### Re: Why aren't bosons included in atomic mass estimates?

Date: Fri Dec 3 06:32:31 1999
Posted By: Georg Hager, Grad student, Theoretical Particle Physics
Area of science: Physics
ID: 942273731.Ph
Message:

Greetings!

You ask why heavy vector bosons (which are responsible for mediating the weak nuclear force) do not seem to play any role for atomic masses. The mass of a hydrogen atom is a little bit smaller than 1GeV, much less than the mass of the heavy vector bosons (about 80GeV). So if weak forces were involved in the composition of an atom, one might conclude that its mass should be much larger.

The answer to this apparent paradox is, unfortunately, somewhat more complicated than just "there are no weak forces in an atom"! Ideed, if there were real W and Z bosons present in the atom, it would have a much larger mass. The reason why I have stressed the word `real' here is that in a physicist's view of the subatomic world, there are actually an infinite number of heavy bosons around in the atom, but they do not obey the usual relationship between energy and momentum: they are virtual particles. (For real particles, the square of the relativistic momentum is the square of the mass; virtual particles do not have to obey this rule)

Let me explain this using a simple example: muon decay. The muon is a lepton that is roughly 200 times heavier than the electron. It is not stable, i.e. it decays with a certian average lifetime into a muon neutrino, an electron, and an electron antineutrino. The `innards' of this decay are the following: The muon decays into a W- vector boson and a muon neutrino, and after that the W boson decays into an electron and an electron antineutrino. I.e., there is a heavy vector boson participating in the process, but the mass of the muon is much much smaller. The reason for this is that the W is a virtual particle in this case. It does carry energy (the difference between the muon and the muon neutrino energies), but this energy is smaller than the rest energy of a W boson, which is allowed because of the virtuality.

The weak interaction does indeed not play a role in the hydrogen atom, but imagine a larger nucleus that is prone to beta decay. There is again a W boson participating in beta decay, but due to above reasons it does not `count' for the atomic mass.

Hope that helps,
Georg.

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