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Posted (edited)

Do not confuse this with the angular momentum of a spinning ball.

It is the angular momentum of a wavefuction. Specifically linear vs angular symmetry/antisymmetric equations with regards to vectors and spinors specifically an internal degree of freedom not a spatial degree of freedom to describe the state of a particle.

Edited by Mordred
Posted
16 minutes ago, Mordred said:

Do not confuse this with the angular momentum of a spinning ball.

It is the angular momentum of a wavefuction. Specifically linear vs angular symmetry/antisymmetric equations with regards to vectors and spinors specifically an internal degree of freedom not a spatial degree of freedom to describe the state of a particle.

No, I equate it to a precessing orbit, am I even close to correct? I know I am expressing this in classical terms.

Posted (edited)

Has nothing to do with orbits. It is an internal degree of freedom of the electron involved in a wavefunction involving vectors.

In QM particles don't orbit the nucleus Swansont mentioned this before in one of your other threads. Throw away any ball like image of the atom you may have.

Edited by Mordred
Posted

The precise answer is:

 

Spin is the part of angular momentum that does not come from position, where angular momentum is the conserved quantity associated with rotational invariance.

 

I don't expect you to understand this as is; it takes quite a bit of study. In order to understand the above, the best thing to do is to learn classical mechanics (up through the Lagrangian and method of least action), then to learn quantum mechanics (including spin, but taking it as a "given"), and then to learn some quantum field theory. 

Posted
5 hours ago, Butch said:

Thanks! More correctly angular momentum not accounted for by a flat orbit?

It is not associated with any orbit whatsoever. Orbital angular momentum is separate. Spin is intrinsic to the particle. 

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