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Lagrangian Mechanics

Equations of motion from the principle of least action — the gateway to analytical mechanics.

· Updated Apr 20, 2026 · 1 min
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Lagrangian Mechanics

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Motivation

Newton’s form F=mr¨\mathbf{F} = m\ddot{\mathbf{r}} is powerful, but for systems with constraints — pendulums, rigid bodies, rotating frames — choosing generalized coordinates qiq_i collapses the equations into something dramatically cleaner. Lagrangian mechanics gives that framework.

Action and the Lagrangian

Describe the state by generalized coordinates q=(q1,,qn)q = (q_1,\dots,q_n). The Lagrangian is the difference between kinetic energy TT and potential VV:

L(q,q˙,t)=T(q,q˙,t)V(q,t).L(q,\dot q,t) = T(q,\dot q,t) - V(q,t).

The action SS is a functional of the path q(t)q(t):

S[q]=t1t2L(q(t),q˙(t),t)dt.S[q] = \int_{t_1}^{t_2} L\big(q(t),\dot q(t),t\big)\,dt.

The principle of least action

Among paths with fixed endpoints q(t1),q(t2)q(t_1), q(t_2), the physical path makes SS stationary:

δS=0.\delta S = 0.

Carrying the variation through gives

δS=t1t2i(LqiddtLq˙i)δqidt=0.\delta S = \int_{t_1}^{t_2} \sum_i \left( \frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot q_i} \right) \delta q_i \, dt = 0.

Since each δqi\delta q_i is independent, the integrand must vanish identically. The Euler–Lagrange equations:

  ddtLq˙iLqi=0.  \boxed{\;\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0.\;}

Example: simple pendulum

Length \ell, mass mm, angle θ\theta:

T=12m2θ˙2,V=mgcosθ.T = \tfrac{1}{2} m \ell^2 \dot\theta^2,\quad V = -m g \ell \cos\theta. L=12m2θ˙2+mgcosθ.L = \tfrac{1}{2} m \ell^2 \dot\theta^2 + m g \ell \cos\theta.

Apply Euler–Lagrange:

ddt(m2θ˙)+mgsinθ=0θ¨+gsinθ=0.\frac{d}{dt}(m \ell^2 \dot\theta) + m g \ell \sin\theta = 0 \quad\Longrightarrow\quad \ddot\theta + \frac{g}{\ell}\sin\theta = 0.

In the small-angle limit sinθθ\sin\theta \approx \theta, this becomes simple harmonic motion with ω=g/\omega = \sqrt{g/\ell}.

Simulation: Interactive Pendulum θ̈ + (g/ℓ) sin θ = 0

Symmetries and conservation: Noether’s theorem

Each continuous symmetry of LL implies a conserved quantity.

SymmetryConserved quantity
Time translationEnergy E=iq˙iLq˙iLE = \sum_i \dot q_i \frac{\partial L}{\partial \dot q_i} - L
Space translationMomentum pi=Lq˙ip_i = \frac{\partial L}{\partial \dot q_i}
RotationAngular momentum

The same logic threads through field theory, quantum mechanics, and the Standard Model.

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