How Does the Electromagnetic Wave Equation Validate Given Solutions?

rmjmu507
Messages
34
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
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.
 
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!
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top