Anchovy said:
OK, a couple of things:
- Does this mean a fermion can be left-handed, right-handed or neither-handed? I never thought about the third option before but if Lorentz boosting can change a reference frame enough to reverse a particle's handedness, then boosting into this particle's own frame I suppose makes its handedness disappear... so that gets you three possible spin projections for a massive particle? But a normal observer could never reach or exceed c, so that leaves only reference frames that can have two possible spin projections for the photon?
You have to distinguish between "handedness=chirality" and "helicity". Handedness is a pretty abstract notion about certain representations of the proper orthochronous Lorentz group. Each representation of the Lorentz group is characterized by two "pseudo-spin" quantum numbers, which are defined through the Lie algebra of the Lorentz group, which consists of 6 basis elements that are the infinitesimal generators of boosts and rotations. For a fermion with spin 1/2 (the spin tells you the representation of the rotation group of a massive particle in its rest frame) there are two representations of the proper orthochronous Lorentz group, called (1/2,0) and (0,1/2). Each of these representations is on two-dimensional spinor space and these spinors are called Weyl fermions. If you like to have a theory that is also invariant under spatial reflections, you have to use both representations and link them together in a direct sum of the spinor spaces. This leads you to Dirac fermions, which are four-component objects consisting of the two types of Weyl fermions. This representation of the Lorentzgroup, containing space reflections in addition to boosts and rotations of the proper orthochronous group, is called ##(1/2,0) \oplus (0,1/2)##. In this way to describe these Dirac spinors a space reflection just exchanges the two Weyl spinors which compose the Dirac spinor, and that's why you call one of these Weyl spinors the left- and the other the right-handed part.
Helicity is the projection of the total (!) angular momentum of a particle to the direction of its momentum. For a massive particle it's obvious that this is not a Lorentz invariant quantum number, because a massive particle always goes with a velocity less than the speed of light, so that you can overtake it, which flips its helicity. That means that a particle which has positive helicity for one observer can have negative helicity for another observer boosted against the first.
This changes for massless particles. They always go with the speed of light in any frame of reference, and you cannot overtake them. Thus all observer agree upon the sign of the helicity (supposed the particle is prepared with definite helicity, i.e., in a specific polarization state). The mathematical formalism shows that for massless Dirac fermions helicity and chirality are the same, but our argument above shows that for massive particles that's not the case: While chirality is a Lorentz invariant quantum number chirality can change by boosts.
Of course particles do not necessarily have definite chirality or helicity, but you have to prepare the particles in such a state if you want to have them in such a state. That's not such an easy task. E.g., the Relativistic Heavy Ion Collider (RHIC) at the Brookhave National Lab does experiments with polarized protons to learn more details about the inner structure of them (in this case about the socalled generalized parton distribution functions and how protons get their spin 1/2 in terms of its "constituents", quarks and gluons).
- This is strange to me. I'm thinking along the lines of the EM wave having the E part oscillating in one plane, and the M part in another... so that's the two transverse modes? But, I must be mistaken in what you're referring to here because, as far as I'm aware, Z boson, say, doesn't have two components oscillating in planes offset by pi/2 like an EM wave... So how can it be said to have more than one transverse mode?
Massless fields with spin ##\geq 1## are not so easy to understand, because they are represented as socalled gauge fields. This you know from classical electrodynamics. From the point of view of the four-vector potential the very same physical situation is represented by a large set of fields, i.e., if ##A_{\mu}## is a four-vector potential describing a physical situation, with any scalar field ##\chi## the vector potential ##A_{\mu}'=A_{\mu}+\partial_{\mu} \chi## discribes exactly the same situation as ##A_{\mu}##.
In classical electromagnetism it's very simple to come to a unique description of the physical situation. You simply don't look at the four-vector potential which is just a mathematical simplification to solve the Maxwell equations, but only at the physically observable electromagnetic field. Relativistically it's described by the Faraday tensor ##F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}##, and thus changing from ##A_{\mu}## to ##A_{\mu}'## doesn't change ##F_{\mu \nu}##. That's gauge invariance: For the physics the concrete gauge of the four-potential is irrelevant and thus the four-potential is only defined up to a gauge transformation, while the observable electromagnetic field ##F_{\mu \nu}## uniquely describes your physical situation.
Now solving the Maxwell equations for the free electromagnetic field, i.e., in absence of charges and currents, you find that the plane waves are a complete basis in the sense that you can describe all electromagnetic fields in terms of Fourier integrals. It turns out that of the original four independent field-degrees of freedom ##A_{\mu}## or the six independent field-degrees of freedom ##F_{\mu \nu}## there's only freedom for two independent field degrees of freedom.
That's easier to see in the non-covariant (3+1)-dimensional formalism, working with ##\vec{E}## and ##\vec{B}## as the electric and magnetic components of the electromagnetic field. You start with the electric field components making the plane-wave ansatz
$$\vec{E}(t,\vec{x})=\text{Re} \left [\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) \right],$$
where ##\vec{E}_0## is an arbitrary complex ##\mathbb{C}^3## constant vector. Of course the field itself is real, i.e., we describe it as the real part only. It's, however simpler to work with the complex field, and we are allowed to do so, because the Maxwell equations are linear real equations. So that we can calculate with the complex field and take the real part at the very end of the calculation.
Now let's see, which constraints the parameters ##\vec{E}_0##, ##\omega##, and ##\vec{k}## follow from Maxwell's equations. The most simple is Gauß's Law for the electric field. Since we assume that there are no charges present we get
$$\vec{\nabla} \cdot \vec{E}=0, \quad \vec{k} \cdot \vec{E}_0=0.$$
That means that for any ##\vec{k} \in \mathbb{R}^3## the amplitude ##\vec{E}_0## is perpendicular to ##\vec{k}##; ##\vec{k}## describes the direction of the wave propation, and that means that electromagnetic waves are transverse. So out of the three field-degrees of freedom ##\vec{E}## only two are independent.
Now we need to fulfill also the other free Maxwell equations (Heaviside-Lorentz units with ##c=1##),
$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \times \vec{B} - \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
These are the Faraday, the Ampere-Maxwell, and Gauß's Law for the magnetic field, respectively.
To disentangle ##\vec{E}## and ##\vec{B}## we take th curl of the Faraday Law:
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})+\vec{\nabla} \times \dot{\vec{B}}=\vec{\nabla} (\vec{\nabla} \cdot \vec{E})-\Delta \vec{E} +\vec{\nabla} \times \dot{\vec{B}}=-\Delta \vec{E} +\vec{\nabla} \times \dot{\vec{B}}=0,$$
where in the last step we have used Gauß's Law for the electric field and ##\rho=0## again.
The time derivative of the Ampere-Maxwell Law reads
$$\vec{\nabla} \times \dot{\vec{B}}=\partial_t^2 \vec{E}. \qquad (*)$$
This we plug in the previous equation, leading to the wave equation
$$\partial_t^2 \vec{E}-\Delta \vec{E}=0,$$
and for our plane-wave ansatz this gives the dispersion relation
$$\omega^2=\vec{k}^2,$$
as it should be.
The magnetic field is now determined from Faraday's Law. With the plane-wave ansatz
$$\vec{B}=\text{Re} \left [\vec{B}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) \right]$$
we find for the complex field
$$\partial_t \vec{B}=-\mathrm{i} \omega \vec{B}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}) = -\vec{\nabla} \times \vec{E}=-\mathrm{i} \vec{k} \times \vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}).$$
This means
$$\vec{B}_0=\frac{\vec{k}}{\omega} \times \vec{E}_0,$$
i.e., the amplitude of the magnetic field is completely determined by that of the electric field, and it's also transverse. The vectors ##\vec{k}##, ##\vec{E}## and ##\vec{B}## build a orthogonal righthanded set of vectors.
All together we have only two independent field-degrees of freedom for the massless vector field.
You can do the same exercise for a massive vector field, leading to the Proca equation. The key point is that a massive vector field is not necessarily a gauge field. The Proca equations lead to three independent field degrees of freedom rather than 2.
