3
$\begingroup$

Wikipedia classifies ELF (Extremely Low Frequency) radiation as between 3 Hz and 300 Hz, and ULF (Ultra Low Frequency) from 300 Hz to 3 kHz. Both ranges can penetrate the earth and sea. What is the physics behind why this is so?

What would be the ideal frequency then to scan up to 500 feet underground for metal detection, a sort of underground metal detection radar?

$\endgroup$
3
$\begingroup$

The physics of the attenuation is that of the physics in conductive materials. The skin depth is a conductor is (see Wikipedia at https://en.m.wikipedia.org/wiki/Skin_effect),

$d = \sqrt(2/\mu_0\sigma\omega)$

See also a simple derivation online at http://farside.ph.utexas.edu/teaching/315/Waves/node65.html Seawater is still a good conductor, as shown there, with conductivity of 5 (ohms-meters)$^{-1}$. The skin depth arises because the charged particles in the medium will tend to cancel the electric field, and eventually do. The higher the conductivity clearly the smaller the skin depths. You have to solve the Maxwell equations with the approximations for a good conductor and you get the wavelength dependence. The article shows the derivation which is basic electromagnetism. It's standard also in textbooks, see it for instance in Jackson. Intuitive interpretation (though better to do the math, you won't misinterpret) is that the electromagnetic energy per unit depth is less at higher wavelengths, the changes are partially scaled by wavelength.

With $\sigma$ the conductivity and $\omega = 2\pi f$, and f the frequency. For lower frequency f the skin depth is higher, greater penetration or distance before attenuation. $\mu_0$ Is the vacuum permeability. Most or almost all the absorption is from the electric field at ELF frequencies so the magnetic properties of the conductor matter little. The derivation of the skin depth is in just about any electromagnetism book, in the most general case also.

The earth is also a conductor, a geological conductor with typically starta that are arranged In a complex way. For lower frequencies like ELF the small scale arrangements have little effect because of wavelengths of a thousand and more kilometers, so an average conductivity is not too bad an approximation. i saw a good reference for the absoprtion through the earth by googling, but can't seem to find it again.

So again low frequencies like ELF propagate better.

By the way, they also propagate pretty good above ground, with the ground and the ionosphere forming a duct. Lots of papaer on the physics and propagation for these waves.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Still what is the reason frequencies being absorbed less? Is it because of less possible interaction with atoms to gain higher energy states(whose possibilities should decrease with lower energy, that is lower frequency)? $\endgroup$ – Tausif Hossain Jul 8 '17 at 5:07
  • $\begingroup$ I edited my answer to describe more of the depth derivation, and reference a specific easy online derivation for the wavelength dependence. The intuitive explanation is approximate because you have to take care of the math right, but higher wavelengths serve as a kind of scale which gives the sqrt(wavelength) as a factor. Much simpler and easier to follow the math, for the sqrt dependence. See the references. $\endgroup$ – Bob Bee Jul 8 '17 at 18:25
  • $\begingroup$ You should also explain what breaks down at higher frequency (i.e. frequency dependence of conductivity itself changes), and why this doesn't occur at ELF. The actual physics behind conductivity should be mentioned I believe. $\endgroup$ – KF Gauss Jul 9 '17 at 18:28
  • $\begingroup$ @user157979. You can go ahead and supplement in whatever I didn't cover. The physics behind conductivity was not the question, and is also standard material covered in textbooks. You need to have electrons or ions which are free or very loosely held by their atoms/molecules. Salt water and similar salts in the earth mixed with water can have ions floating around when those dissolve in water. I don't think the OP asked for any of that, but you never know $\endgroup$ – Bob Bee Jul 10 '17 at 1:19
2
$\begingroup$

The reason for this is frequency attenuation. Electromagnetic radiations, when penetrating through the media, interact with the media and attenuate. The higher-frequency, the faster the attenuation effect. Thus, low-frequency radiations such as VLF, ULF, etc., can penetrate into the medium much deeper than high-frequency ones. However, the drawback is that low-frequency radiations usually give us low-resolution signals.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ -1 You didn't explain the physics of why the attenuation depends of frequency the way it does. $\endgroup$ – KF Gauss Jul 8 '17 at 3:32
  • $\begingroup$ @user157879 I explained that it was due to frequency attenuation. When the radiation interacts with a medium, higher frequency means the interaction occurs more frequently and the effect is stronger. Hence, electromagnetic radiations would not attenuate in the vacuum. Please note that frequency attenuation is different from the decrease in intensity of the radiation as it propagates away from the source, which is inversely proportional to the square of the distance traveled. $\endgroup$ – A Slow Learner Jul 8 '17 at 3:49
  • 1
    $\begingroup$ Your basic understanding is incorrect. Blue light is higher frequency than red light, yet blue light penetrates much farther in water than red light. Obviously frequency is not the whole picture, you need to explain the physics specific to RF radiation. $\endgroup$ – KF Gauss Jul 8 '17 at 4:05
  • $\begingroup$ This is funny: "As light wavelength decreases from red to blue light, so does the ability of light to penetrate water. Blue light penetrates best, green light is second, yellow light is third, followed by orange light and red light. Red light is quickly filtered from water as depth increases." This statement makes no sense from: oceanexplorer.noaa.gov/facts/red-color.html maybe it's just the wording. $\endgroup$ – R. Rankin Jul 8 '17 at 4:39
  • $\begingroup$ @user157879 You are correct about it. Could you provide an explanation for the effect then? $\endgroup$ – A Slow Learner Jul 8 '17 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.