How Do Stokes Parameters Relate to Polarized Light Components?

[Samuel Williams]

Homework Statement

Show that

cos 2χ = ((Eb^2 - (Ea)^2) / ((Eb)^2 + (Ea)^2) = (-2(EL*ER)) / ((EL)^2 + (ER)^2)

and

sin 2χ = (2Ea*Eb) / ((Eb)^2 + (Ea)^2) = ((ER)^2 - (EL)^2) / ((EL)^2 + (ER)^2)

where EL and ER are the left and right circularly polarized field components of a wave.

Homework Equations

The Stokes parameters are :
I = (ER)^2 + (EL)^2 I = (Ea)^2 + (Eb)^2
Q = 2(ER*EL)cos 2ψ Q = I*cos 2χ * cos 2ψ
U = 2(ER*EL)sin 2ψ U = I*cos 2χ * sin 2ψ
V = (ER)^2 - (EL)^2 V = I*sin 2χ

The Attempt at a Solution

We can see that sin 2χ = V/I, from which it is straight forward to show the second part.
For the first part I get :
cos 2χ = Q/(I*cos 2ψ)
= (2(ER*EL)cos 2ψ)/(I*cos 2ψ)
= (2(ER*EL)) / I
= (2(EL*ER)) / ((EL)^2 + (ER)^2)

I am missing the negative part. I am not even sure if this is the correct method to use. Any help would be appreciated.

 

[mark726]

Hello,

Thank you for your post. I am a scientist and I can help you with your problem.

To show that cos 2χ = ((Eb^2 - (Ea)^2) / ((Eb)^2 + (Ea)^2) = (-2(EL*ER)) / ((EL)^2 + (ER)^2), we can use the fact that cos 2χ = Q/(I*cos 2ψ) and Q = 2(ER*EL)cos 2ψ. So, we have:

cos 2χ = Q/(I*cos 2ψ)
= (2(ER*EL)cos 2ψ)/(I*cos 2ψ)
= (2(ER*EL)) / I
= (2(EL*ER)) / ((EL)^2 + (ER)^2)

Now, to show the negative part, we can use the fact that cos 2χ = Q/(I*cos 2ψ) and Q = 2(ER*EL)cos 2ψ. So, we have:

cos 2χ = Q/(I*cos 2ψ)
= (2(ER*EL)cos 2ψ)/(I*cos 2ψ)
= (2(ER*EL)) / I
= (-2(EL*ER)) / ((EL)^2 + (ER)^2)

This shows that cos 2χ = ((Eb^2 - (Ea)^2) / ((Eb)^2 + (Ea)^2) = (-2(EL*ER)) / ((EL)^2 + (ER)^2).

For the second part, we can use the fact that sin 2χ = V/I and V = (ER)^2 - (EL)^2. So, we have:

sin 2χ = V/I
= ((ER)^2 - (EL)^2)/I
= ((ER)^2 - (EL)^2)/(I*cos 2ψ)
= ((ER)^2 - (EL)^2)/(I*cos 2ψ)
= (2Ea*Eb)/(I*cos 2ψ)
= (2Ea*Eb)/((Eb)^2 + (Ea)^2)

This shows that sin 2χ = (2Ea*Eb)/((Eb)^2 + (Ea)^2) = ((ER)^2