Retrieve expression in a less simplified form w/ partial derivs unsolved

Questions
,
1

My print out for rct (ricci curvature tensor) attempts to partially differentiate some terms however for my work i need my solution to be in terms of partial derivatives (hopefully this makes sense). It’s towards Birkhoffs uniqueness theorem if this adds context. thankyou in advance

d = 4 ; y = {t, r, \[Theta], \[Phi]} ; 

G = {{-(E^(2 X[t, r])), 0, 0, 0}, {0, (E^(2 Y[t, r])), 0, 0}, {0, 0, 
    r^2, 0}, {0, 0, 0, r^2  Sin[\[Theta]]^2}};

IG = Inverse[G];

\[CapitalGamma] = 
  Table[1/2*
    Sum[IG[[i, 
       l]]*(D[G[[j, l]], y[[k]]] + D[G[[k, l]], y[[j]]] - 
        D[G[[j, k]], y[[l]]]), {l, 1, d}], {i, 1, d}, {j, 1, d}, {k, 
    1, d}];
Rct = Table[
   D[\[CapitalGamma] [[i, l, j]], y[[k]]] - 
    D[\[CapitalGamma] [[i, l, k]], y[[j]]] + 
    Sum[\[CapitalGamma] [[i, k, a]]*\[CapitalGamma] [[a, l, j]], {a, 
      1, d}] - 
    Sum[\[CapitalGamma] [[i, l, b]]*\[CapitalGamma] [[b, k, j]], {b, 
      1, d}], {i, 1, d}, {j, 1, d}, {k, 1, d}, {l, 1, d}];
rct = Table[Sum[Rct[[c, i, c, j]], {c, 1, d}], {i, 1, d}, {j, 1, d}];
rct
2

Maybe Inactivate[<some expression>, D] is what you are looking for:

\[CapitalGamma] = 
  Inactivate[
   Table[1/2*
     Sum[IG[[i, l]]*(D[G[[j, l]], y[[k]]] + D[G[[k, l]], y[[j]]] - 
         D[G[[j, k]], y[[l]]]), {l, 1, d}], {i, 1, d}, {j, 1, d}, {k, 
     1, d}], D];
Rct = Inactivate[
   Table[D[\[CapitalGamma][[i, l, j]], y[[k]]] - 
     D[\[CapitalGamma][[i, l, k]], y[[j]]] + 
     Sum[\[CapitalGamma][[i, k, a]]*\[CapitalGamma][[a, l, j]], {a, 1,
        d}] - Sum[\[CapitalGamma][[i, l, b]]*\[CapitalGamma][[b, k, 
         j]], {b, 1, d}], {i, 1, d}, {j, 1, d}, {k, 1, d}, {l, 1, d}],
    D];
3

This just keeps throwing up errors unfortunately

4

Well, of course you have to execute the whole code - I posted above the changed part only! OK, it then reads:

ClearAll["Global`*"]
d = 4; y = {t, r, \[Theta], \[Phi]};
G = {{-(E^(2 X[t, r])), 0, 0, 0}, {0, (E^(2 Y[t, r])), 0, 0}, {0, 0, 
    r^2, 0}, {0, 0, 0, r^2 Sin[\[Theta]]^2}};
IG = Inverse[G];
\[CapitalGamma] = 
  Inactivate[
   Table[1/2*
     Sum[IG[[i, l]]*(D[G[[j, l]], y[[k]]] + D[G[[k, l]], y[[j]]] - 
         D[G[[j, k]], y[[l]]]), {l, 1, d}], {i, 1, d}, {j, 1, d}, {k, 
     1, d}], D];
Rct = Inactivate[
   Table[D[\[CapitalGamma][[i, l, j]], y[[k]]] - 
     D[\[CapitalGamma][[i, l, k]], y[[j]]] + 
     Sum[\[CapitalGamma][[i, k, a]]*\[CapitalGamma][[a, l, j]], {a, 1,
        d}] - Sum[\[CapitalGamma][[i, l, b]]*\[CapitalGamma][[b, k, 
         j]], {b, 1, d}], {i, 1, d}, {j, 1, d}, {k, 1, d}, {l, 1, d}],
    D];
rct = Table[Sum[Rct[[c, i, c, j]], {c, 1, d}], {i, 1, d}, {j, 1, d}];
rct

This gives a lengthy result, but with unresolved partial derivatives. Is it this you want?

5

Sorry im really not great with mathematica, and it’s along the right lines, basically my result is the following

ScreenShot2019-04-01at22.06.44.png
ScreenShot2019-04-01at22.06.44.png750×222 14.6 KB

But i was hoping i could have things such as the the 2nd to last line, the 3rd term, ideally id like it to read (sin2?)e?2Y(?1 +e2Y?r?rX+r?rY) instead, i know its correct but for my future calculations it would be nice if in my appendix, the code would print exactly that, thankyou for your time

6

I cannot reproduce your result. When I do not do anything else but execute my code from above I get something like:

partial.png
partial.png830×433 10.2 KB

which does contain the unresolved derivatives you are asking for.
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=partial.png&userId=32203