Incorrect proof that four-current $J^\mu$ is a four-vector

Question

This question is inspired by this answer to a question about proving that $J^\mu$ is a four-vector.

The answer uses the continuity equation $\partial_\mu J^\mu = 0$ and the experimental fact that charge conservation is Lorentz invariant, and concludes that $J^\mu$ must transform like a 4-vector.

This comment by JánLalinský claims that this argument is wrong, and provides as counterexample the 4-tuple [energy density, energy current density], which satisfies the continuity equation but is not a 4-vector.

I don't completely understand why this tuple is not a four-vector, and I don't completely understand why the previous argument is incorrect. Could someone provide a more elaborate explanation for what goes wrong?

It seems to me that $\partial_\mu j^\mu = \text{div }\left(j^\mu\frac{\partial}{\partial x^\mu}\right)$ where $\text{div}$ is the divergence on $4$-dimensions, which is invariant under transformations that preserve the metric since the divergence can be expressed in a coordinate-free way. Since the RHS of $\text{div }\left(j^\mu\frac{\partial}{\partial x^\mu}\right) = 0$ is zero which is also invariant, this requires $j^\mu\frac{\partial}{\partial x^\mu}$ to be invariant so it must be a true vector field (at least locally). This then means that the coefficients $j^\mu$ should transform like the components of a 4-vector.

Clearly something is wrong in my argument, so I would be thankful if someone could resolve my confusion. I am not yet very comfortable with the transformation properties of four-vectors.

To avoid making this a duplicate, I am not asking for any proof that $J^\mu$ is a four-vector, but rather asking why this particular argument does not work.

Answer 1

Let's start with an even simpler example. Let there be a single-component function of coordinates $\phi$, which obeys the equation

$$ \partial_\mu \phi = 0 $$ in all frames. Does this imply value of $\phi$ at some event is a Lorentz scalar, that is, has the same value in all frames? No. It is easy to find counter-examples, e.g. $\phi=\beta$ where $\beta$ is speed of some preferred inertial frame, e.g. speed of the Earth center. This depends on the frame, but does not depend on coordinates, since it is a parameter that is the same for all points of an inertial frame.

On the other hand, there are functions $\phi$ which are Lorentz scalars and obey the above equation, e.g. $\phi = 0$ in all frames. So, we conclude zero derivative of $\phi$ in all frames does not determine how $\phi$ transforms between the frames; it maybe be a Lorentz scalar, it may not be.

Similarly, the fact there is a field $N$ whose components obey the equation $$ \frac{\partial N^\mu}{\partial x^\mu} = 0 $$ in all inertial frames, does not determine how the field components $N^\mu$ transform.

Let the components of field $N$ in all frames be a sum $N^\mu = A^\mu + C^\mu$, where $A^\mu$ are components of a four-vector $A$ which obey the same equation in all frames:

$$ \frac{\partial A^\mu}{\partial x^\mu} = 0, $$ and where the four-tuple $C^\mu = (C,0,0,0)$, with the same constant value in all frames. Then, since components $C^\mu$ do not transform as four-vector components, also components $N^\mu = A^\mu+C^\mu$ do not.

This can be encountered e.g. in gauge transformations of EM potentials. For example, for retarded field of a point particle, we can have electric potential $\phi(x^\mu) = \frac{Kq\gamma_{ret}}{U_\nu(x^\nu-R^\nu)}$ and magnetic vector potential $A^i(x^\mu) = \frac{Kq\gamma_{ret} V^i/c^2}{U_\nu(x^\nu-R^\nu)}$ , where $\gamma_{ret}, R^\nu,V^i,U_\nu$ are retarded gamma, coordinates, velocity and four-velocity of the charged particle. These potentials, by definition, form components of a four-vector: $$ A^\mu=\frac{Kq}{c^2}\frac{U^\mu}{U_\nu(x^\nu-R^\nu)}. $$ and these obey the Lorenz condition (Lorenz gauge) $$ \frac{\partial A^\mu}{\partial x^\mu} = 0 $$ in all frames.

But we can "gauge transform" these potentials into different potentials in such a way that the new potentials, while still having zero four-divergence, are not components of a four-vector.

Let $$ A^{*\mu} = A^\mu + \partial^\mu \lambda $$ where e.g. $\lambda = -Ct$ in all frames, with the same $C$. Then

$$ \phi^* = \phi + C, $$ so the new potential has value $C$ at infinity. The magnetic potential is unchanged by this transformation, $\mathbf A^* = \mathbf A$, so the four-tuple $$ A^{*\mu} = A^\mu + (C, 0,0,0) $$

still has zero four-divergence, but since the time component is affected by $C$, and spatial components are not, these together obviously do not transform as a four-vector.

One can think up many different functions $\lambda$ and corresponding gauge transformations which turn the potentials that form a four-vector into potentials that do not form a four-vector, all while preserving the Lorenz condition.