More specifically, what is in the nature of these particles that they carry a charge, and thus repel objects of similar charge and attract those of opposite charge? Also, why would the charge of an electron and proton have the same magnitude??? Could it be that there are infinite universes, and yet only those where these particles have the same charge could atoms form, and allow enough order for an observer to ask these questions?

This question is different from others about the origin of charge in that it is directly querying about the multiverse theory, and whether this is a possible consequence.


Why do fundamental particles have charge?

The answer on the question from the topic which I have is heuristical and it may not coincide with the answer what you want to get (write, if it doesn't coincide).

We know that particles interact with each other through some interaction law. We also know that apriori we may identify them by such quantities: the $i$-th particle has definite mass, spin etc. We know that for given initial set of particles only definite final sets appear; so that there is some selection rule, which may be interpreted as consequence of conservation law of some quantity. We may parametrize this phenomena by giving to the $i$-th particle number $q_{i}$, so that $$ \sum_{i}q_{i}^{\text{initial}} = \sum_{j}q_{j}^{\text{final}} $$ Quantity $q_{i}$ is called charge.

and thus repel objects of similar charge and attract those of opposite charge

This ability (repel or attract similar charged object) strongly depends on the helicity of mediator of interaction and is promoted by the requirement that the time derivative of spatial components of action, which is just the kinetic term must be with the positive sign (one of implications of the unitarity). Suppose the helicity $s$ boson mediator is described by $A_{\mu_{1}...\mu_{s}}$. Hence the kinetic term in the lagrangian is like $$ \partial_{0}A_{i_{1}...i_{s}}\partial^{0}A^{i_{1}...i_{s}} = (-1)^{s}(\partial_{0}A_{i_{1}...i_{s}})^{2} $$ For even helicity, there is plus one factor in the front, while for odd helicity there is minus. We thus need extra minus sign at the front for an odd helicity. Finally, gauge field is present as the propagator when we calculate the energy of interaction of charges. Extra minus sign for odd helicity gauge bosons which comes from the above statement of the unitarity leads to negative sign of interaction energy with $q_{1}q_{2}$, where $q_{1},q_{2}$ are charges. Hence similar charges are repelled for odd helicity mediator.

For the case of electromagnetic field, we have that corresponding gauge boson is helicity one, so EM interaction leads to repelling of similar charges, while even helicity gauge boson (like gravitons) interaction leads to repelling of similar charges (in fact, there can't be gauge bosons with helicity $>2$ since this violates Lorentz invariance).

Also, why would the charge of an electron and proton have the same magnitude???

From one point of view, the reason is that charge is quantized. In fact, you know that quarks carry electric charges and electron carries it too, but they aren't, naively, related to each other. But in fact the requirement that theory which describes their interaction is unitary leads to the statement that these charges are related to each other by some relations (called anomaly matching conditions). So that, by using these conditions you can calculate the charge of $uud$ quarks bound state, which is a proton. And you surprisingly obtain that it is opposite to the electron charge.

From another point of view, the answer on this question are Ward identities in QED.

Suppose we know apriori that the "bare" charges of electron and proton have the same magnitude. The question: why the proton which is surrounded by clouds of virtual mesons, gluons etc., has the same "physical" charge as the electron, which is surrounded by other virtual particles and doesn't interact as hadron? They come from the quantum requirement of gauge invariance and their result is that quantum corrections from the field renormalization (electron or proton, it doesn't matter) completely reduces quantum corrections from the interaction vertex renormalization (which again depends on the field - electron, proton etc). So that the charge doesn't feel the difference between the nature of the particle which carries it.

  • $\begingroup$ This is not exactly what I was looking for, but it answered most of my question, thank you so much! Most of this is a bit too technical for me, but I think I get the gist of what you are saying. I appreciate the answer! $\endgroup$ – Pliam Mar 10 '16 at 17:31
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    $\begingroup$ I corrected a few spelling and grammar mistakes, for instance you kept writing photon instead of proton. But some words like "coincite" in the last paragraph seem to not be the right word, but I didn't know what you intended. That means the meaning is pretty unclear. I also wish you could be more clear about the energy being "positively defined." As I think that expression is also really vague and totally unclear. $\endgroup$ – Timaeus Mar 10 '16 at 17:33
  • $\begingroup$ @Timaeus : "...I also wish you could be more clear about the energy being "positively defined."..." For example, by starting from the Maxwell pure EM lagrangian $$ L = +F_{\mu \nu}F^{\mu \nu}, $$ we don't obtain the energy density which is positive for all $A_{\mu}$. This means that the theory doesn't have the ground state. $\endgroup$ – Name YYY Mar 10 '16 at 18:17
  • $\begingroup$ @NameYYY I thought the answer would be more clear if you went into more details, but it isn't. At the classical level, a Lagrangian of $+L$ or $-L$ generate the same dynamics. And their Hamiltonian's look different at first, but once you realize the momentums of the two Hamiltonians are opposites of each other then you see they have the same dynamics too. So it seems like you need to state that classically it could be different (and classically you don't have a velocity of a mediator) or state a different reason. $\endgroup$ – Timaeus Mar 10 '16 at 19:58

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