For the nuclear force there is existance of pion is important.why do not consider it in mass of nucleus.


Pions are the excitations of the pion-field. If you excite a field with the right frequency you get a "stable excitation", one that can propagate through space without changing it's shape. This happens when the norm of the momentum four-vector $m^2 = p_\mu p^\mu$ is equal to the mass of the field $m = m_\pi$. Such an excitation is called a real (or on-shell) particle, and is what you commonly imagine when you hear "pion".

If you however do not hit the right frequency, your excitation will quickly decay. It's like when you hit a guitar string but don't get it to ring (it will just make a bzz sound). This happens when $m \neq m_\pi$, and such a particle is called a virtual (or off-shell) particle. Since it does not travel through space-time like a real particle, the $m$ obtained in this way does not correspond to a mass in the kinetic sense.

When we say pions are binding the nucleons, we are not talking about real pions but virtual pions. It's just like in electromagnetism, when two charges are being attracted, you don't "see" light even though the force is mediated by virtual photons, because they are virtual. It's better to think of the virtual pions as a force than an actual particle, in fact we have a name for it, it's called the nuclear force (not exact but you get the idea).

With that in mind I think it should be obvious that the mass difference due to the pions is just another way to call the mass difference due to the nuclear binding energy. If the pions had a different mass, the nuclear force would cause a different binding energy, but you do not "add" the pion mass to the mass of the nucleus as though it were a real particle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.