What Determines the Flavour Content of Elementary Particles?

[Lissajoux]

Homework Statement

To calculate the flavour content of these particles:

\pi^{-},K^{+},\hat{B}^{0},\pi^{0},\hat{\pi}^{0}

*Note that the \hat{} notation should actually be a flat line, which I believe denotes the antiparticle, but I don't know how to do this in latex code. If anyone can let me know for future reference that would be great*

Homework Equations

These are the hints given in the question:

• Pions are the lightest Mesons
  
• Kaons have Strangeness of \pm 1
• B-Particles have Bottomness of \pm 1

The Attempt at a Solution

I'm not sure how to do this.

I know that:

Quarks have flavours of u,d,c,s,t,b. These are abbreviations from Up, Down, Charm, Strange, Top, Bottom. The respective charges are \frac{2}{3},\frac{1}{3},\frac{2}{3},\frac{1}{3},\frac{2}{3},\frac{1}{3}

Leptons have flavours of v_{e},e,v_{\mu},\mu,v_{\tau},\tau and these decrease from lightest to heaviest in this order listed. The respective charges are 0,-1,0,-1,0,-1

Each of the objects must be charge neutral and also colour neutral.

Hence can have either 3 quarks (known as Baryons) or a quark-antiquark pair (known as Mesons). Both Baryons and Mesons are types of Hadrons.

So I've got a bit of the theory, but unsure how to put this together in regards to identifying the flavour content of the particles. A little bit of help and advice would be much appreciated.

 

[nickjer]

You might want to check out:

http://en.wikipedia.org/wiki/Omega_meson

 

[Lissajoux]

Thanks for the link.

So for example on that link, the pion \pi^{+} is listed as having quark content (which is the 'flavour content' right?) of u \ihat{d}. I need it's listed antiparticle \pi^{-} so for this it's flavour content will be the opposite, i.e. \hat{u}d?

 

[nickjer]

Yes, the antiparticle consists of the antiparticles of all the quarks that make it up.

 

[Lissajoux]

OK.

I'm not sure how to explain the results for \pi^{0} and \hat{\pi}^{0}, need to have a little explanation not just state the values and the theory for these two is a bit more complicated it seems just from looking at their flavour contents.

 

[nickjer]

Well for the neutral pion you need 0 charge and lightest possible quarks. For more complicated reasons it is a superposition of u(-u) and d(-d)

 

[nickjer]

You can probably get away with just saying u(-u) for the neutral pion. If it is an introductory course.

 

[Lissajoux]

Yeah it's an introductory course so it's not too in depth particle physics.

So basically for that one its flavour is (u,-u) because it's charge is 0 and flavour must be neutral, right.

 

[nickjer]

That and it is the lightest possible combination you can make.

 

[Lissajoux]

K^{+}=u\hat{s} because kaons have strangeness of \pm1, and since the charge of s is \frac{2}{3} then can balance this with u

.. is this correct? This is what wiki lists it as but I'm slightly confused with the theory and understanding behind the answer.

Alternatively, is it rather that in regards to charge s=-\frac{1}{3} therefore \hat{s}=\frac{1}{3} which combined with u=\frac{2}{3} gives a charge of 0? Unsure if I can just reverse the signs when going between particle and antiparticle.

..but why is it u not c or t? Since these all have the same charge of +\frac{2}{3}. And the same for pairing it with \hat{s} not \hat{d} or \hat{b}

I'm sure there's a simple explanation as to why u and \hat{s} are chosen? I can only think because kaons have strangeness so use s.

In regards to \hat{B}^{0} in a similar way use \hat{b} but again why is this? And why choose to pair it with \hat{d}?

 

[nickjer]

K^{+}=u\hat{s} because kaons have strangeness of \pm1, and since the charge of s is \frac{2}{3} then can balance this with u

The K^+ does not have +/-1 strangeness. It has only +1 strangeness. You were only told that Kaons in general have +/-1 strangeness. But to satisfy the +1 charge, you need an anti-strange quark which gives you strangeness +1. Also the charge of an anti-strange is +1/3 and not +2/3.

Alternatively, is it rather that in regards to charge s=-\frac{1}{3} therefore \hat{s}=\frac{1}{3} which combined with u=\frac{2}{3} gives a charge of 0? Unsure if I can just reverse the signs when going between particle and antiparticle.

1/3+2/3 = +1 and not 0

..but why is it u not c or t? Since these all have the same charge of +\frac{2}{3}. And the same for pairing it with \hat{s} not \hat{d} or \hat{b}

You weren't told it had charm or topness.

I'm sure there's a simple explanation as to why u and \hat{s} are chosen? I can only think because kaons have strangeness so use s.

It doesn't have charm or topness so you chose "u". Also you were told it had strangeness so it must have a strange quark. Then all you do is conserve charge.

In regards to \hat{B}^{0} in a similar way use \hat{b} but again why is this? And why choose to pair it with \hat{d}?

Up and down are the only ones not given crazy quantum number names. You know it must have a bottom quark, so to conserve charge you need a down quark. Just make sure one of the quarks is an antiquark. Unfortunately, I don't have a reason for choosing the bottom quark as an anti-quark. It might just be naming convention.