First-order shell action yields massive and null geodesics

Use the supplied smooth tangent and cotangent bundles under [A1], with the mathematical functionals and selected branches [F1,F2]. In a spacetime chart, variation in $e$ gives $C_m=0$. Variation in $p_a$ gives $x^{\mathrm{\prime }a}=e{g}^{ab}{p}_b$. Variation in $x^a$, integrating the $p_a\delta x^{\mathrm{\prime }a}$ term by parts with $\delta x$ vanishing at endpoints, gives $p_a^{\mathrm{\prime }}=-e{\mathrm{\partial }}_a{C}_m=-\frac{e}{2}{\mathrm{\partial }}_a{g}^{bc}{p}_b{p}_c$. The vertical differential of $C_m$ is $d_p{C}_m=p^{\mathrm{♯}}$; it is nonzero on the massive branch because $g^{-1}(p,p)=-m^2{c}^2<0$, and on the massless branch because $p\mathrm{\ne }0$ and $g^{-1}$ is nondegenerate. Thus the shell is a regular primary constraint by [F7], and [F8] fixes the canonical sign convention. The equations are tensorial and hence agree across overlapping charts.

For $m>0$, the first equation gives $p_a=g_{ab}{x}^{\mathrm{\prime }b}\mathrm{/}e$. The constraint gives $g(x^{\mathrm{\prime }},x^{\mathrm{\prime }})=-e^2{m}^2{c}^2$, so $s=\sqrt{-g(x^{\mathrm{\prime }},x^{\mathrm{\prime }})}=emc$ because $e,m,c>0$. Hence $p_a=mc{g}_{ab}{x}^{\mathrm{\prime }b}\mathrm{/}s$, the fibre derivative of the square-root Lagrangian; this agrees with the exact image result [F9].

Differentiate $p_a=g_{ab}{x}^{\mathrm{\prime }b}\mathrm{/}e$, use the second equation, and use the inverse-metric identity ${\mathrm{\partial }}_a{g}^{bc}{g}_{bi}{g}_{cj}=-{\mathrm{\partial }}_a{g}_{ij}$. Multiplying by $e$ gives $g_{ab}{x}^{\mathrm{\prime }\mathrm{\prime }b}+({\mathrm{\partial }}_c{g}_{ab})x^{\mathrm{\prime }c}{x}^{\mathrm{\prime }b}-\frac{e^{\mathrm{\prime }}}{e}{g}_{ab}{x}^{\mathrm{\prime }b}=\frac{1}{2}({\mathrm{\partial }}_a{g}_{ij})x^{\mathrm{\prime }i}{x}^{\mathrm{\prime }j}\mathrm{.}$ Since $x^{\mathrm{\prime }c}{x}^{\mathrm{\prime }b}$ is symmetric, the difference of the metric-derivative terms is $\frac{1}{2}({\mathrm{\partial }}_c{g}_{ab}+{\mathrm{\partial }}_b{g}_{ac}-{\mathrm{\partial }}_a{g}_{bc})x^{\mathrm{\prime }c}{x}^{\mathrm{\prime }b}=g_{ad}{\mathrm{\Gamma }}^d{}_{cb}{x}^{\mathrm{\prime }c}{x}^{\mathrm{\prime }b}$. Inverting $g_{ab}$ therefore gives ${\mathrm{\nabla }}_{x^{\mathrm{\prime }}}{x}^{\mathrm{\prime }}=\frac{e^{\mathrm{\prime }}}{e}{x}^{\mathrm{\prime }}\mathrm{.}$ This computation uses [F3]–[F5]; it is the coordinate form of the reparameterized geodesic equation.

Substituting $p_a=g_{ab}{x}^{\mathrm{\prime }b}\mathrm{/}e$ into the integrand gives $L_1=g(x^{\mathrm{\prime }},x^{\mathrm{\prime }})\mathrm{/}(2e)-e{m}^2{c}^2\mathrm{/}2=-s^2\mathrm{/}(2e)-e{m}^2{c}^2\mathrm{/}2$, where $s=\sqrt{-g(x^{\mathrm{\prime }},x^{\mathrm{\prime }})}>0$. Its $e$-derivative is $s^2\mathrm{/}(2e^2)-m^2{c}^2\mathrm{/}2$, whose unique positive zero is $e=s\mathrm{/}(mc)$; the second derivative there is $-s^2\mathrm{/}{e}^3\mathrm{\ne }0$. Thus the $p$- and $e$-stationarity equations determine a unique smooth auxiliary pair $(p^{\ast }(x,x^{\mathrm{\prime }}),e^{\ast }(x,x^{\mathrm{\prime }}))$ on the future timelike domain, and the stationary integrand is $-mcs$. For any fixed-endpoint variation $x_{\epsilon }$, choose $p_{\epsilon },e_{\epsilon }$ by these formulas. The chain rule for the varied action has the $x_{\epsilon }$ variation plus terms proportional to its $p$- and $e$-derivatives; those terms vanish at the auxiliary stationary pair. Hence the first variation of the reduced action equals the $x$-part of the first-order variation. Conversely every stationary first-order triple must equal this unique auxiliary pair, so the same identity gives the reverse implication. This proves the equivalence with the square-root action.

For $m>0$, $d\tau =sd\lambda \mathrm{/}c=emd\lambda$, so $e=1\mathrm{/}m$ in proper-time parameter and step 2.2 becomes ${\mathrm{\nabla }}_{U}U=0$; the converse follows by reversing the momentum definition and the same identities.

For $m=0$, [F2] selects $p\mathrm{\ne }0$ future null; the first equation makes $x^{\mathrm{\prime }}=e{p}^{\mathrm{♯}}$, so $x^{\mathrm{\prime }}$ is nonzero future null and the constraint is null. Choose a new parameter σ with $d\sigma \mathrm{/}d\lambda =e\mathrm{/}{e}_0$ for any constant $e_0>0$; the connection transformation law applied to step 2.2 gives ${\mathrm{\nabla }}_{dx\mathrm{/}d\sigma }(dx\mathrm{/}d\sigma )=0$. Since $g(x^{\mathrm{\prime }},x^{\mathrm{\prime }})=0$, the massive proper-time construction cannot be used.