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?

$\endgroup$
  • $\begingroup$ I think $L:TM\to M$ is defined as $X\mapsto |x|^2=g(X,X)$. $\endgroup$ – Olivier Bégassat May 28 '13 at 13:24
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.