I) The Problem
There's a particular metric $[1],[2]$ in general relativity which is written as:
$$ds^{2} = -[c^2-v_{s}^2f(r_{s})^2]dt^2+v_{s}f(r_{s})dtdx+v_{s}f(r_{s})dxdt+ dy^2+dz^2 \tag{1}$$
So my question is:
How can I calculate the Christoffel symbols of this metric using Mathematica or similar software?
II) Important facts before answering to my question and propose code fixing:
Now, everything about that geometry you can find in $[1],[2]$; for the sake of commodity, I'll give important concepts and notation.
Firstly, this metric gives an idea of a "hyperdrive" or a "warp bubble",i.e. a body can move through the point $A$ to point $B$ using just spacetime curvature, given by metric $(1)$; this metric uses the cartesian coordinates $(t,x,y,z)$. I'm saying this because the movement of the bubble could in principle occur in a (spatial) vector direction, but the problem is simplified by assuming a one-dimensional displacement in $x$-axis. Therefore, this implies that the quantity $v_{s}$ (the velocity of the bubble, seen by an external observer) is:
$$v_{s} \equiv v_{s}(t) =: \frac{dx_{s}(t)}{dt} \tag{2}$$
Secondly, the radius from the center of the bubble to the "curvature wall" is given by a function $r_{s}$ which is:
$$r_{s} =: \sqrt{(x-x_{s}(t))^2+y^2+z^2} \tag{3}$$
Finally, the function $f$ is given by:
$$f(r_{s}) =: \frac{tanh[\sigma(r_{s}+R)]-tanh[\sigma(r_{s}-R)]}{2tanh[\sigma R]} \tag{4}$$
Now, $R$ and $\sigma$ are parameters (quantities which we can put in a slider to continuously change); the terms $\sigma(r_{s}+R)$ and $\sigma(r_{s}-R)$ aren't denoting $\sigma(r)$, they are a multiplication of a function ($r_{s}$) by the number (parameter) $\sigma$. The reason for this function can be understood in $[1],[2]$.
III) My (unsuccessful) effort
Now, Christoffel symbols (CS) are "easily" calculated because we have a strong algorithmic process to do it $[3]$. I calculated the Christoffel symbols of this metric "by hand" and my final results turned to be quite right; my outputs are the same as $[2]$, therefore, there's no doubt about the final result of a Mathematica program.
To calculate Christoffel symbols you must declare a metric tensor, calculate the inverse metric tensor and just after that you're able to start the calculations of CS.
The final results which I calculated "by hand" (also in $[2]$) are:
$\Gamma^{0}_{00} = -\frac{\partial_{1}f(r_{s})f(r_{s})v_{s}(t)^3}{c^2}$
$\Gamma^{0}_{01} = \Gamma^{0}_{10} = -\frac{\partial_{1}f(r_{s})f(r_{s})v_{s}(t)^2}{c^2}$
$\Gamma^{0}_{11} = -\frac{v_{s}(t)\partial_{1}f(r_{s})}{c^2}$
$\Gamma^{1}_{00} = \frac{1}{c^2}\Big(\partial_{1}f(r_{s})v_{s}(t)^2f(r_{s})(-c^2+v_{s}(t)^2f(r_{s})^2)+c^2\partial_{0}v_{s}(t)f(r_{s})+c^2\partial_{0}f(r_{s})v_{s}(t)\Big)$
$\Gamma^{1}_{01} = \Gamma^{1}_{10} = \frac{\partial_{1}f(r_{s})f(r_{s})v_{s}(t)^3}{c^2}$
My codes are returning, in fact, the wrong outputs:
FIRST TRY
- SECOND TRY
IV) Some codes from other people which I tried to adapt
The code isn't mine's, is from Hartle's site $[4]$ in the link called "Christoffel Symbols and Geodesic Equations"
$$ * * * $$
$[1]$ https://arxiv.org/abs/gr-qc/0009013
$[2]$ https://arxiv.org/abs/1202.5708
$[3]$ David McMahon's Relativity Demystified
$[4]$ http://web.physics.ucsb.edu/~gravitybook/mathematica.html
