To produce a quark from universe

  • Thread starter Thread starter Quarlep
  • Start date Start date
  • Tags Tags
    Quark Universe
Quarlep
Messages
257
Reaction score
4
Lets think we have a paticle system which consisting two quarks and they are combined.Then here I learned somehing strange.If we separate this two quarks the universe creates two quarks from universe so first we have 1 quark pairs now we have two. In this process virtual quarks (the matter ones) combined the quarks.Here comes my question Whats happened the other two virtual quarks (the antimatter ones).

Thanks
 
Physics news on Phys.org
Not exactly the universe. They pop out from the vacuum as real particles (on-shell particles) because the available energy is enough.
I don't exactly understand the rest of the question, you had 1 pair of quarks, let's say q \bar{q}
and you move them away (since latex doesn't help, allow me to write \bar{q} as q*)
q---q*
q--------q*
q----[q* q]---q*
(qq*)+(qq*)

In [..] I put the popped up quarks... So what do you mean by matter or antimatter? In the first place you can't have a particle consisting of two quarks. The least are the mesons consisting of q \bar{q} and then are the baryons consisted of q q q In the case you try to move away a quark from a baryon, you'll most probably get a meson out.:
qqq
qq---q
qq------q
qq---[q q*]--q
(qqq) + (q*q)

This happens because of the properties of QCD... In electomagnetism, taking two charges away from each other, the energy density between them gets dispersed and so weakens...In QCD though the picture is different, and it looks like a string (no wonder string theory was first proposed to explain strong interactions). The energy density between q and q* gets concentrated around their distance (more correctly the potential becomes that of a string), and if it gets enough [qq*] can be created.
 
ChrisVer said:
I don't exactly understand the rest of the question
I am sorry about my english I am live in Turkey.
ChrisVer said:
So what do you mean by matter or antimatter?
I mean q and q* as that you have wrote it.
I understand that I didn't ask the question right point of view. Let me clear the question let's suppose we have qq than If we separate them then
q...q
q...(q*q)...q
I want to separate qq but I made a mistake which in universe there's no qq.
This is true isn't it there no qq in universe

Thanks
 
When you want to find allowed quark configurations you have to be able to make color singlets.
The fundamental representation of the SU(3) gauge symmetry (the symmetry of QCD) is the \textbf{3}.
Now out of it, in order to make a singlet you have to use: \textbf{3} \otimes \textbf{3} \otimes \textbf{3}= \textbf{10} \oplus \textbf{8} \oplus \textbf{8} \oplus \textbf{1} which contains a singlet \textbf{1} (color neutral).
Having two quarks (3 colors each) can't give a neutral combination: \textbf{3} \otimes \textbf{3} = \textbf{6} \oplus \textbf{3}.

In addition to the fundamental representation \textbf{3} of SU(3), you can also have the antifundamental repr, \bar{\textbf{3}} which together gives you the adjoint representation and a singlet.
\textbf{3} \otimes \bar{\textbf{3}} = \textbf{8} \oplus \textbf{1}
That's why a quark can be combined with an antiquark to give a color neutral combination.

To make this more illustrative, I will also add that the reason you say that the singlets \textbf{1} are color neutral, is because they transform identically under an SU_{color}(3) transformation (that's why they are neutral)
 
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}...

Similar threads

Replies
6
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
7
Views
6K
Replies
12
Views
2K
Replies
35
Views
8K
Back
Top