Charge Invariance and Continuity

Fill a room with hydrogen. In every atom the electron moves at $\alpha c = c/137 = 2.19\times10^{6}\ \mathrm{m\,s^{-1}}$ while the proton sits essentially still. Now suppose — as almost every other property of a particle does — that charge depended on speed: suppose it scaled with $\gamma$ , the mildest guess available. Then the electron's charge would exceed the proton's by $\gamma - 1 = 2.66\times10^{-5}$ , every atom in the room would carry that much net charge, and the gas would tear itself apart. It does not. The proton and electron charges agree to better than one part in $10^{21}$ . The $\gamma$ hypothesis is not merely wrong — it is wrong by sixteen orders of magnitude. So $q$ is a Lorentz scalar: every observer counts the same coulombs on the same body, no $\gamma$ anywhere. Which is strange, because the charge density $\rho$ standing right beside it is emphatically not invariant — $\rho$ is the time slot of a four-vector, it grows by $\gamma$ under a boost, and observers flatly disagree about it. Boost a neutral wire and charge density appears where there was none. Two words get used almost interchangeably here and they are not the same word. Conservation is about time: charge is never created or destroyed, only moved. Invariance is about frames: everybody counts the same total. To total up the charge you must add $\rho$ over all space at one instant — and "one instant" is a slice through spacetime. Two observers slice differently. That they nevertheless agree is a theorem needing both facts: microscopic invariance and continuity. Remove either and "the total charge of the universe" stops being a number and becomes a function of your velocity. Then the payoff: you cannot write electromagnetism down with a leaky source. Charge conservation is not a postulate bolted onto Maxwell — it is forced, by one line of index algebra.

Key takeaways

Signature $(+,-,-,-)$ , $\;\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ — lower index, plus gradient. $\;p\cdot p = +m^2c^2$ .
Charge $q$ is a Lorentz SCALAR. Charge density $\rho$ is not — it is the time slot of the four-current $J^\mu = (c\rho,\ \vec J\,) = \rho_0 U^\mu$ , banked in The Four-Current Density, and it boosts.
Local conservation is one covariant line: $\boxed{\partial_\mu J^\mu = 0}$ , which unpacks — with no minus signs surviving, though the metric has three — to $\boxed{\partial_t\rho + \nabla\!\cdot\!\vec J = 0}$ . Globally, $\dot q_V = -\oint_{\partial V}\vec J\cdot\mathrm d\vec A$ : charge leaves a box only through its walls.
Owned result — MAXWELL FORCES IT. $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ plus antisymmetry gives $\partial_\nu J^\nu = 0$ identically. Conservation is a theorem, and the displacement current is what pays for it.
Owned result — invariance $\ne$ conservation, and global invariance needs BOTH. Charge is the flux of $J^\mu$ through a spacelike slice; two frames slice differently, and the counts agree only because $\partial_\mu J^\mu = 0$ seals the wedge between the slices.
Key idea: charge is the one thing in this chapter a boost cannot touch. Everything else — $\rho$ , $\vec J$ , $\vec E$ , $\vec B$ , the very words "electric" and "magnetic" — is frame-talk. The invariance of $q$ is the fixed point the whole covariant edifice is nailed to, and it is an experimental fact good to $10^{-21}$ .

The core insight

Charge is a counting number; density is a count divided by a volume — and volume is precisely what relativity refuses to hold still. So $\rho \to \gamma\rho_0$ while $\mathrm dV \to \mathrm dV_0/\gamma$ , and the product survives untouched. Everything covariant in electromagnetism rests on that one cancellation: it makes $\rho_0$ a scalar, hence $J^\mu = \rho_0 U^\mu$ a four-vector, hence $\partial_\mu J^\mu$ a scalar — which is what makes "charge is conserved" mean the same thing in every frame.

Where this lives today

Charge conservation is the most rigorously tested conservation law in physics, and it is tested two ways. The first is the neutrality of bulk matter: gas-neutrality experiments — the current best being Bressi and collaborators' 2011 measurement on sulphur hexafluoride, which the Particle Data Group adopts — bound the proton–electron asymmetry below $10^{-21}\,e$ , which is why the electrical universe of Lyttleton and Bondi is a historical curiosity rather than a cosmology. The second is the direct hunt for charge non-conservation, such as an electron decaying to a neutrino and a photon; the underground detectors built for neutrinoless double beta decay double as monitors, and the bound on the electron's lifetime against charge-violating decay now exceeds $10^{26}$ years. Nothing has ever been seen. The reason for that silence is Noether's (1918). Charge conservation is the Noether current of the global phase symmetry $\psi \to e^{i\theta}\psi$ of the charged matter field, and demanding that the phase be free to vary from point to point is precisely what forces $A^\mu$ into existence (Gauge Transformations and the Lorenz Gauge). Charge conservation, gauge invariance and the masslessness of the photon are three faces of one symmetry — which is why violating any of them would be so expensive. Practically, continuity does quiet work everywhere. Every circuit simulator enforces $\sum I = 0$ at each node; every plasma and magnetohydrodynamics code carries $\partial_t\rho + \nabla\!\cdot\!\vec J = 0$ as a hard constraint — and a scheme that violates it does not merely lose accuracy: it manufactures spurious charge, which sources spurious fields, and the run diverges. Antisymmetry is the guardrail in the continuum; on a grid you must build the guardrail yourself.

Origins · 2 minCharge conservation and how this was discovered

Charge conservation

1747–present · Franklin's single fluid · Maxwell's displacement current · Noether's theorem

The law began as accounting. In 1747 Benjamin Franklin proposed in his letters to Peter Collinson that electricity was a single fluid, and in a letter to Cadwallader Colden on 5 June 1747 he stated the conservation principle outright: the electrical fire is "a real Element, or Species of Matter, not created by the Friction, but collected only." Rubbing does not make electricity — it moves it, and a body gaining what Franklin named positive electricity does so at the exact expense of another going negative. That is charge conservation, a century before anyone could write $\partial_t\rho + \nabla\!\cdot\!\vec J = 0$ . The law became structural with James Clerk Maxwell. The displacement current — introduced in "On Physical Lines of Force" (1861–62), built into the completed system of "A Dynamical Theory of the Electromagnetic Field" (1865) — is usually taught as the ingenious step that produced electromagnetic waves, and it did. But its necessity is more prosaic and more inescapable: without it, Ampère's law demands $\nabla\!\cdot\!\vec J = 0$ , which a charging capacitor flatly violates. Conservation of charge was the constraint; light was the dividend. The modern statement is Emmy Noether's (1918): every continuous symmetry of the action yields a conserved current, and the one whose conservation is $\partial_\mu J^\mu = 0$ belongs to the global phase symmetry of the charged matter field. Make that phase local and electromagnetism itself is forced. The covariant packaging — charge and current as one four-vector — is Hermann Minkowski's, from the 1908 Grundgleichungen. Franklin's fluid, Maxwell's constraint, Minkowski's four-vector, Noether's symmetry: four statements of one law, each stronger than the last.

History

1747

— Benjamin Franklin sets out the single-fluid theory of electricity in his letters to Peter Collinson, and in a letter to Cadwallader Colden of 5 June states the conservation principle: the electrical fire is "not created by the Friction, but collected only."

1861–62

— Maxwell introduces the displacement current in "On Physical Lines of Force"; it is required for Ampère's law to be consistent with a charging capacitor, and it is what lets the equations carry waves.

1865

— Maxwell's "A Dynamical Theory of the Electromagnetic Field" presents the completed system and identifies light as an electromagnetic wave.

1908

— Minkowski's "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern" casts charge and current as a single four-vector and continuity as the vanishing of its four-divergence.

1918

— Emmy Noether's "Invariante Variationsprobleme" proves that continuous symmetries generate conserved currents; charge conservation becomes the Noether current of global phase symmetry.

1959

— Lyttleton and Bondi propose an "electrical universe" in Proc. R. Soc. A 252, 313: a proton–electron charge asymmetry of a few parts in 10^18 would let electrostatic repulsion exceed gravity and drive the cosmic expansion. The proposal is testable, and it triggers a generation of precision neutrality experiments.

1960

— J. G. King's search for a small charge carried by molecules (Phys. Rev. Lett. 5, 562) pushes the bound below the Lyttleton–Bondi window, killing the electrical universe.

2011

— Bressi and collaborators measure the neutrality of sulphur hexafluoride (Phys. Rev. A 83, 052101); the Particle Data Group adopts their limit, |q_p + q_e| below 1 × 10^-21 e.

The physical picture

In relativity almost nothing is invariant. Length is not; duration is not; the electric field is not, and neither is the magnetic. Against that it is genuinely surprising that charge simply refuses to move: a proton accelerated to $0.999999c$ carries exactly the charge it had at rest. That is not a theorem of relativity — it is an experimental input. Purcell's version of the evidence is the sharpest: a helium atom and a hydrogen molecule contain the same charged particles, two protons and two electrons. But in H₂ the protons drift slowly, an ångström apart, while in the helium nucleus they are bound with kinetic energies around $1\ \mathrm{MeV}$ — some $4.6\%$ of $c$ . If motion changed charge even slightly, the electronic and nuclear charges could not cancel exactly in both. They do, to $10^{-21}$ . What invariance buys is a four-vector: if $\mathrm dq$ is a scalar and $\mathrm dV_0$ the proper volume carrying it, then $\rho_0 = \mathrm dq/\mathrm dV_0$ is a scalar, and $J^\mu = \rho_0 U^\mu$ — scalar times four-vector — is a four-vector. That construction belongs to The Four-Current Density, which earns $J^\mu = (c\rho, \vec J\,) = \rho_0 U^\mu$ , shows $\rho = \gamma\rho_0$ , and computes $J^\mu J_\mu$ . We take it as given and ask different questions: why is $q$ a scalar, how well do we know it, and what is the covariant statement that charge is never lost? The last has a one-line answer with a surprise in it. The metric carries three minus signs, so one braces for the continuity equation to sprout one. It does not — contracting a lower-index gradient against an upper-index four-vector never invokes the metric at all. And then the question worth staying for: "the total charge" requires a slice through spacetime, and slices are frame-dependent, tilted hyperplanes cutting the charges' world-tubes at different events. Why should two observers agree? They agree because charge is conserved — which is why invariance and conservation, though genuinely different statements, are not the independent axioms the textbooks imply.

Figure

PUNCHLINE — global charge invariance needs BOTH microscopic invariance AND continuity. Two observers slice spacetime differently: Σ is a horizontal now, Σ′ a tilted now. They cut the charge worldlines at different events. The counts agree only because ∂ₘJᵐ = 0 makes the four-dimensional wedge between the slices leak-proof. Delete continuity and 'the total charge' is not a well-defined number at all — it depends on who is slicing

The core equations

The experimental fact, stated as the definition of what kind of object $q$ is:

\boxed{\;q \;=\; \int_{\Sigma}\rho\,\mathrm dV \;=\; \text{the same number in every inertial frame}\;} \qquad (1)

where: $\Sigma$ — a spacelike hypersurface (somebody's "all of space, now"). Physical reading: $q$ is a Lorentz scalar; $\rho$ is not. The scalar is the integral, because the volume shrinks by exactly the factor the density grows by. The four-current, recalled from The Four-Current Density:

J^\mu \;=\; (c\rho,\ \vec J\,) \;=\; \rho_0 U^\mu , \qquad U^\mu = \gamma(c, \vec v\,), \qquad \rho = \gamma\rho_0 \qquad (2)

where: $\rho_0 = \mathrm dq/\mathrm dV_0$ — the proper charge density, a scalar precisely because $\mathrm dq$ is invariant. Physical reading: this is where charge invariance is spent. Take it away and $\rho_0$ is not a scalar, $J^\mu$ is not a four-vector, and every covariant statement below collapses. Local conservation, in the only form worth memorising, and its global partner:

\boxed{\;\partial_\mu J^\mu = 0\;} \iff \boxed{\;\frac{\partial\rho}{\partial t} + \nabla\!\cdot\!\vec J = 0\;} \qquad (3) \qquad\quad \frac{\mathrm d}{\mathrm dt}\int_V \rho\,\mathrm dV = -\oint_{\partial V}\vec J\cdot\mathrm d\vec A \qquad (4)

where: $\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ — the lower-index four-gradient. Physical reading: $\partial_\mu J^\mu$ is a scalar, so a law setting it to zero says the same thing in every frame; charge in a box changes only by flowing through the walls. Note what is absent: minus signs. Contracting lower against upper never touches the metric, so the three minuses of $(+,-,-,-)$ never appear; anyone writing $\partial_t\rho - \nabla\!\cdot\!\vec J$ has quietly used $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ by mistake. And the theorem that makes conservation non-negotiable:

\boxed{\;\partial_\mu F^{\mu\nu} = \mu_0 J^\nu \;\Longrightarrow\; \partial_\nu J^\nu = 0\;} \qquad (5)

where: $F^{\mu\nu} = -F^{\nu\mu}$ — the antisymmetric field tensor. Physical reading: charge conservation is a consequence of electromagnetism, not an assumption of it. Take the four-divergence of the source equation and the left side becomes $\partial_\nu\partial_\mu F^{\mu\nu}$ — a symmetric derivative pair contracted with an antisymmetric tensor, which vanishes identically. Maxwell's equations will not let you write a source that leaks.

Derivation AContinuity in one covariant line, and why none of the metric's minus signs survive

The goal is to show that the scalar equation $\partial_\mu J^\mu = 0$ contains exactly the classical continuity equation — no more, no less — and to settle a sign question that trips up nearly everyone on first contact. In a signature carrying three minus signs one braces for a minus in the unpacking. There is none, and the reason governs every contraction in this chapter.

Step 1. State both index positions. $J^\mu = (c\rho,\ J_x, J_y, J_z)$ is naturally contravariant, being $\rho_0 U^\mu$ with $U^\mu = \mathrm dx^\mu/\mathrm d\tau$ . The four-gradient $\partial_\mu \equiv \partial/\partial x^\mu = \big(\tfrac1c\partial_t,\ \partial_x, \partial_y, \partial_z\big)$ is naturally covariant, with a plus on the spatial part. Why: index position is not a decoration you choose — differentiating by an upper-index coordinate produces a lower-index object, and that is where the spatial plus comes from.

Step 2. Contract, and notice the metric is never invoked. $\;\partial_\mu J^\mu = \partial_0 J^0 + \partial_1 J^1 + \partial_2 J^2 + \partial_3 J^3$ — a lower index summed against an upper one, with $\eta_{\mu\nu}$ appearing nowhere. Why: this is the whole answer to the sign question. The minus signs live inside $\eta$ , and $\eta$ only shows up when you raise or lower. Here nothing is raised or lowered, so nothing flips.

Step 3. Substitute. $\;\partial_0 J^0 = \tfrac1c\,\partial_t(c\rho) = \partial_t\rho$ ; $\;\partial_i J^i = \nabla\!\cdot\!\vec J$ . Hence $\;\boxed{\partial_\mu J^\mu = \partial_t\rho + \nabla\!\cdot\!\vec J = 0}$ . Why: the two factors of $c$ cancel exactly — which is why $J^0$ is defined as $c\rho$ and not $\rho$ , so all four components share the units $\mathrm{A\,m^{-2}}$ . The result is the continuity equation of fluid mechanics, of circuit theory, of probability in quantum mechanics, each time saying the same thing: the stuff is shuffled, never made.

Step 4. Integrate for the global law, and collect the dividend. $\int_V \partial_t\rho\,\mathrm dV = -\oint_{\partial V}\vec J\cdot\mathrm d\vec A$ . Why: the covariant form is strictly stronger than the three-dimensional one, because $\partial_\mu J^\mu$ is a scalar — vanish in one frame, vanish in all. We do not postulate conservation frame by frame; we postulate it once and relativity distributes it. And setting $\partial_t\rho = 0$ leaves $\nabla\!\cdot\!\vec J = 0$ , which over a circuit node is $\sum I_{\rm in} = \sum I_{\rm out}$ : Kirchhoff's current law is the steady-state space part of $\partial_\mu J^\mu = 0$ . *Verified in verify_aq-sr-charge-invariance-continuity.py: $\partial_\mu J^\mu$ built symbolically from $\partial_\mu = (\tfrac1c\partial_t, +\nabla)$ and $J^\mu = (c\rho, \vec J\,)$ with $\rho, \vec J$ arbitrary functions of $(t,x,y,z)$ , asserted equal to $\partial_t\rho + \nabla\!\cdot\!\vec J$ with no sign flips; the $c$ -cancellation checked; and $\partial^\mu J_\mu$ shown to give the identical scalar, as it must.*

Go deeper · Derivation B · 2 minThe total charge is the same on every slice — and that is why it is the same in every frame

Derivation BThe total charge is the same on every slice — and that is why it is the same in every frame

This is the article's owned structural result, and the reason the two words in the title belong on one page. "The total charge" is not self-evident: to compute it you add $\rho$ over all space at one time, and "at one time" means on a spacelike hypersurface — a frame-dependent choice. Two observers in relative motion slice along differently tilted hyperplanes and sample the charge world-tubes at different events. That they nevertheless get the same total is not obvious and it is not free: it is a theorem, and its hypotheses are exactly the two facts this article is about.

Step 1. Write the charge as a flux, not a volume integral. Define $\;q_\Sigma = \tfrac1c\displaystyle\int_\Sigma J^\mu\,\mathrm d\Sigma_\mu$ . On a $t = \text{const}$ slice $\mathrm d\Sigma_\mu = (\mathrm dV,\ \vec 0\,)$ , so $\tfrac1c J^\mu\mathrm d\Sigma_\mu = \tfrac1c(c\rho)\,\mathrm dV = \rho\,\mathrm dV$ , recovering $q = \int\rho\,\mathrm dV$ . Why: the same number, re-expressed covariantly. The gain: a flux of a four-vector through a hypersurface is a scalar, which puts four-dimensional Gauss within reach.

Step 2. Bound a four-dimensional region with two slices: $\Sigma_1$ my simultaneity slice, $\Sigma_2$ yours — tilted — and $\Omega$ the wedge between them, closed at spatial infinity where $\vec J \to \vec 0$ . Why: choosing slices belonging to different frames is what converts a conservation statement into an invariance statement.

Step 3. Apply the four-dimensional divergence theorem. $\;\displaystyle\int_\Omega \partial_\mu J^\mu \,\mathrm d^4x = \oint_{\partial\Omega} J^\mu\,\mathrm d\Sigma_\mu = c\,q_{\Sigma_2} - c\,q_{\Sigma_1}$ , the wall term dying with the current at infinity. Why: exactly as in three dimensions. The minus on $\Sigma_1$ is orientation — the boundary traverses the earlier slice backwards.

Step 4. Impose continuity and conclude. The left side vanishes, so $\;\boxed{q_{\Sigma_1} = q_{\Sigma_2}}$ . Since $\Sigma_1$ and $\Sigma_2$ are the "nows" of two different inertial frames, the frames count the same total — global charge invariance, proved rather than assumed. Why: notice exactly what went in. (i) $J^\mu$ had to be a four-vector, which required microscopic $\mathrm dq$ invariance — the experimental input. (ii) $\partial_\mu J^\mu$ had to vanish — conservation. Remove either and the proof dies.

Step 5. Watch it fail, and fix the logical order. If charge could be created, $\partial_\mu J^\mu = \sigma(x) \ne 0$ inside $\Omega$ , then $c(q_{\Sigma_2} - q_{\Sigma_1}) = \int_\Omega \sigma\,\mathrm d^4x \ne 0$ : the observers disagree, and which total you get depends on how you tilt your slice. Why: "the total charge of the universe" would not be a number — it would be a function of the observer's velocity. The honest order: microscopic invariance is an experimental fact; continuity is a theorem of Maxwell (next); global invariance is a theorem resting on both. *Verified in verify_aq-sr-charge-invariance-continuity.py: the flux formula $\tfrac1c J^\mu\mathrm d\Sigma_\mu = \rho\,\mathrm dV$ on a constant- $t$ slice; and a direct numerical check of slice-independence — a charge packet drifting at $0.6c$ , integrated over the lab slice and over the tilted simultaneity slice of the boosted frame, giving the same total to 10 significant figures — plus the same computation with a deliberately non-conserved source, which is shown to give different totals on the two slices.*

Go deeper · Derivation C · 1 minMaxwell forces conservation: electromagnetism cannot be written down with a leaky source

Derivation CMaxwell forces conservation: electromagnetism cannot be written down with a leaky source

The finest result on this page, and it costs three lines. Charge conservation is usually presented as an independent law that Maxwell's equations happen to respect. That is backwards. Given the inhomogeneous covariant Maxwell equation, conservation is not respected — it is compelled. There is no consistent electromagnetism with a non-conserved source: the field equation refuses, by a piece of index algebra that recurs all over physics.

Step 1. Start from the source equation, unpacked in The Electromagnetic Field Tensor: $\;\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ . Why: this one tensor equation is the two inhomogeneous Maxwell equations — $\nu=0$ gives Gauss's law, $\nu=i$ gives Ampère–Maxwell.

Step 2. Take the four-divergence on the free index. $\;\partial_\nu\partial_\mu F^{\mu\nu} = \mu_0\,\partial_\nu J^\nu$ . Why: $\nu$ is the only index available, and differentiating a tensor equation keeps it covariant.

Step 3. Kill the left-hand side by symmetry. $\partial_\nu\partial_\mu$ is symmetric in $(\mu,\nu)$ — partial derivatives commute — while $F^{\mu\nu}$ is antisymmetric. Relabel the dummies $\mu \leftrightarrow \nu$ : $\;\partial_\nu\partial_\mu F^{\mu\nu} = \partial_\mu\partial_\nu F^{\nu\mu} = -\partial_\nu\partial_\mu F^{\mu\nu}$ . The quantity equals minus itself, so it is zero. Why: this is the whole theorem. It uses no property of the electromagnetic field beyond antisymmetry, and none of the derivative beyond commutation. It is $\nabla\!\cdot\!(\nabla\times\vec A) = 0$ in four-dimensional clothes.

Step 4. Conclude. $\;0 = \mu_0\,\partial_\nu J^\nu \Rightarrow \boxed{\partial_\nu J^\nu = 0}$ . Why: conservation is a theorem of Maxwell's equations, not a postulate added to them. Feed the theory a source that creates charge and there is simply no $F^{\mu\nu}$ satisfying the equations — antisymmetry is a guardrail welded into the geometry of the field. The three-dimensional shadow of this is the displacement current, and Worked Example 2 puts numbers on it. *Verified in verify_aq-sr-charge-invariance-continuity.py: $\partial_\nu\partial_\mu F^{\mu\nu} = 0$ checked symbolically for a general antisymmetric $F^{\mu\nu}(t,x,y,z)$ with six independent arbitrary component functions (not the EM one — the theorem needs only antisymmetry); the $\nu=0$ and $\nu=i$ unpackings into Gauss's law and Ampère–Maxwell; and the divergence of Ampère–Maxwell reproducing $\partial_t\rho + \nabla\!\cdot\!\vec J = 0$ .*

Figure

The neutrality of matter on a log scale — and what the numbers kill. If charge scaled with γ, hydrogen would carry 2.66×10⁻⁵ e. Lyttleton and Bondi noted in 1959 that an excess of a few parts in 10¹⁸ would let electrostatic repulsion beat gravity and drive the cosmic expansion; our own break-even calculation gives 9.0×10⁻¹⁹. The measured bound on |q_p + q_e|/e is 10⁻²¹ — killing the γ hypothesis by 16 orders of magnitude and the electrical universe by three

Limiting cases

Case	Result	Meaning
blob of charge at rest	$J^\mu = (c\rho_0,\ \vec 0\,)$	purely timelike four-current; no current anywhere
boosted along $x$	$\rho = \gamma\rho_0$ , $\;J^x = \gamma\rho_0 v$	density up by $\gamma$ , volume down by $\gamma$ — $q$ untouched
steady state, $\partial_t\rho = 0$	$\nabla\!\cdot\!\vec J = 0$	Kirchhoff's current law: current in = current out at every node
neutral current-carrying wire	$\rho = 0$ , $\vec J \ne \vec 0$ in the lab	$J^\mu J_\mu = -\|\vec J\|^2 < 0$ : spacelike. No frame kills the current, and frames disagree on the sign of $\rho$
$\partial_\mu J^\mu \ne 0$	no $F^{\mu\nu}$ exists	Maxwell's equations become inconsistent. Not "wrong" — unwritable

Common trapρ is not invariant. q is. The words are nearly the same; the objects are not.

The commonest slip in the chapter, because "charge" and "charge density" sound like one idea with a volume attached. They are not the same kind of thing:

$q = \int\rho\,\mathrm dV$ is a Lorentz scalar. Frame-independent. Boost until you are blue and it does not move.
$\rho = J^0/c$ is the time component of a four-vector. Frame-dependent. It boosts like $ct$ does, mixing with $\vec J$ : $\;\rho' = \gamma\big(\rho - \tfrac{v}{c^2}J^x\big)$ , $\;J'^x = \gamma\big(J^x - v\rho\big)$ .

The two coexist because the volume element is also frame-dependent, in exactly the compensating way. A great deal rides on not confusing them: the whole reason a neutral current-carrying wire can push a moving charge is that $\rho' \ne 0$ in the charge's frame even though $\rho = 0$ in the lab. (That story is derived in full in Magnetism as a Relativistic Effect.)

Worked examples

1 · How badly is "charge grows with speed" ruled out?

Propose the mildest possible velocity-dependence, $q(v) = \gamma\,q(0)$ , so that charge transforms as energy does. Test it against the neutrality of hydrogen, and then against the neutrality of the universe.

Step 1 — identify. In hydrogen the proton is essentially at rest while the electron moves, in the Bohr picture, at $v = \alpha c$ , $\alpha = 1/137.036$ . The hypothesis inflates the electron's charge magnitude by $\gamma(\alpha c)$ , so the atom carries net charge $|q_p + q_e| = (\gamma - 1)\,e$ .

Step 2 — the number. $\beta = \alpha = 7.2974\times10^{-3}$ , so $\beta^2 = 5.3252\times10^{-5}$ and $\gamma = 1.0000266$ . Hence $\;\gamma - 1 = \mathbf{2.66\times10^{-5}}$ . (Shortcut: $\gamma - 1 \approx \tfrac12\alpha^2 = 2.663\times10^{-5}$ , agreeing to four figures.)

Step 3 — compare with experiment. The measured bound is $|q_p + q_e|/e < 10^{-21}$ , while the hypothesis predicts $2.66\times10^{-5}$ — excluded by $\;2.66\times10^{-5}/10^{-21} = \mathbf{2.7\times10^{16}}$ , sixteen orders of magnitude. Nor is there a gentler dependence to retreat to: any $\beta^2$ term with an order-unity coefficient dies in the same place.

Step 4 — the boundary number. How small must an asymmetry be to be harmless? Give every hydrogen atom net charge $\varepsilon e$ and compare the electrostatic repulsion of two of them with their gravitational attraction. Both go as $1/r^2$ , so the ratio is distance-free:

\frac{F_{\rm E}}{F_{\rm G}} = \frac{k(\varepsilon e)^2}{G m_{\rm H}^2} = \varepsilon^2\,\frac{k e^2}{G m_{\rm H}^2} = \varepsilon^2 \times \mathbf{1.234\times10^{36}} .

Break-even sits at $\;\varepsilon = 1/\sqrt{1.234\times10^{36}} = \mathbf{9.0\times10^{-19}}$ — an asymmetry in the eighteenth decimal place would make the universe electrostatically self-repulsive. That is precisely the "electrical universe" Lyttleton and Bondi floated in 1959 as a mechanism for cosmic expansion, needing an excess of a couple of parts in $10^{18}$ .

Step 5 — the verdict. The bound is $10^{-21}$ , a thousand times below break-even. At the bound, $F_{\rm E}/F_{\rm G} = (10^{-21})^2 \times 1.234\times10^{36} = \mathbf{1.2\times10^{-6}}$ : the residual electric repulsion between two hydrogen atoms is a million times weaker than their gravitational attraction — and that gravity is already the feeblest force in the catalogue. The electrical universe is dead; the $\gamma$ hypothesis is dead; and charge is a scalar to a precision matched by almost nothing else in physics.

2 · A charging capacitor, and why the displacement current is charge conservation

A parallel-plate capacitor with plate area $A = 100\ \mathrm{cm^2}$ is charged by a steady $I = 1.00\ \mathrm{mA}$ . Show that uncorrected Ampère's law contradicts charge conservation, and that Maxwell's correction is exactly the repair.

Step 1 — identify. Charge piles up on the plate, so $\partial_t\rho \ne 0$ there and $\nabla\!\cdot\!\vec J$ must be non-zero. Any field equation demanding $\nabla\!\cdot\!\vec J = 0$ is in trouble.

Step 2 — the contradiction. From $\nabla\times\vec B = \mu_0\vec J$ , take the divergence: $\nabla\!\cdot\!(\nabla\times\vec B) \equiv 0$ , so the equation demands $\nabla\!\cdot\!\vec J = 0$ — while continuity says $\nabla\!\cdot\!\vec J = -\partial_t\rho \ne 0$ at the plate. The law is not merely inaccurate here; it is inconsistent with charge conservation.

Step 3 — the numbers. Conduction current density onto the plate: $J = I/A = (1.00\times10^{-3})/(1.00\times10^{-2}) = \mathbf{0.100\ \mathrm{A\,m^{-2}}}$ , so $\mathrm d\sigma/\mathrm dt = J = 0.100\ \mathrm{C\,m^{-2}\,s^{-1}}$ . Between the plates $E = \sigma/\varepsilon_0$ , hence $\;\mathrm dE/\mathrm dt = 0.100/(8.8542\times10^{-12}) = \mathbf{1.129\times10^{10}\ \mathrm{V\,m^{-1}\,s^{-1}}}$ .

Step 4 — the repair, and it lands exactly. The displacement term contributes $\varepsilon_0\,\partial_t E = (8.8542\times10^{-12})(1.129\times10^{10}) = \mathbf{0.100\ \mathrm{A\,m^{-2}}}$ — identical to $J$ . The conduction current stops dead at the plate and the displacement current takes over with precisely the same value, so $\vec J + \varepsilon_0\partial_t\vec E$ is divergence-free everywhere: $\nabla\!\cdot\!(\vec J + \varepsilon_0\partial_t\vec E) = \nabla\!\cdot\!\vec J + \partial_t\rho = 0$ , by Gauss's law.

Step 5 — the covariant punchline. $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ contains the displacement current already — it is the $\partial_0 F^{0i}$ term, the piece of the four-divergence a three-dimensional eye does not see — and its consistency condition, proved above from antisymmetry alone, is exactly $\partial_\mu J^\mu = 0$ . Maxwell spent the 1860s discovering by hand a term the four-dimensional formalism cannot omit. It is the clearest illustration in classical physics of what covariance is for: not elegance, but the removal of the very possibility of writing something inconsistent.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §38 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.3 · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §5.4 · Lyttleton & Bondi, Proc. R. Soc. A 252, 313 (1959) · Bressi et al., Phys. Rev. A 83, 052101 (2011).

Provenance. Signature $(+,-,-,-)$ ; $\partial_\mu = (\tfrac1c\partial_t, +\nabla)$ ; SI units. No numeric defect was found in this article's base text, but its framing of conservation and invariance as "logically independent" is scoped down here: they are different statements, but global invariance is a theorem requiring microscopic invariance and continuity, and fails without either — this article's owned structural result. Every printed number is recomputed in verify_aq-sr-charge-invariance-continuity.py, which asserts against this article's own printed digits. The $10^{-21}$ bound is attributed to Bressi et al. (2011), the limit the Particle Data Group adopts, not to the older Dylla–King measurement the PDG declines to use.

Scope. The construction $J^\mu = \rho_0 U^\mu$ , the derivation of $\rho = \gamma\rho_0$ , and the invariant $J^\mu J_\mu$ belong to The Four-Current Density and are recalled, not re-derived. The current-carrying-wire argument belongs to Magnetism as a Relativistic Effect. The field tensor and the full unpacking of $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ belong to The Electromagnetic Field Tensor and The Covariant Maxwell Equations; only the antisymmetry argument is used here.

Builds on. The Four-Current Density · Four-Vectors · Index Notation and the Metric Tensor · Length Contraction. Leads to. The Four-Potential · Gauge Transformations and the Lorenz Gauge · The Covariant Maxwell Equations · The Electromagnetic Stress–Energy Tensor.

Is charge density a Lorentz invariant?

No, and confusing it with charge is the standard error. $\rho$ is the time component of $J^\mu = (c\rho, \vec J\,)$ , so it boosts: $\rho' = \gamma(\rho - \tfrac{v}{c^2}J^x)$ . A blob at rest with density $\rho_0$ is seen by a moving observer to have density $\gamma\rho_0$ — squeezed into a contracted volume. What is invariant is $q = \int\rho\,\mathrm dV$ , because the volume element shrinks by exactly the $1/\gamma$ the density grew by.

What exactly is the difference between charge conservation and charge invariance?

Conservation is about time within one frame: charge is never created or destroyed, only transported — $\partial_\mu J^\mu = 0$ . Invariance is about frames: every observer measures the same coulombs on the same body. They are different claims, and you can imagine a universe with one and not the other. In ours both hold, and once $J^\mu$ is a four-vector the global invariance of $q$ actually follows from conservation.

If Maxwell's equations force charge conservation, why is it ever stated as a separate law?

Two reasons. Historically, conservation was known (Franklin, 1747) long before Maxwell's equations existed, and was used as a constraint to build them — the displacement current was added precisely to make Ampère's law compatible with a charging capacitor. Logically, conservation is a statement about the sources, and it stays true where you never solve for a field at all: circuit theory, plasma transport, semiconductor device physics.

How do the neutrality experiments actually work?

Both classic designs hunt for a residual charge on macroscopically many atoms at once. In a gas-neutrality experiment a known number of molecules of a neutral gas is admitted to or released from an isolated conducting vessel; if each carried $\varepsilon e$ , the vessel would acquire a measurable potential. In a levitation experiment a suspended bulk sample is driven with an oscillating field and the absence of driven motion bounds its charge. Because both interrogate $\sim10^{20}$ atoms at once, they turn a hopeless single-atom measurement into a $10^{-21}$ bound.

Why does the four-current come out spacelike for a current-carrying wire? Is something wrong?

Nothing is wrong — for a neutral wire it must be. $J^\mu J_\mu = c^2\rho^2 - |\vec J|^2$ , and in the lab frame of a neutral wire $\rho = 0$ while $\vec J \ne \vec 0$ , giving $J^\mu J_\mu = -|\vec J|^2 < 0$ . A spacelike four-current means no frame kills the current, while there are frames in which the charge density is non-zero, of either sign. Contrast a beam of electrons all drifting together: that current is timelike, and a rest frame exists in which it is pure charge density and no current.

Is there a connection between charge conservation and gauge invariance?

Yes, and it is the deepest thing on this page. Noether's theorem attaches a conserved current to every continuous symmetry, and the current whose conservation is $\partial_\mu J^\mu = 0$ belongs to the global phase symmetry $\psi \to e^{i\theta}\psi$ of the charged matter field. Demand that the phase be free to vary from point to point — a local symmetry — and you are forced to introduce a compensating field transforming as $A^\mu \to A^\mu + \partial^\mu\chi$ : electromagnetism itself. That is the subject of Gauge Transformations and the Lorenz Gauge. The corollary is that charge conservation is not a law you could break in isolation — breaking it would break gauge invariance, which would give the photon a mass.