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How an em wave propogates?

  1. Jan 1, 2007 #1
    How an em wave propogates??

    i understand that an em wave can be produced due to an oscillating electric field or oscillating magnetic field... but how does this wave move forward at the speed of light??
  2. jcsd
  3. Jan 1, 2007 #2


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    Loosely speaking, when the electric and magnetic fields at some point change, it causes the fields at nearby points to change also, but with a time delay that depends on the distance. This is like when a point on a stretched string moves, it causes nearby points to move also, but with a time delay.

    Mathematically speaking, the electric and magnetic fields each obey the classical differential wave equation,

    <oops... see robphy's posting below for the correct equation :blushing: >

    where [itex]c = 1 / \sqrt{\epsilon_0 \mu_0}[/itex]

    This can be proved from Maxwell's equations for the electric and magnetic fields, as was first done by Maxwell himself.
    Last edited: Jan 1, 2007
  4. Jan 1, 2007 #3

    Gib Z

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    i Think you wanted a visual interpretation rather than mathematical? I like to this of a photon as a sort of charged particle, which is attracted to its own electromagnetic fields that it generates. The fields induce the creation of another one is front, the photon is attracted and propagates. This helps you remember that light is a particle and a wave, but shouldn't be taken too seriously.
  5. Jan 1, 2007 #4


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    I do not think that you analogy is a good one. It simply does not capture the mechanism well at all. Please reread JtBells explanation. It would be difficult to come up with a better one.

    When you have acquired a better understanding of Mathematics you will be able to appreciate the formal mathematical statement. Meanwhile be careful about building incorrect visualizations as you will find that they can become a barrier to gaining a correct understanding.
  6. Jan 1, 2007 #5


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    [tex] \left( \frac {\partial^2 \vec E}{\partial x^2} + \frac {\partial^2 \vec E}{\partial y^2} + \frac {\partial^2 \vec E}{\partial z^2}\right) = {\color{red} \frac {1}{c^2}} \frac {\partial^{\color{red}2} \vec E}{\partial t^{\color{red}2}}[/tex]
  7. Jan 1, 2007 #6
    the thing is, it doesn't really have to oscillate; electromagnetic wave just have to satisfy the equation robphy posted and the physical situation one is in.

    notice that any twice differentiable equation in the form of:
    [tex]E_i=f(x_i-ct) + g(x_i+ct)[/tex]
    satisfy the wave equation.

    the sinusoidal wave is just one simple solution of the wave equation.

    [tex]E_y=\cos\left[ \frac{2\pi}{\lambda}(x+ct) \right][/tex]

    which satisfy the wave equation. you can visualize the electric field pointing in the y-direction changes as x or time varies.
    Last edited: Jan 1, 2007
  8. Jan 2, 2007 #7


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    A somewhat simple minded explanation goes like: Maxwell's equations are not invariant with respect to acceleration of charges. When a charged particle accelerates. the Coloumb (near field in fact) changes. ( We are talking inertial frames here.) The Coloumb does not change instantaneously, rather any change must propagate at c. The accelerative radiation field is just the "delta Coloumb field'. That is, old Coloumb + radiation = new Coloumb. And then, turned around, the radiation field is necessary to keep the Coloumb field as the Coloumb field.

    Note, also, that the radiation from an accelerating charge is not generally monochromatic, but rather has a superposition of frequencies, and, probably, does not even look like a wave.

    I suspect that all the details of Coloumb + radiative adjustment can be worked out for a charge with uniform rotational mostion.
    Reilly Atkinson
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