How light is a neutrino? The answer is closer than ever
nature.com
nature.com
This is the best attempt to date to directly measure the mass of the electron (anti)neutrino. When a nucleus beta decays, it emits an electron and an antineutrino. If the neutrino were massless, the electron could carry away the entire energy of the decay. What they've tried to observe here is the highest energy electrons, to see if they can see the effect of some of that energy going to the neutrino mass. It's an amazingly difficult measurement, since only a very small fraction of those electrons have the highest energies.
Measuring the neutrino mass this way would be relatively unambiguous, which is why it's exciting to see progress.
That said, there have been other experiments that have put stronger constraints on neutrino masses, but only for certain models. Double beta decay experiments look for decays in which the two neutrinos annihilate each other, allowing the full energy to be emitted by the electrons. However, this requires the neutrino to be its own antiparticle. This is allowed, because neutrinos are neutrally charged, and wed call such a particle a Majorana particle. The rate of these neutrinoless double beta decays would tell about the neutrino masses, but only if neutrinos are Majorana particles. The exact measurement would also depend on how well we understand the energy levels of the nuclei involved. So far, the upper limits for neutrino masses from these experiments are on the order of 0.1 eV.
And there are cosmological constraints on the neutrino mass from experiments that look at the cosmic microwave background. Early in the universe's history (like the first second), the mass of neutrinos would have influenced how much matter clumped up due to gravity, which would lead to fluctuations in the microwave background. Modeling this requires us to understand particle physics very well at those early high energies, so there's some uncertainty due to modeling. But again, the limits from these observations are on the order of 0.1 eV.
So we now have one direct observation that's consistent with other, model-dependent observations. The most interesting scenario would be that KATRIN actually observes neutrino mass as it gets more data, implying that our models might be wrong. But even if the experiment doesn't, it's still great to have the extra constraints.
Question - if the model-dependent 0.1eV neutrino gets knocked out because the direct observation (anti)neutrino is say "0.3eV confirmed".
How does this mess with our current understanding of physics?
For cosmology, it would suggest something about our understanding of the big bang or early universe is wrong. I'm less familiar with cosmology, however, so I'm not sure in what way things might change. We'd need a new way to explain why matter is less clumpy than it should be due to the massive neutrinos, or why there weren't as many neutrinos then as we think there should have been.
Photon are also massless but they are still able to carry energy? Why would neutrinos be different here?
If a neutrino has no rest mass, there would be a variety of ways in which the decay energy could be distributed, with some of it going to the electron and some of it going to the neutrino. So over many decays, the electron would be observed to travel at a range of possible speeds. You might see almost all of the energy going into the electron and almost none into the neutrino, or vice versa.
But if some of the decay energy goes into the creation of a massive neutrino, that decreases the upper bound on how much energy is left to go into the electron. So by carefully measuring the extent range of energies of the decay products, you can hope to distinguish the massive and massless cases.
However, experimental evidence suggests that observed neutrinos oscillate between the three neutrino flavours over time & the sane way to allow for that theoretically is for them to to have slightly different small masses. The Wikipedia pages on the Neutrino & Neutrino Oscillation give a good overview.
Hrm. My understanding was that electron/muon/tau neutrinos (bizzarely) don't have a defined mass, since the flavor eigenstates are a mix of the mass eigenstates. Did you mean that they are trying to put bounds on one of the particular mass states?
For comparison, the mass of an electron is approximately 510,998.950 eV and the mass of a proton is 938,272,088. eV.
I was going to write:
> For comparison, the mass of an electron is approximately 500,000 eV and the mass of a proton is 900,000,000 eV.
But all the numbers I wrote are measured experimentally. Both values have a lot of experimentally measured digits! I only removed the part that overlaps with the uncertainty, because the notation with parenthesis is somewhat confusing.
Thank you; this is a style guideline I wish more writers would adopt.
What about this version:
> [an upper bound of 0.9 eV] For comparison, the mass of an electron is approximately 510998.9 eV and the mass of a proton is 938272088.0 eV.
> the mass of an electron is approximately 510,999 eV and the mass of a proton is 938,272,088 eV.
This reads best to me.
neutrino: 0. 8 eV/c^2
electron: 510 000. 0 eV/c^2
proton: 940 000 000. 0 eV/c^2
Of course, if you were to present the masses this way without context, you should add a note about rounding.
In fact, I find it so unintuitive that I didn't even notice the units changed and thought you'd made a mistake in magnitude before I finished your sentence and I went back.
Comparisons should always be done in like units.
(also i would have expected an electron to be... 1 electron volt?)
The electron volt is related to the electric charge of the electron, not to its mass. Similarly, a magnet weights more than its strength turned into mass.
As it happens eV works nicely from meV up to TeV for particles in all the most useful contexts, from chemical bonds to particle experiments. You only start getting insanely huge factors when you stop being subatomic.
Also, the unit used for mass of particle is usually eV/c² or MeV/c² which is more correct (unless in c=1 units) but the c part is sometimes dropped by convention.
eV is a unit of energy, and eV/c^2 is a unit of mass. You can drop the c^2 if you are lazy, and if someone ask you can say that you using a system of units where c=1 and ħ=1. This is fine in the second half of a Physic degree, but you can get in trouble for forgetting the c^2 in the first half or during secondary school.
The technical definition is in a sibling comment, but an eV is similar to the energy that an electron gets or loose when it makes a jump inside a molecule or between two molecules. The exact number depends on the molecule and the jump, it may be x10 bigger or smaller.
For example if you connect a led to a battery, each electron gets like 1.5eV when it pass through the battery, and release that energy as a photon of light in the led. If you ignore a lot of technical details, the electron makes one or two jumps in the battery and a jump in the led, but this is a huge oversimplification.
Note that a led releases like 10^18 photons per second, so each photon with approximately 1eV has a tiny amount of energy.
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On the other hand, if you destroy an electron you get 511000eV of energy. You can't destroy a single electron, but you can make an electron colide with a positron that has the same mass and in total you gen 511000eV+511000eV of energy. So destroying an electron release much more energy than an electron that jumps inside a molecule.
In special relativity there is an equivalence between energy and mass, so if the energy you get while destroying an electron is 511000eV then the mass is 511000eV/c^2. You can also measure the mass directly without destroying the electron, and you get the same number (probably with a bigger uncertainly, I'm not sure how they measured so many digits).
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As an analogy, if you move around and climb some steps and make some jumps, each big movement absorbs or release approximately 1000J.
If someone makes a antimatter clone of you, and you both colide, your mass will be released as energy and they will get like 10^18J+10^18J = 1000000000000000000J+1000000000000000000J.
Changing the units, your mass is like 510^37eV/c^2 and if you colide with your antimatter evil clone, you both will release 510^37eV+5*10^37eV of energy.
No it isn't. At least, not all of them. The mass of a proton varies.
Y'all use a lot of bits for an approximation. In how many digits is the uncertainty?
Electron: 0.51099895000(15) MeV/c^2
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.
Photons are massless particles, but alas, they still gravitate because 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.
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.
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 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.
The Minkowski metric ensures that photons moving at C can not experience local time.
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.
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.
how is that different from 'photons don't have experienced time'
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.
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.
> 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”?
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.
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.
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?
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.
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.
It is still possible that even after 2024, KATRIN will be unable to measure the neutrino’s minimum mass: if the mass is less than 0.2 eV, it could lie outside the experiment’s sensitivity.
The sensitivity's goal is 0.2 eV. So if they report a measurement of 0.2 ± 0.2 eV, they won't be able to rule that the neutrino has a mass. But in fact it would could actually have a mass of 0.4 eV.
Am I missing something? Such as misunderstanding what they mean by sensitivity?
As a sort of counterexample, photons can have circular polarization. But if you measure the linear polarization of a photon in two locations, you may find that it is polarized up and down at one point, but polarized left and right at the other. Does this imply that the photon has mass because its polarization has changed as it propagated? No. It just means that the axis of polarization you measured doesn't line up with the way that the polarization gets propagated.
There's a very similar thing going on with neutrinos. When we measure a neutrino, it collapses into a particular eigenstate with a specific mass. But when the neutrino propagates, it propagates as a mixture that oscillates between the various eigenstates, a little like how a photon propagates with circular polarization.
It turns out that the frequency of these oscillations depends on two things: a parameter that measures the strength of this mixing, and the difference between the squares of the masses of the eigenstates. Since the frequency of oscillations is nonzero this means that the difference between the masses has to be nonzero, which means that at least one neutrino flavor has to have mass. But even if neutrinos had no oscillations this doesn't mean that they are massless --- they could equally well have a mixing coefficient of zero, or just have equal, but nonzero masses.
That is 0.199 eV could be measured even if the actual mass was 0.35 eV.
Unless I'm misunderstanding what they mean by sensitivity.
https://www.katrin.kit.edu/68.php
https://en.wikipedia.org/wiki/KATRIN (the German WP entry is better though if you don't mind using an online translator)
Death by Neutrino