# If particles like the $W$ boson can have a range of masses, can quarks and leptons also have a range of masses?

The reason why the weak nuclear force is weak is because the mediators, the $$W$$ and $$Z$$ bosons, need to take on a ridiculously high mass compared to the mass that they are usually found at (which is very rare). This is only possible because of the time-energy uncertainty given by the uncertainty principal.

But if bosons can have varying masses, why not fermions?

Also if all bosons have varying masses (and for some reason that I do not know, fermions do not have varying masses) does that also mean that mesons do not have a fixed mass? But if mesons have a varying mass doesn't that mean that its constituents, quarks, also don't have a fixed mass?

This idea pretty much an over-simplification that is common on physics-for-the-masses youtube videos, which have a habit of (1) taking virtual particle too seriously, (2) over application of the Heisenberg Uncertainty Principle (HUP).

The idea that virtual particles can borrow an energy, $$E$$, from the vacuum, and thus exist for a time:

$$t \propto \frac{\hbar}E$$

is common, and was the basis of Old Fashioned Perturbation Theory. This leads to diagrams of virtual particles with well define time-ordering, momentum conservation, and no energy conservation. These diagrams are not Feynman diagrams.

In Feynman diagrams, energy and momentum is conserved at all vertices, nothing is mysteriously borrowed from the vacuum, and time-ordering is not always possible. What changes is that the exchanged particle has the wrong rest mass:

$$-Q^2 \equiv p^{\mu}p_{\mu} = p^2 = E^2 - p^2 \ne m^2$$

In a scattering experiments, you'll find that $$Q^2 > 0$$, meaning $$m^2>0$$, and the photon appears space-like.

The amplitude of the diagram contains a virtual particle term, called the propagator, which looks something like:

$$G(p) \propto \frac 1 {p^2 - m^2 +i\epsilon}$$

This is maximized when:

$$p^2 = m^2$$

so when you are doing weak interaction far from the $$W$$ or $$Z$$ mass, the amplitude is small. The interpretation that you are exchanging a low mass weak boson is not wrong, but it is not that useful.

Now in the world of real particles, for instance:

$$\gamma p \rightarrow p\pi^-$$

there is a huge increase in cross-section when the invariant mass of the final state is around $$M=1232\,$$MeV. This is resonant production of the delta baryon:

$$\gamma p \rightarrow \Delta^0 \rightarrow p\pi^-$$

The $$\Delta^0$$ has a lifetime of $$\tau = 5\times 10^{-24}\,$$s, and the cross-section follows the Breit-Wigner distribution, with an amplitude:

$$f(E)\propto \frac 1 {(E^2-M^2)+iM\Gamma}$$

which peaks at $$E = M$$, and has FWHM around $$\Gamma$$. Note that:

$$\Gamma = \frac{\hbar}{\tau}$$

So here you are creating real particles with a definite (per event) mass, but with a variation in mass that is related to lifetime. If the intermediate delta baryon has a mass far from $$M$$, then the reaction is much more unlikely. (Note, however, that the Breit-Wigner resonances is a Cauchy distribution, which famously has an unbounded standard deviation, so there is much more amplitude far from the peak relative to what a Gaussian distribution would predict.)

Given the similarity between the real Breit-Wigner resonance amplitude and the virtual particle propagator, you can see how the interpretation arises.

The reason why the weak nuclear force is weak is because the mediators, the W and Z bosons, need to take on a ridiculously high mass compared to the mass that they are usually found at (which is very rare).

The masses of the Bosons are fixed , see here .

This is only possible because of the time-energy uncertainty given by the uncertainty principal.

The uncertainty principle is an envelope of strict mathematical constraints when solving the relevant Quantum Field Theory equations, connected with the behavior of commutators.

But if bosons can have varying masses, why not fermions?

All particles have fixed masses, the "length" of the four momentum describing them. Your are confusing the virtual states within a Feynman integral with real particle four momentum states.

The real particles are the incoming and outgoing ones, that are studied in the measurements and observations. The "length" of their four momentum is the mass in the table.

The intermediate states are virtual. In the above iconic integral the photon exchanged instead of having mass zero, as when it is real, has a variable mass within the limits of integration for the interaction.

The mass of the virtual exchange particles is in the denominator of the propagator for each interaction described. As the integral gives the crossection for the interaction, the bigger the mass the more repressed the diagram, that is what makes weak interactions weaker, but also the coupling of the weak interaction has an equally large if not larger, role.