Why is the 4-current a tensor rather than a tensor density?

Question

I am trying to understand electromagnetism better in terms of tensors and differential geometry. First recall that (in the Lorenz gauge) the equation of motion for the four-potential $A^\mu$ is

$$(-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2)A^\mu = -\mu_0 J^\mu.$$

My question is how does the 4-current $J^\mu$ transform under a generic change of coordinates? I know that it transforms like a vector but I am confuses as to why since it is ultimately a current density? After all, its components are the charge density, $\rho$, and the current density, J. And, of course, the units of charge density are $C/m^3$ and of current density are $C/(m^2 s)$.

Edit: To spell out my concern about units suppose that I were to implement a global conformal transformation $(t,x,y,z)\mapsto(\bar{t},\bar{x},\bar{y},\bar{z})$ which rescales 1 meter to down 1 foot. Then wouldn't $J^\mu$ be rescaled by $(1\text{ meter}/1\text{ foot})^3$? But this isn't how $J^\mu$ would transform if it were a vector field.

I guess I am confused about how we can combine together a bunch of densities and end up with a tensor instead of a tensor density (which transforms differently than a tensor).

Let's simplify the situation (and actually what follows is the scenario that I really care about). Consider a scalar version of electromagnetism in $n=1+1$ dimensions. The theory is now about the dynamics of a scalar field $\varphi$ given some charge density $\rho$. Note in this theory both of these are scalar fields in their own right and not a component of some other field. Let's say the equation of motion for $\varphi$ is

$$(-\partial_t^2+\partial_x^2)\varphi = -\rho/\epsilon_0$$

where $\rho$ is the scalar charge density which sources this scalar field. Does $\rho$ transform like a scalar or a scalar density? Again I know from this equation that it transforms like a scalar, but I want to know why.

Consider the following bit of hand waving. At any point $p=(t,x)$ define $Q(p)$ to be the total charge to the left of $p$ minus the total charge to the right of $p$. Now $Q$ is clearly a scalar field. Isn't $\rho$ essentially the derivative of $Q$? But then how is $\rho$ a scalar?

Edit: It has been pointed out that $Q$ as I have defined it above depends upon what frame I am using to define it. To fix this, consider the space-like region to the left of $p$ and the space-like region to the right of $p$. Note that these are completely disjoint and can be defined in a coordinate-independent way. Take $Q_\text{alt}(p)$ to be defined as the difference in charges between these two regions. Now define $\rho_\text{alt}$ as the derivative of $Q_\text{alt}$. How does $\rho_\text{alt}$ transform and what makes it different from how $\rho$ transforms?

Answer 1

First of all, the electromagnetic wave equation given in the question is valid (assuming the Lorenz gauge condition) only in flat or at least Ricci-flat spacetime. In general curved spacetime there is an additional term proportional to the Ricci tensor.

The $J$ that appears in this equation is a tensor, not a tensor density. The quantity $J \sqrt{-g}$ is a tensor density and its four components are the charge and current density. $J \sqrt{-g}$ satisfies a continuity equation with partial derivatives, i.e., $\partial_i (J^i \sqrt{-g}) = 0$, which is interpreted as conservation of charge in the local coordinates, while $J$ instead satisfies $\nabla_i J^i = 0$.

Addendum (April 11, 2024): In 1+1 spacetime, it is not correct that $\varphi$ is a scalar field. The electromagnetic potential still has two components, call them $\varphi$ and $A$, which transform as a vector, e.g., $\varphi^2 - A^2$ is invariant under Lorentz transformations (i.e., a scalar, though not a gauge-invariant one), while $\varphi$ and $A$ separately are not. There is no magnetic field, but we would have $E = -\partial_x \varphi - \partial_t A$, so the spacelike component of the potential is still relevant.

The equation $-\partial_t^2 \varphi + \partial_x^2 \varphi = - \rho/\epsilon_0$ is not correct even assuming that the spacetime is flat. The quantity $\varphi$ on the left is a component of a vector, and the quantity on the right also should be. $\rho$ is neither a scalar, nor a component of a vector. $\rho$ is, again, the time component of $J\sqrt{-g}$ where $J$ is a vector.

The OP's attempt to provide a coordinate-independent definition of $\rho$ (which would make it a scalar) is confused. Intuitively, $\rho$ simply cannot be a scalar because it is subject to length contraction. The OP refers to "the space-like region to the left of $p$ and the space-like region to the right of $p$", but these are regions of spacetime, not regions of space. To calculate the amount of charge to the left or right of $p$, it is necessary to take the intersections of these regions with the space-like slice defined by the current coordinate system (i.e., the set of all spacetime points with $t = 0$ or whatever the current time is), which means you end up with something that depends on the coordinate system.

Answer 2

My question is how does the 4-current $J^\mu$ transform under a generic change of coordinates? I know that it transforms like a vector but I am confuses as to why since it is ultimately a current density? After all, its components are the charge density, $\rho$, and the current density, J. And, of course, the units of charge density are $C/m^3$ and of current density are $C/(m^2 s)$.

It is sometimes called current, but it is actually electric current density four-vector. Its spatial components are both a "current density" in the usual physics sense and the whole is a four-vector, these two things do not contradict each other. The description "current density" does not imply this quantity is a tensor density (needed to be multiplied by some correction factor to get a tensor). Maybe the problem is just this misleading language.

Also it seems to me you have a too ambitious expectation of what the concept of four-vector means. The definition is that its components transform the same way as the cartesian coordinates and time of an event when changing inertial frame, that is, via the Lorentz transformation. Changes of scale or units of length are not considered in this definition, and such coordinate transformations are not covered by standard Lorentz transformations! Different four-vector quantities can change differently when changing units of length.

This can be directly seen for $A^\mu$ and $J^\mu$ from the equations

$$(-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2)A^\mu = -\mu_0 J^\mu, $$ where $t,x,y,z$ and any lengths or times are measured in seconds, and

$$(-\partial_{t'}^2+\partial_{x'}^2+\partial_{y'}^2+\partial_{z'}^2)A'^\mu = -\mu_0 J'^\mu. $$ where lengths and times are measured in milliseconds (thousand times smaller units).

Because the derivatives act only on $A$ and not on $J$, these two four-vectors have to transform differently when changing the units of length. $J$ scales as $q/x^3$ and $A$ scales as $q/x$, so $J$ will decrease billion times, while $A$ will decrease only thousand times. But when the units are fixed, and inertial frames are changes, both quantities transform as four-vectors.

Another way to say this is that while the d'Alembert operator

$$ (-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2) $$ produces the same thing in all inertial frames (assuming all use the same units of time and length), and thus the equations above allow for $A$,$J$ to be Lorentz four-vectors, it does not produce the same thing when the coordinate system is changed via change of units, and thus the equations above do not allow for $A$ and $J$ to re-scale the same way.