Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Observer dependent event horizon for Schwarzschild black hole

  1. Nov 23, 2015 #1
    Hello.

    In oral exams my professor likes to ask if Alice and Bob can communicate, if Alice ist just above the event horizon of a schwarzschild black hole and Bob ist just below.

    He wants to hear:
    Communication is possible, because the event horizon is observer dependent. Only an observer resting at infinity distance couldn't see something cross the event horizon.

    Is this correct? I can't understand it.
    Doesn't you draw diagrams like this https://faculty.etsu.edu/gardnerr/planetarium/relat/eventho2.gif exactly to see, that this is not possible?

    I read about apparent horizons, but the lecture is the introduction to general relativity. We talk about an ideal schwarzschild black hole, not a perturbed black hole.

    Would you please help me to understand?
     
  2. jcsd
  3. Nov 23, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    A signal from Bob will never cross the event horizon. In order to communicate, Alice's motion must be such that she enters the event horizon. If she is not willing to do that, they cannot communicate. The point is that, in order not to fall inside the event horizon, Alice will need to accelerate at a high rate. Even if the space-time near the event horizon may be considered locally flat, Alice's acceleration will (locally) create a horizon which Bob will be beyond.
     
  4. Nov 23, 2015 #3

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Another way to see that your professor is wrong is to look at a Penrose diagram. I have a nontechnical introduction to Penrose diagrams in section 11.5 of my book Relativity for Poets: http://www.lightandmatter.com/poets/ .
     
  5. Nov 23, 2015 #4

    Nugatory

    User Avatar

    Staff: Mentor

    This is so basic and so uncontroversial (especially in the context of an introduction to GR) that I'd consider the possibility that you are misunderstanding your professor.
     
  6. Nov 23, 2015 #5
    I have 7 written summaries of oral exams, written by the examined students und collected by the physics students' union.
    The questions asked by the professor and the answers given by the examinees are only as accurate as remembered by the examinee and
    not written very detailed.

    The question is provided in three reports. In the oldest report, the examinee could not answer. Then the professor explains:
    "They can talk to each other because the event horizon is observer-dependent."
    In two more recent, the examinees knew this oldest report. They answered: "Because the event horizon ist observer dependent" and there were no further questions regarding this.

    I'm very sure that I am misunderstanding it. But how can it be meant?
    In https://en.wikipedia.org/wiki/Event_horizon one can find:
    "For the case of the horizon around a black hole, observers stationary with respect to a distant object will all agree on where the horizon is."
    So other observers don't agree? So the event horizon is observer dependent? Huh?

    Is there another way that it would make sense?
    Sorry, I'm very confused.
     
  7. Nov 23, 2015 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No, they all agree. Wikipedia is not a reliable source. The event horizon is a global geometric feature of spacetime for a black hole; all observers agree on where it is, just as all observers on Earth agree on where the Earth's equator is.
     
  8. Nov 24, 2015 #7

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    To try to interpret the wikipedia text, the observers may be able to communicate if the outside observer stops accelerating. However, this will inevitably lead to that observer ending up inside the event horizon, which is probably not the spirit of the question.
     
  9. Nov 24, 2015 #8
    A last question to clarify:

    Is there the same infinite redshift surface for all observers, located at the schwarzschild radius?
    What an observer far away sees is objects that are falling in getting closer and closer to the schwarzschild radius, without ever reaching it, right?
    So the same happens for all observers if things cross the schwarzschild radius? An observer resting one metre above the schwarzschild radius and an observer falling into the black hole but above the schwarzschild radius both see objects that are falling in getting closer and closer to the schwarzschild radius, without ever reaching it?
     
  10. Nov 24, 2015 #9

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    If this is all the information we have, then it is ambiguous.

    If Alice is accelerating upwards and never falls below the event horizon, then two-way communication is not possible, because Bob's messages can never reach Alice while she stays outside the event horizon.

    However, if Alice and Bob are both in freefall radially downwards, then (if they are close enough to each other) it could be possible for them to be in continuous two-way communication with each other. For the short period that Alice is above the event horizon and Bob is below (according to any simultaneity convention), Alice is receiving a message that Bob sent while he was still above the event horizon, while the message that Bob is sending from below the event horizon will be received by Alice only after she has fallen below the event horizon.
     
  11. Nov 24, 2015 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Actually it looks like several questions. :wink:

    Yes.

    Yes, at least in principle. The faraway observer will also see the light from the infalling objects getting more and more redshifted; in practical terms, the light will become unobservable (because it is redshifted to a wavelength that is not detectable by any real device) fairly quickly. So what the faraway observer will actually see, in practice, is infalling objects appearing to get redder and fainter and then disappear, fairly quickly.

    No. The observer who falls in will see objects falling in cross the horizon, because he will cross the horizon himself.
     
  12. Nov 24, 2015 #11
    Pfff, same order of magnitude...


    Okay, that was misleading. The infalling observer was thought to use his rocket to stop before crossing the schwarzschild-radius.

    I think both observers are in freefall radially downward.

    But was does "the event horizon is observer dependent" mean for freely falling observers?
    How would you calculate this "horizon"?
     
    Last edited: Nov 24, 2015
  13. Nov 24, 2015 #12

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think it's just the boundary of the causal past of that observer's world-line. That is, take W to be the set of events that constitute the world-line. Let ##J^-(W)## be the set of all points P such that a timelike or null curve exists from P to a point on W. Then I think the boundary of ##J^-(W)## is what we mean by that observer's horizon.
     
  14. Nov 24, 2015 #13

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The horizon itself is not observer-dependent; but when each observer crosses the horizon is. If Alice is above Bob, she will cross the horizon after him.
     
  15. Nov 24, 2015 #14

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's pretty standard to say that event horizons are observer-dependent. When we say "the" horizon of a black hole, we just mean the horizon as seen by an observer at null infinity.
     
  16. Nov 25, 2015 #15

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    It is? This seems very surprising to me, since the event horizon, as I said before, is a global geometric feature of the spacetime; it's a specific null 3-surface, and which null 3-surface it is is the same for all observers. Can you give examples of references that say this? I don't remember MTW or Wald, for example, ever saying anything like this, and IIRC they say the opposite.

    "Horizon" can also mean "apparent horizon", and apparent horizons are observer-dependent. But event horizons are not, unless I am drastically misreading sources like MTW.
     
  17. Nov 25, 2015 #16

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Off the top of my head, there's Smolin's pop-sci book Three Roads to Quantum Gravity. Also, doesn't every standard treatment of the Rindler coordinates and accelerated observers say the same thing?

    The global geometric feature you're talking about can be described without using the word "observer," but as I pointed out earlier, you can interpret it in terms of an observer at null infinity.

    Maybe we're just differing in terms of how inclusive our terms are. Is there a standard usage according to which "event horizons" are a proper subset of "horizons?"
     
  18. Nov 25, 2015 #17

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No treatment that I'm aware of says that an event horizon is observer-dependent. Lots of them say a Rindler horizon is observer-dependent, but a Rindler horizon is not an event horizon.

    The term "event horizon" has a specific meaning: it is the boundary of the region of spacetime that is not in the causal past of future null infinity. (Wald, IIRC, is the clearest of the major GR textbooks on this definition--see section 12.1.) This definition is obviously not observer-dependent: such a region of spacetime, and its boundary, if it exists in a spacetime at all, is a global geometric feature of that spacetime.

    The term "apparent horizon" means a 3-surface composed of a continuous set of trapped 2-surfaces. A trapped 2-surface is a 2-surface for which the congruence of outgoing null normals has zero expansion. This definition is observer-dependent because which congruence of null curves is the congruence of "null normals" to a 2-surface is observer-dependent: different observers in different states of motion passing through the same 2-surface will pick out different congruences of null curves as the "null normals" to the surface.

    A Rindler horizon is obviously observer-dependent, since it is just the asymptotic surface to the hyperbolic worldline of an observer with constant proper acceleration.

    Those are the only types of "horizons" that I'm aware of that are relevant for GR.
     
  19. Nov 25, 2015 #18

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    I'm not sure this is really a valid interpretation. Future null infinity is a null surface, so, strictly speaking, an "observer", who has to travel on a timelike worldline, can't have future null infinity as his worldline. Even if we view an "observer at infinity" as strictly the result of an idealized limiting process (since future null infinity isn't actually part of spacetime), AFAIK future null infinity isn't the limit of any set of observer worldlines; for example, in Schwarzschild spacetime, it isn't the limit of the worldlines of observers at radius ##r##, as ##r \rightarrow \infty##.
     
  20. Nov 26, 2015 #19

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    @PeterDonis -- Seems like we've gone around a couple of times on this and said what we had to say. In general I think I'm arguing that certain terms and concepts are inherently elastic, whereas you're taking the point of view that they can and should be embodied in a standard and precise way in mathematical definitions, and that the versions you propose are the only right ones.
     
    Last edited: Nov 26, 2015
  21. Nov 26, 2015 #20

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    I do think precise terminology is important; but more than that, in this particular thread, if the term "event horizon" is not being used as I've defined it, then I'm completely confused about what the professor described in the OP is thinking. What does he mean by the term, if not what I defined? In other words, I'm unable to find an interpretation of what's described in the OP under which the professor is saying something true, and is using the term "event horizon" in some way other than what I defined; the only interpretation I can come up with is that the professor is using the term "event horizon" to mean what I defined, but he's confused or misinformed about what its properties are.

    Also, reading back, I see this in your post #12:

    This is a reasonable definition of "an observer's horizon", yes. But note that the horizon defined this way can be observer-dependent in some spacetimes, but not in others, for a similar class of observers. For example, if we consider a class of observers who are (a) accelerated, and (b) following orbits of a timelike Killing vector field of the spacetime they are in, then in Minkowski spacetime, these observers are Rindler observers and their horizons, defined as you define them, are observer-dependent; but in Schwarzschild spacetime, these observers are "hovering" at some finite altitude above the event horizon, and the horizon of all such observers, defined as you define it, is the event horizon--i.e., the same surface, not observer-dependent.

    But again, if we consider a different class of observers in Schwarzschild spacetime, such as the class of observers who free-fall into the hole, then the horizons of these observers are observer-dependent; different observers falling in at different times will have different horizons. And any inertial observer in Minkowski spacetime has no horizon in this sense; the union of the causal past of all events on the observer's worldline is the entire spacetime. So I'm not disputing that there are legitimate senses of the word "horizon" in which the horizon can be observer-dependent; I just don't see any of those senses being used by the professor described in the OP. (And, as I said above, I do think it's important to clarify which sense is being used, since the word "horizon" unqualified is ambiguous.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Observer dependent event horizon for Schwarzschild black hole
Loading...