# Value of g near a black hole

1. Feb 12, 2012

### pawprint

On approach to a simple (non-rotating, uncharged) singularity, does g increase asymptotically near...
a) the singularity,
b) its event horizon,
or c) no?

2. Feb 12, 2012

### bcrowell

Staff Emeritus
At any point in spacetime, g has any value you want it to have. For a free-falling observer, it always equals zero. This is what the equivalence principle is about: http://en.wikipedia.org/wiki/Equivalence_principle

The singularity isn't a point in spacetime. There's no "there" there. Asking the value of g at the singularity is like asking the temperature right now in Narnia. Narnia isn't a place, so "right now in Narnia" doesn't describe a point in spacetime.

That's why we don't talk about g in general relativity. We talk about curvature. Curvature gradually increases as you get closer to the singularity. It doesn't do anything special at the event horizon. It approaches infinity as you approach the singularity.

3. Feb 12, 2012

### Nabeshin

I'd like to expand on bcrowell's answer by talking about what happens to an observer who's trying to stay at rest with respect to the black hole (hover a certain distance away from it). For such an observer, the amount of power he needs to hover at a given radius away increases without bound as his distance drops to the schwarzschild radius.

4. Feb 12, 2012

### pawprint

bcrowell- I'm sorry my question was imprecise. I spent some time editing it to the bone but it looks as though I should have left a little more meat. I was unwise in using the word 'approach'- I meant to suggest that the measurements were taken at progressively closer distances along a 'straight' line ending at the singularity. There is no need to debate the existence of a straight line near a singularity; I seek an intuitive answer rather than a mathematical one.

Thank you Nabeshin. You have grasped my meaning. I take it you mean that b) is the answer within this context.

5. Feb 13, 2012

### pervect

Staff Emeritus
The acceleration required to "hold station" will be equal to
$$\frac{G\,m}{r^2 \sqrt{1-r_s/r}}$$

where G is the gravitational constant, m is the mass of the black hole, r is the Schwarzschild r coordinate, and $r_s$ is the Schwarzschld radius, $r_s= G m / c^2$

The Scwarzchild r coordinate isn't really a distance, because space-time is curved near a black hole, but the surface area of a sphere given by r=constant is 4 pi r^2.

By contrast, for Newtonian gravity, the acceleration requires to hold station is G m / r^2, (but here r is the radius, though it's still true that the surface are of a sphere given by r = constant is 4 pi r^2).

As you can see, the GR expression goes to infinity as r approaches r_s, the Schwarzschild radius.

6. Feb 13, 2012

### Paul.Dent

I can't accept that the event horizon is not "a point in spacetime" We can and do assign coordinates to it, so it must be! However, I do have some grave doubts about its nature.

Here is one issue I am struggling to understand:

I think I am right in saying the gravity gradient at the event horizon is finite and gets smaller, the larger the Black Hole is. So make it large enought, and matter falling in does not necessarily get torn apart while still on the outside, especially if heading in in a straight line plumb dead center. So consider that situation. Now, just before the matter enters, the event horizon's radius is given by thre mass already on the inside; and just after enters, the Black Hole's mass and its radius must have increased.

Here is the \$64000 question: At what point does the Black Hole's event horizon increase its radius? Doe sit come out to meet the falling in mass, like a big snake's mouth opening up" Does that mean the Black Hole gets a bulge on side, that subsides as the matter plummets towards the cental singulaity? I have big problem with the latter, because that implies we a getting information on the outside about events on the inside. I have other reasons to want to believe that a Back Hole can only accrete mass in a sphercially synnetric way. At least in a cylindrically symmetric way, if there is angular momentum invloived. But preferably, only in a spherically symmetric way when there is no angular momentum involved. In the latter case, the matter would cover the event horizon uniformly and then continue its chute towards the central singularity as a uniform sphere of collapsing radius.

7. Feb 14, 2012

### pawprint

This rather depends on the observer. A distant observer will see the infalling mass slow down and finally halt at the event horizon. From her point of view the EH will never expand.

A brave observer entering a sufficiently massive BH may, as you pointed out above, avoid spaghettification. From her point of view the external universe will accelerate and die as she nears the EH. She would believe that the EH expanded slightly as her conveyance passed through, but would be unable to verify this.

Two threads discussing this aspect are-

Over the past few years speculation has grown as to whether BHs can gain any mass in external finite time. I was unable to find the thread(s) discussing this but it is now considered at least an open question. I DO remember the argument is that all infalling matter is re-radiated as energy in exact ballance. I'm sure this has been discussed here (on PF) and I would appreciate someone posting the relevant link(s).

8. Feb 18, 2012

### pawprint

Thank you all. If only I had a mathematically inclined brain I could use those equations. I'm currently faced with variations of g at different heights above the surface of Earth. I'm reduced to expressing my answer in μgals!

9. Feb 18, 2012

### pervect

Staff Emeritus
While it's not a point (technically, it's a null surface), the event horizon is "in space-time". But the singularity at the center of the black hole (which isn't the event horizon) , isn't part of the manifold, i.e. it's not part of space-time. So I think you misunderstood what Ben said earlier, as he was talking about the singularity not being part of space-time, not the event horizon.

The acceleration required to hold station - the "gravity" for a hovering particle approaches infinity as the particle approaches the event horizon. This might lead one to conclude that the gravity gradient also approaches infinity for a particle approaching the event horizon. This is not correct, at last not for a particle that free-falls into the event horizon. The gravity gradient for a free-falling particle is always finite. Explaining exactly why convincingly probalby would require some math - all I can say is that the gravity gradient isn't quite observer independent, but the effect is important only for large accelerations.

Specifically: the Riemann tensor is observer independent quantity (being a true tensor), but the tidal forces you measure with rods and accelerometers are only exactly equal to the Riemann tensor for someone who is not accelerating, for someone who is accelerating there is a (usually very tiny) difference between the observer-independent Riemann tensor components and the actual measurements of the gravity gradient, also known as tidal forces.

Since the acceleration required to hold station at the event horizon is infinite, even a very small observer dependence that's usually ignorable becomes important to know about.

It depends on how you define the event horizon. There's an apparent horizon, and an absolute horizon, as I recall. http://archive.ncsa.illinois.edu/Cyberia/NumRel/AppVsEventHorizonCaption.html and http://en.wikipedia.org/w/index.php?title=Apparent_horizon&oldid=457048068 have some details.

10. Feb 18, 2012

### Nabeshin

If we imagine a hole of mass M, and throwing in some small mass dm, the event horizon smoothly expands from its radius at r=2M to r=2(M+dm) as the mass falls in. If the situation is not axisymmetric, indeed there will be a tidal budge induced on this surface. After the mass falls in, the horizon will ring down, emitting gravitational radiation until it settles back into a symmetric black hole.

No information is transferred here because the event horizon is simply a mathematical surface, not something physical.

11. Feb 18, 2012

### elfmotat

You mean $r_s= 2G m / c^2$, right?

12. Feb 19, 2012

### pervect

Staff Emeritus
Ooops, yes, right...

13. Feb 19, 2012

### George Jones

Staff Emeritus
It is useful to look at the Vaidya metric. See Figure 5.7 on page 134 (pdf page 150) of Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.

Radiation falls into a black hole from v1 to v2, but the left diagram of Figure 5.7 shows that the event horizon starts to grow before the first radiation crosses the event horizon.
Here, "history" means past and future.

14. Feb 19, 2012

### Passionflower

I do not think that is correct.

The total mass will be slightly less than M+dm.

15. Feb 19, 2012

### Nabeshin

Well yes, thank you. My point in using those values was to illustrate that it asymptotes to what you would 'expect' the horizon to be when the additional mass has fallen in, minus of course some losses from gravitational radiation. If I wanted to be a little more analytic, the power radiated by waves will be $\propto dm^2$, while the mass increase is linear, so to first order indeed the mass is M+dm.