Is four-current $J^{\mu}$ necessarily timelike?

Question

In the spacelike case, there is a frame in which the charge density vanishes but current density does not. Additionally, the drift velocity of a free current would propagate superluminally. For this reason I would expect such a current to be impossible.

In the lightlike case, the charge and current densities would come in equal proportions in all frames. The drift velocity would propagate at the speed of light. It's unclear to me whether this is forbidden outright.

Is the four-current necessarily a timelike vector?

Edit: Do Maxwell's equations admit solutions for lightlike or spacelike currents?

Answer 1

Case 1: Current Densities Comprising Charge Carriers of the Same Sign

For currents comprising a lone charged particle, or charge carriers of the same sign, the current four vector must be timelike. But Maxwell's Equations don't impose the limit - relativity does. The limit arises because we postulate it (as I discuss here) to force causality - so that no boost can switch the sign of the time component of a four-vector. Maxwell's equations are not causal: for any given solution you can construct a time-reversed one, by mapping $t\mapsto-t$. When we solve antenna problems, we must explicitly exclude the advanced wave solution as unphysical - nothing in Maxwell's equations rules it out. If you write down and study the Liénard-Wiechert Potentials (well-known, "building block" solutions of Maxwell's Equations) for a supraluminal particle, you'll explicitly see causality violated in that the direction of the Poynting vector is switched for supraluminal observers relative to the direction for subluminal ones. This is because, for an accelerated particle, energy pulses, from a supraluminal standpoint, run inwards to the charged particle and bring about an acceleration of the particle, switching the causal field-particle relationship apparent to the subluminal observer.

Lightlike current densities are excluded because you can't boost such a particle to $c$ with finite energy, even for a charged particle of zero rest mass (we have solid theoretical reasons for believing that they don't exist. See the answers to this question here, especially Marek's and Lubos's answers). This is owing to the Abraham-Lorentz-Dirac force, i.e. the radiation resistance. This is readily seen from the Liénard relativistic generalization

$$P = \frac{\mu_o q^2 a^2 \gamma^6}{6 \pi c}$$

of the Larmor radiation formula (i.e. it diverges like $\gamma^6$) that follows from the Abraham-Lorentz-Dirac formula. Alternatively, use the Larmor formula from the frame momentarily co-moving with the charged particle to show that the same small change in rapidity always takes the same amount of energy (measured from the co-moving frame). Also note that the Liénard-Weichert potentials diverge for lightlike charge motion.

Case 2: Current Densities Comprising Charge Carriers of the Both Signs (Hat Tip: Emilio Pisanty for pointing out a vicious error on my part)

In this case, trivially, we can have a DC current distribution comprising charge carriers of opposite sign such that the charge density at any point is nought. Thus a four current density of the form $(0,\,\vec{J})$, i.e. a spacelike four current density. Another example is a sinusoidally varying with time current density comprising opposite signed charges which is orthogonal to the spatial part of the wave four vector and with a zero charge density so that the continuity equation $J^\alpha\,k_\alpha = 0$ is fulfilled.

Note that neither of these distributions entail faster than light particles, so causality is not threatened by them.

Lightlight distributions can arise as well. Consider again the sinusoidally varying with time case. Lightlike vectors by definition are null (self Minkowski orthogonal) so that put $J_\alpha$ parallel to the wave four co-vector $k_\alpha$ then $J^\alpha\,k_\alpha = 0$ and the current and charge density automatically fulfill the continuity equation.

Answer 2

Suppose the current consists of only negative charges. If $j^μ$ is spacelike, $c^2 ρ^2-j^2<0$. In 2-dim there is a frame where $j^μ= \begin{bmatrix}0\\j\end{bmatrix}$. Under Lorentz boost $Λj=\begin{bmatrix}γ & γβ \\ γβ & γ\end{bmatrix}$ $\begin{bmatrix}0\\j\end{bmatrix}$ $=\begin{bmatrix}γβj\\γj\end{bmatrix}$ $=\begin{bmatrix}cρ^{'}\\j^{'}\end{bmatrix}$, so the sign of $ρ^{'}$ depends on the sign of $β$. We can reverse the sign of the charges by reversing the speed direction. This is a contradiction. On the other hand, if $j^μ$ is timelike, $c^2 ρ^2-j^2>0$. There is a frame where $j^μ= \begin{bmatrix}cρ\\0\end{bmatrix}$. Under boost $Λj=\begin{bmatrix}γ & γβ \\ γβ & γ\end{bmatrix}$ $\begin{bmatrix}cρ\\0\end{bmatrix}$ $=\begin{bmatrix}γcρ\\γβcρ\end{bmatrix}=$ $\begin{bmatrix}cρ^{'}\\j^{'}\end{bmatrix}$. The charge sign never changes.

Answer 3

Timelike and spacelike are characteristics of a pair of points in the Minkowsky space. How can a four-vector possibly timelike or spacelike?

I guess you mean that: whether four-current $J^\alpha$ lies in the light cone. If so, then you call it timelike. If not, then you call it spacelike. We recall that $J^\alpha$ is the product of $\rho$ and four velocity $u^\alpha$. Then, because the moving charged particles cannot exceed speed of light (thus lying in the light cone), and that $\rho$ is real, we see that $J^\alpha$ is timelike indeed.