[proofplan] We compute the first variation of Perelman's $\mathcal L$-length on an arbitrary compact time interval $[a,b]\subset(0,\tau_2]$ with fixed endpoints. The scalar curvature term contributes the spatial gradient of $R$, while the kinetic term is integrated by parts along the time-dependent metric $g(\tau)$. The time dependence of the metric, through $\partial_\tau g=2\operatorname{Ric}$, produces the Ricci term, and differentiating the weight $\sqrt{\tau}$ produces the term $\frac{1}{2\tau}\dot{\gamma}$. Since the resulting integral vanishes against every compactly supported variation field, the coefficient vector field must vanish pointwise. [/proofplan] [step:Compute the first variation on a compact time interval] Fix $0<a<b\leq \tau_2$. Let $\varepsilon>0$, and let $\Gamma:(-\varepsilon,\varepsilon)\times [a,b]\to M$ be a smooth variation of $\gamma|_{[a,b]}$, written $\Gamma(s,\tau)=\gamma_s(\tau)$, with fixed endpoints, so $\gamma_0=\gamma|_{[a,b]}$, $\gamma_s(a)=\gamma(a)$, and $\gamma_s(b)=\gamma(b)$ for all $s$. Define the variation field $X:[a,b]\to TM$ by \begin{align*} X(\tau)=\left.\frac{\partial \gamma_s(\tau)}{\partial s}\right|_{s=0}\in T_{\gamma(\tau)}M. \end{align*} Then $X(a)=X(b)=0$. For each $\tau\in[a,b]$, all gradients, inner products, norms, Ricci tensors, and connections below are taken with respect to $g(\tau)$. Differentiating the scalar curvature term in the spatial direction $X(\tau)$ gives \begin{align*} \left.\frac{\partial}{\partial s}\right|_{s=0} R(\gamma_s(\tau),\tau) = g(\nabla R(\gamma(\tau),\tau),X(\tau)). \end{align*} For the kinetic term, fix $\tau\in[a,b]$ and use the Levi-Civita connection of $g(\tau)$. This connection is torsion-free, so along the two-parameter map $\Gamma$ one has $\nabla_{\partial_s}\partial_\tau\Gamma=\nabla_{\partial_\tau}\partial_s\Gamma$ at time $\tau$; it is also metric-compatible. Hence \begin{align*} \left.\frac{\partial}{\partial s}\right|_{s=0}|\dot{\gamma}_s(\tau)|_{g(\tau)}^2=2g(\nabla_{\dot{\gamma}}X,\dot{\gamma}). \end{align*} Therefore criticality of $\gamma$ gives \begin{align*} 0 = \int_a^b \sqrt{\tau}\left( g(\nabla R,X) + 2g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) \right)\,d\mathcal L^1(\tau). \end{align*} [guided] Fix a compact interval $[a,b]\subset(0,\tau_2]$ so that the factor $\sqrt{\tau}$ is smooth and positive on the whole interval. We vary $\gamma$ through curves with the same endpoints. Thus we take $\varepsilon>0$ and a smooth map $\Gamma:(-\varepsilon,\varepsilon)\times [a,b]\to M$, written $\Gamma(s,\tau)=\gamma_s(\tau)$, with $\gamma_0=\gamma|_{[a,b]}$, $\gamma_s(a)=\gamma(a)$, and $\gamma_s(b)=\gamma(b)$. The associated variation field is the vector field $X:[a,b]\to TM$ along $\gamma$ defined by \begin{align*} X(\tau)=\left.\frac{\partial \gamma_s(\tau)}{\partial s}\right|_{s=0}\in T_{\gamma(\tau)}M. \end{align*} The fixed endpoint condition implies $X(a)=X(b)=0$. Now differentiate the two pieces of the integrand. First, for each fixed $\tau$, the scalar curvature is a smooth function \begin{align*} R(\cdot,\tau):M \to \mathbb{R}. \end{align*} The directional derivative of this function in the direction $X(\tau)$ is its metric gradient paired with $X(\tau)$: \begin{align*} \left.\frac{\partial}{\partial s}\right|_{s=0} R(\gamma_s(\tau),\tau) = g(\nabla R(\gamma(\tau),\tau),X(\tau)). \end{align*} Second, consider the kinetic energy term. For a fixed value of $\tau$, the metric is $g(\tau)$. Its Levi-Civita connection is metric-compatible and torsion-free. Torsion-freeness gives $\nabla_{\partial_s}\partial_\tau\Gamma=\nabla_{\partial_\tau}\partial_s\Gamma$ at this fixed time, so differentiating the velocity field in the variation direction is the same as covariantly differentiating the variation field along $\gamma$. Therefore \begin{align*} \left.\frac{\partial}{\partial s}\right|_{s=0}|\dot{\gamma}_s(\tau)|_{g(\tau)}^2=2g(\nabla_{\dot{\gamma}}X,\dot{\gamma}). \end{align*} The weight $\sqrt{\tau}$ does not depend on $s$, so differentiating the $\mathcal L$-length under the integral sign yields \begin{align*} \left.\frac{d}{ds}\right|_{s=0} \mathcal L[\gamma_s] = \int_a^b \sqrt{\tau}\left( g(\nabla R,X) + 2g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) \right)\,d\mathcal L^1(\tau). \end{align*} Because $\gamma$ is an $\mathcal L$-geodesic, this first variation vanishes for every fixed-endpoint variation, so \begin{align*} 0 = \int_a^b \sqrt{\tau}\left( g(\nabla R,X) + 2g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) \right)\,d\mathcal L^1(\tau). \end{align*} [/guided] [/step] [step:Integrate the kinetic variation by parts using the backward Ricci flow] For the vector fields $X$ and $\dot{\gamma}$ along $\gamma$, the product rule for the time-dependent metric gives \begin{align*} \frac{d}{d\tau}g(X,\dot{\gamma}) = \frac{\partial g}{\partial \tau}(X,\dot{\gamma}) + g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) + g(X,\nabla_{\dot{\gamma}}\dot{\gamma}). \end{align*} Using $\frac{\partial g}{\partial \tau}=2\operatorname{Ric}$, we obtain \begin{align*} g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) = \frac{d}{d\tau}g(X,\dot{\gamma}) - 2\operatorname{Ric}(X,\dot{\gamma}) - g(X,\nabla_{\dot{\gamma}}\dot{\gamma}). \end{align*} Hence the integral of the kinetic variation decomposes as \begin{align*} \int_a^b 2\sqrt{\tau}\,g(\nabla_{\dot{\gamma}}X,\dot{\gamma})\,d\mathcal L^1(\tau)=\int_a^b \left(2\sqrt{\tau}\,\frac{d}{d\tau}g(X,\dot{\gamma})-4\sqrt{\tau}\,\operatorname{Ric}(X,\dot{\gamma})-2\sqrt{\tau}\,g(X,\nabla_{\dot{\gamma}}\dot{\gamma})\right)\,d\mathcal L^1(\tau). \end{align*} Integrating the total derivative term by parts and using $X(a)=X(b)=0$ gives \begin{align*} \int_a^b 2\sqrt{\tau}\,\frac{d}{d\tau}g(X,\dot{\gamma})\,d\mathcal L^1(\tau)=-\int_a^b \frac{1}{\sqrt{\tau}}g(X,\dot{\gamma})\,d\mathcal L^1(\tau). \end{align*} Substituting into the first variation identity yields \begin{align*} 0 = \int_a^b g\left( \sqrt{\tau}\nabla R - \frac{1}{\sqrt{\tau}}\dot{\gamma} - 4\sqrt{\tau}\operatorname{Ric}(\dot{\gamma},\cdot)^\sharp - 2\sqrt{\tau}\nabla_{\dot{\gamma}}\dot{\gamma}, X \right)\,d\mathcal L^1(\tau). \end{align*} [guided] The first variation still contains the derivative of the variation field, namely the term $g(\nabla_{\dot{\gamma}}X,\dot{\gamma})$. To obtain an Euler-Lagrange equation, we must move this derivative off $X$ and onto the coefficient of $X$ by [integration by parts](/theorems/2098). For the vector fields $X$ and $\dot{\gamma}$ along $\gamma$, the metric depends on $\tau$, so the ordinary product rule has an extra term. The time-dependent metric product rule gives \begin{align*} \frac{d}{d\tau}g(X,\dot{\gamma}) = \frac{\partial g}{\partial \tau}(X,\dot{\gamma}) + g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) + g(X,\nabla_{\dot{\gamma}}\dot{\gamma}). \end{align*} Because the metric evolves by backward Ricci flow, $\frac{\partial g}{\partial \tau}=2\operatorname{Ric}$. Solving this identity for the derivative term gives \begin{align*} g(\nabla_{\dot{\gamma}}X,\dot{\gamma}) = \frac{d}{d\tau}g(X,\dot{\gamma}) - 2\operatorname{Ric}(X,\dot{\gamma}) - g(X,\nabla_{\dot{\gamma}}\dot{\gamma}). \end{align*} Substituting this into the kinetic part of the first variation yields \begin{align*} \int_a^b 2\sqrt{\tau}\,g(\nabla_{\dot{\gamma}}X,\dot{\gamma})\,d\mathcal L^1(\tau)=\int_a^b \left(2\sqrt{\tau}\,\frac{d}{d\tau}g(X,\dot{\gamma})-4\sqrt{\tau}\,\operatorname{Ric}(X,\dot{\gamma})-2\sqrt{\tau}\,g(X,\nabla_{\dot{\gamma}}\dot{\gamma})\right)\,d\mathcal L^1(\tau). \end{align*} Now apply the one-dimensional [integration by parts](/theorems/210) formula on $[a,b]$ to the scalar function $\tau \mapsto g(X(\tau),\dot{\gamma}(\tau))$ and the weight $\tau \mapsto 2\sqrt{\tau}$. Since $X(a)=X(b)=0$, the boundary term vanishes. Since $\frac{d}{d\tau}(2\sqrt{\tau})=\frac{1}{\sqrt{\tau}}$, we get \begin{align*} \int_a^b 2\sqrt{\tau}\,\frac{d}{d\tau}g(X,\dot{\gamma})\,d\mathcal L^1(\tau)=-\int_a^b \frac{1}{\sqrt{\tau}}g(X,\dot{\gamma})\,d\mathcal L^1(\tau). \end{align*} Combining this with the scalar curvature term from the first variation gives \begin{align*} 0 = \int_a^b g\left( \sqrt{\tau}\nabla R - \frac{1}{\sqrt{\tau}}\dot{\gamma} - 4\sqrt{\tau}\operatorname{Ric}(\dot{\gamma},\cdot)^\sharp - 2\sqrt{\tau}\nabla_{\dot{\gamma}}\dot{\gamma}, X \right)\,d\mathcal L^1(\tau). \end{align*} This is the key reduction: the first variation is now an integral pairing an arbitrary variation field $X$ against a coefficient vector field along $\gamma$. [/guided] [/step] [step:Use arbitrary compactly supported variations to identify the Euler-Lagrange field] Define the vector field $E:[a,b]\to TM$ along $\gamma|_{[a,b]}$ by \begin{align*} E(\tau)=\sqrt{\tau}\nabla R(\gamma(\tau),\tau)-\frac{1}{\sqrt{\tau}}\dot{\gamma}(\tau)-4\sqrt{\tau}\operatorname{Ric}(\dot{\gamma}(\tau),\cdot)^{\sharp}-2\sqrt{\tau}\nabla_{\dot{\gamma}}\dot{\gamma}(\tau). \end{align*} The preceding step says that \begin{align*} \int_a^b g(E(\tau),X(\tau))\,d\mathcal L^1(\tau)=0 \end{align*} for every smooth variation field $X$ along $\gamma$ with $X(a)=X(b)=0$. Conversely, every smooth compactly supported vector field $X:(a,b)\to TM$ along $\gamma$ arises from a fixed-endpoint variation. Indeed, let $K=\operatorname{supp}X\subset(a,b)$. Since $K$ is compact and $\gamma(K)$ is compact in $M$, choose finitely many coordinate charts whose domains cover $\gamma(K)$, and choose a smooth [partition of unity](/page/Partition%20of%20Unity) in the parameter $\tau$ subordinate to the corresponding open cover of $K$. On each coordinate subinterval, define the local variation by adding $sX(\tau)$ to the coordinate representative of $\gamma(\tau)$; after possibly shrinking $|s|<\varepsilon$, compactness of $K$ ensures that all perturbed coordinate representatives remain in the chart domains. Summing the local coordinate perturbations with the partition of unity gives a smooth variation of $\gamma$ on $[a,b]$ whose variation field is $X$. Because $X$ has compact support in $(a,b)$, this variation equals $\gamma$ near $a$ and $b$, so it has fixed endpoints. Therefore the identity holds for every smooth compactly supported vector field along $\gamma$ on $(a,b)$. We claim that $E(\tau)=0$ for every $\tau\in(a,b)$. If not, choose $\tau_0\in(a,b)$ with $E(\tau_0)\neq 0$. By smoothness and positive-definiteness of $g(\tau_0)$, there is an open interval $I\subset(a,b)$ containing $\tau_0$ and a smooth vector field \begin{align*} Y:I\to TM, \qquad Y(\tau)\in T_{\gamma(\tau)}M, \end{align*} along $\gamma|_I$ such that $g(E,Y)>0$ on $I$. Let \begin{align*} \eta:I\to[0,\infty) \end{align*} be a smooth function with compact support in $I$ and not identically zero, and extend $\eta Y$ by zero to a smooth compactly supported vector field $X$ along $\gamma$ on $(a,b)$. Then \begin{align*} \int_a^b g(E,X)\,d\mathcal L^1(\tau) = \int_I \eta(\tau)g(E(\tau),Y(\tau))\,d\mathcal L^1(\tau) > 0, \end{align*} contradicting the first variation identity. Therefore $E=0$ on $(a,b)$. [/step] [step:Divide by the positive weight and obtain the geodesic equation] Since $\tau>0$ on $(a,b)$, the identity $E(\tau)=0$ gives \begin{align*} 0=\sqrt{\tau}\nabla R-\frac{1}{\sqrt{\tau}}\dot{\gamma}-4\sqrt{\tau}\operatorname{Ric}(\dot{\gamma},\cdot)^\sharp-2\sqrt{\tau}\nabla_{\dot{\gamma}}\dot{\gamma}. \end{align*} Multiplying by $-\frac{1}{2\sqrt{\tau}}$ gives \begin{align*} \nabla_{\dot{\gamma}}\dot{\gamma} - \frac{1}{2}\nabla R + 2\operatorname{Ric}(\dot{\gamma},\cdot)^\sharp + \frac{1}{2\tau}\dot{\gamma} = 0 \end{align*} on $(a,b)$. Because $0<a<b<\tau_2$ were arbitrary, the same identity holds for every $\tau\in(0,\tau_2)$. The curve $\gamma$, the backward Ricci flow metric $g(\tau)$, the scalar curvature $R(\cdot,\tau)$, and the Ricci tensor are smooth up to $\tau=\tau_2$ by the hypotheses implicit in the definition of an $\mathcal L$-geodesic for the flow. Hence the Euler-Lagrange expression is continuous as $\tau\uparrow\tau_2$, and the identity also holds at $\tau=\tau_2$. This is the desired Euler-Lagrange equation for $\mathcal L$-geodesics, with all geometric quantities evaluated using $g(\tau)$. [/step]