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> a photon giving rise to a virtual e-/e+ pair that immediately annihilates back into a photon i'm been wondering about validity of virtual particle trick - wh

> a photon giving rise to a virtual e-/e+ pair that immediately annihilates back into a photon i'm been wondering about validity of virtual particle trick - while it is preserves charge, etc... doesn't it violates entropy principle of the 2nd law, ie. a pair of e-/e+ seems to have less entropy than the original/resulting photon, thus e-/e+ always happy to produce a photon, while photon would never produce e-/e+ pair without "help".

tgb 13y ago
There's actually two photons involved, not just one: http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annih... (If there were just one produced by annihilation, you'd have great difficultly satisfying conservation of momentum and having that photon move at the speed of c (when looking in the center-of-mass frame).)
Steuard 13y ago
The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles (the number 10^23 gets thrown around a lot, though it becomes an awfully compelling "law" even with just a few hundred particles in the mix). But as for your actual example, it's important to distinguish between virtual pair production and longer-lived pair production. A single photon can momentarily "decay" into an e-/e+ pair which then annihilates back into a photon, and in fact quantum field theory tells us that this sort of process is an intrinsic part of how photons travel through space. (Similar things happen for all other particles as well.) But conservation of energy and momentum during this process leads to weird effects like violations of the usual mass/energy/momentum relation of relativity, which is essentially why we call the e-/e+ "virtual" particles (or "off-shell"[1]). For a "real" annihilation between an electron and a positron, you'll always wind up with two photons in the final state (necessary to conserve momentum and energy). By the same token, to produce an e-/e+ pair you'd need two photons, not just one, and those photons would need to have enough energy to provide the mass of both particles and enough kinetic energy to satisfy momentum conservation. I wouldn't invoke the 2nd law here (as I said above), but the point you're making is still valid: because so much of the e-/e+ energy is "used up" in their mass, there is less "phase space" available (fewer possible momentum states) in that configuration than as two photons, so the probability of winding up in that state is correspondingly lower. (Also, an electron and positron are electrically attracted to each other and will naturally tend to meet up to annihilate, but two photons have no such tendency.) I'm not sure that any of that was interesting or useful, but there you go. [1] https://en.wikipedia.org/wiki/On_shell_and_off_shell https://en.wikipedia.org/wiki/On_shell_and_off_shell
- VladRussian2 13y ago
  >The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles All the fundamental laws of physics, including the Schrodinger equation follow the 2nd law, so it is somewhat
  - VladRussian2 13y ago
    sorry, somehow partial text got submitted. >The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles All the fundamental laws of physics, including the Schrodinger equation, follow the 2nd law or in other words they describe application of 2nd law to the given area of physics, i.e. they express "entropy never decreases" principle in terms of the objects of that physical area. >"there is less "phase space" available (fewer possible momentum states)" ie. less entropy :) >an electron and positron are electrically attracted to each other and will naturally tend to meet up to annihilate 2 particles having negative and positive charges - less entropy than the entropy of 2 resulting neutral particles. Attraction (getting closer - increasing entropy) and annihilation (erasing the charge difference - increasing entropy) are in full compliance with, and i'd say direct manifestation of, the 2nd law :)
    - Steuard 13y ago
      My glib objection was not to the notion of there being less entropy in one state than the other, but in calling the result "the 2nd law of thermodynamics" in cases like this. The phase space argument means that it is less likely for the system to wind up in the e-/e+ state than in the 2-photon state, and calling that a manifestation of entropy is entirely sensible. But in general, there's a non-negligible chance that the system will wind up in the e-/e+ state anyway: it's very much a possibility (and becomes even more so when the total energy is greater). I usually reserve the term "2nd law" for cases where the probabilities are tiny enough that it's essentially impossible for the entropy to decrease by any remotely significant amount. That level of certainty typically requires systems of many particles, not just one.

tgb 13y ago

There's actually two photons involved, not just one: http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annih... (If there were just one produced by annihilation, you'd have great difficultly satisfying conservation of momentum and having that photon move at the speed of c (when looking in the center-of-mass frame).)

Steuard 13y ago

The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles (the number 10^23 gets thrown around a lot, though it becomes an awfully compelling "law" even with just a few hundred particles in the mix). But as for your actual example, it's important to distinguish between virtual pair production and longer-lived pair production. A single photon can momentarily "decay" into an e-/e+ pair which then annihilates back into a photon, and in fact quantum field theory tells us that this sort of process is an intrinsic part of how photons travel through space. (Similar things happen for all other particles as well.) But conservation of energy and momentum during this process leads to weird effects like violations of the usual mass/energy/momentum relation of relativity, which is essentially why we call the e-/e+ "virtual" particles (or "off-shell"[1]). For a "real" annihilation between an electron and a positron, you'll always wind up with two photons in the final state (necessary to conserve momentum and energy). By the same token, to produce an e-/e+ pair you'd need two photons, not just one, and those photons would need to have enough energy to provide the mass of both particles and enough kinetic energy to satisfy momentum conservation. I wouldn't invoke the 2nd law here (as I said above), but the point you're making is still valid: because so much of the e-/e+ energy is "used up" in their mass, there is less "phase space" available (fewer possible momentum states) in that configuration than as two photons, so the probability of winding up in that state is correspondingly lower. (Also, an electron and positron are electrically attracted to each other and will naturally tend to meet up to annihilate, but two photons have no such tendency.) I'm not sure that any of that was interesting or useful, but there you go. [1] https://en.wikipedia.org/wiki/On_shell_and_off_shell https://en.wikipedia.org/wiki/On_shell_and_off_shell

VladRussian2 13y ago
>The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles All the fundamental laws of physics, including the Schrodinger equation follow the 2nd law, so it is somewhat
- VladRussian2 13y ago
  sorry, somehow partial text got submitted. >The glib answer here is that there's really no such thing as the 2nd law when you're talking about a small number of individual fundamental particles. The statement "entropy never decreases" is actually a purely statistical claim, and it only rises to its usual level of essentially unbreakable validity when you're dealing with macroscopic numbers of particles All the fundamental laws of physics, including the Schrodinger equation, follow the 2nd law or in other words they describe application of 2nd law to the given area of physics, i.e. they express "entropy never decreases" principle in terms of the objects of that physical area. >"there is less "phase space" available (fewer possible momentum states)" ie. less entropy :) >an electron and positron are electrically attracted to each other and will naturally tend to meet up to annihilate 2 particles having negative and positive charges - less entropy than the entropy of 2 resulting neutral particles. Attraction (getting closer - increasing entropy) and annihilation (erasing the charge difference - increasing entropy) are in full compliance with, and i'd say direct manifestation of, the 2nd law :)
  - Steuard 13y ago
    My glib objection was not to the notion of there being less entropy in one state than the other, but in calling the result "the 2nd law of thermodynamics" in cases like this. The phase space argument means that it is less likely for the system to wind up in the e-/e+ state than in the 2-photon state, and calling that a manifestation of entropy is entirely sensible. But in general, there's a non-negligible chance that the system will wind up in the e-/e+ state anyway: it's very much a possibility (and becomes even more so when the total energy is greater). I usually reserve the term "2nd law" for cases where the probabilities are tiny enough that it's essentially impossible for the entropy to decrease by any remotely significant amount. That level of certainty typically requires systems of many particles, not just one.