Expert: Philip A. Stahl - 1/7/2009

Question
In a previous answer to a question some time ago, dealing with gray atmospheres of stars, you wrote that the effective temperature (T_eff) is equal to the temperature at an optical depth of t= 2/3. Could you show this in more detail and also what would the (gray) surface temp. be in relation to the effective temp.? Thanks! (Note: not a HW question!)

Answer
Hello,

Okay, the key points - as I noted- were to assume: a) radiative equilibrium, and b) "LTE" or local thermodynamic equilibrium. Then, following the suggestions and method given in that problem, one can obtain a relationship between the effective temperature, T_eff, the optical depth, t, and the temperature T:

T^4  =  ½ T_eff^4(1 + 3t/ 2)

Now, allow the optical depth t = 2/3 as per your query. We find:


T^4  =  ½ T_eff^4(1 + 1) = T_eff^4

so, taking the fourth root of both sides:

T = T_eff

or more exactly, T(t=2/3) =  T_eff

which is to say, the temperature at an optical depth of t= 2/3 is equal to the effective temperature.


In the case of the gray surface temperature we are seeking:

T_o =  T(t = 0)

in other words, at the defined surface where the optical depth t = 0:

Then, using the same relation as earlier, we obtain:


T^4  =  ½ T_eff^4(1 + 3t/ 2) and T_o = ½ T_eff^4(1 + 0)

T_o^4  = ½ T_eff^4


take the fourth root of both sides:

T(t= 0) = T_o = 1/ (2)^1/4 T_eff =  0.841 T_eff

Hope this is useful!