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Use metric signature $(-,+,+,+)$ and units $c=1$. Write ${\dot{x}}^{\mu }=d{x}^{\mu }/d\lambda$ in this part. Varying the worldline einbein, without first fixing it, givesThe canonical four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }/e$, 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 isMultiplying by $e$ and then by the inverse metric, and collecting the metric tensor derivatives into the Christoffel symbols, givesThis is a geodesic equation in a possibly nonaffine parameter.

Under a change of parameter, $e^{\prime }({\lambda }^{\prime })=e(\lambda )d\lambda /d{\lambda }^{\prime }$, so $ed\lambda$ is invariant. On the positive branch, the mass shell constraint identifies this invariant with the proper time increment $ds=\sqrt{-g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }}=ed\lambda$. Choosing ${\lambda }^{\prime }=s$ therefore sets $e^{\prime }=1$. 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.