5 ms·

It's more complex than that. If it was that simple, it would emit particles and antiparticles randomly which would not lead to losing mass.

by terryf 8y ago

doubleunplussed 8y ago
What? Particles and antiparticles both have positive mass, as far as we know. Emitting either or both would be a loss of mass.
- terryf 8y ago
  Ok, to be fair I'm not a physicist but it would seem to me that if you get a stream of mixed particles and anti-particles, they would annihilate, thus not producing a mass difference.
  - sbuttgereit 8y ago
    That "annihilation" doesn't lead to non-existence, it does lead to transformation. You get energy rather than matter... Which would still lead to mass loss so long as that energy didn't end up back in the black hole. And very much not a physicist myself...
    - terryf 8y ago
      Right, but the charges of particle-antiparticle pairs do cancel out, so there is no net energy produced. Quick thought experiment: two particle-antiparticle pairs get created: p1-ap1, p2-ap2 In regular space, after a very short amount of time p1 annihilates with ap1 and p2 with ap2. Net result is zero additional matter and energy, otherwise we'd just be creating random matter and/or energy everywhere. If the same happens at the event horizon of a black hole and let's assume purely by chance, p1 and ap2 fall in and p2 and ap1 do not. After a short while, p1 and ap2 annihilate inside and p2 and ap1 outside. The result should be effectively the same as in the previous case, thus zero net energy and/or matter on either side. Thus, also no radiation and mass loss. It'd be nice if an actual physicist chimed in at this point and told me where I went wrong :)
      - sbuttgereit 8y ago
        What is the charge of a photon? For the charge to matter, you're saying that charges infer positive or negative mass. But that's original responder's point.... both the particle and anti-particle both have positive mass regardless of their charge, which is a different attribute. Consider how it is you are conserving energy/mass in your thought experiment. Seems like you might be creating a back door whereby that mass/energy is not conserved in order to preserve charge conservation.
      - JdeBP 8y ago
        Ahem! Net result is photons, which have energy. The actual physicist is Richard Feynman, and here is a Feynman diagram of a particle-antiparticle annihilation. * https://commons.wikimedia.org/wiki/File:Mutual_Annihilation_of_a_Positron_Electron_pair.svg https://commons.wikimedia.org/wiki/File:Mutual_Annihilation_...
        terryf 8y ago
        Well, the virtual p-ap creation happens everywhere, all the time. Why aren't there a large number of photons happening everywhere all the time?
        sbuttgereit 8y ago
        Look at the PBS Space Time video previously linked in other comments. It will help.
        JadeNB 8y ago
        nocturnial's other comment: https://news.ycombinator.com/item?id=19643999 https://news.ycombinator.com/item?id=19643999 .
        Retra 8y ago
        The epistemological reality of those events are not as simple as you assume. For the most part, that annihilation event is a short-hand analogy for a term in an equation. Another way to think of it is that this 'happening everywhere' style of particle-anti-particle annihilation relies on the inherent uncertainty of energy over short time scales, and thus it 'borrows' energy that doesn't exist in order to precede, but under any measurable timescale the energy must be conserved, and so these events are likewise unobservable. So it's not a simple matter to assume they happen, or that they are the same phenomenon as annihilation in general.
    - deleted 8y ago
      [deleted]
- steve76 8y ago
  Massless particles exist, and I believe this goes to the concept that particle and anti-particle pairs exist everywhere, and annihilate each other very very fast. If you put an object into empty space, sunlight will hit it and get hot. Don't have an object in empty space? Something has to be there. It might be part of that region in space, or it might be some property of the object emitting energy. It help me to think of this as a model. It works, whether we have an intuitive understanding of it or not. Light comes on, wall gets bright. Whether it's ripples in a connected medium, or collisions between little projectiles, we know it's always going to be the same. What I'm thinking about is if black holes are not really black and have some light to them, can suns sort of be black holes too? Can we thread particles around it's outer orbit, to gain control of the gravity internal to the star, and also pull out more from what's around it?

It's more complex than that. If it was that simple, it would emit particles and antiparticles randomly which would not lead to losing mass.

by terryf 8y ago

It's more complex than that. If it was that simple, it would emit particles and antiparticles randomly which would not lead to losing mass.

doubleunplussed 8y ago

What? Particles and antiparticles both have positive mass, as far as we know. Emitting either or both would be a loss of mass.

terryf 8y ago
Ok, to be fair I'm not a physicist but it would seem to me that if you get a stream of mixed particles and anti-particles, they would annihilate, thus not producing a mass difference.
- sbuttgereit 8y ago
  That "annihilation" doesn't lead to non-existence, it does lead to transformation. You get energy rather than matter... Which would still lead to mass loss so long as that energy didn't end up back in the black hole. And very much not a physicist myself...
  - terryf 8y ago
    Right, but the charges of particle-antiparticle pairs do cancel out, so there is no net energy produced. Quick thought experiment: two particle-antiparticle pairs get created: p1-ap1, p2-ap2 In regular space, after a very short amount of time p1 annihilates with ap1 and p2 with ap2. Net result is zero additional matter and energy, otherwise we'd just be creating random matter and/or energy everywhere. If the same happens at the event horizon of a black hole and let's assume purely by chance, p1 and ap2 fall in and p2 and ap1 do not. After a short while, p1 and ap2 annihilate inside and p2 and ap1 outside. The result should be effectively the same as in the previous case, thus zero net energy and/or matter on either side. Thus, also no radiation and mass loss. It'd be nice if an actual physicist chimed in at this point and told me where I went wrong :)
    - sbuttgereit 8y ago
      What is the charge of a photon? For the charge to matter, you're saying that charges infer positive or negative mass. But that's original responder's point.... both the particle and anti-particle both have positive mass regardless of their charge, which is a different attribute. Consider how it is you are conserving energy/mass in your thought experiment. Seems like you might be creating a back door whereby that mass/energy is not conserved in order to preserve charge conservation.
    - JdeBP 8y ago
      Ahem! Net result is photons, which have energy. The actual physicist is Richard Feynman, and here is a Feynman diagram of a particle-antiparticle annihilation. * https://commons.wikimedia.org/wiki/File:Mutual_Annihilation_of_a_Positron_Electron_pair.svg https://commons.wikimedia.org/wiki/File:Mutual_Annihilation_...
      - terryf 8y ago
        Well, the virtual p-ap creation happens everywhere, all the time. Why aren't there a large number of photons happening everywhere all the time?
        sbuttgereit 8y ago
        Look at the PBS Space Time video previously linked in other comments. It will help.
        JadeNB 8y ago
        nocturnial's other comment: https://news.ycombinator.com/item?id=19643999 https://news.ycombinator.com/item?id=19643999 .
        Retra 8y ago
        The epistemological reality of those events are not as simple as you assume. For the most part, that annihilation event is a short-hand analogy for a term in an equation. Another way to think of it is that this 'happening everywhere' style of particle-anti-particle annihilation relies on the inherent uncertainty of energy over short time scales, and thus it 'borrows' energy that doesn't exist in order to precede, but under any measurable timescale the energy must be conserved, and so these events are likewise unobservable. So it's not a simple matter to assume they happen, or that they are the same phenomenon as annihilation in general.
  - deleted 8y ago
    [deleted]
steve76 8y ago
Massless particles exist, and I believe this goes to the concept that particle and anti-particle pairs exist everywhere, and annihilate each other very very fast. If you put an object into empty space, sunlight will hit it and get hot. Don't have an object in empty space? Something has to be there. It might be part of that region in space, or it might be some property of the object emitting energy. It help me to think of this as a model. It works, whether we have an intuitive understanding of it or not. Light comes on, wall gets bright. Whether it's ripples in a connected medium, or collisions between little projectiles, we know it's always going to be the same. What I'm thinking about is if black holes are not really black and have some light to them, can suns sort of be black holes too? Can we thread particles around it's outer orbit, to gain control of the gravity internal to the star, and also pull out more from what's around it?