advanced26 min studyLesson 266 in Physics

Four-current and charge conservation

Transform charge and current density together, derive local continuity, and distinguish invariant total charge from observer-dependent density.

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What you’ll understand

  • Transform a four-current with consistent velocity and density units.
  • Use local continuity to relate charge accumulation to boundary flux.
  • Explain how invariant charge can coexist with frame-dependent charge density.

Charge is conserved, but a measured charge density changes when the observer changes. This is not a contradiction because density refers to a spatial volume measured at one observer's simultaneous time. The current density supplies the rest of the relativistic information. Combining both into a four-current makes local conservation transparent and prevents a common error: transforming a density while leaving the associated current unchanged. We will connect the tensor notation to charge flowing through an ordinary laboratory box.

Density and flux form one four-vector

Keep signature (+,−,−,−)(+,-,-,-) and x0=ctx^0=ct. Define Jμ=(cρ,j)J^\mu=(c\rho,\mathbf j), with charge density rho in coulombs per cubic metre and current density j in amperes per square metre. Multiplying rho by c makes all four components share the latter units. Under a passive boost along x, J transforms exactly like any other four-vector. The density equation contains Vjx/c2Vj_x/c^2; the current equation contains VρV\rho. Both combinations have the correct units before the common gamma factor is applied.

For one species moving with ordinary velocity v, its convection current is j=ρv\mathbf j=\rho\mathbf v. If its proper density is ρ0\rho_0, the four-current is Jμ=ρ0uμJ^\mu=\rho_0u^\mu. A mixture need not have any frame where total current vanishes. Oppositely charged populations can carry a net current while their total charge density cancels. Treat each population consistently before summing; using one species' rest density as if it were the net laboratory density is a frequent source of incorrect wire and beam calculations.

ρ′=γ(ρ−Vjxc2),jx′=γ(jx−Vρ),jy′=jy,jz′=jz\rho'=\gamma\left(\rho-\frac{Vj_x}{c^2}\right),\qquad j'_x=\gamma(j_x-V\rho),\qquad j'_y=j_y,\quad j'_z=j_z
The primed observer moves with positive V along x. J has units A/m² in every component; rho and rho prime are measured on their respective observers' simultaneous spatial slices.↔ Scroll wide equations horizontally.

Continuity is a local spacetime statement

The four-divergence ∂μJμ\partial_\mu J^\mu expands to ∂tρ+∇⋅j\partial_t\rho+\nabla\cdot\mathbf j because the c factors cancel in the temporal derivative. Setting it to zero says that charge can accumulate only when more charge enters a region than leaves. Integration over a fixed laboratory volume and the divergence theorem yield the boundary-flux version. The surface normal points outward, so positive outward flux decreases the enclosed charge. This sign is more reliable than a memorized verbal rule about incoming or outgoing current.

A total-charge statement alone would allow charge to vanish at one place and appear instantly elsewhere. Local continuity is stronger: intermediate transport must account for the change. For an isolated system, sufficiently rapid decay of current at spatial infinity gives a time-independent total charge in each inertial frame. Boundary assumptions matter. A finite open region can lose enclosed charge while obeying exact continuity; that does not violate charge conservation. The lost charge has simply crossed the chosen boundary.

∂μJμ=∂tρ+∇⋅j=0,ddt∫Vρ d3x=−∮∂Vj⋅da\partial_\mu J^\mu=\partial_t\rho+\nabla\cdot\mathbf j=0,\qquad \frac{d}{dt}\int_{\mathcal V}\rho\,d^3x=-\oint_{\partial\mathcal V}\mathbf j\cdot d\mathbf a
The integral equation concerns a fixed spatial laboratory volume. Its left side has units C/s and its right side A. A moving boundary requires an additional transport term.↔ Scroll wide equations horizontally.

Total charge needs a spatial slice

Relativistically, total charge is the flux of the four-current through a spacelike hypersurface. With the present normalization it is Q=c−1∫JμdΣμQ=c^{-1}\int J^\mu d\Sigma_\mu. On a constant-time slice this reduces to the familiar integral of rho over volume. Continuity makes the result independent of deformations of the slice when no current escapes through the side boundary. This is how different observers can agree on total isolated charge while disagreeing on the density and on which separated events are simultaneous.

A stationary uniform charge cloud illustrates the density effect simply. An observer moving along positive x sees the charges move along negative x, measures a density enhanced by gamma, and measures a negative x current. The cloud's contracted volume compensates for the density increase, leaving its total charge unchanged. For an infinite cloud, an infinite total charge is not a useful invariant number; the local transformation remains valid. Always distinguish local density calculations from finite-system global charge claims.

Boost a stationary charged cloud

A finite cloud has uniform rest density ρ0=4 nC m−3\rho_0=4\,\mathrm{nC\,m^{-3}} and rest volume 2 m32\,\mathrm{m^3}. A primed observer moves at 0.6c0.6c along x. Find its density, current density, volume, and total charge. Use c=3.00×108 m s−1c=3.00\times10^8\,\mathrm{m\,s^{-1}}.

  1. In the original rest frame jx=0j_x=0 and γ=1.25\gamma=1.25. Therefore ρ′=γρ0=5 nC m−3\rho'=\gamma\rho_0=5\,\mathrm{nC\,m^{-3}}. The charges move in the negative x direction for the primed observer.

  2. The current is jx′=−γVρ0=−(1.25)(1.8×108)(4×10−9)=−0.90 A m−2j'_x=-\gamma V\rho_0=-(1.25)(1.8\times10^8)(4\times10^{-9})=-0.90\,\mathrm{A\,m^{-2}}. Equivalently it is −Vρ′-V\rho'.

  3. Only the dimension parallel to motion contracts, so the simultaneous primed volume is 2/1.25=1.6 m32/1.25=1.6\,\mathrm{m^3}. Its total charge is (5 nC m−3)(1.6 m3)=8 nC(5\,\mathrm{nC\,m^{-3}})(1.6\,\mathrm{m^3})=8\,\mathrm{nC}, equal to the rest-frame value.

The result

The moving observer measures higher density and a negative current, while the cloud still contains exactly 8 nC of charge.

Explore the model

Interactive lab

Interactive lab · spacetime

The same boost mixes temporal and spatial components

For each fresh comparison, Reset and set BOTH β = 0, ct = 1, then Reset and set β = 0.6, ct = 1. Predict that the event's normalized component pair (1, 0.6) becomes (0.8, 0). This is the same algebra used to find the rest frame of a timelike current with jx/(cρ) = 0.6. Inspect paused readouts or Step.

The same boost mixes temporal and spatial componentsA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe instrument transforms a fixed event, not an actual current or a charge cloud. Its normalized pair illustrates four-vector algebra and cannot display density, current, or a contracted volume.

Check your understanding

A fixed laboratory volume has positive net outward charge flux. What does local continuity imply?

Try it yourself

In a one-dimensional slab 0≤x≤2 m0\le x\le2\,\mathrm m, the current density is jx(x)=kxj_x(x)=kx with k=3 A m−3k=3\,\mathrm{A\,m^{-3}}, and the cross-sectional area is 0.5 m20.5\,\mathrm{m^2}. Find the charge-density rate and enclosed-charge rate.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate the current density locally, then compare the result with the current leaving the right face minus that entering the left face.

Compare with the solution

The divergence is ∂xjx=k=3 A m−3\partial_xj_x=k=3\,\mathrm{A\,m^{-3}}, so ∂tρ=−3 C m−3 s−1\partial_t\rho=-3\,\mathrm{C\,m^{-3}\,s^{-1}}. The slab volume is 1 m31\,\mathrm{m^3}, giving dQ/dt=−3 C s−1dQ/dt=-3\,\mathrm{C\,s^{-1}}. The left-face current is zero, while the right-face outgoing current is (6 A m−2)(0.5 m2)=3 A(6\,\mathrm{A\,m^{-2}})(0.5\,\mathrm{m^2})=3\,\mathrm A, confirming the sign and magnitude. This is an instantaneous continuity calculation; the assumed current profile needs suitable sources or evolution to persist.

Take it with you

  • Charge density and current density are observer-dependent parts of one four-current.
  • Local continuity accounts for charge changes through boundary transport.
  • Invariant isolated charge requires consistent hypersurfaces and boundary conditions.

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