The Electromagnetic Field Tensor
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You should be comfortable with the Lorentz transformation, the concept of four-vectors, and the basic definitions of electric and magnetic fields from classical electromagnetism.
Why this matters
Electric and magnetic fields are not separate entities but different facets of a single geometric object.
In classical physics, we treat the electric field \(\mathbf{E}\) and the magnetic field \(\mathbf{B}\) as distinct vector fields. However, special relativity reveals that what one observer identifies as a purely electric field, another moving observer may perceive as a mixture of electric and magnetic components. This unification is elegantly captured by the electromagnetic field tensor, a rank-2 antisymmetric tensor that encodes all six field components into a single covariant object. Understanding this tensor is essential for formulating Maxwell's equations in a way that is invariant under Lorentz transformations, thereby revealing the deep geometric structure of spacetime.
The Geometric Unification
Imagine two observers, Alice and Bob, moving relative to each other. Alice measures a static charge at rest, producing only an electric field. Bob, moving past the charge, sees the charge in motion, which constitutes a current. A moving charge generates a magnetic field. Thus, Bob detects both electric and magnetic fields, while Alice detects only an electric one. Neither is "wrong"; they are simply slicing the four-dimensional spacetime differently. The electromagnetic field tensor \(F^{\mu\nu}\) is the mathematical object that contains the full information of the electromagnetic field in spacetime, independent of any specific observer's frame. It is antisymmetric, meaning \(F^{\mu\nu} = -F^{\nu\mu}\), which ensures that the diagonal components are zero and that the tensor has exactly six independent components—matching the three components of \(\mathbf{E}\) and the three of \(\mathbf{B}\).
Constructing the Tensor
To construct \(F^{\mu\nu}\), we must arrange the electric and magnetic field components into a \(4 \times 4\) matrix that transforms correctly under Lorentz boosts. The standard convention places the electric field components in the first row and column (indices involving time, \(\mu=0\)) and the magnetic field components in the spatial block (indices \(\mu, \nu \in \{1,2,3\}\)). The sign conventions depend on the metric signature, but for the \((+,-,-,-)\) metric, the tensor is defined as:
Here, \(c\) is the speed of light. Notice the factors of \(1/c\) multiplying the electric field components. This ensures that \(F^{\mu\nu}\) has consistent dimensions, as the magnetic field \(B\) has units of \(\mathrm{T}\) (tesla) and \(E/c\) also has units of \(\mathrm{T}\) since \(E\) is in \(\mathrm{V/m}\) and \(c\) in \(\mathrm{m/s}\). The antisymmetry is evident: the upper triangle is the negative of the lower triangle.
Tensor Transformation Laws
Because \(F^{\mu\nu}\) is a tensor, it transforms between inertial frames using the Lorentz transformation matrix \(\Lambda^\mu_{\ \alpha}\). If frame \(S'\) moves with velocity \(v\) along the \(x\)-axis relative to frame \(S\), the components in \(S'\) are given by:
Applying this to the components involving the \(y\)- and \(z\)-directions (perpendicular to the boost) yields the familiar field transformation equations. For example, the electric field component \(E'_y\) in the moving frame is:
and the magnetic field component \(B'_z\) is:
These equations demonstrate that a pure electric field in one frame (\(B=0\)) appears as a mixed field in another. The tensor formalism guarantees that these transformations are consistent and preserve the invariants of the electromagnetic field, such as \(E^2 - c^2 B^2\) and \(\mathbf{E} \cdot \mathbf{B}\).
Covariant Maxwell's Equations
The most powerful application of the field tensor is in writing Maxwell's equations in a compact, covariant form. The inhomogeneous equations (Gauss's law and Ampère's law) can be written as the divergence of the field tensor equaling the current four-vector \(J^\mu\):
Here, \(\partial_\mu = \frac{\partial}{\partial x^\mu}\) is the four-gradient. The homogeneous equations (Faraday's law and Gauss's law for magnetism) are expressed using the dual field tensor \(F^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}\), where \(\epsilon^{\mu\nu\alpha\beta}\) is the Levi-Civita symbol:
This formulation makes it explicit that Maxwell's equations are Lorentz invariant. The tensor \(F^{\mu\nu}\) is not just a bookkeeping device; it is the fundamental geometric object that describes the electromagnetic interaction in spacetime.
The derivation
The Electromagnetic Field Tensor
Assembles the electric and magnetic fields into the antisymmetric rank-2 field tensor whose transformation unifies them as one geometric object.
Open full derivation →A common pitfall is confusing the contravariant tensor \(F^{\mu\nu}\) with the covariant tensor \(F_{\mu\nu}\). While they contain the same physical information, their components differ by signs due to the metric tensor \(g_{\mu\nu}\). For instance, \(F_{0i} = -F^{0i}\) in the \((+,-,-,-)\) convention. Always check the position of indices when applying transformation laws or calculating invariants. C
Worked example
Problem. In frame \(S\), there is a uniform electric field \(\mathbf{E} = 100\ \mathrm{V/m}\) in the \(y\)-direction and no magnetic field (\(\mathbf{B} = 0\)). A frame \(S'\) moves with velocity \(v = 0.6c\) in the \(x\)-direction relative to \(S\). Calculate the electric and magnetic fields in frame \(S'\).
- Identify the non-zero components in \(S\): \(E_y = 100\ \mathrm{V/m}\), \(B_x = B_y = B_z = 0\).
- Use the transformation equations for fields perpendicular to the boost direction (\(y\)- and \(z\)-components): \[ E'_y = \gamma (E_y - v B_z) \] \[ B'_z = \gamma (B_z - \frac{v}{c^2} E_y) \] Since \(B_z = 0\), these simplify to: \[ E'_y = \gamma E_y \] \[ B'_z = -\gamma \frac{v}{c^2} E_y \]
- Calculate the Lorentz factor \(\gamma\): \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.6^2}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25 \]
- Compute \(E'_y\): \[ E'_y = 1.25 \times 100\ \mathrm{V/m} = 125\ \mathrm{V/m} \]
- Compute \(B'_z\): \[ B'_z = -1.25 \times \frac{0.6c}{c^2} \times 100\ \mathrm{V/m} = -1.25 \times 0.6 \times \frac{100}{c}\ \mathrm{T} \] Using \(c \approx 3 \times 10^8\ \mathrm{m/s}\): \[ B'_z = -0.75 \times \frac{100}{3 \times 10^8}\ \mathrm{T} = -2.5 \times 10^{-7}\ \mathrm{T} \]
Units check. \(E'\) is in \(\mathrm{V/m}\), consistent with the input. \(B'\) is in \(\mathrm{T}\). The term \(\frac{v}{c^2} E\) has units \(\frac{\mathrm{m/s}}{\mathrm{m^2/s^2}} \cdot \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{s}}{\mathrm{m}} \cdot \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{V \cdot s}}{\mathrm{m^2}}\). Since \(1\ \mathrm{T} = 1\ \mathrm{V \cdot s / m^2}\), the units are correct.
Three quick tests
- Why is the electromagnetic field tensor antisymmetric?
Answer
It reflects the fact that the electromagnetic field is a two-form in spacetime, and antisymmetry ensures that the diagonal components (which would correspond to self-interaction of the same field component) are zero, leaving exactly six independent degrees of freedom corresponding to \(\mathbf{E}\) and \(\mathbf{B}\). - If \(\mathbf{E} \cdot \mathbf{B} \neq 0\) in one frame, can it be zero in another?
Answer
No. \(\mathbf{E} \cdot \mathbf{B}\) is a Lorentz invariant scalar. If it is non-zero in one inertial frame, it is non-zero in all inertial frames. - What is the physical significance of the term \(\frac{v}{c^2} E\) in the magnetic field transformation?
Answer
It represents the magnetic field induced by the motion of the observer relative to the electric field. It shows that a purely electric field in one frame appears as a combination of electric and magnetic fields in a moving frame, with the magnetic component proportional to the relative velocity.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.