Physics · Advanced university
Euler-Lagrange equations and constraints
For L(q,qdot,t), stationarity of S=∫L dt with fixed endpoints yields d/dt(∂L/∂qdot)−∂L/∂q=0.
Prerequisites
Calculus of variations and Newtonian mechanics.
Study notes
Action
For L(q,qdot,t), stationarity of S=∫L dt with fixed endpoints yields d/dt(∂L/∂qdot)−∂L/∂q=0.
Coordinates
Generalized coordinates describe independent degrees of freedom. Holonomic constraints can reduce their number when handled consistently.
Cyclic coordinate
If L has no dependence on a coordinate, its conjugate momentum is conserved, provided the Euler-Lagrange assumptions apply.
Caution
Stationary action is not always a minimum. Nonconservative forces may require generalized forces or an extended formulation.
Worked examples
What is the usual variational condition on action?
The first variation vanishes; its sign at second order is a separate issue.
Stationarity under allowed variations
For L=3qdot²/2−6q²/2, find the coefficient a in qddot=−a q.
The equation is 3qddot+6q=0, so a=2.
2
Check your understanding
If ∂L/∂q=0, what is conserved?
- ∂L/∂qdot
- L in every driven system
- Time itself
- q in every trajectory
For L=2qdot² and qdot=3, find conjugate momentum.
Derive the harmonic-oscillator equation from L=mqdot²/2−kq²/2.
Which statement about generalized coordinates is correct?
- There must be exactly three
- They need not be Cartesian distances
- They are always momenta
- They must all be angles
Answers and marking guidance
Answers and marking notes
- ∂L/∂qdot
Euler-Lagrange gives zero derivative of the conjugate momentum.
- 12
p=∂L/∂qdot=4qdot=12.
- The velocity derivative is mqdot, its time derivative is mqddot, and the coordinate derivative is −kq. Hence mqddot+kq=0.
Credit both partial derivatives and their signs.
- They need not be Cartesian distances
They parameterize independent configurations under constraints.
Further reading and review references
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