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Evaluating time for falling body to reach event horizon

  1. Dec 2, 2012 #1
    Hi everyone,
    I have read a few different ways of looking at this problem, and it's one of those things where I am happywith the answer, just not how to get there using proper mathematics. My lecturer described this with some complex integrals involving E (but I'm not sure what that is!) but I have found a simpler treatment in a textbook, which revolves around this idea:

    Looking at the Schwarzschild metric (in natural units with c = 1), we can get to

    dt/dr = (1 - 2GM/r)

    so t is the integral of (1 - 2GM/r) from infinity (distant starting point) to 2GM (the event horizon or Schwarzschild radius).


    I know the answer is infinity as to a distant observer a falling object never reaches the event horizon, and I understand the physics of why that happens. I am not sure how to properly evaluate this integral to prove it though. My calculus is pretty poor, so I always try and break these things down as much as possible. I got

    t = ∫(1 - 3GM/r) dr

    t = ∫1 - ∫(2GM/r) dr

    t = ∫1 - 2GM∫(1/r) dr

    t = [r] - 2GM[ln r]

    So when r = 2GM I get

    t1 = 2GM - 2GM ln 2GM or just t = x - x ln x

    and when r = infinity

    t2 = ∞ - 2GM ln ∞

    and infall time t = t2 - t1

    That all looks like total rubbish to me, and certainly doesn't look like a clear answer that dt = ∞ which is what I'd like to get to. Is there a better way to approach this or have I made a stupid mistake/drastic oversimplification somewhere?
     
  2. jcsd
  3. Dec 2, 2012 #2

    HallsofIvy

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    Writing [itex]\infty[/itex] as if it were a number certainly is "rubbish"! Don't do it. If f(x) has anti-derivative F(x), then [itex]\int_a^\infty f(x) dx[/itex] is defined as [itex]\lim_{x\to\infty} F(x)- F(a)[/itex]. In other words, you need to take a limit.

    [itex]\lim_{x\to\infty} x- Cln(x)[/itex] can be done by letting [itex]y= x- Cln(x)= x- ln(x^C)[/itex] so that [itex]e^y= e^{x- ln(x^C)}= e^x/x^C[/itex]. If we apply L'Hopital's rule C times (assuming for the moment that C is an integer) we reduce to [itex]e^x/C[/itex] which has [itex]\infty[/itex] as limit and, since [itex]e^y[/itex] is an increasing function, that tells us that the original limit is [itex]\infty[/itex]. If C is not an integer, it must lie between two integers and comparing to them, we see that the limit is still [itex]\infty[/itex].
     
  4. Dec 2, 2012 #3

    mfb

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    If you start "at infinity" without any velocity, your infalling time "is infinite", that is not meaningful. Consider a large, but finite distance, instead. The solution should correspond to Newtonian gravity (+ maybe corrections of the order of Rs/c).
     
  5. Dec 2, 2012 #4

    Mentz114

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    Whose time are you evaluating ? If a body takes time τ, measured by a comoving clock, to reach the EH, then a distant observer will see this as τ/(1-2m/r) which is infinite at r=2m.
     
  6. Dec 2, 2012 #5
    Thanks for all your responses! I thought I understood the physics, but I wonder now if I have framed the question correctly? I will respond to all your posts to see if I have understood your answers! If you could spare the time for further clarification I would really appreciate it.

    HallsofIvy: I think what you are saying is that rather than trying to evaluate the integral at infinity, we see how it behaves as the variable tends towards infinity? I can see how that gives the correct final result, but this result comes from the functions behaviour at r = ∞. In the "physics" argument, the object is in boring old Minkowski spacetime at very large r, and it is the coordinate singularity that occurs when r = 2GM that gives the infinite time interval. That doesn't seem to be reflected in your mathematical argument, but maybe that is because I am reading it wrong!

    mfb: My understanding is that we have the starting distance at r = ∞, because here the Schwarzschild metric becomes the MInkowski (flat) metric, so we can properly evaluate the behaviour of the test mass as it travels towards the black hole. Your suggestion that the answer should be Newtonian with a correction factor is a bit odd to me (although I am definitely not trying to suggest that you're wrong!) because I thought this situation was used to demonstrate a clear "weirdness" resulting from GR, ie we get a result which is totally counter-intuitive to Newtonian gravity.

    Mentz114: I am trying to calculate the time as measured by a stationary distant observer, and I understand that a comoving clock on the infalling test mass will just keep on ticking happily until it reaches the central (true) singularity. I understand the basic physics at least! I recognise your statement about how a distant observer will see a comoving time τ as τ/(1-2m/r) which is infinite at r=2m. How do we express that mathematically though, when r is constantly changing from ∞ to 2m (or 2GM in my half-natural units!)? I think that's why we need an integral, which takes me neatly around in a circle to where I started . . . :)
     
  7. Dec 2, 2012 #6

    mfb

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    Far away from a black hole, the gravitation is similar to Newton's gravitation.
    It can get weird close to the black hole, if you take the time it needs for the distant observer (see Mentz114).
     
  8. Dec 2, 2012 #7

    Mentz114

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    The proper time can be found from the the integral
    [tex]
    \tau=\int\ d\tau
    [/tex]
    which is finite, but if we reparametrize in terms of t then
    [tex]
    t = \int \frac{d\tau}{dt}dt
    [/tex]
    this integral diverges.
     
  9. Dec 2, 2012 #8

    pervect

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    I'm trying to figure out how you or your textbook got this - the only thing I can imagine i that you're considering the metric of an infalling light ray. If we consider the expression as r->infinity, we get

    dt/dr -> 1

    which is the solution for light. The more usual equation used involve the [add] Energy at infinity, E, a conserved constant of motion in the Schwarzschild geometry.

    So in short, I'm guessing you set ds=0 (which is correct for light) to derive this expression. The fuller treatment with E can handle falling bodies that are not light.

    As far as definining E goes, one can write:

    [tex]\frac{dt}{d\tau} =\frac{E}{1-2M/r} [/tex]

    Here [itex]\tau[/itex] is proper time. This brings up the other question - you are asking about the "time" for the object to reach the black hole. Are you trying to calculate the proper time , [itex]\tau[/itex], or the coordinate time, t? Do you know the difference? (I suspect you should, but I'm not sure of your background, and considering some of the cofusion I've run into on this issue in the past, I thought I should ask.).

    Of course, if you are looking for the proper time [itex]\tau[/itex], you won't find it by calculating the light-like case.
     
    Last edited: Dec 2, 2012
  10. Dec 3, 2012 #9

    Bill_K

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    One of the remarkable features of the Schwarzschild solution is in how many ways it imitates the Newtonian field. In Newtonian theory, radial motion of a test particle is governed by the energy integral,

    E = ½mv2 - GMm/r

    On the other hand, a radial geodesic in Schwarzschild obeys

    (dr/dτ)2 = 2E/m + 2GM/r

    Exactly the same equation except for the replacement of universal time with the particle's proper time.
     
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