According to MTW,

$$F^{\alpha\beta}{}_{;\beta} = 4\pi J^\alpha$$

and we can infer that the four-current must be an ordinary vector field because the left side is tensorial.

But Wikipedia says that the four-current is a vector density and gives the Maxwell Lagrangian density as follows

$$\mathcal{L} = -\frac{1}{4\mu_0} F_{\alpha\beta} F^{\alpha\beta} \sqrt{-g} + A_\alpha J^\alpha$$

which makes it clear that the four-current defined in the way in which the article uses it must be a vector density since there is no $\sqrt{-g}$ factor in the second term. But a definition isn't given.

So which is it? Are there two different conventions in use for the four-current? Which is more common?


2 Answers 2


You could make an argument that the four-current is most naturally defined as a vector density. This is because a vector density uniquely defines a three-form (i.e. a totally antisymmetric tensor $J_{\alpha\beta\gamma}$), and three-forms can be integrated on hypersurfaces without any reference to a metric. So if you think of a current as an object which, when integrated over a hypersurface, gives you a flux through that surface, then it is natural to think of the current as a three-form, or equivalently as a vector density.

Another advantage of treating $\tilde{J}^\mu$ as a vector density (I'll use a tilde to denote a density) is that it satisfies the ordinary conservation law $\partial_\mu \tilde{J}^\mu = 0$, rather than the covariant conservation law $\nabla_\mu J^\mu = 0$. This sort of illustrates the main point of working with the current density: its conservation and coupling to $A_\mu$ are all independent of the metric, so it can be simpler at times to work with.

In the end, you should be able to tell by the context whether someone is working with the vector or the vector density. I think if they are using vector densities, they will probably explicitly mention it, since writing $J^\mu$ usually has the implication that you are talking about a vector, and not a vector density.


Clearly MTW's definition of $J^\alpha$ is a vector field due to the argument given. Note however that in the Wikipedia article you linked, under Summary, $J^\alpha$ is defined with an additional factor $\sqrt{-g}$, making it a vector density. You can easily see this by replacing the ordinary derivative in the definition of $J^\mu$ by a covariant one (which does not change anything, as stated in the article) and pulling the factor $\sqrt{-g}$ outside the covariant derivative (since $\sqrt{-g}_{;\mu}=0$).

I'm not sure which definition is more common, but in my experience I've seen the first one more often, i.e. where $J^\mu$ is an ordinary vector field.


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