3
$\begingroup$

If we have a metric $g_{\mu \nu}$, defined in a Riemannian manifold we can write the equation of the geodesic: $$\frac{d^2x^\mu}{dt^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt}$$ in which $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols. The geodesic in that metric can be obtained also using the Euler - Lagrange equations: $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=0$$ Where the $q_k$ are the generalized coordinates. How can I find the Lagrangian $L$ in order to find the same geodesic obtained using the affine connection symbols?

$\endgroup$
1
  • $\begingroup$ I think $L:TM\to M$ is defined as $X\mapsto |x|^2=g(X,X)$. $\endgroup$ Commented May 28, 2013 at 13:24

1 Answer 1

2
$\begingroup$

The Lagrangian is the standard measurement of a distance. That is $$ L=-\kappa\int ds $$ being $\kappa>0$ a constant, that in your case is $$ S=-\kappa\int\sqrt{g_{\mu\nu}dx^\mu dx^\nu}. $$ Then you are supposed to parametrize the coordinates as $x_\mu=x_\mu(t)$ and the computation becomes a standard one. Notice that, for a Minkowski metric, this reduces to the action of a relativistic free particle.

$\endgroup$
1
  • 2
    $\begingroup$ Or you could do the usual thing and look at energy, instead. That would remove the pesky square root. $\endgroup$ Commented May 28, 2013 at 13:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .