Stokes theorem for smooth differential forms is well-known. If $\alpha$ is a smooth differential $n$-form defined on an $(n+1)$-dimensional compact oriented manifold with boundary, then we have $$\int_Md\alpha=\int_{\partial M}\alpha.$$

In physics, one considers a generalized form of this equation, where the manifold need not be compact, and the differential form need not be smooth everywhere; it may have poles.

For example, in electrostatics one sees a 2-form field generated by a point charge $Q$ $$E=\dfrac{Q}{4\pi\epsilon_0r^3}\left(x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy\right)=\dfrac{Q}{4\pi\epsilon_0r^2}d\Omega_{S^2},$$ where $d\Omega_{S^2}$ is the volume form of the sphere. I suppose this must be understood as a distribution-valued form, because the physicist computes


where $\delta^3$ is the Dirac delta function in $\mathbb{R}^3$. Perhaps this is a weak exterior derivative?

The physics discussions I have seen justify this computation by saying it is necessary to make the Gauss's law hold. But where I come from, we don't conspire to ensure theorems hold because we like them, but rather prove from the definitions that theorems follow logically. Could you say how this theory of distribution valued differential forms looks, and state a version of Stokes' theorem for it? I would also accept a reference.

  • $\begingroup$ You can just puncture at the point charge instead of directly considering poles. This means you only have to worry about noncompactness and not directly about poles. (Among other things, this makes it clearer from the mathematical point of view why one would expect interesting 2-forms; it's because $\mathbb{R}^3$ minus a point is homotopy equivalent to $S^2$.) $\endgroup$ – Qiaochu Yuan Oct 23 '17 at 17:58
  • $\begingroup$ @QiaochuYuan: true, but doesn't it seem like we're missing something that way? The Cauchy-like relationship between the "residue" of $E=Q/r^2$ and the flux integral $\int_{S^2}E$ is gone. If you remove the pole, then we model only the flux, not the charges. If we define charge to be flux, then Gauss's law becomes a tautology. $\endgroup$ – ziggurism Oct 23 '17 at 18:18
  • $\begingroup$ Gauss's law doesn't become a tautology: it's still a nontrivial fact that the integral is the same no matter the size of the sphere, and so forth. In any case, if you know the value of $E$ at every point except the origin then you've completely determined it ("by continuity"), e.g. if we have $E' = \frac{Q'}{r^2}$ then $E = E'$ away from the origin iff $Q = Q'$. $\endgroup$ – Qiaochu Yuan Oct 23 '17 at 18:21
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    $\begingroup$ Perhaps it's a current? $\endgroup$ – md2perpe Oct 23 '17 at 21:21
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    $\begingroup$ See, for example, the discussion in this post. $\endgroup$ – Ted Shifrin Oct 24 '17 at 16:55

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