general relativity - Lagrangian for relativistic massless point particle

For relativistic massive particle, the action is

$$S ~=~ -m_0 \int ds ~=~ -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} ~=~ \int d\lambda \ L,$$ where $ds$ is the proper time of the particle; $\lambda$ is the parameter of the trajectory; and we used Minkowski signature $(+,-,-,-)$. So what is the action for a massless particle?

Answer

I) The equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is

$$\tag{A} \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0,$$

where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is

$$\tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2,$$

where $\lambda(\tau)$ is a Lagrange multiplier. This answer (B) may seem like just a cheap trick. Note however that it is possible by similar methods to give a general action principle that works for both massless and massive point particles in a unified manner, cf. e.g. Ref. 1 and eq. (3) in my Phys.SE here.

II) The corresponding Euler-Lagrange (EL) equations for the action (B) reads

$$\tag{C} 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2,$$

$$\tag{D} 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}.$$

III) The action (B) is invariant under world-line reparametrization

$$\tag{E} \tau^{\prime}~=~f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\qquad \dot{x}^{\mu}~=~\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\qquad$$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the familiar geodesic equation.

References:

1. J. Polchinski, String Theory Vol. 1, 1998; eq. (1.2.5).