Proper time to reach the singularity of a Schwarschild black hole

In summary: I think it should be OK now.In summary, the proper time to fall down to a Schwarzschild coordinate ##r<r_0## is given by the equation ##\tau(r)=-\sqrt{\frac{r_0}{r_S}}\int_{r_0}^{r}\mathrm{d}r^\prime\,\sqrt{\frac{r^\prime}{r_0-r^\prime}}=\sqrt{\frac{r_0}{r_S}}\biggl(\sqrt{r(r_0-r)}+r_0\cos ^{-1}\Bigl(\sqrt{\frac{r}{r_0}}\Bigr)\biggr)##. This shows that the proper time
  • #1
Reggid
43
52
TL;DR Summary
I have a question on how to calculate the proper time to reach the singularity of a Schwarzschild black hole using Schwarzschild coordinates.
When calculating the proper time along a timelike radial geodesic, with the initial condition that object the starts at rest at some Schwarzschild coordinate ##r_0>r_S##, i.e.
[tex]\frac{\mathrm{d}r}{\mathrm{d}\tau}\Bigg|_{r=r_0}=0\;,[/tex]
after using the equations of motion one finds
[tex]\mathrm{d}\tau=-\mathrm{d}r\,\sqrt{\frac{r_0}{r_S}}\sqrt{\frac{r}{r_0-r}}\;.[/tex]
So the proper time to fall down to some Schwarzschild coordinate ##r<r_0## is
[tex]\tau (r)=-\sqrt{\frac{r_0}{r_S}}\int_{r_0}^{r}\mathrm{d}r^\prime\,\sqrt{\frac{r^\prime}{r_0-r^\prime}}=\sqrt{\frac{r_0}{r_S}}\biggl(\sqrt{r(r_0-r)}+r_0\cosh ^{-1}\Bigl(\sqrt{\frac{r}{r_0}}\Bigr)\biggr)\;.[/tex]
From this result one can see that the proper time to reach the event horizon ##\tau(r=r_S)## is finite.

I know that until here everything is fine, but the above integral is also perfectly finite for ##r=0##, such that the proper time to reach the singularity is
[tex]\tau(r=0)=\frac{\pi r_0^{3/2}}{2r_S^{1/2}}\;.[/tex]
Now my question is: is this calculation also reliable beyond the horizon?
On the one hand everything is finite and everything is expressed in terms of invariant proper time (no reference to coordinate time ##t## that diverges at the horizon is needed).
But on the other hand I know that Schwarzschild coordinates only describe the patch of spacetime that lies outside the horizon, so maybe one would first have to go to coordinates that can be extended to the region beyond the horizon to be able to derive that result.
I have the feeling that the result could be correct, but still the way I obtained it is somewhat "sloppy" or not not really OK.

Can somebody help?

Thanks for any answers.
 
Physics news on Phys.org
  • #2
It works because ##r## is definable all the way in (it's the areal radius of the spherical shells reflecting the symmetry of the spacetime), so ##dr/d\tau## is defined everywhere. I agree that using the Schwarzschild coordinate expression of the geodesic equation to get to that point is a bit sketchy because the coordinates don't work at the horizon (although they work above and below it).

I've not done this myself, but I think that formally you could translate everything into a coordinate system that does work at the horizon - e.g. Eddington-Finkelstein coordinates. Since they share a definition of the radial coordinate, ##r##, with Schwarzschild coordinates, that ought to give you the same ##dr/d\tau##.
 
Last edited:
  • Like
Likes Reggid
  • #3
Thank you for your answer.

Yes you are right, using Eddington-Finkelstein coordinates it is actually very easy to see that they give you the same expression for ##\mathrm{d}r/\mathrm{d}\tau## (as they should).
 
  • Like
Likes PeterDonis and Ibix
  • #4
Reggid said:
So the proper time to fall down to some Schwarzschild coordinate ##r<r_0## is
[tex]\tau (r)=-\sqrt{\frac{r_0}{r_S}}\int_{r_0}^{r}\mathrm{d}r^\prime\,\sqrt{\frac{r^\prime}{r_0-r^\prime}}=\sqrt{\frac{r_0}{r_S}}\biggl(\sqrt{r(r_0-r)}+r_0\cosh ^{-1}\Bigl(\sqrt{\frac{r}{r_0}}\Bigr)\biggr)\;.[/tex]
From this result one can see that the proper time to reach the event horizon ##\tau(r=r_S)## is finite.

I'm not sure this integral is good.

##\cosh^{-1}(x)## is only real valued if x>1, but ##r/r_0## is <1

The equation to be integrated passes the sanity check that ##\frac{dr}{d\tau} = \sqrt{E^2 - 1 + 2M/r} = \sqrt{E^2-1 + r_s/r}## for some constant E, but I've got my doubts about the integration process.
 
  • Like
Likes Reggid
  • #5
pervect said:
I'm not sure this integral is good.

cosh−1⁡(x) is only real valued if x>1, but r/r0 is <1

The equation to be integrated passes the sanity check that drdτ=E2−1+2M/r=E2−1+rs/r for some constant E, but I've got my doubts about the integration process.

That's because it should be a ##\cos^{−1}##, not a ##\cosh^{−1}##.

Sorry for the typo (and thanks for finding it)
 

1. What is the singularity of a Schwarschild black hole?

The singularity of a Schwarschild black hole is a point of infinite density and zero volume at the center of the black hole. It is surrounded by the event horizon, which marks the point of no return for anything that enters the black hole.

2. How long does it take for a Schwarschild black hole to reach its singularity?

According to classical physics, a Schwarschild black hole will take an infinite amount of time to reach its singularity. However, in the theory of general relativity, time is relative and can be affected by factors such as gravity. Therefore, the proper time for a black hole to reach its singularity may be different depending on the observer's frame of reference.

3. Can the singularity of a Schwarschild black hole be avoided?

No, according to current understanding, the singularity of a Schwarschild black hole is an unavoidable consequence of the collapse of matter into a black hole. It is a point of infinite density and cannot be escaped or avoided.

4. Is the proper time to reach the singularity of a Schwarschild black hole the same for all black holes?

No, the proper time to reach the singularity of a Schwarschild black hole can vary depending on the mass and spin of the black hole. A more massive black hole will have a longer proper time to reach its singularity compared to a less massive black hole.

5. What happens at the singularity of a Schwarschild black hole?

At the singularity of a Schwarschild black hole, the laws of physics as we know them break down. It is a point of infinite density and gravity, where the laws of general relativity no longer apply. It is currently unknown what exactly happens at the singularity, as it is beyond the reach of our current technology and understanding.

Similar threads

  • Special and General Relativity
Replies
11
Views
168
  • Special and General Relativity
Replies
4
Views
868
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
3
Views
928
  • Special and General Relativity
2
Replies
43
Views
2K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
5
Views
899
Back
Top