# What causes neutrinos to be weakly coupled?

When reading about "active-sterile mixing", which requires some Dirac mass ($$m_D$$) and some Majorana masses ($$M_R$$) to be very small but not zero, the seesaw limit model is discussed ($$M_R \gg m_D$$).

In this paper (Light Sterile Neutrinos: A White Paper), it is mentioned that the active-sterile mixing matrix, when squared can be given by:

$$\Theta^2 \sim \frac{m_{1,2,3}}{m_{4,5,...}} \sim \frac{m_D ^2}{M^2 _R} \tag{1}$$

where $$m_{1,2,3}$$ and $$m_{4,5,...}$$ are the neutrino masses.

My doubt lies in the following statement that is preceded by eq. $$(1)$$:

Given $$m_{1,2,3} \lt 10^{-1}$$ eV, the mostly sterile states are very weakly coupled unless $$m_{4,5,...} \lt 10$$ eV and $$\Theta^2 \gt 10^{-2}$$.

Why were these the values chosen for the mass of the neutrinos? Why are those the conditions for the states to be weakly coupled? What causes them to be weakly coupled, just the value of the masses?

• The basic answers to "why" in physics is "because that is what fits the experiments", theories have to be at least consistent with experimental measurements, and that generates the various limits at the end of the game – anna v May 26 at 3:47

There are two types of neutrinos there: active ones (part of the $$SU(2)_L$$ doublet, carrying the weak charge) and sterile ones (singlets of the SM).