Theoretical question about electromagnetic radiation...

[Jonsy123]

Between a wavelength of 450nm and 451nm, there are theoretically *infinite* number of wavelengths.

 

But, those wavelengths need to be produced by something, and if that something can only produce quantized wavelengths, then there isn't *infinite* number of wavelengths between 450nm and 451nm.

 

For example, In case of blackbody radiation, I think it will be impossible to have the ability to produce an infinite number of wavelengnths, because there isn't an infinite number of electronic excitation and de-excitation configutrations (even if you take into consideration all the atoms and molecules there are), right ?.

 

But, what about other ways to produce electromagnetic radiation ?, is there a way that allows to produce ANY wavelength you want, with no limitation of the frquency/wavelength ?, or, are we faced with a quantization limitation, which will be a characteristic of any wave-producing system we will use ?. If there is such a limitation, and every system that produces electromagnetic radiation, must do it in a quantized way (which is specific to that system), then does it means that there are certain wavelengths in nature, that never happen ?.

[J.C.MacSwell]

Between a wavelength of 450nm and 451nm' date=' there are theoretically *infinite* number of wavelengths.

 

But, those wavelengths need to be produced by something, and if that something can only produce quantized wavelengths, then...

 

For example, In case of blackbody radiation, I think it will be impossible to have the ability to produce an infinite number of wavelengnths, because there isn't an infinite number of electronic excitation and de-excitation configutrations (even if you take into consideration all the atoms and molecules there are), right ?.

 

But, what about other ways to produce electromagnetic radiation ?, is there a way that allows to produce ANY wavelength you want, with no limitation of the frquency/wavelength ?, [b']or, are we faced with a quantization limitation, which will be a characteristic of any wave-producing system we will use ?. If there is such a limitation, and every system that produces electromagnetic radiation, must do it in a quantized way (which is specific to that system), then does it means that there are certain wavelengths in nature, that never happen [/b] ?.

 

Good question.

My guess is that the answer is no for the reason you proposed but I think there may be a lot more to it.

 

For instance has the CMBR been redshifted in a continuous or quantized way?

 

Also for a photon at (say) 450 nm does an infinite number of inertial frames exist thate would equate that photon at every wavelength from (but not including) zero to infinite?

 

Edit: Just a heritical thought to add to my CMBR question but you could possibly build a "tired light" conjecture based on a quantum decay process.

[DV8 2XL]

are we faced with a quantization limitation, which will be a characteristic of any wave-producing system we will use ?. If there is such a limitation, and every system that produces electromagnetic radiation, must do it in a quantized way (which is specific to that system), then does it means that there are certain wavelengths in nature, that never happen ?.

 

yes

[mezarashi]

In semiconductors as in blackbodies, I guess there is an obvious limit in the electromagnetic frequencies that can be emitted depending on the materials electron configuration. However, in other systems like radio wave production, is there really a quantum limit? I've only analyzed such systems using Maxwell's plane wave equations, and thus the notion of a non-continuous spectrum is a bit odd to me in this field. Should someone care to explain why we cannot create any arbitrary wavelength with this respect. *blink*

[swansont]

The Heisenberg Uncertainty Principle means that you would not be able to distinguish between two wavelengths or frequencies that are arbitrarily close together. So at some level we just will never be able to tell.

[MetaFrizzics]

How do we measure frequency when dealing with single photon packets? Frequency measurement is easy with sound, or low frequency devices. But we only deduce frequency indirectly, like we deduce speed, at very high numbers through Moire interference patterns. This I think is beyond Heisenberg's principle, and into physical limitations in the construction of measurement devices.

[Jonsy123]

Another question:

 

Regarding a mirror, let's say we have an aluminum mirror (the most common). Now, theoretically, if you take a specially designed flashlight, which emits almost an infinite type of wavelenghts between 400-700nm, but *not* those very small number of wavelengths that for them, the aluminum has an excitation/de-excitation states for its electrons.

 

So, by shining with such a flashlight on the aluminum mirror, it won't reflect anything, right ?.

 

Question #2:

But, if reflectance always involve absorption of a photon, excitation, and then de-excitation with an ejection of a photon, then, how come when we hit a mirror with light from a certain angle, it will be reflected only at one angle (the same as it hit) ?. I mean, the photon hit an atom at the surface of the mirror from a certain angle, it get absorbed in the atom, and the electron of that atom get excited, then it get de-excited, and that electron needs to emit a photon, but how does the atom "remembers" to which direction it should emit the photon ? (it has to keep a log, and a calculator).

[swansont]

Another question:

 

Regarding a mirror' date=' let's say we have an aluminum mirror (the most common). Now, theoretically, if you take a specially designed flashlight, which emits almost an infinite type of wavelenghts between 400-700nm, but *not* those very small number of wavelengths that for them, the aluminum has an excitation/de-excitation states for its electrons.

 

So, by shining with such a flashlight on the aluminum mirror, it won't reflect anything, right ?.

 

Question #2:

But, if reflectance always involve absorption of a photon, excitation, and then de-excitation with an ejection of a photon, then, how come when we hit a mirror with light from a certain angle, it will be reflected only at one angle (the same as it hit) ?. I mean, the photon hit an atom at the surface of the mirror from a certain angle, it get absorbed in the atom, and the electron of that atom get excited, then it get de-excited, and that electron needs to emit a photon, but how does the atom "remembers" to which direction it should emit the photon ? (it has to keep a log, and a calculator).[/quote']

 

Atomic excitation is very different from what happens in a solid, especially on a good reflector, where you have a band structure.

 

Reflection occurs with the same angle because of conservation of momentum - that's the "log and calculator"

[Jonsy123]

Atomic excitation is very different from what happens in a solid, especially on a good reflector, where you have a band structure.

Wait, so a photon does not interact with each atom of a solid, by means of excitation and de-excitation ?. If this is the case, then using the flashlight I described in the post above, will you still get reflectance from the aluminum mirror ?.

[swansont]

Wait, so a photon does not interact with each atom of a solid, by means of excitation and de-excitation ?. If this is the case, then using the flashlight I described in the post above, will you still get reflectance from the aluminum mirror ?.

 

In a solid, the electronic structure is modified from the single-atom structure. Instead of a single frequency you get a band, which is a continuum of energies. IIRC in the derivation you have to consider the electrons as all part of one system, so you can't have multiple excitations into one state (Pauli exclusion principle) so each atom contributes to a new state of a slightly different energy.

 

Also consider that in a solid there are electronic bonds, so the excitation energy gets modified.

[Jonsy123]

So, in the solid state of the aluminum atoms, your only way to emit light is still by excitation and de-excitation of electrons, so the emitted light is still quantized. But, in comparison to a single aluminum atom, you get much more excitation possibilities, which leads to a wider range of different frequencies which can be emitted... right ?. (You used the words "a continuum of energies", but I assume you didn't mean an *infinite* number of possibilities, but just a higher number than in the single atom).

 

So, I think that in my flashlight situation, you can still flash it on the aluminum mirror and get no reflection, but you will have to take into account all the new excitation possibilities that arise from the fact that the aluminum atoms are at the solid state...

[J.C.MacSwell]

So' date=' there is still a quantization of the emitted light due to excitation and de-excitation of electrons in the aluminum atoms of the solid, but, in comparison to a single aluminum atom, you get much more excitation and de-excitation possibilities, which leads to a wider range of different frequencies which can be emitted... right ?. (You used the words "a continuum of energies", but I assume you didn't mean an *infinite* number of possibilities, but just a higher number than in the single atom).

 

So, I think that in my flashlight situation, you can still flash it on the aluminum mirror and get no reflection, but you will have to take into account all the new excitation possibilities that arise from the fact that the aluminum atoms are at the solid state...[/quote']

 

Also the aluminum has a temperature, so would that not lead to an effective spreading of the excitation /de-excitation possibilities?

[Jonsy123]

Also the aluminum has a temperature, so would that not lead to an effective spreading of the excitation /de-excitation possibilities?

But why would heating cause *more* excitation *possibilities* ?. It seems to me (but I may be wrong) that it will just excite more electrons to an already possible excitation states, but not create new ones...

 

On the other hand, the atoms are in the solid state only because of the current temperature, so the possible excitation states should be a function of the solid's temp.

 

Back to my flashlight idea, I guess that it will work, but the flashlight will have to be designed for a certain temperature of the solid mirror. If the temperature of the solid mirror will change, so will the excitation states of its atoms, and some of the flashlight frequencies might get absorbed and re-emitted now...

[swansont]

(You used the words "a continuum of energies", but I assume you didn't mean an *infinite* number of possibilities, but just a higher

 

You have to consider that any given excitation has a finite frequency width that is inversely related to the state lifetime. A state with a lifetime of 25 ns has a frequency width (full-width at half-maximum) of about 6 MHz.

 

In your example of 450nm-451nm, the frequency width is only about 1.5e12 Hz, IOW less than a million atoms would span that, if the states were spaced accordingly. You'd have a continuous absorption band with a finite number of states. (I'm not a solid-state guy, so this may be way over-simplified on how the states are configured, reflection may be treated differently, and I'm using typical lifetime values from materials I know, not Al. Just showing how it could happen)

[DQW]

Wait, so a photon does not interact with each atom of a solid, by means of excitation and de-excitation ?. If this is the case, then using the flashlight I described in the post above, will you still get reflectance from the aluminum mirror ?.

Absolutely. And at frequencies below the plasma frequency, you will have every one of your incident frequencies be reflected, not just certain special frequencies. And this is not realy because you have a band of energies but because you are exciting collective oscillatory modes known as plasmons - kind of like bouncing a steel ball off a rubber floor.

[J.C.MacSwell]

But why would heating cause *more* excitation *possibilities* ?. It seems to me (but I may be wrong) that it will just excite more electrons to an already possible excitation states' date=' but not create new ones...

 

On the other hand, the atoms are in the solid state only because of the current temperature, so the possible excitation states should be a function of the solid's temp.

 

Back to my flashlight idea, I guess that it will work, but the flashlight will have to be designed for a certain temperature of the solid mirror. If the temperature of the solid mirror will change, so will the excitation states of its atoms, and some of the flashlight frequencies might get absorbed and re-emitted now...[/quote']

 

Due to the relative velocities inherent in "temperature" the some aluminum atoms could "see" some photons as having correct frequencies that the average would not.

I think this is "spreading" of the possibilities over more frequencies not necessarily more "successes" overall.

[Jonsy123]

In your example of 450nm-451nm, the frequency width is only about 1.5e12 Hz, IOW less than a million atoms would span that, if the states were spaced accordingly. You'd have a continuous absorption band with a finite number of states.

When you say a "continuous absorption band", you mean that no matter what the frequency that will hit the solid, as long as it is between 450-451nm, then it will cause excitation ?.

 

I'm not sure if that is what you mean.

[Jonsy123]

Absolutely. And at frequencies below the plasma frequency, you will have every one of your incident frequencies be reflected, not just certain special frequencies.

DQW, did you also affirmed the following sentence ?:

 

a photon does not interact with each atom of a solid, by means of excitation and de-excitation.

[DQW]

DQW, did you also affirmed the following sentence ?:

Yes, I did - in the context of reflection at least. There are processes like x-ray photoemission (of core electrons) that deal with individual atoms, but that's not what we're talking about here.

[DQW]

But why would heating cause *more* excitation *possibilities* ?. It seems to me (but I may be wrong) that it will just excite more electrons to an already possible excitation states, but not create new ones...

That's essentially correct, but not entirely. The most important effect of increasing temperature is that you now populate states that were unoccupied at lower temperatures - specifically some of the states above E(F) in the picture below.

 

 

In any case, this is more or less irrelevant to reflection off a metallic surface.

[Jonsy123]

So, in metalic surface reflectance, the same photon which hit the surface, is the one which bounce back from it ?. Now at least I can understand how it comes out only at a certain direction (like a basketball hitting the floor, only without any energy loss, or maybe there is some energy loss as heat ?, afterall, there IS an imapct...). I don't think it would have been possible, if we were talking about excitation/de-excitation, since the direction information should have been lost in the process...

 

(I guess my flashlight idea won't work afterall).

 

Anyhow, the fact that a metalic surface can reflect *any* wavelengh between 450-451nm, still doesn't mean it will get the chance to do it in practice, since as DV8 2XL affirmed earlier, some wavelengths never happen in nature, due to the quantization limitation which is an inherent characteristic of any system which actually produce radiation.

[swansont]

Anyhow, the fact that a metalic surface can reflect *any* wavelengh between 450-451nm, still doesn't mean it will get the chance to do it in practice, since as DV8 2XL affirmed earlier, some wavelengths never happen in nature, due to the quantization limitation which is an inherent characteristic of any system which actually produce radiation.

 

Even if that were true, you would never actually see a "gap" of a missing wavelength because they aren't infinitely narrow.

[Jonsy123]

But, must they be *infinitely* narrow in order for me to see some gap ?.

 

Also, is there a simple explanation as to how come a metallic surface can reflect a photon, without the excitation/de-excitation mechanism ?. The picture I see is this: a photon is headed toward a metalic surface, at some point, it will hit it (otherwise, it won't change its course). It can either hit an atom's nucleus, or an atom's electron. I assume if it hits something, it will almost always be the electron. At the moment it hit the electron, I understand that the electron does NOT absorb the photon and excite to a higher orbital, but if not, then what happens instead ?, does the photon simply hit the electron, the photon's course is changed, at it continues its journey at that new course ?.

[DQW]

The photon is actually absorbed and re-emitted. It is absorbed by one of the conduction electrons. This transfer of energy and momentum to the conduction electron creates a longitudinal compression wave within the conduction ("free") electrons. So, the impinging photon causes the plasma of free electrons to compress and the flex back (at the natural frequency of the free electron gas) spitting out a photon identical to the incident one.

[mezarashi]

Also' date=' is there a simple explanation as to how come a metallic surface can reflect a photon, without the excitation/de-excitation mechanism ?. The picture I see is this: a photon is headed toward a metalic surface, at some point, it will hit it (otherwise, it won't change its course). It can either hit an atom's nucleus, or an atom's electron. I assume if it hits something, it will almost always be the electron. At the moment it hit the electron, I understand that the electron does NOT absorb the photon and excite to a higher orbital, but if not, then what happens instead ?, does the photon simply hit the electron, the photon's course is changed, at it continues its journey at that new course ?.[/quote']

 

There are actually two types of reflection, diffuse reflection (typical opaque materials) and specular reflection (mirror-like, water reflection). Both phenomena observe the conservation of momentum. I think you're visualization of absorption-emission is a bit too simple. You know about quantum weirdness. It may seem as if the momentum info is lost, but actually it isn't. If the electron emits the photon in any other direction of any other frequency, there will be a high disproportion in equilibrium of the structure, thus it is not probable. Like in direct indirect bandgap semi-conductors, sometimes phonons must be created in order to obey momentum conservation.

 

Back to the topic however, diffuse reflection is the one where you get an absorption followed by a re-emission. This is because the solid's "electron band" vibrates within a certain range of frequencies. This can be due to so many factors like the rotational energy, lattice structure, the hybridization of energy levels, etc.

 

As for specular reflection, the explanation is apparently not easy to explain from a quantum point of view. From the classical equations (Maxwell's), you can derive the Fresnel equations which govern the law of reflection/transmission at a boundary. The quantum explanation as I've been taught, has to do with the most probable path of the photon. But it is true that the photon isn't absorbed, and interacts with the mirror "from a distance" you can say.