25
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Lambda Aug 4 '18 at 16:30
  • 25
    $\begingroup$ 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 anyway—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. $\endgroup$ – dmckee --- ex-moderator kitten♦ Aug 4 '18 at 18:50
  • 5
    $\begingroup$ 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. $\endgroup$ – Mike Scott Aug 5 '18 at 6:50
  • 1
    $\begingroup$ "Light is a particle" and "light is a wave of energy" are not mutually exclusive statements (because particles are waves of energy). There's this idea of wave-particle duality that I think is more confusing than not—its usage often implies that waves and particles are two completely different, mutually exclusive states. A wave and a particle are just two ways of contextualizing the behavior of something. It's easy to say billiard balls act as particles and sound acts as a wave because the functioning of both of these things are almost entirely in their specified state... $\endgroup$ – AmagicalFishy Aug 7 '18 at 14:50
  • $\begingroup$ ... everything implied by being a "particle" is relevant to billiard balls, and everything implied by being a "wave" is relevant to sound. Think of particle and wave instead as general classifications for behavior—and sometimes things don't easily fit into one or the other (i.e. - light). Similarly: We have cars and we have trucks. Trucks are generally big, hold a lot more weight, etc. Cars are smaller. When we see an SUV, we don't say SUVs exhibit "car-truck duality," because that'd be stupid. :) $\endgroup$ – AmagicalFishy Aug 7 '18 at 14:52
54
$\begingroup$

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 $ΔE$, 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 transition energy of the atom, the photon may scatter with the spillover electric fields of the atom or molecule either elastically, or transferring 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.

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Ruslan Aug 4 '18 at 18:14
  • 10
    $\begingroup$ @Ruslan QFT s not the level at which this question is asked $\endgroup$ – anna v Aug 4 '18 at 18:16
  • 1
    $\begingroup$ If it can emit a photon, surely it can absorb a photon? Otherwise there'd be a very definite time asymmetry here. $\endgroup$ – Francis Davey Aug 5 '18 at 11:50
  • 4
    $\begingroup$ @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) $\endgroup$ – anna v Aug 5 '18 at 12:14
  • 2
    $\begingroup$ @MRT as what I think of your answer: it is way beyond what the question asks, which is not asking about ensembles of particles, just about electrons and photons and individual atoms. $\endgroup$ – anna v Aug 5 '18 at 17:46
16
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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), ... $\endgroup$ – Steven Sagona Aug 5 '18 at 0:06
  • 1
    $\begingroup$ "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"), ... $\endgroup$ – Steven Sagona Aug 5 '18 at 0:06
  • 1
    $\begingroup$ 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.) ... $\endgroup$ – Steven Sagona Aug 5 '18 at 0:06
  • $\begingroup$ @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. $\endgroup$ – V.F. Aug 5 '18 at 0:29
  • $\begingroup$ 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. $\endgroup$ – anna v Aug 5 '18 at 3:15
12
$\begingroup$

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.)

|cite|improve this answer
$\endgroup$
2
$\begingroup$

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 atoms (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 electrons of 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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ @PM2Ring you are right, i've edited. The answer was about the general case of a gas of atoms with an electromagnetic field. $\endgroup$ – MRT Aug 7 '18 at 8:13
1
$\begingroup$

I think this is a very good question - in the sense that it is actually hard to answer in a clear and intuitive way. Here's my take, and it is admittedly what Terry Pratchett used to call a good lie (that is, it gives the feeling of being intuitively 'right', but probably isn't factually true):

In physics, we have fields, which are described as something that permeates space; the electric field is one such thing. On the other hand, we have 'masses' or 'charges': some quantity that intereacts with fields of given types; the electric charge interacts with the electric field, for example. What this means is that a charge somehow surrounds itself with a field, and if you change the position of the charge, the field will tend to move with it - but this doesn't happen instantly, the change to the field propagates out from the charge following a wave form. I'm trying to pick my words carefully, that's why it sounds a bit woolly - the way the field changes can be described using sinuses and cosinuses, basically.

This is similar to what happens with a cork floating on water: if you push the cork, the water wobbles, a wave spreads out, then fades away - this is a 'wave particle' if you like, only it spreads evenly in all directions - I don't think we actually know why photons seem to be more particle like than this. However, this example also illustrates how the field can influence the particle: when a wave passes the cork, it wobbles, and the same no doubt happens to electrons.

Some of the previous answers have postulated that free electrons don't absorbe photons, and I feel that is not entirely true: if a photon hits an electron, it will lose some energy, which is transferred to the electron; in my view that means some of the energy was absorbed. We have never seen a photon being fully absorbed by a free electron - and it doesn't fit with current theories, but I don't think the state of experimental physics is able to rule out that it can occur. Also, I don't feel convinced that we can explain why and how a photon can be fully absorbed by a bound electron; quantum mechanics feels rather sketchy on the details - this is perhaps not surprising, since it is a probabilistic/statistical theory: it describes what happens on average, not what necessarily happens to individual particles.

Sorry, if that last part was a bit ranty: I'm a mathematician trying to make sense of the fundamentals of physics, and having a hard time.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – user191954 Aug 5 '18 at 11:41
  • $\begingroup$ It is a tall order to write chapters on my phone. Still I extended my answer quite a bit, although it may disappoint in coolness. $\endgroup$ – my2cts Aug 6 '18 at 10:23
-1
$\begingroup$

Absorption and emission is how we describe the interaction between electrons and the electromagnetic field using quantum field theory. If the photons didn’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).

|cite|improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.