How Does the Electromagnetic Wave Equation Validate Given Solutions?

[rmjmu507]

Homework Statement

Show that the solution $\textbf{E}=E(y,z)\textbf{n}\cos(\omega t-k_xx)$ substituted into the wave equation yields

$\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-k^2E(y,z)$

where $k^2=\frac{\omega^2}{c^2}-k_x^2$

Homework Equations

See above.

The Attempt at a Solution

I plugged the given solution into $\frac{\partial^2 \textbf{E}}{\partial y^2}+\frac{\partial^2 \textbf{E}}{\partial z^2}=\frac{1}{c^2}\frac{\partial^2 \textbf{E}}{\partial t^2}$ and got:

$\textbf{n}\cos(\omega t-k_xx)[\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}]=-\frac{\omega^2}{c^2}E(y,z)\textbf{n}\cos(\omega t-k_xx)$

Now, canceling like terms I get:

$\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-\frac{\omega^2}{c^2}E(y,z)$

But I'm missing a $k_x^2$ term on the RHS, and cannot figure out where this could/would have come from...can someone please explain?

 

[rmjmu507]

I was able to get the $k_x^2$ term by determining $\nabla^2\textbf{E}$ and rearranging, thus obtaining the desired relation.

However, I'm not entirely sure why it's necessary to determine $\nabla^2$. Can someone please explain this to be?

 

[RUber]

You had to evaluate the $\nabla^2$ operator because that is the definition of the wave function. $\nabla^2 \vec{E} = \frac{\partial^2 \vec{E}}{\partial t^2}$ Adding an $x$ dependence into your function for $\vec{E}$ meant you had to fully evaluate the Laplacian.

 

[rmjmu507]

I see...I was considering this equation as only a two-dimensional one...for some reason I was overlooking the x component in the cosine function. Not entirely sure why, perhaps because of the E(y,z) term, but I now realize this is simply a coefficient corresponding to the amplitude.

Thanks!