According to this Wikipedia page W-bosons are involved in beta decay. According to Wikipedia, this occurs when a down quark turns into an up quark and also emits a W− boson. When this happens in a neutron, which consists of 2 down quarks and one up quark, one quark switches from down to up and you have a proton instead. An electron is also emitted, and this is the beta decay.

This brings me to my question, which is as follows: What makes these quarks change from down to up, i.e. when are the W-bosons emitted?


3 Answers 3


The reason for this is the so-called weak interaction (see Wikipedia - Weak interaction - Interaction types.

With very low probabilities (that's why it is named "weak") it causes reactions for example like these:

$$d \to u + W^- \tag{1}$$ $$u \to d + W^+ \tag{2}$$ $$d + W^+ \to u \tag{3}$$ $$u + W^- \to d \tag{4}$$ $$W^- \to e^- + \overline{\nu}_e \tag{5}$$ $$W^+ \to e^+ + \nu_e \tag{6}$$

The $W$ bosons are very heavy. Their rest mass is $80\ \mathrm{GeV}/c^2$. Therefore they cannot be created as permanently existing particles from reaction (1) or (2). That would violate the conservation of energy (including the rest mass energy $mc^2$).

But according to Heisenberg's uncertainty relation $\Delta E \Delta t \geq \hbar/2$ they are allowed to exist for $\approx 10^{-25}\ \mathrm{s}$ as virtual particles. And then they must disappear again through reaction (3), (4), (5) or (6). Actually this is not even enough time for them to leave the nucleon (proton or neutron) where they have been created, and so they get destroyed within that same nucleon.

Therefore a $W$ creating reaction can only happen combined with a $W$ destroying reaction $\approx 10^{-25}\ \mathrm{s}$ later.
By combining reactions (1) and (5) you get $\beta^-$ decay: $n \to p + e^- + \overline{\nu}_e$
By combining reactions (2) and (6) you get $\beta^+$ decay: $p \to n + e^+ + \nu_e$
By combining reactions (1) and (4) you get no change at all, also by combining (2) and (3).

  • $\begingroup$ But if beta decay is when a down quark turns into a up quark and the W-boson “decays” (don’t know if that’s the correct word) into an electron and an antineutrino, why can’t the up quark just turn into a down quark again immediately? $\endgroup$
    – Melvin
    Aug 28, 2019 at 14:14
  • $\begingroup$ @Melvin Do you mean something like $d \to u + W^- \to d$ ? $\endgroup$ Aug 28, 2019 at 14:18
  • 2
    $\begingroup$ @Melvin Of course this can happen all the time. But you will not notice it because at the end nothing has changed. $\endgroup$ Aug 28, 2019 at 14:21
  • 1
    $\begingroup$ @Melvin No. $\beta^-$ decay creates a $e^-$. $\beta^+$ decay creates a $e^+$. $\endgroup$ Aug 28, 2019 at 14:59
  • 2
    $\begingroup$ @jmh perhaps you meant to say the free proton doesn't "decay" into a neutron due to the neutrons larger mass. Obviously the proton does undergo processes that change it to a neutron (otherwise PET scans wouldn't exist), but those aren't free protons of course. $\endgroup$
    – Triatticus
    Aug 28, 2019 at 20:24

This brings me to my question, which is as follows: What makes these quarks change from down to up, i.e. when are the W-bosons emitted?

Physics is the discipline of finding mathematical models for observations, that not only fit the observations (map them) but also predict for new situations successfully. One says then that the model is validated.

So a basic "axiom" for present day theories of physics is that a system relaxes to the lowest energy level if no conservation laws are violated.

Thus, in the case discussed, the decays of the quarks, the higher mass quarks go to lower mass quarks until the lowest mass is reached, if conservation laws allow it.It is the weak interaction, mediated by the Z and W which is involved in weak decays. The decay follows the probabilistic nature of quantum mechanics (see my answer to a similar question)and the conservation laws of the given decay.

  • $\begingroup$ What is the axiom? Are you referring to statistical mechanics? $\endgroup$
    – innisfree
    Aug 28, 2019 at 16:33
  • $\begingroup$ it appears in all physics frameworks in various principles. In thermodynamics it is called en.wikipedia.org/wiki/Principle_of_minimum_energy .. It is implicit in the various laws used to develop the framework. For statistical mechanics en.wikipedia.org/wiki/… $\endgroup$
    – anna v
    Aug 28, 2019 at 17:41
  • $\begingroup$ Energy is conserved so an isolated system cannot "relax" to a lower energy. What is happening is that the state with a proton, antineutrino and electron is much more probable due to a larger phase space. $\endgroup$
    – my2cts
    Aug 28, 2019 at 19:49
  • $\begingroup$ Yes, I see what you are saying. I'm just not sure I'd say that was the cause of beta decay, which is what the OP asked about $\endgroup$
    – innisfree
    Aug 29, 2019 at 3:19
  • $\begingroup$ @innisfree All I am trying to say is that "if energetically allowed, a system will go to the lowest energy level conservation laws allow it", If there were no weak interaction they would not decay of course, so the existence of the weak interaction changes the conservation laws to be obeyed. $\endgroup$
    – anna v
    Aug 29, 2019 at 5:38

The cause of the decay is that the weak interaction couples the neutron to a system of proton + antineutrino + electron. The latter system has a much larger phase space so by statistics the reaction happens in one direction only. The reverse is vanishingly probable.


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