Questions about the Schwarzschild metric

nick41
Messages
4
Reaction score
0
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: \\
\\
1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this? \\
\\
2. What are the geodesics in the Schwarzschild solution? Is there any good way to visualize them all at once?
\\
\\
Every answer would be appreciated
 
Physics news on Phys.org
nick41 said:
1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this?

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.

nick41 said:
2. What are the geodesics in the Schwarzschild solution? Is there any good way to visualize them all at once?

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.
 
nick41 said:
1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this?

There doesn't have to be a black hole at all - the Schwarzschild 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.
 
nick41 said:
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: \\
\\
1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this? \\
\\
2. What are the geodesics in the Schwarzschild solution? Is there any good way to visualize them all at once?
\\
\\
Every answer would be appreciated

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.
 
haushofer said:
For a more formal treatment, see e.g. Townsend's notes on black holes,

http://arxiv.org/pdf/gr-qc/9707012v1.pdf

chapter 5 (Komar integrals).

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).
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top