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If you consider waves in a metal, you can write the index of refraction for the metal as, $$ n^2 = 1 - \frac{\omega_p^2}{\omega^2} $$ I am interested in what will happen if the index is perturbed by some small complex quantity, $$ n^2 = 1 - \frac{\omega_p^2}{\omega^2} - i\epsilon $$ In the low frequency limit, this usually corresponds to an attenuation. To compute quantities that depend on $n$ (such as absorption), I would like to expand $n$ since $\epsilon$ is small.

The function is, $$ n = \sqrt{1 - \frac{\omega_p^2}{\omega^2}-i\epsilon} $$ So I write, $$ n = \sqrt{1 - \frac{\omega_p^2}{\omega^2}}\sqrt{1 - i \frac{\epsilon}{1 - \omega_p^2/\omega^2}} $$ For this problem $\omega_p > \omega$, so I rewrite, $$ n = i \sqrt{\frac{\omega_p^2}{\omega^2}-1}\sqrt{1 + i \frac{\epsilon}{\omega_p^2/\omega^2 - 1}} $$ And, since $\epsilon << 1$, I tried to expand this: $$ \approx i \sqrt{\frac{\omega_p^2}{\omega^2}-1}\left(1 + i \frac{\epsilon}{2(\omega_p^2/\omega^2-1)} \right) = \sqrt{\frac{\omega_p^2}{\omega^2}-1}\left(i - \frac{\epsilon}{2(\omega_p^2/\omega^2-1)} \right)$$

As a sanity check, I tried plugging in some numbers. $\omega_p=2$, $\omega=1$, and $\epsilon=0.01$. The expansion is off by a factor of $-1$. I believe it is due to the branch cut of the complex square root. If I place $i$ back into the square root, $$ \sqrt{-1 - i \frac{\epsilon}{\omega_p^2/\omega^2-1}} $$ is in the third quadrant of the complex plane. This means that the square root should be in the fourth quadrant, where I have a positive real part and negative imaginary part (in contrast to my expansion above, where the opposite is the case).

I spent a long time hunting for this negative, so my question is, is there a cleaner way to expand the function where I won't have this branch cut issue?

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  • $\begingroup$ It's in the third quadrant in which complex plane? About where do you want to expand? You can choose a branch cut for the original sqrt that is far away from the region you want to expand. This is all intuitively of course... I haven't done the calculations. $\endgroup$ – Your Majesty Feb 19 '14 at 23:06
  • $\begingroup$ Can closevoters please keep the outcome of this recent meta discussion in mind, where even real math and physics professors say that migrating every question that contains math away does a disfavor to the site, in particular to advanced / theoretical physics? Do the reviewers really give a damn about what such experts say ...? $\endgroup$ – Dilaton Feb 19 '14 at 23:10
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    $\begingroup$ @LoveLearning, my complex analysis is not very strong, which is why I didn't ask in the math stack exchange. I'm not sure what you mean by "which complex plane"... isn't there only one? I am expanding about $\epsilon=0$. If choosing a branch cut is a feasible option, sure let's do that, but I don't know how to do it. Dilaton, thanks, I've added some information in the question. $\endgroup$ – kordon Feb 19 '14 at 23:40
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    $\begingroup$ @Dilaton, I didn't realize math vs. physics was such a hot topic here. Thanks for all the meta help. $\endgroup$ – kordon Feb 19 '14 at 23:58
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    $\begingroup$ Im on a phone so can't help much but if the question still here tomorrow (and not moved to ms) I can help. $\endgroup$ – Your Majesty Feb 20 '14 at 0:05
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I'm not sure I understand your question correctly, but let me know if the below helps.

Let \begin{array}$ n^2(\omega) &= 1 - \frac{\omega_p^2}{\omega^2} - i\epsilon\\ &=-\Omega-i\epsilon\end{array} where $\Omega:=-\left(1 - \frac{\omega_p^2}{\omega^2}\right)>0$ by assumption.

Then, expanding $n(\omega)$ for small $\epsilon$ we find (using some algebra-software: is this OK with you?) $$\sqrt{-\Omega-i\epsilon} = -i\cdot\mathrm{csgn}(i(-\Omega-i\epsilon))\sqrt{\Omega}+\frac{1}{2}\frac{\mathrm{csgn}(i(-\Omega-i\epsilon))}{\sqrt{\Omega}}\epsilon+\mathcal{O}(\epsilon^2)$$

where the csgn is defined on this page as:

Definition of csgn

In our case we get:

$$\sqrt{-\Omega-i\epsilon} = -i\sqrt{\Omega}+\frac{1}{2}\frac{1}{\sqrt{\Omega}}\epsilon +\mathcal{O}(\epsilon^2)$$ which lies in the fourth quadrant.

Does this answer your question?

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  • $\begingroup$ That's interesting. I'm not sure where the csgn function comes from though. $\endgroup$ – kordon Feb 21 '14 at 19:56
  • $\begingroup$ Okay, I think I figured it out. The csgn function keeps track of the principal value of the square root. $\sqrt{-\Omega-i\epsilon} = -i\sqrt{\Omega+i \epsilon}$. You pull out $-i$ (not $i$) because you need the argument of $-1(\Omega+i\epsilon)$ to be between $[-\pi,\pi]$. $-1 = e^{\pm i \pi}$, but the argument of $\Omega + i \epsilon$ is positive, so you must choose $-1 = e^{-i \pi}$ to stay in the domain of the principal value of the square root (arguments are additive under complex multiplication). The correct series follows. $\endgroup$ – kordon Feb 21 '14 at 20:42
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There is a formula, which is checked by squaring the right side $$\sqrt{a+ib}=\pm \sqrt{(\sqrt{a^2+b^2}+a)/2}\pm i\sqrt{(\sqrt{a^2+b^2}-a)/2}$$, the branches of this formula must be chosen from the condition $$\sqrt{a+ib}=exp(iarg(a+bi)/2+i\pi k),k=0,1$$. Depending on the value $arg(a+bi)/2$, choose the root sign.The exact formula is determined up to a multiplier $\pm 1$. For formula $\sqrt{-1-i\frac{\epsilon}{\omega_p^2/\omega^2-1}}$ has the phase $arg(-1-i\frac{\epsilon}{\omega_p^2/\omega^2-1})/2=-\pi/2-\alpha$ located in the third quarter, therefore we have $$\sqrt{-1-i\frac{\epsilon}{\omega_p^2/\omega^2-1}}=\mp\sqrt{(\sqrt{1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2}-1)/2}\mp i\sqrt{(\sqrt{1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2}+1)/2}=\mp i\sqrt{1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2}(1+ i\frac{\epsilon}{\omega_p^2/\omega^2-1}/2+O(\epsilon^2))$$

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I think you're hitting something equivalent to the classic high school math class riddle: Find the flaw in the equation $$1 = \sqrt{1} = \sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=i^2=-1$$ You do a step near the beginning involving $\sqrt{AB} \rightarrow \sqrt{A}\sqrt{B}$, which is not always true, instead you should replace $\sqrt{AB}\rightarrow\pm\sqrt{A}\sqrt{B}$.

At the very end, you should get an equation that looks like $n=\pm(\text{something})$, and then you pick the sign based on what quadrant you know $n$ is supposed to be in.

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