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Action and the Euler-Lagrange equations

For a trajectory $q(t)$ between times $t_1$ and $t_2$, the action is

$$S[q]=\int_{t_1}^{t_2}L(q,\dot q,t),dt.$$

A physical classical trajectory makes the action stationary with respect to sufficiently small variations that keep the endpoints fixed.

Stationary does not necessarily mean minimum

The condition is

$$\delta S=0,$$

not simply 'the path with the smallest action'. The action may be a minimum, maximum or saddle point depending on the problem.

Euler-Lagrange equation

Applying the stationary-action condition gives

$$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0$$

for each generalized coordinate $q_i$.

For

$$L=\frac12m\dot x^2-V(x),$$

this becomes

$$m\ddot x=-\frac{dV}{dx},$$

which is Newton's second law for a conservative force.

The action principle turns an entire trajectory into the object being varied. That global viewpoint is the classical structure that later appears inside the phase of Feynman's path integral.