# Could we prove that neutrinos have mass by measuring their gravitational signature?

It is now said that neutrinos have mass. If an object has mass then it also emits a gravitational field. I appreciate the neutrinos mass is predicted to be small, but as there are so many produced by our sun and gravity works collectivity should we not be able to detect neutrinos through gravitational differences?

An extract from Wikipedia states:-

"Most neutrinos passing through the Earth emanate from the Sun. About 65 billion ($6.5 \times 10^{10}$) solar neutrinos per second pass through every square centimeter perpendicular to the direction of the Sun in the region of the Earth."

That is a great deal of neutrinos. Would that many not produce a noticeable gravitational signature that we could detect? If we can't detect it does that mean that neutrinos are massless?

The gravitational field of a fast moving particle is from its energy, not its rest-mass. The source of the gravitational field is the energy divided by c2 if you are using unnatural units, or what used to be called "relativistic mass" before that term fell out of favor.

The neutrinos we observe are moving at essentially the speed of light, so we cannot distinguish their gravitational signature from that of a massless particle. Both a neutrino at 1KeV and an exactly massless fermion at 1KeV have pretty much the same gravitational field. The difference is suppressed by the ratio of the mass to the energy, and it is as hard to detect as the difference in the neutrino speed from the speed of light (which is a more direct way to measure the mass, but also impractical). There are not enough neutrinos which are moving slowly compared to the order .01 eV masses, so that the number of nonrelativistic neutrinos is too small to allow their mass to be measured gravitationally.

In short, the answer is just no.

• How is it possible to estimate the number of non relativistic neutrinos? – Lehs Jul 18 '15 at 23:15
• @Lehs: From the manner of their creation, and dissipation. Any beta decay process or supernova happens at scales of KeVs or MeV, and the neutrinos are weakly interacting enough that they don't cool down. Cosmologically produced neutrinos during the big bang are the coolest, and the number of these can be estimated the same way you estimate the nuclear density, from Big Bang models (which are very accurate today). – Ron Maimon Jul 19 '15 at 16:43
• Are there any possibility that the distribution of neutrinos in universe is affected by the Pauli principle? – Lehs Jul 19 '15 at 18:33
• @Lehs: You're asking if the cosmological neutrinos are dense and cold enough to be close to forming a Fermi surface. The answer is probably no, the density and temperature are both known, so you can check explicitly whether there is about one neutrino per typical wavelength at the thermal momentum. I didn't check myself, I don't know the density off the top of my head. – Ron Maimon Jul 28 '15 at 14:51

Neither do we know the speed of the neutrinos. Since mass=energy, anything with energy gravitates as well. Light gravitates in that manner. Neutrinos could have zero mass, but they'd still gravitate if they went at the speed of light (energy=$\frac{m_0c^2}{\sqrt{1-v^2/c^2}}$. If $m_0=0,v=c$, energy is not necessarily zero). Since their speed is disputed as well, knowing their gravitational effects doesn't help. If the "faster than light" thing pans out, they may even have complex mass (which you can see directly from the above equation). So we can't really measure if they have mass if they go faster than light by looking for a signature.

OK, if you're talking about measuring the exact mass-energy of the neutrino (which is unknown atm), that would be theoretically possible, practically impossible. Mass of $$\nu_e<2.2 eV$$ $$\nu_\mu<0.17 eV$$$$\nu_\tau<15.5 MeV$$

Pretty tiny. Multiply by 65 billion and it's still tiny. Taking only the tau neutrinos (IIRC 1/3rd of them) into account, a safe assumption as the others have negligible mass; we get $65\times10^{15} eV/c^2=1.1\times10^{-19} kg$. That may be measurable in a laboratory, but impossible to measure around the Earth. Earth's mountains, irregulatiries in the crust/mantle, and other cosmic rays would create a larger effect. The main issue is that we can't "turn off" the neutrinos to get a control case so that we can eliminate the other effects. If we do this in an accelerator, a few neutrinos have even less mass, and it's even more uncertain. Add quantum mechanics to it and I think you could say that there are no gravitational effects at all (becomes less than the planck length, &c; though mashing GR and QM like this isn't exactly accepted)

Edit: As @dmckee pointed out in the comments, I've used obsolete neutrino masses. The actual masses are much smaller, though that doesn't change the final conclusion.

• BTW---The limits you give for the direct measurement of the heavy neutrino masses are obsolete. The mass differences squared are known (modulo the hierarchy problem) and all the neutrinos are very light. – dmckee --- ex-moderator kitten Feb 26 '12 at 1:46
• @dmckee Hmm, you're right. But it doesn't change the final conclusion.. – Manishearth Feb 26 '12 at 12:16

The neutrino masses are probably on order of $1\text{ eV}$, but we need to consider their total energy (I feel like a dunce having to wait for Ron to point that out in the comments).

The solar neutrinos have energies on order of $1\text{ MeV}$.

So 65 billion neutrinos accordingly have mass of $6.5 \times 10^{16} eV = 65 \times 10^8\text{ GeV}$ which is about the mass of a million atoms of zinc or about $65 \text{ g/mole}* (10^8\text{ atoms})/(6 \times 10^{23}\text{atoms/mole}) \approx 1 \times 10^{-15}\text{ g}$. The volume occupied is 1 square centimeter times $300,000,000$ m (lightspeed times 1 second), or about $30,000$ cubic meters. This gives us a total density of $3 \times 10^{-20} \text{ g}/\text{m}^3$.

So, short answer: In principle yes, in practice no.

• This calculation is faulty, because the energy is the source of the gravitational field, not the mass. If you like, it's only the relativistic mass that is important. You need to consider the total energy of the neutrinos, and for ordinary energies the mass only affects the velocity in a slight way. – Ron Maimon Feb 26 '12 at 1:20
• @RonMaimon: Ah....yes. Moment. – dmckee --- ex-moderator kitten Feb 26 '12 at 1:21
• @dmckee $10^{16} eV\neq10^6GeV$ Where did you get $6*10^{23}$? google.com/search?sourceid=chrome&ie=UTF-8&q=eV%2Fc%5E2 – Manishearth Feb 26 '12 at 1:32
• Or is that avogadro's number? Now I'm confused. Did you convert eV to zinc to grams? – Manishearth Feb 26 '12 at 1:38