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How can a black hole evaporate instantly if speed of gravity is c?

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kamenjar
#19
Jan17-12, 07:29 PM
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Quote Quote by alexg View Post
I'm pretty sure that you get different values for expansion redshift and gravitational redshift. So he concludes the flashlight is down a gravity well.
Ok so he compares the magnitude of the light which is very faint due to gravitational effect and and compares to expansion of space, he concludes that the object is accelerating away from him but is still far more distant than his house.

We can drop this. Distances can be confusing if measured form different reference frames, so we can use time:
a) How long does the light from the EH inside the gravity well take to reach him?
b) How long until the observer notices that the gravitational effect of the BH has dissapeared?

I think that science answers "less than a second" to "b" but think about answer to "a".... It would mean that in case of "a" the black hole never existed. Given my limited knowledge answering "less than a second" to b) would violate some GR speed of information information rules.

So please enlighten me why isn't the answer "billions of years"? You are most likely right in all this, but I don't understand why :)
alexg
#20
Jan17-12, 09:04 PM
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If the observer is 10 light seconds away from the BH, it will take him ten seconds to note that the event horizon has disappeared.

Light does not slow down, and distances do not greatly change.
PeterDonis
#21
Jan17-12, 10:40 PM
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Quote Quote by kamenjar View Post
The part that I don't get though is this change of mass propagating instantly (c) and instantly "straightening" the geodesics. From a reference frame of the black hole itself, yes, but from the observer that is say 1m from the pico black hole, I don't see how this information can reach him in an "instant". I would figure it would take eternity for the effect to disappear.
The change of mass doesn't propagate instantly; it propagates at the speed of light, just as you would expect. Sorry if I wasn't clear about that. Suppose you're hovering at a distance d just above the hole's horizon, and it evaporates a little bit, losing a tiny increment of its mass. It takes a time d (in units where c = 1) for the information that this tiny evaporation has happened to get to you from the hole's horizon (loosely speaking--there are a number of technicalities here that I don't think we need to get into for this discussion). So there is a speed of light time lag for the "change in gravity" to propagate.

However, once the "change in gravity" has propagated outward to someone "hovering" at some distance d above the hole, it causes two things to happen: first, the observed mass of the hole at d is a bit smaller; and second, the light cones at d are tilted inward a bit less. That means that the outgoing information about the change in gravity gets outward a little quicker than it would have if the hole's mass had stayed constant. And each little bit of evaporation makes it quicker still. By the time the hole is almost completely evaporated, outgoing light signals are traveling outward almost as fast as if spacetime were completely flat. So there is no discontinuous change at any time; everything propagates outward at c, but as the light cones tilt less and less, the meaning of "propagate outward at c" changes, so to speak.
PeterDonis
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Jan17-12, 10:45 PM
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Quote Quote by twitch1 View Post
Ive heard that objects at the event horizon appear to be at rest once they reach the event horizon to the distant observer, even though the object being viewed has actually gone over the event horizon and into the black hole.
This is not correct, at least not in either of the senses in which the word "appear" is used in GR:

(1) It is not true that light coming outward from objects at the horizon, to observers "hovering" some distance above the horizon, shows the objects at the horizon to be at rest.

(2) It is not true that there is any valid coordinate chart that the distant observer could use, in which objects at the horizon would be seen to be at rest.

Note that both these statements are true for an evaporating hole, not just for a static "eternal" black hole whose mass never changes.
PeterDonis
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Jan17-12, 10:49 PM
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Quote Quote by kamenjar View Post
Quite correct.
Not really; see my previous post. However, that doesn't completely invalidate the rest of your thought experiment:

Quote Quote by kamenjar View Post
Let's say for the sake of this experiment that the flashlight changed its mind at the last instant before reaching the EH and decided to accelerate to c instantly...
Small nitpick: the flashlight can't accelerate to c; it can only accelerate to a speed very, very close to c (arbitrarily close, but not *exactly* c). The *light* it emits can move outward at c, but the flashlight itself can't.

Quote Quote by kamenjar View Post
...and start traveling towards the observer and away from the black hole that just evaporated as the flashlight took off. My understanding is that even the flashlight would experience tremendous gravity along the way (though the BH already evaporated). Change in gravity is following the flashlight at speed c just behind the flashlight itself.
But the change in gravity is very small, as I showed in an earlier post. If the black hole "just evaporated", then its mass was very tiny when the flashlight turned around. So the change in gravity that is propagating just behind the flashlight (it will eventually overtake the flashlight, but as I said above, that's a small nitpick) is a small one, not a large one; the gravity in the region of spacetime the flashlight is moving through is close to zero, because the hole's mass just before it evaporated was close to zero.
kamenjar
#24
Jan19-12, 06:37 AM
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Quote Quote by PeterDonis View Post
The change of mass doesn't propagate instantly; it propagates at the speed of light...
I think you misunderstood the question or I misunderstood the answer.

Imagine a straight line perpendicular to EH pointing at an observer O:
BH----EH---P1->---P2->------------------------------O
There are a few statements here and one is incorrect:

- If there is an event horizon at point EH (of evaporating or non-evaporating black hole), there may exist such point P2 at which a photon P2 moving at c directly towards O will take 10 Billion years of OBSERVER'S TIME to reach the observer due to curvature of spacetime.

- A photon at P1 will reach the observer later than 10 billion years of OBSERVER'S TIME.

- Information about black hole's mass change will never reach P1 nor P2 at any point in time because P1 and P2 photons move at c.

- P1 and P2, being unaware of the change follow the same "flight path"/curvature and as if BH never evaporated because the information never reached them.

(The fact that P1 does not reach the observer before P2 does is just a supporting fact stating that there is something "behind" P2 that does not reach the observer sooner)

- Hence we can conclude that information that BH evaporated will reach observer O in more than 10 billion years.

- Hence we conclude that we can never get rid of a "micro" black hole if we were to create one and better eject them into outer space after we are done with them.
phinds
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Jan19-12, 09:35 AM
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Quote Quote by kamenjar View Post

- Hence we conclude that we can never get rid of a "micro" black hole if we were to create one and better eject them into outer space after we are done with them.
I followed everything else you said, but this is nonsense. micro black holes DO evaporate, and they do so whether we like it or not, although if we feed them mass, they could perhaps be kept stable. If I recall correctly, a micro black hole moving through the earth would loss mass to Hawking Radiation faster than it would absorb mass from the earth.
PeterDonis
#26
Jan19-12, 10:24 AM
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Quote Quote by kamenjar View Post
Imagine a straight line perpendicular to EH pointing at an observer O:

BH----EH---P1->---P2->------------------------------O

There are a few statements here and one is incorrect:

- If there is an event horizon at point EH (of evaporating or non-evaporating black hole), there may exist such point P2 at which a photon P2 moving at c directly towards O will take 10 Billion years of OBSERVER'S TIME to reach the observer due to curvature of spacetime.
This is incorrect if the black hole is evaporating; there may not be any such point P2. If the BH is "eternal" (i.e., static, its mass never changes), then the statement is correct, there must be some such P2.

Since all the rest of your statements pretty much depend on this one, they're all incorrect in the case of an evaporating BH. But the particular one that's most important is this one:

Quote Quote by kamenjar View Post
- P1 and P2, being unaware of the change follow the same "flight path"/curvature and as if BH never evaporated because the information never reached them.
Nope. In fact, this statement doesn't even follow from the previous incorrect ones. The previous ones were about information traveling from P1 or P2 to O. This one is about information traveling from the EH (or what *was* the EH before the BH lost some mass by evaporation) to P1 or P2. I believe this confusion may be the source of your misunderstanding.

Look at that parenthetical statement I made just above. When the BH evaporates a little bit, it loses a little bit of mass. That means the location of the EH *changes*; the point that was just at the EH before is now slightly above it. And *that* means that light at the original EH, which previously could not escape outward, now *can*. Now carry this same reasoning outward to P1: outgoing light from P1 now escapes outward a little *faster* than it did before, because it's a little farther from the EH. Similarly for P2. And as the BH continues to evaporate, outgoing light from P1 and P2 continues to be able to escape faster. The ultimate result is that, unlike the case of the "eternal" BH, there is a finite limit to how long it can take for information to get out to O from any point above the EH.
kamenjar
#27
Jan19-12, 12:31 PM
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Quote Quote by PeterDonis View Post
This is incorrect if the black hole is evaporating; there may not be any such point P2. If the BH is "eternal" (i.e., static, its mass never changes), then the statement is correct, there must be some such P2.
It doesn't matter if it is evaporating or not. We look at a snapshot in time t = 0. the graph represents the situation at t=0 so nothing can be "changing or evaporating".

Quote Quote by PeterDonis View Post
Nope. In fact, this statement doesn't even follow from the previous incorrect ones. The previous ones were about information traveling from P1 or P2 to O. This one is about information traveling from the EH (or what *was* the EH before the BH lost some mass by evaporation) to P1 or P2. I believe this confusion may be the source of your misunderstanding.
And at t = 1 this information about black hole losing mass (or ANY information, for the matter from that direction) has never reached either of the photons originating from either P1 or P2 because they are moving away form the information at the speed of light.

Quote Quote by PeterDonis View Post
Look at that parenthetical statement I made just above. When the BH evaporates a little bit, it loses a little bit of mass.
That means the location of the EH *changes*; the point that was just at the EH before is now slightly above it.
And how does this relate to the thought experiment? My point is that such change can not rech P1 or P2. Why do you think this info can reach them FTL?
Quote Quote by PeterDonis View Post
And *that* means that light at the original EH, which previously could not escape outward, now *can*.
And this has nothing to do with the argument I am making. Once you can show to me why information travels faster than C (or why this is not really any violation) we can debate what happens in this case.
Quote Quote by PeterDonis View Post
Now carry this same reasoning outward to P1: outgoing light from P1 now escapes outward a little *faster* than it did before, because it's a little farther from the EH. Similarly for P2. And as the BH continues to evaporate, outgoing light from P1 and P2 continues to be able to escape faster. The ultimate result is that, unlike the case of the "eternal" BH, there is a finite limit to how long it can take for information to get out to O from any point above the EH.
And that is the part that I am trying to understand. How can P1 and P2 get this information?
PeterDonis
#28
Jan19-12, 02:09 PM
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Quote Quote by kamenjar View Post
It doesn't matter if it is evaporating or not. We look at a snapshot in time t = 0. the graph represents the situation at t=0 so nothing can be "changing or evaporating".
Your picture is a "snapshot", but your claim that there must be a P2 from which light takes 10 billion years to reach O is not just a claim about the snapshot; it is a claim about the entire spacetime to the future of the snapshot, and for that it does make a difference whether or not the hole is evaporating. See below.

Quote Quote by kamenjar View Post
And at t = 1 this information about black hole losing mass (or ANY information, for the matter from that direction) has never reached either of the photons originating from either P1 or P2 because they are moving away form the information at the speed of light.
Ok, now I see better what is bothering you. Let me take a step back.

First, remember that the term "horizon" can actually be defined in two different ways, one local and one global. The local definition is called the "apparent horizon": it is a point at which outgoing light rays just stay at the same radius. (This is also called a "marginally trapped surface".) The global definition is called the "absolute horizon": it is the boundary of the region of spacetime (if there is one) that cannot send light signals to "infinity" (more precisely, to "future null infinity").

For an "eternal" black hole, which is static and has an unchanging mass, the two horizons coincide: the apparent horizon, where outgoing light rays stay in the same place forever, is also the absolute horizon, the boundary of the region that can't send light signals to infinity. However, as soon as we introduce any kind of change in mass, the two no longer coincide. That means we have to be careful, in any reasoning that includes horizons, to specify which kind of horizon we are talking about.

Second, remember that what we see as the "gravity" of the black hole, which causes the tilting of the light cones, the delaying of outgoing light rays, etc., does *not* "propagate" from the hole. This is why you can feel the hole's gravity even though gravity itself travels at the speed of light (meaning gravity can't "escape" from within the horizon, any more than any other signal can). There was a fairly recent thread about this, which I can't find right now, but in the course of it I linked to this in the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physic...k_gravity.html

The key point is that the field outside the hole, for a real black hole that is formed by a collapsing star or other object, is "propagated" from the collapsing object, *not* from the hole that it forms. So the reason you can feel the hole's gravity is that the collapsing object, some time in the past, left an "imprint" on the spacetime at the radius where you are, which you feel as the hole's gravity.

Third, what actually causes black hole evaporation? The usual explanation involves virtual particle pairs popping out of the quantum vacuum at the horizon, but not annihilating each other a very short time later, as they do in empty space. Instead, one particle goes down the hole, while the other escapes outward to infinity. Since the virtual pair came from the vacuum, its net energy has to be zero; so one particle has to have negative energy and the other positive. It turns out that the only way the process can work is for the negative energy particle to go down the hole, meaning that the hole loses a tiny bit of mass, which is carried away by the positive energy particle.

But if you look at the actual theoretical derivation of black hole evaporation, it turns out that you can't just look at the quantum fields at the horizon. You have to look at field modes propagating *into* the hole from the past; more precisely, propagating in from past null infinity. (I think I've got this right--QM experts, please correct me if needed.) But that means that information about how the hole is going to evaporate is, in principle, predictable from the quantum field modes propagating into the hole in the past--just as the hole's gravity is predictable from the collapsing object that formed it.

Putting all of the above together, we get the following: the future worldlines of photons emitted from P1 and P2, at the instant of your "snapshot", are indeed *not* affected by any signal that starts propagating from EH at the instant of your snapshot (because, as you say, such a signal would have to travel faster than light to catch up). However, that does *not* mean the future worldlines of those photons must be the same as they would be if the black hole were eternal with the mass that it has at the instant of your "snapshot". The fact that the black hole is not eternal is a global fact about the spacetime as a whole, *not* a fact about the local region around EH in the vicinity of your snapshot. And the paths of photons emitted outward from P1 and P2 are determined by information about curvature that propagates from the entire past of your "snapshot", not just from EH.

To see how this works, consider a slightly simpler case: the point EH itself. At the instant of your "snapshot", the point EH is an apparent horizon; outgoing light rays "hover" at the same radius. But an instant later, a point at that radius will no longer be an apparent horizon (since the hole has lost a little mass, so the apparent horizon has moved inward). So an outgoing light ray emitted from EH, at the instant of your snapshot, will in fact escape to infinity! In other words, EH marks an apparent horizon, but *not* an absolute horizon. The absolute horizon at the instant of the "snapshot" is at some smaller radius. But if an outgoing light ray from EH does escape to infinity, then obviously it takes a shorter time than such a ray would have taken if the black hole were eternal (since in that case EH would be an absolute horizon and outgoing light would take an infinite time to escape). But by your reasoning, this could not happen, because the very information that light from EH can actually escape can't propagate outward any faster than the light from EH itself--for example, the light cone at the same radius as EH, but slightly to the future of your "snapshot", must already "know", by the time the light ray from EH gets there, that it is not a trapped surface so that light ray can escape outward a bit. The only way it can "know" that is for the information to already have reached it from the entire past of the "snapshot", not outward from the hole.

So I believe the answer to your question is that the information about how the spacetime to the future of your "snapshot" is going to change, including how the light cones are going to tilt less, light is going to escape faster, and ultimately there is going to be a finite limit on how long it will take light signals to reach O from P1 or P2, is already encoded in the snapshot itself. That information doesn't have to "propagate" outward from the hole, any more than the hole's gravity does; it is already there. So nothing has to "catch up" to photons emitted from P1 or P2.
phinds
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Jan19-12, 02:36 PM
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Gads, that is weird. I'm not suggesting it's WRONG ... I'm just saying it's another one of those many things where my "common sense" makes no sense and makes me want to take up drinking again.
utesfan100
#30
Jan19-12, 03:26 PM
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1) I am confused by the discussion above regarding the distance to the black hole. While the metric indicates the stretching ratio goes to infinity at the horizon, the actual distance integrated over the metric remains finite. Thus a waveform leaving the horizon at velocity c should reach an observer in finite time.

2) Am I to take it that the model of evaporation is an instantaneous, uniform, symmetric burst of radiation traveling at a local speed of c? An object outside the waveform would then see no difference in gravity as, even in retrospect, the observer would observe the gravity from the energy of the radiation that would be the same as the mass of the original source.

Once the wave form hit an observer, they would be instantaneously bathed in the emitted radiation. After this, the curvature from the black hole would be 0, just as there is no field inside a hollow sphere in Newtonian physics.
PeterDonis
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Jan19-12, 05:43 PM
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Quote Quote by utesfan100 View Post
While the metric indicates the stretching ratio goes to infinity at the horizon, the actual distance integrated over the metric remains finite. Thus a waveform leaving the horizon at velocity c should reach an observer in finite time.
For a static black hole that doesn't evaporate, a light signal emitted from anywhere above the horizon (not *at* it) will reach an observer anywhere outside the horizon in some finite time. However, the finite time can be arbitrarily large; it can be made as large as you like by putting the emitter of the waveform closer and closer to the horizon. This is not a matter of the distance from the emitter to the observer; it's a matter of the light cones being tilted inward, so that the light takes longer to cover the distance than it would if the light cones were not tilted. One could say that the light was "moving slower than c", but I don't like that terminology because it invites a lot of confusion about whether GR is consistent with SR; in fact, the rule that "light always moves at c" simply can't be applied as-is in a curved spacetime, because there is no way to uniquely define "relative speed" for spatially separated objects. The rule in GR becomes "light always moves along the light cones", which has a unique, unambiguous definition.

For an evaporating hole, the location of the horizon changes, and the spacetime as a whole is not static, so things get more complicated; but the upshot, as I said in previous posts, is that you can no longer make the finite time arbitrarily large.

Quote Quote by utesfan100 View Post
2) Am I to take it that the model of evaporation is an instantaneous, uniform, symmetric burst of radiation traveling at a local speed of c?
No. Black hole evaporation is a continuous process. The hole starts with some mass M, and very slowly radiates it away, continuously, until it's all gone.


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