1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electromagnetic Wave Equation

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the solution [itex]\textbf{E}=E(y,z)\textbf{n}\cos(\omega t-k_xx)[/itex] substituted into the wave equation yields

    [itex]\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-k^2E(y,z)[/itex]

    where [itex]k^2=\frac{\omega^2}{c^2}-k_x^2[/itex]
    2. Relevant equations
    See above.

    3. The attempt at a solution
    I plugged the given solution into [itex]\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}[/itex] and got:

    [itex]\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)[/itex]

    Now, canceling like terms I get:

    [itex]\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)[/itex]

    But I'm missing a [itex]k_x^2[/itex] term on the RHS, and cannot figure out where this could/would have come from...can someone please explain?
     
  2. jcsd
  3. Oct 5, 2014 #2
    I was able to get the [itex]k_x^2[/itex] term by determining [itex]\nabla^2\textbf{E}[/itex] and rearranging, thus obtaining the desired relation.

    However, I'm not entirely sure why it's necessary to determine [itex]\nabla^2[/itex]. Can someone please explain this to be?
     
  4. Oct 5, 2014 #3

    RUber

    User Avatar
    Homework Helper

    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.
     
  5. Oct 5, 2014 #4
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted