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Challenging problem in GR

  1. Jul 25, 2008 #1
    I'm looking for a reasonably challenging problem that will help me apply concepts of GR.

    Difficulty wise, I'm looking for something that would compare to a QED cross-section calculation.

    Can anyone suggest such a calculation? I'm considering the problem posed in MTW to figure out the number of test particles required to determine the independent components of the Riemann curvature tensor, but I question how useful this would be compared to the time required to do the problem.
     
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  3. Jul 25, 2008 #2

    Mentz114

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    Hi JD,

    My take on finding the curvature tensor from observation.
    1. You need at least as many particles as there are independent components of Rpqrs
    2. A dynamic model incorporating the tidal effects
    3. Make a least squares fit of observed motion to model
    4. Calculate back to R components.

    The more particles observed the smaller the error bars.

    I'm trying to calculate the worldline of a photon in the Schwarzschild space-time ( only considering an equatorial plane ) so maybe you can help me out with that ?

    I don't have MTW, alas.

    M
     
  4. Jul 25, 2008 #3
    Your approach is correct, and it can even be done analytically (although it is difficult because it involves checking if a 20 x 20 determinant vanishes). See

    https://www.physicsforums.com/showthread.php?t=245779
     
  5. Jul 25, 2008 #4
    I guess once you have the metric in hand, the trajectories will be solutions of the Euler Lagrange equations.

    Are you learning GR at the moment Mentz?
     
  6. Jul 25, 2008 #5

    Mentz114

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    JD,

    Good question. I am studying GR, but whether I'm learning anything is moot.

    Yes, the Euler Lagrange extremization gives the geodesic equation. I don't understand the physical meaning of

    [tex]\frac{d^2x^{\mu}}{d\lambda^2} = \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}[/tex] and it's not suitable for null geodesics.

    The method I'm using is to find the null vector ka that is tangent to the null geodesic and is Fermi transported along it. This is give by the conditions

    [tex]k_{m;n}k^n = 0[/tex] and of course it must have length 0.

    See my earlier post here.

    https://www.physicsforums.com/showthread.php?t=244511

    It's easy for a radial photon but more tricky in the equatorial plane. I am making progress, I think.

    M
     
  7. Jul 25, 2008 #6

    gel

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    Why do you say that? It should work for null geodesics too. It's the same as [itex]
    k_{m;n}k^n = 0[/itex] with [itex]k^m=dx^m/d\lambda[/itex].
     
  8. Jul 25, 2008 #7

    Haelfix

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    "Difficulty wise, I'm looking for something that would compare to a QED cross-section calculation."

    Derive the Schwarzschild metric from first principles without looking at a textbook. Thats a standard test problem. Slightly less challenging but dull, given that metric, compute the Riemann tensor and the Ricci scalar.
     
  9. Jul 27, 2008 #8

    Mentz114

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    gel,
    thanks for your remarks. I can only think of one possible affine parameter, and that's [itex]\tau[/itex] which is zero for a null geodesic. In practical computing terms, I can see how I might solve this [itex]k_{m;n}k^n = 0[/itex] but not the geodesic equation.

    M
     
  10. Jul 27, 2008 #9

    gel

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    The geodesic equation is [itex]\frac{d^2x^{\mu}}{d\lambda^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lam bda}\frac{dx^{\beta}}{d\lambda}=0[/itex] (fixing a sign error in the previous post). This is just a second order diff eqn which can be solved by computer using any of the standard techniques for such equations, and gives the path of the particle as a function of a parameter [itex]\lambda[/itex]. The parameter is is not the distance along the curve if you have a null geodesic, but I don't think that matters. If you like you could eliminate lambda and write it in terms of t.

    The same issue would arise using the Fermi transport equation. k is the tangent calculated using an affine parameterization, so you have to introduce the affine parameter and, if you write out the resulting equation, it gives the same geodesic equation.
     
    Last edited: Jul 27, 2008
  11. Jul 27, 2008 #10
    Compute the index of refraction for gravitational waves in a medium of density rho, Young's modulus E and Poisson's ratio nu.
     
  12. Jul 27, 2008 #11

    George Jones

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    One problem with using [itex]t[/itex] is that the computer won't evolve a (lightlike or timelike) geodesic across the event horizon of a black hole.

    I don't remember what pararmeter I used for lightlike geodesics (source is on a hard drive in a laptop that died) in

    https://www.physicsforums.com/showthread.php?p=1813622#post1813622
     
  13. Jul 27, 2008 #12

    Mentz114

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    gel,
    is t an affine parameter ? So

    [tex]\frac{d^2x^{\mu}}{dt^2} = -\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt}[/tex] ?

    In any case, it's a lot easier to solve the Fermi transport equation because it's not DE's, and I only need the vector up to a constant for my purpose.

    M
     
  14. Jul 27, 2008 #13

    gel

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    t is not an affine parameter. You can use change of variables,
    [tex]dx^\mu/d\lambda=(dt/d\lambda)dx^\mu/dt,\ \ d^2x^\mu/d\lambda^2=(d^2t/d\lambda^2)dx^\mu/dt+(dt/d\lambda)^2d^2x^\mu/dt^2[/tex]
    Substitute into the geodesic equation,
    [tex]
    \frac{d^2x^\mu}{dt^2} = -\gamma \frac{dx^\mu}{dt}-\Gamma^\mu_{\alpha\,\beta}\frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}
    [/tex]
    where gamma is given by [itex]\gamma=(dt/d\lambda)^{-2}(d^2t/d\lambda^2)[/itex]. Assuming you have t=x0, then putting mu=0 in the geodesic equation above gives
    [tex]
    \gamma = \Gamma^0_{\alpha\,\beta}\frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}
    [/tex]
    which you can substitute back in to get a 2nd order DE for x1, x2, x3 in terms of t=x0.

    However, as George Jones says, the geodesic won't pass through the event horizon. Instead, it will slow down to a standstill as it approaches the horizon which, of course, is what should happen for the motion of a particle expressed in terms of the time coordinate t. It only passes through the horizon in the limit t -> infinity.

    Not sure what your point about the Fermi transport equation is though. It is still a second order DE. First, it involves the tangent vector which is equal to the derivative of the coordinates along the geodesic (wrt an affine parameter). Second, it involves the derivative of k.
     
  15. Jul 27, 2008 #14

    gel

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    There's other things you can do too. If you're using the Schwarszchild metric, the time invariance and spherical symmetry means that you can integrate out the equations to get some first order diff eqns corresponding to conservation of energy and angular momentum (just as you would with Newton's equations).
    You can then convert to polar coordinates to get first order DEs for radius r & angle theta in terms of time. Then you can convert this to a first order DE for r in terms of theta, which is easy to solve on a computer. It gives you the paths of the particle through space, but elimates the time variable.

    I remember doing just this years ago as part of a computer project. As time has been eliminated you see some of the paths passing through the event horizon.

    I suppose one interesting question you could ask is, what is the effective radius of a black hole as seen by a distant observer? Probably not the same as the actual radius of the event horizon appearing in the Schwarzschild metric.
     
  16. Jul 27, 2008 #15

    Mentz114

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    gel & George,
    thanks a lot. I'll study what you've written with care, especially on the geodesic equation and the spherical symmetry.

    I'm not interested in the event horizon. I'm trying to claculate the frequency shift for light sent between various observers in Schwarzschild ST. Think I've done observers on the same radius, and now looking at the equatorial plane.

    I only need to find the four components of k that satisfy the Fermi equation. This gives me 4 equations in 4 unknowns ( if I'm lucky) and then there is the additional null constraint. No integration or affine parameter required.

    M
     
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