# Difficulties on Mathematica code to solve Christoffel Symbols of a particular metric

I) The Problem

There's a particular metric $$,$$ 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 $$,$$; 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 $$,$$.

III) My (unsuccessful) effort

Now, Christoffel symbols (CS) are "easily" calculated because we have a strong algorithmic process to do it $$$$. I calculated the Christoffel symbols of this metric "by hand" and my final results turned to be quite right; my outputs are the same as $$$$, 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 $$$$) 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: 1. SECOND TRY

IV) Some codes from other people which I tried to adapt

The code isn't mine's, is from Hartle's site $$$$ in the link called "Christoffel Symbols and Geodesic Equations"

$$* * *$$

$$$$ https://arxiv.org/abs/gr-qc/0009013

$$$$ https://arxiv.org/abs/1202.5708

$$$$ David McMahon's Relativity Demystified

$$$$ http://web.physics.ucsb.edu/~gravitybook/mathematica.html

• I am the moderator of Computational Science SE. I accidentally migrated this question here (even though I personally think it belongs here) without making sure. Let me know if it's not OK. I already asked the OP for this on our community. Mar 31, 2020 at 0:47
• Please show us the code text rather than screenshot of it so we can easily test. As to typesetting of code, you may want to read the following post: mathematica.meta.stackexchange.com/q/1584/1871. As to Christoffel symbols, I'm not familiar with the topic, but have you read this post?: mathematica.stackexchange.com/q/8895/1871. Mar 31, 2020 at 2:50
• @AntonMenshov I am not a moderator for Mathematica SE but the post does indeed look like it belongs here. Mar 31, 2020 at 15:17

Several different codes for routine calculations in general relativity were published here. I will give one of them (sorry, I don’t remember where I got it from). This code accurately reproduces your results (where no typos):

Clear [coord, metric, inversemetric, affine, riemann, ricci, scalar, \
einstein, t, x, y, z]

n = 4;
coord = {t, x, y,z};

metric = {{-c^2 + vs[t]^2 f[t, x, y, z]^2, vs[t] f[t, x, y, z], 0,
0}, {vs[t] f[t, x, y, z], 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}};

inversemetric = Simplify[Inverse[metric]];

affine :=
affine = Simplify[
Table[(1/2)*
Sum[inversemetric[[i,
s]]*(D[metric[[s, j]], coord[[k]]] +
D[metric[[s, k]], coord[[j]]] -
D[metric[[j, k]], coord[[s]]]), {s, 1, n}], {i, 1, n}, {j,
1, n}, {k, 1, n}]];

listaffine :=
Table[If[UnsameQ[affine[[i, j, k]],
0], {ToString[\[CapitalGamma][i - 1, j - 1, k - 1]],
affine[[i, j, k]]}], {i, 1, n}, {j, 1, n}, {k, 1, j}];
TableForm[Partition[DeleteCases[Flatten[listaffine], Null], 2],
TableSpacing -> {2, 2}] 