General Relativity easy question

  • Thread starter Replusz
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  • #1
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Homework Statement:
Connection coefficients
Relevant Equations:
part c) of the problems
1571848271121.png


Attempt: I don't know what they mean by effective Lagrangian.
I am aware there is something called 'the lagrangian' that goes as L=g_ab * dx^a/dk * dx^b/dk, but i dont see how this gives me any of the chrostoffel symbols...

cheers
 

Answers and Replies

  • #2
166
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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.
 

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