Classical Field Theory

For a field $\phi(\mathbf{r}, t)$The Lagrangian density $\mathcal{L}$ replaces the discrete Lagrangian $L = \sum_i T_i - V_i$:

$S = \int \mathcal{L}(\phi, \partial_\mu\phi)\,d^4x, \quad \delta S = 0 \implies \frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = 0$

This is the Euler—Lagrange equation for fields.

A real scalar field of mass $m$:

$\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}m^2\phi^2$

The equation of motion: $(\Box^2 + m^2)\phi = 0$ where $\Box^2 = \partial_\mu\partial^\mu = \nabla^2 - \partial^2/\partial t^2$.

Plane wave solutions: $\phi \propto e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}$ with $\omega^2 = k^2 + m^2$ (dispersion relation).

Every continuous symmetry of the action yields a conserved current:

$\partial_\mu j^\mu = 0 \implies Q = \int j^0\,d^3x = \text{const}$

For the complex Klein—Gordon field, the conserved current is:

$j^\mu = i(\phi^*\partial^\mu\phi - \phi\partial^\mu\phi^*)$

With conserved charge $Q = \int(i\phi^*\dot{\phi} - \phi\dot{\phi}^*)\,d^3x$.

The Hamiltonian density:

$\mathcal{H} = \frac{\partial\mathcal{L}}{\partial\dot{\phi}}\dot{\phi} - \mathcal{L} = \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2$

The canonical energy-momentum tensor (symmetric, Belinfante):

$T^{\mu\nu} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi - g^{\mu\nu}\mathcal{L}$

$T^{00} = \mathcal{H}$ (energy density), $T^{0i}$ (momentum density), $T^{ij}$ (stress tensor).

Problem. Derive the equation of motion for a simple pendulum of length $l$ and mass $m$.

Solution. $T = \frac{1}{2}ml^2\dot{\theta}^2$, $V = -mgl\cos\theta$ (taking pivot as reference). $L = T - V = \frac{1}{2}ml^2\dot{\theta}^2 + mgl\cos\theta$.

$\frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta}$, $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} = ml^2\ddot{\theta}$, $\frac{\partial L}{\partial \theta} = -mgl\sin\theta$.

$ml^2\ddot{\theta} + mgl\sin\theta = 0 \implies \ddot{\theta} + \frac{g}{l}\sin\theta = 0$.

$\blacksquare$

Problem. For a 1D harmonic oscillator ($L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2$), find Hamilton’s equations.

Solution. $p = \frac{\partial L}{\partial \dot{x}} = m\dot{x}$, so $H = p\dot{x} - L = \frac{p^2}{2m} + \frac{1}{2}kx^2$.

$\dot{x} = \frac{\partial H}{\partial p} = \frac{p}{m}$, $\dot{p} = -\frac{\partial H}{\partial x} = -kx$.

$\blacksquare$

• Confusing generalised coordinates and Cartesian coordinates. Generalised coordinates ($q_1, \ldots, q_n$) may not have dimensions of length. Fix: The Lagrangian formalism works with any set of independent coordinates.
• Wrong Euler-Lagrange equation. $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0$; the total time derivative is applied to $\partial L/\partial \dot{q}_i$, not to $L$ itself. Fix: Compute $\partial L/\partial \dot{q}_i$ first, then take $d/dt$.
• Ignoring constraints in Lagrangian mechanics. Holonomic constraints reduce degrees of freedom; non-holonomic constraints require Lagrange multipliers. Fix: For holonomic constraints, express the system in terms of independent generalised coordinates.

• Newton’s laws (vector approach) vs Lagrangian mechanics (scalar, energy-based approach).
• Euler-Lagrange equation: $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0$.
• Hamiltonian: $H = \sum p_i \dot{q}_i - L$; Hamilton’s equations give $2n$ first-order ODEs.
• Conservation laws follow from symmetries via Noether’s theorem.