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The geodesic equation used in general relativity is the following: $$ {d^2 x^\mu \over ds^2} =- \Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over ds}{d x^\beta \over ds}. $$ It states that the acceleration of the test particle is a function of the metric (Chistoffel symbol) and the derivative of coordinates with respect to "a scalar parameter of motion s ex.: proper time". So how do you find the trajectory of a particle using a known metric (example the Schwarzschild metric) with the equation above?

Up to now, I haven't done much... I tried to differentiate the components of the Schwarzschild metric with respect to "a scalar parameter of motion"; i chose to differentiate with respect to proper time. But there is no proper time term in the metric I chose... Not on the Wikipedia page anyways. So how do i start?

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    $\begingroup$ Could you show the working you've done so far? E.g. could you try to do it for a Schwarzschild metric so that we can see what's causing you confusion? Otherwise your question's a little confusing: the technique is exactly as for any other second order set of ODEs; perhaps you're having trouble reducing this to a standard initial value problem? $\endgroup$ May 22, 2016 at 14:13

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I will answer this with a simple example. Let us consider the metric for weak gravity, $$ ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2. $$ The $g_{tt}$ metric element is largest by a factor of $c^2$ and we have $$ \Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}. $$ Now let us work with the geodesic equation that is $$ \frac{d^2r}{ds^2} + \Gamma^r_{tt}\left(\frac{dt}{ds}\right)^2 = 0. $$ Now use the fact this is weak gravity with low velocities so that $\frac{dt}{ds} \simeq 1$ and we can replace $ds$ with $dt$ in the second order derivative $$ \frac{d^2r}{dt^2}~+~\frac{GM}{r^2} = 0. $$ This is Newton's law of gravity. In this limit general relativity recovers Newton. This is an elementary start to how to use the geodesic equation.

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  • $\begingroup$ The metric for weak gravity is the Schwarzschild metric right? $\endgroup$
    – Investor
    May 22, 2016 at 17:03
  • $\begingroup$ And why did you use that particular Christoffel symbol? $\endgroup$
    – Investor
    May 22, 2016 at 17:12
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    $\begingroup$ @Investor 1. No. 2. It's the only Christoffel symbol which is "large" in the current approximation scheme. $\endgroup$
    – Ryan Unger
    May 22, 2016 at 18:51
  • $\begingroup$ @0celo7 which means its the only "relevant" symbol? $\endgroup$
    – Investor
    May 22, 2016 at 20:41
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    $\begingroup$ @Investor : Yes. $\endgroup$
    – Ryan Unger
    May 22, 2016 at 21:34

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