Use $\delta F=\epsilon\{F,\varphi\}$ with the canonical <Poisson bracket>. The <first-class constraint> generates
$$
\boxed{\delta x^m=\epsilon p^m,\qquad \delta p_m=0,\qquad \delta e=\dot\epsilon.}
$$
Indeed the variation of the <relativistic particle phase-space action> is
$$
\delta I=\int dt\,\left\{\frac12\frac{d}{dt}
[\epsilon(p^2-M^2)]+(\dot\epsilon-\delta e)\varphi\right\}.
$$
The transformation is therefore a <gauge invariance> when its parameter vanishes at the time endpoints, or suitable boundary conditions remove the total derivative.

On a fixed interval of parameter length $\Delta t$, the <proper-time modulus>
$$
s=\frac1{\Delta t}\int_{t_i}^{t_f}e(t)\,dt
$$
is unchanged by these <gauge transformations>. Choosing $\epsilon(t)=\int_{t_i}^{t}(s-e(u))\,du$ sets $e+\delta e=s$, with $\epsilon(t_i)=\epsilon(t_f)=0$. Thus \b[the nonconstant part of the <worldline einbein> can be fixed, but its constant modulus must still be integrated over]. Fixing $e=1$ as well would remove inequivalent values of the <proper-time modulus>; it is legitimate only if the parameter interval is allowed to vary instead. The name proper-time modulus refers to the Schwinger proper-time parameter; after eliminating $p$, the geometric proper length for a massive on-shell trajectory is $M\int e\,dt$ in this normalization.

For the <gauge fixing> functional $F=e-s$, its variation is $\delta F=\dot\epsilon$. The <Faddeev-Popov determinant> is consequently
$$
\Delta_{\mathrm{FP}}=\det\bigl[\partial_t\delta(t-t')\bigr].
$$
The <Grassmann Gaussian integral> represents this determinant using anticommuting <Faddeev-Popov ghost fields> $b,c$:
$$
\boxed{I_{\mathrm{FP}}=i\int dt\,b\dot c,\qquad
\int\mathcal Db\,\mathcal Dc\,e^{iI_{\mathrm{FP}}}\ \propto\ \det(\partial_t).}
$$
The phase and normalization of the determinant depend on the measure convention. Its domain carries the chosen boundary conditions: on an interval the constant modulus is excluded from the gauge-fixed directions, and on a periodic <worldline> the constant ghost <zero mode in field theory> is removed with the residual gauge volume treated separately. A bare determinant with such zero modes left in would vanish.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 306 / 3 / a / Solution

{title2=$e(t)$}

An einbein specifies a one-dimensional <metric tensor> along a <worldline>. In the <relativistic particle phase-space action> it is the <Lagrange multiplier> for the <mass-shell condition> and transforms as $\delta e=\dot\epsilon$ under the corresponding canonical <gauge transformation>. After eliminating the <momentum>, its geometric interpretation depends on the normalization of the particle <action>.

Worldline einbein

{title2=$I=\int dt\,[p_m\dot x^m-\tfrac e2(p^2+M^2)]$}

The <worldline einbein> imposes the relativistic <mass-shell condition>. With the <Minkowski metric> of signature $(-,+,\ldots,+)$, its <first-class constraint> is $\varphi=(p^2+M^2)/2$.

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 Incoming links: Relativistic particle phase-space action