Difference between revisions of "Radiative Equilibrium"

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Line 11: Line 11:
 
\def\bto{{B_{21}}}
 
\def\bto{{B_{21}}}
 
\def\bot{{B_{12}}}
 
\def\bot{{B_{12}}}
 +
\def\Jbar{{\bar J}}
 
\usepackage{fullpage}
 
\usepackage{fullpage}
 
\usepackage{amsmath}
 
\usepackage{amsmath}
Line 54: Line 55:
 
in state 1, ditto for $n_2$.  Assume we are in thermal, steady-state  
 
in state 1, ditto for $n_2$.  Assume we are in thermal, steady-state  
 
equilibrium, so:
 
equilibrium, so:
$$n_1\bot\_J=n_2\ato+n_2\bto\_J$$
+
$$n_1\bot\Jbar=n_2\ato+n_2\bto\Jbar$$
 
This is because as many atoms need to be
 
This is because as many atoms need to be
 
going from energy state 1 to 2 as visa versa.
 
going from energy state 1 to 2 as visa versa.
 
A second relation is:
 
A second relation is:
$\_J = {n_2\ato\over n_1\bot-n_2\bto}$.
+
$\Jbar = {n_2\ato\over n_1\bot-n_2\bto}$.
 
Using that ${n_2\over n_1}={g_2\over g_1}\eboltz$:
 
Using that ${n_2\over n_1}={g_2\over g_1}\eboltz$:
$$\_J={{\ato\over\bto}\over{g_1\bot\over g_2\bto}\eboltz-1}$$
+
$$\Jbar={{\ato\over\bto}\over{g_1\bot\over g_2\bto}\eboltz-1}$$
  
 
In thermal equilibrium $J_\nu=B_\nu$:
 
In thermal equilibrium $J_\nu=B_\nu$:
$$\begin{aligned}\_J&\equiv\int_0^\infty{J_\nu\phi(\nu)d\nu}\\  
+
$$\begin{aligned}\Jbar&\equiv\int_0^\infty{J_\nu\phi(\nu)d\nu}\\  
 
&=\int_0^\infty{B_\nu\phi(\nu)d\nu}\\  
 
&=\int_0^\infty{B_\nu\phi(\nu)d\nu}\\  
 
&\approx B_\nu(\nu_0)\\  
 
&\approx B_\nu(\nu_0)\\  
 
&={2h\nu_0^3\over c^2(\eboltz-1)}\\ \end{aligned}$$
 
&={2h\nu_0^3\over c^2(\eboltz-1)}\\ \end{aligned}$$
  
Combining this with $\_J$ earlier, we get:
+
Combining this with $\Jbar$ earlier, we get:
 
$$\boxed{g_1\bot=g_2\bto}$$
 
$$\boxed{g_1\bot=g_2\bto}$$
 
\centerline{and}
 
\centerline{and}

Revision as of 14:15, 15 September 2015

Short Topical Videos

Reference Material

Local Thermodynamic Equilibrium (LTE)

Local Thermodynamic Equilibrium means

where is local. Take a sphere at surface temp , with some gas (matter) inside it. Allow to come to thermal equilibrium. Inside the sphere, you see a Planckean spectrum (). Now suppose you take a box of gas (it is both emissive and absorbent) and shoot rays of photons through that box suffer some absorption:

But those photons also pick up some intensity:

But since the photons should not pick up energy going through the box (everything is at the same temperature),

Now suppose you let all the photons out of this sphere. Those photons will no longer be in thermal equilibrium with the gas. Suppose we magically fix the temperature of the gas at this point. The source function will remain the same (it will still be the case that ) because the source function is a property of the matter alone (the absorptive and emissive properties of it). However, .

Suppose we look along a column of this gas. At each point, the source function will be: . Therefore, along that column will be:

Radiative Equilibrium and Relations among Einstein Coefficients

Assume we have many atoms with 2 energy states, and is the # density in state 1, ditto for . Assume we are in thermal, steady-state equilibrium, so:

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: . Using that :

In thermal equilibrium :

Combining this with earlier, we get:

and