# Axisymmetric VS Spherical metrics

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• lmmoreira

#### lmmoreira

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.

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.

Yes. But the coordinates in which the two cases are commonly expressed might not be the same, even if they look the same.

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

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

Where are you getting these expressions from?

• lmmoreira
Yes. But the coordinates in which the two cases are commonly expressed might not be the same, even if they look the same.

You mean that I could use a coordinate transformation for \theta like,

$$d\theta=e^{-\alpha}d\theta,$$

that could make the form of the metric equal?

Where are you getting these expressions from?

The first expression was from Fridolin Weber's book "Pulsars as Astrophysical Laboratories for Nuclear and Particle Physics, 1999" equation 14.14. The second was from an article from Komatsu et al. "Rapidly rotating general relativistic stars, 1989".

You mean that I could use a coordinate transformation for \theta like,

$$d\theta=e^{-\alpha}d\theta,$$

that could make the form of the metric equal?

A general axisymmetric metric is different from a spherically symmetric metric, so you should not expect the forms to be equal always.

However, you might be able to find a transformation (the one you suggest might be part of one that would work) that would make the axisymmetric metric, for the special case ##\omega = 0##, look the same as the spherically symmetric metric.

The first expression was from Fridolin Weber's book "Pulsars as Astrophysical Laboratories for Nuclear and Particle Physics, 1999" equation 14.14. The second was from an article from Komatsu et al. "Rapidly rotating general relativistic stars, 1989".

Unfortunately I don't have either of those books, and I don't have any handy reference online. But both of those books look rather specialized; I would suggest trying to find a more general and basic treatment in a more general textbook. IIRC Misner, Thorne, and Wheeler discuss general axisymmetric metrics as well as spherically symmetric ones.

• lmmoreira
Unfortunately I don't have either of those books, and I don't have any handy reference online. But both of those books look rather specialized; I would suggest trying to find a more general and basic treatment in a more general textbook. IIRC Misner, Thorne, and Wheeler discuss general axisymmetric metrics as well as spherically symmetric ones.

I am working on writing a code to rotating compact objects. The process flows through a series of iterations, starting with a initial guess. I want to use the first guess as solution of Tolman-Oppenheimer-Volkoff (TOV) equations, for the spherical symmetric objects. But the transposition of this guy as written in the axisymmetric way is being particularly difficult to me.

the transposition of this guy as written in the axisymmetric way is being particularly difficult to me.

I'm not aware of any closed-form equation for general axisymmetric spacetimes that corresponds to the TOV equation for spherically symmetric spacetimes. I also don't think there is any known closed-form solution to the Einstein Field Equation for an axisymmetric rotating object made of matter (i.e., with nonzero stress-energy). The Kerr solution is axisymmetric, but it's a vacuum solution.

If you want an example of how an axisymmetric metric reduces to a spherically symmetric metric, you might look at the Kerr solution in Boyer-Lindquist coordinates and compare it with the Schwarzschild solution in Schwarzschild coordinates. These are vacuum solutions, but the way in which taking the "rotating" quantity to zero (in the Kerr case, ##a##, which is the angular momentum per unit mass, is the "rotating" quantity) makes the axisymmetric solution spherically symmetric might be instructive to look at.

• lmmoreira
Hm, but since spherically symmetric spacetimes are special cases of axially symmetric spacetimes, there should be constraints of the parameters of the axisymmetric spacetimes, for which you get spherically symmetric spacetimes. I'm also not aware of a book or paper, where this is treated explicitly.

• lmmoreira
there should be constraints of the parameters of the axisymmetric spacetimes, for which you get spherically symmetric spacetimes

Yes, I gave an example of that in post #6: Kerr spacetime (axisymmetric) with ##a = 0## reduces to Schwarzschild spacetime (spherically symmetric).

• vanhees71
I'm not aware of any closed-form equation for general axisymmetric spacetimes that corresponds to the TOV equation for spherically symmetric spacetimes. I also don't think there is any known closed-form solution to the Einstein Field Equation for an axisymmetric rotating object made of matter (i.e., with nonzero stress-energy). The Kerr solution is axisymmetric, but it's a vacuum solution.
Even for the case of an isolated non-rotating, non-spherical but axisymmetric body of matter, I’m not sure any closed form solutions are known that also satisfy some basic physical plausibility criteria. Synge, p 317 discusses the difficult challenges to finding such a solution.

[edit: one of many issues is that it can’t be any standard fluid stress energy form because these cannot be static without being spherical]

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• vanhees71
Even for the case of an isolated non-rotating, non-spherical but axisymmetric body of matter, I’m not sure any closed form solutions are known that also satisfy some basic physical plausibility criteria. Synge, p 317 discusses the difficult challenges to finding such a solution.

This solution do not exist in a closed form (for what I know). I've been studying the numerical approaches to this problem for some time. There are well know differential equations (and integral equations) that controls the possible solutions but these must be obtained numerically. But anyway, the axial numerical solution should converge to spherical one in the case of null rotation.

• vanhees71
the case of an isolated non-rotating, non-spherical but axisymmetric body of matter

This case seems unrealistic if we want the body to be in hydrostatic equilbrium. If we allow it to not be in hydrostatic equilibrium, that would seem to imply that its self-gravity is too weak to force it into hydrostatic equilibrium, in which case we would not expect its structure to be usefully analyzed using GR in any case.

one of many issues is that it can’t be any standard fluid stress energy form because these cannot be static without being spherical

This might be another way of stating what I am stating above.

• vanhees71
This case seems unrealistic if we want the body to be in hydrostatic equilbrium. If we allow it to not be in hydrostatic equilibrium, that would seem to imply that its self-gravity is too weak to force it into hydrostatic equilibrium, in which case we would not expect its structure to be usefully analyzed using GR in any case.

This might be another way of stating what I am stating above.
Well, to me it is interesting that it is a very hard problem to describe e.g. a cylindrical object in GR. Asteroids are all asymmetric and have measurable surface gravity. An asteroid sized cylinder can certainly exist. Describing the gravity of a cylinder in Newtonian gravity is not a hard problem. In GR, it is apparently unsolved in closed form despite the many symmetries of a static cylinder.

[note: I believe it is not such a hard problem in numerical GR. Thus it is just another example of the different boundary of ‘hard’ between Newtonian gravity and GR:

- Newton, 3 body is hard, GR, two body is hard.
- Newton, any simple rigid shape is a straightforward integral to compute potential field; GR, anything other than a spherical body is very hard. ]

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• vanhees71
In GR, it is apparently unsolved in closed form despite the many symmetries of a static cylinder.

There is a known closed form solution for the exterior of an infinite cylinder, the Levi-Civita spacetime. These papers look at possible sources for an interior region with the Levi-Civita spacetime as the vacuum exterior:

https://arxiv.org/abs/gr-qc/9512029

https://arxiv.org/abs/gr-qc/9607058

• PAllen, vanhees71 and lmmoreira
There is a known closed form solution for the exterior of an infinite cylinder, the Levi-Civita spacetime. These papers look at possible sources for an interior region with the Levi-Civita spacetime as the vacuum exterior:

https://arxiv.org/abs/gr-qc/9512029

https://arxiv.org/abs/gr-qc/9607058
Note, on a quick read, I didn’t see any discussion of energy conditions. Perhaps this is moot for something implausible to start with - an infinite cylinder. Synge’s discussion was about the extreme difficulty of finding a solution of matched interior and exterior for an isolated finite axisymmetric body of everywhere positive energy density and pressure also satisfying energy conditions. Note, one of these papers discusses that an infinite cylinder has no definable mass per the standard GR definitions.

on a quick read, I didn’t see any discussion of energy conditions

I didn't either.

Now, back to practical physics - using generally valid arguments you conclude that for anybody too small to force sphericity via self gravity, the GR correction to Newtonian gravity is negligible, so forget using GR.

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using generally valid arguments you conclude that for anybody too small to force sphericity via self gravity, the GR correction to Newtonian GR is negligible, so forget using GR.

Yes.