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Classical Field Theory | Physics

sources:

  • text: “Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). Wiley.”

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

S=L(ϕ,μϕ)d4x,δS=0    LϕμL(μϕ)=0S = \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 mm:

L=12(μϕ)(μϕ)12m2ϕ2\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}m^2\phi^2

The equation of motion: (2+m2)ϕ=0(\Box^2 + m^2)\phi = 0 where 2=μμ=22/t2\Box^2 = \partial_\mu\partial^\mu = \nabla^2 - \partial^2/\partial t^2.

Plane wave solutions: ϕei(krωt)\phi \propto e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} with ω2=k2+m2\omega^2 = k^2 + m^2 (dispersion relation).

Every continuous symmetry of the action yields a conserved current:

μjμ=0    Q=j0d3x=const\partial_\mu j^\mu = 0 \implies Q = \int j^0\,d^3x = \text{const}

SymmetryConserved Quantity
Time translationEnergy E=Hd3xE = \int\mathcal{H}\,d^3x
Space translationMomentum P=πd3x\mathbf{P} = \int\boldsymbol{\pi}\,d^3x
RotationAngular momentum L=r×πd3x\mathbf{L} = \int\mathbf{r}\times\boldsymbol{\pi}\,d^3x
Phase rotation (ϕeiαϕ\phi \to e^{i\alpha}\phi)Charge QQ

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

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

With conserved charge Q=(iϕϕ˙ϕϕ˙)d3xQ = \int(i\phi^*\dot{\phi} - \phi\dot{\phi}^*)\,d^3x.

12.4 Hamiltonian Density and Energy-Momentum Tensor

Section titled “12.4 Hamiltonian Density and Energy-Momentum Tensor”

The Hamiltonian density:

H=Lϕ˙ϕ˙L=12ϕ˙2+12(ϕ)2+12m2ϕ2\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):

Tμν=L(μϕ)νϕgμνLT^{\mu\nu} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi - g^{\mu\nu}\mathcal{L}

T00=HT^{00} = \mathcal{H} (energy density), T0iT^{0i} (momentum density), TijT^{ij} (stress tensor).

Worked Example 12.1: Noether Current for the Klein--Gordon Field

Consider the infinitesimal phase transformation ϕϕ+δϕ\phi \to \phi + \delta\phi where δϕ=iϵϕ\delta\phi = i\epsilon\phi (a global U(1) transformation).

The change in the Lagrangian density:

δL=Lϕδϕ+L(μϕ)μ(δϕ)=iϵ[Lϕϕμ ⁣(L(μϕ)ϕ)]\delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta\phi) = i\epsilon\left[\frac{\partial\mathcal{L}}{\partial\phi}\phi - \partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\phi\right)\right]

Using the E-L equation L/ϕ=μ(L/(μϕ))\partial\mathcal{L}/\partial\phi = \partial_\mu(\partial\mathcal{L}/\partial(\partial_\mu\phi)):

δL=iϵμ ⁣(L(μϕ)ϕ)=μ(ϵjμ)\delta\mathcal{L} = -i\epsilon\,\partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\phi\right) = \partial_\mu(-\epsilon\,j^\mu)

Where jμ=i[ϕ(μϕ)(μϕ)ϕ]j^\mu = i[\phi^*(\partial^\mu\phi) - (\partial^\mu\phi^*)\phi] (using the complex Klein—Gordon Lagrangian for generality).

By Noether’s theorem: μjμ=0\partial_\mu j^\mu = 0 And the conserved charge:

Q=j0d3x=i(ϕϕ˙ϕ˙ϕ)d3xQ = \int j^0\,d^3x = \int i(\phi^*\dot{\phi} - \dot{\phi}^*\phi)\,d^3x

For a plane wave ϕ=eiωt+ikr\phi = e^{-i\omega t + i\mathbf{k}\cdot\mathbf{r}}: Q2ω>0Q \propto 2\omega > 0 (positive frequency modes have positive charge).

Example 1: Lagrangian of a simple pendulum

Section titled “Example 1: Lagrangian of a simple pendulum”

Problem. Derive the equation of motion for a simple pendulum of length ll and mass mm.

Solution. T=12ml2θ˙2T = \frac{1}{2}ml^2\dot{\theta}^2, V=mglcosθV = -mgl\cos\theta (taking pivot as reference). L=TV=12ml2θ˙2+mglcosθL = T - V = \frac{1}{2}ml^2\dot{\theta}^2 + mgl\cos\theta.

Lθ˙=ml2θ˙\frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta}, ddtLθ˙=ml2θ¨\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} = ml^2\ddot{\theta}, Lθ=mglsinθ\frac{\partial L}{\partial \theta} = -mgl\sin\theta.

ml2θ¨+mglsinθ=0    θ¨+glsinθ=0ml^2\ddot{\theta} + mgl\sin\theta = 0 \implies \ddot{\theta} + \frac{g}{l}\sin\theta = 0.

\blacksquare

Problem. For a 1D harmonic oscillator (L=12mx˙212kx2L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2), find Hamilton’s equations.

Solution. p=Lx˙=mx˙p = \frac{\partial L}{\partial \dot{x}} = m\dot{x}, so H=px˙L=p22m+12kx2H = p\dot{x} - L = \frac{p^2}{2m} + \frac{1}{2}kx^2.

x˙=Hp=pm\dot{x} = \frac{\partial H}{\partial p} = \frac{p}{m}, p˙=Hx=kx\dot{p} = -\frac{\partial H}{\partial x} = -kx.

\blacksquare

  • Confusing generalised coordinates and Cartesian coordinates. Generalised coordinates (q1,,qnq_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. ddtLq˙iLqi=0\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0; the total time derivative is applied to L/q˙i\partial L/\partial \dot{q}_i, not to LL itself. Fix: Compute L/q˙i\partial L/\partial \dot{q}_i first, then take d/dtd/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.
flowchart TD
A[13_Classical Field Theory] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]
  • Newton’s laws (vector approach) vs Lagrangian mechanics (scalar, energy-based approach).
  • Euler-Lagrange equation: ddtLq˙iLqi=0\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0.
  • Hamiltonian: H=piq˙iLH = \sum p_i \dot{q}_i - L; Hamilton’s equations give 2n2n first-order ODEs.
  • Conservation laws follow from symmetries via Noether’s theorem.

Classical field theory extends mechanics from particles to continuous media. A field assigns a value to every point in space, like temperature in a room or displacement in a vibrating string. The Lagrangian density replaces the Lagrangian, integrating over space gives the total Lagrangian. Noether’s theorem connects symmetries to conservation laws: time translation symmetry gives energy conservation, spatial translation gives momentum conservation. The wave equation emerges as the simplest field theory, describing how disturbances propagate through a medium at a finite speed.

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