- #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

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?

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: