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Will 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

Will 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 ago
Has to be the complementary type. In your example, they will repel each other (since both are positively charged).
- willis936 5y ago
  What about a neutron and positron? They could collide with no electrostatic repulsion but have quark-antiquark pairs.
  - fsh 5y ago
    Only hadrons (neutrons, protons, mesons,...) consist of quarks. Positrons are leptons which are elementary particles.
  - db48x 5y ago
    No, 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 ago
      lepton 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 ago
        To 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 ago
        What 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 ago
        As a rule of thumb it’s good enough for most particle interactions.
  - jfengel 5y ago
    The 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 ago
For 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 ago
  Side 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?
  - 5y ago
Xcelerate 5y ago
I’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 ago
  A 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 ago
    Annihilating 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 ago
How 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 ago
  It'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.

Will 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 ago

Has to be the complementary type. In your example, they will repel each other (since both are positively charged).

willis936 5y ago
What about a neutron and positron? They could collide with no electrostatic repulsion but have quark-antiquark pairs.
- fsh 5y ago
  Only hadrons (neutrons, protons, mesons,...) consist of quarks. Positrons are leptons which are elementary particles.
- db48x 5y ago
  No, 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 ago
    lepton 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 ago
      To 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 ago
        What 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 ago
      As a rule of thumb it’s good enough for most particle interactions.
- jfengel 5y ago
  The 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 ago

For 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 ago
Side 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?
- 5y ago

Xcelerate 5y ago

I’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 ago
A 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 ago
  Annihilating 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 ago

How 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 ago
It'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.