The Euler–Lagrange Equations

The principle of least action tells us what nature optimises — the action S = \int L\,dt — but not yet which curve does it. Turning "make this functional stationary" into an actual equation for the motion is a solved problem: it is exactly what the Euler–Lagrange equation does. On this page we borrow that result wholesale and point it at mechanics, and something wonderful falls out: F = ma reappears, all by itself, from a pure statement about a stationary integral.

The equation, translated into mechanics

In the calculus of variations you met the Euler–Lagrange equation for the curve y(x) that makes \int L(x, y, y')\,dx stationary:

\frac{\partial L}{\partial y} - \frac{d}{dx}\!\left(\frac{\partial L}{\partial y'}\right) = 0.

Mechanics is the very same problem wearing physics clothes. Just rename the variables: the independent variable is no longer position x but time t; the unknown curve is no longer y but a generalised coordinate q(t); and the "slope" y' becomes the velocity \dot q = dq/dt. With those renamings the equation reads:

That is the whole engine of Lagrangian mechanics. You never draw a force diagram. You write down two energies, subtract them, and turn the crank.

The crank recovers F = ma

Take the plainest possible system: one particle of mass m moving along a line x in a potential V(x). Its kinetic energy is T = \tfrac12 m\dot x^2, so the Lagrangian is

L = T - V = \tfrac12 m \dot x^2 - V(x).

Now compute the two pieces the Euler–Lagrange equation asks for.

The momentum term. Differentiate L with respect to the velocity \dot x (treating x as a bystander):

\frac{\partial L}{\partial \dot x} = m\dot x.

That is just the ordinary momentum p = m\dot x! This is no accident — in general \partial L/\partial \dot q is called the generalised (or conjugate) momentum for the coordinate q. Its time derivative is

\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot x}\right) = \frac{d}{dt}(m\dot x) = m\ddot x.

The force term. Differentiate L with respect to the position x (now treating \dot x as the bystander):

\frac{\partial L}{\partial x} = -\frac{dV}{dx}.

And -dV/dx is precisely the force F — the force is minus the slope of the potential, exactly as in energy methods. Slot both pieces into the Euler–Lagrange equation \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot x}\right) - \frac{\partial L}{\partial x} = 0:

m\ddot x - \left(-\frac{dV}{dx}\right) = 0 \quad\Longrightarrow\quad m\ddot x = -\frac{dV}{dx} = F.

There it is: m\ddot x = F, Newton's second law, delivered by a machine that never once mentioned a force. Lagrange contains Newton.

Force is minus the slope of the potential

The force term F = -dV/dx is worth seeing. The chart shows a spring-like potential V(x) = \tfrac12 k x^2 (a bowl) together with the force it produces, F(x) = -kx (a straight line through the origin). Slide the stiffness k. Wherever the bowl slopes up to the right, the force points left (negative), always pushing the particle back down toward the bottom of the bowl — the restoring force of the Euler–Lagrange equation, read straight off the geometry.

This is the sense in which the Euler–Lagrange equation is the same physics as F = ma: the \partial L/\partial q term is the force, the \frac{d}{dt}(\partial L/\partial \dot q) term is the rate of change of momentum, and setting them equal is Newton's law dressed in energy.

Why it works in any coordinates — the real superpower

Here is the property that makes the whole framework indispensable. Newton's F = ma is written for Cartesian components; move to polar coordinates and you must hand-derive awkward extra terms (the centrifugal -r\dot\theta^2 and Coriolis 2\dot r\dot\theta pieces) or you get the wrong answer. The Euler–Lagrange equation has no preferred coordinates at all. You may use any generalised coordinates you like — angles, arc-lengths, rotating frames — and the equation \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right) = \frac{\partial L}{\partial q} keeps exactly the same form. All those correction terms emerge automatically when you differentiate a correctly written kinetic energy.

That is why constrained and curvilinear problems that are a nightmare for Newton become routine for Lagrange: choose coordinates that fit the geometry, write T and V in them, and the same equation does the rest.

The single commonest slip is muddling the partial derivatives with the total time derivative. When you take \partial L/\partial \dot q, you treat q, \dot q (and t) as independent slots and differentiate with respect to the velocity slot only — so for L = \tfrac12 m\dot x^2 - V(x) you get m\dot x, and the V(x) term contributes nothing because it has no \dot x in it.

Only afterwards do you apply d/dt, and now everything is allowed to depend on time through the motion: m\dot x becomes m\ddot x. Do the partial first, the total-time derivative second — never mix the two, and never "cancel" a \dot x against a d/dt. Getting the order right is the whole discipline.

A conserved quantity, for free

The Euler–Lagrange form hands you conservation laws almost for nothing. Suppose the Lagrangian does not contain some coordinate q explicitly (only its velocity \dot q appears). Such a q is called a cyclic (or ignorable) coordinate. Then \partial L/\partial q = 0, so the equation collapses to

\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right) = 0 \quad\Longrightarrow\quad \frac{\partial L}{\partial \dot q} = \text{constant}.

The generalised momentum conjugate to a cyclic coordinate is conserved. If L does not depend on x, linear momentum is conserved; if it does not depend on an angle \theta, angular momentum is conserved. Conservation laws, which Newton treats as separate theorems, here drop straight out of a symmetry of the Lagrangian — a first taste of Noether's theorem.