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Quarkly
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- TL;DR Summary
- I ran across this equation in some lecture notes and I'm not able to follow the derivation.
I'm reading through the lecture notes of Wayne Hu regarding the Damping Scale of the CMB. He give the following steps to calculating the damping scale, ##k_D##:$$k_D^{-2}=\int \frac{1}{6(1+R)}\left( \frac {16}{15}+ \frac{R^2}{(1+R)}\right)\frac{1}{\dot\tau} d\eta$$Limiting forms:$$\lim\limits_{R \to 0}k_D^{-2}= \frac{1}{6}\frac{16}{15}\int \frac{1}{\dot\tau} d\eta$$
$$\lim\limits_{R \to \infty}k_D^{-2}= \frac{1}{6}\int \frac{1}{\dot\tau} d\eta$$and finally$$k_D=\frac{\sqrt{6}}{\sqrt{\eta \dot\tau^{-1}}}$$I see roughly what he's doing, but my math is rusty. Could anyone explain in greater detail how he arrived at ##k_D##? For example, why did he discard the limit as ##{R \to 0}##? Why is only ##{R \to \infty}## used in the final formula? Why was he able to extract everything but ##\frac{1}{\dot\tau}## from the integral (since R is also a function of ##\eta##)?
While function ##f(\eta)=\frac{1}{6(1+R(\eta))}\left( \frac {16}{15}+ \frac{R(\eta)^2}{(1+R(\eta))}\right)## is relatively constant, it does change with ##\eta## and looks like this:
and the function ##g(\eta)=\frac{1}{\dot\tau(\eta)}## looks like this:
The bottom axis is ##\eta##.
$$\lim\limits_{R \to \infty}k_D^{-2}= \frac{1}{6}\int \frac{1}{\dot\tau} d\eta$$and finally$$k_D=\frac{\sqrt{6}}{\sqrt{\eta \dot\tau^{-1}}}$$I see roughly what he's doing, but my math is rusty. Could anyone explain in greater detail how he arrived at ##k_D##? For example, why did he discard the limit as ##{R \to 0}##? Why is only ##{R \to \infty}## used in the final formula? Why was he able to extract everything but ##\frac{1}{\dot\tau}## from the integral (since R is also a function of ##\eta##)?
While function ##f(\eta)=\frac{1}{6(1+R(\eta))}\left( \frac {16}{15}+ \frac{R(\eta)^2}{(1+R(\eta))}\right)## is relatively constant, it does change with ##\eta## and looks like this:
and the function ##g(\eta)=\frac{1}{\dot\tau(\eta)}## looks like this:
The bottom axis is ##\eta##.