Euler-Lagrange: Nonlinear Function
Created: July 14, 2022
Modified: July 14, 2022

Euler-Lagrange

This page is from my personal notes, and has not been specifically reviewed for public consumption. It might be incomplete, wrong, outdated, or stupid. Caveat lector.

One-particle system

Let

L(x(t),v(t))=12mv(t)2−U(x(t))\mathcal{L}(x(t), v(t)) = \frac{1}{2}mv(t)^2 - U(x(t))

be the Lagrangian for a system with time-varying position xx and velocity vv, with forces defined by a potential energy function UU.

Naively the position and velocity are separate degrees of freedom. But over time we have the constraint v(t)=x˙(t)v(t) = \dot{x}(t), which we impose with a Lagrange multiplier ρ(t)\rho(t) at each step:

L′(x(t),v(t),ρ(t))=12mv(t)2−U(x(t))+ρ(t)(x˙(t)−v(t)).\mathcal{L}'(x(t), v(t), \rho(t)) = \frac{1}{2}mv(t)^2 - U(x(t)) + \rho(t)\left(\dot{x}(t) - v(t)\right).

The action

S=∫t1t2L′(x(t),v(t),ρ(t))dt\mathcal{S} = \int_{t1}^{t2} \mathcal{L}'(x(t), v(t), \rho(t))dt

has the stationarity conditions

∂S∂ρ=0  ⟹  v(t)=x˙(t)\frac{\partial \mathcal{S}}{\partial \rho} = 0 \implies v(t) = \dot{x}(t)

(recovering the constraint introduced above), and

∂S∂v=0  ⟹  ρ(t)=mv(t)\frac{\partial \mathcal{S}}{\partial v} = 0 \implies \rho(t) = mv(t)

in which the Lagrange multiplier ρ\rho turns out to play the role of the momentum. The final condition for ∂S∂x\frac{\partial \mathcal{S}}{\partial x} is slightly trickier since the action integral includes not just xx but also its time derivative x˙\dot{x}. We remove this dependence using integration by parts to rewrite the relevant term of the integral,

S=∫t1t2ρ(t)dx(t)dtdt−∫t1t2U(x(t))dt+…=∣ρ(t)x(t)∣t1t2−∫t1t2dρ(t)dtx(t)dt−∫t1t2U(x(t))dt+…\begin{align*} \mathcal{S} &= \int_{t1}^{t2} \rho(t)\frac{dx(t)}{dt}dt - \int_{t1}^{t2}U(x(t))dt + \ldots\\ &= \left|\rho(t)x(t)\right|_{t_1}^{t_2} - \int_{t1}^{t2}\frac{d\rho(t)}{dt}x(t)dt - \int_{t1}^{t2}U(x(t))dt + \ldots \end{align*}

omitting (via '…\ldots') the terms that do not depend on xx. Then it followsFormally, take ∂x(t)\partial x(t) to be a perturbation of x(t)x(t) that leaves the endpoints x(t2)x(t_2) and x(t1)x(t_1) unchanged, so that the term ∂∂x∣ρ(t)x(t)∣t1t2\frac{\partial}{\partial x} \left|\rho(t)x(t)\right|_{t_1}^{t_2} vanishes, and apply the fundamental lemma of the calculus of variations. that

∂S∂x=0  ⟹  −U′(x(t))=dρ(t)dt\frac{\partial \mathcal{S}}{\partial x} = 0 \implies -U'(x(t)) = \frac{d\rho(t)}{dt}

which we can read as Newton's third law f=maf = ma, where f=−U′(x)f = -U'(x) is the potential gradient and ma=dρ(t)dtma = \frac{d\rho(t)}{dt} is the time derivative of momentum.

General form

In general form, the preceding derivation establishes the Euler-Lagrange equation,

∂L∂x=ddt∂L∂v.\frac{\partial \mathcal{L}}{\partial x} = \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial v}.

See Wikipedia or other sources for the formal derivation. In the case above:

  1. The left side ∂L∂x=−U′(x)\frac{\partial \mathcal{L}}{\partial x} = -U'(x) is the derivative of the potential, representing the forces on the system.
  2. The right side ddt∂L∂v=ddtρ=ma\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial v} = \frac{d}{dt} \rho = ma is the time derivative of the Lagrange multiplier used to enforce the dynamics.

Thus the Euler-Lagrange equation tells us that the local potential gradient (on the left side) equals the rate of change of the generalized momentum (on the right side), where the generalized momentum is the Lagrange multiplier indicating how strongly the system 'wants' to violate the dynamics constraint.

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