THE LAGRANGIAN

In 1788, at the age of fifty-two, Joseph-Louis Lagrange published Mécanique analytique and boasted in the preface that it contained not a single diagram. Newton's Principia had been a forest of geometry — circles, tangents, secants, shaded triangles — because Newton had thought in shapes. Lagrange wanted to think in symbols. He claimed he could derive all of mechanics from a single scalar function and a handful of partial derivatives, and that if you followed the algebra you would never need to draw a free-body diagram again.

He was right. The function is now called the Lagrangian, and once you know it, the equations of motion of a system fall out mechanically, like turning a crank. It was the first time physics felt like an algorithm.

The rest of this page is what that sentence means — and why, two hundred and fifty years later, every graduate physics student still learns mechanics twice: once Newton's way, and once Lagrange's.

The whole construction starts with one quantity. Take the kinetic energy T, take the potential energy V, and subtract:

That minus sign looks wrong. The conserved quantity we learned to care about in FIG.12 was T + V, the total energy. Why would T − V matter? What could "energy minus energy" even mean?

Classical mechanics offers no intuitive answer. It is a brute algebraic fact that if you minimise the time-integral of T − V — the action of FIG.28 — you recover Newton's laws. In that sense L is a bookkeeping trick, a generating function, chosen because it works and not because it has a physical picture behind it.

The picture shows up later, in two places. In quantum mechanics the path-integral tells you that a particle explores every path between two points, weighted by e^(iS/ℏ) with S = ∫L dt; the classical trajectory is the one where nearby paths interfere constructively, and the Lagrangian is what you were always computing phases with. In general relativity, the matter Lagrangian is what sources the gravitational field — the stress-energy tensor is literally a derivative of L with respect to the metric. Minus signs and all, T − V is the handle the rest of twentieth-century physics grabs the world by.

For now, accept it as a definition. Kinetic minus potential. Integrate it along any candidate trajectory between two fixed endpoints. The trajectory nature actually chooses is the one where the integral is stationary — we proved this in FIG.28, the principle of least action. What we haven't yet done is turn "stationary" into a working equation.

Lagrange and Euler, working in the 1750s and 60s, figured out exactly what equation a stationary action implies. The derivation fits on a napkin. Imagine the true trajectory q(t), and perturb it by a small function η(t) that vanishes at the endpoints. The action S = ∫L(q, q̇, t) dt changes by

Integrate the second term by parts; the boundary terms die because η is zero at the endpoints. What survives is

For δS to vanish for every wobble η, the bracketed quantity must itself be zero everywhere. That condition is the Euler-Lagrange equation:

Stop and notice what this is. It is a second-order differential equation in q(t). The Lagrangian L is a function of position and velocity; take partial derivatives of it, plug them into the formula, and out falls the equation of motion. No forces, no free-body diagrams, no tangent-vector bookkeeping. Just partials.

Whenever L depends on several coordinates q₁, q₂, …, qₙ, you get one Euler-Lagrange equation per coordinate, independently. The machinery scales.

Here is the part that eats Newton's lunch.

Newton's F = ma is written in Cartesian coordinates — x, y, z — because forces naturally live in 3-space. If your system is a bead on a wire, or a pendulum on a rod, or a planet on an orbit, you have to start by parameterising the geometry in Cartesians and then bolt on constraint forces (tensions, normal forces, rod tensions) to keep the motion on the allowed surface. Every first-year student has done the free-body-diagram dance and cursed the vanishing tension terms.

The Lagrangian doesn't care. q in the Euler-Lagrange equation can be any coordinate you like, as long as it actually describes the configuration of the system. For a pendulum, q = θ — the angle — is ideal; the length is fixed, so a single angle pins down the bob. For a bead on a wire, q is the arc-length along the wire. For a double pendulum, q₁ and q₂ are the two angles. These are called generalised coordinates, and the Euler-Lagrange equation works in any of them, unchanged.

The rest of this page is a tour of that fact. One system, one Lagrangian, one crank; Newton's version for comparison, kept deliberately short.

We already solved the simple pendulum in FIG.01 by balancing torques and a restoring force. Lagrange's way is faster.

Pick θ, the angle from vertical, as the generalised coordinate. The bob's velocity is Lθ̇, so the kinetic energy is

The height above the lowest point is L(1 − cos θ), so the potential energy is

The Lagrangian is the difference:

Now turn the crank. ∂L/∂θ̇ = m L² θ̇; its time derivative is m L² θ̈. ∂L/∂θ = −m g L sin θ. Plug into the Euler-Lagrange equation:

The mass dropped out. The rod length and gravity survive. That's the full nonlinear pendulum equation we met in FIG.04 — derived in three lines of partials instead of a page of geometry. Notice in particular that we never wrote down the rod's tension, because θ is the only coordinate we have and constraint forces that act perpendicular to θ simply cannot appear in L. They were eliminated the moment we picked the coordinate.

Attach a second pendulum to the end of the first. Let θ₁ be the angle of the upper rod from vertical, θ₂ the angle of the lower rod, and assume for simplicity that each rod has length L and each bob has mass m.

Try to do this with Newton. You need the position and velocity of each bob in Cartesians, expressed through θ₁ and θ₂ and their derivatives. You need the tensions in both rods. You need to resolve forces along and perpendicular to each rod. You will fill a page with sines and cosines and tangents. The algebra is doable but miserable, and students routinely lose a term in it.

Now try it Lagrange's way.

The upper bob moves on a circle of radius L about the pivot, so T₁ = ½ m L² θ̇₁². The lower bob's velocity is the vector sum of the upper bob's velocity and its own rotation about the upper bob; square it, and the cross term ends up as 2 L² θ̇₁ θ̇₂ cos(θ₁ − θ₂). Heights add the same way. Put it all together:

That's it. Two lines of setup. The Euler-Lagrange machinery then produces two coupled second-order equations for θ̇₁ and θ̇₂ — we'll spare you the explicit form; they live in the lagrangian-mechanics.ts module and they are, bluntly, ugly. But you never had to draw a free-body diagram, you never wrote down a tension, and you never resolved forces along an axis. The coordinates are θ₁ and θ₂ from start to finish.

The trick with the pendulum — pick the angle as your coordinate and the rod tension drops out — is not an accident. It is the general feature of the Lagrangian machinery.

Suppose you have a bead sliding along a wire bent into some fixed shape. Newton's law for the bead asks you to find the constraint force: whatever force the wire exerts on the bead to keep it on the track. That force points perpendicular to the wire, changes direction as the bead moves, and depends on the bead's speed — and for most wire shapes you cannot write it down in closed form. The work of solving the motion is mostly the work of solving for the constraint force first.

Now do it Lagrange's way. Let s be the arc-length along the wire — a single coordinate that fully specifies where the bead is. The bead's speed is ṡ, so T = ½ m ṡ², clean and coordinate-free. The height y(s) is whatever the wire shape prescribes, so V = m g y(s). Plug L = ½ m ṡ² − m g y(s) into Euler-Lagrange, and you get a single equation for s̈(t). The constraint force never appears. It's orthogonal to the coordinate you chose, so its contribution to ∂L/∂s and ∂L/∂ṡ is identically zero. You get the motion for free.

One wire in particular is worth pausing on. In 1659 Huygens, working geometrically, proved that a bead sliding on a cycloid reaches the bottom in a time that is independent of where you drop it from. The cycloid is the tautochrone — the "same-time curve" — and it was what Huygens used to fix the amplitude-dependence of pendulum clocks. His proof took pages of Euclidean geometry. The Lagrangian version is exercise-sized: along the cycloid, the EoM reduces to simple harmonic motion in a well-chosen arc-length coordinate, and the period π√(R/g) drops out.

Every constraint you can pick a coordinate along simply vanishes from L. That is the whole reason the Lagrangian method dominates engineering dynamics today: robot arms, spacecraft attitude, suspension bridges, molecular modes. The coordinate absorbs the constraint, and you are left with one ordinary equation per degree of freedom.

Lagrange's reformulation was the first. It turned mechanics from a geometry problem into an algebra problem. Forty-five years later, in 1833, William Rowan Hamilton sitting in Dublin did it again, and got a second reformulation that looks almost the same — the Euler-Lagrange equation is second-order in q, and Hamilton's version is two first-order equations in q and a new variable p, the momentum conjugate to q.

You will wonder why anyone needs a second version of the same content. The answer is that it exposes a structure the Lagrangian hides: the natural space for mechanics isn't the configuration space of positions — it's the phase space of positions and momenta, and the laws of motion become a geometric flow on that space. Quantum mechanics uses Hamilton's version literally; every modern numerical planet-integrator is Hamiltonian by construction; and chaos, when it appears, is a property of the phase-space flow.

Next: the Hamiltonian.