Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Rigorous Science

Post History

66%
+2 −0
Rigorous Science Can pion production effectively shorten the lifetime of neutrons?

The known charge-conserving decay modes of free neutrons all involve the production of a proton, an electron and an electron antineutrino: $$n\to p^++e^-+\bar{\nu}_{e}$$ This beta decay is why, out...

1 answer  ·  posted 2y ago by HDE 226868‭  ·  edited 2y ago by HDE 226868‭

#6: Post edited by user avatar HDE 226868‭ · 2020-06-20T02:44:52Z (over 2 years ago)
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) (and I think I understand [its application to beta decay](http://hep.ucsb.edu/courses/ph125_02/fgr_schw.pdf)), but I'm not sure if I'm on the right track - and I also have no idea if this sort of approach is valid for a large ensemble of particles.
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
#5: Post edited by user avatar HDE 226868‭ · 2020-06-20T01:36:17Z (over 2 years ago)
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. If we were talking about something classical, I'd expect the rate per unit volume $\Gamma$ to be proportional to the number densities of photons $n_{\gamma}$ and neutrons $n_n$ and the interaction cross-section $\sigma$:
  • $$\Gamma\propto n_{\gamma}n_n\sigma$$
  • but I suspect that the quantum mechanical case is much more complicated. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) (and I think I understand [its application to beta decay](http://hep.ucsb.edu/courses/ph125_02/fgr_schw.pdf)), but the exact computations are beyond me.
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) (and I think I understand [its application to beta decay](http://hep.ucsb.edu/courses/ph125_02/fgr_schw.pdf)), but I'm not sure if I'm on the right track - and I also have no idea if this sort of approach is valid for a large ensemble of particles.
#4: Post edited by user avatar HDE 226868‭ · 2020-06-20T01:34:24Z (over 2 years ago)
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. If we were talking about something classical, I'd expect the rate per unit volume $\Gamma$ to be proportional to the number densities of photons $n_{\gamma}$ and neutrons $n_n$ and the interaction cross-section $\sigma$:
  • $$\Gamma\propto n_{\gamma}n_n\sigma$$
  • but I suspect that the quantum mechanical case is much more complicated. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule), but the exact computations are beyond me.
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. If we were talking about something classical, I'd expect the rate per unit volume $\Gamma$ to be proportional to the number densities of photons $n_{\gamma}$ and neutrons $n_n$ and the interaction cross-section $\sigma$:
  • $$\Gamma\propto n_{\gamma}n_n\sigma$$
  • but I suspect that the quantum mechanical case is much more complicated. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) (and I think I understand [its application to beta decay](http://hep.ucsb.edu/courses/ph125_02/fgr_schw.pdf)), but the exact computations are beyond me.
#3: Post edited by user avatar HDE 226868‭ · 2020-06-20T01:31:50Z (over 2 years ago)
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
  • ---
  • I'm stumped because I have no idea how to figure out the probability of these interactions occurring. If we were talking about something classical, I'd expect the rate per unit volume $\Gamma$ to be proportional to the number densities of photons $n_{\gamma}$ and neutrons $n_n$ and the interaction cross-section $\sigma$:
  • $$\Gamma\propto n_{\gamma}n_n\sigma$$
  • but I suspect that the quantum mechanical case is much more complicated. I assume I'd want to invoke [Fermi's golden rule](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule), but the exact computations are beyond me.
#2: Post edited by user avatar HDE 226868‭ · 2020-06-19T21:29:40Z (over 2 years ago)
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate.
  • The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
  • $$n\to p^++e^-+\bar{\nu}_{e}$$
  • This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.
  • I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
  • $$\gamma+n\to p^++\pi^-$$
  • and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).
  • Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate?
#1: Initial revision by user avatar HDE 226868‭ · 2020-06-19T21:28:21Z (over 2 years ago)
The [known charge-conserving decay modes of free neutrons](http://pdg.lbl.gov/2015/listings/rpp2015-list-n.pdf) all involve the production of a proton, an electron and an electron antineutrino:
$$n\to p^++e^-+\bar{\nu}_{e}$$
This beta decay is why, outside an atom, a neutron is unstable, with a lifetime of 15 minutes. This is actually [quite a long time](https://physics.stackexchange.com/q/31514/56299) compared to the decays of other particles mediated by the weak nuclear force.

I recently learned that there are other ways for a neutron to (any I'm *definitely* abusing terminology here) "decay" into a proton and a charged pion, if it can interact with a photon:
$$\gamma+n\to p^++\pi^-$$
and similarly, a proton interacting with a photon can lead to a neutron and a positively charged pion. Both of these processes are known as [charged pion photoproduction](https://www.jlab.org/exp_prog/proposals/08/PR-08-003.pdf).

Imagine an environment with a large quantity of free neutrons - say, several million. Normally, assuming they avoided interacting with any other particles, they would undergo beta decay after, on average, 15 minutes. But let's say that as soon as the neutrons are released, they're bathed in a sea of high-energy photons, energetic enough to enable pion photoproduction. Handwaving away the precise details of how these photons are produced, could the photons trigger photoproduction to the extent that it would be the predominant "decay" mode over beta decay, or is the reaction simply too unlikely for it to ever dominate.