The original code related stuff that will be used in this post can be found here
Other relevant links from this site related to this discussion are the following: Artes' wonderful answer, another answer by myself, and another one.
pretty sure there are some more related material on this site...
See this recent answer of mine for a thorough explanation of the approach taken here. Here we are moving a bit faster to the end result.
If I understand the OP properly, we are interested in obtaining the functions $A(r)$ and $B(r)$ appearing in the metric.
We set the dimensionality of the spacetime
n = 4;
Now we write the coordinate system
coord = {t, r, θ, φ}
We insert the metric elements. Note that these are the ones with both indices down; $g_{\mu \nu}$. We do it in the following manner:
metric = {{-A[r], 0, 0, 0}, {0, B[r], 0, 0}, {0, 0, r^2, 0}, {0, 0, 0,
r^2 Sin[θ]^2}};
Its inverse is of course $g^{\mu \nu}$
inversemetric = Simplify[Inverse[metric]];
Computation of the Christoffel symbols; indices are $\Gamma^{x}_{xx}$:
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}] ]
Computation of the Riemann tensor; indices are $R^{x}_{xxx}$
riemann := riemann = Simplify[Table[
D[affine[[i, j, l]], coord[[k]] ] -
D[affine[[i, j, k]], coord[[l]] ] +
Sum[affine[[s, j, l]] affine[[i, k, s]] -
affine[[s, j, k]] affine[[i, l, s]],
{s, 1, n}],
{i, 1, n}, {j, 1, n}, {k, 1, n}, {l, 1, n}] ]
Components of the Ricci tensor $R_{xx}$
ricci := ricci =
FullSimplify[
Table[Sum[riemann[[i, j, i, l]], {i, 1, n}], {j, 1, n}, {l, 1,
n}] ]
Now, if you call ricci
you get the output of the above which is a list of lists.
ricci
gives
{{-((Derivative[1][A][r] Derivative[1][B][r])/(4 B[r]^2)) + (
Derivative[1][A][r]/r - Derivative[1][A][r]^2/(
4 A[r]) + (A^\[Prime]\[Prime])[r]/2)/B[r], 0, 0, 0}, {0, (
A[r] (4 A[r] + r Derivative[1][A][r]) Derivative[1][B][r] +
r B[r] (Derivative[1][A][r]^2 - 2 A[r] (A^\[Prime]\[Prime])[r]))/(
4 r A[r]^2 B[r]), 0, 0}, {0, 0,
1/2 (2 - (2 + (r Derivative[1][A][r])/A[r])/B[r] + (
r Derivative[1][B][r])/B[r]^2), 0}, {0, 0, 0, (
Sin[\[Theta]]^2 (-r B[r] Derivative[1][A][r] +
A[r] (2 (-1 + B[r]) B[r] + r Derivative[1][B][r])))/(
2 A[r] B[r]^2)}}
However, we can use the following
ricci[[1, 1]]
to get a single equation. This is the one related to the $\mu=\nu=0$ component of the tensor
-((Derivative[1][A][r] Derivative[1][B][r])/(4 B[r]^2)) + (
Derivative[1][A][r]/r - Derivative[1][A][r]^2/(
4 A[r]) + (A^\[Prime]\[Prime])[r]/2)/B[r]
This is where automatization ends to be honest. Then, one has to do some re-writing manually following the logic outlined here.
DSolve
appears to be helpful. $\endgroup$Part
is also very useful. If you have a list, sayefe={{},{},{},...}
you can look up whatefe[[stuff here]]
is giving you. Also, with all due respect, the OP is lacking crucial details. No definition of the Ricci tensor, Ricci scalar, and the energy-momentum tensor. Not everyone has a physics degree on this site, and maybe you would find more help with more details :-) $\endgroup$