Why does an incandescent solid produce a continuous spectrum rather than emission lines?

[TheExibo]

This whole post was taken from the following URL:

https://www.physicsforums.com/threads/kirchhoffs-three-laws-of-spectroscopy.510370/

And has never been answered. It states the following:

Kirchhoff's Three Laws of Spectroscopy are as follows.

1. An incandescent solid, liquid, or gas under high pressure, emits a continuous spectrum.
2. A hot gas under low pressure emits a "bright-line" or emission-line spectrum.
3. A continuous spectrum source viewed through a cool, low-density gas produces an absorption-line spectrum.

Here is the Wikipedia article I took that from.

How does an incandescent solid, liquid or gas under high pressure NOT have an emission line spectrum? Doesn't each atom this material is made up of emit emission lines as an electron transition happens from a higher energy level to a lower energy level? And and atom with an electron at a lower energy level can absorb a photon of correct frequency and transition to a higher energy level, right? Do these two processes cancel each other out?

Let's take a sample of an incandescent solid. I can understand how the emission photons from atoms at the middle of the sample being absorbed as they travel outwards. But if an atom at the outer edge of the sample emits a photon outwards, that doesn't get absorbed, right? So, shouldn't there be at least very faint emission lines?

And I guess even photons emitted from the center of a gas clump that is diffused can travel out through the clump without hitting other atoms because there is enough spatial separation between the atoms?

And how is light that does NOT have the frequency of an emission line get emitted? Maybe a derivation of black body radiation will help me understand this, but is there a simple qualitative explanation?

Thanks.

 

[mfb]

In solid materials (and to a lesser extent in liquids and dense gases), the electron levels form bands. You cannot consider the atoms individually any more.