Conductivity and loss in electromagnetic waves

It's very hard to follow your question, but you are talking about two regimes, either high or low conductivity medium/frequency combinations, where the key quantity to consider is $sigma/omega epsilon$, where $omega$ is the angular frequency of the waves, $epsilon$ is the (real) permittivity and $sigma$ is the conductivity.

The wave solution to Maxwell's equations in a conductive medium can be written as (for example, for a wave propagating in the x-direction) $$ {bf E} = {bf E_0} exp[i(kx- omega t)], $$ where the wave vector $k$ is a complex number $ k = alpha + i beta$. Thus $${bf E} = {bf E_0} exp(-beta x) exp[i(alpha x - omega t)] .$$

In a common set of of units (i.e. S.I. units), then when $sigma/omega epsilon gg 1$ then we would call that a good conductor. In these circumstances it turns out that $alpha simeq beta$. Whether this is true or not depends on both the properties of the medium and the frequency of the wave. In this scenario, the wave dies away with an exponential scale length of $beta^{-1} = alpha^{-1}$, which is known as the skin depth.

The value of $beta$ depends on the conductivity and frequency of the wave. $$ alpha simeq beta = (omega mu sigma/2)^{1/2}$$

If on the other hand say $sigma/omega epsilon ll 1$, then we would call that an insulator. In that case $beta simeq 0$ and the wave can propagate a long way before being attenuated.

The intermediate case where $sigma/omega epsilon sim 10^{pm 3}$, we might refer to as a semi-conductor or partial insulator. For this case $beta < alpha$ so the wave may die away over many wavelengths (like the case of weak damping in an oscillator). Nevertheless, it will be attenuated.

Looking at your notes, it may be that $alpha$ and $beta$ are switched around compared with my definitions, but I doubt it. It is the size of $beta^{-1}$ (not $alpha^{-1}$ in a partial insulator) that will determine how far the wave gets. In your notes you appear to be looking at $alpha^{-1}$, but because $alpha > beta$ then this will overestimate the attenuation of the wave.

Example - seawater with $epsilon = 81epsilon_0$, $sigma = 4$ S/m and $f=10^6$ Hz.

The ratio $sigma/omega epsilon = 888$, so seawater is a reasonably good conductor at these frequencies. I do not understand why you label it as a good insulator.

In this case $alpha simeq beta = (omega mu sigma/2)^{1/2}$ and the attenuation length of the EM waves electric field amplitude is about 4 m and the power attenuation length is 2 m.

Decreasing $omega$ makes the seawater more conductive, but because the scale length depends on $omega^{-1/2}$, the waves can travel further. If you drop the frequency by a factor of $10^4$ to 100 Hz, then your communication scale length goes up by a factor of 100.

After writing all this, I now think I can see what your question is - why can the waves travel further despite the medium becoming more like a good conductor? To see this we go back to Maxwell's equations and write down the wave equation for this damped wave. Taking the curl of both sides of Faraday's law $$ nabla times (nabla times {bf E}) = - frac{partial}{partial t} nabla times {bf B}$$ Using a well known vector identity, that $nabla cdot {bf E}=0$ in a neutral medium and inserting $nabla times {bf B}$ from Amp`ere's law, we have $$ nabla^2 {bf E} = frac{partial}{partial t} mu {bf J} + mu epsilon frac{partial^2 {bf E}}{partial t^2}$$ Then using ${bf J} = sigma {bf E}$ we write $$ nabla^2 {bf E} - mu sigma frac{partial {bf E}}{partial t} - mu epsilon frac{partial^2 {bf E}}{partial t^2} = 0$$

The middle term is the damping term. It gets bigger for higher $sigma$ (as you might expect), but it also gets higher for larger $partial {bf E}/partial t$. If $E propto exp(iomega t)$ then $partial {bf E}/partial t propto omega {bf E}$. In other words the damping increases linearly with higher frequencies and whilst the material is considered a good conductor - i.e. where the middle term is large compared with the third term, the attenuation length decreases as $omega^{-1/2}$.

However, the third term in the eqution grows as $omega^2$, so as the frequency increases, eventually the object cannot be considered a good conductor and moves to being an insulator and the value of $beta$ tends towards a constant value $$ beta simeq frac{sigma}{2}sqrt{frac{mu}{epsilon}}$$

Another way of thinking about this is that the relative sizes of the the second and third terms determines whether something behaves as a good conductor (when the middle term gets large compared with the third term), but the attenuation length depends on the absolute size of the middle term. The middle term grows with $omega$, but the third term grows as $omega^2$.

A third way of looking at this is to consider the attenuation limit in units of wavelength. The wavelength of the wave is is given by $2pi alpha^{-1}$. When we have a good conductor $alpha = beta$ and so the attenuation length remains a fixed fraction of the wavelength$/2pi$. At increased frequencies $alpha > beta$ and so the damping becomes relatively weaker and the attenuation length increases in terms of number of wavelengths before being damped. However, the wavelength itself is getting shorter, faster, and so the absolute attenuation length decreases.