# The Dirac Equation and the Neutrinos

1. Nov 2, 2007

### Magister

Does the Dirac equation predicts the fact that there are no right handed neutrinos?

2. Nov 2, 2007

### blechman

No. There's nothing that "predicts" such a thing. It's an assumption of the standard model, partially justified by experiment. I say "partially" because even though we have never seen a right-handed neutrino, we know neutrinos have mass. RH neutrinos are not required for this, but that is one way to give them mass.

3. Nov 2, 2007

### Magister

But the Dirac equation does "accept" the fact of the non-existence of the RH neutrinos, doesn´t it?

4. Nov 2, 2007

### Meir Achuz

No. The term "right handed neutrinos" is ambiguous. Now that neutrinos are believed to have mass, the probability of a left handed neutrino in beta decay having right helicity is (1-v^2). The "handedness" of a particle is not related to the Dirac eqluation, but to whether its weak interaction is (V-A) or (V+A).

5. Nov 2, 2007

### Meir Achuz

I just saw this. The DE says nothing about the existence or non-exisence of handeness.
Handedness is a feature of the weak interaction.

6. Nov 2, 2007

### blechman

Be careful: there's "handedness" and there's "chirality". In the massless fermion limit, they are the same thing, but in the (Dirac) massive case, they are not. "Handedness" refers to the spin polarization of the fermion ($\pm\hbar/2$), and has to do with the Lorentz transformation properties of the fermion. "Chirality" has to do with which of the (4) components of the spinor you are dealing with. They're only the same thing in the (Dirac) massless fermion limit. In particular, handedness is not Lorentz invariant for a massive fermion, since you can boost "past" the fermion and change the direction of helicity (massless fermions are moving at the speed of light, so you can't do that for them). However, chirality *is* a Lorentz invariant quantity.

What I think Meir Achuz means is that the weak interactions couple to left-chiral and right-chiral fermions differently (in the latter case, not at all!).

The Dirac equation describes a 4-component fermion, which contains both a Left and a Right chiral fermion. However, it can be decomposed into two separate equations (coupled by a Dirac mass term) that describes 2-component fermions of a given chirality. These equations are sometimes called the Weyl Equations. These are, in some sense, more fundamental than Dirac's equation. When using Weyl's equations, you only need one kind of chirality.

The SM is a chiral theory, meaning that the fermions actually obey Weyl's equations. For this reason, you do not need a R-chiral neutrino. I suppose I was a little sloppy before: I guess that the Dirac equation DOES insist on a R-handed neutrino. But if there is no Dirac mass, then the R-handed neutrino decouples completely and so it might as well not be there (use the Weyl equations). That's what I meant when I said Dirac does not imply R-handed.

Of course, physicists are super-sloppy: we use "handed" and "chiral" interchangably, even when we aren't supposed to!

7. Nov 2, 2007

### Hans de Vries

The DE does describe the handedness as a non-Lorentz invariant
combination of the two chiral components. The chiral terms do have
a Lorentz invariant handedness because they transform light-like
always whether the Dirac electron has mass or not.

$$\begin{array}{|clcccrc|} \hline &&&&&& \\ &\mbox{Momentum:} && J_{Vt}^2\ -\ J_{Vx}^2\ -\ J_{Vy}^2\ -\ J_{Vz}^2\ & = & \ \ (2m)^2 & \\ &&&&&& \\ &\mbox{Spin (axial):}&& J_{At}^2\ -\ J_{Ax}^2\ -\ J_{Ay}^2\ -\ J_{Az}^2\ & = & -(2m)^2 & \\ &&&&&& \\ &\mbox{Left chiral:} && J_{Lt}^2\ -\ J_{Lx}^2\ -\ J_{Ly}^2\ -\ J_{Lz}^2\ & = & \ 0 \quad & \\ &&&&&& \\ &\mbox{Right chiral:}&& J_{Rt}^2\ -\ J_{Rx}^2\ -\ J_{Ry}^2\ -\ J_{Rz}^2\ & = & \ 0 \quad & \\ &&&&&& \\ \hline \end{array}$$

The Chiral terms have definite handedness but they are however only
exactly aligned with the spin (axial current) in frames where V and A
are aligned. So there can be an up to 90 degrees angle between a chiral
component and the spin in a reference frame but never more. The dot
product between the two has a definite sign.

The vector and axial current do of-course tend to line up almost 100%
at relativistic speeds because of the way spin transforms:

$$\vec{J}_A\ =\ \vec{J}_{A}^{\ rest}\ +\ \frac{\gamma^2}{\gamma+1}\ (\ \vec{\beta}\cdot \vec{J}_{A}^{\ rest}\ )\ \vec{\beta}$$

The spin pointer grows always along the velocity with a sign depending
on the dot product. See Jackson (11.159) The notion that the spin is
aligned with the momentum and the chiral components is only true in the
ultra relativistic case.

It's the light like behavior of the chiral terms (whether the Dirac electron
has mass or not) what makes the SM Lorentz invariant. From a path
integral point of view one can consider the square of the amplitude of the
chiral components as the average time which is spend moving in one or
the other direction with the speed of c, with the overall speed being equal
to the speed v of the electron.

Regards, Hans

8. Nov 2, 2007

### Meir Achuz

"Be careful: there's "handedness" and there's "chirality". In the massless fermion limit, they are the same thing, but in the (Dirac) massive case, they are not. "Handedness" refers to the spin polarization of the fermion (LaTeX graphic is being generated. Reload this page in a moment.), and has to do with the Lorentz transformation properties of the fermion."

I hate to argue over words, since I think we are in basic agreement.
That is why I said "ambiguous". I would call your definition of "handedness", helicity,
with handedness used for the two possible chiralities.
I would say that chiral invariance puts fermions into two classes,
right handed and left handed, with the mneomnic that leptons are leftons.

9. Nov 2, 2007

### blechman

I know. It actually took me a while before I finally understood that there are two concepts being described by the same words. Back in grad school, I used to say to myself (and others): "How can you have no 'right-handed' neutrinos when neutrinos have mass, and you can always boost into a frame where it is 'right-handed' (that is, positive helicity)?!" The key is to understand the difference between "helicity" and "chirality", which are only the same in the massless limit.