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Physics · lesson 1 of 3

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.

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Learn the ideas

Prerequisites: Calculus of variations and Newtonian mechanics.

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

Example 1

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

Example 2

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

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

1 What is the usual variational condition on action?

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Stationarity under allowed variations

The first variation vanishes; its sign at second order is a separate issue.

2 For L=3qdot²/2−6q²/2, find the coefficient a in qddot=−a q.

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2

The equation is 3qddot+6q=0, so a=2.

3 If ∂L/∂q=0, what is conserved?

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∂L/∂qdot

Euler-Lagrange gives zero derivative of the conjugate momentum.

4 For L=2qdot² and qdot=3, find conjugate momentum.

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12

p=∂L/∂qdot=4qdot=12.

5 Derive the harmonic-oscillator equation from L=mqdot²/2−kq²/2.

Sample answer and marking guidance

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.

6 Which statement about generalized coordinates is correct?

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They need not be Cartesian distances

They parameterize independent configurations under constraints.

7 Why are endpoint variations set to zero in the standard fixed-endpoint derivation?

Sample answer and marking guidance

Integration by parts produces a boundary term involving δq. Fixed endpoints make that term vanish while leaving interior variations arbitrary.

Credit the boundary term and interior arbitrariness.

8 Why must friction not be silently inserted into a conservative potential?

Sample answer and marking guidance

Generic friction depends on motion and dissipates energy rather than deriving from a position-only conservative potential. A generalized force or appropriate dissipative formulation is needed.

Credit the physical and variational distinction.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.