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Will a free neutron radiate if it is decelerated?
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$begingroup$
In this answer it is said (and I fully agree):
Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.
In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.
Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation
Will a free neutron radiate if it is decelerated?
electromagnetic-radiation photons neutrons
edited Jul 3 at 11:31
Community♦
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
In this answer it is said (and I fully agree):
Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.
In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.
Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation
Will a free neutron radiate if it is decelerated?
electromagnetic-radiation photons neutrons
edited Jul 3 at 11:31
Community♦
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
$endgroup$
$begingroup$
link.springer.com/article/10.1007/BF00892879
$endgroup$
– G. Smith
Jul 2 at 22:09
2
$begingroup$
Note that the neutron consists of three charged particles, one up and two down quarks.
$endgroup$
– infinitezero
Jul 3 at 9:45
$begingroup$
@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
$endgroup$
– HolgerFiedler
Jul 3 at 9:54
2
$begingroup$
I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
$endgroup$
– infinitezero
Jul 3 at 10:00
add a comment |
$begingroup$
In this answer it is said (and I fully agree):
Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.
In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.
Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation
Will a free neutron radiate if it is decelerated?
electromagnetic-radiation photons neutrons
edited Jul 3 at 11:31
Community♦
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
$endgroup$
In this answer it is said (and I fully agree):
Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.
In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.
Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation
Will a free neutron radiate if it is decelerated?
electromagnetic-radiation photons neutrons
electromagnetic-radiation photons neutrons
edited Jul 3 at 11:31
Community♦
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
edited Jul 3 at 11:31
Community♦
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
edited Jul 3 at 11:31
Community♦
1
edited Jul 3 at 11:31
Community♦
1
edited Jul 3 at 11:31
Community♦
1
1
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
asked Jul 2 at 20:41
HolgerFiedlerHolgerFiedler
4,5713 gold badges14 silver badges42 bronze badges
4,5713 gold badges14 silver badges42 bronze badges
$begingroup$
link.springer.com/article/10.1007/BF00892879
$endgroup$
– G. Smith
Jul 2 at 22:09
2
$begingroup$
Note that the neutron consists of three charged particles, one up and two down quarks.
$endgroup$
– infinitezero
Jul 3 at 9:45
$begingroup$
@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
$endgroup$
– HolgerFiedler
Jul 3 at 9:54
2
$begingroup$
I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
$endgroup$
– infinitezero
Jul 3 at 10:00
add a comment |
$begingroup$
link.springer.com/article/10.1007/BF00892879
$endgroup$
– G. Smith
Jul 2 at 22:09
2
$begingroup$
Note that the neutron consists of three charged particles, one up and two down quarks.
$endgroup$
– infinitezero
Jul 3 at 9:45
$begingroup$
@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
$endgroup$
– HolgerFiedler
Jul 3 at 9:54
2
$begingroup$
I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
$endgroup$
– infinitezero
Jul 3 at 10:00
$begingroup$
link.springer.com/article/10.1007/BF00892879
$endgroup$
– G. Smith
Jul 2 at 22:09
$begingroup$
link.springer.com/article/10.1007/BF00892879
$endgroup$
– G. Smith
Jul 2 at 22:09
2
2
$begingroup$
Note that the neutron consists of three charged particles, one up and two down quarks.
$endgroup$
– infinitezero
Jul 3 at 9:45
$begingroup$
Note that the neutron consists of three charged particles, one up and two down quarks.
$endgroup$
– infinitezero
Jul 3 at 9:45
$begingroup$
@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
$endgroup$
– HolgerFiedler
Jul 3 at 9:54
$begingroup$
@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
$endgroup$
– HolgerFiedler
Jul 3 at 9:54
2
2
$begingroup$
I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
$endgroup$
– infinitezero
Jul 3 at 10:00
$begingroup$
I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
$endgroup$
– infinitezero
Jul 3 at 10:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size
$$
mu_n = −9.6623647(23) times 10^−27 rm J,T^−1.
$$
This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.
You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
$endgroup$
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
add a comment |
$begingroup$
We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size
$$
mu_n = −9.6623647(23) times 10^−27 rm J,T^−1.
$$
This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.
You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
$endgroup$
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
add a comment |
$begingroup$
The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size
$$
mu_n = −9.6623647(23) times 10^−27 rm J,T^−1.
$$
This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.
You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
$endgroup$
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
add a comment |
$begingroup$
The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size
$$
mu_n = −9.6623647(23) times 10^−27 rm J,T^−1.
$$
This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.
You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
$endgroup$
The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size
$$
mu_n = −9.6623647(23) times 10^−27 rm J,T^−1.
$$
This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.
You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
answered Jul 2 at 21:29
Andrew SteaneAndrew Steane
8,4721 gold badge11 silver badges45 bronze badges
8,4721 gold badge11 silver badges45 bronze badges
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
add a comment |
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
1
1
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate.
$endgroup$
– Gyromagnetic
Jul 3 at 13:30
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
$begingroup$
@Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect.
$endgroup$
– Andrew Steane
Jul 3 at 14:29
2
2
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero.
$endgroup$
– rob♦
Jul 3 at 16:36
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
$begingroup$
It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron.
$endgroup$
– KF Gauss
Jul 4 at 17:45
add a comment |
$begingroup$
We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
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We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
answered Jul 4 at 14:31
Cang YeCang Ye
605 bronze badges
605 bronze badges
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Cang Ye is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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link.springer.com/article/10.1007/BF00892879
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– G. Smith
Jul 2 at 22:09
2
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Note that the neutron consists of three charged particles, one up and two down quarks.
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– infinitezero
Jul 3 at 9:45
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@infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate.
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– HolgerFiedler
Jul 3 at 9:54
2
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I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation.
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– infinitezero
Jul 3 at 10:00