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MTW 25.29 v.s. 25.38 (first formula)

  1. Sep 22, 2011 #1
    MTW 25.29 v.s. 25.38 (first formula)

    Yesterday I made a quick calculation for a radially free falling observer in a Schwarzschild solution for a given r starting value both starting at rest and at escape velocity.

    I realized that the ratio between those two formulas is always 3/4 Pi. But I have troubles with that because when r->infinity the two formulas should converge.

    Any comments as to where my mistake lies?

    MTW 25.29

    [tex]1/2\,\pi \,R\sqrt {{\frac {R}{2M}}}[/tex]

    MTW 25.38

    [tex]-8/3\,M \left( {\frac {R}{2M}} \right) ^{3/2}[/tex]
  2. jcsd
  3. Sep 23, 2011 #2
    MTW 25.29 is by the way the special case of the more generic:

    [tex]1/2\,\sqrt {{\frac {{R}^{3}}{2M}}} \left( 2\,\sqrt {{\frac {r}{R}}-{
    \frac {{r}^{2}}{{R}^{2}}}}+\pi -\arccos \left( -2\,{\frac {r}{R}}+1
    \right) \right)
  4. Sep 24, 2011 #3
    I don't have access to the part of MTW that derives 25.38, but, if I understand your concern correctly, I'm not sure that anything is wrong. The escape velocity is a critical point.

    Consider that, in the Newtonian case, infall time for an observer starting at the distance R and velocity 0.9*v_escape is finite for all R, and at velocity v_escape it is infinite for all R, even though the two values will converge as R -> infinity. So the ratio of infall times remains equal to infinity. Isn't that worse than having the ratio of 3/4 pi?
  5. Sep 24, 2011 #4
    MTW does not explicitly derive it.

    It is obtained by integrating:

    [tex]{\frac {{\it dr}}{d\tau }}=-\sqrt {{\frac {2M}{r}}}[/tex]

    which can be derived from knowing that a test particle radially falling at escape velocity has an energy E at infinity of exactly 1 and a total angular momentum L of 0 and then plugging it in this 'monster or integration' which is mentioned in MTW (25.27).

    [tex]{\it dr}{\frac {1}{\sqrt {{E}^{2}- \left( 1-{\frac {2M}{r}} \right)
    \left( 1+{\frac {{L}^{2}}{{r}^{2}}} \right) }}}
    Last edited: Sep 24, 2011
  6. Sep 24, 2011 #5


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    I'm confused about a few points. In my copy of MTW (copyright 1973), the first formula at 25.38 is:

    tau = -(2/3) (r/2M)^3/2

    and as I read the derivation of this, its absolute value represents, for any r, the remaining proper time to reach the singularity, given fall to that r from infinity. I presume that's what you mean, but escape velocity confuses me because I would picture that going the other direction (though, of course, the magnitude of speed having fallen from infinity is obviously escape velocity).

    It looks like my my edition likely has a mistake here. There is something much more plausible at the extra M in yours. Without it, for some M, it would purportedly take longer to reach the singularity starting from a faster speed.

    Anyway, the idea of a constant ratio makes sense to me. As long as it takes to fall all the way from some great distance at rest, it takes a fixed factor less to fall with inverse escape velocity. At great distance, it takes so long for an at rest body to build up any speed at all, that one starting with the tiny speed of inverse escape velocity develops a big lead, leading to fixed ratio.
  7. Sep 24, 2011 #6
    In MTW the formula is expressed as tau/2M, just getting tau would require you to multiply the formula with 2M.

    A test observer radially free falling from infinity has a vlocal of escape velocity.

    One can physically determine the local velocity by measuring it wrt to a stationary observer at that location.
    This free falling velocity is:
    [tex]\sqrt {{\frac {2M}{r}}}[/tex]

    A stationary observer has a local velocity of 0.
    Last edited: Sep 24, 2011
  8. Sep 24, 2011 #7


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    oops, right.
    Just terminology: I would call escape velocity, as a vector, pointing the other way.
  9. Sep 24, 2011 #8
    The absolute difference between the local velocity for a free falling particle (from infinity) and a stationary observer decreases with increasing values of r right?

    Thus would it not follow that at infinity the local velocity should approach zero which then is equal to a stationary observer at infinity?
  10. Sep 24, 2011 #9


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    You're taking the limit of the ratio of two things that each go to infinity. At infinity, you have infinity/infinity. For anything less, my argument seems plausible, and your derivation proves it. Accept it - you've made no mistake, you've derived an interesting fact.
  11. Sep 24, 2011 #10
    Yes, I do not disagree with the math, I am just trying to wrap my mind around it. :)

    If we take the limit for r->infinity of:
    [tex]\sqrt {{\frac {2M}{r}}}[/tex]
    we get zero velocity, so how is this zero velocity different from the zero velocity of a stationary observer at infinity?

    So do you think there would be a difference between a free falling observer at infinity and a stationary observer at infinity?

    For instance if we keep the energy argument then the former has an energy at infinity of 1 and the latter has an energy at infinity of <1.
    Last edited: Sep 24, 2011
  12. Sep 24, 2011 #11


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    The thing is, you can never be 'at infinity'. Your result can be stated as follows: the farther the starting point, the smaller the velocity 'head start' you need to beat the stationary faller by a fixed factor. Not at an obvious fact, but it seems perfectly plausible to me.
  13. Sep 24, 2011 #12
    Right, and I am trying to understand why that would not imply that the proper time differential between these two test observers with increasing values of r would become smaller as well.

    As you can see in the graph below that is not the case:

    [PLAIN]http://img684.imageshack.us/img684/431/freefallvszerofall.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  14. Sep 24, 2011 #13


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    It's down to intuition here. Both proper times (to reach r=0) increase without bound as starting r increases. The proper time difference between them also increases without bound. But the ratio stays constant. That feels very plausible to me. Further, I read the section of MTW very carefully and I concur in your analysis. Don't know what else I can say.
  15. Sep 25, 2011 #14
    What is MTW?
  16. Sep 25, 2011 #15


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