Short answer: No.
Long answer:
I believe that in all known (and theorised) particle decays, the total mass of the system necessarily decreases. Say you have particle
P which is stationary and has mass
m0.
P decays into two particles
D1 and
D2, both of mass
m1.
Now, in order for momentum to be conserved, both particles most have equal and opposite velocities of magnitude
v. If both particles have velocities, that means they have kinetic energy AS WELL AS mass energy. So in order for the total energy of the system to be conserved,
Unparseable latex formula: \varepsilon_k_0 + \varepsilon_m_0 = \varepsilon_k_1 + \varepsilon_m_1
, or,
m0c2+m0×0=2×(m1c2+m1v)As you can see, if
m0=2m1, this equation doesn't work unless
v=0, which would make the decay invalid, as the particle isn't really decaying at all.
In conclusion, if the mass of a neutrino were 0, it would be impossible for it to decay into any other particles as the mass of the system must decrease (and you can't have a negative mass!). The only other possible decay path would be into another massless particle, such as a photon or graviton, however neither of these decays would conserve lepton number.