I understand that in beta decay, a quark changes its flavor and emits a $W$-Boson, and this $W$ (is the W Boson virtual, by the way?) quickly decays into an electron and an electron antineutrino or positron and neutrino. What makes the $W$-Boson a gauge boson? I don't see any fundamental force being exchanged, and the W seems to act as an intermediate particle.


Processes in elementary particle physics are described by Quantum Field Theory (QFT). In this theory the fields (eletromagnetic force, weak force etc.) are quantized, so they are presented by particles. The attraction, but also repulsion between particles, even decays are represented by the exchange of a gauge boson. Physicists use for this purpose the famous Feynman diagrams (they are very suggestive diagrams, but actually one has to be careful about the interpretation of these diagrams, each part of such a diagram corresponds to mostly complicated mathematical expressions). So the attraction respectively the repulsion of 2 charged particles is represented by the exchange of photons, and the beta decay is represented by the exchange of another boson, the W-boson.

They are called gauge-bosons since their corresponding force respects some local symmetry --- actually it is a freedom (which is called gauge freedom) in the choice of the corresponding potentials of the force -- in case of a photon it is called U(1), and in case of the W-boson it is SU(2).

But, I think the main message to keep here is that forces between elementary particles are described in QFT by the exchange of a particle, actually always bosons (excluding here SUSY). One can make a little computational exercise (you can look it up in the book of Zee, "QFT in a nutshell" in the first chapter) and show that this approach (exchange of a particle) leads to a decrease (increase in case of repulsion) of the energy of in the interaction participating particles, which is nothing else as a felt force.

  • $\begingroup$ I want to sincerely thank you for your answer. Usually people downvote my questions, knitpick my questions and give me answers that I cannot understand. $\endgroup$ – Indigo2003 Jan 24 '18 at 19:24
  • $\begingroup$ @Indigo2003 You are welcome. $\endgroup$ – Frederic Thomas Jan 25 '18 at 13:35
  • $\begingroup$ Two questions though. First, is the W boson exchanged between two particles? I realize that the W is emitted by a quark in the nucleon, but is it received by another particle? In all other fundamental interaction, gauge bosons are exchanged between two particles, i.e. electromagnetic repulsion and gravitation. Also, What is gauge freedom? $\endgroup$ – Indigo2003 Jan 26 '18 at 16:44
  • $\begingroup$ In $\beta$-decay we have $n\rightarrow p + \overline\nu_e + e^-$, in QFT (Feynman diagrams) the anti-neutrino is considered as an incoming (time backwards running, because it's an antiparticle) neutrino which upon absorption of the $W^-$ turns into an electron. Gauge freedom expresses the observation that potentials (electric magnetic e.g.) can be varied to some extent without changing the fields. If you have further questions, please create another post on SE. $\endgroup$ – Frederic Thomas Jan 27 '18 at 21:39
  • $\begingroup$ @Indigo2003 Before Feynman diagrams, there were time-ordered diagrams, where one diagram has the $q$ emit a W that then decayed, while another had a W spawn a virtual $e\nu$ pair and then get absorbed by the quark. All this was possible because energy wasn't conserved at vertices (uncertainty principle). Feynman unified these into 1 diagram that conserved 4-momentum at all vertices, and here the qW vertex and We$\nu$ vertex have space like separation (the virtual W has negative invariant mass squared). The point: don't fixate on time ordering, it's just an "interaction". $\endgroup$ – JEB Feb 14 '18 at 15:21

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