If neutrinos both move at the speed of light and also have rest mass, what bearing does this have on relativity?
Asked by: Borman


To put it simply, the question is either one or the other. Nobody thinks the answer is both.

Einstein's special relativity predicts that no particle with mass>0 can travel as fast as the speed of light, and it predicts that all particles with mass=0 must travel at the speed of light. This is evident from one of the key equations that results from special relativity: E2 = (m0*c2)2 + p2*c2

Where E is total energy, m0 is rest mass, p is momentum, and c is of course the speed of light. Relativity also shows that E = gamma*m0*c2 (the famous E=mc2, where observed mass m = gamma*m0)and p = gamma*m0*v (where v is velocity and gamma is the dilation factor). Thus p*c2/E = v.

If you put in zero for m0, you get E = p*c from the first equation. Using a little slight of hand (dividing by zero, actually; to do it rigorously, you use a limit) you can turn that into p*c2/E = c. Thus, the velocity of a massless particle is always c.

The neutrino was known from the start to be an extraordinarily light particle, perhaps massless, since its rest mass could never be detected. However, recent results from the Super Kamiokande experiment in Japan and the Sudbury Neutrino Observatory in Canada suggest strong evidence for 'neutrino oscillations,' a phenomenon predicted to occur only if neutrinos have rest mass. So the details are not known, but it seems that neutrinos have a tiny bit of rest mass, and thus travel slightly less than the speed of light.
Answered by: Micah Brodsky, Undergraduate student, University of Washington

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