Changes between Version 1 and Version 2 of u/erica/scratch6

Timestamp:

04/13/16 14:43:50 (9 years ago)

Author:

Erica Kaminski

Comment:

—

u/erica/scratch6

@@ -15 +15 @@
 [[latex($E_{rad}=a T_{gas}^4$)]]
 
-However, in general the energies of the gas and the radiation will not be equal. In radiative equilibrium, the gas energy,
+Recall,
+
+[[latex($\frac{\partial E_{rad}}{\partial t}=(4 \pi B(T)- c E_{rad})$)]]
+
+The left-hand term on the RHS is the blackbody radiation of the gas. It equals, Tgas^4^:
+
+  
+[[latex($\frac{\partial E_{rad}}{\partial t}=(4\pi T_{gas}^4- c E_{rad})$)]]
+
+Thus, there is no coupling when,
+
+[[latex($E_{rad}=\frac{4 \pi}{c} T_{gas}^4$)]]
+
+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,
 
 [[latex($E_{gas}=n K_B T_{gas}$)]]
@@ -27 +42 @@
 [[latex($n K_B=a T_{gas}^3$)]]
 
-Depending on the ratio of !Erad/Egas, will get different rates at which the temperature will rise (right?). This is in addition to the coupling.
-
+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.