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Quick question on Geodesic Equation

  1. Feb 6, 2015 #1
    Starting with the geodesic equation with non-relativistic approximation:

    [tex] \frac{d^2 x^{\mu}}{d \tau^2} + \Gamma_{00}^{\mu} \left( \frac{dx^0}{d\tau} \right)^2 = 0 [/tex]

    I know that ## \Gamma_{\alpha \beta}^{\mu} = \frac{\partial x^{\mu}}{\partial y^{\lambda}} \frac{\partial^2 y^{\lambda}}{\partial x^{\alpha} \partial x^{\beta}}##, so
    [tex]\Gamma_{00}^{\mu} = \frac{\partial x^{\mu}}{\partial y^{\lambda}} \frac{\partial^2 y^{\lambda}}{\partial t^2} [/tex]

    How did they get the relation ##\Gamma_{00}^{\mu} = -\frac{1}{2} g^{\mu \lambda} \frac{\partial g_{00}}{\partial x^{\lambda}}##?

  2. jcsd
  3. Feb 6, 2015 #2


    Staff: Mentor

    Do you have a reference? That's not the formula for the Christoffel symbols that I'm familiar with. The one I'm familiar with, as given, for example, here, reduces to the relation you are asking about immediately for the 0-0 case if you assume that the metric is independent of time (which is what "stationary" means).
  4. Feb 7, 2015 #3
    I think I've seen that somewhere, if I remember correctly it's the christoffel symbols in a general coordinate transform (i.e. non-inertial).
    Just use the ones written in terms of the metric.
  5. Feb 7, 2015 #4
    I got these from my lecturer's notes.

    I tend to get confused by the ever-changing metrics they use. It would be very helpful if I could see the steps leading to ##
    \Gamma_{00}^{\mu} = -\frac{1}{2} g^{\mu \lambda} \frac{\partial g_{00}}{\partial x^{\lambda}} ##.
  6. Feb 7, 2015 #5
    Hi. First, I have never understood why folks like to look at the geodesic equation in a flat-space, non-relativistic limit, as that defeats the beauty of G.R., but, nevertheless, here is how your lecturer obtained that equation, at least according to me:

    Assume a weak gravitational field, where your spacetime metric ##g_{uv}## is perturbed by a Minkowski metric ##\eta_{uv}##. So, we have that:
    ##g_{uv} = \eta_{uv} + h_{uv}##, where ##h_{uv}## is the relativistic part of the metric tensor. If you have done some G.R. already, then, you will see this technique commonly used in gravitational wave theory.

    The full definition of the metric connection (assuming it is torsion-free, and in a coordinate basis) is the Christoffel symbol definition, which is:
    ##\Gamma^{a}_{by} = \frac{1}{2}g^{ad} \left(g_{dy,b} + g_{db,y} - g_{by,d}\right)##.
    Substituting in our definition of ##g_{uv}## from above into this expression, we get that:
    ##\Gamma^{a}_{by} = \frac{1}{2} \eta^{ad} \left(h_{dy,b} + h_{db,y} - h_{by,d}\right)##, this is a first-order approximation of course, as is commonly the case in Newtonian gravity.

    We now just use the above equations to find for example ##\Gamma^{i}_{00}## as:
    ##\Gamma^{i}_{00} = \frac{1}{2} \eta^{ip} \left(h_{0p,0} + h_{0p,0} - h_{00,p}\right)##.
    The key thing is the following, the terms:
    ##h_{0p,0} + h_{0p,0}## must vanish in the Newtonian limit, and using a static approximation, they indeed do. You will notice that both of these terms are derivatives with respect to time. In the Newtonian world, where observers move at speeds v << c, gravitational systems have a very weak time-dependence, compared to their spatial dependence. For example, compare any Newtonian gravitational system's timescales/age to the age of the Sun, or galaxy, it is very, very small on these scales. So, we neglect all time derivatives in this approximation, and obtain:
    ##\Gamma^{i}_{00} = -\frac{1}{2} \eta^{ip} \left(h_{00,p}\right)##.

    Which is precisely the relationship you seek.

    I hope that helps.
    Dr. Ikjyot Singh Kohli
  7. Feb 7, 2015 #6

    Thanks alot for working with me on this, I really appreciate it. But, I think these are too advanced for me. I am trying to work with the notes I have here - Chapters 5 and 6.

    Attached Files:

    • B3..pdf
      File size:
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  8. Feb 7, 2015 #7
    Hi. No problem. I am not familiar with your textbook or set of notes. I am sure, however, that there is no other way to derive the expressions above. I believe your lecturer just skipped some steps, because they may have been too technical, which is unfortunate, but frequently happens.
  9. Feb 7, 2015 #8


    Staff: Mentor

    Yes, this is the definition I am familiar with.

    To get the expression unscientific is asking about, there's a simpler way, if you know that the metric is independent of time (i.e., of ##x^0##), which is the "stationary" assumption referred to in the OP. We have, from the general expression above,

    \Gamma^a{}_{00} = \frac{1}{2}g^{ad} \left(g_{d0,0} + g_{d0,0} - g_{00,d}\right)

    But if the metric is independent of ##x^0##, the first two terms inside the parentheses vanish, so we're left with

    \Gamma^a{}_{00} = - \frac{1}{2}g^{ad} \left( g_{00,d} \right)

    as desired.

    There is for the expression the OP asks about; see above. That expression is valid for any stationary metric; you don't need to make the weak field approximation. (The lecture notes do make that approximation, but later on in the derivation when it's actually needed; they don't make it just to derive the expression above.)
    Last edited: Feb 7, 2015
  10. Feb 7, 2015 #9


    Staff: Mentor

    About halfway down page 11, you'll find the expression for ##\Gamma## given in post #5 (with different labeling of the indexes, but that doesn't matter).
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