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Geodesic equations for the metric $$dS^2 = \left(1 - \dfrac{2GM}{r}\right)\dot t^2 - \dfrac{\dot r^2}{\left(1-\dfrac{2GM}{r}\right)} - r^2\left(\dot \phi^2\right),$$ would be

$$\left(1 - \frac{2GM}{r}\right)\dot t = k$$ $$\left(1 - \frac{2GM}{r}\right)\dot t^2 - \frac{\dot r^2}{\left(1 - \frac{2GM}{r}\right)} - r^2\dot\phi^2 = 1$$ $$r^2\dot\phi^2 = h$$

where h = J(angular momentum)/m and k = E/m and c = 1. I obtain the first and third one but stuck with the second problem. Please Help

Reference: General Relativity by Hobson (Section 9.6)

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Frobenius
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1 Answers1

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The second equation doesn't really come from the Euler-Lagrange equations (though one could probably derive it from them.) It's much easier to see from the definition of the particle's proper time: $$ d\tau^2 = g_{\mu \nu} dx^\mu dx^\nu. $$ Expand out the sum above, divide both sides by $d\tau^2$, and you'll get the equation you want.