Why is the spin in a mirror image opposite?

1. May 30, 2012

nonequilibrium

(without using the intuitive yet eronneous idea of "spin" meaning "spinning")

2. May 30, 2012

Bill_K

In 140 characters or less: Because both spin and orbital angular momentum are types of angular momentum and therefore behave the same way under a reflection.

3. May 30, 2012

nonequilibrium

Why do they behave the same way under a reflection?

4. May 30, 2012

kith

If spin was not an axial vector quantity you would get a wrong total angular momentum when coupling spin with orbital angular momentum. See this wiki article.

5. May 30, 2012

Bill_K

This says the same thing in more mathematical terms. Noether's Theorem says that for any system that can be described by a Lagrangian, corresponding to every continuous symmetry there is a conserved quantity. Angular momentum is the conserved quantity that arises from rotational symmetry. When you write out the general expression for angular momentum, you get Ji = Li + Si where
Li = ∫(∂L/∂∂4ψα)(xik - xkiα dV and Si = ∫ (∂L/∂∂4ψα) Iαβik ψβdV
The first quantity depends explicitly on the choice of the origin of the coordinate system, and represents the orbital angular momentum. The second quantity is independent of the choice of origin and represents the spin. Iαβik is the rotation operator acting on the field components,
ψα(x) → ψ'α(x') = (δαβ + ε Iαβik) ψβ
You can specialize this general result to particles of different spin such as Dirac particles or photons. For Dirac particles, Iαβik are the matrix components of σik.

Anyway, all this mathematics is just for entertainment! The short and simple answer to your question is as I said, Ji = Li + Si, which demands that Li and Si must both be the same kind of quantity. In this case they are both pseudovectors, i.e. under a mirror reflection they behave in the opposite manner to normal vectors.

6. May 30, 2012

nonequilibrium

That's not really true though, think of the $\frac{1}{2 \sqrt{2}} \gamma^\mu \left(1+ \gamma^5\right)$ vertex in the weak interaction force, adding an axial vector to a normal one.

EDIT: it seems the arguments for "S is an axial vector" depend on mirror symmetry, while "S is an axial vector" is used to prove that nature is not mirror symmetric (P-asymmetry in nature). Seems like an inconsistency?

7. May 30, 2012

bahamagreen

I think you all are missing the point...

There is nothing spinning in the mirror image whatsoever. The mirror image is a flat plane from which rays projected from the actual spinning object reflect. It is the translations of these reflections laterally across the flat mirror plane caused by the shifting angles of the ray projections from the actual object to the mirror plane to your eyes that is being seen.

The shifting reflection points are not spinning, but their effect is identical to the same effect of watching an actual object spin - the points on the spinning object shift position.

If you draw some rays from the spinning object to your eyes, then draw some rays from the spinning object to the mirror plane and then on to your eyes this will be quite clear why the image of the object in the mirror appears to spin the opposite direction of the actual object.

8. May 30, 2012

nonequilibrium

bahamagreen, I think you're the one missing the point. In these contexts, "mirror image" is just shorthand for "what you get when you perform (e.g.) $x \mapsto -x, y \mapsto y, z \mapsto z$". We're well aware of how mirrors work...

9. May 30, 2012

Bill_K

Yes, that's the point. If L and S were not both axial vectors, parity would never be conserved.

Also note for a Dirac particle you can have 1 + γ5, and you can have γμ + γμγ5, but you can't do that trick with the spin, because σμν and γ5σμν are the same!

10. May 30, 2012

nonequilibrium

To make sure I understand, is the following correct?
Do you mean that irrelevant to S being an axial vector or not, nature would not be mirror symmetric, however, if S were not axial vector, mirror symmetry would be broken even more manifestly than it is?