# Does rotational energy have effect on gravity/metric?

Intuitively, if energy can be stored in rotational motion, it has to obey $E=mc^2$. Does rotation of typical stellar-sized objects - BHs, pulsars, binaries - have measurable effect on their overall gravity?

(I'm not talking about nearby effects, like frame dragging described by Kerr metric, the effect on measurements from the Earth is of interest)

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## 2 Answers

The source in the Einstein field equations is the stress-energy tensor, not the scalar mass-energy. Adding rotation will affect multiple elements of the stress-energy tensor. You can sometimes get rough estimates of effects in GR by using $E=mc^2$ and pseudo-Newtonian arguments, but sometimes these are way off. As an example where it's way off, two light rays propagating in parallel (not antiparallel) directions experience zero gravitational interaction.

In the case of the distant field, I believe the answer to your conjecture is yes, in the sense that in any asymptotically flat spacetime, the distant field is Newtonian, and its strength is what you would expect based on the Bondi or ADM mass of the ingredients that went in.

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+1 I'll just leave this note here: the Hulse–Taylor binary pulsar may be an example of rotation of binary system having a gravitational effect (all 7.35E24W of it) that we can measure from Earth. –  Stan Liou Aug 16 '11 at 4:35
Stan Liou and Ben Crowell thank you for your information! –  troyaner Jan 12 at 14:10

Yes. Since the Einstein field equations consider the stress energy momentum tensor, it includes momentum density too, with energy density. You may want to learn about the "Kerr-Metric" and more generally, the "Kerr-Newmann metric" which are more general than the Schwarzschild metric". Yes, they can sometimes be measured from the Earth, but I think that is already answered by Ben Crowell and Stan Liou.

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