Relativistic Currents

A conserved four-current combines a density and its flux into a covariant object. Its integral over a spacelike hypersurface gives a conserved charge when boundary flux is controlled. Conservation, positivity, and interpretation are separate questions: the same continuity equation can govern an electric charge of either sign or a nonnegative probability density.

Required background. Four-Vectors defines the current transformation law and covariant divergence.

Helpful background. Causality and Light Cones supplies the geometry of timelike normals and causal vectors.

Local conservation and finite-volume balance

Write $j^\mu=(c\rho,\mathbf j)$. The continuity equation is

$$\partial_\mu j^\mu=0 \quad\Longleftrightarrow\quad \partial_t\rho+\nabla\cdot\mathbf j=0.$$

For a fixed spatial region $V$, the divergence theorem yields

$$\frac{d}{dt}\int_V\rho\,d^3x =-\int_{\partial V}\mathbf j\cdot d\mathbf S.$$

Thus a conserved current does not make the charge inside every finite region constant. Outgoing flux reduces the charge in that region. Total charge is constant if the flux at spatial infinity vanishes, or if another specified boundary condition cancels the net flux. Wave packets with suitable decay and periodic boxes provide different ways to realize this condition.

Charge on a spacelike hypersurface

Let $\Sigma$ be a smooth spacelike hypersurface with future-directed unit normal $n^\mu$, normalized by $n_\mu n^\mu=1$. Write its oriented surface element as $d\Sigma_\mu=n_\mu d\Sigma$. Then define

$$Q_\Sigma=\frac1c\int_\Sigma j^\mu d\Sigma_\mu.$$

For $t=\text{constant}$, $n_\mu=(1,0,0,0)$ and $d\Sigma=d^3x$, so this reduces to $Q=\int\rho\,d^3x$. The factor $1/c$ matches the convention $j^0=c\rho$.

Apply the four-dimensional divergence theorem to the region between two such surfaces and a connecting timelike boundary. With consistent boundary orientations,

$$0=\int_{\mathcal V}d^4x\,\partial_\mu j^\mu =c(Q_{\Sigma_2}-Q_{\Sigma_1}) +\int_{\text{side}}j^\mu d\Sigma_\mu.$$

If the side flux vanishes, the charges agree. This proves independence from the chosen slicing for a sufficiently isolated system. Comparing equal-time integrals in different frames without transforming the integration surface misses precisely this geometric issue.

A tilted plane

For the plane $x^0=\beta x^1$ with $|\beta|<1$, parametrized by $\mathbf x$, the normal and proper surface volume are

$$n_\mu=\gamma(1,-\beta,0,0), \qquad d\Sigma=\frac{d^3x}{\gamma}.$$

Their product gives

$$Q_\Sigma=\int d^3x\left(\rho-\frac{\beta j^1}{c}\right),$$

with the fields evaluated at $t=\beta x^1/c$. The current term and the different evaluation times explain why a tilted-surface integral is not obtained by integrating the original density over an unchanged time slice.

The extra requirement for probability

For a local current to assign nonnegative density to every inertial observer, one needs $j^\mu n_\mu\geq0$ for all future timelike normals. In Minkowski space this is equivalent to $j^\mu$ being future causal or zero. In one frame it can be written as

$$j^0\geq|\mathbf j|, \qquad j^0\geq0.$$

To see sufficiency, transform to the observer’s rest frame, where the contraction is the nonnegative time component of a future causal vector. For necessity, a spacelike current with $|\mathbf j|>j^0$ admits a boost along its spatial direction that makes the measured density negative. A nonzero past-directed causal current already has the wrong sign.

This is a geometric positivity criterion. It does not by itself construct a Hilbert space, specify a position measurement, or guarantee fixed particle number in an interacting relativistic theory.

Klein–Gordon and Dirac currents

The Klein–Gordon Equation derives, for $m>0$, the conventional current

$$j^\mu_{\mathrm{KG}}= \frac{i\hbar}{2m} \left(\phi^*\partial^\mu\phi -(\partial^\mu\phi^*)\phi\right).$$

It is conserved, but its time component is not generally nonnegative. Restricting to positive frequency makes the integrated free KG inner product positive; interference can still make its local density negative. Multiplying the current by a signed charge gives a charge-current convention, not a Born probability interpretation. For massless scalars, a normalization without the arbitrary $1/m$ factor is used.

The Covariant Dirac Equation instead gives

$$j^\mu_D=c\bar\psi\gamma^\mu\psi, \qquad j_D^0=c\psi^\dagger\psi.$$

Its causal property follows without selecting an energy branch. For any unit spatial vector $\mathbf a$, the Hermitian matrix $\mathbf a\cdot\boldsymbol\alpha$ squares to the identity. Its eigenvalues are $\pm1$, so

$$|\mathbf a\cdot\mathbf j_D| \leq c\psi^\dagger\psi=j_D^0.$$

Maximizing over $\mathbf a$ gives $|\mathbf j_D|\leq j_D^0$. The conserved one-particle norm is therefore positive on every spacelike slice under the boundary assumptions above. The electric current $qj_D^\mu$ can have either sign: positivity is a property of the probability current before multiplying by signed charge.

Exercises

1. A current has $j^0=1$ and $j^1=2$ in common units, with the other components zero. Give an observer who measures negative density.

Solution

An $x$ boost gives $j'^0=\gamma(1-2\beta)$. Any $1/2<\beta<1$ makes it negative. Conservation, if imposed separately, would not remove this local spacelike-vector obstruction.

2. Suppose $\rho$ is time independent but there is a nonzero stationary current through a box. What must the net boundary flux be?

Solution

The integral of $\partial_t\rho$ vanishes, so the total outward flux is zero. Current can enter one face and leave another. Local conservation constrains the net flux, not each face separately.

3. Replace $j^\mu$ by $j^\mu+\partial_\nu K^{\mu\nu}$, where $K^{\mu\nu}=-K^{\nu\mu}$ is smooth. Show conservation is preserved and state a condition for the equal-time total charge to remain the same.

Solution

$\partial_\mu\partial_\nu K^{\mu\nu}=0$ because derivatives commute and the tensor is antisymmetric. Since $K^{00}=0$, the charge shift is a spatial boundary integral $c^{-1}\int_{\partial V}K^{0i}dS_i$. It vanishes if that boundary contribution vanishes. A conserved current is not always unique, and local improvements need not preserve pointwise positivity.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — scalar and spinor conserved currents.
• J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998, chapter 11 — covariant conservation laws and charge.
• B. Thaller, The Dirac Equation, Springer, 1992, doi:10.1007/978-3-662-02753-0 — Dirac evolution, norm, and current.