Explain the word particle flavour

  • Thread starter Thread starter wolram
  • Start date Start date
  • Tags Tags
    Explain Particle
wolram
Gold Member
Dearly Missed
Messages
4,410
Reaction score
555
Can someone please explain the word flavour, when used to describe
types of particle, to a none expert.
 
Physics news on Phys.org
Quarks (which make up all matter, save leptons) come in 6 flavors, which is to say that there are 6 distinguishable types of quarks (each with a different set of properties).

flavor, mass (eV)
up, 5 M
down, 7 M
strange, 150 M
charmed, 1.5 G
top, 176 ± 13 G
bottom, 4.8 G

Also, quarks can decay from one flavor to another.

I'll let the high energy folks lead you through the guts of this.
 
Last edited:
Actually, it's the more massive quarks that decay into the less massive ones. We only find the u and d quarks in nature; all others are produced in high energy collisions in accelerators.
 
Just a little precision, you seems to say that the quark masses are fixed, but they are not well known. Here are the last results (see http://pdg.lbl.gov/2004/listings/qxxx.html) :

u : 1.5 MeV --> 4.0 MeV
d : 4 MeV --> 8 MeV
s : 80 MeV --> 130 MeV
c : 1.15 GeV --> 1.35 GeV
b : 4.1 GeV --> 4.4 GeV (MSbar scheme)
t : 178.1^{+10.4}_{-8.3} GeV
 
By Major
Just a little precision, you seems to say that the quark masses are fixed, but they are not well known. Here are the last results (see http://pdg.lbl.gov/2004/listings/qxxx.html) :

u : 1.5 MeV --> 4.0 MeV
d : 4 MeV --> 8 MeV
s : 80 MeV --> 130 MeV
c : 1.15 GeV --> 1.35 GeV
b : 4.1 GeV --> 4.4 GeV (MSbar scheme)
t : GeV
----------------------------------------------------------------------------------------------------
So from this flavour equates to energy level?
I note that the bottom and top quarks are much more massive than the
others, do these change flavour?
 
Last edited:
The first problem when discussing the mass of quarks is that, since they are confined one can not take a single quark and weight it : so does it really make sens at all to speak about their mass ? So it was agreed that, by mass one should mean "the parameter in the kinetic energy term".

The second problem is again related to the ffact that quarks always stick together : while weighting a baryon you will get not only the mass of the constituents but also the binding energy. In the case of quarks, the mass is very small as compared to the binding energy.

From the above we see that u and d quarks are very light indeed.
s quark is still quite light.
c and b are heavy.
t is badly heavy.

There is a very good approximate symmetry called "chiral symmetry" which would be exact if the quarks were massless. Many results can be obtained by studying how this symmetry is broken. This leads to efficient methods for QCD at low energy, in the non-perturbative regime.
 
Thread 'Why is there such a difference between the total cross-section data? (simulation vs. experiment)'
Well, I'm simulating a neutron-proton scattering phase shift. The equation that I solve numerically is the Phase function method and is $$ \frac{d}{dr}[\delta_{i+1}] = \frac{2\mu}{\hbar^2}\frac{V(r)}{k^2}\sin(kr + \delta_i)$$ ##\delta_i## is the phase shift for triplet and singlet state, ##\mu## is the reduced mass for neutron-proton, ##k=\sqrt{2\mu E_{cm}/\hbar^2}## is the wave number and ##V(r)## is the potential of interaction like Yukawa, Wood-Saxon, Square well potential, etc. I first...
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top