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Textbook wisdom in electromagnetism tells you that there is no total electric charge on a compact manifold. For example, consider space-time of the form $\mathbb{R} \times M_3$ where the first factor is time. One defines the total charge via $Q(M_3) = \int_{M_3} \star j$ where $d\star F = \star j$ is the electric current. If $M_3$ has no boundary (e.g. if it is compact) one can use Stokes' theorem to argue that $$ Q(M_3) = \int_{M_3} d\star F = \int_{\partial M_3} \star F = 0.$$

I wonder what happens for general four-manifolds $M_4$, especially in the case that the third Betti number is zero (otherwise one can simply integrate over a three-cycle). Is there a sensible way to define charge in the above sense? Can one argue that it has to vanish if $\partial M_4 = 0$?

JamalS
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jws
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2 Answers2

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Charge integral makes a point only in that the sum of all charges is zero. The difference to the open $R^3$ is, that there are no counter charges at spacelike $\infty$.

For all practical purposes one can distribute the excess charges outside a large sphere. Faraday and Maxwell could rely simply on the polarizabilty of the walls of their laboratory.

Roland F
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The current 1-form is defined as $J = -\rho \mathrm{d}t + J_x \mathrm{d}x + J_y \mathrm{d}y + J_z \mathrm{d}z$. Its Hodge dual is \begin{align} \star J &= - \rho \star \mathrm{d}t + J_x \star \mathrm{d}x + J_y \star \mathrm{d}y + J_z \star \mathrm{d}z \\ &= \rho \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z - J_x \mathrm{d}t \wedge \mathrm{d}y \wedge \mathrm{d}z + J_y \mathrm{d}t \wedge \mathrm{d}x \wedge \mathrm{d}z - J_z \mathrm{d}t \wedge \mathrm{d}x \wedge \mathrm{d}y. \end{align} Now, if you integrate the current 3-form over a certain region of the space, that is, fixing time at a given $t_o$, then the contributions of terms containing the one-form $\mathrm{d}t$ vanish, hence $$ \int_{M_3} \mathrm{d} \star F = \int_{M_3} \star J = \int_{M_3} \rho \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z = Q(M_3), $$ which is identified as the charge inside the region because you are integrating the charge density over space. Invoking Stokes' theorem, $Q(M_3) = 0$ if $\partial M_3 = \emptyset$. Now, can you can generalize this result to an integral over a spacetime region $(t_o, t_1) \times V$, where $[t_o, t_1] \in \mathbb{R}$ and $V \in \mathbb{R}^3$? Consider the quantity $$ \int_{\mathrm{R}^4} \star J, $$ which now, not only does not equal to the charge but it is not well defined, since you can only integrate a 3-form over a 3-dimensional manifold. To fix the integrability problem, we need a way to get a "current 4-form", which can be obtained by the exterior derivative, $\mathrm{d} \star J$. We are now able to integrate this over a spacetime region, $$ \mathrm{d} \star J = \left( \frac{\partial \rho}{\partial t} + \frac{\partial J_x}{\partial x} + \frac{\partial J_y}{\partial y} + \frac{\partial J_z}{\partial z} \right) \mathrm{d}t \wedge \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z. $$ Notice how the term inside the parentheses is reminiscent of the continuity equation in its vector form, $$ \frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0. $$ We also know that the continuity equation must hold, since we have the Maxwell equation $\mathrm{d} \star F = \star J$, and applying the exterior derivative, which satisfies $\mathrm{d}^2 = 0$, $$ \mathrm{d} (\mathrm{d} \star F) = \mathrm{d}\star J \implies \mathrm{d} \star J = 0. $$ Therefore, for an arbitrary $S\in \mathbb{R}^4$, $$ \int_{S} \mathrm{d} \star J = \int_{\mathbb{R}^4} \mathrm{d} \star J = 0. $$ So, even if this is a bit of a stretch, in the case of 3-space you defined charge conservation for a given time, and in the whole 4-dimensional spacetime you generalize this notion to the continuity equation.

plo
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