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> The data still do not rule out the possibility that the mass is zero, says KATRIN member Magnus Schlösser, a particle physicist at the Karlsruhe Institute of Technology. But other lines of evidence, in particular from cosmological observations, show that the neutrino cannot be massless.

I would love if neutrinos were massless, just because it would be so interesting. The only way they would interact with gravity would be through the shape of spacetime itself, which for some reason is a fascinating to me.

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Based on what we've observed, and our current standard model of particle physics, only one type of neutrino can be massless. We've observed neutrinos oscillating flavors (for example, an electron neutrino later interacting as a muon neutrino), and the rate of those oscillations suggest nonzero mass differences between the three different types. So even if one of them is zero, the other two cannot be.

Not a physicist, but I have questions.

> Based on what we've observed, and our current standard model of particle physics, only one type of neutrino can be massless.

Is the standard model complete to the point where we can predict how many types of neutrinos exist and what their properties should be?

I always thought that the standard model as a set of equations (a model) that fits observed data, without venturing far into “why this model is the governing principle for our universe”. That is, it is not able to explain things like “why an electron comes with two heavier varieties”.

Are neutrinos somehow different in a way that we can understand them to the point where we know things like “only one type of neutrino can be massless”?

The oscillation mechanism that we’ve come up with (and that fits data from reactor experiments to astronomical ones pretty well) only works if the mass eigen states of the neutrinos are different from their flavor eigen states, otherwise there is no mixing.

We can measure the mass differences between the neutrinos pretty well through these oscillation experiments, but this also doesn’t tell us which the mass hierarchy. It could be bottom up or the other way round.

In principle, one neutrino could be massless and the mass differences we’ve measured so far would still be correct.

Aside from this, pretty much anything is on the table. Neutrinos being their own anti particles? Maybe. Fourth generation of neutrinos? Could be.

The LEP experiment at CERN (the LHC now inhabits the same tunnels) collided a lot of electrons and positrons to create a lot of Z bosons. The standard model describes these interactions really precisely. And we can observe how often the Z boson decays "invisibly" to particles we can't detect. The rate it does so tells us there are three neutrinos with masses less than the Z boson. So that's established. Could there be more, heavier ones? Possibly.

We observe neutrino oscillations through a variety of channels. We first observed fewer (electron) neutrinos from the sun than expected, suggesting they were oscillating to other flavors. And this has been further observed in neutrinos produced in the atmosphere by cosmic rays, neutrinos produced by decays of particles in beams, and neutrinos from nuclear reactors.

The best explanation, and the one that fits the standard model, is that the pure "flavor" (electron, mu, tau) neutrino states are mixtures of pure "mass" states. And from those different channels, which look at different energies and flavors of neutrinos, we can work out what those mixtures are.

When you go through all the math, it turns out the oscillations depend on the differences of the squares of the masses of the pure mass states. And we observe oscillations that tell us that two of these differences are nonzero. That is, if there are mass states 1, 2, and 3, then we know that (mass 1)^2 - (mass 2)^2 is nonzero, and (mass 3)^2 - (mass 2)^2 is also nonzero. So this implies that at least two of them must have nonzero masses.

To explain that I need to explain symmetry breaking.

Consider a pencil. The equations for a balanced pencil are completely symmetric with the point of symmetry being balanced on its tip. However an actual pencil is never to be found balanced on its tip it is always lying on a side. Therefore a perfectly symmetric theory may describe a real world situation which is not in the least symmetric.

The Standard Model is exactly such a theory. It is in principle completely symmetric, and there is no particular reason in it why an electron would weigh less than a proton rather than more. However it posits a number of free parameters, whose actual values are set by fields, which are therefore fixed throughout the observable universe. Each field is carried by a particle. And we've found all of those particles for the Standard Model, thereby confirming the existence of the fields. And our measurements of the properties of those particles reveals the values of the free parameters, and our theory gets better.

How many parameters? As https://en.wikipedia.org/wiki/Mathematical_formulation_of_th... says, the traditional Standard Model has 19 parameters which we've measured. Thus while the theory itself says nothing about why the electron would have 2 heavier varieties, it does say that there should be 3 varieties on that particle, and the measured parameter values say what the masses are very precisely.

Proposals to include the complexities of the neutrino add another 7 parameters which we've so far not managed to measure. But we're working on it.

>only one type of neutrino can be massless

they would stuck in that type then right? which is not what we see in mid - long baseline experiments, so I think all three types are massive?

We can only produce neutrinos as pure flavor states. These are all mixtures of the mass states.

No, the oscillations only depend on the difference between the squares of the masses. The fact that oscillations between the different flavors is observed just implies that the differences between the masses are all nonzero.

But wouldn't a zero-mass neutrino travel at c, and thus not experience time, and thus not be able to decay?

It's true that a zero mass neutrino would travel at c (just like any massless particle). But neutrino oscillations are not decays. It's more that the neutrino propagates as a mixture of eigenstates. There's nothing stopping one of these eigenstates from having zero mass (as far as we know).

As an analogy, you could think of a photon with circular polarization traveling through a vacuum. If you measured its linear polarization you would sometimes find that it is up-down and other times left-right. The photon is massless but is still able to oscillate between these two linear polarizations because it is propagating with a mixture of these two polarizations.

>The only way they would interact with gravity would be through the shape of spacetime itself, which for some reason is a fascinating to me.

Photons are massless particles, but alas, they still gravitate because it's mass-energy that gravitates, not rest mass.

So… Just like the neutrinos?

Neutrinos aren't massless, but quite like the neutrinos because they're light. Err, unheavy.

No, because neutrinos are believed to have non zero rest mass.

> it's mass-energy that gravitates, not rest mass.

TIL! That's a very fun fact. I never learned that, and I have a physics undergrad! Or I forgot it, which is just as likely this far out.

Ive been around that too and was never explicitly stated but implied by equations once you can calculate with relativity. I mean, they never even formally introduced us to electron volts, I just figured it out myself cause I assumed I was behind the class.

> The only way they would interact with gravity would be through the shape of spacetime itself

I'm not sure what you mean by this. In General Relativity, gravity is "the shape of spacetime", so any gravitational interaction involves the shape of spacetime.

The reason we know that neutrinos aren't massless is that they oscillate between neutrino types. A massless particle must always travel at c, so it doesn't experience time, so it can't decay or change into another particle.

> A massless particle must always travel at c, so it doesn't experience time, so it can't decay or change into another particle.

This is not correct, although it's a common pop science misconception. For example, photons are massless, but they can undergo interactions that, for example, produce particle-antiparticle pairs. If your statement here were true, photons would be unable to undergo any interaction at all.

A correct statement would be, heuristically, that if all three neutrino flavors were massless, they would all have the same mass, namely zero, so they would all oscillate exactly the same way, so any neutrino state that started out as one particular mixture of flavors would stay the same mixture forever. For example, neutrinos that were produced in an interaction like those in the Sun, which only produces electron neutrinos, would stay electron neutrinos forever. But this would also be true if the different neutrino flavors all had nonzero mass, but all the same nonzero mass. The only way for the mixture of neutrino flavors to change as the neutrinos travel is for the different flavors to have different masses. One of those masses could in principle be zero, but only one, not all three.

> photons are massless, but they can undergo interactions

Not by themselves. They have to bump into other particles to interact. Photons in motion do not experience time and can not 'change flavour' or whatever without bumping into something else.

Photon decay is not an observed phenomenon and can only happen if a photon has a non zero mass.

> Not by themselves. They have to bump into other particles to interact.

Not necessarily. There is a very small, but nonzero, probability for photon scattering (due, heuristically, to the small but nonzero probability for photons to become virtual electron-positron pairs). The probability is small, but it is nonzero, and that is sufficient to invalidate your claim.

> Photons in motion do not experience time

This is not correct. A correct statement would be that the concept of "proper time" does not apply to massless objects (objects that move on null worldlines) at all (it's mathematically ill-defined). But that does not mean there are not distinct events on a photon's worldline and that things cannot happen at those events even if the photon is moving by itself.

Can you link me to a source stating that photons moving at C does or does not experience time? I'm not a layman, give it to me straight. I'm happy to work through the math or whatever is needed. I'm asking because I'm very sure that you are wrong here and I would like to be convinced otherwise if possible.

The Minkowski metric ensures that photons moving at C can not experience local time.

> I'm very sure that you are wrong here

What actual textbooks on relativity have you studied? If you haven't studied any, I would strongly suggest doing so before being this confident about your beliefs about relativity. Many pop science sources (including, unfortunately, books by physicists who should know better but who can't help themselves when there are no other experts peer reviewing their work) will say the kinds of things you're saying about photons "not experiencing time", but you won't find a textbook or peer-reviewed paper that says them, because such misstatements are weeded out. Sean Carroll has a set of free online lecture notes [1] that make a good start (they're more focused on GR than SR but they have an introductory section that covers SR). Taylor & Wheeler's Spacetime Physics [2] is a good introductory textbook as well.

The fundamental physical point here is that timelike objects (objects with positive rest mass that follow timelike worldlines) and lightlike objects (objects with zero rest mass that follow null worldlines) are different things, and the concept of "experienced time" or "proper time" (the latter is the correct technical term) only applies to timelike objects. The mathematical basis for this is that timelike worldlines can be parameterized by arc length, and arc length along a timelike worldline corresponds to elapsed time on a clock following that worldline. Null worldlines, however, cannot be parameterized by arc length at all, so the fact that arc length along them is zero does not mean lightlike objects "experience no time", it means that the whole concept of "experienced time" doesn't even work for them: it's mathematically invalid since it requires parameterizing the worldline by arc length.

Another way of seeing the fundamental difference is to look at how Lorentz transformations act on timelike and null vectors. Lorentz transformations hyperbolically rotate timelike vectors: that means the transformation changes which way in spacetime the vector points, without changing its length. But Lorentz transformations do not rotate null vectors: they dilate them, meaning they increase or decrease all components of the vector by the same factor, without changing its direction in spacetime.

This means that a null vector is not a "limiting case" of any set of timelike vectors as far as the Lorentz transformations are concerned; null and timelike vectors are simply two disconnected sets of vectors with respect to Lorentz transformations. Which in turn means that the common pop science image of objects "experiencing less time" as they move faster and faster, until photons moving at the speed of light "experience no time" in the limit, is not correct: it is not a valid description of what is actually happening in the math. Lorentz transformations don't change the length of timelike vectors at all, so they don't change the "experienced time" along them. The apparent "time dilation" of an object that is moving relative to you is due to the angle in spacetime between your worldline and the object's worldline, not to any property of the object's worldline itself. But the "angle" here is a hyperbolic angle, and the hyperbolic angle between your worldline (you being a timelike object) and any null worldline (i.e., any worldline of a light ray or photon) is infinite--in other words, mathematically ill-defined.

In short, pop science authors who make claims like "photons don't experience time" are putting an interpretation on the fact that a photon's worldline has zero length in the Minkowski metric that is not justified by anything in the actual math. They do it, unfortunately, because they believe (quite possibly correctly) that saying things like that, even if they're wrong, will sell more books than trying to teach their readers the actual science.

[1] https://arxiv.org/abs/gr-qc/9712019

[2] https://www.eftaylor.com/spacetimephysics/

You're picking nits, I'm saying 'objects moving at C do not experience local time' and you're saying the same thing in more formal / fancy language without explaining where exactly my statement differs from yours.

> 'm saying 'objects moving at C do not experience local time' and you're saying the same thing

No, I'm not. See below.

> without explaining where exactly my statement differs from yours

You evidently failed to grasp the point of my statement that the concept of "experienced time", or "proper time", is not even well-defined for lightlike worldlines.

Your statement was that photons "do not experience time". And you drew from that the implication that photons cannot undergo any kind of change while propagating freely. That implication is only valid if "do not experience time" means that the concept of "experienced time" is well defined for photons, and the time that they experience is zero.

However, as I explained, the concept of "experienced time" is not well defined for photons. That means you cannot draw any implications either way about whether or not photons can "change" as they propagate freely, based on the fact that the Minkowski length of their worldlines is zero. There simply is no logical implication about "change" for photons from the Minkowski length of their worldlines. To draw any conclusions about whether or not a photon can "change" as it propagates, you have to look at other things.

> the concept of "experienced time" is not well defined for photons.

how is that different from 'photons don't have experienced time'

I explained the issue in the last two paragraphs of my post (the GP to this one). You drew an implication from "photons don't experience time" that is not valid. I explained why it's not valid.

Does polarization (or polarization based filtering) of a photon qualify as an interaction based on time?

Polarization based filtering involves other objects besides the photon being present, so it is not quite the kind of thing we're discussing. We're discussing the possibility of a photon just by itself, with nothing else present, undergoing some kind of change. There is a small probability of this happening; it's small, but it's not zero. So the claim that the photon's worldline being null (i.e., having zero "length" in terms of the metric of spacetime) prevents it from undergoing any change whatever as it travels is not true.

Ah, does doppler effect or relativistic stretching affect the photon wave packet? I was told, that's what causes red/blue shift but aren't photons immune to relativistic effects since they travel at v = c?

> does doppler effect or relativistic stretching affect the photon wave packet?

Yes.

> aren't photons immune to relativistic effects since they travel at v = c?

No. Lorentz transformations still affect photons; they just affect them by dilation (they change all of the components of the photon's wave vector by the same factor, without changing the direction the wave vector points) rather than by hyperbolic rotation.

Only two of the three neutrinos need to be massless, though that would be quite a curious asymmetry, and everyone expects all three to have mass.

A massless particle might not have a restframe or experience proper time, but it still propagates through spacetime, and can definitely decay to other massless particles, at least in theory. After all, moving at the speed of light doesn't preclude it from interacting with ordinary matter either. "Luckily", in our universe there are theoretical reasons for photons to be completely stable (e.g. see https://arxiv.org/abs/hep-th/9508018 ), but there's no such general rule.

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