Why don't radio waves travel through sea water?

Thread Starter

sirch2

Joined Jan 21, 2013
1,029
As I understand it it is because sea water is conductive. But then coax is conductive and radio waves travel along that so is it because with radio waves in air have a differential voltage compared to earth whereas seawater is in contact with the earth and so there is no (or a very small) potential difference between the water and the earth?

If that is the case why do VLF signals work?

Is it also reasonable to think of radio wave in are as being analogous to a very big capacitor and seawater is like shorting out the capacitor?
 

crutschow

Joined Mar 14, 2008
28,170
Any conductive material tends to block radio waves going through them although they can be used to conduct radio waves along wires or waveguides (note they don't go through the material, only along it). Thus a steel building will block radio waves to a large extent. This blocking effect is generally proportional to frequency and to the material conductivity, so lower frequencies are more penetrating than higher frequencies. Thus the VLF frequencies can penetrate low conductivity sea water to a sufficient depth such that submerged submarines can pick up the signal whereas higher frequencies can't.
 

Lestraveled

Joined May 19, 2014
1,946
Change how you look at the world for a moment. Everything is a dielectric media. Space is a great dielectric media, there is very little it does not allow to pass through it. Add air to the formula and some wavelengths begin to get attenuated. Add dry wall and stucco and visual wave lengths are greatly attenuated and RF frequencies are only slightly effected. Add salt water and the higher RF frequencies are knocked out but the lower RF (very low) frequencies are less attenuated. The world is a transmission line. The right frequency will go through anything.
 

Thread Starter

sirch2

Joined Jan 21, 2013
1,029
Thanks for the replies but I suppose I was going for the "why" or "why not" rather than "can they". I mentioned VLF in the OP but why do lower frequencies have more penetrating power? Is it related to the fact that Xc goes up as frequency drops?
 

studiot

Joined Nov 9, 2007
4,998
You can use the formula that the attenuation factor, alpha is


\(\alpha = \sqrt {\frac{{\omega \mu \sigma }}{2}} \)

Where sigma is 4 seimens, and the relative permittivity is 80 for seawater.
 

Thread Starter

sirch2

Joined Jan 21, 2013
1,029
Thanks studiot I've Googled around that but can't really get to the derivation. Still looking for the why of why are radio waves attenuated. What is going on? What causes radio waves of increasing frequency to be more attenuated?
 

studiot

Joined Nov 9, 2007
4,998
I am sorry but I was caught again by this too-clever-by-half new interface.

I had prepared a much more detailed post and double clicked somewhere in the post to set the tex for the formulae and it all disappeared and took me back to the previous thread I was looking at.

I really do hate computer systems that some absent designer thinks he has made foolproof but has really made foolish.
 

studiot

Joined Nov 9, 2007
4,998
The complex exponential solution to the wave equation is very convenient for calculating things about electromagnetic waves.

So for an electric wave travelling in the +x direction into a medium
The Electric field oscillation, Ey, is in the y direction

\(\frac{{{\partial ^2}{E_y}}}{{\partial {x^2}}} - (j\omega \mu \sigma - {\omega ^2}\mu \varepsilon ){E_y} = 0\)

Are you happy with this equation or do I need to go right back to Maxwell's curl equation?

If we replace the coefficient of the second term by gamma squared we have


\({\gamma ^2} = (j\omega \mu \sigma - {\omega ^2}\mu \varepsilon )\)

and the solution is


\({E_y} = {E_0}{e^{ - \gamma x}}\)

However for a conductive medium \(\sigma \gg \omega \varepsilon \)

So \({\gamma ^2} = j\omega \mu \sigma \)

\(\gamma = \sqrt {j\omega \mu \sigma } \)

\(\gamma = (1 + j)\sqrt {\frac{{\omega \mu \sigma }}{2}} \)

putting \(\gamma = \alpha + j\beta \) to separate real and imaginary parts the solution becomes


\({E_y} = {E_0}{e^{ - \alpha x}}{e^{ - j\beta x}}\)

Since we only want the real part


\(\alpha = {\rm Re\nolimits} (\gamma ) = \sqrt {\frac{{\omega \mu \sigma }}{2}} \)

With the figures I gave for seawater typical reduction distances for a factor of 1,000,000 are

1kHz 106m; 10kHz 36m, 100kHz 11m, 1000kHz 3.5m
 
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