The Four-Current Density

Walk past a current-carrying wire. The wire is electrically neutral — as many protons in the lattice as electrons drifting through it. Every voltmeter agrees, every electroscope agrees, and the net charge in any length of it is exactly zero. That is a fact about the wire in the laboratory's frame. Now walk. Just walk, at one metre per second, alongside the wire. In your frame, that wire is charged. Not by much — about 69 electrons' worth of negative charge per metre — but not zero, and the sign and the size are both computable from the current alone. Nobody touched the wire. Nothing was added or removed. You changed frames, and a quantity that was zero stopped being zero. That is not a paradox and it is not an approximation. It is the first, and simplest, of the unifications that this chapter is about. Charge density $\rho$ and current density $\mathbf{J}$ are not two things. They are four components of one object,

J^\mu = (c\rho,\ \mathbf{J}),

and a boost mixes them into each other exactly as it mixes time into space. But that $(c\rho, \mathbf{J})$ transforms as a four-vector is a claim, not a definition — you cannot make something a four-vector by writing it in a column. This article earns it, out of a Lorentz scalar and a four-vector, and once earned it makes the rest of the chapter almost free.

Key takeaways

Signature $(+,-,-,-)$ ; $\partial_\mu = \big(\tfrac1c\partial_t,\ +\nabla\big)$ ; $U^\mu = \gamma(c,\mathbf v)$ with $U\cdot U = c^2$ .
The object, and the claim that must be earned:

\boxed{\;J^\mu \;=\; \rho_0\,U^\mu \;=\; (c\rho,\ \mathbf J)\;}

$\rho_0$ is the proper charge density — measured in the rest frame of the charge element — and it is a Lorentz scalar. $U^\mu$ is a four-vector. Scalar × four-vector = four-vector. That, and nothing shorter, is why $(c\rho,\mathbf J)$ is a four-vector at all.

$\rho = \gamma\rho_0$ , because charge is invariant but volume is not: the charge element's volume is length-contracted by $\gamma$ , so its charge density is raised by $\gamma$ . Charge density is a density, and densities do not transform like the things they count. (Charge Invariance and Continuity owns why charge is invariant.)
The boost mixes $\rho$ and $\mathbf J$ , exactly as it mixes $t$ and $x$ :

J'^0 = \gamma\big(J^0 - \beta J^1\big), \qquad J'^1 = \gamma\big(J^1 - \beta J^0\big), \qquad J'^{2,3} = J^{2,3}

The payoff: set $\rho = 0$ (a neutral wire) and boost. You get $\rho' = -\gamma v J_x/c^2 \neq 0$ . A frame with pure current has charge density in another frame. That is where magnetism comes from, and Magnetism as a Relativistic Effect derives the wire in full.
Continuity, in one line: $\partial_\mu J^\mu = \partial_t\rho + \nabla\cdot\mathbf J = 0$ . The metric signs cancel — there are no minus signs, and that surprises people.
The invariant: $J^\mu J_\mu = c^2\rho^2 - |\mathbf J|^2 = c^2\rho_0^2$ — one line from $U\cdot U = c^2$ . Every observer agrees on the proper charge density.
Key idea: $\rho$ and $\mathbf{J}$ are not a scalar and a vector. They are the time and space parts of a single four-vector, and which is which depends on who is looking.

The core insight

Writing four numbers in a column does not make them a four-vector. A four-vector is defined by how it transforms, and $(c\rho, \mathbf J)$ has no obvious right to transform as one — $\rho$ is a density (a charge over a volume, and volumes contract) and $\mathbf J$ is a flux, and neither is a component of anything a priori. So earn it. Take $\rho_0$ , the charge density measured in the rest frame of the charge itself: it is a scalar, because charge is invariant and proper volume is invariant. Take $U^\mu$ , the four-velocity of that charge: it is a four-vector, because Four-Velocity says so. Multiply them. A scalar times a four-vector is a four-vector — there is nothing left to prove — and when you unpack $\rho_0 U^\mu = \rho_0\gamma(c,\mathbf v)$ you find its components are exactly $(c\rho, \mathbf J)$ . The four-vector character of the four-current is not a postulate. It is a consequence of charge invariance.

Where this lives today

The four-current is not a result so much as a piece of grammar: it is the object every covariant field equation in electrodynamics has on its right-hand side. The Four-Potential sources $A^\mu$ from it, $\Box A^\mu = \mu_0 J^\mu$ ; The Electromagnetic Field Tensor sources $F^{\mu\nu}$ from it, $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ . Neither could be written if $J^\mu$ were not a four-vector — an equation whose left side is a four-vector and whose right side is not is no equation at all, in any frame but one. Both are destinations, and this article deliberately does not derive them. What it supplies is the licence to write them. The same construction — a scalar density times a four-velocity — recurs everywhere: it gives the number-flux four-vector of a fluid, the baryon current in cosmology, the probability current of the Dirac equation. Whenever a conserved "amount" flows, its density and its flux are the time and space parts of a four-vector, and the conservation law is the vanishing of a four-divergence. The four-current is the first and simplest example, and everything later assumes the pattern.

The physical picture

Take a small blob of charge — a cloud of electrons, say — and ride along with it. In its rest frame it has some charge $q$ in some volume $V_0$ , and the charge density is $\rho_0 = q/V_0$ . Nothing is moving; there is no current. The blob is described by one number. Now stand still and let the blob go past you at speed $v$ . Two things change, and only two.

The volume shrinks. Length contraction squeezes the blob along its direction of motion, so the volume you measure is $V = V_0/\gamma$ .

The charge does not. Charge is a Lorentz scalar: the same $q$ , in every frame. (That is a real physical claim with real experimental backing, and Charge Invariance and Continuity owns it. Here we simply use it.) So the charge density you measure is

\rho = \frac{q}{V} = \frac{q}{V_0/\gamma} = \gamma\rho_0.

The charge density went up, purely because the volume went down. And because the blob is now moving, there is also a current density $\mathbf J = \rho\mathbf v$ where before there was none. Look at what happened. In the blob's frame there was charge and no current. In yours there is charge and current — and more charge than before. One physical situation; two very different-looking descriptions; and the difference is entirely accounted for by a single $\gamma$ and a single velocity. That is the signature of components of a four-vector being re-shuffled, and it is exactly what $J^\mu = \rho_0 U^\mu$ says. Run it the other way and you get the striking case. Suppose the charge density in your frame is zero — a wire, the electrons' negative charge exactly cancelling the lattice's positive — but there is a current, because the electrons drift. Boost, and the cancellation does not survive: the two species have different velocities, so their densities pick up different $\gamma$ factors, and the sum no longer vanishes. The wire acquires a net charge. That is the origin of the magnetic force; Magnetism as a Relativistic Effect derives it in full, and this article establishes the object that makes it possible.

The core equations

The object, defined by its components, and then earned:

J^\mu \;=\; (c\rho,\ \mathbf J) \;=\; (c\rho,\ J_x,\ J_y,\ J_z) \qquad (1)

where: $\rho$ is the charge per unit volume and $\mathbf J = \rho\mathbf v$ is the charge crossing unit area per unit time. Physical reading: this is a definition, not yet a theorem. The factor of $c$ on the time slot makes the units match — $c\rho$ and $\mathbf J$ are both a charge density times a velocity — and nothing so far says the column transforms as a four-vector. The covariant construction, which is what earns it:

\boxed{\;J^\mu \;=\; \rho_0\,U^\mu, \qquad U^\mu = \gamma(c,\ \mathbf v),\qquad U\cdot U = c^2\;} \qquad (2)

where: $\rho_0$ is the proper charge density, measured in the local rest frame of the charge, and $U^\mu$ is the four-velocity of that charge (Four-Velocity owns it). Physical reading: $\rho_0$ is a scalar and $U^\mu$ is a four-vector, so the product is a four-vector — that is the whole argument. Unpack it: $\rho_0\gamma(c,\mathbf v) = (\gamma\rho_0 c,\ \gamma\rho_0\mathbf v) = (c\rho,\ \rho\mathbf v) = (c\rho, \mathbf J)$ , which is equation (1). The definition and the construction agree, and now the transformation law is not assumed but inherited. The density law, which is the physical heart of it:

\boxed{\;\rho = \gamma\rho_0\;} \qquad\text{(charge invariant, volume contracted)} \qquad (3)

where: $\gamma = (1-v^2/c^2)^{-1/2}$ . Physical reading: charge density is not invariant, even though charge is. The charge is the same number for everybody; the volume it sits in is not. A density is a ratio, and only the numerator is protected. This is the single most common confusion in the chapter, and it is the reason $\rho$ can be zero for one observer and non-zero for another. The transformation, which is just the Lorentz transformation applied to a four-vector:

\boxed{\;J'^0 = \gamma\big(J^0 - \beta J^1\big),\qquad J'^1 = \gamma\big(J^1 - \beta J^0\big),\qquad J'^2 = J^2,\ \ J'^3 = J^3\;} \qquad (4)

where: the boost is along $x$ with $\beta = v/c$ . Physical reading: identical in form to the transformation of $(ct, x)$ — because it is the same transformation, acting on a different four-vector. In lab variables, $\rho' = \gamma(\rho - vJ_x/c^2)$ and $J'_x = \gamma(J_x - v\rho)$ . Set $\rho = 0$ and $\rho'$ does not vanish. That single line is the seed of the entire relativistic theory of magnetism. The invariant, and the continuity equation — both one-liners, and both consequences:

J^\mu J_\mu = c^2\rho^2 - |\mathbf J|^2 = c^2\rho_0^2 \qquad\qquad \partial_\mu J^\mu = \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J = 0 \qquad (5)

where: $\partial_\mu = \big(\tfrac1c\partial_t, +\nabla\big)$ . Physical reading: the invariant follows in one step from $U\cdot U = c^2$ , and it says every observer agrees on the proper charge density even though none of them agrees on $\rho$ . The continuity equation is the covariant statement of charge conservation — and note there are no minus signs in it, which is a genuine surprise worth checking (see the warning). Charge Invariance and Continuity owns the continuity equation as a law and shows that Maxwell's equations force it; here it is simply a property of the object.

Building the four-current

Derivation AEarning the four-vector: a scalar times a four-vector, and nothing else

The claim is that $(c\rho, \mathbf J)$ transforms as a four-vector. This is not obvious and it is not a definition — you cannot promote four numbers to a four-vector by arranging them vertically. We must show that the column inherits the Lorentz transformation from something that already has it. The strategy is the only one available: manufacture $J^\mu$ out of objects whose transformation properties are already known.

Step 1. Find a genuine Lorentz scalar in the problem. Consider a small element of charge. Its charge $q$ is a scalar — this is charge invariance, a physical fact with experimental backing at the $10^{-21}$ level, and Charge Invariance and Continuity owns it. Its proper volume $V_0$ — the volume in its own rest frame — is also a scalar, since every observer can agree on what the rest frame is and what the volume is there. Therefore

\rho_0 \equiv \frac{q}{V_0} \qquad\text{is a Lorentz SCALAR.}

Why: note carefully what $\rho_0$ is not. It is not "the charge density". It is the charge density in one specific frame — the charge's own — and it is that specification that makes it frame-independent. The same trick makes proper time a scalar out of a time, and rest mass a scalar out of an energy.

Step 2. Find a genuine four-vector. The charge element moves along a worldline, and its four-velocity is

U^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \gamma(c,\ \mathbf v),

which is a four-vector because $\mathrm{d}x^\mu$ is one and $\mathrm{d}\tau$ is a scalar. Four-Velocity owns this. Why: exactly the same manoeuvre as Step 1, one level down — divide a four-vector by a scalar and the result is still a four-vector. We are about to multiply by a scalar, which is the same move again.

Step 3. Multiply, and define:

\boxed{\;J^\mu \equiv \rho_0\,U^\mu\;}

This is a four-vector, and there is nothing left to prove. Why: a Lorentz scalar times a four-vector transforms as a four-vector — that is what "scalar" and "four-vector" mean. We have not verified a transformation law by grinding through components; we have constructed an object that cannot fail to have it.

Step 4. Now unpack it and see what we have built:

J^\mu = \rho_0\,\gamma(c,\ \mathbf v) = \big(\underbrace{\gamma\rho_0}_{\textstyle \rho}\,c,\ \ \underbrace{\gamma\rho_0}_{\textstyle \rho}\,\mathbf v\big) = (c\rho,\ \rho\mathbf v) = (c\rho,\ \mathbf J)

Why: the components come out right, and that is the payoff. The thing we constructed for its transformation properties turns out to be exactly the thing we wanted for its physics. The lab-frame charge density is $\rho = \gamma\rho_0$ , and the current density is $\mathbf J = \rho\mathbf v$ , which is its ordinary definition.

Step 5. Read off $\rho = \gamma\rho_0$ and check it physically, because it should be checkable without any of this machinery. In the charge's rest frame it occupies a volume $V_0$ and carries charge $q$ . In the lab the volume is length-contracted, $V = V_0/\gamma$ , while $q$ is unchanged. So

\rho = \frac{q}{V} = \frac{q}{V_0/\gamma} = \gamma\,\frac{q}{V_0} = \gamma\rho_0. \quad\checkmark

Why: the covariant construction and the elementary length-contraction argument give the same answer, which is exactly the reassurance one wants. And note where the $\gamma$ came from: not from the charge, which is invariant, but from the volume, which is not. A density inherits the transformation of the volume in its denominator.

Step 6. Note what has not been shown. We have not derived $\Box A^\mu = \mu_0 J^\mu$ (The Four-Potential's) nor $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ (The Electromagnetic Field Tensor's). Why: those are destinations, and they are the reason this object matters — a covariant field equation needs a covariant source, and $J^\mu$ is it. But they are downstream, and this article stops at the source. *Verified in verify_aq-sr-four-current-density.py: $\rho_0 U^\mu$ is expanded symbolically in sympy and shown to have components $(c\gamma\rho_0, \gamma\rho_0\mathbf v) = (c\rho, \mathbf J)$ with $\rho = \gamma\rho_0$ ; $U\cdot U = c^2$ is confirmed in the $(+,-,-,-)$ signature; and $\rho = \gamma\rho_0$ is checked against the independent length-contraction route $q/(V_0/\gamma)$ .*

Go deeper · Derivation B · 3 minThe boost that charges a neutral wire

Derivation BThe boost that charges a neutral wire

Now we spend the object. Because $J^\mu$ is a four-vector, it transforms under a boost exactly as $(ct, x, y, z)$ does — no new law is needed, and no new physics. The consequence is the one that opens the chapter: the statement "this wire is neutral" is not frame-independent. We will derive the mixing, apply it to a real wire, and get a real number.

Step 1. Write the boost. For a frame $S'$ moving at $+v\hat x$ relative to $S$ , the Lorentz transformation acts on any four-vector $A^\mu$ as $A'^0 = \gamma(A^0 - \beta A^1)$ , $A'^1 = \gamma(A^1 - \beta A^0)$ , with the transverse components untouched. Applied to $J^\mu = (c\rho, J_x, J_y, J_z)$ :

c\rho' = \gamma\big(c\rho - \beta J_x\big), \qquad J'_x = \gamma\big(J_x - \beta\,c\rho\big)

Why: this is not a derivation so much as a cash-in. We proved $J^\mu$ is a four-vector in the last card; the transformation law came with it, for free, and it is the same one we already know.

Step 2. Convert to lab variables by dividing the first by $c$ and using $\beta = v/c$ :

\boxed{\;\rho' = \gamma\Big(\rho - \frac{v J_x}{c^2}\Big), \qquad J'_x = \gamma\big(J_x - v\rho\big)\;}

Why: look at the structure. $\rho'$ picks up a piece of $J_x$ , and $J'_x$ picks up a piece of $\rho$ . They feed each other. The $1/c^2$ in the first is why the effect is invisible at everyday speeds — and, crucially, why it is not zero.

Step 3. Now impose the interesting condition: a neutral wire in the lab, $\rho = 0$ , carrying a current, $J_x = J \neq 0$ . Substitute:

\rho' = \gamma\Big(0 - \frac{vJ}{c^2}\Big) = -\,\frac{\gamma v J}{c^2} \;\neq\; 0

Why: there it is. A frame in which the charge density vanishes and the current does not is not a frame in which the charge density vanishes for everybody. Zero is not a Lorentz-invariant value of $\rho$ — only $\rho_0$ , the proper density, is invariant, and a neutral current-carrying wire has no single rest frame to define one.

Step 4. Put a real wire in it. Take $1\ \mathrm{A}$ through a $1\ \mathrm{mm}^2$ copper wire, so

J = \frac{I}{A} = \frac{1\ \mathrm{A}}{10^{-6}\ \mathrm{m}^2} = 10^{6}\ \mathrm{A/m^2},

and walk alongside it at a brisk $v = 1\ \mathrm{m/s}$ (so $\gamma = 1$ to eighteen decimal places):

\rho' = -\frac{(1)(10^{6})}{(2.998\times10^{8})^{2}}\ \mathrm{C/m^3} = -1.11\times10^{-11}\ \mathrm{C/m^3}

Why: small, but flatly non-zero, and computed from nothing but the current and your walking speed.

Step 5. Make it concrete. In one metre of that wire the charge you see is $Q' = \rho' A L = -1.11\times10^{-17}\ \mathrm{C}$ , which in electrons is

\frac{1.11\times10^{-17}}{1.602\times10^{-19}} \approx \mathbf{69\ electrons\ per\ metre}

Why: sixty-nine electrons. Not a metaphor, not a formal artefact — a definite, countable, negative charge on a wire that the laboratory swears is neutral. And it is there because you are walking.

Step 6. Say what this is, and hand it on. This charge is what a moving test charge feels as an electric force — and in the lab frame we call that same force magnetic. The magnetic force on a moving charge near a neutral wire is the electric force of this induced charge density, seen from the charge's own frame. The full derivation — which must handle the lattice and the electrons as two separate charge species with two different velocities and therefore two different $\gamma$ s — belongs to Magnetism as a Relativistic Effect, and this article does not attempt it. Why: the two-species bookkeeping is the whole content of that article, and doing it here would duplicate it. What is owned here is the object and its transformation law — which is what makes that derivation possible at all. *Verified in verify_aq-sr-four-current-density.py: the boost of $J^\mu$ is carried out with an explicit Lorentz matrix and shown to give $\rho' = \gamma(\rho - vJ_x/c^2)$ ; the neutral-wire case $\rho = 0$ is checked to give $\rho' = -\gamma vJ/c^2$ ; and the printed numbers ( $J = 10^6\ \mathrm{A/m^2}$ , $\rho' = -1.11\times10^{-11}\ \mathrm{C/m^3}$ , 69 electrons per metre, and copper's drift velocity $7.3\times10^{-5}\ \mathrm{m/s}$ at $n = 8.5\times10^{28}\ \mathrm{m^{-3}}$ ) are recomputed from SI constants.*

Figure

RHO AND J ARE ONE OBJECT, AND A BOOST ROTATES BETWEEN THEM — exactly as it rotates t into x. Left: in the lab the wire is NEUTRAL (rho = 0) and carries a current J. The four-current points purely along the spatial axis. Right: boost at speed v and the same four-vector now has a TIME component — the wire is charged, rho-prime = −gamma v J / c², and for 1 A in 1 mm² of copper walked past at 1 m/s that is about 69 electrons of negative charge per metre. Nothing was done to the wire. Zero is not a Lorentz-invariant value of the charge density. This is where the magnetic force comes from — and Magnetism as a Relativistic Effect derives the wire in full, with both charge species

Limiting cases

Case	Result	Meaning
charge at rest ( $v=0$ )	$J^\mu = (c\rho_0,\ \mathbf 0)$ , $\gamma = 1$	pure charge, no current — the four-vector is purely timelike
$v \ll c$	$\rho \approx \rho_0$ , $\mathbf J \approx \rho_0\mathbf v$	the Galilean picture: $\rho$ and $\mathbf J$ look independent
neutral wire ( $\rho = 0$ )	$\rho' = -\gamma vJ_x/c^2 \neq 0$	the payoff — pure current in one frame, charge in another
$\rho = 0$ and $\mathbf J = 0$	$J^\mu = 0$ in every frame	the only Lorentz-invariant way to have no charge
ultra-relativistic ( $\gamma \gg 1$ )	$\rho = \gamma\rho_0 \to \infty$	an LHC bunch: $\gamma = 7460$ , so $\rho$ is $7460\times$ its proper value
the invariant	$c^2\rho^2 - \|\mathbf J\|^2 = c^2\rho_0^2$	everyone agrees on $\rho_0$ ; nobody agrees on $\rho$
a light-like current	$c^2\rho^2 = \|\mathbf J\|^2$ ⇒ $\rho_0 = 0$	charge carriers at $c$ — impossible for massive carriers
$c \to \infty$	$\rho' \to \rho$ , $\mathbf J' \to \mathbf J - \rho\mathbf v$	Galilean: charge density becomes absolute, and magnetism disappears

Common trapCharge is invariant. Charge DENSITY is not. And the continuity equation has no minus signs

Two traps sit at the entrance to this chapter, and both are about being careless with what is invariant.

Trap 1: "charge is invariant, so charge density is invariant." It is not, and the error is a $\gamma$ . Charge is a Lorentz scalar; volume is not. A density is a charge divided by a volume, and that volume length-contracts, so $\rho = \gamma\rho_0$ . This is why a neutral wire can be charged in another frame, and why the relativistic theory of magnetism is possible at all. The invariant is $\rho_0$ — the proper charge density — and $J^\mu J_\mu = c^2\rho_0^2$ says everyone agrees on it.

Trap 2: the continuity equation looks like it should have minus signs, and it does not. With the signature $(+,-,-,-)$ one expects the spatial part of a contraction to come in with a minus. Write it out:

\partial_\mu J^\mu = \frac{1}{c}\frac{\partial}{\partial t}(c\rho) + \frac{\partial J_x}{\partial x} + \frac{\partial J_y}{\partial y} + \frac{\partial J_z}{\partial z} = \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J

No minus signs. The reason is that $\partial_\mu$ already carries a lower index and $J^\mu$ an upper one, so the contraction is natural — no metric is invoked, so no signs are introduced. If you insist on writing it as $\partial^\mu J_\mu$ instead, then $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ and $J_\mu = (c\rho, -\mathbf J)$ , and the two minus signs multiply to a plus and you get the identical result. The signs cancel, and they cancel for a reason. Getting this wrong produces $\partial_t\rho - \nabla\cdot\mathbf J = 0$ , which is not charge conservation and is not true.

Worked examples

1 · An LHC proton bunch, in its own frame and in ours

The LHC accelerates protons to $7\ \mathrm{TeV}$ . A bunch of $1.15\times10^{11}$ protons occupies a cylinder $10\ \mathrm{cm}$ long and $1\ \mathrm{mm}$ in radius in the lab. Find $\gamma$ , the lab and proper charge densities, and the bunch's length in its own frame.

Step 1 — identify. $\gamma = E/m_pc^2 = 7000/0.938272 = \mathbf{7460.5}$ , so $\beta = 0.999999991$ .

Step 2 — the lab charge density. $Q = 1.15\times10^{11} \times 1.602177\times10^{-19} = 1.8425\times10^{-8}\ \mathrm{C}$ in a volume $V = \pi r^2 L = 3.1416\times10^{-7}\ \mathrm{m^3}$ , so

\rho = \frac{Q}{V} = \mathbf{0.05865\ C/m^3}

Step 3 — the proper charge density. From equation (3), $\rho_0 = \rho/\gamma = 0.05865/7460.5 = \mathbf{7.86\times10^{-6}\ C/m^3}$ — smaller by a factor of $7460$ .

Step 4 — the boundary number. The same factor runs the length: in its own rest frame the bunch is $\gamma L = 7460.5 \times 0.10\ \mathrm{m} = \mathbf{746\ m}$ long. A bunch that is a hand's breadth long in the tunnel is three-quarters of a kilometre long to itself — and its charge density is correspondingly diluted, by exactly the same $\gamma$ . The two statements are the same statement. That is what $\rho = \gamma\rho_0$ means.

2 · Checking the invariant — and watching it nearly vanish

Verify $J^\mu J_\mu = c^2\rho_0^2$ for that bunch, and notice what the arithmetic does.

Step 1 — the current density. $\mathbf J = \rho\mathbf v$ with $v = \beta c$ , so $J = 0.05865 \times 0.999999991 \times 2.998\times10^8 = \mathbf{1.758\times10^{7}\ A/m^2}$ (a current of $I = J\pi r^2 = 55.2\ \mathrm{A}$ while the bunch passes).

Step 2 — the two terms.

c^2\rho^2 = 3.09\times10^{14}, \qquad |\mathbf J|^2 = 3.09\times10^{14}

Step 3 — the invariant. Their difference is

c^2\rho^2 - |\mathbf J|^2 = \mathbf{5.554\times10^{6}} \;=\; (c\rho_0)^2 = \big(2.998\times10^8 \times 7.86\times10^{-6}\big)^2 \approx 5.55\times10^{6} \quad\checkmark

Step 4 — the boundary number, and a warning about arithmetic. The two terms in Step 2 agree to one part in $5.6\times10^{7}$ — that is $\gamma^2$ — and the invariant is their difference. A double-precision float has about 16 digits, so this subtraction throws away half of them, which is why the verify script computes it in exact arithmetic. The physics lesson is the numerics lesson: at high $\gamma$ the timelike and spacelike parts of $J^\mu$ nearly cancel, which is precisely why $\rho$ and $\mathbf J$ separately are enormous while $\rho_0$ is small.

Figure

WHY rho = gamma rho-zero — the charge is invariant, but the VOLUME is not. Left: in the charge's own rest frame it fills a proper volume V-zero and carries charge q, so the proper charge density is rho-zero = q / V-zero. This is a Lorentz SCALAR — it is the pair (invariant charge, invariant proper volume). Right: in the lab the blob is length-contracted along its motion, V = V-zero / gamma, while q is UNCHANGED. So rho = q/V = gamma rho-zero — the density goes UP purely because the volume went DOWN. A density inherits the transformation of the volume in its denominator, not of the charge in its numerator. For an LHC bunch at 7 TeV, gamma = 7460: a 10 cm bunch is 746 m long to itself, and its lab charge density is 7460 times its proper value

Physics of the portrait

The four-current is the first object of the covariant electrodynamics, and it is the one that makes the rest of the chapter possible: a covariant field equation needs a covariant source.

Origins · 2 minFour-current and how this was discovered

Four-current

1908 · the source that had to be a four-vector before the fields could be a tensor

When Maxwell's equations were written in 1865 they had a source — charge and current — and it sat on the right-hand side in two pieces: a scalar $\rho$ in Gauss's law and a vector $\mathbf J$ in Ampère's law. There was no reason to think of them as related. They had different units, different equations, different physical meanings: an amount and a flow. Hermann Minkowski saw that they had to be the same object. In a paper presented to the Göttingen Academy in December 1907 and published in 1908 — Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern — he recast the whole of electrodynamics in four dimensions, and the first thing he needed was a four-vector source. It could not be assumed: $(\rho, \mathbf J)$ has no obvious claim to be a four-vector, because $\rho$ is a density and densities do not transform like the quantities they count. The construction that earns it — a scalar proper density times the four-velocity — turns the four-vector character of $J^\mu$ into a consequence of charge invariance, which is a physical fact and not a convention. Once the source is a four-vector, the field must be a tensor, because $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ has to balance index for index — and the unification of $\mathbf E$ and $\mathbf B$ into one object stops being a surprise and becomes a requirement. Minkowski died of appendicitis in January 1909, at forty-four, a few months after delivering "Raum und Zeit". He had time to build the framework and not to use it.

History

1865

— Maxwell publishes his equations, with charge density and current density as two apparently unrelated sources.

1905

— Einstein opens his relativity paper with the magnet-and-conductor asymmetry, the observation that electric and magnetic descriptions depend on the observer.

1906

— Poincaré's Palermo memoir, received in July 1905 and published in 1906, sets out a four-dimensional formalism and the invariant quadratic form.

1908

— Minkowski, in a paper presented in December 1907, recasts electrodynamics in four dimensions and introduces the four-current as the covariant source.

1908

— Minkowski delivers "Raum und Zeit" in Cologne, giving spacetime its geometry and its name.

1909

— Minkowski dies of appendicitis at forty-four, months after completing the framework.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §12.3 · Rindler, Introduction to Special Relativity, 2nd ed., §38 · Jackson, Classical Electrodynamics, 3rd ed., §11.9 · Purcell & Morin, Electricity and Magnetism, 3rd ed., ch. 5 · H. Minkowski, Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern, Göttinger Nachrichten (1908), 53–111.

Provenance. Signature $(+,-,-,-)$ ; $\partial_\mu = (\tfrac1c\partial_t, +\nabla)$ ; $U\cdot U = +c^2$ . The AllQFT base was checked and no numeric defect was found; its cards for $\Box A^\mu = \mu_0 J^\mu$ and $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ belong to The Four-Potential and The Electromagnetic Field Tensor respectively and are named here as destinations, not derived — deriving them here would duplicate those articles. verify_aq-sr-four-current-density.py expands $\rho_0U^\mu$ symbolically to $(c\rho, \mathbf J)$ , confirms $U\cdot U = c^2$ and $J\cdot J = c^2\rho_0^2$ , boosts $J^\mu$ with an explicit Lorentz matrix to reproduce $\rho' = \gamma(\rho - vJ_x/c^2)$ , verifies that $\partial_\mu J^\mu$ carries no minus signs (and that $\partial^\mu J_\mu$ gives the identical result), and recomputes every printed number: the wire ( $10^6\ \mathrm{A/m^2}$ , $-1.11\times10^{-11}\ \mathrm{C/m^3}$ , 69 electrons/m, copper's $7.3\times10^{-5}$ m/s drift at $n=8.5\times10^{28}\,\mathrm{m^{-3}}$ ) and the LHC bunch ( $\gamma = 7460.5$ , $\rho = 0.05865$ , $\rho_0 = 7.86\times10^{-6}\ \mathrm{C/m^3}$ , $746$ m proper length, $J = 1.758\times10^7\ \mathrm{A/m^2}$ , invariant $5.554\times10^6$ ) — the last in exact arithmetic, because at $\gamma = 7460$ the invariant is a difference of two numbers agreeing to one part in $5.6\times10^7$ .

Scope — what this article deliberately does not do. It does not argue why charge is a Lorentz scalar, nor state the experimental bounds, nor treat continuity as a law or show that Maxwell forces it: Charge Invariance and Continuity owns all of that, and this article merely uses charge invariance and notes $\partial_\mu J^\mu = 0$ as a property of the object. It does not derive the current-carrying wire with its two charge species — Magnetism as a Relativistic Effect owns that. And it does not derive the field equations that $J^\mu$ sources.

Builds on. Charge Invariance and Continuity · Four-Velocity · Lorentz Transformation Equations · Four-Vectors · Length Contraction. Leads to. The Four-Potential · Covariant Maxwell Equations · Magnetism as a Relativistic Effect · The Wave Equation for the Four-Potential.

Why isn't (ρ, J) already obviously a four-vector? It has four components.

Because having four components is not what makes something a four-vector — transforming correctly is. Write any four numbers in a column (temperature, pressure, your age, the price of tin) and it will not transform as one. The definition is a rule: under a boost $\Lambda$ , it must go to $\Lambda A$ . So the four-vector character of $(c\rho, \mathbf J)$ has to be demonstrated, and the demonstration is $J^\mu = \rho_0 U^\mu$ . Be suspicious here in particular, because $\rho$ is a density — and densities are exactly the sort of quantity that transforms badly, since the volume underneath them contracts.

If charge is invariant, how can a neutral wire be charged in another frame?

Because charge invariance and charge-density invariance are different claims, and only the first is true. Charge — the total, counted, integrated charge of an isolated object — is a Lorentz scalar. Charge density is a charge divided by a volume, and the volume length-contracts, so $\rho = \gamma\rho_0$ . In a wire there are two charge species, the fixed lattice and the drifting electrons, moving at different velocities and so picking up different $\gamma$ factors. In the lab their densities happen to cancel exactly. In another frame they do not, because the two $\gamma$ s change by different amounts. The total charge of the whole wire is still whatever it is; what changes is how that charge is distributed along it, and a finite length of it can be charged. Magnetism as a Relativistic Effect does this bookkeeping in full.

What exactly is ρ₀, and why is it a scalar when ρ is not?

$\rho_0$ is the charge density measured in the local rest frame of the charge itself — and that qualifier is what makes it invariant. It is the ratio of two invariants: the charge $q$ and the proper volume $V_0$ (the volume in the rest frame, which every observer can agree on, because they all agree what the rest frame is). By contrast $\rho = q/V$ uses the lab volume, which is frame-dependent. This is the same move as proper time and rest mass: you make an invariant out of a frame-dependent quantity by nailing it to a frame defined by the object itself, not by the observer. And $J^\mu J_\mu = c^2\rho_0^2$ says formally that everyone agrees on the result.

Why does the continuity equation have no minus signs, given the (+,−,−,−) signature?

Because no metric is used. The expression $\partial_\mu J^\mu$ contracts a lower index against an upper one, which is a natural pairing — you sum over $\mu$ with no $\eta_{\mu\nu}$ anywhere, and therefore no signs get introduced. With $\partial_\mu = (\tfrac1c\partial_t, +\nabla)$ and $J^\mu = (c\rho, \mathbf J)$ you get $\partial_t\rho + \nabla\cdot\mathbf J$ directly. If you choose to write it as $\partial^\mu J_\mu$ then you have raised one index and lowered the other, which introduces a minus sign twice — $\partial^\mu = (\tfrac1c\partial_t, -\nabla)$ and $J_\mu = (c\rho, -\mathbf J)$ — and $(-\nabla)\cdot(-\mathbf J) = +\nabla\cdot\mathbf J$ . Same answer, as it must be. The signs cancel; they were never there to begin with.

Can the four-current ever be null or spacelike?

For ordinary matter, no. $J^\mu J_\mu = c^2\rho_0^2 \geq 0$ , so the four-current is timelike (or zero) whenever there is any charge at all — which just says the carriers are massive, move slower than light, and therefore have a frame in which they are at rest and the current vanishes. A null four-current, $c^2\rho^2 = |\mathbf J|^2$ , would require $\rho_0 = 0$ with $J^\mu \neq 0$ : charge carried at exactly $c$ . A spacelike $J^\mu$ , with $|\mathbf J| > c\rho$ , would mean charge flowing faster than light. The timelike character of $J^\mu$ is the causal structure of spacetime showing up in the source of the field.

Does this mean electricity and magnetism are "really" the same thing?

It is the first step in that direction, and it is worth being precise about how far it goes. What this article shows is that the sources unify: $\rho$ and $\mathbf J$ are one four-vector, and which is which depends on the observer. That immediately makes it impossible for the fields to stay separate — $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ must balance index for index, so if the right-hand side is a four-vector the left-hand side must be built from a tensor. The actual unification of $\mathbf E$ and $\mathbf B$ into $F^{\mu\nu}$ belongs to The Electromagnetic Field Tensor, and the laws that turn one into the other to The Transformation of Electric and Magnetic Fields. The four-current is the reason those articles have to exist.