Physic Labs

Analytical mechanics

The Euler–Lagrange equations

The Euler–Lagrange equations derive motion from a Lagrangian and apply to each generalized coordinate.

A system can be described using coordinates adapted to its constraints rather than all Cartesian coordinates. The Euler–Lagrange equations determine how these coordinates evolve when the action is stationary.

ddt∂L∂q˙i−∂L∂qi=0,i=1,…,n\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=0, \qquad i=1,\ldots,n

Definition: Generalized equation of motion

Here qiq_i and q˙i\dot q_i are generalized coordinate and velocity. If L=T−VL=T-V has no explicit time dependence, energy is usually conserved. With nonconservative forces, the generalized form is d(∂L/∂q˙i)/dt−∂L/∂qi=Qi(nc)d(\partial L/\partial\dot q_i)/dt-\partial L/\partial q_i=Q_i^{(nc)}.

Explore a simple pendulum in phase space. Change its initial condition and length; the trajectory follows the Lagrange equation.

Example: the simple pendulum

L=12ml2θ˙2−mgl(1−cos⁡θ),θ¨+glsin⁡θ=0L=\frac12 ml^2\dot\theta^2-mgl(1-\cos\theta), \qquad \ddot\theta+\frac gl\sin\theta=0

Set q=θq=\theta. The kinetic energy is T=12ml2θ˙2T=\tfrac12 ml^2\dot\theta^2 and the potential, zeroed at the bottom, is V=mgl(1−cos⁡θ)V=mgl(1-\cos\theta). Substituting L=T−VL=T-V gives nonlinear motion; only at small angles may we use sin⁡θ≃θ\sin\theta\simeq\theta.

Example: Initial angular acceleration

A pendulum of length l=1l=1 m is released from rest at θ=0.10\theta=0.10 rad. With g=9.8g=9.8 m/s², find its initial angular acceleration.

Solution

θ¨=−(g/l)sin⁡θ=−9.8sin⁡(0.10)≈−0.98\ddot\theta=-(g/l)\sin\theta=-9.8\sin(0.10)\approx-0.98 rad/s²; the negative sign points back toward equilibrium.

Quick check

If LL has no explicit dependence on a coordinate qjq_j, that coordinate is cyclic, and the Euler–Lagrange equation gives dpj/dt=0d p_j/dt=0, where pj=∂L/∂q˙jp_j=\partial L/\partial\dot q_j. This extracts a conserved quantity before solving the equations of motion. For example, in a central potential the azimuthal angle is absent from the Lagrangian, so its conjugate angular momentum is conserved. A valid coordinate change leaves the physics intact, but both velocities and the Lagrangian must be transformed consistently.

In one dimension, L=12mx˙2−V(x)L=\tfrac12m\dot x^2-V(x) gives mx¨=−dV/dxm\ddot x=-dV/dx, Newton’s second law. This equivalence is a useful check: the Lagrangian method does not change the prediction, but organizes it more effectively when many coordinates and constraints are present.

What is a natural generalized coordinate for a simple pendulum?

When is the approximation sin⁡θ≈θ\sin\theta\approx\theta valid?

References

  1. Herbert Goldstein, Charles Poole, John Safko (2002). Classical Mechanics