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Does pair production happen even when the photon is around a neutron?
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$begingroup$
In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.
But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?
nuclear-physics pair-production
asked 9 hours ago
AWanderingMindAWanderingMind
1837
$endgroup$
add a comment |
$begingroup$
In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.
But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?
nuclear-physics pair-production
asked 9 hours ago
AWanderingMindAWanderingMind
1837
$endgroup$
add a comment |
$begingroup$
In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.
But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?
nuclear-physics pair-production
asked 9 hours ago
AWanderingMindAWanderingMind
1837
$endgroup$
In order for a photon to decay into a pair of $e^+ e^-$, it must have at least $E_{gamma}=1.022$ MeV and must be near a nucleus in order to satisfy the conservation of energy-momentum.
But would this happen even if the photon is near a neutron and not necesarily a nucleus? Does the fact that the nucleus is charged have anything to do with this decay? Who acts upon the photon to induce the interaction?
nuclear-physics pair-production
nuclear-physics pair-production
asked 9 hours ago
AWanderingMindAWanderingMind
1837
asked 9 hours ago
AWanderingMindAWanderingMind
1837
asked 9 hours ago
AWanderingMindAWanderingMind
1837
asked 9 hours ago
AWanderingMindAWanderingMind
1837
asked 9 hours ago
AWanderingMindAWanderingMind
1837
1837
add a comment |
add a comment |
2 Answers
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$begingroup$
Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.
In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $sim v/c$ compared to electric effects.
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
$endgroup$
add a comment |
$begingroup$
Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.
In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $sim v/c$ compared to electric effects.
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
$endgroup$
add a comment |
$begingroup$
Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.
In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $sim v/c$ compared to electric effects.
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
$endgroup$
add a comment |
$begingroup$
Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.
In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $sim v/c$ compared to electric effects.
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
$endgroup$
Quantum mechanics says that everything that is not forbidden is compulsory. Any process that doesn't violate a conservation law will happen, with some rate or cross-section. However, this general principle doesn't tell you what the rate is. For example, it's theoretically possible for 124Te to decay into two 62Ni nuclei plus four electrons and four antineutrinos, but to predict the (very small) rate, you need to know the relevant nuclear physics.
In your example, the process probably would go at some rate determined by electromagnetic interactions, because the neutron has a magnetic field. But the rate would presumably be small because the magnetic field of a dipole falls off like $1/r^3$, and magnetic effects are usually down by $sim v/c$ compared to electric effects.
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
answered 7 hours ago
Ben CrowellBen Crowell
56.5k6169327
56.5k6169327
add a comment |
add a comment |
$begingroup$
Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
add a comment |
$begingroup$
Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
add a comment |
$begingroup$
Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Yes, pair production can occur even near a lone neutron. The presence of a mass for the photon to interact with is required for conservation of momentum (further explanation can be found here). To my knowledge, the charge of the nucleus is not significant to the process of pair production, though the probability of pair production increases approximately with atomic number squared. Experimentally, this would be hard to demonstrate as neutrons are hard to control and have a relatively short half life (~10.3 minutes).
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
answered 7 hours ago
Maarten de HaanMaarten de Haan
295
295
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Maarten de Haan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
add a comment |
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
1
1
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
To my knowledge, the charge of the nucleus is not significant to the process of pair production This is not true, and it contradicts your later statement about the $Z^2$ dependence. The charge is relevant because it's a QED process, and the $Z^2$ dependence tells us that this vanishes to first order in QED for the neutron. However, one would expect some higher-order Feynman diagram to provide some nonvanishing rate.
$endgroup$
– Ben Crowell
5 hours ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
$begingroup$
Good point. I tried to find the relevant higher order correction, but I could only find corrections for large mass elements, and not for $Z=0$. I'll have to keep looking. Thanks
$endgroup$
– Maarten de Haan
1 hour ago
add a comment |
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