# Homework Help: Flavour Content of Particles

1. Apr 23, 2010

### Lissajoux

1. The problem statement, all variables and given/known data

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*

2. Relevant 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 $\pm 1$
3. 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.

2. Apr 23, 2010

### nickjer

3. Apr 23, 2010

### Lissajoux

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. $\hat{u}d$?

4. Apr 23, 2010

### nickjer

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

5. Apr 23, 2010

### Lissajoux

OK.

I'm not sure how to explain the results for $\pi^{0}$ and $\hat{\pi}^{0}$, need to have a little explaination 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.

6. Apr 23, 2010

### 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)

7. Apr 23, 2010

### nickjer

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

8. Apr 23, 2010

### 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.

9. Apr 23, 2010

### nickjer

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

10. Apr 24, 2010

### 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 explaination as to why $u$ and $\hat{s}$ are chosen? I can only think becuase 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 $\hat{d}$?

11. Apr 24, 2010

### nickjer

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.

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

You weren't told it had charm or topness.

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.

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.