Whilst reading Light Sterile Neutrinos: A White Paper it is stated on the bottom of p.3 without an explanation that:

"The observed $Z$-boson decay width implies that any additional active neutrinos are quite heavy, $m_\nu \gtrapprox M_Z/2$."

Where was this observation deduced from? Is it from Heisenberg's uncertainty principle somehow?


2 Answers 2


The observation of Z-boson decay doesn't imply active neutrinos to be heavy. The cited white paper says

The observed Z-boson decay width implies that any additional active neutrinos are quite heavy ...

The decay width is related to the number of different ways the Z boson can decay, and also to the strengths of the interactions that lead to those decays. By measuring the decay width and taking note of the number of decays with no detectable decay products (inferred to be neutrinos), the number of active neutrino species with masses $<M_Z/2$ can be inferred. The result of these measurements is that the number of active neutrino species with masses $<M_Z/2$ is three, which is the expected number in the Standard Model. This resuilt shows that there are no additional active neutrino species beyond the three that are expected from the Standard Model.



This is based on the observation of the decay of the $Z^{0}$ into invisible daughter products. When a decay like $Z\rightarrow\nu+\bar{\nu}$ occurs, the neutrino and antineutrino are virtually never observed, so it just looks like the $Z^{0}$ vanishes from the detector apparatus.

The three decays, $Z\rightarrow\nu_{e}+\bar{\nu}_{e}$, $Z\rightarrow\nu_{\mu}+\bar{\nu}_{\mu}$, and $Z\rightarrow\nu_{\tau}+\bar{\nu}_{\tau}$ occur independently. The total rate* of $Z^{0}$ disappearance is proportional into the number of channels available for decays into invisible daughter particles. Thus, it is proportional to the number of neutrino species into which the boson may decay. By measuring experimentally the rate of $Z^{0}$ disappearance, we therefore get a measurement of the number of light neutrino species. According to the Particle Data Group, the best current value for this is $2.984 \pm 0.008$, very close to the expected value of $3$. If there are any other neutrino species that are not sterile (meaning they interaction with the $Z^{0}$ in the usual way), they must be too heavy for the $Z^{0}$ to decay into them, or $m_{\nu}>\frac{1}{2}M_{Z}$.

*The "decay width" is equivalent to the decay rate in this case. The decay width $\Gamma$ is $\hbar/\tau$, where $\tau$ is the mean lifetime of a particle. This is essentially a manifestation of the energy-time uncertainty relation; a long-lived but ultimately unstable state has a small mass uncertainly.


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