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A How to derive equations of motion in GR?

  1. Jul 4, 2017 #1
    Question Background:
    I'm considering the Eddington-Robertson-Schiff line element which is given by
    [tex] (ds)^2 = \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) dt^2 - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (dr^2 + r^2 d\theta^2 + r^2 \sin^2{\theta} \;d\phi^2 ),[/tex]
    where [itex]\mu = GM = \text{const.}[/itex] and [itex] r=|\mathbf{r}|.[/itex]
    I'm interested in determining the equations of motion for such a line element which can be obtained from the least action principle. The classical action [itex]S[/itex] is the integral along the particle trajectory
    [tex] S = \int ds, [/tex]
    which can be equivalently expressed as
    [tex] S = \int \left( \frac{ds}{dt} \right) dt \equiv \int L \; dt. [/tex]
    We can see from the above that
    [tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right]^{1/2}, [/tex]
    where [itex]L[/itex] is the associated Lagrangian over time.
    Problem and question
    The associated equations of motion are given by (Eq. 20)
    [tex]\frac{d^2\mathbf{r}}{dt^2} = \frac{\mu}{r^3} \left[ \left(4 \frac{\mu}{r} - v^2 \right) \mathbf{r} + 4 (\mathbf{r}\cdot \mathbf{\dot{r}} ) \mathbf{\dot{r}}\right]. [/tex]
    I cannot for the life of me obtain this using the Euler-Lagrange equations.
    Attempt at a solution:
    The Euler-Lagrange equations are given by
    [tex] \frac{d}{dt} \left( \frac{\partial L}{\partial \mathbf{\dot{r}}} \right) - \frac{\partial L}{ \partial \mathbf{r}} =0. [/tex]
    I note that the equations of motion should be equivalent for either
    [tex] L = \sqrt{g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu}, [/tex]
    or
    [tex] L = g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu. [/tex]
    Bearing this in mind and working through the process using
    [tex] L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right], [/tex]
    I find
    [tex] \frac{d}{dt} \left( \frac{\partial L}{ \partial \mathbf{\dot{r}}} \right) = -2 \left[ \left( 1 + 2 \frac{\mu}{r} \mathbf{\ddot{r}} \right) - 2 \frac{\mu}{r^3} (\mathbf{r}\cdot \mathbf{\dot{r}}) \mathbf{\dot{r}} \right], [/tex]
    and
    [tex] \left( \frac{\partial L }{\partial \mathbf{r}} \right) = 2\frac{\mu}{r^3} \mathbf{r} - 4 \frac{\mu^2}{r^4} \mathbf{r} + 2 \frac{\mu}{r^3} \mathbf{r} (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}} ). [/tex]
    Clearly, adding these together does not give the desired result. Any suggestions?
     
  2. jcsd
  3. Jul 4, 2017 #2

    Orodruin

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    This is only true if the geodesic is parametrised by an affine parameter. The coordinate ##t## is in general not affine.
     
  4. Jul 4, 2017 #3
    O wow! If this was the issue the whole time I will be very pleased but also annoyed at my ignorance!!
     
  5. Jul 4, 2017 #4

    Hmmm I am confused now! So, normally in GR we can set the Lagrangian = ##\pm c## depending on the signature of the line element. Is this only true when the geodesic is parametrised by an affine parameter also? Even using the ## L = \sqrt{ }##, I can't seem to manage to get the correct expression!!
     
  6. Jul 4, 2017 #5

    Orodruin

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    In general, it will just give you a requirement that gives an affine parametrisation (constant length tangent vector). It tells you nothing about whether or not you found a geodesic.
     
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