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.

  • $\begingroup$ Surely the right one should be closer to $\vec J \to J^{\mu}=\gamma (\rho c,\rho v)$ $\endgroup$
    – Mauricio
    Jun 29, 2023 at 19:24
  • $\begingroup$ 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. $\endgroup$
    – mpc
    Jun 29, 2023 at 19:27
  • $\begingroup$ I agree with you that $J^0 = \gamma \rho c$ and $\vec{J} = \gamma \rho \vec{v}$. $\endgroup$
    – mpc
    Jun 29, 2023 at 19:30
  • $\begingroup$ How are you defining $\rho$? Is it the charge density in the rest frame, or in a frame where the charges are moving? $\endgroup$ Jun 29, 2023 at 19:41
  • $\begingroup$ Thanks for taking a look Michael. I've answered my original question, and in the process I think I've addressed your clarifying question. $\endgroup$
    – mpc
    Jun 29, 2023 at 19:54

1 Answer 1


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."

  • $\begingroup$ that's what I suggested $\endgroup$
    – Mauricio
    Jun 30, 2023 at 18:28

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