Theory question - Blackbody Radiation and Light

[sweetreason]

"Theory" question -- Blackbody Radiation and Light

I am trying to understand the discussion about blackbody radiation in my Modern Physics textbook. (I'll quote it, but it can be found http://phy240.ahepl.org/Chp3-QT-of-Light-Serway.pdf" , page 5 document numbering, 69 textbook numbering).

The book says "light emitted by a small opening [in a heated cavity] is in thermal equilibrium with the walls, because it has been absorbed and re-emitted many times." (the book shows a diagram in which the light has bounced around a bit inside the cavity)

Now, I'm assuming the way that light exchanges heat has to do with the physics 1 equation $\frac{1}{2}mv^2_{av} = \frac{3}{2}k_BT$ (though I am not sure if this equation needs to be modified with a relativistic one). In other words, light has kinetic energy, and when it shines on you the light bounces into your molecules, which moves them around and thus heats them up. By conservation of energy and momentum, though, light would have to give up kinetic energy in order to heat you up. So once the light has reached thermal equilibrium, it must have given up a fair bit of kinetic energy. So how is the light that is in thermal equilibrium with the cavity changed when it exits? Light has no mass, it's energy is purely kinetic. I remembered reading somewhere that light that stopped moving essentially ceases to exist. Is that true? So would the number of photons leaving the cavity be less? Would it follow that the wavelength of the light decreases in accordance with Planck's law? Another bit of confusion here for me arises because of the connection between kinetic energy and velocity. If light gives up kinetic energy, you would at least classically expect it to give up velocity. Does this happen here? The speed of light in materials can be less than c, so this is allowable, right?

 

[Nate810]

The main question I'm asking is how does the light emitted from a heated cavity that has reached thermal equilibrium differ from the light coming in? Is there a decrease in number of photons and thus wavelength or a decrease in velocity, or something else? The key thing to remember when trying to answer this question is that in thermal equilibrium, the energy of the light must remain constant. This means that the number of photons leaving the cavity must be equal to the number of photons entering the cavity. So, while the photon energy may increase due to the increased temperature, the number of photons remains the same. Therefore, there is no change in wavelength or velocity of the light emitted from the cavity.