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Schwartzschild metric difficulty

  1. Apr 12, 2012 #1
    I am trying to understand how the Schwartzschild metric works and coming to the conclusion that if I drop an object that it will not fall to the earth but just stay there when I open my hand. Therefore, I'm confident I have made a mistake, but I don't see where it is.

    Here is the metric directly from Schwartzschild's paper. This agrees with various other sources I've seen although the notation is different.

    ds2 = (1 - α/R)dt2 - dR2/(1 -α/R) - R2(dθ2 + sin2θ d∅)

    where α is the Schwartzschild radius, t is the time coordinate, and R is the radial coordinate. Please note that Schwarzschild uses units where c = 1.

    I'm just interested in dropping a clock, so I set dθ and d∅ to zero, reducing the equation to

    ds2 = A dt2 - dR2/A

    where I have defined A = (1 - α/R) for brevity.

    The traveler along this line element will see himself "at rest" and so ds2 will equal dτ2, where τ is the proper time. dτ will equal f dt for some function f of R that reflects slower ticking of the clock as it falls. Thus, we have

    f2 dt2 = A dt2 - dR2/A

    Rearranging a bit, I get

    dR2/A = A dt2 - f2 dt2

    Solving for dR/dt yields

    dR/dt = (A2 - Af2).5

    I then take the derivative to get the acceleration.

    d2R/dt2 = .5 (A2 - Af2)-.5 (2A dA/dt - A 2f df/dt - f2 dA/dt))

    But, dA/dt = -α/R2 dR/dt, and by hypothesis dR/dt = 0 (we are just dropping an object, whose velocity is initially 0). Therefore, we get

    d2R/dt2 = .5 (A2 - Af2)-.5 (- A 2f df/dt))

    Since f is a function of R, df/dt = df/dR dR/dt = 0 also initially.

    But that means d2R/dt2 = 0 initially as well. So the object does not move!

    What's wrong?
     
  2. jcsd
  3. Apr 12, 2012 #2

    Mentz114

    User Avatar
    Gold Member

    You're using trying to use dynamics, but GR is a kinematic theory. This means that the curvature predefines trajectories that will be followed by test particles with given initial conditions and/or proper accelerations. So it is possible to find the 4-velocity of a dropped object but not the way you're attempting.

    You should look this up in a textbook, or maybe start with this (very good) article in Wiki.

    http://en.wikipedia.org/wiki/Schwarzschild_geodesics
     
  4. Apr 12, 2012 #3
    It might be worth noting that:

    f2 = A - dR2/(dt2A)

    so f is not just a function of R, but a function of R and dt.

    These links show how the Schwarzschild acceleration is obtained:

    http://www.mathpages.com/rr/s6-07/6-07.htm
    http://www.mathpages.com/rr/s6-04/6-04.htm
     
    Last edited: Apr 12, 2012
  5. Apr 12, 2012 #4
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