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Have we as of yet been able to calculate the space-time curvature around a galactic (or idealized) disk?

I have not heard of or been able to find this. Is it even possible to calculate anything meaningful, either analytically or numerically?

What I'm curious about is the shape near the disk itself, either ignoring dark matter or including some halo configuration

2 Answers2

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It is possible to develop a full relativistic theory around a galactic disk. This paper from Bardeen & Wagoner for instance explores the properties of space time around a pressureless rotating disk which is -ignoring the effect of gas- a good first approximation to a spiral galaxy. The metric, in cylindrical coordinates, they use in this case is

$$ ds^2 = e^{2\mu}(dR^2 + dz^2) + R^2 e^{-2\nu}(d\theta - \omega dt)^2 + e^{2\nu}c^2dt^2 $$

In this expression $(R,\theta,z)$ are the usual cylindrical coordinates, $\omega = \omega(R,z)$ the angular frequency and $\nu$ is the equivalent version of the Newtonian potential, $\mu$ can be found in terms of $\nu$ and $\omega$ once they're known. Please see the paper for further details.

These models, however, are mostly useful for rapidly rotating ($\sim c$), very massive disks. This is not the case for rotation supported galaxies, where the field is weak enough to use Newtonian gravity reliably and avoid the complications of GR

caverac
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Galaxies as a whole have relatively weak gravitational fields and can be described using the weak field limit. This links the metric to the gravitational potential, and in this limit the metric is:

$$\mathrm ds^2 \approx -~\left( 1 + \frac{2\phi}{c^2}\right) c^2~\mathrm dt^2 + \frac{1}{1 + 2\phi/c^2}\left(\mathrm dx^2 +\mathrm dy^2 + \mathrm dz^2\right) \tag{1} $$

where $\phi$ is the Newtonian gravitational potential. The problem of finding the metric is then reduced to finding the gravitational potential. I'm not sure how well this is known, but see for example Ergodic Considerations in the Gravitational Potential of the Milky Way by Adi Nusser, which I referenced in my answer to Why isn't the center of the galaxy "younger" than the outer parts?

But I don't think we would normally use a metric when calculating the dynamics of galaxies since it isn't necessary. The deviations from Newtonian gravity are small enough to be described using linearised gravity.

John Rennie
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