The self energy of the electron can be represented in two ways:

  1. the energy required to bring a charge distribution from infinity to the size of the electron (assuming it is a point charge with no other structure)
  2. the work required for the electron to move against its own electric field.

In the second instance, the action of accelerating the electron against its own field could define its mass (also assuming the problems of infinities are eliminated by the vacuum polarization at the quantum level). There does not need to be interaction with any other particles in that the action against its own field produces photons.

Why would a separate Higgs field be needed if the mass could be defined by the electron's own electric field?

  • $\begingroup$ Your are talking of an isolated electron but then say "accelerating the electron against its own field" . What will induce the acceleration if not another field/paraticle $\endgroup$ – anna v Aug 7 '15 at 5:14
  • $\begingroup$ gravity will accelerate an electron, $\endgroup$ – Peter R Aug 7 '15 at 17:19
  • $\begingroup$ Right, so there will be an interaction if there exists a gravitational field, i.e. Feynman diagram. $\endgroup$ – anna v Aug 7 '15 at 17:21
  • $\begingroup$ The standard model and the Higgs field don't deal with gravity. $\endgroup$ – Peter R Aug 7 '15 at 18:44
  • 1
    $\begingroup$ The standard model is not written in stone. Effective quantization of gravitation is already used in cosmological models , and there is continuous work in theories unifying all four fources/interactions. When talking of elementary particles, point particles as far as we can establish experimentally, one has to use quantum mechanics/QFT. Anyhow the acceleration you envisage would break the law of energy conservation, where would it take its energy from to accelerate? $\endgroup$ – anna v Aug 8 '15 at 4:00

The mass of a fundamental particle turns out to be quite an elusive concept, because massless particles act as a source of gravity and they carry momentum. What then is special about mass?

Where mass comes in is in explaining the relationship between the total energy of a particle and its momentum. For any particle we have the expression for the total energy:

$$ E^2 = p^2 c^2 + m^2 c^4 \tag{1} $$

where $m$ is a parameter that we call the rest mass. It's the parameter $m$ that the Higgs mechanism gives us.

I can't see any way that either of the options you mention could usefully describe the value of the parameter $m$ in equation (1). For example, why would the electron, muon and tau have such different rest masses when they are all (as far as we know) point particles with the same charge?

  • $\begingroup$ You say $m$ is what the Higgs mechanism gives us, but what about binding energy? Seems to me I read that the mass of a proton or neutron could be explained simply as the binding energy of the quarks it contains, without invoking Higgs. Maybe you can clarify that? $\endgroup$ – Mike Dunlavey Aug 6 '15 at 17:52
  • $\begingroup$ @MikeDunlavey: my answer applies only to fundamental particles not composite ones. I should probably edit the answer to make this clear ... $\endgroup$ – John Rennie Aug 6 '15 at 18:53
  • $\begingroup$ Thanks. That was a major confusion for me, and maybe others. $\endgroup$ – Mike Dunlavey Aug 6 '15 at 21:30
  • $\begingroup$ I was referring to inertial mass and the "resistance" experienced by an electron working against its own electric field. As far as I know there are no theories out there that can derive the principle of equivalence. $\endgroup$ – Peter R Aug 7 '15 at 17:24

A classical "shell" of charge of radius $R$ will appear to have a field, outside the shell, corresponding to the charge all being at the center of the shell. If the charge on the shell is $q$ then a charge $dq$ being pulled in from infinity will cost an energy $k_e~q~dq / R$; without loss of generality this is the same if $dq$ is spread over an infinitely large sphere that contracts its surface area as it comes inward. Integrating from 0 to Q gives a self-energy of $\frac 12 k_e Q^2 / R$. There used to indeed be a model where the so-called "classical electron radius" $R_e$ was chosen to make this value equal to $m_ec^2$ when $Q = -e$.

The answer you get is well-known and is approximately 3 femtometers (really 2.81794...), where a femtometer is a quadrillionth (10-15) of a meter.

What's the problem? To quote a song by Hank Green, "A quark is a fundamental constituent of matter observed in 1968 through deep inelastic scatter." We were able to do these experiments colliding really high-speed electrons against protons, and there's tons of information online under that title, "deep inelastic scattering", which showed that protons are made up of these three charge-carrying particles called quarks, two with charge $+2/3$ and one with charge $-1/3$.

The interaction of those quarks sets a well-known size for the proton radius, which has been measured as $0.877 \pm 0.005~\text{fm}.$ So, paradoxically, even though we saw these effects, the classical size of the electron is $3.21 \pm 0.02$ times larger than the size of the proton. So if you think of the electron as truly having this fixed size, you cannot readily explain how it was able to probe the structure of something one third of its size.

In fact, at this point there is no good experiment which shows an electron to have any size at all, so its self-energy should by all rights be infinite. Quantum theory almost saves you from this infinity by a rapid shaking called Zitterbewegung that oscillates with the Compton wavelength of the electron, which is huge compared to the sizes we were talking about just now (2426 fm!) but even when you've done the necessary quantum "renormalization" you still have to stop at some distance scale because the quantum theory is still mathematically divergent at small radii, albeit logarithmically.

We don't have any reason to believe that electrons have an internal structure at any particular length scale, but we cannot deal with the self-energies if they are point particles, and we really kind of don't know what the final solution will look like.


I would say (almost) yes... If you are familiar with QFT, then you probably know that the mass $m$ of a given field, say a scalar $\phi$, is introduced via the square term $\frac{m^2}{2} \phi^2$ in the Lagrangian. This however comes at the expense of some of the symmetry of the theory. In particular, if you naively apply this idea within the Standard Model, you will mix Right-Hand with Left-Hand fermions (of $SU(2)$), which "breaks" (explicitly) the EW symmetry. For this reason, unless you don't mind a theory lacking $SU(2)$ symmetry, you need a Higgs field (or another source of spontaneous EW-symmetry breaking)...


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.