Lagrangian Formulation for Lightlike Geodesics

Question

I have found several detailed demonstrations which derive the geodesic equation from the Lagrangian. However, each demonstration states that if we are interested in lightlike geodesics where the standard Largrangian will always be zero, the procedure fails and needs to be modified. I would like to learn about this modification! Will someone please help me understand how to derive the geodesic equation for lightlike geodesics using a Lagrangian technique? For example, in D’Inverno’s introductory GR textbook on page 101 he points out why the straightforward Lagrangian approach must fail for lightlike geodesics and then he simply presents the lightlike result without a derivation. I would like to learn this derivation!

Answer 1

The standard relativistic free particle action is $$ S[x] = - m \int d\tau \sqrt{ - g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu } ~. $$ The equations of motion derived from this action is $$ {\ddot x}^\lambda + \Gamma^\lambda_{\mu\nu}[g(x)] {\dot x}^\mu {\dot x}^\nu = {\dot x}^\lambda \partial_\tau \log \sqrt{ - g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu } $$

This action is of course not suitable to describe massless particles. However, it is possible to modify the Lagrangian a bit which does allow one to do so.

We introduce an auxiliary field $e(\tau)$ and consider the Lagrangian $$ S[e,x] = \frac{1}{2} \int d\tau \left[ \frac{1}{e(\tau) } g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu - m^2 e(\tau) \right] ~. $$ Note that $e(\tau)$ has no time derivatives in the action and therefore appears as an auxiliary field. It is easy to show that the equations of motion derived from $S[e,x]$ is completely equivalent to $S[x]$. The EOM are $$ g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu + m^2 e^2 = 0 ~, \qquad {\ddot x}^\lambda + \Gamma^\lambda_{\mu\nu}[g(x)] {\dot x}^\mu {\dot x}^\nu = {\dot x}^\lambda \partial_\tau \log e $$ Thus, the actions $S[x]$ and $S[x,e]$ are classically equivalent. However, $S[x,e]$ has two advantages over $S[x]$.

1. $S[x,e]$ is quadratic in $x$. The pesky square root in $S[x]$ has been removed. Thus, it is significantly easier to quantize $S[x,e]$.

2. $S[x,e]$ is completely well-defined in the massless limit.

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For the advanced readers, I'll also add that $S[x,e]$ has a very interesting interpretation as a one-dimensional non-linear sigma model. The action has a $\tau$ reparameterization symmetry under which $$ \tau \to \tau' ~, \qquad e(\tau) \to e'(\tau) \quad \text{where}\quad e(\tau) = \frac{d\tau'}{d\tau} e'(\tau')~. $$ This shows that $e(\tau)$ can be thought of as a 1-dimensional metric, $ds^2 = \gamma_{\tau\tau} d\tau^2 = e(\tau)^2 d\tau^2$. We can then write the action as $$ S[e,x] = \frac{1}{2} \int d\tau \sqrt{\gamma} \left[ \gamma^{\tau\tau} g_{\mu\nu}(x) \partial_\tau x^\mu \partial_\tau x^\nu - m^2 \right] ~. $$ This is a $d=1$ NLSM with a "cosmological constant" proportional to $m^2$.

In string theory, the corresponding generalization gives is a $d=2$ NLSM.