# Electric Field One-Form

I know for instance that we can interpret the electric field as the one-form that given a vector gives the change in potential in the direction of the vector, however I'm very unsure about how to write this down. My point is, in the usual definition of electric field we have a charge density $\rho : U \subset \mathbb{R}^3 \to \mathbb{R}$ and we define the electric field to be given by:

$$E(a)=\frac{1}{4\pi\epsilon_0}\iiint_{U}\rho(a') \frac{a-a'}{\left|a-a'\right|^3}dV'$$

Where $a$ is the point of the field and $a'$ is the source point. My problems here are: when we introduce differential forms, the charge density becomes a $3$-form given by $\mathcal{Q}=\rho \ dV$, where $dV$ is the volume $3$-form. There rises the first problem: we are integrating a $3$-form times a vector $a - a'$, which is a little confusing to me.

Also, if we just forget about this detail and use this definition, I'm not sure how to convert this into a differential one-form.

Can someone explain the question of "integrating a form times a vector" and point out how can we proceed in this case ? To get the electric field expresses as a one form ?

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Related: physics.stackexchange.com/q/575/2451 Related question by OP: physics.stackexchange.com/q/54912/2451 –  Qmechanic Apr 2 '13 at 21:08

I think the answer here lies in Hodge duality. I'll present this answer using geometric calculus, which subsumes differential forms.

When you construct the integral with the Green's function $G(a) = a/4\pi a^2$, there's an inherent use of Hodge duality (which I will represent using geometric algebra, by multiplication with the pseudoscalar $i$).

$$iE(a) = \int \frac{\rho(a')}{\epsilon_0} G(a-a') i |dV'|$$

Writing $\rho$ as a 3-covector field lumps the $i$ in the integral into $\rho$, like so:

$$E(a) = i^{-1} \int \left( \frac{[i\rho(a')]}{\epsilon_0} G(a-a') \right) |dV'|$$

So ultimately, you need this $i^{-1}$ to walk back to an electric covector field. In differential forms literature, multiplication by $i$ or $i^{-1}$ is represented by $*$, usually. Otherwise, the term in large brackets is a 2-vector field.

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I feel like this problem disappears if your were to start from the electric potential as a 0-form that reads:

$V(a) = \iiint da' \frac{\rho(a')}{|a-a'|}dV'$

The electric field can them be defined via the external derivative of the potential:

$E(a) \equiv -d_aV(a)$

I am not a mathematician but it seems to me that the form in $a-$space should not be confused with the form in $a'-$space. There must be some kind of tensor product between these two spaces but nothing more harmful than that.

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