[proofplan] The proof is the pointwise tangent-space form of the Benamou-Brenier dynamic formulation. First, the velocity representation theorem for absolutely continuous Wasserstein curves gives, for almost every time, a continuity-equation velocity of minimal $L^2(\rho_t)$ norm equal to the metric derivative. Second, every velocity satisfying the same continuity equation has norm at least that minimal norm. Finally, if a smooth minimizer exists, variations by weighted divergence-free fields force it to lie in the orthogonal complement of the divergence-free subspace; the weighted Helmholtz decomposition identifies this complement with the closure of gradients, and smoothness on the simply connected domain $\mathbb R^n$ upgrades this to an actual smooth gradient field. [/proofplan]

[step:Select the full-measure set where the Wasserstein velocity representation applies]Since $\rho:I\to(\mathcal P_2(\mathbb R^n),W_2)$ is $2$-absolutely continuous, the velocity representation theorem for absolutely continuous Wasserstein curves applies locally to $\rho$ on the interval $I$; in the notation of [citetheorem:9559], there exists a Borel vector field \begin{align*} u:I\times\mathbb R^n\to\mathbb R^n \end{align*} such that $u_t:=u(t,\cdot)\in L^2(\rho_t;\mathbb R^n)$ for $\mathcal L^1$-almost every $t\in I$, the continuity equation \begin{align*} \partial_t\rho_t+\nabla\cdot(\rho_t u_t)=0 \end{align*} holds in the sense of distributions on $I\times\mathbb R^n$, and \begin{align*} \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x)=|\dot\rho_t|_{W_2}^2 \end{align*} for $\mathcal L^1$-almost every $t\in I$. Because $\rho_t=r_t\,\mathcal L^n$ and $t\mapsto r_t$ is differentiable as a distribution-valued map, the space-time continuity equation gives time-slice identities after testing with functions of the form $(s,x)\mapsto \alpha(s)\psi(x)$, where $\alpha\in C_c^\infty(I)$ and $\psi\in C_c^\infty(\mathbb R^n)$. Fubini's theorem applies to the velocity term because [citetheorem:9559] gives that $s\mapsto \|u_s\|_{L^2(\rho_s)}$ is locally square-integrable, hence locally integrable, in time. Moreover, \begin{align*} \left|\int_{\mathbb R^n}\nabla\psi(x)\cdot u_s(x)r_s(x)\,d\mathcal L^n(x)\right|\le \|\nabla\psi\|_\infty\|u_s\|_{L^2(\rho_s)}\rho_s(\operatorname{supp}\psi)^{1/2}\le \|\nabla\psi\|_\infty\|u_s\|_{L^2(\rho_s)}. \end{align*} since $\rho_s$ is a probability measure. Hence the resulting scalar distribution in $s$ is represented by the locally integrable function \begin{align*} s\mapsto \int_{\mathbb R^n}\nabla\psi(x)\cdot u_s(x)\,r_s(x)\,d\mathcal L^n(x). \end{align*} The scalar distributional derivative of the map $s\mapsto \int_{\mathbb R^n}\psi(x)r_s(x)\,d\mathcal L^n(x)$ is therefore represented by the locally integrable velocity term above. Since $t\mapsto r_t$ is differentiable as a distribution-valued map for almost every $t$, this representative agrees with $(\partial_t r_t)(\psi)$ for almost every such $t$. After intersecting over a countable dense family of test functions on each member of a compact exhaustion of $\mathbb R^n$, the identity extends to every $\psi\in C_c^\infty(\mathbb R^n)$ as follows. If $\operatorname{supp}\psi\subset K$, then smooth positivity gives $M_K:=\sup_{x\in K}r_t(x)<\infty$, and the Cauchy-Schwarz inequality bounds the velocity functional by $M_K^{1/2}\|u_t\|_{L^2(\rho_t)}\|\nabla\psi\|_{L^2(K,\mathcal L^n)}$; hence the right-hand side is continuous in the usual test-function topology on $C_c^\infty(K)$. The left-hand side is continuous because $\partial_t r_t\in\mathcal D'(\mathbb R^n)$. Therefore we obtain, for almost every $t\in I$ and every $\psi\in C_c^\infty(\mathbb R^n)$, \begin{align*} (\partial_t r_t)(\psi) = \int_{\mathbb R^n}\nabla\psi(x)\cdot u_t(x)\,r_t(x)\,d\mathcal L^n(x). \end{align*} Thus $u_t\in\mathcal A_t$ for $\mathcal L^1$-almost every $t\in I$ for which the metric derivative and the distributional derivative are represented pointwise.[/step]

[guided]The first task is to reconcile the global continuity equation from Wasserstein theory with the pointwise definition of $\mathcal A_t$. The theorem [citetheorem:9559] applies because $\rho$ is absolutely continuous as a curve in $(\mathcal P_2(\mathbb R^n),W_2)$, exactly as assumed. It gives a Borel velocity field \begin{align*} u:I\times\mathbb R^n\to\mathbb R^n \end{align*} whose time slices $u_t:=u(t,\cdot)$ belong to $L^2(\rho_t;\mathbb R^n)$ for almost every $t$, and whose $L^2(\rho_t)$ norm realizes the metric speed: \begin{align*} \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x)=|\dot\rho_t|_{W_2}^2 \end{align*} for $\mathcal L^1$-almost every $t\in I$. The same theorem also gives the continuity equation \begin{align*} \partial_t\rho_t+\nabla\cdot(\rho_t u_t)=0 \end{align*} in the distributional sense on $I\times\mathbb R^n$. To extract a time-slice identity, we first test the equation against $(s,x)\mapsto \alpha(s)\psi(x)$, where $\alpha\in C_c^\infty(I)$ and $\psi\in C_c^\infty(\mathbb R^n)$. The velocity term is integrable in $s$ on compact subintervals because [citetheorem:9559] gives that $s\mapsto \|u_s\|_{L^2(\rho_s)}$ is locally square-integrable, hence locally integrable, in time. Since $\nabla\psi$ is bounded with compact support and $\rho_s$ is a probability measure, the Cauchy-Schwarz inequality gives \begin{align*} \left|\int_{\mathbb R^n}\nabla\psi(x)\cdot u_s(x)r_s(x)\,d\mathcal L^n(x)\right| \le \|\nabla\psi\|_\infty\|u_s\|_{L^2(\rho_s)}. \end{align*} Fubini's theorem therefore identifies the velocity side with the scalar distribution represented by \begin{align*} s\mapsto \int_{\mathbb R^n}\nabla\psi(x)\cdot u_s(x)\,r_s(x)\,d\mathcal L^n(x). \end{align*} The last point is to identify this scalar derivative with the distribution-valued derivative assumed in the theorem. For the fixed test function $\psi$, distributional differentiability of $t\mapsto r_t$ says that the derivative of $s\mapsto \int_{\mathbb R^n}\psi(x)r_s(x)\,d\mathcal L^n(x)$ is $(\partial_t r_t)(\psi)$ at almost every differentiability time. Hence, after taking a common full-measure set of times for the test functions and extending by continuity in $\psi$, \begin{align*} (\partial_t r_t)(\psi) = \int_{\mathbb R^n}\nabla\psi(x)\cdot u_t(x)\,r_t(x)\,d\mathcal L^n(x) \end{align*} for every $\psi\in C_c^\infty(\mathbb R^n)$ at almost every such $t$. This is exactly the defining condition for $u_t\in\mathcal A_t$. Hence the admissible class is nonempty at almost every time at which the metric derivative and this pointwise distributional equation are both available.[/guided]

[step:Compare every admissible velocity with the minimal Wasserstein velocity]Fix a time $t\in I$ in the full-measure set obtained above. Let $v\in\mathcal A_t$. By definition of $\mathcal A_t$, \begin{align*} (\partial_t r_t)(\psi) = \int_{\mathbb R^n}\nabla\psi(x)\cdot v(x)\,r_t(x)\,d\mathcal L^n(x) \end{align*} for every $\psi\in C_c^\infty(\mathbb R^n)$, so $v$ is a distributional velocity for the curve at time $t$. The pointwise minimal-velocity clause of [citetheorem:9559] applies at this time because the metric derivative exists, $v\in L^2(\rho_t;\mathbb R^n)$, and the preceding identity is exactly the required time-slice distributional continuity equation with density $r_t$. That clause states that every $L^2(\rho_t)$ vector field producing the same distributional derivative as the curve at time $t$ has squared $L^2(\rho_t)$ norm at least $|\dot\rho_t|_{W_2}^2$. Therefore \begin{align*} |\dot\rho_t|_{W_2}^2 \le \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*} Taking the infimum over $v\in\mathcal A_t$ gives \begin{align*} |\dot\rho_t|_{W_2}^2 \le \inf_{v\in\mathcal A_t} \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*}[/step]

[guided]Fix a time $t$ in the full-measure set selected in the previous step, and let $v\in\mathcal A_t$. The definition of $\mathcal A_t$ gives, for every test function $\psi\in C_c^\infty(\mathbb R^n)$, \begin{align*} (\partial_t r_t)(\psi) = \int_{\mathbb R^n}\nabla\psi(x)\cdot v(x)r_t(x)\,d\mathcal L^n(x). \end{align*} Thus $v$ is an $L^2(\rho_t)$ velocity whose distributional divergence realizes the same infinitesimal motion of the measure $\rho_t=r_t\mathcal L^n$ at the chosen time. The pointwise minimal-velocity clause of [citetheorem:9559] is exactly the comparison principle for such time-slice velocities: when the metric derivative exists, every $L^2(\rho_t)$ vector field producing the same distributional derivative as the curve at that time has norm at least the minimal Wasserstein speed. Since $v\in L^2(\rho_t;\mathbb R^n)$ by the definition of $\mathcal A_t$, this principle gives \begin{align*} |\dot\rho_t|_{W_2}^2 \le \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*} Because this inequality holds for every admissible $v\in\mathcal A_t$, taking the infimum over the admissible class yields \begin{align*} |\dot\rho_t|_{W_2}^2 \le \inf_{v\in\mathcal A_t} \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*}[/guided]

[step:Use the minimal velocity to attain the opposite inequality]For the velocity field $u_t\in\mathcal A_t$ selected above, the metric-speed identity from [citetheorem:9559] gives \begin{align*} \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x) = |\dot\rho_t|_{W_2}^2. \end{align*} Since $u_t$ is one admissible element of $\mathcal A_t$, \begin{align*} \inf_{v\in\mathcal A_t} \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x) \le \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x) = |\dot\rho_t|_{W_2}^2. \end{align*} Combining this inequality with the previous step proves \begin{align*} |\dot\rho_t|_{W_2}^2 = \inf_{v\in\mathcal A_t} \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x) \end{align*} for $\mathcal L^1$-almost every $t\in I$.[/step]

[guided]The previous step proves that no admissible velocity can have squared norm below $|\dot\rho_t|_{W_2}^2$. To prove equality, we must exhibit one admissible velocity with exactly that norm. The velocity field $u_t$ selected from [citetheorem:9559] belongs to $\mathcal A_t$ by the time-slice argument in the first step, and the same theorem gives \begin{align*} \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x) = |\dot\rho_t|_{W_2}^2. \end{align*} Since the infimum over $\mathcal A_t$ is bounded above by the value at this particular admissible element, we have \begin{align*} \inf_{v\in\mathcal A_t} \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x) \le \int_{\mathbb R^n}|u_t(x)|^2\,d\rho_t(x) = |\dot\rho_t|_{W_2}^2. \end{align*} Together with the opposite inequality, this proves the metric tensor identity for $\mathcal L^1$-almost every $t\in I$.[/guided]

[step:Derive orthogonality to weighted divergence-free perturbations for a smooth minimizer]Fix a time $t$ for which the preceding equality holds, and assume that the infimum is attained by a vector field \begin{align*} v:\mathbb R^n\to\mathbb R^n \end{align*} with $v\in C^\infty(\mathbb R^n;\mathbb R^n)\cap L^2(\rho_t;\mathbb R^n)$. Define the closed weighted divergence-free subspace \begin{align*} K_t := \left\{ w\in L^2(\rho_t;\mathbb R^n): \int_{\mathbb R^n}\nabla\psi(x)\cdot w(x)\,d\rho_t(x)=0 \text{ for every }\psi\in C_c^\infty(\mathbb R^n) \right\}. \end{align*} If $w\in K_t$ and $s\in\mathbb R$, then $v+s w\in\mathcal A_t$, because the defining weak continuity equation for $v+s w$ is obtained by adding $s$ times the defining identity for $w$ to the defining identity for $v$. Define \begin{align*} Q_w:\mathbb R\to\mathbb R \end{align*} by \begin{align*} Q_w(s):=\int_{\mathbb R^n}|v(x)+s w(x)|^2\,d\rho_t(x). \end{align*} Since $v,w\in L^2(\rho_t;\mathbb R^n)$, expansion of the square gives \begin{align*} Q_w(s)=\int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x)+2s\int_{\mathbb R^n}v(x)\cdot w(x)\,d\rho_t(x)+s^2\int_{\mathbb R^n}|w(x)|^2\,d\rho_t(x). \end{align*} The function $Q_w$ has a minimum at $s=0$, because $v$ minimizes the action over $\mathcal A_t$. Hence $Q_w'(0)=0$, and therefore \begin{align*} \int_{\mathbb R^n}v(x)\cdot w(x)\,d\rho_t(x)=0 \end{align*} for every $w\in K_t$.[/step]

[guided]Fix $w\in K_t$. The point of introducing $K_t$ is that its elements are precisely the $L^2(\rho_t)$ perturbations that do not change the weak continuity equation. Indeed, for every $\psi\in C_c^\infty(\mathbb R^n)$ and every $s\in\mathbb R$, the defining identity for $v\in\mathcal A_t$ and the defining identity for $w\in K_t$ give \begin{align*} (\partial_t r_t)(\psi)=\int_{\mathbb R^n}\nabla\psi(x)\cdot v(x)\,d\rho_t(x)=\int_{\mathbb R^n}\nabla\psi(x)\cdot (v(x)+s w(x))\,d\rho_t(x). \end{align*} Thus $v+s w\in\mathcal A_t$ for every real number $s$. Define the quadratic function $Q_w:\mathbb R\to\mathbb R$ by \begin{align*} Q_w(s):=\int_{\mathbb R^n}|v(x)+s w(x)|^2\,d\rho_t(x). \end{align*} The integral is finite for every $s$ because $v,w\in L^2(\rho_t;\mathbb R^n)$. Expanding the square in the Hilbert space $L^2(\rho_t;\mathbb R^n)$ gives \begin{align*} Q_w(s)=\int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x)+2s\int_{\mathbb R^n}v(x)\cdot w(x)\,d\rho_t(x)+s^2\int_{\mathbb R^n}|w(x)|^2\,d\rho_t(x). \end{align*} Since $v$ minimizes the action over $\mathcal A_t$ and all fields $v+s w$ remain admissible, $Q_w$ has a minimum at $s=0$. Differentiating this polynomial at $0$ gives $Q_w'(0)=0$, hence \begin{align*} \int_{\mathbb R^n}v(x)\cdot w(x)\,d\rho_t(x)=0. \end{align*} Because $w\in K_t$ was arbitrary, $v$ is orthogonal to the whole closed subspace $K_t$, not merely to a smaller class of smooth compactly supported perturbations.[/guided]

[step:Identify the orthogonal complement with gradients and extract a smooth potential]Let \begin{align*} G_t := \overline{\{\nabla\psi:\psi\in C_c^\infty(\mathbb R^n)\}}^{\,L^2(\rho_t;\mathbb R^n)}. \end{align*} By the definition of $K_t$, one has $K_t=G_t^\perp$ in the Hilbert space $L^2(\rho_t;\mathbb R^n)$: the condition defining $K_t$ says precisely that $w$ is orthogonal to every generator $\nabla\psi$, hence to the closed span $G_t$. Since $G_t$ is closed, the Hilbert-space identity $(G_t^\perp)^\perp=G_t$ gives $K_t^\perp=G_t$. Since the previous step proves that the smooth minimizer $v$ is orthogonal to every element of $K_t$, the variational orthogonality places $v$ in $K_t^\perp$. Therefore \begin{align*} v\in G_t. \end{align*} Thus there exists a sequence \begin{align*} (\psi_k)_{k=1}^{\infty}\subset C_c^\infty(\mathbb R^n) \end{align*} such that \begin{align*} \nabla\psi_k\to v \end{align*} in $L^2(\rho_t;\mathbb R^n)$. Because $r_t\in C^\infty(\mathbb R^n;(0,\infty))$, for every compact set $K\subset\mathbb R^n$ there is a constant \begin{align*} m_K:=\inf_{x\in K}r_t(x)>0. \end{align*} Therefore convergence in $L^2(\rho_t;\mathbb R^n)$ implies convergence in $L^2(K,\mathcal L^n;\mathbb R^n)$: \begin{align*} \int_K|\nabla\psi_k(x)-v(x)|^2\,d\mathcal L^n(x)\le \frac{1}{m_K}\int_K|\nabla\psi_k(x)-v(x)|^2r_t(x)\,d\mathcal L^n(x). \end{align*} Hence $\nabla\psi_k\to v$ in $L^2_{\mathrm{loc}}(\mathbb R^n;\mathbb R^n)$. For indices $i,j\in\{1,\dots,n\}$ and every test function $\eta\in C_c^\infty(\mathbb R^n)$, convergence in the local $L^2$ pairing, justified by the Cauchy-Schwarz inequality on $\operatorname{supp}\eta$, gives \begin{align*} \int_{\mathbb R^n}v_i(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x)=\lim_{k\to\infty}\int_{\mathbb R^n}\partial_{x_i}\psi_k(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x). \end{align*} Since mixed partial derivatives of $\psi_k$ commute, \begin{align*} \int_{\mathbb R^n}\partial_{x_i}\psi_k(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}\partial_{x_j}\psi_k(x)\,\partial_{x_i}\eta(x)\,d\mathcal L^n(x). \end{align*} Passing to the limit gives \begin{align*} \int_{\mathbb R^n}v_i(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}v_j(x)\,\partial_{x_i}\eta(x)\,d\mathcal L^n(x). \end{align*} Thus $\partial_{x_j}v_i=\partial_{x_i}v_j$ in the sense of distributions. Since $v$ is smooth, this equality holds pointwise. Equivalently, the smooth one-form $\sum_{i=1}^n v_i\,dx_i$ is closed on the contractible domain $\mathbb R^n$. The smooth Poincare lemma now gives a smooth function \begin{align*} \phi:\mathbb R^n\to\mathbb R \end{align*} such that \begin{align*} v=\nabla\phi. \end{align*}[/step]

[guided]The variational argument only shows that $v$ is orthogonal to weighted divergence-free perturbations. To turn that into a potential, we use the weighted Helmholtz decomposition in the Hilbert space $L^2(\rho_t;\mathbb R^n)$. Define \begin{align*} G_t := \overline{\{\nabla\psi:\psi\in C_c^\infty(\mathbb R^n)\}}^{\,L^2(\rho_t;\mathbb R^n)}. \end{align*} Define the closed weighted divergence-free subspace \begin{align*} K_t := \left\{ w\in L^2(\rho_t;\mathbb R^n): \int_{\mathbb R^n}\nabla\psi(x)\cdot w(x)\,d\rho_t(x)=0 \text{ for every }\psi\in C_c^\infty(\mathbb R^n) \right\}. \end{align*} The identity $K_t^\perp=G_t$ follows directly from the definitions. Indeed, $K_t$ consists exactly of those $w\in L^2(\rho_t;\mathbb R^n)$ that are orthogonal to every compactly supported gradient $\nabla\psi$, so $K_t=G_t^\perp$ because $G_t$ is the closed span of those gradients. Since $G_t$ is a closed subspace of the Hilbert space $L^2(\rho_t;\mathbb R^n)$, the identity $(G_t^\perp)^\perp=G_t$ gives $K_t^\perp=G_t$. The previous step proves \begin{align*} \int_{\mathbb R^n}v(x)\cdot w(x)\,d\rho_t(x)=0 \end{align*} for every $w\in K_t$, so we obtain \begin{align*} v\in G_t. \end{align*} Being in $G_t$ means that there is a sequence \begin{align*} (\psi_k)_{k=1}^{\infty}\subset C_c^\infty(\mathbb R^n) \end{align*} with \begin{align*} \nabla\psi_k\to v \end{align*} in $L^2(\rho_t;\mathbb R^n)$. This is not yet the same as saying $v$ is literally a gradient. The point of the smoothness assumption is that we can upgrade the closed-gradient information into a curl-free condition. Fix a compact set $K\subset\mathbb R^n$. Since $r_t$ is smooth and strictly positive, it has a positive minimum on $K$: \begin{align*} m_K:=\inf_{x\in K}r_t(x)>0. \end{align*} Therefore \begin{align*} \int_K|\nabla\psi_k(x)-v(x)|^2\,d\mathcal L^n(x) \le \frac{1}{m_K} \int_K|\nabla\psi_k(x)-v(x)|^2r_t(x)\,d\mathcal L^n(x), \end{align*} so $\nabla\psi_k\to v$ in $L^2_{\mathrm{loc}}(\mathbb R^n;\mathbb R^n)$. Now gradients have zero curl, and this property passes to local $L^2$ limits in the distributional sense. Let $i,j\in\{1,\dots,n\}$ and let $\eta\in C_c^\infty(\mathbb R^n)$. Since $\nabla\psi_k\to v$ in $L^2$ on the compact support of $\eta$, the Cauchy-Schwarz inequality for the $L^2(\operatorname{supp}\eta,\mathcal L^n)$ pairing gives \begin{align*} \int_{\mathbb R^n}v_i(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x) = \lim_{k\to\infty} \int_{\mathbb R^n}\partial_{x_i}\psi_k(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x). \end{align*} For each $k$, the mixed partial derivatives of the smooth function $\psi_k$ commute, and integration by parts gives \begin{align*} \int_{\mathbb R^n}\partial_{x_i}\psi_k(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}\partial_{x_j}\psi_k(x)\,\partial_{x_i}\eta(x)\,d\mathcal L^n(x). \end{align*} Passing to the limit again yields \begin{align*} \int_{\mathbb R^n}v_i(x)\,\partial_{x_j}\eta(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}v_j(x)\,\partial_{x_i}\eta(x)\,d\mathcal L^n(x). \end{align*} This says $\partial_{x_j}v_i=\partial_{x_i}v_j$ as distributions. Because $v$ is smooth, the same equality holds pointwise. Hence the smooth one-form $\sum_{i=1}^n v_i\,dx_i$ is closed. Since $\mathbb R^n$ is contractible, the smooth Poincare lemma gives a smooth potential \begin{align*} \phi:\mathbb R^n\to\mathbb R \end{align*} such that $v=\nabla\phi$.[/guided]

[step:Substitute the potential into the metric tensor identity]For the time $t$ fixed in the previous steps, the metric-speed identity already proved gives \begin{align*} |\dot\rho_t|_{W_2}^2 = \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*} Since $d\rho_t(x)=r_t(x)\,d\mathcal L^n(x)$ and $v=\nabla\phi$, this becomes \begin{align*} |\dot\rho_t|_{W_2}^2 = \int_{\mathbb R^n}|\nabla\phi(x)|^2r_t(x)\,d\mathcal L^n(x). \end{align*} This is the asserted Benamou-Brenier metric tensor formula.[/step]

[guided]At the chosen time $t$, the metric tensor identity has already shown that the squared metric speed equals the minimum action, and $v$ is assumed to attain that minimum. Therefore \begin{align*} |\dot\rho_t|_{W_2}^2 = \int_{\mathbb R^n}|v(x)|^2\,d\rho_t(x). \end{align*} The previous step produced a smooth potential $\phi:\mathbb R^n\to\mathbb R$ with $v=\nabla\phi$. Since the measure has density $r_t$ with respect to $\mathcal L^n$, we substitute $d\rho_t(x)=r_t(x)\,d\mathcal L^n(x)$ and obtain \begin{align*} |\dot\rho_t|_{W_2}^2 = \int_{\mathbb R^n}|\nabla\phi(x)|^2r_t(x)\,d\mathcal L^n(x). \end{align*} This is the stated formula for the smooth minimizing velocity potential.[/guided]