# Superposition of linearly polarized waves: what is the relationship which is independent of time that describes the electric field components?

Consider an electromagnetic wave propagating in the $z$-direction, which is a superposition of two linearly-polarized waves. The electric field vector in the wave is:

$\mathbf{E} = a\cos(kx-\omega t)\hat{x} +b\cos(kx-\omega t + \delta)\hat{y}$.

My goal is to show that there exists a simple relation between $E_x = a\cos(kx-\omega t)$ and $E_y = b\cos(kx-\omega t + \delta)$ which does not involve $t$, namely, that:

$\frac{E_x^2}{a^2} + \frac{E_y^2}{b^2} - \frac{2E_xE_y}{ab}\cos(\delta) = constant.$

My approach thus far has been to use the fact that $E_x\hat{x} \perp E_y\hat{y}$ to find an expression for the hypotenuse of the right-angle triangle formed by them. To do this, I've tried representing each component in their complex forms, as in:

$E_x = ae^{i(kx - \omega t)}$, and $E_y = be^{i(kx - \omega t +\delta)}$,

and taking the square of their respective magnitudes, but I've not been having much luck and do not know how to proceed from here.

Any help as to how to derive this relation I would much appreciate.

You can write the field components with a difference of phase

$$\vec{E}( \vec{r},t) = \begin{pmatrix} \varepsilon_x e^{i\omega t} \\ \varepsilon_x e^{i(\omega t+ \Delta \varphi)} \end{pmatrix} e^{i\left( \varphi_x + \vec{k}\cdot\vec{r}\right)}$$ Take real part \begin{align} E_x &= \varepsilon_x \cos \omega t \\ E_y &= \varepsilon_y \cos{\left( wt+ \Delta\varphi \right)} \end{align}

then separate $E_x / \varepsilon_x$ and $E_y / \varepsilon_y$

\begin{align} \frac{E_x}{\varepsilon_x} &= \cos \omega t \\ \frac{E_y}{\varepsilon_y} &= \cos{\left( wt+ \Delta\varphi \right)}= \cos\omega t \cos \Delta\varphi- \sin\omega t \sin \Delta\varphi \end{align}

Substituting $E_x / \varepsilon_x$ in $E_y / \varepsilon_y$

\begin{align} \frac{E_y}{\varepsilon_y} &= \frac{E_x}{\varepsilon_x} \cos \Delta\varphi- \sin\omega t \sin \Delta\varphi \\ \frac{E_y}{\varepsilon_x} +\frac{E_x}{\varepsilon_x}\cos \Delta\varphi &= \sin\omega t \sin \Delta\varphi \\ \frac{-\frac{E_y}{\varepsilon_y} +\frac{E_x}{\varepsilon_x}\cos \Delta\varphi}{\sin \Delta\varphi } &= \sin\omega t \end{align}

Then, elevate to 2

$$\left( \frac{E_x}{\varepsilon_x} \right)^2 +\left(\frac{-\frac{E_y}{\varepsilon_y} +\frac{E_x}{\varepsilon_x}\cos \Delta\varphi}{\sin \Delta\varphi } \right)^2= 1$$

and doing a little algebra you can reach your goal.

• Thank-you for your breakdown of the problem. I should have realized that in writing out $\frac{E_x}{a} = \cos(kx - \omega t)$ the $\sin^2(x) + \cos^2(x) = 1$ relation would be of importance. Your help I appreciate. – T. Zaborniak Sep 13 '18 at 22:09