Volume inside a black hole?

[Mr Skeptic]

How much spacetime is there inside the event horizon of a non-spinning uncharged black hole of mass M? I know that a black hole has a Schwarzschild radius of [math]r = \frac{2GM}{c^2}[/math] and a sphere a volume of [math]V = \frac{4}{3} \pi r^3[/math], for a black hole volume of [math]\frac{32}{3} \frac{\pi G^3 M^3}{c^6}[/math].

 

However, I think the above volume would be in our metric rather than accounting for the distortions of spacetime. If you account for spacetime distortion when measuring the volume, what would you get?

Edited April 16, 2010 by Mr Skeptic
units --> metric

[Hendrix]

I'm not sure but I'm curious. My first thought is that if a black hole is infinitly dense, it's volume would be infinitly small i.e. singularity

[J.C.MacSwell]

I'm not sure but I'm curious. My first thought is that if a black hole is infinitly dense, it's volume would be infinitly small i.e. singularity

 

Isn't a black hole considered to be everything inside it's event horizon?

[Simpleton]

In reality, don't all black holes spin?

[Royston]

Isn't a black hole considered to be everything inside it's event horizon?

 

Yeah, it's just the region bounded by the event horizon.

 

In reality, don't all black holes spin?

 

Kerr black holes i.e mass and angular momentum (spin if you like), are the most likely, but that's not really relevant to the thread.

 

IIRC the volume that's bounded by the event horizon, is just the surface area multiplied by ct, where t is the longevity of the black hole, so...

 

[math]V_{bh}=4\pi r^2 ct[/math]

 

Though we still have the problem of [math]V_{bh}\rightarrow\infty[/math]

 

I actually lost sleep thinking about this last night, and wrote a lengthy reply earlier, and then decided to ditch it, so I may come back to this, here's a short version...

 

I wondered if there was a way around [math]V_{bh}\rightarrow\infty[/math] using Kruskal coordinates, which copes with the problem of asymmetry using advanced Eddington-Finkelstein coordinates i.e the modified Schwarzschild solution.

 

In principle, the point where a photon gets expelled (in the other domain), could be a cut off point for the volume, i.e there should be a change in curvature of the outgoing null geodesic (from a straight world line, to one curving outwards.) Although Kruskal coordiantes do away with coordinate singularities, AFAIK there's still a gravitational singularity to worry about, so I'm a bit confused on how this works.

 

However, perhaps an expert could point out the flaw in this, or if it doesn't quite make sense et.c

 

EDIT: Congrats on the promotion to mod Mr. Skeptic

[Farsight]

I don't think there's an answer to this one, because there is no spacetime inside the event horizon. That might sound a bit of a surprise, but think about what a volume is: a distance multiplied by a distance multiplied by a distance. If the coordinate speed of light at the black hole event horizon is zero as measured by observers in the universe at large, there's nothing with which you can calibrate distance.

 

If you switch to the "proper time" of the infalling observer who measures c locally to be 299,792,458 m/s, it takes him forever by our clocks to make a measurement. So you're still a bit stuck.

 

See http://www.mathpages.com/rr/s7-02/7-02.htm where Kevin Brown discusses the "geometric interpretation" (as exemplified by Misner/Thorne/Wheeler's "Gravitation") and the "field interpretation" (as in Weinberg's "Gravitation and Cosmology"). He favoured the former, but the latter gives a black hole picture which is more of a "hole" than the MTW picture, and seems to be gaining more acceptance.

[Royston]

There's no spacetime beyond the event horizon using an unmodified Schwarzschild metric, there is if you use advanced Eddington-Finkelstein coordinates, where the t and r (distance) coordinates are related by...

 

[math]ct'=ct+R_sln\left(\frac{r}{R_s}-1\right)[/math]

 

You're talking about a coordinate singularity, which is what the paper you cited is discussing, (which is flawed reasoning to your argument) in a limited way, what you said is right i.e 'there is no spacetime inside the event horizon', but only when using an unmodified Schwarzschild metric, and that's because it's ill equipped to deal with events at and past the event horizon...time like, becomes space like.

Edited April 2, 2010 by Snail

[Farsight]

I think my argument stands, Snail. See http://casa.colorado.edu/~ajsh/schwp.html and note that "Eddington-Finkelstein coordinates differ from Schwarzschild coordinates only in the relabelling of the time. The problem is that time is a cumulative measure of motion, and from where we're all standing, there just isn't any. The local "observer" cannot adopt a new coordinate system to escape that coordinate c=0, not in all eternity. He will never observe anything, ever.

[Mr Skeptic]

Farsight, if your space is non-flat, then I don't think you can count on the volume being the integral of the surface.

[Spyman]

Here is another view of the answer from Astrophysicists at NASA:

 

The Question

(Submitted August 08, 1997)

 

What is the volume of a black hole?

 

The Answer

Our intuitive sense of volume breaks down in the strong gravitational region in a black hole. So while the "size" of a black hole is given by the radius of its event horizon, it's volume is not determined by the usual 4/3*pi*r³. Instead, relativity makes it more complicated than that. As you pass the event horizon, the spatial direction 'inwards' becomes 'towards the future'-- you WILL reach the center, it's as inevitable as next Monday. The direction outsiders think of as their future becomes a spatial dimension once you are inside. The volume of a black hole, therefore, is its surface area times the length of time the hole exists (using the speed of light to convert from seconds to meters). Since a black hole last practically forever, the black hole's volume is almost infinite. (This is also a way of explaining the fact that you can pour stuff into a black hole forever and never fill it up. Another reason why black holes never fill up is that the radius of the event horizon increases as the mass of the black hole increases.)

 

David Palmer and Jim Lochner

for Ask an Astrophysicist

 

http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970808.html

[Farsight]

Mr Skeptic: agreed. I agree with this more than you might appreciate. There isn't any volume!

 

Spyman: re that answer, IMHO you never reach the centre. Have a read of http://www.mathpages.com/rr/s7-02/7-02.htm and look out for this:

 

"...if the same trajectory is described in terms of Schwarzschild coordinate time, the infalling object traverses through infinite coordinate time in order to reach the event horizon..."

 

It isn't as inevitable as next Monday, it's as inevitable as the end of time.

[Spyman]

Farsight, I think the author is actually arguing against the frozen star concept...

 

"Consider a black hole of mass m. The event horizon has radius r = 2m in Schwarzschild coordinates. Now suppose a large concentric spherical dust cloud of total mass m surrounds the black hole is slowly pulled to within a shell of radius, say, 2.1m. The mass of the combined system is 2m, giving it a gravitational radius of r = 4m, and all the matter is now within r = 4m, so there must be, according to the unique spherically symmetrical solution of the field equations, an event horizon at r = 4m. Evidently the dust has somehow gotten inside the event horizon. We might think that although the event horizon has expanded to 4m, maybe the dust is being held "frozen" just outside the horizon at, say, 4.1m. But that can't be true because then there would be only 1m of mass inside the 4m radius, and the horizon would collapse. Also, this would imply that any dust originally inside 4m must have been pushed outward, and there is no known mechanism for that to happen."

From your own link: http://www.mathpages.com/rr/s7-02/7-02.htm

 

 

Did you read and understand the Chapter 6.4 that was mentioned before the text you quoted ?

 

"Therefore, at r = 2m the curvature of this surface is -1/(4m2), which is certainly finite (and in fact can be made arbitrarily small for sufficiently large m). The only singularity in the intrinsic curvature of the surface occurs at r = 0."

"and we can confirm that the radial coordinate passes smoothly through r = 2m as a function of the proper time t."

From Chapter 6.4: http://www.mathpages.com/rr/s6-04/6-04.htm

[Farsight]

Spyman: yes, I know the author doesn't favour the frozen star interpretation. I think he got that wrong. No, I didn't read chapter 6.4. I'll read it now... huh, proper time, fooey. OK, what's this:

 

This allows us to easily generate geodesic paths in terms of r as a function of t. If we do this, we will notice that the paths invariably go to infinite t as r approaches 2m. Is our 2-dimensional surface actually singular at r = 2m, or are the coordinates simply ill-behaved (like longitude at the North pole)?

 

Sounds promising.

 

The only singularity in the intrinsic curvature of the surface occurs at r = 0.

 

A singularity in curvature is a red herring. That's essentially the rate of change of time dilation, what's important is where it's total.

 

In order to plot r as a function of the proper time t we would like to eliminate t from the two geodesic equations (3).

 

Proper time again. People just don't seem to understand that time is a measure of motion. Infinite time dilation means no more motion.

 

we can confirm that the radial coordinate passes smoothly through r = 2m as a function of the proper time t.

 

There is no proper time. It's an illusion that uses a stopped clock. But what's this?

 

We can also express the Schwarzschild coordinate time t explicitly in terms of alpha by multiplying the two relations

 

Alpha is just some angle, and

 

we must perform a complex integration around the singularity at r = 2m, offsetting the result by ±pi (assuming the path of integration doesn’t make any complete loops around the singularity). This is not surprising, because the t coordinates are discontinuous at r = 2m, so we cannot unambiguously “carry over” the labeling of the t coordinates in the region r > 2m to the region r < 2m.

 

This is performing a hop and a skip to jump over the end of time. Those two little wings at r=2m stop the show:

 

 

This is junk, we don't move in time:

 

Alternatively we could imagine a single universe with two families of particles, whose proper times increase in opposite directions of the coordinate time t. (In fact, John Wheeler once speculated that anti-matter particles might be modeled as particles moving backward in time.)

 

This is good stuff:

 

The existence of two distinct inner regions is perhaps not surprising if we note that an in-falling object requires infinite coordinate time to cross the boundary at r = 2m,

 

It then talks about changing the coordinate system, and mentions Kruskal-Szekeres coordinates. But it still doesn't get past the t=inf at r=2m. Interesting though.

[Deepak Kapur]

How much spacetime is there inside the event horizon of a non-spinning uncharged black hole of mass M? I know that a black hole has a Schwarzschild radius of [math]r = \frac{2GM}{c^2}[/math] and a sphere a volume of [math]V = \frac{4}{3} \pi r^3[/math], for a black hole volume of [math]\frac{32}{4} \frac{\pi G^3 M^3}{c^6}[/math].

 

However, I think the above volume would be in our metric rather than accounting for the distortions of spacetime. If you account for spacetime distortion when measuring the volume, what would you get?

 

I think these equations are derived assuming that normal physical laws operate inside a black hole. At such a high density, I doubt the concept of space is valid. Moreover new theories like the 'String Theory' may altogether dispense with the concept of space-time. It's only a matter of time. I pray to God to give mankind at least 'a single law' that is proof against new theories and inquiries.

[toastywombel]

I think these equations are derived assuming that normal physical laws operate inside a black hole. At such a high density, I doubt the concept of space is valid. Moreover new theories like the 'String Theory' may altogether dispense with the concept of space-time. It's only a matter of time. I pray to God to give mankind at least 'a single law' that is proof against new theories and inquiries.

 

 

String Theory doesn't dispense the concept of space-time. That is simply not true. And actually many of the equations are derived assuming the exact opposite, that the normal laws of physics do not operate inside a black hole.

 

One of the most inclusive of these is the 11-dimensional M-theory, which requires spacetime to have eleven dimensions,[8] as opposed to the usual three spatial dimensions and the fourth dimension of time.

 

http://en.wikipedia.org/wiki/String_theory

 

An open question in fundamental physics is the so-called information loss paradox, or black hole unitarity paradox. Classically, the laws of physics are the same run forward or in reverse (T-symmetry). Liouville's Theorem dictates conservation of phase space volume, which can be thought of as 'conservation of information', so there is some problem even in classical (non-quantum general relativity) physics. In quantum mechanics, this corresponds to a vital property called unitarity, which has to do with the conservation of probability (It can also be thought of as a conservation of quantum phase space volume as expressed by the density matrix).[93]

 

http://en.wikipedia.org/wiki/Black_hole

[Cap'n Refsmmat]

Just wanted to point this out:

 

http://arxiv.org/abs/0801.1734

 

Written by my TA from E&M class.

Edited April 16, 2010 by Cap'n Refsmmat

[Mr Skeptic]

Thanks Cap'n! That seems to answer my question.