Consistency of existence of point charges and energy in fields

Question

In Feynman lectures, Volume 2 chapter 8 (https://www.feynmanlectures.caltech.edu/II_08.html#Ch1-audio) at the very end Feynman remarks

We must conclude that the idea of locating the energy in the field is inconsistent with the assumption of the existence of point charges. One way out of the difficulty would be to say that elementary charges, such as an electron, are not points but are really small distributions of charge

His reasoning for this was $$U=(\epsilon_0/2)\int E^2 dv=\int_0^\infty\frac{q^2}{8\pi r^2\epsilon_0}dr=\infty$$

One may think to calculate the self energy of a point particle by the product of the charge of this particle times the potential caused by this charge itself at its own position that is $r=0$, thus we get $$U=Kq²/0=2\infty$$ Here $K =1/4\pi\epsilon_0$ and the factor of 2 signifies that this is double of the Feynman's calculation.

However this approach is fundamentally flawed which one can see as soon as they realize that a particle cannot apply a force on itself.

But instead of this we can think of a point particle to be made of many infinitesimal charges, $dq$, then we can bring back our potential energy model since here any charge $dq$ would be experiencing force from all the other particles and not itself (though this "self force" is included in the integral below but is of no consequence since it is of second order), thus we can write $$U=(1/2)\int\phi dq = q\phi/2 = \infty$$

Where $\phi$ is the potential of the charge ($\phi=Kq/r$) and is pulled out of the integral since it is constant over an infinitesimal volume we are considering. This is the same result as Feynman's without using the model in which energy is stored in fields (well electric field, that is one can always argue that it's stored in potential field, but then the argument becomes redundant).

So what is Feynman trying to imply/justify here since using both models we get inifinity in the end, how can infinity of electric field model be enough to discredit it?

Note: Though trivial and obvious it is important to realize that $dq≠q $, where $q$ is charge of point particle and that energy is always $+\infty$ no matter the sign of charge.

Answer 1

So what is Feynman trying to imply/justify here?

Exactly what he says:

the idea of locating the energy in the field is inconsistent with the assumption of the existence of point charges.

And your calculation of the energy for "many infinitesimal charges" is incorrect, as the potential will be different. You should have something like a double integral, and the result will not equal infinity.

Answer 2

So what is Feynman trying to imply/justify here since using both models we get inifinity in the end, how can infinity of electric field model be enough to discredit it?

It is not about the electric field model being discredited, but the assumption of point charges or the model validity on small length scales, as obvious from the full quote.

We must conclude that the idea of locating the energy in the field is inconsistent with the assumption of the existence of point charges. One way out of the difficulty would be to say that elementary charges, such as an electron, are not points but are really small distributions of charge. Alternatively, we could say that there is something wrong in our theory of electricity at very small distances, or with the idea of the local conservation of energy. There are difficulties with either point of view. These difficulties have never been overcome; they exist to this day. Sometime later, when we have discussed some additional ideas, such as the momentum in an electromagnetic field, we will give a more complete account of these fundamental difficulties in our understanding of nature.

Infinite energy in problematic per se because once you are interested in work performed in a process, an energy difference, you are doomed when working with non-finite values, you can end up with meaningless stuff like $\infty - \infty$. That should be enough to disregard a model.

Answer 3

Of course, electrons are not point charges. We treat them as such because we have no measuring instruments to determine the diameter. We even get this quite inaccurately for the radius of action. This is because every other subatomic particle as well as photons interact with the electron and it is difficult to use smaller measuring instruments.

Moreover, we are not yet interested in the structure of the electric and magnetic fields of subatomic particles and therefore cannot imagine an internal structure of the particles together with their fields. The whole discussion is questionable as long as we only discuss point-like or extended charges instead of the structuring of their fields.

If someone - because of our ignorance, because of convenient calculations or because it doesn't matter - assumes a point source, that's perfectly fine. At the same time, it should be a common consensus that a source of fields and mass cannot be a geometric point.

Answer 4

I take him to mean that starting out the derivation of the expressions for electrostatic energy with the expression $\frac{q_1q_2}{4\pi \epsilon_0 r_{12}}$, i.e. electrostatic energy equals the work required to bring $two$ charges together, from this and the principle of superposition, you can get $U = \frac{1}{2}\int \rho \,\phi \,dV$, where $\phi$ at a point is made up of the contributions from all other charges than the charge $\rho \, dV$ at that point. (So far in his derivation, even in this expression, there is only energy $between$ charges, and an isolated point charge in the integral therefore does not make sense). It could have ended there with this expression (with energy maybe located where the charges are, i.e. $\phi \, \rho$) and point charges would not give us any headache. But because of the additional fact that we also have that (numericaly at the least) $U = \frac{\epsilon_0}{2}\int E \cdot E \,dV$, and that we can interpret this as there being an energy density $\frac{\epsilon_0}{2}E^2$ associated with the field, and because of the fact that this expression is $actually$ the correct distribution of enery (which is consistent with, but cannot be deduced from electrostatics), because of this it follows that even a single point charge, with no other charges around, will also have energy in its field (so the energy isnt only $between$ charges anymore), namely with density $\frac{q^2}{32\pi^2 \epsilon_0 r^4} $ at a distance r (and the integral of this diverges).