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User:Jaycall/T3 Coordinates/Special Case

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Coordinate Transformations

If the special case q2 = 2 is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

λ1

\equiv

\left( R^2+2z^2 \right)^{1/2}

      and      

λ2

\equiv

\frac{R^2}{\sqrt{2}z}

R2

\equiv

\frac{\lambda_2}{2} \left( - \lambda_2 + \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) = \frac{2{\lambda_1}^2}{\Lambda+1}

      and      

z

\equiv

\frac{1}{2^{3/2}} \left( -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right) = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)

where \Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2} .

From this definition of Λ, we can compute both its partials with respect to the T3 coordinates, and its total time derivative.

\frac{\partial \Lambda}{\partial \lambda_1} = \frac{4 \lambda_1 / {\lambda_2}^2}{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \frac{1}{\lambda_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial \Lambda}{\partial \lambda_2} = - \frac{4 {\lambda_1}^2 / \lambda_2}{\Lambda} = - \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \lambda_1

\dot{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \left( \frac{\dot{\lambda_1}}{\lambda_2} - \lambda_1 \dot{\lambda_2} \right)

Partials of the Coordinates

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

 


\frac{\partial}{\partial R}


\frac{\partial}{\partial z}


\frac{\partial}{\partial \phi}

λ1


\frac{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}


\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}

0

λ2


\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}


-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}

0

λ3

0

0

1

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

 


\frac{\partial}{\partial \lambda_1}


\frac{\partial}{\partial \lambda_2}


\frac{\partial}{\partial \lambda_3}

R


R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}


2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}

0

z


2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}


-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}

0

φ

0

0

1

where \ell \equiv \left( R^2 + 4z^2 \right)^{-1/2} = \frac{1}{\lambda_1} \left( \frac{\Lambda + 1}{2 \Lambda} \right)^{1/2}.

Scale Factors

Furthermore, the scale factors become

h1

=

\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}

h2

=

Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}

h3

=

R = λ3

Useful Relationships

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

R2 + 2z2

=

{\lambda_1}^2

R2 + 4z2

=

2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}

R2 + 8z2

=

4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2

R2 − 2z2

=

- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}

Rz

=

\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell

Additional Partials

Partials of \ell can be taken with respect to the coordinates of either system. They are:

 


\frac{\partial}{\partial R}


\frac{\partial}{\partial z}


\frac{\partial}{\partial \phi}

\ell


-R \ell^3


-4z \ell^3

0

 


\frac{\partial}{\partial \lambda_1}


\frac{\partial}{\partial \lambda_2}


\frac{\partial}{\partial \lambda_3}

\ell


- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)


2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2

0

Partials of the scale factors taken with respect to the T3 coordinates are:

 


\frac{\partial}{\partial \lambda_1}


\frac{\partial}{\partial \lambda_2}


\frac{\partial}{\partial \lambda_3}

h1


\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2


2 {h_2}^2 \ell^3 \lambda_1 \lambda_2

0

h2


2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3


h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2

0

h3


R \ell^2 \lambda_1


2Rz^2 \ell^2 / \lambda_2

0

Conserved Quantity

The conserved quantity associated with the λ2 coordinate is


m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt .

The quantity in brackets needs to be integrated. In terms of Λ, it can be written

{\lambda_2}^2 \Lambda \left( \Lambda - 1 \right) \left[ \frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} \lambda_1 \dot{\lambda_2} - {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}} \frac{\dot{\lambda_1}}{\lambda_2} \right] .

Notice that the thing in square brackets looks very closely related to \dot{\Lambda}. Could this be a hint? If only we could figure out what \frac{\dot{\lambda_1}}{\dot{\lambda_2}} is, maybe we could factor out the \lambda_1 \dot{\lambda_2} - \frac{\dot{\lambda_1}}{\lambda_2}, which appears in \dot{\Lambda}...

If, by some miracle, it should turn out that \frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} = {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}}, factorization would be possible and our integral would read

-\tfrac{1}{4} \int {\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right) \ \dot{\Lambda} \ dt = - \tfrac{1}{4} \int {\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right) \  d\Lambda .

We ought to be able to integrate this, right...? Maybe we could handle the pesky {\lambda_2}^2 with integration by parts...

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