9 ms·
How light is a neutrino? The answer is closer than ever
- gus_massa 5y ago> These data imply an upper bound of 0.9 eV, which goes down to 0.8 eV when combined with the earlier results. For comparison, the mass of an electron is approximately 510,998.950 eV and the mass of a proton is 938,272,088. eV.
- Koshkin 5y agoToo many digits… TL;DR: .5 MeV and .9 GeV, respectively.
- pdonis 5y agoThat's what the GP wrote. Their electron value has a decimal point before the last three digits. The GP values are approximately 510 keV and 938 MeV.
- gus_massa 5y agoI prefer to use the same unit for all numbers because it makes the comparison easier. The bound of mass of the neutrino is really low. Using M and G hides it. 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.
- fknorangesite 5y ago> I prefer to use the same unit for all numbers because it makes the comparison easier. Thank you; this is a style guideline I wish more writers would adopt.
- robotresearcher 5y agoAlternatively, you can use exponential notation to compare numbers of very different sizes while writing them similarly and using the same unit.
- fknorangesite 5y agoYeah. The point is keeping the units consistent.
- deleted 5y ago[deleted]
- _Microft 5y ago511 keV and 938 MeV would have been much clearer than a distinction via comma/point like in 510,998 -> . <- 950 eV and 938,272 -> , <- 088 eV.
- gus_massa 5y agoI agree that the coma/point distinction is confusing. Also, I'd like to cut all the numbers in the same digit, but the next digit of the proton is too dubious. 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.
- cecilpl2 5y agoI would just round to the nearest eV and keep the commas. > 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.
- panda-giddiness 5y agoI know we're just bikeshedding now, but neutrino: 0. 8 eV/c^2 electron: 510 000. 0 eV/c^2 proton: 940 000 000. 0 eV/c^2 (Note: neutrino mass is given as the upper limit.)
- _Microft 5y agoThe first value has one, the others two significant figures in this example. The decimal point is also giving a false sense of certainty when used like this. You can find more about significant figures at Wikipedia: https://en.m.wikipedia.org/wiki/Significant_figures https://en.m.wikipedia.org/wiki/Significant_figures
- panda-giddiness 5y agoYes, I know about significant figures. In fact, I considered adding a note about sig figs before deciding it was too minor of a nitpick to elaborate on. Elsewhere in this thread, people have already listed these masses to much greater precision than almost anyone needs (including me, a practicing physicist); obviously people can infer I'm rounding from context. Of course, if you were to present the masses this way without context, you should add a note about rounding.
- codeflo 5y agoIn contrast to some sibling comments, I think using the same unit is a great idea. It’s just that long numbers with commas are a bit hard to read. At least I think 511 keV and 938 000 keV would have been another clear way to present those numbers.
- throwhauser 5y agoWriting out the digits gives a more visceral impression of the difference in weight, from roughly one, to a six-digit number, to a nine-digit number.
- andrekandre 5y ago(total layperson here but) why is mass measured in electron volts? (also i would have expected an electron to be... 1 electron volt?)
- FranchuFranchu 5y agoMy understanding is that an electron volt is the energy an electron needs to move against an electrical gradient of 1V (imagine like rowing against the current). It is a unit of energy, so at first glance it doesn't really make much sense to use it a a mass unit. However, since e = mc2, you can calculate the mass that is equivalent to 1 eV (which, in kg, turns out to be about 10 to the power of -36). 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.
- ghaff 5y agoBecause eV is not (directly) a mass measure. Though of course it can translate into a mass. To C&P Wikipedia it's "the measure of an amount of kinetic energy gained by a single electron accelerating from rest through an electric potential difference of one volt in vacuum."
- aroberge 5y agoelectron volts is a unit of energy. Using E = mc^2, one expresses masses in the equivalent energy quantity. It turns out that electron-volts is a convenient scale for the (energy equivalent) mass for elementary particles -- much better than trying to express the mass in kilograms.
- TheOtherHobbes 5y agoOr energy in Joules, which are around 10^19 bigger. So you'd have the inverse of that everywhere as an inconvenient factor. 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.
- wiz21c 5y ago
- Maursault 5y ago> the mass of a proton is 938,272,088. eV. No it isn't. At least, not all of them. The mass of a proton varies.
- dtgriscom 5y agoDepends on if it's been dieting recently.
- hbrav 5y agoI assume that is the mass of an unbound proton.
- Maursault 5y agoA proton in a hydrogen atom does not have the same mass as a proton in a Helium atom, nor do any protons in any particular element have the same mass as the protons in any of the other elements. Again, the mass of a proton varies.
- shirleyquirk 5y agoWhy does it vary? Changes in gluon/quark energy due to nucleon-nucleon interactions?
- Maursault 5y agoYes, in nucleons of different sizes, the amount of binding energy changes according to quantum mechanics. The result is the average mass per proton is different in different isotopes. The mass differences are explained by the energy differences. Anyway, idk about you, but this really bothers me.
- robotresearcher 5y agoapproximately <8 significant digit value>, <9 significant digit value>. Y'all use a lot of bits for an approximation. In how many digits is the uncertainty?
- gus_massa 5y agoProton: 938.27208816(29) MeV/c^2 Electron: 0.51099895000(15) MeV/c^2
- treeman79 5y agohttps://what-if.xkcd.com/73/ https://what-if.xkcd.com/73/ Death by Neutrino
- dgb23 5y agoThat’s a very cool book and a nice gift idea.
- deleted 5y ago[deleted]
- 6gvONxR4sf7o 5y ago> 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.
- whatshisface 5y ago>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.
- Koshkin 5y agoSo… Just like the neutrinos?
- whatshisface 5y agoNeutrinos aren't massless, but quite like the neutrinos because they're light. Err, unheavy.
- pja 5y agoNo, because neutrinos are believed to have non zero rest mass.
- 6gvONxR4sf7o 5y ago> 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.
- wholinator2 5y ago
- _Microft 5y agoYou can read more about the experiment here: https://www.katrin.kit.edu/68.php https://www.katrin.kit.edu/68.php https://en.wikipedia.org/wiki/KATRIN https://en.wikipedia.org/wiki/KATRIN (the German WP entry is better though if you don't mind using an online translator)
- SaberTail 5y agoWith the caveat that I'm a few years out from my PhD in the field, this is both really interesting and also maybe not surprising. 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.
- irjustin 5y agoThanks for this! 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?
- JonChesterfield 5y agoDreadful title. "The answer is closer than ever" applies to every single unanswered thing that will ever be answered, with every second that passes.
- peteradio 5y agoBetter than a dreadful article. Also there is an implied lower bound that is not zero which sort of implies that eventually we will converge on that threshold rather than the infinite toil of reducing powers (how much closer to zero is 10^-34 vs 10^-102). If we converge on that threshold without conclusion that has its own benefit of implying a misunderstanding in the way neutrino oscillations work.
- yreg 5y ago"This is the greatest iPhone we have ever made."
- btilly 5y agoI'm wondering whether the following line in the article is wrong: 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?
- cschmid 5y agoI don't know what the issue is. Let's assume for simplicity that the sensitivity is exactly 0.2 eV. Then if you measure something slightly above, like 0.201 eV, you can conclude it has a mass. If you measure something slightly below, 0.199 eV, you don't know.
- btilly 5y agoThe issue is that your measurement can be below the real value. Therefore you can measure something slightly below even if the real value is somewhat above. 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.
- AmericanChopper 5y agoThe way we know for sure that neutrinos have mass is because we know they oscillate between flavours, and in order for them to be capable of undergoing any type of change they must have a mass, because anything without a mass can only travel at the speed of light, and anything travelling at the speed of light has no experience of time and can therefor never undergone any type of change.
- antognini 5y agoThis is sort of true. You are correct that we know that at least some neutrinos have mass because they are observed to oscillate between flavors. But the argument is a bit more subtle than it simply being a consequence of the fact that massless particles don't experience proper time. 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.