- #1
ergospherical
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- 1,276
Treating water as having a conductivity of ##\sigma \sim 5 \ \mathrm{Sm^{-1}} \gg \omega \epsilon_0 \epsilon_{\mathrm{r}}## then Maxwell ##\mathrm{III}## implies\begin{align*}
\nabla \times \mathbf{H} = \sigma \mathbf{E} + \epsilon_0 \epsilon_{\mathrm{r}} \dot{\mathbf{E}} = -i \omega \epsilon_0 \left(\dfrac{i\sigma}{ \omega \epsilon_0} + \epsilon_{\mathrm{r}} \right) \mathbf{E} \equiv \epsilon_0 \epsilon' \dot{\mathbf{E}}
\end{align*}The water has an effective dielectric constant ##\epsilon' \sim \dfrac{i\sigma}{\omega \epsilon_0}## and a refractive index ##n = \sqrt{\mu_{\mathrm{r}} \epsilon'} \sim (1+i)\sqrt{\dfrac{\mu_{\mathrm{r}} \sigma}{2\omega \epsilon_0}}##. A plane electromagnetic wave inside the water has wave-number ##k = \omega n/c \equiv (1+i)/\delta## where the skin depth is given by ##\delta = \sqrt{\dfrac{2}{\mu_{\mathrm{r}} \mu_0 \sigma \omega}}##, and the power decays ##\propto |\mathbf{E}|^2 \propto \mathrm{exp}(-2x/\delta)##. An order of magnitude of power is lost every ##x_*## metres, satisfying\begin{align*}
2x_* / \delta = \log{10} \implies x_* = \dfrac{1}{2} \delta \log{10} \sim \delta
\end{align*}Google tells me that submarines use radio frequencies typically of the order of a few ##\mathrm{kHz}## or tens of ##\mathrm{kHz}##, which results in the power dropping off by an order of magnitude every ##10 \ \mathrm{m}## or so. However, modern submarines are capable of operating at depths of, say, ##\sim 400 \ \mathrm{m}##, at which point the power has decayed by a factor of... ##\sim 10^{40}##. Even at the very lowest limits of radio waves, ##\sim## a couple of ##\mathrm{Hz}##, the power still drops off by an order of magnitude every few hundred metres. Assuming the analysis is vaguely correct, how do submarine engineers get around this issue? Is it merely a case of using very high power radio waves to begin with, so that the signals are still detectable even with this very significant attenuation?
\nabla \times \mathbf{H} = \sigma \mathbf{E} + \epsilon_0 \epsilon_{\mathrm{r}} \dot{\mathbf{E}} = -i \omega \epsilon_0 \left(\dfrac{i\sigma}{ \omega \epsilon_0} + \epsilon_{\mathrm{r}} \right) \mathbf{E} \equiv \epsilon_0 \epsilon' \dot{\mathbf{E}}
\end{align*}The water has an effective dielectric constant ##\epsilon' \sim \dfrac{i\sigma}{\omega \epsilon_0}## and a refractive index ##n = \sqrt{\mu_{\mathrm{r}} \epsilon'} \sim (1+i)\sqrt{\dfrac{\mu_{\mathrm{r}} \sigma}{2\omega \epsilon_0}}##. A plane electromagnetic wave inside the water has wave-number ##k = \omega n/c \equiv (1+i)/\delta## where the skin depth is given by ##\delta = \sqrt{\dfrac{2}{\mu_{\mathrm{r}} \mu_0 \sigma \omega}}##, and the power decays ##\propto |\mathbf{E}|^2 \propto \mathrm{exp}(-2x/\delta)##. An order of magnitude of power is lost every ##x_*## metres, satisfying\begin{align*}
2x_* / \delta = \log{10} \implies x_* = \dfrac{1}{2} \delta \log{10} \sim \delta
\end{align*}Google tells me that submarines use radio frequencies typically of the order of a few ##\mathrm{kHz}## or tens of ##\mathrm{kHz}##, which results in the power dropping off by an order of magnitude every ##10 \ \mathrm{m}## or so. However, modern submarines are capable of operating at depths of, say, ##\sim 400 \ \mathrm{m}##, at which point the power has decayed by a factor of... ##\sim 10^{40}##. Even at the very lowest limits of radio waves, ##\sim## a couple of ##\mathrm{Hz}##, the power still drops off by an order of magnitude every few hundred metres. Assuming the analysis is vaguely correct, how do submarine engineers get around this issue? Is it merely a case of using very high power radio waves to begin with, so that the signals are still detectable even with this very significant attenuation?
