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Confusion between R parallel and R orthogonal in section 10.2 #526

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IceTDrinker opened this issue May 10, 2020 · 7 comments
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Confusion between R parallel and R orthogonal in section 10.2 #526

IceTDrinker opened this issue May 10, 2020 · 7 comments
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@IceTDrinker
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IceTDrinker commented May 10, 2020

Hello,

I re-did the math (as proposed in the 10.2 section on refraction) and there seems to be a confusion on what R parallel and R orthogonal are.

The book defines R parallel as being parallel to the normal.

But from the book when we look at the expression in terms of known quantities we have :

We can now rewrite R′∥ in terms of known quantities:

R′ ∥=η/η′(R+(−Rn)n)

R.n is the component of R along n and we subtract that from R meaning that the right handside of the above equation is tangential (or along the surface) which contradicts the definition of R parallel being parallel to the normal.

I can't recall from past physics lecture if the convention is to consider direction with respect to the surface (which would make sense here as R parallel is parallel to the surface/tangent and R orthogonal is orthogonal to the surface)

Let me know if that makes sense or if I missed something.

Cheers

Edit : typo

@hollasch hollasch added this to the v3.2.0 milestone May 10, 2020
@hollasch
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The associated figure could also be improved, as R and R' are both shown as bidirectional, but the direction of the rays is important to understand the calculations. Is R pointing to the surface or away from it? I imagine that R' is the ray pointing into the refracting medium (downwards), but this should also be made explicit in the diagram.

@IceTDrinker
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Yes you are correct R and R' are not drawn and it could be complicated for people who have never seen wave propagation.

Maybe there was something about Fermat's principle in there ?

@trevordblack
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R and N point in opposing directions,
so R⋅N < 0, hence
-R⋅N > 0

The diagram could have R and R' with correct directions, but refraction is hard. And we should be careful about where what we change

@IceTDrinker
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IceTDrinker commented May 11, 2020

It does not change the fact that R parallel is not parallel to n :)

Edit : yes R and R’ pointing downwards would be the standard way of drawing things

@framigo
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framigo commented May 19, 2020

I agree with this:

the convention is to consider direction with respect to the surface (which would make sense here as R parallel is parallel to the surface/tangent and R orthogonal is orthogonal to the surface)

Here is the other side of the problem:
It is written in the book that R'=−sqrt(1−|R′|2)n.
However, sqrt(1−|R′|2) is a scalar, so R' must be parallel to n, i.e. parallel to n'.

Combined with @IceTDrinker's argument,

R′ ∥=η/η′(R+(−R⋅n)n)
R.n is the component of R along n and we subtract that from R meaning that the right handside of the above equation is tangential (or along the surface) which contradicts the definition of R parallel being parallel to the normal.

, there is clearly a confusion of convention here.

@trevordblack trevordblack self-assigned this May 24, 2020
@trevordblack trevordblack changed the title Confusion bewteen R parallel and R orthogonal in section 10.2 Confusion between R parallel and R orthogonal in section 10.2 May 28, 2020
@trevordblack
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Sorry Folks. I was totally wrong on this one.

You're all correct.

When I worked out the proof to confirm functionality, I used R_perp to represent perpendicular with the surface, and r_parallel to represent parallel to the surface.

After proving it, I realized that it would be more intuitive for the reader to work from the (already available) local normal.

I switched all of the math over, but must have forget to update the subscript.

So, that's pretty embarrassing.

Thanks everyone for pointing this out so others aren't taught incorrectly.
Fix is coming in the new minor release coming sometime this week or next.

@trevordblack
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Solved in #659

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