Wrong proof in textbook of optics?

nikolafmf

In textbook on optics by Pedrotti (and I think Fowles does the same) on page 493 it is said that:

kr=k_rr=k_tr

from which follows

(k-k_r)r=(k-k_t)r=(k_r-k_t)r=0,

where k, k_r and k_t are wave vectors of incident, reflected and transmitted wave.

Than Pedrotti concludes that k_r and k_t must lie in a plane determined by k and r. Must they?

Take, for example this:

k=(-1,0,1), r=(1,0,0), k_r=(-1,5,3), k_t=(-1,2,1).

If we put this vectors in the equation above, we will get 0, but k_r and k_t don't lie in the plane determined by k and r, which is y=0 plane.

So I conclude that Pedrotti is wrong. Am I right or wrong to conclude that? If wrong, why?

Simon Bridge

... so what is ##\vec{r}## ?

I think you may have missed part of the proof out though... merely showing that three vectors have the same extent against a fourth vector does not show that they are all co-planar. All you have to do is rotate one of the first three about the fourth to see this.

nikolafmf

Well, r is vector from the origin the point on the boundary where wave is reflected and transmitted. It is the point where the wave "touches" the boundary.

Well, I think Pedrotti takes only the equations I have written in his argument (Introduction to Optics, third edition, page 493).

And what does that mean three vectors to have the same extent against fourth vector? How rotation helps in proving coplanarity?

Simon Bridge

In the relations you have shown, what are the operations? Are yu multiplying the magnitudes of the vectors, taking the dot-product, what?It is an arbitrary vector then? (Since the choice of origin is arbitrary.) Therefore there is no reason that the rays should be in a plane defined by the incedent ray and some arbitrary vector unless the origin is in some special place.

This sort of setup is normally used when you want to invoke the principle of least time.I don't have that one. You should double-check the context that he's doing this in.

nikolafmf

Operation is dot product.

Yes, origin is arbitrary.

Simon Bridge

What does the dot product of two vectors tell you - geometrically?
From that - what is the first relation in post #1 telling you?... You know from the laws of reflection and transmission that ##\vec{k}##, ##\vec{k}_r##, and ##\vec{k}_t## are coplanar ... so what does ##\vec{r}## have to be in order that the k vectors are in the plane defined by ##\vec{k}## and ##\vec{r}##?

nikolafmf

The first relation is telling me that projection of k on r are equal in magnitude.

Yes, from laws of reflection and transmission it follows they are coplanar, but now I want (and Pedrotti wants) to prove that. It is something should be proved. I don't know what r should be in order other vectors to lie in the plane defined by k and r, but it is irrelevant, since origin can be always taken arbitrary.

nikolafmf

I checked "Principles of Optics" by Born, who says that all wave vectors lie in the plane specified by incident wave vector and vector normal to the boundary, not the position vector. After some consideration I found this true. So, proof by Pedrotti is incorrect.

jtbell

I'm looking at a copy of Pedrotti right now.

We're talking about plane waves here, so ##\vec k##, ##\vec k_r## and ##\vec k_t## must each be the same everywhere on the interface surface (the xy plane in Pedrotti's Figure 23-1). Equations (23-6) and (23-7) have to be true not only for ##\vec r = (1,0,0)## which fits Figure 23.1, but also for ##\vec r = (0,1,0)## and ##\vec r = (1,1,0)## and indeed for any ##\vec r## that has z = 0.

I haven't thought through the logic, but I suspect this is the "missing link" in Pedrotti's argument.

nikolafmf

Yes, it it true. But, again, r is arbitrary, so Pedrotti's conclusion is wrong. Below the equation (23-7) instead of plane determined by k and r it should be "the plane determined by k and the vector normal to the boundary at the interacting point"

jtbell

In the most general case, this is true. However, it seems that Pedrotti is requiring that ##\vec r = 0## lies in the surface, for convenience. Note the argument leading to equation (23-5).

I think it's possible to make the origin completely general, at the cost of introducing another vector.

nikolafmf

Even if r lies on the boundary, what if k has also y component? Than the argument again will not be correct.

nikolafmf

Pedrotti says that the plane of incidence is xz plane. With this constraints, his conclusion is right, yes. But this is not the most general case. In the most general case his conclusion is wrong.

Simon Bridge

... This text is in it's third edition ... what are the odds that a mistake like that in a popular text book used by millions of people all over the World has such a simple mistake in it?
It may be interesting to see if the "error" is in the 2nd ed.

Usually, when something looks like it is missing in the math, it is implied somewhere in the text ... maybe by context. I'm not looking at it - maybe jtbell can see what it could be?

If you feel confident this is a genuine problem with the text, you could try writing the publisher about it (after checking to see if there is a 4th edition).

jtbell

I think that derivation actually is somewhat "fishy." I remember having similar problems with other derivations in Pedrotti when I taught optics from it. In this case, I see the fundamental problem as this:

At each point on the boundary, the phases of the three waves (incident, reflected and transmitted) must be equal. That phase is different at different points on the boundary. Considering only the incident and transmitted wave for simplicity, the basic condition is that ##\phi = \phi_t## at a given location ##\vec r##.

This does not mean that we can say (as Pedrotti does) that (at t = 0) ##\vec k \cdot \vec r = \vec k_t \cdot \vec r##. Each wave has, in general, an initial phase, i.e. the phase at the origin (##\vec r = 0##) at t = 0. So we have instead, ##\vec k \cdot \vec r + \phi_0 = \vec k_t \cdot \vec r + \phi_{0t}##.

Not only is the initial phase different for the two waves, it also depends on the origin of your coordinate system, which is where ##\vec r## starts. If you move the origin, the phase of a wave (##\phi = \vec k \cdot \vec r + \phi_0##) at a given point on the boundary doesn't change, but ##\vec r## changes, so ##\phi_0## must also change in order to keep ##\phi## constant.

nikolafmf

I sent a mail to one of the authors of the book. He was thankful and said that he would add this to his list of needed corrections.

As I said, Born (in his book "Principles of Optics") has different thoughts about the question in which plane the wave vectors must lie. It can be that both are correct.

jtbell

Aha! I looked at Hecht's optics textbook, and it gave me the answer to your original question. First recall what I wrote before:

Rearranging the last equation, we get

$$(\vec k - \vec k_t) \cdot \vec r = \phi_{0t} - \phi_0$$

Note the right side is a constant, for a given choice of origin. Hecht reminded me that in general, if ##\vec a \cdot \vec r = b## where ##\vec a## is a constant vector and b is a constant scalar, then the allowed values of ##\vec r## define a plane normal (perpendicular) to ##\vec a##. In our case, this plane is the reflection/refraction boundary. Therefore, the equation above tells us that ##\vec k - \vec k_t## is normal to the boundary.

The three vectors ##\vec k##, ##-\vec k_t## and ##\vec k - \vec k_t## form a vector-addition triangle, therefore they lie in the same plane, namely the plane of incidence, which is defined by ##\vec k## and the normal to the boundary.

We can make the same argument for ##\vec k## and ##\vec k_r##. Therefore all three vectors ##\vec k##, ##\vec k_t## and ##\vec k_r## lie in the same plane.

Pedrotti chooses the origin of ##\vec r## to lie in the boundary plane, which makes ##\phi_{0t} - \phi_0 = 0## in the equation above. This is a special case of Hecht's more general argument. In fact, Hecht goes on to assume this for simplicity in his further derivation of Fresnel's equations for reflection and refraction.