Radiative Equilibrium

The radiative energy is related to the radiative temperature by (is this always the case?):

E_{rad}=a T_{rad} ^4

(where the radiative energy is the total radiative energy, i.e. the sum of any blackbody radiation, as well as any sources — what else is a source of radiation in the code?)

In radiative equlibrium, the radiative temperature equals the gas temperature,

T_{rad}=T_{gas}

That is to say,

E_{rad}=a T_{gas}^4

Recall,

\frac{\partial E_{rad}}{\partial t}=(4 \pi B(T)- c E_{rad})

The left-hand term on the RHS is the blackbody radiation of the gas. It equals, Tgas⁴:

\frac{\partial E_{rad}}{\partial t}=(4\pi T_{gas}^4- c E_{rad})

Thus, there is no coupling when,

E_{rad}=\frac{4 \pi}{c} T_{gas}^4

That is to say, when the radiative temperature equals the gas temperature.

However, that does not necessarily mean that the gas energy will equal the radiative energy in equilibrium. In general these energies will not be equal. In radiative equilibrium, the gas energy,

E_{gas}=n K_B T_{gas}

will equal the radiative energy,

E_{rad}=a T_{gas}^4

when

n K_B=a T_{gas}^3

Now, depending on the ratio of Erad/Egas, can expect to get different rates at which the temperature will rise or fall due to the radiation (right?). While the coupling rate does tell us how fast tgas will approach trad, there is another factor involved which is how much relative energies these fields have. That is because each can act as a thermal bath for the other. The idea is that if the radiation has >>> energy than the gas, then it can quickly transfer heat into the gas (but note, the gas will not change the temperature of the radiation). However, if the gas >>> than the radiation, then the gas is a bath and the radiation will not change the temperature significantly.

To experiment with this idea, I am taking RUN 2 from (​https://astrobear.pas.rochester.edu/trac/wiki/u/erica/scratch5) and adjusting this ratio. Initially that run had an Egas/Erad~1500 (the simulation was very cold at 10K).

Egas=Erad