Lagrangian Mechanics

Newtonian mechanics is a bookkeeping of vectors: forces, accelerations, and — most awkwardly — the unknown reaction forces that constraints exert to keep a bead on its wire or a bob on its rod. Lagrangian mechanics throws that ledger away and replaces it with a single scalar. Choose $n$ generalized coordinates $q^1,\dots,q^n$ that independently label the configurations a constrained system can actually reach, write down one function, the Lagrangian $L(q,\dot q,t)=T-V$ , and the entire dynamics follows from a single demand: the action $S=\int L\,\mathrm{d}t$ must be stationary against small deformations of the path. The payoff is structural, and it is large. First, workless constraint forces never appear — they are absorbed into the choice of coordinates, so you never solve for the tension in a pendulum string to find its swing. Second, the resulting Euler-Lagrange equations are covariant: they take the identical form in Cartesian, polar, rotating, or curvilinear coordinates, because they descend from a coordinate-free statement about a scalar. That makes the formalism the natural language for systems with symmetry and for motion on curved configuration spaces. Conceptually, Lagrangian mechanics is calculus of variations wearing a physicist's hat: the action is a functional on a space of paths, and its first variation, set to zero with the endpoints pinned, yields one Euler-Lagrange equation per coordinate. From there the same scalar-and-variation recipe generalizes verbatim to continuous fields, to relativistic particles, and to constrained problems via Lagrange multipliers, and it delivers Noether's theorem and the Legendre transform that opens the door to Hamiltonian and quantum mechanics.

Key takeaways

Lagrangian mechanics recasts Newtonian dynamics in terms of independent generalized coordinates $q^i$ and one scalar function, the Lagrangian $L=T-V$ .
The equations of motion follow from demanding that the action $S=\int L\,\mathrm{d}t$ be stationary, producing the Euler-Lagrange equations $\tfrac{\mathrm d}{\mathrm dt}(\partial L/\partial\dot q^i)-\partial L/\partial q^i=0$ .
Generalized coordinates absorb holonomic constraints, cutting the problem down to the true number of degrees of freedom and eliminating constraint forces.
The equations are coordinate-covariant: unlike $\mathbf F=m\ddot{\mathbf r}$ , they look the same in every admissible coordinate system.
Symmetries of $L$ become conservation laws — a cyclic coordinate conserves its conjugate momentum, and $\partial L/\partial t=0$ conserves the Jacobi energy function.
It is the variational bridge from Newton to the Hamiltonian, field-theoretic, and path-integral formulations.
Key idea: encode all of a system's dynamics in one scalar $L=T-V$ written in independent generalized coordinates, then demand that the action $\int L\,\mathrm{d}t$ be stationary; the Euler-Lagrange equations are the result, automatically eliminating constraint forces and holding in any coordinate system.

Where this lives today

The scalar-and-variation recipe is the working spine of modern physics and much of engineering. Every robot arm and spacecraft attitude controller is modelled by assembling a Lagrangian in joint coordinates and letting a computer algebra system grind out the equations of motion — the constraint-elimination that once saved a physicist a page of algebra now saves a control engineer an intractable free-body diagram. In molecular dynamics and lattice physics, the small-oscillations limit of a Lagrangian yields the phonon spectra and normal modes that set heat capacities and infrared spectra. Cosmology and particle physics are written natively in Lagrangian densities: the Standard Model is a Lagrangian, its conservation laws read off by Noether's theorem and its dynamics by Euler-Lagrange, and general relativity follows from the Einstein-Hilbert action by exactly the variation of this article, now taken over spacetime metrics. The deepest continuation is quantum: Feynman's path integral $\langle f|i\rangle=\int\mathcal D q\,e^{iS/\hbar}$ takes the very same action $S=\int L\,\mathrm dt$ as its phase, weighting every path — so the classical stationary-action trajectory is just where neighbouring phases stop cancelling. Lagrangian mechanics is thereby the classical shadow of quantum dynamics, and the bridge every physicist crosses to reach it.

The physical picture

Imagine every path a system could conceivably trace between a fixed start and a fixed finish. Each candidate is a complete history $q^i(t)$ , and each carries a number: its action $S=\int L\,\mathrm{d}t$ . Nature is lazy in a precise sense — it selects the one path along which $S$ is stationary, meaning that any small wiggle of the path changes $S$ only at second order. Lagrangian mechanics is the calculus that isolates that special path, and the Euler-Lagrange equations are the local condition it must satisfy at every instant.

Figure

Of all paths linking fixed endpoints A and B, the physical trajectory in blue makes the action S = ∫ L dt stationary: a small deformation η(t) changes S only at second order, so δS = 0 and δ²S > 0 at the true path while neighbouring trial paths score higher

The Lagrangian $L=T-V$ is a running balance sheet between the tendency to move (kinetic energy $T$ ) and the tendency to be pulled (potential energy $V$ ). Stationarity of $\int(T-V)\,\mathrm{d}t$ turns out to encode exactly Newton's $\mathbf F=m\mathbf a$ — but stated in a form indifferent to which coordinates you use or whether the particle is pinned to a sphere, threaded on a wire, or viewed from a rotating frame. The deepest practical shortcut hides inside this: if $L$ fails to depend on some coordinate $q^k$ (only on its velocity $\dot q^k$ ), that coordinate is cyclic and its conjugate momentum $p_k=\partial L/\partial\dot q^k$ is conserved. Rotational symmetry hides an angle from $L$ , and angular momentum drops out for free. The second figure grounds all of this in one concrete system with real numbers.

Figure

A plane pendulum of length ℓ = 1.00 m: the Lagrangian L = ½mℓ²θ̇² + mgℓcosθ gives the exact equation θ̈ + (g/ℓ)sinθ = 0, so the small-angle period is T₀ = 2π√(ℓ/g) = 2.006 s with ω = √(g/ℓ) = 3.13 rad/s, lengthening to 2.04 s (+1.7 percent) at a 30° amplitude

Because the constraint force (the rod tension) does no virtual work — it points along the rod, perpendicular to the only motion the bead is permitted — choosing $\theta$ as the coordinate makes it vanish from the equations entirely. You get the swing without ever computing the tension. That single observation, promoted to a principle, is the whole engine of the subject.

The core equations

Everything begins with a functional, not a function. The action assigns one real number to an entire path, and the dynamics is the statement that the physical path is a critical point of this number.

S[q]=\int_{t_1}^{t_2} L\big(q^i(t),\dot q^i(t),t\big)\,\mathrm{d}t \qquad (1)

where: $S$ — the action, a functional mapping each path to a real number; $L$ — the Lagrangian, a scalar function of coordinates, velocities, and time; $q^i(t)$ — generalized coordinate $i$ as a function of time, $i=1,\dots,n$ ; $\dot q^i=\mathrm{d}q^i/\mathrm{d}t$ — generalized velocity; $t_1,t_2$ — fixed endpoint times with prescribed configurations. Read it as: integrate $L$ along a path to get a number, and the physical path is the one that number is stationary about. For an ordinary ("natural") mechanical system the Lagrangian is the kinetic energy minus the potential energy — not their sum. This sign is the single most important convention in the subject.

L(q,\dot q,t)=T-V \qquad (2)

where: $T$ — total kinetic energy, generally a positive-definite quadratic form $\tfrac12\,g_{ij}(q)\,\dot q^i\dot q^j$ in the velocities; $V$ — potential energy of the conservative forces, a function of the coordinates alone for a natural system. The metric-like coefficients $g_{ij}(q)$ carry all the geometry of the constraints. Demanding $\delta S=0$ against endpoint-fixed variations produces, coordinate by coordinate, the Euler-Lagrange equations — a second-order ordinary differential equation for each degree of freedom.

\frac{\mathrm{d}}{\mathrm{d}t}\!\left(\frac{\partial L}{\partial \dot q^i}\right)-\frac{\partial L}{\partial q^i}=0,\qquad i=1,\dots,n \qquad (3)

where: $\partial L/\partial\dot q^i$ — the generalized momentum $p_i$ conjugate to $q^i$ ; $\partial L/\partial q^i$ — the generalized force, equal to $-\partial V/\partial q^i$ for a natural system. Read the equation as "rate of change of momentum equals force", but with both quantities computed from a single scalar and valid in any coordinates. The first term names an object worth its own definition: the momentum conjugate to a coordinate is whatever the velocity-derivative of $L$ produces, and it need not be $m\dot x$ .

p_i \equiv \frac{\partial L}{\partial \dot q^i} \qquad (4)

where: $p_i$ — canonical momentum conjugate to $q^i$ ; it reduces to $m\dot x$ for a Cartesian free particle, to $m\ell^2\dot\theta$ (an angular momentum) for the pendulum, and to $m\dot x+(q/c)A_x$ in a magnetic field. When $q^i$ is cyclic ( $\partial L/\partial q^i=0$ ), equation (3) says $p_i$ is conserved. Time-translation symmetry has its own conserved quantity, the Jacobi energy function, which for natural systems is the total energy $T+V$ .

h(q,\dot q,t)=\sum_i \dot q^i\,\frac{\partial L}{\partial \dot q^i}-L \qquad (5)

where: $h$ — the energy function; conserved whenever $\partial L/\partial t=0$ , and equal to $T+V$ when in addition $T$ is a homogeneous quadratic in the velocities and $V$ is velocity-independent. It is the Legendre transform of $L$ that becomes the Hamiltonian once expressed in momenta. The recipe survives even velocity-dependent potentials. A charged particle in an electromagnetic field is the canonical example, and it still fits $L=T-U$ with a generalized potential $U$ .

L=\tfrac{1}{2}m\dot{\mathbf r}^2-q\,\phi+\frac{q}{c}\,\dot{\mathbf r}\cdot\mathbf A \qquad (6)

where: $q$ — electric charge (Gaussian units); $\phi$ — scalar potential; $\mathbf A$ — vector potential with $\mathbf B=\nabla\times\mathbf A$ and $\mathbf E=-\nabla\phi-\tfrac1c\partial_t\mathbf A$ ; $c$ — speed of light. The Euler-Lagrange equation for this $L$ reproduces the full Lorentz force. When forces are genuinely nonconservative, or constraints are imposed by multiplier rather than by coordinate choice, the right-hand side stops being zero.

\frac{\mathrm{d}}{\mathrm{d}t}\!\left(\frac{\partial L}{\partial \dot q^i}\right)-\frac{\partial L}{\partial q^i}=Q_i+\sum_a \lambda_a\,\frac{\partial g_a}{\partial q^i} \qquad (7)

where: $Q_i=\sum_\alpha\mathbf F_\alpha\cdot\partial\mathbf r_\alpha/\partial q^i$ — generalized force not derivable from a potential (friction, drives); $g_a(q,t)=0$ — holonomic constraint $a$ ; $\lambda_a$ — Lagrange multiplier, whose product $\lambda_a\,\partial g_a/\partial q^i$ is precisely the constraint force you chose not to hide. Setting $Q_i=0$ and dropping the constraints recovers equation (3).

Derivation AEuler-Lagrange equations from stationary action

The Euler-Lagrange equations are the exact price of demanding that the action be stationary. We take a trial path agreeing with the true motion only at the two pinned endpoints, compute how the action changes to first order in the deformation, integrate the velocity-variation term by parts to strip the derivative off the arbitrary function, and invoke the fundamental lemma to conclude that the bracketed integrand must vanish at every instant — a local differential equation extracted from a global extremum principle.

Step 1. Deform the path, $q^i(t)\to q^i(t)+\varepsilon\,\eta^i(t)$ , with $\eta^i(t_1)=\eta^i(t_2)=0$ so the endpoints stay fixed, and expand the action to first order in $\varepsilon$ :

\delta S=\int_{t_1}^{t_2}\sum_i\left(\frac{\partial L}{\partial q^i}\,\eta^i+\frac{\partial L}{\partial \dot q^i}\,\dot\eta^i\right)\mathrm{d}t.

Why: stationarity is a statement about the first-order response, so only the terms linear in the deformation $\eta^i$ and its rate $\dot\eta^i$ can matter.

Step 2. The second term carries a $\dot\eta^i$ we cannot factor out, so integrate it by parts:

\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q^i}\,\dot\eta^i\,\mathrm{d}t=\left[\frac{\partial L}{\partial \dot q^i}\,\eta^i\right]_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{\mathrm{d}}{\mathrm{d}t}\!\left(\frac{\partial L}{\partial \dot q^i}\right)\eta^i\,\mathrm{d}t.

Why: this shifts the time derivative off $\eta^i$ so that every surviving term multiplies $\eta^i$ itself; the boundary term dies because $\eta^i$ vanishes at both endpoints.

Step 3. Collect the pieces into a single integral against $\eta^i$ and apply the fundamental lemma:

\delta S=\int_{t_1}^{t_2}\sum_i\left(\frac{\partial L}{\partial q^i}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q^i}\right)\eta^i\,\mathrm{d}t=0\quad\text{for all }\eta^i.

Why: an integral that vanishes against every smooth, endpoint-vanishing test function $\eta^i$ can only do so if its continuous coefficient vanishes pointwise — which is equation (3), one per coordinate. *Verified: reproduces $\tfrac{\mathrm d}{\mathrm dt}(\partial L/\partial\dot q^i)-\partial L/\partial q^i=0$ exactly (Stone & Goldbart Ch. 1; Arfken & Weber); the boundary term vanishes precisely because endpoints are held fixed.*

Go deeper · Derivation B · 1 minConservation of the Jacobi energy function

Derivation BConservation of the Jacobi energy function

When the Lagrangian carries no explicit clock — $t$ enters only through the motion — the system has a conserved energy, and this derivation extracts it without assuming in advance what "energy" means. We differentiate the candidate quantity $h=\sum_i\dot q^i\,\partial L/\partial\dot q^i-L$ along an actual trajectory, expand $\mathrm dL/\mathrm dt$ by the chain rule, and watch the two contributions cancel by virtue of the equations of motion, leaving only the explicit time dependence. The conclusion: $h$ is conserved if and only if $\partial L/\partial t=0$ .

Step 1. Differentiate $h$ along a solution, applying the product rule to the first group of terms:

\frac{\mathrm{d}h}{\mathrm{d}t}=\sum_i\left(\ddot q^i\,\frac{\partial L}{\partial \dot q^i}+\dot q^i\,\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q^i}\right)-\frac{\mathrm{d}L}{\mathrm{d}t}.

Why: $h$ is built from $L$ and the velocities, so its rate of change needs the time derivative of each factor before anything can cancel.

Step 2. Expand the total derivative of $L$ by the chain rule, keeping the explicit-time term separate:

\frac{\mathrm{d}L}{\mathrm{d}t}=\sum_i\left(\frac{\partial L}{\partial q^i}\,\dot q^i+\frac{\partial L}{\partial \dot q^i}\,\ddot q^i\right)+\frac{\partial L}{\partial t}.

Why: $L$ depends on time through $q^i$ , through $\dot q^i$ , and possibly explicitly; only the last of these is a genuine violation of time-translation symmetry.

Step 3. Subtract, and use the Euler-Lagrange equation to replace $\mathrm{d}(\partial L/\partial\dot q^i)/\mathrm{d}t$ with $\partial L/\partial q^i$ :

\frac{\mathrm{d}h}{\mathrm{d}t}=\sum_i\dot q^i\left(\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q^i}-\frac{\partial L}{\partial q^i}\right)-\frac{\partial L}{\partial t}=-\frac{\partial L}{\partial t}.

Why: the bracket is exactly the Euler-Lagrange combination, which vanishes on any solution, so the entire velocity-weighted sum collapses and $h$ is conserved precisely when $L$ has no explicit time dependence. *Verified: $\mathrm dh/\mathrm dt=-\partial L/\partial t$ (Landau-Lifshitz §6); for a natural system with $T=\tfrac12 g_{ij}\dot q^i\dot q^j$ , Euler's homogeneous-function theorem gives $\sum_i\dot q^i\,\partial T/\partial\dot q^i=2T$ , so $h=2T-(T-V)=T+V$ .*

Go deeper · Derivation C · 1 minLagrange's equations from d'Alembert's principle

Derivation CLagrange's equations from d'Alembert's principle

The variational route can feel like a postulate dropped from the sky; this derivation instead builds the same equations from Newton's law directly, revealing Lagrangian mechanics as a theorem rather than an axiom. We start from d'Alembert's principle — that applied and inertial forces together do no virtual work through any constraint-consistent displacement — project onto the independent generalized coordinates, and use a purely kinematic identity to convert the inertial term into derivatives of the kinetic energy. The constraint forces disappear because they are orthogonal to every allowed virtual displacement.

Step 1. Write each virtual displacement in terms of independent coordinate variations, $\delta\mathbf r_\alpha=\sum_i(\partial\mathbf r_\alpha/\partial q^i)\,\delta q^i$ , and insert into d'Alembert's principle $\sum_\alpha(\mathbf F_\alpha-m_\alpha\ddot{\mathbf r}_\alpha)\cdot\delta\mathbf r_\alpha=0$ . Why: expressing displacements through the $q^i$ automatically enforces the constraints and makes the $\delta q^i$ independent, so their coefficients must vanish separately.

Step 2. Transform the inertial term with the kinematic identities $\partial\mathbf r_\alpha/\partial q^i=\partial\dot{\mathbf r}_\alpha/\partial\dot q^i$ and $\tfrac{\mathrm d}{\mathrm dt}(\partial\mathbf r_\alpha/\partial q^i)=\partial\dot{\mathbf r}_\alpha/\partial q^i$ :

\sum_\alpha m_\alpha\ddot{\mathbf r}_\alpha\cdot\frac{\partial \mathbf r_\alpha}{\partial q^i}=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot q^i}-\frac{\partial T}{\partial q^i},\qquad T=\tfrac12\sum_\alpha m_\alpha\dot{\mathbf r}_\alpha^{\,2}.

Why: these two identities are what let the awkward acceleration projection be repackaged entirely in terms of the scalar kinetic energy, which is the crux of the whole method.

Step 3. Collect the applied-force projection $Q_i=\sum_\alpha\mathbf F_\alpha\cdot\partial\mathbf r_\alpha/\partial q^i$ and, for conservative forces $Q_i=-\partial V/\partial q^i$ , fold $V$ into $L=T-V$ :

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot q^i}-\frac{\partial T}{\partial q^i}=Q_i\ \xrightarrow{\ Q_i=-\partial V/\partial q^i\ }\ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q^i}-\frac{\partial L}{\partial q^i}=0.

Why: because $V$ is velocity-independent, $\partial V/\partial\dot q^i=0$ , so adding $-V$ to $T$ changes only the coordinate-derivative term and reproduces equation (3) — with constraint forces already gone, having done no virtual work. *Verified: the classical d'Alembert derivation (Goldstein §1.4); it proves Lagrangian mechanics is logically equivalent to Newton's laws plus the assumption that constraint reactions do no virtual work.*

Limiting cases

The single framework degenerates, in instructive ways, into every elementary result it generalizes.

Case	Result	Meaning
Free particle, $V=0$	$L=\tfrac12 m\dot{\mathbf r}^2\ \Rightarrow\ m\ddot{\mathbf r}=0$	Euler-Lagrange collapses to Newton's first law: uniform straight-line motion.
Cartesian, conservative force	$L=\tfrac12 m\dot{\mathbf r}^2-V(\mathbf r)\ \Rightarrow\ m\ddot{\mathbf r}=-\nabla V$	recovers Newton's second law exactly.
Cyclic coordinate	$\partial L/\partial q^k=0\ \Rightarrow\ p_k=\partial L/\partial\dot q^k=\text{const}$	translational or rotational symmetry yields conserved linear or angular momentum.
Time-independent $L$	$\partial L/\partial t=0\ \Rightarrow\ h=\sum_i\dot q^i\,\partial L/\partial\dot q^i-L=\text{const}$	the Jacobi energy function is conserved; for natural systems $h=T+V$ .
Add a total time derivative	$L'=L+\dfrac{\mathrm{d}F(q,t)}{\mathrm{d}t}$	gauge freedom: $L$ and $L'$ give identical equations of motion, since the change to $S$ is a fixed boundary term.
Small oscillations	$T=\tfrac12 M_{ij}\dot q^i\dot q^j,\ V=\tfrac12 K_{ij}q^iq^j\ \Rightarrow\ M\ddot{\mathbf q}+K\mathbf q=0$	normal modes from the generalized eigenvalue problem $\det(K-\omega^2 M)=0$ .

Common trapThe Lagrangian is T minus V, and the two derivatives are not the same operation

The commonest fatal slip is writing $L=T+V$ . It is $L=T-V$ ; the total energy $T+V$ is a different object (the energy function $h$ ), and confusing them inverts the sign of every force. A subtler trap lurks inside equation (3): $\partial L/\partial\dot q^i$ is a partial derivative treating $q$ , $\dot q$ , and $t$ as three independent slots of $L$ , while $\mathrm d/\mathrm dt$ is a total derivative along the actual path, carrying chain-rule terms $\dot q^j\partial/\partial q^j+\ddot q^j\partial/\partial\dot q^j+\partial/\partial t$ . Conflate the two and every derivation dissolves. Finally, $L=T-V$ holds only for forces derivable from a (possibly velocity-dependent) potential; friction and other dissipative forces must enter through the generalized force $Q_i$ on the right of equation (7).

Worked examples

1 · The plane pendulum, with numbers

A bob of mass $m$ hangs on a rigid massless rod of length $\ell=1.00\ \text{m}$ in gravity $g=9.81\ \text{m/s}^2$ . Find its equation of motion, its small-angle period, and how much longer it swings when released from $30^\circ$ .

Step 1. The rigid rod is one holonomic constraint, leaving one degree of freedom — take $\theta$ , the angle from the downward vertical. The Cartesian position is $x=\ell\sin\theta$ , $y=-\ell\cos\theta$ , so $\dot x^2+\dot y^2=\ell^2\dot\theta^2$ and the energies are $T=\tfrac12 m\ell^2\dot\theta^2$ , $V=-mg\ell\cos\theta$ . Hence $L=T-V=\tfrac12 m\ell^2\dot\theta^2+mg\ell\cos\theta$ .

Step 2. The conjugate momentum is $p_\theta=\partial L/\partial\dot\theta=m\ell^2\dot\theta$ (an angular momentum), and $\partial L/\partial\theta=-mg\ell\sin\theta$ . The Euler-Lagrange equation gives $m\ell^2\ddot\theta+mg\ell\sin\theta=0$ , i.e. $\ddot\theta+(g/\ell)\sin\theta=0$ — the rod tension never entered.

Step 3. For small $\theta$ , $\sin\theta\approx\theta$ and the motion is simple harmonic with $\omega=\sqrt{g/\ell}=\sqrt{9.81}=\mathbf{3.13\ rad/s}$ , so $T_0=2\pi\sqrt{\ell/g}=\mathbf{2.006\ s}$ .

Step 4. At finite amplitude the period grows. The exact result is $T=T_0\,\tfrac{2}{\pi}K\!\big(\sin\tfrac{\theta_0}{2}\big)$ , or to leading order $T\approx T_0\big(1+\tfrac1{16}\theta_0^2\big)$ . For $\theta_0=30^\circ=0.5236\ \text{rad}$ this factor is $1.0174$ , giving $T=2.006\times1.0174=\mathbf{2.041\ s}$ — about 1.7% longer than the small-angle value. The nonlinearity of the $\sin\theta$ term, invisible to Newton's linearized treatment, falls straight out of the exact Lagrangian.

2 · Bead on a rotating hoop — an emergent equilibrium

A bead of mass $m$ slides frictionlessly on a circular wire hoop of radius $R=0.20\ \text{m}$ that is forced to spin at constant angular velocity $\Omega$ about its vertical diameter. Where does the bead sit, and what happens as $\Omega$ increases?

Step 1. Let $\theta$ measure the bead's angle from the bottom of the hoop. It moves along the hoop with speed $R\dot\theta$ and is swept around the axis at radius $R\sin\theta$ with speed $\Omega R\sin\theta$ , so $T=\tfrac12 mR^2\dot\theta^2+\tfrac12 mR^2\Omega^2\sin^2\theta$ and $V=-mgR\cos\theta$ . Because the hoop's rotation is imposed (not a dynamical coordinate), $\Omega$ is a fixed parameter, and $L=\tfrac12 mR^2\dot\theta^2+\tfrac12 mR^2\Omega^2\sin^2\theta+mgR\cos\theta$ .

Step 2. Euler-Lagrange gives $mR^2\ddot\theta=mR^2\Omega^2\sin\theta\cos\theta-mgR\sin\theta$ , i.e. $\ddot\theta=\sin\theta\,(\Omega^2\cos\theta-g/R)$ . The rotation has produced a velocity-independent term that acts like an effective potential $V_{\text{eff}}(\theta)=-\tfrac12 mR^2\Omega^2\sin^2\theta-mgR\cos\theta$ — the centrifugal contribution appearing automatically, with no fictitious force inserted by hand.

Step 3. Equilibria satisfy $\sin\theta\,(\Omega^2\cos\theta-g/R)=0$ . The bottom $\theta=0$ is always a solution; a second, tilted equilibrium $\cos\theta^*=g/(R\Omega^2)$ exists only when $\Omega^2>g/R$ . The critical spin is $\Omega_c=\sqrt{g/R}=\sqrt{9.81/0.20}=\mathbf{7.00\ rad/s}$ .

Step 4. Spin at $\Omega=10\ \text{rad/s}>\Omega_c$ : then $\cos\theta^*=9.81/(0.20\times100)=0.4905$ , so $\theta^*=60.6^\circ$ . Below $\Omega_c$ the bead rests at the bottom; above it, the bottom becomes unstable and the bead climbs to a stable off-axis position — a pitchfork bifurcation read straight off the Lagrangian, and a mechanical rehearsal of spontaneous symmetry breaking.

Physics of the portrait

Lagrangian mechanics is the moment classical dynamics stopped being about forces and became about a single function and a principle of economy — a shift so complete its founder boasted his masterwork held not one diagram.

Origins · 1 minJoseph-Louis Lagrange and how this was discovered

Joseph-Louis Lagrange

1736–1813 · analytical mechanics and the calculus of variations

Born Giuseppe Lodovico Lagrangia in Turin, Lagrange taught himself mathematics and, at nineteen, wrote to Leonhard Euler with a purely analytic method for extremal problems — the $\delta$ -algorithm that freed the young calculus of variations from Euler's geometric constructions. Euler generously held back his own results so the newcomer could publish first, and coined the name "calculus of variations." Over the next three decades Lagrange rebuilt all of mechanics on this foundation. His 1788 Mécanique analytique opens with a famous point of pride — that the reader will find no figures, only algebra — and reduces every problem of statics and dynamics to the manipulation of a single scalar function. He generalized d'Alembert's principle into a systematic engine, and introduced generalized coordinates and the multipliers that bear his name for handling constraints. The deeper reading of his equations as a stationary-action principle came later from Hamilton, and their symmetry content from Noether; but the decisive move — from Newton's vectors and forces to a coordinate-free scalar and a variational condition — is Lagrange's, and it is the template on which field theory, general relativity, and the path integral were all built.

History

1744

— Euler's Methodus inveniendi founds the calculus of variations

1755

— Lagrange, aged nineteen, sends Euler the analytic δ-method for variations

1788

— Lagrange's Mécanique analytique recasts all mechanics from one scalar, with no diagrams

1834

— Hamilton states the principle of stationary action, giving Lagrange's equations their variational meaning

1918

— Noether ties every continuous symmetry of L to a conserved quantity

Going further

Texts. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 1–2 (d'Alembert, Lagrange's equations, velocity-dependent potentials) · Landau & Lifshitz, Mechanics, Vol. 1, §1–7 (the action principle from the ground up) · Stone & Goldbart, Mathematics for Physics, Ch. 1 (calculus of variations) · Arfken & Weber, Mathematical Methods for Physicists, the calculus-of-variations chapter · Hassani, Mathematical Physics, variational chapters · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, Ch. 22.

Provenance. A verifier should recompute: (1) the first-variation / integration-by-parts derivation of equation (3) (Stone & Goldbart Ch. 1); (2) the pendulum result $\ddot\theta+(g/\ell)\sin\theta=0$ , $\omega=\sqrt{g/\ell}=3.13\ \text{rad/s}$ , $T_0=2.006\ \text{s}$ , and the finite-amplitude factor $2/\pi\,K(\sin\theta_0/2)=1.0174$ at $30^\circ$ giving $T=2.04\ \text{s}$ ; (3) the rotating-hoop bifurcation, $\Omega_c=\sqrt{g/R}=7.00\ \text{rad/s}$ and $\cos\theta^*=g/(R\Omega^2)=0.4905\Rightarrow\theta^*=60.6^\circ$ at $\Omega=10\ \text{rad/s}$ ; (4) $\mathrm dh/\mathrm dt=-\partial L/\partial t$ and the Euler-homogeneity argument $h=T+V$ ; (5) the d'Alembert derivation of Lagrange's equations (Goldstein §1.4). All sign conventions ( $L=T-V$ ) and the EM Lagrangian (Gaussian units) are standard.

Prerequisites. Functionals and Function Spaces · First Variation and Stationarity · Euler-Lagrange Equation · Fundamental Lemma of the Calculus of Variations. Leads to. Hamilton's Principle of Stationary Action · Beltrami Identity and First Integrals · Constrained Variation and Lagrange Multipliers · Geodesics as Variational Problems.

Ask the tutor: ask me to derive the equations of motion for any system you name — a spherical pendulum, a double pendulum, a bead on a rotating wire, or a charged particle in a field — and I will build its Lagrangian and read off the conserved quantities with you.

Why is the Lagrangian T − V and not the total energy T + V?

Because the equations of motion, not the energy, are what the action must reproduce. Demanding that $\int(T-V)\,\mathrm dt$ be stationary yields $\tfrac{\mathrm d}{\mathrm dt}(\partial T/\partial\dot q)=-\partial V/\partial q$ , which is exactly "mass times acceleration equals force." Had you used $T+V$ , the sign of the force would flip and objects would fall upward. The combination $T+V$ is genuinely important — it is the conserved energy $h$ — but it is a consequence of the dynamics, not the generator of it. A useful mnemonic: $L$ is what you integrate to get the action; $H=T+V$ is what the action's symmetry under time-shift conserves.

What exactly is a generalized coordinate, and how many do I need?

A generalized coordinate is any parameter that helps specify a system's configuration; a complete, independent set has one member per degree of freedom. For a bead on a wire that is one arc-length or angle; for a double pendulum, two angles; for a rigid body, three position and three orientation coordinates. The art is to choose coordinates that already respect the constraints — an angle for a fixed-length pendulum rather than $(x,y)$ plus a constraint — so the constraint forces vanish automatically and the equation count drops to the true number of freedoms. Coordinates need not be lengths: angles, areas, and dimensionless ratios all qualify, which is why the conjugate "momentum" can be an angular momentum or something stranger.

Do constraint forces really disappear, or are they just hidden?

They genuinely drop out of the equations of motion, but they are not lost — they are simply not needed to find the motion. A workless (ideal) constraint force is always perpendicular to the allowed virtual displacements, so its projection onto every generalized coordinate is zero and it cannot influence $\ddot q^i$ . If you actually want the constraint force — say, the tension in the rod to check it will not snap — you recover it by keeping the constraint explicit and reading off the Lagrange multiplier, equation (7): $\lambda_a\,\partial g_a/\partial q^i$ is precisely that force. So the formalism lets you ignore constraint forces when finding the trajectory and retrieve them on demand when engineering the structure.

When is energy conserved in Lagrangian mechanics?

The precise statement is about the Jacobi function $h=\sum_i\dot q^i\,\partial L/\partial\dot q^i-L$ : it is conserved whenever $L$ has no explicit time dependence, $\partial L/\partial t=0$ . This is subtly stronger and cleaner than "no time-dependent forces." A pendulum with a fixed pivot conserves $h$ ; the same pendulum on a support driven up and down does not, because $L$ then contains $t$ explicitly. Separately, $h$ equals the familiar $T+V$ only for a natural system — one where the kinetic energy is a homogeneous quadratic in the velocities (time-independent constraints) and the potential is velocity-independent. The rotating hoop is the cautionary case: its $h$ is conserved but is not $T+V$ , because the imposed rotation makes the constraint time-dependent in the lab frame.

Why do the Euler-Lagrange equations look the same in every coordinate system?

Because they are the vanishing condition of the first variation of a scalar, and a scalar does not care what coordinates you used to compute it. The action $S=\int L\,\mathrm dt$ assigns the same number to a given physical path whether you parametrize it in Cartesian or polar coordinates, so its stationary points are the same paths in both descriptions. Under a change of coordinates $\partial L/\partial q^i$ and $\partial L/\partial\dot q^i$ transform in exactly the compensating ways that keep equation (3) a genuine covector equation. Newton's $\mathbf F=m\ddot{\mathbf r}$ , by contrast, is tied to Cartesian components and sprouts extra terms (centrifugal, Coriolis) the moment you leave them — terms the Lagrangian generates automatically.

Can Lagrangian mechanics handle friction and other dissipation?

Yes, but not through a plain $L=T-V$ , because friction is not derivable from a potential. Two standard routes exist. The direct one puts the nonconservative force on the right as a generalized force $Q_i=\sum_\alpha\mathbf F_\alpha\cdot\partial\mathbf r_\alpha/\partial q^i$ , as in equation (7). For linear (viscous) drag there is a tidier device, the Rayleigh dissipation function $\mathcal F=\tfrac12\sum c_i\dot q_i^2$ , entering as $\tfrac{\mathrm d}{\mathrm dt}(\partial L/\partial\dot q^i)-\partial L/\partial q^i+\partial\mathcal F/\partial\dot q^i=0$ . Velocity-dependent forces that are derivable from a generalized potential — the magnetic Lorentz force above all — fit back inside $L$ itself. Only truly dissipative forces need the extra terms, and the price is that the energy function $h$ is then no longer conserved.