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Throwing something into a black hole

  1. Dec 30, 2014 #1
    I'm a bit puzzled by the dynamics of things falling into a black hole. If I start with a test mass at infinity, it will fall freely into the black hole and reach the speed of light at the event horizon.
    What happens if I throw something towards the black hole? Will it already reach the speed of light before reaching the event horizon?
    On the other hand - if I start with an object at a certain distance of the BH but that is kept still, it will also fall onto the BH, but be slower than the particle falling from infinity at every point.
    Will the speeds of the three objects as seen from the reference frame of the particle falling from infinity be the same at the event horizon or will it be different? (I assume that one can use a Gullstrand-Painleve metric to calculate this, but my maths abilities are not sufficient to actually do this...)
    And looking from far away - if all particles reach c at the event horizon, what about energy conservation? (After all, the particles started with different energies.)
  2. jcsd
  3. Dec 30, 2014 #2
    I'm no pro in Relativity, but this is what I can gather from all the knowledge I have of black holes:
    The event horizon is not defined as the point at which objects reach the speed of light; rather it's defined as a boundary beyond which there is no return: the curvature of spacetime becomes such that the escape velocity beyond that point become the speed of light, so no matter which direction you take after crossing it, you'll always end up going towards the black hole. The only way "out" is to go back in time, which is essentially impossible. With the frame of reference of any point outside the event horizon of the black hole, all objects that approaches the event horizon appear to slow down and eventually almost stop, and the light that we see from them appears to be redshifted and becomes dimmer. The reason for this is that light loses energy as it comes closer to a black hole due to the strong attractive pull on it, and hence shifts towards longer wavelengths, so that it reduces in luminosity, until the object can no longer be seen.

    Ideally, since a black hole possesses infinite gravity, all objects, no matter how far, experience forces of attraction. However, if you say an object is released from "infinity", then that means (by definition) it will never experience any force of attraction (zero to be exact) and never reach the event horizon (it will reach in infinite time). Also, a frame of reference of a point at "infinity" will mean that you see nothing, since light will never reach you (crossing infinite distance requires infinite time again). I'm guessing what you mean to ask is whether there will be any difference in the velocities reached by objects at different distances from the black hole. If that is so, then yes, there will be variation, just like how meteorites from far away in the solar system hit the Earth at a higher velocity than some other space debris from nearby the atmosphere. You should also recognize that the speed of light is a universal speed limit, and that the relativistic mass of any object tends to infinity as it approaches the speed of light. For objects to reach this velocity, matter must first be converted into light (energy/packets of quanta). However, accelerations required to make an object approach an appreciable percentage of light will only be experienced by objects in close proximity to the black hole(unless the black hole is bigger than anything we know of), so my guess is that objects will never reach the speed of light no matter where they are released from unless they're really close to the event horizon (I don't think it's necessary for this to happen at the event horizon, because colliding matter can get converted to ionized gas and sub-atomic particles fairly quickly in the presence of such strong G-forces, and these particles can spontaneously get converted to energy).
  4. Dec 30, 2014 #3


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    Gold Member

    Uh ... you may want to research that a bit, since you are grossly mistaken.
  5. Dec 30, 2014 #4
    My bad.
  6. Dec 30, 2014 #5
    After thinking about it a bit more, I think there is no problem with having an object at the event horizon that falls with less than c - since the freely-falling particle (which is a valid reference frame) falls with c, no particle can escape (and exactly at the horizon, light would "hover" forever). So the problem I saw is probably not existing at all. Nevertheless, I'd be grateful for a brief confirmation by an expert...
  7. Dec 30, 2014 #6
    Confirmed! ;)
    (by a PHD student)
  8. Dec 30, 2014 #7


    Staff: Mentor

    No, it won't. When it crosses the horizon, the horizon will be moving at the speed of light relative to the falling object; but that's because the horizon is an outgoing lightlike surface. The falling object remains timelike.

    No, it doesn't; see above. If it were falling at speed c, it could not be a "valid reference frame"; there are no valid reference frames that move at c.

    This part is true, but the reason it's true is that phrase in parentheses: at the horizon, light that is moving directly outward "hovers" at the same radius forever. So anything moving slower than light, even if it's moving directly outward, must be actually decreasing the radius it is at with time. But relative to an object freely falling into the black hole, another timelike object could be moving outward at just a little less than c; it would still be trapped inside the horizon, because the horizon, relative to the object freely falling into the hole, is moving outward exactly at c.
  9. Dec 30, 2014 #8

    Thanks a lot - I'm always misphrasing these things because I tend to get confused with the different reference frames.
    I got the basic concept of things falling into the black hole from this site:
    where the black hole is compared to a waterfall (with space falling in at a speed of c at the event horizon, and faster than c inside). However, I have a lot of trouble in understanding what reference frame is used to state this.
  10. Dec 30, 2014 #9


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    To add on to Peter's (as usual) excellent reply, coordinate velocities are not physical; if an observer wants to measure the physical velocity, henceforth just velocity, of some freely falling observer passing the event horizon then of course one needs the local Lorentz frame of this observer at the event for which the worldlines of the two intersect. Since static local Lorentz frames do not exist on the event horizon it doesn't make sense to talk about the velocity of the freely falling observer relative to a static observer; this is of course because the timelike Killing field becomes null on the event horizon.

    Since the coordinate velocity of the freely falling observer is just a rescaling of the velocity relative to the static frame, this explains why one gets the unphysical value of ##c## on the event horizon for the coordinate velocity. One can of course measure the velocity of one freely falling observer at the event horizon relative to the freely falling frame of another coincident freely falling observer and one will always find that this velocity is less than ##c##. Indeed it is easy to prove that the velocity of any observer moving past a local Lorentz frame is always less than ##c## in any spacetime since spacetime is always locally Minkowskian.

    In that link the non-rigid extended reference frame of the Painleve congruence is being used. This is the congruence of observers freely falling radially from rest at infinity. C.f. http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates#A_rain_observer.27s_view_of_the_universe
  11. Dec 30, 2014 #10


    Staff: Mentor

    No. No infalling object (except an ingoing light ray) will reach the speed of light at any point. Timelike objects always remain timelike; objects that move at the speed of light (like light rays) are lightlike. These are fundamentally different types of objects.


    Different. If we suppose that all three objects reach the horizon at the same instant, then at that instant, relative to the object that fell from rest at infinity (call this object A), the object that was thrown inward (object B) towards the hole will be moving inward, and the object that was released from rest at a finite altitude (object C) will be moving outward. Relative to a global coordinate chart, all three objects will be decreasing their radial coordinate ##r## with time: object B will be doing so faster than A, and object C will be doing so slower than A.

    Yes, this is a good chart to use.

    They are traveling at different speeds when they cross the horizon, as above.
  12. Dec 31, 2014 #11
    The equations in this post from 'Exploring Black Holes' by Taylor and Wheeler pretty much address what you're talking about
  13. Dec 31, 2014 #12
    Thanks, that's absolutely great.
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