Maxwell Stress Tensor and Field Momentum

Derivation

Maxwell Stress Tensor and Field Momentum

D-167 Home PU-204 Threads force · energy · fields Depends on Poynting's Theorem and Field Energy, lorentz-force-law

Statement

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.

Why it matters

This derivation is the point at which the electromagnetic field stops being bookkeeping and becomes a mechanical object in its own right. Newton's third law fails for the instantaneous forces between moving charges — the forces two charges exert on each other via retarded fields are not equal and opposite at each instant — so mechanical momentum alone is not conserved in electrodynamics. The resolution is that the field itself carries momentum, with density \(\varepsilon_0\,\mathbf{E}\times\mathbf{B}\); once that is included, total momentum is conserved locally, transported through space by the stress tensor exactly as energy is transported by the Poynting vector.

Practically, the stress tensor is the correct machine for computing electromagnetic forces on anything: radiation pressure on solar sails and laser-trapped particles, forces on capacitor plates and magnet pole faces, tension along and pressure across field lines (Faraday's "tubes of force" made quantitative), and the momentum budget of light in optical tweezers. It is also the spatial part of the relativistic energy–momentum tensor \(T^{\mu\nu}\), the object that gravitates in general relativity.

Assumptions

Microscopic (vacuum) Maxwell equations with total charge and current densities \(\rho\), \(\mathbf{J}\).Inside polarizable media, if you use the macroscopic fields \(\mathbf{D}, \mathbf{H}\) and free charges only, the split between "field momentum" and "matter momentum" becomes convention-dependent — the Abraham–Minkowski ambiguity. The vacuum form used here is unambiguous because all charges, bound and free, sit in \(\rho\) and \(\mathbf{J}\).

The Lorentz force law \(\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}\) gives the entire force on matter.If matter couples to the field by anything other than charge and current (intrinsic magnetization treated as fundamental dipoles, for instance), extra force terms appear and the identification of \(d\mathbf{p}_{\text{mech}}/dt\) with \(\int_V \mathbf{f}\,dV\) must be revisited.

Fields fall off fast enough (or the surface is chosen so) that the surface integral converges.For the total momentum of an infinite system to be finite you need \(\mathbf{E}\times\mathbf{B}\) integrable; radiation fields falling as \(1/r\) give finite momentum flux to infinity — that flux is real (radiated momentum), not an error, but it must be accounted for.

Fields are differentiable (\(C^1\)) throughout \(V\), and the surface \(\mathcal{S}\) does not pass through point charges.At a point charge the self-field stress integrals diverge; the theorem still holds for any surface excluding the singularity, but "the momentum of a point charge's own field" is infinite — the classical self-energy problem, resolved only in renormalized QED.

A single inertial frame; \(\partial_t\) and \(\partial_j\) are ordinary partial derivatives in flat spacetime.In curved spacetime the conservation law becomes \(\nabla_\mu T^{\mu\nu} = 0\) with covariant derivatives, which no longer integrates to a globally conserved four-vector unless the spacetime has the corresponding Killing symmetry.

Derivation

\[ \mathbf{f} = \rho\,\mathbf{E} + \mathbf{J}\times\mathbf{B} \]

Force per unit volume on the charges in \(V\): the Lorentz force law (prior result lorentz-force-law) summed over the charge and current distributions. Then \( d\mathbf{p}_{\text{mech}}/dt = \int_V \mathbf{f}\,dV \) by Newton's second law applied to the matter. A

\[ \rho = \varepsilon_0\,\nabla\cdot\mathbf{E}, \qquad \mathbf{J} = \frac{1}{\mu_0}\nabla\times\mathbf{B} - \varepsilon_0\,\frac{\partial\mathbf{E}}{\partial t} \]

Eliminate the sources in favour of the fields using Gauss's law and the Ampère–Maxwell law. This is the strategic move of the whole derivation: everything on the right is now a field quantity. A

\[ \mathbf{f} = \varepsilon_0(\nabla\cdot\mathbf{E})\,\mathbf{E} + \frac{1}{\mu_0}(\nabla\times\mathbf{B})\times\mathbf{B} - \varepsilon_0\,\frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B} \]

Substitute step 2 into step 1 and distribute. Three terms: an electric divergence term, a magnetic curl term, and a term with an explicit time derivative — the last one is where field momentum will come from. B

\[ \frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B} = \frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right) - \mathbf{E}\times\frac{\partial\mathbf{B}}{\partial t} \]

Product rule for the cross product (valid because \(\times\) is bilinear). We trade the awkward \(\dot{\mathbf{E}}\times\mathbf{B}\) for a total time derivative plus a term we can convert with Faraday's law. B

\[ -\varepsilon_0\,\frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B} = -\varepsilon_0\,\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right) - \varepsilon_0\,\mathbf{E}\times(\nabla\times\mathbf{E}) \]

Insert Faraday's law \( \partial\mathbf{B}/\partial t = -\nabla\times\mathbf{E} \) into the last term of step 4 and multiply through by \(-\varepsilon_0\). The magnetic field's time derivative has been converted into a spatial derivative of \(\mathbf{E}\). B

\[ \mathbf{f} = \varepsilon_0\left[(\nabla\cdot\mathbf{E})\mathbf{E} - \mathbf{E}\times(\nabla\times\mathbf{E})\right] + \frac{1}{\mu_0}\left[(\nabla\cdot\mathbf{B})\mathbf{B} - \mathbf{B}\times(\nabla\times\mathbf{B})\right] - \varepsilon_0\,\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right) \]

Collect steps 3 and 5, using \( (\nabla\times\mathbf{B})\times\mathbf{B} = -\mathbf{B}\times(\nabla\times\mathbf{B}) \) (antisymmetry of the cross product), and add the term \( (\nabla\cdot\mathbf{B})\mathbf{B}/\mu_0 \), which is legal because \(\nabla\cdot\mathbf{B} = 0\) identically — adding zero to make \(\mathbf{E}\) and \(\mathbf{B}\) enter symmetrically. B

\[ \left[\mathbf{A}\times(\nabla\times\mathbf{A})\right]_i = A_j\,\partial_i A_j - A_j\,\partial_j A_i = \tfrac{1}{2}\,\partial_i\!\left(A^2\right) - (\mathbf{A}\cdot\nabla)A_i \]

Index identity, proved from \( [\mathbf{A}\times(\nabla\times\mathbf{A})]_i = \epsilon_{ijk}A_j\epsilon_{klm}\partial_l A_m \) and \( \epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \) (contraction of two Levi-Civita symbols). Summation over repeated indices throughout. C

\[ \left[(\nabla\cdot\mathbf{A})\mathbf{A} - \mathbf{A}\times(\nabla\times\mathbf{A})\right]_i = A_i\,\partial_j A_j + A_j\,\partial_j A_i - \tfrac{1}{2}\,\partial_i A^2 = \partial_j\!\left(A_i A_j - \tfrac{1}{2}\,\delta_{ij}A^2\right) \]

Combine step 7 with \((\nabla\cdot\mathbf{A})A_i = A_i\partial_jA_j\); the first two terms assemble into \(\partial_j(A_iA_j)\) by the product rule run backwards, and \(\partial_i = \delta_{ij}\partial_j\) absorbs the gradient into the same divergence. The whole bracket is a pure divergence — this is the miracle that makes a conservation law possible. C

\[ T_{ij} \equiv \varepsilon_0\!\left(E_iE_j - \tfrac{1}{2}\,\delta_{ij}E^2\right) + \frac{1}{\mu_0}\!\left(B_iB_j - \tfrac{1}{2}\,\delta_{ij}B^2\right) \quad\Longrightarrow\quad f_i = \partial_j T_{ij} - \varepsilon_0\,\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)_i \]

Apply step 8 once with \(\mathbf{A} = \mathbf{E}\) and once with \(\mathbf{A} = \mathbf{B}\), and name the resulting symmetric tensor \(T_{ij}\) — the Maxwell stress tensor. Definition, then substitution; no new physics in this step. A

\[ \mathbf{g} \equiv \varepsilon_0\,\mathbf{E}\times\mathbf{B} = \mu_0\varepsilon_0\,\frac{\mathbf{E}\times\mathbf{B}}{\mu_0} = \frac{\mathbf{S}}{c^2} \]

Identify the time-derivative term as a momentum density: \(\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0\) is the energy flux from poynting-theorem-energy, and \(\mu_0\varepsilon_0 = 1/c^2\). The same vector \(\mathbf{E}\times\mathbf{B}\) carries both energy (as \(\mathbf{S}\)) and momentum (as \(\mathbf{g}\)). B

\[ \frac{d p_{\text{mech},i}}{dt} = \int_V f_i\,dV = \int_V \partial_j T_{ij}\,dV - \frac{d}{dt}\int_V g_i\,dV = \oint_{\mathcal{S}} T_{ij}\,da_j - \frac{d p_{\text{field},i}}{dt} \]

Integrate over the fixed volume \(V\): the divergence theorem converts \(\int_V \partial_j T_{ij}\,dV\) into the flux \(\oint_{\mathcal{S}} T_{ij}\,da_j\) through the boundary (legal for \(C^1\) fields, \(\mathcal{S}\) piecewise smooth), and \(\partial/\partial t\) passes outside the integral because \(V\) is fixed. Rearranging gives the conservation law. B

Result

\[ \frac{d}{dt}\left(\mathbf{p}_{\text{mech}} + \mathbf{p}_{\text{field}}\right)_i = \oint_{\mathcal{S}} T_{ij}\,da_j, \qquad T_{ij} = \varepsilon_0\!\left(E_iE_j - \tfrac{1}{2}\delta_{ij}E^2\right) + \tfrac{1}{\mu_0}\!\left(B_iB_j - \tfrac{1}{2}\delta_{ij}B^2\right), \qquad \mathbf{g} = \varepsilon_0\,\mathbf{E}\times\mathbf{B} = \frac{\mathbf{S}}{c^2} \]

Reading. Momentum is conserved locally in electrodynamics, but only if the field is credited with momentum of density \(\mathbf{g} = \mathbf{S}/c^2\). The rate of change of (mechanical + field) momentum inside any closed surface equals the stress-tensor flux through that surface: \(T_{ij}\,da_j\) is the \(i\)-th component of force transmitted across the area element \(d\mathbf{a}\). Equivalently, \(-T_{ij}\) is the flux density of the \(i\)-component of momentum in the \(j\)-direction. The diagonal elements are pressures/tensions: along a field line the field pulls with tension \(\tfrac{1}{2}\varepsilon_0E^2 + \tfrac{1}{2}B^2/\mu_0\) per unit area; transverse to the lines it pushes with the same magnitude of pressure.

Units check. \([\varepsilon_0 E^2] = (\mathrm{C^2\,N^{-1}\,m^{-2}})(\mathrm{V^2\,m^{-2}}) = \mathrm{J\,m^{-3}} = \mathrm{N\,m^{-2}} = \mathrm{Pa}\), and \([B^2/\mu_0] = \mathrm{T^2}/(\mathrm{T\,m\,A^{-1}}) = \mathrm{T\,A\,m^{-1}} = \mathrm{N\,m^{-2}}\) likewise — so \(T_{ij}\) is a stress and \(\oint T_{ij}\,da_j\) is a force (N). For the momentum density: \([\mathbf{S}/c^2] = (\mathrm{W\,m^{-2}})/(\mathrm{m^2\,s^{-2}}) = \mathrm{kg\,m^{-2}\,s^{-1}} = (\mathrm{kg\,m\,s^{-1}})\,\mathrm{m^{-3}}\), momentum per unit volume. Both sides of the boxed law are newtons. Consistent.

Limiting cases

Statics (\(\partial\mathbf{g}/\partial t = 0\)): the force on everything inside \(\mathcal{S}\) is exactly \(\oint T_{ij}\,da_j\) — electrostatic and magnetostatic forces computed purely from the fields on a surrounding surface, no knowledge of the charge distribution needed.
Uniform field along \(\hat{\mathbf{z}}\): \(T_{zz} = +\tfrac{1}{2}\varepsilon_0E^2\) (tension along the lines), \(T_{xx} = T_{yy} = -\tfrac{1}{2}\varepsilon_0E^2\) (pressure across them), off-diagonals zero — Faraday's picture of field lines as stretched elastic bands that repel sideways.
Plane wave, normal absorption: momentum flux \(=\langle u\rangle = I/c\) per unit area, so radiation pressure \(P = I/c\) on a black surface and \(2I/c\) on a perfect mirror.
Empty volume (\(\rho = 0\), \(\mathbf{J} = 0\) inside \(V\)): \(d\mathbf{p}_{\text{field}}/dt = \oint T_{ij}\,da_j\) — field momentum in a region changes only by stress transmitted through its boundary, a genuine continuity equation for momentum.
Non-relativistic energy transport: for a wave packet of energy \(U\) travelling at \(c\), the total momentum is \(p = U/c\) — recovering the photon relation \(E = pc\) classically, forty years before the photon.

Breaks when

Inside ponderable media using macroscopic fields. Writing the momentum density as \(\varepsilon_0\mathbf{E}\times\mathbf{B}\) (Abraham) or \(\mathbf{D}\times\mathbf{B}\) (Minkowski) gives different answers for "the momentum of light in glass" — differing by exactly the momentum ascribed to the medium's response. The vacuum theorem is exact; the partition into field and matter contributions inside a dielectric is convention until you specify the matter's stress tensor too. Experiments (photon recoil in gases, fibre recoil) probe total momentum and are consistent with both bookkeepings correctly completed.
Surfaces through point charges / classical self-fields. The stress integral over any surface enclosing a single point charge's own field diverges as the surface shrinks; the field momentum of a point charge moving at constant velocity is infinite. Classical electrodynamics has no consistent finite answer (the 4/3 problem of electromagnetic mass); resolution requires Poincaré stresses or QED renormalization.
Field strengths near the Schwinger scale \(E \sim 1.3\times10^{18}\ \mathrm{V/m}\): vacuum polarization makes the effective Lagrangian nonlinear (Euler–Heisenberg); Maxwell's linear equations, and hence this tensor, acquire corrections, and beyond the critical field electron–positron pair creation drains the field entirely.
Strong gravity / curved spacetime. The law \(\partial_\mu T^{\mu\nu} = 0\) must be promoted to \(\nabla_\mu T^{\mu\nu} = 0\); the flat-space surface-flux form fails because momentum components at different points can no longer be added without parallel transport. Gravitational lensing of light is precisely field momentum being exchanged with spacetime geometry.

Failure modes

Sign-convention whiplash. With the convention used here (Griffiths), the force on the enclosed matter is \(+\oint T_{ij}\,da_j\) with the outward normal, and \(-T_{ij}\) is the momentum flux density. Some texts define \(T\) with the opposite sign so that \(+T_{ij}\) is the flux. Mixing conventions flips every force.
Dropping the \(\tfrac{1}{2}\delta_{ij}\) trace terms. Keeping only \(\varepsilon_0E_iE_j\) misses the isotropic pressure part; you then predict (wrongly) zero transverse force from a uniform field and get the capacitor-plate force wrong by a factor and sign structure.
Confusing \(\mathbf{S}/c^2\) with \(\mathbf{S}/c\). \(\mathbf{S}/c^2\) is momentum density (kg m\(^{-2}\) s\(^{-1}\)); \(S/c\) is momentum flux (Pa). Using the wrong one silently multiplies answers by \(c\).
Integrating \(T_{ij}\) over a surface that cuts through charge. The theorem gives the force on everything inside \(\mathcal{S}\); a surface slicing a conductor includes part of its charge and returns the force on that fragment plus surface-term ambiguities. Choose \(\mathcal{S}\) through charge-free space.
Assuming static fields carry no momentum. Static crossed \(\mathbf{E}\) and \(\mathbf{B}\) fields have nonzero \(\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B}\); it is balanced by an equal and opposite "hidden" mechanical momentum in the current-carrying matter, and forgetting either side breaks the momentum ledger (Shockley–James paradox).
Treating \(T_{ij}\) as the force per area on a material surface at that location. \(T_{ij}\,da_j\) is the force transmitted across a mathematical surface in the field; it equals the force on a physical conductor surface only when the surface coincides with the boundary and the field vanishes on one side.

Discussion

The deepest content of the result is the failure and repair of Newton's third law. For two charges in relative motion, the magnetic forces they exert on each other are generally not equal and opposite — momentum is being created or destroyed in the matter sector at each instant. The derivation shows exactly where it goes: into \(\varepsilon_0\mathbf{E}\times\mathbf{B}\), physically real momentum stored in the field between the charges. Feynman's disk paradox makes this vivid: discharge a capacitor at the centre of a magnetized system and the disk starts to rotate, its angular momentum supplied by the \(\mathbf{r}\times\mathbf{g}\) that the static fields were holding all along. Conservation laws in field theory are local ledgers, not action–reaction pairs.

The tensor structure encodes Faraday's geometric intuition exactly. Diagonalize \(T_{ij}\) at a point where only \(\mathbf{E}\) exists: the eigenvalue along \(\hat{\mathbf{E}}\) is \(+\tfrac{1}{2}\varepsilon_0E^2\) (tension), the two transverse eigenvalues are \(-\tfrac{1}{2}\varepsilon_0E^2\) (pressure). Field lines behave like elastic filaments under tension that repel one another laterally: two like charges "push apart" because the bundle of lines between them is compressed; unlike charges "pull together" because the lines connecting them are stretched. Every electrostatic and magnetostatic force diagram can be read off this way, and the pinch effect in plasmas is nothing but the transverse magnetic pressure \(B^2/2\mu_0\).

The relation \(\mathbf{g} = \mathbf{S}/c^2\) is a classical shadow of relativity. Writing it as (momentum density) = (energy flux)/\(c^2\) and reading it in reverse, an energy flux \(\mathbf{S}\) is equivalent to a flow of mass density \(\mathbf{S}/c^2\) — Einstein's box argument for \(E = mc^2\) runs on exactly this identity. In covariant language the pieces assemble into the symmetric energy–momentum tensor \(T^{\mu\nu}\): \(T^{00} = u\) (energy density), \(T^{0i} = S_i/c = cg_i\), \(T^{ij} = -T_{ij}\) (sign per metric convention), and the two separate conservation laws — Poynting's theorem and this derivation — are the time and space components of the single equation \(\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda/c\) coupling field to matter.

Rigorously, \(T^{\mu\nu} = \tfrac{1}{\mu_0}\left(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right)\) is the Belinfante-improved tensor, not the canonical one delivered by Noether's theorem for translations: the canonical tensor is neither symmetric nor gauge-invariant, and the improvement terms (a superpotential \(\partial_\lambda K^{\lambda\mu\nu}\) with \(K\) antisymmetric in \(\lambda\mu\)) shift nothing in the integrated charges. Symmetry of \(T^{\mu\nu}\) is not cosmetic — it is precisely the condition for the angular-momentum current \(x^\mu T^{\nu\lambda} - x^\nu T^{\mu\lambda}\) to be conserved, and \(T^{0i} = T^{i0}\) is the statement \(\mathbf{g} = \mathbf{S}/c^2\) itself: momentum density equals energy flux over \(c^2\) because rotational and boost symmetry demand it. Tracelessness, \(T^{\mu}{}_{\mu} = 0\), reflects the classical scale invariance of the free Maxwell field and is why radiation pressure of isotropic radiation is \(u/3\).

Common misconceptions. Field momentum is not "photon momentum smuggled into classical physics" — it follows from Maxwell's equations alone, with no quantum input; the photon relation \(p = E/c\) is its quantized echo. And radiation pressure does not require absorption "converting energy to force": a perfect mirror absorbs no energy yet feels twice the force of a black absorber, because force is momentum transfer, and reversal transfers twice the momentum.

Worked examples

Example 1 — Radiation pressure on a solar sail at 1 AU. Sunlight of intensity \(I = 1361\ \mathrm{W/m^2}\) strikes a flat, perfectly reflecting square sail of side \(L = 100\ \mathrm{m}\) at normal incidence. Sail-plus-payload mass \(m = 150\ \mathrm{kg}\). Find the radiation force and acceleration.

\[ \langle g \rangle = \frac{\langle S \rangle}{c^2} = \frac{I}{c^2}, \qquad \text{incident momentum flux} = c\,\langle g \rangle = \frac{I}{c} \]

The wave's momentum density is \(\mathbf{S}/c^2\) (step 10); it arrives at speed \(c\), so momentum crosses unit area at rate \(I/c\). Equivalently, \(\langle T_{zz}\rangle\) for a plane wave equals the mean energy density \(\langle u\rangle = I/c\). Symbols first. A

\[ P_{\text{mirror}} = \frac{2I}{c}, \qquad F = P_{\text{mirror}}\,L^2 = \frac{2IL^2}{c} \]

A perfect reflector reverses the normal momentum of the light, so the momentum transferred per unit area per unit time is twice the incident flux. Multiply by the sail area \(L^2\). B

\[ P_{\text{mirror}} = \frac{2(1361\ \mathrm{W/m^2})}{2.998\times10^{8}\ \mathrm{m/s}} = 9.08\times10^{-6}\ \mathrm{Pa} \]

Insert numbers. \(\mathrm{W\,m^{-2}}/(\mathrm{m\,s^{-1}}) = \mathrm{J\,m^{-3}} = \mathrm{Pa}\) — a pressure eleven orders of magnitude below atmospheric. A

\[ F = (9.08\times10^{-6}\ \mathrm{Pa})(100\ \mathrm{m})^2 = 9.08\times10^{-2}\ \mathrm{N}, \qquad a = \frac{F}{m} = \frac{9.08\times10^{-2}\ \mathrm{N}}{150\ \mathrm{kg}} = 6.1\times10^{-4}\ \mathrm{m/s^2} \]

Force = pressure × area; then Newton's second law for the sail. A

\[ F \approx 0.091\ \mathrm{N}, \qquad a \approx 6.1\times10^{-4}\ \mathrm{m/s^2} \]

Reading. Tiny but relentless: sustained for one day this acceleration builds \(\Delta v \approx 53\ \mathrm{m/s}\), and unlike a rocket it costs no propellant — the momentum comes from the field, exactly as the conservation law requires.

Example 2 — Force between capacitor plates via the stress tensor. A parallel-plate capacitor has plate area \(A = 1.0\times10^{-2}\ \mathrm{m^2}\) and a uniform interior field \(E = 1.0\times10^{6}\ \mathrm{V/m}\) along \(\hat{\mathbf{z}}\) (fringing neglected). Find the force on one plate using \(T_{ij}\), not Coulomb's law.

\[ T_{zz} = \varepsilon_0\!\left(E_zE_z - \tfrac{1}{2}\delta_{zz}E^2\right) = \varepsilon_0\!\left(E^2 - \tfrac{1}{2}E^2\right) = \tfrac{1}{2}\,\varepsilon_0 E^2 \]

Evaluate the stress tensor between the plates, where \(\mathbf{E} = E\hat{\mathbf{z}}\) and \(\mathbf{B} = 0\). Only \(T_{zz}\) is needed for the \(z\)-force through a horizontal surface. B

\[ F_z = \oint_{\mathcal{S}} T_{zj}\,da_j = T_{zz}\,A = \tfrac{1}{2}\,\varepsilon_0 E^2 A \]

Enclose the top plate with a surface whose only contribution is a flat sheet of area \(A\) in the gap (field is zero above the plate and fringing fields are neglected on the sides). Statics, so the \(d\mathbf{p}_{\text{field}}/dt\) term vanishes and the surface integral is the force. The positive \(T_{zz}\) with the outward (downward, \(-\hat{\mathbf z}\)) normal gives a downward force: tension along the field lines pulls the plates together. B

\[ |F_z| = \tfrac{1}{2}\left(8.854\times10^{-12}\ \mathrm{F/m}\right)\left(1.0\times10^{6}\ \mathrm{V/m}\right)^2\left(1.0\times10^{-2}\ \mathrm{m^2}\right) \]

Insert numbers only now. \(\mathrm{F\,m^{-1}\cdot V^2\,m^{-2}} = \mathrm{J\,m^{-3}} = \mathrm{Pa}\), and \(\mathrm{Pa\cdot m^2} = \mathrm{N}\): the stress \(\tfrac12\varepsilon_0E^2 = 4.43\ \mathrm{Pa}\) acts on \(10^{-2}\ \mathrm{m^2}\). A

\[ |F| = 4.4\times10^{-2}\ \mathrm{N}\ \text{(attractive)}, \qquad \frac{F}{A} = \tfrac{1}{2}\varepsilon_0E^2 = 4.4\ \mathrm{Pa} \]

Reading. The stress-tensor route reproduces the standard result \(F/A = \sigma^2/2\varepsilon_0\) (with \(\sigma = \varepsilon_0E\)) without ever asking which charges pull on which — the force is read off the field alone on a surface in empty space. The \(\tfrac{1}{2}\) that students fight over in the charge-based calculation (a plate feels the other plate's field, \(E/2\)) is automatic here.

Problems

A pulsed laser delivers a single \(1.0\ \mathrm{mJ}\), \(10\ \mathrm{ns}\) pulse. (a) What total momentum does the pulse carry in vacuum? (b) What average force does it exert while being absorbed by a target? (c) While being retro-reflected?
Solution
(a) Integrate the momentum density over the pulse volume: \(g = S/c^2 = u/c\) (using \(S = uc\) for a travelling wave), and the pulse occupies volume \(A\,c\,\Delta t\), so \(p = \frac{u}{c}\,A\,c\,\Delta t = u\,A\,\Delta t = \frac{U}{c}\), since the pulse energy is \(U = u\,A\,c\,\Delta t\) — i.e. energy \(U\) travelling at \(c\) carries momentum \(p = U/c\). So \(p = (1.0\times10^{-3}\ \mathrm{J})/(2.998\times10^{8}\ \mathrm{m/s}) = 3.3\times10^{-12}\ \mathrm{kg\,m/s}\). (b) Absorption transfers \(p\) in \(\Delta t = 10\ \mathrm{ns}\): \(F = p/\Delta t = 3.3\times10^{-12}/1.0\times10^{-8} = 3.3\times10^{-4}\ \mathrm{N}\) — a third of a millinewton during the pulse. (c) Reflection reverses the momentum, transferring \(2p\): \(F = 6.7\times10^{-4}\ \mathrm{N}\). Note the average force over one second (one pulse per second) is only \(3.3\times10^{-12}\ \mathrm{N}\): radiation forces are large only while the light is actually there.
A region contains a uniform static field \(\mathbf{E} = E\,\hat{\mathbf{z}}\) with \(E = 3.0\times10^{6}\ \mathrm{V/m}\) (about the breakdown field of air). Write out all nine components of \(T_{ij}\), identify the principal stresses, and evaluate them numerically.
Solution
With \(E_x = E_y = 0\), \(E_z = E\): the off-diagonal components all vanish (\(E_iE_j = 0\) for \(i\neq j\)). Diagonals: \(T_{zz} = \varepsilon_0(E^2 - \tfrac12E^2) = +\tfrac12\varepsilon_0E^2\); \(T_{xx} = T_{yy} = \varepsilon_0(0 - \tfrac12E^2) = -\tfrac12\varepsilon_0E^2\). Numerically \(\tfrac12\varepsilon_0E^2 = \tfrac12(8.854\times10^{-12})(3.0\times10^{6})^2 = \tfrac12(8.854\times10^{-12})(9.0\times10^{12}) = 39.8\ \mathrm{Pa} \approx 40\ \mathrm{Pa}\). So the field is in tension \(+40\ \mathrm{Pa}\) along \(\hat{\mathbf{z}}\) (along the field lines) and pushes outward with pressure \(40\ \mathrm{Pa}\) in every transverse direction. Even at the strongest field air can sustain, electrostatic stresses are only \(\sim 4\times10^{-4}\) of atmospheric pressure — which is why electrostatic machines are weak and magnetic machines (where \(B^2/2\mu_0 = 40\ \mathrm{Pa}\) already at \(B = 0.01\ \mathrm{T}\), and \(4\times10^{5}\ \mathrm{Pa}\) at 1 T) dominate engineering.
Two point charges \(+q = +1.0\ \mathrm{\mu C}\) sit on the \(z\)-axis at \(z = \pm d\) with \(d = 0.50\ \mathrm{m}\). Compute the force on the upper charge by integrating the Maxwell stress tensor over the midplane \(z = 0\) (closed at infinity), and check against Coulomb's law.
Solution
On the midplane the \(z\)-components of the two Coulomb fields cancel and the radial components add: at radius \(\rho\), \(E_\rho = 2\cdot\frac{q}{4\pi\varepsilon_0}\cdot\frac{\rho}{(\rho^2+d^2)^{3/2}} = \frac{q\rho}{2\pi\varepsilon_0(\rho^2+d^2)^{3/2}}\), with \(E_z = 0\). Then \(T_{zz} = \varepsilon_0(E_zE_z - \tfrac12E^2) = -\tfrac12\varepsilon_0E_\rho^2\). Enclose the upper charge by the plane \(z=0\) plus a hemisphere at infinity (whose contribution vanishes as \(E^2r^2 \sim r^{-2}\)). The outward normal of the enclosing surface on the plane is \(-\hat{\mathbf{z}}\), so \(F_z = \oint T_{zj}da_j = -\int T_{zz}\,da = +\tfrac12\varepsilon_0\int_0^\infty E_\rho^2\,2\pi\rho\,d\rho = \tfrac{q^2}{4\pi\varepsilon_0}\int_0^\infty \frac{\rho^3\,d\rho}{(\rho^2+d^2)^3}\). Substitute \(u = \rho^2\): \(\int_0^\infty \frac{\rho^3 d\rho}{(\rho^2+d^2)^3} = \tfrac12\int_0^\infty \frac{u\,du}{(u+d^2)^3} = \tfrac12\left[\frac{1}{d^2} - \frac{d^2}{2d^4}\right] = \frac{1}{4d^2}\). Hence \(F_z = \frac{q^2}{16\pi\varepsilon_0 d^2}\), which is exactly Coulomb's law \(\frac{q^2}{4\pi\varepsilon_0(2d)^2}\) for separation \(2d\), directed away from the midplane (repulsion). Numerically: \(F = (8.99\times10^{9})(1.0\times10^{-6})^2/(1.0\ \mathrm{m})^2 = 9.0\times10^{-3}\ \mathrm{N}\). The negative \(T_{zz}\) on the plane is the transverse pressure of the compressed field lines between like charges — the mechanical picture of electrostatic repulsion.
A cube of side \(a = 10\ \mathrm{cm}\) contains uniform crossed static fields \(\mathbf{E} = (1.0\times10^{5}\ \mathrm{V/m})\,\hat{\mathbf{x}}\) and \(\mathbf{B} = (0.10\ \mathrm{T})\,\hat{\mathbf{y}}\). (a) Compute the electromagnetic momentum stored in the cube. (b) Nothing is moving — explain, naming the relevant principle, why this does not violate momentum conservation.
Solution
(a) \(\mathbf{g} = \varepsilon_0\,\mathbf{E}\times\mathbf{B} = \varepsilon_0 EB\,(\hat{\mathbf{x}}\times\hat{\mathbf{y}}) = \varepsilon_0EB\,\hat{\mathbf{z}}\). Magnitude \(g = (8.854\times10^{-12})(1.0\times10^{5})(0.10) = 8.9\times10^{-8}\ \mathrm{kg\,m^{-2}\,s^{-1}}\). Volume \(a^3 = 1.0\times10^{-3}\ \mathrm{m^3}\), so \(p_{\text{field}} = g\,a^3 = 8.9\times10^{-11}\ \mathrm{kg\,m/s}\) along \(\hat{\mathbf{z}}\). (b) A closed static system at rest must have zero total momentum (the centre-of-energy theorem: for any bounded stationary system, total momentum vanishes). The sources maintaining these fields — charged plates and a current loop or magnet — carry an equal and opposite "hidden momentum" \(-\mathbf{p}_{\text{field}}\), a relativistic effect of order \(1/c^2\) residing in the current carriers moving through the electric potential. When the fields are switched off, the field momentum is delivered to the sources as an impulse, and the hidden momentum disappears with the currents; the ledger balances at every instant. Forgetting hidden momentum is the root of the Shockley–James paradox.
Sunlight of intensity \(I = 1361\ \mathrm{W/m^2}\) strikes a perfectly reflecting flat panel at \(60^\circ\) from the normal. (a) Using the momentum-flux interpretation of \(T_{ij}\), show the force per unit panel area is \(P = (2I/c)\cos^2\theta\), directed along the normal. (b) Evaluate it. (c) A comet dust grain is a sphere of radius \(r = 0.50\ \mathrm{\mu m}\) and density \(\rho_m = 2.0\times10^{3}\ \mathrm{kg/m^3}\) that absorbs all light falling on it at 1 AU. Compare radiation force to solar gravity (\(GM_\odot = 1.327\times10^{20}\ \mathrm{m^3/s^2}\), 1 AU \(= 1.496\times10^{11}\ \mathrm{m}\)).
Solution
(a) The incident beam carries momentum flux \(I/c\) along its propagation direction. The panel intercepts, per unit panel area, the light crossing an area \(\cos\theta\) of the beam cross-section, so incident momentum arrives at rate \((I/c)\cos\theta\) per unit panel area. Specular reflection reverses only the normal component of the light's momentum, of magnitude \((I/c)\cos\theta\cdot\cos\theta\) per unit panel area; the tangential component is unchanged (zero tangential force — no shear on a specular mirror). Momentum transfer normal to the panel: \(2\times(I/c)\cos^2\theta\). Hence \(P = (2I/c)\cos^2\theta\,\hat{\mathbf{n}}\) — equivalently the \(nn\)-component of the change in the wave's stress tensor. (b) \(P = 2(1361)/(2.998\times10^{8})\times\cos^2 60^\circ = (9.08\times10^{-6})(0.25) = 2.3\times10^{-6}\ \mathrm{Pa}\). (c) Absorber: \(F_{\text{rad}} = (I/c)\pi r^2 = (1361/2.998\times10^{8})\pi(0.5\times10^{-6})^2 = (4.54\times10^{-6})(7.85\times10^{-13}) = 3.6\times10^{-18}\ \mathrm{N}\). Gravity: \(m = \tfrac43\pi r^3\rho_m = \tfrac43\pi(1.25\times10^{-19})(2.0\times10^{3}) = 1.05\times10^{-15}\ \mathrm{kg}\); \(F_g = GM_\odot m/R^2 = (1.327\times10^{20})(1.05\times10^{-15})/(1.496\times10^{11})^2 = 6.2\times10^{-18}\ \mathrm{N}\). Ratio \(F_{\text{rad}}/F_g \approx 0.6\): for sub-micron grains radiation pressure is comparable to gravity (both scale as \(1/R^2\), so the ratio is distance-independent), which is why comet dust tails are blown antisunward. Grains a few times smaller are expelled from the solar system outright.