The Electromagnetic Stress-Energy Tensor

Sunlight pushes. Not metaphorically — a mirror held in space at Earth's orbit feels a real pressure of about nine micropascals, and in 2019 a spacecraft the size of a loaf of bread unfurled a 32 m² sail and climbed to a higher orbit on nothing but that push. No fuel. The light carried momentum, and gave it up. So the electromagnetic field must be a mechanical object: it holds energy, transports energy, stores momentum, and presses on surfaces. Pre-relativistic physics tracked those four facts with two separate pieces of apparatus — Poynting's theorem for the energy, Maxwell's stress tensor for the force. Relativity says they were never separate:

T^{\mu\nu} \;=\; \frac{1}{\mu_0}\Big[-F^{\mu\alpha}F^{\nu}{}_{\alpha} \;+\; \tfrac14\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\Big].

One symmetric tensor. Its $00$ slot is the energy density. Its $0i$ slots are the Poynting flux and the momentum density — the same numbers, because the tensor is symmetric. Its $ij$ block is the Maxwell stress. Four slices of one object, and the object is quadratic in a field tensor you already know.

Key takeaways

Signature $(+,-,-,-)$ , $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. Dictionary: $F^{0i} = -E^i/c$ , $F^{ij} = -\epsilon^{ijk}B_k$ , $J^\mu = (c\rho,\vec J)$ .
The tensor: $\boxed{T^{\mu\nu} = \mu_0^{-1}\big[-F^{\mu\alpha}F^{\nu}{}_{\alpha} + \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]}$ — symmetric, traceless, gauge-invariant, quadratic in $F$ .
The slots are the physics. $T^{00} = u = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ · $T^{0i} = S^i/c$ · $T^{ij} = -\sigma^{ij}$ . Symmetry $T^{0i} = T^{i0}$ identifies energy flux with momentum density: $\vec g = \vec S/c^2$ .
The trace-subtraction term IS the first invariant, $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ — owned by `aq-sr-electromagnetic-field-invariants`. The $\tfrac14$ is exactly the coefficient making $T^\mu{}_\mu = 0$ , and tracelessness is the fingerprint of a massless field.
Radiation pressure, with a real number. Sunlight at Earth delivers $I = 1361\ \mathrm{W\,m^{-2}}$ , so $P = I/c = \mathbf{4.540\ \mu Pa}$ on a black absorber and $2I/c = \mathbf{9.080\ \mu Pa}$ on a mirror. Feeble — but it never switches off, and a sail is free.
Key idea: the sign is not a convention, it is a measurement. $T^{00} = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ is a sum of squares. It cannot be negative in any frame, in any signature, ever. Get the sign of $T^{\mu\nu}$ wrong and you predict a negative energy density for a static electric field — and the trace will not tell you, because the wrong tensor is traceless too.

The core insight

The stress-energy tensor is the field's mechanical inventory, and it fits on one page because relativity forces energy, energy flux, momentum and stress to be components of a single object. It is also this chapter's sternest lesson in verification. The tensor's structure — symmetric, traceless, quadratic, gauge-invariant — is fixed by arguments that fix everything except the overall sign. And the sign is checked by exactly one quantity: $T^{00}$ , positive because it is a sum of squares. Tracelessness, symmetry and gauge invariance all survive a global sign flip. Only the energy density notices.

Where this lives today

The Maxwell stress tensor is the working tool of every electromagnetic force calculation that would be intolerable by direct summation: the force between the poles of a magnet, the pull on a transformer core, the attraction of a charge to a grounded plane. You draw a surface, integrate $\sigma^{ij}n_j$ , and never touch the sources. Finite-element solvers do exactly this, because a surface integral of a computed field is far better behaved numerically than a volume integral over a singular charge density. Radiation pressure runs from the trivial to the cosmic. Optical tweezers trap a single bacterium using the momentum of focused laser light — Ashkin's 2018 Nobel Prize. Solar sails turn the $9\ \mu\mathrm{Pa}$ of equation (2) into propellant-free propulsion. And inside a star the $p = u/3$ of the third derivation is one of the two pressures holding the thing up against its own gravity — in the most massive stars, the dominant one. The tensor's deepest career is gravitational. The Einstein field equation reads $G^{\mu\nu} = (8\pi G/c^4)\,T^{\mu\nu}$ , and the $T^{\mu\nu}$ on the right is this tensor whenever the matter is electromagnetic. Light gravitates, and it gravitates through the object on this page — which is why tracelessness matters (a photon gas curves spacetime differently from dust of the same energy density), and why $T^{00} > 0$ is not a fussy detail. Feed general relativity a stress-energy tensor with the wrong sign and you get a universe in which light has negative energy, repels matter, and violates every energy condition ever written down.

The physical picture

You already know the field carries energy: $u = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ . You know it carries energy flux: the Poynting vector $\vec S = (\vec E\times\vec B)/\mu_0$ , watts per square metre. And you know it presses on things: the Maxwell stress tensor $\sigma^{ij}$ , which lets you compute the force on a charged conductor by integrating over a surface that encloses it, never touching the charge. Relativity's claim is that these are not three facts but one tensor read in three places. The argument is short. Energy density and momentum density cannot be separate frame-independent notions, because a boost mixes energy into momentum — whatever object holds the energy density must have slots the boost can rotate the momentum density into, and a rank-2 tensor is the smallest object that can do it. The rest is then forced: $T^{\mu\nu}$ must be quadratic in the fields (energy goes as field squared), built from $F^{\mu\nu}$ alone (gauge invariance — energy is measurable, potentials are not), and symmetric (or angular momentum is not conserved). There are exactly two tensors you can build from $F$ with those properties: $F^{\mu\alpha}F^{\nu}{}_{\alpha}$ and $\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$ . Demanding tracelessness fixes their ratio. What it does not fix is the overall sign — and that is where this chapter's bases went wrong.

The core equations

The tensor this page owns:

\boxed{\;T^{\mu\nu} \;=\; \frac{1}{\mu_0}\left[\,-F^{\mu\alpha}F^{\nu}{}_{\alpha} \;+\; \tfrac14\,\eta^{\mu\nu}\,F_{\alpha\beta}F^{\alpha\beta}\right]\;} \qquad (1)

where: $F^{\mu\nu}$ is the field tensor (`aq-sr-electromagnetic-field-tensor`), $F^{\nu}{}_{\alpha} = F^{\nu\beta}\eta_{\beta\alpha}$ , and $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ is the first invariant (`aq-sr-electromagnetic-field-invariants`). Physical reading: quadratic in $F$ , so it does not care how you sign $F$ — flip $F\to-F$ and $T$ is unchanged. The sign in (1) is fixed by the signature, and by the single requirement that $T^{00}$ come out positive. An equivalent form, and the one the electromagnetism track's `aq-em-stress-energy-tensor` prints, is $T^{\mu\nu} = \mu_0^{-1}\big[F^{\mu\alpha}F_{\alpha}{}^{\nu} + \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ — identical, because $F_{\alpha}{}^{\nu} = -F^{\nu}{}_{\alpha}$ . The components, which are the reason anyone cares:

T^{00} = u = \tfrac12\!\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right), \qquad T^{0i} = T^{i0} = \frac{S^i}{c} = c\,g^i, \qquad T^{ij} = -\sigma^{ij} \qquad (2)

where: $\vec S = (\vec E\times\vec B)/\mu_0$ is the Poynting vector, $\vec g = \varepsilon_0\,\vec E\times\vec B = \vec S/c^2$ the momentum density, and $\sigma^{ij} = \varepsilon_0(E^iE^j - \tfrac12\delta^{ij}E^2) + \mu_0^{-1}(B^iB^j - \tfrac12\delta^{ij}B^2)$ the Maxwell stress tensor. Physical reading: the symmetry of $T$ is a physical claim, not a formality. $T^{0i}$ is "the flux of energy in direction $i$ "; $T^{i0}$ is "the density of $i$ -momentum, times $c$ ". Equating them gives $\vec g = \vec S/c^2$ — energy in motion is momentum, which is why a sail works. The trace, and what it means:

T^{\mu}{}_{\mu} \;=\; \eta_{\mu\nu}T^{\mu\nu} \;=\; \frac{1}{\mu_0}\left[-F^{\mu\alpha}F_{\mu\alpha} + \tfrac14\cdot 4\cdot F_{\alpha\beta}F^{\alpha\beta}\right] \;=\; 0 \qquad (3)

where: the $\tfrac14$ and the factor $\eta^\mu{}_\mu = 4$ are what cancel. Physical reading: tracelessness is the fingerprint of masslessness. Dust of rest-mass density $\rho$ has $T^\mu{}_\mu = \rho c^2 \ne 0$ ; the field has none, which is why isotropic radiation obeys $p = u/3$ rather than $p = 0$ . ⚠ But note what (3) does not do: flip the sign of the whole bracket and it still vanishes. Tracelessness cannot detect a sign error. And the local test that can:

\boxed{\;T^{00} \;=\; \tfrac12\!\left(\varepsilon_0E^2 + \frac{B^2}{\mu_0}\right) \;>\; 0\;}\qquad\text{for every non-zero field, in every frame} \qquad (4)

where: nothing. There is no convention in equation (4). Physical reading: $T^{00}$ is a sum of squares with positive coefficients. A static electric field has positive energy density; that is measurable and not negotiable. Equation (4) is the only member of this set that fixes the overall sign of (1) — and it is the one check the bases did not perform.

Building the tensor, and pinning its sign

Derivation AThe two-term structure is FORCED — and then the sign is forced by the energy density

The goal is to construct $T^{\mu\nu}$ from nothing but requirements, and to be precise about which requirement fixes which part of the answer. This matters, because the structural requirements — symmetry, tracelessness, gauge invariance, quadratic in $F$ — fix the tensor up to an overall sign, and every one of them is satisfied by the wrong sign as well as the right one. The sign needs a separate, physical argument, and it gets one.

Step 1. List the candidates. $T^{\mu\nu}$ must be (i) quadratic in the fields, since energy density goes as field squared; (ii) built from $F^{\mu\nu}$ alone, since energy is gauge-invariant and observable while $A^\mu$ is neither; (iii) symmetric, or angular momentum is not conserved. There are exactly two such objects: F^{\mu\alpha}F^{\nu}{}_{\alpha} \qquad\text{and}\qquad \eta^{\mu\nu}\,F_{\alpha\beta}F^{\alpha\beta}. Why: the first is symmetric because relabelling the dummy $\alpha$ swaps $\mu\leftrightarrow\nu$ ; the second because $\eta^{\mu\nu}$ is. There is nothing else — the menu really is this short, which is what makes the construction feel inevitable.

Step 2. Write the general combination and impose tracelessness. Put $T^{\mu\nu} = \mu_0^{-1}\big[a\,F^{\mu\alpha}F^{\nu}{}_{\alpha} + b\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ . Contracting with $\eta_{\mu\nu}$ : T^{\mu}{}_{\mu} = \mu_0^{-1}(a + 4b)\,F_{\alpha\beta}F^{\alpha\beta} = 0 \quad\Longrightarrow\quad b = -\tfrac{a}{4}. Why: tracelessness fixes the RATIO and nothing else. It leaves $a$ completely free — including its sign. That is the whole trap: the check the bases perform is the check that cannot fail.

Step 3. Fix $a$ by computing $T^{00}$ and demanding it be the energy density. With the locked dictionary $F^{0i} = -E^i/c$ we get $F^{0}{}_{i} = F^{0j}\eta_{ji} = +E^i/c$ , so the first contraction is F^{0\alpha}F^{0}{}_{\alpha} = \sum_i\left(-\frac{E^i}{c}\right)\left(+\frac{E^i}{c}\right) = -\frac{E^2}{c^2}. With $\eta^{00} = 1$ , $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ and $b = -a/4$ : T^{00} = \frac{1}{\mu_0}\left[a\left(-\frac{E^2}{c^2}\right) - \frac{a}{4}\cdot 2\left(B^2 - \frac{E^2}{c^2}\right)\right] = -\frac{a}{2}\left(\varepsilon_0E^2 + \frac{B^2}{\mu_0}\right) = -a\,u , using $1/(\mu_0c^2) = \varepsilon_0$ . Demanding $T^{00} = +u$ therefore gives \boxed{\;a = -1,\qquad b = +\tfrac14\;} Why: this is the step the bases skipped. They wrote $a = +1$ — the $(-,+,+,+)$ value — while declaring $(+,-,-,-)$ , and never substituted back to check the energy density. With $a=+1$ the formula returns $T^{00} = -u$ : a negative electromagnetic energy density, which is not a sign convention but a false statement about the world.

Step 4. Confirm the other components fall out right, because a sign error propagates. With $a=-1$ : $T^{0i} = +S^i/c$ and $T^{ij} = -\sigma^{ij}$ . With $a=+1$ both flip — $T^{0i} = -S^i/c$ (energy flowing the wrong way) and $T^{ij} = +\sigma^{ij}$ (tension and pressure exchanged). Why: the sign is global. Get $T^{00}$ right and the Poynting flux and the Maxwell stresses come right for free; get it wrong and everything is wrong at once, silently.

Step 5. Check the divergence — the independent confirmation. Only with $a = -1$ does the tensor satisfy \partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda, whose right-hand side is minus the Lorentz four-force density $f^\nu = F^{\nu\lambda}J_\lambda = (\vec J\cdot\vec E/c,\ \rho\vec E + \vec J\times\vec B)$ . With $a=+1$ the divergence comes out $+F^{\nu\lambda}J_\lambda$ — the field would gain energy by doing work on charges. Why: this is why base 1951 is self-refuting. It prints the $a=+1$ tensor and the correct divergence law and the correct force density, and those three cannot all be true. Only $T^{00} = +u$ closes the books. (The balance law is developed in `aq-sr-field-energy-momentum-conservation`.) *Verified in verify_aq-sr-electromagnetic-stress-energy-tensor.py: $F^{\mu\nu}$ is rebuilt symbolically from $A^\mu = (\phi/c,\vec A)$ ; both tensors ( $a=\pm1$ ) are constructed; $T^{00}_{a=-1} - u = 0$ and $T^{00}_{a=+1} + u = 0$ are proved; $T^{0i} = S^i/c$ and $T^{ij} = -\sigma^{ij}$ are confirmed for all nine spatial components; both tensors are shown to be traceless; and $\partial_\mu T^{\mu\nu} + F^{\nu\lambda}J_\lambda = 0$ is verified component by component for arbitrary $A^\mu(t,x,y,z)$ .*

Go deeper · Derivation B · 2 minOwned result: the symmetry of T is why light pushes — energy flux IS momentum density

Derivation BOwned result: the symmetry of T is why light pushes — energy flux IS momentum density

The single most surprising line of equation (2) is $T^{0i} = T^{i0}$ . It looks like a formality — of course a symmetric tensor has symmetric components — but the two slots were defined by different physics, and their equality is a prediction that had to be checked against experiment. It is the reason a solar sail is possible, and the reason the field must be granted momentum of its own if Newton's third law is to survive.

Step 1. Read $T^{0i}$ as a flux. In a stress-energy tensor, $T^{\mu\nu}$ means "the flux of the $\mu$ -component of four-momentum across a surface of constant $x^\nu$ ". So $T^{0i}$ is the flux of $p^0 = E/c$ in the direction $i$ — the Poynting vector over $c$ : T^{0i} = \frac{S^i}{c},\qquad \vec S = \frac{\vec E\times\vec B}{\mu_0}\ \ [\mathrm{W\,m^{-2}}]. Why: this slot is pure nineteenth-century electromagnetism — Poynting, 1884 — and needs no relativity at all.

Step 2. Read $T^{i0}$ as a density. By the same rule, $T^{i0}$ is the flux of $p^i$ across a surface of constant $x^0 = ct$ — which is to say the density of $i$ -momentum, times $c$ : T^{i0} = c\,g^i . Why: a "flux across a surface of constant time" is just a density. Nothing is assumed about $\vec g$ ; it is defined by this slot.

Step 3. Impose the symmetry the construction already gave us. $T^{0i} = T^{i0}$ forces \frac{S^i}{c} = c\,g^i \qquad\Longrightarrow\qquad \boxed{\;\vec g = \frac{\vec S}{c^2} = \varepsilon_0\,\vec E\times\vec B\;} Why: a physical result obtained from an index symmetry. Energy flux and momentum density were separately defined, separately measured quantities, and relativity says they are one object viewed along different axes. The reason is $E = mc^2$ : energy has inertia, so a stream of energy is automatically a stream of momentum, and the conversion factor can only be $c^2$ .

Step 4. Cash it out as a pressure. A beam of intensity $I = |\vec S|$ striking a black absorber head-on delivers momentum at rate $g\,c = I/c$ per unit area per unit time. A perfect mirror reverses the beam's momentum instead of absorbing it, doubling the transfer: P_{\text{absorb}} = \frac{I}{c}, \qquad P_{\text{reflect}} = \frac{2I}{c}. Why: this is the equation a solar sail is built on, and it is a two-line consequence of an off-diagonal symmetry in a 4×4 matrix. The mirror gets twice the push and no heat — which is why sails are shiny.

Step 5. Note that the field's momentum is not optional. There are ordinary configurations — a charge beside a magnetic dipole — in which the mutual forces are not equal and opposite, and Newton's third law appears to fail. It does not: the missing momentum is in the field, exactly $\vec g = \varepsilon_0\vec E\times\vec B$ per unit volume. Why: refuse the field its momentum and you must give up Newton's third law. Physics kept the conservation law and granted the field an inventory — and this tensor is that inventory. *Verified in verify_aq-sr-electromagnetic-stress-energy-tensor.py: $T^{0i} - T^{i0} = 0$ symbolically for arbitrary fields; $\vec g = \vec S/c^2$ is confirmed; and the plane wave gives $|\vec S| = c\,u$ , so the momentum density is exactly $u/c$ — the photon relation.*

Go deeper · Derivation C · 2 minOwned result: tracelessness, and why isotropic radiation has p = u/3

Derivation COwned result: tracelessness, and why isotropic radiation has p = u/3

Tracelessness is the property that makes $T^{\mu\nu}$ electromagnetic rather than generic. It has a crisp physical meaning — the field has no rest mass — and one very concrete consequence: the equation of state of radiation, which fixes the pressure inside a star and the expansion history of the early universe. It is also, as the warning cards insist, the check that cannot catch the sign error.

Step 1. Compute the trace. Contracting (1) with $\eta_{\mu\nu}$ and using $\eta^{\mu\nu}\eta_{\mu\nu} = 4$ : T^{\mu}{}_{\mu} = \frac{1}{\mu_0}\left[-F^{\mu\alpha}F_{\mu\alpha} + \tfrac14\cdot4\cdot F_{\alpha\beta}F^{\alpha\beta}\right] = 0 . Why: the cancellation is exact and it is a $D=4$ accident. In $D$ spacetime dimensions the same computation gives $T^\mu{}_\mu \propto (D-4)/4$ , non-zero unless $D=4$ . Tracelessness is a feature of gauge fields in four dimensions, not of gauge fields generally.

Step 2. Interpret it. For a fluid of rest-mass density $\rho$ and pressure $p$ , $T^\mu{}_\mu = \rho c^2 - 3p$ . Setting this to zero gives $\rho = 0$ — no rest mass — or, for a gas of massless quanta, $u = 3p$ . Why: the trace is a rest-mass meter. Dust has trace $\rho c^2$ ; radiation has none. This is the classical shadow of the photon being massless, and the same fact that made $k_\mu k^\mu = 0$ in `aq-sr-wave-equation-four-potential`.

Step 3. Get $p = u/3$ by averaging over directions. For a single plane wave along $\hat n$ the tensor collapses to the null dyad $T^{\mu\nu} = u\,n^\mu n^\nu$ , $n^\mu = (1,\hat n)$ — every slot the same number $u$ , with pressure $u$ along the propagation direction. Average isotropically, using $\langle n^in^j\rangle = \tfrac13\delta^{ij}$ : \langle T^{ij}\rangle = \tfrac13 u\,\delta^{ij} \quad\Longrightarrow\quad \boxed{\;p = \frac{u}{3}\;} Why: the three is the three spatial dimensions, nothing more. One beam pushes with pressure $u$ along one axis; an isotropic bath spreads the same energy over three. And $\langle T^\mu{}_\mu\rangle = u - 3\cdot\tfrac13u = 0$ ✓, so $p=u/3$ is not an extra assumption but a restatement of tracelessness.

Step 4. Note the trap, one final time. Replace $T^{\mu\nu}\to -T^{\mu\nu}$ everywhere above. Every equation still holds: the trace is still zero, the null dyad is still rank one, the averaging still gives $|p| = |u|/3$ . The only change is that $u$ becomes negative. Why: tracelessness is blind to the sign, and so is every argument built on it. That is why equation (4) has a box around it, and why this article's verification script asserts $T^{00}>0$ as a numerical inequality rather than a symbolic identity. *Verified in verify_aq-sr-electromagnetic-stress-energy-tensor.py: the trace is computed symbolically for both sign choices and shown to vanish for both; the $D$ -dimensional trace is shown to be proportional to $(D-4)$ ; the plane-wave null dyad $T^{\mu\nu} = u\,n^\mu n^\nu$ is verified slot by slot; and the isotropic average gives $p = u/3$ exactly.*

Limiting cases

Case	$T^{\mu\nu}$	What it says
Static electric field	$T^{00} = \tfrac12\varepsilon_0E^2 > 0$ , $T^{0i} = 0$	energy at rest, no flux. Tension along $\vec E$ , pressure across
Static magnetic field	$T^{00} = B^2/2\mu_0 > 0$ , $T^{0i} = 0$	the same with $\vec B$ — this is how a magnetic bottle confines plasma
Plane wave ( $E = cB$ , $\vec E\perp\vec B$ )	$T^{\mu\nu} = u\,n^\mu n^\nu$ , $n^\mu = (1,\hat n)$	null and rank one. $\lvert\vec S\rvert = cu$ , so $g = u/c$ . Both invariants vanish
Isotropic radiation	$T^{\mu\nu} = \mathrm{diag}(u,\ \tfrac{u}{3},\ \tfrac{u}{3},\ \tfrac{u}{3})$	the radiation equation of state $p = u/3$ . Still traceless
Vacuum, $J^\mu = 0$	$\partial_\mu T^{\mu\nu} = 0$	field energy-momentum conserved on its own
Charges present	$\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$	not conserved alone — see `aq-sr-field-energy-momentum-conservation`
Wrong overall sign	$T^{00} = -u < 0$	still symmetric, still traceless. Only the energy density gives it away

Worked examples

1 · Sunlight on a sail — the whole tensor, in micropascals

The solar constant at Earth's orbit is $I = 1361\ \mathrm{W\,m^{-2}}$ . Compute the energy density, the momentum density, the peak fields, and the pressure on an absorber and on a mirror. Then compute the force on a real spacecraft.

Step 1 — energy density from the flux. For a plane wave, $|\vec S| = c\,u$ (Step 3 of the third derivation), so u = \frac{I}{c} = \frac{1361}{2.99792458\times10^{8}} = \mathbf{4.5398\times10^{-6}\ J\,m^{-3}} . Sunlight at Earth carries about 4.5 microjoules per cubic metre — the energy in a cubic metre of sunbeam would not lift a grain of sand.

Step 2 — momentum density, from the symmetry of $T$ . $g = u/c$ for a plane wave, so $g = \mathbf{1.5143\times10^{-14}\ kg\,m^{-2}\,s^{-1}}$ .

Step 3 — the pressure. Momentum arriving per unit area per unit time is $g\,c = I/c$ . So P_{\text{absorb}} = \frac{I}{c} = \mathbf{4.540\ \mu Pa}, \qquad P_{\text{reflect}} = \frac{2I}{c} = \mathbf{9.080\ \mu Pa} . Compare with atmospheric pressure, $101\,325\ \mathrm{Pa}$ : even a mirror in full sunlight feels $101325/9.080\times10^{-6} = \mathbf{1.12\times10^{10}}$ times less pressure than the air in this room. That ratio is why radiation pressure went unmeasured until 1901, and why it took a torsion balance in a vacuum to find it.

Step 4 — the peak fields, as a sanity check. Using $I = \tfrac12\varepsilon_0 c E_0^2$ , E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \mathbf{1012.6\ V\,m^{-1}}, \qquad B_0 = \frac{E_0}{c} = \mathbf{3.378\ \mu T} . A kilovolt per metre — a substantial field. The fields are not small; the momentum is small, because momentum costs a factor of $c$ .

Step 5 — a real spacecraft. The Planetary Society's LightSail 2 flew a $32\ \mathrm{m^2}$ reflective sail in 2019. At normal incidence, F = P_{\text{reflect}}\times A = (9.080\times10^{-6})(32) = \mathbf{2.906\times10^{-4}\ N} \approx 0.29\ \mathrm{mN}. The weight of a grain of rice. But it acts continuously, for free, forever, and over weeks it measurably raised the spacecraft's apogee. Every term in that calculation came out of a slot in $T^{\mu\nu}$ .

2 · The plane wave is a null dyad — and every slot is the same number

Take a wave travelling along $+z$ with $\vec E = E_0\hat x$ , $\vec B = (E_0/c)\hat y$ . Compute all sixteen components of $T^{\mu\nu}$ .

Step 1 — the invariants vanish. $E = cB$ gives $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2) = 0$ , and $\vec E\perp\vec B$ gives $F_{\alpha\beta}\tilde F^{\alpha\beta} = -(4/c)\vec E\cdot\vec B = 0$ . Both invariants are zero — the signature of a radiation field (`aq-sr-electromagnetic-field-invariants`). So the trace-subtraction term in (1) dies outright, leaving $T^{\mu\nu} = -\mu_0^{-1}F^{\mu\alpha}F^{\nu}{}_{\alpha}$ .

Step 2 — evaluate. With $u = \varepsilon_0E_0^2$ (the electric and magnetic halves being equal for a wave), the only non-zero components are $T^{00} = T^{0z} = T^{z0} = T^{zz} = u$ , every other slot zero. Compactly, \boxed{\;T^{\mu\nu} = u\,n^\mu n^\nu,\qquad n^\mu = (1,0,0,1)\;}

Step 3 — read it. $n^\mu$ is null ( $n_\mu n^\mu = 1 - 1 = 0$ ), so the tensor is rank one: energy density $u$ , energy flux $cu$ along $+z$ , momentum density $u/c$ along $+z$ , pressure $u$ along $z$ , and no transverse stress at all. Light is a dust of energy streaming at $c$ — the whole 4×4 matrix holds exactly one independent number. The trace is $u(n_\mu n^\mu) = 0$ ✓ automatically. And with the wrong overall sign we would have found $T^{\mu\nu} = -u\,n^\mu n^\nu$ : still null, still rank one, still traceless — and carrying negative energy to infinity.

Common trapThe stress-energy sign. Both bases in this chapter print the (−,+,+,+) tensor and label it (+,−,−,−) — and the trace will never tell you.

The AllQFT bases for this article and for `aq-sr-field-energy-momentum-conservation` both print

T^{\mu\nu} = \frac{1}{\mu_0}\Big(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\Big) \qquad\text{labelled ``metric } (+,-,-,-)\text{''}.

In the signature they declare, that tensor has $T^{00} = -u$ . It is the $(-,+,+,+)$ form, mislabelled. Because $T$ is quadratic in $F$ , this has nothing to do with how you sign the field tensor — flip $F\to -F$ and $T$ is unchanged. The sign depends on the signature and on nothing else.

And the base contradicts itself one card later, printing $T^{00} = u$ — the right answer, which does not follow from the formula it just wrote. The energy-momentum base is worse: it prints the wrong $T$ , the correct balance law $\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$ , and the correct force density $f^\nu = F^{\nu\lambda}J_\lambda$ — three statements that cannot simultaneously be true.

Why it survived: the base checks tracelessness, and the wrong tensor is traceless too. So is it symmetric, gauge-invariant and quadratic. Every structural test passes. The one test that fails is $T^{00} > 0$ , and nobody ran it. A gate that cannot fail is not a gate.

Independent corroboration: the live electromagnetism-track article `aq-em-stress-energy-tensor` prints the tensor as $\theta^{\mu\nu} = \mu_0^{-1}\big[F^{\mu\alpha}F_{\alpha}{}^{\nu} + \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ and states $\theta^{00} = u$ — identical to equation (1), since $F_{\alpha}{}^{\nu} = -F^{\nu}{}_{\alpha}$ . Two tracks, one answer.

Common trapTwo factors that are not typos: the c in T⁰ⁱ and the minus sign in Tⁱʲ.

$T^{0i} = S^i/c$ , not $S^i$ . The Poynting vector is measured in $\mathrm{W\,m^{-2}}$ ; every slot of $T^{\mu\nu}$ carries the units of energy density, $\mathrm{J\,m^{-3}}$ . Dividing by $c$ converts a flux into a density — the same factor of $c$ that appears in $J^\mu = (c\rho, \vec J)$ and $A^\mu = (\phi/c, \vec A)$ . If your $T^{\mu\nu}$ has slots with different dimensions, you have dropped a $c$ . $T^{ij} = -\sigma^{ij}$ , with a minus. $\sigma^{ij}n_j$ is defined as the force per unit area transmitted across a surface with normal $\hat n$ ; the stress-energy tensor's spatial block is the momentum flux, which flows the other way. The minus is a genuine relative sign between two standard, entrenched definitions — which is why a field line under tension ( $\sigma$ positive along $\vec E$ ) shows up as a negative $T^{ij}$ . Get it backwards and your capacitor plates repel.

Physics of the portrait

The portrait is the 3×3 block — the oldest part of the tensor.

Origins · 2 minMaxwell stress tensor and how this was discovered

Maxwell stress tensor

1873 · the field as an elastic medium · tension along the lines, pressure across them · the block that became T^{ij}

The Maxwell stress tensor $\sigma^{ij} = \varepsilon_0(E^iE^j - \tfrac12\delta^{ij}E^2) + \mu_0^{-1}(B^iB^j - \tfrac12\delta^{ij}B^2)$ is the oldest part of $T^{\mu\nu}$ and, in its day, the most radical. Maxwell introduced it in the Treatise on Electricity and Magnetism (1873) to answer a question Newtonian action-at-a-distance could not: where, physically, is the force between two charges? His answer: in the space between them. The field is a stressed elastic medium, pulling like a stretched rubber band along the field lines with tension $\tfrac12\varepsilon_0E^2$ and pushing sideways with an equal pressure. Two like charges repel because the lines between them are crowded and press apart; two unlike charges attract because the lines running from one to the other are in tension. The picture is astonishingly literal, and it works: you can compute the force on any charged object by drawing an imaginary closed surface around it and integrating $\sigma^{ij}n_j$ over it — never touching the charge, never summing a Coulomb force. Maxwell believed the stresses lived in a mechanical aether, and on that he was wrong. But the stresses themselves were right, and survived the aether's demolition intact, because they were never about a medium: they were about the field. Minkowski's 1908 four-tensor swallowed the 3×3 block whole, glued the energy density on as $T^{00}$ and the Poynting flux on as $T^{0i}$ , and revealed the three to be one object seen from three angles. Maxwell built the last third of a relativistic tensor thirty-two years before relativity, while trying to describe a substance that does not exist.

History

1873

— Maxwell's "Treatise on Electricity and Magnetism" introduces the stress tensor of the field and predicts that light exerts pressure.

1884

— Poynting publishes "On the Transfer of Energy in the Electromagnetic Field", identifying the energy flux S = E × H.

1901

— Lebedev in Moscow, and independently Nichols and Hull in the United States, measure radiation pressure on a torsion balance.

1908

— Minkowski assembles the energy density, the Poynting flux and the Maxwell stresses into a single four-tensor.

2010

— IKAROS becomes the first spacecraft to fly on solar radiation pressure alone, in interplanetary space.

2019

— LightSail 2 raises its orbit around Earth using a 32 m² reflective sail and no propellant.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §8.2 and §12.3.5 · Jackson, Classical Electrodynamics, 3rd ed., §6.7 and §12.10 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Landau & Lifshitz, The Classical Theory of Fields, 4th ed., §33.

Provenance. Signature $(+,-,-,-)$ , hence $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. Dictionary $F^{0i} = -E^i/c$ , $F^{ij} = -\epsilon^{ijk}B_k$ , $J^\mu = (c\rho,\vec J)$ . This article's base prints the stress-energy tensor with the wrong overall sign — the $(-,+,+,+)$ form under a $(+,-,-,-)$ label, giving $T^{00} = -u$ — then contradicts itself by printing $T^{00} = +u$ on the next card. The correction is derived here from scratch and machine-asserted as an inequality, `assert T00 > 0`, in verify_aq-sr-electromagnetic-stress-energy-tensor.py, along with the proof that both the right and the wrong tensor are traceless, so no trace-based check could ever have caught it. Every number here is recomputed in `mpmath` at 50 digits against the article's own printed digits.

Scope. The field tensor belongs to `aq-sr-electromagnetic-field-tensor`; the invariant $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ in the trace-subtraction term to `aq-sr-electromagnetic-field-invariants`. The balance law $\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$ and its unpacking into Poynting's theorem and momentum conservation are owned by `aq-sr-field-energy-momentum-conservation`; this page derives the tensor and reads off its slots. The electromagnetism track's parallel treatments are `aq-em-stress-energy-tensor` and `aq-em-maxwell-stress-tensor`.

Builds on. `aq-sr-electromagnetic-field-tensor` · `aq-sr-electromagnetic-field-invariants` · `aq-sr-covariant-maxwell-equations` · `aq-sr-four-current-density`. Leads to. `aq-sr-field-energy-momentum-conservation` · `aq-sr-covariant-lorentz-force`.

Does the sign of T^{μν} depend on how I choose to sign the field tensor F^{μν}?

No — and this is the most important thing to understand about the trap. $T^{\mu\nu}$ is quadratic in $F$ : replace $F\to-F$ and every term in (1) is unchanged, because each contains exactly two factors of $F$ . No convention about $F$ can rescue a wrong sign in $T$ . The sign is fixed by the metric signature alone, and pinned by the one physical requirement that $T^{00}$ be the (positive) energy density. If someone tells you the sign of the stress-energy tensor is "just a convention", ask them what the energy density of a charged capacitor is.

The trace-subtraction term uses the first invariant. Is that a coincidence?

It is forced. The second tensor in the menu must be $\eta^{\mu\nu}\times(\text{a Lorentz scalar quadratic in }F)$ , and there are only two such scalars: $F_{\alpha\beta}F^{\alpha\beta}$ and $F_{\alpha\beta}\tilde F^{\alpha\beta}$ . The second is a pseudoscalar — it flips sign under parity — so it cannot appear in a parity-even object like the energy density. That leaves $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ , and no choice about it. `aq-sr-electromagnetic-field-invariants` owns both invariants; this page consumes the first.

Why is T^{ij} equal to MINUS the Maxwell stress tensor rather than plus?

Because the two objects are defined with opposite conventions about which way the momentum goes. $\sigma^{ij}n_j$ is the force per unit area exerted on the material bounded by a surface with outward normal $\hat n$ ; $T^{ij}$ is the flux of $i$ -momentum in the $j$ direction, which is what leaves. Both conventions are standard and entrenched. The practical rule: a field line under tension pulls, which is a positive $\sigma$ along $\vec E$ and a negative $T^{ij}$ . If your capacitor plates come out repelling, you have the sign backwards.

If the field carries momentum, does that not break Newton's third law?

It rescues it. There are ordinary configurations — a point charge beside a magnetic dipole — in which the forces the two objects exert on each other are not equal and opposite. Were the field momentumless, the total momentum would change with nothing to balance it. It does not, because the field holds $\vec g = \varepsilon_0\vec E\times\vec B$ per unit volume, and the ledger closes exactly when you count it. Physics faced a choice between abandoning momentum conservation and granting the field an inventory of its own, and chose the inventory. This tensor is that inventory; `aq-sr-field-energy-momentum-conservation` is the ledger.

Why is a photon gas p = u/3 while ordinary gas pressure has no such universal relation?

Because the photon gas is traceless and ordinary matter is not. $T^\mu{}_\mu = \rho c^2 - 3p$ for a perfect fluid, so $T^\mu{}_\mu = 0$ is the statement $p = u/3$ . Ordinary gas has rest mass, hence a non-zero trace, hence a pressure that depends on temperature and density in a way nothing universal constrains. The equation of state of light is a theorem; the equation of state of air is an experiment.

Radiation pressure is only about 9 micropascals at Earth. Is it ever important?

Constantly. It is one of the two pressures supporting a massive star against gravitational collapse, and in the most massive stars it dominates — the Eddington limit, the luminosity at which radiation pressure balances gravity and the star begins to shed its outer layers, is nothing but $P = I/c$ applied to a whole star. It perturbs the orbits of satellites and asteroids measurably. And it is the entire propulsion system of a solar sail. The pressure is small; the lever arms are astronomical.