# Schwarzschild radius

1. Oct 31, 2006

### MeJennifer

I am a bit confused about the Schwarzschild radius perhaps someone can help me here.

The Schwarzschild radius for a black hole is defined as the distance between the center of mass and the event horizon. Now in GR this distance should not be the arc length of the geodesic but the actual length of the curve right?

How come this distance is identical with the one calculated in Newtonian mechanics (although in this case there is no black hole) and how come it is even finite in GR since the curvature closer to the singularity gets stronger and stronger.

Another question is what is the volume of the region inside the event horizon, clearly it cannot possibly be:

$$\frac{4}{3} \pi r^3$$

where r is the distance betwen the center of mass and the event horizon, since spacetime is curved right?

And am I correct in understanding that the Ricci tensor in this volume, with the exclusion of the actual singularity, and assuming no significant mass/energy is on it's way towards the singularity, is zero?

Last edited: Oct 31, 2006
2. Oct 31, 2006

### lalbatros

MeJennifer,

I did not try, but you could try to calculate the distance and check if you are right.
I think that r is a coordinate, nothing else.
The same geometry can be represented with other coordinates (like Kruskal-...).
At large r, you should expect that the distance if indeed given by the usual formulas with a small correction.
Close to the horizon, I don't expect r will assume the same meaning as usually.

Michel

3. Oct 31, 2006

### dicerandom

The r value of the Schwarzschild radius is the coordinate r, not the "real" r (straight-line-path distance from the center). The coordinate r value does not depend on the curvature of the local spacetime, it's just some grid that an observer out at infinity laid down.

Edit: Somewhat unrelated, but fun. You can get Google to calculate numerical values for a Schwatzschild radius for you, i.e. by googling for "mass of the earth * 2 * G/c^2". Google knows the mass of about anything in the solar system (i.e. sun, jupiter, etc.).

Last edited: Oct 31, 2006
4. Nov 3, 2006

### Ich

The Schwarzschild radius is defined as circumference/2pi.

5. Nov 3, 2006

### pervect

Staff Emeritus
Yep, exactly.

I'll expand on this remark a bit. The Schwarzschild r coordinate can be thought of as being related to the distance, but it is not the distance, it's just a coordinate. This is especially important inside the event horizon, where the r coordinate is time-like rather than space-like.

6. Nov 3, 2006

### robphy

Since Schwarzschild is spherically symmetric, it might be more aesthetically pleasing to define r in terms of the area of a sphere, rather than the circumference of a circle.

7. Nov 19, 2006

### Chris Hillman

Definition and interpretation of Schwazschild radial coordinate

Hi, Jennifer,

No, it's not, and this is rather important to understand. Rather the Schwarzschild radial coordinate is defined so that the surfaces $$r = r_0, t=t_0$$ (in the Schwarzschild coordinate chart) are geometric round spheres (as abstract two dimensional Riemannian manifolds) with surface area $$4 \Pi r^2$$. But the difference $$r_2-r_1$$ where $$r2 > r1 > 2*m$$ is NOT the radial distance between two of these nested spheres.

(Pedantic note: an even better definition, because it is more operational, is formulated in terms of the optical expansion scalar of the radially outgoing null geodesic congruence; the Schwarzschild coordinate is simply the reciprocal of this number, which is defined in a coordinate independent manner. Thus, the notion of Schwarzschild radius does have operational significance independent of what coordinate chart we use to represent the geometry.)

Right, "the volume inside the sphere $$r=r_0$$" is not well-defined in the Schwarzschild geometry. Looking at the fine pictures in Misner, Thorne and Wheeler, Gravitation, 1973 (MTW) should help clarify this.

The Ricci tensor vanishes everywhere in the Schwarzschild spacetime, because this happens to be a vacuum solution in gtr.

Looks like I just learned something from you, incidentally--- how to obtain some TeX-like mathematical markup in this forum. Thanks!

Chris Hillman

Last edited: Nov 19, 2006
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