Problem with proofing polarity of the reflected and transmitted wave.

[yungman]

This is an exercise problem, BUT my question is not the exercise, but the fundamental assumption of the problem. That's the reason I don't post it in the homework section as it has nothing to do with solving the problem. I have issue with the given assumption all together.

This is a case of Normal Incidence where the incident wave travels in +z direction in medium 1. The incident wave hitting the medium 2 boundary on xy plane where z=0. Incident E is in x direction and H in y direction.

This is the original question word by word:
In writing Eq.9.76 and 9.77, I tacitly assumed that the reflected and transmitted waves have the same polarization as the incident wave...along the x direction. Prove that this must be so.

Hint: Let the polarization vectors of the transmitted and reflected waves be:
##\hat n_T=\hat x\cos\theta_T+\hat y \sin\theta_T\;\hbox { and } \;\hat n_R=\hat x\cos\theta_R+\hat y \sin\theta_R##
and prove from the boundary conditions that ##\theta_T=\theta_R##.

##\hbox{(9.76)}\;\Rightarrow\;\tilde E_R(z)=\hat x E_{0R}e^{-jk_1z}\;\hbox { and }\;\tilde B_R(z)=-\hat y \frac 1 {v_1}E_{0R}e^{-jk_1z}##

##\hbox{(9.77)}\;\Rightarrow\;\tilde E_T(z)=\hat x E_{0T}e^{-jk_2z}\;\hbox { and }\;\tilde B_T(z)=\hat y \frac 1 {v_2}E_{0T}e^{-jk_2z}##This is from "Introduction to Electrodynamics" 3rd edition by Dave Griffiths, the gold book for undergrad physics. The pages of relevant are page 384 to 386.

My first issue is: By assuming the polarity of ##\tilde E_T## is ##\hat n_T=\hat x\cos\theta_T+\hat y \sin\theta_T## and ##\tilde E_R## is ##\hat n_R=\hat x\cos\theta_R+\hat y \sin\theta_R##, the author already assuming the ##\tilde E_R## and ##\tilde E_T## are tangential to the boundary. I accept that ##\tilde E_{0T}## has the same polarity as ##\tilde E_{0I}## as this is proven by the boundary condition that tangential E is continuous across the boundary. So this in natural the polarity follows.

BUT I have found nothing in over 5 books explaining why ##\tilde E_{0R}## is the same polarity. In fact, there is no proof I can find that ##\tilde E_{0R}## is tangential to the boundary. By assuming that ##\tilde E_R## is ##\hat n_R=\hat x\cos\theta_R+\hat y \sin\theta_R##, the author already make an important assumption that is not proven yet here.

I might still have more to follow later.

Thanks

Alan

 

[Simon Bridge]

It's set up in the initial conditions isn't it?
Where else could they point given the incidence direction?

 

[yungman]

It's set up in the initial conditions isn't it?
Where else could they point given the incidence direction?

Thanks for the response. It is GIVEN ##\tilde E_I(z)=\hat x E_I(z)##. It is given as a case of normal incidence with boundary on xy plane. That's it.

There is no question that ##\tilde E(z)## is tangential to the boundary plane. I just want to proof of the ##\tilde E_R(z)=\hat x E_R(z)##.

 

[Simon Bridge]

Well... you know the direction of the electric and magnetic fields wrt the wave-vector right?
So I'm not sure what the problem is ... the hint in the exercize just suggests lining up the cartesian axis with the polarity.

Or do you think that the proof is assuming what is to be proved?

 

[yungman]

Well... you know the direction of the electric and magnetic fields wrt the wave-vector right?
So I'm not sure what the problem is ... the hint in the exercize just suggests lining up the cartesian axis with the polarity.

Or do you think that the proof is assuming what is to be proved?

Yes, I think the Hint already made a big leap assumption that the reflected wave is tangential to the boundary. The boundary condition only said about the transmitting E is continuous cross boundary so it remain tangential. I see nothing that the reflected wave of a tangential E is still tangential to the boundary.

To put it in another way, reflection coef ##\Gamma## is a scalar function that give the scalar info of the reflected wave, it does not give the vector of the reflector wave.

Funny, it make a whole world of sense if you consider an optical image, like if you put a pencil in x direction, the reflected image is absolutely in x direction.

BUT this is electric field amplitude, not a physical height. How can you assume it's the same as optical image?

 

[Simon Bridge]

Well you can always attempt the proof another way ;)

The approach in the hint is to set the situation in place and show that it gives results that are consistent with the rest of optics.

Try assuming a different relationship and see what happens.

 

[yungman]

can you give me some guidance?

For me, the proof I can accept is if ## \hat n_R=\hat x \cos \theta_R+\hat y \sin\theta_R+ \hat z f(\theta_R)## and proof that y and z component are both zero.

I was thinking about it. It will go a long way if I can proof the reflected wave of a TEM wave is a TEM wave. That will automatically implies that in the normal incidence case, the reflected wave ##\tilde E_R## is confined to x and y direction only...which is tangential to the boundary plane. Because Snell's Law already proven ##\theta_I=\theta_R##. If ##\theta_I##=0, ##\theta_R##=0 and both waves propagate perpendicular to the boundary plane.

Thanks

 

[yungman]

This is a scanned copy of the solution manual, I don't even know why the author went through the trouble to proof ##\theta_T##=0! So if it is zero, what does that prove the polarity of ##\tilde E_R## and ## \tilde E_T## is same polarity as ##\tilde E_I##?

 

[yungman]

I found a partial prove using equations of J D Jackson p304 (7.37)

##[\epsilon(\vec E_0+\vec E_0'')-\epsilon'\vec E_0']\cdot \hat n=0## (7.37a) for normal E.
##[\hat k \times \vec E_0+\hat k''\times\vec E_0''-\hat k'\times\vec E_0']\cdot \hat n=0## (7.37b) for normal B.
##(\vec E_0+\vec E_0''-\vec E_0')\times\hat n=0## (7.37c) for tangential E.
##\left[\frac 1 {\mu}(\hat k \times \vec E_0+\hat k''\times\vec E_0'')-\frac 1 {\mu'}(\hat k'\times\vec E_0')\right]\times\hat n=0## (7.37d) for tangential B.
(7.37a)##\Rightarrow\;[\epsilon_1(\vec E_{0I}+\vec E_{0R})-\epsilon_2\vec E_{0T}]\cdot \hat z=0## (A).
(7.37b)##\Rightarrow\;[\hat z \times \vec E_{0I}-\hat z \times\vec E_{0R}-\hat z \times\vec E_{0T}]\cdot \hat z=0## (B).
(7.37c)##\Rightarrow\;(\vec E_{0I}+\vec E_{0R}-\vec E_{0T})\times\hat z=0## (C).
(7.37d)##\Rightarrow\;\left[\frac 1 {\mu_1}(\hat z \times \vec E_{0I}-\hat z\times\vec E_{0R})-\frac 1 {\mu_2}(\hat z\times\vec E_{0T})\right]\times\hat z=0## (D).

We know ##\vec E_{0I}=\hat x E_{0I}##.
Let ##\vec E_{0R}=\hat x E_{0R_x}+\hat y E_{0R_y} + \hat z E_{0R_z}##
and ##\vec E_{0T}=\hat x E_{0T_x}+\hat y E_{0T_y} + \hat z E_{0T_z}##

From (C) ##\Rightarrow\; E_{0R_y}=E_{0T_y}##
From (D) ##\Rightarrow\; \frac {E_{0R_y}}{\mu_1}=-\frac {E_{0T_y}}{\mu_2}##
From this, the only possibility is both ##E_{0R_y}=E_{0T_y}##=0.

But I cannot see in these boundary conditions given by J D Jackson can prove ##E_{0R_z}=E_{0T_z}##=0.

Can anyone help please. I am really stuck. So far I proved y components are zero already. If someone can give me prove the reflected and transmitted wave of a Plane Wave is a Plane Wave, then that proves z components are also zero and that will answer my question.

Thanks

 

[Simon Bridge]

You could try it the other way - assume the statement is false and show that this would be inconsistent.

 

[yungman]

There is no proof of the reflected wave and transmitted wave of a Plane Wave is a Plane Wave? I thought this is the ABC of optics and EM! That's all I need. I already proved the reflected wave and transmitted wave of a E=##\hat x E## Plane Wave at normal incidence has NO Y component already!

 

[Simon Bridge]

You are the one doing the proof. You are the one wants to do it the hard way.
I'm not going to serve it up on a platter - it's your work.

You should be careful with your expectations of proofs though - you can prove the math - you can never prove the physics. All you are ever doing with this sort of thing is showing that one thing is consistent with a bunch of other things thought to be true. The best you can manage is to show that the geometry works. Welcome to empiricism.

 

[yungman]

There is no proof of a reflected wave of a Plane Wave is a Plane Wave?

 

[Simon Bridge]

You can direct a good approximation plane wave at a surface and see that there is a good approximation plane wave reflected. That's not "proof" though.

You can put the plane-wave math and the boundary conditions into Maxwell's equations and show that the solution requires a plane wave reflected. That proves that the idea is consistent with Maxwell's equations. Which is what Griffiths appears to be doing.

 

[yungman]

You can direct a good approximation plane wave at a surface and see that there is a good approximation plane wave reflected. That's not "proof" though.

You can put the plane-wave math and the boundary conditions into Maxwell's equations and show that the solution requires a plane wave reflected. That proves that the idea is consistent with Maxwell's equations. Which is what Griffiths appears to be doing.

First of all, I want to make sure you are not offended, I was truly surprised, I thought there must be a very simple explanation that the reflected wave of a plane wave on a flat surface is a plane wave.

If you have the page of either Griffiths, J D Jackson, Balanis, Flanklin, Hyatt & Buck, Kraus that have the derivation to proof this, please tell me the page. I have been looking but have no luck. But books have a way to hide important info somewhere inside and not showing in the chapter heading or index.

 

[Simon Bridge]

I don't have any of those books.
Reflection and refraction of plane waves from/through a dielectric interface is routinely handled at college level so you can just google for course notes and student resources. i.e.
http://ecee.colorado.edu/~ecen3400/Chapter%2022%20-%20Reflection%20and%20Refraction%20of%20Plane%20Waves.pdf

Note: a plane wave in does not guarantee a plane wave out - that will depend on the material properties of the target. You have to define the surface - which is what the boundary conditions are all about.
The above text walks you through the standard variations that students get introduced to, starting from normal incidence to a perfect conductor.

However, since you won't accept the math you've been shown, and you won't do your own math, I cannot help you. Good luck.
Sorry.

 

[yungman]

However, since you won't accept the math you've been shown, and you won't do your own math, I cannot help you. Good luck.
Sorry.

Thanks for the link, at least this one I saw it said reflection from a plane boundary of homogeneous media is a plane wave. This is what I've been looking for as I specified in post #1 that boundary is the xy plane. It said it's too hard to work out the formulas. There is no proof. But at this point, I just have to take their words.

One thing...What math have I been shown here? I see nothing shown to me so far. I did show ALL the math I manage to solve myself, proofing y component is zero, hours in typing out the equations in Latex... If I have any idea how to approach this, I won't be spending hours just to type the detail long equation in Latex? I sure look at the 4 boundary conditions in all different ways.

But anyway, thanks for your time. If I accept the reflected wave is TEM, there will be no z component, I proof there will be no y component, so this answer my question. I still totally disagree the way Griffiths approach this problem all together. As I posted the solution from Griffiths in post #8, I disagree even more. What does ##\theta_T## and ##\theta_R## anything to do with proofing there will be no y components even accepting that reflected and transmitted wave are all TEM wave?!

Thanks

Alan

 

[yungman]

One thing, I can perfectly see in optics that the reflected image of the object behave with the given dimension. For example, if it is a pencil in x direction, I can absolutely see how the reflected image should be.

But this is electric field. What is the physical meaning of it's in x direction vs y direction? It is not a physical dimension that you can use a ruler to measure! That's really the bottom line of the difficulty I have "seeing" the images.