Lagrangian Mechanics Review

Lagrangian mechanics describes classical motion by making an action functional stationary. It is the configuration-space language behind Euler–Lagrange equations, normal-mode expansions, path integrals, and many semiclassical approximations.

The core object is a Lagrangian

$$L(q,\dot q,t),$$

where $q$ denotes generalized coordinates and $\dot q=dq/dt$. The action assigned to a path is

$$S[q] = \int_{t_i}^{t_f} L(q,\dot q,t)\,dt.$$

Classical paths are stationary points of this functional under allowed variations. The full mathematical derivation belongs to Calculus of Variations. This page reviews the mechanics version and the quantum bridges.

Why Quantum Mechanics Needs This

Quantum mechanics is usually introduced through Hamiltonians, but Lagrangian mechanics remains essential for four reasons.

First, path integrals weight histories by $e^{iS[q]/\hbar}$, so the action is the phase-bearing classical quantity.

Second, the semiclassical limit is controlled by stationary phase. The paths satisfying the Euler–Lagrange equations are precisely the saddle points around which the path integral is expanded.

Third, generalized coordinates make constrained systems and curvilinear coordinates natural. This matters before quantization because coordinate choices affect measures, momenta, and operator ordering.

Fourth, field theory is usually written in Lagrangian language. Even in nonrelativistic quantum mechanics, learning actions and stationary variation is preparation for QFT-facing material.

Generalized Coordinates

Generalized coordinates are coordinates $q^i$ on the configuration space of a system. They need not be Cartesian coordinates. Examples include:

• the angle $\theta$ of a pendulum;
• polar coordinates $(r,\theta)$ for planar motion;
• normal-mode amplitudes for small oscillations;
• collective coordinates in approximate or reduced models.

For $n$ generalized coordinates, a path is

$$t\mapsto q(t) = (q^1(t),\ldots,q^n(t)).$$

The velocity coordinates are

$$\dot q^i(t) = \frac{dq^i}{dt}.$$

A Lagrangian is a function on positions, velocities, and time:

$$L = L(q^1,\ldots,q^n,\dot q^1,\ldots,\dot q^n,t).$$

For many elementary systems,

$$L=T-V,$$

kinetic energy minus potential energy. This formula is useful, but it is not the definition of all possible Lagrangians. Magnetic couplings, velocity-dependent potentials, constraints, and field theories can require more general forms.

Action and Stationary Paths

The action is a functional: its input is an entire path and its output is a number. For fixed endpoint data,

$$q(t_i)=q_i, \qquad q(t_f)=q_f,$$

one varies the path by

$$q^i(t) \longmapsto q^i(t)+\epsilon\eta^i(t),$$

with

$$\eta^i(t_i)=\eta^i(t_f)=0.$$

The classical path satisfies

$$\left. \frac{d}{d\epsilon} S[q+\epsilon\eta] \right\rvert_{\epsilon=0} = 0$$

for every allowed variation $\eta^i(t)$.

The word “stationary” is deliberate. The classical path need not minimize the action. It may be a saddle point, especially in real-time mechanics and path-integral applications.

Euler–Lagrange Equations

Stationary action gives one Euler–Lagrange equation for each generalized coordinate:

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot q^i} \right) - \frac{\partial L}{\partial q^i} = 0.$$

Equivalently,

$$\frac{\partial L}{\partial q^i} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q^i} \right) = 0.$$

These equations are second-order ordinary differential equations when $L$ is regular in the velocities. Initial conditions usually specify both $q^i(t_0)$ and $\dot q^i(t_0)$. Boundary-value problems, such as fixed endpoints in a propagator, specify data at two times and can have zero, one, or many classical solutions.

Free Particle Example

For a one-dimensional free particle,

$$L = \frac12m\dot x^2.$$

The Euler–Lagrange equation is

$$\frac{d}{dt} \left( m\dot x \right) = 0,$$

so

$$m\ddot x=0.$$

The classical path is a straight line in time:

$$x_{\mathrm{cl}}(t) = x_i+ \frac{x_f-x_i}{t_f-t_i}(t-t_i)$$

for fixed endpoints $x_i$ and $x_f$.

The classical action along this path is

$$S_{\mathrm{cl}} = \frac{m(x_f-x_i)^2}{2(t_f-t_i)}.$$

This quantity is not just classical bookkeeping. It appears in the phase of the exact free-particle propagator.

Harmonic Oscillator Example

For a harmonic oscillator,

$$L = \frac12m\dot x^2 - \frac12m\omega^2x^2.$$

The partial derivatives are

$$\frac{\partial L}{\partial x} = -m\omega^2x, \qquad \frac{\partial L}{\partial\dot x} = m\dot x.$$

The Euler–Lagrange equation gives

$$m\ddot x+m\omega^2x=0.$$

This is the classical equation whose stable equilibrium becomes the local model behind the Quantum Harmonic Oscillator. Around a stable minimum of a potential, the quadratic approximation to $V(x)$ gives oscillator dynamics.

Curvilinear Example: Central Motion

For a particle in a plane with polar coordinates $(r,\theta)$ and central potential $V(r)$,

$$L = \frac12m \left( \dot r^2+r^2\dot\theta^2 \right) -V(r).$$

The coordinate $\theta$ is cyclic because $L$ does not depend on $\theta$ itself. The Euler–Lagrange equation for $\theta$ gives

$$\frac{d}{dt} \left( \frac{\partial L}{\partial\dot\theta} \right) = 0,$$

or

$$\frac{d}{dt} \left( mr^2\dot\theta \right) = 0.$$

Thus

$$p_\theta = mr^2\dot\theta$$

is conserved. This is angular momentum in planar motion. The example illustrates a general rule: cyclic coordinates produce conserved canonical momenta.

Canonical Momenta

The momentum conjugate to $q^i$ is

$$p_i = \frac{\partial L}{\partial\dot q^i}.$$

This definition agrees with ordinary momentum for simple Cartesian kinetic energy, but it is more general. In polar coordinates, for example,

$$p_\theta=mr^2\dot\theta,$$

not $m\dot\theta$.

When the relation between $\dot q^i$ and $p_i$ can be inverted, one can pass from the Lagrangian to the Hamiltonian by a Legendre transform. The Hamiltonian review page owns that phase-space formulation; here the key point is that Lagrangian mechanics starts from configuration-space paths and their velocities.

Symmetries and Conservation Laws

If a coordinate $q^a$ does not appear explicitly in $L$, then

$$\frac{\partial L}{\partial q^a}=0.$$

The Euler–Lagrange equation gives

$$\frac{d}{dt} \left( \frac{\partial L}{\partial\dot q^a} \right) = 0,$$

so the conjugate momentum $p_a$ is conserved.

This is the simplest Noether-type statement: a continuous symmetry of the action leads to a conserved quantity. The quantum version appears when continuous unitary symmetries have self-adjoint generators, as in Unitary Representations and Generators.

Path-Integral Bridge

In the path-integral representation of a propagator, one writes formally

$$K(q_f,t_f;q_i,t_i) = \int_{q(t_i)=q_i}^{q(t_f)=q_f} \mathcal Dq\, \exp \left( \frac{i}{\hbar}S[q] \right).$$

This expression is not an ordinary integral over a finite-dimensional space. It is a limiting construction or formal representation whose precise meaning depends on context. Its central physics lesson is robust: the action supplies the phase.

When $S$ is large compared with $\hbar$, phases oscillate rapidly. Contributions from nearby paths cancel except near stationary points satisfying

$$\delta S=0.$$

Those stationary paths are classical solutions of the Euler–Lagrange equations. Expanding around them gives the Semiclassical Propagator, fluctuation determinants, and Maslov phases. When the classical action is treated as a function of endpoints, it becomes Hamilton’s principal function; see Hamilton–Jacobi Theory.

For the path-integral construction itself, continue to Why Path Integrals? and Propagators to Path Integrals.

Common Mistakes

• Saying “least action” when the action is only required to be stationary.
• Confusing generalized velocity $\dot q^i$ with ordinary Cartesian velocity components.
• Assuming $L=T-V$ covers every useful Lagrangian.
• Dropping endpoint terms without checking the allowed variations.
• Treating a cyclic coordinate as a coordinate with zero velocity rather than a coordinate absent from $L$.
• Assuming the canonical momentum is always $m\dot q$.
• Forgetting that fixed-endpoint path-integral saddle points are boundary-value solutions, not initial-value trajectories.
• Quantizing a curvilinear-coordinate Hamiltonian without checking measures and operator ordering.

Cross-Links

• Calculus of Variations
• Action Principles
• Hamiltonian Mechanics Review
• Phase Space
• Hamilton–Jacobi Theory
• Functional Derivatives
• Ordinary Differential Equations
• Why Path Integrals?
• Propagators to Path Integrals
• Semiclassical Propagator
• Semiclassical Limit Overview
• Quantum Harmonic Oscillator
• Unitary Representations
• Generators

References

• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
• L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Butterworth-Heinemann, 1976.
• V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, 1989.
• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Derive the Euler–Lagrange equation for the free particle Lagrangian $L=(1/2)m\dot x^2$.

Solution

The derivatives are

$$\frac{\partial L}{\partial x}=0, \qquad \frac{\partial L}{\partial\dot x}=m\dot x.$$

The Euler–Lagrange equation gives

$$\frac{d}{dt}(m\dot x)=0,$$

so

$$m\ddot x=0.$$

2. For $L=(1/2)m\dot x^2-V(x)$, show that the Euler–Lagrange equation gives Newton’s equation.

Solution

Compute

$$\frac{\partial L}{\partial x} = -\frac{dV}{dx}, \qquad \frac{\partial L}{\partial\dot x} = m\dot x.$$

Then

$$\frac{d}{dt} \left( \frac{\partial L}{\partial\dot x} \right) - \frac{\partial L}{\partial x} = 0$$

becomes

$$m\ddot x+\frac{dV}{dx}=0.$$

Thus

$$m\ddot x=-\frac{dV}{dx}.$$

3. A coordinate $q^a$ is cyclic. Prove that its conjugate momentum is conserved.

Solution

Cyclic means

$$\frac{\partial L}{\partial q^a}=0.$$

The Euler–Lagrange equation is

$$\frac{d}{dt} \left( \frac{\partial L}{\partial\dot q^a} \right) - \frac{\partial L}{\partial q^a} = 0.$$

Therefore

$$\frac{d}{dt} \left( \frac{\partial L}{\partial\dot q^a} \right) = 0.$$

Since $p_a=\partial L/\partial\dot q^a$, the conjugate momentum $p_a$ is conserved.

4. Why does the phase $e^{iS/\hbar}$ make stationary action relevant to the semiclassical limit?

Solution

When $S/\hbar$ is large, small changes in the path usually produce rapid changes in the phase. Nearby contributions then cancel by destructive interference. Near a stationary path, the first variation vanishes, so the phase changes only to second order for small variations. These neighborhoods survive the stationary-phase approximation and give the classical paths in the semiclassical propagator.