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Geodesics in schwarzschild solution

  1. Mar 21, 2006 #1
    Hello everybody,

    I was studying the lecture notes about the schwarzschild solution for general relativity. In a particular example they calculate the equations of motion of a particle falling straight into a black hole. But there are some things about the calculation I really don't get.

    First of all they say that because the schwarzschild metric is time independent, so dg/dt = 0 => u_0 = const. Then they show that by using the initial conditions r = R and tau = 0 and t = 0 coincide, that u_0 equals to sqrt( g_00(R) ) = sqrt(1 - 2M/R).

    I've tried to prove the constancy of u_0 from the geodesic equation, but all I got were some nasty coupled differential equations.

    Somehow it is true that constancy of metric in one component direction implies that the four-velocity in that component is constant. But I don't see why that is true. And how can you show from this, that this means that u_0 = sqrt(1-2M/r) ?
  2. jcsd
  3. Mar 21, 2006 #2

    George Jones

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    Are you familiar with the Lagrangian approach to geodesic motion in GR. What happens when a Lagrangian is independent of one of the generalized coordinates?

    In this case, the conserved quantity gives

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

    Take a look at https://www.physicsforums.com/showpost.php?p=621802&postcount=32" of mine. Note that I forgot to "divide" by the [itex]d\tau[/itex]'s in the definition of the Lagrangian.

    Last edited by a moderator: Apr 22, 2017
  4. Mar 21, 2006 #3


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    Oops, I'm afraid that I totally missed the point on my first response :-(. So I've deleted it and will start over.

    Let us suppose that we have some paramaterized infalling trajectory [itex]r(\tau), t(\tau)[/itex] for a particle falling straight into the black hole.

    The "messy diffeqs" you get via the geodesic equations should be, omitting all the zero Christoffel symbols should be eq1 and eq2 below.

    \frac{d^2 t}{d \tau^2} + 2 \Gamma^t{}_{rt} \frac{dt}{d\tau}\frac{dr}{d\tau} = 0

    \frac{d^2 r}{d \tau^2} + \Gamma^r{}_{rr} \frac{dr}{d\tau}\frac{dr}{d\tau} + \Gamma^r{}_{tt} \frac{dt}{d\tau}\frac{dt}{d\tau} = 0

    Now, consider the equation

    g00(r(\tau)) \, \frac{dt}{d\tau} = C

    where C is some constant. We can rewrite this as

    \frac{d g00(r)}{dr} \, \frac{dr}{d\tau} \frac{dt}{d\tau} + g00(r) \,\frac{dt^2}{d\tau^2} = 0

    by making use of the fact that [itex]dC/d\tau[/itex] = 0 and applying the chain rule for differentiation.

    Normalizing the coefficeint for the second derivative by dividing by g00, we get

    \frac{dt^2}{d\tau^2} + \frac {\frac{d g00}{dr}} {g00} \frac{dt}{d\tau} \frac{dr}{d\tau} = 0

    When we substitute g00(r) = -1 + 2m/r and for [itex]\Gamma^t{}_{rt}[/itex], we find then that the messy-looking eq1 is just our eq3.

    eq3 didn't really spring out of thin air, it can be motivated in several ways - see George's posts, or read up on "Killing vectors". Meanwhile, it's good to know that eq3 is really just one of the geodesic equations (i.e eq1) in a simpler form.

    If you like, you can use eq3 to eliminate [itex]dt/d\tau[/itex] entirely from eq2, leaving you a differential equation for r.
    Last edited: Mar 21, 2006
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