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25

Mesons are not elementary, they are composed of quarks. So take a look at their quark content. The charmed eta meson consists of a charm and an anti-charm quark, denoted $c\overline{c}$. An anti charmed eta meson would therefore be an anti-charm and an anti-anti-charm (which is just a charm) quark, i.e. $\overline{c}c$, which is obviously the same as ...


8

Electroweak theory told us where to look for the $W$ and $Z$ gauge bosons. For the Higgs, its mass is a free parameter, hence we didn't know where to look for it. Once you start to look in many places for a particle, you also have to factor in the look-elsewhere effect, which basically means that the more places you look for a particle, the higher the ...


7

The isospin is different. $I=0$ for the $\Lambda^0$ and $I=1$ for the $\Sigma^{0}$. This makes the $\Lambda^0$ an isospin singlet state but the $\Sigma^0$ is part of an isospin triplet. There are quite few other examples e.g. compare a proton (uud with $I=1/2$) with a $\Delta^{+}$ (uud with $I=3/2$).


4

Finally, I am now able to provide an answer to my question (copied from this thesis). The weak charged current interaction is described by the gauge field $W_\mu^\pm$ through the interaction Lagrangian term: \begin{equation} \mathcal{L}_I = -\frac{g}{\sqrt{2}} (\overline{u}_L, \overline{c}_L, \overline{t}_L) \gamma^\mu {W_\mu}^- V_\text{CKM} ...


3

The situation is not symmetric at all: This diagram describes a force between two fermions, but a diagram such as just doesn't exist (in the Standard Model). Fermions can in fact mediate a force between bosons, like in: Such diagrams are highly suppressed loop diagrams though, and the one above would after renormalization be seen as just one ...


3

A similar question is the following. How can $\pi^0$ and $\eta$ in the $SU(3)_F$ meson octet both have the same $SU(3)_F$ flavor content? One could answer that this is because $\pi^0$ is part of an isospin triplet of pions with $I=1$, while $\eta$ is an isospin singlet with $I=0$. Or one may point out that their explicit ket linear combinations of ...


3

[The figure shown in the OP question above ...] is experimental data for the ratio $R = $ [...] as a function of the centre of mass energy $\sqrt{s}$ The so called cross section ratio $$R[~\sqrt{s}~] = \frac{\sigma^{(0)}[~e^+~e^- \rightarrow \text{hadrons}, \sqrt{s}~]}{\sigma^{(0)}[~e^+~e^- \rightarrow \mu^+~\mu^-, \sqrt{s}~]}$$ ?? Actually, ...


2

Is there any possible way to extract all the mass of a quark?probably under extreme Gs or heat or pressure? Not with classical means, like Gs and heat and pressure, because the quarks are elementary particles held together within hadrons because of QCD . Quarks can give up all their energy when quark meets antiquark, as happens in proton antiproton ...


2

In general an atom undergoing fission breaks up into other, smaller nuclei and stable particles: photons (x rays and gamma rays), electrons(beta decays) , alpha particles an other lower mass nuclei. Gluons are never free, because of QCD, and always inside a proton or a neutron. Their spill over attractive force is due to virtual gluons which will never ...


2

I think that this is more about the historical construction of the theory than about the actual interactions. In a lagrangian, two fields A, B interact when there is a product term of both such as AB. So, I see no real fundamental distinction there, even with more complicated expressions. But when one introduces the interaction bosons, it's by the mean of ...


1

The fermion masses result from Yukawa interactions after EWSB: $$ m_f = \frac1{\sqrt 2} y_f v $$ Thus the Yukawa couplings govern lepton and quark masses. Of course the masses should be diagonalized. For the leptons, as there no neutrino masses in the SM, the lepton interactions and masses can be simultaneously diagonlized, whereas differences in up- and ...


1

Defining precisely what are all the quantum numbers is a difficult question because it depends highly on the model under consideration, even for the standard model. In particular any U(1) symmetry leads to a quantum number, and similarly some U(1) subgroup of non-abelian groups that commute with all other interactions can also be associated to quantum ...



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