3Dconfigurations/RiemannEllipsoids: Difference between revisions

From jetwiki
Jump to navigation Jump to search
Line 476: Line 476:
   <td align="left">
   <td align="left">
<math>= 1 \, .
<math>= 1 \, .
</math>
  </td>
</tr>
</table>
<table border="0" align="center" cellpadding="5" width="90%">
<tr>
  <td align="right" width="10%">
<math>2xy:</math>
  </td>
  <td align="left">
<math>
(\cos\psi \cos\phi - \sin\psi \sin\phi \cos\theta)(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
+ (-\sin\psi \cos\phi - \sin\phi \cos\theta \cos\psi)( - \sin\psi \sin\phi + \cos\psi \cos\theta\cos\phi )
+ (\sin\theta\sin\phi)(-\sin\theta \cos\phi)
</math>
  </td>
</tr>
<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\cos\psi \cos\phi (\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
- \sin\psi \sin\phi \cos\theta(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
+ (\sin\psi \cos\phi + \sin\phi \cos\theta \cos\psi)( \sin\psi \sin\phi - \cos\psi \cos\theta\cos\phi )
- \sin^2\theta\sin\phi\cos\phi
</math>
  </td>
</tr>
<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\cos\psi \cos\phi (\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
- \sin\psi \sin\phi \cos\theta(\sin\phi \cos\psi + \sin\psi \cos\theta \cos\phi)
+ (\sin\psi \cos\phi + \sin\phi \cos\theta \cos\psi)( \sin\psi \sin\phi - \cos\psi \cos\theta\cos\phi )
- \sin^2\theta\sin\phi\cos\phi
</math>
</math>
   </td>
   </td>

Revision as of 19:56, 18 June 2021


Riemann (1826 - 1866)

Background

Excerpt from p. 539 of R. Baker, C. Christenson & H. Orde (2004):

"Newton first showed that the departure of the figure of the earth from a sphere is due to its rotation. Jacobi showed in 1834 that gravitational equilibrium of a rotating spheroid is consistent with three distinct axes if the angular momentum exceeds a critical value. Dirichlet had posed and partially analyzed the conditions for a configuration which is an ellipsoid varying with time, such that the motion in an inertial frame, is linear in the coordinates. His results were edited posthumously by Dedekind in 1860. In his published work dated 1861 — five years before his death — Riemann took up this problem of Dirichlet …"

Excerpt from p. 530 of R. Baker, C. Christenson & H. Orde (2004) which, in turn, is taken from R. Dedekind's accounting of The life of Bernhard Riemann:

"In the Easter vacation of 1860 [Riemann] went on a trip to Paris, where he stayed for a month from 26th March; unfortunately the weather was raw and unfriendly and in the last week of his visit there was one day after another of snow and hail which made it almost impossible to see the sights. However, he was delighted with the friendly reception which he received from the Parisian scholars Serret, Bertrand, Hermite, Puiseux and Briot, with whom he spent a pleasant day in the country at Chatenay, along with Bouquet. In the same year, [Riemann] completed his paper on the motion of a fluid ellipsoid …"

Excerpt from pp. 184-185 of EFE:

"Riemann's paper 'Ein Beitrag zu den Untersuchungen über die Bewegung eines flüssigen gleichartigen Ellipsoides,' communicated to Der Königlichen Gesellschaft der Wissenschaften zu Göttingen on December 8, 1860, is remarkable for the wealth of new results it contains and for the breadth of its comprehension of the entire range of problems. In the present writer's [S. Chandrasekhar] view this much neglected paper — [for example] there are no references to it in any of the writings of Poincaré, Darwin, or Jeans — deserves to be included among the other great papers of Riemann that are well known. … In view of Riemann's unique place in science, a critical appraisal of this paper is perhaps justified."

His Published Work on Ellipsoids

Riemann's (1861) work, titled, "Ein Beitrag zu den Untersuchungen über die Bewegung eines flüssigen gleichartigen Ellipsoides" — English translation: "A contribution to the study of the motion of a homogeneous fluid ellipsoid" — can be found in various published collections of his papers:

Our description and detailed analysis of Riemann's (1861) work that follows, draws primarily from the 2004 translation of his collected works.

Derivation

Step 1

At the beginning of his discussion, Riemann denotes by (x,y,z) "the [inertial-frame] coordinates of an element of the fluid body at time t," and he denotes by (ξ,η,ζ) "the coordinates of the point (x,y,z) with respect to a moving coordinate system, whose axes coincide at each instant with the principal axes of the ellipsoid." Drawing from our accompanying discussion of Euler angles, it seems appropriate to associate AXYZ with the vector that points from the origin to the (x,y,z) location of the fluid element as viewed from the inertial reference frame, and to associate Abody with the vector that points from the origin to the same location of the fluid element, but as viewed from Riemann's specified rotating frame. With this in mind, we have the following notation mappings:

AXYZ(AX,AY,AZ)

   

Ainertial(x,y,z)=eX(x)+eY(y)+eZ(z),

Abody(A1,A2,A3)

   

Abody(ξ,η,ζ)=e1(ξ)+e2(η)+e3(ζ).

Again drawing from our accompanying Euler angles discussion, quite generally these two coordinate representations — of the same vector, A — are related to one another via the matrix expression,

Abody=R^Ainertial,

that is,

[ξηζ]

=

[e1eXe1eYe1eZe2eXe2eYe2eZe3eXe3eYe3eZ][xyz]=R^[xyz].

In terms of the trio of Euler angles, the rotation matrix is,

R^(ϕ,θ,ψ)

=

[(cosψcosϕsinϕsinψcosθ)(cosψsinϕ+cosϕsinψcosθ)(sinψsinθ)(sinψcosϕsinϕcosψcosθ)(sinϕsinψ+cosϕcosψcosθ)(cosψsinθ)(sinθsinϕ)(sinθcosϕ)(cosθ)].

This means that — see our accompanying discussion — in terms of the trio of Euler angles, the following three coordinate mappings hold:

ξ

=

x(cosψcosϕsinψsinϕcosθ)+y(sinϕcosψ+sinψcosθcosϕ)+z(sinψsinθ),

η

=

x(sinψcosϕsinϕcosθcosψ)+y(sinψsinϕ+cosψcosθcosϕ)+z(sinθcosψ),

ζ

=

x(sinθsinϕ)+y(sinθcosϕ)+z(cosθ).

Notice the correspondence between this set of coordinate relations and the set marked by Riemann (1861) as equation (2) of §1.

From §1 of Riemann (1861)

Then ξ, η, ζ are known … to be linear expressions in x, y, z,

ξ

=

αx+βy+γz,

η

=

αx+βy+γz,

ζ

=

αx+βy+γz.

The coefficients are the cosines of the angles that the axes of one system form with the axes of the other …

Alternatively, given that,

Ainertial=R^1Abody,

the following additional three mapping relations also must hold:

x

=

ξ(cosψcosϕsinψsinϕcosθ)+η(sinψcosϕsinϕcosθcosψ)+ζ(sinθsinϕ),

y

=

ξ(sinϕcosψ+sinψcosθcosϕ)+η(sinψsinϕ+cosψcosθcosϕ)+ζ(sinθcosϕ),

z

=

ξ(sinψsinθ)+η(sinθcosψ)+ζ(cosθ).

Step 2

Given that (x,y,z) and (ξ,η,ζ) are components of the same position vector, A, it must in general be the case that the square of the length of the vector is the same for both coordinate representations. That is, as Riemann (1861) states in §1 following equation (2), it must be the case that,

ξ2+η2+ζ2=x2+y2+z2.

Let's specifically assess whether or not this holds true when Euler angles are used to relate the components of the two position vector expressions.

ξ2+η2+ζ2

=

[x(cosψcosϕsinψsinϕcosθ)+y(sinϕcosψ+sinψcosθcosϕ)+z(sinψsinθ)]2

 

 

+[x(sinψcosϕsinϕcosθcosψ)+y(sinψsinϕ+cosψcosθcosϕ)+z(sinθcosψ)]2

 

 

+[x(sinθsinϕ)+y(sinθcosϕ)+z(cosθ)]2.

On the right-hand side, the coefficient of various terms is …

x2:

(cosψcosϕsinψsinϕcosθ)2+(sinψcosϕsinϕcosθcosψ)2+(sinθsinϕ)2

 

=cos2ψcos2ϕ2cosψcosϕsinψsinϕcosθ+sin2ψsin2ϕcos2θ+sin2ψcos2ϕ+2sinψcosϕsinϕcosθcosψ+sin2ϕcos2θcos2ψ+sin2θsin2ϕ

 

=(2sinψcosϕsinϕcosθcosψ2cosψcosϕsinψsinϕcosθ)+(sin2ψ+cos2ψ)sin2ϕcos2θ+(sin2+cos2ψ)cos2ϕ+sin2θsin2ϕ

 

=sin2ϕcos2θ+cos2ϕ+sin2θsin2ϕ

 

=1.

y2:

(sinϕcosψ+sinψcosθcosϕ)2+(sinψsinϕ+cosψcosθcosϕ)2+(sinθcosϕ)2

 

=sin2ϕcos2ψ+2sinϕcosψsinψcosθcosϕ+sin2ψcos2θcos2ϕ+sin2ψsin2ϕ2sinψsinϕcosψcosθcosϕ+cos2ψcos2θcos2ϕ+sin2θcos2ϕ

 

=(2sinϕcosψsinψcosθcosϕ2sinψsinϕcosψcosθcosϕ)+(cos2ψ+sin2ψ)sin2ϕ+(sin2ψ+cos2ψ)cos2θcos2ϕ+sin2θcos2ϕ

 

=sin2ϕ+cos2θcos2ϕ+sin2θcos2ϕ

 

=1.

z2:

(sinψsinθ)2+(sinθcosψ)2+(cosθ)2

 

=1.

2xy:

(cosψcosϕsinψsinϕcosθ)(sinϕcosψ+sinψcosθcosϕ)+(sinψcosϕsinϕcosθcosψ)(sinψsinϕ+cosψcosθcosϕ)+(sinθsinϕ)(sinθcosϕ)

 

=cosψcosϕ(sinϕcosψ+sinψcosθcosϕ)sinψsinϕcosθ(sinϕcosψ+sinψcosθcosϕ)+(sinψcosϕ+sinϕcosθcosψ)(sinψsinϕcosψcosθcosϕ)sin2θsinϕcosϕ

 

=cosψcosϕ(sinϕcosψ+sinψcosθcosϕ)sinψsinϕcosθ(sinϕcosψ+sinψcosθcosϕ)+(sinψcosϕ+sinϕcosθcosψ)(sinψsinϕcosψcosθcosϕ)sin2θsinϕcosϕ

See Also

 

Tiled Menu

Appendices: | VisTrailsEquations | VisTrailsVariables | References | Ramblings | VisTrailsImages | myphys.lsu | ADS |