# I How can we visualize spin being invariant?

1. Mar 30, 2016

### idea2000

I have a basic question about what spin invariance means. If we were to use a classical example, if the spin of a basketball was invariant, what would that mean? Would every frame measure the same revs/min? Newtonian mechanics predicts that every frame would measure the same revs/min, but sr predicts that every frame would measure a different revs/min. So, would every frame measure a basketball travelling near the speed of light as having the same revs/min? Or, the same number of rotations, over different periods of time?

2. Mar 30, 2016

### Simon Bridge

Technically the spin quantum number does not analogise well with macroscopic classical angular momentum... certainly not "revs per min".
When I describe the effect using classical terms it will come out as wierd.

Relativistic invariance means it stays the same under the lorentz transform: this means that all inertial observers measure the same thing for the quantity - the most famous example being the speed of light in a vacuum. In the case of light - Bob and Alice observe a pulse of light traversing a system of mirrors, Alice is stationary in the mirror frame and bob is not. Alice and Bob will disagree about the path the light pulse takes, and they will disagree about how long it took, but they will agree about the speed of the light. That is what invariance means.

Spin is basically a fundamental angular momentum - so the analogy would be the basketball angular momentum.
If you have a basketball as a spinning sphere on Alice's fingertip, Bob, who is relativistic in Alice's frame, sees something somewhat odd: his view is length contracted (so the "ball" is an oblate spheroid, flattened in Alice's direction of travel) and time dilated.
The invarience says that even though he disagrees with Alice about the period of the rotation and the mass distribution being rotated, he and Alice will still agree about the value of it's angular momentum.

There's probably a better way to do this...

3. Mar 30, 2016

### Markus Hanke

Spin is a form of intrinsic angular momentum that has nothing at all to do with motion in space; it is not a function of spatial coordinates of any kind, but a fundamental, intrinsic, immutable property of a particle. Thinking of spin as a "rotation" is therefore at best highly misleading.

Consider a flagpole with a flag attached to it at the top, such that, if you rotate the pole about a suitable axis, you also have to twist the pole itself, i.e. rotate the flag at the top. Having a spin of ½ would correspond to an "angle doubling" - rotating the flag pole by 360 degrees about some axis will make it coincide with itself, except that the flag on top will point into the opposite direction. You would have to rotate the whole thing again to make both flagpole and flag coincide with the original version. This means, you have to rotate by 720 degrees to achieve a perfect match, i.e. two full revolutions, not just one.

The "flagpole + flag" analogy is a visualisation for a mathematical object called a spinor. This is an element of a complex vector space, and it is possible to associate rotations of spinors with rotations of 4-vectors in space-time - but the rotation angle will be doubled for our 4-vectors. It is precisely this relationship between rotation angles which is invariant, and hence the same for all observers. It has nothing to do with "revs/min" or anything of that nature; spin is better thought of as a property that determines how the mathematical objects describing our particles ( i.e. spinors ) behave under certain transformations.

All of this also ties in with the concept of orientation entanglement.

4. Mar 30, 2016

### idea2000

Yes, there probably is... =) Thanks for your great reply, btw...

But, please bear with me for just one more second... =) From what I gather from your posting, angular MOMENTUM is what is invariant. So, what we mean when we say that spin is invariant is that angular momentum is invariant between frames, is this correct?

So, each frame could see a different mass distribution, just like a skater pulling her arms in, and spinning faster. But, when we do the math and transform between the two frames, can we at least guarantee that they see the same number of rotations? So, for example, if Alice sees 5 rotations, could Bob see 100 rotations? The reason I am asking is because if one frame sees more rotations than the other, then, can reality still be consistent between the two frames? If reality is consistent, should they both see the same number of rotations? For example, in the train and tunnel paradox, where the train is supposedly longer than the tunnel, and vice versa, there is a way to explain what is going on where reality is still consistent. Is there a way to do this for angular momentum, as well? If we were to use a windmill instead of a basketball, if the windmill were to strike a bug in Alice's frame, then, the windmill should also strike a bug in Bob's frame, or what actually happens in our world would not be consistent. Can we guarantee this consistency when we count the number of rotations? Or is there some other way to explain what is going on?

Last edited: Mar 30, 2016
5. Mar 30, 2016

### Staff: Mentor

If we start the system rotating, let the system rotate for a while, then stop it, all observers will agree about the total number of turns. They will not agree about how much time passed while it was rotating and therefore they will not agree about the rotation rate in revolutions per minute.

This might also be a good time to point out that angular momentum (like total energy and linear momentum) is not frame-independent. It is conserved, meaning that whatever value it has in given frame will never change, but the value itself may be different in different frames.

None of this has anything to do with "spin", which as other posters have said is an intrinsic property of quantum particles.

6. Mar 30, 2016

### Jonathan Scott

Rates of rotation are not relevant. Spin and internal angular momentum within a system are both related to the amount by which the phase of the wave function changes when you follow it all the way round the axis of rotation, which for a system in a stable state is always a multiple of pi radians (a whole number of half-turns), which is effectively just a number. The more the momentum, the more rapidly the phase changes, so the smaller the radius of the path for the same angular momentum.

7. Apr 1, 2016

### Simon Bridge

Yep, I was attempting the analogy in the terms originall tried... basketball rotating. I did warn that it was not good, but I hoped to illustrate how invariance comes about, how to think about it.

Orbital angular momentum (the kind a rotating basketball has) is not, in fact, invariant. I tried to show you "if it were" what that could mean. Spin is a different kind of angular momentum.

Spin is more of a symmetry thing... it can be thought of as related to how far you have to rotate an object before it maps on to itself... ie, in playing cards, the ace of hearts has spin 1 because you have to turn it around one time... the queen of hearts would have spin 2 because you only have to give it a half turn. Which is easy to visualise ... and you should be able to see that all observers will agree how far an object got turned.
Also not perfect... but no analogy is. Remember, we need the concept of intrinsic angular momentum precisely because classical ideas do not work here.

8. Apr 1, 2016

### sweet springs

The question of idea2000 is puzzling me. When we say spin angular momentum of electron is $$\hbar/2$$, in the frame of reference of which an electron is at rest or in motion anyway ? Spin of an electron could be zero in some rotating frame of reference around it?

Last edited: Apr 1, 2016
9. Apr 1, 2016

### Simon Bridge

The "spin" is built in angular momentum, independent of motion.
So the spin of an electron is ħ/2 whether it is at rest of not... a bit like like 511keV of its total energy is always mass, the rest being kinetic.

10. Apr 1, 2016

### Staff: Mentor

Neither. The "spin" of the electron, or any other elementary particle, is an intrinsic property of its wave function (or quantum field, if you want to go to that level of detail). It has to do with how the wave function changes under rotations. This property is independent of any choice of coordinates.

11. Apr 2, 2016

### vanhees71

The spin by definition is invariant, because it's the angular momentum of a massive particle at rest, i.e., it's specified in a (naturally) preferred reference frame. In the same way you make cross sections invariant. It's defined in the lab frame, i.e., where one of particles in the initial state is at rest. That's because in the early days of accelerators they usual setup were fixed-target experiments. That's also why the relative velocity of two particles is defined as the velocity of one of the incoming particles in the rest frame of the other. All these definitions are easily written in a manifest covariant way using Minkowski four-vector/tensor notation.

One should note that "spin" doesn't make sense for a massless particle (although one calls the quantum number also "spin" in this case to indicate that it's described by massless fields in certain representation of the Poincare group like vector fields in the case of photons and gluons in the standard model). Here you have helicity instead, which is defined as the projection of the total angular momentum to the direction of a particle's momentum, which is (for massless particles only!) a Lorentz-invariant number that can take only two values $\pm s$ for a massless particle of "spin" $s \in \{0,1/2,\1,\ldots \}$.

For details, see appendix B in

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

12. Apr 2, 2016

### sweet springs

Thanks. Mass of a particle by definition is invariant because it is the energy(/c^2) of a particle at rest. Mass of the two free particle system is usually more than summation of masses of the two particles by their relative motion with kinetic energy. Similarly, spin of the system of two particles could not be different from the sum of the particle spins ?

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