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An antimatter experiment shows surprises near absolute zero
- Pet_Ant 5y agoWill any anti-matter/matter annhilate or does it have to be the complementary type? Like can a proton and positron annihilate? PS- hate the article title, pure vague clickbait
- czzr 5y agoHas to be the complementary type. In your example, they will repel each other (since both are positively charged).
- willis936 5y agoWhat about a neutron and positron? They could collide with no electrostatic repulsion but have quark-antiquark pairs.
- fsh 5y agoOnly hadrons (neutrons, protons, mesons,...) consist of quarks. Positrons are leptons which are elementary particles.
- db48x 5y agoNo, a positron is a lepton but a neutron is not. Since lepton number must be conserved, these two will not annihilate with each other. Similarly, one has a positive charge and the other neutral, so electric charge would not be conserved either. Are you thinking of a neutron plus an antiproton? Those can annihilate with each other. They are not direct antiparticles of each other, but the neutron is really an up quark and two down quarks, while the antiproton is really two anti–up quarks and an anti–down quark. There are several different ways that they could annihilate with each other. Worse, all hadrons (like neutrons and protons) contain a confined gluon field which you can think of as containing additional quark–antiquark pairs. There’s a chance that the gluon fields of the two particles will interact as if a quark–antiquark pair from one had annihilated with a quark–antiquark pair from the other, further complicating the results. This can cause the creation of other exotic particles as the quarks combine and recombine. You can read more about it on Wikipedia, of course: https://en.wikipedia.org/wiki/Annihilation https://en.wikipedia.org/wiki/Annihilation
- ellis-bell 5y agolepton number does not need to be conserved. it is an approximate symmetry of nature. if lepton number were conserved, neutrinos could not oscillate. quarks, the particles composing protons and neutrons, have fractional charge; these would be the particles that would interact with an electron or positron. the charges wouldn't work out (charge is conserved (as far as we known...)) so there wouldn't be a fundamental electromagnetic interaction between a single quark and a e+/e- (i.e., an annihilation). But there are fundamental weak interactions between quarks and e+/e-; these processes are known as inverse beta decay and are used for pet scans.
- lanstin 5y agoTo quote Wikipedia: “Lepton flavor is only approximately conserved, and is notably not conserved in neutrino oscillation.[6] However, total lepton number is still conserved in the Standard Model.“ The beta decay gives rise to e.g. a positron and a neutrino (or an electron and an anti-neutrino).
- floxy 5y agoWhat is going on in neutron stars? The layman's answer is that the electrons get squeezed into the protons due to the extreme gravity, leaving only neutrons. But I suppose (?) there needs to be a anti-neutrino or similar that comes along to "complete" the reaction? Edit: I guess it is inverse beta decay: https://en.wikipedia.org/wiki/Inverse_beta_decay#Electron_induced https://en.wikipedia.org/wiki/Inverse_beta_decay#Electron_in...
- db48x 5y agoAs a rule of thumb it’s good enough for most particle interactions.
- jfengel 5y agoThe electrostatic repulsion isn't really an issue. If you shoot an electron hard enough, it will overcome the electrostatic repulsion. The bigger issue here is that a positron is a lepton, and a neutron has a 0 lepton count. So they can't react directly while conserving lepton count. However, a neutron can decay into a proton, an electron, and a anti-neutrino. The neutrino is a lepton. So there is a path where the neutron and positron react to form a proton, anti-neutrino, and two gamma rays. I don't know if that actually occurs. There are even more factors to be considered. But it illustrates how you have to go about conserving all of the things, including obscurer factors like leptonicity.
- aardvark179 5y agoFor particles to annihilate everything has to be conserved. A proton and a positron cannot annihilate because they both have positive electric charge, but you also need a bunch of other properties to cancel out as well.
- gus_massa 5y ago> PS- hate the article title, pure vague clickbait I agree. From the article: > Yet physicists have found nothing amiss, no conclusive sign that antimatter particles — which are just the oppositely charged twins of familiar particles — obey different rules. So no surprises. Also, the "near absolute zero" is true, but it's just to make the experiment easier, it's not important from the theoretical point of view. Anyway, it's an interesting experiment in spite of the press article title. > Will any anti-matter/matter anhilate or does it have to be the complementary type? Like can a proton and positron annihilate? There are a lot of conservation rules, for example * proton + positron --> gamma rays is impossible because the electric charge is conserved * proton(+1) + positron (+1) --> gamma rays (0) so +2 charge before and 0 after the annihilation, so it's imposible. There are other number that are conserved, like the number of quarks (a quark is +1 and an anti-quark is -1) so * proton(+3) + positron (0) --> gamma rays (0) And the number of leptons, particles like electron are +1, and particles like positron are -1 * proton(0) + positron (-1) --> gamma rays (0) And there are other examples of numbers that must be conserved. (And there are numbers that are almost conserved and they can change but usually the decay/annihilation is "slower".) It's interesting how this work with changes that are common, for example * neutron(0) --> proton(+1) + electron(-1) + anti-neutrino(0) [charge] * neutron(+3) --> proton(+3) + electron(0) + anti-neutrino(0) [quarks] * neutron(0) --> proton(0) + electron(1) + anti-neutrino(-1) [leptons] or another example * electron(-1) + positron(+1) --> two gamma rays(0) [charge] * electron(0) + positron(0) --> two gamma rays(0) [quarks] * electron(+1) + positron(-1) --> two gamma rays(0) [leptons] Note that as said before there are more numbers that are conserved and more complicated rules. If one of these 3 rules are broken the transformation is impossible (AFAWK). If these 3 rules are ok, then you must check all the other rules.
- cupofpython 5y agoSide question. I understand the positive and negative charges, but when we say "+1" or "-1" what are the units? I never understood that part. an electron is -1 and a proton is +1, but what makes them equal? Why / how do all these things of varying sizes and properties have the same magnitude for charges?
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- Xcelerate 5y agoI’ve always thought “annihilate” was a poorly chosen word to describe what happens here. It makes it sound like stuff is disappearing, and if you consider “stuff” to be only those particles that have rest mass, then fine, the aggregate value of the rest mass property of the system becomes zero, but most of the “substance” of everyday objects isn’t due to rest mass anyway. Also, the notion that rest mass “turns into energy” is similarly silly, as energy is just a scalar value that is conserved over time due to the time translational invariance of the laws of physics. It would be like saying I converted an onion into the number 7; it doesn’t make any sense. The energy of the system is the same after the collision as it was before the collision. A better way to describe matter/anti-matter annihilation is that different types of particles interact in all sorts of different ways, and a collection of particles can turn into other types (and number) of particles during these interactions as long as the complete system obeys certain conservation laws. Matter/anti-matter collision is no more special in this sense than any of the other types of interactions in the Standard Model.
- chasil 5y agoA star that is undergoing a "pair-instability" supernova would express rather dramatic opinions on exactly what is meant by "annihilation" as it is destroyed by positrons. https://en.wikipedia.org/wiki/Pair-instability_supernova https://en.wikipedia.org/wiki/Pair-instability_supernova
- raattgift 5y agoAnnihilating positrons don't really consume an appreciable amount of the star's ordinary matter. Their presence mostly serves to allow a big star to compress into a much smaller space than otherwise, leading to a more powerful thermonuclear detonation. One way to think about the gamma -> lepton-pair is that the parent (the gamma photon) must have at least the same momentum-energy as it's lepton-pair children. That is, the positron and electron each have less momentum than the gamma, and since each half of the pair can go in a different random direction, a gamma whose entire momentum may be outwards might turn into a positron that, for example, heads inwards and an electron that heads outwards, each carrying about half the gamma's original outward momentum. The outward momentum flux is what does the work of keeping the star from imploding gravitationally, and the balance is usually fairly fine; when a substantial amount of that momentum flux is turned in the wrong direction, the implosion is inevitable. There is nothing particularly special about the positrons (indeed, muons and antimuons, and other cascades of particles likely are produced too) other than that they and their partner electrons represent a redirection of mostly-outward momentum. Another way of thinking about it is that a hot gas of gammas becomes a cooler gas of gammas+leptons, and "hot gas rises" is what keeps the star inflated. Cool down the middle, and the star contracts, because there is less hot gas rising. Initially, the sudden central cooling may re-heat the star's middle (compressive heating as outer layers sink inwards), which produces further central cooling (pair-production), which produces yet more heating, and so forth, with the swings from heating to cooling becoming more extreme over time. In some stars undergoing this process, the heating wins, and a thermonuclear detonation completely obliterates the star, scattering its remains to infinity. In other stars other factors help "lift" the outer envelope (e.g. the star's rotation, or the presence of many atomic nuclei heavier than helium) and cooling wins, with a super-dense phase (likely a black hole) holding on to some of the detonating material. More technically, the lepton-pair has more degrees of freedom than the pair's parent gamma, and as one adds more degrees of freedom to a system, one increases the system's heat capacity. This is why one calls the process of pair production "cooling": effective temperature goes down when specific heat capacity goes up. For a nice rabbit hole, visit negative heat capacity in stars: https://en.wikipedia.org/wiki/Heat_capacity#Negative_heat_capacity https://en.wikipedia.org/wiki/Heat_capacity#Negative_heat_ca... with a general attitude of "momentum can condense, or become frozen into structures with large numbers of internal degrees of freedom (DoFs); black holes have enormous numbers of such DoFs inside the horizon, with number of DoFs increasing with black hole mass, which in turn is why more massive black holes are colder than small black holes".
- breezeTrowel 5y agoHow is the article title clickbait? They expected the spectral lines of hybrid atoms to widen the same as regular atoms when immersed in a dense fluid. Instead the spectral lines narrowed. This was so surprising they thought they made a mistake and spent years checking over their work only to conclude that, yes, the finding was in fact real.
- whimsicalism 5y agoIt's not clickbait. And obviously a title about a complex physics topic is going to be vague. It's just classic HN bikeshedding, as usual.
- wadkar 5y agoI don’t understand how an anti-proton can orbit a nucleus without annihilating into pure energy with a proton from the nucleus. What am I missing here? Honestly speaking I am really amazed by these researchers ability to separate and then “shove” an anti-proton in a Helium gas.
- Iwan-Zotow 5y ago> I don’t understand how an anti-proton can orbit a nucleus without annihilating into pure energy with a proton from the nucleus. it means there is bound meta-stable state of combined proton+nucleus interacting system.
- simonh 5y agoIf a proton in the nucleus and orbiting anti-proton don’t collide (to simplify somewhat), then they won’t annihilate.
- mynameismon 5y agoRelated: https://news.ycombinator.com/item?id=30700633 https://news.ycombinator.com/item?id=30700633
- mmmBacon 5y agoI skimmed the article looking for something about the spectroscopy of antiprotonic helium. I’d assume that the spectroscopy of such an atom is very different from normal helium but the article says nothing about how the spectroscopy is different or if the spectroscopy observed is what we expect (thinking of anti Hydrogen here). The article does point to Helium made from a pion though where the atoms spectroscopy was observed.
- mikewarot 5y agoIt took reading it a few times to figure out that you can slam antiprotons into liquid helium and end up with a shelf stable helium/antimatter hybrid atoms. That is the part that amazes me... I had no idea that was even possible. I wonder what other atoms this would work with?
- waterhouse 5y ago> shelf stable helium/antimatter hybrid atoms Well. Depends on the scale you consider to be shelf stable: The antiproton can thus orbit the nucleus for tens of microseconds, before finally falling to its surface and annihilating. I think that is a relatively long time in terms of particle-physics experiments, but it's not something you could stockpile. https://en.wikipedia.org/wiki/Antiprotonic_helium https://en.wikipedia.org/wiki/Antiprotonic_helium
- mikewarot 5y agoIt makes no reference to lifetimes in the article, other than to say the outer electron shields it. The Wikipedia article mentions that it has a high energy, which aligns with the original article, which also states they figured out how to lower the energy. So, mixed signals... it could be stable??
- AnimalMuppet 5y agoNo. The problem is that once the antiproton reaches the ground state, its wavefunction has significant overlap with the nucleus. That means that it's going to encounter the proton in short order.
- gameswithgo 5y agoDoesn't it only last a few microseconds?
- icodestuff 5y agoVery cool. I had no idea antiprotons -- well, an antiproton and an electron -- could orbit a nucleus. Does it fill up the usual quantum state of the first electron, or does it have its own set of quantum states to fill? Can they get a second antiproton to take the place of the second electron? I imagine that should be at least somewhat stable as the electrons of the enclosing container should repel the antiprotons just fine.
- AnimalMuppet 5y ago> Does it fill up the usual quantum state of the first electron, or does it have its own set of quantum states to fill? Not a physicist. But if I understand correctly, it fits the same equations the electron would, except you have to use the proton mass instead of the electron mass, which means it's much closer in. The remaining electron (I think) doesn't have Pauli exclusion with the anti-proton, so it looks like the electron in a hydrogen atom. (Since it's much further out, it sees the effective charge of +1 for the "nucleus" of two protons and an anti-proton.)
- martincmartin 5y agoClassical mechanics has F = ma. Quantum mechanics has Schrodinger's Equation. Given a probability distribution for the location of your particle, and the potential (as in classical mechanics, just the integral of the force), it describes how that probability distribution evolves over time. For something orbiting a nucleus, the only parameters that matter are electric charge and mass. This stuff was figured out in the 1920s, and is taught in a year or two of undergraduate physics. The examples are usually an electron orbiting a proton (the hydrogen atom), but work just as well for muons, or anything else. It's only once you get inside the atom that you get into more advanced stuff.