-1
$\begingroup$

This question already has an answer here:

How do particles with different Quantum Spin Numbers interact differently with other matter? What does spin cause? I understand that electrons etc act as if they are spinning my emitting a magnetic field (I think?) but what does the spin value represent? Does it represent the rate at which a particle would have to spin at to emit an electromagnetic field of x size? If so, what is the multiplier?

(I am new to particle/quantum physics [I am just researching it for fun {I am in 8th Grade}])

$\endgroup$

marked as duplicate by stafusa, John Rennie, Qmechanic Oct 6 '17 at 14:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Spins don't indicate that particles are spinning. That's a horrible misconception. Spin is an intrinsic property of a particle. When particle reactions take place, then several conservation principles are followed, like Lepton number conservation, Barton number conservation, Hypercharge conservation, Isospin conservation, Spin Quantum Number conservation, and so on. The topic is a very broad one and it's difficult for us to teach here. Please do some good research. You may refer to books like "Introduction to Elementary Particle Physics" by D. J. Griffiths for introduction and further reading $\endgroup$ – Wrichik Basu Oct 5 '17 at 21:07
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/1/2451 , physics.stackexchange.com/q/822/2451 and links therein. $\endgroup$ – Qmechanic Oct 5 '17 at 21:21
  • $\begingroup$ @WrichikBasu: I think an Intro to QM text might be more appropriate than particle physics text $\endgroup$ – Kyle Kanos Oct 6 '17 at 9:58
4
$\begingroup$

The spin of particle can roughly be physically thought of as the magnetic moment of the particle (or the strength of the magnetic field generated by the particle). For example, a spin-$0$ particle has no magnetic moment, and a spin-$1/2$ particle has half the magnetic moment of a spin-$1$ particle. Experimentally, what this means is that if I shoot a particle while applying a magnetic field, the particle will deflect a certain amount based on its spin. The greater the spin, the more it gets deflected.

Actually, the situation is a little bit more complicated than this due to quantum mechanical effects; there is something called the Lande g-factor which means that a spin-$1$ particle might not necessarily have twice the response under a magnetic field of that of a spin-$1/2$ particle. This, in part, has to do with the fact that the particle (eg the electron) is not actually spinning (even though we call it spin) and there is deep and complicated physics that goes into this.

Regarding the notion of a "spinning particle", it's a bit more accurate to think of spin as the magnetic moment associated with a particle. "Spinning" or orbiting charge does create a magnetic moment, so we call the quantum property spin, but you should keep in mind that this picture is just a cartoon.

$\endgroup$
  • $\begingroup$ Thanks so much! I did know that spin does not indicate true motion. What do you mean by "if I shoot a particle while applying a magnetic field, the particle will deflect a certain amount"? The particle would deflect a certain amount of what? And why does an external magnetic field need to be applied? $\endgroup$ – Indigo2003 Oct 6 '17 at 4:22
  • $\begingroup$ Imagine a particle moving along; it'll travel in a straight line if there's nothing else in space. Now, if I apply a magnetic field, this will interact with the spin of the particle; the magnetic field will cause the particle to curve instead of moving straight. The amount that it curves tells you how much spin this particle has. Without the magnetic field, no matter what spin it has, all particles will just move straight, without curving. There are, of course, other cool things that spin is important for (Pauli exclusion, superconductivity, etc). $\endgroup$ – Aaron Oct 6 '17 at 15:02
  • $\begingroup$ Ok thanks! I think I get it now. But if it has less spin, is it effected more by the external magnetic field or less by the external magnetic field? In other words, does more spin make a particle more resistant to magnetic fields? $\endgroup$ – Indigo2003 Oct 6 '17 at 15:39
  • $\begingroup$ More spin leads to more deflection. $0$ spin, for example, means it isn't affected at all. As I mentioned in the answer, this entire description should not be taken super literally, but as a rough description of what happens. $\endgroup$ – Aaron Oct 6 '17 at 15:41
0
$\begingroup$

The most striking physical effect of spin is the Pauli exclusion principle (no two identical spin 1\2 particles can occupy the same quantum state) which is responsible for all the chemical properties of atoms. Without spin the world would be very different and life as we know it would not exist.

$\endgroup$

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