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Timelike Geodesics

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the Reissner Nordstrom line element, which I've given in the relevant equations section, I'm looking to show that the time like Geodesics obey the equation again show below.

    2. Relevant equations
    Line Element

    ##ds^2= - U(r)c^2dt^2 +\frac{dr^2}{U(r)} +r^2(d\theta^2 + sin^2(\theta)d\phi^2)##
    ##U(r)=1-\frac{r_s}{r}+\frac{G^2Q^2}{r^2}##

    Equation to Obey
    ##\frac{1}{2} (\frac{dr}{d\tau})^2 +V(r) = \varepsilon##


    3. The attempt at a solution
    I've presumed as we are looking for a ##dr'## the Euler Lagrange equation we would be interested would be

    ##\frac{d}{d\tau}(\frac{\partial L^2}{\partial r'}) - \frac{\partial L^2}{\partial r}##

    If I work this through my answer doesn't really resemble the equation I'm looking for, I get the ##(\frac{dr}{d\tau})^2##, but I can't get the ##\frac{1}{2}## factor, plus I have other terms in the denominator.
    I also have a lot of other terms but they could possibly be grouped in to ##V(r)##.

    I was hoping somebody could confirm whether the method I'm attempting is correct, as then I'll know if I'm incorrectly calculating it or it is something else.

    Many thanks.
     
  2. jcsd
  3. Nov 14, 2014 #2

    stevendaryl

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    You're on the right track, but why don't you first write down what you think [itex]L[/itex] is. Your equations don't define [itex]L[/itex].
     
  4. Nov 14, 2014 #3

    stevendaryl

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    Here are a couple of other hints:

    First, [itex]\tau[/itex] and [itex]s[/itex] are the same thing, so [itex]\frac{ds}{d\tau} = 1[/itex]. So if you take the expression for [itex]s[/itex], this gives you one "constant of the motion".

    Second, if you have a Lagrangian of the form [itex]L(r, \frac{dr}{d\tau}, t, \frac{dt}{d\tau}, \theta, \frac{d\theta}{d\tau}, \phi \frac{d\phi}{d\tau})[/itex], and [itex]L[/itex] doesn't mention [itex]\tau[/itex], then the following quantity is conserved (has the same value for all [itex]\tau[/itex]):

    [itex]H = (\sum_j P_j U^j) - L[/itex]

    where [itex]U^j = \frac{d x^j}{d\tau}[/itex] and [itex]P_j = \dfrac{\partial L}{\partial U^j}[/itex].

    So [itex]H[/itex] gives you a second constant of the motion. So [itex]H = E[/itex], for some constant [itex]E[/itex]
     
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