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I use the code in this thread's answer:

(Calculating Einstein tensor components in Kaluza-Klein model) to get the Einstein tensor components of a four-dimensional Kaluza Klein model. But instead of the diagonal matrix in that thread, I use a perturbed metric 0209156:

$ g_{00}= a^2(t) ( 1+ 2 \phi^{(1)} + \phi^{(2)} )$

$ g_{0i}=0$

$g_{ij}= a^2(t) [( 1- 2 \psi^{(1)}- \psi^{(2)})~\delta_{ij} + \frac{1}{2} D_{ij} \chi_{ij}^{(2)}] $,

Where:

$ D_{ij} =\partial_i \partial_j - \frac{1}{3} \delta_{ij} \partial^k \partial_k $

Here is the entire code I use:

coordList = {t, x, w, z};
coord[i_] := coordList[[i]]

met = a[t]^2 Exp[y] {{-(1 + 2 n \[CapitalPhi][t, x, w, z, y]), 0, 0, 0},
    {0, 1 - 2 n \[CapitalPsi][t, x, w, z, y] +  n 1/2 D[Xi[t, x, w, z, y], {x, 2}] - 
      1/2 n D[Xi[t, x, w, z, y], {w, 2}] - 1/2 n D[Xi[t, x, w, z, y], {z, 2}], 
     n D[D[Xi[t, x, w, z, y], x], w], n D[D[Xi[t, x, w, z, y], x], z]}, {0, n D[D[Xi[t, x, w, z, y], x], w], 1 - 2 n \[CapitalPsi][t, x, w, z, y] + n 1/2 D[Xi[t, x, w, z, y], {x, 2}] -  1/2 n D[Xi[t, x, w, z, y], {w, 2}] -  1/2 n D[Xi[t, x, w, z, y], {z, 2}], 
     n D[D[Xi[t, x, w, z, y], w], z]},{0, n D[D[Xi[t, x, w, z, y], x], z],n D[D[Xi[t, x, w, z, y], w], z],  1 - 2 n \[CapitalPsi][t, x, w, z, y] + n 1/2 D[X[t, x, w, z, y], {x, 2}] - 
      1/2 n D[Xi[t, x, w, z, y], {w, 2}] -1/2 n D[Xi[t, x, w, z, y], {z, 2}]}}; 

 invmetric = Inverse@metric;

m[a_, b_] := metric[[a, b]];

im[a_, b_] := invmetric[[a, b]];

d[a_, f_] := D[f, coord[a]];

dDel[b_, f_] := covd[f, coord[b]];

covd[0, x_] := 0;

d4[f_] := D[f, y]

term1[a_, b_] := dDel[b, d[a, f[t, y]]]/f[t, y];

term2[a_, b_] := d4[f[t, y]] d4[m[a, b]]/f[t, y];

term3[a_, b_] := -d4[d4[m[a, b]]];

term4[a_, b_, g_, d_] := im[g, d] (d4[m[a, g] d4[m[b, d]]]);

term5[a_, b_, g_, d_] := -im[g, d] (d4[m[g, d] d4[m[a, b]]])/2;

term6[a_, b_, g_, d_] := 
  m[a, b] (d4[im[g, d]] d4[m[g, d]] + (im[g, d] d4[m[g, d]])^2)/4;

ein[a_, b_] := -(term1[a, b] + term2[a, b] + term3[a, b] + 
       Sum[term4[a, b, g, d] + term5[a, b, g, d] + 

Where n is the order of perturbation. The problem now the output is huge. How to linearize such output by Series[Exp[n], {n,0,1}], and how to divide it to first order and second order perturbation terms like for instance in 0209156 Equations (30-40)

Edit

Here is an example of a code calculating Einstein tensor from the christoffel symbols, where it uses Series[--, {n,0,1}] for linearizton. And I do not get if it can divide the first order and second order perturbation terms. The idea I can not use this code because I substitute directly by the metric to the Einstein tensor equation. So this code needs to be modified for my calculation.

    x = {t, r, \[Theta], \[Phi]};
    met ={};
    metI=Inverse[met];
    
     Chrest = 
      ParallelTable[
       Normal[Series[
         1/2 (Sum[
            metI[[i, 
              p]] (D[met[[p, j]], x[[k]]] + D[met[[k, p]], x[[j]]] - 
               D[met[[j, k]], x[[p]]]), {p, 1, 4}]), {n, 0, 1}]], {i,1,4}, {j, 1, 4}, {k, 1, 4}];
    
    Reim = ParallelTable[
       Normal[Series[
         D[Chrest[[i, l, j]], x[[k]]] - D[Chrest[[i, k, j]], x[[l]]] + 
          Sum[Chrest[[i, k, m]] Chrest[[m, l, j]], {m, 1, 4}] - 
          Sum[Chrest[[i, l, m]] Chrest[[m, k, j]], {m, 1, 4}], {n, 0, 
          1}]], {i, 1, 4}, {j, 1, 4}, {k, 1, 4}, {l, 1, 4}];
    
     Ricci =  ParallelTable[
       Normal[Series[Sum[Reim[[m, i, m, j]], {m, 1, 4}], {n, 0, 1}]], {i,1, 4}, {j, 1, 4}];
    
    RScal =  Normal[Series[
        Sum[Sum[metI[[i, j]] Ricci[[i, j]], {i, 1, 4}], {j, 1, 4}], {n,0,1}]];
    
     Eins =   ParallelTable[
       Normal[Series[
         Ricci[[i, j]] - 1/2 met[[i, j]] RScal, {n, 0, 1}]], {i, 1, 
        4}, {j, 1, 4}];
    
EinsUD = ParallelTable[
       Normal[Series[
         Sum[metI[[p, i]] Eins[[p, j]], {p, 1, 4}], {n, 0, 1}]], {i, 1, 
        4}, {j, 1, 4}];
    
  For[i = 1, i < 5, i++, For[j = 1, j < 5, j++,
       Eins = 
        ReplacePart[
         Eins, {i, j} -> 
          Normal[Normal[
            Series[Ricci[[i, j]] - 1/2 met[[i, j]] RScal, {n,0,1}]]]]]];
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    $\begingroup$ This might be a good time to download and learn xAct, since they incorporate xPert.m which handles perturbation theory for tensors. Maybe start by going to their site and looking at one of the xPert example notebooks to get an idea if this will meet your needs. $\endgroup$
    – user87932
    Oct 10, 2023 at 19:36
  • $\begingroup$ xAct seems to me to be the most popular option for tensor calculations with mathematica. Other options (that might have more bugs, to be expected particularly with the newer ones) can be found here en.wikipedia.org/wiki/…. I personally prefer to use xAct but I have not really experimented with the options in that list. $\endgroup$ Oct 10, 2023 at 23:37
  • $\begingroup$ I mean a code with no extra MA packages, like the code in the question edit, where it calculates Einstein tensor from the christoffel symbols and make linearization. $\endgroup$
    – Dr. phy
    Oct 11, 2023 at 18:16
  • 1
    $\begingroup$ In the code you posted, you use Series[expr,{n,0,1}]. If you check the documentation, this mean expand in n around the value n=0. But nothing in expr depends on n, so this won't do anything. Series is normally used for expanding around a scalar or a list of scalar variables, not matrices. If you want to compute Christoffel symbols, Riemann tensors, etc. from a metric, without using a package, you might find this post useful: mathematica.stackexchange.com/questions/8895/… $\endgroup$
    – user87932
    Oct 12, 2023 at 18:01
  • $\begingroup$ @jdp. Hi, the metric in met depends on n which is the order of perturbation. $\endgroup$
    – Dr. phy
    Oct 12, 2023 at 19:14

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