A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite?

If the total energy of all three types of neutrinos exceeded an average of 50 eV per neutrino, there would be so much mass in the universe that it would collapse. This limit can be circumvented by assuming that the neutrino is unstable....These indicate that the summed masses of the three neutrinos must be less than 0.3 eV source

Then, if I read this right, rest mass of a neutrino is, on average, $10^{14} h\nu$ and its speed is near (a few nanometres) the speed of light and recently there were also (false) allegations that its speed might exceed the speed of light.

I couldn't find data about the energies recorded at GranSasso on reception of the neutrinos from CERN or from other sources, but surely they are not exceedingly high, else they would have hit the headlines.

So, how come they do not have an almost infinite (or at least enormously huge) relativistic mass/energy which, according to SR, any body with restmass should have in order to approach $c$? Do you have any figures about their relativistic energy and relate it to SR formulas?

So if we plug in to find the total energy of the neutrino we find.

ev∼18keV∼0.03 $m_{electron} Which isn't that big. Ia this answer correct? I posted this question because in my [previous question] (How did Pauli and Fermi deduce the existence of the neutrino?) there was this comment: What makes you think the energy carried away by the neutrino is tiny? The mass of the neutrino is tiny, but its kinetic energy can be of the same scale as that of the electron. – dmckee Now, the energy of an electron is .5 Mega eV, what is then the correct maximum value of the energy of a neutrino? • The finding of neutrinos exceeding the speed of light in the CERN experiment turned out to be faulty wiring. :) Reminds me of when I was learning to wire to NASA standards. – Howard Miller Mar 12 '17 at 3:42 5 Answers So let's say these electrons go really really fast like 0.999999997 times the speed of light. We know the upper bound on the neutrino mass is less than 1 eV (the kinetic energy of taking one electron through a one volt potential difference) from experiments. So if we plug in to find the total energy of the neutrino we find. $$E_{relativistic} = \gamma m_{neutrino}c^2 < \frac{1}{\sqrt{1-(0.999999997)^2}} \text{eV} \sim 13 \text{keV} \sim 0.03 m_{electron}$$ Which isn't that big. And as already stated the mass energy is on the order of at MOST 1eV. So what we find is that these things can travel extremely, extremely fast even though their kinetic energy is still mere percentage points of the next smallest particle, the electron. • @Alb: as I said in my answer the GranSasso measurement of the speed produced the result$c \pm 0.000003c$. That is$c \pm 0.0003$%. – John Rennie Sep 24 '14 at 11:11 The OPERA experiment reported: the speed of neutrinos is consistent with the speed of light within the margin of error The margin of error was around$3 \times 10^{-6}c$, so their result restricted the speed of the neutrinos to be at least$0.999997c$and less than$1.000003c$. For comparison, the speed of the protons in the LHC is$0.999999991 c$, so the OPERA lower limit of$0.999997c$is actually a long way below the speed of light as such measurements go. The most accurate determination of neutrino speed was the detection by Kamiokande of neutrinos from supernova SN 1987a. This placed a lower limit on the speed of$0.999999998c$, which is still comparable with the LHC proton speed. • Hi John, I don't think this actually answers the OP's question. I think he's hung up on the fact that because Neutrinos tend to be so fast, they must be very high energy, neglecting the fact that their insanely low mass is what allows them to be so fast even at relatively low energies. – Brandon Enright Sep 24 '14 at 6:49 • @BrandonEnright: both braxton and I answered v1 of the question, before the edit that specified nearly the speed of light. I'll calculate the Lorentz factor the speeds I quote when I get a moment ... – John Rennie Sep 24 '14 at 7:39 As an ex-theoretical chemist, not claiming any expertise in this field: Even though mass-energy of near light-speed neutrinos, individually, might still be very-very small, there are likely to be many-many more of them (~10^9?) than electrons (used in another answer, above, to make a comparison). Can we therefore make a back-of-the-envelope estimate of the percentage of the universe’s mass represented by neutrinos in the present? Starting with a photon-baryon ratio ~ 10^9, as given in, e.g: http://star-www.st-and.ac.uk/~spd3/Teaching/PHYS3303/obs_cos_lecture3.pdf ...and neutrino-photon ratio 7/4, as in e.g: http://hyperphysics.phy-astr.gsu.edu/hbase/astro/neutemp.html …then if we can take the present baryon-electron number ratio to be a little under 1, say 0.85 (weighted average, given that universe’s atomic components are mostly H, He, Li), we get a neutrino/electron ratio of around 0.85*(7/4)*10^9 or ~ 1.5*10^9. So the comparison with electrons in an earlier answer (above) may not wipe out the neutrino contribution. To get an estimate of the percentage of the universe’s mass represented by neutrinos means looking at the thorny question of neutrino masses and neutrino oscillation. Taking information from: ...where it currently (24th Sept 2014) says: “In 2009, lensing data of a galaxy cluster were analyzed to predict a neutrino mass of about 1.5 eV. All neutrino masses are then nearly equal, with neutrino oscillations of order meV. They lie below the Mainz-Troitsk upper bound of 2.2 eV for the electron antineutrino. The latter will be tested in 2015 in the KATRINexperiment, that searches for a mass between 0.2 eV and 2 eV.” …let’s use a number of 1eV (which presumably incorporates relativistic mass-energy) and assume from the quote above that the three different neutrino types’ masses are about equal (we’re only looking for a ball-park estimate at best). To get an estimate of the percentage of the universe’s mass represented by neutrinos we can therefore use the WMAP estimate: …that atoms represent 4.6% of the current mass/energy of the universe, and use the ratio of neutrino to electron masses. Again using a weighted average, we find that the universe’s mass-energy is about 0.002% electrons (generally moving at sub-light speeds, with rest masses of 0.511 MeV; an earlier post How fast do electrons travel in an atomic orbital? gives v∼e24πϵ0ℏcc=αc i.e. ~ c/137). Thus the universe’s mass-energy stored in neutrinos should be somewhere in the region of 0.002%*(1.5*10^9)*(1eV)/(5.11*10^5 eV) = 5.87% (And if observations such as the lensing data, referred to above, turn out to be reasonably accurate, then the question of whether or not the neutrino's mass-energy is/isn't "(nearly) infinite" is effectively answered.) This says that if over 5% of all the mass-energy in the universe is in the form of neutrinos, that's greater than the 4.6% figure for all atomic matter in the universe! (Can anyone spot a large error in this estimate?) For comparison, WMAP suggests that 13.7 x 10^9 years ago neutrinos comprised 10% of the universe's mass-energy. http://map.gsfc.nasa.gov/media/080998/index.html Their estimate for the present doesn't seem to mention neutrinos at all, unless they're included under dark matter. One last little thought - if the numbers for atomic matter and neutrino matter are so (relatively) close, might observation eventually find them to actually be the same... • It is not that simple! When you make assumptions after assumptions and approximations after approximations and get a crazy result (5% of the energy budget) then its likely due to those assumptions and approximations. Here is one very large error in your estimate: most neutrinos in the universe are not emmited from galaxy clusters (and not with those energies)! The calculation can be done properly (see any cosmology text-book) and if you do it then you will end up with a result that is roughly 1/10000 times smaller today! – Winther Sep 24 '14 at 21:43 • @Winther - It did look rather surprising, and I wasn't aware of the difference between galactic and non-galactic neutrinos, so I appreciate your finding where at least one of the errors lay. – iSeeker Sep 24 '14 at 23:28 • @Winther - I've looked online; will Weinberg's big 2008 'Cosmology' text cover the calculation? There are none more recent that I can see on Amazon that look technical enough (my own copy of Edward Harrison's Cosmology (2000) doesn't seem to cover it, and the only newish book - de Grasse Tyson - seems too thin). – iSeeker Sep 26 '14 at 13:32 • Yes, Weinberg cover this. Another good book (imo) is Dodelson's. I also think there should be some free online reviews about this subject, but I couldn't find them right now (except this but this might be behind a paywall). – Winther Sep 26 '14 at 13:44 • @Winther - Many thanks, Hans - I've ordered a s/h copy of Dodelson. – iSeeker Sep 26 '14 at 14:48 A neutrino has rest-mass and travels at (near)$c$, why isn't its mass/ energy (nearly) infinite? Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite. However, this is not a practical point of view. From the practical point of view, neutrinos have mass-energy as they have (because they got it when they were born (or sometimes in an interaction event afterwards)) and because of this they have speed (that for all currently observable neutrinos is only slightly less than the speed of light). The mass-energy can be anything bigger than the rest mass of neutrino. Neutrino detector using a gallium → germanium transformation has detection threshold of 0.233 MeV and this is the lowest value I have found in the Wikipedia article about neutrino detection. Solar neutrinos have energies up to 18 MeV. Famous, but now known to be mistake, faster-than-light neutrino anomaly has something to do with 28 GeV neutrinos. 6.5 TeV per beam (planed for LHC since early in 2015) can become almost wholly energy of neutrino. (Two protons could stop and emit neutrino and antineutrino, but probability is low, digression 1). The equation for energy of particle with rest mass$m$is$E = \gamma m$where$\gamma = \frac{1}{\sqrt{1-v^2}}$so$v = \sqrt{1-\frac{1}{\gamma^2}} \approx 1 - \frac{1}{2 \gamma^2}$. In the cases about which I write above, the speed is respectively$(1-9.2 \cdot 10^{-14}) c$,$(1-1.5 \cdot 10^{-17}) c$,$(1-3.4 \cdot 10^{-24}) c$and$(1-1.2 \cdot 10^{-28}) c$. On the other hand, "relic background neutrinos are estimated to have (...) temperature 1.9 K ($1.7×10^{−4}$eV) if they are massless, much colder if their mass exceeds 0.001 eV". So even assuming temperature too high for such "heavy" 0.1 eV neutrinos, they are non-relativistic and have average speed (from non-relativistic equations$E = m v^2 / 2$and$E = \frac{3}{2} k T$)$v = \sqrt{3 k T / m} \approx 0.071 c$. That's still about 21000 km/s, but clearly less than the speed of light. So if we plug in to find the total energy of the neutrino we find. ev∼18keV∼0.03$m_{electron} Which isn't that big.

Ia this answer correct?

J.J. must have forgotten about square. I have corrected it and $0.03 m_{electron}$ is now even better approximation, but this says something only about neutrinos with speed $0.999999997 c$.

Now, the energy of an electron is .5 Mega eV, what is then the correct maximum value of the energy of a neutrino?

It is true that "the mass of the neutrino is tiny, but its kinetic energy can be of the same scale as that of the electron" for neutrinos from beta decay and the energy of beta decay is comparable to the mass of electron, so this question makes some sense, but knowing just this, one can only say "of the order of .5 Mega eV". Solar neutrinos are from nuclear reactions and have energies up to 18 MeV - the same order more or less.

You probably wanted some calculations, so I will calculate maximum energy of the neutrinos from decay of tritium. In the first approximation, energy of neutrino is simply difference between rest energy of tritium atom and rest energy of helium-3 atom. (If neutrino takes the whole energy, we could really get neutral helium-3 atom, but probability is low. Digression 2.) Neglecting neutrino mass, its energy is equal to its momentum ($p$), which is equal (with opposite direction) to the helium-3 momentum.

$E_{{^3}\mathrm{He}}^2 = m_{{^3}\mathrm{He}}^2+p^2$

$m_\mathrm{T} = E_{{^3}\mathrm{He}} + p = \sqrt{m_{{^3}\mathrm{He}}^2+p^2} + p$

$m_{{^3}\mathrm{He}}^2 + p^2 = m_\mathrm{T}^2 - 2 m_\mathrm{T} p + p^2$

$p = \frac{m_\mathrm{T}^2 - m_{{^3}\mathrm{He}}^2}{2 m_\mathrm{T}}$

$p = 0.0186 \mathrm{M}e\mathrm{V}$

using

$1 \mathrm{u} = 931.4812 Me\mathrm{V}$

$m_\mathrm{T} = 3.01604927 \mathrm{u}$

$m_{{^3}\mathrm{He}} = 3.01602931 \mathrm{u}$

In this case, we get low energy. From 0.0186 MeV to 18 MeV is something about 0.5 MeV, but writing this I have realised that the rule that nuclear reaction energy is comparable to the mass of electron is not very precise.

Digression 1: Probability is low, but I do not know how low - there are very many events and I do not have any idea, how often this is going to happen - maybe more than once a second, maybe less than once per trillion years. This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with parton energy distribution and probability of reaction. And what does "almost wholly" mean? Something like 99%, in worst case 99.9%, is enough to get $(1-1.2 \cdot 10^{-28}) c$ with 1.2, not 1.3. Anyway, taking one half of this energy should be much more probable and give comparable neutrino speed.

Digression 2: Again I do not know how improbable this is. This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with distribution of speed in decays, but also with something like resonances.

General notes:

I do not know more than Wikipedia says about the 0.320 ± 0.081 eV estimation by Planck collaboration, but here I'm assuming that it is correct. Than, oscillation data show that the mass difference is small, and masses of all three neutrinos are about 0.1 eV, but this is rough approximation from error range itself, so I probably underestimate other errors and write to many digits. All not taking into account the possibility that something is very wrong.

Besides I am using here $c=1$ convention. Masses and energy are both measured in electronovolts, and speed is always compared to the speed of light, so this should not cause any problems.

Suppose a solar neutrino of the right flavor interacts with a chlorine atom in a molecule of perchlorethane (Nobel laureate Ray Davis' Homestake neutrino detector) and changes the chlorine into an atom of argon.

This is a difference in energy (of the atoms themselves) of between 3 and 4 eV, yet the estimated rest mass of a neutrino is supposed to be an order of magnitude (0.3 eV) less than this.

Ray's detector (and the new detector designs that eventually solved the mystery of the missing solar neutrino flux) of course does not measure how much excess kinetic energy is left after the chnlorine atom changes into an argon atom, but evidently the amount of energy added to the neutrino resulting in the energies of the products of this collision is not infinite.