Momentum and the Maxwell Stress Tensor
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You should be comfortable with vector calculus identities, specifically the divergence theorem and the product rule for divergence, and have a working knowledge of Poynting's theorem for energy conservation.
Why this matters
Momentum is not just a property of matter; it is a property of the field.
In the previous lecture, we established that energy is conserved locally, flowing from matter to field and back again. However, Newton's third law, in its simple action-reaction form, fails for electromagnetic interactions because the force on one charge depends on the retarded field of another. To restore the conservation of momentum, we must recognise that the electromagnetic field itself carries momentum. This lecture introduces the Maxwell stress tensor, the mathematical object that describes how momentum flows through space, allowing us to write a local conservation law that unifies mechanical and field momentum.
The Field as a Momentum Carrier
Imagine two charged particles repelling each other. If you look only at the particles, the forces they exert on each other are equal and opposite, so their total mechanical momentum is conserved. But what if the particles are far apart? The force on particle A is determined by the field at A, which was generated by B long ago. The "information" about B's motion travels at the speed of light. During this delay, the simple action-reaction pair seems broken. The resolution is that the field in the space between them is not empty; it is a physical entity that stores momentum. When the field changes, momentum is transferred from the field to the matter, or vice versa. We need a way to calculate how much momentum is inside a volume of space and how much is flowing across its boundary.
From Force Density to Stress
We begin with the Lorentz force density, which tells us the force per unit volume acting on the charges and currents within a region. This is given by \(\mathbf{f} = \rho \mathbf{E} + \mathbf{J} \times \mathbf{B}\). Our goal is to rewrite this expression so that it no longer contains the sources \(\rho\) and \(\mathbf{J}\), but only the fields \(\mathbf{E}\) and \(\mathbf{B}\). We use Maxwell's equations to substitute for \(\rho\) and \(\mathbf{J}\). Specifically, we use Gauss's law \(\rho = \varepsilon_0 \nabla \cdot \mathbf{E}\) and Ampère-Maxwell's law \(\mathbf{J} = \frac{1}{\mu_0} \nabla \times \mathbf{B} - \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\).
Substituting these into the force density expression and using vector calculus identities, we can separate the terms into two distinct parts. One part involves the time derivative of the Poynting vector, representing the rate of change of field momentum. The other part involves spatial derivatives of the fields, which we can combine into a tensor quantity. This tensor is the Maxwell stress tensor, \(T_{ij}\). It describes the flux of the \(i\)-th component of momentum across a surface with normal in the \(j\)-th direction. For example, \(T_{xx}\) is the pressure exerted by the field in the \(x\)-direction on a surface perpendicular to \(x\).
Here, \(\mathbf{g}\) is the momentum density of the field. This equation is the local momentum conservation law. It states that the force on the matter (left side) is equal to the flux of momentum out of the volume (the divergence of stress) minus the rate of change of momentum stored in the field (the time derivative of \(\mathbf{g}\)).
The Maxwell Stress Tensor
The Maxwell stress tensor \(T_{ij}\) is a symmetric \(3 \times 3\) matrix. Its components are defined as:
The term \(\delta_{ij}\) is the Kronecker delta, which is 1 if \(i=j\) and 0 otherwise. The first part of the expression comes from the electric field, and the second from the magnetic field. Notice the structure: it looks like pressure. If the field is purely electric and aligned with the \(x\)-axis, \(E_x = E\), then \(T_{xx} = \varepsilon_0 (E^2 - \frac{1}{2}E^2) = \frac{1}{2}\varepsilon_0 E^2\), which equals the energy density: along the field direction the stress is a tension, as if field lines were stretched strings, while perpendicular to it the stress is a pressure of the same size. If the field is perpendicular to the surface, the stress is positive (tension). If the field is parallel to the surface, the stress is negative (compression). This tensor allows us to calculate the total force on any object by integrating the stress over its surface, without needing to know the internal charge distribution.
Field Momentum and Conservation
The momentum density of the electromagnetic field is given by \(\mathbf{g} = \varepsilon_0 \mathbf{E} \times \mathbf{B}\). This can also be written as \(\mathbf{g} = \frac{\mathbf{S}}{c^2}\), where \(\mathbf{S}\) is the Poynting vector. This relationship is profound: it links the flow of energy (Poynting vector) to the flow of momentum. The total momentum in a volume \(V\) is the sum of the mechanical momentum of the charges and the field momentum:
Integrating the local conservation law over a volume \(V\) bounded by a surface \(\mathcal{S}\), and applying the divergence theorem, we get the global conservation law:
This equation states that the rate of change of the total momentum (matter plus field) is equal to the net flux of momentum through the boundary surface. If the surface is far enough away that the fields vanish, the right-hand side is zero, and the total momentum is conserved. This resolves the apparent violation of Newton's third law: the "missing" momentum is carried by the field.
The derivation
Maxwell Stress Tensor and Field Momentum
Starting from the Lorentz force density on charges and Maxwell's equations, the total electromagnetic force on the matter inside any volume \(V\) can be rewritten entirely in terms of the fields on the boundary and inside \(V\): the \(i\)-th component of the force density is \( f_i = \partial_j T_{ij} - \varepsilon_0\mu_0\,\partial S_i/\partial t \), where \( T_{ij} = \varepsilon_0\left(E_iE_j - \tfrac{1}{2}\delta_{ij}E^2\right) + \tfrac{1}{\mu_0}\left(B_iB_j - \tfrac{1}{2}\delta_{ij}B^2\right) \) is the Maxwell stress tensor and \(\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0\) is the Poynting vector. Integrated over \(V\), this is a local momentum-conservation law: mechanical momentum plus field momentum, with density \( \mathbf{g} = \varepsilon_0\,\mathbf{E}\times\mathbf{B} = \mathbf{S}/c^2 \), changes only by the flux of stress \( \oint_{\mathcal{S}} T_{ij}\,da_j \) through the bounding surface.
A common misconception is that field momentum needs moving charges or a wave. Static fields carry momentum \(\varepsilon_0\,\mathbf{E} \times \mathbf{B}\) whenever \(\mathbf{E}\) and \(\mathbf{B}\) cross, for example a charge sitting near a bar magnet. In Feynman\'s "disc paradox", a disc carrying charges and a current-carrying coil starts to rotate when the current is switched off, with no apparent torque: the angular momentum was stored in the static field and is handed to the disc as the field collapses. (The "hidden momentum" of a magnetic dipole in an electric field is a related but distinct, mechanical, effect.) C
Worked example
Problem. A uniform electric field \(\mathbf{E} = E_0 \hat{\mathbf{x}}\) and a uniform magnetic field \(\mathbf{B} = B_0 \hat{\mathbf{y}}\) exist in a small cubic volume of side length \(L\). Calculate the total electromagnetic momentum contained within this volume.
- Identify the momentum density formula: \(\mathbf{g} = \varepsilon_0 \mathbf{E} \times \mathbf{B}\).
- Compute the cross product: \(\mathbf{E} \times \mathbf{B} = (E_0 \hat{\mathbf{x}}) \times (B_0 \hat{\mathbf{y}}) = E_0 B_0 (\hat{\mathbf{x}} \times \hat{\mathbf{y}}) = E_0 B_0 \hat{\mathbf{z}}\).
- Substitute into the density expression: \(\mathbf{g} = \varepsilon_0 E_0 B_0 \hat{\mathbf{z}}\).
- Integrate over the volume \(V = L^3\). Since the fields are uniform, \(\mathbf{g}\) is constant: \(\mathbf{p}_{\text{field}} = \int_V \mathbf{g} \, dV = \mathbf{g} V = \varepsilon_0 E_0 B_0 L^3 \hat{\mathbf{z}}\).
Units check. \(\varepsilon_0\) is \(\mathrm{F/m} = \mathrm{C^2/(N \cdot m^2)}\). \(E_0\) is \(\mathrm{N/C}\). \(B_0\) is \(\mathrm{T} = \mathrm{N/(A \cdot m)} = \mathrm{N \cdot s/(C \cdot m)}\). \(L^3\) is \(\mathrm{m^3}\). Multiplying: \(\frac{\mathrm{C^2}}{\mathrm{N \cdot m^2}} \cdot \frac{\mathrm{N}}{\mathrm{C}} \cdot \frac{\mathrm{N \cdot s}}{\mathrm{C \cdot m}} \cdot \mathrm{m^3} = \frac{\mathrm{C^2 \cdot N \cdot N \cdot s \cdot m^3}}{\mathrm{N \cdot m^2 \cdot C \cdot C \cdot m}} = \mathrm{N \cdot s} = \mathrm{kg \cdot m/s}\). This is the unit of momentum.
For scale: with \(E_0 = 1.0 \times 10^{5}\ \mathrm{V\,m^{-1}}\), \(B_0 = 1.0\ \mathrm{T}\) and \(L = 0.10\ \mathrm{m}\), \(p_{\text{field}} = 8.854 \times 10^{-12} \times 10^{5} \times 1.0 \times 10^{-3} = 8.9 \times 10^{-10}\ \mathrm{kg\,m\,s^{-1}}\): real, but tiny, which is why field momentum is rarely noticed in the laboratory.
Three quick tests
- What is the direction of the field momentum density if \(\mathbf{E}\) is in the \(x\)-direction and \(\mathbf{B}\) is in the \(z\)-direction?
Answer
\(\mathbf{E} \times \mathbf{B}\) is in the \(-y\)-direction (since \(\hat{\mathbf{x}} \times \hat{\mathbf{z}} = -\hat{\mathbf{y}}\)). - If the electric field is zero, what is the Maxwell stress tensor?
Answer
It reduces to the magnetic part: \(T_{ij} = \frac{1}{\mu_0} (B_i B_j - \frac{1}{2} \delta_{ij} B^2)\). - Does the field momentum depend on the charge density \(\rho\)?
Answer
No, it depends only on the fields \(\mathbf{E}\) and \(\mathbf{B}\), which are determined by the sources but the momentum density expression itself contains no explicit \(\rho\) or \(\mathbf{J}\).
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.