Relation between momentum and mass of quarks

[Hluf]

We say on-shell and off-shell mass of quarks. 1) What is the difference on-shell and off-shell mass of quarks. 2) At lab. center of mass frame for lepton particles p²= -m². Can we apply this equation for quarks.
Thank you!

 

[tiny-tim]

Hi Hluf!

We say on-shell and off-shell mass of quarks. 1) What is the difference on-shell and off-shell mass of quarks.

a quark always has on-shell mass (usually just called "mass")

a quark never has off-shell mass

off-shell mass is a mathematical trick which helps in the calculations for Feynman diagrams

2) At lab. center of mass frame for lepton particles p²= -m². Can we apply this equation for quarks.

(p is the four-momentum)

yes this applies to everything (and in any frame): leptons hadrons and photons

you can regard it as the definition of m (the mass)

 

[Bill_K]

off-shell mass is a mathematical trick which helps in the calculations for Feynman diagrams

If it works once, it's a trick. If it works twice, it's a method!

 

[tiny-tim]

If it works once, it's a trick. If it works twice, it's a method!

But if you don't know how it works, it's magic!

 

[Meir Achuz]

But if you don't know how it works, it's magic!

If you don't know how it works, go to school to learn how.
Quarks can appear as intermediate particles in Feynman diagrams, where they are
'off-shell'.

 

[mfb]

yes this applies to everything (and in any frame): leptons hadrons and photons

you can regard it as the definition of m (the mass)

Well, there is no center of mass frame for (single) photons.

 

[kurros]

a quark always has on-shell mass (usually just called "mass")

a quark never has off-shell mass

Some people would argue the complete opposite. The longer time a particle lives for, the more probably that its mass would be measured to be almost exactly the on-shell mass, but really the whole universe is quantum mechanical and every process a "little bit" virtual, so everything is always off-shell, the exact on-shell mass occupying measure zero of the probability distribution :p.

 

[Vanadium 50]

Yes, some people will make this argument. Some people find quibbling entertaining. It's not a very good argument for promoting understanding of QFT to people new to it, though.

There really is a difference between real and virtual, and to quote Potter Stewart, "I know it when I see it"