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I tried to calculate a stationary dust solution in general relativity where the energy-momentum tensor is $T^{\mu\nu} = \rho c^2 \delta^\mu_0 \delta^\nu_0$. (related question) The question is that I can't find any named solution about this condition. I saw rotating dust, charged dust, null dust, etc. solutions but not stationary dust solution.

Below is my farthest reach about the stationary dust solution.

The ricci tensor is $R=-{8\pi G\over c^2}\rho$, so the field equation is $R_{\mu\nu}={8\pi G\over c^2}\big(T_{\mu\nu}-{1\over 2}\rho c^2g_{\mu\nu}\big)$ with $T_{\mu\nu} = \rho c^2 \delta_\mu^0 \delta_\nu^0$. By spherical symmetry, letting the metric as $-ds^2=A(r)dt^2-B(r)dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2$, $A(r)$ and $B(r)$ should satisfy $${dA\over dr}={(1-k)A(-k^2A^2+2(k^2+2k-1)A+3(k-1)(k-3))\over rk(2k-1)(kA^2-kA-A-k-2)}, B={(kA-2+k)A\over k(2k-1)(A-1)}$$ with $k=1-{4\pi G\rho\over c^2}r^2$.

Qmechanic
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Zjjorsia
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1 Answers1

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Because for dust the pressure $p$ is zero it follows from Einstein's field equations that energy density $\varepsilon$ must be zero as well. This means that there is no static spherical symmetric dust solution.

For a mathematical demonstration of this, you can look at the equations from my other answer here. The steps are simple. Set equation (2) to zero and differentiate it. Then use equation (1) to simplify obtained expression and substitute it into equation (3). The result will be $\varepsilon=0.$

JanG
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