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* [https://www.mediawiki.org/wiki/Special:MyLanguage/Manual:Combating_spam Learn how to combat spam on your wiki]
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==Various Trials & Tests==
{{ SGFfooter }}
 
<ol type="I">
  <li>First (bad) [[User:Jet53man/Tests|MediaWiki tests]]</li>
  <li>Learning how to modify and enhance the [[A2HostingEnvironment|a2Hosting MediaWiki environment]].</li>
</ol>
 
 
I'm just testing to see if I'm properly editing.
<math>
E = mc^2
</math>
 
<div align="center">
<math>H = \int\frac{dP}{\rho}</math> .
</div>
 
Try another way:  {{math|sin &pi; {{=}} 0}}
 
{{math|<VAR>&alpha;</VAR>}}
 
{{ Template:TeX|\alpha \, \! }}
 
{{ &alpha; }}
 
==Lagrangian Representation==
 
===in terms of velocity:===
Among the [[User:Tohline/PGE#Principal_Governing_Equations|principal governing equations]] we have included the
 
<div align="center">
<span id="ConservingMomentum:Lagrangian"><font color="#770000">'''Lagrangian Representation'''</font></span><br />
of the Euler Equation,
 
{{User:Tohline/Math/EQ_Euler01}}
 
[<b>[[User:Tohline/Appendix/References#BLRY07|<font color="red">BLRY07</font>]]</b>], p. 13, Eq. (1.55)
</div>
 
===in terms of momentum density:===
Multiplying this equation through by the mass density {{User:Tohline/Math/VAR_Density01}} produces the relation,
<div align="center">
<math>\rho\frac{d\vec{v}}{dt} = - \nabla P - \rho\nabla \Phi</math> ,
</div>
which may be rewritten as,
<div align="center">
<math>\frac{d(\rho\vec{v})}{dt}- \vec{v}\frac{d\rho}{dt} = - \nabla P - \rho\nabla \Phi</math> .
</div>
Combining this with the [[User:Tohline/PGE/ConservingMass#ConservingMass:Lagrangian|Standard Lagrangian Representation of the Continuity Equation]], we derive,
<div align="center">
<math>\frac{d(\rho\vec{v})}{dt}+ (\rho\vec{v})\nabla\cdot\vec{v} = - \nabla P - \rho\nabla \Phi</math> .
</div>
 
 
==Eulerian Representation==
 
===in terms of velocity:===
 
By replacing the so-called Lagrangian (or "material") time derivative <math>d\vec{v}/dt</math> in the Lagrangian representation of the Euler equation by its Eulerian counterpart (see, for example, the wikipedia discussion titled, "[https://en.wikipedia.org/wiki/Material_derivative Material_derivative]", to understand how the Lagrangian and Eulerian descriptions of fluid motion differ from one another conceptually as well as how to mathematically transform from one description to the other), we directly obtain the
 
<div align="center">
<span id="ConservingMomentum:Eulerian"><font color="#770000">'''Eulerian Representation'''</font></span><br />
of the Euler Equation,
 
{{User:Tohline/Math/EQ_Euler02}}
</div>
 
===in terms of momentum density:===
 
As was done above in the context of the Lagrangian representation of the Euler equation, we can multiply this expression through by {{User:Tohline/Math/VAR_Density01}} and combine it with the continuity equation to derive what is commonly referred to as the,
 
<div align="center">
<span id="ConservingMomentum:Conservative"><font color="#770000">'''Conservative Form'''</font></span><br />
of the Euler Equation,
 
{{User:Tohline/Math/EQ_Euler03}}
 
[<b>[[User:Tohline/Appendix/References#BLRY07|<font color="red">BLRY07</font>]]</b>], p. 8, Eq. (1.31)
</div>
 
The second term on the left-hand-side of this last expression represents the divergence of the "[https://en.wikipedia.org/wiki/Dyadics dyadic product]" or "[https://en.wikipedia.org/wiki/Outer_product outer product]" of the vector momentum density and the velocity vector, and is sometimes written as, <math>~\nabla\cdot [(\rho \vec{v}) \otimes \vec{v}]</math>.
 
===in terms of the vorticity:===
Drawing on one of the standard [https://en.wikipedia.org/wiki/Vector_calculus_identities#Dot_product_rule dot product rule vector identities], the nonlinear term on the left-hand-side of the Eulerian representation of the Euler equation can be rewritten as,
<div align="center">
<math>
(\vec{v}\cdot\nabla)\vec{v} = \frac{1}{2}\nabla(\vec{v}\cdot\vec{v}) - \vec{v}\times(\nabla\times\vec{v})
= \frac{1}{2}\nabla(v^2) + \vec{\zeta}\times \vec{v} ,
</math>
</div>
where,
<div align="center">
<math>
\vec\zeta \equiv \nabla\times\vec{v}
</math>
</div>
is commonly referred to as the [https://en.wikipedia.org/wiki/Vorticity vorticity].  Making this substitution leads to an expression for the,
 
<div align="center">
Euler Equation<br />
<span id="ConservingMomentum:Vorticity"><font color="#770000">'''in terms of the Vorticity'''</font></span>,
 
{{User:Tohline/Math/EQ_Euler04}}
</div>
 
==Double Check Vector Identities==
 
In a subsection of an accompanying chapter titled, [[User:Tohline/AxisymmetricConfigurations/SolutionStrategies#Double_Check_Vector_Identities|''Double Check Vector Identities,'']] we explicitly demonstrate for four separate "simple rotation profiles" that these two separate terms involving a nonlinear velocity expression do indeed generate identical mathematical relations, namely.
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2) \, ;</math>
  </td>
</tr>
</table>
and we explicitly demonstrate that they are among the set of velocity profiles that can also be expressed in terms of the gradient of a "centrifugal potential," <math>~\nabla\Psi</math>.
===Relationship Between State Variables===
 
If the two normalized state variables, <math>~\chi</math> and <math>~z</math>, are known, then the third normalized state variable, <math>~p_\mathrm{total}</math>, can be obtained directly from the [[User:Tohline/SR/PressureCombinations#Total_Pressure|above key expression for the total pressure]], that is,
 
<div align="center">
<math>p_\mathrm{total}(\chi, z) = 8(C_g \chi)^3  z + F(\chi) + \biggl(\frac{8\pi^4}{15}\biggr) z^4 \, ,</math>
</div>
where,
<div align="center">
<math>C_g \equiv \biggl(\frac{\mu_e m_p}{\bar\mu m_u}\biggr)^{1/3} \, .</math>
</div>
 
If it is the two normalized state variables, <math>~\chi</math> and <math>~p_\mathrm{total}</math>, that are known, the third normalized state variable &#8212; namely, the normalized temperature, <math>~z</math> &#8212; also can be obtained analytically.  But the governing expression is not as simple because it results from an inversion of the total pressure equation and, hence, the solution of a quartic equation.  As is [[User:Tohline/SR/Ptot_QuarticSolution#Determining_Temperature_from_Density_and_Pressure|detailed in the accompanying discussion]], the desired solution is,
 
<div align="center">
<math>
z(\chi, p_\mathrm{total}) = \theta_\chi \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] ,
</math>
</div>
 
where,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\theta_\chi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left" bgcolor="yellow">
<math>~\biggl( \frac{3\cdot 5}{2^2 \pi^4} \biggr)^{1/3} C_g\chi \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\phi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left" bgcolor="lightblue">
<math>~ 2^{3/2} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{1/2}
\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2}\, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\lambda</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left" bgcolor="pink">
<math>~
\biggl(\frac{\pi^4}{2\cdot 3^4\cdot 5} \biggr)^{1/3} \biggl[\frac{p_\mathrm{total}-F(\chi)}{(C_g \chi)^{4}}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
It also would be desirable to have an analytic expression for the function, <math>~\chi(z, p_\mathrm{total})</math>, in order to be able to immediately determine the normalized density from any specified values of the normalized temperature and normalized pressure.  However, it does not appear that the [[User:Tohline/SR/PressureCombinations#Total_Pressure|above key expression for the total pressure]] can be inverted to provide such a closed-form expression.
 
 
<span id="Step2"><font color="red"><b>STEP #2</b></font></span>
 
As viewed from the ''tipped'' coordinated frame, the curve that is identified by this intersection should be an
<table border="0" cellpadding="5" align="center">
<tr>
<td align="center" colspan="3"><font color="maroon">'''Off-Center Ellipse'''</font></td>
</tr>
<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{(y')^2  - 2y' y_c + y_c^2}{y^2_\mathrm{max}} \biggr] \, ,</math>
  </td>
</tr>
</table>
 
<span id="Result3">that lies in the</span> x'-y' plane &#8212; that is, <math>~z' = 0</math>.  Let's see if the intersection expression can be molded into this form.
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~1 - \frac{z_0^2}{c^2} - \frac{(x')^2}{a^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(y')^2  \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr]\cos^2\theta  + 2y' \biggl[ \frac{z_0 \sin\theta}{c^2} \biggr]  </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr]\cos^2\theta \biggl\{ (y')^2 - 2y' \biggl[ \frac{-z_0 \sin\theta}{c^2 \cos^2\theta} \biggr]\biggl[\frac{b^2c^2}{c^2 + b^2\tan^2\theta} \biggr]  \biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\kappa^2 \biggl[ (y')^2 - 2y' \underbrace{\biggl( \frac{-z_0 \sin\theta}{c^2 \kappa^2} \biggr)}_{y_c}  \biggr] \, ,</math>
  </td>
</tr>
</table>
 
<table border="1" align="center" cellpadding="10" width="60%" bordercolor="orange">
<tr><td align="center" bgcolor="lightblue">'''RESULT 3'''<br />(same as [[User:Tohline/ThreeDimensionalConfigurations/ChallengesPt2#Result1|Result 1]], but different from [[#Result2|Result 2, below]])
</td></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin\theta}{c^2\kappa^2}
</math>
  </td>
</tr>
</table>
 
</td></tr>
</table>
 
where,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\kappa^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{c^2 \cos^2\theta + b^2 \sin^2\theta}{b^2c^2} \, .
</math>
  </td>
</tr>
</table>
Dividing through by <math>~\kappa^2</math>, then adding <math>~y_c^2</math> to both sides gives,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~(y')^2 - 2y' y_c  + y_c^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\underbrace{\biggl[ \frac{1}{\kappa^2} - \frac{z_0^2}{c^2 \kappa^2} + y_c^2 \biggr]}_{y^2_\mathrm{max}} - \frac{(x')^2}{a^2\kappa^2} \, .</math>
  </td>
</tr>
</table>
Finally, we have,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\frac{1}{y^2_\mathrm{max}} \biggl[ (y')^2 - 2y' y_c  + y_c^2 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" bgcolor="yellow">
<math>~1 - (x')^2 \underbrace{\biggl[ \frac{1}{a^2\kappa^2 y_\mathrm{max}^2} \biggr]}_{ 1/x^2_\mathrm{max} } \, .</math>
  </td>
</tr>
</table>
So &hellip; the intersection expression can be molded into the form of an off-center ellipse if we make the following associations:
 
<table border="1" cellpadding="8" align="center" width="60%"><tr><td align="left">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\sin\theta}{c^2 \kappa^2} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~y_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\kappa^2}\biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~x_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2 \biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, .</math>
  </td>
</tr>
</table>
Note as well that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\kappa^2 = \frac{a^2}{b^2 c^2} \biggl[ c^2 \cos^2\theta + b^2 \sin^2\theta \biggr] \, .</math>
  </td>
</tr>
</table>
 
</td></tr></table>

Revision as of 13:32, 24 July 2021