Graduate Einbein as Lagrange Multiplier: How Does it Work?

[ergospherical]

Let $g_{\mu \nu}(x)$ be a time-independent metric. A photon following a curve $\Gamma$ has action\begin{align*}
I[x,e]= \dfrac{1}{2} \int_{\Gamma} e^{-1}(\lambda) g_{\mu \nu}(x)\dot{x}^{\mu} \dot{x}^{\nu} d\lambda
\end{align*}with $e(\lambda)$ an independent function of $\lambda$ (an einbein). The canonical momentum is $\partial L / \partial \dot{x}^{\mu} = e^{-1}(\lambda) \dot{x}_{\mu}$ which yields a conserved energy $-E \equiv -e^{-1}(\lambda) \dot{t}$ since the Lagrangian does not depend on time. The Hamiltonian is\begin{align*}
H = \dfrac{1}{2} e^{-1}(\lambda) g_{\mu \nu}(x) \dot{x}^{\mu} \dot{x}^{\nu} = \dfrac{1}{2}e(\lambda)\left( - E^2 + \mathbf{p}^2 \right)
\end{align*}From this follows the mass-shell equation $E^2 = \mathbf{p}^2$. Why is this? The einbein field $\dfrac{1}{2}e(\lambda)$ is supposedly acting as a Lagrange multiplier but my variational calculus is rusty. (The next step will be to Legendre transform the cyclic variables only to determine the Routhian).

 

[ergospherical]

Actually I think it makes sense, that $-E^2 + \mathbf{p}^2 \equiv F = 0$ is just a standard constraint between the coordinates and the canonical momentum when re-written.

 

[vanhees71]

You can make $e(\lambda)$ a Lagrange multiplier. Then you have in addition the variation of the action wrt. $e$, and this gives the desired constraint $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$ for the motion of a massless particle.

 

[haushofer]

Why is this? It's per construction, so you can write down an action which also applies to massless particles and is easy to quantize. In string theory I'd call this trick "switching from Nambu-Goto to Polyakov" :P