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

I The Shapiro delay and falling into a black hole

  1. Feb 25, 2017 #1
    I realise that this question has been asked many times on this forum, however, I have yet to come across a satisfying/understandable answer that takes into account gravitational time dilation.

    The speed of light inside a gravitational field is slowed down relative to a distant observer by gravitational time dilation as proven by the Shapiro delay experiment. Therefore the speed of any object is slowed down in a gravitational field as seen by a distant observer. The speed of an object/light is related to the strength of the gravitational field.

    How can anything 'fall' through the event horizon of a black hole within the time span of the entire future of the universe if its speed asymptotically approaches zero as it approaches the event horizon.

    In the proper time of an object falling into a black hole, the object will cross the event horizon in a finite amount of time. For a distant observer, however, the time it takes is infinite. How does one reconcile this paradox?
    Let us also ignore the effect of redshift by presuposing that the distant observer has access to infinitely sensitive observational equipment.

    Logically I know I must be misunderstanding something fundamental as general relativity has been studied by very intelligent physicists for decades and this has not flagged up as an unresolvable paradox.

    Having said all of that, black holes clearly do grow as they come in different sizes. Somehow matter must pass through the event horizon. However, I don't think this necessarily disproves my reasoning.
    If the event horizon of a black hole is determined by the mass / energy within. Is it not the event horizon which grows to swallow/reestablish itself above the approaching object rather than the object travelling through the event horizon?
  2. jcsd
  3. Feb 25, 2017 #2


    User Avatar
    2017 Award

    Staff: Mentor

    Give up the idea of a global reference frame that would give useful results.
  4. Feb 25, 2017 #3
    Imagine that the Alcubierre drive were possible and we could travel through space faster than light. We could theoretically continuously travel between both the reference frames of the in falling object and the distant observer almost instantaneously. Assume that this third person inside the Alcubierre drive spaceship is also in a distant reference frame - the global reference frame.
    Because both frames of reference can now be accessed virtually simultaneously, they must reconcile. I don't understand how giving up a global reference frame helps in any way.
  5. Feb 25, 2017 #4


    User Avatar
    2017 Award

    Staff: Mentor

    Then we can also travel backwards in time and break causality, and not even the question "what caused what" works any more. It will depend on how you travel.
  6. Feb 25, 2017 #5
    I'm not an expert on the Alcubierre drive but I believe all the problems you mentioned have been previously debunked on this very forum. Regardless, I only introduced the Alcubierre drive as a thought experiment. Theoretically the distant observer could also quantum tunnel to and from the in falling object extremely rapidly, again requiring both reference frames to reconcile.
  7. Feb 25, 2017 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    The premise itself is a bit iffy. Unfortunately, even perfectly sound logic will give you misleading results if your premises are wrong.

    There are some foundational issues about the speed of a distant object even being definable. See for instance Baez's remarks in "The Meaning of Einstein's equation" <<link>>.

    I have no idea of your background, but I will mention that I have a certain learned caution for assuming that posters on PF are really familiar with special relativity (SR) and not just jumping into general relativity (GR) without a good understanding of SR. But that's not the main point, so I'll skip on to what is.

    Unfortunately, if you are not already familiar with parallel transport, this explanation may be difficult to follow. And I'm not sure if I have any good non-technical references on the topic. Perhaps another poster can suggest some.

    If I feel ambitious, I may post a bit on parallel transport, but in a separate thread - it would be too derailing to post it to this thread, I think, it's really a separate topic. The point I wanted to mention was that it is dangerous to assume that distant objects have a well-defined velocity as a premise for an argument. So my response would be that even if your logic is sound, there are some issues with your starting premise.

    I'll add something that I hope may be more helpful. It is not necessary to assume that the speed of light is slowed down at all by gravity. In fact, if you look at the purely local speeds, the local speed of light is not affected at all, it is in fact a premise of both SR and GR that the local speed of light is constant.

    So it is not necessary (and in my opinion not desirable) to formulate GR in the manner you have in your premise. It's better (and also more standard) to formulate GR as a geometric theory in which the local speed of light is always constant and equal to "c". By not introducing the concept of global speeds, but restricting ourself to local speeds, we can avoid the pitfalls Baez mentions.

    The standard approach to GR and the complexities introduced by curvature is similar to how we handle the problems of making a map of the curved surface of the Earth. No map drawn on a flat sheet of paper of the globe can be perfectly to scale, but if we consider a small enough area, we can draw an approximate map that's good enough to cover some local region. So we can utilize a street atlas with a lot of flat maps, to plan a route in a city, without learning about spherical geometry. If we wanted to plan an ocean voyage, though, we would probably have to make the effort to learn a bit about spherical geometry first.

    Similar remarks can be made about GR. If we look at a small enough region, we can avoid having to learn about Riemannian geometry.

    It's really, really helpful if one appreciates the geometric formulation of SR first, before one tackled GR. In the geometric formulation of SR, we view space-time as having a unified geometry, a non-Euclidean geometry, called a Lorentian geometry. So studying how to go from Euclidean geometry to Lorentzian geometry is a way to learn SR that is a good step for learning GR. The next larger step goes from Lorentzian geometries (which are flat) to Riemannian and pseudo-Riemannian geometries, which are in general not flat.
  8. Feb 25, 2017 #7


    User Avatar
    2017 Award

    Staff: Mentor

    It cannot. Quantum tunneling cannot lead to superluminal information transfer. Quantum field theory is local, and while some interpretations of it are not local, none of them can allow superluminal information transfer (because that is an interpretation-independent process).

    pervect explained the problem with the global reference frame nicely.
  9. Feb 25, 2017 #8
    That may be so. I'm not very familiar with quantum tunnelling and was probably a little too hasty in my reply. Even so, as I understand it, quantum tunnelling is instantaneous and there is no upper limit on the distance that can be traversed. What fundamentally stops an entire person from tunnelling immense distances fully intact?

    I'm a second year physics undergrad. We have only lightly covered special relativity but I'll try my best.
    Thank you for the link, it's an interesting read. But your right about it being confusing for someone who hasn't heard about parallel transport before. I am confused.
    As far as I understand, each vector makes sense in describing the velocity of a particle in the context of its own local region of space. To compare vectors you first need to convert one of the vectors from one language to another(one context to another). To do this you parallel transport which can be ambigous because the conversion you get depends on the method you used(path you took).

    What I don't understand is why there exists a need for a conversion in the first place. Isn't the whole point of a vector to convey all the necessary information about an object simply, by using a standardised coordinate system?
    Using your analogy of the map of the earth: You wouldn't need to parallel transport across the surface of the map to map out an intercontinental journey if your coordinate system was in a higher dimension.
    Equivalently, using the analogy of the ball in the link you posted: Parallel transport wouldn't be needed if you had a 3 dimensional coordinate system to describe the curvature of the 2 dimensional surface of the ball.

    So I guess what I am trying to say is, could general relativity be made less ambiguous by introducing a higher reference frame? A reference frame of some sort beyond 4D space time.

    I'm not sure I understand what you mean by 'it is not necessary to assume the speed of light slows down'. Do you mean necessary to show that the in falling object crosses the event horizon? I know it does, I said as much my self. But by sticking with local reference frames are we not just choosing to ignore the perspective of the outside observer and the resulting inconsistency in events?
    Just because introducing a global reference frame complicates things doesn't make unworthy of investigating

    Thank you both for your input thus far.
  10. Feb 26, 2017 #9


    Staff: Mentor

    That's because gravitational time dilation, as a concept, only applies to static observers (observers who are at a constant altitude above the horizon), and there are no static observers at or below the horizon. So you are trying to apply a concept that simply has no meaning in the domain in which you are trying to apply it.

    Your understanding is incorrect. Where did you get this from?

    Yes, but with a key additional aspect that you did not state: a vector only describes something in a particular small neighborhood of spacetime. A vector is "attached" to a particular event in spacetime (a point in space at an instant of time), and only has meaning in the immediate neighborhood of that event (i.e., at points/times infinitesimally close to the event the vector is attached to). Having a "standardised coordinate system" does not change that--all a coordinate system does is attach an agreed upon set of 4 numbers to every event, so we can unambiguously refer to events and know which ones we are talking about. It doesn't let you compare vectors attached to events that are separated from each other.

    It's not a matter of the coordinate system being in a higher dimension: it's a matter of the Earth being embedded in a higher dimensional space which happens to be flat (at least in the Earth's vicinity, to a good enough approximation). So in that higher dimensional space, parallel transport is unambiguous (since the space is flat), and you can compare vectors at widely separated points. But that's not because parallel transport isn't required; it's because you are working in a flat space.

    There is no analogue of this for spacetime. The spacetime of our universe is not embedded in any higher dimensional flat space. (Even in theories like string theory that postulate extra dimensions, the higher dimensional space that includes those extra dimensions is not flat.) So there is no way to avoid the ambiguities in parallel transporting vectors between distant events.
  11. Feb 26, 2017 #10
    Please look up the Shapiro delay. It's in the title of this thread.

    It may well be. I deduced it from the limited understanding I have of quantum mechanics. A wavefunction represents a probability distribution of finding a particle in a particular place along that wavefunction. Wavefunctions span the entire universe. When you are not observing the particle, it is in a superposition of states. When you begin to observe it it can appear at any position depending on the probability distribution of the wavefunction. What stops it from materialising on the other side of the universe?
    If my understanding is wrong can you explain why?

    That's what I alluding to when I suggested using a higher dimension. I should have made that more clear. If the higher dimension is not flat, why invoke it at all?

    Why? Is it impossible to invent a new theory that embeds general relativity in a flat higher dimensional reference frame?
    I dont know. I'm asking.
  12. Feb 26, 2017 #11


    User Avatar

    Staff: Mentor

    You misunderstand Peter's point about the Shapiro delay. He's not saying it doesn't exist, he's saying that gravitational time dilation is only a meaningful concept for static observers so won't help visualize what's going on with someone falling towards and through the event horizon.

    You are thinking of non-relativistic quantum mechanics, QM formulated without considering the effects of relativity. That's the only kind you'll encounter in the first few years of a college-level physics degree program so that's the only kind that most people ever hear about. The proper relativistic formulation of quantum mechanics, which you probably won't ever meet unless you're in a PhD program, doesn't have these universe-spanning relativity-defying global-frame wave functions. The apparent contradiction you're seeing is the result of trying to use a theory that ignores relativistic effects in a situation where they are to large to ignore.
  13. Feb 26, 2017 #12
    So far no one has even attempted to answer the question. Everyone has decided to concern themselves with explaining the conventions in examining general relativity.
    I'm not interested in the conventional approach, the question is much simpler than this.
    I understand that you don't need a global reference frame to "predict" what happens in any particular reference frame. I am bringing into the question the validity of this very assumption. The penrose diagram also describes what happens to a person falling into a black hole. But only from the perspective of the falling person. It does not work at all at describing what an observer would see because they are two very different reference frames. It completely neglects time dilation.

    In one reality (the distant observer's), the in falling person never crosses the horizon. In another (the in falling person's), he does. Which is correct? There is a concept called the relativity of simultaneity which explains how events that are simultaneous in one reference frame may not be in another. The end result however must be agreed upon. And crossing the event horizon is a very clear event because it has very clear implications. If the person crosses, he will never get out. If he does not cross he will.
    In fact you could design a test to prove whether or not there grows a virtually infinite time difference between the two reference frames. When the in falling person gets extremely close to the black hole, he fires rockets to get himself back out. If from the outside observer's perspective it takes him quintillions upon sextillions of years (just a very large number) to go down and come back up but only minutes from the infalling person's perspective, you have proven the existence paradox.

    Can you or can you not run this test?

    Noted, thanks.
  14. Feb 26, 2017 #13


    User Avatar
    Science Advisor

    Of course. We've done it by taking clocks to the top of a mountain, waiting, and coming back down. The clocks disagree with ones left in the lab. That's less extreme than your experiment, but it's the scenario you are talking about.

    The result is not paradoxical. One clock took a longer route through spacetime than the other, that's all. It's not really any different from two cars starting at the same factory and ending at the same scrapyard but having different odometer readings.
  15. Feb 26, 2017 #14
    It is paradoxical because unlike the route up and down a mountain, the route to the event horizon is infinitely long.
  16. Feb 26, 2017 #15


    User Avatar
    Science Advisor

    There is no timelike route touching the event horizon and returning to the outside. That's pretty much the definition of an event horizon.
  17. Feb 26, 2017 #16


    User Avatar

    Staff: Mentor

    It's not correct to say that "in one reality (the distant observer's), the in falling person never crosses the horizon". What's going on here is that the outside observer is using a particular (and locally sensible to him) convention for assigning time values to distant events; and this convention doesn't assign any time value to the event "infaller crosses horizon" so there's no way to describe that event in those coordinates. That doesn't mean the event didn't happen, it means that we've made a poor choice of coordinates for analyzing this problem.

    Let's consider one way that the outside observer could be assigning time values to events on the infaller's worldline. The method I'm going to use ("radar coordinates") is different from the Schwarzschild coordinates that you've been assuming throughout this thread, but it is equivalent to them in the local patch of spacetime around the observer, has the same problem with assigning a time to the horizon-crossing event, aand is easier to visualize. You are free to choose some other method (but be sure that you can explain how it's implemented - it doesn't work to, for example, assert that you and the infaller can both just look at the same big clock in the sky!) but no matter which one you choose you'll conclude that the infaller does cross the horizon and light from that event doesn't reach the outside observer.

    Let's say that I am the outside observer and at time ##T## according to my wristwatch I send a radio message to the infaller and the infaller immediately replies (he might be receiving then transmitting a message back, or we might just be bouncing my signal back from a mirror). I receive the reply at time ##T+\Delta{T}##. This tell me that he received the signal at the same time (using this particular convention for defining "at the same time" - your point about relativity of simultaneity is well-taken) that my wristwatch read ##T+\frac{\Delta{T}}{2}##; the distance out is the same as the distance back so the time out has to be equal to the time back.

    Now consider what happens as I send a continuous stream of such signals. The ##\Delta{T}## values become longer and longer, and I never stop receiving replies even if I wait forever (although the replies may be redshifted beyond the ability of my receiver to detect them, they never stop). However, there is a moment when I send the signal that reaches the infaller just as he passes through the horizon - his reply never makes it back to me, so I don't have a ##\Delta{T}## value and the ##T+\Delta{T}## convention stops working as a means of assigning times to events on the infaller's worldline. However, that's telling us something about the behavior of light signals between me and the infaller, not about what's happening to the infaller. If I want to describe events on the infaller's worldline, I have to choose some other convention for assigning times.
    Last edited: Feb 26, 2017
  18. Feb 26, 2017 #17


    User Avatar

    Staff: Mentor

    It is not infinitely long. It has a finite length that is easily calculated.

    The apparent infinity on the inbound path is the result of trying to use Schwarzschild coordinates, which have a coordinate singularity at the event horizon so do not apply in this situation (at a handwaving level, using them across the event horizon is a mathematical error akin to dividing both sides of an equation by zero). Kruskal, Eddington-Finkelstein, and Painleve coordinates will all work and will give you the right answer. (The radar coordinates I described in my previous post won't work, but I deliberately chose them because they have the same sort of pathology as Schwarzschild coordinates).
  19. Feb 26, 2017 #18
    I think you missed my point by about a million miles here.
    I meant simply just to the horizon in the first place.

    Is the theory that explains why the light signals are delayed not exactly the same theory that explains the motion of the falling person? Regardless of what analogy you want to use - either there is more space so the light has more distance to travel or the speed of the beam itself slows down - the same should apply to the falling observer. The closer to the event horizon you are, the slower you would travel out and the slower you would travel in.

    By changing the coordinate system all you're doing is changing reference frames and ignoring time dilation again.
    Clearly the Schwarzschild coordinates singularity has real physical significance because it predicts the increase in signal delay that you mentioned.

    But only using a coordinate system that ignores time dilation? I'm getting out of my depth here.
  20. Feb 26, 2017 #19


    User Avatar
    Science Advisor

    Then your point is not correct. From any point outside the event horizon, there exist timelike paths to the event horizon that have finite proper time along them.
  21. Feb 26, 2017 #20
    Could you please give an example of this? I'm not actually very familiar with timelike paths.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: The Shapiro delay and falling into a black hole
  1. Shapiro Delay (Replies: 1)

  2. Shapiro delay (Replies: 12)

  3. Shapiro delay (Replies: 15)