Strange geodesic in Schwartzschild metric

  1. The following curve is geodesic in Schwardschild metric:
    [tex] \tau \mapsto [(1-2m/r_0)^{-1/2}\tau,r_0,0,0][/tex].
    The tangent vector is: [tex] [(1-2m/r_0)^{-1/2},0,0,0] [/tex], its length is 1 and its
    product with killing vector [tex]\partial_t [/tex] is equal: [tex] (1-2m/r_0)^{1/2} = \textrm{const}[/tex]. So the body lays at rest in gravitational field - why it's possible??
    In newtonian limit it's impossible - the body which does not rotate around a star cannot
    have constant radious.
  2. jcsd
  3. Ich

    Ich 1,931
    Science Advisor

    Constant energy is necessary for a geodesic, not sufficient. Plug this tangent vector in the equation of motion, you'll get [itex]\ddot r \neq 0[/itex].
    Btw., it's Schwarzschild.
  4. George Jones

    George Jones 6,470
    Staff Emeritus
    Science Advisor
    Gold Member

    This not a geodesic. If

    [tex]\mathbf{u} = \left( u^t , u^r, u^\theta, u^\phi, \right) = \left( \left( 1 - \frac{2m}{r_0} \right)^{-\frac{1}{2}}, 0, 0, 0 \right),

    then the 4-acceleration is given by

    \mathbf{a} &= \nabla_{\mathbf{u}} \mathbf{u} \\
    &= u^\alpha \nabla_{\partial_\alpha} \left( u^\beta \partial_\beta \right) \\
    &= u^\alpha \left( \nabla_{\partial_\alpha} \left( u^\beta \right) \partial_\beta + u^\beta \nabla_{\partial_\alpha} \left( \partial_\beta \right) \right) \\
    &= \left( u^t \right)^2 \Gamma^\mu {}_{tt} \partial_\mu

    which is non-zero.
    Last edited: Feb 22, 2010
  5. George Jones

    George Jones 6,470
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    I was waiting for comments before finishing this off.


    [tex]0 = \Gamma^t {}_{tt} = \Gamma^\theta {}_{tt} = \Gamma^\phi {}_{tt}[/tex]


    [tex]\Gamma^r {}_{tt} = \left( 1 - \frac{2m}{r_0} \right) \frac{m}{r_0^2}[/tex]


    [tex]\mathbf{a} = \left( 0, \frac{m}{r_0^2}, 0, 0 \right)[/tex]

    with magnitude

    [tex]a = \left( 1 - \frac{2m}{r_0} \right)^{-\frac{1}{2}} \frac{m}{r_0^2}[/tex]

    Taking [itex]r_0[/itex] to be much larger that the Schwarzschild radius, and restoring [itex]c[/itex] and [itex]G[/itex] gives

    [tex]a = \frac{Gm}{r_0^2}[/tex].

    Consequently, such a hovering observer experiences normal Newtonian weight.
  6. Mentz114

    Mentz114 4,507
    Gold Member

    George, that's very instructive, thank you. I'm still reading Lee's book and I've bookmarked this thread.
  7. A very confusing thing here is the use of [tex]m[/tex] to denote both half of the Schwarzschild redius and the mass of gravitating body! I think in textbooks whose authers prefer using the notation [tex]2m[/tex] instead of [tex]r_s[/tex] to symbolize the Schwarzschild redius, they later use


    where [tex]M[/tex] is the mass of mass of gravitating body. But I respect George's style and accept it as another alternative.:wink:

  8. DrGreg

    DrGreg 1,986
    Science Advisor
    Gold Member

    There is a convention that many authors of advanced texts use, as well as choosing units of distance and time such that c=1, they also choose units of mass such that G=1. It can cause confusion to persons unfamiliar with it.
  9. But you didn't notice that George put a G in the last equation which means the convention that I probably seem to have forgotten leads to


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