Lagrangian and Hamiltonian mechanics

Perturb the path by $q(t)+\epsilon\,\eta(t)$ with $\eta(t_1)=\eta(t_2)=0$ and require $\frac{d}{d\epsilon}S[q+\epsilon\eta]\big|_{\epsilon=0}=0$ for every such $\eta$. Expanding to first order in $\epsilon$ and integrating the $\dot\eta$ term by parts (the boundary term vanishes because $\eta(t_1)=\eta(t_2)=0$) leaves $\int \left(\frac{\partial L}{\partial q_k} - \frac{d}{dt}\frac{\partial L}{\partial \dot q_k}\right)\eta_k\,dt = 0$ for every $\eta$, which forces the bracket to vanish identically.

Step 1 (set up the variation). Let $q(t)$ be the true path and perturb it by $q(t)+\epsilon\,\eta(t)$ for a small parameter $\epsilon$, where $\eta(t)$ is an arbitrary smooth function satisfying $\eta(t_1)=\eta(t_2)=0$ so the perturbed path shares the same endpoints as $q(t)$. Hamilton's principle requires $S[q+\epsilon\eta]$ to be stationary at $\epsilon=0$ for every admissible $\eta$, i.e. $\frac{d}{d\epsilon}S[q+\epsilon\eta]\big|_{\epsilon=0}=0$.

Step 2 (differentiate under the integral). Differentiating $S[q+\epsilon\eta]=\int_{t_1}^{t_2}L(q+\epsilon\eta,\dot q+\epsilon\dot\eta,t)\,dt$ with respect to $\epsilon$ and setting $\epsilon=0$ gives, by the chain rule applied term by term, $$\frac{d}{d\epsilon}S[q+\epsilon\eta]\Big|_{\epsilon=0}=\int_{t_1}^{t_2}\sum_k\left(\frac{\partial L}{\partial q_k}\eta_k+\frac{\partial L}{\partial \dot q_k}\dot\eta_k\right)dt$$

**Step 3 (integrate the $\dot\eta_k$ term by parts).** For each $k$, integration by parts turns $\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q_k}\dot\eta_k\,dt$ into a boundary term minus a bulk term: $$\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q_k}\dot\eta_k\,dt = \left[\frac{\partial L}{\partial \dot q_k}\eta_k\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_k}\right)\eta_k\,dt$$

Step 4 (the boundary term vanishes). Because $\left[\frac{\partial L}{\partial \dot q_k}\eta_k\right]_{t_1}^{t_2}=0$ by the fixed-endpoint condition $\eta(t_1)=\eta(t_2)=0$, only the bulk integral survives, and substituting back into Step 2 gives $$\int_{t_1}^{t_2}\sum_k\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}\right)\eta_k(t)\,dt=0$$

Step 5 (fundamental lemma of the calculus of variations). This integral vanishes for every smooth $\eta_k$ vanishing at the endpoints. If the bracket $\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}$ were nonzero and continuous at some interior point $t_0$, it would keep a fixed sign on some small interval around $t_0$; choosing $\eta_k$ to be a smooth bump function supported on that interval (positive where the bracket is positive, and zero elsewhere) would make the integral strictly nonzero, a contradiction. Hence the bracket must vanish identically, giving $\frac{d}{dt}\frac{\partial L}{\partial \dot q_k} - \frac{\partial L}{\partial q_k} = 0$ for every $k$ — the Euler–Lagrange equation.

Differentiating $L(q(s),\dot q(s),t)=L(q,\dot q,t)$ (invariance) with respect to $s$ at $s=0$ and using the Euler–Lagrange equations to rewrite $\partial L/\partial q_k$ as $\frac{d}{dt}\left(\partial L/\partial \dot q_k\right)$ turns the result into $\frac{d}{dt}\sum_k \frac{\partial L}{\partial \dot q_k}\frac{\partial q_k}{\partial s}=0$: exactly the statement that $I$ has zero time derivative, i.e. is conserved.

Step 1 (differentiate the invariance identity). Invariance of $L$ under the transformation means $L(q(s),\dot q(s),t)=L(q,\dot q,t)$ holds identically in $s$. Differentiating both sides with respect to $s$ and evaluating at $s=0$ (using $q_k(0)=q_k$) gives, by the chain rule, $$\sum_k\left(\frac{\partial L}{\partial q_k}\frac{\partial q_k}{\partial s}+\frac{\partial L}{\partial \dot q_k}\frac{\partial \dot q_k}{\partial s}\right)\bigg|_{s=0}=0$$

Step 2 (use the Euler–Lagrange equation on the actual trajectory). Because $q(t)$ solves the Euler–Lagrange equations, $\frac{\partial L}{\partial q_k}=\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}$. Substituting this into the first term of Step 1 replaces $\partial L/\partial q_k$ by a total time derivative.

**Step 3 (commute the $s$- and $t$-derivatives).** Since $s$ and $t$ are independent variables, mixed partial derivatives of $q_k(s,t)$ commute: $\frac{\partial \dot q_k}{\partial s}=\frac{d}{dt}\frac{\partial q_k}{\partial s}$. This lets the second term of Step 1 be rewritten with a total time derivative as well.

Step 4 (recognize a total derivative). With both terms now built from $\frac{d}{dt}(\cdot)$, the product rule runs in reverse: $\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}\cdot\frac{\partial q_k}{\partial s}+\frac{\partial L}{\partial \dot q_k}\cdot\frac{d}{dt}\frac{\partial q_k}{\partial s}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_k}\frac{\partial q_k}{\partial s}\right)$. Substituting into the identity of Step 1 collapses it to a single total time derivative equal to zero: $$\frac{dI}{dt}=\frac{d}{dt}\sum_k\frac{\partial L}{\partial \dot q_k}\frac{\partial q_k}{\partial s}\bigg|_{s=0}=0$$

Step 5 (conclude conservation). A quantity whose time derivative vanishes identically along every solution of the equations of motion is constant along that solution. Hence $I=\sum_k \frac{\partial L}{\partial \dot q_k}\frac{\partial q_k}{\partial s}\big|_{s=0}$ is conserved, which is Noether's theorem.

The phase-space velocity field $(\dot q,\dot p) = (\partial H/\partial p, -\partial H/\partial q)$ is divergence-free: $\sum_k\left(\frac{\partial \dot q_k}{\partial q_k}+\frac{\partial \dot p_k}{\partial p_k}\right) = \sum_k\left(\frac{\partial^2 H}{\partial q_k \partial p_k} - \frac{\partial^2 H}{\partial p_k \partial q_k}\right)=0$, because mixed partial derivatives commute. A divergence-free flow preserves volume, by the same reasoning as the divergence theorem for an incompressible fluid.

Step 1 (the phase-space flow). Hamilton's equations $(\dot q,\dot p)=\left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right)$ define a velocity field on phase space; every point $(q,p)$ moves along this field, so a region $\Omega(t_1)$ is carried by the flow to a region $\Omega(t_2)$. The Reynolds transport theorem gives the rate of change of the enclosed volume $V(t)=\int_{\Omega(t)}dq\,dp$ as the flux of the divergence: $$\frac{d}{dt}V(t)=\int_{\Omega(t)}\nabla\cdot(\dot q,\dot p)\,dq\,dp$$

Step 2 (compute the divergence). Directly from Hamilton's equations, $$\sum_k\left(\frac{\partial}{\partial q_k}\frac{\partial H}{\partial p_k}+\frac{\partial}{\partial p_k}\left(-\frac{\partial H}{\partial q_k}\right)\right)=\sum_k\left(\frac{\partial^2 H}{\partial q_k\partial p_k}-\frac{\partial^2 H}{\partial p_k\partial q_k}\right)=0$$ because mixed second partial derivatives of the smooth function $H$ commute (equality of mixed partials, Schwarz's theorem): $\partial^2H/\partial q_k\partial p_k=\partial^2H/\partial p_k\partial q_k$, so the two terms cancel exactly.

Step 3 (zero divergence forces zero rate of change). Substituting the Step 2 result into the transport formula of Step 1 gives $$\frac{d}{dt}V(t)=0$$ at every instant $t$, for every choice of initial region $\Omega(t_1)$.

Step 4 (integrate to get equal volumes). A function of $t$ whose derivative is identically zero is constant, so $V(t)=V(t_1)$ for all $t$: the phase-space volume of the evolving region never changes, exactly as claimed.