How Do B and Gamma Values Affect Polarization in Electromagnetic Waves?

[Rubiss]

Homework Statement

I'm currently trying to understand linear and circular polarization of electromagnetic plane waves. Let's say I have an electric field given by

\vec{E}=Acos(kx-\omega t)\hat{x}+Bcos(kx-\omega t - \gamma)\hat{y}

A is given and nonzero. I want to find what values of B and gamma that can make the wave linear or circularly polarized.

Homework Equations

I can also write the electric field as

\vec{E}=Re \big((A\hat{x}+Be^{-i\gamma}\hat{y})e^{i(kx-\omega t)}\big)

The Attempt at a Solution

For the linear polarization, I'm thinking I can either make B=0, or gamma equal to n*pi if B is not equal to zero. Can anyone comment on my thinking?

For circular polarization, I'm thinking I need to have A=B and gamma equal to n*pi/2 for odd n. I will have right handed circular polarization if n=3,7,11,... and have left handed circular polarization if n=1,5,9,... Is this thinking correct?

Is there an easier way to do this that I am not seeing?

 

[rude man]

Are you sure it's not cos(kz - wt) etc.? Because if not this is not a possible e-m wave.

 

[rude man]

Assuming propagation in the z direction, you can solve this by writing the E field at z = 0, writing the x and y components of E and eliminating t between them. The result is y(x,phi,A,B) which then makes it obvious the restrictions on B and phi you must impose to effect linear or circular polarization. You could even go further and do the same for elliptical polarization for extra credit ...

 

[rude man]

Homework Statement

I'm currently trying to understand linear and circular polarization of electromagnetic plane waves. Let's say I have an electric field given by

\vec{E}=Acos(kx-\omega t)\hat{x}+Bcos(kx-\omega t - \gamma)\hat{y}

A is given and nonzero. I want to find what values of B and gamma that can make the wave linear or circularly polarized.

Homework Equations

I can also write the electric field as

\vec{E}=Re \big((A\hat{x}+Be^{-i\gamma}\hat{y})e^{i(kx-\omega t)}\big)

OK. I never did respond to you directly. Here goes:

The Attempt at a Solution

For the linear polarization, I'm thinking I can either make B=0, or gamma equal to n*pi if B is not equal to zero. Can anyone comment on my thinking?

Right.

For circular polarization, I'm thinking I need to have A=B and gamma equal to n*pi/2 for odd n. I will have right handed circular polarization if n=3,7,11,... and have left handed circular polarization if n=1,5,9,... Is this thinking correct?

Almost right. B can be + or -A. The choice of sign determines right or left circular polarization. That makes it tantamount to choosing n for γ the way you did.

Is there an easier way to do this that I am not seeing?

Hard to say since you didn't tell us how you got your results in the first place ...