# Questions about the Schwarzschild metric

1. Nov 5, 2012

### nick41

Hello everybody! I have some questions concerning the structure of the Schwarzschild metric, which is given by
$$ds^2=-(1- \frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+ r^2(d\theta^2+ \sin^2(\theta) d\phi^2)$$
where we set $c=1$. I would like to know the following: \\
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1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this? \\
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2. What are the geodesics in the Schwarzschild solution? Is there any good way to visualize them all at once?
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2. Nov 5, 2012

### Staff: Mentor

Put simply, M is the mass you will obtain if you put a small test object in orbit about the hole, measure its orbital parameters, and apply Kepler's Third Law.

Probably not; there are too many different kinds. A couple of types that are commonly considered are:

(1) Purely radial timelike geodesics, particularly infalling ones; these represent the worldlines of objects that are free-falling into the hole.

(2) Tangential timelike geodesics, which represent the worldlines of objects that are orbiting the hole.

3. Nov 5, 2012

### Staff: Mentor

There doesn't have to be a black hole at all - the Schwarzchild solution works just fine in the empty space around any stationary spherically symmetric mass. You only get a black hole if the object is so dense that r=2M lies above its surface.

4. Nov 5, 2012

### pervect

Staff Emeritus
As others have pointed out, if you look at the orbits of bodies far away from the black hole, where gravity is nearly Newtonian, you can find that the Newtonian limit gives M as the mass of the black hole.

The geodesics of the Schwarzschild solution are just the various orbits of unaccelerated test particles. These include orbits that go around the black hole, go into it, or fly by it, as long as the test particle in question a) isn't accelerating like a rocket and b) isn't so big that its own gravity perturbs the metric and c) you don't have significant emission of gravitational radiation, which would also cause the test particle to accelerate.

5. Nov 6, 2012

6. Nov 6, 2012

### pervect

Staff Emeritus
Note though that while in the exterior region the Komar formula gives you the Schwarzschild mass parameter M, it doesn't give you the same distribution of mass in the interior region as the fairly well known integral, valid only in Schwarzschild coordinates

$m(r) = \int \rho(r) 4 \pi r^2 dr$

does. The difference won't matter with a Schwarzschild black hole, but it will matter with a a Schwarzschild non-black hole solution in the interior region (not the exterior region though).

The easiest way of convincing oneself of this is direct calculation, the particular case that convinced me was the "box of light" case. I'm not sure I've seen much comment on this difference, though the fact that Townsend mentions specifically showing it's equal in the exterior region is suggestive of the issues that arise in the interior region.

In the box of light case, the Schwarzschild formula will suggest a mass that increases monotonically as r increases. In the idealized "box of light" case, where the box is a sphere of exotic matter with no mass but the tension required to hold the box together, you get a picture where the contribution of the exotic matter shell to the total mass is actually negative, obviously different from the monotonic increase in mass predicted by the Schwarzschild formula.

The total exterior mass remains the same in both cases, but the details of how it's distributed (and in particular, the contribution of the shell to the mass) varies with the two aproaches.

This may sound funky, but the Komar approach correlates better to the gravitational fields one would measure by measuring the 4-acceleration of a static observer. This is because you can cast the Komar approach into a rather Gauss-law like integral (see Wald, for example, around pg 288).