I've heard that differential forms are related to densities, however I'm still a little confused about that. I thought on the case of charge density and I came to that: let $U\subset\mathbb{R}^3$ be a region of $3$-space, and let $\rho : U \to \mathbb{R}$ give the charge density at every point of $U$. I can then create the $3$-form $\omega = \rho \ dx \wedge dy \wedge dz$, which in my understanding gives me the approximate amount of charge enclosed by a volume determined by $3$ vectors when they're given.

So, if I give the vectors $v, u, w$, the value of $dx \wedge dy \wedge dz(v,u,w)$ should be the volume enclosed by those vectors, and hence $\omega(u, v, w)$ should be an approximation of the charge enclosed. Is this correct?

My only problem is: in this point of view, the form isn't giving me the density, the density itself is being given by a scalar field, while the form gives me the charge instead of the density.

Is this correct? The form is always meant to give the charge instead of the density? The density should always be regarded as a scalar field?

  • 1
    $\begingroup$ Integrating over a volume of a charge density will give you a charge. To get mass you would need the density itself (rather than the charge density). So unless you know some relation between charge and mass, e.g. that only electrons inhabit the space and thus go charge density->electron density->mass density, you don't have the right information. Is it a different kind of density you are expecting? $\endgroup$
    – Phil H
    Commented Mar 11, 2013 at 14:28

2 Answers 2


By 'density' in this case I think you just mean "something on a manifold that can be integrated to give you a scalar". By this definition, on an $n$-manifold, a density would be an $n$-form (since if you integrate over a form of lower dimension you get zero). So in your 3d case, take 3 smooth functions $f,g,h:M^3\to \mathbb{R}$, the form

$df\wedge dg\wedge dh$

is a density. Now, in your example you are integrating over a scalar field multiplied by a 3-form, which is again a 3-form, which can be integrated over a 3-manifold to get you the change in the region. But the scalar field $\rho:M^3 \to\mathbb{R}$ is NOT a density (not a 3-form), so it cannot be integrated over to find the total charge. The charge density is

$\omega=\rho dx\wedge dy\wedge dz,$

and $\rho$ just tells us how 'big' this should be.

In other words, the mathematical term 'density' can be stated as '$n$-form on $M^4$', whereas the colloquial 'density' for 'something per unit length/area/volume' is shorthand for what we really mean ($n$-form).


Expanding on cduston's point, you get this sort of nonintuitive behavior (that the "true" charge density is the pseudoscalar field $\omega$ and not the $\rho$ from Maxwell's equations) as a consequence of our definition of the derivative $\nabla$ (or $d$ or $\delta$ in differential forms parlance). This all goes back to how we originally defined the gradient of a scalar field: as the covector in the direction of greatest increase.

But what if we had a different definition instead? What if we defined the "derivative" of a scalar field as the tangent $(n-1)$-form on a level surface of the scalar field--that is, the Hodge dual of the cotangent one-form? The structure of Maxwell's equations would not be radically changed, but the charge density appearing in those equations would naturally be a 3-form.

Hodge duality lies at the heart of the issue here; in the end, it's not unjustified to gloss over that duality in many cases.

Anyway, the trivector $\omega$ is really the charge density. $\rho$, which is the scalar magnitude of $\omega$, is not really the charge density, even though we often refer to it as such. Some of this terminology may be related to a common piece of mathematical voodoo in which factors of metric determinants get lumped into functions so that curved space volume integrals look like flat space integrals instead. Such functions, with metric determinants lumped into them, are also referred to as "densities".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.