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PU-204 / L28 · lecture

The Field Lagrangian and Gauge Invariance

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BEFORE YOU START

You should be comfortable with the four-vector formalism, the definition of the electromagnetic field tensor \( F_{\mu\nu} \), and the basic principles of the Euler–Lagrange equations for fields.

Why this matters

Gauge invariance is not merely a mathematical curiosity; it is the fundamental symmetry that dictates the structure of electromagnetism.

In previous lectures, we viewed Maxwell’s equations as empirical laws governing electric and magnetic fields. Here, we elevate our perspective to the Lagrangian formulation, where the dynamics of the field are derived from a single scalar quantity: the action. This approach reveals that the requirement of local gauge invariance—our freedom to redefine the potential without altering the physical fields—strictly enforces the conservation of electric charge. It unifies the source-free and source-dependent Maxwell equations into a coherent relativistic framework, demonstrating that the very existence of the photon as a massless spin-1 particle is a direct consequence of this symmetry.

The Principle of Least Action for Fields

Classical mechanics is built on the principle of stationary action. For a single particle, we minimise the action \( S = \int L \, dt \). For a field, which is a function of space and time, we generalise this to the action \( S = \int \mathcal{L} \, d^4x \), where \( \mathcal{L} \) is the Lagrangian density. The physical state of the electromagnetic field is the configuration that makes this integral stationary. This is a powerful unification: instead of solving four coupled differential equations (Maxwell’s equations) directly, we define a single density function. The equations of motion then emerge naturally as the condition for the action to be invariant under small variations of the field. This variational principle ensures that the resulting laws are automatically consistent with special relativity, as the action is a Lorentz scalar.

Constructing the Gauge-Invariant Lagrangian

To construct \( \mathcal{L} \), we must respect two constraints: Lorentz invariance and gauge invariance. The field strength tensor \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \) is the only object built from the potential \( A_\mu \) that is invariant under the gauge transformation \( A_\mu \to A_\mu + \partial_\mu \chi \). The simplest Lorentz scalar formed from \( F_{\mu\nu} \) is \( F_{\mu\nu}F^{\mu\nu} \). Thus, the free-field part of the Lagrangian density is proportional to \( -\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} \). The interaction with matter is described by the coupling term \( -J^\mu A_\mu \), where \( J^\mu \) is the four-current. The total Lagrangian density is therefore:

\[ \mathcal{L} = -\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} - J^\mu A_\mu \]

Notice that \( \mathcal{L} \) depends on \( A_\mu \) and its first derivatives. This structure is crucial because it allows us to apply the Euler–Lagrange equation for fields, \( \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\nu)} \right) - \frac{\partial \mathcal{L}}{\partial A_\nu} = 0 \).

Recovering Maxwell’s Equations

Applying the Euler–Lagrange equation to our Lagrangian yields the inhomogeneous Maxwell equations. The term \( \frac{\partial \mathcal{L}}{\partial A_\nu} \) is simply \( -J^\nu \). The term involving the derivative requires careful index manipulation. Since \( F_{\mu\nu} \) is antisymmetric, the variation leads to \( \partial_\mu F^{\mu\nu} \). Setting the variation to zero gives:

\[ \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \]

This single tensor equation encapsulates both Gauss’s law and the Ampère–Maxwell law. The homogeneous equations, \( \nabla \cdot \mathbf{B} = 0 \) and \( \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0 \), are not derived from the action but are identities satisfied by the definition of \( F_{\mu\nu} \) in terms of \( A_\mu \). Specifically, the Bianchi identity \( \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \) holds automatically because \( F_{\mu\nu} \) is defined as a curl of the potential. This distinction is vital: the inhomogeneous equations are dynamical laws, while the homogeneous equations are kinematic constraints.

Gauge Invariance and Charge Conservation

The most profound consequence of this formulation is the link between symmetry and conservation. Consider the action \( S = \int \mathcal{L} \, d^4x \). Under a local gauge transformation \( A_\mu \to A_\mu + \partial_\mu \chi \), the field strength \( F_{\mu\nu} \) remains unchanged. However, the interaction term changes by \( -\int J^\mu \partial_\mu \chi \, d^4x \). Integrating by parts, this becomes \( +\int (\partial_\mu J^\mu) \chi \, d^4x \) (ignoring boundary terms). For the action to be invariant for arbitrary \( \chi \), we must have \( \partial_\mu J^\mu = 0 \). This is the continuity equation. Thus, local gauge invariance of the action is equivalent to the local conservation of electric charge. This is a specific instance of Noether’s second theorem, which states that local symmetries imply identities among the equations of motion.

The derivation

◆ DERIVATION EMBED · D-177 · not re-derived here

Field Lagrangian and Gauge Invariance

Starting from the covariant field Lagrangian density \( \mathcal{L} = -\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} - J^\mu A_\mu \), with \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \), we derive the inhomogeneous Maxwell equations \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) as the Euler–Lagrange field equations, recover the homogeneous pair as an identity of the potential formulation, and prove that invariance of the action under the local gauge transformation \( A_\mu \to A_\mu + \partial_\mu \chi \) holds if and only if the coupled current is conserved, \( \partial_\mu J^\mu = 0 \).

\[ \partial_\mu F^{\mu\nu} = \mu_0 J^\nu, \qquad \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0, \qquad \partial_\mu J^\mu = 0 \]
Open full derivation →
Tier 3 · watch out

A common misconception is that the Lagrangian density itself is gauge invariant. The free term \( -F_{\mu\nu}F^{\mu\nu}/4\mu_0 \) is, but the coupling \( -J^\mu A_\mu \) changes by \( -J^\mu\partial_\mu\chi \), which equals the total derivative \( -\partial_\mu(J^\mu\chi) \) only when \( \partial_\mu J^\mu = 0 \). So \( \mathcal{L} \) is invariant only up to a total derivative, which leaves the action and the field equations unchanged, and this works only for a conserved current. C

Worked example

Problem. Consider a static, spherically symmetric charge distribution with total charge \( Q \). Using the Lagrangian formulation, verify that the resulting electric field satisfies Gauss’s law and calculate the electrostatic energy stored in the field outside a sphere of radius \( R \).

  1. For a static charge, \( J^0 = c\rho \) and \( \mathbf{J} = 0 \). The relevant component of the Euler–Lagrange equation is \( \nu = 0 \): \( \partial_\mu F^{\mu 0} = \mu_0 J^0 \). In 3D vector notation, this is \( \nabla \cdot \mathbf{E} = \rho / \epsilon_0 \).
  2. For a point charge \( Q \) at the origin, \( \rho = Q \delta^3(\mathbf{r}) \). The solution to Gauss’s law is \( \mathbf{E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{\mathbf{r}} \).
  3. The energy density of the electric field is \( u_E = \frac{1}{2} \epsilon_0 E^2 \). The total energy outside radius \( R \) is \( U = \int_R^\infty u_E \, dV \).
  4. Substitute \( E^2 = \frac{Q^2}{16\pi^2 \epsilon_0^2 r^4} \) and \( dV = 4\pi r^2 dr \). The integral becomes \( U = \int_R^\infty \frac{1}{2} \epsilon_0 \frac{Q^2}{16\pi^2 \epsilon_0^2 r^4} 4\pi r^2 dr = \frac{Q^2}{8\pi \epsilon_0} \int_R^\infty \frac{dr}{r^2} \).
  5. Evaluate the integral: \( \int_R^\infty r^{-2} dr = \left[ -\frac{1}{r} \right]_R^\infty = \frac{1}{R} \).
\[ U = \frac{Q^2}{8\pi \epsilon_0 R} \]

Units check. \( Q^2 / (\epsilon_0 R) \) has units \( \mathrm{C^2 / (F/m \cdot m)} = \mathrm{C^2 / F} = \mathrm{J} \). The result is consistent with the standard expression for electrostatic self-energy.

For scale: a charge \(Q = 1.0\ \mathrm{nC}\) spread over a sphere of radius \(R = 0.10\ \mathrm{m}\) stores \(U = Q^2/(8\pi\varepsilon_0 R) = 8.99 \times 10^{9} \times (1.0 \times 10^{-9})^2/(2 \times 0.10) = 4.5 \times 10^{-8}\ \mathrm{J}\) in the field outside it.

CHECK YOURSELF

Three quick tests

  1. Why is the term \( F_{\mu\nu}F^{\mu\nu} \) preferred over \( A_\mu A^\mu \) in the Lagrangian?
    Answer\( A_\mu \) is not gauge invariant, whereas \( F_{\mu\nu} \) is. The Lagrangian must be constructed from gauge-invariant quantities to ensure the physics is independent of the potential choice.
  2. What is the physical significance of the Bianchi identity \( \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \)?
    AnswerIt encodes the homogeneous Maxwell equations (\( \nabla \cdot \mathbf{B} = 0 \) and \( \nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t \)), which are kinematic identities arising from the definition of \( F_{\mu\nu} \) in terms of the potential \( A_\mu \).
  3. If the current \( J^\mu \) were not conserved, what would happen to the gauge invariance of the action?
    AnswerThe action would change by a term proportional to \( \int (\partial_\mu J^\mu) \chi \, d^4x \), which is non-zero for arbitrary \( \chi \). Thus, local gauge invariance would be broken, and the theory would lose its consistency with the principle of least action.
PROBLEMS

Problem set for PU-204 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-177 · Threads: force, energy, light