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General Relativity easy question
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[QUOTE="Antarres, post: 6251795, member: 643619"] Well, there are two ways of getting the connection coefficients. One way is getting them via the formula where they are defined as Christoffel symbols of the second kind: $$\Gamma^\mu_{\nu\rho} = \frac{1}{2}g^{\mu\sigma}(\partial_\nu g_{\sigma\rho} + \partial_\rho g_{\nu\sigma} - \partial_\sigma g_{\nu\rho})$$ The second way is by variation of action: $$I = \frac{1}{2}\int g_{\mu\nu}\dot{x^\mu}\dot{x^\nu}d\lambda$$ The Lagrangian in this action is so called geodesic Lagrangian, and by variating the action ##\delta I = 0##, you find geodesic equations from which you can read connection coefficients. You variate with respect to every component ##x^\mu##. Since in general relativity, free particles travel along geodesics, this is what is meant by effective Lagrangian for free particles(I think, at least, I always called it geodesic Lagrangian, but there should be no other). Of course, ##\dot{x^\mu} \equiv \frac{dx^\mu}{d\lambda}## where ##\lambda## is parameter of the geodesic. Sometimes the action is also defined as: $$I = \int \sqrt{g_{\mu\nu}\dot{x^\mu}\dot{x^\nu}}d\lambda$$ but these two actions have same equations of motion. The second action is precisely the line element form, but it's pretty much the same, and I like using the one without the square root. [/QUOTE]
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