What Determines the Flavour Content of Elementary Particles?

[Lissajoux]

Homework Statement

To calculate the flavour content of these particles:

[tex]\pi^{-},K^{+},\hat{B}^{0},\pi^{0},\hat{\pi}^{0}[/tex]

*Note that the [tex]\hat{}[/tex] 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 [itex]\pm 1[/itex]
• B-Particles have Bottomness of [itex]\pm 1[/itex]

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 [itex]\frac{2}{3},\frac{1}{3},\frac{2}{3},\frac{1}{3},\frac{2}{3},\frac{1}{3}[/itex]

Leptons have flavours of [itex]v_{e},e,v_{\mu},\mu,v_{\tau},\tau[/itex] 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 [itex]\pi^{+}[/itex] is listed as having quark content (which is the 'flavour content' right?) of [itex]u \ihat{d}[/itex]. I need it's listed antiparticle [itex]\pi^{-}[/itex] so for this it's flavour content will be the opposite, i.e. [itex]\hat{u}d[/itex]?

 

[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 [itex]\pi^{0}[/itex] and [itex]\hat{\pi}^{0}[/itex], 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]

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

.. 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 [itex]s=-\frac{1}{3}[/itex] therefore [itex]\hat{s}=\frac{1}{3}[/itex] which combined with [itex]u=\frac{2}{3}[/itex] gives a charge of [itex]0[/itex]? Unsure if I can just reverse the signs when going between particle and antiparticle.

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

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

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

 

[nickjer]

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

The [tex]K^+[/tex] 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 [itex]s=-\frac{1}{3}[/itex] therefore [itex]\hat{s}=\frac{1}{3}[/itex] which combined with [itex]u=\frac{2}{3}[/itex] gives a charge of [itex]0[/itex]? 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 [itex]u[/itex] not [itex]c[/itex] or [itex]t[/itex]? Since these all have the same charge of [itex]+\frac{2}{3}[/itex]. And the same for pairing it with [itex]\hat{s}[/itex] not [itex]\hat{d}[/itex] or [itex]\hat{b}[/itex]

You weren't told it had charm or topness.

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

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

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.