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Question

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.

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. — yesterday
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 anywayÃ¢Â€Â”to 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. — yesterday
Light is neither a wave nor a particle. It is a quantum field, which has behaviours that are in some ways similar to what we see as particles and waves in our macroscopic environment. — 18 hours ago

score 28 · Accepted Answer · 2018-08-05 15:00:40Z

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 (within a small $ÃŽÂ”E$, the width of the energy level ) 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.

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. — yesterday
If it can emit a photon, surely it can absorb a photon? Otherwise there'd be a very definite time asymmetry here. — 13 hours ago
@FrancisDavey note the "if accelerated" . this means there is a third party in the equation except an electron and a photon, a field accelerating the electron. It is actually the accelerating field that keeps energy and momentum conservation in the set up. The field can accelerate and decelerate (synchrotron and brehmstrahlung radiation) — 13 hours ago
@annav not at all: you said that the energy of the photon has to COINCIDE with the difference in energy between the two atomic energy levels and this is not literally true. If you put a gas of atoms in a box, you will not observe an emission spectrum like a Dirac delta function, but you will have 3 different contributions of Lorentian and Gaussian nature. If the spectrum had been a delta, it would have been only a unique exact frequency that can excite the atom, but if you observe Lorentian and Gaussian contribution, it means that there is a little band of frequencies that can exite the atom. — 11 hours ago

V.F. 4,6761417 · Answer 2 · 2018-08-04 15:44:33Z

up vote
12
down vote

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 yesterday

V.F.

4,6761417

I don't think this is a helpful answer in it's current state. And I think there is a common trend to answer "why" question with this type of unhelpful, dismissive language. Yes, you can't answer "why" to the most fundamental laws, but you can answer why up /until/ you get to a fundamental law. You can't answer "why are Newtons laws true?", but you can answer almost any other why question related to it: "Q: Why does an apple fall to the ground" (Ans: Newtons Laws& Gravity), "Q: Why should we believe gravity is correct?" (Scientific method and induction), ...
â€“Â Steven Sagona
yesterday

"Q: "Q: Why does a pendulum absorb the energy of a push" (Ans: A framework that discusses energy conservation is derivable from Newton's laws, and has its own set of helpful intuitions). In this case there are many "why's" than can be answered, but (despite getting many upvotes and an accepted answer) this answer answers none of them. For example, the photoelectric effect could be explained and how energy is quantized in quantum mechanics (giving a first notion of a "photon"), ...
â€“Â Steven Sagona
yesterday

how the atom also has quantized energy levels due to a very complicated interaction between many electrons and a nucleus system, intuition of how this electron-nucleus system can be thought of as an effective dipole. Additionally, even if you want to dismiss why question as "just axioms", then at least point out /what/ axioms or laws produce this event. (I think in this case it's tricky! You probably need to use field theory and some assumptions to combine maxwells equations and schrodinger's equations.) ...
â€“Â Steven Sagona
yesterday

@StevenSagona Thank you for taking time and expressing your thoughts about my answer. You are definitely entitled to your opinions and interpretations - so are other people. I am sorry if my answer came across as dismissive - that was certainly not my intention.
â€“Â V.F.
yesterday

please note that electrons and photons belong to the quantum mechanical framework and obey different laws than classicale laws. The analogy is very gross dependent on general conservation laws that hold in both frames.
â€“Â anna v
22 hours ago

Â |Â
show 2 more comments

Peter Mortensen 1,86011223 · Answer 3 · 2018-08-06 01:27:31Z

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 do 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.)

Peter Mortensen 1,86011223 · Answer 4 · 2018-08-05 19:30:28Z

Absorption and emission is how we describe the interaction between electrons and the electromagnetic field using quantum field theory. If the photons didnÃ¢Â€Â™t scatter off electrons they would not interact.

Basically if you couple light fields and matter fields and quantize you must get a process where the quanta of the fields (electrons and photons) must scatter (absorb and re-emit).

Peter Mortensen 1,86011223 · Answer 5 · 2018-08-05 19:31:16Z

Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electromagnetic field (i.e. photons) and also a gas of electrons (i.e. fermions).

As you probably know, the photons are continuously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the thermal equilibrium. It is important to notice that this process of continuous absorption and re-emission of the photons by the walls of the box (i.e. by the matter!) is always present when you put together matter and light. This is fundamental if you want to reach the Planck distribution, because 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 to for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas because 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, because there is a continuous re-equilibrium process of light and matter that live together in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorption and emission of photons by the matter. It means that photons are moving, but if they are moving it means that there is a non-zero
scattering-probability --> interaction.

score -2 · Answer 6 · 2018-08-05 14:03:24Z

The question is "why do electrons bother to absorb and re-emit light and not just let it pass all the time? "

Nearly all of the time atoms do not absorb and reemit light. This only happens for photons that are resonant with an excitation frequency of the atom. Non-resonant photons are not absorbed and certainly not reemitted. At very high temperatures, such as in a plasma when thermal radiation occurs close in frequency to atomic excitation energies, the conditions can be right for absorption and reemission.

Also, absorption and emission are not the only ways by which light interacts with matter. Light can be scattered by molecules, elastically by Rayleigh scattering and inelastically by Compton scattering. In dielectrics such as window glass reflection and transmission occur by the mixing of light with electronic excitations of the dielectric. Some of this light is also absorbed by inelastic scattering with such excitations.

In general no relativistic field theory is required unless very accurate atomic energy levels are required, heavy atoms are involved or esoteric effects such as vacuum polarisation kick in.

A lot of these ultra-short answers simply point out an objection to the premise of the question, without attempting to constructively describe the way stuff works. They seem kind of half-hearted and mechanical to me (whether or not they are), particularly when there are other answers which tell me some really cool stuff. — 14 hours ago

score 28 · Accepted Answer · 2018-08-05 15:00:40Z

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 (within a small $ÃŽÂ”E$, the width of the energy level ) 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.

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. — yesterday
If it can emit a photon, surely it can absorb a photon? Otherwise there'd be a very definite time asymmetry here. — 13 hours ago
@FrancisDavey note the "if accelerated" . this means there is a third party in the equation except an electron and a photon, a field accelerating the electron. It is actually the accelerating field that keeps energy and momentum conservation in the set up. The field can accelerate and decelerate (synchrotron and brehmstrahlung radiation) — 13 hours ago
@annav not at all: you said that the energy of the photon has to COINCIDE with the difference in energy between the two atomic energy levels and this is not literally true. If you put a gas of atoms in a box, you will not observe an emission spectrum like a Dirac delta function, but you will have 3 different contributions of Lorentian and Gaussian nature. If the spectrum had been a delta, it would have been only a unique exact frequency that can excite the atom, but if you observe Lorentian and Gaussian contribution, it means that there is a little band of frequencies that can exite the atom. — 11 hours ago

V.F. 4,6761417 · Answer 8 · 2018-08-04 15:44:33Z

up vote
12
down vote

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 yesterday

V.F.

4,6761417

I don't think this is a helpful answer in it's current state. And I think there is a common trend to answer "why" question with this type of unhelpful, dismissive language. Yes, you can't answer "why" to the most fundamental laws, but you can answer why up /until/ you get to a fundamental law. You can't answer "why are Newtons laws true?", but you can answer almost any other why question related to it: "Q: Why does an apple fall to the ground" (Ans: Newtons Laws& Gravity), "Q: Why should we believe gravity is correct?" (Scientific method and induction), ...
â€“Â Steven Sagona
yesterday

"Q: "Q: Why does a pendulum absorb the energy of a push" (Ans: A framework that discusses energy conservation is derivable from Newton's laws, and has its own set of helpful intuitions). In this case there are many "why's" than can be answered, but (despite getting many upvotes and an accepted answer) this answer answers none of them. For example, the photoelectric effect could be explained and how energy is quantized in quantum mechanics (giving a first notion of a "photon"), ...
â€“Â Steven Sagona
yesterday

how the atom also has quantized energy levels due to a very complicated interaction between many electrons and a nucleus system, intuition of how this electron-nucleus system can be thought of as an effective dipole. Additionally, even if you want to dismiss why question as "just axioms", then at least point out /what/ axioms or laws produce this event. (I think in this case it's tricky! You probably need to use field theory and some assumptions to combine maxwells equations and schrodinger's equations.) ...
â€“Â Steven Sagona
yesterday

@StevenSagona Thank you for taking time and expressing your thoughts about my answer. You are definitely entitled to your opinions and interpretations - so are other people. I am sorry if my answer came across as dismissive - that was certainly not my intention.
â€“Â V.F.
yesterday

please note that electrons and photons belong to the quantum mechanical framework and obey different laws than classicale laws. The analogy is very gross dependent on general conservation laws that hold in both frames.
â€“Â anna v
22 hours ago

Â |Â
show 2 more comments

Peter Mortensen 1,86011223 · Answer 9 · 2018-08-06 01:27:31Z

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 do 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.)

Peter Mortensen 1,86011223 · Answer 10 · 2018-08-05 19:30:28Z

Absorption and emission is how we describe the interaction between electrons and the electromagnetic field using quantum field theory. If the photons didnÃ¢Â€Â™t scatter off electrons they would not interact.

Basically if you couple light fields and matter fields and quantize you must get a process where the quanta of the fields (electrons and photons) must scatter (absorb and re-emit).

Peter Mortensen 1,86011223 · Answer 11 · 2018-08-05 19:31:16Z

Imagine a cubic box at thermal equilibrium with a room a temperature $T$ and let's think that in this box there is an electromagnetic field (i.e. photons) and also a gas of electrons (i.e. fermions).

As you probably know, the photons are continuously absorbed and emitted by the walls of the box and they tend to reach the Planck frequency distribution at the thermal equilibrium. It is important to notice that this process of continuous absorption and re-emission of the photons by the walls of the box (i.e. by the matter!) is always present when you put together matter and light. This is fundamental if you want to reach the Planck distribution, because 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 to for sure understand that these photons are moving into this box. During the motion they will scatter with the atoms of the gas because 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, because there is a continuous re-equilibrium process of light and matter that live together in order to reach the thermal equilibrium of photons given by the Planck distribution. This process is made of absorption and emission of photons by the matter. It means that photons are moving, but if they are moving it means that there is a non-zero
scattering-probability --> interaction.

score -2 · Answer 12 · 2018-08-05 14:03:24Z

The question is "why do electrons bother to absorb and re-emit light and not just let it pass all the time? "

Nearly all of the time atoms do not absorb and reemit light. This only happens for photons that are resonant with an excitation frequency of the atom. Non-resonant photons are not absorbed and certainly not reemitted. At very high temperatures, such as in a plasma when thermal radiation occurs close in frequency to atomic excitation energies, the conditions can be right for absorption and reemission.

Also, absorption and emission are not the only ways by which light interacts with matter. Light can be scattered by molecules, elastically by Rayleigh scattering and inelastically by Compton scattering. In dielectrics such as window glass reflection and transmission occur by the mixing of light with electronic excitations of the dielectric. Some of this light is also absorbed by inelastic scattering with such excitations.

In general no relativistic field theory is required unless very accurate atomic energy levels are required, heavy atoms are involved or esoteric effects such as vacuum polarisation kick in.

A lot of these ultra-short answers simply point out an objection to the premise of the question, without attempting to constructively describe the way stuff works. They seem kind of half-hearted and mechanical to me (whether or not they are), particularly when there are other answers which tell me some really cool stuff. — 14 hours ago

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