# What exactly is spin?

1. May 13, 2007

### Hoof47

I am interested in quantum mechanics, but my maths isn't up to understanding some of the concepts. I've tried to look at spin, but all explanations are mathematical. Is it possible for it to be explained in words?

2. May 13, 2007

### cristo

Staff Emeritus
Spin is a type of "intrinsic" angular momentum-- i.e. a type of angular momentum that a particle has, regardless of its motion in space. marlon has posted in his blog an https://www.physicsforums.com/blogs/marlon-13790/what-is-spin-152/ [Broken] to this, which you may wish to look at.

A common misunderstanding is for one to interpret a paricle's spin as the "rotation about its own axis." However, spin is a purely quantum mechanical property, and it therefore makes no sense to interpret it in any classical sense.

Last edited by a moderator: May 2, 2017
3. May 13, 2007

Are all the terms associated with spin all analogous? E.g. "intrinsic angular momentum"; a particle has neither angular momentum or "spin" [EDIT: meaning rotation], right? Are all these just naming conventions?

I'm assuming the term "spin", etc., is due to the Stern-Gerlach experiment, where a particle's "spin" is measured in a way similar to what could be done for classical spinning object. True?

Since particles do deflect in a magnetic field as if they had angular momentum, but they don't actually have any, what is going on?

Last edited: May 13, 2007
4. May 13, 2007

### jostpuur

I've had some trouble with this "purely quantum mechanical" argument. Classical electromagnetic field has internal angular momentum, that is not cross product of location and momentum, and is pretty much spin angular momentum then.

When I asked what was spin a long time ago, one already graduated student responded to me that it is "angular momentum without rotation". That was quite good. But you can have internal angular momentum that is not result of any kind of rotation also in classical context.

Last edited by a moderator: May 2, 2017
5. May 13, 2007

### lightarrow

Particles with spin *do* have angular momentum and magnetic momentum.
But they don't "spin".
If you wonder how could this be, you're not alone.

6. May 13, 2007

### Sojourner01

Am I the only person who doesn't have a problem with writing it off as "a quantum mechanical property that a human being is not physiologically able to conceptualise"? In much the same way that I'm not able to comprehend what a dog dreams about - because I'm not a dog - I have no problem with the lack of realistic interpretation of spin, because I'm not a quantum mechanical object, I'm a macroscopic object, and I perceive macroscopic things.

So the answer to the question is, spin is spin. It doesn't need to go any deeper than that.

7. May 13, 2007

But...it isn't *real* momemtum, is it (i.e. no rotation -> no velocity)? I'm wondering if it's just a model to account for deflection in a mag field?

(And you meant magnetic moment?)

Last edited: May 13, 2007
8. May 13, 2007

9. May 13, 2007

If someone will tell me definitively that there is no explanation for how a particle is deflected in the Stern-Gerlach thing other than via a set of functions which have familiar sounds like "spin" and "angular momemtum", then I can be content, or at least as content as I am with duality (not!).

10. May 13, 2007

### jostpuur

This is where I asked about the matter:

My understanding of the Dirac's field has been rather "changing" lately, as result of discussions here in PF, and the OP of my thread probably isn't relevant for this matter anymore, but pay attention to the later comments on the classical electromagnetic field too.

11. May 13, 2007

### Tomsk

My tutor explained it to me as a dynamical variable associated with the rotational symmetry of the particle. (Or of space, I can't remember!) Linear momentum is a variable associated with the translational symmetry of space, energy is linked to time in a similar way, and angular momentum comes from the rotational symmetry of space. So because a particle looks the same from every angle, there is a spin associated with it. But I think it is only noticable when there is something that breaks that symmetry, like a B field. This probably hasn't helped much, tbh I don't get it myself! :/

12. May 13, 2007

BTW, there is another thread with a lot of responses on this subject, in it I saw this:

https://www.physicsforums.com/showpost.php?p=1320028&postcount=29"

saying,
Wikipedia has this to say:

There is no explanation how to make the leap from seeing a metal bar rotate to knowing it's because of the "spin angular momentum" of its [electrons/atoms?]. Heck, for all I know, spin/intrinsic angular momentum is defined by this effect.

Last edited by a moderator: Apr 22, 2017
13. May 13, 2007

Ok, so you're talking about the internal angular momentum of an EM field, so at least there's not some wacky compound I didn't know about.

But, come on! It's bad enough that an electron has angular momentum with rotation, but now EM fields have angular momentum without mass!?

What the hell, I guess QM is re-defining all my CM terms according to its own fancy. Sigh, at least some of the QM fathers had the sense to use non-conflicting terms, like "flavor".

14. May 13, 2007

### Hootenanny

Staff Emeritus
Photons are massless and yet have linear momentum...

15. May 13, 2007

### Staff: Mentor

Well, EM fields carry energy and linear momentum, so they pretty much have to have angular momentum too, based on the definition of angular momentum as $\vec r \times \vec p$.

16. May 13, 2007

Yeh, I remember now (I wasn't making the connection from photon momentum to fields...), but is that supposed to make it better? Anyway, that was a rhetorical question venting moment[um]ary outrage.

But now that you bring it up,
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photonMass.html" [Broken]

Says,

So, light does have "relativistic mass", but no rest/invariant mass.
It does go on to tell about a theoretical upper limit for the rest mass of light between 7 × 10-17 eV and 3 × 10-27 eV. I'm not sure why they bother to state the rest mass at zero, and then go on to give the estimate.

But anyway, now I feel better.

Last edited by a moderator: May 2, 2017
17. May 13, 2007

### Hootenanny

Staff Emeritus
That is an experimentally derived upper limit, from observations. The theoretical mass of the photon is defined as zero. Period.

18. May 13, 2007

### jostpuur

This is getting slightly off topic, but I could make a remark, that electric and magnetic fields can have rest mass too. The plane waves which travel at the speed of light are massless, but static electric fields around some charge distributions are not.

Last edited: May 13, 2007
19. May 13, 2007

### lightarrow

You can define angular momentum without the need to have a rotating mass: think about light, as jtbell wrote. What is striking in that case is that *statics* electric and magnetic fields can have angular momentum: you only need a circulation of the Poyinting vector.
Yes I meant magnetic moment.

20. May 13, 2007

### lightarrow

Example: if you shoot a very strong light pulse in a specific direction, you receive a recoil in the opposite direction. If you want to hold true the law of momentum conservation (higly advisable!) you are forced to conclude that light too carries momentum. That is a more general definition than p = m*v (for non zero rest mass and not relativistically) or p = m*v*gamma (for non zero rest mass and relativistically).
The more general formula which comes out is:
p = Sqrt[(E/c)^2 - (mc)^2]
where E is the total energy and m is the rest mass. For a photon:
p = Sqrt[(E/c)^2 - (0)^2] = E/c
To have angular momentum, you only have to compute $$\vec r \times \vec p$$