This is the first time I try to apply the calculus of differential forms to make some computation so sorry if I say something very silly. My try was the following: $M=\mathbb{R}^3$, and $E\in\mathfrak{X}(M)$ the electric field due to a point charge on the origin we have using the spherical coordinate system

$$E = \dfrac{q}{4\pi\epsilon_0} \dfrac{1}{r^3} \dfrac{\partial}{\partial r},$$

we want to compute the divergence of $E$ using differential forms. Considering then the metric tensor $g = dr^2+r^2d\theta^2+ r^2\sin^2\phi d\phi^2$ we can associate a $1$-form to $E$. Before doing that we are going to use basis $dr$, $r d\theta$, $r\sin\theta d\phi$ so that the metric tensor has the identity matrix at each point. Now, the form associated with $E$, has components $E_i$ given by

$$E_i = g_{ij}E^j,$$

then since $g_{ij} = 0$ for $i\neq j$, $g_{11}=1$ and since $E^j = 0$ for $j \neq 1$ there remains just $E_1 = E^1$ so that we consider the $1$-form

$$E^\flat = \dfrac{q}{4\pi\epsilon_0} \dfrac{1}{r^3}dr,$$

but this is a $1$-form, since we want divergence we need a $2$-form, so we want to take the hodge star operator. In the basis we are working with it maps $dr\mapsto r^2\sin\theta d\theta\wedge d\phi$ so that we have

$$\ast E^\flat = \dfrac{q}{4\pi \epsilon_0} \dfrac{1}{r^3}r^2\sin\theta d\theta\wedge d\phi$$

now we take the exterior derivative of this which by definition is

$$d(\ast E^\flat) = \dfrac{q}{4\pi \epsilon_0}d\left(\dfrac{1}{r^3}\right)\wedge r^2\sin\theta d\theta\wedge d\phi = \dfrac{q}{4\pi \epsilon_0} \dfrac{-3}{r^2}dr\wedge rd\theta\wedge r\sin\theta d\phi,$$

but this isn't right. We know that the result should be zero. So certainly I made some huge mistake there. What's the problem with this computation?


I see two serious errors off the bat. First, in your formula for the metric, you have $\sin\phi$ where you should have $\sin\theta$. But that was a typo you corrected in the next line. More substantively, your formula for $E$ should have $1/r^2$. Then everything works as it should.

EDIT: Oops, one more. You need $d$ of everything once you have the $2$-form!

  • $\begingroup$ Thanks for your help @TedShifrin! I see, when I picked that basis I already took care of the extra $r$ in the denominator right? Also, I confused myself when applying $d$ and considered $r$ and $r\sin\theta$ as part of the differentials $d\theta$ and $d\phi$. Taking $d$ of everything made it work all fine. Thanks again. $\endgroup$
    – Gold
    Mar 14 '14 at 2:27
  • 1
    $\begingroup$ Yes, $dr$ is dual to $\hat{\mathbf r}$, not $\mathbf r$. $\endgroup$ Mar 14 '14 at 2:36

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