What is the significance of the newly discovered pentaquark at LHCb?

[Michel_vdg]

And how would you have 1.66 charge? or 0.66?

Quarks have fractional electric charge values of either 1⁄3 or 2⁄3 times the elementary charge ... so when you have 5 of them within this new 'proton' than what is the total charge or what are the individual charges ... I used those numbers more in a figuratively speaking sense.

 

[ChrisVer]

Do you think you could obtain a colorless bound state by just adding arbitrarily 5 quarks? (Try $3 \otimes 3 \otimes 3 \otimes 3 \otimes 3$)

 

[Michel_vdg]

Do you think you could obtain a colorless bound state by just adding arbitrarily 5 quarks?

Oh, I didn't know if they were saying that the Pentaquark was 'colorless' ... it says in the article:

"The quarks could be tightly bound, or they could also be loosely bound in meson-baryon molecule, in which color-neutral meson and baryon feel a residual strong force similar to the one that binds nucleons together within nuclei. " http://lhcb-public.web.cern.ch/lhcb-public/Welcome.html#Penta

Edit: But thus that mean that the positive proton charge is just 1 when it is colorless even with 5 quarks, just like a normal proton with only 3 quarks?

 

[ChrisVer]

even in that case (which OK somehow tells you that it's colorless since it implies $3 \otimes 3 \otimes 3 \otimes 3 \otimes \bar{3}$) how would you add the charges of a hadron and meson (none has fractional charge) to obtain fractional total charge in the end? I hope this helped.

PS. But it can have more positive charge, like +2...

 

[Michel_vdg]

even in that case (which OK somehow tells you that it's colorless since it implies $3 \otimes 3 \otimes 3 \otimes 3 \otimes \bar{3}$) how would you add the charges of a hadron and meson (none has fractional charge) to obtain fractional total charge in the end? I hope this helped.

I don't know ... those numbers were more to ask the question ... see the edit that I added:

"... thus that mean that the positive proton charge is just 1 when it is colorless even with 5 quarks, just like a normal proton with only 3 quarks?"

It also comes down to a previous question and mfb's reply where he says that it would act just like a 'regular' proton.

 

[mfb]

It has to be color-neutral to be a hadron. This also implies that it has to have an integer charge. The discovered particle has a charge of 1 (or -1 for its antiparticle).

Are you asking about elastic collisions via the electromagnetic interaction? Those are independent of the substructure of the hadron, that should be obvious.

 

[ChrisVer]

"... thus that mean that the positive proton charge is just 1 when it is colorless even with 5 quarks, just like a normal proton with only 3 quarks?"

I may be misinterpreting your question, but as I mentioned you can have other charges too...
In the case of the discovered particle $u u d \bar{c}c$ this happens to be a +1 charged particle.
I don't see the reason why the $u u d \bar{c} s$ cannot exist (which would have a charge 0)

As for the 5 quarks, I wouldn't call it similar to proton (or a proton excited state)... it's a little different.