First variation of the length of a timelike curve: stationarity is the geodesic equation

(Differentiability of $L$.) The integrand $\rho (s,t):=\sqrt{-g_{H(s,t)}(T(s,t),T(s,t))}$ is strictly positive and continuous on $(-\epsilon ,\epsilon )\times [a,b]$ (indeed $C^1$): $H$ and $T={\mathrm{\partial }}_{t}H$ are $C^1$ in both variables, the metric components are smooth, and the square root is smooth on $(0,\mathrm{\infty })$. Hence [L4] applies and $L^{\mathrm{\prime }}(0)={\int }_a^b{\mathrm{\partial }}_s\rho (0,t)dt$.

(Derivative of the integrand.) By [L1] applied to $V=W=T$ with the derivative in the $s$-direction (the pairing identity holds for the two-parameter family by the same component computation as for one parameter), ${\mathrm{\partial }}_s\rho (0,t)=-\frac{g(D_{s}T,T)}{\sqrt{-g(T,T)}}{\mid }_{s=0}$.

(Commutator.) On the chart $U$, writing $H^a=H^a(s,t)$ and using [L2], $D_s{T}^a-D_t{J}^a={\sum }_{b,c}{\mathrm{\Gamma }}^a{}_{bc}(J^b{T}^c-T^b{J}^c)=0$ since ${\mathrm{\partial }}_s{\mathrm{\partial }}_t{H}^a={\mathrm{\partial }}_t{\mathrm{\partial }}_s{H}^a$ by [L3] and $\mathrm{\Gamma }$ is symmetric in its lower indices. Hence $D_{s}T=D_{t}J$ at $s=0$, where $J={\mathrm{\partial }}_{s}H$, and $T={\gamma }^{\mathrm{\prime }}$.

(First variation formula.) Substituting steps 2.1 and 2.2 into step 1.1, and using $g({\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})=-c^2$, gives the formula of claim 1.

(Integration by parts.) By [L1], $\frac{d}{dt}g(J,{\gamma }^{\mathrm{\prime }})=g(D_{t}J,{\gamma }^{\mathrm{\prime }})+g(J,D_t{\gamma }^{\mathrm{\prime }})$ and both terms are continuous, so the derivative of the $C^1$ function $t\mapsto g(J,{\gamma }^{\mathrm{\prime }})$ is integrable; [L5] gives ${\int }_a^b(g(D_{t}J,{\gamma }^{\mathrm{\prime }})+g(J,D_t{\gamma }^{\mathrm{\prime }}))dt=g(J,{\gamma }^{\mathrm{\prime }}){\mathrm{\mid }}_a^b=0$ because $J(a)=J(b)=0$. Multiplying the first variation formula by $c$ and using this identity gives claim 2, $L^{\mathrm{\prime }}(0)=\frac{1}{c}{\int }_a^{b}g(J,D_t{\gamma }^{\mathrm{\prime }})dt$.

($\Rightarrow$ direction of claim 3.) If $D_t{\gamma }^{\mathrm{\prime }}=0$ on $[a,b]$, the integrand of claim 2 vanishes identically for the variation at hand, so $L^{\mathrm{\prime }}(0)=0$; since $H$ was arbitrary, every such variation is stationary.

(Converse.) If $D_t{\gamma }^{\mathrm{\prime }}$ is nonzero at an endpoint, continuity makes it nonzero at an interior point as well; hence, if $D_t{\gamma }^{\mathrm{\prime }}$ is not identically zero on $[a,b]$, choose $t_0\in (a,b)$ with $A(t_0):=D_t{\gamma }^{\mathrm{\prime }}(t_0)\mathrm{\ne }0$. The field $A$ is continuous because $\gamma$ is $C^2$ and the Christoffel symbols are smooth. By [L7], choose a chart coordinate vector field with sign $Z=\pm {\mathrm{\partial }}_j$ along $\gamma$ so $g(Z(t_0),A(t_0))>0$; by continuity, this pairing is positive on a neighborhood of $t_0$. Choose a closed interval $[u,v]\subseteq (a,b)$ with $t_0$ in its interior on which the pairing stays positive. Then choose $0<r<R$ with $[t_0-r,t_0+r]\subseteq [u,v]$ and $(t_0-R,t_0+R)\subseteq (u,v)$, and let $\phi \ge 0$ be the bump of [L6] equal to $1$ on $[t_0-r,t_0+r]$ with support in $(t_0-R,t_0+R)$. Put $J(t):=\phi (t)Z(t)$; its chart components are smooth and it vanishes at $a,b$. Define $H(s,t):=x^{-1}(x(\gamma (t))+s{J}^{\mathrm{c}\mathrm{h}}(t))$ on a sufficiently small $\mathrm{\mid }s\mathrm{\mid }$, where $J^{\mathrm{c}\mathrm{h}}$ denotes the chart components of $J$. This is a $C^2$ variation, and [L7] ensures $g(J,D_t{\gamma }^{\mathrm{\prime }})>0$ wherever $\phi >0$.

(Admissibility of the constructed variation.) $H$ fixes the endpoints because $J(a)=J(b)=0$; it is $C^2$; and for all sufficiently small $\mathrm{\mid }s\mathrm{\mid }$ its fibers are future timelike and lie in $U$: the perturbation $s{J}^{\mathrm{c}\mathrm{h}}$ vanishes outside $[u,v]$, so on the compact set $[u,v]$ the image stays in a fixed compact subset of the chart domain for small $\mathrm{\mid }s\mathrm{\mid }$, and the family of velocities $T(s,t)$ depends continuously on $(s,t)$ with $T(0,t)={\gamma }^{\mathrm{\prime }}(t)$ future timelike, so by compactness of $[a,b]$ the timelike condition persists for small $\mathrm{\mid }s\mathrm{\mid }$.

(Nonstationarity.) For this variation, claim 2 and the choice of $J$ give $L^{\mathrm{\prime }}(0)=\frac{1}{c}{\int }_a^b\phi (t)g_{\gamma (t)}(Z(t),D_t{\gamma }^{\mathrm{\prime }}(t))dt>0$: the pairing is positive wherever $\phi >0$, while $\phi$ is nonnegative and is $1$ on the nondegenerate interval $[t_0-r,t_0+r]$, so the integrand is continuous, nonnegative and strictly positive on that interval; hence the integral is strictly positive by [L5]. Therefore a curve with $D_t{\gamma }^{\mathrm{\prime }}\mathrm{\ne }0$ admits a variation with $L^{\mathrm{\prime }}(0)\mathrm{\ne }0$, and stationarity for every variation forces $D_t{\gamma }^{\mathrm{\prime }}=0$, proving claim 3.