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Deriving radial velocity as observed from infinity

  1. Jan 12, 2009 #1
    In Wheeler and Taylor's 'Exploring Black Holes', on pages 3-12 and 3-13, the bookkeeper measure of radial velocity (i.e. radial velocity as measured from infinity) is derived. Basically the equation for 'Energy in Schwarzschild geometry' is established-

    [tex]\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}=1[/tex]

    The book states-

    'From the energy equation and the Schwarzschild metric, we can find an expression for [itex]dr/dt[/itex], the rate of the change of the r-coordinate with far-away time t for a stone starting from rest at a very great distance. To obtain this derivative, square terms on either side of the right-hand equality, multiply through by [itex]d\tau^2[/itex], and equate it to the Schwarzschild metric equation for [itex]d\tau^2[/itex] in the case of radial fall [itex](d\phi=0)[/itex]:

    [tex]\left(1-\frac{2M}{r}\right)^2dt^2=d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\frac{dr^2}{\left(1-\frac{2M}{r}\right)}[/itex]

    Divide through by [itex]dt^2[/itex], solve for [itex]dr/dt[/itex], and take the square root to obtain

    [tex]\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\left(\frac{2M}{r}\right)^{1/2}[/tex]


    I'd appreciate if someone could show the process of derivation between the second and third equation.
     
    Last edited: Jan 12, 2009
  2. jcsd
  3. Jan 12, 2009 #2

    tiny-tim

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    Hi stevebd1! :smile:

    Ignore the dtau2 in the middle, and the coefficient of dt2 becomes (1 - 2M/r)(1 - 2M/r - 1) :wink:
     
  4. Jan 15, 2009 #3
    Hi tiny-tim

    Thanks for the response. I'm probably missing something elementary here but could you shed some light on how you arrived at that coefficient for dt2?
     
  5. Jan 15, 2009 #4

    tiny-tim

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    Yup … in
    rearrange to [tex]\left[\left(1-\frac{2M}{r}\right)^2\ -\ \left(1-\frac{2M}{r}\right)\right]dt^2\ =\ -\frac{dr^2}{\left(1-\frac{2M}{r}\right)}[/tex]

    which is [tex]\left[\left(1-\frac{2M}{r}\right)\left(1-\frac{2M}{r}\right\ -\ 1)\right]dt^2\ =\ -\frac{dr^2}{\left(1-\frac{2M}{r}\right)}[/tex] :smile:
     
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