Schwarzschild radius of a black hole

In summary, 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.
  • #1
MeJennifer
2,008
6
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:

[tex]\frac{4}{3} \pi r^3 [/tex]

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:
Physics news on Phys.org
  • #2
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
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:
  • #4
The Schwarzschild radius for a black hole is defined as the distance between the center of mass and the event horizon.
The Schwarzschild radius is defined as circumference/2pi.
 
  • #5
Ich said:
The Schwarzschild radius is defined as circumference/2pi.

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
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
Definition and interpretation of Schwazschild radial coordinate

Hi, Jennifer,

MeJennifer said:
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.

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

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

Right, "the volume inside the sphere [tex]r=r_0[/tex]" 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.

MeJennifer said:
And am I correct in understanding that the Ricci tensor in (the region [tex]0 < r < r_0[/tex]) with the exclusion of the actual singularity, and assuming no significant mass/energy is on it's way towards the singularity, is zero?

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:

Related to Schwarzschild radius of a black hole

1. What is the Schwarzschild radius of a black hole?

The Schwarzschild radius is a measure of the size of the event horizon of a non-rotating black hole. It represents the point at which the escape velocity exceeds the speed of light, making it impossible for anything, including light, to escape the gravitational pull of the black hole.

2. How is the Schwarzschild radius calculated?

The Schwarzschild radius is calculated using the mass of the black hole and the gravitational constant. The formula is: Rs = 2GM/c^2, where Rs is the Schwarzschild radius, G is the gravitational constant, M is the mass of the black hole, and c is the speed of light.

3. What happens at the Schwarzschild radius?

At the Schwarzschild radius, the curvature of space-time becomes infinite, and time and space become distorted. This is known as the event horizon, and it marks the point of no return for anything that crosses it.

4. Can the Schwarzschild radius change?

The Schwarzschild radius is directly proportional to the mass of the black hole, so it can change if the mass of the black hole changes. As matter falls into a black hole, its mass increases, and the Schwarzschild radius will also increase.

5. How does the Schwarzschild radius relate to the size of a black hole?

The Schwarzschild radius is the size of the event horizon of a black hole. However, the actual size of a black hole can vary depending on its properties, such as its spin and charge. The event horizon is the point at which the gravitational pull is strong enough to trap light, but the size of the black hole itself can extend beyond this boundary.

Similar threads

  • Special and General Relativity
Replies
4
Views
595
  • Special and General Relativity
Replies
26
Views
678
  • Special and General Relativity
Replies
1
Views
351
  • Special and General Relativity
Replies
30
Views
3K
  • Special and General Relativity
2
Replies
43
Views
2K
  • Special and General Relativity
Replies
4
Views
606
  • Special and General Relativity
Replies
17
Views
2K
Replies
3
Views
1K
  • Special and General Relativity
Replies
18
Views
2K
  • Special and General Relativity
2
Replies
46
Views
2K
Back
Top