[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.

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

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.

[Bored chemist]

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?

[alancalverd]

Problem with the above images of molten metals is one of dynamic range. It's clear that the image receptor is saturated and therefore unable to resolve any spectral information. A dense filter might resolve differences between  two metals at the same temperature precisely because they aren't "black" at any temperature, but you wouldn't be able to see anything else in the picture.

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.

[hamdani yusuf]

We can all see the difference.
But that difference is between the reflectance spectra, not absorbance or emission..

What happens to the unreflected light?

[Bored chemist]

We can all see the difference.
But that difference is between the reflectance spectra, not absorbance or emission..

What happens to the unreflected light?

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.

There are two sorts of people in the world.
One sort can extrapolate from incomplete data.

[hamdani yusuf]

We can all see the difference.
But that difference is between the reflectance spectra, not absorbance or emission..

What happens to the unreflected light?

Let me help you to answer my question, just in case you are having difficulty.
It's absorbed. Which means that absorbed spectra between gold and silver in room temperature are different.

[Bored chemist]

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?