# Spin 4-vector

1. Jul 18, 2012

### Heirot

I have a problem understanding the Lorentz transformation of the spin. The spin 4-vector is defined in the rest frame of the particle as

$s^{\mu} = (0, \vec{s})$

and then boosted in any other frame according to

$s'^{\mu} = (\gamma \vec{\beta} \cdot \vec{s}, \vec{s} + \frac{\gamma^2}{1+\gamma}(\vec{\beta} \cdot \vec{s}) \vec{\beta})$

I have a couple of question concerning this.

1) How can spin transform as a 4-vector, when the angular momentum transforms as a 4-tensor with two indices?
2) How can I interpret the zeroth component of the spin 4-vector in an arbitrary inertial reference frame?
3) It is often said that one cannot separate the total angular momentum J into the orbital angular momentum L and spin S in a covariant way. In light of this (which I do not completely understand), what to make of this spin 4-vector?

Thank you!

2. Jul 18, 2012

### bcrowell

Staff Emeritus
There is no difference between the transformation properties of spin angular momentum and orbital angular momentum under a Lorentz boost. Both are really 2-index tensors. Both of them can also be expressed as 1-index tensor densities (not tensors), which is what you're talking about here. Note that there is nothing quantum-mechanical about this topic, so there can't be any difference between spin and orbital angular momentum. The "spin" can, if you like, be the spin of a baseball or a black hole.

It's not a 4-vector, it's a tensor density. The most straightforward way to define a one-index angular momentum tensor density is like $L^a=\epsilon_{abc}r^bp^c$, where $\epsilon$ is the tensor-density version of the Levi-Civita symbol (not the tensorial version defined in some books, so we don't need to throw in separate factors of $\sqrt{-|g|}$). In the object's rest frame, the terms with indices bc being timelike-spacelike vanish, and the yz, zx, and xy terms correspond to the three components of the angular momentum 3-vector. If index a is timelike, then I believe the timelike component of $L^a$ simply ends up being equal to the sum of the components of the 3-vector. Maybe someone else could check me on this.

This may help: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.6 [Broken]

Last edited by a moderator: May 6, 2017
3. Jul 19, 2012

### Heirot

So, you are saying that the 1-index and 2-index description of spin are equivalent?

My main concern is this: If we agree that spin should be a 3-vector in every inertial frame of reference, and we know that is is given by $\vec{s}$ in some frame, what is the transformation property of this physical 3-vector?

I.e.

$\vec{s}' = f(\vec{\beta},\vec{s})$

what is $f$?

4. Jul 19, 2012

### bcrowell

Staff Emeritus
Yes, although the 1-index thing can only be interpreted as angular momentum if it's *spin* angular momentum, i.e., if it's constructed in the object's rest frame.

No, it's not a 3-vector, it's a tensor density.

No, only in the object's rest frame.

It transforms as the three spacelike components of a one-index tensor density.

5. Jul 19, 2012

### Heirot

Surely you would agree that neither the electric or magnetic field are geometrically 3-vectors, yet they have well defined Lorentz transformation properties?

6. Jul 19, 2012

### bcrowell

Staff Emeritus
Huh? Both E and B are 3-vectors.

They have well-defined transformation properties, but they don't transform as tensors. (They don't even have the right number of components to be one-index tensors in 4-dimensional spacetime.)

7. Jul 20, 2012

### Bill_K

As long as we're only considering Lorentz transformations, the distinction between tensors and tensor densities is irrelevant, since √g = 1 in Minkowski coordinates. Also, to say a physical quantity "is" a tensor density belies the fact that you can construct tensors from tensor densities and vice versa just by multiplying by √g. Thirdly, there's an ambiguity in common language: e.g. we call T00 the energy density even though it's a component of an ordinary tensor and √g T00 is really the density.

Are the 1-index and 2-index description of spin equivalent?

Not by a long shot. From the energy-momentum tensor Tμν we can construct a three index quantity mμνσ = xμTνσ - xνTμσ which satisfies the equation mμνσ = 0, implying that the six quantities Mμν = ∫mμνσd∑σ are conserved. The space-space components are the orbital angular momentum r x p, while the "extra three" time components are also conserved quantities: Mi0 = x0pi - p0xi = (ct)p - (E/c)r. They relate the momentum of a system to the motion of its center of mass.

The Pauli-Lubanski 4-vector on the other hand, Wμ = ½ εμνστMνσPτ is a quantity of interest, but is not even an angular momentum, it doesn't have the right units. It is conserved if M and p are. It vanishes for the orbital part of M, and therefore relates only to the spin part. It has one "extra" component W0 = εijkMijpk, which is proportional to the particle's helicity.

8. Jul 20, 2012

### strangerep

Rindler [1] introduces it this way:
$S_\mu$ now has the correct dimensions. (Rindler's $L^{\nu\rho}$ is the angular momentum 4-tensor.)

[1] W. Rindler, "Introduction to Special Relativity", 2nd Ed.

9. Jul 20, 2012

### Heirot

I find it very strange that the total angular momentum is a 4-tensor with two indices, while the (Pauli-Lubanski) spin vector is a 4-vector when they should both transform the same way. It's like the PL spin vector carries the information about the spin of the particle as measured in its rest frame (due to contraction with the particle's 4-velocity).

Is there any sense in asking - what is the spin of the particle in motion as measured by the lab frame? Is this related to the fact that one cannot separate orbital and spin angular momentum in an arbitrary frame of reference?

10. Jul 20, 2012

### vanhees71

It doesn't make sense to ask this question, because the distinction between spin and orbital angular momentum is not a Lorentz invariant concept. Thus, for massive particles, one defines the spin of the particle referring to its (local) rest frame. The spin generates rotations of states of a particle at rest. As you correctly say, the Pauli-Lubanski vector (operator) is indeed precisely the spin operator in this sense. Of course, total angular momentum is a welldefined Lorentz invariant object, and the corresponding operator is given from Noether's theorem by the behavior of the fields under rotations. The total angular-momentum is given by the space-space components of an antisymmetric 2nd-rank tensor. These you can map to a three-pseudovector, thereby leaving the manifest covariant notation (which in general is not too good an idea ;-)).

For massless particles, you do not have spin at all but only helicity. One nevertheless talks about "spin", meaning the corresponding quantum number 0, 1/2, 1,... For $s>0$ Helicity can take only two values, namely $\lambda=\pm s$. Helicity is defined as the projection of the total angular momentum to the momentum direction of the particle, and for massless particles this is a Lorentz invariant concept.

11. Jul 20, 2012

### Heirot

Can you please refer me to some literature which further discusses the imposibility of distinction between spin and orbital angular momentum in a Lorent covariant way?

12. Jul 20, 2012

### Bill_K

I would be interested in this also. Especially since I have just given above an unambiguous expression for the orbital angular momentum density.

13. Jul 20, 2012

### bcrowell

Staff Emeritus
When you define a quantity like area or angular momentum as the cross product of two three-vectors, you're exploiting a special fact about three-dimensional space. There is no cross product in a four-dimensional space. It is in some sense "wrong" to want these things to be one-index creatures, and "right" to want them to have two indices.

14. Jul 20, 2012

### Heirot

I completely agree with you. I should rephrase my original question as: how to define J, L and S, all being four-tensors with two indices and corresponding to total, orbital and spin angular momentum, respectively, in an arbitrary frame of reference?

15. Jul 21, 2012

### strangerep

For readers who have a copy of Misner, Thorne & Wheeler, take a look at Box 5.6.
Bill_K's earlier expression for orbital angular momentum density is a special case of what MTW describe in their part A.

Oh heck, I'll just type out an extract...
Later, in part D, they specialize this to
Following MTW's part D a bit further, they explain...

Working in the system's rest frame, this has components
$$S^{0j} = 0 ~,~~~~ S^{jk} = \dots$$
Then define a 3-vector
$$S^i ~:=~ \epsilon^i_{~jk} S^{jk}$$
Finally, define "intrinsic angular momentum 4-vector" $S^\mu$ to be that 4-vector whose components in the rest frame are $(0, S^i)$. This means that
$$S^{\mu\nu} = U_\alpha S_\beta \epsilon^{\alpha\beta\mu\nu} ~,$$ where $U_\beta := P_\beta/M$ is the 4-velocity of the center of mass.

Also,
$$U_\beta S^\beta = 0 ~.$$

Now part E...
The first term is $S^{\mu\nu}$ (intrinsic), and the second term is $L^{\mu\nu}$ (orbital).

Unfortunately, MTW don't bother to show how these two terms don't transform independently under Lorentz transformations in general.

Last edited: Jul 21, 2012
16. Jul 21, 2012

### vanhees71

This derivation by MTW shows that the splitting of total angular momentum in a "spin" and an "orbital angular momentum" makes use of a specific frame, namely the rest frame of the particle. Then this definition is written in covariant form. That's clear definition and that's the only natural meaning you can give to such a concept. With "natural" I mean that it is defined by a physically distinguished reference frame, namely the rest frame of the object in question.

For massless particles, however such a splitting doesn't make sense. Take photons as an example. As far as I know there is no Lorentz and gauge invariant split of the angular momentum in an orbital and an intrinsic part. Here only "helicity" makes sense in a "natural" way, and this should be viewed as the substitute for the intrinsic "spin" part for massive particles.

17. Jul 21, 2012

### Bill_K

Ok, clear enough. The definition is like Jμν = ∫jμνσσ where jμνσ = sμνσ + ℓμνσ. And while the total quantity jμνσ is conserved, jμνσ = 0, implying Jμν is hypersurface independent, the same is not true of sμνσ and ℓμνσ separately. So they ARE hypersurface dependent.

18. Jul 21, 2012

### Heirot

OK, thing are beginning to clarify...

Let us first note the similarities between the four-vector of momentum $P^{\mu}$ and four-tensor of angular momentum $J^{\mu \nu}$. Both have two substantially different ingredients. Symbolically we can write $P^{\mu} = (E, \vec{p})$ and $J^{\mu \nu} = (\vec{J}, \vec{K})$, meaning that the momentum four-vector is composed of energy and momentum three-vector, while the angular momentum four-tensor is composed of angular momentum three-vector and "boost" three-vector. In one specific frame of reference, namely the rest frame, we have $E \to m, \vec{p} \to \vec{0}$ so that $P^{\mu} = (m, \vec{0})$ and $P^{\mu}P_{\mu}=m^2$. On the other hand, for the angular momentum in the rest frame, $\vec{J} \to \vec{S}, \vec{K} \to \vec{0}$ so that $J^{\mu \nu} = (\vec{S}, \vec{0})$ and $J^{\mu \nu} J_{\mu \nu} = \vec{S}^2$.

Therefore, the decomposition of angular momentum into orbital and spin part is meaningful in the same sense as the decomposition of the total energy into rest energy and kinetic energy. But this composition is frame dependent and therefore not covariant.

Also, the question, what is the spin three-vector of a particle measured in the lab frame is the same as asking what is the mass of the particle measured in the lab frame. The point is that lab frame can measure only the total energy (rest plus kinetic) and total angular momentum (orbital plus spin) and then the mass and the spin of the particle can be determined from the Lorentz transformation to the rest frame.