Limiting the mass of the neutrino for a relativistic case

Question

I came across a question that states

What mass would a neutrino need to still be relativistic today (T = 2.37K) ?

So for a particle to be relativistic we need $pc \gg mc^2$

Well Neutrino was relativistic in the early universe, so I took the time when the neutrino decoupled which is approximately $\approx 1 MeV$

So I did something like

$$\frac{E_{now}}{E_{dec}} = \frac{kT_{now}}{T_{dec}} = \frac{8.617\times 10^{-5} eV K^{-1} \times 2.73K}{1Mev} = 2.35 \times 10^{-10}$$

But I am kind of stuck here since we need some value for the neutrino mass I guess ? Or my approach is completely wrong (?) In general, how can we solve this kind of problem? What makes the transition from Non-Relativistic to the relativistic case? The temperature of the universe right..? For instance when the temperature of the universe was larger than the $1 MeV$ we would call protons relativistic

Answer 1

So for relativistis case $$pc > mc^2$$ We also know that at the reheating thr temperature of the neutrinos and photons evolved differently.

$$T_{\tau} = T_{\gamma}/1.40$$

We know that $T_{CMB}^{now}= 2.73 K$ so

$T_{CNB}^{now}= 1.95 K$

By using $E = kT$

$E_{\tau} = 1.68 \times 10^{-4} eV$$

So particle will be relativistic today if

$$m_{tau} < 1.68 \times 10^{-4} eV$$