[Eternal Student (OP)]

Hi.

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.

[Bored chemist]

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..

[evan_au]

police car in it with the lights flashing

If 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...

[Bored chemist]

the filament is acting like a black body radiator.

The blue lamp isn't.

[Eternal Student (OP)]

Hi and thanks for everyone's time.

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.


Fabry-Perot.jpg (74.49 kB . 783x480 - viewed 1319 times)

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.

[Bored chemist]

Hi and thanks for everyone's time.

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.

Your etalon is not very different in one way from a bit of blue glass.
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

[hamdani yusuf]

police car in it with the lights flashing

If 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...

The blue color is then produced by filtering out lower frequency light. Which makes the spectrum no longer resemble black body radiation.

[Eternal Student (OP)]

Hi.

@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.

[Bored chemist]

it can be enough that they will reduce the amplitude of light found between the mirrors 

They 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

[Eternal Student (OP)]

Hi.

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.

[Bored chemist]

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.

[Bored chemist]

The diagram you cited is misleading; it forgets where the light goes.

etalon.png (54.24 kB . 201x224 - viewed 1226 times)

[Eternal Student (OP)]

Hi.

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.

[Bored chemist]

With a laser the power bouncing back and to between the mirrors will be much higher than the output power.
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?

[hamdani yusuf]

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.

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.

[Bored chemist]

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.

[hamdani yusuf]

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.

To equate Kirchhoff's law with conservation of energy, some assumptions are required.
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.

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

[hamdani yusuf]

Here's how molten silver looks like, around 4:35 time stamp.

Here's how molten gold looks like, around 5:30 time stamp.

It doesn't look bluer than molten silver.

[Eternal Student (OP)]

Hi.

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.

   

....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).

- - - - - - -

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.

[hamdani yusuf]

What's the material of the cavity wall?
Is other type of energy transfer, such as conduction and convection controlled?