[Bored chemist]

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.

Imagine thinking that you can do spectroscopy by looking at a TV picture.

Obviously, it should look bluer; it's a bit hotter.

[Bored chemist]

The system is in equilibrium

That assumption has been made tacitly or explicitly  throughout the thread.
So, yes, fluorescent things are odd but gold doesn't normally fluoresce.

[Bored chemist]

 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?

[Eternal Student (OP)]

Hi.

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.

[hamdani yusuf]

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.

[hamdani yusuf]

The system is in equilibrium

That assumption has been made tacitly or explicitly  throughout the thread.
So, yes, fluorescent things are odd but gold doesn't normally fluoresce.

Kirchoff's law only applies in more restrictive conditions than conservation of energy.

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.

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/

It doesn't say that emission spectrum of a material is proportional to its absorption spectrum.

Water is a good absorber in microwave, but ice is not. AFAIK, they are not good emitter of microwave.

[Bored chemist]

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 391 times).
So... what is it?

[hamdani yusuf]

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 391 times).
So... what is it?

You are confusing between emission and absorption spectrum, because you believe that they must be the same.

[Bored chemist]

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 391 times).
So... what is it?

You are confusing between emission and absorption spectrum, because you believe that they must be the same.

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.

So, what's the metal?

[hamdani yusuf]

The absorption spectrum is the same as the emission spectrum.

How do you know?
What's your reference?

Take a look at this.

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

[Bored chemist]

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"?

[hamdani yusuf]

If someone quotes newton's laws do you say "what's the reference"?

Principia.

[Bored chemist]

What's the metal?

[hamdani yusuf]

The conservation of energy.
You keep trying or pretend that Kirchhoff's law doesn't exist.

This is not Kirchhoff's law.

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.

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

[hamdani yusuf]

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.

[Bored chemist]

What's the metal?

I don't know. There's not enough information to distinguish one from the others.

But you said there was.

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.

[Bored chemist]

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.

OK.
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)

[Bored chemist]

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.

Let's apply some less basic logic.
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.

[hamdani yusuf]

What's the metal?

I don't know. There's not enough information to distinguish one from the others.

But you said there was.

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.

They can distinguish the difference in absorption spectrum.
Is this false?
Can't you see the difference?

[hamdani yusuf]

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.

Your conclusion was, molten gold emits more blue light than  molten silver, because gold absorbs more blue light when they are cool.