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I am guessing that the angular momentum of a particle measured by different inertial observers (no gravity) is different. At the same time, since quantum spin is an intrinsic property,

  1. Is it true that spin angular momentum is lorentz invariant ?
  2. If so, how does it reconcile with the fact that being an angular momentum, it should change ?
  3. given some unknown particle's total angular momentum, is there any way to partition it into spin and other angular momentum ?

I am thinking (1) is true but am undecided about 2. and 3. Any help is appreciated.

Maybe for 2., all the lorentz transformation is obligated to do is change the total angular momentum but not necessarily its contributors so spin itself could stay constant. In this case, wouldn't 3. be affirmative ?.

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – G. Smith Dec 4 '19 at 6:18
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    $\begingroup$ The magnitude of the spin is a Lorentz-invariant, but the spin itself is a four-vector (or antisymmetric two-index four-tensor) that transforms according to the usual Lorentz transformation rules. $\endgroup$ – G. Smith Dec 4 '19 at 6:22
  • $\begingroup$ @G.Smith so does spin 4 vector change independently of the total angular momentum 4 vector? $\endgroup$ – lineage Dec 4 '19 at 6:38
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    $\begingroup$ Not only can it have independent dynamics, it can exist independently. You can have spin without any orbital angular momentum. $\endgroup$ – G. Smith Dec 4 '19 at 6:41
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The spin of a particle is defined by its intrinsic angular momentum and as such is defined in the particle's rest frame. So your spin four vector in the rest frame must take the form $$S^\mu = (0, \mathbf S) $$ Where the 3-vector $\mathbf S$ is your familiar non-relativistic spin.

The four vector will transform precisely thus, in the vector representation of the Lorentz group, so in an arbitrary frame it will have components $$S^\mu = (s, \mathbf S'), $$ where since the length of the vector is invariant we must have $-s^2 + \mathbf S^{\prime 2}= \mathbf S^2$. In fact we can show that the zeroth component of the spin vector is proportional to the helicity of the particle (an invariant): $s = h|\mathbf p|$ where $\mathbf p$ is the (non-invariant) three momentum of the particle in whatever frame you're in and h is its helicity. Helicity is essentially the projection of the particle's spin in the direction of its motion - - as such for massless particles spin coincides with helicity.

We can give an explicit form for the spin vector following Pauli-Lubanski that has the properties mentioned above. Let (their notation) $$W^\mu := \frac{1}{2} \varepsilon^{\mu}{} _{\nu \alpha \beta} P^\nu M^{\alpha \beta } $$ With $P^\nu$ the components of the four momentum and $M^{\alpha \beta }$ the (matrix valued) generators of the Lorentz group (for vectors this is just the orbital angular momentum but for the spin 1/2 representation there is an additional piece involving the $\gamma$ matrices). Note that in the rest frame, $P^\nu = (t, \mathbf 0)$ and the anti-symmetry of the Levi-Civita tensor implies that $W^0=0.$ we can get a more general relation than this by noting that in all reference frames $P_\mu W^\mu =0$ so the spin vector is covariantly orthogonal to the particle velocity.

In fact the Pauli-Lubanski (pseudo-)vector is a Casimir of the Poincaré group and is used extensively in the classification of irreducible representations of the Lorentz group.

Finally OP asked about orbital (L) and spin (S) contributions - the Lorentz generators are built up from the sum of an orbital term and a spin term, $$M^{\mu \nu} = x^\mu p^\nu - x^\nu p^\mu + S^{\mu \nu} $$ where the final term depends upon the model (for the Dirac field it's proportional to $ [\gamma^\mu, \gamma^\nu] $). The generators transform in the two tensor rep of the Lorentz group which ensures that the spin vector transforms in the fundamental rep - I.e. As a four vector.

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  • $\begingroup$ So your spin four vector in the rest frame must take the form Sμ=(0,S) This logic doesn't work. If we apply it to other quantities, such as force or electric field, it leads to a false result. $\endgroup$ – Ben Crowell Dec 5 '19 at 21:44
  • $\begingroup$ @lux thanks for the detailed answer (explaining concepts i was hitherto unaware of). However can you please explain the reasoning in the context of the specific questions i have asked. More precisely, I understand that answer to 1 is false since $S$ is a 4 vector but how does it transform when seen from the perspective of being an ordinary angular momentum? Would the total $L+S$ transform or they transform separately? $\endgroup$ – lineage Dec 6 '19 at 13:48
  • $\begingroup$ The spin vector is defined in the rest frame as stated - what is spin if not the angular momentum observed when it's orbital part vanishes? See en.m.wikipedia.org/wiki/Relativistic_angular_momentum#Four-spin to see that this is not my definiton but the accepted one. I agree it does not work for other quantities but I do no suggest applying the same logic to force or electric fields. $\endgroup$ – lux Dec 6 '19 at 14:50
  • $\begingroup$ As for the separation into orbital and spin contributions I have now edited the answer $\endgroup$ – lux Dec 6 '19 at 14:50

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