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Determination of Riemann curvature tensor from tidal forces

  1. Jul 19, 2008 #1

    Given a large number of test particles N, it should be possible to determine the Riemann curvature tensor by tracking their motion as they undergo geodesic deviation.

    Is there a minimum number N that will achieve this in any situation, or does it vary from problem to problem?

    How difficult is it to compute this least N?
  2. jcsd
  3. Jul 19, 2008 #2
    I have a partial answer to your question. I found this reasoning in https://www.amazon.com/Gravitation-Physics-Charles-W-Misner/dp/0716703440" so don't hesitate to read the page 72 if you find my explanation too messy, or simply wrong.

    The geodesic deviation equation is

    [tex]\frac{D^{2}\xi_{\alpha}}{d\tau^{2}} = -R_{\alpha\beta\gamma\delta}u^{\beta}\xi^{\gamma}u^{\delta}[/tex] in any coordinate system.

    The idea is to use the method of least squares to determine the components of the Riemann tensor.

    We define the 'error' between the observed value and the predicted value for the Kth particle by

    [tex]\ddot{\xi}_{\alpha}^{\phantom{\alpha}K} + R_{\alpha\beta\gamma\delta}u^{\beta,K}\xi^{\gamma,K}u^{\delta,K} = \delta a_{\alpha}^{\phantom{\alpha}K}[/tex]

    The expression you now have to minimize is the sum of the squares of the [itex]\delta a_{\alpha}^{\phantom{\alpha}K}[/itex]

    [tex]S = \sum_{K}\eta^{\alpha\beta}\delta a_{\alpha}^{\phantom{\alpha}K}\delta a_{\beta}^{\phantom{\beta}K}[/tex]

    In this expression, everything is regarded as known, except the components [itex]R_{\alpha\beta\gamma\delta}[/itex]. Minimize with respect to these components. It gives a system of 20 equations (the 20 independant components out of the 256).

    It should be possible to determine the minimum number of particles by checking for which value of K the determinant of the 20 equations system 'stops to vanish'. However, this is not very elegant and might prove difficult to achieve (I haven't tried).
    Last edited by a moderator: May 3, 2017
  4. Jul 22, 2008 #3
    Interesting. I will try to do this after trying the Lorentz force version.

    By the way shouldn't the expression for the sum of the square involve a [itex]g^{\alpha\beta}[/itex] rather than [itex]\eta^{\alpha\beta}[/itex].
  5. Jul 22, 2008 #4
    I'm a little bit confused by this box in MTW.

    In the first column they seem to make it out as though 2 particles is insufficient to determine all the components of [itex]F^{\alpha\beta}[/itex] (by boosting to one particle's frame and then rotating spatial axes so that one axis is aligned with the motion of the second particle). Does this mean that 3 is sufficient?
  6. Jul 27, 2008 #5
    Yes, the box is a bit confusing. At the end of the box, the author says that with 2 particles the determinant vanishes. I suppose it means that for 3 particles the system becomes solvable (the determinant is different from 0).

    For the [tex]g^{\alpha\beta}[/tex] instead of the [tex]\eta^{\alpha\beta}[/tex] I think you're right.
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