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Understanding Black Holes

  1. Dec 22, 2014 #1
    Hello. If at the event horizon..the escape velocity exceeds the speed of light..or time at that 'point' comes to
    a halt..how can anything get past or thru the horizon and make the journey to the center of a black hole?

    Thanks for any and all your responses.
    Bye
    SC
     
  2. jcsd
  3. Dec 22, 2014 #2
    Hey SC,

    You should always be mindful of which reference frame you are viewing a given physical situation.
    By virtue of the equivalence principle the physical laws are independent
    of the reference frame but what different observers 'sees' can be different depending on the
    reference frame.

    For the case of falling into a black hole consider two observers
    1. An observer freely falling into a black hole
    2. An observer who is at rest relative to the black hole but very far away such that spacetime
    around this observer is flat.

    For observer 1. their reference frame ticks away at their proper time and they experience nothing
    out of the ordinary, no slowing of time, nothing, as they cross the event horizon.
    (assuming that the black hole is massive enough such that the tidal forces at the horizon
    is negligible).

    For observer 2. they observe something different, as observer 1. approaches the horizon
    obs. 2. sees obs. 1's time slow due to time dilation from the fact that obs 1. is in a deeper
    gravitational potential. What happens is that as obs. 1. asymptotes to the horizon
    they are observed to slow down until they are 'frozen in time' (according to obs 2.).
    And also, due to gravitational redshift, they also get redder and fainter. You can calculate
    this by considering what happens to light signals emitted by obs. 1. and observed by obs. 2.

    But remember that obs. 1. actually just passes straight through the horizon, experiencing
    *their* proper time.

    I hope I explained it ok and didn't fumble my words too much!

    Best,
    Sebastian
     
    Last edited: Dec 22, 2014
  4. Dec 22, 2014 #3
    Hello. Thank you. At the bottom I think you meant observer 1. That's okay. Now I will have to think about this a bit.
    Bye
    SC
     
  5. Dec 22, 2014 #4

    phinds

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    I'm sure this was supposed to say "obs 1", not "obs 2" (who is the remote observer)
     
  6. Dec 22, 2014 #5
    Thanks! I've changed this now :D
     
  7. Dec 22, 2014 #6
    Hello. To continue in this thread..Question: does a material object approaching a black hole increase it's velocity towards it? I am assuming it will get closer and closer to the speed of light. No mass can reach the speed of light..but black holes can get more massive by pulling in material. My real question is then..is there anything located past the event horizon that is not radiation? Hmm..just trying to learn the nature of black hole's a little.
     
  8. Dec 22, 2014 #7

    PeterDonis

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    Pop science presentations often give this description, but you have to be careful how you interpret it. It is true that, once something is at or inside the horizon of a black hole, it can never escape. One could describe this as the "escape velocity" being equal to the speed of light at the horizon, and being greater than the speed of light inside the horizon. But that doesn't prevent the horizon or the region inside it from existing. It just means spacetime there doesn't work the way you are used to.

    It doesn't. Skhaaan's description is basically correct; the only thing I would add to it is that observer 2, since he is spatially separated from observer 1, has to be careful interpreting his observations of observer 1. The fact that he sees observer 1's clock slow down, and his light signals get redder and redder, does not mean observer 1 has "really" had his clock slow down, and it does not prevent observer 1 from reaching and falling inside the horizon. Again, all this really means is that spacetime at or inside a black hole's horizon doesn't work the way you are used to.

    By falling in. ;)

    This question assumes that the "velocity" of the object is a unique, well-defined property. It isn't. In order to make the question well-defined, we have to specify how the velocity is measured, and by which observer(s).

    Here's one way to do it, which makes the answer to your question "yes": suppose we have a fleet of observers all "hovering" at constant altitudes above the hole's horizon, but a different altitude for each observer. Each observer measures the velocity of the infalling object as it passes by him (so it's a local measurement and has a unique, well-defined result). The "hovering" observers closer to the hole's horizon will measure the infalling object's velocity to be larger; in the limit as the "hovering" observer's altitude above the horizon approaches zero, the velocity he measures for the infalling object approaches the speed of light.

    However, this method of measuring the "velocity" of the infalling object can be interpreted another way. Notice that, as the "hovering" observers get closer to the horizon, they have to accelerate harder and harder to maintain altitude (in the limit as the altitude above the horizon goes to zero, the acceleration required goes to infinity). So from the viewpoint of the infalling object, the reason its velocity relative to the "hovering" observers gets closer and closer to the speed of light is that those observers are accelerating towards him harder and harder. In the limit, when the infalling object passes the horizon itself, the horizon is moving outward, relative to the object, at the speed of light. See further comments below.

    Yes; any object that falls through the horizon will be located "past" (i.e., inside) it, at least until it hits the singularity at the center and is destroyed. See further comments below.

    Yes. See my description above of the infalling object passing "hovering" observers closer and closer to the horizon, and finally passing the horizon itself. The reason the object's velocity relative to the horizon is the speed of light is that the horizon is moving outward at the speed of light; it is the lightlike "object". The infalling object remains perfectly normal throughout.
     
  9. Dec 28, 2014 #8
    Hello. Sorry about any delay's but I am busy. Not a great response..but I am continuing. I can not see at the moment how a 'finite' sized horizon could be 'moving' out at the speed of light?......I understand the rest of it. My thinking has been that a material object does not have to be travelling or is not caused to be travelling at or near the speed of light when it crosses the horizon. I talked earlier about there being a need to be travelling faster than the speed of light in order to enter or escape. I quote unquote think that it is not so much the density of the hole..or its size..it's just at and beyond the event horizon..the space-time continuum warps such that nothing can escape the area no matter it's speed. What's the real point of the question? I at the moment do not believe in a 'singularity' or an infinite density in a finite sized black hole. The space time continuum is just warped enough to prevent anything form escaping. I suppose what the real question is maybe..is there a difference?

    Thanks for any and all responses
    Bye
    SC
     
  10. Dec 28, 2014 #9
    That is true, due to the curvature of space time inside the black hole nothing can avoid the singularity. Also, time and space do a strange kind of flipping, probably depending on whether or not the hole has considerably huge spin or electrical charge. But the singularity becomes a time-like coordinate whether than a spacial coordinate upon passing the outer horizon; you must now inevitably hit the singularity, as it is literally your future. Maybe it switches back to a spacial coordinate as you pass the inner horizon, but if the hole is a charged one or a normal schwarzschild one, you'll be in bad shape.

    The interesting part is for rotating Kerr type black holes; in those kinds you would have to actually try to fall into the singularity, as its spin would be pushing material outwards from it, depending on the angle it falls in from. Not all singularities have to be infinitely small; in a spinning black hole, it would take on the shape of a ring with definite volume, although its thickness would still be zero. The singularity is pretty much the easy way out that people take when they notice their equations break down, so perhaps it could possibly have a finite but still inconceivably small size; but that's beyond my understanding.
     
  11. Dec 28, 2014 #10

    PeterDonis

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    "Movement" is relative. The horizon is moving outward at the speed of light relative to an observer falling into the black hole. Globally, the horizon "stays in the same place"; it doesn't "move" at all. But since the horizon is a lightlike surface, it's impossible to "see" it staying in the same place; the only way to observe light from the horizon is to be at the horizon, and the only way to do that is to fall into the hole (and you will only be at the horizon for an instant).

    Correct. This is similar to the way that, in special relativity, light rays are moving at the speed of light relative to all material objects. The horizon is a surface composed of outgoing light rays (where "outgoing" here describes a direction, not a state of motion--as above, motion is relative), so any material object that crosses the horizon will observe those light rays to be moving at the speed of light.

    Yes. To be more precise, at the event horizon, light rays that are moving radially outward "stay at the same place" (globally speaking), so they never get any further outward; and material objects must fall inward (since they can't move at the speed of light). Inside the horizon, everything, both material objects and light rays, must fall inward. So nothing can escape outside the horizon once it is at or inside the horizon.

    The singularity is a feature of the classical GR model of black holes. Most physicists appear to believe (as I do) that the presence of the singularity in the model is a sign that the model breaks down at the center of a black hole; some other theory (presumably some form of quantum gravity) must take over and tell us what actually happens there.

    However, whether or not there is actually a singularity at the center of a black hole is a different question from whether or not there is a black hole--i.e., whether or not there is an event horizon. There could be an event horizon, and therefore a region of spacetime from which nothing can escape once it falls in, even if there isn't an actual singularity at the center.
     
  12. Dec 28, 2014 #11

    PeterDonis

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    There is a kernel of truth in here, but the way you state it is rather confused. The singularity is not a coordinate, and does not "become a timelike coordinate" inside the (outer) horizon. In some coordinate charts, the coordinates usually called ##t## and ##r## "switch roles" (timelike to spacelike or vice versa) inside the horizon, but that's a peculiarity of those particular coordinate charts; there are other charts in which the "switch" does not happen.

    The kernel of truth is that, once you are inside the horizon of a non-rotating, uncharged black hole, the singularity is in fact "your future" (at least in the classical GR model of a non-rotating black hole--see my previous post for caveats about whether that model actually matches reality all the way down to the singularity). This is because, in a non-rotating, uncharged black hole, the singularity is spacelike--it's like a moment of time, not like a place in space. And the moment of time that is the singularity is to the future of any other moment of time along any timelike worldline inside the horizon.

    Note, however, that I carefully said a "non-rotating, uncharged" black hole in the previous paragraph. In a black hole that is either charged or rotating or both, the singularity is not spacelike; it's timelike, so it's like a place in space, not a moment of time. In these holes, there are also two horizons (outer and inner), not one, and the singularity is inside the inner horizon. Inside the inner horizon for these holes, timelike worldlines no longer have to fall inward; they can stay at the same radius or move outward. However, at least in the idealized classical model for these holes, if you are inside the inner horizon and move outside it again, you do not emerge into the same region of spacetime; you emerge into a different region, with a different outer and inner horizon, and beyond that different outer horizon (out of which you can now emerge, so it's really a "white hole", not a black hole) is a different exterior universe. AFAIK, nobody really believes that all this is physically realistic; I believe the general view is that the inner horizon in these models is where classical GR breaks down and something else (like quantum gravity) has to take over.
     
  13. Dec 28, 2014 #12
    Thanks for correcting me! I didn't think calling it a coordinate would really make a difference at first, but I guess you learn something new each day you're on here; it all comes together so clearly now.
     
  14. Dec 29, 2014 #13
    New evidence, has been introduced, that says that black holes do not exist. http://theweek.com/speedreads/index...ays-she-has-proof-that-black-holes-dont-exist Hawking has recently said that his ideas were blunderous as well, so what gives.
     
  15. Dec 29, 2014 #14

    PeterDonis

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  16. Dec 29, 2014 #15
    Where does your information, for what happens on the inside of a non rotating black hole come from?
     
  17. Dec 29, 2014 #16

    PeterDonis

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    From the classical GR model of a non-rotating, uncharged black hole, i.e., Schwarzschild spacetime. That's the model we're using in this thread.

    If you want to start a discussion about how quantum effects might change the model we use, you should start a new thread. (But please first search PF for previous threads on that topic, of which there are many--I linked to one in my previous post. There's no point in rehashing discussions that have already been had.)
     
  18. Dec 29, 2014 #17
    Well then, where did Schwarzschild get the information for his black hole model?
     
  19. Dec 29, 2014 #18

    PeterDonis

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    By solving the Einstein Field Equation for a spherically symmetric vacuum spacetime. Any textbook on GR will explain it--a good online one is Sean Carroll's:

    http://preposterousuniverse.com/grnotes/
     
  20. Dec 29, 2014 #19
    Hello. Just a small point. I saw a time-lapse video once (ie. about 10 years duration) showing the 'literal orbital movements' of stars at the center of our galaxy. Stars were seen to speed up and slow down with a focal point that was clearly dark. Very convincing about there being a black hole there!
     
  21. Dec 30, 2014 #20

    Drakkith

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    Black holes were found as particular solutions to Einstein's Field Equations long, long before we ever found observational evidence for them. I believe Schwarzschild produced the first solution for a non-rotating, spherically symmetric mass in 1915. If the mass is great enough, the equations produce a black hole as the solution.
     
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