[hamdani yusuf (OP)]

In particular, there are some extra assumptions that seem essential and must be added to the mathematical model to correctly predict that very small and peculiarly shaped cavities will still ultimately produce the usual BB spectrum.

Can you specify what are they?

[hamdani yusuf (OP)]

I smell a philosopher! Indeed, only God can answer the question "why", and as there is no God, the question is meaningless.  But if you want to know "how", I refer to the three hon gents I mentioned earlier.

Learning is basically information collection, filtration, and compression process.
When we find a physical observation, we can accept that as the natural process of involved objects in that particular situation. Alternatively, we can try to find underlying basic principles which can be used to reproduce that observation, given some reasonable model and assumptions.
Keppler's laws of planetary motions essentially compressed Tycho's table of planetary observations. Newton's universal gravitation compressed them even further.

[Bored chemist]

Incidentally, hot hydrogen atoms do have a strong absorption for red light.
It's the hydrogen alpha line looked at from the other point of view.

How hot or cold does it take to produce absorption spectrum?

How strongly do you want it to absorb?

Hot enough to significantly populate the first excited state.
How come you don't know that?

[hamdani yusuf (OP)]

How strongly do you want it to absorb?

Hot enough to significantly populate the first excited state.
How come you don't know that?

Strong enough to be detected by common phone camera unambiguously.
Can it be done under room temperature?

Your statement suggests that there's a minimum temperature limit to show the absorption spectrum, contrary to the pictures which suggest that there's a maximum temperature limit instead. What's your reference?

[hamdani yusuf (OP)]

Incidentally, hot hydrogen atoms do have a strong absorption for red light.
It's the hydrogen alpha line looked at from the other point of view.

How hot or cold does it take to produce absorption spectrum?

http://www.4college.co.uk/as/el/how.gif

https://www.daviddarling.info/images/types_of_spectra.jpg


https://sites.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect6/figure05-14.jpg

Instead of absorption, the pictures look more like scattering effect.
I've tried to find the video version of the experiment shown in the pictures above. Unfortunately I can't find it on YouTube. I can only find some experiment of hydrogen emission spectrum using hydrogen lamp.
But I'll take it as an opportunity to make the video myself. Let me hear your suggestions and advices.

To produce hydrogen gas, I think I can use electrolysis of dilute sodium hydroxide solution. Note that pure water doesn't conduct electricity well, while salt water produces dangerous chlorine gas as byproduct.

Torricelli's method using tall tube can be used to create partial vacuum or low pressure gas.

Sunlight focussed by a convex lens or concave mirror can be used as the bright light source.

[Bored chemist]

Sodium carbonate is less of a hazard than the hydroxide but in either case you will be here until doomsday trying to scrub salt spray out of the hydrogen or you will get a yellow flame.

How strongly do you want it to absorb?

Hot enough to significantly populate the first excited state.
How come you don't know that?

Strong enough to be detected by common phone camera unambiguously.
Can it be done under room temperature?

Your statement suggests that there's a minimum temperature limit to show the absorption spectrum, contrary to the pictures which suggest that there's a maximum temperature limit instead. What's your reference?

You have not understood the pictures.
You seem to have read "cooler gas" as if it says "cool gas". Quite often, the "cool" gas is an oxyacetylene flame.
The only requirement there is that it's cooler than the hot black-body which is providing the radiation.

Your statement suggests that there's a minimum temperature limit to show the absorption spectrum

As I already pointed out.
It depends.
It depends on the answer to this

How strongly do you want it to absorb?

And you failed to answer it. Why is that?

This tells you about flame absorption spectra.
https://www.agilent.com/en/support/atomic-spectroscopy/atomic-absorption/flame-atomic-absorption-instruments/how-does-aas-work-aas-faqs

[Eternal Student]

Hi.

Thanks for your time and suggestion @Bored chemist .     I've spent a bit of time considering your idea.    I've also written two replies but none of them were short.   This is the third attempt with brevity as high priority.

You could make it out of small cube shaped BB (for which the maths is known)

    The general idea of building a new (bigger) shape by joining some shapes which you already know will produce a BB spectrum together (face to face etc.) is reasonably sound.

And you can then rescale the problem - effectively using a different unit of length-

    Yes, there are some places in the usual mathematical method for deriving the BB spectrum formula where scale is fairly arbitrary.   The main problem is just that you could be magnifying the errors in the approximations that are made in that method.   
   
     The standard mathematical method for deriving the Planck formula pivots around assuming two things that are actually contradictory (an assumption for the first half, a different one for the second half):   1.  Start by assuming you will only have only a discrete set of supported modes.  You need this to estimate a "Density of modes" function.
    2.  Now you assume you have a continuous distribution of supported frequencies which is given by this "density of modes" function.   Stop assuming you had a discrete distribution, throw that away.

    Those two steps or assumptions are only compatible or consistent enough  when  L (the size of the cavity) is large or λ (the wavelength of a supported mode) is small.   While we can argue that scale is arbitrary at some places in the mathematics, we do want to stay in some sensible region (of physical length L and wavelengths λ) where we can believe that both halves of the method will be reasonably true simultaneously.   (i.e. both assumptions are reasonable representations of what Nature is actually doing).
- - - - - - - - - - -

Can you specify what they are?    (extra assumptions for small cavities etc.)

    One of the things you'd like to do is to reduce the discrete nature of the frequencies for the supported modes and make them more spread out or continuous immediately.
    The spectrum we observe from experiment (even for small cavities) does tend to be fairly smooth and continuous rather than having well defined spikes like a discrete spectrum.  So it is clear that Nature is not doing exactly what the mathematical model assumed (about fitting standing waves in the cavity etc. - as in step1 or the first half of the method).    One adjustment we can make to the mathematical model is to relax the boundary conditions for the standing waves slightly   (the boundary conditions are that the E and B fields are 0 at the walls,  which was represented in your videos by insisting the wave returns to 0 displacement at the walls).   This modification is reasonably justified on physical grounds....
    Specifically, it seems that Nature does NOT see the wall of a cavity as a sharply defined hard wall.   For example, not all of the radiation is stopped or bounced back at the inner edge of the wall,  radiation of some frequencies may penetrate the wall slightly before being absorbed or bounced back etc.   Indeed that does happen and is verifiable in other experiments - e.g. X rays can penetrate most materials to some depth.
    We can fairly easily imagine that if we allow some penetration, then if a wave of precisely λ wavelength would have fitted inside the cavity, then we can actually assume some continuous range of wavelengths around λ are supported.
    We can do even more than this to further justify some more supported wavelengths:  We don't need to consider photons in the model but it's just easier to have this in your mind for a moment and imagine an e-m wave as some collection of photons where each photon could get bounced back at different places into the wall.   It provides some physical grounds to justify having a wave that is incident on the wall be reflected in small pieces and interesting ways at more than just one depth into the wall.  With the principle of superposition we can then maintain standing waves inside the cavity with a wavelength that is much longer than (twice) the length of the cavity.   This helps to mitigate the cut-off or maximum wavelength for supported modes normally governed by the length of the cavity (see above, λ_max = 2L ).
       This post is already too long so I've got to stop.   Basically, for very small cavities, we need to adjust assumptions in the mathematical model.   One of the most sensible adjustments is to allow more complicated support of modes (standing waves) inside the cavity (for example allowing penetration into the wall).

Best Wishes.

[hamdani yusuf (OP)]

And you failed to answer it. Why is that?

It looks like you fail to find my answer.

Strong enough to be detected by common phone camera unambiguously.

[hamdani yusuf (OP)]

This tells you about flame absorption spectra.
https://www.agilent.com/en/support/atomic-spectroscopy/atomic-absorption/flame-atomic-absorption-instruments/how-does-aas-work-aas-faqs

The sample shown in the article is Pb. Turning it into gas requires high temperature. If the element is already gaseous in room temperature, like hydrogen, the flame doesn't seem to be necessary.
The flame is not shown in the pictures I posted either.

[hamdani yusuf (OP)]

Specifically, it seems that Nature does NOT see the wall of a cavity as a sharply defined hard wall.   For example, not all of the radiation is stopped or bounced back at the inner edge of the wall,  radiation of some frequencies may penetrate the wall slightly before being absorbed or bounced back etc.   Indeed that does happen and is verifiable in other experiments - e.g. X rays can penetrate most materials to some depth.

Also, if the wall is not a good electric conductor, radio wave can penetrate it. I discussed the limitations of photon model in explaining electromagnetic radiation in another thread.

[Bored chemist]

And you failed to answer it. Why is that?

It looks like you fail to find my answer.

Strong enough to be detected by common phone camera unambiguously.

Opops!
Sorry, I missed that.
But I'm not sure it would be practically possible to detect with a phone camera, even under optimal conditions.

[Bored chemist]

This tells you about flame absorption spectra.
https://www.agilent.com/en/support/atomic-spectroscopy/atomic-absorption/flame-atomic-absorption-instruments/how-does-aas-work-aas-faqs

The sample shown in the article is Pb. Turning it into gas requires high temperature. If the element is already gaseous in room temperature, like hydrogen, the flame doesn't seem to be necessary.
The flame is not shown in the pictures I posted either.

Heating has three effects.
One is converting the material to vapour- which is not important for hydrogen.
Another is converting the material into atoms (rather than molecules).
That's important for hydrogen, but not with (for example) mercury.

And the third thing heating it does is raise some of the atoms to their first excited state.
The ground state of hydrogen atoms doesn't absorb visible light.

[hamdani yusuf (OP)]

Instead of absorption, the pictures look more like scattering effect.

It's unfortunate that the diagram doesn't mention temperature requirements for the depicted phenomena to be observed. Although the spectrum looks like Balmer series, which implies that the gas is hydrogen.

[hamdani yusuf (OP)]

Something like this.

I see no photons.

The video shows em radiation caused by macroscopic motion.
I didn't mention any photon. I didn't expect anyone to see it either.

[hamdani yusuf (OP)]

Instead of absorption, the pictures look more like scattering effect.

It's unfortunate that the diagram doesn't mention temperature requirements for the depicted phenomena to be observed. Although the spectrum looks like Balmer series, which implies that the gas is hydrogen.

Before I saw this diagram, I wondered where does the absorbed energy go. I guessed that some are transformed into heat, which is then dissipated to the environment.

[Bored chemist]

Instead of absorption, the pictures look more like scattering effect.

There's some similarity but...
There are two big differences.
Absorption warms up the absorber.
The timescales are different.
Scattering is much faster.


A less obvious difference is the spectrum. Shorter wavelengths are much more strongly scattered and all wavelengths are scattered at least slightly.

[Bored chemist]

Before I saw this diagram, I wondered where does the absorbed energy go. I guessed that some are transformed into heat, which is then dissipated to the environment.

Some is.

[hamdani yusuf (OP)]

The second one is quantization of atomic radiation frequency, which is observed in spectral line emission. Balmer discovered empirical formula to describe the spectral line emissions of the hydrogen atom. Bohr interpreted it as the evidence for the existence of atomic orbitals.

In atomic physics, the Bohr model or Rutherford?Bohr model of the atom, presented by Niels Bohr and Ernest Rutherford in 1913, consists of a small, dense nucleus surrounded by orbiting electrons. It is analogous to the structure of the Solar System, but with attraction provided by electrostatic force rather than gravity. In the history of atomic physics, it followed, and ultimately replaced, several earlier models, including Joseph Larmor's solar system model (1897), Jean Perrin's model (1901),[2] the cubical model (1902), Hantaro Nagaoka's Saturnian model (1904), the plum pudding model (1904), Arthur Haas's quantum model (1910), the Rutherford model (1911), and John William Nicholson's nuclear quantum model (1912). The improvement over the 1911 Rutherford model mainly concerned the new quantum mechanical interpretation introduced by Haas and Nicholson, but forsaking any attempt to explain radiation according to classical physics.
https://en.m.wikipedia.org/wiki/Bohr_model

Bohr's model was proposed because then classical physicists thought that accelerating electrons must radiate em wave. But experiments with superconductor shows that electric current can flow in circular motion without radiating away its energy. It's generally accepted that superconductivity is created by forming Cooper pairs.
IMO, it can happen because the radiation from one electron is cancelled out by its pair.

[Bored chemist]

Bohr's model was proposed because then classical physicists thought that accelerating electrons must radiate em wave.

Bohr's model was proposed in spite of classical physics saying that accelerating electrons must radiate em wave.

[Bored chemist]

But experiments with superconductor shows that electric current can flow in circular motion without radiating away its energy.

You don't need superconductors to run into that problem.

An ordinary  magnet should be impossible for much the same reason.