Analytical mechanics

Hamilton's principle of stationary action

The physical trajectory makes the action S = ∫L dt stationary under small variations that keep both endpoints fixed.

Newtonian mechanics asks which force produces an acceleration. Analytical mechanics asks a different question: among possible paths joining the same endpoint states, which makes a time-integrated quantity stationary? That quantity is the action.

S[q]=\int_{t_1}^{t_2} L(q,\dot q,t)\,dt, \qquad \delta S=0

Definition: Action and stationarity

For a suitable conservative system, the Lagrangian is $L=T-V$ (kinetic minus potential energy). The action $S$ has units J·s. The condition $\delta S=0$ means its first-order variation vanishes; it does not assert that $S$ is always a minimum—the path may be a minimum, maximum, or saddle point.

Drag the path's midpoint to vary the trajectory. Compare its action with the stationary path between the same endpoints.

From principle to equation of motion

Vary the path as $q(t)\to q(t)+\delta q(t)$ with $\delta q(t_1)=\delta q(t_2)=0$ . Integrating the term containing $\delta\dot q$ by parts moves the derivative off the variation.

\delta S=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial\dot q}\right)\delta q\,dt=0

Because $\delta q(t)$ is arbitrary in the interior, its coefficient must vanish, giving the Euler–Lagrange equation. For $L=\tfrac12m\dot x^2-V(x)$ , it becomes $m\ddot x=-dV/dx$ , Newton's second law.

Example: Free particle

A particle of mass $m$ travels from $x=0$ to $x=2$ m in $2$ s with no force acting. Find its classical path.

Solution

With $V=0$ , Euler–Lagrange gives $m\ddot x=0$ . The velocity is constant, so $x(t)=t$ m when $t$ is measured in seconds.

Quick check

The stationary condition becomes concrete by varying a path as $q(t)\mapsto q(t)+\epsilon\eta(t)$ with $\eta(t_1)=\eta(t_2)=0$ . Expanding to first order and integrating by parts gives $\delta S=\int_{t_1}^{t_2}[\partial L/\partial q-d(\partial L/\partial\dot q)/dt]\eta(t)\,dt$ . Since $\eta$ is arbitrary in the interior, the bracket must vanish: stationary action yields the Euler–Lagrange equation. Fixed endpoints matter; if an endpoint is free, an additional natural boundary condition must be imposed.

What variational condition characterizes the classical path?

In $S=\int Ldt$ , what are the units of action?

References

Herbert Goldstein, Charles Poole, John Safko (2002). Classical Mechanics