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From this PhD dissertation, page 29:

Reduce the formula $$ T(0, t) = T_{eq} + \frac{(1 − A)}{(1+i)\sqrt{\pi fKC\rho} + 4\epsilon\sigma T^{4}_{eq}}\epsilon_{1}e^{i2\pi ft} $$

down to $$ T(0, t) = T_{eq} + \frac{(1 − A)}{4\epsilon\sigma T^{3}_{eq}} \frac{1}{1+2\Theta +2\Theta^{2}}\epsilon_{1}e^{i(2\pi ft + \phi_{th})} $$

using the fact that $$ \Theta = \frac{\sqrt{\pi fKC\rho}}{4\pi\epsilon\sigma T^{3}_{eq}} $$ and $$ \tan{\phi_{th}} = -\frac{\Theta}{1 + \Theta} $$

I can't for the life of me get this. I suspect it is a lot of work, some pointers would be appreciated.

EDIT: So I think this may be incorrect. In my opinion $$ \Theta = \frac{\sqrt{\pi fKC\rho}}{4\epsilon\sigma T^{4}_{eq}} $$ With this it can be easily shown that $$ T(0, t) = T_{eq} + \frac{(1 − A)}{4\epsilon\sigma T^{4}_{eq}} \frac{1}{\sqrt{1+2\Theta +2\Theta^{2}}}\epsilon_{1}e^{i(2\pi ft - \phi_{th})} $$ which is somewhat different, but also makes more sense to me from a physics perspective (minus phase factor and quadratic square rooted). Also how he went from T^4 to T^3 is really beyond me (it's thermal radiation so should really radiate according to T^4 imo).

sunra
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  • Shorn of the clutter, you are aiming to show that $$\frac1{(1+i)\Theta+\sqrt\pi}=\frac{\exp(-i\arctan(\Theta/(1+\Theta)))} {1+2\Theta+2\Theta^2}$$ or at least something similar (I'm not sure that is quite right). My tip is to work on $\exp(-i\arctan(\Theta/(1+\Theta)))$. – Angina Seng Jun 07 '20 at 12:36
  • Thanks, I followed your advice, just working on the denominator as that's all that changes, but I'm thinking there may be an error here in his working. Posted and edit. – sunra Jun 07 '20 at 16:22

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