# Variational principle for a point particle (massive or massless) in curved space

We know that for a point particle, the action is

$$S[x,e] ~=~ \frac{1}{2}\int_{\lambda_A}^{\lambda_B} d\lambda\left[e^{-1}(\lambda)~g_{\mu\nu}(x(\lambda))~\dot{x}^\mu(\lambda)~\dot{x}^\nu(\lambda) -(mc)^2e(\lambda)\right] ,$$

with signature convention $$(-,+,+,+)$$. It was mentioned on some website as a I googled that $$e$$ and $$x$$ are the dynamical variables and from them we should get the Euler-Lagrange equations.

I was wondering how to start since just a few minutes ago I first encountered this einbein variable (which I didn't know was a variable in the first place)!

1. The einbein field $$e(\lambda)\neq 0$$ is not a dynamical field because there is no $$\dot{e}(\lambda)$$ present. It is a so-called auxiliary field or generalized Lagrange multiplier. Its EL eq. simplifies to $$(mce)^2~\approx~-g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu} . \tag{1}$$ [Here the $$\approx$$ symbol means equality modulo eom.] Here $$m$$ is the rest mass of the point particle. See also this related Phys.SE post and links therein.
2. In the massive case $$m>0$$, we can integrate out the $$e$$ field, which means to replace it in the action $$S[x,e]$$ by its eom $$e~\approx~\pm\frac{1}{mc}\sqrt{-g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}} ,\tag{2}$$ which has two branches. The resulting action is \begin{align} S_{\pm}[x]~:=~&S\left[x,e=\pm\frac{1}{mc}\sqrt{ \ldots}\right]\cr ~=~&\mp mc \int \!d\lambda~ \sqrt{- g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}}~ \left\{ \begin{array}{c} < \cr > \end{array}\right\}~0. \end{align}\tag{3} The $$S_{+}[x]$$ branch is the standard square-root action for a massive point particle, cf. e.g. this and this Phys.SE posts. The minimum of $$S_{+}[x]$$ (and the maximum of $$S_{-}[x]$$) is obtained for timelike geodesic curves $$x^{\mu}(\lambda)$$. Often we throw away the $$S_{-}[x]$$ branch since we are mostly interested in the minimum. This can be encoded into the variational principle by imposing that the einbein field $$e(\lambda)>0$$ is positive.
3. In the massless case $$m=0$$, eq. (1) becomes $$g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu} ~\approx~0, \tag{4}$$ which is the equation of motion (EOM) for a massless particle, cf. e.g. this Phys.SE post.