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The event horizon (maybe dumb question)

  1. Oct 23, 2012 #1
    The event horizon

    Wouldn't it be possible for light to orbit a black hole inside the event horizon?

    In other words, the event horizon, while being a point of no return, is not necessarily a point of ultimate doom.
     
    Last edited: Oct 23, 2012
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  3. Oct 23, 2012 #2

    PeterDonis

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    Re: The event horizon

    No. Inside the horizon, even light that is pointed directly outward actually moves inward, towards the singularity.

    Light can orbit the black hole *outside* the horizon, at a radius 1.5 times the horizon radius.

    Yes, it is. Anything that once gets inside the horizon is doomed to eventually hit the singularity.
     
  4. Oct 24, 2012 #3
    I am most likely beyond my scope of knowledge, but I am under the understanding that an object's orbital velocity required to maintain orbit is less than the escape velocity. In other words, traveling in a direction exactly opposite of the force of gravity originating from the singularity seems to me like an inefficient means to avoid ultimately falling into into the singularity. Wouldn't a velocity at a right angle to the force of gravity be more efficient? I guess what I am asking is :

    If what you say is true, why does the orbital velocity needed to maintain an orbit inside the event horizon somehow exceed the escape velocity?
     
    Last edited: Oct 24, 2012
  5. Oct 24, 2012 #4

    Nabeshin

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    This is the intuition you have from Newtonian mechanics -- close to a black hole, this intuition breaks down completely. The accurate description is given by General Relativity, and is as PeterDonis said above.
     
  6. Oct 24, 2012 #5

    PeterDonis

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    For an ordinary material object at a sufficient distance from the body it's orbiting, this is true. But there is a minimum radius that the orbiting object (if it's an ordinary material object) can be from the body it's orbiting, for a stable orbit to be possible. For an ordinary material object, that minimum radius is three times the horizon radius. (Just as a check, the orbital velocity at this radius is half the speed of light, and the escape velocity at this radius is 1/sqrt(3) times the speed of light, which is larger, so the rule does indeed hold.)

    For light, as I said before, a stable orbit is possible at 1.5 times the horizon radius, and *only* there. Yes, the "orbital velocity" of the light at this radius is greater than the escape velocity at this radius. Light works differently than ordinary material objects.

    If you are at a radius large enough for a stable orbit to be possible (see above), then yes, it is more efficient to move tangentially (I assume that by "efficient" you mean "expending less rocket power/fuel/energy"), since once you've achieved a stable orbit you can stay in it indefinitely without expending any more energy. Even down to the radius at which light can orbit the hole (1.5 times the horizon radius), although you will have to expend rocket power to maintain altitude, it will take less rocket power to do so if you are moving tangentially.

    However, below 1.5 times the horizon radius (but still above the horizon), this is no longer true; instead, it takes *more* rocket power to maintain altitude if you are moving tangentially. In this region, the most efficient way to maintain altitude is to hover, not moving sideways at all. Some books and papers refer to this as "centrifugal force reversal".

    Inside the horizon, efficiency no longer really matters if you are trying to avoid the singularity, since there is *no* way to maintain a constant altitude at all. See further comments below.

    There is no "escape velocity" in any real sense inside the event horizon. Some books and papers will say that the "escape velocity" inside the horizon is greater than the speed of light, but that just amounts to saying that escape is impossible, since nothing can move faster than the speed of light.

    In fact, not only is escape impossible from inside the horizon, you can't even orbit the hole inside the horizon. You can't stay at the same radius by *any* means whatsoever. In order to stay at the same radius, by whatever means, you would have to move faster than the speed of light.

    If you want to at least make the time you will experience prior to hitting the singularity as long as possible, it turns out that the best way to do that is to just let yourself free-fall into the hole. Any expenditure of rocket power will shorten the time you experience prior to hitting the singularity. But you'll still hit it, no matter what you do.
     
  7. Oct 24, 2012 #6

    tom.stoer

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    Re: The event horizon

    And light can stay "frozen" on the horizon.

    The horizon is a light-like 2-surface, i.e. a surface on which light-like motion is possible. A photon emitted radially outwards directly at the horizon stays on this horizon-surface. Or the other way round: the horizon can be defined as the set of all light rays with fixed radial distance from the singularity.

    For the Schwarzschild metric with coordinates (t,r,Ω) there are light-like solutions with r(t)=rs=const and Ω(t)=const.
     
  8. Oct 24, 2012 #7
    Re: The event horizon

    As far as I am aware, existing gravitational theory (i.e. GR) cannot answer this question. The mathematics (and physics) of GR breaks down at the event horizon. What happens to anything inside the event horizon is unknown.
     
  9. Oct 24, 2012 #8
    Re: The event horizon

    That would seem to be an unwarranted conclusion. Based on GR, looking at the event horizon and just outside, one can say that escape is impossible once the event horizon is reached. What happens inside though is not described by GR as per my understanding.
     
  10. Oct 24, 2012 #9

    Nabeshin

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    Re: The event horizon

    This is incorrect -- GR describes the situation inside the event horizon just fine (what it doesn't describe is the single point of the gravitational singularity).

    You might be thinking that in the schwarzschild coordinates, the metric is singular at the horizon. But this is just a coordinate artifact -- in Kruskal-Szekeres coordinates, for example, it's easy to see that you can describe the entire spacetime (- the singularity) just fine.
     
  11. Oct 24, 2012 #10
    Re: The event horizon

    This is absolutely news to me, so thanks for the info.

    If I understand correctly, it is a limitation of the Schwarzchild solution, and not of GR, that events within the event horizon cannot be described. In general, GR can be used to define what happens within the even horizon exactly (except at the singularity itself which remains out of bounds for GR also)?

    Very helpful in correcting a wrong idea I had been carrying for a long time.
     
  12. Oct 24, 2012 #11

    PeterDonis

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    Re: The event horizon

    Yes, I should have clarified that the light orbiting the hole at 1.5 times the Schwarzschild radius is moving purely tangentially. The light that is "frozen" at the horizon is moving purely radially (outward).

    And zero tangential motion, as you note next when you say Ω(t)=const.

    (I know you know all this; I want to make sure the OP understands the distinction.)
     
  13. Oct 24, 2012 #12

    PeterDonis

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    Re: The event horizon

    No. The "Schwarzschild solution" describes events inside the horizon just fine.
     
  14. Oct 24, 2012 #13
    I greatly appreciate all of your responses. Basically what it comes down to is I have a lot of learning to do regarding General Relativity. Thanks.
     
  15. Oct 25, 2012 #14
    Re: The event horizon

    Can that be explained a bit more please - that is, how the light pointing directly outward is actually moving towards the singularity?
     
  16. Oct 25, 2012 #15
    In Newtonian physics it can't. 'Pointing' takes on a different meaning inside an BH event horizon.

    In GR when you pass inward and cross the event Horizon, at distance r = 2M from the center singularity, the mathematics changes such that the radial distance [r] becomes time [the sign changes to minus, meaning time] and is such that time flows inward towards the singularity....so everything inside the horizon moves in that 'direction'...to the future....nothing can move 'out' to the past.
     
  17. Oct 25, 2012 #16

    PeterDonis

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    Re: The event horizon

    Naty1's answer is basically correct (but see the response to him that I'm about to post for some clarifications), but may not be very easy to picture. Here's another way of looking at it, based on the river model of black holes:

    http://arxiv.org/abs/gr-qc/0411060

    The idea of the river model is that space is flowing inward towards the singularity, like a river. Anything that moves outward has to move against the river flow. Inside the horizon, the river is flowing inward faster than light, so even light that is moving directly outward is getting pulled inward by the river flow faster than it can "swim upstream" against it.
     
  18. Oct 25, 2012 #17

    PeterDonis

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    A clarification: this is only true in Schwarzschild coordinates. In other coordinate charts (such as the Painleve chart, which is the basis for the river model that I referred to in my previous post), the radial r coordinate remains spacelike all the way to the singularity.

    Another way of looking at this is that the term "radially outward" doesn't fully pin down a particular spacetime direction by itself; you also have to pick a particular set of surfaces of constant time (or, equivalently, a particular time coordinate). At a given event, for a given observer following a particular infalling worldline, there are actually *three* [edit: at least three--see below] possible choices for those surfaces:

    (1) The surfaces of constant Schwarzschild time. These surfaces are timelike, so vectors pointing "radially outward" in them are also timelike (and pointing "into the past").

    (2) The surfaces of constant Painleve time. These surfaces are spacelike, so vectors pointing "radially outward" in them are also spacelike.

    (3) The surfaces of constant time in the local inertial frame of the infalling observer. These surfaces are also spacelike, as are vectors pointing radially outward in them. What's more, light that is pointed radially outward *does* move outward in this frame. (Viewed from a global chart such as Painleve coordinates, what is happening is that the outward-moving light "swims upstream" relative to the infalling observer, so the observer sees it move outward relative to him.)

    Another clarification: robinpike was asking about outward-moving *light*, and light follows null curves, not timelike (or spacelike) ones. So the meaning of "light moving radially outward" at a given event depends on which *null* vectors at that event are pointing "radially outward". In the case we're discussing, this ends up meaning more or less the same thing as what we've been saying, but it's well to bear the distinction in mind.

    [Edit: Actually there are more than three choices, because there are more than three possible coordinate charts. At least two other global ones deserve mention, Eddington-Finkelstein and Kruskal. Both of these have surfaces of constant "time" that are different from the three I described above.]
     
    Last edited: Oct 25, 2012
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