# Tag Info

0

For one dimensional crystal, when we have periodic boundary condition, we may think that the atoms are arranged in a circle, and so each lattice point will be equal to other lattice point and the translation symmetry is conserved. And for three dimensional crystal, due to periodic boundary condition, each atom will feel the same in the lattice. They are in ...

2

You're question is interesting because it is connected to the notion of elementary particle. As mentioned by anna v, the elementary particles (fermions) of the standard model have very specific properties under the symmetry of the standard model ($SU(2)_L\times U(1)_Y \times SU(3)_c$): they lie in the fundamental representation of the group, which in ...

11

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Here I will only consider elementary, non composite particles. All the hadronic resonances are composite particles of quark antiquark combinations as well as the neutron . The standard model of particle ...

4

Inmediately is not really true, there is some proportionality. Sorry I am not answering directly about "the standard model", this is quarks and leptons. But they will fit the general pattern, you will see. Let me first consider all the "particles" listed in the particle data group file. Most of the particles decaying via photons have a half-live about ...

2

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Nobody has an issue calling the electron a particle. Ditto for a neutron. It's stable in a nucleus, and the fact that a free neutron decays in circa 15 minutes doesn't much matter. It's similar for a muon, ...

4

I think the most direct answer to this would be the fact that a heavier particle can decay into many different lighter particles for different reactions. The abundance of occurence of these relations are const. Again the same heavy particle can be created in multiple types of collisions of various different lighter particles. Thus we cannot say that the ...

24

Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2$\mu$s. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter ...

2

Your question seems quite general, but perhaps you're confused about what "decay" is. When we say something "decays" we don't always mean that it's somehow "breaking up" into it's constituent parts. In fact, we hardly ever do. The heavier particles aren't really "transient interplay of the stable forms", unless I misunderstand, and that isn't something that ...

Top 50 recent answers are included