nice math proof arctan(x)+arctan(1/x)=sign(x)pi/2

[johnsmi]

Hi everybody, im new here and like physics very much
I came across a post here about the proof for: arctan(x)+arctan(1/x)=sign(x)pi/2
and wanted to share a differnt point of view (not completely scientific but could be aranged)
OK. so the proof uses the Brewster angle in Optics.

Brewster summarization:
Brewster's angle is the angle where the wave is completely transfered from
one matter (n_1) to another (n_2): tan(Theta_B)=n_2/n_1
=>Theta_B=arctan(n_2/n_1)
Now Imagine two waves one from n_1 hitting the surface at brewster angle and another on the other side of the surface at it's own brewster angle (the fraction of n's is inverse)
it is pretty "clear" (unexplained here but true) that the first one will continue at the same angle as the second one hit the surface and therefore according to snell's law the angle between the first hitting wave and it's transfered wave will be pi/2. (same for the second wave)
===>arctan(x)+arctan(1/x)=sign(x)pi/2
Yeah, I know its not complete and it does not work for negative x's (no negative refractive index as far as I know) but it's a little 'out of the box'

[romolo]

Hi John,

I'm new here, too. I like the idea of using an application to prove a pure math theorem, but I don't think I have a very good visual of your explanation. Can you post a diagram so I can see what's going on?

[johnsmi]

Here is a sketch:
http://img9.imageshack.us/img9/3328/45971616gg3.jpg

Two different matters (refractive indexes n1 & n2)
The xy plane is the border between matter 1 and 2.

K1,2 are the wave vectors and ThetaB12,21 are the Brewster angles for waves from 1 to 2 and from 2 to 1 respectively.

Now try and read the last post and see if it makes it any clearer.
If not I don't mind trying again.

[The Dagda]

I'm afraid there are negative refractive indexes, now all you have to do is generalise it. :tongue2:

http://en.wikipedia.org/wiki/Negative_refractive_index#Negative_refractive_inde x

Superlens

The first superlens with a negative refractive index provided resolution three times better than the diffraction limit and was demonstrated at microwave frequencies at the University of Toronto by A. Grbic and G.V. Eleftheriades.[9] Subsequently, the first optical superlens (an optical lens which exceeds the diffraction limit) was created and demonstrated in 2005 by Xiang Zhang et al. of UC Berkeley, as reported that year in the April 22 issue of the journal Science,[10] but their lens did not rely on negative refraction. Instead, they used a thin silver film to enhance the evanescent modes through surface plasmon coupling. This idea was first suggested by John Pendry in Physical Review Letters.

[johnsmi]

Yeah your right I forgot.

In one of my courses (EM fields) we actually tried to build (theoretically) a matter made of little dielectric balls and calculated their properties in order to get a negative refractive index.
http://skullsinthestars.files.wordpress.com/2008/08/negativerefraction.jpg

And here is what it would look like: (the one on the right)
http://www.nature.com/nature/journal/v445/n7126/images/445346a-i2.0.jpg

By the way that is one of the ways they intend to build invisible cloaks where the EM waves detour the object

[The Dagda]

I wonder if anyone would be interested in a proof in all cases?

Never the less it is interesting. I always groan when my maths program throws out the signum function though. :eek:

[johnsmi]

you prefer looking at it as a step function?:wink:

[The Dagda]

you prefer looking at it as a step function?:wink:

Given a preference, not looking at it at all would be favourite. :smile:

[johnsmi]

Here's an article about the metamaterials in progress
http://www.sciencedaily.com/releases/2008/08/080811092450.htm