--- title: The Four-Current and Four-Potential module: Covariant Electromagnetism moduleNumber: 4 lessonNumber: 1 order: 401 summary: > Charge density and current combine into a single four-vector whose divergence is charge conservation. The scalar and vector potentials combine likewise into the four-potential, whose gauge freedom fixes to the Lorenz condition, reducing Maxwell's equations for the potentials to a single wave equation sourced by the four-current. topics: [Covariant Electromagnetism] draft: false sources: - book: Schutz ref: "Ch. 4 — Perfect Fluids in Special Relativity; §4.2 the number–flux four-vector and the four-current construction" - book: Carroll ref: "Lecture Notes on General Relativity §1 — Special Relativity and Flat Spacetime (classical field theory, electromagnetism), arXiv:gr-qc/9712019" --- Electromagnetism is the theory that forced relativity. Maxwell's equations single out one speed, $c = 1/\sqrt{\mu_0 \epsilon_0}$, with no reference to any frame, and reconciling that with mechanics is what produced the Lorentz transformation.[^carroll1] The reward for building the tensor machinery of the earlier modules is that the whole theory collapses into a few index equations in which Lorentz invariance is manifest. This module rebuilds it. The starting point is the source of the fields: charge and current. Throughout, the metric signature is $(-,+,+,+)$, coordinates are $x^\mu = (ct, x, y, z)$ so that $x^0 = ct$, and the derivative operator is $\partial_\mu = \partial/\partial x^\mu = \left(\tfrac{1}{c}\partial_t,\ \nabla\right)$. Indices are raised and lowered with $\eta_{\mu\nu} = \operatorname{diag}(-1,+1,+1,+1)$, following the [four-vector conventions](/relativity/spacetime-and-the-lorentz-group/four-vectors-and-index-notation) of Module 2. ## Charge and current as one object Charge density $\rho$ and current density $\vec J$ are not separate quantities in different frames. A static line of charge, seen by a moving observer, is a current; a stationary charge cloud, boosted, carries both a compressed density and a flow. The transformation that ties them together is the same one that ties $ct$ to $\vec x$, so they must be the components of a four-vector. Take a cloud of charge with **proper charge density** $\rho_0$, the density measured in the local rest frame of the charge. In that frame there is no current. Give the cloud four-velocity $U^\mu = \gamma(c, \vec u)$, where $\vec u$ is its ordinary velocity and $\gamma = (1 - u^2/c^2)^{-1/2}$. The **four-current** is $$ J^\mu = \rho_0\, U^\mu = \big(\gamma \rho_0 c,\ \gamma \rho_0 \vec u\big). $$ Its components in a general frame are the ordinary charge and current densities: $$ J^\mu = (c\rho,\ \vec J), \qquad \rho = \gamma \rho_0, \qquad \vec J = \rho \vec u. $$ The time component $J^0 = c\rho$ carries the density; the space components carry the flow. The factor $\gamma$ in $\rho = \gamma\rho_0$ is length contraction of the charge distribution: the same charge occupies a volume shrunk by $1/\gamma$ along the motion, so the density measured by a moving observer is larger by $\gamma$. Charge itself is a Lorentz invariant — every observer counts the same number of elementary charges — but charge _density_ is not, because volume is not. > **Definition (Four-current).** For a charge distribution with proper density > $\rho_0$ and four-velocity $U^\mu$, the four-current is $J^\mu = \rho_0 U^\mu = > (c\rho, \vec J)$. It is a four-vector: under a Lorentz transformation > $\Lambda^\mu{}_\nu$ its components mix as $J'^\mu = \Lambda^\mu{}_\nu J^\nu$, > exactly as $x^\mu$ does. The cleanest illustration is the one-dimensional case. A wire at rest carries charge per unit length $\lambda_0$ and no current. View it from a frame moving at speed $v$ antiparallel to the wire, so the charges stream past at $+v$. The spacing between charges contracts by $1/\gamma$, so the linear density becomes $\lambda = \gamma \lambda_0$, and a current $I = \lambda v = \gamma \lambda_0 v$ appears where there was none. Density and current are the same physical charge seen through two frames. $$ % caption: A line of charge at rest has only a density; boosting to a frame in % which the charges stream past contracts their spacing, raising the density to % gamma times its rest value and producing a current where there was none. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % rest frame \node[black!70] at (-1.6,2.0) {rest frame}; \draw[black] (-0.3,2.0) -- (6.3,2.0); \foreach \x in {0,1,2,3,4,5,6} \node at (\x,2.0) {$+$}; \node[black, anchor=west] at (6.6,2.0) {density only}; % moving frame \node[black!70] at (-1.6,0.4) {moving frame}; \draw[black] (-0.3,0.4) -- (6.3,0.4); \foreach \x in {0.4,1.1,1.8,2.5,3.2,3.9,4.6} \node at (\x,0.4) {$+$}; \draw[->, acc, very thick] (5.0,0.4) -- (6.2,0.4) node[above, midway, black!70] {v}; \node[black, anchor=west] at (6.6,0.4) {density and current}; \end{tikzpicture} $$ > **Worked example (Four-current of a proton beam).** A proton beam has number > density $n_0 = 1.0 \times 10^{14}\ \text{m}^{-3}$ in its own rest frame and > moves at $\beta = 0.90$. Its proper charge density is $\rho_0 = n_0 e = (1.0 > \times 10^{14})(1.6 \times 10^{-19}) = 1.6 \times 10^{-5}\ \text{C/m}^3$. With > $\gamma = (1 - 0.81)^{-1/2} = 2.29$, the lab density is $\rho = \gamma\rho_0 = > 3.67 \times 10^{-5}\ \text{C/m}^3$ and the current density is $J = \rho v = > (3.67 \times 10^{-5})(0.90)(3.0 \times 10^8) = 9.9 \times 10^{3}\ > \text{A/m}^2$. The four-current is $J^\mu = (c\rho,\ J, 0, 0)$ with $c\rho = 1.1 > \times 10^{4}$ in the same units as $J$, confirming that the time and space > parts are comparable in magnitude once the beam is relativistic. ## Charge conservation as a divergence The statement that charge is neither created nor destroyed is the continuity equation, $$ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0. $$ Charge that leaves a region must flow across its boundary. In four-vector form this is the vanishing four-divergence of the current: $$ \partial_\mu J^\mu = \frac{1}{c}\frac{\partial (c\rho)}{\partial t} + \nabla \cdot \vec J = \frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0. $$ The four-divergence $\partial_\mu J^\mu$ is a Lorentz scalar: its value is the same in every frame. Charge conservation is therefore not a separate law in each frame but a single invariant statement. If charge is conserved for one observer, the tensor equation $\partial_\mu J^\mu = 0$ guarantees it for all. The geometric content is a flux balance. Integrate $\partial_\mu J^\mu = 0$ over a four-dimensional region of spacetime and apply the divergence theorem: the net flux of $J^\mu$ through the boundary vanishes. For a box that is a spatial volume $V$ swept from time $t_1$ to $t_2$, the flux through the two spacelike faces is the total charge in $V$ at each time, and the flux through the timelike sides is the current leaving through the walls. Setting the two equal gives $$ Q(t_2) - Q(t_1) = -\int_{t_1}^{t_2}\!\!\oint_{\partial V} \vec J \cdot \hat n \,\d A \,\d t, $$ the integral form of charge conservation. $$ % caption: Continuity as a flux balance on a spacetime box: the change in charge % between the two spacelike faces equals the current carried out through the % timelike walls, which is the integrated four-divergence set to zero. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % box (a spatial slab drawn as a slanted prism) \draw[black, fill=acc!10] (0,0) rectangle (4,2.6); \draw[black] (0,2.6) -- (1.1,3.4) -- (5.1,3.4) -- (4,2.6); \draw[black] (4,0) -- (5.1,0.8) -- (5.1,3.4); \draw[black] (0,0) -- (1.1,0.8) -- (5.1,0.8); \draw[black] (1.1,0.8) -- (1.1,3.4); % faces \node[black!70] at (2.0,1.3) {charge $Q$ inside}; % charge in through bottom (past) \draw[->, very thick] (2.0,-1.0) -- (2.0,-0.1) node[right, midway, black!70] {charge at $t_1$}; % charge out through top (future) \draw[->, very thick] (2.9,2.9) -- (2.9,3.9) node[right, midway, black!70] {charge at $t_2$}; % current through walls \draw[->, black, thick] (4.05,1.3) -- (5.3,1.3) node[right, black!70] {current out}; \draw[->, black, thick] (-0.05,1.3) -- (-1.3,1.3) node[left, black!70] {current in}; \end{tikzpicture} $$ ## The four-potential The electric and magnetic fields derive from potentials. The scalar potential $\phi$ and the vector potential $\vec A$ produce the fields through $$ \vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t}, \qquad \vec B = \nabla \times \vec A. $$ These two potentials assemble into the **four-potential** $$ A^\mu = \left(\frac{\phi}{c},\ \vec A\right), \qquad A_\mu = \eta_{\mu\nu} A^\nu = \left(-\frac{\phi}{c},\ \vec A\right). $$ That $A^\mu$ is a genuine four-vector is not obvious from the definitions of $\phi$ and $\vec A$ alone; the justification is that the field equations written in terms of $A^\mu$ take the form of a four-vector equation, which the next section establishes. The four-potential is the fundamental variable of the covariant theory. The fields $\vec E$ and $\vec B$ are extracted from its derivatives. > **Definition (Four-potential).** The four-potential is $A^\mu = (\phi/c, \vec > A)$. The scalar potential sits in the time slot, the vector potential in the > space slots. The observable fields are built from $\partial^\mu A^\nu - > \partial^\nu A^\mu$, a combination that turns out to be gauge invariant. ## Gauge freedom The potentials are not unique. The fields $\vec E$ and $\vec B$ are unchanged by the **gauge transformation** $$ \vec A \to \vec A + \nabla \chi, \qquad \phi \to \phi - \frac{\partial \chi}{\partial t}, $$ for any scalar function $\chi(t, \vec x)$. In four-vector form this is a single shift, $$ A^\mu \to A^\mu + \partial^\mu \chi, $$ with $\partial^\mu = (-\tfrac{1}{c}\partial_t,\ \nabla)$. The magnetic field is unchanged because $\nabla \times \nabla \chi = 0$; the electric field is unchanged because the added pieces $-\partial_t \nabla \chi$ and $\nabla \partial_t \chi$ cancel. Different four-potentials related by $\partial^\mu \chi$ describe identical physics. $$ % caption: A whole family of four-potentials, differing by the gradient of an % arbitrary scalar, produces the same electric and magnetic fields; the physical % content lives in the gauge-invariant combination of derivatives. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % three potentials mapping to one field \node[draw, black, minimum width=1.7cm] (p1) at (0,2.4) {potential $A$}; \node[draw, black, minimum width=1.7cm] (p2) at (0,1.2) {shifted $A$}; \node[draw, black, minimum width=1.7cm] (p3) at (0,0.0) {shifted $A$}; \node[draw, acc, very thick, fill=acc!12, minimum width=1.9cm] (f) at (5.0,1.2) {same E and B}; \draw[->, black] (p1) -- (f); \draw[->, black] (p2) -- (f); \draw[->, black] (p3) -- (f); \node[black, anchor=west] at (1.7,2.05) {shift by gradient of a scalar}; \end{tikzpicture} $$ Gauge freedom is a redundancy, and it can be used to simplify the equations. The **Lorenz gauge** is the condition $$ \partial_\mu A^\mu = \frac{1}{c^2}\frac{\partial \phi}{\partial t} + \nabla \cdot \vec A = 0, $$ a manifestly Lorentz-invariant statement (it is the vanishing four-divergence of $A^\mu$). Any potential can be brought to Lorenz gauge: if $\partial_\mu A^\mu = f \ne 0$, choose $\chi$ solving $\Box \chi = -f$, and the shifted potential satisfies the condition. Even within Lorenz gauge a residual freedom remains — any $\chi$ with $\Box \chi = 0$ preserves it — but the condition is enough to decouple the field equations. > **Definition (Lorenz gauge).** The Lorenz gauge condition is $\partial_\mu A^\mu > = 0$. Because it equates a Lorentz scalar to zero, a potential in Lorenz gauge > in one inertial frame is in Lorenz gauge in every inertial frame. This is what > makes it the natural gauge for a manifestly covariant formulation, unlike the > Coulomb gauge $\nabla \cdot \vec A = 0$, which singles out a frame. ## The sourced wave equation Writing Maxwell's equations for the potentials, in Lorenz gauge, gives a single equation. The two inhomogeneous Maxwell equations, Gauss's law $\nabla \cdot \vec E = \rho/\epsilon_0$ and the Ampère–Maxwell law $\nabla \times \vec B = \mu_0 \vec J + \tfrac{1}{c^2}\partial_t \vec E$, when expressed through the potentials and simplified with $\partial_\mu A^\mu = 0$, both reduce to the same form. Define the d'Alembertian, the four-dimensional wave operator, $$ \Box \equiv \partial^\mu \partial_\mu = -\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \nabla^2. $$ Then the potentials satisfy $$ \Box A^\mu = -\mu_0 J^\mu. $$ Component by component, this is one wave equation for the scalar potential and one for each component of the vector potential: $$ \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = -\frac{\rho}{\epsilon_0}, \qquad \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\vec A = -\mu_0 \vec J. $$ The check on the time component uses $A^0 = \phi/c$ and $J^0 = c\rho$: $\Box (\phi/c) = -\mu_0 c\rho$ gives $\Box \phi = -\mu_0 c^2 \rho = -\rho/\epsilon_0$, which is the wave equation above, and reduces to Poisson's equation $\nabla^2\phi = -\rho/\epsilon_0$ in the static limit. > **Result.** In Lorenz gauge, Maxwell's equations for the potentials become the > single four-vector wave equation $\Box A^\mu = -\mu_0 J^\mu$. Each component is > a wave equation with a source: charges and currents generate potentials that > propagate at $c$. The sign of the source depends on the wave operator's sign > convention; here $\Box = \partial^\mu\partial_\mu = \nabla^2 - > c^{-2}\partial_t^2$. The solution is the retarded potential: the field at a point is set by the source at the **retarded time** $t - |\vec x - \vec x\,'|/c$, the earlier moment from which a signal at speed $c$ reaches the point. Disturbances in charge and current propagate outward as waves at $c$, the light speed built into the wave operator. That electromagnetism predicts waves at $c$ in every frame is the puzzle Module 1 opened with, now written as a manifestly invariant equation. $$ % caption: The four-current sources the four-potential through a wave equation; % a localized disturbance in charge and current radiates outward at speed c, the % single speed built into the d'Alembertian. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % source \fill (0,0) circle (2.2pt); \node[black!70, anchor=north] at (0,-0.25) {source $J$}; % expanding wavefronts \draw[acc, thick] (0,0) circle (1.0); \draw[acc!70, thick] (0,0) circle (1.9); \draw[acc!45, thick] (0,0) circle (2.8); \draw[->, black] (0,0) -- (2.8,0) node[right, midway, above, black!70] {speed $c$}; \node[black, anchor=west] at (2.0,2.2) {potential $A$ radiating}; \end{tikzpicture} $$ ## Summary - Charge density and current density are the components of one four-vector, the four-current $J^\mu = (c\rho, \vec J) = \rho_0 U^\mu$. Density and current are frame-dependent shadows of the same charge; a boost turns a static density into a current. - Charge conservation is the single invariant statement $\partial_\mu J^\mu = 0$, the vanishing four-divergence, equivalent to the continuity equation and to a flux balance on any spacetime region. - The scalar and vector potentials combine into the four-potential $A^\mu = (\phi/c, \vec A)$, defined only up to a gauge shift $A^\mu \to A^\mu + \partial^\mu \chi$ that leaves the fields untouched. - The Lorenz gauge $\partial_\mu A^\mu = 0$ is Lorentz invariant and reduces the field equations for the potentials to the wave equation $\Box A^\mu = -\mu_0 J^\mu$, whose solutions propagate at $c$. The [next lesson](/relativity/covariant-electrodynamics/the-electromagnetic-field-tensor) builds the gauge-invariant field strength $F^{\mu\nu}$ from the derivatives of $A^\mu$ and reads $\vec E$ and $\vec B$ off its components. [^carroll1]: Carroll, _Lecture Notes on General Relativity_, §1 (classical field theory and electromagnetism), arXiv:gr-qc/9712019. The four-current and four-potential constructions and the Lorenz-gauge wave equation follow Schutz, _A First Course in General Relativity_, Ch. 4.