2
$\begingroup$

Related: x-ray interaction with atmosphere

I know that electromagnetic waves of particular frequencies reflect from the ionosphere. And the light (which from one perspective is an electromagnetic wave) also reflects from the.. let's say snow. These types of reflection have the same nature or different mechanisms are involved here?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Light is an EM wave from all perspectives, not just one. $\endgroup$
    – Manishearth
    Commented Feb 13, 2012 at 11:06
  • $\begingroup$ It's also particles (phatons), innit? $\endgroup$
    – Stanislav Bashkyrtsev
    Commented Feb 13, 2012 at 11:37
  • $\begingroup$ All EM waves are photons. It's not just light that has particle nature, all EM waves have that nature. OK, I see what you're saying. I though you were segregating light from the rest of the EM spectrum from some esoteric perspective. $\endgroup$
    – Manishearth
    Commented Feb 13, 2012 at 12:07

4 Answers 4

Reset to default
2
$\begingroup$

Electromagnetic radiation will reflect from any change in refractive index.

For light reflecting off snow you have an air/ice boundary, and the refractive index of air is about 1.0 while ice is about 1.3, so you get reflection. Actually you get multiple reflections which is why you get a diffuse scattering and why ice looks white.

For RF waves reflecting off the ionosphere, the ionosphere refractive index is changed by the free elections in it, so there is a refractive index mismatch between the very low pressure air below the ionosphere and the ionosphere itself and this causes reflection. It's a bit hand-waving to claim it's just that the ionosphere has a different refractive index, but that's a good starting point. If you want to look in more detail see http://ecjones.org/physics.html and many other easily Googlable articles.

So I suppose the mechanisms are basically the same.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ "ionosphere has a different refractive index" is correct, not hand-waving. Higher conductivity means larger imaginary part of refractive index - link. $\endgroup$
    – Steve Byrnes
    Commented Jul 27, 2012 at 12:05
1
$\begingroup$

It was such a simple question, but you guys gave such complicated answers:(

It is really not that hard to answer: in both cases, we can model the reflection as light (an electromagnetic wave) passing through a medium with changing index of refraction. In both cases, we can analyze it as waves or rays, but it is better to analyze both as waves. In both cases, waves are reflected when the index of refraction changes (as Rennie already pointed out). In that sense, it is the same mechanism.

Now true, technically, what happens in the ionosphere is a bit more complicated, with radio waves being bent gradually back to earth rather than reflected at a sharply defined boundary with a discontinuity in index of refraction, but that is ignored when actual users of such radio communication (such as radio amateurs) work in terms of the (virtual) height of the relevant layer of the ionosphere (layers D through F2). See http://www.electronics-radio.com/articles/radio/basic_radio/propagation/ionospheric-hf-propagation.php for a simplified explanation of ionospheric reflection, http://www.qsl.net/zl1bpu/IONO/iono101.htm for more details.

Oh, there is one other aspect in which they are different: since ice and air are both dielectrics with ice having the higher index, the reflection is phase inverting. We don't usually care, because the phase is lost in the diffuse reflections anyway.

But since air in the ionosphere is conducting, we have to model it as a reflection where one medium is dielectric and the other conducting, so the treatment of phase is different. If it were a good conductor, the reflection would be phase preserving; but since the ionosphere is rather weakly conducting, the full electromagnetic wave treatment of this case is rather difficult, and is not found in elementary E&M texts: one has to resort to odd or old sources like Frankel to see it fully analyzed. Yet as the QSL site reference above mentions, one can at times hear the effects of phase distortion in HF signals reflected off the ionosphere. So the difference is relevant.

Come to think of it, my own answer turned out to be more complicated than I expected when staring it, too;) But I hope it has answered your question more fully.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Probably to understand this I need to understand what refraction index is from the perspective of EM wave. Everywhere it's just described 'light goes through medium blah-blah', but it's not described why/how this medium changes light as EM wave on the level of particles/fields. It looks to be described here: farside.ph.utexas.edu/teaching/315/Waves/node39.html But need to re-read that several times to understand at least something :) $\endgroup$
    – Stanislav Bashkyrtsev
    Commented Aug 2, 2012 at 14:02
0
$\begingroup$

First of all, in order to describe this kind of phenomenon the EM "perspective" is enough, but if you want to look at the interaction between light and one atom you need the description given by quantum electrodynamics. So light is not an EM wave from all perspectives!

The reflection due to the ionosphere and the one you observe with snow are the same and it is called diffusive reflection. Because snow and ionosphere are not ordered/regular medium in fact you observe a reflection in all directions. With a mirror, for which atoms are regularly spaced, reflection appears only in one direction which explain Snell-Descartes.

I hope my answer is clear!

Improve this answer
$\endgroup$
4
  • $\begingroup$ But.. EM wave reflects from ionosphere not in all directions. E.g. at this page you can see k'' wave that is mirrored by the plasma:ecjones.org/physics.html which is used for radio waves propagation actually. $\endgroup$
    – Stanislav Bashkyrtsev
    Commented Feb 13, 2012 at 11:44
  • $\begingroup$ So light is not an EM wave from all perspectives!?? Why so?? $\endgroup$
    – Vineet Menon
    Commented Feb 13, 2012 at 15:46
  • $\begingroup$ The right description of what is called EM wave since Maxwell is given by Quantum Electrodynamics (read The Strange Theory of Light and Matter, Feynman). As you maybe know, a great debate during 17th century was about the nature of light : is it a wave or made of particles? Newton thought that light it is composed of particles in order to explain reflection ; and Huygens thought that it is a wave. Indeed you can't explain interferences with classical particles. $\endgroup$
    – PanAkry
    Commented Feb 13, 2012 at 16:25
  • $\begingroup$ During the 19th century the theory exposed by Maxwell confirmed that light is a wave and all phenomena were explained with this theory. At the beginning of the 20th century, Planck and Einstein explained the black body and the photoelectric effect only with help of photons, which contradict Maxwell's theory! Then the complete theory called Quantum Electrodynamics was achieved by Feynman, Schwinger and Tomonaga. Nowadays this theory explains all phenomena which involve matter and light (Lamb shift, stimulated emission), which is not the case of Maxwell theory. $\endgroup$
    – PanAkry
    Commented Feb 13, 2012 at 16:38
0
$\begingroup$

I am here because of the bright spots on Ceres. It does not have the dense atmosphere of the Ionsphere, so I will quote you a reference on the phenomena of Reflection.

This is covered by Philip M. Morse "Chpt 7 Hndbk of Physics Vibrations of Elastic Bodies; Wave propagation in Elastic Solids Sec 4. Reflection from a Plane Interface, Surface Waves"

'(if the free surface were the surface of the earth, these would be SH waves) is reflected without change of type and with amplitude -1. In other words P and SV waves interchange on reflection but SH waves remain SH waves.'

The terminology is on 3-99 he is using the physical analogy of Seismic waves.

Compressional waves, or longitudinal waves = P

Shear or transverse = S

Shear Waves in a horizontal plane = SH

The analogy is apt and he covers Rayleigh reflection and terms them "attenuated" not scattered.

In Part 6 Optics Sec. 6 Reflection and Refraction at a Plane Boundary

E. U. Condon of Colorado University

Covers Snell's Law and Brewsters Angle.

Here the application of boundary conditions is tied back to Maxwell's Field equations

of Section 3 eqn. 1.17

Normal components of vectors D and B Tangential components of E and H **

Improve this answer
$\endgroup$
1
  • $\begingroup$ It would be preferable if you eg. write down the equation, to which you are referring. You can see this help post to learn how to write your equations in $\LaTeX$. $\endgroup$
    – Gonenc
    Commented Jul 10, 2015 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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