# Circular Polization type in an EM wave.

1. Dec 18, 2011

### Silversonic

My bad for the spelling fail in the title.

1. The problem statement, all variables and given/known data

I am utterly confused with the mathematics in a section of my notes, I swear that it's wrong - and it related to a piece of homework which I cannot understand.

The electric field vector in a z-directed monochromatic EM wave is given by;

http://img213.imageshack.us/img213/4768/58814275.jpg [Broken]

It says that if $\alpha$ = $\pi$/2, $\theta$ = 0 and $E_{0,x}$ = $E_{0,y}$

Then

$E_{x}$ = $E_{R}$cos(kz-wt),

$E_{y}$ = $E_{R}$sin(kz-wt)

and $E_{R}$ = $\frac{1}{\sqrt{2}}E_{0,x}$

3. The attempt at a solution

I do not see how the component in y direction is right. Taking the parameters $\alpha$ = $\pi$/2, $\theta$ = 0, $E_{0,x}$ = $E_{0,y}$

we get

$E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + E_{0,x} e^{i(\pi/2)} \hat{y}]$

$e^{i(\pi/2)} = i$

$E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}]$

Subbing this in the equation for overall E.

$E = Re(\frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}]e^{i(kz-wt)})$

$E_{R}$ = $\frac{1}{\sqrt{2}}E_{0,x}$

$E = Re(E_{R}[\hat{x} + i\hat{y}]e^{i(kz-wt)})$

$E = Re(E_{R}[\hat{x} + i\hat{y}][cos(kz-wt)+isin(kz-wt)])$

$E = E_{R}[\hat{x} cos(kz-wt) - \hat{y}sin(kz-wt)]$

So then surely for $\alpha$ = $\pi$/2

We have

$E_{x}$ = $E_{R}cos(kz-wt)$

$E_{y}$ = -$E_{R}sin(kz-wt)$

So the $E_{y}$ direction is minus instead?

Similarly for $\alpha$ = $-\pi$/2 We'd have;

$E_{x}$ = $E_{R}cos(kz-wt)$

$E_{y}$ = $E_{R}sin(kz-wt)$

Is this correct?

$\alpha$ = $\pi$/2 should correspond to left circular polarization, and $\alpha$ = $-\pi$/2 to right.

Last edited by a moderator: May 5, 2017