4 ms·

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 “stuf

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".

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".