Classical Mechanics · Upper Division

Lagrangian Mechanics

Instead of forces, Lagrangian mechanics asks: what path minimizes the action? This single variational principle contains all of classical mechanics — and generalizes naturally to quantum fields and general relativity.

Prerequisites

Newton's laws (Ch. 3) \cdot Energy (Ch. 4) \cdot Calculus (Ch. 22) \cdot Differential equations (Ch. 2

Learning Goals

Choose generalized coordinates appropriate to a constrained system and count the degrees of freedom.
$Construct the Lagrangian L = T - V and derive the equations of motion using the Euler-Lagr$ ange equation.
Identify cyclic coordinates and read off the corresponding conserved momenta directly.
State Noether's theorem and connect each continuous symmetry to its conserved quantity.
Apply the Lagrangian method to multi-body systems such as the Atwood machine and double pendulum.

L.1 Generalized Coordinates

The power of Lagrangian mechanics lies in the freedom to choose any coordinates that describe the system's configuration — not just Cartesian x, y, z. These aregeneralized coordinates qᵢ: angles, arc lengths, normal mode amplitudes, whatever parametrizes the space of possible configurations.

A system with N particles in 3D has 3N degrees of freedom. Constraints (a bead on a wire, a pendulum on a rod) reduce this number. The number of independent generalized coordinates equals the number of degrees of freedom after constraints are imposed.

Definition L.1 — The Lagrangian

The Lagrangian L of a mechanical system is the difference between kinetic energy T and potential energy V:

L(q, q̇, t) = T - V

where q = (q_{1}, q_{2}, \cdots) are the generalized coordinates and q̇ = dq/dt their time derivative

L.2 The Euler-Lagrange Equation

The principle of stationary action(Hamilton's principle) states that the actual path taken by a system between two configurations is the one for which the action S = ∫L dt is stationary (first variation vanishes). Applying variational calculus yields:

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = 0 \qquad \text{(Euler-Lagrange equation, for each }i\text{)}

(L.1)

This is one equation per degree of freedom. Remarkably, it automatically handles constraints — there is no need to compute constraint forces. The equation is alsocovariant: it takes the same form in any coordinate system.

Example L.1 — Simple Pendulum

Derive the $equation of motion for a pendulum (mass m, length l) using the Lagrangian. Use the angle \theta$ eralized coordinate.

Kinetic energy:

T = \frac{1}{2}m(l\thetȧ)^{2} = \frac{1}{2}ml^{2}\thetȧ^{2}

Potential energy:

V = -mgl cos \theta (taking pivot as origin, downward positive

Lagrangian:

L = T - V = \frac{1}{2}ml^{2}\thetȧ^{2} + mgl cos \theta

∂L/∂θ̇:=

ml^{2}\thetȧ \qquad \to d/dt(ml^{2}\thetȧ) = ml^{2}\thetä

∂L/∂θ:=

-mgl sin \theta

E-L equation:

ml^{2}\thetä + mgl sin \theta = 0 \quad \to \quad \thetä = -(g/l) sin \theta ✓

Example L.2 — Atwood Machine

$Two masses m_{1} and m_{2} hang over a frictionless pulley. Find the acceleration using one gene$ $ralized coordinate x (extension of m_{1} below its initial position$ .

Constraint:

When m_{1} descends by x, m_{2} rises by x. Both speeds = |ẋ|

T = \frac{1}{2}(m_{1}+m_{2})ẋ^{2}

V = -m_{1}gx + m_{2}gx = (m_{2}-m_{1})gx \qquad (taking initial position as reference

L = \frac{1}{2}(m_{1}+m_{2})ẋ^{2} - (m_{2}-m_{1})gx

E-L:

d/dt[(m_{1}+m_{2})ẋ] - [-(m_{2}-m_{1})g] = 0 \quad \to \quad ẍ = (m_{1}-m_{2})g/(m_{1}+m_{2}) ✓

L.3 Conserved Quantities and Noether's Theorem

If the Lagrangian does not depend explicitly on a coordinate qᵢ (a cyclic coordinate), then the Euler-Lagrange equation gives d/dt(∂L/∂q̇ᵢ) = 0 — the conjugate momentum pᵢ = ∂L/∂q̇ᵢ is conserved.

Theorem L.1 — Noether's Theorem (1915)

For every continuous symmetry of the action, there is a corresponding conserved quantity:

Time translation symmetry \quad \to \quad conservation of energy (Hamiltonian H

Spatial translation symmetry \quad \to \quad conservation of linear momentum

Rotational symmetry \quad \to \quad conservation of angular momentum

This is the deepest explanation of conservation laws in physics — they are not axioms but consequences of the symmetry structure of nature.

The Hamiltonian H = Σ(pᵢq̇ᵢ) − L equals the total energy for natural systems (no explicit time dependence, no velocity-dependent potentials). The Hamiltonian formulation replaces 2nd-order ODEs in q with 1st-order Hamilton's equations in phase space (q,p): q̇ = ∂H/∂p, ṗ = −∂H/∂q.

Example L.3 — Central Force — Angular Momentum Conservation

$A particle moves under a central force V(r) only. Use polar coordinates (r, \phi). Show that$ angular momentum is conserved.

T in polar:

T = \frac{1}{2}m(ṙ^{2} + r^{2}\phi̇^{2}

L = \frac{1}{2}m(ṙ^{2} + r^{2}\phi̇^{2}) - V(r

φ is cyclic:

\partialL/\partial\phi = 0 \to d/dt(\partialL/\partial\phi̇) = 0 \to d/dt(mr^{2}\phi̇) = 0

Conserved:ℓ =

mr^{2}\phi̇ = const — this is the angular momentum! No force calculation needed

Definition L.2 — Common Traps

Generalized coordinates are not necessarily distances: angles, mode amplitudes, and constraints can all serve as coordinates.
The Lagrangian is not usually energy: $L = T - V, while total mechanical energy is T + V for natural systems$
Constraints reduce degrees of freedom: count coordinates after applying constraints, not before.
Cyclic means absent from L: if q does not appear, its conjugate momentum is conserved.
Stationary action is broader than minimum action: the physical path makes the first variation vanish; it need not be a true minimum.

Exercises — L.1–L.3 Lagrangian Mechanics

A bead slides without friction on a straight wire inclined at angle \alpha. Find the equation o

f motion using x (distance along wire) as the generalized coordinate.

m/s²

Straightforward

Write the Lagrangian for a double pendulum (m_{1},l_{1}) and (m_{2},l_{2}). What makes this system cha

otic at large amplitudes?

Intermediate

3.Find the Lagrangian for a spherical pendulum. Identify the cyclic coordinate and its conserved quantity. What condition gives uniform circular motion (conical pendulum)?

Intermediate

4.Explicitly derive the conserved quantity associated with rotational symmetry around the z-axis using Noether's theorem. Show that it equals the z-component of angular momentum.

Challenging

Key Takeaways

Generalized coordinates q describe configuration in any convenient parameterization.
$Lagrangian L = T - V; stationary action \to Euler-Lagrange equations d/dt(\partialL/\partialq̇) = \partialL/\partialq$ .
$Cyclic coordinate (\partialL/\partialq = 0) \to conjugate momentum p = \partialL/\partialq̇ is conserved$ .
Noether's theorem: every continuous symmetry corresponds to a conserved quantity.
$Time symmetry \to energy; space translation \to momentum; rotation \to angular momentum$ .
Constraints are handled automatically — no need to compute constraint forces.