# Is there a size?

1. Apr 26, 2005

### UrbanXrisis

my physics teacher said that there is an actual size for the black hole and this size can be calculated. I thought a black hole is a point mass.

2. Apr 26, 2005

### chroot

Staff Emeritus
A classical black hole is a point mass, but that point is surrounded by a mathematical surface called the event horizon. The event horizon is the "point of no return," in the sense that anything that ventures inside the event horizon cannot come back out. The event horizon has a radius, usually known as the Schwarzschild radius:

$$r_s = \frac{2 G M}{c^2}$$

This is the figure most people use when describing the "size" of a black hole.

- Warren

3. Apr 26, 2005

### turbo

Warren's explanation is spot-on. Let me point out in common English the dichotomy that he laid out. In mathematical terms, the BH singularity is a dimensionless point. Pretty neat, isn't it, taking 3-D physical objects and embedding them in a dimensionless entity? In the physical model, the effects of a BH in our universe extend to the event horizon, which has a radius that can be calculated, and any object that ventures to that radius is committed to join the BH. The size of a BH varies infinitely between none and some depending on whether you are a mathemetician or a physicist.

4. Apr 26, 2005

### Marjan

That is classical mistake people make. BH has surface, ofcourse the problem can not be solved in SM. You need more fancy tools like quantum information theory!

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5. Apr 26, 2005

### chroot

Staff Emeritus
Uh, what? You apparently don't know what you're talking about.

The classical black hole is modelled entirely by the general theory of relativity.

- Warren

6. Apr 27, 2005

### Marjan

Yes, but is BH modelled entirely by the GR enough to describe real BH? I know that SM isn't at all. We wouldn't need QG if GR would be enough!

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7. Apr 27, 2005

### mapper

What I would like to know is if there is a min or max density a black hole can get. If so what happens next? Also, what happens if a black hold meets another one? If black holes truly do exists, then this would also be a possible scenario.

8. Apr 27, 2005

### chroot

Staff Emeritus
Of course, but such complexity is not necessary to describe the event horizon. Please attempt to keep responses at a level appropriate for the original poster.

- Warren

9. Apr 27, 2005

### chroot

Staff Emeritus
No.
The two coalesce, and generate a lot of gravitational radiation. The merger of two black holes is actually a very well-studied scenario. The various gravitational-wave detectors (LIGO, etc.) are, in fact, looking for neutron star mergers or black hole mergers, since these are the most prolific generators of gravitational radiation in the universe.

- Warren

10. Apr 27, 2005

### mapper

I would think that there is a min. =/

11. Apr 27, 2005

### JesseM

There'd be no max density according to general relativity, but there might be one according to quantum gravity, probably on the order of one planck mass per planck volume.

12. May 2, 2005

### Phobos

Staff Emeritus
density = mass/volume
under GR, the black hole singularity has zero volume...not much you can do with that.

However, for BHs formed from the cores of "dead" stars, there is a minimum MASS that is required to be able to crunch the matter down to a singularity.

13. May 6, 2005

### kleinwolf

I know nothing about QG, but some things can be said about Schwarzschild solution :

In fact the Schwarzschild metric has 2 singularities : r=r_s and r=0.

If r is smaller than r_s but greater than 0 the metric is not singular.

Curiously : when going inside this radius (let say just looking at the metric), then time becomes space, and the radial coordinate becomes time !!

The other thing is that I think there is a coordinate system called Kruskal coordinates which are singular only at r=0.

Interprete this as : Schwarzschild radius is a singularity of coordinates, but not of space-time.

However, r=0 is singularity of symmetry....sthg like that.

14. May 6, 2005

### chroot

Staff Emeritus
Yes, kleinwolf, there are coordinate systems like the Finkelstein-Eddington coordinates which are only singular at the singularity, not at the event horizon.

- Warren

15. May 6, 2005

### ohwilleke

If I'm not mistaken while density is ill defined for a black hole itself, you can approach the problem from the other direction.

For example, a spherically symmetric object cannot exceed a certain density (with GR you really need to be looking at some sort of mass-pressure combination I imagine) or it will become a black hole. The maximum non-black hole density would corrospond to probably the most intuitive way to define a minimum black hole density, since every black hole candidate had that density at some point prior to becoming a black hole.

16. May 6, 2005

### ohwilleke

Incidentally, while black holes may marry, they do not divorce.

17. May 7, 2005

### kleinwolf

Consider a point mass at r=0, hence spherical symmetry.

a) Consider this is static, plug the standard spherical metric in Einstein's equ., solve it for the vacuum...get the Schwarzschild solution after a LONG calculation :

$$g_{00}(r)=1-\frac{2GM}{rc^2}$$

this corresponds to the squared time-dilatation factor.

b) I use the equivalence principle : gravitation is equiv. to an accelerating frame...of course the accel. depends on r (Newtonian potential) What is the time-dilation due to that change of frame squared ?

$$1-\frac{v^2}{c^2}$$ (*)

But what is the speed ?

1) It depends on r and I remark that it is just the classical speed :

$$\frac{1}{2}mv^2=\frac{GMm}{r}\Rightarrow v^2=\frac{2GM}{r}$$

Plug into (*)...you get the Schwarzschild coefficient....Is this a pure coincidence that works only in this case ?

Because if not, then I could say :

2) I Just compute the speed with the relativistic formula :

$$mc^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)=\frac{GMm}{r}$$

$$g1_{00}=1-\frac{v^2}{c^2}=\frac{1}{(1+\frac{GM}{rc^2})^2}$$

This has no singularity outer r=0...but look far away from r=0 leads to :

$$g1_{00}->1-\frac{2GM}{rc^2}+\frac{3G^2M^2}{r^2c^4}-...$$

The first approx. gives the Schwarzschild solution again....

Coincidence again ? Or is it somehow a physical calculation ?

Last edited: May 7, 2005
18. May 7, 2005

### Chronos

That is meaningless. You are substituting identities.

19. May 7, 2005

### kleinwolf

You mean i do things of the type

F(a)=b and a=g(c) hence F(g(c))=b ?