If Maxwell's equations are linear....

[Charles Link]

I posted previously about this topic a couple of years ago, and it really came across like a ton of bricks, but now that I have established some credibility, perhaps it will be read with some interest.

In the course of my career, (in the years 1986-1990), on two occasions I discussed with two very intelligent physics PhD.'s a problem that surfaces in regards to interference patterns. The first PhD. knew very little Optics and (around the year 1986) was asking me, "how can you possibly get an interference pattern? Are you telling me if I have one solution to Maxwell's equations, and a second solution, that the sum is not a solution?" I didn't have a good answer for him at the time. I told him there seems to be something non-linear going on, but I couldn't put my finger on it.

A couple years later, around 1990, I was working on the problem of reflections in sections of r-f cables with slightly different characteristic impedances with another older physics PhD., and the problem involved the Schellnukopf equations (Edit: A google shows these are the "Schoelkopf" equations after R.J. Schoelkopf). We were basically solving for what is the Fabry-Perot effect in Optics at normal incidence. After observing some of the effects we were seeing, the results seemed to be non-linear. The existence of a second beam seemed to affect what the first one did. I asked the older PhD. , "Are you sure these equations are linear". His reply was a very quick, "Of course they are linear". I really couldn't argue it further with him at this point.

About 20 years later, in 2008, it finally dawned on me the solution to this dilemma: In analyzing both of these systems, there is an energy equation=energy is conserved= and that equation is second order in the E-field parameter. The result is that there is a non-linear process in the E-field parameter going on in these systems that does make interference patterns not only possible, but quite commonplace. The systems are linear in the E-fields, but they are not linear in how the energy redistributes itself. The energy distribution does not need to obey linear principles, because the energy equations are second order in the E-field parameter.

Anyway, hopefully at least a couple of you find this good reading.

 

[hutchphd]

How do these considerations differ from "squaring" the wavefn in quantum mechanics? I guess I don't really see what is the surprise?

 

[Charles Link]

How do these considerations differ from "squaring" the wavefn in quantum mechanics? I guess I don't really see what is the surprise?

It's the exact same thing as the squaring in Q.M.
The effect with a Michelson interferometer is very surprising. If you just have a single beam, the beamsplitter does a 50-50 energy split to both receivers. The presence of the second beam (from the second mirror) can cause all of the energy to go to one receiver or the other , depending on the relative phases of the two beams, which is determined by the path distance difference.

 

[hutchphd]

I agree wholeheartedly it is a surprise which is why we expend our hours at this silliness. I should have inquired as to why it was unexpected by you since the context is the same. Very good!
I'm still surprised by the interposed 45deg polarizer result and yet I can do the formalism...and don't get me started on soap bubbles!

 

[Charles Link]

An important feature for the 45 degree beamsplitter is that it is a reflection from just one face, (air-material interface), with an AR (anti-reflection) coating on the other face making it asymmetric. Another important feature is the Fresnel reflection coefficient is $\rho=\frac{n_1-n_2}{n_1+n_2}$ resulting in a $\pi$ phase shift (= factor of -1) for the external reflection as opposed to the internal one.
I actually wrote up a whole Insights article on this topic, see https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/, before someone informed me that Schwinger solved this whole thing with a matrix formalism, sometime around 1930.
(See also my Edit of post 1, where it is actually the "Schoelkopf" equations, and not "Schellnukopf").

 

[Klystron]

I have always felt comfortable with RF and EM equations as you describe in your Insights article and many related posts. I had several years experience using RF Fresnel lenses as a young tech and am somewhat dismayed that lenses do not receive the attention on the forums as reflective antennae. The physicist insisting on solving emf problems using linear equations was likely oversimplifying if not ignoring the underlying physics.

If I can offer an audio analogy, people often describe and seem to understand sound in simplistic linear terms ignoring logarithmic relations and the 3-dimensional nature of sound. A similar analogy involves the scales used to measure earthquake intensity and describe seismic waves to the public. News reports tend to describe damage and subjective shaking effects in lieu of actual measurements to compare events thereby avoiding, as you say "a little algebra".

 

[PhilDSP]

I'd have to say the Maxwell Equations are inherently non-linear. We tend to forget that when we're working with media that is isotropic, such as a vacuum. In that case $\epsilon$ and $\rho$ are simply scalar constants and the situation is one of linearity. However, in media that is anisotropic $\epsilon$ and $\rho$ become tensors where non-linearity becomes the rule.

 

[Charles Link]

I'd have to say the Maxwell Equations are inherently non-linear. We tend to forget that when we're working with media that is isotropic, such as a vacuum. In that case $\epsilon$ and $\rho$ are simply scalar constants and the situation is one of linearity. However, in media that is anisotropic $\epsilon$ and $\rho$ become tensors where non-linearity becomes the rule.

The subject of my discussion here is not non-linear Optics. I am discussing the very ideal case of completely linear behavior from Maxwell's equations, and treating only media that can be approximated as being completely linear.

 

[PhilDSP]

But don't the EM waves interacting with either a beam splitter or a dielectric interface essentially mean that you're dealing with media and therefore have multiple values of $\rho$ for example? That's where I believe I see the non-linearities arise.

 

[Charles Link]

But don't the EM waves interacting with either a beam splitter or a dielectric interface essentially mean that you're dealing with media and therefore have multiple values of $\rho$ for example? That's where I believe I see the non-linearities arise.

The Fresnel reflection coefficients $\rho=\pm \sqrt{R}$ where $R$ is the energy reflection coefficient for a single beam. With two beams present $R$ is no longer a good number, but the $\rho's$ remain good. For a 50-50 beamsplitter, $R=\frac{1}{2}$ and $\rho=\pm \frac{1}{\sqrt{2}}$.
The problem is solved by treating the individual E fields with their respective $\rho's$ and $\tau's$, and summing the resulting E fields. The system is non-linear in the sense that the energy solutions don't superimpose. Energy is completely conserved, and as magic as it may seem, when two coherent beams are incident on a single beamsplitter, all of the emerging energy, i.e. 100% of the incident energy, can wind up in a single direction. In that case the beamsplitter combines the two beams.
Work the problem with the Fresnel coefficients, keeping track of the phase change of $\pi$ for the external reflection, ($\rho$ gets a minus sign), and you will see how it works. The composite $\tau=\frac{1}{\sqrt{2}}$ in both cases. (There are two air-material interfaces to cross for the transmission. The beam also crosses the interface with the AR coating. Thereby the composite $\tau=\frac{1}{\sqrt{2}}$ is actually quite simple).

 

[Charles Link]

@PhilDSP I strongly urge you to work through this with the Fresnel coefficients in its entirety. See also the Insights, https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/ where I believe I worked it out for the reader. It is actually surprisingly simple once you see what is going on, as well as quite amazing, but it can be very elusive and not completely obvious if you don't actually sit down and work out the case of two beams of arbitrary phase incident on a single 50-50 beamsplitter from two directions, with an AR (anti-reflection) coating on one face of the beamsplitter, so that all reflection occurs from the one air material interface. One of these reflections is external, the other internal.

 

[PhilDSP]

Thanks so much for your attention and suggestions Charles! My immediate inclination is to approach the problem from a bit different mathematical foundation. I hope I'll have time to see it through to complement your observations. Of course we all have different conceptual foundations and have to proceed from there. Vive la difference!

 

[Charles Link]

Thanks so much for your attention and suggestions Charles! My immediate inclination is to approach the problem from a bit different mathematical foundation. I hope I'll have time to see it through to complement your observations. Of course we all have different conceptual foundations and have to proceed from there. Vive la difference!

I saw the Optics equations for working multiple layers of thin films=the Fabry-Perot effect in 1979. Something seemed to be missing, but I couldn't put my finger on it until 2008-2009.$\\$
In 2008-2009 I discovered by myself that two beams incident on a single interface at normal incidence, (have the interface do a 50-50 energy split by proper choice of $n_1$ and $n_2$), from opposite directions exhibits the Fabry-Perot effect with complete energy conservation. That case is even much simpler to treat than the 45 degree beamsplitter, because computing the $\tau_{composite}$ at 45 degrees (the product of the $\tau ′s$ across two air-material interfaces) actually takes a considerable amount of work, even though it turns out to be a very simple $\tau_{composite}=\frac{1}{\sqrt{2}}$, whereas the $\tau_{12}=\frac{2n_1}{n1+n_2}$ at normal incidence is quite simple, and requires almost no effort to compute it. $\\$ Note: For these Optics problems you can use simplified intensity units: Intensity$I=nE^2$ $\\$ e.g. For a single beam coming from the left: $\\$ $I_{incident}=n_1 E_{incident}^2$ $\\$ $I_{reflected}=n_1(ρ_{12}E_{incident})^2$,and $\\$ $I_{transmitted}=n_2(τ_{12}E_{incident})^2$.

 

[PhilDSP]

Oh. Thanks for that additional insight!

 

[Charles Link]

For the case of two beams, you take $E_{emerging\, left}=E_{1 \, reflected}+E_{2 \, transmitted}$ and $E_{emerging \, right}=E_{1 \, transmitted}+E_{2 \, reflected}$. $\\$ You then of course write $\\$ $I_{emerging \, left}=n_1 E_{emerging \, left}^2$ and $\\$ $I_{emerging \, right}=n_2 E_{emerging \, right}^2$. $\\$ Don't forget to include a possible phase term between $E_{1 \, incident }$ and $E_{2 \, incident}$, and use the Fresnel coefficients to compute the various terms. $\\$ Meanwhile the incident energies are just $\\$ $I_{1 \, incident}=n_1 E_{1 \, incident}^2$ and $\\$ $I_{2 \, incident }=n_2 E_{2 \, incident}^2$. $\\$ Now set $I_{1 \, incident}=I_{2 \, incident}$ and $\\$ choose $n_1$ and $n_2$ so that energy split for single beam $R=\frac{(n_1-n_2)^2}{(n_1+n_2)^2}=\frac{1}{2}$. $\\$ Finally $I_{emerging \, left}$ and $I_{emerging \, right}$ will be a function of the relative phases, with the sum being conserved and equal to the sum of the incident intensities. $\\$ Note $\rho_{12}=\frac{n_1-n_2}{n_1+n_2}$ and $\rho_{21}=-\rho_{12}$.

 

[Charles Link]

Notice that the single beam energy reflection coefficient $R=(\rho_{12})^2=(\rho_{21})^2$.
With $n_2>n_1$, and a 50-50 energy split, $\rho_{12}=-\frac{1}{\sqrt{2}}$ , and $\rho_{21}=+\frac{1}{\sqrt{2}}$.
The minus sign is equivalent to a $\pi$ phase shift because $e^{i \pi}=-1$. For a quick qualitative look, the beam that gets reflected off of the higher index surface undergoes a $\pi$ phase change.
For very simplistic purposes, you can do a qualitative analysis that doesn't require the complete algebra.
You can do the case of incident beams (normal incidence) that are in phase, and then compare with the case of incident beams that are $\pi$ out of phase.
For the first case,(incident beams in phase), for $n_2 >n_1$, all of the emerging energy goes to the right. (This is because the beam that is incident from the left undergoes a $\pi$ phase change upon reflection. Meanwhile the transmitted beams and the beam incident from the right that gets reflected don't undergo any phase changes).
For the $\pi$ relative phase case, all of the emerging energy goes to the left.
(This is for incident beams of equal energy with $n_1$ and $n_2$ such as to make a 50-50 energy split for a single incident beam).

 

[Charles Link]

@PhilDSP Please see the latest additions above.

 

[Charles Link]

Note: Posts 10, 13, 15, and 16 above basically explain the science of Interferometry.

 

[tade]

@Charles Link Basically what it means is that Maxwell's equations applied to E and B vector fields are distributive, but the same does not apply for the amplitude-squared?

 

[Charles Link]

@tade Very good ! It's a slightly different type of distributive law, but I think you got the basic idea.

 

[Charles Link]

Just a general comment: $\\$
This solution (and the resulting interference) for the case where two coherent waves of arbitrary phase are incident from opposite directions onto an interface at normal incidence (the extension to other angles like 45 degrees is really straightforward) using the Fresnel coefficients is such an important fundamental physics concept that really everyone should have a working knowledge of it. $\\$
For some reason, it doesn't seem to be emphasized in the basic curriculum, and the result is that for many, how the various interferometers work remains a puzzle, and unnecessarily so. $\\$ With a Michelson interferometer, the beamsplitter is used twice: The first pass splits the beam, and then the beams are recombined, with the resulting distribution of energy being phase-dependent. $\\$ And with the Michelson interferometer, it also has two beams incident on a single interface=one face of the beamsplitter, with the other face having an AR (anti-reflection) coating. It is also of interest that the waves incident on the beamsplitter after returning from the mirrors do not interact with the initial incident wave that is getting split by the beamsplitter. The two processes are independent of each other. $\\$ In a simplified version of the Michelson interferometer, the mirrors can be replaced by two mutually coherent plane wave sources, and the resulting interference is the same as that of a plane wave type Michelson, where the initial plane wave source is split by the beamsplitter.

 

[Charles Link]

The question even comes up by students of Optics, is it possible to combine/overlay two macroscopic size beams together into a single beam and conserve energy? $\\$ That is exactly what happens in the case of the plane-wave Michelson interferometer, if you adjust the phases of the two incident beams properly to get all of the energy to arrive at the receiver with complete constructive interference there, and complete destructive interference at the other point. $\\$ The algebra for the process is done by the Fresnel coefficients. It is important to note when doing the energy conservation calculation, that the beam intensity is proportional to the square of the electric field amplitude. Basically, in simplified units, intensity $I=nE^2$, where $n$ is the index of refraction of the medium. $\\$ To show the simplified arithmetic, two beams each with amplitude $E_o$ get recombined into a single beam with $E_T=\frac{E_o}{\sqrt{2}}+\frac{E_o}{\sqrt{2}} =\sqrt{2} E_o$ where the $\frac{1}{\sqrt{2}}$ factors are the Fresnel coefficients (one for reflection, and the other for transmission). Finally, we compute $I_T=E_T^2=2E_o^2$, and energy is conserved. At the other point, the Fresnel reflection coefficient picks up a minus sign if it is the external reflection, so that $\rho=-\frac{1}{\sqrt{2}}$ and $E_T =0$ at the other location.

 

[PAllen]

Somehow, this all seems obvious. Just take water waves. The wave equation in terms of displacement ( a signed quantity) is additive. Amplitudes are not.

 

[Charles Link]

Somehow, this all seems obvious. Just take water waves. The wave equation in terms of displacement ( a signed quantity) is additive. Amplitudes are not.

For the case of two waves incident on the interface from opposite directions at the same time, the result with the interference is that you can get no energy emerging in one of the directions and all of the emerging energy going in the other. This IMO is quite remarkable. With two beams, the 50-50 energy beamsplitter no longer splits the energy in such a fashion as it does for a single incident beam. $\\$ Note: In order to get such a thing with light waves, the beamsplitter surface needs to be optically flat. $\\$ The ability to combine two beams with a beamsplitter as mentioned in post 22 is also something of interest, that I think undergraduate physics students would especially find interesting.

 

[Charles Link]

@vanhees71 Might you have an opinion on this? Perhaps I underestimate them, but I believe most upper level undergraduate physics students who are taking an Optics course have not seen the problem of two waves incident on a single interface from opposite directions, and also have not treated the Michelson interferometer in that fashion. $\\$ Our upper level undergraduate Optics class (University of Illinois at Urbana-Champaign) in 1976 treated the Michelson as two images (sources) that interfere with each other. The complete mathematical details were omitted. We treated the case of the diffuse source Michelson, (where you get ring patterns of interference). The plane wave (Twyman-Green) Michelson was not really treated at all in that course.

 

[PAllen]

Let me clarify. I think the scenarios discussed in this thread are fascinating, and correct treatment is not obvious just based on elementary use of Maxwell’s equation. Much of the details were new to me. The only aspect I was commenting on was that constructive and destructive interference are elementary undergrad physics, and these already imply that energy cannot be remotely additive, since it is classically determined by amplitude (a nonadditive scalar quantity for waves).

 

[Charles Link]

Let me clarify. I think the scenarios discussed in this thread are fascinating, and correct treatment is not obvious just based on elementary use of Maxwell’s equation. Much of the details were new to me. The only aspect I was commenting on was that constructive and destructive interference are elementary undergrad physics, and these already imply that energy cannot be remotely additive, since it is classically determined by amplitude (a nonadditive scalar quantity for waves).

Hi @PAllen Very glad to hear it. As a graduate student, I was actually the T.A. (teaching assistant) for the upper level undergraduate Optics course at UCLA (1979-1980). Although I think I did a very good job of teaching the material to the students back then, I think I could have done an even better job if I could have presented this material on the Michelson interferometer to the students, but I didn't figure it out myself until about 30 years later in 2009. $\\$ As elusive as this solution was to me before I finally recognized it in 2009, it is actually very simple. I urge the reader to work through the case themselves of two plane waves of equal intensity ($I=nE^2$) incident at normal incidence onto an interface from opposite directions where $\rho=\pm \frac{1}{\sqrt{2}}$, ($R=\frac{1}{2}$ for a single beam), with arbitrary phase. I think many will be surprised at how simple the results are for the emerging energy distribution.$\\$
@Dale Did you ever work through the complete algebra?=give it a try: $\\$ $E_1=A \cos(\omega t)$, and $E_2= B \cos(\omega t+\phi)$. $\\$ The transmission coefficients are $\tau_{12}==\frac{2n_1}{n_1+n_2}$, and $\tau_{21}=\frac{2n_2}{n_1+n_2}$, and $n_1$ and $n_2$ have $\rho=\frac{n_1-n_2}{n_1+n_2}=\pm \frac{1}{\sqrt{2}}$. $\\$ We also have $I_1=n_1A^2=I_2=n_2B^2$.$\\$ The rest is algebra=computing the resultant emerging E-fields and the resulting intensity of each. (I actually worked some of the algebra in the Insights article below for $n_1=1$ and $n_2=n_o$, (basically it's just the ratio $\frac{n_2}{n_1}=n_o$ that matters), but I urge you to try it with the phase term as well. See: https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/)