Why do electrons absorb and re-emit photons?
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Up to a certain time, I was told photons a.k.a. light was just a wave of energy. Then I was told, no, light is actually a particle. And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? (An electron would also be unstable by absorbing the energy and thus it re-emits it but in the first place why does it absorb it?)
*Note:- A similar question was asked earlier (How does an electron absorb or emit light?) but my question is not the same. The earlier asked question was how does it happen and I ask why does it happen.
quantum-mechanics photons atomic-physics
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up vote
6
down vote
favorite
Up to a certain time, I was told photons a.k.a. light was just a wave of energy. Then I was told, no, light is actually a particle. And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? (An electron would also be unstable by absorbing the energy and thus it re-emits it but in the first place why does it absorb it?)
*Note:- A similar question was asked earlier (How does an electron absorb or emit light?) but my question is not the same. The earlier asked question was how does it happen and I ask why does it happen.
quantum-mechanics photons atomic-physics
2
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
2
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Up to a certain time, I was told photons a.k.a. light was just a wave of energy. Then I was told, no, light is actually a particle. And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? (An electron would also be unstable by absorbing the energy and thus it re-emits it but in the first place why does it absorb it?)
*Note:- A similar question was asked earlier (How does an electron absorb or emit light?) but my question is not the same. The earlier asked question was how does it happen and I ask why does it happen.
quantum-mechanics photons atomic-physics
Up to a certain time, I was told photons a.k.a. light was just a wave of energy. Then I was told, no, light is actually a particle. And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time? (An electron would also be unstable by absorbing the energy and thus it re-emits it but in the first place why does it absorb it?)
*Note:- A similar question was asked earlier (How does an electron absorb or emit light?) but my question is not the same. The earlier asked question was how does it happen and I ask why does it happen.
quantum-mechanics photons atomic-physics
edited 2 hours ago
psmears
26915
26915
asked 9 hours ago
![](https://lh5.googleusercontent.com/-e82tiWnvVKo/AAAAAAAAAAI/AAAAAAAAACc/YvT-b8RwMGA/photo.jpg?sz=32)
![](https://lh5.googleusercontent.com/-e82tiWnvVKo/AAAAAAAAAAI/AAAAAAAAACc/YvT-b8RwMGA/photo.jpg?sz=32)
Saptarshi sarma
16317
16317
2
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
2
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago
add a comment |Â
2
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
2
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago
2
2
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
2
2
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
5
down vote
accepted
But why do electrons bother to absorb and re-emit light and not just
let it pass all the time? (An electron would also be unstable by
absorbing the energy and thus it re-emits it but in the first place
why does it absorb it?)
A similar question could be asked about macro objects, say, a pendulum.
If you push a pendulum it is going to up and then it goes down. So, why, you could ask, does it bother to go up, if it is going down afterwards? Why does it absorb the energy of a push instead of just ignoring it?
I guess a simplistic answer is that it absorbs the energy because it gets a direct hit and it's not up to the pendulum to decide whether it should take it or just ignore it.
add a comment |Â
up vote
8
down vote
And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time?
There is a basic misunderstanding in your question.
An electron is an elementary particle of fixed mass. It can scatter off a photon, ( which is also an elementary particle); if accelerated it can emit a photon ,but it does not absorb it, because the electron's mass is fixed , and if it were able to absorb a photon , at the electron's center of mass, the mass would have to change, which contradicts observations and special relativity for elementary particles.
The terms "absorption" and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom, when the atom absorbs a photon. So it is not the electron that absorbs the photon , but the atom.
The atom has energy levels, and if the photon energy coincides with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon ( not the electron). So the answer to "why", above , is "because the photon has the appropriate energy to transfer the electron to an empty energy level".
If the photon energy does not coincide with a transition energy of the atom, the photon may scatter with the spill over electric fields of the atom or molecule either elastically, or transfering energy and a lower energy photon continues on its way.
The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can.
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
add a comment |Â
up vote
0
down vote
Absorption and emission is how we describe the interaction between electrons and the electro magnetic field using quantum field theory. If the photons didnâÂÂt scatter off electrons they would not interact. Basically if you couple light fields and matter fields and quantize you must get process where the quanta of the fields (electrons and photons) must scatter( absorb and re-emit)
add a comment |Â
up vote
0
down vote
Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electro-magnetic field (i.e. photons) and also a gas of electrons (i.e. Fermions). As you probably know, the photons are continously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the termal equilibrium. Is importante to notice that this process of continous absorption and re-emission of the photons by the walls of the box (i.e by the matter!) is always present when you put togheter matter and light, this is fundamental if you want to reach the Planck distribution, beacuase the latter has got chemical-potential $mu=0$ (i.e. the energy cost in order to produce (or killing) a photon is practically zero). If this framework is clear, now you have for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas beacuase the cross section elctron-photon is not zero: this scattering process characterize the interaction between photons and electrons and so, the result excitation of the atoms.
The answer of your question can be:
they interact with the electrons beacause there is a continuos re-equilibrium process of light and matter that live togheter in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorbtion and emission of photons by the matter. It means that photons are moving, but if they are moving it means that ther is non zero
scattering-probability --> interaction.
add a comment |Â
up vote
0
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It's really down to two questions: why do electrons interact with photons, and why do atoms absorb photons?
Why interact with photons?
One can understand why electrons interact with photons by considering relativistic quantum field theory. In order to combine quantum mechanics with special relativity, you have to think of reality as consisting of "quantum fields". A field is something that has a value at every location, for example $Phi(x,t)$ might be a (time-dependent) field, the value of the function signifying the value at every point in space (and every time t). A classical, non-quantum, field simply has a value at every location - you can think of it as the height of some system, say the deviation from equilibrium of an oscillator, at every point in space. A quantum field instead has a quantum system at every point in space; you can think of it as having a quantum harmonic oscillator at every point in space. The state of the point-like system, i.e the deviation of this oscillator from equilibrium, is the "height" of the field at that point in space.
Now a core principle of quantum mechanics is that the phase of the quantum state does not matter. In order to carry this principle into a quantum field, the equations describing the physics of the system, known as the Lagrangian, has to not change if we change the phases of the states of the points in space. This requirement is known as "gauge symmetry". Now it so happens that it's rather difficult to build a gauge-symmetric Lagrangian using only standard expressions like derivatives. Instead, in order to maintain gauge-symmetry one has to introduce another quantum field, known as the gauge-field. This is the only way to maintain gauge symmetry, i.e. to maintain the requirement that the phase of a quantum state has no physical meaning.
So if you try to build laws of physics (a Lagrangian) to describe a simple matter field (e.g. an electron's field), you need to introduce an additional "gauge" field that interacts with it. The waves in the matter field will be the matter particles, such as electrons. The waves in the gauge field will be force-carrying particles, such as photons.
To summarize then, the reason an electron interacts with photons is that an electron is really a wave in a quantum (relativistic) field, and these waves have to interact with waves in the (gauge) electromagnetic field, which we call photons, in order for the electron's field to be a quantum field (i.e. for the phase of the point-like states to lack any physical meaning).
Why atoms absorb photons?
Anna v beautifully explained already why an elementary electron cannot absorb a photon - it has to scatter it instead, as the electron's energy and hence mass cannot increase in its rest frame. But why is it that atoms absorb photons?
The important point here is that you cannot turn the electromagnetic interaction "off" for one effect while keeping in "on" for another. If you build an equation describing an electron that's attracted to a positive nucleus by the electromagnetic force, then this same system will also be affected by waves in the electromagnetic field.
So the same equations that describe the stable orbits (the electron levels/orbitals) due to the electromagnetic interaction with the potential energy of the nucleus, also describe a response to an electromagnetic wave (usually dealt with only as a perturbation off the stable state). And this interaction with the waves amounts to annihilating a normal-mode of the wave (annihilating a photon), while at the same time increasing in energy to maintain energy conservation. (Or conversely creating a normal-mode wave while dropping in energy.)
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up vote
-1
down vote
Absorption and emission are not the only ways by which light interacts with matter. Reflection and transmission, in which light and electronic excitations mix to give the characteristic reflection and transmission, are also possible.
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
But why do electrons bother to absorb and re-emit light and not just
let it pass all the time? (An electron would also be unstable by
absorbing the energy and thus it re-emits it but in the first place
why does it absorb it?)
A similar question could be asked about macro objects, say, a pendulum.
If you push a pendulum it is going to up and then it goes down. So, why, you could ask, does it bother to go up, if it is going down afterwards? Why does it absorb the energy of a push instead of just ignoring it?
I guess a simplistic answer is that it absorbs the energy because it gets a direct hit and it's not up to the pendulum to decide whether it should take it or just ignore it.
add a comment |Â
up vote
5
down vote
accepted
But why do electrons bother to absorb and re-emit light and not just
let it pass all the time? (An electron would also be unstable by
absorbing the energy and thus it re-emits it but in the first place
why does it absorb it?)
A similar question could be asked about macro objects, say, a pendulum.
If you push a pendulum it is going to up and then it goes down. So, why, you could ask, does it bother to go up, if it is going down afterwards? Why does it absorb the energy of a push instead of just ignoring it?
I guess a simplistic answer is that it absorbs the energy because it gets a direct hit and it's not up to the pendulum to decide whether it should take it or just ignore it.
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
But why do electrons bother to absorb and re-emit light and not just
let it pass all the time? (An electron would also be unstable by
absorbing the energy and thus it re-emits it but in the first place
why does it absorb it?)
A similar question could be asked about macro objects, say, a pendulum.
If you push a pendulum it is going to up and then it goes down. So, why, you could ask, does it bother to go up, if it is going down afterwards? Why does it absorb the energy of a push instead of just ignoring it?
I guess a simplistic answer is that it absorbs the energy because it gets a direct hit and it's not up to the pendulum to decide whether it should take it or just ignore it.
But why do electrons bother to absorb and re-emit light and not just
let it pass all the time? (An electron would also be unstable by
absorbing the energy and thus it re-emits it but in the first place
why does it absorb it?)
A similar question could be asked about macro objects, say, a pendulum.
If you push a pendulum it is going to up and then it goes down. So, why, you could ask, does it bother to go up, if it is going down afterwards? Why does it absorb the energy of a push instead of just ignoring it?
I guess a simplistic answer is that it absorbs the energy because it gets a direct hit and it's not up to the pendulum to decide whether it should take it or just ignore it.
answered 7 hours ago
V.F.
4,5251416
4,5251416
add a comment |Â
add a comment |Â
up vote
8
down vote
And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time?
There is a basic misunderstanding in your question.
An electron is an elementary particle of fixed mass. It can scatter off a photon, ( which is also an elementary particle); if accelerated it can emit a photon ,but it does not absorb it, because the electron's mass is fixed , and if it were able to absorb a photon , at the electron's center of mass, the mass would have to change, which contradicts observations and special relativity for elementary particles.
The terms "absorption" and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom, when the atom absorbs a photon. So it is not the electron that absorbs the photon , but the atom.
The atom has energy levels, and if the photon energy coincides with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon ( not the electron). So the answer to "why", above , is "because the photon has the appropriate energy to transfer the electron to an empty energy level".
If the photon energy does not coincide with a transition energy of the atom, the photon may scatter with the spill over electric fields of the atom or molecule either elastically, or transfering energy and a lower energy photon continues on its way.
The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can.
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
add a comment |Â
up vote
8
down vote
And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time?
There is a basic misunderstanding in your question.
An electron is an elementary particle of fixed mass. It can scatter off a photon, ( which is also an elementary particle); if accelerated it can emit a photon ,but it does not absorb it, because the electron's mass is fixed , and if it were able to absorb a photon , at the electron's center of mass, the mass would have to change, which contradicts observations and special relativity for elementary particles.
The terms "absorption" and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom, when the atom absorbs a photon. So it is not the electron that absorbs the photon , but the atom.
The atom has energy levels, and if the photon energy coincides with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon ( not the electron). So the answer to "why", above , is "because the photon has the appropriate energy to transfer the electron to an empty energy level".
If the photon energy does not coincide with a transition energy of the atom, the photon may scatter with the spill over electric fields of the atom or molecule either elastically, or transfering energy and a lower energy photon continues on its way.
The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can.
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
add a comment |Â
up vote
8
down vote
up vote
8
down vote
And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time?
There is a basic misunderstanding in your question.
An electron is an elementary particle of fixed mass. It can scatter off a photon, ( which is also an elementary particle); if accelerated it can emit a photon ,but it does not absorb it, because the electron's mass is fixed , and if it were able to absorb a photon , at the electron's center of mass, the mass would have to change, which contradicts observations and special relativity for elementary particles.
The terms "absorption" and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom, when the atom absorbs a photon. So it is not the electron that absorbs the photon , but the atom.
The atom has energy levels, and if the photon energy coincides with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon ( not the electron). So the answer to "why", above , is "because the photon has the appropriate energy to transfer the electron to an empty energy level".
If the photon energy does not coincide with a transition energy of the atom, the photon may scatter with the spill over electric fields of the atom or molecule either elastically, or transfering energy and a lower energy photon continues on its way.
The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can.
And electrons in an atom absorb and re-emit it. But why do electrons bother to absorb and re-emit light and not just let it pass all the time?
There is a basic misunderstanding in your question.
An electron is an elementary particle of fixed mass. It can scatter off a photon, ( which is also an elementary particle); if accelerated it can emit a photon ,but it does not absorb it, because the electron's mass is fixed , and if it were able to absorb a photon , at the electron's center of mass, the mass would have to change, which contradicts observations and special relativity for elementary particles.
The terms "absorption" and absorbs are not usable with free electrons. It is the bound electrons in an atomic system, which may change energy levels in the atom, when the atom absorbs a photon. So it is not the electron that absorbs the photon , but the atom.
The atom has energy levels, and if the photon energy coincides with the transition energy of kicking an electron to an empty energy level, then the atom can absorb the photon ( not the electron). So the answer to "why", above , is "because the photon has the appropriate energy to transfer the electron to an empty energy level".
If the photon energy does not coincide with a transition energy of the atom, the photon may scatter with the spill over electric fields of the atom or molecule either elastically, or transfering energy and a lower energy photon continues on its way.
The relevant thought to keep is that an elementary particle cannot absorb a photon. Composite ones as atoms, molecules and lattices, can.
answered 7 hours ago
![](https://i.stack.imgur.com/JA9pM.jpg?s=32&g=1)
![](https://i.stack.imgur.com/JA9pM.jpg?s=32&g=1)
anna v
148k7138425
148k7138425
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
add a comment |Â
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
1
1
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
Actually the terms "absorption" and "emission" are often used in the meaning of transition to a virtual state and subsequent decay from it. This would work also for a free electron.
â Ruslan
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
@Ruslan QFT s not the level at which this question is asked
â anna v
5 hours ago
add a comment |Â
up vote
0
down vote
Absorption and emission is how we describe the interaction between electrons and the electro magnetic field using quantum field theory. If the photons didnâÂÂt scatter off electrons they would not interact. Basically if you couple light fields and matter fields and quantize you must get process where the quanta of the fields (electrons and photons) must scatter( absorb and re-emit)
add a comment |Â
up vote
0
down vote
Absorption and emission is how we describe the interaction between electrons and the electro magnetic field using quantum field theory. If the photons didnâÂÂt scatter off electrons they would not interact. Basically if you couple light fields and matter fields and quantize you must get process where the quanta of the fields (electrons and photons) must scatter( absorb and re-emit)
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Absorption and emission is how we describe the interaction between electrons and the electro magnetic field using quantum field theory. If the photons didnâÂÂt scatter off electrons they would not interact. Basically if you couple light fields and matter fields and quantize you must get process where the quanta of the fields (electrons and photons) must scatter( absorb and re-emit)
Absorption and emission is how we describe the interaction between electrons and the electro magnetic field using quantum field theory. If the photons didnâÂÂt scatter off electrons they would not interact. Basically if you couple light fields and matter fields and quantize you must get process where the quanta of the fields (electrons and photons) must scatter( absorb and re-emit)
answered 9 hours ago
Shane P Kelly
934312
934312
add a comment |Â
add a comment |Â
up vote
0
down vote
Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electro-magnetic field (i.e. photons) and also a gas of electrons (i.e. Fermions). As you probably know, the photons are continously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the termal equilibrium. Is importante to notice that this process of continous absorption and re-emission of the photons by the walls of the box (i.e by the matter!) is always present when you put togheter matter and light, this is fundamental if you want to reach the Planck distribution, beacuase the latter has got chemical-potential $mu=0$ (i.e. the energy cost in order to produce (or killing) a photon is practically zero). If this framework is clear, now you have for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas beacuase the cross section elctron-photon is not zero: this scattering process characterize the interaction between photons and electrons and so, the result excitation of the atoms.
The answer of your question can be:
they interact with the electrons beacause there is a continuos re-equilibrium process of light and matter that live togheter in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorbtion and emission of photons by the matter. It means that photons are moving, but if they are moving it means that ther is non zero
scattering-probability --> interaction.
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Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electro-magnetic field (i.e. photons) and also a gas of electrons (i.e. Fermions). As you probably know, the photons are continously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the termal equilibrium. Is importante to notice that this process of continous absorption and re-emission of the photons by the walls of the box (i.e by the matter!) is always present when you put togheter matter and light, this is fundamental if you want to reach the Planck distribution, beacuase the latter has got chemical-potential $mu=0$ (i.e. the energy cost in order to produce (or killing) a photon is practically zero). If this framework is clear, now you have for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas beacuase the cross section elctron-photon is not zero: this scattering process characterize the interaction between photons and electrons and so, the result excitation of the atoms.
The answer of your question can be:
they interact with the electrons beacause there is a continuos re-equilibrium process of light and matter that live togheter in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorbtion and emission of photons by the matter. It means that photons are moving, but if they are moving it means that ther is non zero
scattering-probability --> interaction.
add a comment |Â
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up vote
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Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electro-magnetic field (i.e. photons) and also a gas of electrons (i.e. Fermions). As you probably know, the photons are continously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the termal equilibrium. Is importante to notice that this process of continous absorption and re-emission of the photons by the walls of the box (i.e by the matter!) is always present when you put togheter matter and light, this is fundamental if you want to reach the Planck distribution, beacuase the latter has got chemical-potential $mu=0$ (i.e. the energy cost in order to produce (or killing) a photon is practically zero). If this framework is clear, now you have for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas beacuase the cross section elctron-photon is not zero: this scattering process characterize the interaction between photons and electrons and so, the result excitation of the atoms.
The answer of your question can be:
they interact with the electrons beacause there is a continuos re-equilibrium process of light and matter that live togheter in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorbtion and emission of photons by the matter. It means that photons are moving, but if they are moving it means that ther is non zero
scattering-probability --> interaction.
Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electro-magnetic field (i.e. photons) and also a gas of electrons (i.e. Fermions). As you probably know, the photons are continously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the termal equilibrium. Is importante to notice that this process of continous absorption and re-emission of the photons by the walls of the box (i.e by the matter!) is always present when you put togheter matter and light, this is fundamental if you want to reach the Planck distribution, beacuase the latter has got chemical-potential $mu=0$ (i.e. the energy cost in order to produce (or killing) a photon is practically zero). If this framework is clear, now you have for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas beacuase the cross section elctron-photon is not zero: this scattering process characterize the interaction between photons and electrons and so, the result excitation of the atoms.
The answer of your question can be:
they interact with the electrons beacause there is a continuos re-equilibrium process of light and matter that live togheter in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorbtion and emission of photons by the matter. It means that photons are moving, but if they are moving it means that ther is non zero
scattering-probability --> interaction.
answered 7 hours ago
![](https://lh5.googleusercontent.com/-7kFH0PnWvbc/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7oLX1rYP7W1mxKnvgHgbKmq5YruIQ/mo/photo.jpg?sz=32)
![](https://lh5.googleusercontent.com/-7kFH0PnWvbc/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7oLX1rYP7W1mxKnvgHgbKmq5YruIQ/mo/photo.jpg?sz=32)
MRT
114
114
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It's really down to two questions: why do electrons interact with photons, and why do atoms absorb photons?
Why interact with photons?
One can understand why electrons interact with photons by considering relativistic quantum field theory. In order to combine quantum mechanics with special relativity, you have to think of reality as consisting of "quantum fields". A field is something that has a value at every location, for example $Phi(x,t)$ might be a (time-dependent) field, the value of the function signifying the value at every point in space (and every time t). A classical, non-quantum, field simply has a value at every location - you can think of it as the height of some system, say the deviation from equilibrium of an oscillator, at every point in space. A quantum field instead has a quantum system at every point in space; you can think of it as having a quantum harmonic oscillator at every point in space. The state of the point-like system, i.e the deviation of this oscillator from equilibrium, is the "height" of the field at that point in space.
Now a core principle of quantum mechanics is that the phase of the quantum state does not matter. In order to carry this principle into a quantum field, the equations describing the physics of the system, known as the Lagrangian, has to not change if we change the phases of the states of the points in space. This requirement is known as "gauge symmetry". Now it so happens that it's rather difficult to build a gauge-symmetric Lagrangian using only standard expressions like derivatives. Instead, in order to maintain gauge-symmetry one has to introduce another quantum field, known as the gauge-field. This is the only way to maintain gauge symmetry, i.e. to maintain the requirement that the phase of a quantum state has no physical meaning.
So if you try to build laws of physics (a Lagrangian) to describe a simple matter field (e.g. an electron's field), you need to introduce an additional "gauge" field that interacts with it. The waves in the matter field will be the matter particles, such as electrons. The waves in the gauge field will be force-carrying particles, such as photons.
To summarize then, the reason an electron interacts with photons is that an electron is really a wave in a quantum (relativistic) field, and these waves have to interact with waves in the (gauge) electromagnetic field, which we call photons, in order for the electron's field to be a quantum field (i.e. for the phase of the point-like states to lack any physical meaning).
Why atoms absorb photons?
Anna v beautifully explained already why an elementary electron cannot absorb a photon - it has to scatter it instead, as the electron's energy and hence mass cannot increase in its rest frame. But why is it that atoms absorb photons?
The important point here is that you cannot turn the electromagnetic interaction "off" for one effect while keeping in "on" for another. If you build an equation describing an electron that's attracted to a positive nucleus by the electromagnetic force, then this same system will also be affected by waves in the electromagnetic field.
So the same equations that describe the stable orbits (the electron levels/orbitals) due to the electromagnetic interaction with the potential energy of the nucleus, also describe a response to an electromagnetic wave (usually dealt with only as a perturbation off the stable state). And this interaction with the waves amounts to annihilating a normal-mode of the wave (annihilating a photon), while at the same time increasing in energy to maintain energy conservation. (Or conversely creating a normal-mode wave while dropping in energy.)
add a comment |Â
up vote
0
down vote
It's really down to two questions: why do electrons interact with photons, and why do atoms absorb photons?
Why interact with photons?
One can understand why electrons interact with photons by considering relativistic quantum field theory. In order to combine quantum mechanics with special relativity, you have to think of reality as consisting of "quantum fields". A field is something that has a value at every location, for example $Phi(x,t)$ might be a (time-dependent) field, the value of the function signifying the value at every point in space (and every time t). A classical, non-quantum, field simply has a value at every location - you can think of it as the height of some system, say the deviation from equilibrium of an oscillator, at every point in space. A quantum field instead has a quantum system at every point in space; you can think of it as having a quantum harmonic oscillator at every point in space. The state of the point-like system, i.e the deviation of this oscillator from equilibrium, is the "height" of the field at that point in space.
Now a core principle of quantum mechanics is that the phase of the quantum state does not matter. In order to carry this principle into a quantum field, the equations describing the physics of the system, known as the Lagrangian, has to not change if we change the phases of the states of the points in space. This requirement is known as "gauge symmetry". Now it so happens that it's rather difficult to build a gauge-symmetric Lagrangian using only standard expressions like derivatives. Instead, in order to maintain gauge-symmetry one has to introduce another quantum field, known as the gauge-field. This is the only way to maintain gauge symmetry, i.e. to maintain the requirement that the phase of a quantum state has no physical meaning.
So if you try to build laws of physics (a Lagrangian) to describe a simple matter field (e.g. an electron's field), you need to introduce an additional "gauge" field that interacts with it. The waves in the matter field will be the matter particles, such as electrons. The waves in the gauge field will be force-carrying particles, such as photons.
To summarize then, the reason an electron interacts with photons is that an electron is really a wave in a quantum (relativistic) field, and these waves have to interact with waves in the (gauge) electromagnetic field, which we call photons, in order for the electron's field to be a quantum field (i.e. for the phase of the point-like states to lack any physical meaning).
Why atoms absorb photons?
Anna v beautifully explained already why an elementary electron cannot absorb a photon - it has to scatter it instead, as the electron's energy and hence mass cannot increase in its rest frame. But why is it that atoms absorb photons?
The important point here is that you cannot turn the electromagnetic interaction "off" for one effect while keeping in "on" for another. If you build an equation describing an electron that's attracted to a positive nucleus by the electromagnetic force, then this same system will also be affected by waves in the electromagnetic field.
So the same equations that describe the stable orbits (the electron levels/orbitals) due to the electromagnetic interaction with the potential energy of the nucleus, also describe a response to an electromagnetic wave (usually dealt with only as a perturbation off the stable state). And this interaction with the waves amounts to annihilating a normal-mode of the wave (annihilating a photon), while at the same time increasing in energy to maintain energy conservation. (Or conversely creating a normal-mode wave while dropping in energy.)
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's really down to two questions: why do electrons interact with photons, and why do atoms absorb photons?
Why interact with photons?
One can understand why electrons interact with photons by considering relativistic quantum field theory. In order to combine quantum mechanics with special relativity, you have to think of reality as consisting of "quantum fields". A field is something that has a value at every location, for example $Phi(x,t)$ might be a (time-dependent) field, the value of the function signifying the value at every point in space (and every time t). A classical, non-quantum, field simply has a value at every location - you can think of it as the height of some system, say the deviation from equilibrium of an oscillator, at every point in space. A quantum field instead has a quantum system at every point in space; you can think of it as having a quantum harmonic oscillator at every point in space. The state of the point-like system, i.e the deviation of this oscillator from equilibrium, is the "height" of the field at that point in space.
Now a core principle of quantum mechanics is that the phase of the quantum state does not matter. In order to carry this principle into a quantum field, the equations describing the physics of the system, known as the Lagrangian, has to not change if we change the phases of the states of the points in space. This requirement is known as "gauge symmetry". Now it so happens that it's rather difficult to build a gauge-symmetric Lagrangian using only standard expressions like derivatives. Instead, in order to maintain gauge-symmetry one has to introduce another quantum field, known as the gauge-field. This is the only way to maintain gauge symmetry, i.e. to maintain the requirement that the phase of a quantum state has no physical meaning.
So if you try to build laws of physics (a Lagrangian) to describe a simple matter field (e.g. an electron's field), you need to introduce an additional "gauge" field that interacts with it. The waves in the matter field will be the matter particles, such as electrons. The waves in the gauge field will be force-carrying particles, such as photons.
To summarize then, the reason an electron interacts with photons is that an electron is really a wave in a quantum (relativistic) field, and these waves have to interact with waves in the (gauge) electromagnetic field, which we call photons, in order for the electron's field to be a quantum field (i.e. for the phase of the point-like states to lack any physical meaning).
Why atoms absorb photons?
Anna v beautifully explained already why an elementary electron cannot absorb a photon - it has to scatter it instead, as the electron's energy and hence mass cannot increase in its rest frame. But why is it that atoms absorb photons?
The important point here is that you cannot turn the electromagnetic interaction "off" for one effect while keeping in "on" for another. If you build an equation describing an electron that's attracted to a positive nucleus by the electromagnetic force, then this same system will also be affected by waves in the electromagnetic field.
So the same equations that describe the stable orbits (the electron levels/orbitals) due to the electromagnetic interaction with the potential energy of the nucleus, also describe a response to an electromagnetic wave (usually dealt with only as a perturbation off the stable state). And this interaction with the waves amounts to annihilating a normal-mode of the wave (annihilating a photon), while at the same time increasing in energy to maintain energy conservation. (Or conversely creating a normal-mode wave while dropping in energy.)
It's really down to two questions: why do electrons interact with photons, and why do atoms absorb photons?
Why interact with photons?
One can understand why electrons interact with photons by considering relativistic quantum field theory. In order to combine quantum mechanics with special relativity, you have to think of reality as consisting of "quantum fields". A field is something that has a value at every location, for example $Phi(x,t)$ might be a (time-dependent) field, the value of the function signifying the value at every point in space (and every time t). A classical, non-quantum, field simply has a value at every location - you can think of it as the height of some system, say the deviation from equilibrium of an oscillator, at every point in space. A quantum field instead has a quantum system at every point in space; you can think of it as having a quantum harmonic oscillator at every point in space. The state of the point-like system, i.e the deviation of this oscillator from equilibrium, is the "height" of the field at that point in space.
Now a core principle of quantum mechanics is that the phase of the quantum state does not matter. In order to carry this principle into a quantum field, the equations describing the physics of the system, known as the Lagrangian, has to not change if we change the phases of the states of the points in space. This requirement is known as "gauge symmetry". Now it so happens that it's rather difficult to build a gauge-symmetric Lagrangian using only standard expressions like derivatives. Instead, in order to maintain gauge-symmetry one has to introduce another quantum field, known as the gauge-field. This is the only way to maintain gauge symmetry, i.e. to maintain the requirement that the phase of a quantum state has no physical meaning.
So if you try to build laws of physics (a Lagrangian) to describe a simple matter field (e.g. an electron's field), you need to introduce an additional "gauge" field that interacts with it. The waves in the matter field will be the matter particles, such as electrons. The waves in the gauge field will be force-carrying particles, such as photons.
To summarize then, the reason an electron interacts with photons is that an electron is really a wave in a quantum (relativistic) field, and these waves have to interact with waves in the (gauge) electromagnetic field, which we call photons, in order for the electron's field to be a quantum field (i.e. for the phase of the point-like states to lack any physical meaning).
Why atoms absorb photons?
Anna v beautifully explained already why an elementary electron cannot absorb a photon - it has to scatter it instead, as the electron's energy and hence mass cannot increase in its rest frame. But why is it that atoms absorb photons?
The important point here is that you cannot turn the electromagnetic interaction "off" for one effect while keeping in "on" for another. If you build an equation describing an electron that's attracted to a positive nucleus by the electromagnetic force, then this same system will also be affected by waves in the electromagnetic field.
So the same equations that describe the stable orbits (the electron levels/orbitals) due to the electromagnetic interaction with the potential energy of the nucleus, also describe a response to an electromagnetic wave (usually dealt with only as a perturbation off the stable state). And this interaction with the waves amounts to annihilating a normal-mode of the wave (annihilating a photon), while at the same time increasing in energy to maintain energy conservation. (Or conversely creating a normal-mode wave while dropping in energy.)
answered 5 hours ago
PhysicsTeacher
1184
1184
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Absorption and emission are not the only ways by which light interacts with matter. Reflection and transmission, in which light and electronic excitations mix to give the characteristic reflection and transmission, are also possible.
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
add a comment |Â
up vote
-1
down vote
Absorption and emission are not the only ways by which light interacts with matter. Reflection and transmission, in which light and electronic excitations mix to give the characteristic reflection and transmission, are also possible.
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Absorption and emission are not the only ways by which light interacts with matter. Reflection and transmission, in which light and electronic excitations mix to give the characteristic reflection and transmission, are also possible.
Absorption and emission are not the only ways by which light interacts with matter. Reflection and transmission, in which light and electronic excitations mix to give the characteristic reflection and transmission, are also possible.
answered 7 hours ago
my2cts
2,606313
2,606313
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
add a comment |Â
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
Why the minus sign? Come forward, anonymous.
â my2cts
5 hours ago
add a comment |Â
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2
No doubt people who study this stuff are smart, but everything about their explanations in this area seem off. Why does the reemitted photon travel in the same direction? Why does the wavelength decrease? These and many other questions make the current theory suspect to me.
â Lambda
6 hours ago
2
No doubt the process is explained at great length and in great detail in textbooks on electromagnetism (to be sure that is in the classical picture, but that is the more natural one for these problems anywayâÂÂto follow a in-depth explanation in the photon picture takes more background). And no doubt each assertion of the theory in those books is grounded in experimental truth, and has been for more than one hundred years. If you don't actually understand the current theory then it is a bit presumptuous to declare it suspect.
â dmckeeâ¦
4 hours ago
is not the actual, single electron excitation that "absorbs and reemits photons" but the nucleon-electron systems we call atoms
â lurscher
1 hour ago
the simplest answer would be: atoms have enough external legs to allow absorption and emission of light-like momentum quanta
â lurscher
1 hour ago