Energy-Momentum Conservation of Fields

Take a copper wire carrying a current. It gets warm. Ask where the energy comes from, and the obvious answer — "it flows along the wire, with the current" — is wrong. Compute the Poynting vector at the wire's surface and it points radially inward. The energy enters through the side of the wire, out of the empty space around it. Integrate it over the surface and you get exactly $I^2R$ , to the last decimal. The battery does not push energy down the copper; it sets up fields in the space outside the copper, and those fields carry the energy in from the sides. That is Poynting's theorem, and this page's claim is that it is one component of something bigger:

\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda .

The $\nu = 0$ slot is Poynting's theorem. The $\nu = i$ slots are the momentum balance, radiation pressure included. And the right-hand side is not zero — the field's energy-momentum is not conserved on its own. It leaks, at exactly the rate the Lorentz force delivers energy and momentum to matter. Add the matter back and the ledger closes.

Key takeaways

Signature $(+,-,-,-)$ , $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. $J^\mu = (c\rho,\vec J)$ , $F^{0i} = -E^i/c$ .
The law: $\boxed{\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda}$ . The right-hand side is minus the Lorentz four-force density, $f^\nu = F^{\nu\lambda}J_\lambda = (\vec J\cdot\vec E/c,\ \rho\vec E + \vec J\times\vec B)$ .
$\nu = 0$ IS Poynting's theorem: $\partial_t u + \nabla\cdot\vec S = -\vec J\cdot\vec E$ . Field energy falls wherever the field does work on charges.
$\nu = i$ IS momentum conservation: $\partial_t g^i - \partial_j\sigma^{ij} = -(\rho\vec E + \vec J\times\vec B)^i$ . Radiation pressure lives here.
The field alone is NOT conserved, and that is the point. $\partial_\mu T^{\mu\nu}_{\text{field}} \ne 0$ whenever charges are present. Only the sum closes: $\partial_\mu(T^{\mu\nu}_{\text{field}} + T^{\mu\nu}_{\text{matter}}) = 0$ .
Key idea: energy and momentum conservation are the same equation, read in different slots. Poynting's theorem (1884) and Maxwell's stress balance (1873) were discovered twenty-four years and one dimension apart. Minkowski's tensor shows they are $\nu = 0$ and $\nu = i$ of a single four-vector identity — and the wire above proves it with a number: $\oint\vec S\cdot d\vec A = \mathbf{0.5348\ W} = I^2R$ , exactly.

The core insight

A conservation law with a source on the right is not a failed conservation law — it is a transfer law, and it is more informative than the conserved version. $\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$ says precisely where the energy went: into the charges, at the rate the Lorentz force does work. The field is an accountant that never loses a joule, but does hand joules over. Sum the field's tensor and the matter's tensor and the divergence vanishes identically — which is Noether's theorem for spacetime translation invariance, arriving in classical electromagnetism a decade before Noether.

Where this lives today

Every power engineer's intuition that "energy flows in the wires" is, strictly, wrong — and at high frequency the error stops being academic. In a coaxial cable the energy travels in the dielectric between the conductors, not in the copper, which is why the geometry of that gap sets the impedance and why a damaged dielectric ruins a cable that looks electrically intact. In a waveguide there is no centre conductor at all and the energy is manifestly in the empty volume. The Poynting vector is the working tool for antenna gain, for specific-absorption-rate limits on mobile phones, and for the thermal design of every high-current busbar. The momentum half runs from optical tweezers to the Abraham–Minkowski controversy — the century-old question of what the momentum of light inside a dielectric is, which was only resolved by recognising that the two candidate answers correspond to two different (and both correct) splits of the total momentum between field and matter. The split is a choice; the total, which is what equation (1) governs, is not. And the structure generalises. In general relativity the matter stress-energy tensor obeys $\nabla_\mu T^{\mu\nu} = 0$ — the same statement, with the covariant derivative — and it is not an independent postulate there either: it follows from the Bianchi identity applied to the Einstein tensor, in a derivation that is line-for-line the analogue of the one on this page. Conservation of energy has never been an axiom in either theory. It has always been a corollary of the field equations, and the reason is always the same: a geometric identity.

The physical picture

The stress-energy tensor of the previous article (`aq-sr-electromagnetic-stress-energy-tensor`) is the field's inventory: energy density in $T^{00}$ , energy flux and momentum density in $T^{0i}$ , stresses in $T^{ij}$ . An inventory is only useful if you can audit it, and the audit is the four-divergence. For an isolated conserved quantity the audit reads $\partial_\mu T^{\mu\nu} = 0$ : whatever leaves a region crosses its boundary. For the electromagnetic field that is false whenever charges are present, and the failure is exactly quantified:

\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda .

The right-hand side is the four-force density the field exerts on matter, with a minus sign. Read it as a ledger entry: every joule the field loses is a joule the charges gain. The field is not leaking into nowhere; it is paying the charges. Two things make this worth a page of its own. First, unpacking $\nu = 0$ gives Poynting's theorem and unpacking $\nu = i$ gives Maxwell's momentum balance — two nineteenth-century results, discovered independently, in different decades, by different people, which turn out to be one equation. Second, the identity is derived, not postulated: it follows from the covariant Maxwell equations plus the Bianchi identity, with no further physics injected. Conservation of energy is not an extra axiom of electromagnetism. It is a theorem about $F^{\mu\nu}$ .

Figure

THE ENERGY DOES NOT FLOW ALONG THE WIRE. It flows IN FROM THE SIDES. A copper wire of radius 1.0 mm carrying 10 A has an axial electric field E_z = V/L = 53.48 mV/m (it must, to push the current through the resistance) and an azimuthal magnetic field B_φ = μ₀I/2πa = 2.0 mT at its surface. The Poynting vector S = E × H is the cross product of those two — and E is along the wire while B wraps around it, so S points RADIALLY INWARD, straight into the copper. Magnitude: 85.11 W/m². Multiply by the lateral surface of one metre of wire, 62.83 cm², and you get 0.5348 W — which is exactly I²R = (10)²(5.348 mΩ). The battery does not shove energy down the copper. It builds fields in the empty space AROUND the copper, and the energy arrives sideways, through the surface. This is the ν = 0 component of ∂_μT^{μν} = −F^{νλ}J_λ, and it is Poynting's theorem.

The core equations

The law this page owns:

\boxed{\;\partial_\mu T^{\mu\nu} \;=\; -\,F^{\nu\lambda}J_\lambda \;=\; -f^\nu\;} \qquad (1)

where: $T^{\mu\nu}$ is the stress-energy tensor of `aq-sr-electromagnetic-stress-energy-tensor`, and $f^\nu = F^{\nu\lambda}J_\lambda$ is the Lorentz four-force density. Physical reading: the field's energy-momentum is not conserved when charges are present — it changes at exactly the rate the field does work on, and pushes, the matter. The minus sign is the ledger: what the field loses, the charges gain. The four-force density, unpacked:

f^\nu \;=\; F^{\nu\lambda}J_\lambda \;=\; \left(\frac{\vec J\cdot\vec E}{c},\ \ \rho\vec E + \vec J\times\vec B\right) \qquad (2)

where: $J_\lambda = (c\rho, -\vec J)$ is the four-current with its index lowered. Physical reading: the time component is power density and the space components are force density — the Lorentz force per unit volume, which is what you would have written down anyway. $f^0 = \vec J\cdot\vec E/c$ is the Ohmic heating rate over $c$ ; $\vec f = \rho\vec E + \vec J\times\vec B$ is the force per cubic metre. Note that $\vec J\times\vec B$ does no work, which is why it appears in $\vec f$ but not in $f^0$ — magnetic forces never heat anything. The $\nu = 0$ component, which has a name:

\frac{\partial u}{\partial t} + \nabla\cdot\vec S = -\,\vec J\cdot\vec E \qquad\text{(Poynting's theorem)} \qquad (3)

where: $u = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ and $\vec S = (\vec E\times\vec B)/\mu_0$ . Physical reading: a continuity equation with a sink. Field energy in a region drops either because energy flows out across the boundary ( $\nabla\cdot\vec S$ ) or because the field does work on charges ( $\vec J\cdot\vec E$ ). In a resistive wire the second term is the Ohmic heating, and the first term is negative — energy flows in. The $\nu = i$ components, which also have a name:

\frac{\partial g^i}{\partial t} \;-\; \partial_j\sigma^{ij} \;=\; -\big(\rho\vec E + \vec J\times\vec B\big)^i \qquad\text{(momentum balance)} \qquad (4)

where: $\vec g = \varepsilon_0\vec E\times\vec B = \vec S/c^2$ is the field momentum density and $\sigma^{ij}$ the Maxwell stress tensor. Physical reading: this is Newton's second law for the field. The momentum stored in the field changes because momentum flows across the boundary (the stress term) or because the field pushes on charges. Radiation pressure is this equation with the source term set to a mirror.

Deriving the balance law

Derivation AFrom Maxwell and Bianchi to the four-divergence — conservation is a THEOREM, not an axiom

The goal is to compute $\partial_\mu T^{\mu\nu}$ directly, using nothing but the two covariant Maxwell equations, and to watch the Lorentz force density appear on the right-hand side without ever having been put in. This matters: it means that energy conservation in electromagnetism is not a separate postulate to be checked, but an algebraic consequence of the field equations. It also means a wrong-signed $T^{\mu\nu}$ cannot possibly satisfy it — which is how the base for this very article convicts itself.

Step 1. Differentiate the tensor. From $T^{\mu\nu} = \mu_0^{-1}\big[-F^{\mu\alpha}F^{\nu}{}_{\alpha} + \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ , the product rule gives \mu_0\,\partial_\mu T^{\mu\nu} = -(\partial_\mu F^{\mu\alpha})F^{\nu}{}_{\alpha} - F^{\mu\alpha}\,\partial_\mu F^{\nu}{}_{\alpha} + \tfrac12 F^{\alpha\beta}\,\partial^\nu F_{\alpha\beta} . Why: three terms, and each is about to be handled by a different piece of standard machinery. The first will use the inhomogeneous Maxwell equation; the last two will annihilate each other using the Bianchi identity.

Step 2. Kill the first term with Maxwell. The inhomogeneous equation $\partial_\mu F^{\mu\alpha} = \mu_0 J^\alpha$ (owned by `aq-sr-covariant-maxwell-equations`) turns it into -(\partial_\mu F^{\mu\alpha})F^{\nu}{}_{\alpha} = -\mu_0 J^\alpha F^{\nu}{}_{\alpha} = -\mu_0\,F^{\nu\lambda}J_\lambda . Why: there is the Lorentz force density, and we did not put it there. It fell out of the divergence of the field's own energy-momentum. That is the whole content of the derivation.

Step 3. Kill the remaining two with Bianchi. The homogeneous Maxwell equation is the cyclic identity $\partial^\nu F^{\alpha\beta} + \partial^\alpha F^{\beta\nu} + \partial^\beta F^{\nu\alpha} = 0$ , automatic once $F = dA$ . Contracting it with $\tfrac12 F_{\alpha\beta}$ and using the antisymmetry of $F$ shows \tfrac12 F^{\alpha\beta}\partial^\nu F_{\alpha\beta} = F^{\mu\alpha}\,\partial_\mu F^{\nu}{}_{\alpha}, so the second and third terms of Step 1 cancel exactly. Why: this is where the "no magnetic monopoles" half of Maxwell earns its keep. Without the Bianchi identity there would be a leftover term and the balance law would not close. Both halves of Maxwell are needed, and they enter through completely different doors.

Step 4. Collect. \boxed{\;\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda\;} Why: conservation of energy is now a theorem about $F^{\mu\nu}$ . Set $J^\mu = 0$ and the field's energy-momentum is exactly conserved. Turn the charges back on and the field pays them, at precisely the Lorentz rate.

Step 5. Now run the same computation with the wrong-signed tensor, $T^{\mu\nu}_{\text{bad}} = -T^{\mu\nu}$ , as printed by this article's own base. Every step goes through with a global sign flip, giving \partial_\mu T^{\mu\nu}_{\text{bad}} = +F^{\nu\lambda}J_\lambda . Why: the base is self-refuting, and one can say exactly how. It prints the wrong tensor, the correct balance law (1), and the correct force density (2) — three claims that are mutually inconsistent. A field obeying its equations would gain energy by doing work on charges. Only $T^{00} = +u$ makes the three consistent. See the warning card. *Verified in verify_aq-sr-field-energy-momentum-conservation.py: $F^{\mu\nu}$ is rebuilt symbolically from $A^\mu = (\phi/c,\vec A)$ with arbitrary functions of $(t,x,y,z)$ ; $J^\nu$ is defined by $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ rather than assumed; and $\partial_\mu T^{\mu\nu} + F^{\nu\lambda}J_\lambda = 0$ is confirmed component by component for the correct tensor, and shown to fail by exactly a factor of $-1$ for the base's.*

Go deeper · Derivation B · 2 minOwned result: the nu = 0 slot IS Poynting's theorem — and it says the energy comes in sideways

Derivation BOwned result: the nu = 0 slot IS Poynting's theorem — and it says the energy comes in sideways

Poynting published his theorem in 1884, twenty-one years before relativity, as a three-dimensional statement about $\vec E$ , $\vec B$ and $\vec J$ . The claim here is that it is not an independent law at all but the time component of equation (1). Unpacking it is a matter of substituting components — but the content is startling, and the wire in the figure is the reason.

Step 1. Take $\nu = 0$ in equation (1). The left-hand side is \partial_\mu T^{\mu 0} = \frac{1}{c}\frac{\partial T^{00}}{\partial t} + \partial_i T^{i0} = \frac{1}{c}\frac{\partial u}{\partial t} + \partial_i\!\left(\frac{S^i}{c}\right), using $T^{00} = u$ , $T^{i0} = S^i/c$ and $\partial_0 = \tfrac1c\partial_t$ . Why: the $1/c$ 's are bookkeeping — they will cancel against the $1/c$ in $f^0$ .

Step 2. Take the right-hand side. From equation (2), $-f^0 = -\vec J\cdot\vec E/c$ . Why: this is the power the field delivers to the charges, per unit volume, divided by $c$ .

Step 3. Equate and multiply through by $c$ : \boxed{\;\frac{\partial u}{\partial t} + \nabla\cdot\vec S = -\,\vec J\cdot\vec E\;} Why: this is Poynting's theorem, and it arrived as one slot of a tensor equation. A continuity equation for field energy, with a sink wherever the field does work.

Step 4. Apply it to a resistive wire, and read the surprise. In steady state $\partial_t u = 0$ , so $\nabla\cdot\vec S = -\vec J\cdot\vec E$ — the divergence of the Poynting flux is negative inside the copper. Integrating over a length of wire and using the divergence theorem, \oint \vec S\cdot d\vec A = -\int \vec J\cdot\vec E\ \mathrm{d}V = -I^2R < 0 . The net Poynting flux out of the wire is negative — which means it flows in. Why: the energy enters through the side surface, from the field in the empty space around the wire. $\vec E$ is axial (it must be, to drive current against resistance) and $\vec B$ is azimuthal, so $\vec S = \vec E\times\vec H$ is radial and inward. The copper is not a pipe carrying energy; it is a sink absorbing energy that travelled through the surrounding vacuum. This is the single most counter-intuitive true statement in classical electromagnetism, and the numbers in the worked example confirm it to eight figures. *Verified in verify_aq-sr-field-energy-momentum-conservation.py: the $\nu=0$ extraction is done symbolically from the tensor; and for the worked example the surface integral $\oint\vec S\cdot d\vec A$ is computed independently of $I^2R$ and the two are shown to agree to 40 digits in `mpmath`.*

Go deeper · Derivation C · 1 minOwned result: the nu = i slots are momentum balance — and only field-plus-matter is conserved

Derivation COwned result: the nu = i slots are momentum balance — and only field-plus-matter is conserved

The spatial components of (1) are Maxwell's momentum balance, which he wrote in 1873 using the stress tensor. Their content is Newton's third law rescued: the field carries momentum, and once you count it, the total is conserved. This derivation closes the ledger by adding the matter's own stress-energy tensor and showing the sum is divergence-free.

Step 1. Take $\nu = i$ . Using $T^{0i} = c\,g^i$ and $T^{ji} = -\sigma^{ji}$ , \partial_\mu T^{\mu i} = \frac{1}{c}\partial_t(c\,g^i) + \partial_j(-\sigma^{ji}) = \frac{\partial g^i}{\partial t} - \partial_j\sigma^{ij}, and the right-hand side of (1) is $-(\rho\vec E + \vec J\times\vec B)^i$ . Hence \boxed{\;\frac{\partial g^i}{\partial t} - \partial_j\sigma^{ij} = -\big(\rho\vec E + \vec J\times\vec B\big)^i\;} Why: Newton's second law, for the field. Field momentum changes because momentum flows across the boundary or because the field pushes on matter.

Step 2. Identify the matter side. The charges obey the Lorentz force law: their momentum density $\vec p_{\text{mech}}$ satisfies $\partial_t\vec p_{\text{mech}} = \rho\vec E + \vec J\times\vec B$ (this is `aq-sr-covariant-lorentz-force`, in density form). Why: the same quantity appears on both sides of the ledger, with opposite signs. That is what makes it a transfer.

Step 3. Add them. \frac{\partial}{\partial t}\big(g^i + p^i_{\text{mech}}\big) = \partial_j\sigma^{ij} . The source terms cancel identically, leaving a pure divergence. Why: total momentum — field plus matter — is conserved, with the Maxwell stress as its flux. Integrate over all space with fields falling off fast enough and the right-hand side vanishes: the total momentum is constant.

Step 4. State it covariantly. Defining the matter stress-energy tensor $T^{\mu\nu}_{\text{matter}}$ so that $\partial_\mu T^{\mu\nu}_{\text{matter}} = +f^\nu$ , \boxed{\;\partial_\mu\big(T^{\mu\nu}_{\text{field}} + T^{\mu\nu}_{\text{matter}}\big) = 0\;} Why: this is the real conservation law, and it is the reason equation (1) is not a defect. A four-divergence that vanishes gives four conserved charges — total energy and total momentum — and by Noether's theorem those are exactly the consequences of spacetime being translation-invariant. Classical electromagnetism was already doing Noether's theorem thirty years before Noether proved it. *Verified in verify_aq-sr-field-energy-momentum-conservation.py: the $\nu=i$ extraction is confirmed symbolically; $\vec J\times\vec B$ is shown to contribute nothing to $f^0$ (magnetic forces do no work); and the cancellation of the source terms in the field+matter sum is checked component by component.*

Limiting cases

Case	Balance law	What it says
Vacuum, $J^\mu = 0$	$\partial_\mu T^{\mu\nu} = 0$	field energy-momentum strictly conserved; radiation propagates without loss
Steady current in a resistor	$\nabla\cdot\vec S = -\vec J\cdot\vec E$	Poynting flux flows inward; $\oint\vec S\cdot d\vec A = -I^2R$
Electrostatics, $\vec J = 0$	$-\partial_j\sigma^{ij} = -\rho E^i$	Maxwell stress balances the electrostatic force — compute forces by surface integral
Magnetic force only	$f^0 = 0$	$\vec J\times\vec B$ does no work: it changes momentum, never energy
Light on a mirror	$\partial_t g + \nabla\cdot(-\sigma) = -f$	radiation pressure: the momentum the field loses, the mirror gains
Field + matter	$\partial_\mu(T^{\mu\nu}_{\text{field}} + T^{\mu\nu}_{\text{matter}}) = 0$	the real conservation law. Nothing is lost, only transferred
Wrong-signed $T$	$\partial_\mu T^{\mu\nu} = +F^{\nu\lambda}J_\lambda$	the field would gain energy by heating a resistor. Absurd — see the warning

Worked examples

1 · The wire — where the energy actually comes from

A copper wire of radius $a = 1.0\ \mathrm{mm}$ carries $I = 10\ \mathrm{A}$ . Take $\rho_{\text{Cu}} = 1.68\times10^{-8}\ \Omega\,\mathrm{m}$ . Show that the Poynting flux entering one metre of it equals the Ohmic dissipation, exactly.

Step 1 — the resistance and the dissipation. Cross-section $A = \pi a^2 = 3.1416\ \mathrm{mm^2}$ , so for $L = 1\ \mathrm{m}$ : R = \frac{\rho L}{A} = \mathbf{5.3476\ m\Omega}, \qquad P = I^2R = \mathbf{0.53476\ W} . That is the number to reproduce, by a completely different route.

Step 2 — the axial electric field. To drive $10\ \mathrm{A}$ through that resistance the wire must carry a voltage drop $V = IR = 53.476\ \mathrm{mV}$ over its metre, so there is an electric field inside and at the surface of the copper, pointing along it: E_z = \frac{V}{L} = \mathbf{53.476\ mV\,m^{-1}} . This field is the one people forget. A "perfect conductor" would have none — but a real resistive wire must have one, or no current would flow.

Step 3 — the azimuthal magnetic field. Ampère's law at the surface gives B_\varphi = \frac{\mu_0 I}{2\pi a} = \frac{(4\pi\times10^{-7})(10)}{2\pi(1.0\times10^{-3})} = \mathbf{2.0\ mT} .

Step 4 — the Poynting vector. $\vec E$ is axial, $\vec B$ is azimuthal, so $\vec S = \vec E\times\vec B/\mu_0$ is radial, and the right-hand rule puts it pointing inward. Its magnitude at the surface: |\vec S| = \frac{E_z B_\varphi}{\mu_0} = \frac{(53.476\times10^{-3})(2.0\times10^{-3})}{4\pi\times10^{-7}} = \mathbf{85.110\ W\,m^{-2}} .

Step 5 — the audit. The lateral surface of one metre of wire is $2\pi a L = 62.832\ \mathrm{cm^2}$ . The power flowing in through it is P_{\text{in}} = |\vec S|\times 2\pi a L = (85.110)(6.2832\times10^{-3}) = \mathbf{0.53476\ W} . Compare Step 1: $I^2R = 0.53476\ \mathrm{W}$ . The same number, to five figures — and to forty, in the verification script. The energy heating this wire did not travel through the copper. It travelled through the air, and entered through the skin. Poynting's theorem is not a bookkeeping identity; it is a claim about where energy physically is, and it is right.

2 · A kilowatt laser on a mirror — the same law, the other slots

A $1.00\ \mathrm{kW}$ laser is focused to a spot of radius $1.0\ \mathrm{mm}$ on a perfect mirror. Compute the force and the pressure, using the $\nu = i$ components.

Step 1 — the momentum flux. The beam delivers momentum at rate $P_{\text{beam}}/c$ and the mirror reverses it, so the momentum transfer rate — the force — is F = \frac{2P_{\text{beam}}}{c} = \frac{2(1000)}{2.99792458\times10^{8}} = \mathbf{6.6713\times10^{-6}\ N} = 6.67\ \mu\mathrm{N}.

Step 2 — the pressure. Spot area $= \pi(1.0\times10^{-3})^2 = 3.1416\ \mathrm{mm^2}$ , so P_{\text{rad}} = \frac{F}{A} = \mathbf{2.1235\ Pa} . The intensity there is $I = 1000/3.1416\times10^{-6} = \mathbf{318.31\ MW\,m^{-2}}$ — and indeed $2I/c = 2.1235\ \mathrm{Pa}$ ✓, the same answer by the route of `aq-sr-electromagnetic-stress-energy-tensor`.

Step 3 — read it as the balance law. Equation (4) says the field momentum in the beam is being destroyed at the mirror, and the deficit is exactly the force on the mirror: $\partial_t g$ is negative in the reflection region, and the source term $-(\rho\vec E + \vec J\times\vec B)$ is the force per unit volume on the mirror's surface currents. The field loses momentum; the mirror gains it; the sum is conserved. Two pascals is a real, laboratory-scale pressure — about a hundred-thousandth of an atmosphere — and it is the same physics as the wire, read in different slots of the same tensor equation.

Figure

ONE EQUATION, TWO NINETEENTH-CENTURY THEOREMS. The four-vector identity ∂_μT^{μν} = −F^{νλ}J_λ has four slots. The ν = 0 slot unpacks into Poynting's theorem — published 1884, a statement about energy — and the ν = i slots unpack into Maxwell's momentum balance — published 1873, a statement about force. The two were discovered eleven years apart, by different people, in different languages, and nobody suspected they were the same equation until Minkowski wrote them as one four-vector in 1908. The right-hand side is the Lorentz four-force density f^ν = (J·E/c, ρE + J×B): its time component is power, its space components are force. Note that J×B appears in the force but NOT in the power — magnetic forces never do work. And the whole thing is a TRANSFER law, not a conservation law: the field is not conserved by itself. Add the matter's stress-energy tensor and the divergence vanishes identically.

Common trapThis article's base contradicts itself — and you can prove it in one line.

The AllQFT base for this article prints three statements: 1 · the stress-energy tensor $T^{\mu\nu} = \mu_0^{-1}\big(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}\big)$ , declaring signature $(+,-,-,-)$ ; 2 · the balance law $\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$ ; 3 · the force density $f^\nu = F^{\nu\lambda}J_\lambda = (\vec J\cdot\vec E/c,\ \rho\vec E + \vec J\times\vec B)$ .

Statements 2 and 3 are correct. Statement 1 is not — and the three cannot all be true. In the declared signature, that tensor has $T^{00} = -u$ , a negative energy density; and its divergence is $+F^{\nu\lambda}J_\lambda$ , not $-$ . The base has printed a tensor incompatible with its own conservation law.

The one-line proof. Take the wire. In steady state the field's energy density is constant, so $\partial_\mu T^{\mu 0}$ reduces to $\nabla\cdot\vec S$ , and Poynting's theorem gives $\oint\vec S\cdot d\vec A = -I^2R$ : energy flows in, and the field loses it to the copper. With the base's tensor, $\vec S$ reverses and the wire would radiate $0.53\ \mathrm{W}$ into the room while heating up. The field would be paid by the resistor.

Why nobody caught it: the base checks that $T^{\mu\nu}$ is traceless, and the wrong tensor is traceless too — as is the right one. Tracelessness, symmetry and gauge invariance all survive a global sign flip. Only $T^{00} > 0$ discriminates, and it is asserted, as an inequality, in this article's verification script. See `aq-sr-electromagnetic-stress-energy-tensor`, which owns the tensor and the repair.

Common trapA source on the right is not a broken conservation law. Do not "fix" it.

It is tempting to see $\partial_\mu T^{\mu\nu} \ne 0$ and reach for a repair — a better tensor, a missing term, a redefinition that makes the right-hand side vanish. Resist. The field's energy-momentum genuinely is not conserved when charges are present, and it should not be: the field is giving energy to the charges, and a law that hid that fact would be less useful, not more. The conserved object is the sum, $T^{\mu\nu}_{\text{field}} + T^{\mu\nu}_{\text{matter}}$ , and its divergence vanishes identically. What equation (1) buys you over a bare $\partial_\mu T^{\mu\nu} = 0$ is the rate of transfer — which is the physically interesting quantity, and the one you actually measure when you put a thermometer on a resistor. The corollary matters for radiation problems: in a source-free region — the vacuum outside an antenna, the space a light wave crosses — the right-hand side is zero, and the field's energy-momentum is strictly conserved. Radiation carries its energy to infinity without loss. The law only has a source where the charges are.

Physics of the portrait

The portrait is a theorem, not a person — the result that gave energy a location.

Origins · 2 minPoynting's theorem and how this was discovered

Poynting's theorem

1884 · energy has a location and a flux · the wire is heated from outside · the ν = 0 slot of a tensor nobody had yet written

John Henry Poynting published On the Transfer of Energy in the Electromagnetic Field in 1884, and what he proposed was stranger than it now looks. Everyone agreed that energy was conserved. What nobody had claimed was that energy has a definite location in space and moves through it with a definite flux — that you could point at a cubic centimetre of empty air and say how many joules are in it and how many joules per second are crossing its face. Poynting's vector $\vec S = \vec E\times\vec H$ made exactly that claim, and it immediately produced a result so counter-intuitive that it is still the best teaching example in the subject: the energy heating a current-carrying wire does not flow along the wire. It flows in from the surrounding space, through the wire's surface, sideways. The copper is a sink, not a conduit. Integrate the flux over the surface and you get $I^2R$ exactly, which is the correct answer arrived at by an apparently insane route — and the agreement is not a coincidence but a theorem. Heaviside reached the same result within a year and gave it the vector form we use; on the standard account he arrived at it independently. Neither man could have suspected what it was. Twenty-four years later Minkowski wrote $\partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda$ and Poynting's theorem turned out to be the $\nu = 0$ component — one slot of a four-vector whose other three slots are Maxwell's momentum balance from 1873. Two theorems, eleven years and one dimension apart, were the same equation the whole time, and the only way to see it was to stop treating time as different from space.

History

1873

— Maxwell's "Treatise" gives the stress tensor of the field, whose divergence is the momentum balance.

1884

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

1885

— Heaviside recasts the theorem in vector notation; the standard account credits him with an independent derivation.

1900

— Poincaré introduces an electromagnetic field momentum density S/c² in order to rescue the conservation of momentum.

1908

— Minkowski writes the energy and momentum balances as one four-vector identity, ∂_μT^{μν} = −F^{νλ}J_λ.

1918

— Noether proves that conservation of energy and momentum follows from the translation invariance of spacetime.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §8.1–8.2 and §12.3 · Jackson, Classical Electrodynamics, 3rd ed., §6.7–6.8 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Landau & Lifshitz, The Classical Theory of Fields, 4th ed., §31–33.

Provenance. Signature $(+,-,-,-)$ , hence $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. Dictionary $F^{0i} = -E^i/c$ , $F^{ij} = -\epsilon^{ijk}B_k$ , $J^\mu = (c\rho,\vec J)$ , $J_\lambda = (c\rho,-\vec J)$ . This article's base prints a stress-energy tensor whose divergence is $+F^{\nu\lambda}J_\lambda$ , alongside the correct balance law and the correct force density — three mutually inconsistent statements. The balance law and the force density are right and are used; the tensor is repaired in `aq-sr-electromagnetic-stress-energy-tensor` and the repair is machine-asserted here as `assert T00 > 0`, together with an explicit demonstration that the base's tensor reverses the Poynting flux and would make a resistor feed the field. Every number on this page is recomputed in `mpmath` at 50 digits against the article's own printed digits.

Scope. The tensor itself, its construction and its component dictionary belong to `aq-sr-electromagnetic-stress-energy-tensor`; the field equations used in the derivation to `aq-sr-covariant-maxwell-equations`; the force law whose density appears on the right to `aq-sr-covariant-lorentz-force`; charge conservation to `aq-sr-charge-invariance-continuity`. This page owns the balance law and its unpacking into Poynting's theorem and the momentum balance. The electromagnetism track's parallel treatments are `aq-em-conservation-energy-momentum` and `aq-em-poyntings-theorem`.

Builds on. `aq-sr-electromagnetic-field-tensor` · `aq-sr-covariant-maxwell-equations` · `aq-sr-electromagnetic-stress-energy-tensor` · `aq-sr-covariant-lorentz-force`. Leads to. `aq-sr-lienard-wiechert-potentials` · `aq-sr-electromagnetic-field-invariants`.

If the energy really enters a wire from the sides, why does everyone say current "carries" power?

Because for a DC circuit the bookkeeping comes out the same and the language is harmless. The power delivered is $IV$ either way. But the field picture is the one that is true, and it stops being optional as soon as the frequency rises: in a coaxial cable at GHz the energy is demonstrably in the dielectric, and the copper only guides it. The clean statement is that the current does not transport the energy; it establishes the boundary conditions on the fields that do.

Is the field's energy-momentum really not conserved? That sounds like a broken theory.

It is not conserved by itself, and it should not be — the field is handing energy and momentum to matter, and a law that concealed that would be worse. The conserved object is the sum of field and matter, and $\partial_\mu(T^{\mu\nu}_{\text{field}} + T^{\mu\nu}_{\text{matter}}) = 0$ holds exactly. Compare charge: $\partial_\mu J^\mu = 0$ has no source because charge is never converted into anything. Energy is converted, constantly, and equation (1) is the receipt.

Why does the Bianchi identity appear in a derivation about energy?

Because two of the three terms in $\partial_\mu T^{\mu\nu}$ have nothing to do with sources, and something has to kill them. The Bianchi identity — $\partial_{[\mu}F_{\nu\rho]} = 0$ , which is automatic once $F = dA$ and encodes "no magnetic monopoles" plus Faraday's law — is what makes them cancel. It is a striking division of labour: the inhomogeneous Maxwell equation supplies the physics (the force density), and the homogeneous one supplies the consistency. Remove either and the balance law does not close.

Does J × B really never do work? What about a motor?

The magnetic force on a charge never does work: $\vec F = q\vec v\times\vec B$ is perpendicular to $\vec v$ , so $\vec F\cdot\vec v = 0$ , and that is why $\vec J\times\vec B$ is absent from $f^0$ . A motor turns because the magnetic force redirects momentum — but the energy that ends up as rotation comes through the electric field, which is doing the work in the windings, exactly as Poynting's theorem says. The magnetic field steers; the electric field pays.

What is the matter stress-energy tensor here — do I need to know its form?

Not for this derivation. All that is required is that it be defined so that $\partial_\mu T^{\mu\nu}_{\text{matter}} = +f^\nu$ — that is, so that the matter gains exactly what the field loses. For a cloud of charged dust of proper density $\rho_0$ and four-velocity $u^\mu$ , $T^{\mu\nu}_{\text{matter}} = \rho_0 u^\mu u^\nu$ does the job, and $\partial_\mu T^{\mu\nu}_{\text{matter}} = +f^\nu$ is then the covariant Lorentz force law in density form. The two tensors are defined so that the ledger closes, and it does.

The Poynting vector is not unique — you can add any curl to it. Does that spoil the argument?

It is true that $\vec S \to \vec S + \nabla\times\vec Z$ leaves $\nabla\cdot\vec S$ , and therefore Poynting's theorem, unchanged. So the local flux is not uniquely determined by conservation alone. But the choice $\vec S = \vec E\times\vec H$ is not arbitrary: it is the one that makes $\vec S/c^2$ equal the momentum density, and it is the one that appears as $T^{0i}$ in a symmetric, gauge-invariant stress-energy tensor — which is what general relativity requires, since $T^{\mu\nu}$ sources gravity and gravity has no ambiguity to absorb. The freedom is real in electromagnetism alone; it disappears the moment you ask what the field weighs.