The Euler-Lagrange Equations

Every serious robot runs these equations in its inner loop. The manipulator equation of robotics, $M(q)\ddot q + C(q,\dot q)\dot q + g(q) = \tau$, is the Euler–Lagrange system (7) for the robot's Lagrangian — $M(q)$ the velocity Hessian, $C$ the chain-rule cross-terms, $g$ the potential gradient, $\tau$ the generalized forces from the motors. Modern dynamics engines (MuJoCo, Drake, Isaac) evaluate exactly this structure thousands of times per second for humanoids with 30-plus degrees of freedom, and the reinforcement-learning pipelines training 2026's robot policies differentiate through it, backpropagating rewards through the mass matrix. When a humanoid catches its balance, a pendulum-on-a-cart problem — literally the worked example above, with the sign of gravity flipped — is being solved in real time. The same structure has become a machine-learning prior. Lagrangian and Hamiltonian neural networks learn the scalar $L$ from trajectory data and then obtain forces by automatic differentiation of the learned function — building conservation laws and covariance into the model instead of hoping data teaches them. And at the opposite end of fundamentality, quantum field theory runs the identical crank with fields in place of coordinates: $\partial_\mu\big(\partial\mathcal L/\partial(\partial_\mu\phi)\big) = \partial\mathcal L/\partial\phi$ is equation (1) with the time derivative promoted to four spacetime derivatives. Maxwell's equations, the Klein–Gordon equation, and the Standard Model's field equations are all outputs of the same two-partial-derivatives-and-a-crank recipe you used on the Atwood machine.

Start with what "stationary action" can possibly say about a single instant. Hamilton's principle grades whole histories; a differential equation acts here and now. The bridge is the freedom to wiggle the path locally. Take the true path and deform it only inside a tiny time window — a small bump, zero everywhere else. Stationarity against that one bump is a statement about the path only inside the window. Slide the window anywhere: you get a condition at every instant separately. That per-instant condition — the price of flatness against localized bumps — is the Euler–Lagrange equation. The global principle was never mystical; it is a differential equation wearing an integral coat.

The second thing to see before any algebra is why these equations deserve their reputation: two independent foundations converge on them. From above, Hamilton's stationary action plus the calculus of variations. From below, Newton's laws filtered through d'Alembert's principle — constraint forces do no virtual work, so project the dynamics onto the directions the constraints allow. Same equations, letter for letter. If you distrust variational reasoning, the equations are still ordinary Newtonian mechanics; if you distrust virtual-work bookkeeping, they are still the flatness condition of a functional. The double derivation is the certificate that nothing physical has been smuggled in or lost.

Last, the practical superpower: covariance. Newton's law is a vector equation; bend the coordinates and you must transform every component, sprouting centrifugal and Coriolis corrections by hand. The action is a scalar built from a scalar $L$; its flat spot cannot depend on how paths are labelled. So the stationarity condition has the identical form in every chart — and all the coordinate-bending corrections are generated automatically by the derivatives of $L$. Choose the angle of a pendulum, the arc length along a wire, the separation of a two-body pair: constraints vanish into the choice of coordinates, ideal constraint forces (which do no work along allowed motions) never enter $L$, and the equations arrive already projected onto the directions in which the system can actually move.

Here is the machine, stated first, derived immediately after. For a system with $n$ degrees of freedom described by independent generalized coordinates $q_1,\dots,q_n$ and Lagrangian $L(q,\dot q,t)$ — for ordinary mechanics, $L=T-V$ built as in the Lagrangian-function article — the motion obeys

where: $\partial L/\partial\dot q_i$ and $\partial L/\partial q_i$ are partial derivatives taken treating $q$, $\dot q$, $t$ as independent slots of the function $L$; the outer $\mathrm{d}/\mathrm{d}t$ is a total derivative along the path, acting through $q(t)$, $\dot q(t)$, and any explicit $t$. That distinction — partials inside, total outside — is where most first attempts die, and it is worth a moment: the total derivative produces terms in $\ddot q$, in $\dot q$, and in $\partial^2L/\partial\dot q_i\partial t$, all by the chain rule. The equations acquire physical eyes through two definitions. Name the things being differentiated:

where: $p_i$ — the generalized (conjugate) momentum of coordinate $q_i$; $Q_i$ — the generalized force. Equation (1) then reads $\dot p_i = Q_i$: rate of change of momentum equals force, Newton's second law in coordinate-free clothing. When $q_i$ is a Cartesian position, $p_i = m\dot q_i$; when $q_i$ is an angle, $p_i$ is an angular momentum — the formalism reshapes momentum to fit the coordinate, a story told fully in the generalized-forces-and-momenta article. The variational origin needs the action. Each admissible history between pinned endpoints is graded by

where: $S$ — the action functional (units J·s), whose anatomy has its own article; here it is the object whose flatness we now cash out.

The same equations have a second, fully Newtonian pedigree. Starting from d'Alembert's principle — constraint forces do no virtual work — and projecting Newton's second law onto the generalized coordinates, one arrives (the full derivation lives in the d'Alembert article, whose property this route is) at

where: $T$ — the kinetic energy in generalized coordinates; $Q_i$ — the generalized force of the applied forces only, constraint forces having dropped out. For monogenic forces, $Q_i = -\partial V/\partial q_i$ with $V$ velocity-independent, and defining $L = T - V$ absorbs the right side: equation (5) becomes equation (1). Keep (5) in your kit even after you trust (1): it is the form that welcomes friction and other nonpotential forces on the right-hand side. Two structural facts complete the toolbox. First, the Lagrangian producing a given dynamics is not unique:

where: $F(q,t)$ — any smooth function. The addition shifts the action of every path by the same endpoint constant, so the flat spot never moves — the gauge freedom exploited constantly in electromagnetism. Second, the equations are second-order ODEs in disguise, and one should know when the disguise can be removed. Expanding the total derivative in (1) by the chain rule:

where: the matrix $W_{ij} = \partial^2L/\partial\dot q_i\partial\dot q_j$ — the velocity Hessian, physically the mass matrix — multiplies the accelerations. Everything on the right is known once $(q,\dot q,t)$ are given, so the system can be solved for $\ddot q$ exactly when $W$ is invertible. For a genuine kinetic energy $T=\tfrac12\sum M_{ij}(q)\dot q_i\dot q_j$, $W = M$ is positive-definite and the initial-value problem is well-posed; degenerate Hessians (Lagrangians linear in some velocity) signal constraints and lead toward Hamiltonian constraint theory.

Which form answers which question

1 · The Atwood machine in three lines (confidence builder)

Masses $m_1 = 3.0\ \mathrm{kg}$ and $m_2 = 2.0\ \mathrm{kg}$ hang from an ideal (massless, frictionless) pulley. Find the acceleration and the speed of $m_1$ after $1.0\ \mathrm{s}$ from rest. ($g = 9.81\ \mathrm{m/s^2}$)

Step 1. Spend the constraint on the coordinate. The string is inextensible, so if $m_1$ has descended by $x$, $m_2$ has risen by $x$: one degree of freedom. Both masses move at speed $|\dot x|$, so $T = \tfrac12(m_1+m_2)\dot x^2$, and taking $x$ downward-positive for $m_1$, $V = -m_1gx + m_2gx$. Hence $L = \tfrac12(m_1{+}m_2)\dot x^2 + (m_1{-}m_2)gx$.

Step 2. Crank equation (1): $\partial L/\partial\dot x = (m_1{+}m_2)\dot x$ and $\partial L/\partial x = (m_1{-}m_2)g$, so $(m_1{+}m_2)\ddot x = (m_1{-}m_2)g$.

Step 3. Numbers: $\ddot x = \dfrac{(3.0-2.0)(9.81)}{3.0+2.0} = \dfrac{9.81}{5} = \boxed{1.96\ \mathrm{m/s^2}}$, constant. From rest, after $1.0\ \mathrm{s}$: $v = \boxed{1.96\ \mathrm{m/s}}$, having fallen $0.98\ \mathrm{m}$.

Step 4. Notice what never appeared: the tension. The Lagrangian route sacrificed it by design — ideal constraint forces do no work along allowed motions. If the tension itself is wanted, one line of after-market Newton delivers it, $T_{\mathrm{str}} = m_1(g-\ddot x) = 3.0(9.81-1.96) = 23.5\ \mathrm{N}$ — or keep the constraint explicit and let a Lagrange multiplier produce it, the subject of that article.

2 · Pendulum on a free cart (exam level)

A cart of mass $M = 2.0\ \mathrm{kg}$ rolls without friction on a horizontal rail. From it hangs a pendulum: mass $m = 0.50\ \mathrm{kg}$ on a light rod of length $\ell = 1.0\ \mathrm{m}$. Find the equations of motion, the conserved momentum, and the small-oscillation frequency and period. Compare with a fixed-pivot pendulum.

Step 1. Coordinates: cart position $x$, pendulum angle $\theta$ from the vertical. The bob sits at $(x+\ell\sin\theta,\, -\ell\cos\theta)$, so its velocity squared is $\dot x^2 + 2\ell\dot x\dot\theta\cos\theta + \ell^2\dot\theta^2$. Assemble the scalar:

Step 2. The $x$ equation. $x$ is absent from $L$ (cyclic!), so equation (1) collapses to conservation of its conjugate momentum:

— the total horizontal momentum of cart plus bob: the rail exerts no horizontal force, and the formalism noticed by inspection.

Step 3. The $\theta$ equation. $\partial L/\partial\dot\theta = m\ell\dot x\cos\theta + m\ell^2\dot\theta$; the total derivative (chain rule — the warning above earns its keep) gives $m\ell\ddot x\cos\theta - m\ell\dot x\dot\theta\sin\theta + m\ell^2\ddot\theta$; and $\partial L/\partial\theta = -m\ell\dot x\dot\theta\sin\theta - mg\ell\sin\theta$. The $\dot x\dot\theta\sin\theta$ terms cancel, leaving $\ell\ddot\theta + \ddot x\cos\theta + g\sin\theta = 0$.

Step 4. Small oscillations about the ground state ($\theta,\dot\theta$ small, start from $p_x=0$ so $\dot x = -\tfrac{m\ell}{M+m}\dot\theta$ to first order). Substituting into the $\theta$ equation with $\cos\theta\to1$, $\sin\theta\to\theta$: $\ell\ddot\theta\big(1-\tfrac{m}{M+m}\big) + g\theta = 0$, i.e.

where: the factor $(M{+}m)/M > 1$ is the recoil correction — the pivot backs away from the bob, shortening the effective swing. Numbers: $\omega^2 = \dfrac{9.81\times2.5}{2.0\times1.0} = 12.26\ \mathrm{s^{-2}}$, so $\omega = \boxed{3.50\ \mathrm{rad/s}}$ and $T_{\mathrm{osc}} = \boxed{1.79\ \mathrm{s}}$.

Step 5. Compare: a fixed-pivot pendulum of the same length has $\omega_0 = \sqrt{g/\ell} = 3.13\ \mathrm{rad/s}$, period $2.01\ \mathrm{s}$. The free cart raises the frequency by $\sqrt{1+m/M} = \sqrt{1.25} = 1.118$ — about $12\%$ faster. Sanity limits: $M\to\infty$ recovers the fixed pivot; $M\to0$ sends $\omega\to\infty$ (a massless cart cannot anchor any swing). Attempting this with $\mathbf F=m\mathbf a$ means rod forces on two bodies and an accelerating pivot; with the crank it is two partial derivatives per coordinate.

References. Morin, Introduction to Classical Mechanics, Ch. 6 · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 1–2 · Landau & Lifshitz, Mechanics Vol. 1, §1–2 · Taylor, Classical Mechanics, Ch. 7

Provenance. Equations cross-checked against Goldstein (§1.4, §2.3), Landau–Lifshitz Mechanics (§1–2), Morin Ch. 6, and Taylor Ch. 7. A verifier should: (1) reproduce the integration-by-parts derivation and the d'Alembert derivation independently; (2) confirm the polar-coordinate equations $m\ddot r-mr\dot\theta^2=-V'(r)$ and conservation of $mr^2\dot\theta$; (3) verify the total-derivative gauge argument; (4) check the energy-function identity $\mathrm{d}h/\mathrm{d}t=-\partial L/\partial t$. No empirical numbers are asserted; all results are symbolic and textbook-standard. Worked-example numbers on this page (Atwood $\ddot x = 1.96\ \mathrm{m/s^2}$, string tension $23.5\ \mathrm{N}$; cart-pendulum $\omega = 3.50\ \mathrm{rad/s}$, $T_{\mathrm{osc}} = 1.79\ \mathrm{s}$ versus $2.01\ \mathrm{s}$ fixed-pivot, recoil factor $\sqrt{1.25}=1.118$) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: The Action Functional · Hamilton's Principle of Stationary Action · The Lagrangian Function · Generalized Coordinates and Configuration Space · d'Alembert's Principle · Degrees of Freedom Leads to: Cyclic Coordinates and Conservation of Conjugate Momenta · The Energy Function and Jacobi Integral · Symmetries and Continuous Transformations · Generalized Forces and Generalized Momenta