# Electromagnetic waves are transverse, but in which direction?

Consider a linearly polarized harmonic plane wave with a scalar amplitude of $$A$$, propagating along a line in the x-plane at $$45^°$$ to the $$x$$-axis with the $$xy$$-plane as its plane of vibration. We assume that $$k_x$$ and $$k_y$$ are both positive. The wave propagates in vacuum. So first, we know that $$\vec{E}(\vec{r},t)=\vec{E}(x,y,t)=A\cos(k_xx+k_yy-\omega t)\vec{u}$$ since $$\vec{k}\cdot \vec{r}=k_xx+k_yy$$. We need to determine $$\vec{u}$$ the unit vector that indicates the direction of polarization. Thanks to the Maxwell equations, we know that $$\vec{E}$$ is transverse to the direction of propagation, which is: $$\vec{k}\cdot \vec{E}=0$$. So my question is:

• do I take $$\vec{u}=\frac{1}{\sqrt{2}}\hat{x}-\frac{1}{\sqrt{2}}\hat{y}$$ or $$\vec{u}=-\frac{1}{\sqrt{2}}\hat{x}+\frac{1}{\sqrt{2}}\hat{y}$$ since both satisfy $$\vec{u}\cdot \vec{k}=0$$ i.e. $$\vec{E}$$ is transverse to the direction of propagation. Also, I would like to know where would be localized the wave-front, the plane of this wave?

Any help would be appreciated!

The two different possibilities you have given for the polarization vector are only different by an overall minus sign. You could use either one of these and you would still be describing a valid plane wave solution to Maxwell's equations. The only difference is that these two solutions are shifted from eachother by a phase of $$\pi$$ in either space or time. Behold:

\begin{align} \vec{u}_1 =& \frac{1}{\sqrt{2}} \hat{x} - \frac{1}{\sqrt{2}}\hat{y}\\ \vec{u}_2 =& -\frac{1}{\sqrt{2}} \hat{x} + \frac{1}{\sqrt{2}}\hat{y} \end{align}

\begin{align} \vec{E}_1(\vec{r},t) =& A \cos(\vec{k}\cdot{\vec{r}} - \omega t) \vec{u}_1\\ \vec{E}_2(\vec{r},t) =& A \cos(\vec{k}\cdot{\vec{r}} - \omega t) \vec{u}_2\\ =& A \cos(\vec{k}\cdot{\vec{r}} - \omega t) (-1) \vec{u}_1\\ =& A\cos(\vec{k}\cdot{\vec{r}} - \omega t + \pi) \vec{u}_1 \end{align}

because $$-\cos(\theta) = \cos(\theta+\pi)$$. Let $$T = \frac{2\pi}{\omega}$$ be the temporal period and let $$\lambda = \frac{2\pi}{k}$$ be the wavelength (spatial period) with $$k=|\vec{k}|$$.

We can see that $$\omega \frac{T}{2} = \pi$$ so we can write

\begin{align} E_2(\vec{r}, t) =& A\cos\left(\vec{k}\cdot{\vec{r}} - \omega \left(t-\frac{T}{2}\right)\right) \vec{u}_1\\ =& E_1\left(\vec{r}, t-\frac{T}{2}\right) \end{align}

Alternatively, let $$\delta \vec{r} = \hat{k} \frac{\lambda}{2}$$. With this we have $$\vec{k}\cdot \delta{\vec{r}} = \pi$$ so that

\begin{align} E_2(\vec{r}, t) =& A\cos\left(\vec{k}\cdot\left({\vec{r}} + \delta\vec{r}\right) - \omega t\right) \vec{u}_1\\ =& E_1\left(\vec{r} + \delta\vec{r}, t\right) \end{align}

Note that $$\delta \vec{r}$$ simply advances the wave along its propagation direction by a half wavelength.

So we see that changing the polarization vector from pointing one way or the other corresponds to the exactly same plane wave, just shifted in space or time by half a period. Since the absolute phase of a plane wave is typically not important we might say that these two solutions represent the same plane wave.

$$\vec{E}(\vec{r},t)=\vec{E}(x,y,t)=A\cos(k_xx+k_yy-\omega t)\vec{u}$$

does not match the description, nor is it suitably general.

First, to be propagating along the 45-degree line described:

$$k_x = k_y \equiv \frac k {\sqrt 2}$$

The general solution consistent with the description allows an arbitrary phase, so:

$$\vec{E}(x,y,t)=A\cos(\frac k{\sqrt 2}(x+y)-\omega t + \phi)\vec{u}$$

With:

$$\vec k = \frac 1 {\sqrt 2}(k, k, 0)$$

and

$$\vec u = (u_x, u_y, u_z)$$

there are two normalized solutions to $$\vec k \cdot \vec u=0$$:

$$\vec u = \frac 1 {\sqrt 2}(1, -1, 0)$$

and

$$\vec u = (0, 0, 1)$$

Either one can be multiplied by a phase (including $$e^{i\pi}=-1$$), so the sign is arbitrary. (You can put it in $$\phi$$ or in the sign of $$\vec u$$, and still describe the same wave).

Note that when you average over time and/or consider an arbitrary origin: linear polarization is not a vector. Rather, it is a tensor alignment without direction. Hence, rotating a linear polarizer by 180 degrees leaves its effects unchanged.