Post by: Eternal Student on 09/03/2023 01:22:57
Blackbody radiation is usually considered (theoretically modelled and suitable equations derived) by looking at the radiation that can exist inside a cavity or an "oven". The interior of the oven can support various modes of radiation within it, each mode has a particular frequency, those frequencies ultimately assumed to be caused (generated by) charged particles oscillating at that frequency in the walls of the oven.
For a cuboidal oven you can have modes of radiation within the oven with this general form:
ψ = A. Sin (Kx x) . Sin (Ky y) .Sin (Kz z).
[Equation 1]
Where A = amplitude (a constant) and Kx, Ky and Kz are also constants but not quite as arbitrary. One of the boundary conditions that must be satisfied is that ψ = 0 at the walls of the oven (since these waves do not continue through the walls and propagate away from the oven). We really are looking for stationary wave solutions that fit inside the oven. Hence only Kx = p.π / Lx , Ky = q π/Ly and Kz = r π / Lz for some integers p,q,r will be allowed (Lx being the length of the oven in the x direction etc.)
That's a typical way in which the blackbody radiation is modelled and you can find more details in most textbooks or online resources. I won't take up any more of your time here.
So, taking everything at face value, it would seem that.... if you have a big oven then you can support e-m radiation at most wavenumbers k or wavelengths λ = 2π/k. This is just because π/Lx , π/ Ly etc. will be small numbers and so the permitted wave vectors are separated by only a small amount.
For a small oven, however, the permitted wavenumbers are more widely separated. You'd still get the overall shape given by the usual blackbody spectrum curve but you may begin to notice that you only have small discrete chunks of that spectrum present.
I don't know... I'm asking not telling. I've just presented some of the background so that you can hopefully see it isn't a silly question to ask.
Taking it to a more extreme situation.... what if you don't even have a cuboidal shaped oven? Imagine an arbitrary blob with protruding lumps shaped oven. I haven't spent much time trying the mathematics, it's just apparent that easy and separable solutions like [equation 1] are out of the question - that's never going to satisfy ψ=0 at the walls. It would seem that a blob shaped oven would support considerably less modes of radiation within it.
So, does the spectrum of the interior of an oven actually show any changes when you alter the shape and size of that interior space? For example, does the spectrum seem to become more discrete and less continuous with very small interior spaces? Does a cube shaped oven have a better and cleaner match to a blackbody radiation curve than the interior of a kidney bean shaped oven?
None of this is urgent, I'm of the opinion that the usual way blackbody radiation is presented in the textbooks (very roughly as the above) is just far too simplified. I am pretty sure that the shape of an oven does not affect the radiation spectrum, I'm just not sure I could explain why.
Best Wishes.
Post by: evan_au on 09/03/2023 07:15:45
Quote from: OP
One of the boundary conditions that must be satisfied is that ψ = 0 at the walls of the oven ...since these waves do not continue through the wallsI understand that historically, black body radiation was calibrated with a block of graphite (already pretty black, visually), with a sphere carved out in the center, and a hole bored through from the outside, so you could observe the radiation inside the cavity.
- You heat up the graphite block to a certain temperature (presumably with no oxygen present), and measure the spectrum of the radiation visible at the small hole.
Because graphite is a fairly effective black body, it actually absorbs any radiation impinging on the surface of the sphere
- getting hotter in the process
- and generating black-body radiation back into the cavity
- You measure the spectrum after this has settled down to an equilibrium
Because the graphite is absorptive, you don't need "ψ = 0 at the walls", since the radiation acts as if it is propagating off to infinity
- When in fact it gets absorbed within a few mm of the wall
- So there are no standing waves in the cavity
- This is quite unlike a metal-walled microwave oven, where the E field is zero at the conductive walls, and standing waves are definitely present (causing uneven heating of the food)
So I suggest that a reflective metal oven is a much worse representation of black body radiation than a cavity carved in graphite.
See the diagram here:
https://www.physics.utoronto.ca/~phy293lab/experiments/blackbody.pdf
Post by: Zer0 on 09/03/2023 23:33:49
https://www.thenakedscientists.com/forum/index.php?topic=66414.0
(thou it's a very long cake, supposedly with no cherries atop)
Post by: Eternal Student on 10/03/2023 01:09:26
The article you linked to includes this statement:
Classical physics suggested that all modes had an equal chance of being produced, and that the
number of modes went up proportional to the square of the frequency. (From page 2, just above fig.1)
I would have to ask what were they considering as a "mode"? If it was a particular frequency of radiation that could exist within the oven, then EVERY frequency is possible and no special conditions or restrictions apply at any frequency (assuming as they state, you don't need standing waves, or ψ = 0 at the walls). The second half of the sentence then seems to have no explanation - why would the number of modes available increase with the square of the frequency?
Compare with these diagrams (and brief explanations):
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c2 (LATE EDITING: I think that link goes exactly where needed):
That idea shown in the diagram is NOT the only way people think about fitting different waves in a box so that ψ = 0 at the walls - but it's an online resource that I know people can access so that's why I've given it.
and then this shows how the number of modes is determined:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c2
Notice that the number of modes varies ~ 1/λ3. Invert to obtain relationship to frequency ~f3 and then instead of considering all modes from f=0 to f=our value, we usually consider a density (the number of modes per unit frequency at f=our value) so that is ~ f 2 as mentioned in the article you linked to. (They were a bit sloppy not mentioning that it was a density). Anyway, it seems more common to consider the density of modes in terms of wavelength so it is a density per unit wavelength at a given wavelength which has the form ~ 1/λ4 .
In all the derivations of blackbody radiation that I have seen (classical based on Rayleigh-Jeans OR quantum based on Planck etc.) the density of modes permitted is ALWAYS of that form ~ 1/λ4 (it seems more common to use wavelength instead of frequency). It is only the allocation of Energy in each mode that changes.
So, there just really ought to be some reason why the density of modes varies with wavelength. In most texts (I can find multiple references if required) that is explained purely on the basis of requiring standing waves inside the oven.
None-the-less, I agree with the general idea presented in the article you sighted. I've been thinking for quite a while that it isn't the electromagnetic waves inside the oven that are the explanation for the density of modes. It does seem reasonable to consider a situation much as you have outlined with the graphite oven - waves of any frequency can be generated, they won't be stationary, they will travel but they will be dealt with at the walls regardless. To say this another way, why on earth should we require standing waves?
Under that assumption, the explanation for the density of modes must be something else...... I'll keep the speculation to a minimum but the most obvious candidate is the particles in the walls of the oven, they are the thing that would generate the radiation of each frequency. They could have more modes (of oscillation) when their oscillation frequency is higher - but I've tried and still can't see a good reason to explain why.
Overall, I'm seriously wondering if we should just abandon the notion of there being "modes" (of radiation in the oven) entirely. If you don't require standing waves then there doesn't seem to be any sensible way of defining what a "mode" should be. I'm going to have another good look through various bits of literature about Blackbody radiation and see what's what.
For the moment, if anyone has any information and especially any relevant practical investigations showing that the spectrum of radiation inside an oven is affected by the shape of the oven, please let me know. Otherwise I'm going to assume that everyone is of the opinion that the shape of the oven makes no difference at all.
So I suggest that a reflective metal oven is a much worse representation of black body radiation than a cavity carved in graphite.Yes, that seems absolutely certain. Obviously the walls were not blackbodies, they reflected something but by definition a blackbody must absorb all radiation that hits it. So it is clearly just the cavity and not the walls that we are talking about here. Such a metal oven really should have a spectrum that is very different to a blackbody spectrum - there was nothing in the walls (an atom) that absorbed the radiation you are describing as being reflected and so there were no particles in the walls that could possibly emit such radiation into the cavity. In short there should be a clear deficit in radiation of the frequency (frequencies) that the walls reflect. The only way radiation of those frequencies could have got into the cavity would be when the oven door was opened.
If we remove the part about the walls being reflective, so we just have an oven with (dull) conductive metal walls then your comment is more interesting. That is precisely the sort of thing I would like to know about, especially if anyone has done a suitable experiment and analysed the spectrum inside a metal oven compared to a graphite oven.
Microwave ovens are complicated and seem only partially related to this discussion - it's not the walls of the oven that are being raised to some temperature to heat that oven - but yes, I see some of your point about the standing waves.
Best Wishes.
Post by: Eternal Student on 10/03/2023 01:13:20
Post by: evan_au on 10/03/2023 06:51:48
Quote from: Eternal Student
The only way radiation of those frequencies could have got into the cavity would be when the oven door was openedIf you are looking at the spectrum excited by the cavity magnetron, it spans a range of about 10MHz (it jumps around between different modes inside the magnetron).
- So there are a number of modes that might be excited by the magnetron
- To minimise uneven heating due to standing waves, microwave ovens often have a rotating platform for the food, or else a mode stirrer (like a metal fan, but usually out of sight)
https://www.researchgate.net/figure/Magnetron-spectrum-without-PLL-control-and-with-full-heater-power-68-W_fig4_3074637
Post by: evan_au on 10/03/2023 08:04:10
- Cheap semiconductor laser pointers continually jump between these modes
- More expensive lasers for telecommunications often have a second optical cavity coupled to the first; only one mode is able to resonate in both cavities, reducing interference due to the different speed of light of different modes when traversing an optical fiber.
This video shows mode changes every few seconds in a green laser.
This page shows a typical laser spectrum, averaged over time - but in practice, not all of these resonances are there at the same time.
https://www.researchgate.net/figure/The-laser-spectrum-at-t-6-nsec-Axial-cavity-modes-are-visible-in-the-laser-spectrum_fig4_321257044
Post by: Bored chemist on 10/03/2023 11:45:01
I think we are looking at the idea that wavelengths comparable with the length of the side of the box will be perturbed by the size and shape of the box.
If I get a metal box (like a microwave oven) and put a tungsten light bulb in it. the light doesn't get out.
If I make a small (say 1mm) hole in the wall, the light gets out and I can measure its spectrum. The box hardly makes a difference to that.
I can, also measure the infra red from the lamp.
In principle, the bulb also emits microwave and radio wave radiation.
But I can't see them- not because the box changes the spectrum of the lamp, but because the long wavelengths will not pass through the small hole.
The longer than about 1mm wavelengths are filtered out
So, I can only observe the spectrum of what's happening in the box through a filter which I know removes anything with a wavelength longer than 1mm.
If I make the hole bigger (comparable with the size of the box) in order to look at wavelengths comparable with the size of the box then...
it isn't a box any more.
Post by: alancalverd on 10/03/2023 11:52:53
Much the same applies to wind instruments. A flute extracts a fairly pure sine wave from the exciting edge tone, whereas a conical-bore instrument like a saxophone carries a lot more overtones.
Post by: Bored chemist on 10/03/2023 13:36:57
Waveguides are often silver plated or even lined with superconductors to improve the sharpness of their frequency response.
A black body is (pretty much definitively) not.
The analogy isn't getting a resonance from a flute; it is trying to play a tune on an exhaust silencer. (I might pay good money to watch someone try to do that.)
If the wall opposite absorbs all the radiation which hits it, there's no way for the radiation to "know" how far away it is.
It's not a good approximation if one of the walls is a good mirror.
I did some pointless research on this as a student. You can change the half life of fluorescence by putting the emitter near a mirror.
But you don't make black bodies out of mirrors so it's beside the point.
Post by: Eternal Student on 10/03/2023 15:40:10
Thank you to everyone who has spent some time here. I am busy reading and thinking about what has been said.
For the moment, if anyone has any information and especially any relevant practical investigations showing that the spectrum of radiation inside an oven is affected by the shape of the oven, please let me know.Overall, especially with the comments from @evan_au about lasers, it seems that the shape of an oven / cavity almost certainly can influence the spectrum of radiation within it.
(sampling the spectrum is a problem....) because the long wavelengths will not pass through the small hole.Yes. I was going to say they would need to put a spectrum analyser inside the oven - but I'm fairly sure a "spectrum analyser" was actually the name for a piece of hi-fi audio kit that we all wanted back in the 1980's. Anyway, it's that sort of equipment that would be wanted regardless of the name it may have. I'm aware that analysing the spectrum of all e-m radiation is actually extremely difficult. Radio waves aren't a problem, an aerial and a simple electrical circuit can catch those but shorter wavelengths are a problem and electrical equipment is not fast enough to rectify a ray of light shone onto an aerial. Spectrum analysers in the lab are often based on refracting the e-m radiation (say through a prism) and looking at the rainbow of rays that emerge. Such pieces of equipment can be too big to get in the oven and very likely to melt etc. Obviously the equipment must be such that it doesn't significantly interfere with the properties of the oven (e.g. it had better be a blackbody itself, very small and avoid removing too much of the radiation during the sampling process). You may be unable to control the spectrum anlayser remotely or get data from it, radio won't get through a typical oven wall and you must not flood the cavity with e-m radiation of that range because that will obviously interfere with the results, so the device may need to operate automatically and record the data for later retrieval. Overall quite a demanding set of criteria for the spectrum analyser. I'm not much of a practical scientist but I'll guess that such equipment isn't available for standard issue from the storeroom? (question mark because you've got to be hopeful).
Best Wishes.
LATE EDITING: Someone added the tag "microwave oven" to this thread. I have reduced that to "oven". Microwave ovens are not a good example of blackbody radiation existing inside the cavity and are only relevant when properties like standing waves were being discussed.
Post by: alancalverd on 10/03/2023 16:21:10
*mathematicians and theoretical physicists may like cubes, but engineers ask awkward questions about what happens in the corners, so physicists prefer to use spherical shells.
Post by: evan_au on 10/03/2023 21:57:51
- This allows selection of one mode amongst many possibilities
- By selecting different modes in different devices, you can have many different devices transmitting on the same optical fiber, increasing the fiber capacity by a factor of 10 to 100: "Wavelength Division Multiplexing"
- If the independent wavelengths are closer together, then better control over wavelength accuracy and stability is needed (ie more cost)
https://en.wikipedia.org/wiki/Wavelength-division_multiplexing
Today, Erbium-Doped Fiber Amplifiers (EDFA) are used to amplify a wide range of wavelengths (other exotic elements like Thulium, Praseodymium and Ytterbium are also usable).
- This is effectively a laser without a cavity, and it is able to amplify all wavelengths in a Wavelength Division Multiplexing system.
- See EDFA at https://en.wikipedia.org/wiki/Optical_amplifier
Post by: Bored chemist on 11/03/2023 11:53:23
if we look inside an ideal carbon spherical shell* at red heat, the tiniest surface imperfection will alter the Planck resonance spectrum by adding or subtracting wavelengths equal to arbitrary multiples of the height of the imperfection,No, it won't.
That's why (or because, depending on how you look at it) they use graphite; it's black.
You can not get cavity resonances if the walls are black.
Post by: alancalverd on 11/03/2023 17:45:41
Post by: Zer0 on 11/03/2023 20:33:17
Best Wishes.
LATE EDITING: Someone added the tag "microwave oven" to this thread. I have reduced that to "oven". Microwave ovens are not a good example of blackbody radiation existing inside the cavity and are only relevant when properties like standing waves were being discussed.
Thanks for Correcting it & providing an Explanation for the same.
Post by: Bored chemist on 11/03/2023 22:32:49
So Planck was wrong after all.No, he was right, which is why he didn't say what shape or size the cavity was; he knew it didn't matter as long as it was black.
Post by: hamdani yusuf on 12/03/2023 14:13:44
Blackbody radiation is usually considered (theoretically modelled and suitable equations derived) by looking at the radiation that can exist inside a cavity or an "oven". The interior of the oven can support various modes of radiation within it, each mode has a particular frequency, those frequencies ultimately assumed to be caused (generated by) charged particles oscillating at that frequency in the walls of the oven.The effect of standing wave to radiation spectrum is only significant when the oven walls are highly reflective. Otherwise, the effect of standing wave, and shape of the oven, would be miniscule.
Candle soot is highly absorbtive/emissive, even when it's only applied to a flat surface. It absorbs light and reemit electromagnetic radiation close to black body spectrum.
Post by: Eternal Student on 12/03/2023 15:42:04
@evan_au, your point about lasers and fibre optic cables is understood and appreciated. I have glanced at the various links you provided but not spent hours on it - it seems slightly tangent. Overall, I can appreciate that they (laser and fibre experts) make a lot of use of resonance effects from cavities and chambers.
@alancalverd and @Bored chemist , the effect of imperfections on the surface of a chamber is not all that clear to me. For a graphite wall it may not be important but I don't know, that's half the point of this thread and the original question.
So Planck was wrong after all.Yes, it looks like it, although possibly not directly about the imperfections in the graphite wall but over some things, yes.
Based on the information I have found recently Planck made a number of assumptions which were later revised, some by Planck himself, others by people who came much later. The modern presentation of the derivation of blackbody radiation that appears in textbooks is, to my best knowledge and belief, often an awkward mesh of the original ideas and some new ones to produce something based on more reasonable and often much more simply explained assumptions.
Examples:
1. Planck assumed radiation of (angular) frequency ω would only be generated by harmonic oscillators in the walls that oscillate at the same same frequency.
2. Most sources of information state: He also assumed the radiation was generated by transitions between energy levels of those oscillators.
3. One of the key assumptions, which every source of information agrees on, is that the energy levels (of the oscillators) were quantised and came in packets of size ħω.
Those assumptions get re-worked and re-phrased in many modern textbooks. With modern science it is possible to understand that classical electromagnetism is not always the best way to go. While you would have wanted an oscillator to be a charged particle oscillating at frequency ω to generate e-m radiation of (angular) frequency ω, if you understand that the things generating the radiation were atoms and not just arbitrary charged particle oscillators, then assumption 2 can point you in a different direction. Radiation of a given frequency ω could be generated by a transition of energy levels in many different ways: It could have been an electron jumping just one electron orbital, or a different atom having an electron jumping two orbitals all in one go, or a different atom jumping 3, etc. etc. The requirement for there to be a thing in the walls that is an oscillator with frequency ω starts to look a bit unnecessary or abstract. You can do reasonably well replacing the density of modes (of radiation supported in the oven) with the notion of there being more ways that radiation of frequency ω can be generated when ω increases.
Another common variation on the assumptions made is that you just assume the radiation (in the oven) must have quantised packets of energy, ħf. (Just to be clear, this is different from assuming the oscillators in the walls were quantised). Then the Boltzmann distribution can be applied directly to each mode of radiation instead of applying it to the oscillators in the walls. This seems to be the way many of the modern texts have decided to go and it has some advantages, in particular you don't need many (if any) assumptions about oscillators existing in the walls. However, to apply the Boltzmann distribution you do really want a "mode" of radiation to be a thing that exists. You will want to treat a "mode" of radiation as being a small system all of its own that is in thermal equilibrium with a larger system at temperature T (the larger system being the rest of the oven including all other modes of radiation EXCEPT the one mode you are working with at the time).
The effect of standing wave to radiation spectrum is only significant when the oven walls are highly reflective. Otherwise, the effect of standing wave, and shape of the oven, would be miniscule.Possibly correct and almost exactly what I had though originally. However, the following two things must be noted:
1. There is some need to define what a "mode" of radiation should be, especially with the assumptions made in most modern textbook derivations. A "mode" is usually explained as being a frequency of radiation that can be sustained within the oven, i.e. a standing wave.
2. Comments made by @evan_au et.al. there do seem to be examples where standing waves become very important and do effectively eliminate other frequencies of radiation (lasers, fibre optics etc.)
Best Wishes.
Post by: Bored chemist on 12/03/2023 18:11:55
The emission line from a laser can be subject to "mode hopping" because, as its temperature changes, the wavelengths reinforced by resonance also chance. The Fabry Perot resonator depends critically on the size and shape of the cavity.
With a black body cavity, the light can't know how far way the other side is, and the spectrum depends on the temperature.
If I get a box with mirror walls and put a police car in it with the lights flashing, that light is still blue.
Post by: Eternal Student on 13/03/2023 01:45:40
As (almost) always, the answer is "it depends".Yes, seems very likely.
If I get a box with mirror walls and put a police car in it with the lights flashing, that light is still blue.You've answered that yourself.
1. Yes, it should be blue for a low powered light bulb shining through some slightly absorbing media and with imperfect reflective walls etc. That is to say in our everyday experience.
2. However, it should be very different under idealised conditions OR when the experiment is adjusted to reduce the effects of the real world noise. Make the chamber a vaccum, boost the blue light output, use good mirrors and wait a few milliseconds (depending on how big the chamber is and how long light takes to reach the walls and start reflecting after you switched the light bulb on). Then you do (or should) have a situation like the Fabry-Perot interferometer that you mentioned.
Overall, I've probably not said anything you (Bored Chemist) didn't know but I've just turned the presentation upside down. My last few lines end with the impression that the theory matters and everday experience does not always win out.
Best Wishes.
Post by: Bored chemist on 13/03/2023 08:51:36
Then you do (or should) have a situation like the Fabry-Perot interferometer that you mentioned.Not quite.
There's no population inversion (as you would find in a laser) and thus no mechanism for amplifying the light that happens to"fit" between the mirrors.
So there's no reason for any particular wavelength to be favoured.
Odd things would happen because the emitter would overheat, but that's a different kettle of fish and not strongly dependent on the shape or size of the enclosure..
Post by: evan_au on 13/03/2023 09:14:24
Quote from: bored chemist
police car in it with the lights flashingIf it is an old-style police car with an incandescent filament in its blue light, the filament is acting like a black body radiator.
- But Tungsten is probably not quite as good as an idealised black body as Graphite...
Post by: Bored chemist on 13/03/2023 16:10:58
the filament is acting like a black body radiator.The blue lamp isn't.
Post by: Eternal Student on 13/03/2023 18:47:34
There's no population inversion (as you would find in a laser)...Sorry, I'm fairly sure that is not really required. Lasers are often used just to provide a good clean monochromatic source of light but there's nothing about the laser or population inversion that needs to be involved other than that. This is most easily seen just by having the laser outside the interferometer and tilted at some small angle to the mirrored surfaces so that there is no path along which any kind of return ray of light could be passed back into the laser and influence it.
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Image based on: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fabry.html
Shine a monochromatic source of light into one side of a Fabry- Perot interferometer and then set the gap correctly and you get interesting results - exactly what the source of that monochromatic light was doesn't matter. Depending on the gap and angle of incidence of the incoming ray, you can have all, some or none of the light being emitted out of the other side and brought to focus on the screen, along with there being no, some, or lots of light to be found inside the gap between the mirrors. In situations where the light is effectively eliminated, the interferometer itself can get hot (the air inside it and/or the mirrored walls of it will get hot) the laser (if that was the original source) should not be affected by what is happening in the interferometer.
Best Wishes.
Post by: Bored chemist on 13/03/2023 19:30:21
Hi and thanks for everyone's time.Your etalon is not very different in one way from a bit of blue glass.There's no population inversion (as you would find in a laser)...Sorry, I'm fairly sure that is not really required. Lasers are often used just to provide a good clean monochromatic source of light but there's nothing about the laser or population inversion that needs to be involved other than that. This is most easily seen just by having the laser outside the interferometer and tilted at some small angle to the mirrored surfaces so that there is no path along which any kind of return ray of light could be passed back into the laser and influence it.
[ Invalid Attachment ]Image based on: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fabry.html
Shine a monochromatic source of light into one side of a Fabry- Perot interferometer and then set the gap correctly and you get interesting results - exactly what the source of that monochromatic light was doesn't matter. Depending on the gap and angle of incidence of the incoming ray, you can have all, some or none of the light being emitted out of the other side and brought to focus on the screen, along with there being no, some, or lots of light to be found inside the gap between the mirrors. In situations where the light is effectively eliminated, the interferometer itself can get hot (the air inside it and/or the mirrored walls of it will get hot) the laser (if that was the original source) should not be affected by what is happening in the interferometer.
Best Wishes.
It transmits some wavelengths better than others.
But it's different in another way.
It doesn't (ideally) absorb energy.
And it can't change the wavelength of light.
so, if your cavity resonates at 500 nm and you feed in light at 501 what you get out is at 501nm.
There's a problem with the definition of a perfect mirror already discussed at futile length in this thread.
https://www.thenakedscientists.com/forum/index.php?topic=82373.0
Post by: hamdani yusuf on 13/03/2023 22:03:49
The blue color is then produced by filtering out lower frequency light. Which makes the spectrum no longer resemble black body radiation.Quote from: bored chemistpolice car in it with the lights flashingIf it is an old-style police car with an incandescent filament in its blue light, the filament is acting like a black body radiator.
- But Tungsten is probably not quite as good as an idealised black body as Graphite...
Post by: Eternal Student on 15/03/2023 12:46:43
@Bored chemist said something like this:
Etalons / F-P interferometers don't change the frequency of light.
---> Agreed but it can be enough that they will reduce the amplitude of light found between the mirrors if frequency ≠ a resonance just because there will be a lot of destructive interference.
Recall that a blackbody spectrum can have some blue light, it just can't be too much.
Moving to a general example instead of worrying about the F-P interferometer too much.....
Suspend a strong γ emitting radioisotope inside an oven and you would have thought the spectrum of radiation inside the oven would be seriously perturbed. I mean you've got plenty of high energy γ frequency radiation in there and it doesn't seem to ask or be concerned about whether you can fit a standing wave of that frequency inside the oven cavity.
However, the usual blackbody spectrum calculations (where only standing waves are considered as modes) only claims to predict the spectrum inside the oven AT EQUILIBRIUM. You're not at an equilibrium situation at the moment. Assume the walls of the oven are capable of fully absorbing the γ rays and wait a while. Two things happen:
(i) The walls get warm because they are absorbing γ rays. This begins to shift the predicted blackbody spectrum to somewhere where some measurable intensity of γ frequencies is expected.
(ii) Now that the walls are producing some γ rays, you can get some destructive interference with those produced by the radioisotope, lowering the overall intensity of γ frequencies.
Those two things together can get the final spectrum, at equilibrium, back in reasonable agreement with a blackbody spectrum.
Now to my mind, it still doesn't really explain WHY we do the calculations for a blackbody spectrum considering "modes" as being ONLY that which can be supported as a standing wave but it does seem to suggest that you CAN make those assumptions and you will still get the right spectrum in the end.
That was rushed and may not have linked smoothly with previous comments, sorry. I seem to have some other things to do and may not respond or put "likes" on future comments for a while, sorry.
Best Wishes.
Post by: Bored chemist on 15/03/2023 13:08:48
it can be enough that they will reduce the amplitude of light found between the mirrorsThey don't absorb it; where has it gone?
Now that the walls are producing some γ rays, you can get some destructive interference with those produced by the radioisotope, lowering the overall intensity of γ frequencies.Again; where has the energy gone?
Interference doesn't remove energy from the system
Post by: Eternal Student on 15/03/2023 17:32:46
They don't absorb it; where has it gone?Heat. The Etalons get hot when light is shone in one side but none is transmitted out of the other side. They can run much cooler if the gap is adjusted so that all the light is transmitted out through the other side.
----
I haven't done the experiment, or found any theory, for the oven / blackbody cavity situation but you'd guess for something similar.
Best Wishes.
Post by: Bored chemist on 15/03/2023 18:50:24
he Etalons get hot when light is shone in one side but none is transmitted out of the other side.In principle, they reflect it.
A prefect etalon doesn't absorb radiation. of course, in practice they do a bit.
At wavelengths where they absorb, they are a black body and their shape is irrelevant.
Post by: Bored chemist on 15/03/2023 18:54:35
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Post by: Eternal Student on 16/03/2023 20:44:45
Ummm.. yes some will be some light transmitted out that side. An earlier post mentioned that there was no return path to the laser, but there could be paths coming back to the left just not aimed at the original source, much as shown in your digram. So there is no disagreement that there could be some light coming out of the left side of the etalon.
Overall, the relevance or connection between etalons and the interior space of an oven is somewhat oblique or marginal. I'm not sure worrying about etalons is vital to the original issue. I was only considering that the interior of the oven has some comparisons with the interior of an etalon. In particular, changing the gap between mirrors of an etalon does seriously influence the amplitude of a given frequency of light that would be found inside the etalon. As you stated, it does not change that frequency of light to a new frequency - but simply reducing the amplitude can be important (as outlined in a previous post with a gamma-emitter in the oven).
Best Wishes.
Post by: Bored chemist on 16/03/2023 21:42:57
The power builds up until the loss through the partially reflecting mirror equals the input power.
But without a source inside the resonator inside an oven I can't see why that would happen.
The same mirror does a fine job of stopping radiation getting into the cavity.
Essentially, it's a bunch of mirrors.
They change the direction of light, but not the wavelength.
If you make the reflectance wavelength dependent- e.g. you use gold instead of silver then yes, you will change the spectrum; but that's not because of the shape of the cavity- it's because of the colour.
And, even then, I'm not sure it affects the spectrum. The gold will heat up. And it will do a better job of emitting blue light than red (because it does a better job of absorbing blue than red- That's Kirchhoff's law). Eventually, it will heat up until it's glowing as brightly as the walls.
If the walls of the oven are perfectly reflective then they must have zero absorbance and thus zero emissivity.
If they have some spectrally preferential absorbance then they have exactly the same preferential emittance.
So, if there's an enhanced emission from the south wall of the oven, but it's cancelled out when the light hits the north side of the oven and is preferentially absorbed,
Fundamentally, how does one wall know where the other wall is in order to change its emission spectrum?
Post by: hamdani yusuf on 17/03/2023 22:37:20
If you make the reflectance wavelength dependent- e.g. you use gold instead of silver then yes, you will change the spectrum; but that's not because of the shape of the cavity- it's because of the colour.The Kirchhoff's law is not generally true. A fluorescent matter can absorb ultraviolet light while emitting longer wavelength, e.g. green. But when white light is shone on it, it doesn't necessarily absorb the green nor emit ultraviolet.
And, even then, I'm not sure it affects the spectrum. The gold will heat up. And it will do a better job of emitting blue light than red (because it does a better job of absorbing blue than red- That's Kirchhoff's law). Eventually, it will heat up until it's glowing as brightly as the walls.
Post by: Bored chemist on 18/03/2023 00:10:45
The Kirchhoff's law is not generally true.In the limit, it's a restatement of the conservation of energy. It's true.
But when white light is shone on it, it doesn't necessarily absorb the green nor emit ultraviolet.Nobody said it did.
Post by: hamdani yusuf on 18/03/2023 03:25:18
To equate Kirchhoff's law with conservation of energy, some assumptions are required.The Kirchhoff's law is not generally true.In the limit, it's a restatement of the conservation of energy. It's true.But when white light is shone on it, it doesn't necessarily absorb the green nor emit ultraviolet.Nobody said it did.
The system is in equilibrium, which means no energy is transferred into nor out from it, other than electromagnetic radiation. Hence, the system isn't undergoing other forms of energy transfer/transformation, such as photoelectric effect, nuclear reaction, electric current, photochemistry, heat conduction, convection, phase changing, etc.
But the following statement is an overreaching stretch of Kirchhoff's law.
And it will do a better job of emitting blue light than red (because it does a better job of absorbing blue than red- That's Kirchhoff's law).
Good absorption of radiation in a certain frequency range doesn't necessarily mean good emission in that same range. Some materials can absorb UV and turn it into heat, which then radiate in mostly infrared light. It's not necessarily a good UV emitter.
(https://aerospace.basf.com/uploads/1/1/8/0/118080269/figure-basf-tinuvin-carboprotect-1_orig.png)
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Figure 1. Absorbance and transmittance of UV absorbers. Solutions of 20 mg/l (absorbance, left) and 80 mg/l (transmittance, right) in chloroform. Method: Perkin Elmer UV/VIS/NIR spectrophotometer Lambda 650. HPT = high performance triazine; BZT = benzo-triazole; the numbers in the graph on the right-hand side indicate the wavelengths where 50% transmittance is observed.
https://aerospace.basf.com/uv-absorber-technology.html
Post by: hamdani yusuf on 18/03/2023 04:16:46
Here's how molten gold looks like, around 5:30 time stamp.
It doesn't look bluer than molten silver.
Post by: Eternal Student on 18/03/2023 04:28:23
I've not had a lot of time to read the new posts. At a glance quite a lot is tangential to the original question - which is fine, I'm not troubled at all - it's just that I can post this bit without needing to address those other comments because there isn't any conflict etc.
I'm starting to find a few papers which suggest the shape of cavity does matter and can influence the spectrum of radiation within it. Sadly, a lot of them are behind paywalls. Best I can do is pull some details out of the abstracts because I'm not going pay for access or get to a good academic library for at least a month.
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....Under the assumption that the cavity is a spherical one, the intensity of the blackbody radiation at some frequency is obtained and found to be uniform only in a small region around the center of the cavity. With the help of the theorem of equipartition, the intensity, or the spectrum of the blackbody radiation, is then expressed as a function of the temperature of the cavity and shown to satisfy the familiar Rayleigh–Jeans’ law. Some other properties of the blackbody radiation are also discussed.Taken from: "Non-uniform distribution of low-frequency blackbody radiation inside a spherical cavity", Journal of the Optical Society of America A Vol. 37, Issue 9, pp. 1428-1434 (2020)
Other than the obvious issue they mention with a spherical cavity, it would also be interesting to see how they derived the Rayleigh-Jeans law. That is usually derived classically and it is only an approximation to the Planck's law (the usual description of Blackbody radiation).
- - - - - - -
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By taking into account all of the standing electromagnetic wave frequencies inside cubical and spherical cavities, generalized expressions for the spectral and total radiation from cubical and spherical blackbodies are derived. It is found that the Stefan–Boltzmann law becomes valid only when χT≫hc/k, where χ denotes the length of a cube edge or the diameter of the sphere, T is the blackbody temperature, h is Planck’s constant, c is the speed of light, and k is Boltzmann’s constant. When χT≲hc/k, the radiated power per unit area is less than that predicted by the SB law.Taken from: "Blackbody radiation from cubes and spheres with application to rapid solidification of microspheres", Journal of Applied Physics 56, 1347 (1984)
Basically, the size of the cavity does seem to matter. If they were radiating like Blackbodies following Planck's law then the Stefan-Boltzmann law would follow.
- - - - - - - - - - - -
There's a few other examples but overall, it seems that -
(i) Very small cavities will deviate from the usual spectrum and,
(ii) the low frequency radiation in some larger but non-cubical shaped cavities seems to show some deviation.
Best Wishes.
Post by: hamdani yusuf on 18/03/2023 04:50:34
Is other type of energy transfer, such as conduction and convection controlled?
Post by: Bored chemist on 18/03/2023 11:47:57
Here's how molten silver looks like, around 4:35 time stamp.Imagine thinking that you can do spectroscopy by looking at a TV picture.
Here's how molten gold looks like, around 5:30 time stamp.
It doesn't look bluer than molten silver.
Obviously, it should look bluer; it's a bit hotter.
Post by: Bored chemist on 18/03/2023 11:52:58
The system is in equilibriumThat assumption has been made tacitly or explicitly throughout the thread.
So, yes, fluorescent things are odd but gold doesn't normally fluoresce.
Post by: Bored chemist on 18/03/2023 11:55:37
Basically, the size of the cavity does seem to matter.That's interesting.
My question stands.
Fundamentally, how does one wall know where the other wall is in order to change its emission spectrum?
Post by: Eternal Student on 18/03/2023 16:46:39
What's the material of the cavity wall?The articles are described as concerning blackbody radiation and could be theoretical BUT they could also involve some experimentation. I can't answer further questions, as stated earlier, those articles are paywalled and I'm not going to a suitable library for a while.
- - - - - - - - - - - - -
Fundamentally, how does one wall know where the other wall is in order to change its emission spectrum?1. Why does it need to? The amount of radiation of frequency f can grow if that frequency can be supported as a standing wave inside the cavity.
Conversely if it can't be supported as a standing wave then it can't grow. Only an amount of radiation equal to (proportional to) the rate of production times the flight time before it hits a wall can exist inside the cavity. Making the cavity bigger (moving the walls further away) shouldn't matter because it's the energy density (energy per unit volume of the cavity) that is considered for the blackbody spectrum calculations and you'd only be making the volume bigger in proportion to the increased flight time. So a classical interpretation may argue that frequencies not capable of forming standing waves never become significant.
2. For all I know, you were starting a discussion about more general aspects of Quantum Mechanics. A photon is a QM particle and as such its properties may not be local. To say this a different way, if we use something like QFT then a photon should be an excitation in the underlying photon field. If the space available or boundary conditions are such that no solution exists that looks like a photon of frequency f located at a given place then nothing that looks like a photon of frequency f can be found in that place. It wouldn't matter if the restrictive boundary conditions were located on the other side of the galaxy.
The issue you were describing is only an issue if you consider a photon as being something that an atom can create inherently or independantly of the rest of the universe.
Best Wishes.
Post by: hamdani yusuf on 20/03/2023 04:13:31
Imagine thinking that you can do spectroscopy by looking at a TV picture.Cameras and TV screens were designed to represent human vision. They are good enough to distinguish the absorption spectrum of gold from silver. If their difference in emission spectrum of gold and silver is proportional to the difference in absorption spectrum, then cameras and TV screens should be able to distinguish them as well.
Post by: hamdani yusuf on 20/03/2023 04:38:40
Kirchoff's law only applies in more restrictive conditions than conservation of energy.The system is in equilibriumThat assumption has been made tacitly or explicitly throughout the thread.
So, yes, fluorescent things are odd but gold doesn't normally fluoresce.
the system isn't undergoing other forms of energy transfer/transformation, such as photoelectric effect, nuclear reaction, electric current, photochemistry, heat conduction, convection, phase changing, etc.
Metamaterials and nanotechnology can produce anomalous spectra of absorption and emission.
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Kirchoff's Law of RadiationIt doesn't say that emission spectrum of a material is proportional to its absorption spectrum.
At a given temperature, the ratio of the emissive power of a body to its absorptive power is constant and is equal to the emissive power of a black body at the same temperature.
Alternative statement of Kirchhoff's law: At any given temperature, the emissivity of a body is equal to its coefficient of absorption.
https://www.toppr.com/ask/content/concept/kirchoffs-law-of-heat-radiation-and-its-theoretical-proof-209979/
Water is a good absorber in microwave, but ice is not. AFAIK, they are not good emitter of microwave.
Post by: Bored chemist on 20/03/2023 08:56:56
Cameras and TV screens were designed to represent human vision. They are good enough to distinguish the absorption spectrum of gold from silver.Really?
If that's relevant you can tell me what metal is being cast here.
casting.png (199.58 kB . 386x245 - viewed 1105 times).So... what is it?
Post by: hamdani yusuf on 20/03/2023 09:27:46
You are confusing between emission and absorption spectrum, because you believe that they must be the same.Cameras and TV screens were designed to represent human vision. They are good enough to distinguish the absorption spectrum of gold from silver.Really?
If that's relevant you can tell me what metal is being cast here.
So... what is it?
Post by: Bored chemist on 20/03/2023 12:45:47
The material is not fluorescing, undergoing a chemical reaction or exhibiting nuclear decay.You are confusing between emission and absorption spectrum, because you believe that they must be the same.Cameras and TV screens were designed to represent human vision. They are good enough to distinguish the absorption spectrum of gold from silver.Really?
If that's relevant you can tell me what metal is being cast here.
So... what is it?
So Kirchhoff's law applies.
The absorption spectrum is the same as the emission spectrum.
So, what's the metal?
Post by: hamdani yusuf on 20/03/2023 13:00:15
The absorption spectrum is the same as the emission spectrum.How do you know?
What's your reference?
Take a look at this.
(https://www.researchgate.net/profile/Roman-Sobolewski/publication/48308616/figure/fig3/AS:339872926715927@1458043428951/Schematic-diagram-of-absorption-and-emission-spectra-of-Ti-3-as-impurities-in-a.png)
1. Schematic diagram of absorption and emission spectra of Ti 3+ as impurities in a sapphire.
https://www.researchgate.net/figure/Schematic-diagram-of-absorption-and-emission-spectra-of-Ti-3-as-impurities-in-a_fig3_48308616
Post by: Bored chemist on 20/03/2023 13:03:28
How do you know?The conservation of energy.
You keep trying or pretend that Kirchhoff's law doesn't exist.
But it's beside the point.
You said that a camera and monitor allowed you to do spectroscopy in the context of the emissions from molten metal.
Well- can you?
What's the metal
What's your reference?If someone quotes newton's laws do you say "what's the reference"?
Post by: hamdani yusuf on 20/03/2023 13:50:42
If someone quotes newton's laws do you say "what's the reference"?Principia.
Post by: Bored chemist on 20/03/2023 17:24:25
Post by: hamdani yusuf on 20/03/2023 17:31:56
The conservation of energy.This is not Kirchhoff's law.
You keep trying or pretend that Kirchhoff's law doesn't exist.
The material is not fluorescing, undergoing a chemical reaction or exhibiting nuclear decay.
So Kirchhoff's law applies.
The absorption spectrum is the same as the emission spectrum.
This is.
Kirchoff's Law of Radiation
At a given temperature, the ratio of the emissive power of a body to its absorptive power is constant and is equal to the emissive power of a black body at the same temperature.
Alternative statement of Kirchhoff's law: At any given temperature, the emissivity of a body is equal to its coefficient of absorption.
https://www.toppr.com/ask/content/concept/kirchoffs-law-of-heat-radiation-and-its-theoretical-proof-209979/
The absorption spectrum is the same as the emission spectrum for low pressure gases. But for other type of materials, it doesn't generally apply, as shown in a counterexample here.
Take a look at this.
(https://www.researchgate.net/profile/Roman-Sobolewski/publication/48308616/figure/fig3/AS:339872926715927@1458043428951/Schematic-diagram-of-absorption-and-emission-spectra-of-Ti-3-as-impurities-in-a.png)
1. Schematic diagram of absorption and emission spectra of Ti 3+ as impurities in a sapphire.
https://www.researchgate.net/figure/Schematic-diagram-of-absorption-and-emission-spectra-of-Ti-3-as-impurities-in-a_fig3_48308616
Post by: hamdani yusuf on 20/03/2023 17:38:21
What's the metal?I don't know. There's not enough information to distinguish one from the others.
Let's apply some basic logic.
Cameras can distinguish between absorption spectrum of gold and silver.
Cameras cannot distinguish between emission spectrum of gold and silver.
Conclusion: emission spectrum can be different from absorption spectrum.
Post by: Bored chemist on 20/03/2023 17:42:32
But you said there was.What's the metal?I don't know. There's not enough information to distinguish one from the others.
They are good enough to distinguish the absorption spectrum of gold from silver. If their difference in emission spectrum of gold and silver is proportional to the difference in absorption spectrum, then cameras and TV screens should be able to distinguish them as well.
Post by: Bored chemist on 20/03/2023 17:50:12
Kirchoff's Law of RadiationOK.
At a given temperature, the ratio of the emissive power of a body to its absorptive power is constant and is equal to the emissive power of a black body at the same temperature.
So, it's true at any temperature.
So that means it's true when the object is red hot- mainly emitting red and IR. And it's also true when the object is so hot that it's emitting blue light.
For it to be true regardless of what wavelengths are being emitted, it must be true at all wavelengths individually.
So the spectra must be the same.
(Unless you pick something weird like a fluorescent material)
So why have you chosen to pretend that Ti doped sapphire is similar to molten metal?
(Incidentally, if you don't disperse the Ti+++ ions into a sapphire lattice, you get Ti2O3 which is black)
Post by: Bored chemist on 20/03/2023 17:56:10
Let's apply some less basic logic.What's the metal?I don't know. There's not enough information to distinguish one from the others.
Let's apply some basic logic.
Cameras can distinguish between absorption spectrum of gold and silver.
Cameras cannot distinguish between emission spectrum of gold and silver.
Conclusion: emission spectrum can be different from absorption spectrum.
Correct conclusion:
The camera doesn't have good enough colour rendering when faced with something that bright.
It is further doomed to failure because the temperatures are different.
Just because a particular piece of equipment is not sensitive enough to detect an effect, that does not mean that the effect is not there. It means you ae not using the right equipment.
All the spectroscopists know about Kirshhoff, so there's no reason to keep citing the details.
Everyone knows that an absorption spectrum typically looks like an emission spectrum.
So, they only publish examples of the interesting cases where the rule doesn't apply- for example- with fluorescence.
Unfortunately, that means that, if you look at the publications, they will mislead you into thinking that the the spectra are usually different.
Post by: hamdani yusuf on 21/03/2023 12:05:04
They can distinguish the difference in absorption spectrum.But you said there was.What's the metal?I don't know. There's not enough information to distinguish one from the others.They are good enough to distinguish the absorption spectrum of gold from silver. If their difference in emission spectrum of gold and silver is proportional to the difference in absorption spectrum, then cameras and TV screens should be able to distinguish them as well.
Is this false?
Can't you see the difference?
(https://www.ino.com/blog/wp-content/uploads/2020/10/Gold-Silver.jpg)
Post by: hamdani yusuf on 21/03/2023 13:13:11
So, it's true at any temperature.Your conclusion was, molten gold emits more blue light than molten silver, because gold absorbs more blue light when they are cool.
So that means it's true when the object is red hot- mainly emitting red and IR. And it's also true when the object is so hot that it's emitting blue light.
Post by: Bored chemist on 21/03/2023 13:22:09
They can distinguish the difference in absorption spectrum.But you said there was.What's the metal?I don't know. There's not enough information to distinguish one from the others.They are good enough to distinguish the absorption spectrum of gold from silver. If their difference in emission spectrum of gold and silver is proportional to the difference in absorption spectrum, then cameras and TV screens should be able to distinguish them as well.
Is this false?
Can't you see the difference?
(https://www.ino.com/blog/wp-content/uploads/2020/10/Gold-Silver.jpg)
We can all see the difference.
But that difference is between the reflectance spectra, not absorbance or emission..
I don't think any of us can see the relevance of your original pictures of the molten metals.
Post by: Bored chemist on 21/03/2023 13:26:44
Your conclusion was, molten gold emits more blue light than molten silver, because gold absorbs more blue light when they are cool.My view is that the electronic band structures don't change much on heating to the melting point(s).
And we know that silver is a good reflector.
We know (from the conservation of energy) that that means it can not be a good absorber.
And we know (from Kirchhoff's work) that it can not be a good emitter.
For red light, the same is true of gold.
But, for blue light we know that gold is not a good reflector.
It is a relatively good absorber and we therefore know it is a relatively good emitter.
None of this is anything other than 19th century physics.
Why are you arguing about stuff that has been known for all that time?
Post by: alancalverd on 21/03/2023 14:07:25
Also worth noting that digital image receptors (cameras, phones, etc) use fixed filters to replicate human visual response in addition to any removable physical or computational filters that the user might add.
Post by: hamdani yusuf on 21/03/2023 15:49:21
We can all see the difference.What happens to the unreflected light?
But that difference is between the reflectance spectra, not absorbance or emission..
Post by: Bored chemist on 21/03/2023 17:55:05
We can all see the difference.What happens to the unreflected light?
But that difference is between the reflectance spectra, not absorbance or emission..
And we know that silver is a good reflector.There are two sorts of people in the world.
We know (from the conservation of energy) that that means it can not be a good absorber.
One sort can extrapolate from incomplete data.
Post by: hamdani yusuf on 22/03/2023 13:21:21
Let me help you to answer my question, just in case you are having difficulty.We can all see the difference.What happens to the unreflected light?
But that difference is between the reflectance spectra, not absorbance or emission..
It's absorbed. Which means that absorbed spectra between gold and silver in room temperature are different.
Post by: Bored chemist on 22/03/2023 17:11:42
Let me help you to answer my question,I did.
just in case you are having difficulty.I wasn't.
absorbed spectra between gold and silver in room temperature are different.That's exactly why I referred to the use of gold.
Let me know when you catch up.
My view is that the electronic band structures don't change much on heating to the melting point(s).
And we know that silver is a good reflector.
We know (from the conservation of energy) that that means it can not be a good absorber.
And we know (from Kirchhoff's work) that it can not be a good emitter.
For red light, the same is true of gold.
But, for blue light we know that gold is not a good reflector.
It is a relatively good absorber and we therefore know it is a relatively good emitter.
None of this is anything other than 19th century physics.
Why are you arguing about stuff that has been known for all that time?