I'm trying to derive Hamilton's Principle from Lagrange's Equations, as I've heard they're logically equivalent statements, and am stuck on a final step. For simplicity, assume we're dealing with a system with only one generalized coordinate $q$, so that there is only one equation of motion of the system:
$$\frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) – \frac{\partial L_q}{\partial q} \circ \Lambda_t = 0 \quad,$$
where $L_q:\mathbb{R}^3 \to \mathbb{R}$ is the Lagrangian of the system expressed in terms of $q$, $\dot{q}$ and time $t$, $\Lambda_t(t) = (q_t(t), \dot{q_t}(t),t)$, where $q_t:\mathbb{R} \to \mathbb{R}$ is an actual trajectory, so a solution to the differential equation.
Working back from the steps taken to arrive at this equation from the variational principle, I did the following: let $\delta q_t:\mathbb{R} \to \mathbb{R}$ be a deviation from the actual trajectory, so a smooth function that satisfies $\delta q_t (t_1) = 0, \delta q_t(t_2) = 0$, for some given time instants $t_1$ and $t_2$. Multiplying the differential equation through by $\delta q_t$ and integrating from $t_1$ to $t_2$, we get
$$\int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t – \frac{\partial L_q}{\partial q} \circ \Lambda_t \phantom{,} \delta q_t \right] = 0 \quad.$$
Since by hypothesis $\delta q_t (t_1) = 0, \delta q_t(t_2) = 0$, the following holds
$$\int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \delta q_t \right) = 0$$
$$\Rightarrow \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t = \int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \delta q_t \right) – \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t \right] = \int_{t_1}^{t_2} \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \dot{\delta q_t} \quad.$$
So the full integral becomes
$$\int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t + \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \dot{\delta q_t} \right] = 0 \quad,$$
which we can recognize as being the statement that the variation of some functional of the form
$$A(q_t) = \int_{t_1}^{t_2} L_q(q_t(t), \dot{q_t}(t), t) \phantom{,} dt$$
must be equal to zero for all admissible $\delta q_t$, so
$$\delta A(q_t, \delta q_t) = 0 \quad.$$
So starting from Lagrange's Equation for the system, we logically arrive at the implication that the variation of the functional $A$ (the action functional) must be zero at the system's true trajectory, for such $\delta q_t$. But this doesn't on its own imply that $q_t$ is in fact an extremum of $A$, which is what Hamilton's Principle states, and from which $\delta A(q_t, \delta q_t) = 0$ follows. Where's the missing piece that lets us conclude that $q_t$ is in fact an extremum of $A$, and not only a point at which its variation vanishes? Is it a mathematical fact about the form of the functional itself that allows us to conclude that?
Best Answer
Hamilton’s principle is often sloppily stated. It is not a “least-action” principle, nor is it a “principle of extremal action”. It is a principle of “stationary” action.
For example, on the round sphere, take two non-antipodal points, and consider the longer arc of the great circle which joins these two points (i.e the longer geodesic). This path is a saddle point for the length functional on the sphere (the length functional is the action in this context).