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I'm working with perturbation in differential geometry. I got a huge matrix for Christoffel symbol, and I need to simplify it using the consideration that $\phi^2$, $\psi^2$, $\phi\partial_{\mu}\phi$, $\psi\partial_{\mu}\phi$,$\psi\partial_{\mu}\psi$, $\phi\partial_{\mu}\psi$ ($\mu=0,1,2,3$, where $0$ is temporal derivative and $1,2,3$ are spatial derivative), $\phi\psi$ are equal to zero.

My code

crd = {t, x1, x2, x3};
metric = {{-n[t]^2 (1 + 2*\[CapitalPhi][crd]), 0 , 0, 0}, {0, 
A[t]^2 (1 - 2*\[CapitalPsi][crd]), 0, 0}, {0, 0, 
A[t]^2 (1 - 2*\[CapitalPsi][crd]), 0}, {0, 0, 0, 
A[t]^2 (1 - 2*\[CapitalPsi][crd])}};
MatrixForm[metric];
inversemetric = {{-(1 - 2*\[CapitalPhi][crd])/n[t]^2, 0 , 0, 
0}, {0, (1 + 2*\[CapitalPsi][crd])/a[t]^2, 0, 0}, {0, 
0, (1 + 2*\[CapitalPsi][crd])/a[t]^2, 0}, {0, 0, 
0, (1 - 2*\[CapitalPsi][crd])/a[t]^2}};

christoffel =  Expand[FullSimplify[Table[ 1/2 Sum[inversemetric[[b, s]] (D[metric[[d, s]], crd[[c]]] + D[metric[[s, c]], crd[[d]] ] - D[metric[[c, d]], crd[[s]] ] ), {s, 1, 4}], {b, 1, 4}, {c, 1, 4}, {d, 1, 4}]]];

christoffel[[1, 1, 1]]

$\partial_{0}\Phi-2 \partial _0\Phi \Phi +\frac{n'(t)}{n(t)}-\frac{4 \Phi ^2 n'(t)}{n(t)}$

It needs to be:

$\partial_{0}\Phi +\frac{n'(t)}{n(t)}$

and

christoffel[[1, 2, 2]]

$\frac{4 \Phi \Psi A(t) A'(t)}{n(t)^2}-\frac{2 \Phi A(t) A'(t)}{n(t)^2}-\frac{2 \Psi A(t) A'(t)}{n(t)^2}+\frac{A(t) A'(t)}{n(t)^2}+\frac{2 \partial _0\Psi \Phi A(t)^2}{n(t)^2}-\frac{\partial _0\Psi A(t)^2}{n(t)^2}$

It needs to be

$-\frac{2 \Phi A(t) A'(t)}{n(t)^2}-\frac{2 \Psi A(t) A'(t)}{n(t)^2}+\frac{A(t) A'(t)}{n(t)^2}-\frac{\partial _0\Psi A(t)^2}{n(t)^2}$

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  • $\begingroup$ Can you provide the inversemetric? When I try to execute your code (using inversemetric=Inverse[metric];), I don't obtain the same results with cristoffels. $\endgroup$ – Chanto Sep 8 '18 at 18:41
  • $\begingroup$ @Chanto,sadly "inverse matrix=Inverse[metric]" don't give my inverse metric correctly, because it is a perturbed metric, so I wrote a "invesermetric" matrix myself. It was a lame trick, I know... $\endgroup$ – Kamog Sep 8 '18 at 18:46
  • $\begingroup$ Nevertheless (physics aside), if you include the inversematrix here, I might be able to help you out. $\endgroup$ – Chanto Sep 8 '18 at 20:22
  • $\begingroup$ @Chanto, I'm sorry for that, I forgot to put the inverse matrix above! I just corrected it! Thank you! $\endgroup$ – Kamog Sep 8 '18 at 20:39
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I found a way using a parameter "e" multiplying my perturbation:

metric = {{-n[t]^2 (1 + 2*e*\[CapitalPhi][crd]), 0 , 0, 0}, {0,A[t]^2 (1 - 2*e*\[CapitalPsi][crd]), 0, 0}, {0, 0, A[t]^2 (1 - 2*e*\[CapitalPsi][crd]), 0}, {0, 0, 0, A[t]^2 (1 - 2*e*\[CapitalPsi][crd])}};

Now I can use:

Series[ ..., {e,0,1}], 

Example, to calculate Christofell symbol:

christoffel = Expand[FullSimplify[Series[Table[1/2 Sum[inversemetric[[\[Rho], \[Sigma]]] (D[metric[[\[Sigma], \[Mu]]], crd[[\[Nu]]]] +         D[metric[[\[Sigma], \[Nu]]], crd[[\[Mu]]] ] - D[metric[[\[Mu], \[Nu]]], crd[[\[Sigma]]] ] ), {\[Sigma], 1, 4}], {\[Rho], 1, 4}, {\[Mu], 1, 4}, {\[Nu], 1, 4}], {e, 0, 1} ]]];
christofell[[1,1,1]]

$\partial_{0}\Phi + \frac{n'}{n}$

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