Basically my main question is the one in the heading of my post, because apparently "Gravitation" by Misner, Thorne and Wheeler takes it directly as true and valid (a common place for null geodesics).

We know that the rest mass of neutrinos is very very low, although we are sure it is not entirely zero. According to Wikipedia (sorry, I don't know where to read more about this topic), their lowest rest mass can be around $\sim$ 0.1 - 1 eV/$c²$. So they don't have null mass, unlike photons which are regarded as particles of mass zero.

In General Relativity we can regard null geodesics as paths followed by photons, stuck onto the surface of null cones. So why is it valid (or maybe not) that we can take as an approximation that neutrinos follow those same null geodesics?

I can imagine that, because their rest mass is extremely small, it is more straightforward to consider that the energy of neutrinos can be simply $E\approx pc$, just like photons. And I can see that since their speed is nearly $c$, then their line element must be very close to that of a null event: $ds =c d\tau \approx 0$. So then, their proper time could be approximated too as null: $d\tau\approx 0$? And can we then assume they also lay in the null cone like photons, because an affine parameter for their geodesics will be always very close to zero?

Furthermore, I can imagine ultra relativistic electrons or muons going pretty close to $c$, however I know maybe it will not be true that their proper time is $d\tau \approx 0$.

  • 1
    $\begingroup$ Since the rest mass of neutrinos is 0.1-1 eV then most neutrinos (from the big bang) are moving with a speed much less than $c$ now. I guess your question is really whether ultra relativistic particles follow null geodesics. $\endgroup$
    – ProfRob
    Commented Dec 30, 2023 at 9:37
  • 3
    $\begingroup$ MTW was first published in 1973, when it was common to assume that neutrinos are massless. $\endgroup$
    – PM 2Ring
    Commented Dec 30, 2023 at 10:06
  • $\begingroup$ @ProfRob So then not all neutrinos are ultra relativistic? What is a good average speed for neutrinos? Don't take necessarily and solely that as their rest mass, PM2 Ring says CMB neutrinos are lighter. And for example, I find the absolute error of neutrinos speed vs. light speed to be under 0.000000001 after googling a little. In common sense I can understand then that they can be in a lightcone (or very close to it) but I wonder how good as an approximation that is. $\endgroup$
    – omivela17
    Commented Dec 30, 2023 at 20:43
  • $\begingroup$ @PM2Ring I get you but in the 2017 reprint they still include that same sentence, and I don't still see if there would be any trouble in stating they can be treated as null particles, because their proper time is merely 0 due to their Lorentz factor. $\endgroup$
    – omivela17
    Commented Dec 30, 2023 at 20:59
  • $\begingroup$ @omivela17 I didn't say that CNB neutrinos are lighter. Rob & I both said they are slow, moving at non-relativistic speeds. We don't yet know the exact neutrino masses, but neutrino mass is complicated because of oscillation. $\endgroup$
    – PM 2Ring
    Commented Dec 30, 2023 at 22:11

1 Answer 1


As ProfRob mentions, most of the neutrinos (and antineutrinos) in the universe decoupled from the rest of the matter in the early moments of the Big Bang, when the universe was around 1 second old. These neutrinos now form the cosmic neutrino background (CNB). At the time of decoupling, the CNB neutrinos had a mean kinetic energy around 2.5 MeV, equivalent to ~17.4 billion K, but due to expansion their temperature is now ~1.95 K, and their mean kinetic energy is now estimated to be $10^{-6}$ to $10^{-4}$ eV.

Neutrinos are notoriously difficult to detect. Our best neutrino detectors, using the Ga → Ge → Ga reaction sequence (aka, the "Alsace-Lorraine" technique), have a threshold of 233 keV. And even then only a tiny fraction of the neutrinos passing through the detector actually trigger the reaction sequence. A well-known heuristic is that a light-year of lead would only absorb ~50% of the neutrinos passing through it (assuming the neutrinos have energies typical of those we can detect).

So neutrinos are effectively invisible unless their kinetic energy is at least on the order of a million times greater than their rest mass. That is, we can't detect neutrinos unless their Lorentz $\gamma$ is >1000000. The trajectories of such neutrinos are very close to the light-cone.

For example, the neutrinos emitted by the core-collapse supernova have energies in the range 10-30 MeV. We detected 19 neutrinos from the supernova SN1987A located in the Large Magellanic Cloud, at a distance of ~168,000 light-years. If we assume that neutrinos have a rest mass of 0.1 eV, a 20 MeV neutrino from SN1987A would arrive at Earth about 66 microseconds after a photon that left SN1987A at the same time.

Wikipedia mentions that there are prospects of detecting CNB neutrinos, using tritium. I suppose we might be able to detect them by running the Alsace-Lorraine reaction in a particle accelerator like the LHC. But even if we can detect their presence, that won't give us any data on their trajectory.

  • $\begingroup$ For you, what is a good approximation for a Lorentz factor $gamma$ to make a particle go very close to the light cone? $\endgroup$
    – omivela17
    Commented Dec 30, 2023 at 20:30
  • $\begingroup$ @omivela17 All the neutrinos that we can currently detect are highly ultra-relativistic, so their trajectories are virtually indistinguishable from null geodesics. We can only estimate their speed and $\gamma$ from the kinetic energy because we don't know the rest mass. But we can calculate the speed from the gamma using $(v/c)^2+(1/\gamma)^2=1$ $\endgroup$
    – PM 2Ring
    Commented Dec 30, 2023 at 22:16

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