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When we solve the Einstein Field Equations

$G_{\mu\nu} = 8\pi T_{\mu\nu}$

one way of doing it is by specifying a symmetry (and thus a general form of the metric) and then specifying the stress-energy-tensor $T_{\mu\nu}$. Most examples found in text are for vacuum solutions (vanishing $T_{\mu\nu}$) or homogeneous isotropic $T_{\mu\nu}=\textrm{diag}(\rho,p,p,p)$.

Is it physically meaningful to solve the field equations for a radially-dependent stress-energy tensor? For example

$T_{\mu\nu}=\textrm{diag}(\rho(r),p(r),p(r),p(r))$

where there is a radially-dependent mass-energy density $\rho(r)$. For example, a mass-energy density that resembles a Gaussian or Poisson distribution, where the highest concentration is at the center. And if the mass-energy density is radially-dependent, is the above expression for $T_{\mu\nu}$ correct? Or will the momentum flux and shear now be non-zero so that $T_{\mu\nu}$ is no longer a diagonal matrix?

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Is it physically meaningful to solve the field equations for a radially-dependent stress-energy tensor?

Yes. For example, the interior Schwarzschild metric (which has nothing to do with the metric inside a Schwarzschild black hole) has a diagonal stress-energy tensor with a uniform energy density but a radially-dependent pressure.

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  • $\begingroup$ Oh thanks. And can it be rightfully assumed that the tensor remains a diagonal matrix like what I wrote up there? $\endgroup$
    – jboy
    Oct 19 '20 at 15:23
  • $\begingroup$ Yes. I mentioned that this radially-dependent stress-energy tensor is diagonal. $\endgroup$
    – G. Smith
    Oct 19 '20 at 15:59

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