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Yet another question about black holes

  1. Sep 16, 2012 #1
    A comparison of a large star and a black hole: Both have identical masses.

    Correct me if I'm wrong but gravitationally, they would be the same because we assume that the gravitational force acts from the center of mass.

    So here's the question: If both the black hole and the star have the same mass and the same gravitational pull, why does the black hole have an event horizon?

    I know that it has something to do with the density of the stellar object but I was hoping that someone could explain this to me a little better.
     
  2. jcsd
  3. Sep 16, 2012 #2
    Hi.

    My guess is that if the two celestial bodies are covered by curtains and we feel only the same gravitational effects outside them thus we cannot know which is BH and which is ordinaly star. When we remove curtains we find that star is big and sparse and BH is compact and dense like 1 kg of cotton and 1 kg of iron. Who claims that cotton and iron should be same volume ? GR says too much compact body make horizon around it.
     
  4. Sep 16, 2012 #3
    I understand that despite the difference in the densities, the gravitational forces would be identical to a point. My question has to do with why one has an escape velocity < c and why the other has an escape velocity > c.
     
  5. Sep 16, 2012 #4

    Drakkith

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    Let us say the event horizon is the "surface" of the black hole and the photosphere is the "surface" of the star. What is the major difference between the two? Diameter. The black hole is TINY, while the star is hundreds of thousands if not millions of kilometers across. We can get MUCH closer to the center of the black hole than we can to the star before we cross the boundary. If we could go into the star we would see that there is material BEHIND us now that isn't pulling us towards the center anymore. So for the same distance from the center, the black hole has more mass pulling you in to the center.
     
  6. Sep 16, 2012 #5

    cepheid

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    The star isn't compact (dense) enough. Recall that gravity gets stronger the closer you get. So the surface representing the boundary at which escape v becomes > c would be *inside* the object (except that once you're inside the star the gravitational pull starts to lessen, because now some of the mass is outside of your radius). The strongest gravity you could ever experience for the star while still being outside it would be the surface gravity of the star, and for any stellar radius, this surface gravity is such that escape velocity is < c.

    For a black hole, for which all of the mass is essentially crushed down to a single point, you can actually get close enough i.e. to within a radius of the object where escape v > c.
     
  7. Sep 17, 2012 #6
    These explanations were exactly what I was looking for. Thanks!
     
  8. Sep 24, 2012 #7
    The gravity is given by the mass of the gravitational body and the distance between the point you measure and its point-mass (the point in its centre), as you know. (and G, but this is besides the point) More precisely, directly proportional with the former and inversely proportional with the later.

    In the case of the star, because it is too extended (large radius), the distance between its centre and this point on its surface is too large to generate a gravity sufficient to stop light. Now, you may get inside the star towards its centre, but the net gravity does not increase anymore, because you've got mass around pulling in all directions. This was just to explain why there's no black hole inside. :P

    But the matter of the black hole is that compact that it fits - or is smaller than - the Schwarzschild radius, which is the distance from the center to the point where, in respect to the given mass of the object, the gravity is that large that light cannot escape.

    And don't confuse the event horizon with the surface of the gravitational body of the black hole. If the radius of the body is equal to its corresponding mass' Schwarzschild radius (which is also the radius of the sphere determined by the event horizon), then they coincide. But this body may be even more compact, which means the radius of its matter can be smaller than its Schwarzschild radius, however no one outside can see that because its even horizon (the "surface" at the distance of its Schwarzschild radius) is around it and lets nothing escape. So the black hole is this entire phenomenon: not only the body inside, but also this "black aura" around it.
     
    Last edited: Sep 24, 2012
  9. Sep 24, 2012 #8
    Is this always the case, that the radius of the body is necessarily zero? (after its collapse ends, of course) Intuitively yes, just there's always a catch. :)
     
  10. Sep 24, 2012 #9

    Chronos

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    A black hole is formed anytime the radius of a gravitating body is less than its Schwarzschild limit [which is a function of mass]. This does not require a singularity, although there is currently no explanation what would prevent a singularity from forming at such high densities.
     
  11. Sep 24, 2012 #10
    Yeah, cool, thank you Chronos! Minutes before I found an answer on a different thread that ringed the bell:
    Edit: basically the same thing you said.
     
  12. Sep 26, 2012 #11
    The event horizon is where escape velocity increases past c. There is no such boundary for stars, because when you go inside a star, the mass you have passed screens out the gravitational force. the force actually approaches zero as you approach the center of a massive star. You can get arbitrarily close to a black hole's singularity and there is still no screening of the gravitational force. Therefore, you will ultimately pass a radius as which the escape velocity is greater than the speed of light.

    Study GAUSS'S LAW in the context of mass and gravity (instead of charge and electromagnetism), and take a look at the equations for escape velocity and gravitational force (Newton's law will do just fine for these purposes) and you will see the difference between the two scenarios you posed.
     
  13. Sep 28, 2012 #12
    This is a question that has puzzled me for some time regarding black holes.

    I have searched for information on black holes many times and never found this question directly addressed.

    How can a singularity exist when matter that falls into a black hole “stops” at the event horizon? I understand that from the perspective of the matter which has entered the event horizon that time hasn’t stopped but from the perspective of the outside universe it has.

    So that all the matter that has ever fallen into a black hole is still suspended at the event horizon, meaning that in finite time no matter will ever reach the singularity.

    So you might say that the matter in a black hole will reach the singularity in the infinite future.

    However, given that black holes evaporate by Hawking Radiation, all black holes have finite lifespans, and will evaporate before any matter can reach the singularity.

    As far as I can tell, from the perspective of the matter which has reached the event horizon the black hole would instantly evaporate the moment the event horizon is reached.

    It appears that singularities are not possible.

    The answer may be obvious, and I simply lack the understanding to see it.
     
    Last edited: Sep 28, 2012
  14. Sep 28, 2012 #13

    cepheid

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    Welcome to PF,

    You have to remember that before the black hole forms, it doesn't exist, and therefore, there is no event horizon and no severely curved spacetime, with all the associated weird effects. The black hole forms when the object in question loses all forms of support against gravity and undergoes a catastrophic collapse in which all the matter in it is crushed down to a singularity. This, I think, is the way to think about your question.
     
  15. Sep 28, 2012 #14
    I had considered this also. In the process of collapse, the event horizon forms when the gravitational field becomes strong enough that the escape velocity >c. The radius at the instant of formation will be >0, and the instant that the event horizon forms all matter which is inside it is frozen. This frozen matter would be small and dense. Extremely dense. Absurdly dense. But not infinitely dense. So again, no singularity.
     
    Last edited: Sep 28, 2012
  16. Sep 28, 2012 #15

    cepheid

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    I'm not entirely sure that this is true, but I don't have the technical knowledge of general relativity in the context of black holes to comment. Another thing is, I personally consider the idea of "infinite density" to be unphysical. In other words, for me, a "singularity" is not a physical object, it is a mathematical problem in which a theory gives an infinite (i.e. invalid) answer for some quantity. So I would say that maybe we don't fully understand exactly what is going on in the interior of black holes. Just my two cents, take it with the appropriate grain of salt.
     
  17. Sep 28, 2012 #16

    cepheid

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    Another thing is, in the frame of reference of the infalling observer, time passes at the normal rate, and all indications from theory are that such an observer would continue to fall in, be torn apart by tidal forces, and then smash into the "singularity" (or whatever is in there). Either this happens or it doesn't, the only cause for disagreement can be over when it happens. So I'm pretty dubious about your claim that matter would get "frozen" at the event horizon. This is really not how gravity works. You keep falling in until you die. For what it is worth, here is the relevant section of the Wikipedia article on black holes:

    http://en.wikipedia.org/wiki/Black_hole#Singularity
     
  18. Sep 28, 2012 #17
    The reason I'm sure that the event horizon forms before the singularity is the same reason that was given above about why a large star doesn't have an event horizon. As you move beneath the surface the force of gravity lessens. So as the large star collapses, the gravitational field at the center would remain effectively 0 as the density increased. The radius where the event horizon forms will be near the surface (possibly at the surface) because that is where the gravitational force is greatest.

    I agree with you that infinite density would be unphysical. Which is why I began to ponder how such an impossibility could in fact exist. I keep hearing physicists say that physics breaks down at the singularity and that quantum gravity will provide the answers that neither quantum machanics nor general relativity can.

    But as I thought about it and read about it I was struck with the question I have asked. It appears that singularities are mathematical artifacts, and can't actually happen in nature (by my understanding). And yet physicists continue to talk about them as if they exist so I'm probably missing something.

    I don't claim to have thought of something nobody has thought of before. Undoubtedly this question has been asked and answered long before I stumbled on it.
     
  19. Sep 28, 2012 #18
    But that is a misunderstanding. The event horizon forms before the singularity, though matter actually falls in even after that moment, and actually very quickly, only the light it emits for an observer outside makes it appear it stops at the event horizon.

    Think of a vacuum cleaner (aspirator) whose mouth you place near the mouth of a tube that periodically releases, at the same time, a fly and a popcorn puff, each fly always flying away from the vacuum. Now, as you increase the power of the vacuum, the flies will clear away slower and slower, and when the speed of the absorbed air will be equal to the speed of the flies, the flies that get out will stay still in place, while the puffs will be swept inside faster than ever before. Now, think that these flies are the photons of the light which you, an observer, will use to detect that matter that goes into the black hole, while the puff is the actual matter. It's basically the animated photographic image of the matter that freezes at the event horizon, not the matter itself. And BTW, the light emitted towards the black hole will sink into it, only the one emitted away from it will contribute to this phenomenon.

    I hope I'm not missing or distorting something, please correct me if the case.
    ---

    Darth, I relate to your concerns about treating these theoretical expectations as facts (another example is the dark matter), however reasonable researchers do not do it, but treat them as they are, simply models which remain to be confirmed or infirmed when we will have the means to. In fact, I am myself skeptical about the possibility of the existence of the black holes altogether and have reasons to believe so, still when I discuss the current most adopted theory - the General Relativity - it can be done with no problem. Within this model these phenomena make sense.
     
  20. Sep 28, 2012 #19
    It isn't a misunderstanding as much as a difference in perspective.

    I understand that from the perspective of the infalling matter time doesn't stop at the event horizon, nor does the fall inward toward the center. From the inside perspective it is time outside that changes, and the speed of time becomes infinite.

    But whether you look at it from the outside in or the inside out there is still an infinite difference in the speed of time. Given that black holes do not have infinite lifespans, the singularity will still never form, as the black hole will completely evaporate in less than infinite time.

    As far as I can tell the pilot of a spaceship that flew into a large black hole would notice nothing particularly unusual until contact with the event horizon, at which point the passage of time in the outside universe would be so fast that the black hole would instantly evaporate, converting him and his ship to Hawking Radiation.
     
  21. Sep 29, 2012 #20
    No, no. Time does not stop relative to an outside observer at the event horizon; time stops relative to an outside observer at the singularity. At that point, it makes little difference, as said observer cannot see the in-falling observer.
     
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