# Electrodynamics question (probably not of general interest) **URL:** \ **Category:** Factual Questions\ **Created:** [August 5, 2002, 10:01pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485 "2002-08-05T22:01:24Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [August 5, 2002, 10:01pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/1 "2002-08-05T22:01:24Z")
While paging through my old E&M text I ran across the following problem (I have a guess as to what the answer might be but I’m far from sure about it) If an EM plane wave passes from an optically dense medium into a less dense one, and is incident at greater than the critical angle, then there is no refraction and you get 100% reflection. But (and here’s my problem) the fields in the less dense medium are not zero and you therefore have an evanescent wave which exponentially decays perpendicular to the interface but propagates and transmits energy parallel to the interface. But if there is energy transmitted parallel to the interface then how can there be 100% reflection? Wouldn’t this energy subtract from the reflected wave? My guess There can be no such thing as a plane wave since this would imply infinite energy. Thus this is a bogus statement by my textbook. ---
**Author:** ![Balduran](https://avatars.discourse-cdn.com/v4/letter/b/cc9497/32.png) [@Balduran](https://boards.straightdope.com/u/Balduran)\ **Post date:** [August 6, 2002, 1:41am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/2 "2002-08-06T01:41:45Z")
It is an interesting problem. According to my E&M text, “Even though (the exponentially decaying) fields exist on the other side of the interface, there is no energy flow through the surface. Hence total internal reflection occurs … The lack of energy flow can be verified by calculating the time-averaged normal component of the Poynting vector just inside the surface” So there is no energy flow across the interface, but you were asking about parallel to the interface. My guess is this will imply that all energy will be reflected back. This is because all the energy travelling parallel to the interface will eventually be absorbed and re-emitted by the media (recall the harmonic oscillator model of waves travelling through a media). The re-emitted waves will be emitted radially in all directions by the oscillators along the interface. The superposition of these waves will give wavefronts travelling away from and across the interface. Since there is no energy flow across the surface as discussed above, all waves will eventually be reflected back. That is just my guess as to the reason, but you can also show mathematically “the ratios of incident field amplitude and reflected field amplitude have modulus unity (although there is a phase change)” source: **Classical Electrodynamics** J.D. Jackson ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [August 6, 2002, 2:06am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/3 "2002-08-06T02:06:24Z")
> [@](#): > > \*Originally posted by Balduran \* > \*\*It is an interesting problem. According to my E&M text, “Even though (the exponentially decaying) fields exist on the other side of the interface, there is no energy flow through the surface. Hence total internal reflection occurs … The lack of energy flow can be verified by calculating the time-averaged normal component of the Poynting vector just inside the surface” > > So there is no energy flow across the interface, but you were asking about parallel to the interface. My guess is this will imply that all energy will be reflected back. This is because all the energy travelling parallel to the interface will eventually be absorbed and re-emitted by the media (recall the harmonic oscillator model of waves travelling through a media). The re-emitted waves will be emitted radially in all directions by the oscillators along the interface. The superposition of these waves will give wavefronts travelling away from and across the interface. Since there is no energy flow across the surface as discussed above, all waves will eventually be reflected back. > > That is just my guess as to the reason, but you can also show mathematically “the ratios of incident field amplitude and reflected field amplitude have modulus unity (although there is a phase change)” > > source: **Classical Electrodynamics** J.D. Jackson \*\* I remember reading somewhere, and darnit! I can’t remember where, that if another section of the same dielectric as the original is placed next to the one in which total reflection is happening, the “evanescent wave” that **Ring** wrote of enters the second dielectric. ---
**Author:** ![Balduran](https://avatars.discourse-cdn.com/v4/letter/b/cc9497/32.png) [@Balduran](https://boards.straightdope.com/u/Balduran)\ **Post date:** [August 6, 2002, 2:46am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/4 "2002-08-06T02:46:04Z")
**David** , I think you are correct. In fact, I believe that principle is used a lot in various fibre optic and other opto-electronic devices. One such application is called **distributed feedback** which enables you to send information on more than one carrier wavelength down an optical fibre therefore increasing the so-called _bandwidth_. The EM analogue to quantum tunneling perhaps? Maybe there is some fibre guy here who can confirm that. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [August 6, 2002, 3:17am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/5 "2002-08-06T03:17:30Z")
> [@](#): > > \*Originally posted by David Simmons \* > \*\* > > I remember reading somewhere, and darnit! I can’t remember where, that if another section of the same dielectric as the original is placed next to the one in which total reflection is happening, the “evanescent wave” that **Ring** wrote of enters the second dielectric. \*\* Yep, but it must be placed closer than a wavelength. If so the evanescent wave in effect jumps the gap and turns into a regular EM wave with a non zero Poynting vector and the energy of the original reflected wave decreases by the same amount. This phenomenon is almost an exact EM duplicate of quantum mechanical tunneling. **Balduran** I think what you say is correct but a _true_ plane wave would have to have infinite energy so I think were talking about things like dividing infinities of different orders etc. In other words a _true_ plane wave can’t really exist. ---
**Author:** ![Balduran](https://avatars.discourse-cdn.com/v4/letter/b/cc9497/32.png) [@Balduran](https://boards.straightdope.com/u/Balduran)\ **Post date:** [August 6, 2002, 3:14pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/6 "2002-08-06T15:14:04Z")
Why can’t a true plane wave exist? Can’t you use some dielectric tricks like lenses to create one from a radially expanding wave (ignoring edge effects in the propagation)? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [August 6, 2002, 5:46pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/7 "2002-08-06T17:46:02Z")
> [@](#): > > But if there is energy transmitted parallel to the interface then how can there be 100% reflection? Wouldn’t this energy subtract from the reflected wave? When the wave first arrives, there isn’t 100% reflection during the transient period, when the fields in the evanescent region build up to their steady-state levels. The 100% reflection refers to steady-state. It’s not much different than if you look at the reflection coefficient referenced to some plane in front of a perfectly reflecting plane. In the steady state, you have 100% reflection, but there’s a delay between when the front of the wave arrives at the plane, and when the reflection from the conductor arrives, and for that short time, you have no reflection. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [August 6, 2002, 5:54pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/8 "2002-08-06T17:54:53Z")
> [@](#): > > But if there is energy transmitted parallel to the interface then how can there be 100% reflection? Wouldn’t this energy subtract from the reflected wave? When the wave first arrives, there isn’t 100% reflection during the transient period, when the fields in the evanescent region build up to their steady-state levels. The 100% reflection refers to steady-state. It’s not much different than if you look at the reflection coefficient referenced to some plane in front of a perfectly reflecting plane. In the steady state, you have 100% reflection, but there’s a delay between when the front of the wave arrives at the plane, and when the reflection from the conductor arrives, and for that short time, you have no reflection. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [August 6, 2002, 7:43pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/9 "2002-08-06T19:43:19Z")
> [@](#): > > \*Originally posted by ZenBeam \* > \*\*When the wave first arrives, there isn’t 100% reflection during the transient period, when the fields in the evanescent region build up to their steady-state levels. The 100% reflection refers to steady-state. It’s not much different than if you look at the reflection coefficient referenced to some plane in front of a perfectly reflecting plane. In the steady state, you have 100% reflection, but there’s a delay between when the front of the wave arrives at the plane, and when the reflection from the conductor arrives, and for that short time, you have no reflection. \*\* But if you have a true plane wave the transmission of energy parallel to the interface isn’t a transient phenomenon. Here’s what Griffiths says about the steady state condition: > [@](#): > > “What we get is a so called evanescent wave…this is a wave propagating in the x direction (parallel to the interface!) and attenuated in the z direction…no energy is transmitted in the z direction, only in the x direction” > [@](#): > > _Originally posted by Balduran_ > Why can’t a true plane wave exist? Can’t you use some dielectric tricks like lenses to create one from a radially expanding wave (ignoring edge effects in the propagation)? Whenever I want the straight scoop on a science subject I always do a search to see if Matti Meron (Ph.D. physicist - University of Chicago) has posted anything on the subject. Here’s what he says: > [@](#): > > **Questioner:** Does the plane wave exist ? > > \*\*Meron’s answer: \*\* As a physical entity? No, it needs infinite room to accommodate it, takes infinite time to generate and carries infinite energy. ---
**Author:** ![Ring](https://avatars.discourse-cdn.com/v4/letter/r/6a8cbe/32.png) [@Ring](https://boards.straightdope.com/u/Ring)\ **Post date:** [August 6, 2002, 8:25pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/10 "2002-08-06T20:25:21Z")
**Balduran** I just notice that you suggested using a lens to create a plane wave. Unfortunately even using a perfect lens this won’t work, because the wavelike nature of light gives it a transverse momentum spread whenever it goes through a finite aperture. In other words the output will still spread. Please keep in mind I’m talking about a _true_ plane wave not an approximate or idealistic one used for modeling purposes. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [August 7, 2002, 4:03am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/11 "2002-08-07T04:03:54Z")
> [@](#): > > \*Originally posted by Ring \* > > > [@](#): > > > > \*Originally posted by David Simmons \* > > > > \*\* > > I remember reading somewhere, and darnit! I can’t remember where, that if another section of the same dielectric as the original is placed next to the one in which total reflection is happening, the “evanescent wave” that Ring wrote of enters the second dielectric. > > Yep, but it must be placed closer than a wavelength. If so the evanescent wave in effect jumps the gap and turns into a regular EM wave with a non zero Poynting vector and the energy of the original reflected wave decreases by the same amount. This phenomenon is almost an exact EM duplicate of quantum mechanical tunneling. > > \*\* I found the reference. It was in _Thirty Years That Shook Physics_ by George Gamow. He says that the total reflection doesn’t take place at a geometric line but over “several wavelengths” and if the second dielectric is placed within that distance some of the energy will be captured and will proceed through the second piece in the same direction as the original. And he also said that it is an analog of quantum tunneling, although he didn’t use that terminology. This sounds to me as if it would lead to a certain amount of cross-talk in fiber optic cables. _I.e._, if one fiber is right next to another, some small leakage from one at the reflecting surface should enter into the adjacent one and vice versa. Fiber optic cable designers? ---
**Author:** ![Balduran](https://avatars.discourse-cdn.com/v4/letter/b/cc9497/32.png) [@Balduran](https://boards.straightdope.com/u/Balduran)\ **Post date:** [August 7, 2002, 4:26am UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/12 "2002-08-07T04:26:37Z")
I will concede the point **Ring**. **David** you are correct about there being crosstalk (which I think actually refers to the direction parallel to the wave propagation). However, from what I recall, you actually design for it in some cases such as the afore-mentioned distributed feedback. In the direction parallel to wave propagation this is also used. It is possible to design a system in such a way to have the energy pretty much completely oscillates between two fibres. I think this is used for amplification purposes (you can reroute the signal from the current fibre onto another one that is designed to amplify the signal and then put it back into the original fibre). I don’t think eliminating crosstalk is too big of a deal. Like quantum tunneling, the larger the “barrier”, the shorter the decay distance so it is just a matter of putting an extra layer of dielectric around the fibre (known as cladding). ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 7, 2002, 10:53pm UTC](https://boards.straightdope.com/t/electrodynamics-question-probably-not-of-general-interest/122485/13 "2002-08-07T22:53:20Z")
Also, higher-quality optical fibers don’t rely on total reflection at the edge, but have a variable index of refraction such that the light is gradually bent back onto course, resulting in less loss.