The Euler-Lagrange Equation and Conserved Momenta
Applying the Euler-Lagrange equation to a Lagrangian returns the equation of motion, and a coordinate missing from the whole Lagrangian gives a conserved momentum when no nonconservative generalized force acts in it. Conservation laws stop being separate rules and become consequences of symmetry.
What a learner can do afterwards
- Applies the Euler-Lagrange equation to recover a known equation of motion
- Spots a coordinate absent from the whole Lagrangian and writes down the conserved conjugate momentum it implies
- Connects a symmetry of the Lagrangian to the quantity it conserves
1 · Read
The Euler Lagrange equation turns a Lagrangian into equations of motion, one per generalised coordinate. A dot marks a time derivative: q̇ is the rate of q and q̈ its acceleration. For a coordinate q it reads d/dt(∂L/∂q̇) − ∂L/∂q = 0. Differentiate L with respect to the rate, take the time derivative, and subtract the derivative with respect to the coordinate itself. Applied to a falling mass or a pendulum it reproduces the Newton result, which is the first consistency check to run. With a nonconservative generalised force Q, the right side becomes Q.
A coordinate absent from the whole Lagrangian is called cyclic, and its conjugate momentum ∂L/∂q̇ is conserved. The recipe keeps that derivative constant, so you only need to spot which coordinates the Lagrangian skips. Each such skip is a symmetry of the description. The conservation holds when no nonconservative generalised force acts in that coordinate; with such a force Q, the right side becomes Q and the momentum no longer coasts. Free polar motion with V = 0 has T = ½m(ṙ² + r²φ̇²), which depends on r, so the radial momentum need not be conserved; a straight radial trajectory is the special case where mṙ stays constant.
Pendulum: script L = ½ml²θ̇² + mgl cos θ. Differentiate with respect to the rate: ∂L/∂θ̇ = ml²θ̇, whose time derivative is ml²θ̈. Differentiate with respect to the angle: ∂L/∂θ = −mgl sin θ. Euler-Lagrange sets the difference to zero: ml²θ̈ + mgl sin θ = 0, so θ̈ + (g/l) sin θ = 0. Small swings: sin θ ≈ θ, so θ̈ + (g/l)θ = 0.
In an inertial Cartesian system with the natural L = T − V, if L skips x then no force acts along x and p_x = mẋ coasts unchanged. Rotation symmetry gives conserved angular momentum. A pendulum angle is not cyclic under gravity, so its angular momentum about the pivot is not conserved.
Energy check: dL/dt = Σᵢ(q̇ᵢ ∂L/∂qᵢ + q̈ᵢ ∂L/∂q̇ᵢ) + ∂L/∂t. Replace ∂L/∂qᵢ with d/dt(∂L/∂q̇ᵢ) from the unforced Euler-Lagrange equation, and the sums collapse into d/dt(Σᵢ q̇ᵢ ∂L/∂q̇ᵢ). So d/dt(Σᵢ q̇ᵢ ∂L/∂q̇ᵢ − L) = −∂L/∂t. An autonomous Lagrangian has ∂L/∂t = 0, so the generalized energy E = Σᵢ q̇ᵢ ∂L/∂q̇ᵢ − L is conserved. E = T + V when L = T − V with T homogeneous quadratic in the rates and V rate-independent.
Apply the Euler Lagrange equation to each coordinate, and spot the coordinates the whole Lagrangian skips for conserved conjugate momenta.
Three readings, six flashcards and no extra questions were written for this Euler-Lagrange topic. Two Khan Academy videos on momentum conservation are linked from their original sources. No video transcript is copied.
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