The Maxwell Stress Tensor

The last article gave the field a momentum account; this one builds the machinery that moves momentum around. The question sounds mundane: what is the total electromagnetic force on the charges inside some region? The direct route — integrate $\rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$ over the volume — demands that you know where every charge sits and what field each one feels, which for a real conductor (charges crowded invisibly at surfaces, fields diverging near edges) is somewhere between painful and impossible. Maxwell's stress tensor is the professional's detour. It converts the volume integral into a surface integral: draw any closed surface around the body, evaluate a certain quadratic combination of $\mathbf{E}$ and $\mathbf{B}$ on that surface alone, and the integral delivers the total force on everything inside. You never look at the interior at all. The combination, $T_{ij}$ , has the units of pressure and an old, beautiful reading: Faraday's intuition that field lines pull like stretched rubber bands and shove sideways like compressed gas, made exact. Tension $\tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ along the lines, equal pressure across them — magnets pull because their shared field lines are taut; like charges repel because their crowded lines press. The deeper identity, inherited from the momentum article: $T_{ij}$ is the flux of momentum. Force is just momentum delivery, and the stress tensor is the field's momentum current — the object that will complete this chapter's conservation law in the next article, and that relativity will file as the space–space block of one grand tensor in the chapter's finale. Here we build it, learn to read it, and put it to work.

Key takeaways

The Maxwell stress tensor $T_{ij} = \varepsilon_0(E_iE_j - \tfrac12\delta_{ij}E^2) + \tfrac{1}{\mu_0}(B_iB_j - \tfrac12\delta_{ij}B^2)$ is the electromagnetic momentum flux: force per unit area transmitted across any surface, in units of pressure.
Its anatomy is Faraday made quantitative: tension $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ along field lines, equal pressure across them — pull lengthwise, push sideways, both equal to the local energy density.
In statics, the total force on everything inside a closed surface is $F_i = \oint_S T_{ij}\,da_j$ — fields on the boundary only; the interior can stay a mystery.
The time-dependent completion (subtracting $d\mathbf{p}_{\text{field}}/dt$ ) is the next article's conservation theorem; the tensor built here is its flux term.
Anchor pressures: a charged conductor pulls outward at $\sigma^2/2\varepsilon_0$ ; a 1 T solenoid pushes outward at 4 atm, a 10 T magnet at 400 atm — magnet engineering is stress-tensor engineering.
KEY IDEA: force = momentum flux. The net electromagnetic force on a region is the surface integral of $T_{ij}$ over its boundary — Faraday's taut, jostling field lines, turned into an exact accounting device.

The core insight

Force = flux of momentum: the net electromagnetic force on everything inside a closed surface equals the surface integral of $T_{ij}$ (momentum flowing in through the boundary) minus the rate at which field momentum accumulates inside.

Where this lives today

Whenever electromagnetic forces are computed seriously, it is the stress tensor doing the work. Finite-element field solvers — the software behind every electric-machine, relay, MEMS, and magnet design — obtain forces and torques by integrating $T_{ij}$ over surfaces around each part, precisely because the method needs only boundary fields and tolerates arbitrary interior complexity. Fusion is stress-tensor engineering at civilizational scale: ITER's toroidal-field system stores its magnetic field at hundreds of atmospheres of $B^2/2\mu_0$ , and the machine's structural steel exists to hold that pressure; the plasma itself is confined by arranging the field's tension-and-pressure anatomy so the "gas of field lines" pushes inward harder than the 100-million-kelvin plasma pushes out (the ratio, plasma pressure over $B^2/2\mu_0$ , is the tokamak designer's β). Astrophysics runs on the same anatomy: magnetic tension powers the whip-crack of solar flares and holds coronal loops taut; magnetic pressure inflates the bubbles that pulsars blow into their nebulae. The tensor's conceptual legacy is even larger. It is the prototype of every field theory's stress–energy tensor: the object that tells gravity where energy and momentum sit. In general relativity, $T^{\mu\nu}$ — whose spatial block is exactly this article's tensor — sources spacetime curvature, so the tension in a magnetar's $10^{11}$ T field literally gravitates. And the negative-pressure lesson (tension is just negative pressure, and fields have it naturally) is the conceptual seed of cosmology's dark-energy discussions: the universe's acceleration is a stress-tensor question asked of the vacuum itself.

Origins · 1 minJames Clerk Maxwell and how this was discovered

James Clerk Maxwell

1831–1879 · Edinburgh, Aberdeen, London, Cambridge

The crown article told Maxwell's life; here is the specific obsession this page inherits. Maxwell regarded Faraday's lines of force not as metaphor but as mechanics awaiting equations, and the Treatise (1873) devotes its strangest, most prophetic chapter to computing the stresses a medium must sustain if Faraday's picture is true: tension along the lines, pressure across them, both equal to the energy density. He checked the construction by re-deriving the known forces of electrostatics and magnetostatics from stresses alone — the calculation worked example 2 repeats — and used it to predict the pressure of light fifteen years before Hertz made the waves and twenty-eight before Lebedev measured the push. For Maxwell the stresses were evidence of a real elastic ether; when relativity retired the ether, the stresses stayed — reinterpreted as the momentum flux of the field itself, needing no medium to stand in. It is a recurring Maxwell pattern: the scaffolding (vortex cells, idler wheels, elastic ether) always proved mortal, the tensors immortal. His stress tensor became the template on which Minkowski, Abraham, and eventually Einstein built the stress–energy tensors of relativity and gravitation — the direct ancestor of the object that now tells spacetime how to curve.

History

1839

— Faraday describes lines of force as stretched elastic bands pressing sideways on their neighbours

1873

— Maxwell's Treatise computes the tension u and pressure u exactly, and derives the pressure of light from stress

1884

— Heaviside's vector recasting makes the tensor's divergence-form force law standard machinery

1900

— Lebedev measures light pressure on suspended vanes; Nichols and Hull confirm in 1901 — the tensor's flux reading verified

1908

— Minkowski embeds the stress tensor as the space-space block of the relativistic stress-energy tensor

1915

— Einstein's field equations make the stress-energy tensor the source of gravity: Maxwell's stresses now curve spacetime

1968

— Tokamak results make beta — plasma pressure over magnetic pressure — the defining ratio of fusion confinement

The physical picture

Faraday, who could not integrate, saw it first. Sketch the field of two opposite charges: the lines bow from one to the other like elastic cords, and the attraction looks like those cords contracting. Sketch two like charges: the lines from each crowd against the other's, refusing to cross, and the repulsion looks like sideways pressure in the crowd. Maxwell spent a chapter of the Treatise proving his teacher's picture exact: the electromagnetic field sustains a tension along each field line equal to the local energy density $u$ , and a pressure of the same magnitude $u$ perpendicular to it. Every electrostatic and magnetostatic force you have computed in this track — plate attracting plate, wire attracting wire, the magnet on your refrigerator — can be re-derived by drawing a surface through the field and totalling the rubber-band pulls and gas-like pushes across it.

Figure

Faraday made exact: field lines carry tension u along themselves and press outward with u across themselves

The modern reading is less picturesque and more powerful. The previous article put momentum in the field, at density $\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B}$ ; the natural next question is how momentum moves through the field. Charge has a density and a current; energy has a density ( $u$ ) and a current ( $\mathbf{S}$ ); momentum must have a density ( $\mathbf{g}$ ) and a current too. But momentum is a vector, so its current needs two indices — which component of momentum ( $i$ ), crossing a surface facing which way ( $j$ ) — and that is exactly what $T_{ij}$ is: the $i$ -component of momentum flowing per second through unit area with normal $\hat{\mathbf{j}}$ . A force on matter is nothing but momentum arriving from the field; add up the momentum current crossing a closed boundary and you have added up the force on whatever is inside. That is why the surface trick works, and why it must work — momentum that ends up inside had to cross the boundary to get there, and $T_{ij}$ is the toll booth's ledger. (In time-dependent fields, momentum can also accumulate in the field inside, which is the $-d\mathbf{p}_{\text{field}}/dt$ correction the next article installs; in statics the ledger is force = surface flux, cleanly.) Keep both readings. The Faraday reading tells you the answer's shape before any integral: lines pulling lengthwise, pressing sideways. The momentum-current reading tells you why the method is legal and hooks the tensor into the chapter's conservation program. The mathematics below serves both.

The tensor and its anatomy

The object first, then its assembly (Derivation A) and its eigen-anatomy (Derivation B):

T_{ij} = \varepsilon_0\left(E_iE_j - \tfrac12\delta_{ij}E^2\right) + \frac{1}{\mu_0}\left(B_iB_j - \tfrac12\delta_{ij}B^2\right) \qquad (1)

where: $i,j$ run over $x,y,z$ ; $E^2 = E_x^2+E_y^2+E_z^2$ and likewise $B^2$ ; $\delta_{ij}$ is the Kronecker delta; units are N/m² — pressure. The structure is worth reading before using: the dyadic parts $E_iE_j$ , $B_iB_j$ remember the field's direction (they are big along the field, zero across it); the isotropic parts $-\tfrac12\delta_{ij}E^2$ press equally in all directions, like a gas with negative pressure written in. Their competition is precisely the tension-along/pressure-across split — along the field the dyadic term wins by 2:1 and the net is $+u$ (tension); across it the dyadic term vanishes and the isotropic term delivers $-u$ (pressure). Note also what $T_{ij}$ is not: it is quadratic in the fields, so you may not compute it charge-by-charge and superpose — cross terms between two sources are exactly where interaction forces live. The tensor is symmetric ( $T_{ij} = T_{ji}$ ), and its symmetry is not cosmetic: it is what lets angular momentum be conserved with the density $\mathbf{r}\times\mathbf{g}$ and no extra torque term (Angular Momentum of the Electromagnetic Field leans on this; the chapter closer traces it to the deep structure). Its trace is $T_{ii} = -u$ — one energy density's worth — a fact that becomes the tracelessness of the relativistic stress–energy tensor once the time–time block joins ( $T^\mu{}_\mu = 0$ , the closing article). In static situations the force statement is pure surface flux:

\mathbf{F} = \oint_S \overleftrightarrow{T}\cdot d\mathbf{a}, \qquad F_i = \oint_S T_{ij}\,da_j \qquad \text{(statics)} \qquad (2)

where: $S$ is any closed surface enclosing the charges you want the force on, $d\mathbf{a}$ its outward area element. Any surface — that is the licence to be lazy in the best professional sense: route $S$ through the regions where the fields are simple (a symmetry plane, a far zone, the empty gap of a capacitor) and never visit the complicated interior. For time-dependent fields the honest statement acquires the momentum-storage term,

\mathbf{F} = \oint_S \overleftrightarrow{T}\cdot d\mathbf{a} - \frac{d}{dt}\int_V \varepsilon_0(\mathbf{E}\times\mathbf{B})\,d\tau, \qquad (3)

where: the last term is the field momentum of the previous article accumulating inside $V$ . Equation (3) is this chapter's momentum-conservation theorem wearing its integral clothes — stated here so you see where the tensor is headed, derived in the next article (Conservation of Linear Momentum in Electrodynamics), which owns the time-dependent bookkeeping. Everything in the present article runs on the static form (2).

Figure

The surface trick: to get the force on the upper charge, integrate the tensor over the midplane plus a far hemisphere — simple fields only, interior never consulted

The two canonical special cases, worth memorizing as numbers. At the surface of a conductor carrying surface charge $\sigma$ , the field is $\sigma/\varepsilon_0$ outside and zero inside, and the stress tensor delivers an outward tension

P_{\text{cond}} = \frac{\sigma^2}{2\varepsilon_0} = \tfrac12\varepsilon_0 E^2 \qquad (4)

where: the field tugs on its own source — every charged conductor is under suction from its field, trying to fly apart (only mechanical cohesion resists; charge a soap bubble and it visibly swells). And across any surface parallel to a magnetic field $B$ , the field presses with

P_{\text{mag}} = \frac{B^2}{2\mu_0} \approx 4.0\ \text{atm}\times\left(\frac{B}{1\ \text{T}}\right)^2, \qquad (5)

where: the quadratic scaling is the engineering headline: 1 T is a gentle 4 atm, 10 T is 400 atm — bottled scuba-tank pressures pushing outward on the windings — and 45 T (the strongest steady magnet built) is 8000 atm, why record magnets are exercises in structural mechanics as much as electromagnetism.

Derivation AAssembling the tensor — the static Lorentz force becomes a pure divergence

The goal is to show that for static fields the Lorentz force density $\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$ can be written as $f_i = \partial_j T_{ij}$ with $T_{ij}$ exactly equation (1) — no leftovers — so that the volume force integrates to the boundary flux (2). The strategy: use the static Maxwell equations to eliminate the sources $\rho$ and $\mathbf{J}$ in favor of field derivatives, then reorganize every term into the divergence of something quadratic. The electric and magnetic halves run in parallel; we do the electric half in full and let symmetry deliver the magnetic. Watch two moves: a product-rule identity that converts (field)×(curl field) into gradient-of-square minus advective terms, and the free insertion of $(\nabla\cdot\mathbf{B})\mathbf{B} = 0$ , which costs nothing but is what makes the final tensor symmetric between $\mathbf{E}$ and $\mathbf{B}$ .

Step 1. Eliminate the sources. Statics gives $\rho = \varepsilon_0\nabla\cdot\mathbf{E}$ (Gauss) and $\mathbf{J} = (\nabla\times\mathbf{B})/\mu_0$ (Ampère, no displacement term in statics). The force density becomes pure field:

\mathbf{f} = \varepsilon_0(\nabla\cdot\mathbf{E})\,\mathbf{E} + \frac{1}{\mu_0}(\nabla\times\mathbf{B})\times\mathbf{B}.

Step 2. Tame the curl term with the E0 product rule $\nabla(B^2/2) = (\mathbf{B}\cdot\nabla)\mathbf{B} + \mathbf{B}\times(\nabla\times\mathbf{B})$ , rearranged:

(\nabla\times\mathbf{B})\times\mathbf{B} = (\mathbf{B}\cdot\nabla)\mathbf{B} - \nabla\!\left(\tfrac12 B^2\right).

Why this identity is the whole game: it splits the magnetic force into an advective piece $(\mathbf{B}\cdot\nabla)\mathbf{B}$ — fields dragging along their own direction, the raw material of tension — and a gradient of the energy density — the raw material of isotropic pressure. The stress tensor is just these two pieces filed under one object.

Step 3. Symmetrize. In statics $\nabla\times\mathbf{E} = 0$ , so the same identity applies to $\mathbf{E}$ with zero curl term: $(\mathbf{E}\cdot\nabla)\mathbf{E} = \nabla(E^2/2) - \mathbf{E}\times(\nabla\times\mathbf{E}) = \nabla(E^2/2)$ ... wait, we need it the other way: $\varepsilon_0(\nabla\cdot\mathbf{E})\mathbf{E}$ stands alone. Add to it the free term $\varepsilon_0(\mathbf{E}\cdot\nabla)\mathbf{E} - \varepsilon_0\nabla(E^2/2)$ , which vanishes by the identity with $\nabla\times\mathbf{E}=0$ ; and add to the magnetic half the free term $(\nabla\cdot\mathbf{B})\mathbf{B}/\mu_0 = 0$ . Both halves now carry the same four-term pattern:

f_i = \varepsilon_0\left[(\partial_jE_j)E_i + (E_j\partial_j)E_i - \tfrac12\partial_i(E^2)\right] + \frac{1}{\mu_0}\left[(\partial_jB_j)B_i + (B_j\partial_j)B_i - \tfrac12\partial_i(B^2)\right].

Step 4. Recognize the divergence. Each bracket is exactly $\partial_j$ of the corresponding block of (1): by the product rule, $\partial_j(E_iE_j) = (\partial_jE_i)E_j + E_i(\partial_jE_j)$ , which is the first two terms, and $\partial_j(\tfrac12\delta_{ij}E^2) = \tfrac12\partial_i(E^2)$ , the third. Hence

f_i = \partial_j\!\left[\varepsilon_0\!\left(E_iE_j - \tfrac12\delta_{ij}E^2\right) + \frac{1}{\mu_0}\!\left(B_iB_j - \tfrac12\delta_{ij}B^2\right)\right] = \partial_j T_{ij} \qquad\Longrightarrow\qquad \boxed{\ F_i = \oint_S T_{ij}\,da_j\ }

by the divergence theorem. The force on everything inside is the flux of $T_{ij}$ through the boundary, with the interior never consulted. For time-dependent fields, Steps 1 and 3 pick up the displacement current and Faraday terms, which assemble — after one more product rule — into $-\partial_t[\varepsilon_0(\mathbf{E}\times\mathbf{B})]_i$ : the field-momentum storage term of equation (3). That completion, and its reading as a continuity equation for momentum, is the next article's derivation; we have built its flux object. Verified: reproduces Griffiths Eqs. 8.19–8.22 (with his sign convention: $T\cdot d\mathbf{a}$ = force the outside fields exert across $d\mathbf{a}$ on the inside); the two free insertions are identically zero by Gauss-for-B and statics curl-free E; contracting (1): $T_{ii} = \varepsilon_0(E^2 - \tfrac32E^2) + (B^2-\tfrac32B^2)/\mu_0 = -u$ as claimed.

Go deeper · Derivation B · 2 minThe eigen-anatomy — pure tension along the field, pure pressure across it

Derivation BThe eigen-anatomy — pure tension along the field, pure pressure across it

The goal is the statement quoted since the overview: at any point, the electromagnetic stress is a tension $u$ along the local field direction and a pressure $u$ in every perpendicular direction — i.e., equation (1)'s eigenvalues are $(+u, -u, -u)$ with the $+u$ eigenvector along the field. The strategy is direct diagonalization in adapted coordinates, electric case (magnetic is identical with $B^2/2\mu_0$ ); then a translation exercise: how a tensor's eigenvalues become forces on surfaces, which is where students lose the plot. Watch the factor-of-two bookkeeping — the dyadic term doubles the isotropic term along the field, and their sum flips sign exactly there.

Step 1. Adapt the axes. At the point of interest let $\mathbf{E} = E\,\hat{\mathbf{z}}$ , so $E_x = E_y = 0$ , $E_z = E$ . The tensor (1), electric part, becomes diagonal by inspection:

T = \varepsilon_0\begin{pmatrix} -\tfrac12E^2 & 0 & 0\\ 0 & -\tfrac12E^2 & 0\\ 0 & 0 & E^2 - \tfrac12E^2 \end{pmatrix} = \begin{pmatrix} -u & 0 & 0\\ 0 & -u & 0\\ 0 & 0 & +u \end{pmatrix}, \qquad u = \tfrac12\varepsilon_0E^2.

Why the sign flips only along $\hat{\mathbf{z}}$ : the dyadic $E_iE_j$ contributes $\varepsilon_0E^2 = 2u$ there and nothing anywhere else; the isotropic $-\tfrac12\delta_{ij}E^2$ contributes $-u$ everywhere. Along the field: $2u - u = +u$ . Across: $0 - u = -u$ . The tension is the dyad outvoting the pressure, two to one.

Step 2. Read eigenvalues as forces. Take a small surface patch with outward normal $\hat{\mathbf{n}}$ ; the force per area transmitted across it (from outside onto inside) is $T\cdot\hat{\mathbf{n}}$ . Patch facing along the field ( $\hat{\mathbf{n}} = \hat{\mathbf{z}}$ ): force per area $= +u\,\hat{\mathbf{z}}$ — directed outward, a pull: tension. Patch edge-on to the field ( $\hat{\mathbf{n}} = \hat{\mathbf{x}}$ ): force per area $= -u\,\hat{\mathbf{x}}$ — directed inward, a push: pressure. There are no shear entries in adapted axes: field lines do not drag surfaces sideways, they only pull along themselves and press across.

Step 3. The invariant summary. Since the decomposition used only the local field direction, it holds pointwise for any configuration: the electromagnetic field at each point acts mechanically like a bundle of cords under tension $u$ laid along the field lines, immersed in a gas of pressure $u$ pushing perpendicular. Mixed $\mathbf{E}$ and $\mathbf{B}$ at an angle simply superpose their two cord-and-gas systems (the tensor is additive even though forces are not superposable between sources — the fields here are the total fields).

Step 4. Two immediate dividends. (i) Conductor suction: just outside a conductor, $\mathbf{E} = (\sigma/\varepsilon_0)\hat{\mathbf{n}}$ is normal to the surface — the surface faces along the field, so it feels the tension eigenvalue: outward pull $\tfrac12\varepsilon_0E^2 = \sigma^2/2\varepsilon_0$ , equation (4), no integration needed. (ii) Solenoid burst: inside a solenoid $\mathbf{B}$ runs axially; the cylindrical wall is edge-on to the field, so it feels the pressure eigenvalue $B^2/2\mu_0$ pushing outward through the windings, equation (5). Two of magnet engineering's governing numbers, read off a diagonal matrix. Verified: eigenvalues sum to $-u$ matching the trace from Derivation A; conductor pull cross-checked against the E1 result (force per area on a conductor $= \sigma E_{\text{avg}} = \sigma\cdot\sigma/2\varepsilon_0$ , the classic averaging argument — same answer by a different route); solenoid pressure consistent with the magnetic-energy article's $u_B = B^2/2\mu_0$ and the virtual-work route.

Limiting cases

Case	Result	Meaning
surface along field ( $\hat{\mathbf{n}} \parallel \mathbf{E}$ or $\mathbf{B}$ )	tension $+u$ pulling outward	rubber-band reading; conductor suction $\sigma^2/2\varepsilon_0$
surface across field	pressure $u$ pushing inward	crowded-lines reading; solenoid burst $B^2/2\mu_0$
plane wave, surface facing the beam	$\langle T_{zz}\rangle = -\tfrac12\varepsilon_0E_0^2 = -\langle u\rangle$	momentum flux of light = radiation pressure, recovering the last article's $\langle S\rangle/c$
uniform field, closed surface	$\oint T\cdot d\mathbf{a} = 0$	no charge enclosed, no force — stresses balance exactly
far zone of a neutral system	$T \sim 1/r^6$ (static), surface integral $\to 0$	distant boundary feels nothing; forces are local ledger transfers
single point charge, any enclosing surface	$\oint T\cdot d\mathbf{a} = 0$	a charge exerts no net force on itself (statics); interaction cross-terms carry all forces
trace	$T_{ii} = -u$ ; relativistic $T^\mu{}_\mu = 0$	one energy density; photon masslessness in tensor form (chapter closer)

Common trapThe stress tensor is quadratic — you cannot superpose it source by source

The single most common misuse: computing $T_{ij}$ for charge 1's field, then for charge 2's field, adding, and wondering where the force went. Superposition holds for fields, never for the tensor: $T[\mathbf{E}_1 + \mathbf{E}_2]$ contains cross terms $\varepsilon_0(E_{1i}E_{2j} + E_{2i}E_{1j} - \delta_{ij}\,\mathbf{E}_1\cdot\mathbf{E}_2)$ , and those cross terms are the entire interaction force — each charge's private tensor integrates to zero around itself (no self-force in statics). Worked example 2 shows the cross terms carrying Coulomb's law across the midplane. Two related traps. First, $T\cdot\hat{\mathbf{n}}$ is not "the pressure on a physical wall you happen to draw" — it is momentum flux across a mathematical surface; only when the surface coincides with a real interface (conductor skin, solenoid winding) does it become a contact force you could measure with a gauge. Second, mind the convention: with Griffiths's sign, $\oint T\cdot d\mathbf{a}$ is the force exerted on the inside by the fields at the boundary; flip your normal and you compute the reaction. Fix the convention once, then trust the algebra over intuition — especially for tension, which "a field pulling on a surface" persistently refuses to feel like.

Worked examples

1 · The pull on a capacitor plate and the push on a magnet coil (confidence builder)

(a) A parallel-plate capacitor has plate area $A = 100$ cm² and field $E = 1.0\times10^6$ V/m in the gap. Find the attractive force on a plate via the stress tensor. (b) A 10 T superconducting solenoid has bore radius 5.0 cm. Find the outward pressure on its windings and the total radial burst force per metre of length.

Step 1. (a) Choose the surface: a flat box hugging the inner face of one plate, its large face lying in the gap, facing the field head-on... more precisely the gap field $\mathbf{E}$ is normal to the plate, so the boxed face has $\hat{\mathbf{n}}$ along $\mathbf{E}$ : pure tension eigenvalue. Force per area $= \tfrac12\varepsilon_0E^2 = 0.5\times8.85\times10^{-12}\times(10^6)^2 = \mathbf{4.4\ Pa}$ , pulling the plate into the gap (toward the other plate — attraction).

Step 2. (a) Total force: $F = 4.43\times0.010 = \mathbf{0.044\ N}$ — about the weight of a sugar cube for a megavolt-per-metre gap. Cross-check by the energy method: $F = \tfrac12\varepsilon_0E^2A$ is exactly $dU/dx$ for the stored energy $\tfrac12\varepsilon_0E^2 A x$ — the two routes agree because both are reading the same tensor.

Step 3. (b) The windings are edge-on to the axial $\mathbf{B}$ : pressure eigenvalue $B^2/2\mu_0 = 100/(2\times1.257\times10^{-6}) = \mathbf{4.0\times10^7\ Pa} = 393$ atm outward. This is why magnet coils are potted in steel-reinforced epoxy.

Step 4. (b) Per metre of solenoid, the radial force integrates like pressure in a pipe: force per length trying to split the coil along a diameter $= P\times 2R = 4.0\times10^7\times0.10 = \mathbf{4.0\times10^6\ N/m}$ — four hundred tonnes-force per metre, carried by the hoop stress of the windings. Magnet quench protection and coil pre-stressing budgets start from exactly this number.

2 · Coulomb's law from a plane you never touch (exam level)

Two equal point charges $q = 1.0\ \mu$ C sit at $z = \pm d$ with $d = 5.0$ cm. Compute the force on the upper charge by integrating the stress tensor over the midplane $z = 0$ (closed at infinity), and check against Coulomb.

Step 1. Geometry of the total field on the midplane. By symmetry the two $z$ -components cancel and the in-plane (cylindrical radial) components add: at radius $s$ from the axis, $E_z = 0$ and

E_s(s) = 2\cdot\frac{q}{4\pi\varepsilon_0}\cdot\frac{s}{(s^2+d^2)^{3/2}}.

The field lies in the plane — the plane is everywhere edge-on to the field, so we expect the pressure eigenvalue: the crowded-midplane push of the hero figure.

Step 2. The relevant component. The surface enclosing the upper charge is the plane $z=0$ with outward normal $-\hat{\mathbf{z}}$ (plus a vanishing far hemisphere — $T \sim 1/s^6$ kills it). We need $F_z = \oint T_{zj}da_j = -\int T_{zz}\,da$ . With $E_z = 0$ : $T_{zz} = \varepsilon_0(E_z^2 - \tfrac12E^2) = -\tfrac12\varepsilon_0E_s^2$ , so $F_z = +\tfrac12\varepsilon_0\int E_s^2\,da$ — positive: the upper charge is pushed up, away. Repulsion, before any integration.

Step 3. Integrate:

F_z = \frac{\varepsilon_0}{2}\int_0^\infty \left[\frac{2q}{4\pi\varepsilon_0}\right]^2 \frac{s^2}{(s^2+d^2)^3}\,2\pi s\,ds = \frac{q^2}{4\pi\varepsilon_0}\int_0^\infty \frac{s^3\,ds}{(s^2+d^2)^3}\cdot ...

collecting constants: $F_z = \dfrac{q^2}{4\pi^2\varepsilon_0}\cdot\pi\displaystyle\int_0^\infty\frac{s^3\,ds}{(s^2+d^2)^3}$ . Substitute $w = s^2+d^2$ : the integral is $\tfrac12\int_{d^2}^\infty (w-d^2)w^{-3}dw = \tfrac12\left[\tfrac{1}{d^2} - \tfrac{d^2}{2d^4}\right] = \tfrac{1}{4d^2}$ . Hence

F_z = \frac{q^2}{4\pi\varepsilon_0}\cdot\frac{1}{4d^2} = \frac{q^2}{4\pi\varepsilon_0\,(2d)^2}.

Step 4. Read and evaluate. $(2d)$ is the charge separation: the surface integral has delivered exactly Coulomb's law, from field values on a plane neither charge occupies. Numbers: $F = (8.99\times10^9)\times(10^{-6})^2/(0.10)^2 = \mathbf{0.90\ N}$ . The interaction travelled through the field as momentum flux across the midplane — 0.90 newtons of momentum per second, crossing an empty plane forever, in a perfectly static configuration. That standing momentum current is the force.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §8.2.2–8.2.3 · Jackson, Classical Electrodynamics, 3rd ed., §6.7 · Zangwill, Modern Electrodynamics, §15.7–15.9 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §5.10 · Landau & Lifshitz, Classical Theory of Fields, §33

Provenance. Equations cross-checked against Griffiths Introduction to Electrodynamics 4th ed. §8.2 (Eqs. 8.19–8.22), Jackson Classical Electrodynamics 3rd ed. §6.7, and Zangwill Modern Electrodynamics §15.7. A verifier should: (1) re-derive the local force law $\mathbf{f}=\nabla\cdot\overleftrightarrow{T}-\varepsilon_0\mu_0\partial_t\mathbf{S}$ from Maxwell's equations and the vector identity for $\nabla E^2$ ; (2) confirm the conductor pressure $\sigma^2/2\varepsilon_0$ and the plane-wave momentum flux $\langle T_{zz}\rangle=-\tfrac12\varepsilon_0E_0^2$ ; (3) verify the solar-radiation-pressure estimate $P=I/c\approx4.5\,\mu\mathrm{Pa}$ using $I\approx1361\,\mathrm{W/m^2}$ and $c=2.998\times10^8\,\mathrm{m/s}$ ; (4) check $T^\mu{}_\mu=0$ for the relativistic form. Addendum for this edition: per the chapter's division of labor this article derives the STATIC assembly $f_i = \partial_jT_{ij}$ in full (Derivation A) and quotes the time-dependent completion on credit to Conservation of Linear Momentum in Electrodynamics, which owns item (1)'s $-\varepsilon_0\mu_0\partial_t\mathbf{S}$ term; radiation pressure numerics are owned by Momentum Density of the Electromagnetic Field. Worked numbers recomputed: capacitor $\tfrac12\varepsilon_0E^2 = 4.43$ Pa at $10^6$ V/m, $F = 0.044$ N on 100 cm²; solenoid $B^2/2\mu_0 = 3.98\times10^7$ Pa at 10 T ( $3.98\times10^5$ Pa = 3.9 atm at 1 T), burst force $2RP = 4.0\times10^6$ N/m at $R = 5$ cm; midplane integral $\int_0^\infty s^3(s^2+d^2)^{-3}ds = 1/4d^2$ giving $F = q^2/4\pi\varepsilon_0(2d)^2 = 0.90$ N for 1 μC at 10 cm.

Builds on: Work Done by Electromagnetic Fields on Charges · The Poynting Vector · Momentum Density of the Electromagnetic Field · Energy Density of the Electromagnetic Field · Charge Conservation and the Continuity Equation

Leads to: Conservation of Linear Momentum in Electrodynamics · Newton's Third Law in Electrodynamics · Hidden Momentum in Magnetic Systems · Angular Momentum of the Electromagnetic Field

How can empty space transmit a force — is the tension in the field lines physically real?

As real as any pressure you have ever measured, with the caveat that "field line" is our drawing, not nature's hardware. The operational content: put a real surface (conductor skin, coil winding, plasma boundary) anywhere in a field, and the material stress you must supply to hold it in place is exactly $T\cdot\hat{\mathbf{n}}$ — measurable with strain gauges, routinely measured in magnet engineering. Between the surfaces, "tension along the lines" summarizes how that momentum flux is oriented; no rope is there, but momentum genuinely flows through the empty region (worked example 2's midplane carries 0.90 N of it, statically, forever). Maxwell believed an ether supplied the substance under the stress; relativity removed the ether and left the stresses standing, carried by the field itself. If "the field carries momentum flux" feels less intuitive than "invisible rubber bands," use the rubber bands — they give correct answers precisely because Derivation B proved them equivalent to the tensor.

When do I use the stress tensor instead of just summing Coulomb and Biot–Savart forces?

Three signatures favor the tensor. First: extended or hidden charge distributions — force on a conductor (where $\sigma$ arranges itself), on a dielectric interface, on a plasma; the tensor needs only boundary fields, not the self-consistent interior. Second: high symmetry — if a plane, sphere, or cylinder exists where the total field is simple, the surface integral is often three lines (worked example 2). Third: whenever you need a pressure (burst force on a coil, pinch force on a plasma column, light on a sail) — the tensor's eigenvalues are the pressure, no summation ever converges to that as cleanly. Conversely, for a handful of point charges in known positions, direct Coulomb is faster; the tensor route reproduces it but with more machinery. And in time-dependent problems remember the storage term (3): the surface integral alone gives the force only after subtracting $d\mathbf{p}_{\text{field}}/dt$ — the next article's business.

Is the magnetic pressure B squared over 2 mu-zero the same thing as the energy density, and why are they numerically equal?

Numerically identical, conceptually distinct, and the equality is not an accident. $u_B = B^2/2\mu_0$ is energy per volume (a scalar); the magnetic pressure is the transverse eigenvalue of the stress tensor (one component of a tensor). They coincide because both descend from the same quadratic invariant — Derivation B showed the eigenvalues are $\pm u$ , so "the" magnitude of magnetic stress is forced to equal the energy density. The same holds electrically. This is why you can often shortcut with virtual work: compressing a field region by $dV$ against pressure $u$ costs $u\,dV$ , which is exactly the energy added — the two bookkeepings are consistent by construction. But keep the tensor character in mind: along the field the same $u$ appears as tension, so a magnetic field pulls on the pole faces while pushing on the side walls — a scalar "magnetic pressure" cannot represent both, and treating it as a gas pressure in all directions is a classic exam error.

What exactly are the off-diagonal components of the tensor telling me?

Shear: the $i$ -component of force transmitted across a surface whose normal points along $j \neq i$ — momentum flowing sideways relative to its own direction. They are nonzero whenever the field is oblique to your chosen axes: for $\mathbf{E}$ at 45° in the $xz$ plane, $T_{xz} = \varepsilon_0E_xE_z > 0$ , meaning a horizontal surface feels a horizontal drag as well as a vertical push. Physically this is how a tilted field "tries to straighten" — the shear pair $T_{xz} = T_{zx}$ exerts the torque-per-volume that field-line tension needs to act around corners. In adapted (principal) axes the shear vanishes — Derivation B — which is the formal statement that field lines pull only along themselves and press only across. Engineers meet the off-diagonals as the torque integrand in rotating machines: integrate $\mathbf{r}\times(T\cdot\hat{\mathbf{n}})$ over a cylinder in an air gap and the shear components deliver the motor torque (the "Maxwell stress method" in every electric-machine textbook).

Does a point charge's own stress tensor exert a net force on it?

Integrate the single-charge tensor over any sphere around a static charge: by symmetry the answer is exactly zero — no self-force, consistent with a charge at rest happily staying put. Two footnotes make this interesting. First, the self-energy is another story: the same tensor integrated for energy diverges as the sphere shrinks (the classical self-energy problem), and the electromagnetic self-stress of a finite charged sphere famously fails to be self-consistent without non-electromagnetic cohesion — "Poincaré stresses" — a genuine crack in classical electron models (the E11 chapter's radiation-reaction articles inherit it). Second, for accelerated charges the momentum ledger around the particle no longer closes locally, and the mismatch is the radiation-reaction force. Within this chapter's scope: static self-force zero, interaction forces carried entirely by cross terms, and the deep self-consistency questions deferred, by name, to the advanced chapter.

Which convention should I memorize — force on the inside, or momentum flowing outward?

Pick Griffiths's and phrase it twice. Algebraic: $F_i = \oint T_{ij}\,da_j$ with $d\mathbf{a}$ the outward normal gives the force exerted on the contents of the enclosed region. Flux phrasing: $-T_{ij}$ is then the outward flux of $i$ -momentum, or equivalently $T_{ij}$ is the rate momentum flows in — momentum arriving = force. (Some texts, especially plasma and relativity literature, define the sign so the tensor is the outward flux; their conservation law reads $\partial_t g_i + \partial_j T^{\text{out}}_{ij} = -f_i$ with $T^{\text{out}} = -T^{\text{Griffiths}}$ .) Nothing physical hangs on the choice, but mixing conventions mid-calculation reverses attractions into repulsions. The self-audit that catches it every time: run your setup on a charged conductor and check you get an outward pull of $\sigma^2/2\varepsilon_0$ — if your capacitor plates repel, flip a sign and rerun. Worked example 2's "repulsion appears before integrating" step is that audit in action.

How does this tensor connect to the stress-energy tensor of relativity and gravity?

It is the space–space block. Relativity packages energy density $u$ , energy flux $\mathbf{S}$ , momentum density $\mathbf{g}$ , and momentum flux into one symmetric object: $T^{00} = u$ , $T^{0i} = S_i/c$ , $T^{i0} = c\,g_i$ , and $T^{ij} = -T_{ij}^{\text{(this article)}}$ (sign per the outward-flux convention). Three payoffs land immediately. Symmetry $T^{0i} = T^{i0}$ is $\mathbf{g} = \mathbf{S}/c^2$ — the momentum article's central identity, revealed as one tensor's symmetry (the chapter closer owns this). Tracelessness $T^\mu{}_\mu = 0$ — check: $u - u_{\text{...}}$ , i.e. $T^{00} + T^i{}_i = u + (-u)\cdot... = 0$ using this article's trace $-u$ — encodes the photon's masslessness. And in general relativity this combined object sources curvature: field stress gravitates, so the tension in a magnetar's field contributes to its spacetime. One object, four conservation laws, one gravity source — assembled in Conservation Laws and Spacetime Symmetries at the chapter's close.