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I want to deduce the special form of $$w^2 = \frac{k_z^2}{2a^2}\left[b + \sqrt{b^2 - 4a^2T}\right]$$ from the equation $$ \left(1+ \frac{k_z}{wk_x}\right){k_x}^2 = \left(T -\frac{ w^2a^2}{k_z^2}\right)\left(\frac{k_z^2}{w^2} - 1\right)$$ Also it is given that $k_x\gg k_z.$ And $b=T+a^2+k_z^2$. So we have $\frac{k_z}{k_x}\approx0$. And using this I tried to solve for the desired form, but I was unable to solve it.

Any help towards this will be appreciated. Thank you.

caverac
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user533186
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1 Answers1

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Under the assumption $k_z / k_x\approx 0$ the second equation becomes

$$ k_x^2 = \left(T - \frac{w^2a^2}{k_z^2} \right)\left(\frac{k_z^2}{w^2} - 1\right) $$

Multiply by $w^2$ and expand

$$ k_x^2 w^2 = Tk_z^2 - Tw^2 - a^2 w^2 + \frac{w^4 a^2}{k_z^2} $$

Rearranging

$$ w^4 \frac{a^2}{k_z^2} - \underbrace{(T + a^2 + k_x^2)}_{\color{blue}{b}} w^2 + Tk_z^2 = 0 $$

So you end up with

$$ \frac{a^2}{k_z^2} x^2 - b x + T k_z^2 = 0 $$

where I called $x = w^2$. This is just a quadratic equation in $x = w^2$.

\begin{eqnarray} w^2 &=& \frac{b\pm\sqrt{b^2 - 4 (a^2/k_z^2)(Tk_z^2)}}{2a^2/k_z^2} \\ &=& \frac{k_z^2}{2a^2} \left[ b\pm\sqrt{b^2 - 4 a^2T}\right] \end{eqnarray}

caverac
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