= Solution
Use <metric signature> $(-,+,+,+)$ and units $c=1$. Write $\dot x^\mu=dx^\mu/d\lambda$ in this part. Varying the <worldline einbein>, without first fixing it, gives
$$
0=\frac{\partial L}{\partial e}=\frac m2\left(-\frac{g_{\mu\nu}\dot x^\mu\dot x^\nu}{e^2}-1\right),\qquad \boxed{g_{\mu\nu}\dot x^\mu\dot x^\nu=-e^2.}
$$
The canonical four-momentum is $p_\mu=mg_{\mu\nu}\dot x^\nu/e$, so the same <mass shell> constraint is the <mass shell> $g^{\mu\nu}p_\mu p_\nu=-m^2$.
The <Euler-Lagrange equation> for the trajectory is
$$
\frac{d}{d\lambda}\left(\frac{g_{\mu\nu}\dot x^\nu}{e}\right)-\frac1{2e}\partial_\mu g_{\alpha\beta}\dot x^\alpha\dot x^\beta=0.
$$
Multiplying by $e$ and then by the <inverse metric>, and collecting the <metric tensor> derivatives into the <Christoffel symbols>, gives
$$
\boxed{\ddot x^\mu+\Gamma^\mu_{\alpha\beta}\dot x^\alpha\dot x^\beta=\frac{\dot e}{e}\dot x^\mu.}
$$
This is a <geodesic equation> in a possibly nonaffine parameter.
Under a change of parameter, $e'(\lambda')=e(\lambda)d\lambda/d\lambda'$, so $e\,d\lambda$ is invariant. On the positive branch, the <mass shell> constraint identifies this invariant with the <proper time> increment $ds=\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}=e\,d\lambda$. Choosing $\lambda'=s$ therefore sets $e'=1$. The trajectory equation is then the affinely parametrized <geodesic equation> and its tangent has norm minus one. This is <proper-time gauge for a massive worldline einbein>. The <mass shell> constraint must still be imposed: fixing $e$ in the action before varying would lose it. A fixed parameter interval can retain a <proper-time modulus>; here changing to the actual proper-time interval is permitted.
Back to article page
