I was recently reading a paper https://arxiv.org/abs/0912.0225 which describes Coulomb's law in a closed universe. I've seen the argument from several sources that the total electric charge in a closed universe must be zero (and the same formula has been derived elsewhere); however, I noticed that given the electric field of a test charge at the “North Pole” of the three-sphere:


{Where we are using the fairly standard angular coordinates on a three-sphere universe ( $S^{3}$ with Radius R): $\chi,\theta,\phi$.}

If we Consider someone someone measuring the field of this charge at some distance $d_{0}$ (in say meters) $d_{0}=R\chi_{0}$

Then the electric field is:


Now let us double the radius of the universe, the same observer, the same distance $(d_{0}=2R(\frac{1}{2}\chi_{0}))$ from the same charge now measures the electric field:$$\overrightarrow{E}_{2R}(d_{0})=\frac{q}{4\pi\epsilon_{0}(2R)^{2}sin^{2}(\frac{d_{0}}{2R})}\hat{\chi}$$

Which is already strange in that we might claim (somewhat naively) that the charge appears to be different! But consider now (letting the scale parameter R vary with time as is the case for our physical universe) that this must also be indicative of an induced magnetic field:

$$\overrightarrow{\Delta}\times\overrightarrow{B}=\mu_{0}\epsilon_{0}\frac{\partial\overrightarrow{E}}{\partial t}=\frac{q\mu_{0}}{4\pi}\frac{-1}{[(R^{2}sin^{2}(\frac{d_{0}}{R})]^{2}}\left[2Rsin^{2}(\frac{d_{0}}{R})-R^{2}2sin(\frac{d_{0}}{R})cos(\frac{d_{0}}{R})\frac{d_{0}}{R^{2}}\right]\frac{\partial R}{\partial t}\hat{\chi}$$

$$=\frac{-q\mu_{0}}{2\pi}\left[\frac{1}{R^{3}sin^{2}(\frac{d_{0}}{R})}-\frac{cos(\frac{d_{0}}{R})}{sin^{3}(\frac{d_{0}}{R})}\frac{d_{0}}{R^{4}}\right]\frac{\partial R}{\partial t}\hat{\chi}$$

Where $\frac{\partial R}{\partial t}$ can directly related to the Hubble factor. This seems to indicate that any charge would have an intrinsic magnetic moment in a closed expanding universe.

Note there is no unique orientation for the magnetic moment either, if you were to pick a direction arbitrarily, it would satisfy the above equation.

Finally, it is worth noting that the Pauli matrices are a natural choice for the representation of $S^3$ (since $SU(2)$ is diffeomorphic to the three-sphere).

My question: Would these effects still persist in a more rigorous treatment? Is anyone aware of how to proceed?

I thought it was really curious at the least and was wondering if anyone's seen a more rigorous way to go about this. I found this excellent paper, but wasn't sure how to apply it: http://www.numdam.org/item/AIHPA_1990__53_3_319_0


It occured to me that the same type of effect should also happen in a negatively curved spacetime. Where difference in the initial equation would simply be $sin(\chi)\longrightarrow sinh(\chi)$. I have therefore edited the question title appropriately.

  • $\begingroup$ The integral form of Gauss's law fails in a curved spacetime. There is a good discussion of this in Misner, Thorne, and Wheeler, p. 457. $\endgroup$ – user4552 Jul 20 '17 at 23:46
  • $\begingroup$ @BenCrowell Agreed, it fails utterly at the antipodal point of the charge, though for any physically achievable distance r<<R it does work just fine. one can map $S^3$ to the one point (bohr) compactification of $R^3$ and say it works everywhere except infinity. $\endgroup$ – R. Rankin Jul 20 '17 at 23:55
  • $\begingroup$ @BenCrowell See pg 331 of all Alertz (referenced at the bottom of the question). $\endgroup$ – R. Rankin Jul 21 '17 at 0:13
  • $\begingroup$ See here: wrap.warwick.ac.uk/89265/1/WRAP_Theses_Toala_Enriquez_2016.pdf $\endgroup$ – user97261 Aug 21 '17 at 4:25
  • $\begingroup$ @BenCrowell I was just browsking MTW and the Gauss's law seems to work just fine, you just can't utilize it over the $\it{entire}$ universe. Charge is still defined locally no problemo, and they refer to the non-existence of such a quantity for the entire closed universe as "a happy salvation". I'm going to reword the title to stimulate answers hopefully $\endgroup$ – R. Rankin Jan 4 '19 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.