Difference between revisions of "User:Tohline/Appendix/Ramblings/T3Integrals"

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(→‎Integrals of Motion in T3 Coordinates: look at angular momentum component)
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</div>
</div>


Hence, it must be true that for our T3 coordinate system,
Hence, it must be true that,


<div align="center">
<table align="center" border="1" cellpadding="10" width="50%">
<tr>
  <td align="center">
<font color="darkblue">
For T3 Coordinates
</font>
  </td>
</tr>
<tr>
  <td align="center">
<math>
<math>
\lambda_2 \frac{d h_2}{dt} = - A(h_1 {\lambda}_1) .
\lambda_2 \frac{d h_2}{dt} = - A(h_1 {\lambda}_1) .
</math>
</math>
</div>
  </td>
</tr>
</table>


Looking now at the "<math>\hat{e}_2</math>" component of the acceleration (which we will set equal to zero), and assuming no motion in the <math>3^\mathrm{rd}</math> component direction, we have,
Looking now at the "<math>\hat{e}_2</math>" component of the acceleration (which we will set equal to zero), and assuming no motion in the <math>3^\mathrm{rd}</math> component direction, we have,
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\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = -  A(h_1 \dot{\lambda}_1)  
\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = -  A(h_1 \dot{\lambda}_1)  
</math><br /><br />
</math><br /><br />
or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates<br /><br />
<math>
<math>
\Rightarrow ~~~~~ h_2\frac{d(\dot{\lambda}_2)}{dt} - A(h_1 {\lambda}_1) \frac{\dot{\lambda}_2 }{\lambda_2} = -  A(h_1 \dot{\lambda}_1)
\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} (h_1 \dot{\lambda}_1) \biggl[ \biggl(\frac{\lambda_2}{h_1 \lambda_1}\biggr) \frac{dh_2}{dt} \biggr]
</math>
</math>
<br /><br />
<br /><br />
<math>
 
\Rightarrow ~~~~~ h_2\frac{d\dot{\lambda}_2}{dt}  = A\biggl[ (h_1 {\lambda}_1)\frac{\dot{\lambda}_2 }{\lambda_2} -  (h_1 \dot{\lambda}_1)\biggr]
</math>
<br /><br />
<math>
\Rightarrow ~~~~~ \frac{d\dot{\lambda}_2}{dt}  = A \lambda_1 \biggl( \frac{h_1}{h_2} \biggr) \biggl[ \frac{\dot{\lambda}_2 }{\lambda_2} -  \frac{\dot{\lambda}_1 }{\lambda_1} \biggr] = \lambda_1 \biggl( \frac{h_1}{h_2} \biggr) \biggl[ \frac{\dot{\lambda}_2 }{\lambda_2} -  \frac{\dot{\lambda}_1 }{\lambda_1} \biggr] \biggl[\frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} -
\frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2}\biggr].
</math>
</div><br />
</div><br />



Revision as of 19:30, 24 May 2010

Whitworth's (1981) Isothermal Free-Energy Surface
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Integrals of Motion in T3 Coordinates

Motivated by the HNM82 derivation, in an accompanying chapter we have introduced a new T2 Coordinate System and have outlined a few of its properties. Here we offer a modest redefinition of the second radial coordinate in an effort to bring even more symmetry to the definition of the position vector, <math>\vec{x}</math>.


Definition

By defining the dimensionless angle,

<math> \Zeta \equiv \sinh^{-1}\biggl( \frac{qz}{\varpi} \biggr) , </math>

the two key "T3" coordinates will be written as,

<math> \lambda_1 </math>

<math>\equiv</math>

<math>\varpi \cosh\Zeta = ( \varpi^2 + q^2z^2 )^{1/2}</math>

      and      

<math> \lambda_2 </math>

<math>\equiv</math>

<math>\varpi [\sinh\Zeta ]^{1/(1-q^2)} = \biggl[\frac{\varpi^{q^2}}{qz}\biggr]^{1/(q^2-1)}</math>

Here are some relevant partial derivatives:

 

<math> \frac{\partial}{\partial x} </math>

<math> \frac{\partial}{\partial y} </math>

<math> \frac{\partial}{\partial z} </math>

<math>\lambda_1</math>

<math> \frac{x}{\lambda_1} </math>

<math> \frac{y}{\lambda_1} </math>

<math> \frac{q^2}{\lambda_1} </math>

<math>\lambda_2</math>

<math> \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) x </math>
<math> =\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{x}{\varpi^2} \biggr) </math>

<math> \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) y </math>
<math> =\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{y}{\varpi^2} \biggr) </math>

<math> - \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\zeta} \biggr]^{q^2/(q^2-1)} \frac{q}{\varpi^{q^2}} </math>
<math> =- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \frac{1}{z} </math>

<math>\lambda_3</math>

<math> - \frac{y}{\varpi^{2}} </math>

<math> + \frac{x}{\varpi^{2}} </math>

<math> 0 </math>

The scale factors are,

<math>h_1^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\lambda_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> \lambda_1^2 \ell^2 </math>

 

 

<math>h_2^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\lambda_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> (q^2-1)^2 \biggl(\frac{\varpi z \ell}{\lambda_2} \biggr)^2 </math>

 

 

<math>h_3^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> \varpi^2 </math>

 

 

where,        <math>\ell \equiv (\varpi^2 + q^4 z^2)^{-1/2}</math>.


The position vector is,

<math>\vec{x}</math>

<math>=</math>

<math> \hat{i}x + \hat{j}y + \hat{k}z </math>

<math>=</math>

<math> \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2) . </math>

Vector Derivatives

For orthogonal coordinate systems, the time-rate-of-change of the three unit vectors are given by the expressions,

<math> \frac{d}{dt}\hat{e}_1 </math>

<math> = </math>

<math> \hat{e}_2 A + \hat{e}_3 B </math>

<math> \frac{d}{dt}\hat{e}_2 </math>

<math> = </math>

<math> - \hat{e}_1 A + \hat{e}_3 C </math>

<math> \frac{d}{dt}\hat{e}_3 </math>

<math> = </math>

<math> - \hat{e}_1 B - \hat{e}_2 C </math>

where,

<math> A </math>

<math> \equiv </math>

<math> \frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} - \frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2} </math>

<math> B </math>

<math> \equiv </math>

<math> \frac{\dot{\lambda}_3}{h_1} \frac{\partial h_3}{\partial \lambda_1} - \frac{\dot{\lambda}_1}{h_3} \frac{\partial h_1}{\partial \lambda_3} </math>

<math> C </math>

<math> \equiv </math>

<math> \frac{\dot{\lambda}_3}{h_2} \frac{\partial h_3}{\partial \lambda_2} - \frac{\dot{\lambda}_2}{h_3} \frac{\partial h_2}{\partial \lambda_3} </math>

Position and Velocity Vectors

In general for an orthogonal coordinate system, the velocity vector can be written as,

<math> \vec{v} = \hat{e}_1 (h_1 \dot{\lambda}_1) + \hat{e}_2 (h_2 \dot{\lambda}_2) +\hat{e}_3 (h_3 \dot{\lambda}_3) . </math>

So, in general, the time-rate-of-change of the velocity vector is,

<math>\frac{d\vec{v}}{dt}</math>

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] + (h_1 \dot{\lambda}_1)\frac{d\hat{e}_1}{dt} + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] + (h_2 \dot{\lambda}_2)\frac{d\hat{e}_2}{dt} + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] + (h_3 \dot{\lambda}_3)\frac{d\hat{e}_3}{dt} </math>

 

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] + (h_1 \dot{\lambda}_1)\biggl[ \hat{e}_2 A + \hat{e}_3 B \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] + (h_2 \dot{\lambda}_2)\biggl[ - \hat{e}_1 A + \hat{e}_3 C \biggr] + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] + (h_3 \dot{\lambda}_3)\biggl[ - \hat{e}_1 B - \hat{e}_2 C \biggr] </math>

 

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt} - A(h_2 \dot{\lambda}_2) - B(h_3 \dot{\lambda}_3) \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) - C(h_3 \dot{\lambda}_3) \biggr] + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt} + B(h_1 \dot{\lambda}_1) + C(h_2 \dot{\lambda}_2) \biggr] </math>

Now, for the T3 coordinate system the position vector has a similar form, specifically,

<math> \vec{x} = \hat{e}_1 (h_1 {\lambda}_1) + \hat{e}_2 (h_2 {\lambda}_2) . </math>

By analogy, then, the time-rate-of-change of the position vector is,

<math>\frac{d\vec{x}}{dt}</math>

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] + (h_1 {\lambda}_1)\frac{d\hat{e}_1}{dt} + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] + (h_2 {\lambda}_2)\frac{d\hat{e}_2}{dt} </math>

 

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] + (h_1 {\lambda}_1)\biggl[ \hat{e}_2 A \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] + (h_2 {\lambda}_2)\biggl[ - \hat{e}_1 A \biggr] </math>

 

<math>=</math>

<math> \hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt} - A(h_2 {\lambda}_2) \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt} + A(h_1 {\lambda}_1) \biggr] </math>

Looking at the "<math>\hat{e}_2</math>" component of this expression, we have,

<math> \hat{e}_2 \cdot \frac{d\vec{x}}{dt} = \frac{d(h_2 {\lambda}_2)}{dt} + A(h_1 {\lambda}_1) . </math>

But we also know that,

<math> \hat{e}_2 \cdot \vec{v} = h_2 \dot{\lambda}_2 . </math>

Hence, it must be true that,

For T3 Coordinates

<math> \lambda_2 \frac{d h_2}{dt} = - A(h_1 {\lambda}_1) . </math>

Looking now at the "<math>\hat{e}_2</math>" component of the acceleration (which we will set equal to zero), and assuming no motion in the <math>3^\mathrm{rd}</math> component direction, we have,

<math> \hat{e}_2 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) =0 </math>

<math> \Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = - A(h_1 \dot{\lambda}_1) </math>

or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates

<math> \Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = (h_1 \dot{\lambda}_1) \biggl[ \biggl(\frac{\lambda_2}{h_1 \lambda_1}\biggr) \frac{dh_2}{dt} \biggr] </math>


Another option is to look at the equivalent angular momentum element, <math>h_2^2 \dot{\lambda}_2</math>:


<math> \hat{e}_2 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) =0 </math>

<math> \Rightarrow ~~~~~ h_2 \frac{d(h_2 \dot{\lambda}_2)}{dt} = - A(h_1 h_2 \dot{\lambda}_1) </math>

<math> \Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} - h_2 \dot{\lambda}_2 \frac{dh_2}{dt} = - A(h_1 h_2 \dot{\lambda}_1) </math>

<math> \Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} + h_2 \dot{\lambda}_2 \biggl[A h_1 \biggl( \frac{\lambda_1}{\lambda_2} \biggr) \biggr] = - A(h_1 h_2 \dot{\lambda}_1) </math>

<math> \Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} = - A h_1 h_2 \lambda_1 \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr] = - h_1 h_2 \lambda_1 \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr]\biggl[\frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} - \frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2}\biggr] . </math>

I'm not yet quite sure what to do with all of this.

See Also

 

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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