- #1

schrodingerscat11

- 89

- 1

## Homework Statement

Hi! The entire problem is this:

(a) Two plane-polarized harmonic plane waves having the same propagation constant are polarized, respectively, along two perpendicular directions. Show that if the phases of the two waves are different, their superposition yields generally an elliptically polarized plane wave.

(b) Show that the time-average Poynting vector of an elliptically polarized plane wave is equal to the sum of the time-average, Poynting vectors of the two orthogonal plane-polarized waves into which it can be decomposed.

## Homework Equations

__Plane waves__

Def: a constant-frequency wave whose wavefronts (surfaces of constant phase) are infinite parallel planes of constant peak-to-peak amplitude normal to the phase velocity vector (Wikipedia).

[itex]A(x,t)=A_ocos(kx-\omega t +\phi)[/itex]

[itex]A(\mathbf{r},t)=A_o cos(\mathbf{k} \cdot \mathbf{r}-\omega t +\phi)[/itex]

[itex]A(\mathbf{r},t)=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t +\phi)}[/itex]

where

[itex]A(x,t)[/itex] is the wave height at position x and t.

[itex]A_o[/itex] is the amplitude

[itex]k[/itex] is the wave number

[itex]\phi[/itex] is the phase constant

[itex]\omega[/itex] is the angular frequency

Propagation constant:

Propagation constant:

[itex]\frac{A_o}{A_x}=e^{\gamma x} [/itex]

[itex]\gamma=\alpha+i\beta[/itex]

[itex]\beta=k=\frac{2\pi}{\lambda}[/itex]

where

[itex]A_x[/itex] and [itex]A_o[/itex] are the amplitude at position x and the amplitude at source of propagation, respectively.

[itex]\gamma[/itex] is the propagation constant

[itex]\alpha[/itex] is the attenuation constant

[itex]\beta[/itex] is the phase constant

__Equation of an ellipse:__

[itex]\frac{x^2}{a}+\frac{y^2}{b}=1[/itex]

whose parametric equations are

[itex]x=a ~ cos ~t[/itex]

[itex]y=b ~sin ~t[/itex]

## The Attempt at a Solution

So far these are the things that I am not sure:

- I now know that plane waves have mathematical forms as given above. My question is how will they change if they become
*harmonic*? - I assume that plane polarization means that if [itex]\mathbf{A}(\mathbf{r},t)[/itex] is a vector, the disturbance is along a certain direction only. That is,[itex]\mathbf{A}(\mathbf{r},t)=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t +\phi)}\mathbf{\hat{x}}[/itex] is said to be plane polarized along the x direction. Right?
- If the propagation constant is the same, I assume the phase constant is also the same which means that k is the same for both plane waves. Also by the definition of propagation constant above, the amplitude of the two plane waves are equal any time. Right?
- I am utterly confused on which among these quantities are complex and which are real. Hence, I don't know how to manipulate the exponential parts or if I can apply Euler's formula to simplify these.

Let the first plane wave be

[itex]\mathbf{A_1}(\mathbf{r},t)=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t +\phi)}\mathbf{\hat{x}}=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}e^{\phi}\mathbf{\hat{x}}[/itex]

and the second plane wave be

[itex]\mathbf{A_2}(\mathbf{r},t)=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t +\psi)}\mathbf{\hat{y}}=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}e^{\psi}\mathbf{\hat{y}}[/itex]

Taking their superposition:

[itex]\mathbf{A}=\mathbf{A_1}+\mathbf{A_2}[/itex]

[itex]\mathbf{A}=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}e^{\phi}\mathbf{\hat{x}}+A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}e^{\psi}\mathbf{\hat{y}}[/itex]

[itex]\mathbf{A}=A_o e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}(e^{\phi}\mathbf{\hat{x}}+e^{\psi}\mathbf{\hat{y}})[/itex]

[itex]1=\frac{A_o}{\mathbf{A}}e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t )}(e^{\phi}\mathbf{\hat{x}}+e^{\psi}\mathbf{\hat{y}})[/itex]

I want to recast this to the form of equation of an ellipse (see relevant equations above) but I'm stuck.

Thank you very much.