Speed of neutrinos

Question

Everyone knows it is close to $c$, but how close? What are the recent results?

Answer 1

Your question is equivalent to asking what the absolute mass of the neutrinos is, and the answer is currently unknown. We do have decent values for the differences of the squared masses for all three required neutrino states (one pair separated by about $7.7 \times 10^{-5}\text{ eV}^2$, and another one standing off from them by about $\pm 2.4 \times 10^{-3}\text{ eV}^2$ (but note the sign ambiguity a problem known as the mass hierarchy question)). This puts a lower bound on the mass of the most massive state at about $0.05\text{ eV}$, but puts no non-trivial bound on the lowest neutrino mass.

Supernova neutrinos may eventually be able to answer this by time-of-flight, but it depends on the theoretical understanding of exactly what is going on inside the exploding star.

There is some data from SN 1987a, but it is not of sufficient quality to provide even a rough answer.

The best we can say at this time is that they are known to be quite small, indeed. (And that is another problem to keep the theorists busy...)

Answer 2

It is technically impossible to measure the speed of such a particle directly; and it all depend on "which" neutrino you are talking about.

The speed is related to the momentum and the momentum to the energy. So you can have a neutrino of some MeV of total energy, another one of some GeV, etc.

But in any cases, the answer will be "very very close" to c.

In all cases neutrinos can be considered as ultra relativistic: their total energy is much higher than their rest energy: $E_t >> m_0 c^2$.

In that case, this relation holds: $E = p c$ where $ p = \gamma m_0 v$. The problem is that $m_0$ is not know with a high precision, it is of the order of the eV.

You can do the math inserting a plausible energy for a solar neutrino, or for a neutrino coming from a high energy collision process, taking any reasonable value for the rest mass.

Edit: To fix the ideas, the rest mass can be something of the order of 0.1 eV and a typical solar neutrino can be of the order of 10 MeV. That's a $\gamma = 10^8$ !

Answer 3

Experimental measurements of neutrino speed have been performed. But thus far, they have all found that the neutrinos were going so close to the speed of light that no difference was detectable within the precision of the experiments.

A couple measurements are described here: http://en.wikipedia.org/wiki/Neutrino#Speed

Answer 4

A paper by Wie et al. (2016) uses the coincidence between gamma rays from Gamma ray bursts and individual 3-30 TeV neutrinos to estimate the difference in propagation speed between neutrinos and photons.

They find $$ \frac{|v-c|}{c} < 2.5 \times 10^{-18} $$

They say this is about 7 orders of magnitude more precise than any previously published limit. However, it does rely on the genuine coincidencce between the GRBs and these individual neutrino events, which does seem to be in question or even doubtful in some cases.

The mass limit this places on the neutrinos appears to be a rather uninteresting $<6.8$ keV, but it is a direct measurement of speed rather than an extrapolation of the speed from a mass limit and special relativity.

Answer 5

The energy of a particle is gamma mc^2. Neutrinos, say electron neutrinos, are produced in about 1-10 MeV units of energy. We are not as certain about the absolue masses of neutrinos, but have measured estimates of the relative masses of neutrinos. From the CKM and PMNS matrix and data input on neutrino oscillations the e-neutrino is estimated at about 1ev or so. So that gamma is very large, gamma ~ 10^7. Then compute from gamma^2 = 1/(1 - v^2) which gives a v (or v/c) =~ .9999998.

There is the funny matter of the P_- = (1 - gamma^5)/2, which projects the helicity or spin in the opposite direction of the momentum --- left handedness and parity violation. A massive neutrino means that in principle one could boost to a frame where the neutrino is right handed. The small mass and large gamma is a sort of obstruction to that, though it is not absolute.

Answer 6

Edit: This answer is obviously dated, the result referred to being due to a timing error (details here). Se the other answers for more realistic values

The answer as of today (2011/09/23) seems to be "measured slightly faster than c" ! (Press release and arXiv). With a big question mark, of course.

More specifically, the measurement result, (from the arXiv paper) is $$\frac{v-c}c = (2.48 \pm 0.28 ({\text{stat}})\pm 0.30 (\text{syst}))\times 10^{-5} $$