Lagrangian for relativistic massless point particle

Question

For relativistic massive particle, the action is $$\begin{align}S ~=~& -m_0 \int ds \cr ~=~& -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} \cr ~=~& \int d\lambda \ L,\end{align}$$ 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 1

1. OP's square root action is not differentiable along null/light-like directions, which makes it ill-suited for a massless particle. So we have to come up with something else. One equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is that its tangent should be null/light-like $$ \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, \tag{A}$$ where dot denotes differentiation wrt. the world-line (WL) parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo EOM.] This suggests that a possible action is $$ S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2, \tag{B} $$ 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 answer here.

2. More importantly, the corresponding Euler-Lagrange (EL) equations for the action (B) are the null/light-like condition $$ 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2, \tag{C}$$ and the geodesic equations $$ 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu},\tag{D} $$ as they should be.

3. The action (B) is invariant under WL reparametrization $$ \begin{align} \tau^{\prime}~=~&f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\cr \dot{x}^{\mu}~=~&\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\end{align}\tag{E} $$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the more familiar affinely parametrized geodesic equations.

References:

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

Answer 2

It is conceptually possible to have a massless charged particle, although there are none that we know of. It is not true that the Lorentz force has to equal mass times acceleration. The momentum of a massless particle is a quantity indepenedent of its speed as all massless particles travel at the speed of light. The momentum $p$ is instead equal to $E/c$, the energy divided by the speed of light.

Answer 3

For a massless particle we can not have a centre-of-mass frame.

Unfortunately I cannot yet add comments. Are you studying Classical Field Theory (CFT) or Quantum Field Theory (QFT)? My guess is CFT since this looks like a line out of a few lectures into a CFT course when you start to make find equations of motion.

In that case, for the (massless) photon, $A_{\mu}(x)$ say, we use the Maxwell Lagrangian, which is Lorentz invariant, and is given (in Heaviside-Lorentz units) by $$ \mathcal{L_{Max}} =-\frac{1}{4} \int d^4x \mathcal{F_{\mu \nu}} \mathcal{F^{\mu \nu}} $$ where $$\mathcal{F_{\mu \nu}} = \partial_{\mu}A_{\nu}(x) - \partial_{\nu}A_{\mu}(x) $$

Answer 4

Particle with zero mass has to have also zero electric charge, otherwise the Lorentz formula for EM force acting on it cannot be used to find its acceleration according to $$ m\mathbf a = q\mathbf{E}_{ext} + q\frac{\mathbf v}{c} \times \mathbf B_{ext}. $$ However, particle with zero mass and zero charge has trivial equation of motion $$ 0=0 $$ and has zero effect on the EM forces on other particles. It seems like a vacuous concept.