Difference between revisions of "User:Tohline/ThreeDimensionalConfigurations/ChallengesPt4"

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</tr>
</tr>
</table>
</table>
where,
where the position-dependent &#8212; and, hence also, the time-dependent &#8212; length scale,
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


Line 426: Line 426:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
Line 436: Line 436:
y_\mathrm{max}^4 (x')^2  
y_\mathrm{max}^4 (x')^2  
\biggr]^{1 / 2} \, .
\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>STEP #4</b></font>
As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,
<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(\dot\varphi t)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~y' - y_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}\sin(\dot\varphi t) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\dot{x}'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~\dot{y}' </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .</math>
  </td>
</tr>
</table>
This means that the (dimensional) velocity vector is,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \dot{x}' + \boldsymbol{\hat\jmath'} \dot{y}'
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr) \dot\varphi \biggr]
+
\boldsymbol{\hat\jmath'} \biggl[ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\boldsymbol{u'} \cdot \boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr) \dot\varphi \biggr]^2
+
\biggl[ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr]^2
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\dot\varphi^2}{x_\mathrm{max}^2 y_\mathrm{max}^2}
\biggl[ (y_c - y')^2 x_\mathrm{max}^4
+
(x')^2 y_\mathrm{max}^4
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(u_0')^2 \dot\varphi^2 \, .
</math>
</math>
   </td>
   </td>

Revision as of 21:33, 2 May 2021

Challenges Constructing Ellipsoidal-Like Configurations (Pt. 4)

This chapter extends the accompanying chapters titled, Construction Challenges (Pt. 1), (Pt. 2), and (Pt. 3). The focus here is on firming up our understanding of the relationships between various "tilted" Cartesian coordinate frames.

Whitworth's (1981) Isothermal Free-Energy Surface
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The Plan

Intersection Expression

STEP #1

First, we present the mathematical expression that describes the intersection between the surface of an ellipsoid and a plane having the following properties:

  • The plane cuts through the ellipsoid's z-axis at a distance, <math>~z_0</math>, from the center of the ellipsoid;
  • The line of intersection is parallel to the x-axis of the ellipsoid; and,
  • The line that is perpendicular to the plane and passes through the z-axis at <math>~z_0</math> is tipped at an angle, <math>~\theta</math>, to the z-axis.

As is illustrated in Figure 1, we will use the line referenced in this third property description to serve as the z'-axis of a Cartesian grid that is tipped at the angle, <math>~\theta</math>, with respect to the body frame; and we will align the x' axis with the x-axis, so it should be clear that the z'-axis lies in the y-z plane of the ellipsoid.

Figure 1

<math>~x</math>

<math>~=</math>

<math>~x' \, ,</math>

<math>~y</math>

<math>~=</math>

<math>~ y' \cos\theta - z'\sin\theta \, ,</math>

<math>~(z - z_0)</math>

<math>~=</math>

<math>~ z' \cos\theta + y'\sin\theta \, .</math>

Primed Coordinates

<math>~x'</math>

<math>~=</math>

<math>~x \, ,</math>

<math>~y'</math>

<math>~=</math>

<math>~ y \cos\theta + (z - z_0) \sin\theta \, ,</math>

<math>~z'</math>

<math>~=</math>

<math>~ (z-z_0) \cos\theta - y \sin\theta \, .</math>


As has been shown in our accompanying discussion, we obtain the following,

Intersection Expression

<math>~1 - \frac{x^2}{a^2} </math>

<math>~=</math>

<math>~y^2 \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr] + y \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] + \frac{z_0^2}{c^2} \, , </math>

as long as z0 lies within the range,

<math>~z_0^2</math>

<math>~\le</math>

<math>~c^2 + b^2\tan^2\theta \, .</math>

Rewriting this "intersection expression" in terms of the tipped (primed) coordinate frame gives us,

<math>~1 - \frac{(x')^2}{a^2} </math>

<math>~=</math>

<math>~(y' \cos\theta - z' \sin\theta)^2 \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr] + (y' \cos\theta - z' \sin\theta) \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] + \frac{z_0^2}{c^2} \, . </math>

STEP #2

As viewed from the tipped coordinated frame, the curve that is identified by this intersection should be an

Off-Center Ellipse

<math>~1</math>

<math>~=</math>

<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 </math>

 

<math>~=</math>

<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>

that lies in the x'-y' plane — that is, <math>~z' = 0</math>. Let's see if the intersection expression can be molded into this form.

<math>~1 - \frac{z_0^2}{c^2} - \frac{(x')^2}{a^2} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\kappa^2 \biggl[ (y')^2 - 2y' \underbrace{\biggl( \frac{-z_0 \sin\theta}{c^2 \kappa^2} \biggr)}_{y_c} \biggr] \, ,</math>

where,

<math>~\kappa^2</math>

<math>~\equiv</math>

<math>~ \frac{c^2 \cos^2\theta + b^2 \sin^2\theta}{b^2c^2} \, . </math>

Dividing through by <math>~\kappa^2</math>, then adding <math>~y_c^2</math> to both sides gives,

<math>~(y')^2 - 2y' y_c + y_c^2</math>

<math>~=</math>

<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>

Finally, we have,

<math>\frac{1}{y^2_\mathrm{max}} \biggl[ (y')^2 - 2y' y_c + y_c^2 \biggr]</math>

<math>~=</math>

<math>~1 - (x')^2 \underbrace{\biggl[ \frac{1}{a^2\kappa^2 y_\mathrm{max}^2} \biggr]}_{ 1/x^2_\mathrm{max} } \, .</math>

So … the intersection expression can be molded into the form of an off-center ellipse if we make the following associations:

<math>~\frac{y_c}{z_0}</math>

<math>~=</math>

<math>~- \frac{\sin\theta}{c^2 \kappa^2} \, ,</math>

<math>~y_\mathrm{max}^2</math>

<math>~=</math>

<math>~\frac{1}{\kappa^2}\biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, ,</math>

<math>~x_\mathrm{max}^2</math>

<math>~=</math>

<math>~a^2 \biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, .</math>

Note as well that,

<math>~\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)^2</math>

<math>~=</math>

<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>


Lagrangian Trajectory and Velocities

We presume that the off-center ellipse that is defined by the intersection expression identifies the trajectory of a Lagrangian fluid element. If this is the case, there are a couple of ways that the velocity — both the amplitude and its vector orientation — can be derived.

STEP #3

If the intersection expression identifies a Lagrangian trajectory, then the velocity vector must be tangent to the off-center ellipse at every location. At each <math>~(x', y')</math> coordinate location, the slope of the above-defined off-center ellipse is,

<math>~\frac{dy'}{dx'}</math>

<math>~=</math>

<math>~ \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr)^2 \frac{x'}{(y_c - y')} \, . </math>

From this expression we deduce that the x'- and y'- components of the velocity vector are, respectively,

<math>~\frac{\boldsymbol{\hat\imath'} \cdot \boldsymbol{u'} }{ [\boldsymbol{u'}\cdot \boldsymbol{u'}]^{1 / 2} }</math>

<math>~=</math>

<math>~ \frac{1}{u'_0} \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) (y_c - y') \, , </math>

      and,      

<math>~\frac{\boldsymbol{\hat\jmath'} \cdot \boldsymbol{u'} }{ [\boldsymbol{u'}\cdot \boldsymbol{u'}]^{1 / 2} }</math>

<math>~=</math>

<math>~ \frac{1}{u'_0} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x' \, , </math>

where the position-dependent — and, hence also, the time-dependent — length scale,

<math>~u'_0</math>

<math>~\equiv</math>

<math>~ \biggl\{ \biggl[ \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) (y_c - y') \biggr]^2 + \biggl[ \frac{1}{|u'|} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x' \biggr]^2 \biggr\}^{1 / 2} </math>

 

<math>~=</math>

<math>~ \frac{1}{x_\mathrm{max} y_\mathrm{max}} \biggl[ x_\mathrm{max}^4 ( y_c - y')^2 + y_\mathrm{max}^4 (x')^2 \biggr]^{1 / 2} \, . </math>

STEP #4

As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,

<math>~x'</math>

<math>~=</math>

<math>~x_\mathrm{max}\cos(\dot\varphi t)</math>

      and,      

<math>~y' - y_c</math>

<math>~=</math>

<math>~y_\mathrm{max}\sin(\dot\varphi t) \, ,</math>

<math>~\dot{x}'</math>

<math>~=</math>

<math>~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi </math>

      and,      

<math>~\dot{y}' </math>

<math>~=</math>

<math>~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .</math>

This means that the (dimensional) velocity vector is,

<math>~\boldsymbol{u'}</math>

<math>~=</math>

<math>~ \boldsymbol{\hat\imath'} \dot{x}' + \boldsymbol{\hat\jmath'} \dot{y}' </math>

 

<math>~=</math>

<math>~ \boldsymbol{\hat\imath'} \biggl[ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr) \dot\varphi \biggr] + \boldsymbol{\hat\jmath'} \biggl[ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr] </math>

<math>~\Rightarrow~~~\boldsymbol{u'} \cdot \boldsymbol{u'}</math>

<math>~=</math>

<math>~ \biggl[ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr) \dot\varphi \biggr]^2 + \biggl[ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr]^2 </math>

 

<math>~=</math>

<math>~\frac{\dot\varphi^2}{x_\mathrm{max}^2 y_\mathrm{max}^2} \biggl[ (y_c - y')^2 x_\mathrm{max}^4 + (x')^2 y_\mathrm{max}^4 \biggr] </math>

 

<math>~=</math>

<math>~ (u_0')^2 \dot\varphi^2 \, . </math>

Various Coordinate Frames

Tipped Orbit Planes

Summary

In a separate discussion, we have shown that, as viewed from a frame that "tumbles" with the (purple) body of a Type 1 Riemann ellipsoid, each Lagrangian fluid element moves along an elliptical path in a plane that is tipped by an angle <math>~\theta</math> about the x-axis of the body. As viewed from the (primed) coordinates associated with this tipped plane, by definition, z' = constant and dz'/dt = 0, and the planar orbit is defined by the expression for an,

Off-Center Ellipse

<math>~1</math>

<math>~=</math>

<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c(z')}{y_\mathrm{max}} \biggr]^2 \, .</math>

Tipped Orbit Frame (yellow, primed)

Tipped Orbital Planes

Given that b/a = 1.25 and c/a = 0.4703 for our chosen Example Type I Ellipsoid, we find that, <math>~\theta = - 1.18122 ~\mathrm{rad} = -67.68^\circ</math>.

Notice that the offset, <math>~y_c</math>, is a function of the tipped plane's vertical coordinate, <math>~z'</math>. As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,

<math>~x'</math>

<math>~=</math>

<math>~x_\mathrm{max}\cos(\dot\varphi t)</math>

      and,      

<math>~y' - y_c</math>

<math>~=</math>

<math>~y_\mathrm{max}\sin(\dot\varphi t) \, ,</math>

<math>~\dot{x}'</math>

<math>~=</math>

<math>~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi </math>

      and,      

<math>~\dot{y}' </math>

<math>~=</math>

<math>~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .</math>

We have determined that (numerical value given for our chosen example Type I ellipsoid),

<math>~\tan\theta</math>

<math>~=</math>

<math>~ - \frac{b^2 \beta \Omega_2}{c^2 \gamma \Omega_3} = -2.43573\, , </math>

where, as has also been specified in an accompanying discussion,

<math>~\beta</math>

<math>~=</math>

<math>~ - \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \frac{\zeta_2}{\Omega_2} = +1.13451 </math>

      and,      

<math>~\gamma</math>

<math>~=</math>

<math>~ - \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\Omega_3} = +1.80518\, . </math>

We also have determined that,

<math>~\biggl[\frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr]^4</math>

<math>~=</math>

<math>~ \frac{a^4 (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)}{b^4 c^4(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2)} ~~~\Rightarrow ~~~\frac{x_\mathrm{max}}{y_\mathrm{max}} = 1.26218 \, , </math>

<math>~{\dot\varphi}^4 </math>

<math>~=</math>

<math>~ \frac{a^4}{b^4 c^4} \biggl(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2 \biggr) (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2) ~~~\Rightarrow ~~~ \dot\varphi = 1.59862\, , </math>

<math>~\frac{y_c}{z_0}</math>

<math>~=</math>

<math>~ -\sin\theta ~~~\Rightarrow~~~~ \frac{y_c}{z_0} = -0.92507 \, .</math>

Demonstration

In order to transform a vector from the "tipped orbit" frame (primed coordinates) to the "body" frame (unprimed), we use the following mappings of the three unit vectors:

<math>~\boldsymbol{\hat\imath'}</math>

<math>~\rightarrow</math>

<math>~\boldsymbol{\hat\imath} \, ,</math>

<math>~\boldsymbol{\hat\jmath'}</math>

<math>~\rightarrow</math>

<math>~\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta \, ,</math>

<math>~\boldsymbol{\hat{k}'}</math>

<math>~\rightarrow</math>

<math>~-\boldsymbol{\hat\jmath}\sin\theta + \boldsymbol{\hat{k}}\cos\theta \, .</math>

Given that, by design in our "tipped orbit" frame, there is no vertical motion — that is, <math>~\dot{z}' = 0</math> — mapping the (primed coordinate) velocity to the body (unprimed) coordinate is particularly straightforward. Specifically,

<math>~\boldsymbol{u'}</math>

<math>~=</math>

<math>~ \boldsymbol{\hat\imath'} \dot{x}' + \boldsymbol{\hat\jmath'} \dot{y}' + \boldsymbol{\hat{k}'} \cancelto{0}{\dot{z}'} </math>

 

<math>~~~\rightarrow~~</math>

<math>~ \boldsymbol{\hat\imath} \dot{x}' + [\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \dot{y}' </math>

 

<math>~=</math>

<math>~ \boldsymbol{\hat\imath} \biggl\{ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \dot\varphi \biggr\} + [\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \biggl\{ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr\} \, . </math>

Recognizing, as before, that the relevant coordinate mapping is,

<math>~x</math>

<math>~=</math>

<math>~x' \, ,</math>

<math>~y</math>

<math>~=</math>

<math>~ y' \cos\theta - z'\sin\theta \, ,</math>

<math>~(z - z_0)</math>

<math>~=</math>

<math>~ z' \cos\theta + y'\sin\theta \, .</math>

Primed Coordinates

<math>~x'</math>

<math>~=</math>

<math>~x \, ,</math>

<math>~y'</math>

<math>~=</math>

<math>~ y \cos\theta + (z - z_0) \sin\theta \, ,</math>

<math>~z'</math>

<math>~=</math>

<math>~ (z-z_0) \cos\theta - y \sin\theta \, .</math>

we have,

<math>~\boldsymbol{u'}</math>

<math>~~~\rightarrow~~~</math>

<math>~ \boldsymbol{\hat\imath} \dot\varphi \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)\biggl\{y_c - y\cos\theta - (z - z_0)\sin\theta\biggr\} + \boldsymbol{\hat\jmath} \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \biggr\{ x\cos\theta \biggr\} + \boldsymbol{\hat{k}} \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \biggr\{ x\sin\theta \biggr\} \, . </math>

Therefore, as viewed from the body frame, the individual velocity components are,

<math>~\boldsymbol{\hat\jmath} \cdot \boldsymbol{u'}</math>

<math>~=</math>

<math>~ \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x\cos\theta </math>

 

<math>~=</math>

<math>~x \cdot \biggl\{ \frac{b^4 c^4(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2)}{a^4 (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)} \frac{a^4}{b^4 c^4} \biggl(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2 \biggr) (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2) \biggr\}^{1 / 4} \frac{{c^2 \gamma \Omega_3} }{[c^4 \gamma^2 \Omega_3^2 + b^4\beta^2 \Omega_2^2]^{1 / 2}} </math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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