# 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, $\vec{x}$.

## Definition

By defining the dimensionless angle,

$\Zeta \equiv \sinh^{-1}\biggl( \frac{qz}{\varpi} \biggr) ,$

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

 λ1 $\equiv$ $\varpi \cosh\Zeta = ( \varpi^2 + q^2z^2 )^{1/2}$ and λ2 $\equiv$ $\varpi [\sinh\Zeta ]^{1/(1-q^2)} = \biggl[\frac{\varpi^{q^2}}{qz}\biggr]^{1/(q^2-1)}$

### Relevant Partial Derivatives

Here are some relevant partial derivatives:

 $\frac{\partial}{\partial x}$ $\frac{\partial}{\partial y}$ $\frac{\partial}{\partial z}$ λ1 $\frac{x}{\lambda_1}$ $\frac{y}{\lambda_1}$ $\frac{q^2 z}{\lambda_1}$ λ2 $\frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) x$ $=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{x}{\varpi^2} \biggr)$ $=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{x}{\varpi} \biggr)$ $\frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) y$ $=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{y}{\varpi^2} \biggr)$ $=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{y}{\varpi} \biggr)$ $- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\zeta} \biggr]^{q^2/(q^2-1)} \frac{q}{\varpi^{q^2}}$ $=- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \frac{1}{z}$ $=- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz^{q^2}} \biggr]^{1/(q^2-1)}$ λ3 $- \frac{y}{\varpi^{2}}$ $+ \frac{x}{\varpi^{2}}$ 0

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

 $\frac{\partial}{\partial \varpi}$ $\frac{\partial}{\partial z}$ $\frac{\partial}{\partial \phi}$ λ1 $\frac{\varpi}{\lambda_1}$ $\frac{q^2 z}{\lambda_1}$ 0 λ2 $\frac{q^2}{q^2-1} \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)}$ $-\frac{1}{q^2-1} \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)}$ 0 λ3 0 0 1

Furthermore, the inverted partials are

 $\frac{\partial}{\partial \lambda_1}$ $\frac{\partial}{\partial \lambda_2}$ $\frac{\partial}{\partial \lambda_3}$ ${\varpi}$ $\varpi \ell^2 \lambda_1$ $(q^2-1) q^2 \varpi z^2 \ell^2 / \lambda_2$ 0 z $q^2 z \ell^2 \lambda_1$ $- (q^2-1) \varpi^2 z \ell^2 / \lambda_2$ 0 φ 0 0 1

### Scale Factors

The scale factors are,

 $h_1^2$ = $\biggl[ \biggl( \frac{\partial\lambda_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial z} \biggr)^2 \biggr]^{-1}$ = $\lambda_1^2 \ell^2$ $h_2^2$ = $\biggl[ \biggl( \frac{\partial\lambda_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial z} \biggr)^2 \biggr]^{-1}$ = $(q^2-1)^2 \biggl(\frac{\varpi z \ell}{\lambda_2} \biggr)^2$ $h_3^2$ = $\biggl[ \biggl( \frac{\partial\lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial z} \biggr)^2 \biggr]^{-1}$ = $\varpi^2$ where,        $\ell \equiv (\varpi^2 + q^4 z^2)^{-1/2}$.

### Direction Cosines

The following table contains expressions for the nine direction cosines.

 Direction Cosines for T3 Coordinates $\gamma_{ni} = h_n\frac{\partial\lambda_n}{\partial x_i}$ i n $\ell x$ $\ell y$ $q^2 \ell z$ $\frac{q^2 x z \ell}{\varpi}$ $\frac{q^2 y z \ell}{\varpi}$ $- \varpi \ell$ $- \frac{y}{\varpi}$ $+\frac{x}{\varpi}$ 0 where: $\ell \equiv (\varpi^2 + q^4 z^2)^{-1/2}$

### Orthogonality Condition

Next, let's use the example orthogonality condition derived elsewhere in connection with our overview of direction cosines. Specifically, let's see if Equation DC.02 is satisfied.

 $\frac{\partial\lambda_1}{\partial \varpi} \cdot \frac{\partial\lambda_2}{\partial \varpi} + \frac{\partial\lambda_1}{\partial z} \cdot \frac{\partial\lambda_2}{\partial z}$ = $\frac{\varpi}{\lambda_1} \biggl[ \frac{q^2}{q^2-1} \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)} \biggr] - \frac{q^2z}{\lambda_1} \biggl[ \frac{1}{q^2-1} \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)} \biggr]$ = $\frac{q^2}{(q^2-1)\lambda_1} \biggl\{ \varpi\biggl[ \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)} \biggr] - z \biggl[ \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)} \biggr] \biggr\}$ = $\frac{q^2}{(q^2-1)\lambda_1} \biggl[ q^{-1/(q^2-1)} \varpi^{1 + 1/(q^2-1)} z^{-1/(q^2-1)} - q^{-1/(q^2-1)} \varpi^{q^2/(q^2-1)} z^{1-q^2/(q^2-1)} \biggr]$ = 0.

Hence, the key orthogonality condition defined by Equation DC.02 is satisfied. MF53 also gives us relationships that should apply between the various direction cosines if the coordinate system is orthogonal. Let's check a few cases to see whether γmn = Mmn, where "Mmn is the minor of γmn in the determinant | γmn | ":

$M_{11} = \gamma_{22}\gamma_{33} - \gamma_{23}\gamma_{32} = - (-\varpi\ell)\frac{x}{\varpi} = + \ell x .$

$M_{12} = \gamma_{23}\gamma_{31} - \gamma_{21}\gamma_{33} = -\varpi\ell \biggl(-\frac{y}{\varpi}\biggr) = +\ell y .$

$M_{13} = \gamma_{21}\gamma_{32} - \gamma_{22}\gamma_{31} = \biggl( \frac{q^2 x z \ell}{\varpi} \biggr) \frac{x}{\varpi} - \biggl( \frac{q^2 y z \ell}{\varpi} \biggr) \biggl( -\frac{y}{\varpi} \biggr) = q^2 \ell z.$

$M_{31} = \gamma_{12}\gamma_{23} - \gamma_{13}\gamma_{22} = \ell y (-\varpi\ell) - q^2 \ell z \biggl(\frac{q^2 yz\ell}{\varpi}\biggr) = - \frac{y \ell^2}{\varpi} \biggl( \varpi^2 + q^4 z^2 \biggr) = - \frac{y}{\varpi} .$

$M_{33} = \gamma_{11}\gamma_{22} - \gamma_{12}\gamma_{21} = \ell x \biggr(\frac{q^2 y z \ell}{\varpi} \biggr) - \ell y \biggr(\frac{q^2 x z \ell}{\varpi} \biggr) = 0 .$

All of these beautifully obey the relationship, γmn = Mmn.

### Position Vector

The position vector is,

 $\vec{x}$ = $\hat{i}x + \hat{j}y + \hat{k}z$ = $\hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2) .$

## Vector Derivatives

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

 $\frac{d}{dt}\hat{e}_1$ = $\hat{e}_2 A + \hat{e}_3 B$ $\frac{d}{dt}\hat{e}_2$ = $- \hat{e}_1 A + \hat{e}_3 C$ $\frac{d}{dt}\hat{e}_3$ = $- \hat{e}_1 B - \hat{e}_2 C$

where,

 A $\equiv$ $\frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} - \frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2}$ B $\equiv$ $\frac{\dot{\lambda}_3}{h_1} \frac{\partial h_3}{\partial \lambda_1} - \frac{\dot{\lambda}_1}{h_3} \frac{\partial h_1}{\partial \lambda_3}$ C $\equiv$ $\frac{\dot{\lambda}_3}{h_2} \frac{\partial h_3}{\partial \lambda_2} - \frac{\dot{\lambda}_2}{h_3} \frac{\partial h_2}{\partial \lambda_3}$

Another way of expressing this involves Christoffel symbols and is most easily written using index notation.

$\frac{d}{dt} \hat{e}_c = \frac{h_a}{h_c} \ \Gamma^a_{bc} \dot{\lambda}_b \ \hat{e}_a \ \ (a \ne c)$

Here, we have been admittedly slopping with the placement and notation of indices in order to best accommodate the notation we have been using up to here. The b index is summed over all the coordinates. The a index is summed over all coordinates EXCEPT the c coordinate. The c index is NOT summed over because it is a free index, meaning that it can equal any of the coordinates depending on which unit vector you want to differentiate.

Writing this out for each of the individual unit vectors, and striking through terms that are automatically zero when dealing with an orthogonal coordinate system, produces

 $\frac{d}{dt} \hat{e}_1$ = $\overbrace{\frac{h_2}{h_1} \left( \Gamma^2_{11} \dot{\lambda}_1 + \Gamma^2_{21} \dot{\lambda}_2 + \cancel{\Gamma^2_{31} \dot{\lambda}_3} \right)}^{A} \hat{e}_2 + \overbrace{\frac{h_3}{h_1} \left( \Gamma^3_{11} \dot{\lambda}_1 + \cancel{\Gamma^3_{21} \dot{\lambda}_2} + \Gamma^3_{31} \dot{\lambda}_3 \right)}^{B} \hat{e}_3$ $\frac{d}{dt} \hat{e}_2$ = $\overbrace{\frac{h_1}{h_2} \left( \Gamma^1_{12} \dot{\lambda}_1 + \Gamma^1_{22} \dot{\lambda}_2 + \cancel{\Gamma^1_{32} \dot{\lambda}_3} \right)}^{-A} \hat{e}_1 + \overbrace{\frac{h_3}{h_2} \left( \cancel{\Gamma^3_{12} \dot{\lambda}_1} + \Gamma^3_{22} \dot{\lambda}_2 + \Gamma^3_{32} \dot{\lambda}_3 \right)}^{C} \hat{e}_3$ $\frac{d}{dt} \hat{e}_3$ = $\overbrace{\frac{h_1}{h_3} \left( \Gamma^1_{13} \dot{\lambda}_1 + \cancel{\Gamma^1_{23} \dot{\lambda}_2} + \Gamma^1_{33} \dot{\lambda}_3 \right)}^{-B} \hat{e}_1 + \overbrace{\frac{h_2}{h_3} \left( \cancel{\Gamma^2_{13} \dot{\lambda}_1} + \Gamma^2_{23} \dot{\lambda}_2 + \Gamma^2_{33} \dot{\lambda}_3 \right)}^{-C} \hat{e}_2$

where, quite generally, the 27 Christoffel symbols are,

 $\Gamma^1_{11}$ = $\frac{\partial_1 h_1}{h_1}$ $\Gamma^1_{12} = \Gamma^1_{21}$ = $\frac{\partial_2 h_1}{h_1}$ $\Gamma^1_{13} = \Gamma^1_{31}$ = 0 $\Gamma^1_{22}$ = $-\frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1}$ $\cancel{\Gamma^1_{23}} = \cancel{\Gamma^1_{32}}$ = 0 $\Gamma^1_{33}$ = $-\frac{h_3}{h_1} \frac{\partial_1 h_3}{h_1}$ $\Gamma^2_{11}$ = $-\frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2}$ $\Gamma^2_{12} = \Gamma^2_{21}$ = $\frac{\partial_1 h_2}{h_2}$ $\cancel{\Gamma^2_{13}} = \cancel{\Gamma^2_{31}}$ = 0 $\Gamma^2_{22}$ = $\frac{\partial_2 h_2}{h_2}$ $\Gamma^2_{23} = \Gamma^2_{32}$ = 0 $\Gamma^2_{33}$ = $-\frac{h_3}{h_2} \frac{\partial_2 h_3}{h_2}$ $\Gamma^3_{11}$ = 0 $\cancel{\Gamma^3_{12}} = \cancel{\Gamma^3_{21}}$ = 0 $\Gamma^3_{13} = \Gamma^3_{31}$ = $\frac{\partial_1 h_3}{h_3}$ $\Gamma^3_{22}$ = 0 $\Gamma^3_{23} = \Gamma^3_{32}$ = $\frac{\partial_2 h_3}{h_3}$ $\Gamma^3_{33}$ = 0

Notice that it is the Christoffel symbols labeled with all three of the coordinate indices that are automatically zero for orthogonal coordinate systems.

For additional details surrounding the Christoffel symbols (the significant role they play in the field equations, and how they can be calculated) visit this page coming soon.

## Time-Derivative of Position and Velocity Vectors

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

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

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

 $\frac{d\vec{v}}{dt}$ = $\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] + (h_1 \dot{\lambda}_1)\frac{d\hat{e}_1}{dt} + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] + (h_2 \dot{\lambda}_2)\frac{d\hat{e}_2}{dt} + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] + (h_3 \dot{\lambda}_3)\frac{d\hat{e}_3}{dt}$ = $\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] + (h_1 \dot{\lambda}_1)\biggl[ \hat{e}_2 A + \hat{e}_3 B \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] + (h_2 \dot{\lambda}_2)\biggl[ - \hat{e}_1 A + \hat{e}_3 C \biggr] + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] + (h_3 \dot{\lambda}_3)\biggl[ - \hat{e}_1 B - \hat{e}_2 C \biggr]$ = $\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt} - A(h_2 \dot{\lambda}_2) - B(h_3 \dot{\lambda}_3) \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) - C(h_3 \dot{\lambda}_3) \biggr] + \hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt} + B(h_1 \dot{\lambda}_1) + C(h_2 \dot{\lambda}_2) \biggr]$

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

$\vec{x} = \hat{e}_1 (h_1 {\lambda}_1) + \hat{e}_2 (h_2 {\lambda}_2) .$

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

 $\frac{d\vec{x}}{dt}$ = $\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] + (h_1 {\lambda}_1)\frac{d\hat{e}_1}{dt} + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] + (h_2 {\lambda}_2)\frac{d\hat{e}_2}{dt}$ = $\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] + (h_1 {\lambda}_1)\biggl[ \hat{e}_2 A + \hat{e}_3 B \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] + (h_2 {\lambda}_2)\biggl[ - \hat{e}_1 A + \hat{e}_3 C \biggr]$ = $\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt} - A(h_2 {\lambda}_2) \biggr] + \hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt} + A(h_1 {\lambda}_1) \biggr] + \hat{e}_3 \biggl[B(h_1 {\lambda}_1) + C(h_2 {\lambda}_2) \biggr]$

## Derived Identity for T3 Coordinates

Looking at the "$\hat{e}_2$" component of this last expression, we have,

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

But we also know that,

$\hat{e}_2 \cdot \vec{v} = h_2 \dot{\lambda}_2 .$

Hence, it must be true that,

 For T3 Coordinates $\lambda_2 \frac{d h_2}{dt} = - A(h_1 {\lambda}_1)$ or, equivalently, $A = - \frac{\lambda_2}{h_1 \lambda_1} \frac{d h_2}{dt}$

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 A, 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 A. So this must be a correct identity in the context of T3 coordinates.)

## Implications of Equation of Motion

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

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

$\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = - A(h_1 \dot{\lambda}_1)$

or, inserting the relation derived above for A in terms of dh2 / dt for T3 coordinates,

EOM.01

$\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = \biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt}$

$\Rightarrow ~~~~~ \frac{d(h_2 \lambda_1 \dot{\lambda}_2)}{dt} = \dot{\lambda}_1 \frac{d(h_2 \lambda_2)}{dt} .$

Or, equivalently (but perhaps more perversely),

EOM.02

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

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

Note that Equation EOM.01 is equivalent to the simplest form of the conservation that we derived — actually, Jay Call derived it — back in January, 2009. Specifically,

 $\lambda_1 \frac{d(h_2 \dot{\lambda}_2)}{dt} = \lambda_2 \dot{\lambda}_1 \frac{dh_2}{dt}$

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?"

Let's look at the 1st component of the equation of motion.

$\hat{e}_1 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_1 \dot{\lambda}_1)}{dt} - A(h_2 \dot{\lambda}_2) = - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1}$

$\Rightarrow ~~~~~ \frac{d(h_1 \dot{\lambda}_1)}{dt} = A(h_2 \dot{\lambda}_2) - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1} = - \biggl[ \frac{h_2 \lambda_2 }{h_1\lambda_1} \biggr]\dot{\lambda}_2 \frac{dh_2}{dt} - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1}$

## Logarithmic Derivatives of Scale Factors

Given the collection of expressions detailed in our derivation up to this point, the logarithmic derivatives of the scale factors are:

 $\frac{\partial\ln h_1}{\partial\ln\lambda_1}$ = $- \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1 } \biggr)^2$ $\frac{\partial\ln h_1}{\partial\ln\lambda_2}$ = $+ \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1 } \biggr)^2$ $\frac{\partial\ln h_2}{\partial\ln\lambda_1}$ = $+ (q h_1^2)^2$ $\frac{\partial\ln h_2}{\partial\ln\lambda_2}$ = $- ( qh_1^2 )^2$

This means, for example, that,

$\frac{\partial\ln h_1}{\partial\ln\lambda_2} = - \frac{\partial\ln h_1}{\partial\ln\lambda_1} ,$

and,

$\frac{\partial\ln h_2}{\partial\ln\lambda_2} = - \frac{\partial\ln h_2}{\partial\ln\lambda_1} .$

It also means that,

$\frac{d\ln h_2}{dt} = \biggl(\frac{\partial\ln h_2}{\partial\ln\lambda_1} \biggr) \frac{\dot{\lambda}_1}{\lambda_1} + \biggl(\frac{\partial\ln h_2}{\partial\ln\lambda_2} \biggr) \frac{\dot{\lambda}_2}{\lambda_2} = ( q h_1^2)^2 \biggl[\frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr]$

$\Rightarrow ~~~~~ \frac{1}{h_1^4}\frac{d\ln h_2}{dt} = \frac{d\ln(\lambda_1/\lambda_2)^{q^2}}{dt} ;$

and,

$\frac{d\ln h_1}{dt} = \biggl(\frac{\partial\ln h_1}{\partial\ln\lambda_1} \biggr) \frac{\dot{\lambda}_1}{\lambda_1} + \biggl(\frac{\partial\ln h_1}{\partial\ln\lambda_2} \biggr) \frac{\dot{\lambda}_2}{\lambda_2} = - \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1} \biggr)^2 \biggl[\frac{\dot{\lambda}_1}{\lambda_1}- \frac{\dot{\lambda}_2}{\lambda_2} \biggr]$

$\Rightarrow ~~~~~ - \biggl( \frac{\lambda_1}{h_1 h_2 \lambda_2} \biggr)^2 \frac{d\ln h_1}{dt} = \frac{d\ln(\lambda_1/\lambda_2)^{q^2}}{dt}.$

Hence, the logarithmic time-derivatives of the two key scale factors can be related to one another via the expression,

T3 Coordinates

$(h_1 \lambda_1)^2 \frac{d\ln h_1}{dt} + (h_2 \lambda_2)^2 \frac{d\ln h_2}{dt} = 0 .$

Also, plugging the relevant logarithmic derivatives of the key scale factors directly into the definition of A given in our discussion of vector derivatives, above, we can confirm the relationship derived earlier between A and dh2 / dt. We also can relate A directly to dh1 / dt. Specifically,

T3 Coordinates

$A = - \biggl( \frac{h_2 \lambda_2}{h_1 \lambda_1} \biggr) \frac{d\ln h_2}{dt} = + \biggl( \frac{h_1 \lambda_1}{h_2 \lambda_2} \biggr) \frac{d\ln h_1}{dt} .$

It must therefore also be true that,

T3 Coordinates

$(h_1\lambda_1)^2 \frac{\partial \ln h_1}{\partial\lambda_j} = - (h_2\lambda_2)^2 \frac{\partial \ln h_2}{\partial\lambda_j}$