Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's say we have two photons, whose momentum vectors point to opposite directions. Also spin angular momentum vectors of the photons point to opposite directions. (Sum of spins is zero)

Now we change into a frame where the momentum vectors point into almost same direction. Do the spin vectors point into almost same direction in this frame? (Is the sum of spins not zero in this frame?)

(Photons were far away from us, moving to the left and to the right, then we accelerated towards the photons)

(Momentum and spin vectors are either parallel or anti-parallel in massless particles, according quantum-mechanics)

I mean: Can acceleration of an observer change spins of particles that the observer observes?

share|improve this question
add comment

3 Answers

Yes, what you are suggesting is exactly what is happening, but that is if you have an expression which transforms like an axial vector which you can identify with the the spin of the photon. The inherent spin property of photons ($1\hbar$) and electrons ($\tfrac12\hbar$) is of course reference frame independent.

Maybe without realizing you brought forward here the issue of "what is the expression which represents the spin of the electromagnetic field", This field can not be expressed in a gauge invariant way because it contains the vector potential $A^\mu$.

The spin density in the Lorentz gauge would be:

$ {\cal C}^\mu ~~=~~ \epsilon_o\,\tfrac12\varepsilon^{\,\mu\nu\alpha\beta} F_{\alpha\beta}A_\nu ~~=~~ \epsilon_o\,\varepsilon^{\,\mu\alpha\beta\gamma} A_\alpha\partial_\beta A_\gamma $

Which (in vacuum) is equal to.

$ {\cal C}^\mu ~~=~~ \left( \begin{array}{c c c c} ~ 0 &-\tfrac1c\,\mathsf{H}_x &-\tfrac1c\,\mathsf{H}_y &-\tfrac1c\,\mathsf{H}_z \\ \tfrac1c\,\mathsf{H}_x & ~~~ 0 & \ \ ~~\mathsf{D}_z & ~-\mathsf{D}_y \\ \tfrac1c\,\mathsf{H}_y & ~-\mathsf{D}_z & ~~~ 0 & \ \ ~~\mathsf{D}_x \\ \tfrac1c\,\mathsf{H}_z & \ \ ~~\mathsf{D}_y & ~-\mathsf{D}_x & ~~~ 0 \end{array} \right) \left( \begin{array}{c} \ \ A_0 \\ -A_x \\ -A_y \\ -A_z \end{array} \right) $

From this expression one can already see that it transforms like an axial vector. If you go through the trouble of calculating the $A^\mu$ field of a circulating charge using Liénard Wiechert (like i did here) then you get indeed the required $1\hbar$ ratio with the momentum density for circular polarized photons and $0\hbar$ for linear polarized photons.

The latter expression is equivalent to the electron's spin density found via the Gordon decomposition of the axial Dirac current of the electron. In this case the matrix is given by the Magnetization Polarization tensor of the Dirac field while the column vector is given by the dynamic momentum of the electron. (The phase change rates minus the phase induced by the $A^\mu$ field, $\partial_\mu-ieA_\mu$).

share|improve this answer
1  
By the way, @Hans, what enables us to associate axial current with spin density? –  Murod Abdukhakimov Feb 28 '12 at 22:38
add comment

My understanding is that Spin $S$ can be defined as the residual angular momentum in the rest frame. So, you will measure a different angular momentum $J$.

share|improve this answer
add comment

I don't know if what i'm saying something correct since my QM knowledge is limited.

Spin for a photon is a binary quantity associated to the particle via a tensor product. It's not merely and arrow pointing up or down: It is a new property that objects have but it is defined in a space completely different from the position space. In fact one have functions for angular momentum, like the one used for orbitals for atoms and molecules, but not for spin.

So when one make a change of reference it should be specified if it is in the angular momentum space or spin space: one change of reference does not necessary imply the other.

I understand that acceleration involves relativistic consideration that i did not make.

share|improve this answer
add comment

protected by Qmechanic Mar 18 at 8:17

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.