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GR - Find the orbit of the planet

  1. Feb 25, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    Find the orbit of a planet in a grav. field corresponding to the metric

    [tex]d\tau^2=(1+\alpha^2(x^2+y^2))dt^2-(dx^2+dy^2+dz^2)[/tex]

    in the newtonian limit with the initial conditions at t=0, x=R, dx/dt=[itex]\beta[/itex],0=y=z=dy/dt=dz/dt where alpha and beta are cositive constants.

    2. Relevant equations
    The Levi-Civita connexion and the geodesic equations

    3. The attempt at a solution

    I calculates the connexion coefficients and wrote down the 4 geodesic equations with the proper time as the parameter:

    [tex]\frac{d^2t}{d\tau^2}+\frac{2\alpha^2x}{(1+\alpha^2(x^2+y^2))} \frac{dt}{d\tau}\frac{dx}{d\tau}+\frac{2\alpha^2y}{(1+\alpha^2(x^2+y^2))}\frac{dt}{d\tau}\frac{dy}{d\tau}=0[/tex]

    [tex]\frac{d^2x}{d\tau^2}+\alpha^2x\left(\frac{dt}{d\tau}\right)^2=0[/tex]

    [tex]\frac{d^2y}{d\tau^2}+\alpha^2y\left(\frac{dt}{d\tau}\right)^2=0[/tex]

    [tex]\frac{d^2z}{d\tau^2}=0[/tex]

    I can solve the 4th equation of course:

    [tex]z(\tau)=a\tau +b[/tex]

    and if I understand correctly, the newtonian limit means that

    [tex]\frac{dx^i}{d\tau}<<\frac{dt}{d\tau}[/tex]

    But I don't see how that can be applied here to simplify the equations.
     
    Last edited: Feb 25, 2007
  2. jcsd
  3. Feb 25, 2007 #2

    nrqed

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    In the Newtonian limit you can set t = tau so [itex] \frac{dt}{d \tau} = 1 [/itex].

    And, of course, [itex] \frac{d^2t}{d \tau^2} =0 [/itex].
    So the equations become trivial to solve.
     
    Last edited: Feb 25, 2007
  4. Feb 25, 2007 #3

    quasar987

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    Are you sure?

    [tex]d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}[/tex]

    And here, [itex]g_{ii}=-1[/itex], [itex]g_{00}=1+\alpha^2(x^2+y^2)[/itex] so

    [tex]d\tau^2=(g_{00}-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2)dt^2[/tex]

    and in the small speed limit, this only reduces to

    [tex]d\tau^2\approx g_{00}dt^2[/tex]

    which is not just dtau=dt because g00 is not just 1.
     
  5. Feb 25, 2007 #4

    nrqed

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    You are right. Sorry, I had misread the spacetime interval and thought that the (1+alpha^2(x^2+y^2)) term went with the spatial part. So you have an expression for dt/dtau to replace in all the equations. Using the chain rule you can also get an expression for the second derivative d^2 t/dtau^2.

    Sorry for my mistake.
     
  6. Feb 25, 2007 #5

    Mentz114

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    Use the weak-field approximation, which gives a straight Newtonian potential of -
    [tex] - \frac{1}{2}c^2( 1 + g_{00})[/tex]

    You can derive this by assuming the

    [tex] g_{ij} = \eta_{ij} + h_{ij} [/tex] where [tex] \eta[/tex] is the flat space metric and h
    is a small perturbation.
     
    Last edited: Feb 25, 2007
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