3
$\begingroup$

The mass of an electron includes the mass of the electric field it creates. If electrons have a radius, it's known to be small enough that the mass of the electric field would be more than the total mass, so talking about the mass of just the electron would be negative, and possibly infinite.

The mass of the color charge field generated by a single quark is infinite. This is why it's impossible to have a single quark. So when they talk about the mass of a quark, what does that mean? The best guess I've come up with is using one third the mass of a proton for the mass of an up quark, and subtracting twice that from two up quarks and an X quark to find the mass of an X quark.

$\endgroup$
2
  • 1
    $\begingroup$ I'm no expert but "The mass of an electron includes the mass of the electric field it creates" sounds wrong to me. I think you're thinking of the black-hole model for electrons which would make them super-extremal (a darn good argument for why they aren't black holes). $\endgroup$ Commented Nov 12, 2014 at 6:51
  • $\begingroup$ I'm thinking of the classical electron radius, and the fact that electrons are known to be smaller than that. The mass of the electric field is not just significant. It's orders of magnitude larger than the total. $\endgroup$
    – DanielLC
    Commented Mar 13, 2015 at 1:27

1 Answer 1

6
$\begingroup$

Statements like "The mass of the electron includes the mass of the electric field it generates" have to be taken veeery carefully.

In quantum field theory we can calculate the backreaction of the electric field our electron creates on itself, which is the mass shift I think you talk about. For these calculations to make sense, we need to make use of renormalization.

The issue is that we get a lot of infinities as results. In order to sensibly deal with these, we introduce so-called "counter terms", that have the same structure and the opposite sign. The problem is that we still substract infinities from each other and we don't know the result.

What we do know however, is the electron rest mass as measured in experiments - this is $m_e = 511\, \mathrm{keV}$. So we put this in our calculation: substract from the infinity we calculate another infinity such that for an electron at rest the mass is 511 keV. We can then go on to calculate how this changes for electrons with large speeds (since it does! this is called running mass) and find a great agreement with experiments at particle colliders.

The issue of quark masses is even more complicated. Quarks do not exist freely, only in bound states and we cannot calculate the binding energy precisely. We can measure the mass of heavy quarks by looking at the total energy of their decay products, but for the $u$, $d$ and $s$ quarks this does not work well.

Here, there exist several definitions of mass. For the so-called "current mass" (or "pole mass") one tries to extrapolate the mass for the $u$ and $d$ quark from other measurements to get a result basically for the mass parameter in the lagrangean. On the other hand there are also "constituent masses", where one does what you describe: Take the mass of the proton and divide at amongs two $u$ and one $d$ quark, such that this approximately also fits the masses of the $\pi$ mesons - but this is crude at best, since you include binding energy in the mass definition which has actually very little to do with the mass of the bound particles.


tl;dr

There are two definitions for quark masses. One refers to the mass parameter in the Lagrangean, the other tries to include the binding energy in hadrons. The first is very hard to measure for the light quarks ($u$, $d$, $s$), while the second combines effects into a mass parameter that have nothing to do with each other.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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