Why doesn't the atom absorb heat energy when it is low?

[DanMP]

I think Wikipedia is completely wrong here.

So, why it is still unchanged?

It is important to note that the time-energy uncertainty principle is not Heisenberg's uncertainty principle, since time is not an observable in QM.

It is not Heisenberg's uncertainty principle, but it is something worth considering, as you may see in this article.

QM doesn't work that way. The interaction of the photon with the atom (or molecule) will result in the atom and the electromagnetic field being in a superposition of non-excited atom + one photon and excited atom + no photon.

In my opinion QM is not quite complete, at least in explaining things. I found something to support this in Nature:

What Is Real?: The Unfinished Quest for the Meaning of Quantum Physics Adam Becker Basic: 2018.

All hell broke loose in physics some 90 years ago. Quantum theory emerged — partly in heated clashes between Albert Einstein and Niels Bohr. It posed a challenge to the very nature of science, and arguably continues to do so, by severely straining the relationship between theory and the nature of reality. Adam Becker, a science writer and astrophysicist, explores this tangled tale in What Is Real?.
...
What Is Real? is an argument for keeping an open mind. Becker reminds us that we need humility as we investigate the myriad interpretations and narratives that explain the same data.

So why not keeping an open mind and investigate the Fizeau experiment and the Sagnac effect as I suggested?

I did point out that this is a semi-classical explanation. The optical wave is here a classical electromagnetic wave, which is being scattered by molecules. There are no "new" photons coming out.

So how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

 

[Herman Trivilino]

How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

New photons are produced in the medium.

 

[Nugatory]

So, why is [wikipedia] still unchanged?

For the same reason that Sisyphus's rock is still at the bottom of the hill
And seriously, kidding aside, there's a reason why Wikipedia is not in general an acceptable source under the forum rules.
(Although to be fair the anonymous Wikipedian who wrote that section may not have intended to make the suggestion that you took away, that "there is a process that takes time...").

In my opinion QM is not quite complete, at least in explaining things. I found something to support this in Nature:

That is a fairly non-controversial position, as quantum mechanics has never claimed to explain things in the sense that you're using the term here. Nonetheless, you would be better served by Becker's book than by Skibba's review of it.

So why not keeping an open mind and investigate the Fizeau experiment and the Sagnac effect as I suggested?

The first step in the open-mind review is to understand the best current theory. It's difficult to improve on something when you don't know what it does and does not do well.

 

[DanMP]

The first step in the open-mind review is to understand the best current theory. It's difficult to improve on something when you don't know what it does and does not do well.

That's why I asked the question:

how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

New photons are produced in the medium.

Ok, but from one photon entering the medium (air, water), we don't usually get more, similar photons at the other end, as wikipedia suggests ...

 

[Nugatory]

Ok, but from one photon entering the medium (air, water), we don't usually get more, similar photons at the other end, as wikipedia suggests ...

I'm not seeing that suggestion in that Wikipedia article (which only mentions "photons" in one place, unrelated to this discussion).

 

[Lord Jestocost]

So how is the "non-classical" explanation? How are the photons traveling trough air & water in such a way that the light slows down and no new/extra photons are produced?

As Peter W. Milonni writes in “Answer to Question #21 [‘‘Snell’s law in quantum mechanics,’’ Steve Blau and Brad Halfpap, Am. J. Phys. 63 (7), 583 (1995)]” (American Journal of Physics 64, 842 (1996)):

“In the quantum-mechanical description of a plane wave incident on a dielectric medium, each photon has a probability amplitude to be scattered by anyone atom. The complete probability amplitude for a photon to be found at any point inside or outside the medium is the amplitude for it to get there without any scattering, plus the sum over all the possible paths by which it can get there via single- and multiple-atom scattering. The result of this superposition of all possible probability amplitudes is an amplitude that is nonzero both inside and outside the medium.”

 

[DanMP]

I'm not seeing that suggestion in that Wikipedia article (which only mentions "photons" in one place, unrelated to this discussion).

This is from Wikipedia (the underline and the comment in red are mine):

At the atomic scale, an electromagnetic wave's phase velocity is slowed in a material because the electric field creates a disturbance in the charges of each atom (primarily the electrons) proportional to the electric susceptibility of the medium. (Similarly, the magnetic field creates a disturbance proportional to the magnetic susceptibility.) As the electromagnetic fields oscillate in the wave, the charges in the material will be "shaken" back and forth at the same frequency.[1]:67 The charges thus radiate their own electromagnetic wave [photons, right?] that is at the same frequency, but usually with a phase delay, as the charges may move out of phase with the force driving them (see sinusoidally driven harmonic oscillator). The light wave traveling in the medium is the macroscopic superposition (sum) of all such contributions in the material: the original wave plus the waves radiated by all the moving charges. This wave is typically a wave with the same frequency but shorter wavelength than the original, leading to a slowing of the wave's phase velocity. Most of the radiation from oscillating material charges will modify the incoming wave, changing its velocity. However, some net energy will be radiated in other directions or even at other frequencies (see scattering).

 

[DanMP]

“In the quantum-mechanical description of a plane wave incident on a dielectric medium, each photon has a probability amplitude to ...

Ok, but how is this related to my question? It doesn't say anything about the speed of light in the medium.

 

[Lord Jestocost]

It doesn't say anything about the speed of light in the medium.

In case one wants to understand how the apparent speed of light in media comes about, I recommend to thoroughly read chapter 31 “The Origin of the Refractive Index” in “The Feynman Lectures on Physics, Volume I “(http://www.feynmanlectures.caltech.edu/I_31.html).

 

[Nugatory]

This is from Wikipedia (the underline and the comment in red are mine):

[1]:67 The charges thus radiate their own electromagnetic wave [photons, right?]

Not right. A light wave is not made up of photons the way you're thinking.

 

[Herman Trivilino]

This is from Wikipedia (the underline and the comment in red are mine):
[...]The charges thus radiate their own electromagnetic wave [photons, right?]

You need large collections of photons for the electromagnetic wave approximation to be valid.

 

[DanMP]

In case one wants to understand how the apparent speed of light in media comes about, I recommend to thoroughly read chapter 31 “The Origin of the Refractive Index” in “The Feynman Lectures on Physics, Volume I “(http://www.feynmanlectures.caltech.edu/I_31.html).

Thank you!

I just read it. It was interesting, but still a classical explanations ... with a peculiar way of seeing the electrons in the atoms:

Each of the electrons in the atoms of the plate will feel this electric field and will be driven up and down (we assume the direction of E0 is vertical) by the electric force qE. To find what motion we expect for the electrons, we will assume that the atoms are little oscillators, that is, that the electrons are fastened elastically to the atoms, which means that if a force is applied to an electron its displacement from its normal position will be proportional to the force.
You may think that this is a funny model of an atom if you have heard about electrons whirling around in orbits. But that is just an oversimplified picture. The correct picture of an atom, which is given by the theory of wave mechanics, says that, so far as problems involving light are concerned, the electrons behave as though they were held by springs. So we shall suppose that the electrons have a linear restoring force which, together with their mass m, makes them behave like little oscillators, with a resonant frequency ω0.

This is not the modern view of the atom, and apparently not consistent with the views expressed here.

 

[Nugatory]

Thank you!

I just read it. It was interesting, but still a classical explanations ... with a peculiar way of seeing the electrons in the atoms:

Yes of course it is - you entered this thread in #33 looking for a classical explanation. Much of the subsequent discussion has been about trying to stop you from confusing yourself by unnecessarily introducing quantum misconceptions into the discussion.

This is not the modern view of the atom

Of course not. It is an accurate and thorough classical analysis of a problem that can be treated classically and for which the methods of qantum electrodynamics are overkill. You don't have to be satisfied by it, but if you aren't willing to learn quantum electrodynamics, there's nothing better on offer except this and other semiclassical descriptions.

 

[Herman Trivilino]

... with a peculiar way of seeing the electrons in the atoms

Not peculiar to people who work in that area. In fact, the casual learner is better served by this simple model of matter than the more complex models presented everywhere from chemistry textbooks to popular science books.

 

[Lord Jestocost]

This is not the modern view of the atom, and apparently not consistent with the views expressed here.

This is a complete different story and refers to the acceleration of an atom as a whole.

 

[cmb]

. If the photon energy doesn't match this energy difference, then the atom's energy level doesn't change. We describe that by saying that the photon is not absorbed.

There is something about this thread I don't like!

So ... if this is the whole story ...
firstly, photons from a black body radiator should never be able to excite an atom, because the 'analogue' energy from a BB emitter will never exactly match the energy levels between electron states, there will always be a little bit left over.
secondly, hydrogen electrons have, as a maximum, an energy state of ~13 eV , so any photons with more than 14 eV cannot ionise hydrogen!

I don't accept those corollaries, please can we look at this proposition that incoming energy has to be exactly the electron band levels, but also if an atom does absorb energy higher than its band energy, what happens to the rest?

 

[phinds]

Thanks to everyone for your opinion.

Just so you are clear, no one here has given you an opinion, they have given you facts. Science does not work on opinions.

 

[sophiecentaur]

because the 'analogue' energy from a BB emitter will never exactly match the energy levels between electron states,

You seem to be wanting a classical and a quantum argument to apply at the same time "Exactly" does not apply. The "analogue" energy you are describing can be regarded as consisting of many different photons and not a continuum. There is a finite spread of energy levels amongst the H atoms and there are a finite number of photons with energy that fall within this spread of atomic energy levels. So the absorption is not of a single energy but in a restricted band of energies. Different substances have different widths of absorption lines and you can look at this as different bandwidths of receiver.

if an atom does absorb energy higher than its band energy, what happens to the rest?

The surplus energy goes into Kinetic Energy of the released electron.

 

[cmb]

You seem to be wanting a classical and a quantum argument to apply at the same time "Exactly" does not apply. The "analogue" energy you are describing can be regarded as consisting of many different photons and not a continuum. There is a finite spread of energy levels amongst the H atoms and there are a finite number of photons with energy that fall within this spread of atomic energy levels. So the absorption is not of a single energy but in a restricted band of energies. Different substances have different widths of absorption lines and you can look at this as different bandwidths of receiver.

The surplus energy goes into Kinetic Energy of the released electron.

I don't think quantum theory allows for a 'band' of energies for electron states. What is the range of this 'band'?

I am aware that the energy bands can change depending on electromagnetic effects (Zeeman/Stark), is that what you mean?

In regards the idea that a freed electron has any 'spare energy' manifest as kinetic energy I understand the "mechanical logic" of it but I don't find this convincing for the following 'philosophical' reason; if the state of the freed electron was defined by both the atom from which it came and the energy that impinged on that atom, then it naturally implies that the electron and the atom's nucleus were like separate components of a two-component system, but that turns my understanding of quantum physics around because surely it is the 'atom' that defines the energy levels of the electron, not the nucleus of the atom as a distinct object from the electron?

 

[Dale]

What is the range of this 'band'?

The range of the band for an individual transition depends on the characteristic time of the transition. For a transition that happens quickly the linewidth is broad, and for a transition that is very unlikely the linewidth is narrow.

When there are many transitions involved it gets more complicated with overlapping transitions forming broad bands, as described.

 

[sophiecentaur]

I don't think quantum theory allows for a 'band' of energies for electron states. What is the range of this 'band'?

I am aware that the energy bands can change depending on electromagnetic effects (Zeeman/Stark), is that what you mean?

We all learn about 'The Hydrogen atom' and how the photon interaction only occurs for Exactly one frequency. However . . .
All the atoms in a region of gas have slightly different Energy levels due to the Pauli Exclusion Principle. They all have a slight effect on each others' fields. If the atoms are in motion then they will absorb photons with different frequencies in the same way that the emitted photons are affected. Line broadening is greater in high pressure gas because the 'bandwidth' is greater.
Edit: Plus what @Dale said.

 

[Dale]

Line broadening is greater in high pressure gas because the 'bandwidth' is greater.

Oh, yeah, I completely forgot about Doppler broadening too.

 

[sophiecentaur]

Oh, yeah, I completely forgot about Doppler broadening too.

There are so many additions to the elementary hydrogen atom model yet that is the model that nearly 'everyone' quotes when linking for explanations. The introductions to QM at school should really contain caveats very early on.

 

[cmb]

But distribution of atoms with different energies still doesn't affect this outcome, I think this is a 'red herring'. 'A' photon interacts with 'an' atom. 'That' atom has energy levels. How broad are those?

A photon may well have the right energy to interact with an atom in the middle of a volume of stuff, but the question is; will it?

What we started with in this thread was that if a photon wasn't the right amount of energy then it wouldn't interact with the atom. My query is that an 'analogue' level, say photon energy from BB, will never be exactly the level in 'an atom'. If there were a million atoms it'd simply mean it was different to a million of them. Either the statements earlier in the thread are wrong and BB radiation can interact with atoms, or the statements earlier were right and they can't. It can't be both.

If the excitation state for hydrogen is, let's say, exactly 2eV for the sake of the point, then will a photon of 2.1eV interact with it? If not, then 2.01eV. If not then 2.001eV. If not then ... stop me when it will interact ...

 

[Dale]

'A' photon interacts with 'an' atom. 'That' atom has energy levels. How broad are those?

That is what my response in post 72 talked about. In a solid, however, it is not usually true that a photon interacts with an individual atom. It interacts with the whole lattice.

If the excitation state for hydrogen is, let's say, exactly 2eV for the sake of the point, then will a photon of 2.1eV interact with it? If not, then 2.01eV. If not then 2.001eV. If not then ... stop me when it will interact ...

See here for details:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node428.html
For the hydrogen 2p to 1s transition the linewidth is about 0.4 E-6 eV.

say photon energy from BB, will never be exactly the level in 'an atom'. ... Either the statements earlier in the thread are wrong and BB radiation can interact with atoms, or the statements earlier were right and they can't. It can't be both.

There is a third option, which is that your claim that BB radiation will never be the right level to be absorbed is wrong. Thus the statements are right and yet BB radiation can interact with atoms.

If I understand your reasoning it is simply that in a continuous distribution the probability of any single real number is 0. But the natural or intrinsic linewidth is finite, so this argument simply doesn’t apply.

 

[cmb]

There is a third option, which is that your claim that BB radiation will never be the right level to be absorbed is wrong. Thus the statements are right and yet BB radiation can interact with atoms.

If I understand your reasoning it is simply that in a continuous distribution the probability of any single real number is 0. But the natural or intrinsic linewidth is finite, so this argument simply doesn’t apply.

Thanks for the response and the additional information, which is good.

Yes, of course, if there were a 3rd option with a margin of uncertainty then that resolves the thing, but the statements above didn't say that. It's like I put But you've resolved that now, thanks.

I am still curious as to what happens to the 'excess/lacking' energy after an interaction? What I mean is that if you have a load of photons all bombarding a substance with a 'real' energy level right at the bottom of that margin and no photons above the median excitation level, where does the system get the extra energy from? Like a quantum version of Maxwell's demon?

 

[Dale]

Any extra energy can either go back into the EM field (inelastic scattering) or it can go into KE.

Edit: oops, on rereading I see you were actually asking about missing energy. There isn’t any missing energy either. Energy is uncertain. There is no “real” energy level until it is measured.