# Experiments to find right handed neutrinos

The question of how neutrino masses arise has not been answered conclusively. In the Standard Model of particle physics, fermions only have mass because of interactions with the Higgs field (see Higgs boson). These interactions involve both left- and right-handed versions of the fermion (see chirality). However, only left-handed neutrinos have been observed so far.

An then I have found a link therein (on Wikipedia) which lead to this book:

But so far I have found nothing searhing for right-handed neutrinos, or any experiments to look for them.

Why are right hand neutrinos unaffected by all forces except gravity

If the mass of neutrino is not zero, why we cannot find right-handed neutrinos and left-handed anti-neutrinos?

Question:

1. Are there any recent experiments that are aimed at finding right-handed neutrinos?
• Hmmm ... righthanded matter type neutrinos require that neutrinos have Dirac rather than Majorana nature. Neutrinoless double beta decay would be a signature for Majorana nature (and therefore rules out right handed neutrinos) and discovery or near-discover scale experiments are underway as we type. Jun 8, 2018 at 0:00
• thank you do you have any link to any such kid of experiment (or their preparation)? Jun 8, 2018 at 0:13
• I have a colleague working on COURE, and they have some competitors. Jun 8, 2018 at 0:16

The simplest way to incorporate massive neutrinos into the SM is to make them Dirac particles much like quarks, but with itty-bitty couplings to the Higgs field. Let’s take this much on faith.

The SM predicts that right-handed neutrinos will interact with Higgs scalars but not with W, Z, or photons. Interactions with scalars flip helicities. I cannot think of any experimentally feasible process involving incoming neutrinos and virtual Higgs exchange (e.g., ${{\nu }_{R}}+p\to {{\nu }_{L}}+p$) that would have a telltale signature. Besides being very feeble due to the itty-bitty couplings, such processes would be mistaken for Z exchange, even though there would be a subtle difference in the angular distributions of products.

Although The question already has an answer I feel like I should contribute some complementary information.

As you already pointed out right-handed neutrinos do not interact via the usual gauge interactions i.e. electromagnetic, strong and weak forces. That's why they are often called "sterile". And as Bert Barrois pointed out interactions via virtual Higgs are possible but in practice hard to detect.

However, there are experiments that search for right-handed sterile neutrinos. Those that come to my mind are experiments investigating neutrino oscillations. During the investigation of the oscillation properties of neutrinos produced at nuclear reactors the so-called Reactor Antineutrino Anomaly (RAA) has been observed i.e. the measured neutrino flux in short distance to the reactors was lower than expected (if one assumes the usual oscillation of neutrinos between three flavors). Those "missing" neutrinos can be explained by the oscillation of reactor neutrinos into a fourth neutrino type. For more detail on the RAA see e.g. this link .

The red line in this graph shows the expected data for the usual 3 flavor oscillations. As you see the tendency of the neutrino events close to the reactor is to be lower than expected. However, this anomaly could in principle still be statistical. Experiments that want to further analyze this with respect to the possible existence of right-handed neutrinos are e.g. the STEREO experiment.

That being said I want to add something regarding the comments about Majorana Neutrinos. The statement that neutrinos being Majorana neutrinos would rule out the existence of right-handed neutrinos is wrong.

Let me explain this.

In the standard model Lagrangian a Dirac particle $$\Psi$$ consisting of two chiral fields $$\Psi_R$$ and $$\Psi_L$$ has a mass term which looks like $$\mathcal{L}_D=m\overline{\Psi_R}\Psi_L$$ where m is generated by the coupling to the Higgs and its non-zero vacuum expectation value (VEV). That's why for neutrinos being Dirac particles we need a right-handed neutrino to form such a mass term.

The reason that we need the Higgs to explain the existence of fermion masses is that such a mass term needs to be a single with respect to the standard model gauge group SU3 x SU2$$_L$$ x U1$$_Y$$. This basically means the overall charges of the term need to be zero if you add up all fields inside. That, by the way, is the reason why we need the Higgs field. The weak force (represented by the SU2$$_L$$ group) only couples to left-handed particles. Therefore right-handed fields are not charged under it while left-handed fields are. This is solved by using the Higgs field to compensate the charge of the left-handed Field and therefore allow for mass terms to existing.

Now let's have a look at Majorana particles. Majorana particles are their own antiparticles $$\Psi=\Psi^C$$

The interesting thing is that this particle-antiparticle conjugation $$C$$ flips the chirality of a field $$(\Psi_R)^C=(\Psi^C)_L$$ Hence we can use this to form a Majorana mass term which looks like $$\mathcal{L}_m=\frac{1}{2}m\overline{(\Psi_L)^C}\Psi_L\\=\frac{1}{2}m\Psi_L^TC\Psi_L$$

not that one can do the same for right-handed fields $$\Psi_R$$

$$\mathcal{L}_m==\frac{1}{2}m\Psi_R^TC\Psi_R$$

Now again due to left-handed fields being equally charged such a team cannot occur. (It could if we impose another second Higgs field to compensate again...)

However, as you pointed out already right-handed neutrinos are not charged under ANY of the standard model gauge group. Hence nothing forbids us to write down such a Majorana mass term for right-handed neutrinos. This means that NATURALLY right-handed neutrinos ARE MAJORANA particles. This model is well established as the seesaw type 1 model.

• Just a pedantic comment: it is not that since the field $\Psi$ is Majorana, then you're allowed to include a term $\mathcal L _m$ in the lagrangian. Indeed $\mathcal L _m$ is an honest Lorentz invariant structure for any Weyl field $\Psi _R$. It's rather the other way round: the fact that your lagrangian includes a term such as $\mathcal L _m$ implies that $\Psi _R$ describes a Majorana fermion. Sep 22, 2018 at 10:39