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Black Holes What?

  1. Oct 13, 2014 #1
    I was thinking about this and either I have a misunderstanding of black holes or they are simply not how the standard model proposes them to be.

    Lets start out by setting a few a statements from the standard model that you agree with.

    If you disagree about any of these points please comment so we can fix that misunderstanding early.

    As soon as an object hits the event horizon according to the standard model it will appear as if it is locked in time right there.

    Well if that's true why do we just see a black hole? Aren't we supposed to see all the matter that's inside the black hole right at the event horizon?
     
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  3. Oct 13, 2014 #2

    PeterDonis

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    No, it won't. To someone outside the horizon, it won't appear at all; light emitted by the object when it is just crossing the horizon stays at the horizon forever, it never gets any further out, so it never reaches someone who is further out.

    To someone who falls through the horizon, they see the object cross the horizon just as they cross the horizon themselves; that is, the light emitted by the object when it crossed the horizon stays at the horizon, so that's where the observer will see it.

    Meaning, why do we see nothing from inside the black hole? I'll assume that's what you mean; see below.

    No. No one outside the horizon ever sees anything that's at or inside the horizon. See above.
     
  4. Oct 13, 2014 #3

    Matterwave

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    When an object moves closer to the horizon, the light from this object gets more and more red-shifted until you can no longer detect this light. Even if you want to take the time-dilation approach, an object which is "frozen in time" won't be something that you can see. Imagine an object that emits flashes of light every 1s (according to its own clock), you, far from the horizon, begin by seeing 1 flash per second. As this object gets closer and closer to the horizon, you begin seeing fewer and fewer flashes per second. You start off maybe seeing 1 flash every 2 seconds, and then every 3 seconds, and then every hour...and when the object gets "to" the horizon, you just stop seeing flashes. That last flash will never reach you.
     
  5. Oct 13, 2014 #4
    Just beyond the event horizon by 1m time will move extremely slow so couldn't you see it then?
     
  6. Oct 13, 2014 #5

    Matterwave

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    Yep, if the object is outside the horizon, you could still technically "see" it, but the light would be very very red-shifted (so perhaps you would see only radio waves or some such).
     
  7. Oct 13, 2014 #6

    Ok so why can't we see all the objects even with very very red shifted light.

    I mean we can see galaxies extremely far away even though the light is very red shifted as well
     
  8. Oct 13, 2014 #7

    Matterwave

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    As long as the object is outside the event horizon, we can technically "see" it (depending on limits to our detection apparatus), but it gets more and more red-shifted until we can't see it as it passes through the event horizon.
     
  9. Oct 13, 2014 #8
    Correct so why can't we see mass outside the event horizon? The gravitational 1m outside of the event horizon is still VERY great
     
  10. Oct 13, 2014 #9

    Matterwave

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    How much time-dilation there is is directly proportional to how much red-shift there is. The more time-dilated, the more redshift. By the time you reach "freezing time" kind of dilation, you're at redshifts where all the visible light has been shifted to well outside the visible range.
     
  11. Oct 13, 2014 #10
    It may not be in the very tiny visible range, but can't we still measure it and with software convert it into visible images.
     
  12. Oct 13, 2014 #11

    Nugatory

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    Not only does it get redder, it gets weaker, because a red shift reduces the amount of energy carried by the light per unit time. No matter how good our detection equipment, the light from the infalling object will eventually (actually this happens quite quickly) be dimmer than the equipment can detect.

    If the light is going to become infinitely dim, then you would need infinitely sensitive equipment to detect it - and there's no such thing.
     
  13. Oct 13, 2014 #12

    Matterwave

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    Realize that it also gets dimmer since its light is losing energy being red-shifted. But sure, you can use a radio telescope to perhaps see a very bright object as it gets close to a black hole.

    EDIT: sniped by nug
     
  14. Oct 13, 2014 #13

    PAllen

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    All other parts of you post have been well addressed, but I didn't notice this one being addressed as I would.

    Imagine you are watching a ball of luminescent, transparent, dust collapsing into a BH. The point of this conceit is to allow seeing throughout the body as it collapses. First, ask carefully, what you see before it is close to horizon. You see matter throughout the body, but the light you image from the center was emitted earlier than the light from the surface closest to you. This is unsurprising and you still image as seeing throughout the body. As it gets close to the horizon, the whole body (without any dramatic change in appearance - especially if it is isolated so you don't see extreme lensing of bodies behind the collapsing body) grows dimmer, redder. At all times, until it is blacker than anything else in the universe except another BH, you image (at longer and longer wavelengths) the whole body, matter throughout. As was the case well away from the horizon, the image of matter inside represents earlier emission such that this this light escaped the surface closest to you just before horizon formation. Thus, in this highly idealized collapse, you simply see the whole body vanishing throughout, fairly quickly.

    To get an idea how black an isolated BH would be (if there is nearby matter, infall, even at slow rates, makes a BH bright), you need to define some 'blackest possible' state. For example, a box at 10-15 degrees above absolute zero that is perfect opaque. Then, in quite a short time, for any given definition you pick, a BH will become a trillion times blacker than that within a remarkably short period of time.
     
  15. Oct 13, 2014 #14

    I'll try to answer the question without almost any physics:

    Yes, objects appear to freeze at the event horizon.

    When an object appears to freeze, it appears to stop doing anything. If an object is emitting light, it appears to stop emitting light, when it appears to freeze.

    It seems possible that a frozen object would still reflect light. But there appears to be no reflection because:

    A system consisting of an object and light approaching that object seems to become frozen at the event horizon. Light-object distance was changing, that change seems to stop happening, because light-object system seems to freeze at the even horizon.
     
    Last edited: Oct 13, 2014
  16. Oct 13, 2014 #15

    PeterDonis

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    No, it isn't, because the object isn't "frozen". The light emitted by the object before it reaches the horizon is being redshifted by the hole's gravity. The object itself falls through the horizon; that's why it can't reflect light.
     
  17. Oct 14, 2014 #16
    Thanks for the answers guys!
     
  18. Oct 15, 2014 #17
    I'm not sure that is quite right. In the frame of a falling object as it cross the horizon, it can be reflecting light. In the frame of the distant observer, you could imagine the object frozen on the horizon reflecting infinitely redshifted light. However, the math at the horizon from the frame of a distant observer can be become a little confusing more maybe even ill defined, so maybe it is better to talk about light reflecting just before crossing the horizon, which would be too redshifted and faint to see.

    I have a thought experiment that helped me a little exploring this kind of question from a different angle.. Imagine we have a powerful telescope observing from a distance a clock as it approaches the horizon of a symmetrical Black Hole. Now imagine what would happen if the observed light was not actually redshifted and was intense enough to see. As the clock approaches the horizon imagine seeing the clocks time slow down to an apparent stand still. You might expect to also observe the clock to stop falling and to stay stuck at the horizon, however, counter intuitively, I don't think this is the case. Because of the geometry of space is so curved, the closer an object gets to the horizon, the more light reflecting from it will diverge. I would image his may create the appearance that the clock is continuing to fall into the hole, even while it's time will remain frozen. If the light diverges to the point it is evenly spread around the horizon, then it will appear that the clock has fallen into a singular point at the center. The clock could be imagined as frozen across the entire surface of the BH, but giving the optical illusion that it is a a singular point a the center.

    Of course, once you take into account light intensity and redshfts, all the above because invisible or perhaps even nonsense.
     
  19. Oct 15, 2014 #18

    PAllen

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    Peter is right and you are not. Imagine an external observer sending a signal. It reaches the surface at some event. The reception of the signal by the surface is one event, and the interaction is frame invariant. It occurs either before the surface crosses the horizon, or at the horizon, or inside, but whichever it is invariant. In particular, I have explained, in many threads here, that for a given external observer world line, sending a sequence of signals to the surface of catastrophically collapsing body, there is a precise event on the external world line whose signal reaches the body at horizon crossing. Signals sent after this event by the external observer reach the surface inside the horizon. These features are frame invariant. Signals sent after this critical event on the external observer world line cannot be reflected, in principle.

    Relating this to an image of a frozen clock, imagine the last visible time on a collapsing surface clock is 3:00 pm - this is the time the clock is seen to freeze at for the external observer. A signal sent after the critical external event will arrive, say, at 3:01 PM at the surface clock, but the external observer can never see this arrival, nor can the signal be returned or reflected because this clock time is inside the horizon.
     
    Last edited: Oct 15, 2014
  20. Oct 15, 2014 #19

    PeterDonis

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    Yes, I should have been clearer: once an object reaches the horizon, it can't reflect light that is visible outside the horizon. It can still reflect light, yes, but that reflected light will only be visible to other observers inside the horizon (and since all such observers must be falling towards the singularity, there are limits to which observers can see reflected light from which others).

    There are other issues with your thought experiment as well, but I'll address those in a separate post if I get a chance.

    I think you mean, such signals can't be reflected back to the observer outside the horizon, correct? A light signal could still reach the falling object inside the horizon, be reflected by it, and be seen by some other observer inside the horizon. It just can't get back outside the horizon once it's inside.

    (Of course, if the signal is sent after another critical time on the external observer's worldline, it will not reach the falling object before the latter hits the singularity; in that case, yes, there is no reflection, period. The ingoing light signal hits the singularity, not the falling object.)
     
  21. Oct 15, 2014 #20

    PAllen

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    Yes, of course, that is what I meant.
     
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