13 ms·

Particle seen switching between matter and antimatter at CERN

203 points by birriel 5y ago

akmarinov 5y ago
Nice! ELI5 what implications does that have?
- Y_Y 5y ago
  No. Five-year-olds can't do particle physics. It may as well imply nothing at all.
  - BobbyJo 5y ago
    Then just don't respond homie.
  - burundi_coffee 5y ago
    Relevant xkcd: https://xkcd.com/1364/ https://xkcd.com/1364/ Some things are indeed too complicated to be explained in simple terms and analogies.
- jerf 5y ago
  The article really already has that in it, better than you're going to get from anyone else.
  - echelon 5y ago
    That simply is not true. HN is filled with subject matter experts that can break down complex topics better than popsci journalists, and provide commentary on how this fits within the framework of current and future research.
    - krapp 5y ago
      HN is filled with people who believe themselves to be subject matter experts, and who refuse to engage with posted articles as they consider leaving the sanctum sanctorum of HN to expose themselves to the plebian knowledge and interests of others to be a waste of time. In other words, with people who wear their ignorance as a badge of elitism. Not unrelated, HN has a few actual subject matter experts who are often aghast at how wrong HN tends to be about any field or subject not directly related to programming, and how often their confidence is in inverse proportion to their knowledge. You can safely assume that while there may be actual, credible particle physicists on HN, their voices are likely to be drowned out by armchair expositors of bullshit ranting about dark matter being a hoax perpetrated by cultural Marxists or somesuch. Do the work yourself. Read the article yourself. Enrich your own mind and satisfy your own intellectual curiosity, don't expect HN to do it for you, or to be better than the mainstream in terms of general expertise, or you'll be misled and disappointed more often than not.
      - echelon 5y ago
        You're projecting here. I only stepped in to defend someone asking for an explanation from the HN community. I'm not putting aside skepticism and intellectual rigour when I read the comments here. I also think you're wrong about the members of this community. There are a tremendously wide assortment of backgrounds represented here. I come to Hacker News mostly for, not in spite of, the comments. I think telling people to not ask questions is dangerous. The person I originally responded to was not doing that exactly, but it felt borderline enough to remind them that the community here is filled with experts and teachers. It's also worth pointing out that we shouldn't silence curiosity. Attacking what you perceive as a lazy mind can only harm the desire to learn.
        krapp 5y ago
        I'm not objecting to asking questions, I'm objecting to preferring the HN commentariat over at least engaging with the posted content. Asking questions after reading the article is fine - that's part of creating fruitful discussion. But HN does have a problem in that no one believes reading the article is worth the effort, and everyone prefers to just read the comments. The end result of that is a negative feedback loop of insularity and incuriosity, more people only engaging with the subject (or what they believe the subject is based on their reading of the title) on a superficial level rather than real intellectual discussion.
      - jodrellblank 5y ago
        "In Comments: Be kind. [...] Please don't sneer, including at the rest of the community. - https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html" HN and Reddit are filled with people sneering at how dumb everyone else on HN and Reddit are. It makes for tedious reading, and a cringeworthy "I'm superior because I see how dumb everyone else is" status grab. If something is wrong, correct it or downvote it or ignore it; don't conjour up some imaginary mind-reading fantasy about how great the author must think they are and then sneerily tear down the strawman.
        krapp 5y ago
  - AnimalMuppet 5y ago
    That article didn't clear up a whole lot for me personally. Maybe some others are like me. I also read the actual CERN article, which this article copied almost entirely. But after reading twice, what seems to be going on is that a particle has a slightly different mass from its antiparticle. (That is the most unbelievable sentence I have written so far this year! It's just... wow. I didn't think that was even theoretically possible.)
    - dukwon 5y ago
      > a particle has a slightly different mass from its antiparticle This is not the case, and that would violate CPT symmetry. There are two states with different masses (the mass eigenstates) but they are not antiparticles of eachother. They are linear superpositions of the flavour eigenstates: |D1> = p|D⁰> + q|D̅⁰> |D2> = p|D⁰> − q|D̅⁰> where p and q are complex coefficients. The difference in mass between D1 and D2 is directly proportional to the oscillation frequency. If p/q ≠ 1, you have CP violation in this mixing. The amount of CP violation in this measurement was found to be consistent with zero.
      - AnimalMuppet 5y ago
        Thanks. That's quite informative. But... having flavor eigenstates not being mass eigenstates is almost as mind-blowing to me. How is that possible? And, by "p/q ≠ 1", I presume you meant their norms?
        dukwon 5y ago
        If you think of the Hamiltonian as a 2x2 matrix operating on the flavour eigenstate, there are off-diagonal terms arising from the amplitude of spontaneously changing from particle to antiparticle. The leading-order diagrams look like this: https://ppd.fnal.gov/experiments/e871/public/images/Kaon_mixing.gif https://ppd.fnal.gov/experiments/e871/public/images/Kaon_mix... You get CP violation even if |p/q|=1 but Arg(p/q)≠0.
- treeman79 5y ago
  Spin two toy tops in opposite directions. When they bump into each other they both stop. All fun stops. This experiment says when they bump into each other ones spinning a certain direction always have a little spin to it left over. A tiny bit of fun still exists. That little bit of fun is all physical matter in the universe.
- adrian_b 5y ago
  The title is confusing, because the whole particle does not switch between matter and antimatter, only its internal components switch. The hadrons, i.e. the particles made of components bound together by the so-called strong nuclear forces are partitioned into 3 groups: particles made of 3 quarks, particles made of 3 anti-quarks and particles made of 1 quark together with 1 anti-quark. Particles made from more quarks than in these 3 groups are possible in principle, but they would disintegrate extremely quickly, i.e. in times many orders of magnitude less than 1 nanosecond. The reasons why only those 3 kinds of hadrons are normally encountered is that exactly like the electric forces attempt to neutralize the electric charge and prevent the negative charges to separate from the positive charges, the strong nuclear forces also try to neutralize a similar quantity, with the difference that this quantity is 2-dimensional, unlike the electric charge, which is 1-dimensional. The condition of neutralizing that 2-dimensional quantity can be satisfied by the 3 kinds of quark combinations listed above. The lightest hadron made of 3 quarks is the proton and the lightest hadron made of 3 anti-quarks is the anti-proton. The proton and the similar hadrons can be considered as matter and the anti-proton and similar hadrons can be considered as anti-matter. None of these hadrons made of either 3 quarks or 3 anti-quarks can switch between matter and anti-matter and this remains true for their compounds, like nuclei or anti-nuclei. On the other hand, the 3rd kind of hadrons, which are usually named mesons, and which are made from 1 quark and 1 anti-quark, are in fact neither matter nor anti-matter. The article linked here refers to a particle of this kind, where the total number of quarks is +1 -1 = 0. So in mesons it is possible that the pair of 1 quark + 1 anti-quark will transform in another pair of 1 quark + 1 anti-quark, but where the quarks are different types of quarks than before the transformation. This transformation is possible because the total number of quarks is 0, both before and after the transformation, so any such transformation is possible as long as the new types of quarks are such so that the electrical charge remains the same. The compositions of mesons can oscillate between different pairs quark + anti-quark, but the heavier mesons decay extremely fast so it is exceedingly difficult to observe such oscillations. The linked article reports the experimental observation of such an oscillation, which was known as possible since decades ago, but it was not seen in experiments due to technical difficulties. So in these transformations some quarks and anti-quarks appear and disappear, but the total quantity of matter and anti-matter is the same before and after, which is why the title is misleading.
- steve76 5y ago
  Water drips out of a faucet, and makes a sink wet. If you are nature, and have no internal mind, and can interact on extreme limits such as exists/non-exist, you can drip dryness from the sink, up through the faucet, through the pipes and into the water tower. Nature always takes the easiest route. If that route includes changing "drip wet down" or "drip dry up" it will take it. Electricity works with positive and negative. At very high temperatures in very precise regions, a third pole leads to better calculations. Better calculations led to better applications, such as plasma, measuring and manipulating the Sun, quantum biology where the molecules of genes are abandoned for a much higher resolution in the subatomic particle interactions, or giving your nervous system a therapeutic sunburn with a positron emitter.
brabel 5y ago
> if ever a matter and antimatter particle come into contact, they will annihilate each other in a burst of energy. > To complicate things, some particles, such as photons, are actually their own antiparticles. I believe "a burst of energy" means a bunch of photons? So, we still end up with the same amount of energy as before, but all particles are now photons? Like in the heat death of the universe?
- leephillips 5y ago
  I don’t know about the heat death of the universe, but yes, the mass-energy is conserved, and the energy that comes out is in the form of photons.
  - arbitrage 5y ago
    mass-energy is only conserved in static spaces. since the universe is expanding, it doesn't necessarily hold all the time.
    - leephillips 5y ago
      Do you think that the cosmic expansion of spacetime is remotely relevant to the experiment that we are talking about?
- Filligree 5y ago
  Nah, it's symmetric. An electron and a positron can annihilate, producing a pair of photons. Or two photons can annihilate, producing an electron and a positron. Or... Well, in fact the symmetry works for any rotation, not just 180 degrees. An electron and a photon can "annihilate", producing an electron* and a photon going the other way. This is also known as "A photon bouncing off a mirror". *: The careful reader will note there's no positron in this one. That's because positrons are what happen when an electron hits a photon so hard it bounces backwards in time -- that is, it's an electron going backwards in time. For one definition of 'time', at least. To make the mirror version work, the photon needs to have lower energy than would be involved with matter annihilation. Otherwise your gamma-ray photon will tear through the mirror instead of bouncing back, probably making a nice hole. This doesn't mean the rotation doesn't work, it's just that what I was describing wasn't really a "mirror" as commonly considered. But also, because momentum (or, more usefully, rapidity) is defined in terms of the angle the particle makes with time, rotations other than 180 degrees don't conserve momentum. So it's not a gamma-ray photon anymore.** **: Photons, obviously, have a fixed 'angle with relation to time'. You can take that as referring to the angle their waves make with time, instead.
Shank 5y ago
The official announcement may also be relevant: https://home.cern/news/news/physics/lhcb-measures-tiny-mass-difference-between-particles https://home.cern/news/news/physics/lhcb-measures-tiny-mass-... Edit: Official seminar: https://indico.cern.ch/event/1027299/ https://indico.cern.ch/event/1027299/
- dukwon 5y ago
  The LHCb collaboration's own summary of the result: https://lhcb-public.web.cern.ch/Welcome.html#Deltamc https://lhcb-public.web.cern.ch/Welcome.html#Deltamc
  - lamontcg 5y ago
    Yeah this is much better than the title article. K-Kbar mixing was observed decades ago. The headline here should be about the first measurement of a mass asymmetry between matter and antimatter.
    - dukwon 5y ago
      > The headline here should be about the first measurement of a mass asymmetry between matter and antimatter. The mass difference is not between particle and antiparticle, but between the mass eigenstates, which are not antiparticles of eachother. It's also not the first Δm to be measured. In fact, it was the last remaining one.
      - lamontcg 5y ago
        So if the mass eigenstates aren't different then you don't get mixing between the flavor eigenstates, so this is expected? And by measuring the frequency of the FCNCs then that is what gives them the difference between the mass eigenstates? (bear with me, its been 25 years since I decided to become a programmer instead...)
        dukwon 5y ago
macksd 5y ago
>> the Big Bang should have produced matter and antimatter in equal amounts, and over time that all would have collided and annihilated, leaving the cosmos a very empty place. Obviously that didn’t happen Why do we expect the Big Bang would have created little more matter than we observe? How do we know that almost all matter wasn't annihilated, and what's left really isn't just a rounding error after ~50% annihilated ~50%?
- laurowyn 5y ago
  I'm not reading that as the matter/antimatter didn't annihilate each other after the big bang. Rather, that matter and antimatter were not produced in equal amounts - if they were produced in equal quantities, why do our observations show such a large amount of matter and not equal amounts of antimatter? production and annihilation are balanced processes, so why the heavy sway to one side? Th obvious suggestion is that if there was equal amounts of antimatter near us, we would likely be annihilated already. But that doesn't explain why so little can be found remotely either. The potential for matter and antimatter to alternate is an interesting proposal that could explain our little island of stability in the universe. Or it could be something else entirely. I'm a software engineer, not a physicist.
- eigenket 5y ago
  Two reasons: 1. Because there isn't even close to the amount of radiation around we would see from such incredible amounts of anihlillation 2. Matter/antimatter annihilation is symmetric - it removes the same amount of matter and antimatter from the equation so however much stuff we started with we expect to still see the same amount of matter and antimatter and the antimatter doesn't seem to be around Its worth emphasising that there is some matter/antimatter asymmetry in the standard model, they don't behave as exact "mirror opposites" particularly in the context of some kaon physics, but there isn't even close to enough asymmetry to explain the amount of matter and the lack of antimatter we observe.
  - fb13 5y ago
    Thanks for the information. Does dark matter ever enter the discussion in terms of "there isn't even close to enough asymmetry to explain the amount of matter and the lack of antimatter we observe."
_Microft 5y ago
Now we know how much mass a minus-sign has ... 10^-41kg ;)
- deleted 5y ago
  [deleted]
- _Microft 5y ago
  1bit of information stored at 300K has a mass of 10^-38kg according to [0]. That’s a few orders of magnitude off but almost uncannily close. [0] https://aip.scitation.org/doi/10.1063/1.5123794 https://aip.scitation.org/doi/10.1063/1.5123794
  - zzt123 5y ago
    Hold up. The article shows a mass difference of 10^-38 for the matter/antimatter states. That’s not a few orders of magnitude off your number, that’s literally your exponent.
    - _Microft 5y ago
      The article gives the mass difference in grams, not kilograms. That’s where the factor of 10^-3 to the number in my top comment comes from.
      - zzt123 5y ago
        Ahhhh. Tricky units!
  - throw0101a 5y ago
    This reminds me of a 2004 weblog post by one of the co-creators of ZFS, Jeff Bonwick, on why 128 bits was chosen: > Some customers already have datasets on the order of a petabyte, or 2^50 bytes. Thus the 64-bit capacity limit of 2^64 bytes is only 14 doublings away. Moore's Law for storage predicts that capacity will continue to double every 9-12 months, which means we'll start to hit the 64-bit limit in about a decade. Storage systems tend to live for several decades, so it would be foolish to create a new one without anticipating the needs that will surely arise within its projected lifetime. > If 64 bits isn't enough, the next logical step is 128 bits. That's enough to survive Moore's Law until I'm dead, and after that, it's not my problem. But it does raise the question: what are the theoretical limits to storage capacity? > Although we'd all like Moore's Law to continue forever, quantum mechanics imposes some fundamental limits on the computation rate and information capacity of any physical device. In particular, it has been shown that 1 kilogram of matter confined to 1 liter of space can perform at most 10^51 operations per second on at most 10^31 bits of information [see Seth Lloyd, "Ultimate physical limits to computation." Nature 406, 1047-1054 (2000)]. A fully-populated 128-bit storage pool would contain 2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2^140 bits) / (10^31 bits/kg) = 136 billion kg. > That's a lot of gear. > To operate at the 10^31 bits/kg limit, however, the entire mass of the computer must be in the form of pure energy. By E=mc^2, the rest energy of 136 billion kg is 1.2x10^28 J. The mass of the oceans is about 1.4x10^21 kg. It takes about 4,000 J to raise the temperature of 1 kg of water by 1 degree Celcius, and thus about 400,000 J to heat 1 kg of water from freezing to boiling. The latent heat of vaporization adds another 2 million J/kg. Thus the energy required to boil the oceans is about 2.4x10^6 J/kg * 1.4x10^21 kg = 3.4x10^27 J. Thus, fully populating a 128-bit storage pool would, literally, require more energy than boiling the oceans. * https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-high https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-hi...
deleted 5y ago
[deleted]
azinman2 5y ago
> That means that if ever a matter and antimatter particle come into contact, they will annihilate each other in a burst of energy. Dumb q: if you could get a bunch of anti-matter, could it be harvested to generate electricity?
- eigenket 5y ago
  Yes, but the only ways we know to generate antimatter require a lot of energy. Theoretically it could be an incredibly energy dense "battery" for spaceships or whatever though.
  - User23 5y ago
    It’s expensive too[1]. “Right now, antimatter is the most expensive substance on Earth, about $62.5 trillion a gram ($1.75 quadrillion an ounce).” [1] https://science.nasa.gov/science-news/science-at-nasa/1999/prop12apr99_1 https://science.nasa.gov/science-news/science-at-nasa/1999/p...
    - h2odragon 5y ago
      Of course they really nail you on the shipping and handling fees.
      - dukwon 5y ago
        They're genuinely working on being able to deliver it in the back of a van: https://home.cern/news/news/physics/cern-approves-two-new-experiments-transport-antimatter https://home.cern/news/news/physics/cern-approves-two-new-ex...
    - opportune 5y ago
      Humans use positrons all the time though, it just turns out (anti)leptons have extremely low mass relative to their usefulness.
- Filligree 5y ago
  You'd need some way to capture the gamma-ray photons and pions, but yes. I'd recommend a wall of plasma. ...then use that to run a steam engine. Sigh. We'll never escape those.
mdoms 5y ago
> This subatomic particle is normally made up of a charm quark and an up antiquark, while its antimatter equivalent consists of a charm antiquark and an up quark. Q: if these subatomic particles consist of a quark and an anti-quark, why don't they self-annihilate?
- mkaic 5y ago
  as someone who isn't well versed in particle physics i'm also wondering this. can anyone with more experience weigh in?
- dukwon 5y ago
  Mesons have 'annihilation' decay modes, certainly, but they can also decay in other ways.
- ganzuul 5y ago
  Charm, Up and Top are all different. They share some properties but not all and therefore annihilation can not take place.
  - dukwon 5y ago
    Sure it can, but it has to proceed via the weak interaction, and it's suppressed by the GIM mechanism. For example, D⁰→γγ (c̅u or cu̅ annihilation) is rare and not yet observed, but it's allowed.
    - deleted 5y ago
      [deleted]
- gus_massa 5y ago
  There are six quarks: down, up, strange, charm, bottom, top. There are six anti-quarks: down, up, strange, charm, bottom, top. Using the strong force, a quark can annihilate only an antiquark of the same type, so the number of charm quark minus the number of charm antiquarks is "almost" conserved. If you start with a system with one charm quark like the particle here, you "allays" will have a system with one charm quark, or 2 charm quarks and on charm antiquark, or ... but "never" a system with no charm quarks like a few photons after annihilation. It's more complicated, because the weak force can annihilate one quark with a different antiquark. (The rules are somewhat complicated, and some annihilations are more usual than others.) So after some time, you can be lucky and a weak interaction can destroy your charm quark and a up antiquark, so the number of charm quark minus the number of charm antiquarks changes "slowly". Here "slowly" means that to see an effect of the weak interaction you must wait like 10E-12 seconds instead of the 10E-24 seconds you must wait until you see an effect of the strong interaction. The numbers change for each decay and are difficult to calculate and may be much longer. So everything in scare quotes means it's just an aproximation. For example the neutron->proton+electron+antineutrino decay use the weak force and you must wait like 10 minutes that is a lot for an unstable particle. The particle discussed in this article has a life of almost a millisecond, that is very long for a particle but short for us. (Electromagnetism can annihilate only an antiquark of the same type, so it's not important here and it's much fainter.) --- On the other hand, you are right. This article is about the decay of a particle D0 into a Ks0 and pions. After some time (probably a few milliseconds after the experiment finalized) the Ks0 decays in more pions. And the new pions and the other pions after some time decay into photons and muons and neutrinos, and after some time the muons decay into electrons and neutrinos. The number of electrons and antielectrons (positrons) are equal, so they annihilate each other and you get more photons. Also, the number of neutrinos is equal to the number of antineutrinos, so they theoretically can annihilate each other, but in practice they just escape. So the global reaction is an annhihilation: real: D0 ---long-story---> photons + neutrinos + antineutrinos optimistic theoretical possibility: D0 ---longer-difficult-story---> photons
smlss_sftwr 5y ago
Aha, sneaky little particle thought it could make a clean escape without being noticed, did it?
ephimetheus 5y ago
I’d love it if they could stop trying to assign these results to individual institutes (Oxford in this case). This is an LHCb publication and that’s that c
- gnufx 5y ago
  Did the Oxford group actually not do the relevant analysis as the university suggested? (I conclude from the lack of mention in my current and previous universities' news -- wot, no Tara Shears? -- that at least their LHCb groups weren't significantly involved!)
  - dukwon 5y ago
    2 of the 10 people who worked on the analysis are at Oxford. The rest are affiliated to CERN, Milano-Bicocca, EPFL and Manchester
- 867-5309 5y ago
  >according to Oxford physicists analyzing data from the Large Hadron Collider >data gathered during the Large Hadron Collider’s second run, by physicists at Oxford University >Sources: Oxford University, CERN
- ineedasername 5y ago
  Consider it like any other resource for scientific exploration. If it was a radio telescope array and a group from Oxford made a discovery while analyzing the massive amounts of data those generate, the array facility wouldn't get top credit for the discovery.
  - dukwon 5y ago
    That might be how it works in astronomy, but not in particle physics. The collaborations which build and operate the detectors are also the people who get to analyse the data. There is "open data" but it's released after a long embargo period and is much less complete. (In fact, you won't find any LHCb open data yet because CERN won't give us enough storage to host it.) Officially, all qualified members sign all papers. The names of the people who worked on an analysis are not publicised. There is already a preference to only work on analysis, because it's cheap and easy (no need to send people to CERN or kit-out expensive labs), but that doesn't help much in the way of actually collecting data, so "service work" shouldn't be discouraged by attributing extra credit to people doing analysis. Institutes sometimes write press releases such as this where they exaggerate their involvement. In this instance, only 2 out of 10 people in the analysis group are affiliated to Oxford. It's very disingenuous not to acknowledge that they're part of a larger collaboration.
ineedasername 5y ago
Others have even been seen to exist as a weird mixture of both states at the same time, thanks to the quantum quirk of superposition This makes the Schrodinger's Box experiment much more dangerous. How much energy would be released in the annihilation of Cat with Anti-Cat? I'm actually honestly interested in that last question.
- frutiger 5y ago
  > I'm actually honestly interested in that last question. The answer to your question can be calculated with a formula you already know. Perhaps the most famous formula of the modern era: E = mc²
  - ineedasername 5y ago
    Ah, very true. But then (not having much knowledge on this area of physics) I also don't know if it would be the energy of the mass of one cat, or two cats.
    - ViViDboarder 5y ago
      I think that would depend. If half the particles in the cat turn to antimatter and the other half stay as matter and annihilate each other, it’d be the mass of one cat.
- danbruc 5y ago
  About 43 megatons of TNT for one kilogram of cat annihilating with one kilogram of anti-cat. So about 200 megatons of TNT, four Tsar Bombas, for an average cat and an average anti-cat of 4.5 kilograms each.
JProthero 5y ago
I've often heard it said that a particular prediction of Quantum Electrodynamics [1] may be the most precise experimentally confirmed value in physics [2], with an accuracy in the region of one part in a trillion (10^12). I haven't heard the same claim made for the LIGO experiment yet, but I understand it is capable of detecting distance changes smaller than one part in a billion trillion (10^21) [3]. If this LHCb result is confirmed, given that it involves the detection of a mass difference of one part in a hundred million quadrillion quadrillion (10^38), would it qualify as a new precision record in physics? I'd be grateful if anyone familiar with the field could comment. [1] https://en.wikipedia.org/wiki/Electron_magnetic_moment https://en.wikipedia.org/wiki/Electron_magnetic_moment [2] https://en.wikipedia.org/wiki/Precision_tests_of_QED https://en.wikipedia.org/wiki/Precision_tests_of_QED [3] https://en.wikipedia.org/wiki/LIGO#Observations https://en.wikipedia.org/wiki/LIGO#Observations
- dtgriscom 5y ago
  This isn't a ratio of one part in 10^38; it's an absolute difference between the particles' masses of 1x10^-38 grams. The ratio implied depends on the mass of each particle. I presume this is far less than a gram, which means the radio is also far less.
- murkt 5y ago
  From LHCb results summary https://lhcb-public.web.cern.ch/Welcome.html#Deltamc https://lhcb-public.web.cern.ch/Welcome.html#Deltamc : (m1-m2)/(D0mass) = 3x10^-15 10^-38 is an absolute measurement, not the ratio.
  - JProthero 5y ago
    Thanks for the replies. To clarify — is this not a measurement of the two particles' masses to ~38 significant figures?
    - mkl 5y ago
      None of those zeros are significant figures.

203 points by birriel 5y ago

akmarinov 5y ago

Nice! ELI5 what implications does that have?

Y_Y 5y ago
No. Five-year-olds can't do particle physics. It may as well imply nothing at all.
- BobbyJo 5y ago
  Then just don't respond homie.
- burundi_coffee 5y ago
  Relevant xkcd: https://xkcd.com/1364/ https://xkcd.com/1364/ Some things are indeed too complicated to be explained in simple terms and analogies.
jerf 5y ago
The article really already has that in it, better than you're going to get from anyone else.
- echelon 5y ago
  That simply is not true. HN is filled with subject matter experts that can break down complex topics better than popsci journalists, and provide commentary on how this fits within the framework of current and future research.
  - krapp 5y ago
    HN is filled with people who believe themselves to be subject matter experts, and who refuse to engage with posted articles as they consider leaving the sanctum sanctorum of HN to expose themselves to the plebian knowledge and interests of others to be a waste of time. In other words, with people who wear their ignorance as a badge of elitism. Not unrelated, HN has a few actual subject matter experts who are often aghast at how wrong HN tends to be about any field or subject not directly related to programming, and how often their confidence is in inverse proportion to their knowledge. You can safely assume that while there may be actual, credible particle physicists on HN, their voices are likely to be drowned out by armchair expositors of bullshit ranting about dark matter being a hoax perpetrated by cultural Marxists or somesuch. Do the work yourself. Read the article yourself. Enrich your own mind and satisfy your own intellectual curiosity, don't expect HN to do it for you, or to be better than the mainstream in terms of general expertise, or you'll be misled and disappointed more often than not.
    - echelon 5y ago
      You're projecting here. I only stepped in to defend someone asking for an explanation from the HN community. I'm not putting aside skepticism and intellectual rigour when I read the comments here. I also think you're wrong about the members of this community. There are a tremendously wide assortment of backgrounds represented here. I come to Hacker News mostly for, not in spite of, the comments. I think telling people to not ask questions is dangerous. The person I originally responded to was not doing that exactly, but it felt borderline enough to remind them that the community here is filled with experts and teachers. It's also worth pointing out that we shouldn't silence curiosity. Attacking what you perceive as a lazy mind can only harm the desire to learn.
      - krapp 5y ago
        I'm not objecting to asking questions, I'm objecting to preferring the HN commentariat over at least engaging with the posted content. Asking questions after reading the article is fine - that's part of creating fruitful discussion. But HN does have a problem in that no one believes reading the article is worth the effort, and everyone prefers to just read the comments. The end result of that is a negative feedback loop of insularity and incuriosity, more people only engaging with the subject (or what they believe the subject is based on their reading of the title) on a superficial level rather than real intellectual discussion.
    - jodrellblank 5y ago
      "In Comments: Be kind. [...] Please don't sneer, including at the rest of the community. - https://news.ycombinator.com/newsguidelines.html https://news.ycombinator.com/newsguidelines.html" HN and Reddit are filled with people sneering at how dumb everyone else on HN and Reddit are. It makes for tedious reading, and a cringeworthy "I'm superior because I see how dumb everyone else is" status grab. If something is wrong, correct it or downvote it or ignore it; don't conjour up some imaginary mind-reading fantasy about how great the author must think they are and then sneerily tear down the strawman.
      - krapp 5y ago
- AnimalMuppet 5y ago
  That article didn't clear up a whole lot for me personally. Maybe some others are like me. I also read the actual CERN article, which this article copied almost entirely. But after reading twice, what seems to be going on is that a particle has a slightly different mass from its antiparticle. (That is the most unbelievable sentence I have written so far this year! It's just... wow. I didn't think that was even theoretically possible.)
  - dukwon 5y ago
    > a particle has a slightly different mass from its antiparticle This is not the case, and that would violate CPT symmetry. There are two states with different masses (the mass eigenstates) but they are not antiparticles of eachother. They are linear superpositions of the flavour eigenstates: |D1> = p|D⁰> + q|D̅⁰> |D2> = p|D⁰> − q|D̅⁰> where p and q are complex coefficients. The difference in mass between D1 and D2 is directly proportional to the oscillation frequency. If p/q ≠ 1, you have CP violation in this mixing. The amount of CP violation in this measurement was found to be consistent with zero.
    - AnimalMuppet 5y ago
      Thanks. That's quite informative. But... having flavor eigenstates not being mass eigenstates is almost as mind-blowing to me. How is that possible? And, by "p/q ≠ 1", I presume you meant their norms?
      - dukwon 5y ago
        If you think of the Hamiltonian as a 2x2 matrix operating on the flavour eigenstate, there are off-diagonal terms arising from the amplitude of spontaneously changing from particle to antiparticle. The leading-order diagrams look like this: https://ppd.fnal.gov/experiments/e871/public/images/Kaon_mixing.gif https://ppd.fnal.gov/experiments/e871/public/images/Kaon_mix... You get CP violation even if |p/q|=1 but Arg(p/q)≠0.
treeman79 5y ago
Spin two toy tops in opposite directions. When they bump into each other they both stop. All fun stops. This experiment says when they bump into each other ones spinning a certain direction always have a little spin to it left over. A tiny bit of fun still exists. That little bit of fun is all physical matter in the universe.
adrian_b 5y ago
The title is confusing, because the whole particle does not switch between matter and antimatter, only its internal components switch. The hadrons, i.e. the particles made of components bound together by the so-called strong nuclear forces are partitioned into 3 groups: particles made of 3 quarks, particles made of 3 anti-quarks and particles made of 1 quark together with 1 anti-quark. Particles made from more quarks than in these 3 groups are possible in principle, but they would disintegrate extremely quickly, i.e. in times many orders of magnitude less than 1 nanosecond. The reasons why only those 3 kinds of hadrons are normally encountered is that exactly like the electric forces attempt to neutralize the electric charge and prevent the negative charges to separate from the positive charges, the strong nuclear forces also try to neutralize a similar quantity, with the difference that this quantity is 2-dimensional, unlike the electric charge, which is 1-dimensional. The condition of neutralizing that 2-dimensional quantity can be satisfied by the 3 kinds of quark combinations listed above. The lightest hadron made of 3 quarks is the proton and the lightest hadron made of 3 anti-quarks is the anti-proton. The proton and the similar hadrons can be considered as matter and the anti-proton and similar hadrons can be considered as anti-matter. None of these hadrons made of either 3 quarks or 3 anti-quarks can switch between matter and anti-matter and this remains true for their compounds, like nuclei or anti-nuclei. On the other hand, the 3rd kind of hadrons, which are usually named mesons, and which are made from 1 quark and 1 anti-quark, are in fact neither matter nor anti-matter. The article linked here refers to a particle of this kind, where the total number of quarks is +1 -1 = 0. So in mesons it is possible that the pair of 1 quark + 1 anti-quark will transform in another pair of 1 quark + 1 anti-quark, but where the quarks are different types of quarks than before the transformation. This transformation is possible because the total number of quarks is 0, both before and after the transformation, so any such transformation is possible as long as the new types of quarks are such so that the electrical charge remains the same. The compositions of mesons can oscillate between different pairs quark + anti-quark, but the heavier mesons decay extremely fast so it is exceedingly difficult to observe such oscillations. The linked article reports the experimental observation of such an oscillation, which was known as possible since decades ago, but it was not seen in experiments due to technical difficulties. So in these transformations some quarks and anti-quarks appear and disappear, but the total quantity of matter and anti-matter is the same before and after, which is why the title is misleading.
steve76 5y ago
Water drips out of a faucet, and makes a sink wet. If you are nature, and have no internal mind, and can interact on extreme limits such as exists/non-exist, you can drip dryness from the sink, up through the faucet, through the pipes and into the water tower. Nature always takes the easiest route. If that route includes changing "drip wet down" or "drip dry up" it will take it. Electricity works with positive and negative. At very high temperatures in very precise regions, a third pole leads to better calculations. Better calculations led to better applications, such as plasma, measuring and manipulating the Sun, quantum biology where the molecules of genes are abandoned for a much higher resolution in the subatomic particle interactions, or giving your nervous system a therapeutic sunburn with a positron emitter.

brabel 5y ago

> if ever a matter and antimatter particle come into contact, they will annihilate each other in a burst of energy. > To complicate things, some particles, such as photons, are actually their own antiparticles. I believe "a burst of energy" means a bunch of photons? So, we still end up with the same amount of energy as before, but all particles are now photons? Like in the heat death of the universe?

leephillips 5y ago
I don’t know about the heat death of the universe, but yes, the mass-energy is conserved, and the energy that comes out is in the form of photons.
- arbitrage 5y ago
  mass-energy is only conserved in static spaces. since the universe is expanding, it doesn't necessarily hold all the time.
  - leephillips 5y ago
    Do you think that the cosmic expansion of spacetime is remotely relevant to the experiment that we are talking about?
Filligree 5y ago
Nah, it's symmetric. An electron and a positron can annihilate, producing a pair of photons. Or two photons can annihilate, producing an electron and a positron. Or... Well, in fact the symmetry works for any rotation, not just 180 degrees. An electron and a photon can "annihilate", producing an electron* and a photon going the other way. This is also known as "A photon bouncing off a mirror". *: The careful reader will note there's no positron in this one. That's because positrons are what happen when an electron hits a photon so hard it bounces backwards in time -- that is, it's an electron going backwards in time. For one definition of 'time', at least. To make the mirror version work, the photon needs to have lower energy than would be involved with matter annihilation. Otherwise your gamma-ray photon will tear through the mirror instead of bouncing back, probably making a nice hole. This doesn't mean the rotation doesn't work, it's just that what I was describing wasn't really a "mirror" as commonly considered. But also, because momentum (or, more usefully, rapidity) is defined in terms of the angle the particle makes with time, rotations other than 180 degrees don't conserve momentum. So it's not a gamma-ray photon anymore.** **: Photons, obviously, have a fixed 'angle with relation to time'. You can take that as referring to the angle their waves make with time, instead.

Shank 5y ago

The official announcement may also be relevant: https://home.cern/news/news/physics/lhcb-measures-tiny-mass-difference-between-particles https://home.cern/news/news/physics/lhcb-measures-tiny-mass-... Edit: Official seminar: https://indico.cern.ch/event/1027299/ https://indico.cern.ch/event/1027299/

dukwon 5y ago
The LHCb collaboration's own summary of the result: https://lhcb-public.web.cern.ch/Welcome.html#Deltamc https://lhcb-public.web.cern.ch/Welcome.html#Deltamc
- lamontcg 5y ago
  Yeah this is much better than the title article. K-Kbar mixing was observed decades ago. The headline here should be about the first measurement of a mass asymmetry between matter and antimatter.
  - dukwon 5y ago
    > The headline here should be about the first measurement of a mass asymmetry between matter and antimatter. The mass difference is not between particle and antiparticle, but between the mass eigenstates, which are not antiparticles of eachother. It's also not the first Δm to be measured. In fact, it was the last remaining one.
    - lamontcg 5y ago
      So if the mass eigenstates aren't different then you don't get mixing between the flavor eigenstates, so this is expected? And by measuring the frequency of the FCNCs then that is what gives them the difference between the mass eigenstates? (bear with me, its been 25 years since I decided to become a programmer instead...)
      - dukwon 5y ago

macksd 5y ago

>> the Big Bang should have produced matter and antimatter in equal amounts, and over time that all would have collided and annihilated, leaving the cosmos a very empty place. Obviously that didn’t happen Why do we expect the Big Bang would have created little more matter than we observe? How do we know that almost all matter wasn't annihilated, and what's left really isn't just a rounding error after ~50% annihilated ~50%?

laurowyn 5y ago
I'm not reading that as the matter/antimatter didn't annihilate each other after the big bang. Rather, that matter and antimatter were not produced in equal amounts - if they were produced in equal quantities, why do our observations show such a large amount of matter and not equal amounts of antimatter? production and annihilation are balanced processes, so why the heavy sway to one side? Th obvious suggestion is that if there was equal amounts of antimatter near us, we would likely be annihilated already. But that doesn't explain why so little can be found remotely either. The potential for matter and antimatter to alternate is an interesting proposal that could explain our little island of stability in the universe. Or it could be something else entirely. I'm a software engineer, not a physicist.
eigenket 5y ago
Two reasons: 1. Because there isn't even close to the amount of radiation around we would see from such incredible amounts of anihlillation 2. Matter/antimatter annihilation is symmetric - it removes the same amount of matter and antimatter from the equation so however much stuff we started with we expect to still see the same amount of matter and antimatter and the antimatter doesn't seem to be around Its worth emphasising that there is some matter/antimatter asymmetry in the standard model, they don't behave as exact "mirror opposites" particularly in the context of some kaon physics, but there isn't even close to enough asymmetry to explain the amount of matter and the lack of antimatter we observe.
- fb13 5y ago
  Thanks for the information. Does dark matter ever enter the discussion in terms of "there isn't even close to enough asymmetry to explain the amount of matter and the lack of antimatter we observe."

_Microft 5y ago

Now we know how much mass a minus-sign has ... 10^-41kg ;)

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_Microft 5y ago
1bit of information stored at 300K has a mass of 10^-38kg according to [0]. That’s a few orders of magnitude off but almost uncannily close. [0] https://aip.scitation.org/doi/10.1063/1.5123794 https://aip.scitation.org/doi/10.1063/1.5123794
- zzt123 5y ago
  Hold up. The article shows a mass difference of 10^-38 for the matter/antimatter states. That’s not a few orders of magnitude off your number, that’s literally your exponent.
  - _Microft 5y ago
    The article gives the mass difference in grams, not kilograms. That’s where the factor of 10^-3 to the number in my top comment comes from.
    - zzt123 5y ago
      Ahhhh. Tricky units!
- throw0101a 5y ago
  This reminds me of a 2004 weblog post by one of the co-creators of ZFS, Jeff Bonwick, on why 128 bits was chosen: > Some customers already have datasets on the order of a petabyte, or 2^50 bytes. Thus the 64-bit capacity limit of 2^64 bytes is only 14 doublings away. Moore's Law for storage predicts that capacity will continue to double every 9-12 months, which means we'll start to hit the 64-bit limit in about a decade. Storage systems tend to live for several decades, so it would be foolish to create a new one without anticipating the needs that will surely arise within its projected lifetime. > If 64 bits isn't enough, the next logical step is 128 bits. That's enough to survive Moore's Law until I'm dead, and after that, it's not my problem. But it does raise the question: what are the theoretical limits to storage capacity? > Although we'd all like Moore's Law to continue forever, quantum mechanics imposes some fundamental limits on the computation rate and information capacity of any physical device. In particular, it has been shown that 1 kilogram of matter confined to 1 liter of space can perform at most 10^51 operations per second on at most 10^31 bits of information [see Seth Lloyd, "Ultimate physical limits to computation." Nature 406, 1047-1054 (2000)]. A fully-populated 128-bit storage pool would contain 2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2^140 bits) / (10^31 bits/kg) = 136 billion kg. > That's a lot of gear. > To operate at the 10^31 bits/kg limit, however, the entire mass of the computer must be in the form of pure energy. By E=mc^2, the rest energy of 136 billion kg is 1.2x10^28 J. The mass of the oceans is about 1.4x10^21 kg. It takes about 4,000 J to raise the temperature of 1 kg of water by 1 degree Celcius, and thus about 400,000 J to heat 1 kg of water from freezing to boiling. The latent heat of vaporization adds another 2 million J/kg. Thus the energy required to boil the oceans is about 2.4x10^6 J/kg * 1.4x10^21 kg = 3.4x10^27 J. Thus, fully populating a 128-bit storage pool would, literally, require more energy than boiling the oceans. * https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-high https://blogs.oracle.com/bonwick/128-bit-storage:-are-you-hi...

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azinman2 5y ago

> That means that if ever a matter and antimatter particle come into contact, they will annihilate each other in a burst of energy. Dumb q: if you could get a bunch of anti-matter, could it be harvested to generate electricity?

eigenket 5y ago
Yes, but the only ways we know to generate antimatter require a lot of energy. Theoretically it could be an incredibly energy dense "battery" for spaceships or whatever though.
- User23 5y ago
  It’s expensive too[1]. “Right now, antimatter is the most expensive substance on Earth, about $62.5 trillion a gram ($1.75 quadrillion an ounce).” [1] https://science.nasa.gov/science-news/science-at-nasa/1999/prop12apr99_1 https://science.nasa.gov/science-news/science-at-nasa/1999/p...
  - h2odragon 5y ago
    Of course they really nail you on the shipping and handling fees.
    - dukwon 5y ago
      They're genuinely working on being able to deliver it in the back of a van: https://home.cern/news/news/physics/cern-approves-two-new-experiments-transport-antimatter https://home.cern/news/news/physics/cern-approves-two-new-ex...
  - opportune 5y ago
    Humans use positrons all the time though, it just turns out (anti)leptons have extremely low mass relative to their usefulness.
Filligree 5y ago
You'd need some way to capture the gamma-ray photons and pions, but yes. I'd recommend a wall of plasma. ...then use that to run a steam engine. Sigh. We'll never escape those.

mdoms 5y ago

> This subatomic particle is normally made up of a charm quark and an up antiquark, while its antimatter equivalent consists of a charm antiquark and an up quark. Q: if these subatomic particles consist of a quark and an anti-quark, why don't they self-annihilate?

mkaic 5y ago
as someone who isn't well versed in particle physics i'm also wondering this. can anyone with more experience weigh in?
dukwon 5y ago
Mesons have 'annihilation' decay modes, certainly, but they can also decay in other ways.
ganzuul 5y ago
Charm, Up and Top are all different. They share some properties but not all and therefore annihilation can not take place.
- dukwon 5y ago
  Sure it can, but it has to proceed via the weak interaction, and it's suppressed by the GIM mechanism. For example, D⁰→γγ (c̅u or cu̅ annihilation) is rare and not yet observed, but it's allowed.
  - deleted 5y ago
    [deleted]
gus_massa 5y ago
There are six quarks: down, up, strange, charm, bottom, top. There are six anti-quarks: down, up, strange, charm, bottom, top. Using the strong force, a quark can annihilate only an antiquark of the same type, so the number of charm quark minus the number of charm antiquarks is "almost" conserved. If you start with a system with one charm quark like the particle here, you "allays" will have a system with one charm quark, or 2 charm quarks and on charm antiquark, or ... but "never" a system with no charm quarks like a few photons after annihilation. It's more complicated, because the weak force can annihilate one quark with a different antiquark. (The rules are somewhat complicated, and some annihilations are more usual than others.) So after some time, you can be lucky and a weak interaction can destroy your charm quark and a up antiquark, so the number of charm quark minus the number of charm antiquarks changes "slowly". Here "slowly" means that to see an effect of the weak interaction you must wait like 10E-12 seconds instead of the 10E-24 seconds you must wait until you see an effect of the strong interaction. The numbers change for each decay and are difficult to calculate and may be much longer. So everything in scare quotes means it's just an aproximation. For example the neutron->proton+electron+antineutrino decay use the weak force and you must wait like 10 minutes that is a lot for an unstable particle. The particle discussed in this article has a life of almost a millisecond, that is very long for a particle but short for us. (Electromagnetism can annihilate only an antiquark of the same type, so it's not important here and it's much fainter.) --- On the other hand, you are right. This article is about the decay of a particle D0 into a Ks0 and pions. After some time (probably a few milliseconds after the experiment finalized) the Ks0 decays in more pions. And the new pions and the other pions after some time decay into photons and muons and neutrinos, and after some time the muons decay into electrons and neutrinos. The number of electrons and antielectrons (positrons) are equal, so they annihilate each other and you get more photons. Also, the number of neutrinos is equal to the number of antineutrinos, so they theoretically can annihilate each other, but in practice they just escape. So the global reaction is an annhihilation: real: D0 ---long-story---> photons + neutrinos + antineutrinos optimistic theoretical possibility: D0 ---longer-difficult-story---> photons

smlss_sftwr 5y ago

Aha, sneaky little particle thought it could make a clean escape without being noticed, did it?

ephimetheus 5y ago

I’d love it if they could stop trying to assign these results to individual institutes (Oxford in this case). This is an LHCb publication and that’s that c

gnufx 5y ago
Did the Oxford group actually not do the relevant analysis as the university suggested? (I conclude from the lack of mention in my current and previous universities' news -- wot, no Tara Shears? -- that at least their LHCb groups weren't significantly involved!)
- dukwon 5y ago
  2 of the 10 people who worked on the analysis are at Oxford. The rest are affiliated to CERN, Milano-Bicocca, EPFL and Manchester
867-5309 5y ago
>according to Oxford physicists analyzing data from the Large Hadron Collider >data gathered during the Large Hadron Collider’s second run, by physicists at Oxford University >Sources: Oxford University, CERN
ineedasername 5y ago
Consider it like any other resource for scientific exploration. If it was a radio telescope array and a group from Oxford made a discovery while analyzing the massive amounts of data those generate, the array facility wouldn't get top credit for the discovery.
- dukwon 5y ago
  That might be how it works in astronomy, but not in particle physics. The collaborations which build and operate the detectors are also the people who get to analyse the data. There is "open data" but it's released after a long embargo period and is much less complete. (In fact, you won't find any LHCb open data yet because CERN won't give us enough storage to host it.) Officially, all qualified members sign all papers. The names of the people who worked on an analysis are not publicised. There is already a preference to only work on analysis, because it's cheap and easy (no need to send people to CERN or kit-out expensive labs), but that doesn't help much in the way of actually collecting data, so "service work" shouldn't be discouraged by attributing extra credit to people doing analysis. Institutes sometimes write press releases such as this where they exaggerate their involvement. In this instance, only 2 out of 10 people in the analysis group are affiliated to Oxford. It's very disingenuous not to acknowledge that they're part of a larger collaboration.

ineedasername 5y ago

Others have even been seen to exist as a weird mixture of both states at the same time, thanks to the quantum quirk of superposition This makes the Schrodinger's Box experiment much more dangerous. How much energy would be released in the annihilation of Cat with Anti-Cat? I'm actually honestly interested in that last question.

frutiger 5y ago
> I'm actually honestly interested in that last question. The answer to your question can be calculated with a formula you already know. Perhaps the most famous formula of the modern era: E = mc²
- ineedasername 5y ago
  Ah, very true. But then (not having much knowledge on this area of physics) I also don't know if it would be the energy of the mass of one cat, or two cats.
  - ViViDboarder 5y ago
    I think that would depend. If half the particles in the cat turn to antimatter and the other half stay as matter and annihilate each other, it’d be the mass of one cat.
danbruc 5y ago
About 43 megatons of TNT for one kilogram of cat annihilating with one kilogram of anti-cat. So about 200 megatons of TNT, four Tsar Bombas, for an average cat and an average anti-cat of 4.5 kilograms each.

JProthero 5y ago

I've often heard it said that a particular prediction of Quantum Electrodynamics [1] may be the most precise experimentally confirmed value in physics [2], with an accuracy in the region of one part in a trillion (10^12). I haven't heard the same claim made for the LIGO experiment yet, but I understand it is capable of detecting distance changes smaller than one part in a billion trillion (10^21) [3]. If this LHCb result is confirmed, given that it involves the detection of a mass difference of one part in a hundred million quadrillion quadrillion (10^38), would it qualify as a new precision record in physics? I'd be grateful if anyone familiar with the field could comment. [1] https://en.wikipedia.org/wiki/Electron_magnetic_moment https://en.wikipedia.org/wiki/Electron_magnetic_moment [2] https://en.wikipedia.org/wiki/Precision_tests_of_QED https://en.wikipedia.org/wiki/Precision_tests_of_QED [3] https://en.wikipedia.org/wiki/LIGO#Observations https://en.wikipedia.org/wiki/LIGO#Observations

dtgriscom 5y ago
This isn't a ratio of one part in 10^38; it's an absolute difference between the particles' masses of 1x10^-38 grams. The ratio implied depends on the mass of each particle. I presume this is far less than a gram, which means the radio is also far less.
murkt 5y ago
From LHCb results summary https://lhcb-public.web.cern.ch/Welcome.html#Deltamc https://lhcb-public.web.cern.ch/Welcome.html#Deltamc : (m1-m2)/(D0mass) = 3x10^-15 10^-38 is an absolute measurement, not the ratio.
- JProthero 5y ago
  Thanks for the replies. To clarify — is this not a measurement of the two particles' masses to ~38 significant figures?
  - mkl 5y ago
    None of those zeros are significant figures.