Volume of a black hole using the Schwarzschild metric

[happyparticle]

Homework Statement

Calculate the volume using the Schwarzschild metric

Relevant Equations

$\text -c^2d\tau^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$

Hi,

I'm wondering if it is possible to calculate the volume of a black hole using the Schwarzschild metric.

After computing the volume I get the follow integral:

$$V = 4 \pi \int_0^r \frac{1}{\sqrt{ (1- \frac{r_s}{r'})}} r'^2 dr$$

This integral diverges at the upper bound. The only way I found to compute the integral analytically is if $$r \gg r_s$$, then I can perform a Taylor's expansion.

However, in this case the volume is not really that of a black hole.

Is there something I don't understand?

Thank you

 

[PeroK]

$r < r_s$ is a timelike coordinate. For fixed Schwarzschild $t$, you are not calculating a spatial volume.

 

[weirdoguy]

Is there something I don't understand?

Basically, since "space" is not an invariant concept in this context, "volume" is not a meaningfull concept either.

 

[happyparticle]

$r < r_s$ is a timelike coordinate. For fixed Schwarzschild $t$, you are not calculating a spatial volume.

I can just add a minus sign inside the square root.


I'm not sure to understand exactly why I can't calculate the volume.

 

[Ibix]

I can just add a minus sign inside the square root.

That isn't the point. The point is that to calculate the volume you need to integrate over the three spatial coordinates, but the coordinate you have labelled $r$ is not a spatial coordinate inside the black hole. You need to integrate over the coordinates you have labelled $t,\phi,\theta$.

Note that the integral with respect to $t$ is trivial - all you need to make sure you have right is the limits.

 

[happyparticle]

I suppose to upper bound would be $ct$.

 

[Ibix]

I suppose to upper bound would be $ct$.

No, why would they be that?

Is there ever a point when you can't increase $t$? Or decrease it?

 

[happyparticle]

I'm not sure to understand. From the metric I would say at $r_s$, but this is a radius not a time.

I understand the part that inside the black hole $t$ is now a spatial coordinate, but I don't understand how a time would be the upper limit of my volume.

 

[Ibix]

I'm not sure to understand. From the metric I would say at $r_s$, but this is a radius not a time.

I understand the part that inside the black hole $t$ is now a spatial coordinate, but I don't understand how a time would be the upper limit of my volume.

So you're saying there's no limit to the $t$ coordinate. Fair enough. As noted, the integral is trivial and therefore you have your answer to the internal volume.

To see why this is so, take a look at a Schwarzschild black hole in a better coordinate system - Kruskal-Szekeres coordinates. There's a diagram at Wikipedia drawn by our own DrGreg. You're integrating along a surface of constant $r$ inside region II (so one of the blue hyperbolae) which is infinite in extent.

 

[happyparticle]

I'll have to think about all that. I can't understand that the expression for the volume has 2 variables : $r , t$ and $t$ goes to infinity inside the black hole, but the volume is clearly not infinite. I think I'm more confuse.

For instance, I must know the volume to compute the density.

 

[PeroK]

Homework Statement: Calculate the volume using the Schwarzschild metric
Relevant Equations: $\text -c^2d\tau^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$

Try a coordinate transformation ($r \leftrightarrow t$) to give:
$$\text -c^2d\tau^2=-\left(1-\frac{2GM}{tc^2}\right)c^2dr^2+\left(1-\frac{2GM}{tc^2}\right)^{-1}dt^2+t^2(d\theta^2+\sin^2\theta\,d\phi^2)$$

 

[Ibix]

but the volume is clearly not infinite

With space defined the way you are doing, yes the interior volume is infinite - your spatial surfaces never intersect the horizon. In these coordinates, the timelike extent of the interior is bounded in the future by the singularity and in the past by the white hole singularity. The spacelike extent is infinite in one direction and closed and boundaryless in the others.

In these coordinates, space inside the black hole can be thought of as a hypercylinder. Fixed $t,r$ gives you a sphere of area $4\pi r^2$, and varying your remaining spatial coordinate moves you along an infinite chain of identical spheres - just like moving along a drinking straw moves you along an infinite chain of identical circles. Varying the timelike coordinate moves you on to spheres of differing area - getting smaller as you move into the future. The axial extent of these surfaces remains infinite - so you have a variable $r$ that relates to the area of the spheres, but its value is irrelevant to the volume because there are always infinitely many such spheres in your space. Until you get to the singularity (which is in the future, not in the center) and the maths fails, anyway.

 

[PeroK]

I'm not sure to understand.

You are missing the point. You have a mental model of a regular Euclidean sphere. A black hole might have formed by a collapsing star, which was approximately a solid sphere, and must have collapsed into a smaller sphere of "black hole stuff". It has a certain spatial volume (approximately $\frac 4 3 \pi r_s^3$), a mass of $M$ and a resulting density.

However, you have a mathematical model (the Schwarzschild metric for $r < r_s$) which is nothing like that. It's not Euclidean; it's a not a Euclidean sphere; it's vacuum (not black hole stuff); it has no definable spatial volume; the $r$ coordinate is timelike and the $t$ coordinate is spacelike; it has no definable density.

What you must do is start with the mathematical model and try to rebuild your intuition (as far as that is possible) based on the mathematics of GR. You must abandon the Euclidean geometrical thinking. It's not compatible with GR.

 

[happyparticle]

However, you have a mathematical model (the Schwarzschild metric for r<rs) which is nothing like that. It's not Euclidean; it's a not a Euclidean sphere; it's vacuum (not black hole stuff);

Thus, the Schwarzschild metric does not allow to define a spatial volume inside the Schwarzschild radius?

All you and Ibix say seems right, but I have no clue why I'm tasked to calculate the volume using the Schwarzschild metric.

Is it possible to use the Schwarzschild metric to approximate a volume ?

 

[WWGD]

I'm by no means an expert, but don't we use the Kerr metric if the black hole is rotating? Is it assumed it's not rotating?

 

[happyparticle]

It it indeed not rotating.

 

[Ibix]

Thus, the Schwarzschild metric does not allow to define a spatial volume inside the Schwarzschild radius?

Sure - it's just infinite.

All you and Ibix say seems right, but I have no clue why I'm tasked to calculate the volume using the Schwarzschild metric.

Probably to hammer home that using your intuition in GR is a bad idea unless it's well-trained intuition.

Is it possible to use the Schwarzschild metric to approximate a volume ?

You've got an exact volume. Why would you need to approximate?

You can use other definitions of space and get a finite volume, but it'll be time-varying because your definition of space will be so. And there's no reason to prefer one definition over another unless you've been told to.

 

[PeroK]

All you and Ibix say seems right, but I have no clue why I'm tasked to calculate the volume using the Schwarzschild metric.

If you are studying GR, perhaps it's assumed that you have the maturity by now to deal with a problem that has no solution - or a surprising result.

 

[julian]

I vaguely remember a paper by Rovelli titled How big is a black hole? (https://arxiv.org/abs/1411.2854). I'm busy at the moment, so I can't go through it right now, but it might be interesting.

 

[Ibix]

I vaguely remember a paper by Rovelli titled How big is a black hole? (https://arxiv.org/abs/1411.2854). I'm busy at the moment, so I can't go through it right now, but it might be interesting.

Clever. He finds a way to pick a spacelike surface to integrate over that crosses the horizon and is reasonably non-arbitrary. Thus he gets a finite answer. It's not the answer to the problem in this thread, but is interesting.

 

[happyparticle]

You've got an exact volume. Why would you need to approximate?

I have to compare the density of a supermassive black hole with another black hole and I can't really perform that with a infinite volume.

 

[Ibix]

I have to compare the density of a supermassive black hole with another black hole and I can't really perform that with a infinite volume.

Please state the exact question as written. As you are paraphrasing it, the answer is "this question doesn't make sense".

 

[Orodruin]

The point is that there is no good answer. You first need to define a spacelike hypersurface to define what you even mean by ”volume”. Preferably one which is invariant under translations by a time-like Killing field to actually call it a volume of something persistent. Such a Killing field does not exist inside the horizon. The coordinate surface of some arbitrary time-like coordinate is going to be just that - arbitrary.

(I am not going to delve into the paper by Rovelli here as that is doing something different)

 

[happyparticle]

Please state the exact question as written. As you are paraphrasing it, the answer is "this question doesn't make sense".

Yes sure.

Calculate the volume of a sphere in the Schwarzschild metric. Based on this calculation, compute the density of a black hole.

 

[weirdoguy]

And from which book does this come from? Or did you made it up yourself? I'm asking because authors of GR book know very well that density of a black hole is not a well defined concept.

 

[Orodruin]

And from which book does this come from? Or did you made it up yourself? I'm asking because authors of GR book know very well that density of a black hole is not a well defined concept.

However, authors of “modern physics” or high-schook books many times won’t in my experience … and yet they feel they should cover just a little bit of GR for legitimacy …

 

[Ibix]

Calculate the volume of a sphere in the Schwarzschild metric. Based on this calculation, compute the density of a black hole.

For a surface not matching the spherical symmetry of the spacetime, there's quite a lot of work to be done to define the word "sphere". For a surface that does match it, the answer is "this question doesn't make sense".

Consider using a different book.

 

[weirdoguy]

However, authors of “modern physics” or high-schook books many times won’t in my experience …

Oh yes, that's my experience too.

 

[happyparticle]

After further research I found this answer. I think this is what I'm looking for. However, I'm not sure to understand what exactly is the first-order excess volume.

 

[Orodruin]

Yea, no. That’s just wrong.

 

[weirdoguy]

I think this is what I'm looking for

No you're not. And again: from wich textbook did you get this exercise? Autors name, and the name of the book please.

 

[happyparticle]

No you're not. And again: from wich textbook did you get this exercise? Autors name, and the name of the book please.

I didn't get this exercise from a textbook. This exercise is one of multiple exercises I found to get used with the Scharwarzshield metric. I tought this problem would be a good practice. However, it is possible that I just don't understand the question.

It is possible that the density of the black hole is $$\rho = \frac{3}{32 \pi} \frac{c^6}{G^3 M^2}$$ as explained [link to vixra removed], meaning that the volume is "simply" $$V = \frac{4 \pi}{3 r_s^3}$$. Which makes sense in my opinion. $r_s$ is the radius of Scharwarzshield, which is the radius at the horizon of the black hole. Hence, $$V = \frac{4 \pi}{3 r_s^3}$$ would be the volume of the black hole. However, I'm not sure how to get this expression from the Scharwarzshield metric.

 

[renormalize]

This exercise is one of multiple exercises I found to get used with the Scharwarzshield metric. I tought this problem would be a good practice.

But where precisely was it "found"? Or did you invent the exercise yourself?

 

[Orodruin]

I didn't get this exercise from a textbook. This exercise is one of multiple exercises I found to get used with the Scharwarzshield metric. I tought this problem would be a good practice. However, it is possible that I just don't understand the question.

It is possible that the density of the black hole is $$\rho = \frac{3}{32 \pi} \frac{c^6}{G^3 M^2}$$ as explained [vixra link], meaning that the volume is "simply" $$V = \frac{4 \pi}{3 r_s^3}$$. Which makes sense in my opinion. $r_s$ is the radius of Scharwarzshield, which is the radius at the horizon of the black hole. Hence, $$V = \frac{4 \pi}{3 r_s^3}$$ would be the volume of the black hole. However, I'm not sure how to get this expression from the Scharwarzshield metric.

ViXra is a well-known crackpot site. You should not trust material posted there. Correspondingly, what you just said is nonsense.

A black hole has no density per se. The Schwarzschild solution is a vacuum solution to the EFEs. There is no stress-energy anywhere.

 

[PeterDonis]

as explained here,

It would have saved a lot of time if you had given this reference at the start of the thread--as you're supposed to do anyway for a homework thread.

As @Orodruin has said, your reference is nonsense. A black hole does not have a well-defined volume. Depending on how you pick a spacelike hypersurface in the hole's interior, you can get pretty much any finite answer you like, or an infinite answer, when integrating over the surface. But none of those answers have any physical meaning that's anything like the "volume" of an ordinary object. In short, the question is not well-defined.

Thread closed.