Quantum Field Theory Course

Total: 440+ pages covering classical field theory through the Standard Model

Part I, Chapter 1 | Page 6 of 8

The Electromagnetic Field

Maxwell's equations from a Lagrangian and gauge invariance

1.19 The Electromagnetic 4-Potential

The electromagnetic field is described by the 4-vector potential:

$$A^\mu = (\phi, \vec{A})$$

where φ is the scalar potential and A is the vector potential. The electric and magnetic fields are:

$$\vec{E} = -\nabla \phi - \frac{\partial \vec{A}}{\partial t}, \quad \vec{B} = \nabla \times \vec{A}$$

Field Strength Tensor

The electromagnetic field strength tensor Fμν is defined as:

$$\boxed{F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu}$$

This is antisymmetric: Fμν = -Fνμ. In matrix form:

$$F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}$$

The tensor Fμν encodes both E and B fields in a covariant way.

1.20 The Maxwell Lagrangian

The Lagrangian for the free electromagnetic field is:

$$\boxed{\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}}$$

Expanding the field strength:

$$F_{\mu\nu}F^{\mu\nu} = 2(\vec{B}^2 - \vec{E}^2)$$

Therefore:

$$\boxed{\mathcal{L} = \frac{1}{2}(\vec{E}^2 - \vec{B}^2)}$$

This is exactly the classical electromagnetic Lagrangian density! Note the relative sign: electric field energy is positive, magnetic field energy enters with opposite sign in the Lagrangian (but both contribute positively to the Hamiltonian).

With Sources

Including a 4-current Jμ = (ρ, j), the Lagrangian becomes:

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - J_\mu A^\mu$$

The second term is the interaction between the field and the sources:

$$J_\mu A^\mu = -\rho \phi + \vec{j} \cdot \vec{A}$$

1.21 Deriving Maxwell's Equations

Apply the Euler-Lagrange equations to Aμ. The Lagrangian is:

$$\mathcal{L} = -\frac{1}{4}(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu) - J_\mu A^\mu$$

Computing the variational derivative:

$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\nu)} = -F^{\mu\nu}$$
$$\frac{\partial \mathcal{L}}{\partial A_\nu} = -J_\nu$$

The Euler-Lagrange equation:

$$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) - \frac{\partial \mathcal{L}}{\partial A_\nu} = 0$$

gives:

$$\boxed{\partial_\mu F^{\mu\nu} = J^\nu}$$

These are Maxwell's equations in covariant form! Explicitly, ν = 0 gives:

$$\nabla \cdot \vec{E} = \rho \quad \text{(Gauss's law)}$$

and ν = i gives:

$$\nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{j} \quad \text{(Ampère-Maxwell law)}$$

The other two Maxwell equations (no magnetic monopoles, Faraday's law) are automatically satisfied by the definition Fμν = ∂μAν - ∂νAμ. They can be written as:

$$\partial_\mu \tilde{F}^{\mu\nu} = 0$$

where $\tilde{F}^{\mu\nu}$ is the dual tensor.

1.22 Gauge Invariance

The electromagnetic field has a profound symmetry: gauge invariance.

Gauge Transformation

The physics is unchanged under the transformation:

$$\boxed{A^\mu \to A^\mu + \partial^\mu \Lambda}$$

where Λ(x) is an arbitrary scalar function. This is called a gauge transformation. Under this transformation:

$$F^{\mu\nu} \to F^{\mu\nu} + \partial^\mu \partial^\nu \Lambda - \partial^\nu \partial^\mu \Lambda = F^{\mu\nu}$$

since partial derivatives commute. Therefore Fμν (and hence E and B) are gauge invariant.

Physical Consequences

  • The potential Aμ has redundant degrees of freedom
  • Only gauge-invariant quantities are observable
  • Different gauges correspond to different coordinate choices in field space
  • Common gauge choices:
    • Lorenz gauge: ∂μAμ = 0
    • Coulomb gauge: ∇·A = 0
    • Temporal gauge: A0 = 0

Gauge invariance is the prototype for all gauge theories, including the Standard Model of particle physics (QED, QCD, electroweak theory).

1.23 Energy-Momentum Tensor for EM Field

The canonical energy-momentum tensor for the electromagnetic field is:

$$T^{\mu\nu}_{\text{can}} = -F^{\mu\lambda}\partial^\nu A_\lambda + \frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$$

However, this tensor is not gauge invariant and not symmetric. The correct, physical tensor is the symmetric energy-momentum tensor:

$$\boxed{T^{\mu\nu} = -F^{\mu\lambda}F^\nu_{\phantom{\nu}\lambda} + \frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}}$$

This is gauge invariant and symmetric: Tμν = Tνμ.

Energy and Momentum Densities

The energy density is:

$$T^{00} = \frac{1}{2}(\vec{E}^2 + \vec{B}^2)$$

The momentum density (Poynting vector / c2):

$$T^{0i} = (\vec{E} \times \vec{B})^i = S^i$$

The stress tensor:

$$T^{ij} = -E^iE^j - B^iB^j + \frac{1}{2}\delta^{ij}(\vec{E}^2 + \vec{B}^2)$$

This is the Maxwell stress tensor, well-known from classical electrodynamics.

Key Concepts (Page 6)

  • • Field strength tensor: Fμν = ∂μAν - ∂νAμ
  • • Maxwell Lagrangian: $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$
  • • Maxwell equations: ∂μFμν = Jν
  • • Gauge invariance: Aμ → Aμ + ∂μΛ
  • • Energy density: $\frac{1}{2}(\vec{E}^2 + \vec{B}^2)$
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