Volume of a black hole using the Schwarzschild metric

[weirdoguy]

I think this is what I'm looking for

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

 

[happyparticle]

No you're not. And again: from which 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.