6 ms·

This just says he came to the conclusion that there should be spontaneous emission and estimated it. Coming up with how it actually happens (particle-antipartic

by comboy 8y ago

This just says he came to the conclusion that there should be spontaneous emission and estimated it. Coming up with how it actually happens (particle-antiparticle pair) is another story, so it seems a bit unfair to Hawking to say that Feynman got it first.

coldtea 8y ago
What other thing could be the emission? Besides, isn't particle-antiparticle pair obvious, if one knows about their spontaneous appearance?
- nocturnial 8y ago
  The particle-antiparticle explanation is just an analogy and doesn't really cover what actually happens. I thought this video from PBS Space Time explains it really well while keeping it accessible for non-physicists: https://www.youtube.com/watch?v=qPKj0YnKANw https://www.youtube.com/watch?v=qPKj0YnKANw
  - stcredzero 8y ago
    And that video is good background info for the one on the Unruh Effect, which I found truly mind-blowing! https://www.youtube.com/watch?v=7cj6oiFDEXc https://www.youtube.com/watch?v=7cj6oiFDEXc Any time we accelerate, we produce a distant effective event horizon behind us! Also, "The Unruh" would make for a good name for aliens in Star Control!
- 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.
- mannykannot 8y ago
  I am always skeptical of retrospective claims of obviousness (partly from the number of times I have seen it made in math textbooks.)
  - CalChris 8y ago
    This happens in patents where the law distinguishes between obviousness (section 103 rejection) and novelty (section 102 rejection). You can invent something which is in fact novel but still obvious.
  - JadeNB 8y ago
    > I am always skeptical of retrospective claims of obviousness (partly from the number of times I have seen it made in math textbooks.) I agree with you about your claim, but not about your justification. I believe that "historically obvious" things ("that's the first thing I would have thought of in that situation; how did it take them so long?") are usually subject to considerable ex post facto bias. On the other hand, when a mathematics textbook says that something is obvious, it means, or should mean: we've specifically set up the presentation to this point so that there's a unique best way to assemble the material so far, and that unique best way will accomplish the next step. This isn't always true when it's claimed, but it is possible in textbooks in a way that it isn't (or that has only a small probability) in history, since textbooks are consciously organised and history isn't.
scrooched_moose 8y ago
Also, this says it was a year before Hawking published his paper. Presumably he'd been working on it for quite a while before he published so I'm not sure it's fair to say Feynman was "first".
mannykannot 8y ago
This isn't about priority, and that is clearly something that Feynman did not care about. It is about Feynman's astonishing ability to make fruitful connections.
jpmattia 8y ago
> Coming up with how it actually happens (particle-antiparticle pair) is another story, so it seems a bit unfair to Hawking to say that Feynman got it first. It is analogous to evolution of how stimulated and spontaneous emission were explained microscopically in the first place: For the case of general photon behavior, Einstein showed that the statistics of photons phenomenologically demanded that there be a stimulated process; and then (several decades later) QED provided a microscopic description of how stimulated and spontaneous processes work. For the case of black holes, it seemed that Feynman noticed that a stimulated-like process was occurring, which (again, probably because of statistics) phenomenologically demands that there be an associated spontaneous process; and then (merely one year later) Hawking provided a microscopic description of how the spontaneous process works. To be honest, I've never heard anyone talk about stimulated emission from black holes in the context of Hawking radiation, but perhaps it is well-known to those in the field.

by comboy 8y ago

coldtea 8y ago

What other thing could be the emission? Besides, isn't particle-antiparticle pair obvious, if one knows about their spontaneous appearance?

nocturnial 8y ago
The particle-antiparticle explanation is just an analogy and doesn't really cover what actually happens. I thought this video from PBS Space Time explains it really well while keeping it accessible for non-physicists: https://www.youtube.com/watch?v=qPKj0YnKANw https://www.youtube.com/watch?v=qPKj0YnKANw
- stcredzero 8y ago
  And that video is good background info for the one on the Unruh Effect, which I found truly mind-blowing! https://www.youtube.com/watch?v=7cj6oiFDEXc https://www.youtube.com/watch?v=7cj6oiFDEXc Any time we accelerate, we produce a distant effective event horizon behind us! Also, "The Unruh" would make for a good name for aliens in Star Control!
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.
mannykannot 8y ago
I am always skeptical of retrospective claims of obviousness (partly from the number of times I have seen it made in math textbooks.)
- CalChris 8y ago
  This happens in patents where the law distinguishes between obviousness (section 103 rejection) and novelty (section 102 rejection). You can invent something which is in fact novel but still obvious.
- JadeNB 8y ago
  > I am always skeptical of retrospective claims of obviousness (partly from the number of times I have seen it made in math textbooks.) I agree with you about your claim, but not about your justification. I believe that "historically obvious" things ("that's the first thing I would have thought of in that situation; how did it take them so long?") are usually subject to considerable ex post facto bias. On the other hand, when a mathematics textbook says that something is obvious, it means, or should mean: we've specifically set up the presentation to this point so that there's a unique best way to assemble the material so far, and that unique best way will accomplish the next step. This isn't always true when it's claimed, but it is possible in textbooks in a way that it isn't (or that has only a small probability) in history, since textbooks are consciously organised and history isn't.

scrooched_moose 8y ago

Also, this says it was a year before Hawking published his paper. Presumably he'd been working on it for quite a while before he published so I'm not sure it's fair to say Feynman was "first".

mannykannot 8y ago

This isn't about priority, and that is clearly something that Feynman did not care about. It is about Feynman's astonishing ability to make fruitful connections.

jpmattia 8y ago

> Coming up with how it actually happens (particle-antiparticle pair) is another story, so it seems a bit unfair to Hawking to say that Feynman got it first. It is analogous to evolution of how stimulated and spontaneous emission were explained microscopically in the first place: For the case of general photon behavior, Einstein showed that the statistics of photons phenomenologically demanded that there be a stimulated process; and then (several decades later) QED provided a microscopic description of how stimulated and spontaneous processes work. For the case of black holes, it seemed that Feynman noticed that a stimulated-like process was occurring, which (again, probably because of statistics) phenomenologically demands that there be an associated spontaneous process; and then (merely one year later) Hawking provided a microscopic description of how the spontaneous process works. To be honest, I've never heard anyone talk about stimulated emission from black holes in the context of Hawking radiation, but perhaps it is well-known to those in the field.