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Trying to understand Electromagnetic waves

  1. Feb 15, 2008 #1
    The problem i have with understanding waves is that people just start drawing graphs of the sine function or some equally appaling function that oogles me over.

    The problem with this is that people begin to correlate a graph as such the Electric field to be something spatial. They assume that something like an awesome wave *moves* through space. I used to think it in the same way.. and well.. it never got me anywhere.

    So, here is what I tried to think it like: A wave is basically an 'influence' that oscillates at many different places in space as a function of time.

    What it means is.. if there is an e.m. wave with an Electromagnetic component [itex]E = E_o\sin{(kx - \omega t)}[/itex] (i'm taking the plane wave solution for simplicity). Now, i keep a charge '[itex]q[/itex]' at say '[itex]l[/itex]' and the wave reaches that point at a particular time '[itex]\tau[/itex]. Can i say that the charge at that particular instant will feel a force given by [itex]F = qE_o\sin{(kl - \omega \tau)}[/itex] ?

    Also, let's say we have a pivoted charge at '[itex]l[/itex] which is pivoted to some intrument such that when the charge experiences any force.. the instrument provides an opposing force instantaneously such that the charge doesn't move at all and also records the force provided.

    If i tabulate the data from this instrument, will i get a periodic reading of results? Is this what a wave is all about? Am i thinking right or i am making a mistake?

    Can i also do the same thing assuming a small current element on the y-axis and the magnetic field component being given by another plane wave equation?
  2. jcsd
  3. Feb 15, 2008 #2


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    Staff: Mentor


    Yes. Furthermore, if you place a second charge slightly "downstream" (larger x) from the first one, the force on it oscillates with the same period and frequency as the first one, but with a slight time delay [itex]\Delta t = (k / \omega) \Delta x = \Delta x / c[/itex] where [itex]c = \omega / k[/itex] is the propagation speed of the wave. For example, the maxima of the second charge's oscillation occur [itex]\Delta t[/itex] later than the maxima of the first charge's oscillation.
  4. Feb 15, 2008 #3
    well.. thanks a lot.. jtbell. It means that i had pretty much the right idea 'bout e.m. waves :D

    P.S: @jtbell: is that guy in your avatar Mr. Garrison?
  5. Feb 15, 2008 #4


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    Staff: Mentor

    It's actually from the standard collection of avatars that comes with this version of vBulletin. I just happened to pick that one because it looks like a nerdy middle-aged guy which is what I am (except I also have a beard and I'm not bald... yet). I thought it looked kind of familiar, and now I know why! :rofl:

    (I watch South Park occasionally because one of our local TV stations shows it late at night.)
  6. Feb 15, 2008 #5
    Also.. does the plane wave solution account for the speed of the wave. For example, how do i assert in my equation that at [itex]t = 0[/itex], [itex]x = 0[/itex]. Because, if i keep my object say 10 lightminutes away.. then it will experience a force 10 minutes later.

    But, at t=0 and x=lightminutes, the electric field is non-zero. How do I get around this?
  7. Feb 17, 2008 #6

    Claude Bile

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    A plane wave solution is infinite in extent with respect to time. You can't really define a beginning or an end of such a wave. Instead you need to use waveforms that are non-periodic (and therefore finite in extent with respect to time).

    Such waveforms are generally termed "wave packets", and have a characteristic duration that depends on the spectral (frequency) bandwidth of the wave packet.

    Due to the non-periodic nature of wave packets (or any modulated wave), we need to define another velocity, the group velocity in additional to the "usual" wave velocity (The v = f[itex]\lambda[/itex]) which is termed the "phase velocity".

    The difference between the group and phase velocity of a wave is that the phase velocity gives the speed of the peaks and troughs in the wave, while the group velocity gives the speed that the envelope (or modulation) travels at.

    In summary, to "get around" the causality problem you need to use a wave that has a finite duration.

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