Consider a frame of coordinates F, in movement with a velocity V with respect to the lab. Assume that we prepare in the lab a beam of spin 1 bosons, polarized along a direction Q which makes an angle θ with the direction of movement of the frame F.
What will be the direction of polarization of the beam from the point of view of an observer in F?
In the classical mechanics the angular momentum is defined as L = r×p where r is the position of the object and p its linear momentum. To calculate L according to the frame F given r and p in the lab, is not a problem. But how to proceed with the spin. It's an intrinsic property, it is quantized.
I would appreciate advice.
Pauli-Lubansky pseudo vector is defined as Wμ=1/2 εμνρσ Jμρ Pσ where J and P are angular momentum and linear momentum operators. WμWμ is a Lorentz invariant operator which commutes with momentum operator. Similarly PμPμ operator is also Lorentz invariant operator. These two operators can be used to label irreducible representations of the Lorentz group. Because any elementary particle state should be a representation of the Lorentz group, we can label these states by their rest mass and intrinsic spin which are Lorentz invariant quantities; that is these numbers are universally valid for all those observers who are connected via Lorentz transformations.
PS: by the term "Lorentz invariant operator" what we mean is that it commutes with generators of Lorentz transformation. "Intrinsic spin" angular momentum is the angular momentum of a free particle as opposed to the orbital angular momentum of a particle under the influence of an external force field due to it's presence inside an atom or in an atomic nucleus.
Note: If we change z --> -z i.e. a reflection then a spin up state will become a spin down state.
Actually in the symmetric phase (before Higgs mechanism) fermions were indeed mass-less due to a chiral nature of the standard model. Left and right handed projections had different transformation properties under weak isospin and hypercharge. In principle both the left and right handed projections are allowed in a quantum field theory and hence they can pair up in the Lagrangian, unless such pairing is forbidden by extra discrete symmetries. The coefficients of the term ψ̅LψR are arbitrary complex numbers.
Mike, thank you, but from home I don't have access to Rev. Mod. Phys. Do you have the possibility to send me a copy of Wigner's article?
With thanks in advance,
Sofia
Dear Mike, would you please send the expected copy also to me. Thanks Hanno
Sophia:
For the case of the case of the electron, which should give you some insight into the spin angular momentum of other particles, see www.tachyonmodel.con, Section 1, and especially Section 1.6.
I have not updated this for the proton or other particles,
as yet.
Mike, I don't possess that book, and I don't feel like buying a book for one article. If you have the PDF of that article, can't you send me a copy?
With thanks in advance,
Sofia
Dear Mike, I sent you my e-mail by a personal message.
Many, many thanks,
Sofia
For the electron we can define two different angular momenta: One with respect to the center of mass and another with respect to the center of charge. Dirac spin operator represents the angular momentum with respect to the center of charge and it transforms under a Poincare transformation llike the space-space part of an antisymmetric tensor.
See attached my contribution:
Kinematical Theory of elementary spinning particles
Research Kinematical Theory of Elementary Spinning Particles. Lecture Notes
“…For the electron we can define two different angular momenta: One with respect to the center of mass and another with respect to the center of charge…”
For the electron (and for any other fundamental particle that has rest mass) we indeed can define two different angular momentums – one with respect to 4-th, i.e. coordinate time, axis, and another with respect to some 3D spatial direction, e.g. to Z-axis in the Matter’s [5]4D Euclidian spacetime; and at that seems indeed both momentums are equal to 1, when their “spatial sum” is equal to ½.
When it seems rather probably that at corresponding rotation of something in the particle, what [the something] has corresponding linear momentum, it is impossible to define a rotation of this something around “the center of mass” or “the center of charge” – as well to define what are those centers …
Cheers
What can be shown is that if an elementary particle has a center of mass(CM) and a center of charge(CC) which are different points, then the CC moves around the CM in circles at the speed of light. This relative motion produces the magnetic moment and contributes to the spin of the particle. Other part of the spin is coming from the rotation of the particle. The existence of two separate centers allows us to define two angular momenta, one about each point. The angular momentum w.r.t. the CC transforms like the space-space part of a four-dimensional angular momentum tensor while the angular momentum w.r.t. the CM does not. Both angular momenta satisfy different dynamical equations. The spin w.r.t. the CC satisfy the same dynamical equation than Dirac's spin operator in the quantum case, while the CM spin satisfy Bargmann-Michel-Telegdi dynamical equation.
Best regards
@Martin Rivas:
“…What can be shown is that if an elementary particle has a center of mass(CM)..”
to define a center of mass [around what the masses rotate with an angular momentum] seems necessary to have at least two masses – or a spatially distributed one mass; so what physically are those masses/mass when the showing above? And what [values] are those “respect to CM” and “respect to CC” angular momentums?
Cheers
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