# How to show $z$-component of EM plane wave in free space is 0?

I know that for a uniform plane wave propagating in the z-direction in free space, there should be no z-component, however, I am having trouble proving this. Assuming $$\vec{E} = E_x (z,t) \hat{x} + E_y (z,t) \hat{y} + E_z (z,t) \hat{z}$$ and taking the divergence, we have $$\delta_z E_z (z,t) = 0 \implies E_z (z,t) = c_1 + E_z (t)$$ (or would it be $$c_1 E_z (t)$$?). Then, plugging this into the wave equation, we get $$\delta^2_t E_z (t) = 0 \implies E_z (t) = c_1 t + c_2 \frac{t^2}{2}$$ (or $$c_1 + c_2 t$$ for $$c_1 E_z (t)$$). This should be really easy. Where am I making a mistake?

## 1 Answer

If you are searching for plane wave solutions, then you know that $$E_z$$ must be in the form $$E_z=E_{z0}e^{i(kz-\omega t)}.$$ Applying the $$\partial/\partial z$$ operator, you get $$\frac{\partial}{\partial z}E_z=ikE_{z0}e^{i(kz-\omega t)},$$ which must be zero at all points, according to Gauss' law. As the complex exponencial is certainly not zero everywhere, the only way to satisfy Gauss' law in this situation is by making $$E_z=0$$.

• Ahh, so I guess my mistake was thinking that $E_z (z,t)$ was equivalent to a plane wave, so I completely ignored the plane wave aspect. – pmac May 19 at 19:14