# Which direction does boosted spin point?

1. Dec 20, 2012

### Zoot

The Pauli-Lubanski spin 4-vector for a particle in its rest frame is given by Sμ = (0, Sx, Sy, Sz). (Note this is the version from Rindler, which defines S in terms of the 4-velocity instead of 4-momentum) From the 3 spatial components, it is easy to construct a unit vector that points along the direction of spin. (I realize that since quantum mechanics doesn't allow us to simultaneously know Sx, Sy, and Sz these represent expectation values). Now suppose we boost this particle to another frame using a standard Lorentz transformation. We pick up a time component S0, and the spatial components change also. My question is, can we still use the 3 boosted spatial components Sx', Sy', Sz' to determine the 3-direction that the spin points in the boosted frame?

As an example, suppose an electron is "spin-up" along the z-axis in its rest frame (meaning 100% chance of measuring spin up along z). Would its spin vector be S = (0, 0, 0, h/2) where h = h bar?

As another example, suppose the electron's spin in the rest frame points in some arbitrary direction such that S = (0, Sx, Sy, Sz) and we do a boost along the x axis to get
S = (γβSx, γSx, Sy, Sz). If the boost is LARGE, so the x component dominates, does this indicate that the spin now points along the x direction?

2. Dec 21, 2012

### bcrowell

Staff Emeritus
Yes.

It might seem counterintuitive if the direction of the spin preferentially aligns itself with the direction of motion for a randomly chosen orientation of the spin in the rest frame. But this is exactly what we see, for example, with relativistic aberration of light rays emitted from a moving point source.

In the limit of large boosts, your electron acts like a massless particle, and the timelike component of S is what people usually define as the helicity. But the x and t components are equal up to a sign.

I'm not convinced that it's valid in general in SR to define a notion of spatial direction by writing a four-vector of the form (0,x,y,z), where x, y, and z make a unit vector. I think it would depend on what exactly how you defined "spatial direction" and what phenomenon you were talking about.

However, in the case of the Pauli-Lubanski vector W there may be a justification for this. Is W conserved (not just its square but the vector itself)? If so, then particle would be able to exchange W with a measuring device. If the particle is in motion relative to the measuring device, then the result of the measurement would definitely show the aberration effect you describe.

 Deleted an incorrect remark about classical versus quantum.

Last edited: Dec 21, 2012
3. Dec 21, 2012

### Zoot

Thank you for your help Ben. It is assuring to know that I'm at least on the right track here.

My hope is that the spin angular 4-momentum Sμ is similar to 4-momentum pμ in the sense that the spatial components describe the 3-momentum in all frames, not just the rest frame. It seems like all the 4-vectors I'm familiar with have spatial components related to some well-known 3-vector. So the question remains: Do the 3 spatial components of Wμ (or Sμ) represent the spin angular momentum 3-vector in ALL frames, not just the rest frame?

4. Dec 21, 2012

### pervect

Staff Emeritus
If you imagine a telescope along the spin axis in the rest frame, it will be pointing in some spatial direction. The spin vector S and the 4-vector of the telescope should be the same - and the direction that the telescope poits is given by the 3-vector part of S.

Now, if you boost the assemblage, mathematicaly we can transform the 4-vector that's the direction that the telescope is pointing via the Lorentz transform. But it seems to me that because of the relativity of simultaneity, the spatial components of this vector won't necessarily be the "direction" the telescope is pointing at, because the simultaneity conventions are different.

I think you really want to consider the "world sheet" - the assemblage of worldlines - traced out by the telescope. Then you slice the worldsheet by the new convention of "simultaneity" in the boosted frame, and that this gives the direction you think the telescope is pointing in.

I'd check these results by comparing them to the formula for aberration - it should give the identical result.

Wiki gives http://en.wikipedia.org/w/index.php?title=Relativistic_aberration&oldid=484933593

my textbook gives a similar complex-looking formula for the aberration angle.

There's also the semi-philosophical question of "is the direction of the telescope really waht you want to know". I think that's what your question is about, but I could have misinterpreted the question.

5. Dec 21, 2012

### Zoot

Pervect: Interesting points made. Maybe what I should be asking is: What are the 3 components of the spin angular momentum 3-vector in the boosted frame? I'm hoping this is a valid question analogous to "What are the components of the 3-momentum of a particle in a boosted frame?" I know the answer to the latter question, they are simply the 3 spatial components of the 4-momentum in that frame. But I'm not sure what the answer is to the former question.

Regarding your telescope analogy, I see a "paradox" of sorts that arises. Suppose the electron's spin and telescope in the rest frame are pointing up along a direction at 45 degrees from the horizontal (x axis). Now we do a LARGE boost along x. Due to Lorentz contraction, the telescope loses it's length along x, so it appears to be pointing vertically straight up along the z axis? But we know that the electron's spin axis will now point in the horizontal direction along x, because the x component is multiplied by γ. So in this sense the telescope and spin direction have moved in opposite directions?

Here's what I believe to be the solution to the paradox: Model the spin angular momentum as being generated by a spinning disk. The disk's orientation looking at it edge-on is \ while S points along / . After a boost along x, the disk now appears vertical, and S points horizontal as required.

Last edited: Dec 21, 2012
6. Dec 21, 2012

### bcrowell

Staff Emeritus
There are two separate questions we could ask here here: (1) the question quoted above, and (2) the question I thought you were asking originally, which was whether the direction of the spatial part of W is the same as the direction of the spin 3-vector.

Conjecture #1 is much stronger, and I think the answer is that it's false. The spacelike part of W in the boosted frame has the wrong magnitude to be interpreted as the spin 3-vector.

Hmm...I think this argument may actually prove that even the weaker conjecture #2 is false. Don't the usual equations for relativistic aberration of light simply come from taking a lightlike energy-momentum four-vector and boosting it? If so, then it seems to me that a telescope that is aligned with an electron's spin, in their common rest frame, is misaligned with it in any other frame. This is because the Pauli-Lubanski vector is spacelike (for a massive particle). When a lightlike vector and a spacelike vector have parallel spatial parts in one frame, I don't think they'll have parallel spatial parts in other frames.

I don't think this is generally valid. E.g., the relativistic velocity 4-vector has a spacelike part that isn't the same as the velocity 3-vector.

7. Dec 21, 2012

### bcrowell

Staff Emeritus
I think the answer may be that this is just not a well-defined question. Fundamentally, there is no reason to expect there to be a relativistic angular momentum vector with nice properties. Classically, the most natural thing to construct is the rank-2 tensor $L^{ab}=p^ar^b$.

Yeah, this seems to be the same conclusion I reached in #6.

I don't think this works, because you can't model a quantum-mechanical spin using mechanical rotation, and the Pauli-Lubanski vector is a purely quantum-mechanical thing. (This was what I originally didn't understand and got wrong in my #2, which I edited.) Since I was confused about this, here's some more detail on what I think is the correct story. Let $W_a=(1/2)\epsilon_{abcd}L^{bc}p^d$. If the angular momentum L came from motion of material particles, $L^{bc}=r^bp^c$, then we'd have $W_a=(1/2)\epsilon_{abcd}r^bp^cp^d$. For a massive particle, this vanishes in the rest frame because c and d have to be a repeated index. Since it vanishes in the rest frame, it vanishes in all frames. This seems to show that, at least for massive particles, the classical P-L vector is zero and of no interest. It's purely a quantum-mechanical thing, which can only be nonzero for intrinsic spin.

I think this purely quantum-mechanical nature of the thing resolves the paradox. If the spin came from a classical object like a gyroscope, you could duct-tape it to the telescope in the rest frame. To simplify, this would be like duct-taping two pencils together in their rest frame, so that they were essentially two identical line segments. If an observer in the rest frame says that the world-strips swept out by these line segments are identical, then every other observer says so as well.

But the "duct-tape" argument fails when the spin is quantum-mechanical and therefore can't be manifested as a line segment with some spatial extent.

8. Dec 21, 2012

### pervect

Staff Emeritus
I have to agree that in 4d (i.e. relativity) it's more reasonable to think of classical spin as a bi-vector, an anti-symmetric rank 2 tensor. In 3d the dual of a bivector is a vector, thats how we get away with talking about "spin vectors".

I don't know how to make sense of asking "how does a bivector point". Conceptually, it's more of a plane. And it's dual is a plane as well.

Quantum mechanically, the "natural" representation of spin state is a pair of complex numbers, $\left< \alpha,\beta \right>$. This gives the probability of measuring "up" and "down" spin.

Measuring a spin state is done by using a spin operator. The various components of the spin are defined by the complex paulii matrices, which are 2x2 complex matrices. Of course, the Paulii matrices for the chosen direction of "up" and "down is the identity matrix.

So, speaking accoding to QM, spin still isn't a 3-vector, it's a complex 2 -vector.

There is a mapping from the QM to the classical, but in the case of a fermion (say an electron), you have to rotate a fermion through 720 degrees to restore it to the same state - rotating it through 360 degrees inverts the wavefunction.

Measuring the spin component in a given direction quauntum mechanically requires one of the Paulii matrices.

Now, what happens when we combine relativity and QM? I believe you wind up with a relativistic 2x2 matrix, but its now called a spinor. But I haven't worked with them much.

So, in conclusion, there's a bunch of mathematical descriptions of spin I've talked about, but I can only make sense of "how it points" in the 3d, nonrelativistic case.

9. Dec 21, 2012

### Zoot

I'll think carefully about what you guys are saying. I think the best strategy for me will be to go back to basics, remove the quantum mechanics, and start by considering a classical system with angular momentum like a spinning disk or 2 masses attached by a rod, etc. spinning about the origin at some orientation. I can then characterize this system by the relativistic angular momentum tensor Lαβ and the classical Pauli-Lubanski spin 4-vector (Rindler's version which depends on 4-velocity of CM frame, not 4-momentum, so that S has units of angular momentum). Then I can consider boosting this object and see how S transforms. I think this analysis will answer many of my questions, which are not necessarily quantum mechanical, but more about how S transforms and how to interpret the physical meaning of the components.

10. Dec 21, 2012

### pervect

Staff Emeritus
I'm not sure if this will help - but while you can't really define a plane by a direction, given a time basis vector, you can define a plane as the wedge product of your given time basis vector and another vector. There are an infinite number of other vectors you chose, you can specify one unique one that's orthogonal to your time basis vector (and hence will be space-like). I think that this would be the closest you could come to defining spin as a vector, but it requires specifying a time basis vector.

I'm not sure if you're familar with "wedge products", it's an anti-symmeterized tensor product.

11. Dec 22, 2012

### andrien

the definition given by cartan for spinors has an indeterminacy in sign i.e. both sign(+ and -) represent same states.
I have not heard anywhere that spinors are a consequence of relativity.May be it first arose in dirac work which combines QM with relativity so it is supposed to come from mix of them.But it is not so,they were already known before dirac.

12. Dec 22, 2012

### pervect

Staff Emeritus
A quick read of wiki indicates you're probably right - Cartan introduced them first in 1913, and they can be used in 3d Euclidean or 4d Lorentzian spaces.