# If Gravity travels at c

1. May 25, 2007

### OSalcido

How does a blackholes gravity escape its own event horizon? thanks

2. May 25, 2007

### lalbatros

It does not: no internal structure of a BH can be revealed, only a few constant of motion: no hair theorem.

3. May 25, 2007

### OSalcido

I mean how are we able to measure the weight of a black hole and why is surrounding material able to feel the black holes gravity if gravity only travels at c and should be too slow to escape its own gravity.

in other words, why is the black holes gravity not limited to its own event horizon

4. May 25, 2007

### cristo

Staff Emeritus
What do you mean by "a black hole's gravity"? In GR, we not longer think of gravity as a force, but rather that massive objects create curvatures in spacetime. This means that the path that an unaccelerated observer takes (a geodesic) is no longer a straight line, but is curved. This is how "gravity affects" observers in GR.

Last edited: May 25, 2007
5. May 25, 2007

### MeJennifer

OSalcido, gravitationally speaking a black hole is not all that different from a star with the same mass. But in case of a star the surface of the star is outside of the event horizon so it is not a black hole.

When a star collapses to a black hole there is no change in the gravitational field from the perspective of someone far away from it.

6. May 25, 2007

### OSalcido

oh I thought I had heard something about gravitons. nevermind about that then

so if gravity is just a curvature of space, why is it limited to c?

7. May 25, 2007

### OSalcido

yes, i am aware of this. but I am referring to the event horizon

8. May 25, 2007

### MeJennifer

Then you should realize that gravity does not need to "escape".
The spacetime was already curved before the star collapsed to a black hole.

9. May 25, 2007

### OSalcido

Yes I admit I am unclear on that. I thought gravitons had to escape, i understand now about the curvature. I am still not clear on what exactly is travelling at c in regards to gravity though and why this curvature of space should not be felt instantenously?

10. May 25, 2007

### cristo

Staff Emeritus
You say instantaneously, which makes it seems that you are still thinking of gravity as a force. Consider a universe with only one observer, and one black hole. At infinity, the observer's geodesic is a straight line, as the spacetime is asymptotically flat. As the observer falls closer to the black hole, his geodesic deviates from a straight line due to the curvature that the black hole has caused in the spacetime.

I'm not sure exactly what you mean when you say "gravity travels at c". Are you talking about gravitational waves, or are you still thinking of gravity as a force?

11. May 25, 2007

### OSalcido

well I saw this video and it said that if the sun were to be erased from the center of the solar system, Earth would still be orbiting seemingly nothing for 8 minutes because that is how long it takes for the loss of gravity to reach earth at the speed of light. my question is , if gravity is just curves in space, why the earth wouldnt just instantly shoot straight off at the deletion of the sun instead of orbiting basically nothing for 8 minutes

12. May 25, 2007

### cesiumfrog

Nicely worded riddle. My answer is:

Because "gravity" doesn't "travel at c", only changes in gravity travel at c. So if no gravitons can escape to cause those changes, when the star collapses beneath the event horizon, the gravity outside gets left just as it was before.

13. May 25, 2007

### Staff: Mentor

This is question is answered in the Usenet Physics FAQ:
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html
Purely in terms of general relativity, there is no problem here. The gravity doesn't have to get out of the black hole. ... If a star collapses into a black hole, the gravitational field outside the black hole may be calculated entirely from the properties of the star and its external gravitational field before it becomes a black hole.

Nevertheless, the question in this form is still worth asking, because black holes can have static electric fields, and we know that these may be described in terms of virtual photons.

14. May 25, 2007

### Staff: Mentor

Like a bowling ball on a trampoline, if you remove the object, the curvature does not get removed instantly, it propagates at the speed of waves on the medium - in this case, the speed of light.

15. May 25, 2007

### DaveC426913

I agree with the first part but not the second.

Yes, changes in gravity propogate at c.

But if they couldn't propogate outside a BH then it would never change its mass or gravitational effect once it was created. Yet it does.

16. May 25, 2007

### cesiumfrog

No it doesn't. It only changes if something else approaches from outside! But don't take my trapped-graviton-no-hair-theorem too seriously, as the question it aimed to alleviate didn't really make sense in GR.

17. May 25, 2007

### Chris Hillman

Elaboration of "news"

I was just about to mention http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html when I saw someone else beat me to it.

Well, the trouble with that video (where did you see it, BTW?) is that by conservation of mass-energy, you can no more envision a scenario where the Sun "disappears" in gtr than you can in Newtonian gravitation! But a minor modification which is physically acceptable makes the same point: suppose the Sun suddenly explodes and redistributes its mass distribution in an asymmetrical manner. Then the "bad news" propagates outward at speed c; you see the explosion at the same time that your gravitational wave detector detects gravitational radiation resulting from the sudden change in the distribution of mass-energy.

Exactly.

Let's elaborate somewhta on the physics FAQ essay by Matt McIrvin quoted above.

Recall, first, that a realistic black hole is formed by the gravitational collapse of matter, and second that the simplest model of the exterior gravitational field of a black hole (the Schwarzschild vacuum solution) is the same used in the simplest models of the exterior gravitational field of a star having the same mass.

This should immediately suggest that the case of a static stellar model, the static exterior field is a "fossil" of the orginal formation of the star by slowly concentrating matter in the distant past. Now the Sun has been very stable for a long time, so this field isn't rapidly changing to a first approximation, so very little gravitational radiation is coming from the Sun. But in a scenario where the Sun suddenly blows apart into two pieces or something like that, the field suddenly changes very rapidly and very dramatically, which causes intense radiation.

More realistic scenarios of this kind involve the collision and merger of two neutron stars, or a neutron star falling into a black hole, etc.

In all these cases, what is trying to "climb out of the well" is the news that something dramatic has happened. The original curvature of spacetime a la Schwarzschild or Kerr happened slowly and long ago during the formation of the original mass m object, which was initially a star, so no news need climb out to explain the existence of this field.

18. Jun 6, 2007

### Brinx

I have a question related to this matter - I think it's sufficiently similar to the OP's question to simply add it to this thread.

Consider a non-rotating black hole, with a nice spherically symmetric event horizon. If one drops some mass 'into' the black hole from a specific direction, what will be the resulting gravitational field of the now slightly bigger black hole? Before you say 'simply spherically symmetric again, only a bit stronger', let me elaborate.

I'd think that the last position from which the gravity field of the infalling mass is 'updated' for an external observer lies at the event horizon. That, after all, is the place from which the changes in the gravitational field brought about by the infalling mass can still be propagated outward. After the mass passes the event horizon, its position can no longer be measured, by any means including via gravitational waves. So, then we get a slightly asymmetric gravitational field as a result: the initial spherically symmetric gravitational field of the black hole (centered, presumably, at the center/singularity of the black hole), plus the small addition of the gravitational field of the infalling mass (or rather, the 'fossil field' of the mass from the moment it passes the event horizon), of which the center resides at the point on the event horizon through which the mass fell.

The sum of these fields isn't spherically symmetric anymore, it'd seem.

Does my line of reasoning make sense? Is the conclusion of a resulting non-spherically-symmetric gravitational field, if valid, problematic in any sense?

19. Jun 6, 2007

### cesiumfrog

That's actually a difficult question:
- it *will* end up being still spherically symmetric (google "no hair theorem"), albeit slowly moving in the direction of the momentum you threw in.
- you will never actually see the small piece of mass reach the event horizon (due to time dilation).
- as the mass approaches, the "event horizon" isn't well defined (mathematically, it's a global rather than local concept).
- the finer details still aren't all known, basically because the precise equations are so difficult to extract accurate solutions from.

There is a closely related question, of what happens to the electric field if a charge is thrown into a black hole. In that case I think the finer details can all be worked out, and you see that as an electric charge approaches the black hole, the event horizon practically takes on a like charge, so externally it makes little difference whether the charge is "just outside" the horizon or even "in the singularity".

Last edited: Jun 6, 2007
20. Jun 6, 2007

### Chris Hillman

Dropping a smallish object into a Schwarzchild hole

Let's say uncharged and isolated (no other mass present), so that we can model the initial situation using a portion of the Schwarzschild vacuum solution.

It is often helpful to first study the analogous Newtonian question. Here, there is a tidal tensor which can be compared directly with that electrogravitic or tidal tensor used in gtr. In Newtonian theory it is, in a vacuum region, simply by $E_{jk} = \Phi_{,j \, ,k}$ where $\Phi$ is the Newtonian gravitational potential http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=28781955 (or more generally, by the traceless part of the Hessian). So answering the question comes down to how the potential changes. In a vacuum region, the potential is just a harmonic function, and since we are dealing with an isolated system (central object plus smallish mass) it vanishes asymptotically. As the smaller object approaches its gravitational field distorts that of the big one. Since Newtonian gravitation is governed by a linear field equation, the Poisson equation, the net potential is simply the sum of the two original potentials. So at this point, you might want to add two point mass Newtonian potentials and animate some contour plots, then follow suit for the Hessian.

In general relavity, as the two objects near each other, we expect strong fields to bring nonlinearity into play, so that the predictions of gtr will differ greatly from those of Newtonian gravitation in various ways. However, it should already be clear at this point that one thing you can expect during the "approach phase" is that the field may retain approximal axial symmetry but certainly won't be spherically symmetrical.

In both Newtonian gravitation and gtr it is useful to have some way of describing precisely the "shape" of the field. In the case of an isolated system, as in our situation, the answer is multipole moments. If you haven't seen these before, you might take a moment (no pun intended) to read about the elementary kinds of moments used everywhere in mathematics, including weak-field gtr, which is an approximate theory governed by the "linearized EFE". In fully nonlinear gtr a more sophisticated kind of multipole moment is appropriate (there is a standard notion, given by competing mathematical definitions which turn out to be largely equivalent).

One of the most essential differences between Newtonian gravitation and gtr is that gtr is a true relativistic field theory of gravitation, and thus it predicts the existence of gravitational radiation. In particular, when the multipole moments of an isolated system are appropriately time varying, gravitational radiation will be emitted from the system. This condition is satisfied in our situation.

Now, there is a principle, discovered long ago by Richard Price and others, to the effect that when a smallish object falls into a black hole, after the collision and merger, the field may be briefly distorted (in particular, the horizon might be distorted and wrinkled), but the hole "radiates away" these distortions in the form of gravitational radiation which moves out at the speed of light and leaves behind an "smooth and symmetrized" region. The result is that after all the fuss has died down, you have a somewhat larger hole, but if no angular momentum was introduced during the merger, it will still be modeled by a Schwarzschild vacuum.

I am oversimplifying above, incidently. In fact much more can be said about the time sequence which the radiation will follow, how strong it will be, and so on. A huge amount of work has been done on such problems, so quite a bit is known by now about what gtr predicts in such merger scenarios.

Which is not static in this situation.

Comparing with what I said above, sounds about right.

Well, much is known, but not everything which might happen is well understood yet. Merger scenarios continue to be the subject of intense research. Because of these typically have little or no symmetry, much of the work involves numerical simulations, and it turns out to be highly nontrivial to concoct numerical integration schemes which remain stable sufficiently long to model the "approach, merger, radiative smoothing" scenario sketched above.

One thing which is thought to be "well understood" from simulations is the idea that in some merger scenarios, the gravititational radiation emitted from the system might be asymmetric, basically because the smallish object, roughly speaking, will suddenly plunge into the (distorted) horizon of the hole. In some situations this radiation can carry off very substantial amounts of energy and momentum. The result is that the new hole should be "kicked" by the recoil of the outgoing gravitational radiation! As a result, one should apparently expect to find holes speeding out of their parent galaxies as the result of such mergers. But as someone recently mentioned, a current mystery is that this hasn't yet been observed.

In principle, the event horizon would generally be well defined but it is always defined in "teleological" fashion. That is, knowing the locus of the event horizon requires roughly speaking knowledge of the entire future history. Thus, numerical simulations have to use "horizon detectors" based on local (as in local neighborhood) notions. One fact worth mentioning is that in the static Schwarzschild vacuum, the world sheet of the horizon (static and spherically symmetric!) happens to be generated by a Killing congruence. This is related to the fact that the covariant derivative of certain scalar curvature invariants vanishes at the horizon, which is an "infinitesimal" criterion. Such a thing is called a "Killing horizon". But in your scenario, the true event horizon would not be given this way.

See the website in my sig for links to on-line resources and citations of relevant books. In particular, the monograph by Hawking and Ellis, The Large Scale Structure of Space-Time, contains a classic discussion of how the horizon behaves during merger of two black holes, and the monograph by Frolov and Novikov, Black Hole Physics, contains an introduction to such topics as quasinormal modes, which govern how a hole responds to small perturbations. But you might want to start with the excellent survey articles in Black Holes and Relativistic Stars, ed. by Robert Wald. See section 4.7.3 of the article by Rees for recoil, see the article by Teukolsky for numerical simulation of mergers, and see the article by Kip Thorne for the radiation predicted in merger scenarios.

HTH

Last edited: Jun 6, 2007
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