Consider the Einstein Field equations $$R_{\mu\nu}-\frac12Rg_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}.$$ Typically, the stress energy tensor $T_{\mu\nu}$ is assumed to be a perfect fluid, which implies that $T_{\mu\nu}$ takes the form $$T_{\mu\nu}=(\rho+p)U_{\mu}U_{\nu}+pg_{\mu\nu},$$ which implies that $T_{00}=\rho$ and $T_{ii}=p$. Furthermore the Einstein field equations can be rearranged to solve for the stress energy tensor if you first specify a metric. Why is it that in nearly all cases we assume the stress energy tensor takes on the form of a perfect fluid and or dust, and what happens when we are trying to model something that is not a perfect fluid, i.e How do we know what stress energy tensor to choose given a certain situation. For example a wormhole, wormholes are modeled by anisotropic fluids.
1 Answers
"Why is it that in nearly all cases we assume the stress energy tensor takes on the form of a perfect fluid and or dust?"
Because in that case the non-linear Einstein's field equations reduce to ordinary linear differential equations: a homogeneous second-order on $\sqrt{g_{tt}}$ and a non-homogeneous first order on $g_{rr}^{-1}$. See for example https://physics.stackexchange.com/a/679431/281096.
"How do we know what stress energy tensor to choose given a certain situation?"
A good orientation is to choose the stress-energy tensor which is compatible with Special Relativity Theory (SRT). Otherwise, with some restrictions (energy conditions), you are free to define it. You may like to read the answer to similar question on this platform, too, see https://physics.stackexchange.com/a/90323/281096 .
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