2
$\begingroup$

My question pertains to entire General Relativity, but to be specific, I'll restrict to black holes.

The black hole solution of the Einstein equations might be represented in a number of different metrics. The most commonly used is the metric written in the Schwarzschild, a.k.a "curvature", coordinates. As well known, these coordinates do not cover the entire manifold, and to treat the full spacetime, one then moves to Kruskal-Szekeres coordinates which do not have spurious singularity at horizon. There are also Eddington-Finkelstein coordinates, which are regular at horizon as well. Another form is the isotropic coordinates, in which the spatial part is conformally flat.

As far as theory is concerned, I think I don't have trouble understanding these forms of the metric and which features they highlight. After all, the major principle of GR is general covariance, so that all coordinate systems are equally "good". What obscures me is how to apply this knowledge in practice. To be specific, let's consider observation of the supermassive black hole in the center of the Milky Way, as currently being undertaken by the Event Horizon Telescope. Which form of the metric should be used to describe what we should see in the direction of that black hole from the Earth (e.g., the expected black hole "shadow")? My intuition is that Finkelstein or Kruskal coordinates are not the best choice here. But what about curvature and isotropic coordinates? Which of them is suitable for this purpose? In fact, a priori, I find the latter more "physical" or logical than traditional curvature coordinates, since in isotropic coordinates both $r$ and $dr$ are equally affected by gravity.

$\endgroup$
  • $\begingroup$ Remember it's not just what what you are describing but what you want to compute. Numerical schemes in GR for example have strong implications on what coordinates to choose. $\endgroup$ – JamalS Nov 17 '17 at 15:18
  • 1
    $\begingroup$ It sounds like you're asking which coordinates to use. These are all the same metric. $\endgroup$ – user4552 Jan 25 at 1:46
2
$\begingroup$

This question is quite old, but let me give it a shot. In general, as you pointed out, all coordinate charts are "equally good", as long as the physical processes that you're interested in take place in the domain of your chart.

If you want to calculate the orbit of body around a black hole, the precession or mercury, the gravitational redshift and so on, then Schwarzschild coordinates are fine. If you want to find out the fate of someone falling in the black hole, then the coordinate singularity at the horizon is troublesome. The difficulty is not insormountable, as you can try to show yourself that a body reaches the horizon for a finite value of your affine parameter (e.g. proper time), even in Schwarzschild coordinates. In other situations this obviously can't be done if the region of the manifold isn't even covered by the chart, for example if you want to consider more complicated processes in rotating and/or charged black holes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "If you want to calculate the orbit of body around a black hole, the precession or mercury, the gravitational redshift and so on, then Schwarzschild coordinates are fine" - but isotropic coordinates are also fine for this, aren't they? But the results of the calculation will differ from those obtained in Schwarzschild coordinates. So how do I decide which of them to choose for the calculations (specifically, if I observe the black hole from the Earth)? $\endgroup$ – Maximko Jan 26 at 4:09
  • $\begingroup$ They are, and if by "different results" you mean "different representations of the same result" then yes, for example geodesics will be the same, but look different in different coordinates. There's no recipe to find out the "best coordinates", one tries with various charts and finds out. $\endgroup$ – user35319 Jan 26 at 13:48
1
$\begingroup$

There is no recipe for this, just try them all (and usually there are not so many) and pick the one that makes the calculations simpler.

If your system is symmetric, you may want to use that symmetry to reduce the complexity. For instance if the system is spherically symmetric use spherical coordinates, so that if the motion is radial and the problem has been reduce to 2d.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.