Computing Christoffel symbols of the second kind [duplicate]

Question

I want to compute the Christoffel-symbol for a given metric. I am using the code here, but I am missing something.

The Chrisfoffel-symbol formula is

$\Gamma^{\mu}_{\phantom{\mu}\nu\sigma}=\frac{1}{2}g^{\mu\alpha}\left\{\frac{\partial g_{\alpha\nu}}{\partial x^{\sigma}}+\frac{\partial g_{\alpha\sigma}}{\partial x^{\nu}}-\frac{\partial g_{\nu\sigma}}{\partial x^{\alpha}}\right\}\quad$

The metric is given to be

$g_{\mu \nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & r^2+b^2 & 0 & 0 \\ 0 & 0 & (r^2+b^2)\sin^2(\theta) & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $

The provided solution is:

$\Gamma^{1}_{22}=-r$

$\Gamma^{1}_{33}=-r\sin^2(\theta)$

$\Gamma^{2}_{21}=\frac{r}{b^2+r^2}$

$\Gamma^{2}_{33}=-\cos(\theta)\sin(\theta)$

$\Gamma^{3}_{31}=\frac{r}{b^2+r^2}$

$\Gamma^{3}_{32}=\cot(\theta)$

The code I'm using is

xx = {t, x, \[Theta], \[Phi]};

g  = { {1,0,0,0},
{0,r^2+b^2,0,0},
{0,0,(r^2+b^2)Sin[\[Theta]]^2,0},
{0,0,0,-1}};

inversemetric = Simplify[Inverse[metric]];

ChristoffelSymbol[g_, xx_] := 
Block[{n, ig, res}, 
n = 4; ig = InverseMetric[ g]; 
res = Table[(1/2)*Sum[ ig[[i,s]]*(-D[ g[[j,k]], xx[[s]]] + D[ g[[j,s]], xx[[k]]] 
+ D[ g[[s,k]], xx[[j]]]), {s, 1, n}], {i, 1, n}, {j, 1, n}, {k, 1, n}];
Simplify[ res]
]

But I do not get the desired answer.

What am I missing? Besides, I'd like to learn how could I display the answer once I know how to actually get it.

Note I also checked Artes' solution here but I do not get how to run the code either.

---

EDIT

After playing around a bit with the Christoffel symbols (which is much more fun when you use Mathematica ;)) I've realized of several features:

1. If the metric is diagonal then the only way to get a non-zero Christoffel symbol is when any of the indices appears at least twice.

2. If the metric is diagonal we cannot have any index appearing three times yielding a non-trivial Christoffel symbol. The reason is because $g_{rr}$ is independent of $r$, $g_{\theta \theta}$ is independent of $\theta$, $g_{\phi \phi}$ is independent of $\phi$ and $g_{tt}$ is independent of $t$, which implies $\partial_{\mu} g_{\nu \rho}=0$ when $\mu=\nu=\rho$

3. Based on 1. and 2. we conclude that (when the metric is diagonal) all non-trivial Christoffel symbols must show repeated indices exactly twice.

Answer 1

The code you provided is a definition for a function to compute the Christoffel symbol (and Inverse to compute the inverse metric, I do not know "InverseMetric")

ChristoffelSymbol[g_, xx_] := 
 Block[{n, ig, res}, n = Length[xx]; ig = Inverse[g];
  res = Table[(1/2)*
     Sum[ig[[i, s]]*(-D[g[[j, k]], xx[[s]]] + D[g[[j, s]], xx[[k]]] + 
         D[g[[s, k]], xx[[j]]]), {s, 1, n}], {i, 1, n}, {j, 1, n}, {k,
      1, n}];
  Simplify[res]]

Then you have to define coordinates xx and the compontents metric with respect to the coordinate basis.

(* The coordinates *)
xx = {r, \[Theta], \[Phi], t};

(* The metric *)
g = {{1, 0, 0, 0}, {0, r^2 + b^2, 0, 0}, {0, 
    0, (r^2 + b^2) Sin[\[Theta]]^2, 0}, {0, 0, 0, -1}};

(* The Christoffel *)
sol = ChristoffelSymbol[g, xx] (* This calls the function! *);

sol[[1, 2, 2]]
(* -r *)

sol[[1, 3, 3]]
(* -r Sin[\[Theta]]^2 *)

sol[[2, 2, 1]]
(* r/(b^2 + r^2) *)

sol[[2, 3, 3]]
(* -Cos[\[Theta]] Sin[\[Theta]] *)

sol[[3, 3, 1]]
(* r/(b^2 + r^2) *)

sol[[3, 3, 2]]
(* Cot[\[Theta]] *)

Edited the answer with the correct coordinates.

Note that you get some coefficients "twice" since the Christoffel symbols are symmetric.

Union@Flatten@With[{n = Length[xx]}, Table[sol[[i, j, k]] == sol[[i, k, j]], {i, n}, {j, n}, {k, n}]]
(* {True} *)