I've always wondered why the number of protons in the Universe exactly matches the number of electrons. They are such different particles with totally different cross sections. So, first of all, is this true? And, secondly, how could it be that they have been produced in the same amount after the Big-Bang?
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3 Answers
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It has been confirmed true to a high level of precision, since it only takes a tiny unbalance in charge would overwhelm gravity and tear apart the galaxy (or clusters of galaxies on a larger scale). There are more charged particles than simply electrons and protons, but since they account for almost all charged particles the universe, equality of charge is equivalent to equality of electrons and protons.
As to why there is charge conservation: we assume that there was perfect conservation of charge, even with all the crazy high energy physics that happened just after the big bang. Why so? If there wasn't, you could generate perpetual energy by allowing a proton to fall into a deep negative charge potential well, converting it into a negatively charged particle, allowing it to fly away, converting it back, etc. The energy required/released by converting a particle would not depend on the electric potential since conversion would be a local process that knows nothing about the global structure of the electrostatic potential you have set up.
However, what about initial charge distribution? It turns out that in the early universe there was a time of inflation for which the universe expanded without the soup of mass/energy being diluted; the expansion would have diluted any charge imbalance.
Note: the only way in which energy can be created/destroyed is by having the universe expand. However, energy will always be conserved in a closed system of constant size even if the universe expands.
answered Nov 5, 2013 at 16:48
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$\begingroup$ Couldn't an acceleration due to charge imbalance exist up to, and including, the (still unknown) dark-energy acceleration? $\endgroup$ Commented Nov 5, 2013 at 16:52
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$\begingroup$ Not sure, maby charge would have been more concentrated in the past and would have produced a very have acceleration. See adsabs.harvard.edu/full/1979ApJ...227....1B $\endgroup$ Commented Nov 5, 2013 at 16:55
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$\begingroup$ Seems the last sentence (v.1) is self-conflicting. $\endgroup$– Chin YehCommented Nov 5, 2013 at 16:56
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$\begingroup$ Fixed last sentence to make it clearer. $\endgroup$ Commented Nov 5, 2013 at 17:04
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$\begingroup$ I still don't get the last sentence. Does the size mean not the physical size when the universe expands? Or perhaps a separate sentence explaining size would do. $\endgroup$– Chin YehCommented Nov 5, 2013 at 17:13
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There is another conceptual problem of having a non-zero average charge density. If Gauss law for the divergence of the electric field is still valid, the electric field cannot be uniform: $\mathrm{div} \vec{E}=\rho/\epsilon_0$.
It is still possible to have a spherically symmetrical solution $\vec{E}=\vec{r}\rho/\epsilon_0$, i.e. electric field must grow linearly with the distance to the coordinate origin. But the universe then would have to have a center, which, as far as we know, it doesn't.
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$\begingroup$ In cosmology, where is “the origin”? $\endgroup$– rob ♦Commented Oct 12, 2021 at 0:05
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$\begingroup$ There is no origin, that's the problem with having a non-zero charge density. With non-zero charge density there has to be a special point. $\endgroup$ Commented Oct 12, 2021 at 0:06
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$\begingroup$ Gravitational charge (a.k.a. mass) has no such restriction. This seems like a weakness in your model of the electric field, not like evidence against a small nonzero electric charge spread over cosmological distance scales. $\endgroup$– rob ♦Commented Oct 12, 2021 at 0:13
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$\begingroup$ Gravitational field does not follow the same equations as the electromagnetic field. Mass can perfectly have a non-zero density. You just have to pick coordinates associated with the expanding universe. After you picked the coordinates, your electric field is fixed by Maxwell equations in curved coordinates. It still turns out proportional to $\vec{r}$, and there is no solution where the charge is static in the expanding coordinates $\endgroup$ Commented Oct 12, 2021 at 1:21
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$\begingroup$ For a more careful analysis, see this related question. $\endgroup$– rob ♦Commented Oct 12, 2021 at 2:08
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Accepting that long ago there would likely be equal numbers for the two polarities. Once one galaxy ever borrowed electrons from another, then it could coat a protostar with them thus presenting a negative electric field at the center of that host. That field would drive electrons away from atoms there and neutralize naked protons so that they could cluster. Consequential fusion would be devoid of the parity occurring with fusion within plasma. Hence a positron could be lost to converted energy and a durable negative bias could begin.
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$\begingroup$ As far as I understand the question, the question is exactly whether there were equal number of protons and electrons at the very beginning. "Accepting that long ago there would likely be equal numbers for the two polarities" is not a valid assumption in this case. $\endgroup$– GonencCommented Oct 14, 2016 at 14:18
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1$\begingroup$ "Consequential fusion would be devoid of the parity..." - why so? "A positron could be lost to converted energy..." - You seem to suggest that charge is not a conserved quantity. That doesn't sound like main stream science to me. $\endgroup$– FlorisCommented Oct 14, 2016 at 18:07
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$\begingroup$ Emmy noether provided an appropriate clause with her disclosure of conservation of charge. $\endgroup$ Commented Oct 15, 2016 at 7:17
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$\begingroup$ @gonene Please advise how you disagree with the assumption that things started out with no macroscopic electric charge. I was only trying to clarify the possibility of a large growing negative electric charge today! $\endgroup$ Commented Oct 15, 2016 at 7:52