# EM wave propagation direction.

1. Oct 2, 2011

### Lavabug

1. The problem statement, all variables and given/known data
Given the following EM wave propagating in vacuum, find the direction along which the electric field oscillates and the direction of propagation of the wave:

$$\vec{E} = (-3\hat{i} + 3\sqrt{3}\hat{j}) 10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]}$$

Btw how can I prevent the thread template from reappearing every time I click preview post? It is quite annoying.

3. The attempt at a solution

I understand the electric field oscillates along the direction $-1\hat{i} \sqrt{3}\hat{j}$, by looking at the amplitide. But how do I determine the direction of propagation from the phase argument? Do I just treat the x and y in the phase as if they were unit vectors of $\vec{k}$?

2. Oct 2, 2011

### vela

Staff Emeritus
Pretty much. You know a wave propagating in the k direction has a phase given by $i(\vec{k}\cdot\vec{r}-\omega t)$, so just pick out the components of k.

3. Oct 2, 2011

### Lavabug

Thanks, I see it now. One more question, if I am asked to find the associated $\vec{H}$ field by calculating $\frac{1}{\mu_0 c} \hat{u}\times \vec{E}$, where û is the normalized wave vector, how would I split up the components of E for the determinant? Would it be like:

$$E_x = -3\cdot10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]}$$

$$E_y = 3\sqrt{3}\cdot10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]}$$

for example? Or do I also have to split up the phase/exponent for each component of E?

4. Oct 2, 2011

### vela

Staff Emeritus
You can keep the exponential part separate; it's part of the amplitude of the electric field. I would write the electric field as the product of a magnitude and a unit vector:
$$\vec{E} = \left(-\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\right)(3\times10^4~\mathrm{V/m})e^{i(\vec{k}\cdot\vec{r}-\omega t)}$$The only part you need for the determinant is the unit vector. The rest just scales the result of the cross product.