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Appendix/PolytropicBinaries

2024-07-04T21:59:12Z

Jet53man:

__FORCETOC__


=Polytropic Models of Close Binary Star Systems=

Over the past half-a-dozen years, Patrick Motl, Mario D'Souza, and Wes Even have used the Hachisu SCF technique to construct 3D equilibrium models of synchronously rotating, tidally distorted binary polytropes. To date, four of these models have been used extensively as initial states for our dynamical simulations of binary mass-transfer. Various properties of these four SCF-code-generated models are summarized in the following table; the listed parameters are:
<table border="0" align="center" cellpadding="2">
<tr>
<td align="right">
<math>q \equiv M_d/M_a</math>  :
</td>
<td align="left">
System mass ratio
</td>
</tr>

<tr>
<td align="right">
<math>M</math>  :
</td>
<td align="left">
Mass
</td>
</tr>

<tr>
<td align="right">
<math>a</math>  :
</td>
<td align="left">
Binary separation
</td>
</tr>

<tr>
<td align="right">
<math>\Omega</math>  :
</td>
<td align="left">
Orbital angular velocity
</td>
</tr>

<tr>
<td align="right">
<math>J_\mathrm{tot}</math>  :
</td>
<td align="left">
Total angular momentum
</td>
</tr>

<tr>
<td align="right">
<math>\rho^\mathrm{max}</math>  :
</td>
<td align="left">
Maximum (central) density
</td>
</tr>

<tr>
<td align="right">
<math>K_\mathrm{n}</math>  :
</td>
<td align="left">
Constant in the polytropic equation of state, {{Math/EQ_Polytrope01}}
</td>
</tr>

<tr>
<td align="right">
<math>V</math>  :
</td>
<td align="left">
Volume occupied by the star or by the Roche Lobe (RL) surrounding the star
</td>
</tr>

<tr>
<td align="right">
<math>R\equiv [3V/(4\pi)]^{1/3}</math>  :
</td>
<td align="left">
Mean stellar radius
</td>
</tr>

<tr>
<td align="right">
<math>f_\mathrm{RL} \equiv V/V_\mathrm{RL}</math>  :
</td>
<td align="left">
Roche-lobe filling factor
</td>
</tr>

</table>

<table align="center" border="1" cellpadding="8" width="95%">
<tr>
<td align="center" colspan="15">
'''<font color="darkblue">
Properties of (<math>n=3/2</math>) Polytropic Binary Systems
</font>'''
</td>
</tr>
<tr>
<td colspan="1" align="center">
'''Model'''
</td>
<td align="center" colspan="5">
'''Binary System'''
</td>
<td align="center" colspan="4">
'''Accretor'''
</td>
<td align="center" colspan="5">
'''Donor'''
</td>
</tr>

<tr>
<td colspan="1" align="center">
 
</td>
<td align="center" colspan="1">
<math>q</math>
</td>
<td align="center" colspan="1">
<math>M_\mathrm{tot}</math>
</td>
<td align="center" colspan="1">
<math>a</math>
</td>
<td align="center" colspan="1">
<math>\Omega</math>
</td>
<td align="center" colspan="1">
<math>J_\mathrm{tot}</math>
</td>

<td align="center" colspan="1">
<math>M_a</math>
</td>
<td align="center" colspan="1">
<math>\rho^\mathrm{max}_a</math>
</td>
<td align="center" colspan="1">
<math>K^a_{3/2}</math>
</td>
<td align="center" colspan="1">
<math>R_a</math>
</td>

<td align="center" colspan="1">
<math>M_d</math>
</td>
<td align="center" colspan="1">
<math>\rho^\mathrm{max}_d</math>
</td>
<td align="center" colspan="1">
<math>K^d_{3/2}</math>
</td>
<td align="center" colspan="1">
<math>R_d</math>
</td>
<td align="center" colspan="1">
<math>f_\mathrm{RL}</math>
</td>
</tr>

<tr>
<td colspan="1" align="center">
'''Q13'''
</td>
<td align="center" colspan="1">
1.323
</td>
<td align="center" colspan="1">
0.0309
</td>
<td align="center" colspan="1">
0.8882
</td>
<td align="center" colspan="1">
0.2113
</td>
<td align="center" colspan="1">
<math>1.40\times 10^{-3}</math>
</td>

<td align="center" colspan="1">
0.0133
</td>
<td align="center" colspan="1">
1.0000
</td>
<td align="center" colspan="1">
0.0264
</td>
<td align="center" colspan="1">
0.2672
</td>

<td align="center" colspan="1">
0.0176
</td>
<td align="center" colspan="1">
0.6000
</td>
<td align="center" colspan="1">
0.0372
</td>
<td align="center" colspan="1">
0.3509
</td>
<td align="center" colspan="1">
0.968
</td>
</tr>

<tr>
<td colspan="1" align="center">
'''Q07'''
</td>
<td align="center" colspan="1">
0.70000
</td>
<td align="center" colspan="1">
0.02371
</td>
<td align="center" colspan="1">
0.83938
</td>
<td align="center" colspan="1">
0.20144
</td>
<td align="center" colspan="1">
<math>8.938\times 10^{-4}</math>
</td>

<td align="center" colspan="1">
0.013945
</td>
<td align="center" colspan="1">
1.0000
</td>
<td align="center" colspan="1">
0.02732
</td>
<td align="center" colspan="1">
0.2728
</td>

<td align="center" colspan="1">
0.009761
</td>
<td align="center" colspan="1">
0.6077
</td>
<td align="center" colspan="1">
0.02512
</td>
<td align="center" colspan="1">
0.2888
</td>
<td align="center" colspan="1">
0.998
</td>
</tr>

<tr>
<td colspan="1" align="center">
'''Q05'''
</td>
<td align="center" colspan="1">
0.500
</td>
<td align="center" colspan="1">
<math>9.216\times 10^{-3}</math>
</td>
<td align="center" colspan="1">
0.8764
</td>
<td align="center" colspan="1">
0.1174
</td>
<td align="center" colspan="1">
<math>1.97\times 10^{-4}</math>
</td>

<td align="center" colspan="1">
<math>6.143\times 10^{-3}</math>
</td>
<td align="center" colspan="1">
1.0000
</td>
<td align="center" colspan="1">
0.016
</td>
<td align="center" colspan="1">
0.2067
</td>

<td align="center" colspan="1">
<math>3.073\times 10^{-3}</math>
</td>
<td align="center" colspan="1">
0.235
</td>
<td align="center" colspan="1">
0.016
</td>
<td align="center" colspan="1">
0.2689
</td>
<td align="center" colspan="1">
0.898
</td>
</tr>

<tr>
<td colspan="1" align="center">
'''Q04'''
</td>
<td align="center" colspan="1">
0.4085
</td>
<td align="center" colspan="1">
0.02399
</td>
<td align="center" colspan="1">
0.8169
</td>
<td align="center" colspan="1">
0.2112
</td>
<td align="center" colspan="1">
<math>7.794\times 10^{-4}</math>
</td>

<td align="center" colspan="1">
0.01703
</td>
<td align="center" colspan="1">
1.0000
</td>
<td align="center" colspan="1">
0.03119
</td>
<td align="center" colspan="1">
0.2918
</td>

<td align="center" colspan="1">
0.006957
</td>
<td align="center" colspan="1">
0.71
</td>
<td align="center" colspan="1">
0.01904
</td>
<td align="center" colspan="1">
0.2453
</td>
<td align="center" colspan="1">
0.996
</td>
</tr>

<tr>
<td align="left" colspan="15">
References:
* Model '''Q13''' (<math>q = 1.323</math>): Table 4 in [http://iopscience.iop.org/0004-637X/643/1/381/pdf/63230.web.pdf publication DMTF06]
* Model '''Q07''' (<math>q = 0.700</math>): First page of the [http://www.phys.lsu.edu/~tohline/clayton/q07.pdf accompanying PDF document]. <font color="red">NOTE: In this PDF document, Roche-lobe volumes appear to be too large by factor of 2.</font>
* Model '''Q05''' (<math>q = 0.500</math>): Table 5 in [http://iopscience.iop.org/0004-637X/643/1/381/pdf/63230.web.pdf publication DMTF06]
* Model '''Q04''' (<math>q = 0.4085</math>): Table 1 in [http://iopscience.iop.org/0004-637X/670/2/1314/pdf/71427.web.pdf publication MFTD07]
</td>
</tr>
</table>

All of the parameter values listed in these tables are specified in dimensionless ''polytropic units'', defined as follows:

<table border="1" cellpadding="20" align="center" width="75%">
<tr>
<td align="center">
<font color="blue"><b>Polytropic Units</b></font>
</td>
</tr>
<tr>
<td align="left">
Here, ''Polytropic Units'' are defined such that the radial extent of the computational grid for the self-consistent-field (SCF) model, <math>R_\mathrm{edge}</math>,
the maximum density of one binary component, <math>\rho^\mathrm{max}_\mathrm{Accretor}</math>, and the gravitational
constant, <math>G</math>, are all unity, that is,
<div align="center">
<math>G = \rho^\mathrm{max}_\mathrm{Accretor} = R_\mathrm{edge} = 1</math>.
</div>
In an [http://www.phys.lsu.edu/~tohline/clayton/PolytropicUnits.pdf accompanying PDF document], we explain how to convert from this set of dimension code units to real (''e.g.,'' cgs) units.
</td>
</tr>
</table>

{{ SGFfooter }}

Appendix/Ramblings/ForOuShangli

2024-07-04T18:21:28Z

Jet53man: /* Wiki_edits */

=For Shangli Ou=

In order to download any one of (a selected subset of) the COLLADA-based model files referenced below, go to …

<table border="3" align="center" cellpadding="10"><tr><td align="center" bgcolor="red">
<table border="0" cellpadding="8" align="center">
<tr>
<td align="center">
[[Appendix/Ramblings/COLLADA/RiemannSType|<font color="white" size="+1">DOWNLOADABLE 3D MODELS</font>]]
</td>
</tr>
</table>
</td></tr></table>

==EFE Diagram (Review)==

<table border="0" cellpadding="12" align="left"><tr><td align="center">'''Figure 1'''</td></tr><tr><td align="center">[[File:EFEdiagram4.png|left|400px|EFE Diagram identifying example models from Ou (2006)]]</td></tr></table>
In the context of our broad discussion of ellipsoidal figures of equilibrium, the label "EFE Diagram" refers to a two-dimensional parameter space defined by the pair of axis ratios (b/a, c/a), ''usually'' covering the ranges, 0 ≤ b/a ≤ 1 and 0 ≤ c/a ≤ 1. The classic/original version of this diagram appears as Figure 2 on p. 902 of {{ Chandrasekhar65_XXVfull }}; a somewhat less cluttered version appears on p. 147 of Chandrasekhar's [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

The version of the EFE Diagram shown here, on the left, highlights four model ''sequences'', all of which also can be found in the original version:
<ul>
<li>''Jacobi'' sequence — the smooth curve that runs through the set of small, dark-blue, diamond-shaped markers; the data identifying the location of these markers have been drawn from §39, Table IV of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>]. The small red circular markers lie along this same sequence; their locations are taken from our own determinations, as detailed in [[ThreeDimensionalConfigurations/JacobiEllipsoids#Table2|Table 2]] of our accompanying discussion of Jacobi ellipsoids. All of the models along this sequence have <math>f \equiv \zeta/\Omega_f = 0</math> and are therefore solid-body rotators, that is, there is no internal motion when the configuration is viewed from a frame that is rotating with frequency, <math>\Omega_f</math>.</li>
<li>''Dedekind'' sequence — a smooth curve that lies precisely on top of the ''Jacobi'' sequence. Each configuration along this sequence is ''adjoint'' to a model on the ''Jacobi'' sequence that shares its (b/a, c/a) axis-ratio pair. All ellipsoidal figures along this sequence have <math>1/f = \Omega_f/\zeta = 0</math> and are therefore stationary as viewed from the ''inertial'' frame; the angular momentum of each configuration is stored in its internal motion (vorticity).</li>
<li>The X = -1 ''self-adjoint'' sequence — At every point along this sequence, the value of the key frequency ratio, <math>\zeta/\Omega_f</math>, in the ''adjoint'' configuration <math>(f_+)</math> is identical to the value of the frequency ratio in the ''direct'' configuration <math>(f_-)</math>; specifically, <math>f_+ = f_- = -(a^2+b^2)/(ab)</math>. The data identifying the location of the small, solid-black markers along this sequence have been drawn from §48, Table VI of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].</li>
<li>The X = +1 ''self-adjoint'' sequence — At every point along this sequence, the value of the key frequency ratio, <math>\zeta/\Omega_f</math>, in the ''adjoint'' configuration <math>(f_+)</math> is identical to the value of the frequency ratio in the ''direct'' configuration <math>(f_-)</math>; specifically, <math>f_+ = f_- = +(a^2+b^2)/(ab)</math>. The data identifying the location of the small, solid-black markers along this sequence have been drawn from §48, Table VI of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].</li>
</ul>

Riemann S-type ellipsoids all lie between or on the two (self-adjoint) curves marked "X = -1" and "X = +1" in the EFE Diagram. The yellow circular markers in the diagram shown here, on the left, identify four Riemann S-type ellipsoids that were examined by {{ Ou2006 }} and that we have also chosen to use as examples.

==Example 3D Interactive Animations==

For each model described below, note that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>f \equiv \frac{\zeta_\mathrm{EFE}}{\Omega_\mathrm{EFE}}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- \frac{\lambda_\mathrm{EFE}}{\Omega_\mathrm{EFE}} \biggl[ \frac{1+(b/a)^2}{b/a} \biggr] \, .</math>
</td>
</tr>
</table>

For additional details, see the [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS#The_COLLADA_Code_.26_Initial_3D_Scene|accompanying chapter titled, "Riemann Meets COLLADA & Oculus Rift S"]].

===b41c385===
The model that we have chosen to use in our first successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 2a</b><br /> <br />
[[File:B41c385EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.41</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 2b</b><br />
[[File:COLLADA3rdViewpoint.png|300px|EFE Model b41c385]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.385</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.547874</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.079886</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.079886</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.547874</math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.415418</math>
</td>
<td align="center">
<math>~f = - 19.53923</math>
</td>
</tr>
</table>

The subscript "EFE" on Ω and λ means that the relevant frequency is given in units that have been adopted in [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], that is, in units of <math>~[\pi G\rho]^{1 / 2}</math>. In Figure 2a, the yellow circular marker, that has been placed where the pair of purple dashed lines cross, identifies the location of this model in the "c/a versus b/a" ''[[ThreeDimensionalConfigurations/RiemannStype#Fig2|EFE Diagram]]'' that appears as Figure 2 on p. 902 of {{ Chandrasekhar65_XXV }}; essentially the same diagram appears in §49 (p. 147) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

In a [[ThreeDimensionalConfigurations/JacobiEllipsoids#Jacobi_Ellipsoids|separate chapter]] we have discussed various properties of uniformly rotating, ''Jacobi'' ellipsoids; they are equilibrium configurations that lie along the sequence that runs from "M<sub>2</sub>" (on the b/a = 1, Maclaurin sequence) to the origin of this diagram. Our chosen model lies off of — just above — the Jacobi-ellipsoid sequence, which means that it is not rotating as a solid body. Instead, as we focus first on the ''direct'' (as opposed to the ''adjoint'') configuration, we appreciate that while the ellipsoid is spinning prograde (counter-clockwise) with a frequency given by |Ω<sub>EFE</sub>|, each Lagrangian fluid element inside as well as on the surface of the ellipsoid is traveling retrograde (clockwise) along an elliptical path with a frequency given by |λ<sub>EFE</sub>|.

===b90c333===

The model that we have chosen to use in our second successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 3a</b><br /> <br />
[[File:B90c333EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.90</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 3b</b><br />
[[File:COLLADAb90c333NewModel.png|300px|EFE Model b90c333]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.333</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.447158</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.221411</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.221411</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.447158 </math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = + 0.995805</math>
</td>
<td align="center">
<math>~f = + 4.061607 </math>
</td>
</tr>
</table>

===b74c692===

The model that we have chosen to use in our third successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 4a</b><br /> <br />
[[File:B74c692EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.74</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 4b</b><br />
[[File:COLLADAb74c692NewModel.png|300px|EFE Model b74c692]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.692</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.638747</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.217773</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.217773</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.638747 </math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.713019</math>
</td>
<td align="center">
<math>~f = - 6.13413 </math>
</td>
</tr>
</table>

===b28c256===

The model that we have chosen to use in our second successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 5a</b><br /> <br />
[[File:B28c256EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.28</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 5b</b><br />
[[File:COLLADAb28c256OldModel.png|300px|EFE Model b28c256]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.256</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.456676</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.020692</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.020692</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.456676</math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.174510</math>
</td>
<td align="center">
<math>~f = - 85.0007</math>
</td>
</tr>
</table>

=Catalog of Local 3D Model Files <font size="-1">(cannot be downloaded)</font>=
==3Dviewers==
===AutoRiemann===

<ul>
<li>
Trial.Ou_b28c256<br />
''Generally speaking ...'' Translucent surface; one Lagrangian (red surface arrow) fluid element; no axis arrow; NO WALL LABELs.
<ul>
<li>
<font color="maroon">Ou_b28c256.Dir.Inertial.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Dir.Rot.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Adj.Inertial.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Adj.Rot.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
Cylinder<br />
''Generally speaking ...'' Translucent surface; one Lagrangian (red surface arrow) fluid element; no axis arrow; PROBLEM RESETING CLOCK.
<ul>
<li>
<font color="maroon">Dir.Inertial_b28c256.labeled.dae</font> (12/22/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Dir.Rot_b28c256.labeled.dae</font> (12/22/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
BESTmodels<br />
''Generally speaking ...'' Translucent surface; six Lagrangian (red arrow markers) fluid elements.
<ul>
<li>
<font color="maroon">b28c256/b28c256.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b41c385/b41c385.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b74c692/b74c692.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b90c333/b90c333.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
TypeI<br />
Translucent surface; just '''Type I tilted''' figure motion (no Lagrangian fluid elements).
<ul>
<li>
<font color="maroon">TypeIa10.GREAT.dae</font> (01/04/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 1.25~~~~~c/a = 0.470</math><br />'''<font color="red">DIRECT figure</font>''' (only) motion - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
TypeI/Lagrange<br />
Depicts motion of (3) Lagrangian fluid elements across an opaque surface; no figure motion; three separate, tilted fluid orbits are identified by small yellow markers.
<ul>
<li>
<font color="maroon">TL15.lagrange.dae</font> (01/26/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 1.25~~~~~c/a = 0.470</math><br />'''<font color="red">LAGRANGIAN element</font>''' (only) motion - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

==Lenovo==

===KhronosGroup's COLLADA_to_gltf/COLLADA2GLTF-v2===

<ul>
<li>
ModelsB41C385<br />
Fundamentally this model is identical to the one developed earlier (and [[#b41c385|presented above]]) that shows how the ''DIRECT'' flow version of model "b41c385" appears when viewed from either the inertial or rotating reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">MultiFluidElements/MultiLagrange26.dae</font> [translucent purple] (05/28/2020) and <font color="orange">/output/MultiLagrange26.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b41c385DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">InertialFrame/Inertial34.dae</font> [opaque purple] (05/29/2020) and <font color="orange">/output/Inertial34.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b41c385DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
ModelsB90C333<br />
Fundamentally this model is identical to the one developed earlier (and [[#b90c333|presented above]]) that shows how the ''DIRECT'' flow version of model "b90c333" appears when viewed from either the inertial or a rotating reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, prograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">FastRot79.dae</font> [translucent purple] (06/03/2020) and <font color="orange">/output/FastRot79.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b90c333DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">FastInertial80.dae</font> [opaque purple] (06/03/2020) — Also available is <font color="orange">../output/FastInertial80.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b90c333DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
ModelsB74C692<br />
Fundamentally this model is identical to the one developed earlier (and [[#b74c692|presented above]]) that shows how the ''DIRECT'' flow version of model "b74c692" appears when viewed from the rotating reference frame: The ellipsoidal surface is translucent (purple); the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">TestMulti74.dae</font> (06/03/2020) — Also available is <font color="orange">../output/TestMulti74.glb</font> (06/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">InertialB74C692aa.dae</font> [opaque purple] (06/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
(no subdirectory)<br />
Fundamentally this model is identical to the one developed earlier (and [[#b28c256|presented above]]) that shows how the ''DIRECT'' flow version of model "b28c256" appears when viewed from the inertial reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (very slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">Pencil94.dae</font> [translucent purple surface] (06/04/2020) and <font color="orange">/output/Pencil94.glb</font> (06/04/2020) — both files also copied (01/30/2021) to LSU physics website with filename … COLLADA/b28c256DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">PencilInertial96.dae</font> [opaque purple surface] (06/04/2020) and <font color="orange">/output/PencilInertial95.glb</font> (06/04/2020) — both files also copied (01/30/2021) to LSU physics website with filename … COLLADA/b28c256DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

==Workfolder==
===Wiki_edits===

<ul>
<li>
CoordinateSystems<br />
Fundamentally this model is identical to the one developed earlier (and [[#b41c385|presented above]]) that shows how the ''DIRECT'' flow version of model "b41c385" appears when viewed from the inertial reference frame: The ellipsoidal surface is translucent (purple); the equatorial plane is identified by a series of small yellow markers; and a single, red marker depicts the (slow, retrograde) Lagrangian motion of one fluid element that lies in the equatorial plane.

But here, in addition across the "northern" hemisphere of the ellipsoid, a sequence of small yellow markers is used to identify the location of one <math>~\lambda_3</math> coordinate curve as defined within our so-called [[Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates|"T8" coordinate system]].
<ul>
<li>
<font color="maroon">Lambda3Curve03.dae</font> (12/26/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

=See Also=

* Discussion of [[ThreeDimensionalConfigurations/RiemannStype|Ou's Riemann-Like Ellipsoids]]

* [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS|Riemann Meets COLLADA & Oculus Rift S]]: Example <b>(b/a, c/a) = (0.41, 0.385)</b>
** [[Appendix/Ramblings/VirtualReality#Virtual_Reality_and_3D_Printing|Virtual Reality and 3D Printing]]
** [[Appendix/Ramblings/OculusRiftS|Success Importing Animated Scene into Oculus Rift S]]
** [[Appendix/Ramblings/RiemannMeetsOculus|Carefully (Re)Build Riemann Type S Ellipsoids Inside Oculus Rift Environment]]
** Other Example S-type Riemann Ellipsoids:
*** <b>[[Appendix/Ramblings/RiemannB90C333|(b/a, c/a) = (0.90, 0.333)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB74C692|(b/a, c/a) = (0.74, 0.692)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB28C256|(b/a, c/a) = (0.28, 0.256)]]</b>

{{ SGFfooter }}

Appendix/Ramblings/ForOuShangli

2024-07-04T18:17:01Z

Jet53man: /* Wiki_edits */

=For Shangli Ou=

In order to download any one of (a selected subset of) the COLLADA-based model files referenced below, go to …

<table border="3" align="center" cellpadding="10"><tr><td align="center" bgcolor="red">
<table border="0" cellpadding="8" align="center">
<tr>
<td align="center">
[[Appendix/Ramblings/COLLADA/RiemannSType|<font color="white" size="+1">DOWNLOADABLE 3D MODELS</font>]]
</td>
</tr>
</table>
</td></tr></table>

==EFE Diagram (Review)==

<table border="0" cellpadding="12" align="left"><tr><td align="center">'''Figure 1'''</td></tr><tr><td align="center">[[File:EFEdiagram4.png|left|400px|EFE Diagram identifying example models from Ou (2006)]]</td></tr></table>
In the context of our broad discussion of ellipsoidal figures of equilibrium, the label "EFE Diagram" refers to a two-dimensional parameter space defined by the pair of axis ratios (b/a, c/a), ''usually'' covering the ranges, 0 ≤ b/a ≤ 1 and 0 ≤ c/a ≤ 1. The classic/original version of this diagram appears as Figure 2 on p. 902 of {{ Chandrasekhar65_XXVfull }}; a somewhat less cluttered version appears on p. 147 of Chandrasekhar's [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

The version of the EFE Diagram shown here, on the left, highlights four model ''sequences'', all of which also can be found in the original version:
<ul>
<li>''Jacobi'' sequence — the smooth curve that runs through the set of small, dark-blue, diamond-shaped markers; the data identifying the location of these markers have been drawn from §39, Table IV of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>]. The small red circular markers lie along this same sequence; their locations are taken from our own determinations, as detailed in [[ThreeDimensionalConfigurations/JacobiEllipsoids#Table2|Table 2]] of our accompanying discussion of Jacobi ellipsoids. All of the models along this sequence have <math>f \equiv \zeta/\Omega_f = 0</math> and are therefore solid-body rotators, that is, there is no internal motion when the configuration is viewed from a frame that is rotating with frequency, <math>\Omega_f</math>.</li>
<li>''Dedekind'' sequence — a smooth curve that lies precisely on top of the ''Jacobi'' sequence. Each configuration along this sequence is ''adjoint'' to a model on the ''Jacobi'' sequence that shares its (b/a, c/a) axis-ratio pair. All ellipsoidal figures along this sequence have <math>1/f = \Omega_f/\zeta = 0</math> and are therefore stationary as viewed from the ''inertial'' frame; the angular momentum of each configuration is stored in its internal motion (vorticity).</li>
<li>The X = -1 ''self-adjoint'' sequence — At every point along this sequence, the value of the key frequency ratio, <math>\zeta/\Omega_f</math>, in the ''adjoint'' configuration <math>(f_+)</math> is identical to the value of the frequency ratio in the ''direct'' configuration <math>(f_-)</math>; specifically, <math>f_+ = f_- = -(a^2+b^2)/(ab)</math>. The data identifying the location of the small, solid-black markers along this sequence have been drawn from §48, Table VI of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].</li>
<li>The X = +1 ''self-adjoint'' sequence — At every point along this sequence, the value of the key frequency ratio, <math>\zeta/\Omega_f</math>, in the ''adjoint'' configuration <math>(f_+)</math> is identical to the value of the frequency ratio in the ''direct'' configuration <math>(f_-)</math>; specifically, <math>f_+ = f_- = +(a^2+b^2)/(ab)</math>. The data identifying the location of the small, solid-black markers along this sequence have been drawn from §48, Table VI of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].</li>
</ul>

Riemann S-type ellipsoids all lie between or on the two (self-adjoint) curves marked "X = -1" and "X = +1" in the EFE Diagram. The yellow circular markers in the diagram shown here, on the left, identify four Riemann S-type ellipsoids that were examined by {{ Ou2006 }} and that we have also chosen to use as examples.

==Example 3D Interactive Animations==

For each model described below, note that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>f \equiv \frac{\zeta_\mathrm{EFE}}{\Omega_\mathrm{EFE}}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- \frac{\lambda_\mathrm{EFE}}{\Omega_\mathrm{EFE}} \biggl[ \frac{1+(b/a)^2}{b/a} \biggr] \, .</math>
</td>
</tr>
</table>

For additional details, see the [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS#The_COLLADA_Code_.26_Initial_3D_Scene|accompanying chapter titled, "Riemann Meets COLLADA & Oculus Rift S"]].

===b41c385===
The model that we have chosen to use in our first successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 2a</b><br /> <br />
[[File:B41c385EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.41</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 2b</b><br />
[[File:COLLADA3rdViewpoint.png|300px|EFE Model b41c385]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.385</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.547874</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.079886</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.079886</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.547874</math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.415418</math>
</td>
<td align="center">
<math>~f = - 19.53923</math>
</td>
</tr>
</table>

The subscript "EFE" on Ω and λ means that the relevant frequency is given in units that have been adopted in [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], that is, in units of <math>~[\pi G\rho]^{1 / 2}</math>. In Figure 2a, the yellow circular marker, that has been placed where the pair of purple dashed lines cross, identifies the location of this model in the "c/a versus b/a" ''[[ThreeDimensionalConfigurations/RiemannStype#Fig2|EFE Diagram]]'' that appears as Figure 2 on p. 902 of {{ Chandrasekhar65_XXV }}; essentially the same diagram appears in §49 (p. 147) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

In a [[ThreeDimensionalConfigurations/JacobiEllipsoids#Jacobi_Ellipsoids|separate chapter]] we have discussed various properties of uniformly rotating, ''Jacobi'' ellipsoids; they are equilibrium configurations that lie along the sequence that runs from "M<sub>2</sub>" (on the b/a = 1, Maclaurin sequence) to the origin of this diagram. Our chosen model lies off of — just above — the Jacobi-ellipsoid sequence, which means that it is not rotating as a solid body. Instead, as we focus first on the ''direct'' (as opposed to the ''adjoint'') configuration, we appreciate that while the ellipsoid is spinning prograde (counter-clockwise) with a frequency given by |Ω<sub>EFE</sub>|, each Lagrangian fluid element inside as well as on the surface of the ellipsoid is traveling retrograde (clockwise) along an elliptical path with a frequency given by |λ<sub>EFE</sub>|.

===b90c333===

The model that we have chosen to use in our second successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 3a</b><br /> <br />
[[File:B90c333EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.90</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 3b</b><br />
[[File:COLLADAb90c333NewModel.png|300px|EFE Model b90c333]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.333</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.447158</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.221411</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.221411</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.447158 </math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = + 0.995805</math>
</td>
<td align="center">
<math>~f = + 4.061607 </math>
</td>
</tr>
</table>

===b74c692===

The model that we have chosen to use in our third successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 4a</b><br /> <br />
[[File:B74c692EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.74</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 4b</b><br />
[[File:COLLADAb74c692NewModel.png|300px|EFE Model b74c692]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.692</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.638747</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.217773</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.217773</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.638747 </math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.713019</math>
</td>
<td align="center">
<math>~f = - 6.13413 </math>
</td>
</tr>
</table>

===b28c256===

The model that we have chosen to use in our second successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 5a</b><br /> <br />
[[File:B28c256EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.28</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 5b</b><br />
[[File:COLLADAb28c256OldModel.png|300px|EFE Model b28c256]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.256</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.456676</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.020692</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.020692</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.456676</math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.174510</math>
</td>
<td align="center">
<math>~f = - 85.0007</math>
</td>
</tr>
</table>

=Catalog of Local 3D Model Files <font size="-1">(cannot be downloaded)</font>=
==3Dviewers==
===AutoRiemann===

<ul>
<li>
Trial.Ou_b28c256<br />
''Generally speaking ...'' Translucent surface; one Lagrangian (red surface arrow) fluid element; no axis arrow; NO WALL LABELs.
<ul>
<li>
<font color="maroon">Ou_b28c256.Dir.Inertial.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Dir.Rot.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Adj.Inertial.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Ou_b28c256.Adj.Rot.dae</font> (12/01/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
Cylinder<br />
''Generally speaking ...'' Translucent surface; one Lagrangian (red surface arrow) fluid element; no axis arrow; PROBLEM RESETING CLOCK.
<ul>
<li>
<font color="maroon">Dir.Inertial_b28c256.labeled.dae</font> (12/22/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">Dir.Rot_b28c256.labeled.dae</font> (12/22/2019)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
BESTmodels<br />
''Generally speaking ...'' Translucent surface; six Lagrangian (red arrow markers) fluid elements.
<ul>
<li>
<font color="maroon">b28c256/b28c256.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b28c256/b28c256.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b41c385/b41c385.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon">b41c385/b41c385.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41 ~~~~~c/a = 0.385</math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b74c692/b74c692.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b74c692/b74c692.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74 ~~~~~c/a = 0.692 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
 
<hr>
 
<li>
<font color="maroon">b90c333/b90c333.DI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.DRot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.AI.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
<li>
<font color="maroon"> b90c333/b90c333.ARot.dae</font> (01/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90 ~~~~~c/a = 0.333 </math><br />'''<font color="red">ADJOINT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
TypeI<br />
Translucent surface; just '''Type I tilted''' figure motion (no Lagrangian fluid elements).
<ul>
<li>
<font color="maroon">TypeIa10.GREAT.dae</font> (01/04/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 1.25~~~~~c/a = 0.470</math><br />'''<font color="red">DIRECT figure</font>''' (only) motion - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
TypeI/Lagrange<br />
Depicts motion of (3) Lagrangian fluid elements across an opaque surface; no figure motion; three separate, tilted fluid orbits are identified by small yellow markers.
<ul>
<li>
<font color="maroon">TL15.lagrange.dae</font> (01/26/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 1.25~~~~~c/a = 0.470</math><br />'''<font color="red">LAGRANGIAN element</font>''' (only) motion - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

==Lenovo==

===KhronosGroup's COLLADA_to_gltf/COLLADA2GLTF-v2===

<ul>
<li>
ModelsB41C385<br />
Fundamentally this model is identical to the one developed earlier (and [[#b41c385|presented above]]) that shows how the ''DIRECT'' flow version of model "b41c385" appears when viewed from either the inertial or rotating reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">MultiFluidElements/MultiLagrange26.dae</font> [translucent purple] (05/28/2020) and <font color="orange">/output/MultiLagrange26.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b41c385DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">InertialFrame/Inertial34.dae</font> [opaque purple] (05/29/2020) and <font color="orange">/output/Inertial34.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b41c385DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
ModelsB90C333<br />
Fundamentally this model is identical to the one developed earlier (and [[#b90c333|presented above]]) that shows how the ''DIRECT'' flow version of model "b90c333" appears when viewed from either the inertial or a rotating reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, prograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">FastRot79.dae</font> [translucent purple] (06/03/2020) and <font color="orange">/output/FastRot79.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b90c333DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">FastInertial80.dae</font> [opaque purple] (06/03/2020) — Also available is <font color="orange">../output/FastInertial80.glb</font> (06/04/2020) — both files also copied (01/31/2021) to LSU physics website with filename … COLLADA/b90c333DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.90~~~~~c/a = 0.333</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
ModelsB74C692<br />
Fundamentally this model is identical to the one developed earlier (and [[#b74c692|presented above]]) that shows how the ''DIRECT'' flow version of model "b74c692" appears when viewed from the rotating reference frame: The ellipsoidal surface is translucent (purple); the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">TestMulti74.dae</font> (06/03/2020) — Also available is <font color="orange">../output/TestMulti74.glb</font> (06/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">InertialB74C692aa.dae</font> [opaque purple] (06/02/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.74~~~~~c/a = 0.692</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

<ul>
<li>
(no subdirectory)<br />
Fundamentally this model is identical to the one developed earlier (and [[#b28c256|presented above]]) that shows how the ''DIRECT'' flow version of model "b28c256" appears when viewed from the inertial reference frame: The ellipsoidal surface is purple [either translucent or opaque, as indicated]; the equatorial plane is identified by a series of small yellow markers; and 9 red markers depict the (very slow, retrograde) Lagrangian motion of fluid elements that lie in the equatorial plane.
<ul>
<li>
<font color="maroon">Pencil94.dae</font> [translucent purple surface] (06/04/2020) and <font color="orange">/output/Pencil94.glb</font> (06/04/2020) — both files also copied (01/30/2021) to LSU physics website with filename … COLLADA/b28c256DirRotating.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">ROTATING</font>''' frame</div>
</td></tr></table>
</li>
</ul>
<ul>
<li>
<font color="maroon">PencilInertial96.dae</font> [opaque purple surface] (06/04/2020) and <font color="orange">/output/PencilInertial95.glb</font> (06/04/2020) — both files also copied (01/30/2021) to LSU physics website with filename … COLLADA/b28c256DirInertial.
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.28~~~~~c/a = 0.256</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

==Workfolder==
===Wiki_edits===

<ul>
<li>
CoordinateSystems<br />
Fundamentally this model is identical to the one developed earlier (and [[#b41c385|presented above]]) that shows how the ''DIRECT'' flow version of model "b41c385" appears when viewed from the inertial reference frame: The ellipsoidal surface is translucent (purple); the equatorial plane is identified by a series of small yellow markers; and a single, red marker depicts the (slow, retrograde) Lagrangian motion of one fluid element that lies in the equatorial plane.

But here, in addition across the "northern" hemisphere of the ellipsoid, a sequence of small yellow markers is used to identify the location of one <math>~\lambda_3</math> coordinate curve as defined within our so-called [[Appendix/Ramblings/ConcentricEllipsodalT8Coordinates#Concentric_Ellipsoidal_.28T8.29_Coordinates|"T8" coordinate system]].
<ul>
<li>
<font color="maroon">Lambda3Curve03.dae</font> (12/26/2020)
<table border="1" cellpadding="8" align="center" width="50%"><tr><td align="left">
<div align="center"><math>~b/a = 0.41~~~~~c/a = 0.385</math><br />'''<font color="red">DIRECT</font>''' flow - as viewed from '''<font color="red">INERTIAL</font>''' frame</div>
</td></tr></table>
</li>
</ul>
</li>
</ul>

=See Also=

* Discussion of [[ThreeDimensionalConfigurations/RiemannStype|Ou's Riemann-Like Ellipsoids]]

* [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS|Riemann Meets COLLADA & Oculus Rift S]]: Example <b>(b/a, c/a) = (0.41, 0.385)</b>
** [[Appendix/Ramblings/VirtualReality#Virtual_Reality_and_3D_Printing|Virtual Reality and 3D Printing]]
** [[Appendix/Ramblings/OculusRiftS|Success Importing Animated Scene into Oculus Rift S]]
** [[Appendix/Ramblings/RiemannMeetsOculus|Carefully (Re)Build Riemann Type S Ellipsoids Inside Oculus Rift Environment]]
** Other Example S-type Riemann Ellipsoids:
*** <b>[[Appendix/Ramblings/RiemannB90C333|(b/a, c/a) = (0.90, 0.333)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB74C692|(b/a, c/a) = (0.74, 0.692)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB28C256|(b/a, c/a) = (0.28, 0.256)]]</b>

{{ SGFfooter }}

ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS

2024-07-02T23:08:45Z

Jet53man: /* Khronos Group's COLLADA2GLTF Converter */

__FORCETOC__


=Riemann Meets COLLADA & Oculus Rift S=
{| class="RiemannMeetsCOLLADA" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
|-
! style="height: 125px; width: 125px; background-color:white;" |[[H_BookTiledMenu#Equilibrium_Structures_2|<b>Riemann<br />meets<br />COLLADA<br />&<br />Oculus Rift S</b>]]
|}

If you want to jump straight to the fun part, copy the ASCII text file that is contained below within the [[#Scroll1|scrollable box]] named, <b>Inertial17.dae</b>, and read the accompanying instructions regarding how to view this example model on a Mac. A second file — presenting the same example model, but as viewed from a rotating frame of reference — can be extracted from a separate [[#Scroll2|scrollable box]] named, <b>MultiLagrange18.dae</b>.
 <br />
 <br />
 <br />
 <br />

==Example Model==

===Key Physical Parameters===

The model that we have chosen to use in our first successful construction of a COLLADA-based, 3D and interactive animation has the following properties; this model has been selected from [[ThreeDimensionalConfigurations/RiemannStype#Table2|Table 2 of our accompanying discussion of Riemann S-type ellipsoids]]:

<table border="0" cellpadding="5" align="center">

<tr>
<td align="center" rowspan="6">
<b>Figure 1a</b><br /> <br />
[[File:B41c385EFEdiagram02.png|325px|EFE Parameter Space]]
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" colspan="3">
<math>~\frac{b}{a} = 0.41</math>
</td>
<td align="center" rowspan="6" width="2%">
 
</td>
<td align="center" rowspan="6" bgcolor="lightgrey">
<b>Figure 1b</b><br />
[[File:COLLADA3rdViewpoint.png|300px|EFE Model b41c385]]
</td>
</tr>

<tr>
<td align="center" colspan="3">
<math>~\frac{c}{a} = 0.385</math>
</td>
</tr>

<tr>
<td align="center">
''Direct''
</td>
<td align="center" width="2%" rowspan="4">
 
</td>
<td align="center">
''Adjoint''
</td>
</tr>

<tr>
<td align="center">
<math>~\Omega_\mathrm{EFE} = 0.547874</math>
</td>
<td align="center">
<math>~\Omega_\mathrm{EFE} = - 0.079886</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\lambda_\mathrm{EFE} = 0.079886</math>
</td>
<td align="center">
<math>~\lambda_\mathrm{EFE} = - 0.547874</math>
</td>
</tr>

<tr>
<td align="center">
<math>~f = - 0.415418</math>
</td>
<td align="center">
<math>~f = - 19.53923</math>
</td>
</tr>
</table>

The subscript "EFE" on Ω and λ means that the relevant frequency is given in units that have been adopted in [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], that is, in units of <math>~[\pi G\rho]^{1 / 2}</math>. In Figure 1a, the yellow circular marker, that has been placed where the pair of purple dashed lines cross, identifies the location of this model in the "c/a versus b/a" ''[[ThreeDimensionalConfigurations/RiemannStype#Fig2|EFE Diagram]]'' that appears as Figure 2 on p. 902 of [http://adsabs.harvard.edu/abs/1965ApJ...142..890C S. Chandrasekhar (1965)]; essentially the same diagram appears in §49 (p. 147) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

In a [[ThreeDimensionalConfigurations/JacobiEllipsoids#Jacobi_Ellipsoids|separate chapter]] we have discussed various properties of uniformly rotating, ''Jacobi'' ellipsoids; they are equilibrium configurations that lie along the sequence that runs from "M<sub>2</sub>" (on the b/a = 1, Maclaurin sequence) to the origin of this diagram. Our chosen model lies off of — just above — the Jacobi-ellipsoid sequence, which means that it is not rotating as a solid body. Instead, as we focus first on the ''direct'' (as opposed to the ''adjoint'') configuration, we appreciate that while the ellipsoid is spinning prograde (counter-clockwise) with a frequency given by |Ω<sub>EFE</sub>|, each Lagrangian fluid element inside as well as on the surface of the ellipsoid is traveling retrograde (clockwise) along an elliptical path with a frequency given by |λ<sub>EFE</sub>|.

===The COLLADA Code & Initial 3D Scene===

The following scrollable box titled, <b>Inertial17.dae</b>, contains an <xml>-formatted [https://www.khronos.org/collada/ COLLADA] code that, when properly executed, uses the interactive, 3D animation capabilities of video game drivers to illustrate, in a quantitatively precise fashion, the dynamical flows that are associated with our chosen example model of an equilibrium S-type Riemann ellipsoid. On a Macintosh computer platform, all of the features of this animated and interactive model can be accessed by simply opening the file using Apple's in-house "Preview" application. Figure 1b provides a snapshot of the Preview app's window immediately (TIME = 0.0) after the COLLADA code has been read into the app. More elements of this 3D visual scene will be identified and discussed below, but for now we simply point out that the center of the opaque, purple ellipsoid coincides with the origin of the principal underlying "laboratory" coordinate frame — that is, at (X<sub>0</sub>, Y<sub>0</sub>, Z<sub>0</sub>) = (0, 0, 0) — and the ellipsoid has the desired axis ratios (b/a, c/a) = (0.41, 0.385).

<div align="center" id="Scroll1">
<b>Inertial17.dae</b>     …     a COLLADA code containing 7968 lines of <xml>-formatted ASCII text
</div>

<div style="height: 400px; width: 100%; overflow: scroll;">
<pre>
<?xml version="1.0" encoding="UTF-8" standalone="no" ?>
<COLLADA xmlns="http://www.collada.org/2005/11/COLLADASchema" version="1.4.1">
<asset>
<contributor>
<author>jet53man</author>
<authoring_tool>Raw XML</authoring_tool>
</contributor>
<created>2020-04-28</created>
<modified>2020-05-20</modified>
<unit meter="0.0254" name="inch" />
<up_axis>Z_UP</up_axis>

</asset>
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<profile_COMMON>
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<diffuse>
<color>0.5 0.0 0.99 1.0</color>
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<transparency>
<float sid="transparency">1.0</float>
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</profile_COMMON>
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<input semantic="POSITION" source="#EquatorialPlane_Positions" />
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<input semantic="INPUT" source="#First_rotation_euler_Z-input"/>
<input semantic="OUTPUT" source="#First_rotation_euler_Z-output"/>
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<sampler id="Frame_rotation_euler_Z-sampler">
<input semantic="INPUT" source="#First_rotation_euler_Z-input"/>
<input semantic="OUTPUT" source="#Frame_rotation_euler_Z-output"/>
</sampler>
<channel source="#First_rotation_euler_Z-sampler" target="EntireEllipsoid/rotationZ.ANGLE"/>
<channel source="#Frame_rotation_euler_Z-sampler" target="Inertial/rotationZ.ANGLE"/>
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<source id="Lagrange1_location_X-input">
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0.000 0.000 -1.00 0.00 -0.7071 0.7071 0.000 0.000 0.7071 0.7071 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.9239 0.3827 0.000 0.000 0.3827 0.9239 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -1.0000 0.0000 0.000 0.000 0.0000 1.0000 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.9239 -0.3827 0.000 0.000 -0.3827 0.9239 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.7071 -0.7071 0.000 0.000 -0.7071 0.7071 0.000 0.00 0.000 0.000 0.000 1.000
</float_array>
<technique_common>
<accessor source="#Cube_quaternion_Z-output-array" count=" 147" stride="16">
<param name="TRANSFORM" type="float4x4"/>
</accessor>
</technique_common>
</source>
<sampler id="Cube_quaternion_Z-sampler">
<input semantic="INPUT" source="#Cube_quaternion_Z-input"/>
<input semantic="OUTPUT" source="#Cube_quaternion_Z-output"/>
</sampler>
<channel source="#Cube_quaternion_Z-sampler" target="Jetman/transformation"/>
</animation>

<animation id="HourHand_euler_Z">
<source id="HourHand_euler_Z-input">
<float_array id="HourHand_euler_Z-input-array" count=" 147">
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75
4.00 4.25 4.50 4.75 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75
8.00 8.25 8.50 8.75 9.00 9.25 9.50 9.75 10.00 10.25 10.50 10.75 11.00 11.25 11.50 11.75
12.00 12.25 12.50 12.75 13.00 13.25 13.50 13.75 14.00 14.25 14.50 14.75 15.00 15.25 15.50 15.75
16.00 16.25 16.50 16.75 17.00 17.25 17.50 17.75 18.00 18.25 18.50 18.75 19.00 19.25 19.50 19.75
20.00 20.25 20.50 20.75 21.00 21.25 21.50 21.75 22.00 22.25 22.50 22.75 23.00 23.25 23.50 23.75
24.00 24.25 24.50 24.75 25.00 25.25 25.50 25.75 26.00 26.25 26.50 26.75 27.00 27.25 27.50 27.75
28.00 28.25 28.50 28.75 29.00 29.25 29.50 29.75 30.00 30.25 30.50 30.75 31.00 31.25 31.50 31.75
32.00 32.25 32.50 32.75 33.00 33.25 33.50 33.75 34.00 34.25 34.50 34.75 35.00 35.25 35.50 35.75
36.00 36.25 36.50
</float_array>
<technique_common>
<accessor source="#HourHand_euler_Z-input-array" count=" 147" stride="1">
<param name="TIME" type="float"/>
</accessor>
</technique_common>
</source>
<source id="HourHand_euler_Z-output">
<float_array id="HourHand_euler_Z-output-array" count=" 147">
270.0
268.125
266.25
264.375
262.5
260.625
258.75
256.875
255.0
253.125
251.25
249.375
247.5
245.625
243.75
241.875
240.0
238.125
236.25
234.375
232.5
230.625
228.75
226.875
225.0
223.125
221.25
219.375
217.5
215.625
213.75
211.875
210.0
208.125
206.25
204.375
202.5
200.625
198.75
196.875
195.0
193.125
191.25
189.375
187.5
185.625
183.75
181.875
180.0
178.125
176.25
174.375
172.5
170.625
168.75
166.875
165.0
163.125
161.25
159.375
157.5
155.625
153.75
151.875
150.0
148.125
146.25
144.375
142.5
140.625
138.75
136.875
135.0
133.125
131.25
129.375
127.5
125.625
123.75
121.875
120.0
118.125
116.25
114.375
112.5
110.625
108.75
106.875
105.0
103.125
101.25
99.375
97.5
95.625
93.75
91.875
90.0
88.125
86.25
84.375
82.5
80.625
78.75
76.875
75.0
73.125
71.25
69.375
67.5
65.625
63.75
61.875
60.0
58.125
56.25
54.375
52.5
50.625
48.75
46.875
45.0
43.125
41.25
39.375
37.5
35.625
33.75
31.875
30.0
28.125
26.25
24.375
22.5
20.625
18.75
16.875
15.0
13.125
11.25
9.375
7.5
5.625
3.75
1.875
0.0
-1.875
-3.75
</float_array>
<technique_common>
<accessor source="#HourHand_euler_Z-output-array" count=" 147" stride="1">
<param name="ANGLE" type="float"/>
</accessor>
</technique_common>
</source>
<sampler id="HourHand_euler_Z-sampler">
<input semantic="INPUT" source="#HourHand_euler_Z-input"/>
<input semantic="OUTPUT" source="#HourHand_euler_Z-output"/>
</sampler>
<channel source="#HourHand_euler_Z-sampler" target="HourHand/rotationZ.ANGLE"/>
</animation>
</library_animations>
<library_controllers/>
<library_visual_scenes>
<visual_scene id="ID1">
<node id="EntireEllipsoid" name="OculusRift_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 3.0</rotate>
<scale sid="scale">5.0 5.0 5.0</scale>
<node id="First" name="First_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Second" name="Second_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 180.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Third" name="Third_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 180.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Fourth" name="Fourth_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 180.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 180.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>

<node id="CylinderPointer" name="instance_cylinder">
<translate sid="location">-1.0 0.0 0.0</translate>
<rotate sid="rotationX">1 0 0 0</rotate>
<rotate sid="rotationY">0 1 0 0</rotate>
<rotate sid="rotationZ">0 0 1 180</rotate>
<scale sid="scale">0.05 0.025 0.025</scale>
<instance_node url="#CylPointer" />
</node>

<node id="Lagrange01" name="instance_Lagrange01">
<translate sid="location">1.0000 0.0000 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">0.075 0.075 0.075</scale>
<instance_node url="#Primary_MarkerID3" />
</node>
<node id="Marker01" name="instance_marker01">
<matrix>0.025 0.0 0.0 1.0000 0.0 0.025 0.0 -0.0000 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker02" name="instance_marker02">
<matrix>0.025 0.0 0.0 0.9921 0.0 0.025 0.0 -0.0514 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker03" name="instance_marker03">
<matrix>0.025 0.0 0.0 0.9686 0.0 0.025 0.0 -0.1020 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker04" name="instance_marker04">
<matrix>0.025 0.0 0.0 0.9298 0.0 0.025 0.0 -0.1509 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker05" name="instance_marker05">
<matrix>0.025 0.0 0.0 0.8763 0.0 0.025 0.0 -0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker06" name="instance_marker06">
<matrix>0.025 0.0 0.0 0.8090 0.0 0.025 0.0 -0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker07" name="instance_marker07">
<matrix>0.025 0.0 0.0 0.7290 0.0 0.025 0.0 -0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker08" name="instance_marker08">
<matrix>0.025 0.0 0.0 0.6374 0.0 0.025 0.0 -0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker09" name="instance_marker09">
<matrix>0.025 0.0 0.0 0.5358 0.0 0.025 0.0 -0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker10" name="instance_marker10">
<matrix>0.025 0.0 0.0 0.4258 0.0 0.025 0.0 -0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker11" name="instance_marker11">
<matrix>0.025 0.0 0.0 0.3090 0.0 0.025 0.0 -0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker12" name="instance_marker12">
<matrix>0.025 0.0 0.0 0.1874 0.0 0.025 0.0 -0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker13" name="instance_marker13">
<matrix>0.025 0.0 0.0 0.0628 0.0 0.025 0.0 -0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker14" name="instance_marker14">
<matrix>0.025 0.0 0.0 -0.0628 0.0 0.025 0.0 -0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker15" name="instance_marker15">
<matrix>0.025 0.0 0.0 -0.1874 0.0 0.025 0.0 -0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker16" name="instance_marker16">
<matrix>0.025 0.0 0.0 -0.3090 0.0 0.025 0.0 -0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker17" name="instance_marker17">
<matrix>0.025 0.0 0.0 -0.4258 0.0 0.025 0.0 -0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker18" name="instance_marker18">
<matrix>0.025 0.0 0.0 -0.5358 0.0 0.025 0.0 -0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker19" name="instance_marker19">
<matrix>0.025 0.0 0.0 -0.6374 0.0 0.025 0.0 -0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker20" name="instance_marker20">
<matrix>0.025 0.0 0.0 -0.7290 0.0 0.025 0.0 -0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker21" name="instance_marker21">
<matrix>0.025 0.0 0.0 -0.8090 0.0 0.025 0.0 -0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker22" name="instance_marker22">
<matrix>0.025 0.0 0.0 -0.8763 0.0 0.025 0.0 -0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker23" name="instance_marker23">
<matrix>0.025 0.0 0.0 -0.9298 0.0 0.025 0.0 -0.1509 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker24" name="instance_marker24">
<matrix>0.025 0.0 0.0 -0.9686 0.0 0.025 0.0 -0.1020 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker25" name="instance_marker25">
<matrix>0.025 0.0 0.0 -0.9921 0.0 0.025 0.0 -0.0514 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker26" name="instance_marker26">
<matrix>0.025 0.0 0.0 -1.0000 0.0 0.025 0.0 -0.0000 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>

<node id="Marker28" name="instance_marker28">
<matrix>0.025 0.0 0.0 -0.9921 0.0 0.025 0.0 0.0514 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker29" name="instance_marker29">
<matrix>0.025 0.0 0.0 -0.9686 0.0 0.025 0.0 0.1020 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker30" name="instance_marker30">
<matrix>0.025 0.0 0.0 -0.9298 0.0 0.025 0.0 0.1509 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker31" name="instance_marker31">
<matrix>0.025 0.0 0.0 -0.8763 0.0 0.025 0.0 0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker32" name="instance_marker32">
<matrix>0.025 0.0 0.0 -0.8090 0.0 0.025 0.0 0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker33" name="instance_marker33">
<matrix>0.025 0.0 0.0 -0.7290 0.0 0.025 0.0 0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker34" name="instance_marker34">
<matrix>0.025 0.0 0.0 -0.6374 0.0 0.025 0.0 0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker35" name="instance_marker35">
<matrix>0.025 0.0 0.0 -0.5358 0.0 0.025 0.0 0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker36" name="instance_marker36">
<matrix>0.025 0.0 0.0 -0.4258 0.0 0.025 0.0 0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker37" name="instance_marker37">
<matrix>0.025 0.0 0.0 -0.3090 0.0 0.025 0.0 0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker38" name="instance_marker38">
<matrix>0.025 0.0 0.0 -0.1874 0.0 0.025 0.0 0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker39" name="instance_marker39">
<matrix>0.025 0.0 0.0 -0.0628 0.0 0.025 0.0 0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker40" name="instance_marker40">
<matrix>0.025 0.0 0.0 0.0628 0.0 0.025 0.0 0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker41" name="instance_marker41">
<matrix>0.025 0.0 0.0 0.1874 0.0 0.025 0.0 0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker42" name="instance_marker42">
<matrix>0.025 0.0 0.0 0.3090 0.0 0.025 0.0 0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker43" name="instance_marker43">
<matrix>0.025 0.0 0.0 0.4258 0.0 0.025 0.0 0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker44" name="instance_marker44">
<matrix>0.025 0.0 0.0 0.5358 0.0 0.025 0.0 0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker45" name="instance_marker45">
<matrix>0.025 0.0 0.0 0.6374 0.0 0.025 0.0 0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker46" name="instance_marker46">
<matrix>0.025 0.0 0.0 0.7290 0.0 0.025 0.0 0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker47" name="instance_marker47">
<matrix>0.025 0.0 0.0 0.8090 0.0 0.025 0.0 0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker48" name="instance_marker48">
<matrix>0.025 0.0 0.0 0.8763 0.0 0.025 0.0 0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker49" name="instance_marker49">
<matrix>0.025 0.0 0.0 0.9298 0.0 0.025 0.0 0.1509 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker50" name="instance_marker50">
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<instance_node url="#MarkersID3" />
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<node id="Marker51" name="instance_marker51">
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<instance_node url="#MarkersID3" />
</node>

</node> 
<node id="Inertial" name="instance_inertial"> 
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 3.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<node id="Zup_Pointer" name="instance_zupPointer">
<translate>0.0 0.0 5.0</translate>
<rotate sid="rotationX">1 0 0 0.0</rotate>
<rotate sid="rotationY">0 1 0 0.0</rotate>
<rotate sid="rotationZ">0 0 1 0.0</rotate>
<scale sid="scale">0.125 0.5 1.5</scale>
<instance_node url="#ZupPointer" />
</node>

<node id="MidPlane" name="instance_MidPlane">
<translate>0.0 0.0 0.0</translate>
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<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 -90.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1 270.0</rotate>
<scale sid="scale">0.75 0.75 0.75</scale>
<instance_node url="#HourID3" />
</node> 
</node>

</node> 
</visual_scene>
</library_visual_scenes>
<scene>
<instance_visual_scene url="#ID1" />
</scene>
</COLLADA>

</pre>
</div>

<table border="1" cellpadding="10" align="center" width="100%">
<tr>
<td align="left">
<b>Instructions for Viewing the Example Model on a Mac:</b>
<br />
<ol>
<li>Click, then drag your cursor down through the scrollable box to highlight and ''select'' all 7968 lines of the COLLADA code.</li>
<li>Make a copy (⌘-c) of the entire code.</li>
<li>Paste (⌘-v) the code into an empty, plain text file (using, for example, Apple's <font type="Courier">TextEdit</font> app) and save it with the name, <b>Inertial17.dae</b>. [NOTE: ".dae" is the file name suffix that is universally associated with an executable COLLADA file.]</li>
<li>Open this file using Apple's ''Preview'' app; the static 3D model of our (opaque purple) ellipsoid should appear at TIME = 0.0 — see Figure 1b.</li>
<li>To view the 3D (TIME = 0.0) model from different angles, place your mouse cursor anywhere near the center of the Preview app's image, then click and drag the cursor in different directions.</li>
<li>The play/pause buttons — shown at the bottom of Figure 1b — become visible if the cursor is simply placed near the bottom of the image being displayed by the Preview app; the animation will start (on a continuous loop) when the play button is clicked.</li>
</ol>

<b>Instructions for Viewing the Example Model on a Windows 10 PC:</b>
<br />We have not yet figured out how to similarly (at no cost) view COLLADA-based (.dae) 3D scenes on a PC. On my Lenovo system, the Windows 10 startup menu includes a "3D Viewer" app, but this app does not recognize .dae-formatted files. However, the app appears to read .glb-formatted files satisfactorily. As is [[#Inside_the_Oculus_Rift_S_Environment|discussed below]], the Oculus Rift S system will import our COLLADA-based models only if each model file has been translated from a ".dae" format to a ".glb" format. Hence, after this translation has been made principally to facilitate the importation of our model into the Oculus Rift S virtual-reality environment — see our description, below, of the [[#Khronos_Group.27s_COLLADA2GLTF_Converter|Khronos Group's COLLADA2GLTF Converter]] — we have also been able to straightforwardly view the 3D scene on our Lenovo PC using its "3D Viewer" app.
</td>
</tr>
</table>

At TIME = 0.0 (see, again, Figure 1b), our purple ellipsoid is oriented such that its longest axis (of length, 2a) aligns with the X<sub>0</sub>-axis and such that the small red cubic "fluid marker" at the tip of the ellipsoid lies in the plus-X direction — initially at location (X<sub>0</sub>, Y<sub>0</sub>, Z<sub>0</sub>) = (+a, 0, 0). At the opposite end of the ellipsoid — initially at location (X<sub>0</sub>, Y<sub>0</sub>, Z<sub>0</sub>) = (-a, 0, 0) — we have attached a stubby black pointer; as the ellipsoid spins, this black pointer will remain affixed to the ellipsoid to aid in counting spin cycles, whereas the red fluid marker will not. The ellipsoid's shortest axis (length, 2c) aligns with the Z<sub>0</sub>-axis, with straight up being associated with the positive Z direction. The ellipsoid's intermediate axis (length, 2b) initially aligns with the lab's Y<sub>0</sub>-axis, with the positive direction being defined by the standard "righthand rule."

In addition to displaying an opaque purple Riemann ellipsoid, our COLLADA-defined 3D visual scene contains (see, again, Figure 1b):
<ul>
<li>A square gray "frame" that lies in the laboratory frame's equatorial, Z<sub>0</sub> = 0.0, plane, just outside of and "framing" the ellipsoid; and a small gray solid rectangle that hovers above the ellipsoid and is affixed to the plus-Z<sub>0</sub> axis, also intended to provide 3D orientation and registration queues.</li>
<li>A couple of lines of alpha-numeric characters that has been attached to an imaginary laboratory wall to label/quantify the geometric properties of this ellipsoid.</li>
<li>A clock that has been hung on the same imaginary laboratory wall.</li>
<li>A yellow-dotted ellipse (comprised of 50 individual small yellow markers) that lies in the equatorial plane and is attached to the ellipsoid, marking the trajectory that will be followed by the red cubic "fluid marker" during the animation sequence.</li>
</ul>

===Animation===

When the "play" button is clicked on the Preview app's control panel (again, reference Figure 1b), the COLLADA-directed scene-animation begins. Three different characteristic timescales are relevant:
[[File:ClockImage01.png|175px|right|Clock Image]]<ol>
<li>
We use the motion of the hands on the wall-mounted clock to identify the natural oscillation frequency of the physical system, namely, <math>~\Omega_0 \equiv [\pi G \rho]^{1 / 2}</math>. More specifically, by declaration, one oscillation period of the red "minute" hand — that is to say, one wall-clock "hour" — is <math>~P_0 = 2\pi[\pi G \rho]^{-1 / 2} = [4\pi/(G \rho)]^{1 / 2}</math>. One full cycle of the wall-clock's black "hour" hand is therefore 12P<sub>0</sub>. For pedagogical purposes, in all of our COLLADA-driven animations we have equated P<sub>0</sub> with a COLLADA TIME interval of 4.0, which means that during each animation, one wall-clock hour passes every (approximately) 4 seconds of real time.<sup>&Dagger;</sup></li>
<li>The purple ellipsoid spins in a prograde direction about its shortest axis — that is, about the Z<sub>0</sub>-axis — on a timescale set by the value of <math>~\Omega_\mathrm{EFE}</math>, which has already been normalized to Ω<sub>0</sub>. Specifically, as viewed from the inertial frame of reference, the ellipsoid will complete one full spin in a time, <math>~P_\mathrm{spin} = |\Omega_\mathrm{EFE}|^{-1} = [0.5479]^{-1} = 1.825</math> wall-clock hours, which is approximately 7.3 seconds of real time.</li>
<li>As viewed from a frame of reference that is spinning with the ellipsoid, each Lagrangian fluid element inside as well as on the surface of the ellipsoid moves in a retro-grade direction along an elliptical trajectory that has the same ellipticity as does the equatorial surface of the 3D ellipsoid. Each and every Lagrangian fluid element completes one full elliptical orbit in a time, <math>~P_\mathrm{Lagrange} = \lambda_\mathrm{EFE}^{-1} = [0.0799]^{-1} = 12.52</math> wall-clock hours, which is about 50 seconds of real time.
</ol>

In the COLLADA-based code (<b>Inertial17.dae</b>) that has been provided above in the scrollable box, we have instructed the animation sequence to follow the system's evolution through five complete spins of the purple ellipsoid. (Unless the ''pause'' button is clicked, the animation will actually loop repeatedly, but the time on the wall-mounted clock will reset to 0<sup>h</sup>0<sup>m</sup> after every fifth spin period.) In practice, this animation sequence lasts (5 × 1.825) × 4 = 36.5 seconds of real time.

Figure 2 displays a pair of still-frame images of our example (purple) ellipsoidal configuration after the ellipsoid has completed precisely five (counter-clockwise) spin cycles. (The snapshots have been taken at the same point in time, but from two different camera viewing angles.) In the left-hand image (Figure 2a), the time on the clock appears to be about 9<sup>h</sup>8<sup>m</sup>. This means that as the ellipsoid has completed five spin cycles, the "minute" hand of the clock has completed approximately [9 + 8/60] ≈ 9.13 cycles. Dividing this number by 5 gives a value of <math>~P_\mathrm{spin} \approx 1.827</math>, which is indeed the spin period we identified earlier. This illustrates that the proper physical timescales governing oscillations of the wall-mounted clock and of the spinning ellipsoid have been implemented in a quantitatively accurate manner within the COLLADA code.

<div align="center">
<table border="1" align="center" cellpadding="8">
<tr>
<th align="center">Figure 2a</th>
<th align="center">Figure 2b</th>
</tr>
<tr>
<td align="left" bgcolor="lightgrey">
[[File:COLLADA1stViewpoint.png|300px|EFE Model b41c385]]
</td>
<td align="left" bgcolor="lightgrey">
[[File:COLLADA4thViewpoint.png|300px|EFE Model b41c385]]
</td>
</tr>
</table>
</div>

In Figure 1b, above, we have used a small red cube to mark the Lagrangian fluid element that is initially (TIME = 0.0) located at the end of the longest axis of the ellipsoid (in the +X direction). In Figure 2b we can see where this fluid element is located after the ellipsoid has completed five complete spins. After moving in a retrograde (clockwise) direction along its assigned elliptical trajectory while the ellipsoid completed five full spins, we see that the red-tagged Lagrangian fluid element has completed a little less than 3/4 of one full orbit; let's call it 73% of one orbit. We therefore infer from the COLLADA-driven animation that <math>~0.73 P_\mathrm{Lagrange}\approx~5 P_\mathrm{spin}</math>, that is, <math>~P_\mathrm{Lagrange}\approx 6.849 P_\mathrm{spin}</math>. This is indeed a proper reflection of the physical properties of our chosen example Riemann S-type ellipsoid, as its mathematically determined equilibrium frequency ratio is, <math>~|\Omega_\mathrm{EFE}|/|\lambda_\mathrm{EFE}| = 6.8582</math>. This, too, illustrates that the proper physical timescales governing oscillations of our chosen example Riemann ellipsoid have been implemented in a quantitatively accurate manner within the COLLADA code.

===Rotating Reference Frame===

<table border="0" align="right" cellpadding="8"><tr><td align="center">
<table border="0" align="center"><tr><td align="center" bgcolor="lightgray"> <br /><b>Figure 3</b><br />
[[File:COLLADA1stRotatingView.png|300px|Rotating Frame at TIME zero]]
</td></tr></table>
</td></tr></table>
The following scrollable box titled, <b>MultiLagrange18.dae</b>, contains another <xml>-formatted [https://www.khronos.org/collada/ COLLADA] code. It is identical in most respects to the COLLADA code named <b>Inertial17.dae</b> that has been discussed above in considerable detail. But there are two key differences:
<ol type="A">
<li>As Figure 3 illustrates, while the surface of the ellipsoid is still colored purple, it has been made partially transparent to reveal the location of three — rather than only one — Lagrangian fluid elements. All three are initially positioned along the plus-X<sub>0</sub> axis, but two are located inside, rather than on the surface of, the ellipsoid.</li>
<li>This new, "MultiLagrange" COLLADA animation presents the system's evolution as viewed from a frame of reference that is spinning with the frequency of the ellipsoid, Ω<sub>EFE</sub>. As before, the COLLADA-driven animation is carried through five full ellipsoid spins, but from the perspective of this new rotating frame, the ellipsoid remains stationary and it is the components of the "laboratory" that actually spin; and they spin ''clockwise'' rather than counter-clockwise.</li>
</ol>

From the point of view of this rotating reference frame, it is much easier to watch and diagnose the behavior of the fluid motion interior to the surface of the ellipsoid. Note, in particular, the relative speed and alignment of the three fluid elements at different points along their coordinated orbits. Because Riemann S-type ellipsoids are uniform-density configurations and they are composed of an incompressible fluid, at all points in time and in space (within the volume of the ellipsoid) the divergence of the velocity must be zero.

<div align="center" id="Scroll2">
<b>MultiLagrange18.dae</b>     …     a COLLADA code containing 7994 lines of <xml>-formatted ASCII text
</div>

<div style="height: 400px; width: 100%; overflow: scroll;">
<pre>
<?xml version="1.0" encoding="UTF-8" standalone="no" ?>
<COLLADA xmlns="http://www.collada.org/2005/11/COLLADASchema" version="1.4.1">
<asset>
<contributor>
<author>jet53man</author>
<authoring_tool>Raw XML</authoring_tool>
</contributor>
<created>2020-04-28</created>
<modified>2020-05-10</modified>
<unit meter="0.0254" name="inch" />
<up_axis>Z_UP</up_axis>

</asset>
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</profile_COMMON>
</effect>
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<profile_COMMON>
<technique sid="COMMON">
<lambert>
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<input semantic="INPUT" source="#First_rotation_euler_Z-input"/>
<input semantic="OUTPUT" source="#First_rotation_euler_Z-output"/>
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<sampler id="Frame_rotation_euler_Z-sampler">
<input semantic="INPUT" source="#First_rotation_euler_Z-input"/>
<input semantic="OUTPUT" source="#Frame_rotation_euler_Z-output"/>
</sampler>
<channel source="#First_rotation_euler_Z-sampler" target="EntireEllipsoid/rotationZ.ANGLE"/>
<channel source="#Frame_rotation_euler_Z-sampler" target="Inertial/rotationZ.ANGLE"/>
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<param name="TIME" type="float"/>
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<channel source="#Lagrange1_location_Y-sampler" target="Lagrange01/location.Y"/>
<channel source="#Lagrange1_location_X-sampler" target="Lagrange02/location.X"/>
<channel source="#Lagrange1_location_Y-sampler" target="Lagrange02/location.Y"/>
<channel source="#Lagrange1_location_X-sampler" target="Lagrange03/location.X"/>
<channel source="#Lagrange1_location_Y-sampler" target="Lagrange03/location.Y"/>
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0.000 0.000 -1.00 0.00 -0.3827 0.9239 0.000 0.000 0.9239 0.3827 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.7071 0.7071 0.000 0.000 0.7071 0.7071 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.9239 0.3827 0.000 0.000 0.3827 0.9239 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -1.0000 0.0000 0.000 0.000 0.0000 1.0000 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.9239 -0.3827 0.000 0.000 -0.3827 0.9239 0.000 0.00 0.000 0.000 0.000 1.000
0.000 0.000 -1.00 0.00 -0.7071 -0.7071 0.000 0.000 -0.7071 0.7071 0.000 0.00 0.000 0.000 0.000 1.000
</float_array>
<technique_common>
<accessor source="#Cube_quaternion_Z-output-array" count=" 147" stride="16">
<param name="TRANSFORM" type="float4x4"/>
</accessor>
</technique_common>
</source>
<sampler id="Cube_quaternion_Z-sampler">
<input semantic="INPUT" source="#Cube_quaternion_Z-input"/>
<input semantic="OUTPUT" source="#Cube_quaternion_Z-output"/>
</sampler>
<channel source="#Cube_quaternion_Z-sampler" target="Jetman/transformation"/>
</animation>

<animation id="HourHand_euler_Z">
<source id="HourHand_euler_Z-input">
<float_array id="HourHand_euler_Z-input-array" count=" 147">
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75
4.00 4.25 4.50 4.75 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75
8.00 8.25 8.50 8.75 9.00 9.25 9.50 9.75 10.00 10.25 10.50 10.75 11.00 11.25 11.50 11.75
12.00 12.25 12.50 12.75 13.00 13.25 13.50 13.75 14.00 14.25 14.50 14.75 15.00 15.25 15.50 15.75
16.00 16.25 16.50 16.75 17.00 17.25 17.50 17.75 18.00 18.25 18.50 18.75 19.00 19.25 19.50 19.75
20.00 20.25 20.50 20.75 21.00 21.25 21.50 21.75 22.00 22.25 22.50 22.75 23.00 23.25 23.50 23.75
24.00 24.25 24.50 24.75 25.00 25.25 25.50 25.75 26.00 26.25 26.50 26.75 27.00 27.25 27.50 27.75
28.00 28.25 28.50 28.75 29.00 29.25 29.50 29.75 30.00 30.25 30.50 30.75 31.00 31.25 31.50 31.75
32.00 32.25 32.50 32.75 33.00 33.25 33.50 33.75 34.00 34.25 34.50 34.75 35.00 35.25 35.50 35.75
36.00 36.25 36.50
</float_array>
<technique_common>
<accessor source="#HourHand_euler_Z-input-array" count=" 147" stride="1">
<param name="TIME" type="float"/>
</accessor>
</technique_common>
</source>
<source id="HourHand_euler_Z-output">
<float_array id="HourHand_euler_Z-output-array" count=" 147">
270.0
268.125
266.25
264.375
262.5
260.625
258.75
256.875
255.0
253.125
251.25
249.375
247.5
245.625
243.75
241.875
240.0
238.125
236.25
234.375
232.5
230.625
228.75
226.875
225.0
223.125
221.25
219.375
217.5
215.625
213.75
211.875
210.0
208.125
206.25
204.375
202.5
200.625
198.75
196.875
195.0
193.125
191.25
189.375
187.5
185.625
183.75
181.875
180.0
178.125
176.25
174.375
172.5
170.625
168.75
166.875
165.0
163.125
161.25
159.375
157.5
155.625
153.75
151.875
150.0
148.125
146.25
144.375
142.5
140.625
138.75
136.875
135.0
133.125
131.25
129.375
127.5
125.625
123.75
121.875
120.0
118.125
116.25
114.375
112.5
110.625
108.75
106.875
105.0
103.125
101.25
99.375
97.5
95.625
93.75
91.875
90.0
88.125
86.25
84.375
82.5
80.625
78.75
76.875
75.0
73.125
71.25
69.375
67.5
65.625
63.75
61.875
60.0
58.125
56.25
54.375
52.5
50.625
48.75
46.875
45.0
43.125
41.25
39.375
37.5
35.625
33.75
31.875
30.0
28.125
26.25
24.375
22.5
20.625
18.75
16.875
15.0
13.125
11.25
9.375
7.5
5.625
3.75
1.875
0.0
-1.875
-3.75
</float_array>
<technique_common>
<accessor source="#HourHand_euler_Z-output-array" count=" 147" stride="1">
<param name="ANGLE" type="float"/>
</accessor>
</technique_common>
</source>
<sampler id="HourHand_euler_Z-sampler">
<input semantic="INPUT" source="#HourHand_euler_Z-input"/>
<input semantic="OUTPUT" source="#HourHand_euler_Z-output"/>
</sampler>
<channel source="#HourHand_euler_Z-sampler" target="HourHand/rotationZ.ANGLE"/>
</animation>
</library_animations>
<library_controllers/>
<library_visual_scenes>
<visual_scene id="ID1">
<node id="EntireEllipsoid" name="OculusRift_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 5.0</rotate>
<scale sid="scale">5.0 5.0 5.0</scale>
<node id="First" name="First_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Second" name="Second_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 180.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Third" name="Third_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 180.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>
<node id="Fourth" name="Fourth_instance_EFE">
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 180.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 180.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.0 1.0 1.0</scale>
<instance_node url="#ID3_EFE" />
</node>

<node id="CylinderPointer" name="instance_cylinder">
<translate sid="location">-1.0 0.0 0.0</translate>
<rotate sid="rotationX">1 0 0 0</rotate>
<rotate sid="rotationY">0 1 0 0</rotate>
<rotate sid="rotationZ">0 0 1 180</rotate>
<scale sid="scale">0.05 0.025 0.025</scale>
<instance_node url="#CylPointer" />
</node>

<node id="Lagrange01" name="instance_Lagrange01">
<translate sid="location">1.0000 0.0000 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">0.06 0.06 0.06</scale>
<instance_node url="#Primary_MarkerID3" />
</node>
<node id="Shrink02" name="instance_Shrink02">
<scale sid="scale">0.8 0.8 0.8</scale>
<node id="Lagrange02" name="instance_Lagrange02">
<translate sid="location">1.0000 0.0000 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">0.075 0.075 0.075</scale>
<instance_node url="#Primary_MarkerID3" />
</node>
</node>
<node id="Shrink03" name="instance_Shrink03">
<scale sid="scale">0.6 0.6 0.6</scale>
<node id="Lagrange03" name="instance_Lagrange03">
<translate sid="location">1.0000 0.0000 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">0.1 0.1 0.1</scale>
<instance_node url="#Primary_MarkerID3" />
</node>
</node>
<node id="Marker01" name="instance_marker01">
<matrix>0.025 0.0 0.0 1.0000 0.0 0.025 0.0 -0.0000 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker02" name="instance_marker02">
<matrix>0.025 0.0 0.0 0.9921 0.0 0.025 0.0 -0.0514 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker03" name="instance_marker03">
<matrix>0.025 0.0 0.0 0.9686 0.0 0.025 0.0 -0.1020 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker04" name="instance_marker04">
<matrix>0.025 0.0 0.0 0.9298 0.0 0.025 0.0 -0.1509 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker05" name="instance_marker05">
<matrix>0.025 0.0 0.0 0.8763 0.0 0.025 0.0 -0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker06" name="instance_marker06">
<matrix>0.025 0.0 0.0 0.8090 0.0 0.025 0.0 -0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker07" name="instance_marker07">
<matrix>0.025 0.0 0.0 0.7290 0.0 0.025 0.0 -0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker08" name="instance_marker08">
<matrix>0.025 0.0 0.0 0.6374 0.0 0.025 0.0 -0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker09" name="instance_marker09">
<matrix>0.025 0.0 0.0 0.5358 0.0 0.025 0.0 -0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker10" name="instance_marker10">
<matrix>0.025 0.0 0.0 0.4258 0.0 0.025 0.0 -0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker11" name="instance_marker11">
<matrix>0.025 0.0 0.0 0.3090 0.0 0.025 0.0 -0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker12" name="instance_marker12">
<matrix>0.025 0.0 0.0 0.1874 0.0 0.025 0.0 -0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker13" name="instance_marker13">
<matrix>0.025 0.0 0.0 0.0628 0.0 0.025 0.0 -0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker14" name="instance_marker14">
<matrix>0.025 0.0 0.0 -0.0628 0.0 0.025 0.0 -0.4092 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker15" name="instance_marker15">
<matrix>0.025 0.0 0.0 -0.1874 0.0 0.025 0.0 -0.4027 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker16" name="instance_marker16">
<matrix>0.025 0.0 0.0 -0.3090 0.0 0.025 0.0 -0.3899 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker17" name="instance_marker17">
<matrix>0.025 0.0 0.0 -0.4258 0.0 0.025 0.0 -0.3710 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker18" name="instance_marker18">
<matrix>0.025 0.0 0.0 -0.5358 0.0 0.025 0.0 -0.3462 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker19" name="instance_marker19">
<matrix>0.025 0.0 0.0 -0.6374 0.0 0.025 0.0 -0.3159 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
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<node id="Marker20" name="instance_marker20">
<matrix>0.025 0.0 0.0 -0.7290 0.0 0.025 0.0 -0.2807 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker21" name="instance_marker21">
<matrix>0.025 0.0 0.0 -0.8090 0.0 0.025 0.0 -0.2410 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker22" name="instance_marker22">
<matrix>0.025 0.0 0.0 -0.8763 0.0 0.025 0.0 -0.1975 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker23" name="instance_marker23">
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<instance_node url="#MarkersID3" />
</node>
<node id="Marker24" name="instance_marker24">
<matrix>0.025 0.0 0.0 -0.9686 0.0 0.025 0.0 -0.1020 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker25" name="instance_marker25">
<matrix>0.025 0.0 0.0 -0.9921 0.0 0.025 0.0 -0.0514 0.0 0.0 0.025 0.0000 0.0 0.0 0.0 1.0</matrix>
<instance_node url="#MarkersID3" />
</node>
<node id="Marker26" name="instance_marker26">
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<instance_node url="#MarkersID3" />
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<node id="Marker28" name="instance_marker28">
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<node id="Marker30" name="instance_marker30">
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<instance_node url="#MarkersID3" />
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<node id="Marker31" name="instance_marker31">
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<instance_node url="#MarkersID3" />
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<node id="Marker32" name="instance_marker32">
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<instance_node url="#MarkersID3" />
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<node id="Marker33" name="instance_marker33">
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<instance_node url="#MarkersID3" />
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<node id="Marker34" name="instance_marker34">
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<instance_node url="#LettersID3" />
</node>
</node>

</node>

<node id="EntireClock" name="instance_Clock">
<translate>7.0 0.0 5.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 0.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.00 1.00 1.00</scale>
<node id="firstclock" name="instance_1"> 
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 -90.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1.0 0.0</rotate>
<scale sid="scale">1.00 1.00 1.00</scale>
<instance_node url="#ID3" />
</node> 
<node id="Jetman" name="instance_Arrow"> 
<matrix sid="transformation"> 0.000 0.000 -1.00 0.00 -1.0000 0.0000 0.000 0.000 0.0000 1.0000 0.000 0.00 0.000 0.000 0.000 1.000</matrix>
<instance_node url="#ArrowID3" />
</node> 
<node id="HourHand" name="instance_Hour"> 
<translate>0.0 0.0 0.0</translate>
<rotate sid="rotationX">1.0 0.0 0.0 0.0</rotate>
<rotate sid="rotationY">0.0 1.0 0.0 -90.0</rotate>
<rotate sid="rotationZ">0.0 0.0 1 270.0</rotate>
<scale sid="scale">0.75 0.75 0.75</scale>
<instance_node url="#HourID3" />
</node> 
</node>

</node> 
</visual_scene>
</library_visual_scenes>
<scene>
<instance_visual_scene url="#ID1" />
</scene>
</COLLADA>

</pre>
</div>

==Inside the Oculus Rift S Environment==

===Introduction===

The [https://www.oculus.com/rift-s/ Oculus Rift S] will import user-constructed 3D objects and allow you to place those objects in a high-quality, interactive virtual reality (VR) environment. But the Oculus system does not accept COLLADA-formatted files directly. Instead, it is set up to import files that are in a ".glb" format, which is the binary version of a "GL Transmission Format" ([https://www.khronos.org/gltf/ glTF]) file. My original expectation was that, after I learned how to construct some intricate and astrophysically interesting 3D visual scenes using COLLADA — successfully accomplished (!) as illustrated above — all I would need to do is find a community tool that could translate a COLLADA (.dae) file into glTF (.glb).

Here are a few discoveries I made along the way:
<ul>
<li>The Windows 10 startup menu includes a "3D Viewer" app but, at least on my Lenova PC, it does not recognize .dae-formatted files.</li>
<li>The Windows 10 startup menu includes a "Paint 3D" application. Using this app, it was very easy to "draw" some simple 3D geometric objects — such as a cube or a torus — and it was straightforward to export each object to a .glb-formatted file.</li>
<li>[https://www.adobe.com/products/photoshop.html Adobe Photoshop] can read in a COLLADA (.dae) file and it is capable of exporting a 3D scene in .glb format.</li>
<li>[https://blackthread.io BlackThread.io] offers a browser-based tool, named "glTF Converter", that can ostensibly translate .dae-formatted files into .glb-formatted files.</li>
<li>The [https://www.khronos.org Khronos Group]'s "COLLADA to glTF" conversion tool can be located under [https://github.com/KhronosGroup/COLLADA2GLTF/ the group's name on GitHub].</li>
</ul>

===Khronos Group's COLLADA2GLTF Converter===

Among this group of potentially useful tools, we have had — by far — the most success/luck using the <font color="red">Khronos Group's COLLADA2GLTF</font> converter. We downloaded and installed version 2 of this Khronos Group conversion tool on our Mac Pro. The executable, named <b>COLLADA2GLTF-bin</b>, was installed on our Mac at the end of a directory tree named …/Dropbox/Lenovo/COLLADA_to_gltf/COLLADA2GLTF-v2/. In order to translate a specific COLLADA-formatted (''filename.dae) file:
<ul>
<li>From inside the Mac's "Terminal" application, we changed directories until we were sitting in the above-specified Mac (Dropbox) directory</li>
<li>We moved (or copied) the chosen COLLADA-formatted file (<b>''filename''.dae</b>) into the same directory
<li>At the terminal prompt, we typed the line command:   <b>./COLLADA2GLTF-bin -i ''filename''.dae -b</b></li>
<li>The .glb-formatted output file was stored as <b>…/Dropbox/Lenovo/COLLADA_to_gltf/COLLADA2GLTF-v2/output/''filename''.glb</b></li>
</ul>

<table border="1" cellpadding="8" align="center">
<tr>
<td align="center">Example Mac-Pro Line Commands + Responses</td>
</tr>
<tr><td align="left">
<pre>
command: pwd
response: ../Dropbox/Lenovo/COLLADA_to_gltf/COLLADA2GLTF-v2
command: ls *.dae
response: Inertial17.dae
command: ./COLLADA2GLTF-bin -i Inertial17.dae -b
response: Converting Inertial17.dae -> output/Inertial17.glb
Time: 48.166 ms
</pre>
</td></tr>
</table>

<font color="red">CAUTION!</font> While we found this COLLADA2GLTF converter to be a very useful tool, we also discovered that porting a model from one visualization environment to another can be a challenging and frustrating endeavor. It is difficult enough to build a (error-free) COLLADA-formatted model that will properly display a desired animated scene in the Mac's Preview app. But it is naive to expect that that same COLLADA code will survive a translation to glTF and seamlessly construct the identical animated scene within the Oculus Rift S environment. In a pair of accompanying chapters — [[Appendix/Ramblings/OculusRiftS|this one]], followed by [[Appendix/Ramblings/RiemannMeetsOculus#Rebuild_Riemann_S-Type_Ellipsoid_Inside_Oculus_Rift_S_Environment|another one]] — we have chronicled the steps that we have taken to generate our first few (error-free) COLLADA-based models that display properly in both environments. Experience has taught us that … (1) When interpreting COLLADA's <xml>-formatted instructions, the Mac's Preview app is considerably more forgiving than is the Oculus system; (2) almost any code that display's properly in the Mac's Preview app will survive the translation to glTF, in the sense that the COLLADA2GLTF converter seldom complains; but, (3) nevertheless, frequently the Oculus system won't display the model because it detects an error(s), but it almost never suggests what the error(s) might be. When embarking on the development of a new interactive and animated 3D visual scene, the best strategy is a time-consuming one: Start with a COLLADA code that works in both environments — we provide two such codes herein; make only incremental changes and ensure that the model displays properly in the Mac's Preview app; convert the COLLADA-based code to glTF (.glb); don the Oculus Rift S headset & hand controllers, and attempt to import this new model into your "Oculus Home".

===Import to Oculus Rift S Environment===

We purchased a Lenovo PC (Windows 10) gaming laptop to drive the Oculus Rift S system; more specifically, it is Lenovo's Legion Y740 15" with NVIDIA GeForce GTX 1660Ti 6GB. Two connections were required: (1) Mini DisplayPort; and (2) USB. We downloaded and installed the Oculus program files under the directory <b>C:\Program Files\Oculus</b>. We also mounted Dropbox on the Lenovo, which gave us immediate access to the model files that were created using the Mac Pro. As an example, we located the model file named <b>Inertial17.glb</b> and placed a copy of the file in the directory <b>… This PC\Documents\Oculus Home\_Import</b>. Any (error free) ".glb"-formatted file that is located in this directory is available for import into the Oculus' 3D virtual-reality environment.

When we want to view one of our models, we first start the Oculus application on the Lenovo then click on "Devices" to check whether the headset and two hand controllers are in proper communication with the PC (GREEN buttons) or not (RED buttons). We don the headset & controllers, thereby entering the Oculus Rift S VR environment. [Occasionally we forget to start the Oculus app before putting on the headset. When this happens, the app seems to sense that we have donned the headset and it starts itself!] The following sequence of steps allows us to place the example model, <b>Inertial17</b>, into our ''Oculus Home''.

<table border="1" align="left" width="100%" cellpadding="8">
<tr>
<th align="center" colspan="2"><font size="+1">Inside the ''Oculus Rift S'' Home Environment</font></th>
</tr>
<tr>
<td align="center">[[File:HomeImage01sm.png|200px|Oculus Home (view 1)]]</td>
<td align="left">The image shown here, immediately to the left, is a snapshot of the first scene that is displayed in our ''Oculus Home'' shortly after we have donned the headset and the pair of hand controllers. The shallow, curved desk immediately in front of us presents ten or so clickable panels (menu choices). In the image, we have highlighted the "Home" button by pointing to it with the index finger of our right hand.
<ul>
<li><font color="red">Click</font> on the "Home" button; the interactive panels in this scene will disappear.</li>
<li>Wait (~ 10 seconds) and our stored "Home" environment will load.</li>
<li>"Menu" button vs. "Oculus" button</li>
</ul>
</td>
</tr>
<tr>
<td align="center">[[File:HomeImage02sm.mod.png|200px|Oculus Home (view 2)]]</td>
<td align="left">
<ul>
<li><font color="red">Click</font> the "Menu" button on the left-hand controller; it sits just below the "Y" and "X" buttons;</li>
</ul>

The image shown here, immediately to the left, is a snapshot of a portion of the scene that is displayed after we have clicked the "Menu" button on our left-hand controller. As illustrated, we have pointed to — and thereby have highlighted in blue — the top menu item, "<font color="blue">Places</font>," and we are in the process of sliding our aim down to the next menu item, "Inventory."
<ul>
<li><font color="red">Click</font> the "Inventory" menu item.</li>
</ul>
</td>
</tr>
<tr>
<td align="center">[[File:HomeImage03sm.mod.png|200px|Oculus Home (view 3)]]</td>
<td align="left">
As is shown here, immediately to the left, the menu panel is updated after the "Inventory" button has been clicked, as directed above.
<ul>
<li><font color="red">Click</font> the "My Imports" menu item.</li>
</ul>

After highlighting then clicking "<font color="blue">My Imports</font>," a three-dimensional display case (a shadow box) containing up to 4 × 3 objects appears, as is illustrated here in the screenshot on the left. Each object in this display case is generated from one of the (error free) ".glb"-formatted files that is located in the aforementioned Lenovo directory <b>… This PC\Documents\Oculus Home\_Import</b> — including files that reside in subdirectories of this parent _Import directory.
</td>
</tr>
<tr>
<td align="left" colspan="2">
NOTE:
<ul>
<li>The display case can hold more than twelve objects, but it will only display twelve at a time. In order to see more, you need to point to (any) one of the displayed objects then use the hand-controller's toggle button to scroll the shelves of the display case up or down.</li>
<li>If Oculus sees a (error free) ".glb" file that it has never seen before — for example, a new model file that you have just placed in the PC's <b>…\_Import</b> directory — it will display a panel asking you to "<font color="blue">Confirm</font>" that you have permission to load this material.</li>
<li>If the Oculus determines that one of your ".glb" files contains an error, it may try to display the faulty object in the display case and attach some sort of warning message. More often than not, however, it will simply ignore the file and leave you wondering why the object does not appear in the display case.
</ul>
</td>
</tr>
<tr>
<td align="center">[[File:HomeImage04sm.mod.png|200px|Oculus Home (view 4)]]</td>
<td align="left">
<ul>
<li>Point at an object in the display case, then <font color="red">Grab</font> it (using the controller's middle-finger "grip" switch) and place it somewhere in your ''Home'' room.</li>
</ul>
In the image shown here, immediately on the left, we are pointing at, and are about to "grab", the model we have named, <b>MultiLagrange18</b>. We could have, just as well, pointed at and grabbed the model sitting on the shelf next to it named, <b>Inertial17</b>.

<br />
It is worth mentioning that after you have ''grabbed'' an object — and before you have placed it in the room — you can do several things with it.
<ol><li>Using the hand controller's ''thumbstick'', you can move the object farther from (or closer to) you by pushing the thumbstick away from you (or by pulling it toward you).</li>
<li>Using the controller's index-finger ''trigger'', you can toggle between a mode that lets you spin the object and a mode that let's you resize it. In either mode, the direction or amplitude of the spin/resizing is determined by pushing the ''thumbstick'' to the left or right varying amounts.</li>
</ol>
</td>
</tr>
<tr>
<td align="center">[[File:HomeImage05sm.png|200px|Oculus Home (view 5)]]</td>
<td align="left">
The image shown here, immediately on the left, is a snapshot of our Oculus ''Home'' shortly after our model named, <b>Inertial17</b>, has been resized (enlarged) then placed in the room.
</td>
</tr>
<tr>
<td align="left" colspan="2">
NOTE: The steps required to remove a model from the "<font color="blue">My Imports</font>" Inventory — that is, from the display case — are not easy to guess. Here we identify what appears to be the required sequence of steps, using our model, <b>MultiLagrange18</b>, as an example.
<ul>
<li>Assuming the <b>MultiLagrange18</b> model is one of the objects that is on display in the ''Oculus Home'' room, grab it with the right-hand controller and use the index-finger "trigger" to toggle to the mode that lets you spin the object (the resize mode does not seem to work for this task).</li>
<li>Press the "B" button; this should return the object to the "My Imports" display case.</li>
<li>Point your index finger at the compartment on the "My Imports" display case that contains the (miniaturized) <b>MultiLagrange18</b> model, then use your thumb to press the "B" controller button. [Be careful not to squeeze the "grip" button with your middle finger instead, as you will end up unintentionally grabbing the object again.] A small panel containing bibliographical information about the <b>MultiLagrange18</b> file will appear immediately in front of your face; an even smaller accompanying panel will ask if you want to "Remove Upload", that is, do you want to remove the file from the inventory.</li>
<li>Point your index finger at the smaller of these two panels and click on it (using the middle finger's "grip" button) to indicate you want the file removed. The panels will disappear but, confusingly, the inventory display box will not change in any way to signify that your request is being implemented.</li>
<li>Remove the headset; sit down in front of the Lenovo PC; locate the <b>MultiLagrange18.glb</b> that has been stored in the Oculus' _Import directory and trash it!

The next time you don the Oculus Rift S headset and look at the "My Imports" display case, the <b>MultiLagrange18</b> will no lonber be among the group of displayed objects.
</ul>
</td>
</tr>
</table>

<p> </p>

<table border="1" align="center" cellpadding="0" width="710px" cellpadding="5">
<tr>
<th align="center" colspan="2"><font size="+1">Additional Photos from Inside the ''Oculus Rift S'' Home Environment</font></th>
</tr>
<tr>
<td align="center">[[File:B41c385 Oculus03.jpg|350px|Third Oculus Rift Selfie]]</td>
<td align="center">[[File:B41c385 Oculus02.jpg|350px|Second Oculus Rift Selfie]]</td>
</tr>
<tr>
<td align="center" colspan="2">[[File:B41c385 Oculus01.jpg|700px|Third Oculus Rift Selfie]]</td>
</tr>
<tr>
<td align="left" colspan="2">
Two of our COLLADA-driven animated models can be identified in each of the three frames of this image montage: In all three cases, the one on the right is model <b>Inertial17</b> and the one on the left is model <b>MultiLagrange18</b>. The bottom (selfie) visual scene illustrates that we can add to our Oculus home environment other items that are straightforwardly selected from a large Oculus objects inventory.
</td>
</tr>
</table>

==Footnotes==

<sup>&Dagger;</sup>Adopting the characteristic mass <math>~(M_\mathrm{Ch} = \mu_e^{-2} \cdot 1.14205 \times 10^{34} ~\mathrm{g})</math> and characteristic length <math>~(\ell_1 = \mu_e^{-1} \cdot 7.71395 \times 10^{8} ~\mathrm{cm})</math> that are associated with a [[SSC/Structure/WhiteDwarfs#White_Dwarfs|white dwarf]] equation of state, we can reasonably state that the characteristic mean density of a spherically symmetric white dwarf is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~{\bar\rho}_\mathrm{WD}</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
\frac{3 M_\mathrm{Ch}}{4\pi \ell_1^3} = \mu_e \cdot 5.93972 \times 10^{6}~\mathrm{g}~{\mathrm{cm}}^{-3} \, .
</math>
</td>
</tr>
</table>

Drawing from our [[#Animation|above discussion of timescales that arise in the context of our COLLADA-driven animations]], we therefore appreciate that the characteristic oscillation period associated with a white dwarf whose molecular weight of electrons is <math>~\mu_e = 2</math>, proves to be,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~[P_0]_\mathrm{WD}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{4\pi}{G{\bar\rho}_\mathrm{WD}} \biggr]^{1 / 2} = 3.98~\mathrm{s}\, .
</math>
</td>
</tr>
</table>

By presenting animation scenes in which one wall-clock "hour" equates to approximately 4 seconds of real time, we can reasonably assert that our animated models fairly represent how a Riemann-ellipsoid-like white dwarf should behave in real time.

=See Also=

* Discussion of [[ThreeDimensionalConfigurations/RiemannStype|Ou's Riemann-Like Ellipsoids]]

* [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS|Riemann Meets COLLADA & Oculus Rift S]]: Example <b>(b/a, c/a) = (0.41, 0.385)</b>
** [[Appendix/Ramblings/VirtualReality#Virtual_Reality_and_3D_Printing|Virtual Reality and 3D Printing]]
** [[Appendix/Ramblings/OculusRiftS|Success Importing Animated Scene into Oculus Rift S]]
** [[Appendix/Ramblings/RiemannMeetsOculus|Carefully (Re)Build Riemann Type S Ellipsoids Inside Oculus Rift Environment]]
** Other Example S-type Riemann Ellipsoids:
*** <b>[[Appendix/Ramblings/RiemannB90C333|(b/a, c/a) = (0.90, 0.333)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB74C692|(b/a, c/a) = (0.74, 0.692)]]</b>
*** <b>[[Appendix/Ramblings/RiemannB28C256|(b/a, c/a) = (0.28, 0.256)]]</b>

{{ SGFfooter }}

Appendix/Ramblings/PowerSeriesExpressions

2024-06-28T23:46:57Z

Jet53man: /* Broadly Used Mathematical Expressions (shown here without proof) */

__FORCETOC__

=Approximate Power-Series Expressions=

==Broadly Used Mathematical Expressions (shown here without proof)==
===Binomial===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1 \pm x)^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2
~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4
~~\pm ~~ \cdots
</math>
     for <math>~(x^2 < 1)</math>
</td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
<th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
</th>
</tr>
<tr>
<td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 §1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(a+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</td>
</tr>
</table>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\binom{n}{k}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</td>
</tr>
</table>

----

'''Our Example:'''  Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+b)^{n}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
</math>
</td>
</tr>
</table>

</td>
</tr>
</table>

Note, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~(1+x)^{-1}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-3}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~(1+x)^{-4}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
</td>
</tr>
</table>

See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]

===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
</math>
</td>
</tr>
</table>
</div>

==Expressions with Astrophysical Relevance==
===Polytropic Lane-Emden Function===
We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ Math/EQ_SSLaneEmden01 }}
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>
</table>
</div>

First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
</math>
</td>
</tr>
</table>
</div>

Left-hand-side of Lane-Emden equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots
</math>
</td>
</tr>
</table>
</div>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively):
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3
+ \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4
~~+ ~~ \cdots
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~F</math>
</td>
<td align="center">
<math>~\equiv</math>
</td>
<td align="left">
<math>~
a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .
</math>
</td>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\Theta_H^n</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</td>
</tr>
</table>
</div>
Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~2\cdot 3 b</math>
</td>
<td align="center">
<math>~-1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3 c</math>
</td>
<td align="center">
<math>~-na</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~2^2\cdot 5 d</math>
</td>
<td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>
</td>
</tr>
</table>
</div>

By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds:

* Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged.
* The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>.
* The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>~h</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:
<div align="center" id="PolytropicLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

===Isothermal Lane-Emden Function===



Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the [[SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~e^{-w} \, . </math>
</td>
</tr>
</table>
</div>

A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</td>
</tr>
</table>
</div>

Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, .
</math>
</td>
</tr>
</table>
</div>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e)
+ r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots
</math>
</td>
</tr>
</table>
</div>

Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr]
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
\times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr]
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl\{
\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24}
- dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\biggl[
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
\biggr]
\times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr)
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr]
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
+ \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr]
+ bcr^5\biggl[1 -ar \biggr]
+ r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr)
</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~2a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~6b</math>
</td>
<td align="center">
<math>~1</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~2^2\cdot 3c</math>
</td>
<td align="center">
<math>~-a</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
</td>
<td align="center">
<math>~\frac{a^2}{2} - b</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
</td>
</tr>
</table>
</div>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~e^{-w}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
1 -d r^4 -e r^5 -f r^6
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</td>
</tr>
</table>
</div>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{3}:</math>
</td>
<td align="center">
<math>~30e</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~e=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{4}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
</td>
<td align="center">
<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>
</td>
<td align="left">
<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{5}:</math>
</td>
<td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
</td>
<td align="center">
<math>~-e</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~g = 0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{6}:</math>
</td>
<td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
</td>
<td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr)
= -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7}
</math>
</td>
</tr>
</table>
</div>

Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center">
<tr><th align="center">For Spherically Symmetric Configurations</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

See also:
* Equation (377) from §22 in Chapter IV of [[Appendix/References#C67|C67]].

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.

===Displacement Function for Polytropic LAWE===
The [[SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} +
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\theta</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots
</math>
</td>
</tr>
</table>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
\frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr]
\, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
- (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } -
\frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl\{ \mathfrak{F}
+ 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2}
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>
</td>
</tr>
</table>
</div>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math> and, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>.

====Displacement Finite at Center====
Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr)
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~ -
(n+1) \biggl[ \mathfrak{F}
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>
</td>
</tr>
</table>
</div>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>.
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{-1}:</math>
</td>
<td align="center">
<math>~24a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{0}:</math>
</td>
<td align="center">
<math>~(12b + 48b)</math>
</td>
<td align="center">
<math>~-(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{1}:</math>
</td>
<td align="center">
<math>~[36c+72c-2a(n+3)]</math>
</td>
<td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\xi^{2}:</math>
</td>
<td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
</td>
<td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(\xi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

===Displacement Function for Isothermal LAWE===
The [[SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,
</math>
</td>
</tr>
</table>
</div>
where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~w(r)
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
</td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{dw}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
</td>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
</math>
</td>
</tr>
</table>
</div>

Let's now adopt a power-series expression for the displacement function of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\frac{d^2x}{dr^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2b + 6cr + 12dr^2 + \cdots
</math>
</td>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] \biggl( 1 + ar + br^2 + cr^3 + dr^4 \biggr) \, .
</math>
</td>
</tr>
</table>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
2b + 6cr + 12dr^2
+ 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]
- \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]
</math>
</td>
<td align="center">
<math>~\approx</math>
</td>
<td align="left">
<math>~
- \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr)
- \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr)
</math>
</td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">Term</td>
<td align="center">LHS</td>
<td align="center">RHS</td>
<td align="center">Implication</td>
</tr>

<tr>
<td align="right">
<math>~r^{-1}:</math>
</td>
<td align="center">
<math>~4a</math>
</td>
<td align="center">
<math>~0</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{0}:</math>
</td>
<td align="center">
<math>~2b + 8b</math>
</td>
<td align="center">
<math>~- \frac{\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{1}:</math>
</td>
<td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
</td>
<td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~c=0</math>
</td>
</tr>

<tr>
<td align="right">
<math>~r^{2}:</math>
</td>
<td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
</td>
<td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>
</td>
<td align="left">
<math>~\Rightarrow ~~~
28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] ~
\Rightarrow~
d = \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</td>
</tr>
</table>
</div>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x(r)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

===Maclaurin Spheroid Index Symbols===
In our accompanying discussion of the equilibrium properties of models along the [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Maclaurin spheroid sequence]], we find the "Index Symbols" expressions,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>e</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2}</math>
      (always positive).
</td>
</tr>
</table>

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations <math>(e \ll 1)</math>, and (b) in the case of an infinitesimally thin disk <math>(c/a \ll 1)</math>.

====Nearly Spherical Configurations====

On p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
</math>
      for,      
<math>
\biggl(e^2 < 1, -\tfrac{\pi}{2} < \sin^{-1}e < \tfrac{\pi}{2}\biggr) \, .
</math>
</td>
</tr>
</table>
Also, from the [[#Binomial|above binomial-theorem expression]], we have,

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>(1 - e^2)^{1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \tfrac{1}{2}e^2 + \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}\biggr]e^4
- \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}\biggr]e^6
+ \biggl[\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)(\tfrac{1}{2}-3)}{4!}\biggr]e^8
- \cdots
</math>
     for <math>~(e^4 < 1)</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
</math>
</td>
</tr>
</table>
So we can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_1 \equiv (1 - e^2)^{1 / 2}\biggl[ \frac{\sin^{-1}e}{e}\biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 - \biggl[ \frac{1}{2} \biggr]e^2 - \biggl[\frac{1}{2^3}\biggr]e^4
- \biggl[\frac{1}{2^4}\biggr]e^6
- \biggl[\frac{5}{2^7 }\biggr]e^8
- \cdots
\biggr\}
\times ~ \biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]e^8
\biggr\}
-\frac{e^2}{2}
\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]e^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{e^4}{2^3}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]e^4
\biggr\}
- \frac{e^6}{2^4}\biggl\{
1 + \biggl[\frac{1}{2\cdot 3}\biggr]e^2
\biggr\}
- \biggl[\frac{5}{2^7}\biggr]e^8
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + e^2 \biggl[ \frac{1}{2\cdot 3} - \frac{1}{2} \biggr]
+ e^4 \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3} - \frac{1}{2^3} \biggr]
+ e^6 \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7} - \frac{1}{2}\cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5}
- \frac{1}{2^3} \cdot \frac{1}{2\cdot 3} - \frac{1}{2^4}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ e^8 \biggl[\frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9} - \frac{1}{2} \cdot \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
-\frac{1}{2^3} \cdot \frac{1\cdot 3 }{2\cdot 4\cdot 5} - \frac{1}{2^4} \cdot \frac{1}{2\cdot 3} - \frac{5}{2^7}\biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{3^2 - 2\cdot 5 - 3\cdot 5}{2^3 \cdot 3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 3\cdot 5^2 - 3^2\cdot 7 - 5\cdot 7 - 3\cdot 5\cdot 7}{2^4 \cdot 3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 5^2 \cdot 7^2 - 2^2 \cdot 3^2\cdot 5^2 - 2\cdot 3^3 \cdot 7 - 2^2\cdot 3\cdot 5\cdot 7
- 3^2 \cdot 5^2 \cdot 7 }{2^7 \cdot 3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - e^2 \biggl[ \frac{1}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>
</table>

Hence,

<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>e^2 A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_1 - (1-e^2)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{2}{3} \biggr]
- e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{2}{3}
- e^2 \biggl[\frac{2}{3 \cdot 5}\biggr]
- e^4 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
- e^6 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\biggl(\frac{e^2}{2}\biggr) A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 - \mathcal{F}_1
</math>
</td>
</tr>

<tr>
<td align="right">
 

</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
e^2 \biggl[ \frac{1}{3} \biggr]
+ e^4 \biggl[\frac{2}{3 \cdot 5}\biggr]
+ e^6 \biggl[ \frac{ 2^3}{3 \cdot 5 \cdot 7}\biggr]
+ e^8 \biggl[\frac{ 2^4 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{10})
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~ A_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{2}{3}
+ e^2 \biggl[\frac{2^2}{3 \cdot 5}\biggr]
+ e^4 \biggl[ \frac{ 2^4}{3 \cdot 5 \cdot 7}\biggr]
+ e^6 \biggl[\frac{ 2^5 }{3^2 \cdot 5 \cdot 7} \biggr]
+ \mathcal{O}(e^{8}) \, .
</math>
</td>
</tr>
</table>

This looks okay, in the sense that <math>(2A_1 + A_3) = 2</math>.

====Infinitesimally Thin Axisymmetric Disk====
As <math>e \rightarrow 1</math> — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{c}{a}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(1 - e^2)^{1 / 2} \ll 1 \, .
</math>
</td>
</tr>
</table>

Recognizing as well that,

<table align="center" border="0" cellpadding="5">

<tr>
<td align="right">
<math>
\sin^{-1} e
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\cos^{-1}(1 - e^2)^{1 / 2} = \cos^{-1}\biggl(\frac{c}{a}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow ~~~ \mathcal{F}_2 \equiv \frac{\sin^{-1} e}{e^3} \cdot (1 - e^2)^{1 / 2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{c}{a}\biggr)\biggl[\cos^{-1}\biggl(\frac{c}{a}\biggr)\biggr]\biggl[ 1 - \frac{c^2}{a^2}\biggr]^{-3 / 2} \, ,
</math>
</td>
</tr>
</table>

the expressions for the pair of relevant index symbols may be rewritten as,
<table align="center" border=0 cellpadding="3">
<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\frac{A_3}{2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 \, .
</math>
</td>
</tr>
</table>

Pulling again from p. 457 of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\cos^{-1}\biggl(\frac{c}{a}\biggr)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} - \frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
      for,      
<math>
\biggl(\frac{c^2}{a^2} < 1, 0 < \cos^{-1}\biggl(\frac{c}{a}\biggr) < \pi\biggr)\, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="8"><tr><td align="left">
<div align="center"><b>LAGNIAPPE:</b></div>

According to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{1}{e} = \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{1}{2}(-\tfrac{1}{2}-1)(-\tfrac{1}{2}-2)(-\tfrac{1}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5}{2^3 \cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5\cdot 7}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{c}{a}\biggl\{1 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^2
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times ~\biggl\{
1 + \frac{1}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2}\biggl\{
1 + \biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{\biggl(\frac{c}{a}\biggr) + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\frac{1}{2} \biggl\{\biggl(\frac{c}{a}\biggr)^3 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl[\frac{3}{2^3 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^5 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
+ \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{5}{2^4 }\biggr] \biggl\{\biggl(\frac{c}{a}\biggr)^7 + \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
- ~\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2^2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3}{2^3 }\biggr] \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^4 }\biggr] \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

(continue expression simplification)

<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\frac{\sin^{-1}e}{e} = \cos^{-1}\biggl(\frac{c}{a}\biggr) \biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-1 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr) - \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3 }{2^3\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[ \frac{5}{2^4\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \frac{1}{2} \biggl(\frac{c}{a}\biggr)^3 - \biggl[\frac{1}{2^2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{5}{2^5 \cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl[\frac{3}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^5 - \biggl[\frac{1}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{5}{2^4 }\biggr] \biggl(\frac{c}{a}\biggr)^7 - \biggl[\frac{5}{2^5\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^9

- \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl\{
\biggl[\frac{1}{2\cdot 3}\biggr]
+ \frac{1}{2}
\biggr\}\biggl(\frac{c}{a}\biggr)^3
- \biggl\{
\biggl[\frac{3 }{2^3\cdot 5}\biggr]
+ \biggl[\frac{1}{2^2\cdot 3}\biggr]
+ \biggl[\frac{3}{2^3 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^5

- \biggl\{
\biggl[ \frac{5}{2^4\cdot 7}\biggr]
+ \biggl[\frac{ 3 }{2^4 \cdot 5}\biggr]
+ \biggl[\frac{1}{2^4}\biggr]\
+ \biggl[\frac{5}{2^4 }\biggr]
\biggr\}\biggl(\frac{c}{a}\biggr)^7
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl\{
\biggl[ \frac{5 \cdot 7}{2^7\cdot 3^2}\biggr]
+ \biggl[ \frac{5}{2^5 \cdot 7}\biggr]
+ \biggl[\frac{3^2 }{2^6\cdot 5}\biggr]
+ \biggl[\frac{5}{2^5\cdot 3}\biggr]
+ \biggl[\frac{ 5\cdot 7}{2^7 }\biggr]
\biggr\} \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} + \frac{\pi}{2}\biggl[\frac{1}{2}\biggr] \biggl(\frac{c}{a}\biggr)^2 + \frac{\pi}{2}\biggl[\frac{3}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \frac{\pi}{2}\biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \frac{\pi}{2}\biggl[\frac{ 5\cdot 7}{2^7 }\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- \biggl(\frac{c}{a}\biggr)
- \biggl[\frac{2}{3}\biggr] \biggl(\frac{c}{a}\biggr)^3

- \biggl[\frac{2^3 }{3\cdot 5}\biggr] \biggl(\frac{c}{a}\biggr)^5

- \biggl[ \frac{2^4}{5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^7

- \biggl[ \frac{2^7}{3^2 \cdot 5 \cdot 7}\biggr] \biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

Referring again to the [[#Binomial|above binomial-theorem expression]], we find for <math>(c^2/a^2 < 1)</math>,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl[ 1 - \frac{c^2}{a^2} \biggr]^{-3 / 2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{-\tfrac{3}{2}(-\tfrac{3}{2}-1)(-\tfrac{3}{2}-2)(-\tfrac{3}{2}-3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)}{2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)}{3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{\tfrac{3}{2}(\tfrac{3}{2}+1)(\tfrac{3}{2}+2)(\tfrac{3}{2}+3)}{4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^2 \cdot 2!}\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{3\cdot 5 \cdot 7}{2^3\cdot 3!}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{3\cdot 5 \cdot 7 \cdot 9}{2^4 \cdot 4!}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
</math>
</td>
</tr>
</table>

We therefore can write,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\mathcal{F}_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
- \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
\times \biggl\{
1 + \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2 + \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+ \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
- \biggl[ \frac{1 \cdot 3\cdot 5 \cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\biggr]\biggl(\frac{c}{a}\biggr)^{10}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \frac{3}{2}\biggl(\frac{c}{a}\biggr)^2
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+ \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^6
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
\biggr\}
+ \biggl[\frac{5 \cdot 7 \cdot 9}{2^7}\biggr]\biggl(\frac{c}{a}\biggr)^8
\biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
- \biggl[\frac{1}{2\cdot 3}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
- \biggl[\frac{1}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{3^2 }{2^4 \cdot 5}\biggr]\biggl(\frac{c}{a}\biggr)^8
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{3\cdot 5}{2^3 }\biggr]\biggl(\frac{c}{a}\biggr)^6
- \biggl[\frac{5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{1}{2\cdot 3}
+ \frac{3}{2}\biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{1\cdot 3 }{2\cdot 4\cdot 5} + \frac{1}{2^2} + \frac{3\cdot 5}{2^3 }\biggr] \biggl(\frac{c}{a}\biggr)^6
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{1 \cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}
+ \frac{3^2 }{2^4 \cdot 5}
+ \frac{5}{2^4 }
+ \frac{5 \cdot 7}{2^4}\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{5 \cdot 7 \cdot 9}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>
Once again from the binomial theorem,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\biggl(1 - \frac{c^2}{a^2} \biggr)^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots </math>
</td>
</tr>
</table>
which gives us,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>A_1 = \mathcal{F}_2 - \biggl(\frac{c}{a}\biggr)^2 \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1}</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
- ~\biggl\{
\biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- 2\biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[\frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr) \, .
</math>
</td>
</tr>
</table>

And,
<table align="center" border=0 cellpadding="3">

<tr>
<td align="right">
<math>\frac{A_3}{2} = \biggl(1 - \frac{c^2}{a^2}\biggr)^{-1} - \mathcal{F}_2 </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl\{
1 + \biggl(\frac{c}{a}\biggr)^2 + \biggl(\frac{c}{a}\biggr)^4 + \biggl(\frac{c}{a}\biggr)^6 + \biggl(\frac{c}{a}\biggr)^8 + \biggl(\frac{c}{a}\biggr)^{10} + \cdots
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>
-~ \biggl\{
\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
- \biggl(\frac{c}{a}\biggr)^2
+ \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
- \biggl[\frac{5}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
+
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
- \biggl[\frac{11}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
+
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
- \biggl[ \frac{ 93}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
+
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
1
-\frac{\pi}{2} \biggl(\frac{c}{a}\biggr)
+ 2\biggl(\frac{c}{a}\biggr)^2
- \pi \biggl[\frac{3}{2^2}\biggr] \biggl(\frac{c}{a}\biggr)^3
+ \biggl[\frac{8}{3} \biggr]\biggl(\frac{c}{a}\biggr)^4
-
\pi \biggl[\frac{3\cdot 5}{2^4 }\biggr]\biggl(\frac{c}{a}\biggr)^5
+ \biggl[\frac{2^4}{ 5} \biggr] \biggl(\frac{c}{a}\biggr)^6
-
\pi\biggl[\frac{5 \cdot 7}{2^5}\biggr] \biggl(\frac{c}{a}\biggr)^7
+ \biggl[ \frac{ 2^7}{5 \cdot 7}
\biggr]\biggl(\frac{c}{a}\biggr)^8
-
\pi \biggl[\frac{3^2\cdot 5 \cdot 7}{2^8}\biggr]\biggl(\frac{c}{a}\biggr)^9
\biggr\}
+ \mathcal{O}\biggl(\frac{c^{10}}{a^{10}}\biggr)\, .
</math>
</td>
</tr>
</table>

Notice that, to the highest order retained in these expressions, we find as expected that, <math>(A_1 + A_3/2) = 1</math>.

====Frequency (temporary)====

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
<math>
\omega_0^2
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr]
</math>
</td>
</tr>
<tr>
<td align="right">
<math>
\Rightarrow ~~~ \Omega^2_\mathrm{Mc} \equiv \frac{\omega_0^2}{\pi G \rho }
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3}
- \frac{6(1-e^2)}{e^2}
\, ,
</math>
</td>
</tr>
</table>

==Taylor Series (Hunter77)==

===First (Unsuccessful) Try===

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{''}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \frac{3^2}{2} \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}\biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
3 f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{15}{2^2} \biggr] f_3
+ \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{9}{4} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\}\biggl[- 2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-3 \biggr]f_0
+ \biggl[\frac{3^3 }{2} \biggr] f_1
- 3^3 f_2
+ \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3
+ \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{''}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 - 2(\Delta) f_3^'
- \frac{2^2}{3} (\Delta)^3 f_3^{'''} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
- \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{'''} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'
+ \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{'''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[- \frac{5}{6} \biggr] \biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0
+ \biggl[5\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3
+ \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{''}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'''} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
2f_0
+ \biggl[- \frac{3^3}{2^2}\biggr] f_1
+ \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{2}{3^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{
\biggl[\frac{2^2}{3^2} \biggr] f_0
- f_1
+ f_3 \biggl[\frac{5}{3^2} \biggr]
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
\biggr\} \biggl[-\frac{3}{2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}\biggr] f_1
+ \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr]
+ \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0
+ \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1
+ \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
- f_0
+ \biggl[\frac{3^3}{2^2}\biggr] f_1
- 3^3 f_2
+ \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2}{3^2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
\biggl[-\frac{2}{3} \biggr] f_0
+ \biggl[4\biggr] f_1
+ \biggl[- 2\cdot 5 \biggr] f_2
+ \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3
+ \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -\frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{24}\biggl\{
\biggl[-\frac{1}{3^2} \biggr]f_0
+ \biggl[\frac{1}{2} \biggr] f_1
- f_2
+ \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
\biggr\} \biggl[ -2^2\cdot 3 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2} \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3} \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+ \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2} \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+ \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3
+ \biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Result:

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Definitely WRONG!
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ \frac{17}{3} (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
<th align="center">
Somewhat Improved
</th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+ 4 (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
</td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

===Second Try===

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term, the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

First:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
f_4
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

This expression works very well for a parabola.

----

Second:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv}
\biggr\} \biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+ \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above. Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

----

Third:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1
+ f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3
+ \biggl[\frac{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

And,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
\biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}
</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 3 \biggr] f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3} \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

----

Finally, then:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~f_4</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3} \biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2^2} \biggl\{
\biggl[ \frac{2}{3} \biggr] f_0 + \biggl[- 4 \biggr] f_1
+ \biggl[ 2\cdot 5 \biggr] f_2 + \biggl[- \frac{2^2 \cdot 5}{3} \biggr] f_3
+ \biggl[ 4 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr] f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{2\cdot 3} \biggr] f_0 + \biggl[- 1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3} \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1 - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\biggl[\frac{1}{3} \biggr] f_0
+ \biggl[- 2\biggr] f_1
+ \biggl[ 6 \biggr] f_2
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~
f_4
</math>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</td>
</tr>
</table>
</td></tr></table>
</div>

{{ SGFfooter }}

Appendix/Mathematics/EulerAngles

2024-06-14T22:19:20Z

Jet53man: /* Using Matrix Notation */

__FORCETOC__


=Euler Angles</font>=
Here we will follow quite closely the [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf online class notes] prepared by [https://phas.ubc.ca/~berciu/ Professor Mona Berciu], Department of Physics & Astronomy, University of British Columbia.

==Basic Relations==

In terms of the (black) unit vectors of the <math>(X, Y, Z)</math> Cartesian coordinate system shown in the left panel of Figure 1, we can specify the (red) vector, <math>\vec{A}</math>, by the expression,
<div align="center">
<math>\vec{A} = \vec{e}_X (A_X) + \vec{e}_Y (A_Y) + \vec{e}_Z (A_Z)</math>
</div>
where the three coefficients, <math>(A_X, A_Y, A_Z)</math>, give the length (and <math>\pm</math> direction) of each vector component. Alternatively, in terms of the (black) unit vectors of the <math>(x_1, x_2, x_3)</math> Cartesian coordinate system shown in the right panel of Figure 1, the same (red) vector is specified by the expression,
<div align="center">
<math>~\vec{A} = \vec{e}_1 (A_1) + \vec{e}_2 (A_2) + \vec{e}_3 (A_3) \, .</math>
</div>

<table border="1" align="center" cellpadding="3">
<tr><td align="center" colspan="2">'''Figure 1'''</td></tr>
<tr>
<td align="center">
[[File:BerciuFig1a.png|250px|Berciu Figure 1a]]
</td>
<td align="center">
[[File:BerciuFig1bAgain.png|250px|Berciu Figure 1b]]
</td>
</tr>
</table>

It is clear from the nature of unit vectors and vector dot-products that the value of the coefficient, <math>A_1</math> — which explicitly appears in the second of these two expressions — may be obtained from the dot product, <math>\vec{e}_1 \cdot \vec{A}</math>. The same must be true if we insert, for <math>\vec{A}</math>, the first of the two expressions; that is to say,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>A_1 = \vec{e}_1 \cdot \vec{A}</math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\vec{e}_1 \cdot\vec{e}_X (A_X) + \vec{e}_1 \cdot\vec{e}_Y (A_Y) + \vec{e}_1 \cdot\vec{e}_Z (A_Z) \, .</math>
</td>
</tr>
</table>
Analogously, we can write,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>A_2 = \vec{e}_2 \cdot \vec{A}</math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\vec{e}_2 \cdot\vec{e}_X (A_X) + \vec{e}_2 \cdot\vec{e}_Y (A_Y) + \vec{e}_2 \cdot\vec{e}_Z (A_Z) \, ,</math>
</td>
</tr>
<tr>
<td align="right">
<math>A_3 = \vec{e}_3 \cdot \vec{A}</math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\vec{e}_3 \cdot\vec{e}_X (A_X) + \vec{e}_3 \cdot\vec{e}_Y (A_Y) + \vec{e}_3 \cdot\vec{e}_Z (A_Z) \, .</math>
</td>
</tr>
</table>
<span id="FormMatrix">This set of three relations</span> therefore provides a generic way to express one set of Cartesian vector projections in terms of the other. When written in ''matrix form'', the set becomes,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
\vec{e}_1 \cdot\vec{e}_X & \vec{e}_1 \cdot\vec{e}_Y & \vec{e}_1 \cdot\vec{e}_Z \\
\vec{e}_2 \cdot\vec{e}_X & \vec{e}_2 \cdot\vec{e}_Y & \vec{e}_2 \cdot\vec{e}_Z \\
\vec{e}_3 \cdot\vec{e}_X & \vec{e}_3 \cdot\vec{e}_Y & \vec{e}_3 \cdot\vec{e}_Z
\end{bmatrix}
\cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}
= \hat{R} \cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix} \, .
</math>
</td>
</tr>
</table>

Alternatively, we could "dot" the unit-vector triplet <math>(\vec{e}_X, \vec{e}_Y, \vec{e}_Z)</math> into the vector, <math>\vec{A}</math>, in which case it would be easy to demonstrate that mapping the other direction is accomplished via the relation,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\hat{R}^{-1} \cdot
\begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix} \, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\hat{R}^{-1}</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
\vec{e}_X \cdot\vec{e}_1 & \vec{e}_X \cdot\vec{e}_2 & \vec{e}_X \cdot\vec{e}_3 \\
\vec{e}_Y \cdot\vec{e}_1 & \vec{e}_Y \cdot\vec{e}_2 & \vec{e}_Y \cdot\vec{e}_3 \\
\vec{e}_Z \cdot\vec{e}_1 & \vec{e}_Z \cdot\vec{e}_2 & \vec{e}_Z \cdot\vec{e}_3
\end{bmatrix} \, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8">
<tr><td align="left">

Adopting the shorthand notation used in [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's class notes], we will define,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\vec{A}_\mathrm{body} \equiv \begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix}</math>
</td>
<td align="center">
      and,      
</td>
<td align="left">
<math>\vec{A}_\mathrm{XYZ} \equiv \begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix} \, .</math>
</td>
</tr>
</table>
To be clear, these are not different vectors. They are, rather, two different coordinate representations of the same vector as illustrated in Figure 1. Using this notation, mapping from one representation to the other is accomplished via the compact expressions,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\vec{A}_\mathrm{body} = \hat{R} \cdot \vec{A}_{XYZ}</math>
</td>
<td align="center">
      or,      
</td>
<td align="left">
<math>\vec{A}_\mathrm{XYZ} = \hat{R}^{-1} \cdot \vec{A}_\mathrm{body} \, .</math>
</td>
</tr>
</table>

</td></tr>
</table>

<span id="Transpose">'''Matrix Transpose:'''</span>  By definition, a matrix <math>\hat{M}</math> is the ''transpose'' of the matrix <math>\hat{N}</math> — that is, <math>\hat{M} = \hat{N}^T</math> — if <math>m_{ij} = n_{ji}</math> for all matrix-element indices, <math>i, j</math>. We recognize, therefore, that
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\hat{R}^{-1} = \hat{R}^T </math>
</td>
<td align="center">
      and,      
</td>
<td align="left">
<math>\hat{R} = [\hat{R}^{-1}]^T \, .</math>
</td>
</tr>
</table>
Specifically, this means that all three diagonal elements (for which, <math>i=j</math>) are the same in <math>\hat{R}</math> and <math>\hat{R}^{-1}</math>; but the locations of paired off-diagonal elements are swapped.

==A Sequence of Rotations==

It is quite generally true that we can transition/map/migrate from one set of orthogonal unit vectors — such as the <math>(\vec{e}_X, \vec{e}_Y, \vec{e}_Z)</math> ''inertial/laboratory'' frame illustrated by the black arrows in the left panel of Figure 1 — to any other set of orthogonal unit vectors — such as the <math>(\vec{e}_1, \vec{e}_2, \vec{e}_3)</math> ''body'' frame illustrated by the black arrows in the right panel of Figure 1 — by carrying out three rotations. These are traditionally referred to as ''Euler angles''. As is [[#The_Order_of_Rotations|emphasized below]], care must be taken when choosing the order in which rotations are carried out. One fairly standard sequence of rotations is illustrated in Figure 2:

<table border="1" align="center" cellpadding="3">
<tr><td align="center" colspan="3">'''Figure 2'''</td></tr>
<tr>
<td align="center">
[[File:BerciuFig2a.png|250px|Berciu Figure 2a]]
</td>
<td align="center">
[[File:BerciuFig2bAgain.png|250px|Berciu Figure 2b]]
</td>
<td align="center">
[[File:BerciuFig2c.png|250px|Berciu Figure 2c]]
</td>
</tr>

<tr>
<td align="center">
Rotation #1
</td>
<td align="center">
Rotation #2
</td>
<td align="center">
Rotation #3
</td>
</tr>
</table>

===First Rotation===
As depicted in the left-most panel of Figure 2, rotate the triplet of unit vectors about the <math>Z</math> (''i.e.,'' <math>\vec{e}_Z</math>) axis by an angle, <math>\phi</math>. The result is the ''green'' coordinate system labeled, <math>(1', 2', 3')</math>. Note: (a) The 3' axis remains aligned with the inertial-frame Z-axis; and (b) we will refer to the 1' axis as the ''line of nodes''. Given that the 3' axis is aligned with the Z-axis — that is, <math>\vec{e}_{3'} = \vec{e}_Z</math> — it is easy to recognize that the other two mappings are:
<div align="center"><math>\vec{e}_{1'} = \vec{e}_X \cos\phi + \vec{e}_Y \sin\phi \, ,</math>     and     <math>\vec{e}_{2'} = \vec{e}_Y \cos\phi - \vec{e}_X \sin\phi</math>.</div>
Hence, the corresponding rotation matrix is,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\hat{R}_{3}(\phi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\vec{e}_{1'} \cdot\vec{e}_X & \vec{e}_{1'} \cdot\vec{e}_Y & \vec{e}_{1'} \cdot\vec{e}_Z \\
\vec{e}_{2'} \cdot\vec{e}_X & \vec{e}_{2'} \cdot\vec{e}_Y & \vec{e}_{2'} \cdot\vec{e}_Z \\
\vec{e}_{3'} \cdot\vec{e}_X & \vec{e}_{3'} \cdot\vec{e}_Y & \vec{e}_{3'} \cdot\vec{e}_Z
\end{bmatrix}}
=
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>
Note that the subscript, 3, has been attached to <math>\hat{R}</math> in order to indicate that rotation was about the "third" axis.

===Second Rotation===
As depicted in the center panel of Figure 2, rotate the triplet of unit vectors about the ''line of nodes'' by an angle, <math>\theta</math>. The result is the ''light-blue'' coordinate system labeled, (1", 2", 3"). Note: (a) The 1" axis is aligned with the 1' axis (''line of nodes''); and (b) the values of the first pair of rotation angles, <math>(\phi, \theta)</math>, have been chosen, here, to ensure that the 3" axis aligns with the <math>\vec{e}_3</math> unit vector. Given that, <math>\vec{e}_{1''} = \vec{e}_{1'}</math>, the other two mappings are,
<div align="center">
<math>\vec{e}_{2''} = \vec{e}_{2'} \cos\theta + \vec{e}_{3'} \sin\theta </math>     and     <math>\vec{e}_{3''} = \vec{e}_{3'} \cos\theta - \vec{e}_{2'} \sin\theta</math>.
</div>
The corresponding rotation matrix is, then,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\hat{R}_{1}(\theta)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\vec{e}_{1''} \cdot\vec{e}_{1'} & \vec{e}_{1''} \cdot\vec{e}_{2'} & \vec{e}_{1''} \cdot\vec{e}_{3'} \\
\vec{e}_{2''} \cdot\vec{e}_{1'} & \vec{e}_{2''} \cdot\vec{e}_{2'} & \vec{e}_{2''} \cdot\vec{e}_{3'} \\
\vec{e}_{3''} \cdot\vec{e}_{1'} & \vec{e}_{3''} \cdot\vec{e}_{2'} & \vec{e}_{3''} \cdot\vec{e}_{3'}
\end{bmatrix}}
=
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>
In this case the subscript, 1, has been attached to <math>\hat{R}</math> in order to indicate that rotation was about the "first" axis.

===Third Rotation===
As depicted in the right-most panel of Figure 2, rotate the triplet of unit vectors about the 3" axis by an angle, <math>\psi</math>. The result is the desired black coordinate system labeled, <math>(\vec{e}_1, \vec{e}_2, \vec{e}_3)</math>. Given that this step — like the first — calls for rotation about the "third" axis, we can immediately deduce that the relevant rotation matrix is,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\hat{R}_{3}(\psi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

===Other Properties of Particular Note===
In her [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf online class notes], Professor Berciu points out:
<ol>
<li><math>\hat{R}_3(\phi_1)\cdot \hat{R}_3(\phi_2) = \hat{R}_3(\phi_1 + \phi_2)</math>. This means that if you rotate, first, by angle <math>\phi_2</math> followed by a rotation by angle <math>\phi_1</math> '''about the same axis(!)''', it is as if you carry out a single rotation by the angle, <math>(\phi_1 + \phi_2)</math>.</li>
<li>In flipping the angle of rotation from positive to negative, the rotation matrix flips to its transpose. That is to say, for example, <math>\hat{R}_3(-\phi) = \hat{R}_3^{-1}(\phi) = \hat{R}^{T}_3(\phi)</math>.</li>
</ol>

===Simple Numerical Example===

Suppose we set <math>(\phi, \theta, \psi) = (30^\circ, 25^\circ, 15^\circ)</math>, which are roughly consistent with the trio of rotation angles displayed in Figure 2. The corresponding trio of rotation matrices are:
<table border="0" align="center" cellpadding="2">
<tr>
<td align="right">
<math>\hat{R}_{3}(\phi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.8660 & 0.5000 & 0 \\
-0.5000 & 0.8660 & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
<td align="center">,      </td>

<td align="right">
<math>\hat{R}_{1}(\theta)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
1 & 0 & 0 \\
0 & 0.9063 & 0.4226 \\
0 & -0.4226 & 0.9063
\end{bmatrix}}
</math>
</td>
<td align="center">,      </td>

<td align="right">
<math>\hat{R}_{3}(\psi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.9659 & 0.2588 & 0 \\
-0.2588 & 0.9659 & 0 \\
0 & 0 & 1
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

==Combined Transformations==

Here, we combine the three rotations in the order defined by Figure 2 — and use the corresponding trio of ''Euler angles'' — to relate the three vector components of <math>\vec{A}_\mathrm{body}</math> to the three vector components of <math>\vec{A}_\mathrm{XYZ}</math>. We first review how this can be accomplished via a standard — but rather ''brute force'' — unit-vector substitution approach; then we show how the more elegant matrix rotation technique can be employed to achieve the same result.

===Brute Force===

Bringing forward the results of our [[#A_Sequence_of_Rotations|above discussion]], we recognize that the following unit-vector relationships result from each of the three specified rotations:
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">Rotation #1:</td>
<td align="left">
<math>\vec{e}_{1'} = \vec{e}_X \cos\phi + \vec{e}_Y \sin\phi</math>     and    
<math>\vec{e}_{2'} = \vec{e}_Y \cos\phi - \vec{e}_X \sin\phi</math>     and    
<math>\vec{e}_{3'} = \vec{e}_Z \, ;</math>
</td>
</tr>

<tr>
<td align="right">Rotation #2:</td>
<td align="left">
<math>\vec{e}_{2''} = \vec{e}_{2'} \cos\theta + \vec{e}_{3'} \sin\theta </math>     and    
<math>\vec{e}_{3''} = \vec{e}_{3'} \cos\theta - \vec{e}_{2'} \sin\theta</math>     and    
<math>\vec{e}_{1''} = \vec{e}_{1'} \, ;</math>
</td>
</tr>

<tr>
<td align="right">Rotation #3:</td>
<td align="left">
<math>\vec{e}_{x_1} = \vec{e}_{1''} \cos\psi + \vec{e}_{2''} \sin\psi</math>     and    
<math>\vec{e}_{x_2} = \vec{e}_{2''} \cos\psi - \vec{e}_{1''} \sin\psi</math>     and    
<math>\vec{e}_{x_3} = \vec{e}_{3''} \, .</math>
</td>
</tr>
</table>

Hence, we have,

<table border="0" align="center" cellpadding="0">

<tr>
<td align="right">
<math>\vec{e}_{x_1}</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_{1''} \cos\psi + \vec{e}_{2''}\sin\psi
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\cos\psi [ \vec{e}_{1'} ] + \sin\psi [ \vec{e}_{2'} \cos\theta + \vec{e}_{3'} \sin\theta ]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\cos\psi [ \vec{e}_X \cos\phi + \vec{e}_Y \sin\phi ]
+ \sin\psi \cos\theta[ \vec{e}_Y \cos\phi - \vec{e}_X \sin\phi]
+ \sin\psi\sin\theta [\vec{e}_Z ]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_X[\cos\psi \cos\phi - \sin\phi \sin\psi \cos\theta]
+
\vec{e}_Y [ \cos\psi \sin\phi + \cos\phi \sin\psi \cos\theta ]
+
\vec{e}_Z [\sin\psi\sin\theta ] \, .
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\vec{e}_{x_2}</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_{2''} \cos\psi - \vec{e}_{1''}\sin\psi
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\cos\psi [\vec{e}_{2'} \cos\theta + \vec{e}_{3'} \sin\theta ] - \sin\psi [\vec{e}_{1'} ]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
- \sin\psi [\vec{e}_X \cos\phi + \vec{e}_Y \sin\phi ]
+ \cos\psi \cos\theta [\vec{e}_Y \cos\phi - \vec{e}_X \sin\phi ]
+ \cos\psi \sin\theta [\vec{e}_{Z} ]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_X [- \sin\psi \cos\phi - \sin\phi \cos\psi \cos\theta ]
+ \vec{e}_Y [- \sin\phi \sin\psi + \cos\phi \cos\psi \cos\theta ]
+ \vec{e}_{Z} [\cos\psi \sin\theta ] \, .
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\vec{e}_{x_3}</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_{3''}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_{3'} \cos\theta - \vec{e}_{2'} \sin\theta
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_{Z} [ \cos\theta ] - \sin\theta [ \vec{e}_Y \cos\phi - \vec{e}_X \sin\phi ]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">=</td>
<td align="left">
<math>
\vec{e}_X[\sin\theta \sin\phi ]
- \vec{e}_Y [ \sin\theta \cos\phi]
+ \vec{e}_{Z} [ \cos\theta ] \, .
</math>
</td>
</tr>
</table>

We can therefore write,
<table border="0" align="center" cellpadding="8">

<tr>
<td align="right">
<math>\hat{R}</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
\vec{e}_{x_1} \cdot\vec{e}_X & \vec{e}_{x_1} \cdot\vec{e}_Y & \vec{e}_{x_1} \cdot\vec{e}_Z \\
\vec{e}_{x_2} \cdot\vec{e}_X & \vec{e}_{x_2} \cdot\vec{e}_Y & \vec{e}_{x_2} \cdot\vec{e}_Z \\
\vec{e}_{x_3} \cdot\vec{e}_X & \vec{e}_{x_3} \cdot\vec{e}_Y & \vec{e}_{x_3} \cdot\vec{e}_Z
\end{bmatrix}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
(\cos\psi \cos\phi - \sin\phi \sin\psi \cos\theta)
& ( \cos\psi \sin\phi + \cos\phi \sin\psi \cos\theta )
& (\sin\psi\sin\theta ) \\

(- \sin\psi \cos\phi - \sin\phi \cos\psi \cos\theta )
& (- \sin\phi \sin\psi + \cos\phi \cos\psi \cos\theta )
& (\cos\psi \sin\theta ) \\

(\sin\theta \sin\phi )
& ( -\sin\theta \cos\phi)
& ( \cos\theta )
\end{bmatrix} \, .
</math>
</td>
</tr>
</table>

===Using Matrix Notation===

<table border="1" align="right" cellpadding="1">
<tr><td align="center">Astrophysics Example</td></tr>
<tr><td align="center">[[File:EulerAnglesExample02.jpeg|250px|Euler Angles Example02]]</td></tr>
</table>
With more finesse, we can write the general rotation matrix that links the ''body frame'', <math>(x_1, x_2, x_3)</math>, to the ''inertial/laboratory'' frame, <math>(X, Y, Z)</math>, as the product of the three rotations about the corresponding axes:
<table border="0" align="center" cellpadding="3">
<tr>
<td align="right">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}} \, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">
[https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's online class notes], bottom of p. 3
</td>
</tr>
</table>

<span id="RotationMatrix">Carrying out the matrix multiplications</span>, starting from the right, gives,
<table border="0" align="center" cellpadding="3">
<tr>
<td align="right">
<math>\hat{R}(\phi, \theta, \psi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi \cos\theta & \cos\theta \cos\phi & \sin\theta \\
\sin\theta\sin\phi & -\sin\theta \cos\phi & \cos\theta
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)
& (\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
& (\sin\psi \sin\theta) \\
(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)
& ( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
& \sin\theta \cos\psi \\
\sin\theta\sin\phi & -\sin\theta \cos\phi & \cos\theta
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

This precisely matches the Euler-angle expression for the rotation matrix, <math>\hat{R}</math>, that we derived above using a more brute force approach.

----

For the [[#Simple_Numerical_Example|'''simple numerical example''' established above]], these three derivation steps give, respectively,
<table border="0" align="center" cellpadding="2">
<tr>
<td align="right">
<math>\hat{R}(\phi, \theta, \psi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.9659 & 0.2588 & 0 \\
-0.2588 & 0.9659 & 0 \\
0 & 0 & 1
\end{bmatrix}}
\cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & 0.9063 & 0.4226 \\
0 & -0.4226 & 0.9063
\end{bmatrix}}
\cdot
{\begin{bmatrix}
0.8660 & 0.5000 & 0 \\
-0.5000 & 0.8660 & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.9659 & 0.2588 & 0 \\
-0.2588 & 0.9659 & 0 \\
0 & 0 & 1
\end{bmatrix}}
\cdot
{\begin{bmatrix}
0.8660 & 0.500 & 0 \\
-0.4532 & 0.7849 & 0.4226 \\
0.2113 & -0.3660 & 0.9063
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.7192 & 0.6861 & 0.1094 \\
-0.6619 & 0.6287 & 0.4082 \\
0.2113 & -0.3660 & 0.9063
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

----

<span id="Rinverse">Following</span> the [[#Transpose|steps provided above]], we recognize that the inverse — or transpose — of this rotation matrix is,
<table border="0" align="center" cellpadding="3">
<tr>
<td align="right">
<math>\hat{R}^{-1} = \hat{R}^T</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)
&(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)
& (\sin\theta\sin\phi) \\

(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
& ( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
& (-\sin\theta \cos\phi) \\

(\sin\psi \sin\theta) & (\sin\theta \cos\psi) & (\cos\theta)
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">

From the last row of the column labeled "Proper Euler angles" in [https://en.wikipedia.org/wiki/Euler_angles#Rotation_matrix Wikipedia's discussion of the ''rotation matrix''], we find,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>~Z_1 X_2 Z_3</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
(c_1c_3 - c_2s_1s_3) & (-c_1s_3 - c_2 c_3s_1) & (s_1s_2) \\
(c_3s_1 + c_1c_2s_3) & (c_1c_2c_3 - s_1s_3) & (-c_1s_2) \\
(s_2s_3) & (c_3s_2) & (c_2)
\end{bmatrix}
</math>
</td>
</tr>
</table>
The equivalent expression can be found at the top of p. 4 of [https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Professor Berciu's online class notes]; it reads,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>~\hat{R}(\phi, \theta, \psi)</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\begin{bmatrix}
(\cos\phi \cos\psi - \cos\theta \sin\phi \sin\psi) & (-\cos\phi \sin\psi - \cos\theta \cos\psi \sin\phi) & (\sin\phi \sin\theta) \\
(\cos\psi \sin\phi + \cos\phi \cos\theta \sin\psi) & (\cos\phi \cos\theta \cos\psi - \sin\phi \sin\psi) & (-\cos\phi \sin\theta) \\
(\sin\theta \sin\psi) & (\cos\psi \sin\theta) & (\cos\theta)
\end{bmatrix}
</math>
</td>
</tr>
</table>

These two rotation-matrix expressions are equivalent to one another, but they do not match our derived expression for <math>\hat{R}</math>. Instead, then match our expression for <math>\hat{R}^{-1}</math>. <font color="red">It is not (yet) clear to us why this is the case.</font>

----

For the [[#Simple_Numerical_Example|'''simple numerical example''' established above]], the rotation matrix on the right-hand-side of both of these expressions gives,
<table border="0" align="center" cellpadding="2">
<tr>
<td align="left">
<math>
{\begin{bmatrix}
0.7192 & -0.6619 & 0.2113 \\
0.6861 & 0.6287 & -0.3660 \\
0.1094 & 0.4082 & 0.9063
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>

</td></tr></table>

==Extra Illustrations==

For the most part, we have focused our discussion on the case where three simple rotations are carried out '''in the order''' depicted in Figure 2. This sequence — first <math>\phi</math> (about the original <math>Z</math> axis), second <math>\theta</math> (about the line of nodes), third <math>\psi</math> (about the new, tilted <math>\boldsymbol{x_3}</math> axis) — has been denoted by the mathematical expression,
<div align="center">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi) \, .</math><br />
[https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Berciu's online class notes], bottom of p. 3
</div>
It is extremely important to appreciate that the '''order''' in which the rotations are carried out matters! On the right-hand-side of our adopted expression, the matrix that specifies the first rotation <math>(\phi)</math> is placed farthest to the right, while the matrix that specifies the third/last rotation <math>(\psi)</math> is placed farthest to the left. In the paragraph that follows, we explicitly demonstrate that a different mapping arises if the first and third rotation matrices are swapped. But, as we shall illustrate, once the '''order''' of rotations has been specified, it doesn't matter whether the steps that are taken to combine matrix elements begins with multiplication of the first and second rotation matrices, or begins with multiplication of the second and third matrices.

===The Order of Rotations Matters===

Let's swap the order of the rotations that are identified in Figure 2 as #1 <math>(\phi)</math> and #3 <math>(\psi)</math>. The resulting combined matrix expression is,

<table border="0" align="center" cellpadding="3">

<tr>
<td align="right">
<math>\hat{R}_3(\phi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\psi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi \cos\theta & \cos\theta \cos\psi & \sin\theta \\
\sin\theta \sin\psi & -\sin\theta \cos\psi & \cos\theta
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
(\cos\phi \cos\psi - \sin\phi \sin\psi \cos\theta) & (\cos\phi \sin\psi + \sin\phi \cos\theta \cos\psi) & (\sin\phi \sin\theta) \\
(-\sin\phi \cos\psi - \cos\phi \sin\psi \cos\theta) & (-\sin\phi \sin\psi + \cos\phi \cos\theta \cos\psi) & (\sin\theta \cos\phi) \\
(\sin\theta \sin\psi) & (-\sin\theta \cos\psi) & (\cos\theta)
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>
This does '''not''' match the [[#Using_Matrix_Notation|expression that was derived earlier]] for <math>\hat{R}(\phi,\theta,\psi)</math>; nor does it match [[#Rinverse|the inverse]], <math>\hat{R}^{-1}</math>. This illustrates that the '''order of specified rotations''' does matter. It is worth pointing out that the ''diagonal elements'' of all of these rotation-matrix expressions are identical, but the off-diagonal terms differ.

===Steps Taken to Combine Matrix Elements===

In our [[#Using_Matrix_Notation|above discussion]], we carried out the multiplication of three rotation matrices — strategically combining the elements of various columns and rows — by starting from the right, that is, by first multiplying the 2<sup>nd</sup> and 3<sup>rd</sup> matrices. Here, beginning with the same expression for <math>\hat{R}(\phi, \theta,\psi)</math>, we multiply the three matrices by starting from the left, that is, by first multiplying the 1<sup>st</sup> and 2<sup>nd</sup> matrices.

<table border="0" align="center" cellpadding="3">

<tr>
<td align="right">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi & 0 \\
-\sin\psi & \cos\psi & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
\cos\psi & \sin\psi \cos\theta & \sin\psi \sin\theta \\
-\sin\psi & \cos\psi \cos\theta & \cos\psi \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \cdot
{\begin{bmatrix}
\cos\phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
(\cos\psi \cos\phi - \sin\psi\cos\theta \sin\phi) & (\cos\psi \sin\phi + \sin\psi \cos\theta \cos\phi) & (\sin\psi \sin\theta) \\
(-\sin\psi \cos\phi - \cos\psi \cos\theta \sin\phi) & (-\sin\psi \sin\phi + \cos\psi \cos\theta \cos\phi) & (\cos\psi \sin\theta) \\
(\sin\theta \sin \phi) & (-\sin\theta \cos\phi) & (\cos\theta)
\end{bmatrix}} \,.
</math>
</td>
</tr>
</table>

This is precisely the same result that we obtained previously, illustrating that once the '''order''' of the multiplications has been established, it doesn't matter how pairs of matrices are grouped in order to carry out the row-column multiplications.

==Switching Coordinate Representations of a Vector==
Now, given that,
<div align="center">
<math>\vec{A}_\mathrm{body} = \hat{R} \cdot \vec{A}_{XYZ} \, ,</math>
</div>
the following three mapping relations must hold:

<table border="0" align="center" cellpadding="0">

<tr>
<td align="right">
<math>A_1</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_X(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)
+ A_Y(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
+ A_Z(\sin\psi \sin\theta) \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_2</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_X(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)
+ A_Y( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
+ A_Z (\sin\theta \cos\psi) \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_3</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_X (\sin\theta\sin\phi)
+ A_Y (-\sin\theta \cos\phi)
+ A_Z (\cos\theta) \, .
</math>
</td>
</tr>
</table>

Alternatively, given that,
<div align="center">
<math>\vec{A}_{XYZ} = \hat{R}^{-1} \cdot \vec{A}_\mathrm{body} \, ,</math>
</div>

the following additional three mapping relations must hold:

<table border="0" align="center" cellpadding="0">

<tr>
<td align="right">
<math>A_X</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_1(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)
+ A_2(-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)
+ A_3(\sin\theta\sin\phi) \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_Y</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_1(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
+ A_2( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
+ A_3 (-\sin\theta \cos\phi) \, ,
</math>
</td>
</tr>

<tr>
<td align="right">
<math>A_Z</math>
</td>
<td align="center">=</td>
<td align="left">
<math>
A_1 (\sin\psi \sin\theta)
+ A_2 (\sin\theta \cos\psi)
+ A_3 (\cos\theta) \, .
</math>
</td>
</tr>
</table>

----

Extending the [[#Simple_Numerical_Example|'''simple numerical example''' established above]], let's assume that the (red) vector that appears in both panels of Figure 1 has the ''inertial-frame'' (left panel) components,
<div align="center">
<math>(A_X, A_Y, A_Z) = (0.8, 0.8, 0.9) \, .</math>
</div>
The components of this same (red) vector as viewed from the ''body'' frame (right panel of Figure 1) should be,
<table border="0" align="center" cellpadding="8">

<tr>
<td align="right">
<math>\begin{bmatrix}A_1 \\ A_2 \\ A_3 \end{bmatrix}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\hat{R} \cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.7192 & 0.6861 & 0.1094 \\
-0.6619 & 0.6287 & 0.4082 \\
0.2113 & -0.3660 & 0.9063
\end{bmatrix}}
\cdot
\begin{bmatrix}0.8 \\ 0.8 \\ 0.9 \end{bmatrix}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\begin{bmatrix} 1.2227 \\ 0.3408 \\ 0.6919 \end{bmatrix} \, .
</math>
</td>
</tr>
</table>

Out of curiosity, what do we obtain if we use the inverse of the rotation matrix? Well …
<table border="0" align="center" cellpadding="8">

<tr>
<td align="right">
<math>\begin{bmatrix}?_1 \\ ?_2 \\ ?_3 \end{bmatrix}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\hat{R}^{-1} \cdot
\begin{bmatrix}A_X \\ A_Y \\ A_Z \end{bmatrix}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
0.7192 & -0.6619 & 0.2113 \\
0.6861 & 0.6287 & -0.3660 \\
0.1094 & 0.4082 & 0.9063
\end{bmatrix}}
\cdot
\begin{bmatrix}0.8 \\ 0.8 \\ 0.9 \end{bmatrix}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\begin{bmatrix} 0.2360 \\ 0.7224 \\ 1.2298 \end{bmatrix} \, .
</math>
</td>
</tr>
</table>

=Riemann Type I Ellipsoids=

==Tipped Coordinate System==
In an accompanying discussion of Riemann Type I ellipsoids, we have presented a [[ThreeDimensionalConfigurations/RiemannTypeI#Try_Again|methodical derivation of orbital parameters]] in a "tipped" reference frame. In the context of our present discussion of Euler angles, it seems clear that this separate coordinate-frame transformation involved only one nonzero Euler angle; specifically, <math>(\phi, \theta, \psi) = (0, \theta, 0)</math>. In terms of Euler angles, then, the transformation between the ellipsoid's unprimed frame and the primed "tipped" frame is given by the [[#FormMatrix|above expressions]] …

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}_\mathrm{tipped}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\hat{R} \cdot
\begin{bmatrix}x \\ y \\ (z - z_0) \end{bmatrix} \, ,
</math>
</td>
</tr>
</table>

and,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\begin{bmatrix}x \\ y \\ (z - z_0) \end{bmatrix}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>
\hat{R}^{-1} \cdot
\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}_\mathrm{tipped} \, ,
</math>
</td>
</tr>
</table>
where,

<table border="0" align="center" cellpadding="3">

<tr>
<td align="right">
<math>\hat{R}(\phi, \theta, \psi) = \hat{R}_3(\psi) \cdot \hat{R}_1(\theta) \cdot \hat{R}_3(\phi)</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \cdot
{\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{bmatrix}} \, ,
</math>
</td>
</tr>
</table>
and,

<table border="0" align="center" cellpadding="3">

<tr>
<td align="right">
<math>\hat{R}^{-1}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta
\end{bmatrix}} \, .
</math>
</td>
</tr>
</table>
That is to say,
<table border="1" align="center" cellpadding="8"><tr>
<td align="center">

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>x'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>x</math></td>
</tr>
<tr>
<td align="right"><math>y'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>y \cos\theta + (z - z_0)\sin\theta</math></td>
</tr>
<tr>
<td align="right"><math>z'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>- y \sin\theta + (z - z_0)\cos\theta</math></td>
</tr>
</table>
</td>

<td align="center">
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>x</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>x'</math></td>
</tr>
<tr>
<td align="right"><math>y</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>y' \cos\theta - z'\sin\theta</math></td>
</tr>
<tr>
<td align="right"><math>(z - z_0)</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>y' \sin\theta + z'\cos\theta</math></td>
</tr>
</table>
</td>
</tr>
</table>
which is identical to the coordinate transformations presented in the [[#Try_Again|top two panels of the table in our accompanying ''methodical derivation of orbital parameters]].

Similarly, the three components of the velocity field as viewed from the "tipped" plane can be obtained from the velocity components in the untipped plane <math>(u_1, u_2, u_3)</math> via the expressions,
<table border="1" align="center" cellpadding="8"><tr>
<td align="center">

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>u_1^'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>u_1</math></td>
</tr>
<tr>
<td align="right"><math>u_2^'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>u_2 \cos\theta + u_3\sin\theta</math></td>
</tr>
<tr>
<td align="right"><math>u_3^'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>- u_2 \sin\theta + u_3\cos\theta</math></td>
</tr>
</table>
</td>
</tr>
</table>

==Velocity in Tipped Plane==
Now, according to EFE — see eq. 154 of Chapter 7, §51 (p. 156) —
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~u_1</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y + \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z</math>
</td>
</tr>

<tr>
<td align="right">
<math>~u_2</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~+\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x</math>
</td>
</tr>

<tr>
<td align="right">
<math>~u_3</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x</math>
</td>
</tr>
</table>

Hence, in the "tipped" plane we find,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>u_1^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z
- \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y
</math>
</td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \biggl[z_0 + y'\sin\theta + z'\cos\theta\biggr]
+ \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 \biggl[ z' \sin\theta - y'\cos\theta \biggr]
\, ,
</math>
</td>
</tr>
<tr>
<td align="right"><math>u_2^'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>\cos\theta\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x
- \sin\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x</math></td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\biggl\{\cos\theta\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3
- \sin\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\}x'
\, ,
</math>
</td>
</tr>
<tr>
<td align="right"><math>u_3^'</math></td>
<td align="center"><math>=</math></td>
<td align="left"><math>- \sin\theta\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x -
\cos\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x</math></td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left"><math>
\biggl\{ - \sin\theta\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3
- \cos\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\} x'
\, .
</math>
</td>
</tr>
</table>
Ideally, we would like to choose the orientation of our "tipped" plane such that there are no vertical motions as viewed from that plane, that is, such that <math>u_3^' = 0</math>. This means that,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\sin\theta\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
- \cos\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2
</math>
</td>
</tr>
<tr>
<td align="right"><math>\Rightarrow ~~~ \tan\theta</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
- \frac{\zeta_2}{\zeta_3}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]\frac{c^2}{b^2} \, .
</math>
</td>
</tr>
</table>
Given, as well, that we are only analyzing motion in the <math>z' = 0</math> plane, we have for the other two velocity components,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>u_2^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\cos\theta \biggl\{\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3
- \tan\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\}x'
</math>
</td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\cos\theta \biggl\{\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3
+ \frac{\zeta_2}{\zeta_3}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]\frac{c^2}{b^2}
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\}x'
</math>
</td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\biggl[ \frac{b^2}{a^2 + b^2} \biggr]\zeta_3 \cos\theta \biggl\{1
+ \frac{\zeta_2^2}{\zeta_3^2}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]^2\frac{c^4}{b^4}
\biggr\}x'
</math>
</td>
</tr>
<tr>
<td align="right"><math>u_1^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\cos\theta \biggl\{
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \biggl[\frac{z_0}{\cos\theta} + y'\tan\theta + \cancelto{0}{z'}\biggr]
+ \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 \biggl[ \cancelto{0}{z'} \tan\theta - y'\biggr]
\biggr\}
</math>
</td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] z_0 \zeta_2
- \zeta_3 \cos\theta \biggl\{
\biggl[ \frac{a^2}{a^2 + c^2} \biggr]
\frac{\zeta_2^2}{\zeta_3^2}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]\frac{c^2}{b^2}
+ \biggl[ \frac{a^2}{a^2 + b^2} \biggr]
\biggr\}y'
</math>
</td>
</tr>
<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] z_0 \zeta_2
- \biggl[ \frac{a^2}{a^2 + b^2} \biggr]\zeta_3 \cos\theta \biggl\{
\frac{\zeta_2^2}{\zeta_3^2}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]^2\frac{c^2}{b^2}
+ 1
\biggr\}y'
\, .
</math>
</td>
</tr>
</table>

==Rewrite Velocity-Component Expressions==
After defining the term,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\Lambda^2</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\zeta_2^2}{\zeta_3^2}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]^2\frac{c^2}{b^2} \, ,</math>
</td>
</tr>
</table>
the expressions for the two "tipped" velocity components may be rewritten as,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr] u_2^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>\biggl\{b^2
+ \Lambda^2 \cdot c^2
\biggr\}x' \, ,
</math>
</td>
</tr>

<tr>
<td align="right"><math>\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr] u_1^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr]
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] z_0 \zeta_2
- \biggl\{ a^2 + \Lambda^2 \cdot a^2
\biggr\}y'
\, .
</math>
</td>
</tr>
</table>
Or,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\biggl\{\frac{1+\Lambda^2}{b^2 + \Lambda^2 \cdot c^2 }\biggr\}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr] u_2^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
( 1 + \Lambda^2 )x'
\, ,
</math>
</td>
</tr>

<tr>
<td align="right"><math>\biggl[ \frac{a^2 + b^2}{a^2 \zeta_3 \cos\theta}\biggr] u_1^'</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr]
\biggl[ \frac{z_0 \zeta_2}{a^2 + c^2} \biggr]
- ( 1 + \Lambda^2 )y'
</math>
</td>
</tr>

<tr>
<td align="right"> </td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
( 1 + \Lambda^2 )(y_0 - y')
\, ,
</math>
</td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>y_0</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>
\frac{1}{1+\Lambda^2}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr]
\biggl[ \frac{z_0 \zeta_2}{a^2 + c^2} \biggr]
\, .
</math>
</td>
</tr>
</table>

==Lagrangian Particle Motion==

Now, if we assume that each Lagrangian particle executes a closed elliptical orbit ''in the plane'' of the tipped coordinate system (''i.e.,'' <math>~z' = \dot{z}' = 0</math> ), but whose orbit-center may be shifted by an amount, <math>~y_0</math>, away from the z'-axis, we expect …
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~x'</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~x_\mathrm{max}\cos(\varphi t)</math>
</td>
<td align="center">      and,      
<td align="right">
<math>~y' - y_0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~y_\mathrm{max}\sin(\varphi t) \, ,</math>
</td>
</tr>
</table>
which implies,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~\dot{x}'</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~- x_\mathrm{max}~ \varphi \cdot \sin(\varphi t)</math>
</td>
<td align="center">      and,      
<td align="right">
<math>~\dot{y}' </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~y_\mathrm{max}~\varphi \cdot \cos(\varphi t) \, .</math>
</td>
</tr>
</table>
That is to say, we expect that the two velocity components will exhibit the form, respectively,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>u_1^'</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[\frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \varphi \cdot (y_0 - y') \, ,
</math>
</td>

<td align="center">      and,      </td>

<td align="right">
<math>u_2^'</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[\frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr] \varphi \cdot x' \, .
</math>
</td>

</tr>
</table>
This pair of expressions will only match the underlying Riemann Type I flow if,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\biggl\{\frac{1+\Lambda^2}{b^2 + \Lambda^2 \cdot c^2 }\biggr\}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr] </math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{x_\mathrm{max}}{y_\mathrm{max}}
\, ,
</math>
      and,
</td>
</tr>

<tr>
<td align="right"><math>\biggl[ \frac{a^2 + b^2}{a^2 \zeta_3 \cos\theta}\biggr] </math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\frac{y_\mathrm{max}}{x_\mathrm{max}} \, .
</math>
</td>
</tr>
</table>
That is,

<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>\biggl\{\frac{1+\Lambda^2}{b^2 + \Lambda^2 \cdot c^2 }\biggr\}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr] </math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl[ \frac{a^2 \zeta_3 \cos\theta}{a^2 + b^2}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~
\frac{1}{a^2}\biggl\{\frac{1+\Lambda^2}{b^2 + \Lambda^2 \cdot c^2 }\biggr\}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr]^2 </math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
1 \, .
</math>
</td>
</tr>
</table>
Let's try the numerical values from the [[ThreeDimensionalConfigurations/RiemannTypeI#Example_b1.25c0.470|example Type I ellipsoid used in an accompanying discussion]], namely, <math>b/a = 1.25, c/a = 0.4703, \zeta_2 = -2.2794, \zeta_3 = -1.9637</math>.

Hence,

<table border="0" align="center" cellpadding="8">

<tr>
<td align="right"><math>\tan\theta</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
- \frac{\zeta_2}{\zeta_3}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]\frac{c^2}{b^2}
= - 0.3448 ~~~ \Rightarrow ~~~ \theta = -0.3320
\, ,
</math>
</td>
</tr>

<tr>
<td align="right"><math>\Rightarrow ~~~ \cos\theta</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
- 0.9454
\, .
</math>
</td>
</tr>
</table>
And,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\Lambda^2</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\zeta_2^2}{\zeta_3^2}\biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr]^2\frac{c^2}{b^2}
= \tan^2\theta \cdot \frac{b^2}{c^2}
= 0.8399
\, .</math>
</td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>\Rightarrow ~~~
\frac{1}{a^2}\biggl\{\frac{1+\Lambda^2}{b^2 + \Lambda^2 \cdot c^2 }\biggr\}\biggl[ \frac{a^2 + b^2}{\zeta_3 \cos\theta}\biggr]^2 </math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
\biggl\{ 1.0524 \biggr\} \biggl[ 1.3803 \biggr]^2
=
2.005
\, .
</math>
</td>
</tr>
</table>

=See Also=
<ul>
<li>[https://en.wikipedia.org/wiki/Euler_angles Wikipedia Chapter on Euler Angles]</li>
<li>[https://phas.ubc.ca/~berciu/TEACHING/PHYS206/LECTURES/FILES/euler.pdf Online class notes] from [https://phas.ubc.ca/~berciu/ Professor Mona Berciu], Department of Physics & Astronomy, University of British Columbia</li>
</ul>

 <br />
{{SGFfooter}}

2024-04-20T17:35:14Z

Jet53man: /* Overview */

__FORCETOC__

<font color="red"><b>
NOTE to Eric Hirschmann & David Neilsen...
</b></font>
I have moved the earlier contents of this page to a new Wiki location called
[[Apps/RiemannEllipsoidsCompressible|Compressible Riemann Ellipsoids]].

=Rotating Reference Frame=
==Overview==
<span id="InertialFrame">Among</span> the [[PGE#Principal_Governing_Equations|principal governing equations]] we have included the
 <br />
 <br />

<div align="center">
<span id="ConservingMomentum:Lagrangian"><font color="#770000">'''Lagrangian Representation'''</font></span><br />
of the inertial-frame Euler Equation,

{{Template:Math/EQ_Euler01}}

[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], p. 671, Appendix Eq. (1E-6)<br />
[<b>[[Appendix/References#BLRY07|<font color="red">BLRY07</font>]]</b>], p. 13, Eq. (1.55)
</div>
Alternatively, a rewrite of the LHS gives what we refer to as the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the intertial-frame Euler Equation
</td>
</tr>
<tr>
<td align="right">
<math>
\frac{\partial \vec{v}}{\partial t} + (\vec{v} \cdot \nabla)\vec{v}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-\frac{1}{\rho}\nabla P - \nabla \Phi \, .
</math>
</td>
</tr>
</table>
At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with an angular velocity
<div align="center">
<math>\vec{\Omega} = \hat\imath ~\Omega_1 + \hat\jmath ~\Omega_2 + \hat{k}~\Omega_3 \, .</math>
</div>

----

<table border="0" align="center" width="90%" cellpadding="8"><tr><td align="left">Often it suffices to align <math>\vec\Omega</math> with the z-axis of the chosen coordinate system — in which case, <math>\Omega_1 =\Omega_2 =0</math> — and to set <math>d\vec\Omega/dt = 0</math>, in which case the nonzero component of the frame's angular velocity, <math>\Omega_3</math>, is independent of time.</td></tr></table>

----

In what follows we show that, when viewed from this rotating reference frame, we have what will be referred to as the
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the ''rotating-frame'' Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>
\frac{d\vec{u}}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, ,
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">{{ Rossner67 }}, §II, Eq. (1)<br />
[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], Appendix 1.D, §3, (p. 664) Eq. (1D-42)</td>
</tr>
</table>
where the difference between the rotating-frame velocity, <math>\vec{u}</math>, and the inertial-frame velocity, <math>\vec{v}</math>, is given by the expression,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\vec{v} - \vec{u}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\vec\Omega \times \vec{x}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\hat\imath (\Omega_2 z - \Omega_3 y)
+ \hat\jmath (\Omega_3 x - \Omega_1 z)
+ \hat{k} (\Omega_1 y - \Omega_2 x))
\, .
</math>
</td>
</tr>
</table>

As above, a rewrite of the LHS gives what we will refer to as the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Eulerian Representation'''</font><br />of the ''rotating-frame'' Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>
\frac{\partial\vec{u}}{dt} + (\vec{u} \cdot \nabla)\vec{u}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, .
</math>
</td>
</tr>
</table>

Along the way, and being guided by Chandrasekhar's presentation in EFE, we appreciate that it can be useful to highlight a ''hybrid'' representation of the Euler Equation that involves a mixture of the velocity variables, <math>\vec{u}</math> along with <math>\vec{v}</math>; specifically,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>-\frac{1}{\rho}\nabla P - \nabla \Phi</math>
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)[\vec{u} + \vec\Omega \times \vec{x}]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t}
+ (\vec{v}\cdot \nabla)\vec{u}
+ (\vec{v}\cdot \nabla)[ \vec\Omega \times \vec{x}]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ \biggl[ v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \biggr]
\biggl[
\hat\imath (\Omega_2 z - \Omega_3 y)
+ \hat\jmath (\Omega_3 x - \Omega_1 z)
+ \hat{k} (\Omega_1 y - \Omega_2 x))
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ v_x
\biggl[
\hat\jmath (\Omega_3 )
- \hat{k} (\Omega_2 )
\biggr]
+ v_y
\biggl[
-\hat\imath (\Omega_3)
+ \hat{k} (\Omega_1 )
\biggr]
+ v_z
\biggl[
\hat\imath (\Omega_2 )
- \hat\jmath (\Omega_1)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ \underbrace{\hat\imath (\Omega_2 v_z - \Omega_3 v_y)
+ \hat\jmath (\Omega_3 v_x - \Omega_1 v_z)
+\hat{k} (\Omega_1 v_y - \Omega_2 v_x)}_{\vec\Omega\times\vec{v}} \, .
</math>
</td>
</tr>
</table>

That is, along the way we derive a,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Hybrid (Eulerian) Representation'''</font><br />of the rotating-frame Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\vec\Omega \times \vec{v}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]
\, .
</math>
</td>
</tr>
</table>

<font color="red"><b>NOPE!</b></font>    Alternatively,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>-\frac{1}{\rho}\nabla P - \nabla \Phi</math>
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + \biggl[(\vec{u} + \vec\Omega \times \vec{x})\cdot \nabla \biggr]\vec{v}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t}
+ \biggl[\vec{u} \cdot \nabla \biggr]\vec{v}
+ \biggl\{\biggl[\hat\imath(\Omega_2 z - \Omega_3 y)
+ \hat\jmath(\Omega_3 x - \Omega_1 z) + \hat{k}(\Omega_1 y - \Omega_2 x)\biggr]
\cdot \nabla \biggr\}\vec{v}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t}
+ \biggl[\vec{u} \cdot \nabla \biggr]\vec{v}
+ \biggl\{\biggl[ (\Omega_2 z - \Omega_3 y)\frac{\partial}{\partial x}
+ (\Omega_3 x - \Omega_1 z)\frac{\partial}{\partial y} + (\Omega_1 y - \Omega_2 x) \frac{\partial}{\partial z}\biggr]
\biggr\}\vec{v}
</math>
</td>
</tr>
</table>

==Coordinate Transformation==

===Traditional Presentation===
At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with a uniform (''i.e.,'' time-independent) angular velocity <math>~\Omega_f</math>. In order to transform any one of the [[PGE#Principal_Governing_Equations|principal governing equations]] from the inertial reference frame to such a rotating reference frame, we must specify the orientation as well as the magnitude of the angular velocity vector about which the frame is spinning, <math>{\vec\Omega}_f</math>; and the <math>~d/dt</math> operator, which denotes Lagrangian time-differentiation in the inertial frame, must everywhere be replaced as follows:
<div align="center">
<math>
\biggl[\frac{d}{dt} \biggr]_{inertial} \rightarrow \biggl[\frac{d}{dt} \biggr]_{rot} + {\vec{\Omega}}_f \times .
</math>

[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (11)

</div>

Operating on the fluid element's position vector, <math>\vec{x}</math>, we obtain the transformation,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\frac{d\vec{x}}{dt}\biggr|_\mathrm{inertial}</math></td>
<td align="center"><math>\rightarrow</math></td>
<td align="right"><math>\frac{d\vec{x}}{dt}\biggr|_\mathrm{rotating} + \vec\Omega \times \vec{x} \, ,</math></td>
</tr>
</table>
that is,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\vec{v}_\mathrm{inertial}</math></td>
<td align="center"><math>\rightarrow</math></td>
<td align="right"><math>\vec{v}_\mathrm{rot} + \vec\Omega \times \vec{x} \, .</math></td>
</tr>
<tr>
<td align="center" colspan="3" align="center">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (15)
</td>
</tr>
</table>

Performing this transformation implies, for example, that
<div align="center">
<math>
\vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}}_f \times \vec{x} ,
</math>
</div>
and,
<div align="center">
<math>
\biggl[ \frac{d\vec{v}}{dt}\biggr]_{inertial} = \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} + {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
</math>

<math>
= \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - \frac{1}{2} \nabla \biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]
</math>
</div>
(If we were to allow <math>{\vec\Omega}_f</math> to be a function of time, an additional term involving the time-derivative of <math>{\vec\Omega}_f</math> also would appear on the right-hand-side of these last expressions; see, for example, Eq.~1D-42 in [[Appendix/References|BT87]].) Note as well that the relationship between the fluid [[PGE/RotatingFrame#WikiVorticity|vorticity]] in the two frames is,
<div align="center">
<math>
[\vec\zeta]_{inertial} = [\vec\zeta]_{rot} + 2{\vec\Omega}_f .
</math>
</div>

===Chandrasekhar's Approach===

Here we draw extensively from Chapter 4, §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

====Transformation Matrix====

Following Chandrasekhar, we let <math>\vec{X}</math> represent the inertial-frame position vector of a fluid element, in which case <math>d\vec{X}/dt</math> is the inertial-frame velocity <math>(\vec{v})</math> of that fluid element, and the acceleration, <math>d\vec{v}/dt</math>, that appears on the LHS of the [[PGE/Euler#Lagrangian_Representation|Lagrangian representation of the (intertial-frame) Euler equation]] may be rewritten as the second time-derivative of <math>\vec{X}</math>, namely,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the (inertial-frame) Euler Equation</td>
</tr>
<tr>
<td align="right">
<math>\frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{inertial} \, .</math>
</td>
</tr>
</table>

Chandrasekhar uses the matrix, <math>\mathbf{T}(t)</math>, to represent the (time-dependent) linear transformation that relates <math>\vec{X}</math> to the corresponding moving-frame position vector, <math>\vec{x}</math>. Specifically, he sets,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{x}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{T}\vec{X} \, .</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (1)</td>
</tr>
</table>
Applying the same transformation to the inertial-frame velocity, <math>d\vec{X}/dt</math>, gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{T}\frac{d\vec{X}}{dt} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (14a)</td>
</tr>
</table>
which Chandrasekhar refers to as the velocity in the <b>inertial frame</b> that has been <font color="darkgreen">"resolved along the instantaneous coordinate axes of the moving frame."</font> And applying this transformation to the inertial-frame acceleration gives the term, <math>\mathbf{T} [d^2\vec{X}/dt^2]</math>, which Chandrasekhar describes as representing <font color="darkgreen">"… the acceleration in the <b>inertial frame</b> resolved, however, along the instantaneous directions of the coordinate axes of the moving frame."</font> Applying the transformation to both sides of the Lagrangian representation of the Euler equation gives,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>\mathbf{T} \frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{moving} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, combination of Eqs. (16) & (17)</td>
</tr>
</table>
where, as Chandrasekhar clarifies, the gradients on the RHS must be <font color="darkgreen">"… evaluated in the coordinates of the moving frame."</font>

====Rotating-Frame Euler Equation====

=====Foundation=====
Suppose the 3-component vector, <math>\vec\Omega</math>, represents a general time-dependent rotation of the <math>(x_1, x_2, x_3)</math>-frame with respect to the inertial frame. In this context, Chandrasekhar introduces a (3 × 3) matrix, <math>\mathbf{\Omega^*}</math>, whose nine components can be expressed in terms of the three components of <math>\vec\Omega</math> via the relations,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{ij}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{ijk}\Omega_k \, .</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (6a)</td>
</tr>
</table>
Alternatively, we may write,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{ik}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{ikj}\Omega_j = -~\epsilon_{ijk}\Omega_j\, .</math>
</td>
</tr>
</table>

<table border="1" align="center" width="60%" cellpadding="8"><tr><td align="left">

Both of these expressions make use of the three-element [https://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition Levi-Civita tensor], <math>\epsilon_{ijk}</math>. Its six nonzero component values are …
<table border="1" align="center" cellpadding="5">
<tr>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
<td rowspan="4" bgcolor="lightgrey"> </td>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
</tr>
<tr>
<td align="center">123</td>
<td align="center" rowspan="3">+1</td>
<td align="center">132</td>
<td align="center" rowspan="3">-1</td>
</tr>
<tr>
<td align="center">312</td>
<td align="center">321</td>
</tr>
<tr>
<td align="center">231</td>
<td align="center">213</td>
</tr>
</table>
Hence, the six nonzero components of the matrix, <math>\mathbf{\Omega^*}</math>, are,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{12}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_3\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{13}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_2\, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>(\Omega^*)_{21}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_3\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{23}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_1\, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>(\Omega^*)_{31}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_2\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{32}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_1\, .</math>
</td>
</tr>
</table>

----

<table border="0" align="center" cellpadding="10" width="30px">

<tr>
<td align="center" colspan="3" bgcolor="white"><math>\mathbf{\Omega^*}</math><br /><font size="-1">(3 × 3 matrix)</font></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_3</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_2</math></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_3</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_1</math></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_2</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_1</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
</tr>
</table>

</td></tr></table>

<span id="Product">For later use,</span> we note as well that for an arbitrary vector — call it, <math>\vec{Q}</math> — the individual components of the product, <math>\mathbf{\Omega^*} \vec{Q}</math>, are given by the expression,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\mathbf{\Omega^*}\vec{Q} )_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(\Omega^*)_{ij}Q_j
=
(\epsilon_{ijk}\Omega_k)Q_j
\, .
</math>
</td>
</tr>
</table>
Compare, for example, Eqs. (17) and (19) in §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

Now, if the motion of the moving frame relative to the inertial frame is specified entirely by the vector <math>\vec\Omega</math>, Chandrasekhar proves that any time-dependent vector defined in the inertial frame — call it <math>\vec{F}</math> — will obey the following operator relation:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr]\vec{F}
\, .
</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (11)</td>
</tr>
</table>

=====Lagrangian Representation In Terms of Inertial-Frame Velocities=====

For example, if we set <math>\vec{F} = d\vec{X}/dt</math>, we find,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr] \frac{d\vec{X}}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \mathbf{T} \frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\frac{d\vec{X}}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{d\vec{U}}{dt} - \mathbf{\Omega^*}\vec{U} \, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eqs. (13) & (16)</td>
</tr>
</table>
This allows us to write the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the (transformed) inertial-frame velocity, <math>\vec{U}</math></td>
</tr>
<tr>
<td align="right">
<math>\frac{d\vec{U}}{dt} - \mathbf{\Omega^*}\vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d\vec{U}}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{\Omega^*}\vec{U}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating}
\, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (17)</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
Appreciating [[#Product|from above]] that <math>(\mathbf{\Omega^*}\vec{Q})_i = \epsilon_{ijk}\Omega_k Q_j</math>, in component form this version of the Euler equation reads,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{imk}\Omega_k U_m -\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} \, .</math>
</td>
</tr>
</table>
That is,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">Component #1:   </td>
<td align="right">
<math>\frac{dU_1}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_3 U_2 - \Omega_2 U_3
-\frac{1}{\rho} \frac{\partial p}{\partial x_1} - \frac{\partial\Phi}{\partial x_1} </math>
</td>
</tr>

<tr>
<td align="right">Component #2:   </td>
<td align="right">
<math>\frac{dU_2}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_1 U_3 - \Omega_3 U_1
-\frac{1}{\rho} \frac{\partial p}{\partial x_2} - \frac{\partial\Phi}{\partial x_2} </math>
</td>
</tr>

<tr>
<td align="right">Component #3:   </td>
<td align="right">
<math>\frac{dU_3}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_2 U_1 - \Omega_1 U_2
-\frac{1}{\rho} \frac{\partial p}{\partial x_3} - \frac{\partial\Phi}{\partial x_3} </math>
</td>
</tr>
</table>

Notice as well that the individual components of the cross product of <math>\vec{U}</math> and <math>\vec\Omega</math> can be represented by the same summation expression, that is,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\vec{U} \times \vec\Omega)_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{imk}\Omega_k U_m \, .</math>
</td>
</tr>
</table>

This allows us to rewrite the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the (transformed) inertial-frame velocity, <math>\vec{U}</math></td>
</tr>

<tr>
<td align="right">
<math>\frac{d\vec{U}}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\vec\Omega \times \vec{U}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating}
\, .
</math>
</td>
</tr>
</table>
in what is perhaps more recognizable notation.

</td></tr></table>

=====Lagrangian Representation In Terms of Rotating-Frame Velocities=====

<span id="Utou">Alternatively,</span> setting <math>\vec{F} = \vec{X}</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr]\vec{X}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \mathbf{T}\frac{d\vec{X}}{dt}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}\vec{X}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\vec{x}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\vec{u} - \mathbf{\Omega^*}\vec{x} \, ,
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eqs. (12) & (15)</td>
</tr>
</table>

where, adopting Chandrasekhar's notation, the variable,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{u}</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{d\vec{x}}{dt} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (14b)</td>
</tr>
</table>
denotes the fluid velocity as measured <font color="darkgreen">"… with respect to an observer [that is] at rest in the moving frame."</font> This allows us to write the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the rotating-frame velocity, <math>\vec{u}</math></td>
</tr>

<tr>
<td align="right">
<math>
\frac{d}{dt}\biggl[ \vec{u} - \mathbf{\Omega^*}\vec{x} \biggr]
- \mathbf{\Omega^*}\biggl[ \vec{u} - \mathbf{\Omega^*}\vec{x} \biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} </math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
Again appreciating [[#Product|from above]] that <math>(\mathbf{\Omega^*}\vec{Q})_i = \epsilon_{ijk}\Omega_k Q_j = -\epsilon_{ijk}\Omega_j Q_k</math>, in component form this version of the Euler equation reads,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d}{dt}\biggl[ u_i + \epsilon_{ijk}\Omega_j x_k \biggr] </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}\Omega_k \biggl[ u_m + \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d u_i}{dt} + \epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}u_m\Omega_k
+
\epsilon_{imk}\Omega_k \biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
Now, if we …
<ol type="a">
<li>Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
-~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr] \, ;
</math>
</td>
</tr>
</table>
note also that we have set <math>dx_j/dt \rightarrow u_j</math>;
</li>
<li>In the first term on the RHS, replace the index, "m", with the index, "j":

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk} u_m \Omega_k
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Inside the square brackets of the second term on the RHS, replace the "jk" indices with "h&ell;" in order to avoid confusion, then swap the "h&ell;" indices of the two variables, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[ \epsilon_{mh\ell}\Omega_h x_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h\Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
- ~ \epsilon_{ikm}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
</ol>

the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d u_i}{dt} -~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k
+
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~\frac{d u_i}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\underbrace{2\epsilon_{ijk} u_j \Omega_k}_{[2\vec{u}\times \vec\Omega]_i}
+
\underbrace{\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]}_{[\vec\Omega \times(\vec{x}\times\vec\Omega)]_i}
+
\underbrace{\epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) \biggr]}_{[\vec{x} \times (d\vec\Omega/dt)]_i}
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
</td></tr></table>

We therefore can rewrite in a more familiar ''vector'' formulation, the
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the rotating-frame velocity, <math>\vec{u}</math></td>
</tr>

<tr>
<td align="right">
<math>
\frac{d\vec{u}}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">{{ Rossner67 }}, §II, Eq. (1)<br />
[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], Appendix 1.D, §3, (p. 664) Eq. (1D-42)</td>
</tr>
</table>
----

The RHS of this equation matches the RHS of Eq. (1) from {{ Rossner67full }} after making the notation switch, <math>\Phi \rightarrow -~\mathfrak{B}</math>, and after acknowledging that <math>\nabla P/\rho \rightarrow \nabla(P/\rho)</math> when the mass-density is spatially uniform. The referenced equation from [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>] contains all of the terms shown here, except there, the effects of pressure are ignored.

----

====Component Form====

<font color="red">EXAMPLE #1:</font> Inertial-frame velocities, <math>\vec{U}</math>, as viewed in the inertial frame.

Adding the Euler equation,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\rho\frac{d\vec{U}}{dt}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi \, ,
</math>
</td>
</tr>
</table>
to the product of the inertial-frame velocity and the equation of continuity,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\vec{U}~\frac{d\rho}{dt}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\rho\vec{U}(\nabla\cdot \vec{u})\, ,
</math>
</td>
</tr>
</table>
gives,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\frac{d(\rho\vec{U})}{dt} + \rho\vec{U}(\nabla\cdot \vec{u})</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi
</math>
</td>
</tr>
<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{\partial(\rho\vec{U})}{\partial t} + \nabla\cdot [(\rho\vec{U})\vec{u}]</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi
</math>
</td>
</tr>
</table>

<span id="ComponentForm">In component form,</span> the relation between <math>\vec{U}</math> and <math>\vec{u}</math> reads,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="left">
<math>
U_i
=
u_i - (\Omega^*)_{ik}x_k
=
u_i + \overbrace{\epsilon_{ijk} \Omega_j x_k}^{[~\vec{\Omega} \times \vec{x}~]_i}
\, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="1">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (21)</td>
</tr>
</table>

and the rotating-frame Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
U_i
=
u_i - (\Omega^*)_{ik}x_k
=
u_i + \overbrace{\epsilon_{ijk} \Omega_j x_k}^{[~\vec{\Omega} \times \vec{x}~]_i}
\, .
</math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>

</math>
</td>
</tr>

<tr>
<td align="center" colspan="1">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (21)</td>
</tr>
</table>

===Part B===
Drawing from Chapter 4, §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] — where the Cartesian components of the inertial-frame velocity <math>(\vec{v}_\mathrm{inertial})</math> are represented by <math>U_i</math> and the Cartesian components of the rotating-frame velocity <math>(\vec{v}_\mathrm{rot})</math> are represented by <math>u_i</math> — we begin by restating the [[PGE/Euler#in_terms_of_velocity|Lagrangian representation of the intertial-frame Euler equation]]:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{inertial}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} \, .</math>
</td>
</tr>
</table>

The LHS of this (Euler) equation transform as follows:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{inertial}</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{rot} - \epsilon_{imk}\Omega_k U_m\, ,</math>
</td>
</tr>
</table>
where we also recognize that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_i</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_i + \epsilon_{ijk}\Omega_j x_k \, .</math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">

Both of these expressions make use of the three-element [https://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition Levi-Civita tensor], <math>\epsilon_{ijk}</math>. Its six nonzero component values are …
<table border="1" align="center" cellpadding="5">
<tr>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
<td rowspan="4" bgcolor="lightgrey"> </td>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
</tr>
<tr>
<td align="center">123</td>
<td align="center" rowspan="3">+1</td>
<td align="center">132</td>
<td align="center" rowspan="3">-1</td>
</tr>
<tr>
<td align="center">312</td>
<td align="center">321</td>
</tr>
<tr>
<td align="center">231</td>
<td align="center">213</td>
</tr>
</table>

Hence, for example, transforming the x-component <math>(i=1)</math> of <math>\vec{U}</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_1</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_1 + \epsilon_{1jk}\Omega_j x_k = u_1 + \epsilon_{123}\Omega_2 x_3 + \epsilon_{132}\Omega_3 x_2
= u_1 + \Omega_2 z -\Omega_3 y \, ;</math>
</td>
</tr>
</table>
transforming the y-component <math>(i=2)</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_2</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_2 + \epsilon_{2jk}\Omega_j x_k = u_2 + \epsilon_{231}\Omega_3 x_1 + \epsilon_{213}\Omega_1 x_3
= u_2 + \Omega_3 x -\Omega_1 z \, ;</math>
</td>
</tr>
</table>
and transforming the z-component <math>(i=3)</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_3</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_3 + \epsilon_{3jk}\Omega_j x_k = u_3 + \epsilon_{312}\Omega_1 x_2 + \epsilon_{321}\Omega_2 x_1
= u_3 + \Omega_1 y -\Omega_2 x \, .</math>
</td>
</tr>
</table>

These are the same three components that arise from the vector expression (from above),
<div align="center">
<math>
\vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}} \times \vec{x} \, ;
</math>
</div>
we therefore recognize that, <math>\vec{\Omega} \times \vec{x} = \epsilon_{ijk}\Omega_j x_k</math>. We note as well that, <math>\vec{\Omega} \times \vec{x} = -\epsilon_{ijk}\Omega_k x_j</math>.

</td></tr></table>

Therefore, as viewed from the rotating frame of reference, the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{rot} - \epsilon_{imk}\Omega_k U_m</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d}{dt}\biggl[ u_i + \epsilon_{ijk}\Omega_j x_k \biggr] </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}\Omega_k \biggl[ u_m + \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d u_i}{dt} + \epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}u_m\Omega_k
+
\epsilon_{imk}\Omega_k \biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
Now, if we …
<ol type="a">
<li>Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
-~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
</tr>
</table>
note also that we have set <math>dx_j/dt \rightarrow u_j</math>;
</li>
<li>In the first term on the RHS, replace the index, "m", with the index, "j":

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk} u_m \Omega_k
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Inside the square brackets of the second term on the RHS, replace the "jk" indices with "h&ell;" in order to avoid confusion, then swap the "h&ell;" indices of the two variables, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[ \epsilon_{mh\ell}\Omega_h x_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h\Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
- ~ \epsilon_{ikm}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
</ol>

the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d u_i}{dt} -~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k
+
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~\frac{d u_i}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\underbrace{2\epsilon_{ijk} u_j \Omega_k}_{[2\vec{u}\times \vec\Omega]_i}
+
\underbrace{\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]}_{[\vec\Omega \times(\vec{x}\times\vec\Omega)]_i}
+
\underbrace{\epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) \biggr]}_{[\vec{x} \times (d\vec\Omega/dt)]_i}
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>

==Continuity Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the continuity equation presented [[PGE/ConservingMass#Continuity_Equation|elsewhere]], we obtain the:

<div align="center">
<font color="#770000">'''Lagrangian Representation'''</font><br />
of the Continuity Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{d\rho}{dt} \biggr]_{rot} + \rho \nabla \cdot {\vec{v}}_{rot} = 0</math> ;
</div>

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Continuity Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{\partial\rho}{\partial t} \biggr]_{rot} + \nabla \cdot (\rho {\vec{v}}_{rot}) = 0</math> .
</div>

==Euler Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the Euler equation presented [[PGE/Euler#Euler_Equation|elsewhere]], we obtain the:

<div align="center">
<font color="#770000">'''Lagrangian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math> ;
</div>

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math> ;
</div>

<div align="center">
Euler Equation<br />
written <font color="#770000">'''in terms of the Vorticity'''</font> and<br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}v_{rot}^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .
</div>

==Centrifugal and Coriolis Accelerations==

Following along the lines of the discussion presented in Appendix 1.D, §3 of [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], in a rotating reference frame the Lagrangian representation of the Euler equation may be written in the form,
<div align="center">
<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi + {\vec{a}}_{fict} </math>,
</div>
where,
<div align="center">
<math>
{\vec{a}}_{fict} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) .
</math>
</div>
So, as viewed from a rotating frame of reference, material moves as if it were subject to two ''fictitious accelerations'' which traditionally are referred to as the,
<div align="center">
<font color="#770000">'''Coriolis Acceleration'''</font>

<math>
{\vec{a}}_{Coriolis} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} ,
</math>
</div>

(see the related [[PGE/RotatingFrame#WikiCoriolis|Wikipedia discussion]]) and the
<div align="center">
<font color="#770000">'''Centrifugal Acceleration'''</font>

<math>
{\vec{a}}_{Centrifugal} \equiv - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
= \frac{1}{2} \nabla\biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]
</math>
</div>
(see the related [[PGE/RotatingFrame#WikiCentrifugal|Wikipedia discussion]]).

==Nonlinear Velocity Cross-Product==

In some contexts — for example, our discussion of [[Apps/RiemannEllipsoidsCompressible|Riemann ellipsoids]] or the analysis by [[Apps/Korycansky_Papaloizou_1996|Korycansky & Papaloizou (1996)]] of nonaxisymmetric disk structures — it proves useful to isolate and analyze the term in the "vorticity formulation" of the Euler equation that involves a nonlinear cross-product of the rotating-frame velocity vector, namely,
<div align="center">
<math>
\vec{A} \equiv ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot} .
</math>
</div>

NOTE: To simplify notation, for most of the remainder of this subsection we will drop the subscript "rot" on both the velocity and vorticity vectors.

===Align Ω<sub>f</sub> with z-axis===
Without loss of generality we can set <math>{\vec\Omega}_f = \hat{k}\Omega_f</math>, that is, we can align the frame rotation axis with the z-axis of a Cartesian coordinate system. The Cartesian components of <math>{\vec{A}}</math> are then,
<div align="left">
<math>
\hat{i}: ~~~~~~ A_x = \zeta_y v_z - (\zeta_z + 2\Omega) v_y ,
</math><br />

<math>
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x - \zeta_x v_z ,
</math><br />

<math>
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</div>
where it is understood that the three Cartesian components of the vorticity vector are,
<div align="center">
<math>
\zeta_x = \biggl[\frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} \biggr] ,
~~~~~~ \zeta_y = \biggl[ \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} \biggr] ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</div>
In turn, the curl of <math>\vec{A}</math> has the following three Cartesian components:
<div align="left">
<math>
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr],
</math><br />

<math>
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = \frac{\partial}{\partial z}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math><br />

<math>
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr] - \frac{\partial}{\partial y}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] .
</math>
</div>

===When v<sub>z</sub> = 0 ===

If we restrict our discussion to configurations that exhibit only planar flows — that is, systems in which <math>v_z = 0</math> — then the Cartesian components of <math>{\vec{A}}</math> and <math>\nabla\times\vec{A}</math> simplify somewhat to give, respectively,
<div align="left">
<math>
\hat{i}: ~~~~~~ A_x = - (\zeta_z + 2\Omega) v_y ,
</math><br />

<math>
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x ,
</math><br />

<math>
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</div>

and,
<div align="left">
<math>
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x \biggr],
</math><br />

<math>
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial}{\partial z}\biggl[(\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math><br />

<math>
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x \biggr] + \frac{\partial}{\partial y}\biggl[ (\zeta_z + 2\Omega) v_y \biggr] ,
</math>
</div>
where, in this case, the three Cartesian components of the vorticity vector are,
<div align="center">
<math>
\zeta_x = - \frac{\partial v_y}{\partial z} ,
~~~~~~ \zeta_y = \frac{\partial v_x}{\partial z} ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</div>

=Related Discussions=
* <span id="WikiVorticity">Wikipedia discussion of [http://en.wikipedia.org/wiki/Vorticity vorticity].</span>
* <span id="WikiCoriolis">Wikipedia discussion of [http://en.wikipedia.org/wiki/Coriolis_effect#Formula Coriolis Effect].</span>
* <span id="WikiCentrifugal">Wikipedia discussion of [http://en.wikipedia.org/wiki/Centrifugal_force#Derivation_using_vectors Centrifugal acceleration].</span>

{{ SGFfooter }}

PGE/RotatingFrame

2024-04-19T16:48:23Z

Jet53man: /* Overview */

__FORCETOC__

<font color="red"><b>
NOTE to Eric Hirschmann & David Neilsen...
</b></font>
I have moved the earlier contents of this page to a new Wiki location called
[[Apps/RiemannEllipsoidsCompressible|Compressible Riemann Ellipsoids]].

=Rotating Reference Frame=
==Overview==
<span id="InertialFrame">Among</span> the [[PGE#Principal_Governing_Equations|principal governing equations]] we have included the
 <br />
 <br />

<div align="center">
<span id="ConservingMomentum:Lagrangian"><font color="#770000">'''Lagrangian Representation'''</font></span><br />
of the inertial-frame Euler Equation,

{{Template:Math/EQ_Euler01}}

[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], p. 671, Appendix Eq. (1E-6)<br />
[<b>[[Appendix/References#BLRY07|<font color="red">BLRY07</font>]]</b>], p. 13, Eq. (1.55)
</div>
Alternatively, a rewrite of the LHS gives what we refer to as the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the intertial-frame Euler Equation
</td>
</tr>
<tr>
<td align="right">
<math>
\frac{\partial \vec{v}}{\partial t} + (\vec{v} \cdot \nabla)\vec{v}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
-\frac{1}{\rho}\nabla P - \nabla \Phi \, .
</math>
</td>
</tr>
</table>
At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with an angular velocity
<div align="center">
<math>\vec{\Omega} = \hat\imath ~\Omega_1 + \hat\jmath ~\Omega_2 + \hat{k}~\Omega_3 \, .</math>
</div>

----

<table border="0" align="center" width="90%" cellpadding="8"><tr><td align="left">Often it suffices to align <math>\vec\Omega</math> with the z-axis of the chosen coordinate system — in which case, <math>\Omega_1 =\Omega_2 =0</math> — and to set <math>d\vec\Omega/dt = 0</math>, in which case the nonzero component of the frame's angular velocity, <math>\Omega_3</math>, is independent of time.</td></tr></table>

----

In what follows we show that, when viewed from this rotating reference frame, we have what will be referred to as the
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the ''rotating-frame'' Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>
\frac{d\vec{u}}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, ,
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">{{ Rossner67 }}, §II, Eq. (1)<br />
[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], Appendix 1.D, §3, (p. 664) Eq. (1D-42)</td>
</tr>
</table>
where the difference between the rotating-frame velocity, <math>\vec{u}</math>, and the inertial-frame velocity, <math>\vec{v}</math>, is given by the expression,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\vec{v} - \vec{u}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\vec\Omega \times \vec{x}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\hat\imath (\Omega_2 z - \Omega_3 y)
+ \hat\jmath (\Omega_3 x - \Omega_1 z)
+ \hat{k} (\Omega_1 y - \Omega_2 x))
\, .
</math>
</td>
</tr>
</table>

As above, a rewrite of the LHS gives what we will refer to as the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Eulerian Representation'''</font><br />of the ''rotating-frame'' Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>
\frac{\partial\vec{u}}{dt} + (\vec{u} \cdot \nabla)\vec{u}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, .
</math>
</td>
</tr>
</table>

Along the way, and being guided by Chandrasekhar's presentation in EFE, we appreciate that it can be useful to highlight a ''hybrid'' representation of the Euler Equation that involves a mixture of the velocity variables, <math>\vec{u}</math> along with <math>\vec{v}</math>; specifically,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>-\frac{1}{\rho}\nabla P - \nabla \Phi</math>
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)[\vec{u} + \vec\Omega \times \vec{x}]</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t}
+ (\vec{v}\cdot \nabla)\vec{u}
+ (\vec{v}\cdot \nabla)[ \vec\Omega \times \vec{x}]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ \biggl[ v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z} \biggr]
\biggl[
\hat\imath (\Omega_2 z - \Omega_3 y)
+ \hat\jmath (\Omega_3 x - \Omega_1 z)
+ \hat{k} (\Omega_1 y - \Omega_2 x))
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ v_x
\biggl[
\hat\jmath (\Omega_3 )
- \hat{k} (\Omega_2 )
\biggr]
+ v_y
\biggl[
-\hat\imath (\Omega_3)
+ \hat{k} (\Omega_1 )
\biggr]
+ v_z
\biggl[
\hat\imath (\Omega_2 )
- \hat\jmath (\Omega_1)
\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}
+ \underbrace{\hat\imath (\Omega_2 v_z - \Omega_3 v_y)
+ \hat\jmath (\Omega_3 v_x - \Omega_1 v_z)
+\hat{k} (\Omega_1 v_y - \Omega_2 v_x)}_{\vec\Omega\times\vec{v}} \, .
</math>
</td>
</tr>
</table>

That is, along the way we derive a,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Hybrid (Eulerian) Representation'''</font><br />of the rotating-frame Euler Equation</td>
</tr>

<tr>
<td align="right">
<math>\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla)\vec{u}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\vec\Omega \times \vec{v}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]
\, .
</math>
</td>
</tr>
</table>

<font color="red"><b>NOPE!</b></font>    Alternatively,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>-\frac{1}{\rho}\nabla P - \nabla \Phi</math>
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t} + \biggl[(\vec{u} + \vec\Omega \times \vec{x})\cdot \nabla \biggr]\vec{v}</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center"><math>=</math> </td>
<td align="left">
<math>\frac{\partial\vec{v}}{\partial t}
+ \biggl[\vec{u} \cdot \nabla \biggr]\vec{v}
+ \biggl\{\biggl[\hat\imath(\Omega_2 z - \Omega_3 y)
+ \hat\jmath(\Omega_3 x - \Omega_1 z) + \hat{k}(\Omega_1 y - \Omega_2 x)\biggr]
\cdot \nabla \biggr\}\vec{v}
</math>
</td>
</tr>
</table>

==Coordinate Transformation==

===Traditional Presentation===
At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with a uniform (''i.e.,'' time-independent) angular velocity <math>~\Omega_f</math>. In order to transform any one of the [[PGE#Principal_Governing_Equations|principal governing equations]] from the inertial reference frame to such a rotating reference frame, we must specify the orientation as well as the magnitude of the angular velocity vector about which the frame is spinning, <math>{\vec\Omega}_f</math>; and the <math>~d/dt</math> operator, which denotes Lagrangian time-differentiation in the inertial frame, must everywhere be replaced as follows:
<div align="center">
<math>
\biggl[\frac{d}{dt} \biggr]_{inertial} \rightarrow \biggl[\frac{d}{dt} \biggr]_{rot} + {\vec{\Omega}}_f \times .
</math>

[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (11)

</div>

Operating on the fluid element's position vector, <math>\vec{x}</math>, we obtain the transformation,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\frac{d\vec{x}}{dt}\biggr|_\mathrm{inertial}</math></td>
<td align="center"><math>\rightarrow</math></td>
<td align="right"><math>\frac{d\vec{x}}{dt}\biggr|_\mathrm{rotating} + \vec\Omega \times \vec{x} \, ,</math></td>
</tr>
</table>
that is,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right"><math>\vec{v}_\mathrm{inertial}</math></td>
<td align="center"><math>\rightarrow</math></td>
<td align="right"><math>\vec{v}_\mathrm{rot} + \vec\Omega \times \vec{x} \, .</math></td>
</tr>
<tr>
<td align="center" colspan="3" align="center">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (15)
</td>
</tr>
</table>

Performing this transformation implies, for example, that
<div align="center">
<math>
\vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}}_f \times \vec{x} ,
</math>
</div>
and,
<div align="center">
<math>
\biggl[ \frac{d\vec{v}}{dt}\biggr]_{inertial} = \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} + {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
</math>

<math>
= \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - \frac{1}{2} \nabla \biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]
</math>
</div>
(If we were to allow <math>{\vec\Omega}_f</math> to be a function of time, an additional term involving the time-derivative of <math>{\vec\Omega}_f</math> also would appear on the right-hand-side of these last expressions; see, for example, Eq.~1D-42 in [[Appendix/References|BT87]].) Note as well that the relationship between the fluid [[PGE/RotatingFrame#WikiVorticity|vorticity]] in the two frames is,
<div align="center">
<math>
[\vec\zeta]_{inertial} = [\vec\zeta]_{rot} + 2{\vec\Omega}_f .
</math>
</div>

===Chandrasekhar's Approach===

Here we draw extensively from Chapter 4, §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

====Transformation Matrix====

Following Chandrasekhar, we let <math>\vec{X}</math> represent the inertial-frame position vector of a fluid element, in which case <math>d\vec{X}/dt</math> is the inertial-frame velocity <math>(\vec{v})</math> of that fluid element, and the acceleration, <math>d\vec{v}/dt</math>, that appears on the LHS of the [[PGE/Euler#Lagrangian_Representation|Lagrangian representation of the (intertial-frame) Euler equation]] may be rewritten as the second time-derivative of <math>\vec{X}</math>, namely,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the (inertial-frame) Euler Equation</td>
</tr>
<tr>
<td align="right">
<math>\frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{inertial} \, .</math>
</td>
</tr>
</table>

Chandrasekhar uses the matrix, <math>\mathbf{T}(t)</math>, to represent the (time-dependent) linear transformation that relates <math>\vec{X}</math> to the corresponding moving-frame position vector, <math>\vec{x}</math>. Specifically, he sets,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{x}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{T}\vec{X} \, .</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (1)</td>
</tr>
</table>
Applying the same transformation to the inertial-frame velocity, <math>d\vec{X}/dt</math>, gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{T}\frac{d\vec{X}}{dt} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (14a)</td>
</tr>
</table>
which Chandrasekhar refers to as the velocity in the <b>inertial frame</b> that has been <font color="darkgreen">"resolved along the instantaneous coordinate axes of the moving frame."</font> And applying this transformation to the inertial-frame acceleration gives the term, <math>\mathbf{T} [d^2\vec{X}/dt^2]</math>, which Chandrasekhar describes as representing <font color="darkgreen">"… the acceleration in the <b>inertial frame</b> resolved, however, along the instantaneous directions of the coordinate axes of the moving frame."</font> Applying the transformation to both sides of the Lagrangian representation of the Euler equation gives,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>\mathbf{T} \frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{moving} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, combination of Eqs. (16) & (17)</td>
</tr>
</table>
where, as Chandrasekhar clarifies, the gradients on the RHS must be <font color="darkgreen">"… evaluated in the coordinates of the moving frame."</font>

====Rotating-Frame Euler Equation====

=====Foundation=====
Suppose the 3-component vector, <math>\vec\Omega</math>, represents a general time-dependent rotation of the <math>(x_1, x_2, x_3)</math>-frame with respect to the inertial frame. In this context, Chandrasekhar introduces a (3 × 3) matrix, <math>\mathbf{\Omega^*}</math>, whose nine components can be expressed in terms of the three components of <math>\vec\Omega</math> via the relations,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{ij}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{ijk}\Omega_k \, .</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (6a)</td>
</tr>
</table>
Alternatively, we may write,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{ik}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{ikj}\Omega_j = -~\epsilon_{ijk}\Omega_j\, .</math>
</td>
</tr>
</table>

<table border="1" align="center" width="60%" cellpadding="8"><tr><td align="left">

Both of these expressions make use of the three-element [https://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition Levi-Civita tensor], <math>\epsilon_{ijk}</math>. Its six nonzero component values are …
<table border="1" align="center" cellpadding="5">
<tr>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
<td rowspan="4" bgcolor="lightgrey"> </td>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
</tr>
<tr>
<td align="center">123</td>
<td align="center" rowspan="3">+1</td>
<td align="center">132</td>
<td align="center" rowspan="3">-1</td>
</tr>
<tr>
<td align="center">312</td>
<td align="center">321</td>
</tr>
<tr>
<td align="center">231</td>
<td align="center">213</td>
</tr>
</table>
Hence, the six nonzero components of the matrix, <math>\mathbf{\Omega^*}</math>, are,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\Omega^*)_{12}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_3\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{13}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_2\, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>(\Omega^*)_{21}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_3\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{23}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_1\, ;</math>
</td>
</tr>

<tr>
<td align="right">
<math>(\Omega^*)_{31}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_2\, ;</math>
</td>

<td align="center">     </td>

<td align="right">
<math>(\Omega^*)_{32}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\Omega_1\, .</math>
</td>
</tr>
</table>

----

<table border="0" align="center" cellpadding="10" width="30px">

<tr>
<td align="center" colspan="3" bgcolor="white"><math>\mathbf{\Omega^*}</math><br /><font size="-1">(3 × 3 matrix)</font></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_3</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_2</math></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_3</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_1</math></td>
</tr>

<tr>
<td align="center" width="10px" bgcolor="lightblue"><math>\Omega_2</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>-~\Omega_1</math></td>
<td align="center" width="10px" bgcolor="lightblue"><math>0</math></td>
</tr>
</table>

</td></tr></table>

<span id="Product">For later use,</span> we note as well that for an arbitrary vector — call it, <math>\vec{Q}</math> — the individual components of the product, <math>\mathbf{\Omega^*} \vec{Q}</math>, are given by the expression,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\mathbf{\Omega^*}\vec{Q} )_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
(\Omega^*)_{ij}Q_j
=
(\epsilon_{ijk}\Omega_k)Q_j
\, .
</math>
</td>
</tr>
</table>
Compare, for example, Eqs. (17) and (19) in §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>].

Now, if the motion of the moving frame relative to the inertial frame is specified entirely by the vector <math>\vec\Omega</math>, Chandrasekhar proves that any time-dependent vector defined in the inertial frame — call it <math>\vec{F}</math> — will obey the following operator relation:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr]\vec{F}
\, .
</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (11)</td>
</tr>
</table>

=====Lagrangian Representation In Terms of Inertial-Frame Velocities=====

For example, if we set <math>\vec{F} = d\vec{X}/dt</math>, we find,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr] \frac{d\vec{X}}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \mathbf{T} \frac{d^2\vec{X}}{dt^2}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\frac{d\vec{X}}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\frac{d\vec{U}}{dt} - \mathbf{\Omega^*}\vec{U} \, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eqs. (13) & (16)</td>
</tr>
</table>
This allows us to write the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the (transformed) inertial-frame velocity, <math>\vec{U}</math></td>
</tr>
<tr>
<td align="right">
<math>\frac{d\vec{U}}{dt} - \mathbf{\Omega^*}\vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d\vec{U}}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\mathbf{\Omega^*}\vec{U}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating}
\, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (17)</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
Appreciating [[#Product|from above]] that <math>(\mathbf{\Omega^*}\vec{Q})_i = \epsilon_{ijk}\Omega_k Q_j</math>, in component form this version of the Euler equation reads,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{imk}\Omega_k U_m -\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} \, .</math>
</td>
</tr>
</table>
That is,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">Component #1:   </td>
<td align="right">
<math>\frac{dU_1}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_3 U_2 - \Omega_2 U_3
-\frac{1}{\rho} \frac{\partial p}{\partial x_1} - \frac{\partial\Phi}{\partial x_1} </math>
</td>
</tr>

<tr>
<td align="right">Component #2:   </td>
<td align="right">
<math>\frac{dU_2}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_1 U_3 - \Omega_3 U_1
-\frac{1}{\rho} \frac{\partial p}{\partial x_2} - \frac{\partial\Phi}{\partial x_2} </math>
</td>
</tr>

<tr>
<td align="right">Component #3:   </td>
<td align="right">
<math>\frac{dU_3}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\Omega_2 U_1 - \Omega_1 U_2
-\frac{1}{\rho} \frac{\partial p}{\partial x_3} - \frac{\partial\Phi}{\partial x_3} </math>
</td>
</tr>
</table>

Notice as well that the individual components of the cross product of <math>\vec{U}</math> and <math>\vec\Omega</math> can be represented by the same summation expression, that is,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>(\vec{U} \times \vec\Omega)_i</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\epsilon_{imk}\Omega_k U_m \, .</math>
</td>
</tr>
</table>

This allows us to rewrite the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the (transformed) inertial-frame velocity, <math>\vec{U}</math></td>
</tr>

<tr>
<td align="right">
<math>\frac{d\vec{U}}{dt} </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-~\vec\Omega \times \vec{U}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating}
\, .
</math>
</td>
</tr>
</table>
in what is perhaps more recognizable notation.

</td></tr></table>

=====Lagrangian Representation In Terms of Rotating-Frame Velocities=====

<span id="Utou">Alternatively,</span> setting <math>\vec{F} = \vec{X}</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>0</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[
\mathbf{T}\frac{d}{dt}
-
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}
\biggr]\vec{X}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \mathbf{T}\frac{d\vec{X}}{dt}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\mathbf{T}\vec{X}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \vec{U}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{d}{dt} - \mathbf{\Omega^*}\biggr)\vec{x}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\vec{u} - \mathbf{\Omega^*}\vec{x} \, ,
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eqs. (12) & (15)</td>
</tr>
</table>

where, adopting Chandrasekhar's notation, the variable,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\vec{u}</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{d\vec{x}}{dt} \, ,</math>
</td>
</tr>
<tr>
<td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (14b)</td>
</tr>
</table>
denotes the fluid velocity as measured <font color="darkgreen">"… with respect to an observer [that is] at rest in the moving frame."</font> This allows us to write the,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the rotating-frame velocity, <math>\vec{u}</math></td>
</tr>

<tr>
<td align="right">
<math>
\frac{d}{dt}\biggl[ \vec{u} - \mathbf{\Omega^*}\vec{x} \biggr]
- \mathbf{\Omega^*}\biggl[ \vec{u} - \mathbf{\Omega^*}\vec{x} \biggr]</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} </math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
Again appreciating [[#Product|from above]] that <math>(\mathbf{\Omega^*}\vec{Q})_i = \epsilon_{ijk}\Omega_k Q_j = -\epsilon_{ijk}\Omega_j Q_k</math>, in component form this version of the Euler equation reads,

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d}{dt}\biggl[ u_i + \epsilon_{ijk}\Omega_j x_k \biggr] </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}\Omega_k \biggl[ u_m + \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d u_i}{dt} + \epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}u_m\Omega_k
+
\epsilon_{imk}\Omega_k \biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
Now, if we …
<ol type="a">
<li>Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
-~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr] \, ;
</math>
</td>
</tr>
</table>
note also that we have set <math>dx_j/dt \rightarrow u_j</math>;
</li>
<li>In the first term on the RHS, replace the index, "m", with the index, "j":

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk} u_m \Omega_k
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Inside the square brackets of the second term on the RHS, replace the "jk" indices with "h&ell;" in order to avoid confusion, then swap the "h&ell;" indices of the two variables, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[ \epsilon_{mh\ell}\Omega_h x_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h\Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
- ~ \epsilon_{ikm}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
</ol>

the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d u_i}{dt} -~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k
+
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~\frac{d u_i}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\underbrace{2\epsilon_{ijk} u_j \Omega_k}_{[2\vec{u}\times \vec\Omega]_i}
+
\underbrace{\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]}_{[\vec\Omega \times(\vec{x}\times\vec\Omega)]_i}
+
\underbrace{\epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) \biggr]}_{[\vec{x} \times (d\vec\Omega/dt)]_i}
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
</td></tr></table>

We therefore can rewrite in a more familiar ''vector'' formulation, the
<table border="0" align="center" cellpadding="5">
<tr>
<td align="center" colspan="3"><font color="#770000">'''Lagrangian Representation'''</font><br />of the rotating-frame Euler Equation<br />in terms of the rotating-frame velocity, <math>\vec{u}</math></td>
</tr>

<tr>
<td align="right">
<math>
\frac{d\vec{u}}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
2\vec{u}\times \vec\Omega + \vec\Omega \times (\vec{x}\times \vec\Omega)
+ \vec{x}\times \frac{d\vec\Omega}{dt}
- ~\biggl[\frac{1}{\rho}\nabla P + \nabla\Phi \biggr]_\mathrm{rotating} \, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="3">{{ Rossner67 }}, §II, Eq. (1)<br />
[<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], Appendix 1.D, §3, (p. 664) Eq. (1D-42)</td>
</tr>
</table>
----

The RHS of this equation matches the RHS of Eq. (1) from {{ Rossner67full }} after making the notation switch, <math>\Phi \rightarrow -~\mathfrak{B}</math>, and after acknowledging that <math>\nabla P/\rho \rightarrow \nabla(P/\rho)</math> when the mass-density is spatially uniform. The referenced equation from [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>] contains all of the terms shown here, except there, the effects of pressure are ignored.

----

====Component Form====

<font color="red">EXAMPLE #1:</font> Inertial-frame velocities, <math>\vec{U}</math>, as viewed in the inertial frame.

Adding the Euler equation,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\rho\frac{d\vec{U}}{dt}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi \, ,
</math>
</td>
</tr>
</table>
to the product of the inertial-frame velocity and the equation of continuity,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\vec{U}~\frac{d\rho}{dt}</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\rho\vec{U}(\nabla\cdot \vec{u})\, ,
</math>
</td>
</tr>
</table>
gives,
<table border="0" align="center" cellpadding="5">
<tr>
<td align="right"><math>\frac{d(\rho\vec{U})}{dt} + \rho\vec{U}(\nabla\cdot \vec{u})</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi
</math>
</td>
</tr>
<tr>
<td align="right"><math>\Rightarrow ~~~ \frac{\partial(\rho\vec{U})}{\partial t} + \nabla\cdot [(\rho\vec{U})\vec{u}]</math></td>
<td align="center"><math>=</math></td>
<td align="left">
<math>
-~\nabla P - \rho\nabla\Phi
</math>
</td>
</tr>
</table>

<span id="ComponentForm">In component form,</span> the relation between <math>\vec{U}</math> and <math>\vec{u}</math> reads,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="left">
<math>
U_i
=
u_i - (\Omega^*)_{ik}x_k
=
u_i + \overbrace{\epsilon_{ijk} \Omega_j x_k}^{[~\vec{\Omega} \times \vec{x}~]_i}
\, .
</math>
</td>
</tr>

<tr>
<td align="center" colspan="1">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (21)</td>
</tr>
</table>

and the rotating-frame Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
U_i
=
u_i - (\Omega^*)_{ik}x_k
=
u_i + \overbrace{\epsilon_{ijk} \Omega_j x_k}^{[~\vec{\Omega} \times \vec{x}~]_i}
\, .
</math>
</td>
<td align="center"><math>=</math></td>
<td align="left">
<math>

</math>
</td>
</tr>

<tr>
<td align="center" colspan="1">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chap. 4, §25, Eq. (21)</td>
</tr>
</table>

===Part B===
Drawing from Chapter 4, §25 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] — where the Cartesian components of the inertial-frame velocity <math>(\vec{v}_\mathrm{inertial})</math> are represented by <math>U_i</math> and the Cartesian components of the rotating-frame velocity <math>(\vec{v}_\mathrm{rot})</math> are represented by <math>u_i</math> — we begin by restating the [[PGE/Euler#in_terms_of_velocity|Lagrangian representation of the intertial-frame Euler equation]]:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{inertial}</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} \, .</math>
</td>
</tr>
</table>

The LHS of this (Euler) equation transform as follows:
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{inertial}</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{rot} - \epsilon_{imk}\Omega_k U_m\, ,</math>
</td>
</tr>
</table>
where we also recognize that,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_i</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_i + \epsilon_{ijk}\Omega_j x_k \, .</math>
</td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">

Both of these expressions make use of the three-element [https://en.wikipedia.org/wiki/Levi-Civita_symbol#Definition Levi-Civita tensor], <math>\epsilon_{ijk}</math>. Its six nonzero component values are …
<table border="1" align="center" cellpadding="5">
<tr>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
<td rowspan="4" bgcolor="lightgrey"> </td>
<td align="center"><math>ij k</math></td>
<td align="center"><math>\epsilon_{ijk}</math></td>
</tr>
<tr>
<td align="center">123</td>
<td align="center" rowspan="3">+1</td>
<td align="center">132</td>
<td align="center" rowspan="3">-1</td>
</tr>
<tr>
<td align="center">312</td>
<td align="center">321</td>
</tr>
<tr>
<td align="center">231</td>
<td align="center">213</td>
</tr>
</table>

Hence, for example, transforming the x-component <math>(i=1)</math> of <math>\vec{U}</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_1</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_1 + \epsilon_{1jk}\Omega_j x_k = u_1 + \epsilon_{123}\Omega_2 x_3 + \epsilon_{132}\Omega_3 x_2
= u_1 + \Omega_2 z -\Omega_3 y \, ;</math>
</td>
</tr>
</table>
transforming the y-component <math>(i=2)</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_2</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_2 + \epsilon_{2jk}\Omega_j x_k = u_2 + \epsilon_{231}\Omega_3 x_1 + \epsilon_{213}\Omega_1 x_3
= u_2 + \Omega_3 x -\Omega_1 z \, ;</math>
</td>
</tr>
</table>
and transforming the z-component <math>(i=3)</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>U_3</math>
</td>
<td align="center">
<math>\rightarrow</math>
</td>
<td align="left">
<math>u_3 + \epsilon_{3jk}\Omega_j x_k = u_3 + \epsilon_{312}\Omega_1 x_2 + \epsilon_{321}\Omega_2 x_1
= u_3 + \Omega_1 y -\Omega_2 x \, .</math>
</td>
</tr>
</table>

These are the same three components that arise from the vector expression (from above),
<div align="center">
<math>
\vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}} \times \vec{x} \, ;
</math>
</div>
we therefore recognize that, <math>\vec{\Omega} \times \vec{x} = \epsilon_{ijk}\Omega_j x_k</math>. We note as well that, <math>\vec{\Omega} \times \vec{x} = -\epsilon_{ijk}\Omega_k x_j</math>.

</td></tr></table>

Therefore, as viewed from the rotating frame of reference, the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{dU_i}{dt}\biggr|_\mathrm{rot} - \epsilon_{imk}\Omega_k U_m</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i} </math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d}{dt}\biggl[ u_i + \epsilon_{ijk}\Omega_j x_k \biggr] </math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}\Omega_k \biggl[ u_m + \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow ~~~ \frac{d u_i}{dt} + \epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{imk}u_m\Omega_k
+
\epsilon_{imk}\Omega_k \biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>
Now, if we …
<ol type="a">
<li>Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{ijk}\biggl[\biggl( \frac{d\Omega_j}{dt} \biggr) x_k + \Omega_j \biggl( \frac{dx_k}{dt} \biggr) \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
-~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
</tr>
</table>
note also that we have set <math>dx_j/dt \rightarrow u_j</math>;
</li>
<li>In the first term on the RHS, replace the index, "m", with the index, "j":

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk} u_m \Omega_k
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Inside the square brackets of the second term on the RHS, replace the "jk" indices with "h&ell;" in order to avoid confusion, then swap the "h&ell;" indices of the two variables, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\biggl[ \epsilon_{mjk}\Omega_j x_k \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[ \epsilon_{mh\ell}\Omega_h x_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
<li>Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well:

<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>
\epsilon_{imk}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h\Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
- ~ \epsilon_{ikm}\Omega_k\biggl[-~ \epsilon_{mh\ell} x_h \Omega_\ell \biggr]
</math>
</td>
<td align="center">
<math>~~\rightarrow ~~</math>
</td>
<td align="left">
<math>
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr] \, ;
</math>
</td>
</tr>
</table>
</li>
</ol>

the Euler equation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
<td align="right">
<math>\frac{d u_i}{dt} -~ \epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) + u_j \Omega_k \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\epsilon_{ijk} u_j \Omega_k
+
\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow~~~\frac{d u_i}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\underbrace{2\epsilon_{ijk} u_j \Omega_k}_{[2\vec{u}\times \vec\Omega]_i}
+
\underbrace{\epsilon_{ikm}\Omega_k\biggl[\epsilon_{mh\ell} x_h \Omega_\ell \biggr]}_{[\vec\Omega \times(\vec{x}\times\vec\Omega)]_i}
+
\underbrace{\epsilon_{ijk}\biggl[x_j \biggl( \frac{d\Omega_k}{dt} \biggr) \biggr]}_{[\vec{x} \times (d\vec\Omega/dt)]_i}
-\frac{1}{\rho} \frac{\partial p}{\partial x_i} - \frac{\partial\Phi}{\partial x_i}
\, .
</math>
</td>
</tr>
</table>

==Continuity Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the continuity equation presented [[PGE/ConservingMass#Continuity_Equation|elsewhere]], we obtain the:

<div align="center">
<font color="#770000">'''Lagrangian Representation'''</font><br />
of the Continuity Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{d\rho}{dt} \biggr]_{rot} + \rho \nabla \cdot {\vec{v}}_{rot} = 0</math> ;
</div>

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Continuity Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{\partial\rho}{\partial t} \biggr]_{rot} + \nabla \cdot (\rho {\vec{v}}_{rot}) = 0</math> .
</div>

==Euler Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the Euler equation presented [[PGE/Euler#Euler_Equation|elsewhere]], we obtain the:

<div align="center">
<font color="#770000">'''Lagrangian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math> ;
</div>

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math> ;
</div>

<div align="center">
Euler Equation<br />
written <font color="#770000">'''in terms of the Vorticity'''</font> and<br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}v_{rot}^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .
</div>

==Centrifugal and Coriolis Accelerations==

Following along the lines of the discussion presented in Appendix 1.D, §3 of [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], in a rotating reference frame the Lagrangian representation of the Euler equation may be written in the form,
<div align="center">
<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi + {\vec{a}}_{fict} </math>,
</div>
where,
<div align="center">
<math>
{\vec{a}}_{fict} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) .
</math>
</div>
So, as viewed from a rotating frame of reference, material moves as if it were subject to two ''fictitious accelerations'' which traditionally are referred to as the,
<div align="center">
<font color="#770000">'''Coriolis Acceleration'''</font>

<math>
{\vec{a}}_{Coriolis} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} ,
</math>
</div>

(see the related [[PGE/RotatingFrame#WikiCoriolis|Wikipedia discussion]]) and the
<div align="center">
<font color="#770000">'''Centrifugal Acceleration'''</font>

<math>
{\vec{a}}_{Centrifugal} \equiv - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
= \frac{1}{2} \nabla\biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]
</math>
</div>
(see the related [[PGE/RotatingFrame#WikiCentrifugal|Wikipedia discussion]]).

==Nonlinear Velocity Cross-Product==

In some contexts — for example, our discussion of [[Apps/RiemannEllipsoidsCompressible|Riemann ellipsoids]] or the analysis by [[Apps/Korycansky_Papaloizou_1996|Korycansky & Papaloizou (1996)]] of nonaxisymmetric disk structures — it proves useful to isolate and analyze the term in the "vorticity formulation" of the Euler equation that involves a nonlinear cross-product of the rotating-frame velocity vector, namely,
<div align="center">
<math>
\vec{A} \equiv ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot} .
</math>
</div>

NOTE: To simplify notation, for most of the remainder of this subsection we will drop the subscript "rot" on both the velocity and vorticity vectors.

===Align Ω<sub>f</sub> with z-axis===
Without loss of generality we can set <math>{\vec\Omega}_f = \hat{k}\Omega_f</math>, that is, we can align the frame rotation axis with the z-axis of a Cartesian coordinate system. The Cartesian components of <math>{\vec{A}}</math> are then,
<div align="left">
<math>
\hat{i}: ~~~~~~ A_x = \zeta_y v_z - (\zeta_z + 2\Omega) v_y ,
</math><br />

<math>
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x - \zeta_x v_z ,
</math><br />

<math>
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</div>
where it is understood that the three Cartesian components of the vorticity vector are,
<div align="center">
<math>
\zeta_x = \biggl[\frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} \biggr] ,
~~~~~~ \zeta_y = \biggl[ \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} \biggr] ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</div>
In turn, the curl of <math>\vec{A}</math> has the following three Cartesian components:
<div align="left">
<math>
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr],
</math><br />

<math>
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = \frac{\partial}{\partial z}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math><br />

<math>
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr] - \frac{\partial}{\partial y}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] .
</math>
</div>

===When v<sub>z</sub> = 0 ===

If we restrict our discussion to configurations that exhibit only planar flows — that is, systems in which <math>v_z = 0</math> — then the Cartesian components of <math>{\vec{A}}</math> and <math>\nabla\times\vec{A}</math> simplify somewhat to give, respectively,
<div align="left">
<math>
\hat{i}: ~~~~~~ A_x = - (\zeta_z + 2\Omega) v_y ,
</math><br />

<math>
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x ,
</math><br />

<math>
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</div>

and,
<div align="left">
<math>
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x \biggr],
</math><br />

<math>
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial}{\partial z}\biggl[(\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math><br />

<math>
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x \biggr] + \frac{\partial}{\partial y}\biggl[ (\zeta_z + 2\Omega) v_y \biggr] ,
</math>
</div>
where, in this case, the three Cartesian components of the vorticity vector are,
<div align="center">
<math>
\zeta_x = - \frac{\partial v_y}{\partial z} ,
~~~~~~ \zeta_y = \frac{\partial v_x}{\partial z} ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</div>

=Related Discussions=
* <span id="WikiVorticity">Wikipedia discussion of [http://en.wikipedia.org/wiki/Vorticity vorticity].</span>
* <span id="WikiCoriolis">Wikipedia discussion of [http://en.wikipedia.org/wiki/Coriolis_effect#Formula Coriolis Effect].</span>
* <span id="WikiCentrifugal">Wikipedia discussion of [http://en.wikipedia.org/wiki/Centrifugal_force#Derivation_using_vectors Centrifugal acceleration].</span>

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