Changes between Version 9 and Version 10 of u/erica/scratch3

Timestamp:

04/03/16 15:37:58 (9 years ago)

Author:

Erica Kaminski

Comment:

—

u/erica/scratch3

@@ -65 +65 @@
 Depending on the value of R, we will get different values of [[latex($\lambda$)]]. So, we start by approximating R. 
 
-Note that 
+Note that,
 
-[[latex($\frac{\nabla E}{E}=\frac{d(\ln E}{E}$)]]
+[[latex($\frac{\nabla E}{E}=\frac{d(\ln E)}{dx}$)]]
+
+so if we make the approximation, 
+
+[[latex($\frac{\nabla E}{E}\approx -\frac{1}{h}$)]]
+
+(by dimensional arguments and assuming the gradient is negative), we have: 
+
+[[latex($\frac{d(\ln E)}{dx}=-\frac{1}{h}$)]]
+
+which integrates to:
+
+[[latex($\ln E = \ln E_0 - \frac{x}{h}$)]]
+
+or,
+
+[[latex($E = E_0 e^{ - \frac{x}{h}}$)]]
+
+From this equation, it is clear that h is the scale-height. Thus, by writing R as:
+
+[[latex($R \approx |-\frac{\tau_R}{L \rho}| \approx \frac{\tau_R}{h \rho}$)]]
+
+we can set h to be the distance between the sink and box side (h=L), and imagine it as the scale height for the radiation. 
+
+