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General Relativity - Motion in Newtonian Limit

  1. Jun 25, 2013 #1
    General Relativity -- Motion in Newtonian Limit

    1. The problem statement, all variables and given/known data

    Assume that the metric of space-time in a weak gravitational field, [itex] \frac{\phi}{c^2} << 1 [/itex] is [itex] ds^2 = (1 + \frac{2\phi}{c^2})c^2 dt^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2 [/itex] for some arbitrary scalar function phi.
    Use the variational principle to derive the equation of motion of particles, in the approximation that the velocity is small compared to c. Compare to the equations of Newtonian gravity.

    Note that my teacher has been using the squared version of the metric in our actionals to derive things, so I wanted to follow his example. This however means that our parameterization [itex] \tau [/itex] is not arbitrary and that we need to solve for it from the metric.


    I have no trouble solving for my spacial components as soon as I make the assumption that [itex] \frac{dt}{d\lambda} \approx 1 [/itex]. However I'm having trouble solving for this quantity from my euler-lagrange equations. I could really use some help in figuring out what approximations I am allowed to make here. Specifically I derive the equation

    [itex] \frac{d}{d\tau} [ c^2 \frac{dt}{d\tau} + 2\phi(x) \frac{dt}{d\tau}] - \frac{\partial \phi}{\partial t} (\frac{dt}{d\tau})^2 = 0[/itex]

    which we can divide through and by c squared and do a tiny bit of algebra to find something which looks ready to approximate--

    [itex] \frac{d}{d\tau} [ \frac{dt}{d\tau} +\frac{2\phi}{c^2} \frac{dt}{d\tau}] = \frac{1}{c^2} \frac{\partial \phi}{\partial \tau} (\frac{dt}{d\tau}) [/itex]

    clearly this must solve such that [itex] \frac{dt}{d\tau} = 1 + O(c^{-1}) [/itex] but getting there is a bit of a headache for me.. I was wondering what approximations I might make to reach this final form.

    Edit: My best thought right now is to throw away the right hand side, and for whatever arbitrary reason integrate so that [itex] \frac{dt}{d\tau} = constant - \frac{2\phi}{c^2} [/itex] where our second term is negligible and can be neglected.. however I have no idea what to set the constant or what the justification for throwing away the RHS might be.
     
    Last edited: Jun 26, 2013
  2. jcsd
  3. Jun 26, 2013 #2
    Oh common,it is not that tough.Just do the differentiation on the left,the differentiation of second term on left will give two terms,drop one term accounting for [itex]\phi[/itex]/c2<<1,and one other term will be subtracted from the right(we can not set (1/c2)∂[itex]\phi[/itex]/∂t to zero)..See now
     
  4. Jun 26, 2013 #3
    Thats right, I already realized we could do a little bit of algebra and write [itex] \frac{d^2t}{d\tau^2} = - \frac{1}{c^2} \frac{dt}{d\tau} \frac{d\phi}{d\tau} [/itex] but again I wasn't sure what to do with this term, I considered integration by parts and using our fixed boundary conditions on lambda, but I couldn't get anything too meaningful from that.
     
  5. Jun 26, 2013 #4
    Assume dt/dτ=y.Also dτ will cancel.
     
    Last edited: Jun 26, 2013
  6. Jun 26, 2013 #5
    After substituting I have [itex] \frac{dy}{d\tau} = \frac{-1}{c^2} y(\tau) \frac{d\phi}{d\tau} [/itex]. None of the ODE tables on wikipedia have a solution for a differential equation written as [itex] y' = F'(y) y [/itex].. what am I missing here?

    Edit: After looking at the equation I suppose I could write [itex] \frac{y'}{y} = F'(y) [/itex] and solve for [itex] y =A e^{ F(y(\tau))} [/itex]

    Thank you
     
    Last edited: Jun 26, 2013
  7. Jun 27, 2013 #6
    you can as well now use the expansion of ex and neglect higher order terms in [itex]\phi[/itex]/c2 to get result what you wanted in op.
     
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