Black Hole: Knock on the horizon

[Ibix]

Are you thinking that second term evaluates to 0 or otherwise goes away? Or did I or Mathematica make a mistake somewhere?

No, you're right. I did think the log term went away, but I'd misread the result on my tiny phone screen.

 

[Dale]

I'd misread the result on my tiny phone screen.

That has happened to me more times than I would care to admit.

 

[Dale]

I just tried to work this out myself and got a slightly different answer. So I have $$g_{rr}=\frac{1}{1-\frac{R_S}{r}}$$ which integrates (according to Mathematica) to $$\int_{R_S}^R \sqrt{g_{rr}} \ dr = \sqrt{R \left(R-R_S\right)}+\frac{1}{2} R_S \log \left(\frac{-R_S+2 \sqrt{R \left(R-R_S\right)}+2
R}{R_S}\right) $$
Are you thinking that second term evaluates to 0 or otherwise goes away? Or did I or Mathematica make a mistake somewhere?

@mef to illustrate the issue about the coordinate system, I performed the same calculation in isotropic coordinates. We have $$g_{rr}=\frac{\left(2 r+\frac{R_S}{2}\right){}^4}{16 r^4}$$ which integrates to $$\int_{R_S}^R \sqrt{g_{rr}} \ dr = R-R_S+\frac{\left(R-R_S\right) R_S}{16 R}+\frac{1}{2} R_S \log \left(\frac{R}{R_S}\right)$$

Note that the results are different, but still finite.

 

[cianfa72]

If you mean a spacelike 3-surface formed by a continuous series of such spacelike 2-spheres, yes, those also exist inside the horizon. They just aren't surfaces of constant Schwarzschild $t$ coordinate, since those are timelike (more precisely, they have one timelike and 2 spacelike linearly independent tangent vectors) inside the horizon.

Since the $t$ coordinate is spacelike inside the horizon and the Schwarzschild metric in Schwarzschild coordinate chart is diagonal, then any 3-surface of constant coordinate time $t= \text{const}$ inside the horizon cannot be spacelike (there are not 3 linearly independent spacelike 4-vectors part of the orthogonal complement of a spacelike vector in the tangent space at each point).

 

[PeterDonis]

Since the $t$ coordinate is spacelike inside the horizon and the Schwarzschild metric in Schwarzschild coordinate chart is diagonal, then any 3-surface of constant coordinate time $t= \text{const}$ inside the horizon cannot be spacelike

Yes, you are agreeing with what I said.