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The Schwarzschild solution from general relativity defines the spacetime of a spherically symmetric body in a vacuum. During this derivation, the stress-energy tensor is set to zero and the solutions are obtained by solving the resultant differential equations.

As a thought experiment, imagine we have 2 stars (each isolated from each other), one is rotating on its axis extremely quickly, and the other is not rotating at all. In all other respects, they are equal. Would not the energy from rotation cause additional spacetime distortion in comparison to the non-rotating star? It seems like this should manifest as a non-zero stress-energy tensor. Is this reasonable? If so, how would one account for this rotation? I imagine that the speed of rotation relative to the mass of the star is negligible, but I am still curious...

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Yes it does. A rotating mass drags spacetime around with it so the spacetime geometry around it changes from the Schwarzschild geometry to the Kerr geometry.

This is not purely theoretical as it has been experimentally observed for the Earth by the Gravity Probe B experiment. The rotation of the spacetime around the Earth is known as frame dragging, or the Lense-Thirring effect, and this is what Gravity Probe B measured (although for the Earth the effect is so small it was just barely observable).

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  • $\begingroup$ Although, the OP seems to suggest that the metric of a rotating planet should have some non-zero stress energy tensor even outside the planet. This is not necessary, right? For example, Kerr is a vacuum solution, etc. $\endgroup$
    – user87745
    Commented May 27, 2020 at 16:27
  • $\begingroup$ Thank you!! I will look into the Kerr solution and see the process of obtaining a solution. That may answer my non-zero stress energy tensor question... $\endgroup$
    – user41178
    Commented May 27, 2020 at 16:29
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    $\begingroup$ @user41178 as Dvij says, outside the planet the stress-energy tensor is zero in both cases. However for the region inside the planet the stress-energy tensors of a stationary and a rotating planet are different. $\endgroup$
    – John Rennie
    Commented May 27, 2020 at 16:32
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    $\begingroup$ @user41178 Outside, the rotation affects the metric and the Riemann curvature but not the zero Ricci curvature or zero stress tensor. $\endgroup$
    – G. Smith
    Commented May 27, 2020 at 16:56

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