Does the $\frac{4}{3}$ problem of classical electromagnetism remain in quantum mechanics?

Question

In Volume II Chapter $28$ of the Feymann Lectures on Physics, Feynman discusses the infamous $\frac43$ problem of classical electromagnetism. Suppose you have a charged particle of radius $a$ and charge $q$ (uniformly distributed on the surface). If you integrate the energy density of the electromagnetic field over all space outside the particle, you'll get the total electromagnetic energy, which is an expression proportional to $c^2$. The energy divided by $c^2$ is what we usually call the mass, so if we calculate the "electromagnetic mass" in this manner we'll get $m = \frac{1}{2}\frac{1}{4\pi\epsilon_0}\frac{q^2}{ac^2}$. If, on the other hand, you took the momentum density of the electromagnetic field and integrated it over all space outside the particle, you'd get the total electromagnetic momentum, which turns out (for $v<<c$) to be proportional to the velocity of the particle. The constant of proportionality of momentum and velocity is what we call mass, so if we calculated the electromagnetic mass in this way we would get $m = \frac{2}{3}\frac{1}{4\pi\epsilon_0}\frac{q^2}{ac^2}$, which is $\frac{4}{3}$ times the value we got before! That is the $\frac43$ problem.

Feynman claims that this fundamental issue remains when we move to quantum electrodynamics. Was he right, and if so has the situation changed since the $1960's$ when he was writing? I've seen claims on the Internet (I don't have the links) that the $\frac43$ problem is still there in QED, but instead of $\frac{4}{3}$ the coefficient is something closer to $1.$ Is that true, and if so what's the coefficient? All of this is of course related to issues of self-energy and renormalization.

Any help would be greatly appreciated.

Answer 1

Problem outline

In the cited reference, Feynman starts his argument by stating that the energy for an inertial sphere of radius $a$ and uniform charge $q$ is given by

$$ U_{elec} = \int_{\mathbb{R}^3} dV \, u_{elec} = \int_{\mathbb{R}^3} dV \, \left( \frac{\epsilon_0 E}{2} \right) = \frac{1}{2a} \frac{q^2}{4 \pi \epsilon_0} \equiv \frac{e^2}{2a} \, \qquad \qquad(*) $$

where $E$ and $u_{elec}$ are its electric field and density and the last equation defines the symbol $e^2$. If we now consider the sphere to be moving with constant speed $v \ll c$, its momentum density is obtained from the Poynting vector:

$$ \vec{p} = \int_{\mathbb{R}^3} dV \vec{S} = \int_{\mathbb{R}^3} dV \left( \epsilon_0 \vec{E} \times \! \vec{B} \right) = \left( \frac{2}{3} \frac{e^2}{a c^2} \right) \vec{v} \equiv m_e \vec{v} \, , \qquad \qquad(**)$$

where the last equation defines the electromagnetic mass $m_e$. We can also associate a second electric mass $m_{e}'$ to the field $U_{elec}$ by using special relativity (SR):

$$ E = m c^2 \Longrightarrow \frac{U_{elec}}{c^2} = m_{e}' \Longleftrightarrow \frac{e^2}{2a c^2} = m_{e}' = \frac{3}{4} m_{e} \Longrightarrow U_{elec} = \frac{4}{3} m_{e} c^2 \, , $$

that is, the "relativistic" mass $m_e'$ is not the same as the "electrical" mass $m_e$, the absurd being explicit in the above equations.

The loophole

Do we have the right of invoking SR in a derivation's single step? Although we used Einstein's famous formula to arrive at the $\frac{4}{3}$ problem, notice that we have not used SR earlier: $(*)$ and $(**)$ are not even Lorentz invariant! If we adapt the calculations to include SR, $U_{elec}$ and $\vec{p}$ transform as

\begin{align} U_{elec}' &= \gamma \left( U_{elec} + \vec{p} \cdot \vec{v} \right) \\ \vec{p}' &= \gamma \left( \vec{p} + \dfrac{U_{elec} \vec{v}}{c^2} \right) + (\gamma-1)\vec{v} \times \left( \vec{v} \times \vec{p} \right) \, . \end{align}

That is, for a particle at rest we have $\vec{p} = 0$ and, introducing velocity in a relativistically meaningful way by changing into a moving reference frame with velocity $\vec{v}$,

\begin{align} U_{elec}' &= \gamma \, U_{elec} = \gamma \left( \frac{e^2}{2a} \right) = \gamma m_e' c^2 \\[8pt] \vec{p}' &= \gamma \left( \dfrac{U_{elec}}{c^2} \right) \vec{v} = \gamma \left( \frac{e^2}{2ac^2} \right) \vec{v} = \gamma m_e' \vec{v} \, , \end{align}

where now, lo and behold, everything fits and there is absolutely no $\frac{4}{3}$ paradox. It was just a matter of moving between reference frames correctly.

What did Feynman mean?

Being one of the most brilliant physicists in history, Richard Feynman would never misuse SR as necessary to arrive at the $\frac{4}{3}$ paradox. In fact, Abraham and Lorentz, the discoverers, only did it because they didn't really know how to transform between reference frames, since SR had not been finished at the time. Let us then quote what Feynman really said:

There are difficulties associated with the ideas of Maxwell’s theory which are not solved by and not directly associated with quantum mechanics. You may say, “Perhaps there’s no use worrying about these difficulties. Since the quantum mechanics is going to change the laws of electrodynamics, we should wait to see what difficulties there are after the modification.” However, when electromagnetism is joined to quantum mechanics, the difficulties remain.

Indeed, the $\frac{4}{3}$ problem is solved not by quantum mechanics (QM), but by the careful use of SR. Abraham and Lorentz used $(*)$ and $(**)$ indiscriminately because they didn't know moving electrons are deformed into ellipsoids according to SR. The beautiful article by Rohrlich, of which this answer is a resume, shows that the error in neglecting the electron's deformation amounts to exactly $-\frac{1}{3} m_e' c^2$, which would correct their result and also get rid of the $\frac{4}{3}$ problem. I believe Feynman was talking about other problems that migrate from classical to quantum electrodynamics (QED), not this apparent paradox. The problem of self-energy, for instance, requires the electron in QED to have either a rather non-intuitive internal structure or interact with stuff that has negative mass, otherwise it would explode (apart from renormalization). This problem migrates from classical electrodynamics erroneously due to the postulation of a "Poincaré stress" that would not only stabilize the electron, but also correct the $\frac{1}{3}$ factor missing in the $\frac{4}{3}$ problem. Fortunately, many physicists helped to dissociate the problem of electronic stability from the $\frac{4}{3}$ problem, which is not really a problem, since it doesn't actually show up in QED. The problem of the electron's internal structure is, however, pretty real.

Answer 2

In the modern view of quantum field theory, the $4/3$ problem, as well as related questions about point charges, are issues of regularization.

The problem is that you want to deal with a singular quantity, namely the energy and momentum of an ideal point charge, but it is impossible to say anything definite because all these quantities are divergent. In order to get a finite answer, you modify the system by regularizing it in some way, so the answers come out finite. Generically, regularization can and will break symmetries, and we generally try our best to choose a regularization scheme which preserves the symmetries we care about most.

The philosophical reason this works is because we know that our theories are only valid up to a certain cutoff scale $\Lambda$, above which new phenomena appear. Below the cutoff, the effects of these new phenomena can be parametrized in terms of an effective Lagrangian. Hence we can get the same low-energy predictions in any regularization scheme that yields the same effective Lagrangian, even if these schemes put dramatically different physics above the cutoff scale.

In this case, Feynman is regularizing the classical theory by cutting off the field integrals at $\Lambda \sim 1/a$. Unfortunately, this regularization scheme does not preserve Lorentz invariance, and furthermore Lorentz invariance isn't restored even in the continuum limit $\Lambda \to \infty$. This isn't an uncommon thing. For instance, lattice QCD has to address this issue as well, because a lattice regularization clearly breaks Lorentz invariance.

The reason that lattice QCD works anyway, in the RG language, is that their systems are engineered so that only extra irrelevant operators can be produced, whose effects vanish in the continuum limit $a \to 0$. Intuitively, the problem with Feynman's regularization is that it allows extra relevant operators, which mess up the effective Lagrangian no matter what $a$ is, though this language isn't exact since Feynman isn't dealing with a quantum field theory.

The modern resolution of the $4/3$ problem in quantum field theory is essentially by fiat. An unregularized theory is ill-defined; we cannot speak of unregularized QED because such a theory doesn't even exist. Instead, we need a regulator in mind from the start. Since we know Lorentz invariance is a symmetry of our world, we choose a regulator which recovers Lorentz invariance in the continuum limit.

Answer 3

Apart from H. Poincaré, the first physicist who made correct relativistic "derivation" was E. Fermi.

Here I would like to focus on the physical effect of the famous self-action: it is simply a self-induction that "slows down" or "resists" to any acceleration (change in time of a constant current). Aparently it is not a desired effect - it is not a small radiative resistance at all, whatever size of electron is.

Note, the EMF "moves" fine with the electron - according to the Maxwell equation, so there is no need to take it into account "again" in the electron (mechanical) equations. The radiative resistance must be guessed from different physical ideas rather than from "self-action".

In QED this problem stays and it is "solved" by modifying the (wrong) equation solutions rather than by modifying the wrong equations.

My research results are still in an embrionic state, but I proceed from another physical idea that catches the radiative resistance well.

Answer 4

Regarding the problem described In Volume II Chapter 28 of the Feynman Lectures on Physics, the 4/3 problem reason is that one of the terms that affect the motion of the particle is not taken into account. That term is the destruction of the field energy in the front of the charged sphere and its creation in the back. While the overall field energy is not changing, if you evaluate the energy locally, you can see that in the front, the v·E term means the energy is being taken from the field to the particle and in the back, the opposite happens, since this happens all the time its effect is equivalent to a backward movement. When you add it to the normal movement caused by the momentum, you fin that you require a mass m_e = (1/2)e^2/a and there is no more inconsistences.

I placed a small document on the topic here: https://www.slideshare.net/SergioPL81/adding-a-shift-term-to-solve-the-43-problem-in-classical-electrodinamics