"Theory" question -- Blackbody Radiation and Light

sweetreason

I am trying to understand the discussion about blackbody radiation in my Modern Physics textbook. (I'll quote it, but it can be found here , 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 [itex] \frac{1}{2}mv^2_{av} = \frac{3}{2}k_BT [/itex] (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?