5
$\begingroup$

I'm currently studying particle physics and find the experimental fact that massless particles are always left-handed quite surprising.In fact, the need to add projectors in the lagrangian seems not very fundamental but rather a pragmatic ad-hoc thing theoretical physicists have come up with.

The question is: Does any current unifying theory explain this asymmetry? Maybe there is a theory where left and right are treated equivalently or, at least, there is an explanation to why left-handed particles are "prefered" by nature.

Or maybe this is just the way the universe works and I just have to get used to it... Anyway, I'd like to know what's the insight on chirality in unifying theories.

Improve this question
$\endgroup$
5
  • $\begingroup$ Apparently, $E_8\times E_8$ heterotic string theory would do it once they work it out. Other than Type IIA superstring theories are chiral. $\endgroup$ – Conifold Apr 12 '17 at 3:39
  • $\begingroup$ +1 "... the need to add projectors in the lagrangian seems not very fundamental but rather a pragmatic ad-hoc thing theoretical physicists have come up with." $\endgroup$ – HolgerFiedler Apr 12 '17 at 4:11
  • 1
    $\begingroup$ It does look bizarre because parity transformations go from an inertial frame to what appears to be another reasonable inertial frame (according to every day life of EM + gravity). So Cohen and Glashow had a neat idea where they proposed we started with the wrong symmetry. They call this approach "very special relativity" arxiv.org/abs/hep-ph/0601236 (Which is not a unifying theory, and so not an answer, but I thought the change of view would be relevant.) $\endgroup$ – PPenguin Apr 12 '17 at 7:25
  • 3
    $\begingroup$ But you don't need to add projectors. If you accept that fundamental objects are Weyl spinors, and that a Dirac spinor is two distinct Weyl spinors, coupling by e.g., mass terms, it isn't so ad hoc. $\endgroup$ – innisfree Apr 12 '17 at 9:37
  • $\begingroup$ Is it like the question can be rephrased has "what is the origin of parity violation"? $\endgroup$ – sam Apr 12 '17 at 17:52
4
$\begingroup$

I believe Pati-Salam model is what you're looking for.

Quick overview: It is very different from $SU(5)$ unification. The quark-lepton unification is achieved by adding a fourth color labeled "lilac" for leptons, so e.g. electrons are lilac down-quarks. The strong force symmetry group naturally enlarges to $SU(4)$. At the same time, a Pati-Salam partner $SU(2)_R$ for the $SU(2)_L$ field is introduced, which restores chirality.

UPD: Pati-Salam is just one example of a left-right symmetric GUT. Another models also exist, as mentioned by innisfree@.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Why not any model where a left-right symmetric GUT group is spontaneously broken to SM? Why Pati-Salam? $\endgroup$ – innisfree Apr 12 '17 at 9:39
  • $\begingroup$ @innisfree no reason, it is just the most well-known example (and I believe the first ever constructed). You are right, any other model would suffice as an example. $\endgroup$ – Prof. Legolasov Apr 12 '17 at 10:10
  • $\begingroup$ Ah ok, sure. You could add that in your answer, it's useful info imho $\endgroup$ – innisfree Apr 12 '17 at 10:12
3
$\begingroup$

Massless particles are not all left-handed. A photon, for example, has two perfectly good polarisations.

A Dirac fermion can always be written in terms of a left-handed weyl spinor and the complex conjugate of a left handed spinor (which is to say, a right-handed spinor).

It is often convenient to adopt this convention, but that's all it is. The only reason for preferring left over right is that the weak interactions couple to the left-handed particles. This is particularly noticeable for neutrinos (which are not massless). But a right-handed electron exists and has different hypercharge and weak isospin to the left-handed electron to which it is coupled by the Higgs.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Have you ever pay attention the fact that accelerated in the same manner electrons (in an antenna rod) all emit photons of the same spin? Otherwise the common radio wave wouldn't have such an oscillating electric AND magnetic field component. $\endgroup$ – HolgerFiedler Apr 13 '17 at 4:35
  • $\begingroup$ I'm really not sure how this comment relates to the chiral representations of the Lorentz group to which the OP refers. $\endgroup$ – rwold Apr 15 '17 at 14:08
  • $\begingroup$ rwold Photons are polarised if all of them undergoing n alignment of their electric field component. By this the positive axis could point in two opposite directions only So for an antenna radiation the photons in one half wave are all directed in one direction and in the other half wave are directed in the opposite direction. Now the point: the direction of the magnetic field component is strongly correlated to the direction of the electric field. That means that only one of the two spin directions takes place. $\endgroup$ – HolgerFiedler Apr 15 '17 at 14:27
  • $\begingroup$ I really don't think this is related to the question. This is a property of photons in a particular circumstance, whilst the question assumes that all massless particles are intrinsically left-handed. $\endgroup$ – rwold Apr 17 '17 at 12:31
  • $\begingroup$ It was a remark only to your statement that "A photon, for example, has two perfectly good polarisations." $\endgroup$ – HolgerFiedler Apr 17 '17 at 13:22
1
$\begingroup$

In the Standard Model (SM) of particle physics, fermions are taken to be left handed. The reason is to incorporate the parity violation in weak interactions. Which is a fact of nature. If one check then in will be seen that $V-A$ type current can account for the inclusion of Left handed fermions ($V+A$ type current may also possible, but its ruled but by experiment).

Minimal extension of SM is the Left-Right symmetric model, proposed by Rabindra Mohapatra and Goran Senjanovic back late seventies, based on gauge group $SU(2)_{L}\times SU(2)_{R}\times U(1)_{B-L}$. In this model the fermion fields are assigned to the doublets

$$L_{L}^{i}= \begin{pmatrix}\nu \\ e^- \end{pmatrix}_{L}^{i}\qquad L_{R}^{i}= \begin{pmatrix}\nu \\ e^- \end{pmatrix}_{R}^{i} \qquad Q_{L}^{i}= \begin{pmatrix} u \\ d \end{pmatrix}_{L}^{i} \qquad Q_{R}^{i}= \begin{pmatrix} u \\ d \end{pmatrix}_{R}^{i}$$

It is obvious from the field representations that both left and right handed fermions entered in the game with both hands. The transformation between $L$ and $R$ fields accomplished parity, and impose parity invariance before spontaneous breaking down $U(1)_{e}$.

This model has interesting features and anomaly free, and looks more symmetric compared to SM itself. But despite intense search, no data has found yet supporting this model.

An interesting paper published by Roni Harnik et al recently. Where authors considered the possibility of an universe without weak interactions. They provided theoretical arguments that it is indeed possible to have a stable universe with nucleosynthesis, matter domination, structure formation, in the absence of weak interactions, which is responsible for parity violation.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Interesting to know from a historical point of view, but it looks like this doesn't attempt to explain the asymmetry (?) $\endgroup$ – Helen Apr 12 '17 at 17:40
  • $\begingroup$ @Helen he mentions the proposed extension $SU(2)_L\times SU(2)_R\times ...$ which spontaneously breaks down to the SM . $\endgroup$ – Kosm Apr 12 '17 at 17:44
  • $\begingroup$ What do you mean fermions are taken to be left-handed? Sure, only LH participate in weak interactions, but e.g. RH charged leptons exist and interact vectorially with the photon. $\endgroup$ – innisfree Apr 13 '17 at 1:29
  • $\begingroup$ @Kosm, I see. But this sounds more like description than "explanation". $\endgroup$ – Helen Apr 13 '17 at 4:15
0
$\begingroup$

The theory presented in Schmelzer, I.: A Condensed Matter Interpretation of SM Fermions and Gauge Fields, Found. Phys. vol. 39, 1, p. 73-107 (2009), arXiv:0908.0591, predicts the fermionic sector, the SM gauge group, and its action on the fermions. See http://ilja-schmelzer.de/matter for a popular presentation.

Given that it predicts the gauge group only as a maximal group fulfilling some assumptions, and the left-right reversed action would be another maximal one, it does not exactly answer the question why the left-handed. It could have been the right-handed as well. But they are above chiral.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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