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May 20, 2020 at 4:55 comment added vikrant If all we need is curvature to feel the effect of gravity, then where does gravitational radiation comes into the picture?
Mar 3, 2011 at 3:01 comment added Daniel Sure, you're probably right. But, i'd rather not use any more than i have to (in terms of assumptions or extra estructure): if you can answer with some sort of 'minimum' set of assumptions, why 'complicate' the problem — and have to deal with the consequences of said 'complications' later? ;-)
Mar 3, 2011 at 1:13 comment added Daniel Grumiller @Daniel: you do not need to quantize gravity or actually calculate some S-matrix elements to conceptually understand how and why (Coulomb-)gravitons "escape" a black hole. I think this is sufficient for the purpose of answering this question...
Feb 15, 2011 at 14:17 comment added Daniel @Daniel Grumiller: In more general grounds, you don't need to agree with nor anyone else. But, in this particular case, the reason is pretty straightforward: because it's possible to answer the original question without having to talk about a possible quantization of GR. To appropriately and scientifically answer the original question, taking gravitons into account, sooner or later one would have to compute the scattering of light by gravitons (and vice-versa), which would immensely complicate an otherwise simpler answer.
Feb 10, 2011 at 1:40 comment added Daniel Grumiller Nogwater specifically asked for an explanation in terms of gravitons, so why should we agree on not bringing them into the answer?
Nov 18, 2010 at 10:51 comment added Vagelford I would like to point out that the existence of an event horizon is not a high curvature effect. The event horizon depends on the causal structure of the space-time while the curvature gives you the strength of the tidal forces. You can have an event horizon with low curvature. For example, in the case of a Schwarzschild black hole, where the event horizon is at radius $R_s=2M,$ if you calculate the relevant curvature components which are $\propto\frac{M}{r^3},$ you can see that on the horizon the curvature scales like $\frac{1}{M^2}.$ So, the bigger the mass, the smaller the curvature.
Nov 17, 2010 at 18:43 comment added Daniel @Nogwater: Even though DavidZ already answered, i just want to offer my 2¢: things that happen inside a black hole (eg, mass increase) affect the curvature of spacetime, and this can be measured from the outside. You can choose to understand this via the Holographic Principle, but you don't necessarily need to: it's vanilla Differential Geometry (and Topology).
Nov 17, 2010 at 18:30 comment added David Z @Nogwater: As far as we know, the mass itself is at the singularity, for some definition of "is" (namely that the proper time between crossing the event horizon and reaching the singularity is finite). But, speaking in vague terms, the information of how much mass there is gets "imprinted" on the horizon, it does not "fall" down to the singularity along with the mass. In more precise form, this is called the holographic principle.
Nov 17, 2010 at 1:40 comment added Nogwater So... if I'm following, flipped light-cones means that the information about the mass doesn't come from within the event horizon (causality doesn't flow that way). Instead the gravity comes from event horizon, which sounds like it fits with @Vagelford's answer. The mass isn't at some singularity at the center of the black hole, but is distributed around the edge? A sphere of mass "creates" the same gravitational field as that same mass at a point from the point of view of someone outside the sphere, right?
Nov 16, 2010 at 20:27 comment added Daniel @MarkE: I could open my magic toolbox and talk about holonomies and their relation with orbits in GR (ie, with closed geodesics). And, by mapping the holonomies of a space you can get information about its curvature. So, if you're orbiting a black hole, you can gather all the information about its curvature. (That's why i made that comment about global properties of spaces: they are very non-intuitive.)
Nov 16, 2010 at 20:25 comment added Daniel @MarkE: only if you were God. Look, the bottom-line is that we're dealing with classical GR, and not Quantum Gravity nor its effects. And, within the framework of classical GR, it's simply not possible for you to change any of the properties (charge, mass, angular momentum) of a black hole from the inside of it. A black hole is simply a "sink" of gravitational fields.
Nov 16, 2010 at 20:05 comment added Mark Eichenlaub But then doesn't information go from inside the black hole to outside it? I could send morse code by turning my mass on and off, right?
Nov 16, 2010 at 19:50 comment added Daniel The thing to note is that curvature is not something that lives only inside the Black Hole: this is a property of spacetime as a whole, and that's what counts. Global, topological, properties are very non-intuitive things. ;-)
Nov 16, 2010 at 19:46 comment added Daniel If "The Hand of God" (and we're <a href="en.wikipedia.org/wiki/… talking about Maradona</a> here ;-) did something like this, and we were only thinking about classical GR, we can say the following: it would be possible to measure such change, in the sense that the curvature of spacetime would change and we'd be able to see that the black hole got more massive this way (the curvature increased).
Nov 16, 2010 at 19:40 comment added Mark Eichenlaub Suppose, very hypothetically of course, that some extra mass were suddenly created inside the black hole. Would the spacetime curvature outside the black hole change? I realize this is an unphysical process, but if the hand of God reached down and created a large lump of stuff just inside the event horizon, what do the equations of GR tell us about whether we would we be able to tell about the event from outside the event horizon?
Nov 16, 2010 at 19:32 history answered Daniel CC BY-SA 2.5