The Field Tensor and Covariant Maxwell Equations
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You must be comfortable with the Minkowski metric, four-vectors, and the distinction between covariant and contravariant indices. You should also recall the standard Maxwell equations in SI units from Electromagnetism I.
Why this matters
Maxwell's equations are not just a set of four vector laws; they are a single geometric object.
In the previous lecture, we established that special relativity demands that physical laws look the same in all inertial frames. The electric and magnetic fields, which appear distinct in one frame, mix into each other in another. This suggests they are components of a deeper entity. By packaging the electromagnetic field into a tensor, we reveal that the four separate Maxwell equations are actually just two compact tensor equations. This unification is not merely aesthetic; it guarantees that charge conservation is a mathematical necessity rather than an additional assumption, and it provides the direct bridge to the relativistic formulation of mechanics and quantum field theory.
The Geometric Unification
Imagine you are describing a rotation in three-dimensional space. You might list the angle and the axis, or you might use a matrix that maps one vector to another. In electromagnetism, we have been treating the electric field \(\vec{E}\) and magnetic field \(\vec{B}\) as separate entities. However, just as a rotation matrix is a single object that contains all the information about how vectors change, the electromagnetic field is a single object that contains all the information about how charges and currents interact with spacetime.
The key insight is that the "time" and "space" parts of the field are not fundamentally different. The electric field is essentially the time-space part of the interaction, while the magnetic field is the space-space part. When we switch to a moving reference frame, the "time" axis tilts relative to the "space" axes. Consequently, what was purely electric in one frame becomes a mixture of electric and magnetic components in another. The tensor formalism captures this mixing automatically. It tells us that \(\vec{E}\) and \(\vec{B}\) are not two separate forces, but two faces of the same coin, determined by the observer's motion through spacetime.
Constructing the Field Tensor
To build the tensor, we first assemble the scalar potential \(\phi\) and vector potential \(\vec{A}\) into a four-vector \(A^\mu\). Following the convention where the time component is first, we define \(A^\mu = (\phi/c, A_x, A_y, A_z)\). The factor of \(c\) ensures all components have the same units.
The electromagnetic field tensor \(F^{\mu\nu}\) is defined as \(F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu\), using the metric \(\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\) and \(x^0 = ct\). This construction guarantees that the tensor is antisymmetric, meaning \(F^{\mu\nu} = -F^{\nu\mu}\). The diagonal elements are zero. The off-diagonal elements contain the field components. Specifically, the time-space block (indices 0 and \(i\)) contains the electric field components, and the space-space block (indices \(i\) and \(j\)) contains the magnetic field components.
This matrix form makes the unification explicit. The electric field appears in the first row and column, while the magnetic field occupies the lower-right \(3 \times 3\) block. The antisymmetry is visually apparent: the matrix is skew-symmetric.
Covariant Maxwell Equations
With the tensor defined, we can write Maxwell's equations in a form that is explicitly invariant under Lorentz transformations. We introduce the four-current \(J^\mu = (c\rho, \vec{J})\), which combines charge density and current density into a single four-vector.
The first of the two covariant equations represents the sourced equations: Gauss's law and the Ampère-Maxwell law. It states that the divergence of the field tensor equals the four-current, scaled by the permeability of free space.
When you expand this single equation for \(\nu = 0\), you recover Gauss's law. When you expand it for \(\nu = 1, 2, 3\), you recover the Ampère-Maxwell law. The second covariant equation represents the homogeneous equations: the absence of magnetic monopoles and Faraday's law of induction. It is a cyclic identity involving three derivatives of the tensor.
This equation is a mathematical identity that holds because \(F_{\mu\nu}\) is derived from a potential \(A_\mu\). It ensures that the magnetic field lines are always closed loops and that a changing magnetic field induces an electric field in a way consistent with relativity.
Automatic Charge Conservation
One of the most profound results of this formulation is that charge conservation is not an independent law; it is a direct consequence of the antisymmetry of the field tensor. In the standard vector formulation, we often state the continuity equation \(\partial \rho / \partial t + \nabla \cdot \vec{J} = 0\) as a separate requirement for consistency.
In the tensor formulation, we take the four-divergence of the sourced equation \(\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\). Because partial derivatives commute, \(\partial_\nu \partial_\mu F^{\mu\nu}\) is the contraction of a symmetric tensor (\(\partial_\nu \partial_\mu\)) with an antisymmetric tensor (\(F^{\mu\nu}\)). The contraction of a symmetric and an antisymmetric tensor is always zero.
Thus, \(\partial_\nu J^\nu = 0\) follows automatically. This demonstrates that the structure of the electromagnetic field enforces the conservation of charge. It is a beautiful example of how geometric constraints in the theory lead to physical conservation laws.
The derivation
The Electromagnetic Field Tensor
The scalar and vector potentials are assembled into a four-vector \( A^\mu = (\phi/c,\ \vec{A}) \), and the six field components into the antisymmetric field tensor \( F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \), whose time–space block is \( \vec{E}/c \) and whose space–space block is \( \vec{B} \). With the four-current \( J^\mu = (c\rho,\ \vec{J}) \), Maxwell's four vector equations collapse into two manifestly Lorentz-covariant tensor equations: the sourced equation \( \partial_\mu F^{\mu\nu} = \mu_0 J^\nu \) (Gauss + Ampère–Maxwell) and the identity \( \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \) (no monopoles + Faraday), with charge conservation \( \partial_\nu J^\nu = 0 \) following automatically from antisymmetry.
Be careful with the placement of indices. The equation \(\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\) uses a raised index on \(F\) and a lowered index on \(\partial\). If you mistakenly write \(\partial_\mu F_{\mu\nu}\), you will get the wrong sign for the magnetic field terms because the metric signature affects how indices are raised and lowered. Always check your metric convention (\(+---\) or \(-+++\)) when converting between tensor components and \(\vec{E}, \vec{B}\). C
Worked example
Problem. Using the matrix above, expand the \(\nu = 1\) component of \(\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\) and show that it is the \(x\)-component of the Ampère–Maxwell law. Then find \(F^{01}\) for \(E_x = 3.0 \times 10^{5}\ \mathrm{V\,m^{-1}}\).
- Read off the column \(\nu = 1\): \(F^{01} = -E_x/c\), \(F^{11} = 0\), \(F^{21} = B_z\), \(F^{31} = -B_y\) (antisymmetry: \(F^{21} = -F^{12}\), \(F^{31} = -F^{13}\)).
- Sum over \(\mu\), with \(\partial_0 = \frac{1}{c}\frac{\partial}{\partial t}\): \(\partial_\mu F^{\mu 1} = -\frac{1}{c^2}\frac{\partial E_x}{\partial t} + \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}\).
- Recognise \(\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} = (\nabla \times \mathbf{B})_x\) and \(J^1 = J_x\), so \((\nabla \times \mathbf{B})_x - \frac{1}{c^2}\frac{\partial E_x}{\partial t} = \mu_0 J_x\).
- With \(1/c^2 = \mu_0\varepsilon_0\) this is \((\nabla \times \mathbf{B})_x = \mu_0 J_x + \mu_0\varepsilon_0\,\partial E_x/\partial t\), the Ampère–Maxwell law, displacement current included.
- Numerically \(F^{01} = -E_x/c = -3.0 \times 10^{5}/(3.00 \times 10^{8}) = -1.0 \times 10^{-3}\ \mathrm{T}\): written this way, a strong laboratory electric field is comparable to a millitesla magnetic field.
Units check. Every entry of \(F^{\mu\nu}\) is in tesla: \(E/c\) has units \(\mathrm{V\,m^{-1}}/(\mathrm{m\,s^{-1}}) = \mathrm{V\,s\,m^{-2}} = \mathrm{T}\). Each term of the expanded equation is in \(\mathrm{T\,m^{-1}}\), matching \(\mu_0 J_x\) in \(\mathrm{(T\,m\,A^{-1})(A\,m^{-2})}\).
Three quick tests
- Why is the diagonal of the field tensor \(F^{\mu\nu}\) always zero?
Answer
Because the tensor is antisymmetric, \(F^{\mu\mu} = -F^{\mu\mu}\), which implies \(2F^{\mu\mu} = 0\). - If \(\vec{E} = 0\) and \(\vec{B} \neq 0\) in one frame, can \(\vec{E}\) be zero in all frames?
Answer
No, unless \(\vec{B}\) is parallel to the relative velocity of the frames. Generally, a pure magnetic field in one frame appears as a mixture of electric and magnetic fields in another. - What physical law is automatically satisfied by the identity \(\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0\)?
Answer
The absence of magnetic monopoles (\(\nabla \cdot \vec{B} = 0\)) and Faraday's law of induction.
Problem set for PU-204 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.