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First integral of Lagrangian

  1. Nov 13, 2011 #1
    I am trying to find the equations of motion for a test particle in the schwarzschild metric. However, I cannot find the correct first integral for the Lagrangian.

    The Schwarzschild metric is:
    [tex]ds^2 = -\left(1-\frac{2M}{r}\right)\,dt^2+\left(1-\frac{2M}{r}\right)^{-1}\,dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2[/tex]
    If we look at the action, and parameterise it using [itex]s=\frac{\tau}{m}[/itex],
    [tex]S=\int\,ds = \int_{\tau_i}^{\tau_f} \sqrt{g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}m^2}\left(\frac{1}{m}\,d\tau\right)[/tex]
    If we DEFINE our lagrangian to be [itex]L=\frac{1}{2}m^2g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}[/itex], then
    [tex]S = \int_{\tau_i}^{\tau_f} \frac{\sqrt{2L}}{m}\,d\tau[/tex]
    Due to reparameterisation invariance, we can set once again set [itex]\tau=s[/itex], and in that case we get [itex]\frac{\sqrt{2L}}{m}=1\Rightarrow L = \frac{m^2}{2}[/itex].

    However, in the book I'm reading, "Black Holes, White Dwarfs, and Neutron stars (Shapiro)", they have [itex]L=-\frac{m^2}{2}[/itex]. This is correct, since you need it to find the equation of motion for the test particle. However, I can't seem to get the minus sign. How does it come about?

  2. jcsd
  3. Nov 13, 2011 #2


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    With the signature you're using, gμνdxμ/dτ dxν/dτ is negative along a timelike curve, so you want to integrate ds = √-gμνdxμ/dτ dxν/dτ dτ
  4. Nov 13, 2011 #3
    Thanks Bill_K, but I'm not exactly sure I understand.

    By signature I'm guessing you mean
    instead of

    But when we take the action, why do we need to integrate over a timelike curve? What's wrong with integrating over a spacelike curve? After all, isn't the definition

    [tex]S = \int\,ds[/tex]
  5. Nov 13, 2011 #4


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    Particles don't move along space-like curves. They must move on time-like curves.
  6. Nov 14, 2011 #5
    ah of course. cheers
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