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PU-105 / L26 · lecture

Maxwell's Equations in Covariant Form

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the definition of the electromagnetic field tensor \( F^{\mu\nu} \) and the four-current \( J^\nu \), as introduced in the previous lecture on the relativistic origin of magnetic force.

Why this matters

Maxwell's equations are the first physical laws to be fully consistent with special relativity.

For decades, physicists struggled to reconcile the fixed speed of light in Maxwell's equations with the variable speeds of Galilean mechanics. The resolution was not to modify Maxwell's equations, but to modify our understanding of space and time. By expressing electromagnetism in the language of four-vectors and tensors, we reveal that electric and magnetic fields are not separate entities, but different components of a single geometric object. This covariant formulation ensures that the laws of physics hold true in every inertial frame, eliminating the need for ad-hoc transformation rules for fields.

The Geometric Unification

In classical electrodynamics, we treat the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \) as distinct vector fields. However, an observer moving relative to a charge will measure a different mix of electric and magnetic effects than a stationary observer. This suggests that \( \mathbf{E} \) and \( \mathbf{B} \) are not fundamental, independent quantities, but rather projections of a deeper structure onto different spatial and temporal axes.

Special relativity provides the framework for this unification. We package the six components of the electric and magnetic fields into a single antisymmetric object called the electromagnetic field tensor, \( F^{\mu\nu} \). Similarly, the charge density \( \rho \) and current density \( \mathbf{J} \) are combined into the four-current \( J^\nu \). The beauty of this approach is that the four separate Maxwell equations in three-dimensional space and time collapse into just two compact tensor equations. These equations are "covariant," meaning they maintain the same mathematical form under Lorentz transformations. If the equation is true in one inertial frame, the tensor nature of the objects involved guarantees it is true in all inertial frames.

The Inhomogeneous Equations

The inhomogeneous Maxwell equations describe how sources (charges and currents) generate fields. In standard vector calculus, these are Gauss's law and the Ampère-Maxwell law. In covariant form, they are combined into a single equation involving the divergence of the field tensor.

The equation is given by:

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

Here, \( \partial_\mu \) is the four-gradient operator, defined as \( (\frac{1}{c}\frac{\partial}{\partial t}, \nabla) \). The index \( \nu \) runs from 0 to 3. When \( \nu = 0 \), this equation yields Gauss's law, \( \nabla \cdot \mathbf{E} = \rho / \epsilon_0 \). When \( \nu = 1, 2, 3 \), it yields the Ampère-Maxwell law, \( \nabla \times \mathbf{B} - \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t} = \mu_0 \mathbf{J} \). The constant \( \mu_0 \) appears here to maintain consistency with SI units, where the four-current \( J^\nu \) is typically defined as \( (c\rho, \mathbf{J}) \).

This single tensor equation encapsulates the source-field relationship. Because \( F^{\mu\nu} \) and \( J^\nu \) are tensors, the equation is automatically Lorentz invariant. We do not need to derive how \( \mathbf{E} \) and \( \mathbf{B} \) transform individually; the tensor transformation law handles it all at once.

The Homogeneous Equations

The homogeneous Maxwell equations describe the intrinsic properties of the fields in the absence of sources. These are Faraday's law of induction and the absence of magnetic monopoles (Gauss's law for magnetism).

In covariant form, these are expressed using the dual field tensor \( \star F^{\mu\nu} \), or more elegantly, using the antisymmetrized derivative of the field tensor itself. The equation is:

\[ \partial_{[\alpha} F_{\beta\gamma]} = 0 \]

Here, the square brackets denote antisymmetrization over the indices \( \alpha, \beta, \gamma \). This means the sum over all permutations of the indices, with a sign change for each swap. This single equation is equivalent to the two familiar vector equations: \( \nabla \cdot \mathbf{B} = 0 \) and \( \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0 \).

The physical significance of this equation is profound. It states that the electromagnetic field is "source-free" in the sense that it has no magnetic charges. Mathematically, it implies that the field tensor \( F_{\mu\nu} \) can be derived from a four-potential \( A_\mu \) via \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \). This potential formulation is crucial for quantum electrodynamics, where the photon is the quantum of the potential field.

Charge Conservation as a Consequence

One of the most elegant results of the covariant formulation is that charge conservation is not an independent assumption, but a direct mathematical consequence of the field equations. In standard electrodynamics, we often state the continuity equation \( \nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0 \) as a separate law required for consistency.

In the tensor formalism, we can derive this immediately. Take the divergence of the inhomogeneous equation with respect to the index \( \nu \):

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

On the left-hand side, we have the contraction of two partial derivatives, which are symmetric under the exchange of indices \( \mu \) and \( \nu \), with the field tensor \( F^{\mu\nu} \), which is antisymmetric. The contraction of a symmetric tensor with an antisymmetric tensor is always zero. Therefore, the left-hand side vanishes identically.

This leaves us with:

\[ \partial_\nu J^\nu = 0 \]

Expanding this four-divergence gives \( \frac{1}{c}\frac{\partial}{\partial t}(c\rho) + \nabla \cdot \mathbf{J} = 0 \), which simplifies to the standard continuity equation. This demonstrates that local charge conservation is built into the geometric structure of spacetime and the definition of the electromagnetic field.

The derivation

◆ DERIVATION EMBED · D-106 · not re-derived here

Covariant Form of Maxwell's Equations

The four Maxwell equations, together with the Lorentz force and charge conservation, are equivalent to two four-tensor equations: the inhomogeneous pair ∂μFμν = μ0Jν, and the homogeneous pair ∂μ(⋆F)μν = 0 (equivalently ∂[αFβγ] = 0). Here Fμν is the electromagnetic field tensor built from E and B, Jν = (cρ, J) is the four-current, and ∂μ = (∂ct, ∇). Both equations transform as tensors under the Lorentz group, so their validity in one inertial frame guarantees validity in every inertial frame.

∂μFμν = μ0Jν       ∂[αFβγ] = 0  (⟺ ∂μ(⋆F)μν = 0)
Open full derivation →
Tier 3 · watch out

A common source of confusion is the placement of indices and the metric signature. In this lecture, we use the metric signature \( (+, -, -, -) \). If you switch to the \( (-, +, +, +) \) signature, the signs in the field tensor components and the definition of the dual tensor will change. Always verify the metric convention before comparing your tensor components with standard vector forms. The physical predictions remain identical, but the algebraic signs in intermediate steps can differ significantly. C

Worked example

Problem. Consider a frame \( S' \) moving with velocity \( v = 0.6c \) relative to frame \( S \) along the \( x \)-axis. In frame \( S \), there is a uniform electric field \( E_y = 100\ \mathrm{V/m} \) and no magnetic field (\( \mathbf{B} = 0 \)). Calculate the electric and magnetic field components in frame \( S' \).

  1. Identify the non-zero components of the field tensor \( F^{\mu\nu} \) in frame \( S \). Since \( E_y \neq 0 \) and \( \mathbf{B} = 0 \), the relevant components are \( F^{01} = -E_x/c = 0 \), \( F^{02} = -E_y/c = -100/c \), and \( F^{03} = -E_z/c = 0 \). The magnetic components \( F^{ij} \) are all zero.
  2. Apply the Lorentz transformation to the tensor components. For a boost along \( x \), the transformation matrix \( \Lambda^\mu_{\ \nu} \) has \( \Lambda^0_{\ 0} = \gamma \), \( \Lambda^0_{\ 1} = -\gamma \beta \), \( \Lambda^1_{\ 0} = -\gamma \beta \), \( \Lambda^1_{\ 1} = \gamma \), and \( \Lambda^2_{\ 2} = 1 \), where \( \beta = v/c = 0.6 \) and \( \gamma = 1/\sqrt{1-\beta^2} = 1/\sqrt{1-0.36} = 1/0.8 = 1.25 \).
  3. Calculate the transformed component \( F'^{02} \). Using \( F'^{\mu\nu} = \Lambda^\mu_{\ \alpha} \Lambda^\nu_{\ \beta} F^{\alpha\beta} \), we focus on \( F'^{02} = \Lambda^0_{\ 0} \Lambda^2_{\ 2} F^{02} + \Lambda^0_{\ 1} \Lambda^2_{\ 2} F^{12} \). Since \( F^{12} = 0 \) (no \( B_z \)), we have \( F'^{02} = \gamma (1) (-100/c) = -125/c \).
  4. Transformed magnetic component: \( F'^{12} = \Lambda^1_{\ 0} \Lambda^2_{\ 2} F^{02} + \Lambda^1_{\ 1} \Lambda^2_{\ 2} F^{12} = (-\gamma\beta)(1)(-100/c) + 0 = 75/c \). With the SI convention \( F^{12} = -B_z \), this gives \( B'_z = -75/c \approx -2.5 \times 10^{-7}\ \mathrm{T} \), the same as the direct transformation \( B'_z = \gamma(B_z - vE_y/c^2) = -\gamma vE_y/c^2 \).
\[ E'_y = 125\ \mathrm{V/m}, \quad B'_z = -\frac{75}{c}\ \mathrm{T} \]

Units check. The electric field is in \( \mathrm{V/m} \). The magnetic field term \( \frac{v E}{c^2} \) has units \( \frac{(\mathrm{m/s})(\mathrm{V/m})}{(\mathrm{m/s})^2} = \frac{\mathrm{V}}{\mathrm{m/s}} = \frac{\mathrm{V \cdot s}}{\mathrm{m}} = \mathrm{Wb/m^2} = \mathrm{T} \). The units are consistent.

CHECK YOURSELF

Three quick tests

  1. Why does the contraction \( \partial_\nu \partial_\mu F^{\mu\nu} \) vanish?
    AnswerBecause \( \partial_\nu \partial_\mu \) is symmetric in \( \mu, \nu \) while \( F^{\mu\nu} \) is antisymmetric.
  2. What physical law does the equation \( \partial_{[\alpha} F_{\beta\gamma]} = 0 \) represent?
    AnswerFaraday's law and the absence of magnetic monopoles.
  3. If \( \mathbf{E} = 0 \) and \( \mathbf{B} \neq 0 \) in one frame, can an observer find a frame where \( \mathbf{B} = 0 \)?
    AnswerNo, if \( \mathbf{E} \cdot \mathbf{B} = 0 \) and \( B^2 - E^2/c^2 > 0 \), the field is magnetic-like and cannot be transformed away.
PROBLEMS

Problem set for PU-105 →

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

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-106 · Threads: force, energy, light