How can two electrons repel if it's impossible for free electrons to absorb or emit energy?
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There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
electrons wavefunction-collapse
add a comment |
up vote
3
down vote
favorite
There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
electrons wavefunction-collapse
3
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
electrons wavefunction-collapse
There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
electrons wavefunction-collapse
electrons wavefunction-collapse
edited 1 hour ago
asked 4 hours ago
user6760
2,23511734
2,23511734
3
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago
add a comment |
3
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago
3
3
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
4
down vote
It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
add a comment |
up vote
1
down vote
Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
|
show 2 more comments
up vote
0
down vote
To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
add a comment |
up vote
0
down vote
Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
add a comment |
up vote
4
down vote
It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
add a comment |
up vote
4
down vote
up vote
4
down vote
It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
answered 4 hours ago
knzhou
40.5k11113194
40.5k11113194
add a comment |
add a comment |
up vote
1
down vote
Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
|
show 2 more comments
up vote
1
down vote
Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
|
show 2 more comments
up vote
1
down vote
up vote
1
down vote
Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
answered 2 hours ago
niels nielsen
15.3k42649
15.3k42649
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
|
show 2 more comments
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
– user6760
2 hours ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
– Bill Alsept
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@BillAlsept: I'm referring to free electron not valence electron
– user6760
1 hour ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
– Bill Alsept
47 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
@BillAlsept: of course 1 free electron interacts with one atom
– user6760
40 mins ago
|
show 2 more comments
up vote
0
down vote
To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
add a comment |
up vote
0
down vote
To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
add a comment |
up vote
0
down vote
up vote
0
down vote
To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
answered 1 hour ago
Bruce Greetham
1,1751416
1,1751416
add a comment |
add a comment |
up vote
0
down vote
Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
add a comment |
up vote
0
down vote
Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
add a comment |
up vote
0
down vote
up vote
0
down vote
Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
answered 37 mins ago
Emilio Pisanty
81.6k21194406
81.6k21194406
add a comment |
add a comment |
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3
"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago
@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago
Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
1 hour ago