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To add to the other answer: Your intuition is right in a way. The fact that the $W$ and $Z$ bosons are so heavy is the reason for the weakness of the interaction. For example, $\pi^+$ mesons can decay over the weak interaction, the process is described by the following Feynman diagram: According to the Feynman rules, the probability amplitude of such a ...
When e.g. a neutron decays, there is no "real" W-boson inside, in the sense that it could be detected at every point. Instead, the decay of the neutron involves a "virtual" W-boson, a W-boson that only exists for a very short time. Quantum mechanics allows the energy conservation law to be violated by $\Delta E$ for a very short time $\Delta t$ as long as $\... 1 The forces entering the standard model all are symmetric, with zero mass exchange particles at high energies before symmetry breaking. It is one of the beauties of the standard model. Any proposed model for further complexity , preons etc has to predict phenomena which are outside the standard model and at the same time, incorporate the phenomena which ... 10 The problem with "weak charges" is that electroweak symmetry is spontaneously broken. Before the symmetry breaking, electroweak symmetry is described by an$SU(2)_L \times U(1)_Y$gauge group.This amounts to three charges: weak hypercharge$Y$for$U(1)_Y$and weak isospin (total isospin$T$and third component$T_3$) for the$SU(2)_L$. Some examples of ... 7 You're mixing a few things up here. When you say "three" for the strong force, you're counting the number of colors of quarks, but when you guess "three" for the weak force, you're counting the number of force carriers. These are two different things. For example, if you counted the number of gluons (the force carriers for the strong force), you'd get ... 4 I would say two, which is pleasantly consistent with the$SU(2)$structure of the weak force. One is the coupling strength with the$Z$boson, and one is the weak isospin which is raised and lowered by the$W^\pm$. 0 Thanks everyone for lots of good responses - I'm going to summarize them here for future convenience. It looks like there's no consensus on this issue, but here are some takeaways on two variants of my question: If the weak interaction were to suddenly "turn off" with the universe in its current condition, then solar fusion would stop and we'd all be ... 0 Without the weak force, the asymmetry between matter and anti matter, on which we all depend, (as we definitely do not want any anti matter near us), may not have occured in the primordial universe and therefore would not be found in today's universe. This scenario is an extrapolation of the work of Cronin and Fitch,who found that for electically neutral ... 1 I would tentatively say no. It could be a matter of geophysics. Without weak interactions there would be no weak decay of Potassium-40. This means the interior of the Earth would be cold, and as a result there would be no tectonic activity nor would there likely be a strong geomagnetic field. Tectonic activity cycles carbon and other elements. Without the ... 0 It's pure group theory. In the normal electroweak theory, the Yukawa interaction (which is what always produces the fermionic mass terms after the symmetry breaking - after the Higgs gets a vev) is simply (schematically) $$h_2 \cdot L_{L2} \cdot L_{R1}$$ where the Higgs$h_2$and the left-handed leptons$L_L$are doublets (contracted with one another) and ... 1 @Michael Brown is right. The SM has 12 exactly conserved charges. All local invariances, a fortiori also imply global invariances, if you ignore (for the sake of argument) the spacetime variability of transformation parameters/angles. So SU(3) has 8, not 3 conserved charges, RG, BG, .... The group has 8 generators. Likewise, SU(2) has 3, not 2 conserved ... 1 The CKM, PMNS matrices are mathematically absolutely analogous, except that the values and even hierarchies of all the parameters are entirely different in the two cases. (Also, we don't know whether right-handed neutrinos exist and whether the effective Majorana masses may be derived from Dirac masses or something else.) But both matrices may be reduced to ... 0 Following the convention of Peskin & Schroeder- transformation rule of Dirac spinors$\psi(x,t)$under$\cal{C}$and$\cal{P}$are given by, \begin{eqnarray} \cal{C}\psi(t,x)\cal{C}^{\dagger} & = & -i(\bar{\psi}\gamma^{0}\gamma^{2})^{\textbf{T}}\\ \cal{C}\bar{\psi}(t,x)\cal{C}^{\dagger} & = & -i(\gamma^{0}\gamma^{2}\psi)^{\textbf{T}}\\ \... 2 I am answering a question after some clarifications. If the gauge field maps the electron to a positron, it really connects the left-handed 2-spinor and right-handed 2-spinor in the Dirac's electron field into one multiplet. But that field is also a part of the$SU(2)_W\$ doublet with the neutrinos. So the theory you are proposing wants to extend the ...