Weak decay, decay rate, colour-supressed and colour-allowed

[binbagsss]

My textbook says:

One of the decays occurs via a $b \to c$ quark transition and is 'colour-allowed'.(Left of diagram).

The other decay has a of $b \to c$ and is 'colour-suppressed'.(Right of diagram)

I'm unsure of what is meant by these terms, it doesn't really explain.

I've attached the Feynman diagrams .

Could anyone explain these to me or send me a good link?
Thanks very much.

 

[Orodruin]

The left diagram works automatically as long is colour is conserved in the upper W vertex. The right only works if the upper vertex connects to quarks of the same colour as the original b quark.

 

[binbagsss]

The left diagram works automatically as long is colour is conserved in the upper W vertex. The right only works if the upper vertex connects to quarks of the same colour as the original b quark.

Thanks. I've tried to have a look into what you are saying, but my textbook doesn't explain it.

I don't understand what you've said because I don't think I understand some basic concepts- I don't understand why the right diagram can't be drawn in the same form as the left in terms of the W boson. Why is this?

 

[mfb]

I don't understand why the right diagram can't be drawn in the same form as the left in terms of the W boson. Why is this?

There is no b->s transition that could emit a W, and also no W -> c cbar process. Both would violate electric charge conservation.

 

[binbagsss]

The left diagram works automatically as long is colour is conserved in the upper W vertex. The right only works if the upper vertex connects to quarks of the same colour as the original b quark.

Ok, so I know that colour must be conserved at a vertex. And that quarks come in 3 colours. As the W boson is colourless, am I correct in thinking that in the left diagram, the anti bottom quark and anti charm quark must be of the same colour, as must the anti down and up, and on the right diagram the anti bottom and anti charm must be, and the anti strange and charm?

 

[mfb]

Right.

And to form the hadrons, the right diagram has the additional requirement that charm and anti-charm have to have opposite colors (also down and anti-strange but that is equivalent to the previous requirement). The left diagram has this automatically as the original two quarks have opposite colors.

 

[binbagsss]

Right.

And to form the hadrons, the right diagram has the additional requirement that charm and anti-charm have to have opposite colors (also down and anti-strange but that is equivalent to the previous requirement). The left diagram has this automatically as the original two quarks have opposite colors.

Opposite as in anti colour/colour?

 

[mfb]

Yes.

 

[binbagsss]

Ahh okay thanks I think I've got it. Mesons have the additional criteria that the quark and antiquark have to be the same colour. (For a baryon does anything go? )

So in the right diagram I've concluded all quarks are of the same type of colour.
But the left diagram there is more freedom of choice as the upper vertex two quarks can be either r,g,b, whereas all other quarks in that diagram are of the same colour?

 

[Orodruin]

For a baryon does anything go?

No, a baryon has to have one quark of each colour.