Transformation from cartesian to cylindrical coordinates

[Otacon23]

Homework Statement

I'm trying to get to grips with Godel's 1949 Paper on Closed Time-like Curves (CTCs). Currently I'm trying to confirm his transformation to cylindrical coordinates using maple but seem to keep getting the wrong answer.

Homework Equations

The line element in cartesian coordinates is: $dS^2 = a^2[dx_0^2-dx_1^2+\frac{1}{2}$e^$2x_1$$dx_2^2-dx_3^2+2$e^$x_1$$dx_0dx_2]$ using the following substitutions:
$e$^$x_1$$= \cosh(2r) + \cos(\phi)\sinh(2r)$
$x_2e$^$x_1$$= \sqrt{2}\sin(\phi)sinh(2r)$
$\tan(\frac{\phi}{2} + \frac{x_0 - 2t}{2\sqrt{2}}) = e$^$2r$$\tan(\frac{\phi}{2})$
$x_3 = 2y$
I need to show that the line element is $dS^2 = 4a^2[dt^2 -dr^2 -dy^2 + (\sinh^4(r) - \sinh^2(r))d\phi^2 + 2\sqrt{2}\sinh^2(r)d\phi dt]$

The Attempt at a Solution

I started by differentiating these four equations to obtain expressions for $dx_0, dx_1, dx_2, dx_3$

$dx_0 = 2e$^$-2r$$\sqrt{2}cos^2(\frac{\phi}{2} + \frac{x_0 - 2t}{2\sqrt{2}})(\frac{1}{2}\sec^2(\frac{\phi}{2})d\phi - 2\tan(\frac{\phi}{2})dr) + 2dt - \sqrt{2}d\phi$

$dx_1 =2e$^$-x_1$$(sinh(2r) + cos(\phi)cosh(2r))dr - e$^$-x_1$$sin(\phi)sinh(2r)d\phi$

$dx_2 = e$^$-x_1$$sinh(2r)(\sqrt{2}cos(\phi) +x_2\sin(\phi))d\phi + 2e$^$-x_1$$(\sqrt{2}sin(\phi)\cosh(2r)-x_2\sinh(2r)-cos(\phi)\cosh(2r))dr$

$dx_3 = 2dy$

In maple I then defined the variable 'X' to be
$X := a^2[dx_0^2-dx_1^2+\frac{1}{2}$e^$2x_1$$dx_2^2-dx_3^2+2$e^$x_1$$dx_0dx_2] - 4a^2[dt^2 -dr^2 -dy^2 + (\sinh^4(r) - \sinh^2(r))d\phi^2 + 2\sqrt{2}\sinh^2(r)d\phi dt];$

Then defined 'XX' to be
$XX:=(subs(dx_0 = ..., dx_1 = ..., dx_2 = ..., dx_3 = ..., X));$
Then simplified the result hoping it to be zero.
$simplify(XX);$
Instead of zero I receive 4 pages or so of trash, so basically can anyone point out where or how I have gone askew?

Many thanks!

 

[Otacon23]

I tried letting all the variables equal to a random constant and managed to get $'XX'$ to equal 0. It seems that I get a non-zero value of $'XX'$ whenever I let any of $t, r, \phi$ remain a variable and not a constant, which leads me to believe that my substitution and hence my differentiation is incorrect. If anyone has the time to double check any of my differentiations I would really appreciate it!
Maths Love xx

 

[Otacon23]

Narrowed it down to there just being a problem with $dx_0$
Anyone care to look?

 

[haruspex]

$dx_0 = 2e$^$-2r$$\sqrt{2}cos^2(\frac{\phi}{2} + \frac{x_0 - 2t}{2\sqrt{2}})(\frac{1}{2}\sec^2(\frac{\phi}{2})d\phi - 2\tan(\frac{\phi}{2})dr) + 2dt - \sqrt{2}d\phi$

You have an x₀ on the RHS side there. Assuming you want rid of that, I get
$dx_0 = 2dt - \sqrt{2}d\phi + \sqrt{2}\frac{2 sin(\phi)dr+d\phi}{e^{-2r}cos^2(\phi/2)+e^{2r}sin^2(\phi/2))}$
The denominator could also be written $cosh(2r) - sinh(2r)cos(\phi)$