One of the most interesting problem in underwater communication is how deep the device (i.e. submarine) could go to still be able to communicate with on-shore data centers.
The attenuation in a medium is usually explained away as a function of how conductive the medium is. http://www.diva-portal.org/smash/get/diva2:649690/FULLTEXT01.pdf see fig 2.3
My question is: What is the fundamental reason as to why freshwater attenuates signals less than seawater? Is it because freshwater is more conductive than seawater?
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The thesis you refer to puts values of 0.02 S on the conductivity of freshwater and 4 S for seawater. I'm assuming that the S here stands for the SI unit of Siemens per metre, the value of $\sim 5$ Siemens per metre is one I have used for seawater in the past.
Submarines communicate with frequencies as low as 100 Hz.To test for a "good conductor" we compare conductivity $\sigma$ with $\omega \epsilon_0 \epsilon_r$, where $\omega$ is the angular frequency, and $\epsilon_r$ is the relative permittivity (about 80 for water). This comparison suggests both freshwater and seawater are good conductors for waves at this frequency. (Indeed you would not try mixing AC electricity with your bathwater, right?)
Given that, there is an easy expression for the characteristic length, the "skin depth", that an EM wave will travel before being attenuated by $1/e$: i.e. $ l = \sqrt{2/\mu_0 \sigma \omega}$. At these frequencies, the skin depth in freshwater is 355m, whilst for seawater it is 25m.
OK, but I guess you knew all that, and your question is why does conductivity affect the skin depth in this way when you might intuitively have thought that better conductors would somehow allow EM waves to pass more freely?
The way to think about it is in term of Ohm's law and the dissipation of energy by currents. The power dissipated by a current in a circuit is of course $I^2 R$, but how is this related to electric field and conductivity? If we let current density $J$ be current per unit area, so that $I=JA$, and resistance $R = l/\sigma A$, where $l$ is distance travelled through the medium; then, further noting that $J = \sigma E$ in a linear conductor, we finally get a power dissipation of $J^2 A^2 l/\sigma A = \sigma E^2 A l$. i.e. the power dissipated per unit volume is proportional to $\sigma$. A higher conductivity yields higher currents and more Ohmic heating. The heat is provided by the electromagnetic energy flux of the waves. Thus the Poynting vector of the wave decays as it travels through the conducting medium; the missing power (per unit volume) is equal to $\sigma E^2$.