Difference between revisions of "User:Tohline/Appendix/Ramblings/T3Integrals"
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Hence, it must be true that | Hence, it must be true that, | ||
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<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> | ||
</ | </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 ~~~~~ | \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 /> | ||
</div><br /> | </div><br /> | ||
Revision as of 19:30, 24 May 2010
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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{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{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>\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
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