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- TL;DR Summary
- I made a question on how to find the spherically symmetric metric as a case of the axisymmetric one.

Hello. I expect this question is not repeated. I look from it in the forum but I found nothing.

I am confused on how an axisymmetric spacetime (generated by a rotating object) can manifest the spherically symmetric case. The axisymmetric spacetime should describe objects with any angular velocity (supposing a rigit body rotation) including a null rotation. So the spherical spacetime should be a specific case of this bigger one.

My question is on the comparison of both spacetime metrics, for the spherical one we have

$$

ds^2=-e^{-2\Phi}dt^2 +e^{-2\Lambda}dr^2 +r^2 d\theta^2 +r^2 sin^2(\theta) d\phi^2,

$$

and for the axisymmetric case

$$

ds^2=-e^{2\nu}dt^2 +e^{2\alpha}(dr^2 +r^2 d\theta^2) +e^{2\beta} r^2 sin^2(\theta)( d\phi- \omega dt)^2,

$$

I know it is written in two different ways but the idea is that there is a same function multiplying dr and d\theta in the axisymmetric case and it doesn't happen in the spherical case. We could set \omega as zero and then set \beta=0 also. But I don't know how to do it for the \alpha.

I am confused on how an axisymmetric spacetime (generated by a rotating object) can manifest the spherically symmetric case. The axisymmetric spacetime should describe objects with any angular velocity (supposing a rigit body rotation) including a null rotation. So the spherical spacetime should be a specific case of this bigger one.

My question is on the comparison of both spacetime metrics, for the spherical one we have

$$

ds^2=-e^{-2\Phi}dt^2 +e^{-2\Lambda}dr^2 +r^2 d\theta^2 +r^2 sin^2(\theta) d\phi^2,

$$

and for the axisymmetric case

$$

ds^2=-e^{2\nu}dt^2 +e^{2\alpha}(dr^2 +r^2 d\theta^2) +e^{2\beta} r^2 sin^2(\theta)( d\phi- \omega dt)^2,

$$

I know it is written in two different ways but the idea is that there is a same function multiplying dr and d\theta in the axisymmetric case and it doesn't happen in the spherical case. We could set \omega as zero and then set \beta=0 also. But I don't know how to do it for the \alpha.