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Time dilation at the event horizon of a black hole

  1. Nov 3, 2015 #1
    Dear all,

    As far as I understand, for a distant observer, time stands still at the event horizon of a black hole, right? In particular, nothing will ever pass the EH. Instead, everything that approaches the BH will get stuck at the EH and stay there forever from the perspective of the distant observer.

    That leads me to a couple of questions:

    1. Under these conditions, are the usual depictions of a BH actually right? Should we really assume that a massive black hole must have a huge EH from the perspective of a distant observer? Or would it be possible that (from our perspective) Sagittarius A* is only a rather small BH with lots of stuff piled up at its EH. After all, the gravitational effects of the BH should not change too much if the majority of its mass is piled up at the EH, should they?

    2. The light that a distant observer receives from objects near the EH would be extremely red-shifted. Now if we change perspectives, for an observer near the EH, the rest of the observable universe would be extremely blue-shifted. Actually, the observer near the EH would be bombed with intense gamma rays from everywhere else, wouldn't he? Does he even have a chance to survive this gamma ray inferno until he gets spaghettified by the gravitation of the BH?

    3. Not only would the observer near the EH be bombed with gamma rays, he would also see the rest of the observable universe age in extremely fast motion, wouldn't he? He would probably see a firework of supernova explosions and maybe even the big freeze. Question: Can we calculate, how much time will elapse outside of the BH's gravitational field before an observer near the EH actually passes the EH?

    Sorry in advance for any misconceptions.

    Best regards,
    Robert
     
  2. jcsd
  3. Nov 3, 2015 #2

    bcrowell

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    This may be helpful: http://physics.stackexchange.com/a/146852/4552
     
  4. Nov 3, 2015 #3
    Are you implying that the curvature of spacetime is only an Illusion?

    Edit: As far as I understand, gravitational time dilation is not symmetric like the twin paradox in SRT. Time in the gravitational field "really" runs slower.

    Best regards,
    Robert
     
    Last edited: Nov 3, 2015
  5. Nov 3, 2015 #4

    PeterDonis

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    No. This has been discussed in many previous threads. For example, these:

    https://www.physicsforums.com/threads/does-an-event-horizon-ever-exist.683076/

    https://www.physicsforums.com/threads/infinite-time-dilation-at-the-surface-of-a-black-hole.718648/

    No. The correct statement is that the coordinates you are implicitly assuming for the distant observer, namely exterior Schwarzschild coordinates, do not cover the horizon and the region of spacetime inside it. Therefore, those coordinates cannot be used to make statements about what happens at or inside the horizon.

    Also, the "time" assigned by those coordinates is not a physical thing, it's just a convention. To investigate what actually happens, physically, you need to look at invariants. There are no invariants which say that things falling into the BH will get stuck at the EH. In fact, the invariants say the opposite: that things fall right through the horizon, and onward to the singularity.

    The area of a black hole's horizon is an invariant; it doesn't depend on the observer, or on which coordinates you adopt.

    Yes, they would, because for mass to "pile up" near the EH, it would have to be highly accelerated, not in free fall, and the additional energy that produced that acceleration would greatly increase the externally measured mass of the hole. Free-falling objects, as above, don't pile up at the horizon; they fall right through.

    If he were highly accelerated so that he was "hovering" above the horizon but very near it, yes, he would. But not if he were free-falling into the hole; in that case he would see incoming radiation as redshifted, not blueshifted, and he would hit the singularity and be destroyed before he had seen very much at all.

    No, he's not. He's just implying that the curvature of spacetime does not work the way you were thinking it works.
     
  6. Nov 3, 2015 #5
    Does not fit, because it refers to what happens "inside a black hole" [sic], whereas I made very clear that I am referring to what happens outside of the EH.

    What are you referring to? I never made any statements about what happens inside the EH.

    Now, this is very interesting, because it is something that I really might have misunderstood so far.

    So you are saying that the gravitational time dilation only occurs when I leave the geodetic lines of curved spacetime, but as long as I am traveling along a geodetic line, no gravitational time dilation will occur even if I am near the EH of a black hole?

    Again, very interesting. I guess the redshift would be a result of the increasing relative speed to the light sources, right?

    So if he's hovering above the EH, blueshift occurs. If he falls freely toward the EH, redshoft occurs. And I guess there would also be something in between where no frequency shift occurs?

    Just to clarify the terminology: As long as I am travelling on a geodetic line, I am *not* accelerating even though my relative speed increases?

    Something else: When you say that he would hit the singularity, then *you* are making assumptions about what happens inside the EH, don't you?

    Have you read the post he was referring to?

    Best regards,
    Robert
     
  7. Nov 3, 2015 #6

    PeterDonis

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    But your incorrect understanding of what happens outside the EH depends on your incorrect belief that things "pile up" at the EH instead of falling through. The only way to correct that belief is for you to correct your understanding of what happens at and inside the EH, so that you realize that things that fall into the hole do not, in fact, "pile up" at the EH--they fall through it.

    I'm saying something more drastic than that. I'm saying that the concept of "gravitational time dilation" doesn't even make sense for someone free-falling into the hole. There is no invariant way to compare their "rate of time flow" with that of a distant observer. That can only be done for someone who is "hovering" above the horizon (and therefore, as you say, traveling on a non-geodesic path), so that they are at rest relative to the distant observer.

    That's a heuristic way of looking at it, but you have to be careful with it. The concept of "relative speed" between spatially separated objects is not really well-defined in a curved spacetime. In this case, it happens to give the right qualitative answer, but that doesn't always happen.

    There would be some such in between state of motion, yes. It would be accelerated, but not enough to "hover", so someone in this state of motion would fall through the horizon. I think they would have to accelerate harder and harder as they fell to keep the observed frequency shift zero, but I have not done the calculation.

    In GR, the term "acceleration" unqualified normally means proper acceleration, i.e., what is measured by an accelerometer. If you are in free fall, you have zero proper acceleration.

    As I said above, the concept of "relative speed" between spatially separated objects doesn't really make sense in a curved spacetime.

    No, I'm not making "assumptions". I'm telling you what GR predicts. GR is the only confirmed theory we have that covers this regime. There are speculations about various ways that quantum gravity effects might change things from what GR predicts, but that's all they are, speculations.
     
  8. Nov 4, 2015 #7
    Hi Peter,

    First of all thank you very much for your enlightening reply.

    If things "pile up" from a distant oberver's perspective, that does not necessarily rule out that they can be "falling through" from their own perspective, does it? My belief (which might certainly be wrong) was that they fall through from their own perspective, but from the distant observer's persepctive they pile up like slow-moving traffic in front of a bottleneck (with the subtile difference that it's not the cars themselves that are slowing down, but rather the "rate of time flow").

    Two questions:

    1. Why does the comparison have to be "invariant" when I am referring to the distant observer's perspective?

    2. Under the assumption that the comparison really would have to be invariant:
    If there is no invariant comparison, then how can a distant oberserver ever say with any certainty that an object has already passed the EH? If their rate of time flow cannot be compared at all, then the statement that something is stuck at the EH (from the perspective of the distant observer) is not more true or false than the statement that it has fallen through, isn't it?

    O.k., I am fine with that. So unless specified otherwise zero accelaration means free fall.

    Spatially separated? Just for clarification: As long as an object is located *outside* of the EH, there can be distant observers that are *not* spatially separated from the object. Do you agree?

    Best regards,
    Robert
     
  9. Nov 4, 2015 #8

    Nugatory

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    You might want to give this paper a quick read: http://arxiv.org/abs/0804.3619
     
  10. Nov 4, 2015 #9

    PeterDonis

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    They don't. Light emitted by infalling objects takes longer and longer to get back out to the distant observer; but that is an effect of light propagation, and the distant observer has to correct for it if he really wants to know what is happening to the infalling objects. Once the distant observer corrects for that effect, he realizes that infalling objects are not "piling up".

    All of the actual physics is contained in invariants, so if there is no invariant comparison, then you're not talking about actual physics, you're talking about somebody's choice of labels that has no physical meaning.

    He can't, if you're looking for certainty, because the word "already" implies an invariant comparison, and there isn't one.

    However, there are other invariants that are still meaningful. Here is one: there will be some definite time, on the distant observer's clock, at which, if he emits a light signal going inward, towards the infalling object, it will not reach the infalling object until after the infalling object has crossed the horizon--where "after" here is to be taken in the geometric sense, i.e., the ingoing light ray will intersect the infalling object's worldline at a point inside the horizon. That being the case, it is perfectly reasonable for the distant observer to say, after that definite time on his clock, that the infalling object has "already" crossed the horizon. But this "already" is still an interpretation: the only actual physics is what I just described about the behavior of light rays and where they intersect the infalling object's worldline.

    There will also be a somewhat later time, on the distant observer's clock, at which, if he emits a light signal going inward, it won't be able to reach the infalling object's worldline at all--it will hit the singularity instead. So at any time after that time, it is perfectly reasonable for the distant observer to say that the infalling object has "already" hit the singularity. But again, that "already" is an interpretation; the actual physics is just what I described about the behavior of light rays and which ones can and can't reach the infalling object's worldline. instead of hitting the singularity.

    No, because "stuck" implies that the object is "at rest" at the EH, and nothing can be at rest at the EH. Not being able to make invariant comparisons of rate of time flow does not mean you can say anything you like; it just means you have to look at other invariants if you want to understand the physics.

    If the observers are "distant", that implies that they are not near the horizon, so they would be spatially separated from any object that was near the horizon. The only way for two objects to not be spatially separated is for them to be right next to each other--to be, from the standpoint of geometry, at the same point in spacetime.
     
  11. Nov 5, 2015 #10
    O.k., so the observer's observation differs from what really happens?

    But the observer cannot specify when the light signal will actually reach the EH respectively the infalling object?

    Even if the observer does not see what really happens, describing what the observer sees is not the same thing as saying anything I like. Or do you imply that the observer's observations are arbitrary?

    [/QUOTE]

    O.k., I guess I made a translation error here.

    Best regards,
    Robert
     
  12. Nov 5, 2015 #11

    PeterDonis

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    The observer's observation is time delayed, because the light has to travel from where it was emitted to him. This is always true; light travels at a finite speed everywhere, not just when climbing away from a black hole. It just happens that, in the case of a black hole, the curvature of spacetime around the hole means the light takes a very long time to get out, so the time delay is much larger than we intuitively expect based on our experience in very weak gravity fields.

    Why not? He controls when he emits the signal, and he knows the spacetime geometry around the hole, so he knows exactly what worldline the light will follow, and he knows what worldline the infalling object will follow. So computing when he has to emit the signal to reach the object at any specified point on its worldline is just a matter of straightforward geometry.

    When you use the word "when", or "already", or anything like that, referring to events that aren't on the observer's worldline, you aren't describing what the observer sees. You're describing an abstract construction that the observer makes--an assignment of coordinates to events.
     
  13. Nov 5, 2015 #12
    O.k., but then he could at least say with certainty when the infalling object has *reached* (although not passed) the EH, couldn't he?

    But I did not use the words "when" or "already" in the statement that you were originally referring to. I rather said that the infalling objects get stuck at the EH from the observer's perspective which is a pretty close description of what the oberver would see, isn't it? So even if it is only an illusion, it is not just an arbitrary statement.

    The "when" and "already" came later into the game, which statements are incorrect from your point of view and which not.[/QUOTE]
     
  14. Nov 5, 2015 #13

    PeterDonis

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    If he can say with certainty when the object has reached the EH, he can say with certainty when the object has passed it, since the one implies the other.

    But, as I said before, being able to "say with certainty" anything about "when" events happen that are not on the distant observer's worldline, according to the distant observer's clock, requires adopting a convention for matching up distant events with times on his clock. There is no invariant fact of the matter about such things; it's just a convention.

    However, there is an invariant fact of the matter about when a light signal sent by the distant observer will reach the infalling object, according to the infalling object's clock. Computing that is what I referred to in my previous post as a matter of simple geometry.

    I don't think so, because the term "get stuck at the EH from the observer's perspective" does not imply that it's an optical effect which the observer has to correct for. It implies that the observer has a "perspective" from which he can say what is happening, and what is happening from that "perspective" is that objects are getting stuck at the EH.

    Also, "getting stuck at the EH" isn't really a good description of what the observer would actually see, because it ignores the fact that the light signals coming from the infalling object would be more and more redshifted as well as taking longer and longer to get to the distant observer. Since the observer can't detect light of arbitrarily long wavelength, the light signals would eventually become undetectable; and "eventually" is actually fairly quick for a black hole of reasonable size. So a more apt description of what the observer actually sees is that the light from the infalling object would appear fainter and fainter and then vanish. How much of an apparent "slowing down" of the object's motion the distant observer would see would depend on how long a light wavelength he could detect; for any reasonable detector with current technology, the light would become undetectable well before any significant apparent slowdown was seen.
     
  15. Nov 6, 2015 #14
    O.k, but then the observer can only say *that* the infalling object will eventually cross the EH, but he cannot specify *when*. Which might be perfectly o.k.

    But how can we say that a black hole has a certain mass then? Because that statement that it *has* a certain mass implies that this mass has already passed the EH, doesn't it?

    But isn't perspective always something that we have to correct for? If we say "from A's perspective it is like this", doesn't that already imply it might be different from somone else's perspective?

    O.k. This is a good point. If the observer does not even see something getting stuck but he only sees the infalling objects fainting away, then of course my description was simply wrong.
     
  16. Nov 6, 2015 #15

    PeterDonis

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    He can't specify "when" in any absolute sense, yes.

    No. The mass of the hole is a property of the spacetime around it, and it can be measured without even having to know what has or hasn't crossed the EH. You just put a test object into orbit about the hole and measure its orbital parameters.

    Also, an infalling object does not have to actually cross the EH for its mass to be observed as part of the mass of the hole. It only has to be closer to the EH than the test object whose orbital parameters we are using to measure the hole's mass. So, for example, if the test object is a million kilometers from a black hole of a few solar masses (and whose horizon radius is therefore a few kilometers), and a large asteroid falls into the hole, the asteroid's mass will affect the test object's orbit as soon as it is closer to the hole than a million kilometers. If we imagine the asteroid falling past the test object, the test object's orbit will change as soon as it falls past, and the mass that we infer from the orbit will be the increased mass (including the asteroid's mass) from that point on.
     
  17. Nov 6, 2015 #16
    O.k., I initially thought so, but when I wrote a few posts ago that it wouldn't change the gravitational effect of the BH if the objects got stuck at the EH, you disagreed. But apparently you disagreed only with respect to the "got stuck", because in your understanding that would mean that the objects are hovering above the EH which implies they are accelerating away from the BH. My originally thought that it does not make a difference with respect to the gravitational effects if the infalling objects have already crossed the EH or not, seems to be correct however.

    But then again, this only seems to be valid if measure the mass of a BH based on its gravitational effects. However, doesn't this mean that the mass of the BH depends on the test object that we use to calculate its mass? So if we select a test object that is farer apart from the BH, we will calculate a higher mass than if we select a test object that is closer to the BH?
     
  18. Nov 6, 2015 #17

    PeterDonis

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    That's because we were talking about a different scenario. For objects to be "stuck" at the EH, as opposed to falling in, they have to be highly accelerated, and that acceleration requires a lot of extra energy, which would appear to a distant observer as the black hole having additional mass, over and above the mass of the "stuck" objects.

    Here, we are talking about an object free-falling into the hole, and when its mass can be viewed as part of the mass of the hole by a distant observer. There's no acceleration involved and therefore no extra energy, just the mass of the infalling object.

    No. The orbital parameters of an object in a free-fall orbit are independent of its mass. That is one way in which GR is the same as Newtonian gravity. (The only caveat is that the object has to be small enough that its own effect as a "source" of gravity is negligible. The term "test object" is used to denote this condition.)

    No. Consider calculating the mass of the Sun by measuring the orbital parameters of the planets. They are all at very different distances from the Sun, but the mass we get for the Sun by measuring their orbital parameters is the same in all cases.

    What will be different if we take test objects at different distances is "when" the change in the black hole's mass will be observed. But let's now rephrase that without the word "when". What we have, heuristically, is a spacetime geometry with two regions: a region where the black hole has mass M, and a region where the black hole has mass M + m, where m is the mass of the infalling asteroid. (There is also, strictly speaking, a "transition" region between these two regions, where the hole's mass increases from M to M + m, but we can assume that region is very thin so it can be treated as just a boundary surface between the two other regions.)

    The two test objects each have worldlines that cross from one region to the other, and the points at which they cross are the points at which the test objects' orbits change. How we assign a "time" to those orbital changes, and therefore to the change in the mass of the hole as seen by each test object, depends on what convention for "time" we adopt. But the geometric facts, that the worldlines each cross the boundary at particular (different) points, are the same regardless. And the fact that the two crossing points are different for the two test objects is really what we are saying when we say that the two objects "see" the mass of the hole increase at different "times".
     
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