The Euler–Lagrange equation
~22 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the action functional S[q] = ∫ L(q, q̇, t) dt (L05) and with taking partial derivatives of a multivariable function.
Why this matters
One equation replaces "write down all the forces."
Instead of tracking every push and pull on a system, Lagrangian mechanics asks a single question: of all the paths the system could take between two moments, which one does it actually follow? The answer is the path that makes the action stationary — and the Euler–Lagrange equation is the exact condition that path must satisfy. Learn it once and it applies to pendulums, planets, fields, and eventually the Standard Model. It is the grammar that the rest of PU-201 is written in.
From "stationary" to an equation
Saying the action is stationary means that if you nudge the path a little — replace q(t) with q(t) + ε η(t) for a small bump η that vanishes at both endpoints — the action does not change to first order in ε. That single demand, δS = 0 for every allowed bump, is enough to pin down a differential equation the true path obeys everywhere along its length.
The derivation
The Euler–Lagrange equation
Requiring the action to be stationary under all endpoint-fixed variations yields a local condition on the Lagrangian:
Reading the result
The equation is a balance. The term ∂L/∂q acts like a generalized force — how the Lagrangian changes as you move the coordinate. The term d/dt(∂L/∂q̇) is the rate of change of the generalized momentum p = ∂L/∂q̇. Setting their difference to zero is Newton's second law in disguise: for L = ½mq̇² − V(q) it collapses straight to mq̈ = −V′(q). But it holds in any coordinates — angles, arc lengths, field amplitudes — with no need to resolve forces along axes.
Stationary does not mean minimum. The action is only guaranteed to be a critical point of S, not a minimizer — beyond the first conjugate point the true path can be a saddle. "Principle of least action" is a historical misnomer; "principle of stationary action" is the honest name. Whether a given trajectory is a genuine minimum is a second-order question decided by the Jacobi condition. C
Three quick tests
1 · Why must the variation η vanish at the endpoints? 2 · Show that for a Lagrangian with no explicit q, the momentum ∂L/∂q̇ is conserved. 3 · What are the units of each term in the equation?
Problem set for PU-201 →
Apply the Euler–Lagrange equation to the pendulum, the sliding-block-on-wedge, and a double pendulum. Full problem set lives on the unit page.