Polarization states via relativistic effect

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  • #1
nomadreid
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In the following article

http://books.google.com/books?id=u1...&resnum=1&ved=0CBMQ6AEwAA#v=onepage&q&f=false

it states,
"massless guage particles...can have only two independent states of polarization....but the ...massive guage particles ....have three independent states of polarization....(the difference in the number of polarization states for massive and massless particles of the same spin is a relativistic effect.)" (my emphasis)

Could someone briefly explain, not too technically, what relativistic effect is being referred to, and how it accounts for the extra polarization state?
 

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  • #2
Bill_K
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A particle with mass has a rest frame. In that frame, assuming the particle has spin S, there are 2S+1 spin states. Thus a massive vector boson has three spin states.

A massless particle does not have a rest frame. The spin projection along its direction of motion is the same in every frame. Hence there will be only one spin state. Except - if the particle's interactions conserve parity the particle's mirror image will also exist, and it will have the opposite spin. Thus a parity-conserving massless particle has two spin states.
 
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nomadreid
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Thank you, Bill_K. That clears up my question.
 
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Actually I don't quite get it!
What if the massless photon has perpendicular polarization?
 
  • #5
nomadreid
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karlzr, it would help if you could be precise as to what difficulty this presents.
 
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karlzr, it would help if you could be precise as to what difficulty this presents.
I try to consider what would happen if a perpendicular spin vector was lorentz boosted to another frame, but there is a related question I feel confused.
Since spin exists only in quantum world, [tex]S_x[/tex],[tex]S_y[/tex],[tex]S_z[/tex] don't commute, why can we write spin into a four-vector form? and what is [tex]S_0[/tex]?
 
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bcrowell
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What if the massless photon has perpendicular polarization?
The photon does have perpendicular polarization.

We observe that electromagnetic waves are always polarized transversely, never longitudinally. Such a constraint can only apply to a wave that propagates at c. If it applied to a wave that propagated at less than c, we could move into a frame of reference in which the wave was at rest. In this frame, all directions in space would be equivalent, and there would be no way to decide which directions of polarization should be permitted.

A massless particle does not have a rest frame. The spin projection along its direction of motion is the same in every frame. Hence there will be only one spin state. Except - if the particle's interactions conserve parity the particle's mirror image will also exist, and it will have the opposite spin. Thus a parity-conserving massless particle has two spin states.
I'm intrigued by this part of your argument, but I don't quite understand it. You say, "Hence there will be only one spin state." You seem to be saying that if there is a certain type of massless particle T (T=photons, say), then by default we should expect all particles of that type to have the same helicity. It seems to me that this should be a much weaker statement. Say T has spin 1, and I propose four laws of physics: (1) All particles of type T have helicity h=+1. (2) All particles of type T have h=-1. (3) All particles of type T have either h=+1 or h=-1. (4) Particles of type T exist with h=-1, 0, or +1. I don't see any obvious way to rule out any of these based on simple arguments about Lorentz invariance.

In the second part of your argument, "if the particle's interactions conserve parity...," you talk about the particle's interactions, but the connection to what follows isn't clear to me, because you never again mention it interacting with anything. I'm sure what you're saying is valid, but it's a little unclear to me because it seems like you must be omitting a step that's obvious to you but not to me.
 
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bcrowell
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I try to consider what would happen if a perpendicular spin vector was lorentz boosted to another frame, but there is a related question I feel confused.
Since spin exists only in quantum world, [tex]S_x[/tex],[tex]S_y[/tex],[tex]S_z[/tex] don't commute, why can we write spin into a four-vector form? and what is [tex]S_0[/tex]?
By the way, to prevent the line breaks you can use itex tags instead of tex.

Even in Newtonian mechanics, it would be simpler in some sense to think of angular momentum as a tensor rather than a vector. It's only a coincidence that in three-dimensional Euclidean space we can reduce the tensor to something that transforms like a vector.

This may be helpful: http://physics.stackexchange.com/qu...ngular-momentum-in-special-general-relativity
 
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The photon does have perpendicular polarization.

We observe that electromagnetic waves are always polarized transversely, never longitudinally. Such a constraint can only apply to a wave that propagates at c. If it applied to a wave that propagated at less than c, we could move into a frame of reference in which the wave was at rest. In this frame, all directions in space would be equivalent, and there would be no way to decide which directions of polarization should be permitted.
I try to interpret your statement. For a massive particle, a positive helicity state can always be transformed to a negative helicity or zero helicity after a lorentz tranformation. So there must be three polorizations (as for this point, I believe this result should also be achieved through a vector or tensor transformation [itex]S'=\Lambda S[/itex], where [itex]\Lambda[/itex] is the transformation matrix, but how?).
However, this is not the case for massless particles like photon. A photon with positive helicity could never be transformed to negative helicity state. But I can see why there is no photon with zero helicity? could you please give a more detailed explanation? Thanks for your time!




By the way, to prevent the line breaks you can use itex tags instead of tex.

Even in Newtonian mechanics, it would be simpler in some sense to think of angular momentum as a tensor rather than a vector. It's only a coincidence that in three-dimensional Euclidean space we can reduce the tensor to something that transforms like a vector.
……
Thanks!
But still, can we put noncommutative quantities such as [itex]s_x[/itex], [itex]s_y[/itex], [itex]s_z[/itex] together into the same tensor? After all, they can never be determined simultaneously.
 
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bcrowell
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as for this point, I believe this result should also be achieved through a vector or tensor transformation [itex]S'=\Lambda S[/itex], where [itex]\Lambda[/itex] is the transformation matrix, but how?).
You actually need two [itex]\Lambda[/itex]'s in there, because S is a rank-2 tensor, as described in the stackexchange link I posted above.

But I can see why there is no photon with zero helicity?
I don't know. I think this corresponds to the fact that there are no longitudinally polarized photons, but I don't have a good explanation for why there are no longitudinally polarized photons. If you don't get a good answer here, maybe try the qm or particle physics forums.

But still, can we put noncommutative quantities such as [itex]s_x[/itex], [itex]s_y[/itex], [itex]s_z[/itex] together into the same tensor? After all, they can never be determined simultaneously.
The issue isn't that they don't commute, the issue is that the collection of the three of them makes a three-vector, not a relativistic four-vector. That's why the rank-2 tensor works better in relativity -- it transforms like a tensor.
 

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