2
$\begingroup$

The "Schwarzschild solution" in "Schwarzschild metric" looks like:

$$\text -c^2d\tau^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$

Using this metric you get certain values for the radius of the "Schwarzschild radius", the "photon sphere" and the "least stable circular orbit", among other things. When the Jet Propulsion Laboratory or other institutions are dealing with orbital dynamics in the solar system, they instead use the "isotropic metric" or at least a first order expansion of the isotropic metric to calculate relativistic effects on the orbits. Now the Schwarzschild solution in isotropic coordinates looks like:

$$ -c^2d\tau^2 = -\left(\frac{1-GM/2rc^2}{1+GM/2rc^2}\right)^2c^2dt^2 +(1+GM/2rc^2)^4[dr^2 +r^2( d\theta^2 +\sin^2\theta d\phi^2)]$$

Now in isotropic coordinates the "Schwarzschild radius", the "photon sphere", the "least stable circular orbit" among other things take other values.

Question 1: Now that we have the measurements from the "Event Horizon Telescope" on the black hole, can something be said about what coordinates that fits the data better?

Question 2: It might be that both metrics fit the data equally well if we change the mass of the central black hole. Do we get different values on the mass of the central black hole if we fit the data using the isotropic coordinates instead of the Schwarzschild coordinates?


According to the Schwarzschild solution in Schwarzschild coordinates the photon sphere is located at an "r-parameter" of $3GM/c^2$. According to this post when using the Schwarzschild solution in Schwarzschild coordinates the apparent radius ($r_{obs}$) of something residing in a spherically symmetric gravitational potential when viewed from infinity is: $$r_{obs}=r(1-\frac{2GM}{rc^2})^{-0.5} $$ According to this formula we will observe the photon sphere at a radial distance of $r_{obs}=3\sqrt{3}GM/c^2$. Now I assume astronomers estimate the "observed radial distance", "$r_{obs}$" from the center of the M87 black hole to the photon sphere around the M87 black hole by measuring the angular size of the observed ring together with some estimate of the distance from earth to M87. If you use the formula for the photon sphere radius in Schwarzschild coordinates together with the formula for gravitational lensing enlargement you end up with a formula for the central mass of M87 as: $M=\frac{r_{obs}c^2}{{G3\sqrt{3}}}$.

Now if isotropic coordinates are used instead you will get another expression for the "r-parameter" of the photon sphere (I do not know what that expression looks like) and another expression for the "gravitational lensing enlargement effect" that makes objects in a spherically symmetric gravitational field look larger when observed from infinity.

What i specifically would like to know in the second question above is how much the mass of the central body will "change" if you keep $r_{obs}$ constant but you use isotropic coordinates instead of Schwarzschild coordinates and in both cases equate the "r-coordinate" with radial distance.


I am a bit confused about the concept that you can do various coordinate transformations of the "r-parameter" in the Schwarzschild solution and still get a solution that fits reality.

$\endgroup$
  • $\begingroup$ Regarding your final paragraph, as en.wikipedia.org/wiki/Coordinate_conditions says, the description of the world as given by the laws of physics does not depend on our choice of coordinate systems. However, it is often useful to fix upon a particular coordinate system, in order to solve actual problems or make actual predictions. Also see en.wikipedia.org/wiki/… $\endgroup$ – PM 2Ring Apr 13 at 4:15
6
$\begingroup$

One of the things that can be confusing for beginners to GR is that there is no physical significance to the choice of coordinates we make. For example when studying static black holes we can use Schwarzschild coordinates, isotropic coordinates, Gullstrand-Painlevé coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates and lots of others that I can't think of offhand.

But regardless of what coordinates we choose it's the same metric, just written in different ways. You ask about the Schwarzschild metric and the isotropic metric as if they were different things, but it's the same metric just written in different coordinates. It would be better to call them the static black hole metric written in Schwarzschild coordinates and the static black hole metric written in isotropic coordinates but, well, life is too short and even experienced physicists will just say Schwarzschild metric and isotropic metric to save time.

The point is that the coordinates are not directly observable. We choose some coordinate system and then do a calculation to compute something that is observable e.g. the trajectories of photons. And the results will come out the same. After all the photons don't care what coordinate system a physicist chooses to compute their paths.

The reason we have all these different coordinates is that it's often the case that some calculations are easier in some coordinates than others. For example the Schwarzschild coordinates are the closest to our intuitive grasp of the geometry around the black hole. If we were considering simple questions like the time dilation for an observer hovering near the black hole I'd use Schwarzschild coordinates to do this because they make the calculation easy. But Schwarzschild coordinates don't work well at the event horizon. If I was explaining why light can't escape from a black hole I'd use Gullstrand-Painlevé coordinates.

To return to your question, I'm not familiar with the method the JPL team use for computing trajectories in the Solar System so I can't comment on why the calculations would be easier in the isotropic coordinates than in the Schwarzschild coordinates. Presumably the computational method they use is easier in isotropic coordinates, but the observables that they calculate will come out the same regardless of which coordinates are used.

In any case neither coordinates are useful for the EHT calculations because Messier 87* is rotating not static and the spacetime geometry around it is described by the Kerr metric not the Schwarzschild metric. The Kerr metric can also be written in lots of different coordinate systems - I don't know which coordinate system the EHT team used.

$\endgroup$
  • $\begingroup$ Assume one is able to determine the distance to the particular black hole in Messier 87 and its angular size in a way that is fairly indepenent on the coordinate choice (I do not know if this is possible). Now according to this answer we will see the ring of light as being at a radial distance of $\sqrt{27}GM/c^2$. Is the perceived radial distance to the light ring the same if we use "isotropic coordinates" or "Schwarzschild coordinates"? (ignoring rotation at the moment) $\endgroup$ – Agerhell Apr 13 at 10:14
  • 1
    $\begingroup$ @Agerhell you need to be careful what you mean by radial distance. In both the Schwarzschild and isotropic coordinates the $r$ parameter is not the radial distance, or at least it's not the distance you'd measure if you let a measuring tape down towards the centre of the black hole. $\endgroup$ – John Rennie Apr 13 at 10:20
  • 1
    $\begingroup$ What we measure experimentally on Earth is the angular width of the ring, and clearly that is independent of whatever coordinates were used to do the calculation of the light trajectories. $\endgroup$ – John Rennie Apr 13 at 10:21
  • $\begingroup$ JPL uses the "post-Newtonian expansion". They take a first order expansion of the spherical solution in isotropic coordinates to get what the call the "Schwarzschild isotropic one-body point mass metric", then they use a schedule I do not really follow to get relativistic perturbation terms to the classical Newtonian gravitational acceleration. Essentially they take the isotropic "r-coordinate" and treat it as the radial distance. Here is a thread on the physics site:[physics.stackexchange.com/questions/469181/… $\endgroup$ – Agerhell Apr 16 at 16:51

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.