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B Confused about how long it takes to cross an event horizon

  1. Dec 8, 2016 #1
    If there were 2 observers, 1 at rest with CMB, and the other falling into a black hole, how much time passes for the observer at rest with CMB while the other falls past the event horizon?

    I ask because I keep reading that time dilation (due to gravity) will cause the object falling in to appear to never fall past the event horizon from the perspective of the outside observer.

    I don't understand how anything has had time to fall past any event horizon because an infinite amount of time would pass for an observer at a distance, and an infinite amount of time has not passed in the universe

    But things have clearly fallen into black holes in a universe with a finite history and observers (or objects) at rest and not falling in.

    What am I missing?
     
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  3. Dec 8, 2016 #2

    phinds

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    Never happens

    ? What happens according to an outside observer has no impact on what ACTUALLY happens to the infalling object.

    There are TONS of threads on this. I suggest a forum search if you want more discussion and a good place to start is with the links at the bottom of this thread.
     
  4. Dec 8, 2016 #3

    PAllen

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    Actually I beg to differ. the question of how how long it takes for an infaller to cross the horizon per an external obsever at rest with the CMB is ill formed. You can ask a physical question: "Does the external observer ever (literally) see the infaller inside the horizon?" with the answer being trivially no because signals can't get from inside to outside per the very definition of horizon. However, asking when something happens at distance is normally NOT taken to be a matter of seeing but of simultaneity convention. Then, the answer depends entirely on the convention, and may be never, or may be any time after the event on the external world line that corresponds to a signal emissions that will reach the infaller exactly at horizon crossing (depending purely on coordinate choice). This is because any time after this event on the external world line is connected only by spacelike paths to the horizon crossing, and is thus an allowable definition of simultaneity.
     
  5. Dec 8, 2016 #4

    phinds

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    OK. I was being simplistic and taking the point of view of just "seeing".

    EDIT: also, what I have expressed is stated in dozens of threads here on PF.
     
    Last edited: Dec 8, 2016
  6. Dec 8, 2016 #5

    PAllen

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    On most of those threads, the above point is made in response, in some form, by someone. By analogy to cosmology, we say the image of a distant galaxy we see corresponds to its distant past. We then make extrapolations per some model on its current state and location, per some useful coordinate system. Similarly, in the BH case, the image we see of an infaller is from its past. Where it is 'now' then depends on which of several useful coordinate systems we choose for modeling. For some of these there is a perfectly well defined time of horizon crossing, establishing simultaneity between horizon crossing and some event on an external world line.
     
  7. Dec 8, 2016 #6

    phinds

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    No argument with that.
     
  8. Dec 8, 2016 #7
    Yes I have read many of those threads, and a lot of other material in the subject. I think I have my head around what an external observer sees, and when and why they see it that way (sort of) and I know my question was poorly formed (I tried hard)

    I am trying to get me head around what actually happens and when. If I threw a baseball into a black hole from a great distance away, and waited around and observed the black hole at some later time (time for me), would I ever notice the event horizon getting larger because of the added mass of the baseball? Would an unrelated observer see a change in the mass of the BH, and calculate a change in the size of the event horizon?

    If I am understanding this response, the answer is 'it depends'...

    Perhaps I need to get a better understanding GR

    Thanks for the replies, its been helpful
     
  9. Dec 8, 2016 #8

    Nugatory

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    In principle: Yes, and fairly quickly too.
    In practice: You might try calculating the Schwarzschild radius of a black hole of ten solar masses, and then compare with the Schwarzschild radius of a black hole of ten solar masses plus another 150 grams for the baseball. Even a very careful observer could be forgiven for not noticing the increase in the size (strictly speaking, the surface area) of the event horizon.... but that doesn't mean it didn't happen.
     
  10. Dec 9, 2016 #9
    I don't know how the size of the event horizon could be easily measured, that's why I changed the problem a little bit:

    Let's say we drop a charged baseball into a neutral black hole. During some time we can change the path of the baseball by creating electric fields. After some time we can no longer send any information to the ball, so electric fields created after that time do not have an effect on the ball, so the electric fields have an effect on the black hole instead, so we could say the black hole has absorbed the charge. Right?


    Now let's say we drop a baseball, which has some mass, into a black hole. During some time we can change the path of the baseball by creating non-uniform gravity fields, I mean a tidal force is stretching the ball - black hole system. After some time we can no longer send any information to the ball, so gravity fields created after that time do not have an effect on the ball, so the gravity fields have an effect on the whole black hole - ball system instead, so we could say the black hole has absorbed the gravitating mass of the ball. Right?
     
  11. Dec 9, 2016 #10

    PeterDonis

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    There is such a time, but it's not the time when the ball crosses the hole's horizon. After the ball crosses the hole's horizon, we can no longer receive any information from the ball, but we can still send information to the ball. If the hole stayed neutral, then the ball could receive information from us until it hit the singularity at ##r = 0## and was destroyed. If the hole becomes charged (which is what would happen if the ball were charged), then if we stick with idealized solutions and ignore all the reasons why they are unphysical, the ball could receive information from us until it crossed the inner Cauchy horizon of the hole (whereas we could no longer receive information from the ball after it crossed the outer horizon).

    They have an effect on it anyway, because they contain energy, so turning on an electric field in the hole's vicinity will add some energy (mass) to the hole.

    However, if you are thinking of the electric fields being able to deflect the path of the hole the way they would deflect the path of the ball, that's much too simplistic. The hole is not an ordinary object; it's spacetime geometry. So you have to figure out how the electric fields will affect spacetime geometry, which again brings into play the fact that the fields contain energy.

    This will stretch/squeeze the ball, but that's not necessarily the same as deflecting its path.

    This is also too simplistic. The hole, again, is not an ordinary object; it's spacetime geometry. You can't just stretch and squeeze it the way you stretch and squeeze an ordinary object.

    Same comment as above.

    According to what you said just before this, that was already true before the ball fell in--you said "a tidal force is stretching the ball - black hole system". So I don't understand your logic here.

    Also please see my comment above.

    We can say that without having to put any other gravity fields into the picture. We can just put test objects in orbit about the hole and watch the orbital parameters change when the ball falls in.
     
  12. Dec 10, 2016 #11

    timmdeeg

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    Supposed, we drop the ball from a stationary position at time ##t_1##. The ball reaches the event horizon at his proper time ##t_2##. We will see our ball somehow redshifted at our time ##t_2##. At what time would we notice that the black hole's mass has increased?
     
  13. Dec 10, 2016 #12

    PAllen

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    It depends a bit on how you define it. Suppose the infalling body is big enough that some hypersensitive GW detector can detect GW from its merger with the BH. Then after GW ring down (a well defined t3 along the world line of some external observer), all of the following would be true:

    - The behavior of orbiting bodies would be identical to an ancient BH with the same mass, charge, and angular momentim.
    - The image of the BH against background stars would become indistinguishable from an ancient BH of the same parameters.
     
  14. Dec 10, 2016 #13

    PeterDonis

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    Let me first give a case where we know the answer exactly. Suppose that instead of a ball, it's a spherically symmetric shell of matter that is falling into the hole. In that case, as soon as the shell passes us--as soon as it is at a smaller radial coordinate than we are--we will notice that the hole's mass has increased. In other words, the mass we measure by orbital parameters is the mass of everything that is below the orbit.

    In the case of the ball, that exact answer can't be quite right because the matter falling into the hole is not spherically symmetric around the hole. But it still gives us a good heuristic, that once the infalling matter is below you, it can potentially contribute to the mass you see. So if you yourself are dropping matter into the hole, the matter you drop in, such as a ball, should start contributing to the hole's mass that you measure right after you drop it, at least heuristically.

    This brings up an important point: when we say "the mass of the hole", that mass is not really localized--it's a global property of the spacetime, not a local property of some piece of matter. Strictly speaking, this is true for the mass of any object; but black holes bring up the point more forcefully because they are vacuum--there is no matter inside them.
     
  15. Dec 10, 2016 #14

    PeterDonis

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    Adding GWs to the picture makes things a bit more complicated than what I described in my last post, because now there are three masses involved. Heuristically, they are:

    (1) The mass of the hole before we drop the ball;

    (2) The (larger than #1) mass of the hole after we drop the ball, but before any GWs emitted when the ball falls in have passed us;

    (3) The (smaller than #2, but most likely still larger than #1) final mass of the hole after all GWs that were emitted by the infall process and ringdown have passed us on their way out to infinity.

    #3 is smaller than #2 because the GWs carry away energy, and that energy comes from the mass of the hole. And, for the same basic reason that we see the mass of the hole increase as soon as an infalling mass is below us, we will see the mass decrease as soon as outgoing energy (whether it's carried by GWs or anything else) is above us. Again, this won't be an exact answer because the outgoing GWs won't be spherically symmetric around the hole, but it should be a good enough heuristic approximation for this discussion.
     
  16. Dec 11, 2016 #15

    timmdeeg

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    I'm not sure if I understand you correctly.
    The mass ##m## is falling in and crosses the event horizon at proper time t2. Then the mass of the black hole previously ##M## is now ##M+m## minus the small loss of mass carried away due to emitted gravitational waves. Is it this what you call "ancient BH"?

    Now let ##m## emit light during falling and its merger with the BH. It seems to follow from this that an external observer far away will receive both the light and the gravitational waves emitted during merger with the same high redshift and at (his time) t3 >>> t2. Is this reasoning correct? Is it this event which you call "GW ring down"?
     
  17. Dec 11, 2016 #16

    timmdeeg

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    This seems to follow from Birkhoff's theorem. So the r-coordinate of said shell makes no difference to an orbit outside of it, there will be no change anymore e.g. when the shell reaches the horizon and since GW will not be emitted.
    And I think this is true, at least in principle, if I don't drop the ball but instead it crosses my orbit opposite of the BH, right?
     
  18. Dec 11, 2016 #17

    PAllen

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    By ancient BH I simply mean one that formed a very long time ago, all at once . Then, for a BH into which some mass m infalls, once GW ring down is complete (a quite short amount of time for the external observer), the augmented BH is already indistinguishable in all ways from one with the same total mass, charge, and angular momentum that formed eons ago. This mass will be smaller than just adding infalling m, due to the GW. I bring in GW to have an external observation that corresponds to the infall body being merged with the BH with full symmetry restored. Ring down is a well known technical term for the process of the horizon settling down to a stationary state with the symmetry characterized only by mass, charge and angular momentum. Until this settling occurs, vibration of horizon emits GW, which then carry away the corresponding excess energy.

    Red shift will not apply to the GW since their generation is not a local process. Light emitted before full quiescence of the BH, which is typically long after the infaller has not only crossed the horizon, but may have already reached the singularity (or whatever is really there instead), will be ever more red shifted. This light all comes from emission events outside of the merged horizon, and will continue to leak, in principle. However, by the time of detection of final ring down, the degree of red shift and faintness will mean such emissions are no longer detectable, in practice. If you want to propose infinitely sensitive detectors, then such leaked light from ancient near horizon events continues forever for all BH.
     
  19. Dec 11, 2016 #18

    timmdeeg

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    Ok, I understand.
    This answers my question, thanks for explaining.
     
  20. Dec 11, 2016 #19

    PeterDonis

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    The mass of the hole does not change in an instant; when it is seen to change depends on where the observer is. The key point is that the mass "of the hole" is a property of the spacetime geometry, and the spacetime geometry does not change everywhere all at once.

    You mean, if it crosses your orbit on the opposite side of the BH from where you are? In that case it will take some time for you to measure the increase in mass: heuristically, the change in spacetime geometry has to propagate from where the ball crossed your orbit to where you are on your orbit, and that can't happen instantaneously.
     
  21. Dec 11, 2016 #20

    timmdeeg

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    Ok, thanks for clarifying and your continuous help.
     
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