diff --git a/_psets/2.md b/_psets/2.md index 90ad9f8b11f6c0a100638d2269f8b6411776133d..c2ca630409a7e0d3ce86150dcb5450c697321aaa 100644 --- a/_psets/2.md +++ b/_psets/2.md @@ -416,9 +416,16 @@ $$ So our result above for $$p(\tau)$$ follows because $$E = k T \log(\tau / \tau_0)$$, so $$dE / d\tau = k T / \tau$$. - +Now we must show that $$S(f) \propto 1/f$$. $$ \begin{align*} +S(f) &= \int_0^\infty p(\tau) \frac{2 \tau}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\ +&= \int_0^\infty \frac{2 p(E) k T}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\ +&= \frac{1}{2 \pi f} \int_0^\infty \frac{2 p(E) k T}{1 + \tau^2} \mathrm{d} \tau \\ +&= \frac{p(E) k T}{\pi f} \int_0^\infty \frac{1}{1 + \tau^2} \mathrm{d} \tau \\ +&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \frac{sec^2{\theta}}{1 + tan^2(\theta)} \mathrm{d} \theta \\ +&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \mathrm{d} \theta \\ +&= \frac{p(E) k T}{2 f} \end{align*} $$