Wave Attenuation in Two Limits

Recall from above that:

$$\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$} = -i \om... ...lon_b + i \frac {\sigma}{\omega} \right ) \mbox{\boldmath$E$}.$$ (9.128)

Then:

$$k^2 = \frac{\omega^2}{v^2} = \mu\epsilon\omega^2 = \mu \eps... ... \omega^2 \left( 1 + i\frac{\sigma}{\omega \epsilon_b} \right)$$ (9.129)

Also $k = \beta + i\frac{\alpha}{2}$ so that

$$k^2 = \left(\beta^2 - \frac{\alpha^2}{4}\right) + i\alpha\b... ... \omega^2 \left( 1 + i\frac{\sigma}{\omega \epsilon_b} \right)$$ (9.130)

Oops. To determine $\alpha$ and $\beta$, we have to take the square root of a complex number. How does that work again? See the appendix on Complex Numbers...

In many cases we can pick the right branch by selecting the one with the right (desired) behavior on physical grounds. If we restrict ourselves to the two simple cases where $\omega$ is large or $\sigma$ is large, it is the one in the principle branch (upper half plane, above a branch cut along the real axis. From the last equation above, if we have a poor conductor (or if the frequency is much higher than the plasma frequency) and $\alpha \ll \beta$, then:

$$\beta \approx \sqrt{\mu\epsilon_b}\omega$$ (9.131)

$$\alpha \approx \sqrt{\frac{\mu}{\epsilon_b}}\sigma$$ (9.132)

and the attenuation (recall that $\mbox{\boldmath$E$} = \mbox{\boldmath$E$}_0 e^{-\frac{\alpha}{2}} e^{i\beta \hat{\mbox{\scriptsize\boldmath$n$}}\cdot\mbox{\scriptsize\boldmath$E$}}$) is independent of frequency.

The other limit that is relatively easy is a good conductor, $\sigma \gg \omega\epsilon_b$. In that case the imaginary term dominates and we see that

$$\beta \approx \frac{\alpha}{2}$$ (9.133)

or

$$\beta \approx \sqrt{\frac{\mu\sigma\omega}{2}}$$ (9.134)

$$\alpha \approx \sqrt{2 \mu\sigma\omega}$$ (9.135)

Thus

$$k = (1 + i)\sqrt{\frac{\mu\sigma\omega}{2}}$$ (9.136)

Recall that if we apply the $\mbox{\boldmath$\nabla$}$ operator to $\mbox{\boldmath$E$}e^{ik(\mbox{\scriptsize\boldmath$n$}\cdot\mbox{\scriptsize\boldmath$x$} - i\omega t}$ we get:

$$\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$E$} = 0$$

$$ik \mbox{\boldmath$E$}_0\cdot\hat{\mbox{\boldmath$n$}} = 0$$

$$\mbox{\boldmath$E$}_0\cdot\hat{\mbox{\boldmath$n$}} = 0$$ (9.137)

and

$$-\frac{\partial \mbox{\boldmath$B$}}{\partial t} = \mbox{\boldmath$\nabla$}\times \vec{E}$$

$$i\omega\mu\mbox{\boldmath$H$}_0 = i(\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$E$}_0)(1 + i)\sqrt{\frac{\mu\sigma\omega}{2}}$$

$$\mbox{\boldmath$H$}_0 = \frac{1}{\omega}\sqrt{\frac{\sigma\omega}{\mu}} (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$E$}_0)\frac{1}{\sqrt{2}}(1 + i)$$

$$= \frac{1}{\omega}\sqrt{\frac{\sigma\omega}{\mu}} (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$E$}_0) e^{i\pi/4}$$ (9.138)

so $\mbox{\boldmath$E$}_0$ and $\mbox{\boldmath$H$}_0$ are not in phase (using the fact that $i = e^{i\pi/2}$).

In the case of superconductors, $\sigma \to \infty$ and the phase angle between them is $\pi/4$. In this case $\mbox{\boldmath$H$}_0 \gg \mbox{\boldmath$E$}$ (show this!) and the energy is mostly magnetic.

Finally, note well that the quantity $\left(\frac{\alpha}{2}\right)^{-1} = \delta$ is an exponential damping length that describes how rapidly the wave attenuates as it moves into the conducting medium. $\delta$ is called the skin depth and we see that:

$$\delta = \frac{2}{\alpha} = \frac{1}{\beta} = \sqrt{ \frac{2}{\mu\sigma\omega} }$$ (9.139)

We will examine this quantity in some detail in the sections on waveguides and optical cavities, where it plays an important role.