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 z}{\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>

Alternatively, partials can be taken with respect to the cylindrical coordinates, <math>\varpi</math>, <math>z</math> and <math>\phi</math>. (Incidentally, I have reversed the traditional order of the <math>\phi</math> and <math>z</math> coordinates in an attempt to parallelize structure between cylindrical and T3 coordinates since <math>\lambda_3 \equiv \phi</math>.)

	<math> \frac{\partial}{\partial \varpi} </math>	<math> \frac{\partial}{\partial z} </math>	<math> \frac{\partial}{\partial \phi} </math>
<math>{\lambda_1}</math>	<math> \frac{\varpi}{\lambda_1} </math>	<math> \frac{q^2 z}{\lambda_1} </math>	<math> 0 </math>
<math>\lambda_2</math>	<math> \frac{q^2}{q^2-1} \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)} </math>	<math> -\frac{1}{q^2-1} \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)} </math>	<math> 0 </math>
<math>\lambda_3</math>	<math> 0 </math>	<math> 0 </math>	<math> 1 </math>

Furthermore, the inverted partials are

	<math> \frac{\partial}{\partial \lambda_1} </math>	<math> \frac{\partial}{\partial \lambda_2} </math>	<math> \frac{\partial}{\partial \lambda_3} </math>
<math>{\varpi}</math>	<math> \varpi \ell^2 \lambda_1 </math>	<math> (q^2-1) q^2 \varpi z^2 \ell^2 / \lambda_2 </math>	<math> 0 </math>
<math>z</math>	<math> q^2 z \ell^2 \lambda_1 </math>	<math> (q^2-1) \varpi^2 z \ell^2 / \lambda_2 </math>	<math> 0 </math>
<math>\phi</math>	<math> 0 </math>	<math> 0 </math>	<math> 1 </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> \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>

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

Derived Identity for T3 Coordinates

Looking at the "<math>\hat{e}_2</math>" component of this last 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>

or, equivalently,

<math> A = - \frac{\lambda_2}{h_1 \lambda_1} \frac{d h_2}{dt} </math>

At some point, this identity needs to be checked by taking various partial derivatives of the scale factors and plugging them into the generic definition of <math>A</math>, given above. (Actually, this shouldn't be necessary because in January, 2009, we derived the same equation-of-motion result shown below while using the uglier expression for <math>A</math>. So this must be a correct identity in the context of T3 coordinates.)

Implications of Equation of Motion

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} = \biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt} </math>

<math> \Rightarrow ~~~~~ \frac{d(h_2 \lambda_1 \dot{\lambda}_2)}{dt} = \dot{\lambda}_1 \frac{d(h_2 \lambda_2)}{dt} . </math>

Or, equivalently (but perhaps more perversely),

<math> \frac{\dot{\lambda}_2}{\lambda_2} \biggl[ \frac{1}{h_2 \dot{\lambda}_2} \frac{d(h_2 \dot{\lambda}_2)}{dt} \biggr] = \frac{\dot{\lambda}_1}{\lambda_1} \biggl[ \frac{1}{h_2} \frac{dh_2}{dt} \biggr] </math>

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

Note that one of these last few expressions is equivalent to the simplest form of the conservation that we derived — actually, Jay Call derived it — back in January, 2009. Specifically,

<math> \lambda_1 \frac{d(h_2 \dot{\lambda}_2)}{dt} = \lambda_2 \dot{\lambda}_1 \frac{dh_2}{dt} </math>

The 64-thousand dollar question is, "Can we turn any of these expressions into a form which states that the total time-derivative of some function equals zero?"

User:Tohline/Appendix/Ramblings/T3Integrals

Contents

Integrals of Motion in T3 Coordinates

Definition

Vector Derivatives

Time-Derivative of Position and Velocity Vectors

Derived Identity for T3 Coordinates

Implications of Equation of Motion

See Also

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