# Correcting the "wrong" definition of relativistic electric current density

In several standard references (I'm using conventions from David Tong's graduate electrodynamics notes), we take charge density $$\rho$$ as a primitive and then define the current density $$\vec{J}$$ by the relation $$I = \int_S \vec{J} \cdot d \vec{A}$$ where $$S$$ is arbitrary oriented surface (not necessarily a closed one) and $$I$$ is the total signed charge per unit time passing through $$S$$. By considering the case where $$S$$ is a small flat surface normal to one of the coordinate axes, and considering the universe's collection of moving charged particles as having a velocity field $$\vec{v}(\vec{x},t)$$ in addition to its charge density $$\rho(\vec{x},t)$$ this implies

$$\vec{J} = \rho \vec{v}.$$

As David Tong correctly points out, this can't be the "right" answer in some sense, since this can't be the spatial part of a 4-vector. By analogy with 3-momentum and 4-momentum of a massive particle we can quickly guess the corrected relation

$$\vec{J} = \rho \gamma \vec{v}$$

where $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$. But this doesn't explain what's wrong with the original derivation. One of two things must be true:

1. The definition we started with isn't the right one.

2. There are subtle Galilean assumptions we make in deducing $$\vec{j} = \rho \vec{v}$$ from the given definition.

I can't tell which is the culprit. On one hand, the definition I start with here readily implies

$$\frac{d \rho}{dt} = - \nabla \cdot \vec{J}$$

by considering $$S$$ a closed surface with enclosed charge $$Q$$, and we know this conservation law is correct. On the other hand, the derivation of $$\vec{J} = \rho \vec{v}$$ from the starting definition seems very straightforward without hidden assumptions.

• Surely the right one should be closer to $\vec J \to J^{\mu}=\gamma (\rho c,\rho v)$ Jun 29, 2023 at 19:24
• I'm not sure what you're asking. In the context of the question, I'm using $\vec{J}$ to denote the spatial component of the relativistic 4-current density $J^\mu$. My point here is that the naive result $\vec{J} = \rho \vec{v}$ seems to be incorrect, and I'm not sure where the error comes from in deducing that result.
– mpc
Jun 29, 2023 at 19:27
• I agree with you that $J^0 = \gamma \rho c$ and $\vec{J} = \gamma \rho \vec{v}$.
– mpc
Jun 29, 2023 at 19:30
• How are you defining $\rho$? Is it the charge density in the rest frame, or in a frame where the charges are moving? Jun 29, 2023 at 19:41
• Thanks for taking a look Michael. I've answered my original question, and in the process I think I've addressed your clarifying question.
– mpc
Jun 29, 2023 at 19:54

I'm leaving the original question here for posterity, but in posting this question I rubber-duck debugged myself and I have found the root cause of my deep confusion: somehow I got it in my head that the exact expression for the relativistic 4-current is $$J^\mu = (\gamma \rho c, \gamma \rho \vec{v})$$ where $$\rho, \vec{v}$$ are both as measured in the same inertial frame that is measuring $$J^\mu$$ to begin with (some sources take $$\rho$$ as measured in the rest frame of the charges at $$\vec{x}$$, which of course changes the final expression). To clarify matters, with those definitions the correct expression is
$$J^\mu = (\rho c, \rho \vec{v}).$$
The analogy with single-particle mechanical 4-momentum $$p^\mu = (\gamma m c, \gamma m \vec{v})$$ is misleading because the invariant mass $$m$$ is a Lorentz scalar while $$\rho$$ (under my conventions) is not.
I think the root cause of my confusion is that David Tong's notes say that the result $$\vec{J} = \rho \vec{v}$$ holds "neglecting relativistic effects." I suspect this is just a mistake on his part, or else he means "the definition of $$\rho$$ is subtle with relativistic effects matter."