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Energy at zero axis in an Electromagnetic field?

  1. Jul 16, 2015 #1
    Can someone correct my understanding on electromagnetic waves please; I've clearly got the wrong end of a stick somewhere, but can't figure out where my misunderstanding is! Thanks.

    If I draw a classic EM wave with the electric field on the up and down axis and the magnetic field at right angles, then both of the sine waves cross the time (or distance) axis at the same points. What I really don't get is "where is the energy" at this point? I can comprehend that a moving electric field generates a magnetic field and vice-versa and that this is how the wave propagates over a distance, but if the heights above and below the axis represent field strength and therefore (possible mistake) a level of energy, then what's happening at the point they both cross the axis, and how does the next phase of the wave get "kick started"?

    Don't shoot me... I know I've missed something, I just can't see what!

    Thanks
    Matt
     
  2. jcsd
  3. Jul 16, 2015 #2

    Drakkith

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    Staff: Mentor

    It's not a 'moving' electric field that generates a magnetic field, it's a 'changing' electric field. Same for the magnetic field. How this causes the wave to propagate, I don't know.
     
  4. Jul 16, 2015 #3
    Thanks Drakkith. I'm perfectly happy to accept that there is a causal relationship between the two components that causes propagation, the thing I'm struggling to get my head around is what's happening at the point both the two components cross the zero point on the time/distance axis?

    Does the height of the wave in any way relate to energy? If it does, then I'm confused in one way; if it doesn't then how do you get a larger electric and magnetic field at one location compared on a line to another if the total energy is the same... or not??!?!?! Brain just exploded :-(
     
  5. Jul 16, 2015 #4

    Drakkith

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    Staff: Mentor

    I've yet to take an E&M course, but I believe that when the graphs of the fields are passing the X axis, the change in the fields is also the highest, so you should get a continuous transfer of energy over one period of the wave.

    Sure. The higher the amplitude of the wave, the more energy it has. Remember though that you're looking at a graph. A real EM wave has no height. Its components only have a magnitude and direction.
     
  6. Jul 16, 2015 #5
    <Slaps head hard> Thanks Drakkith, you've just spotted where my thinking fell down. Ironic really.... I've just been posting on a thread (you were on) about the dangers of confusing the math with the reality, and that's exactly what I've done here! Much appreciated
     
  7. Jul 17, 2015 #6
    Hi mgkii:

    A very, very important point to understand the Field Energy of a propagation EM wave is that, this energy is a kind of 'potential energy'. The actual power of a EM-wave on a given time is not only dependent on the exact field amplitude at this time, but also dependent on 'the potential that the amplitude can be how much higher'. As an analogy, if a ball is falling from a table and bounce from the floor back and forth, we can not state that at the time the ball hit the floor the ball has zero energy. As the matter of fact, if the collision from the ball to the floor is perfect elastic(without any energy loss), the ball will bounce forever, and the total energy (kinetic+potential) will be conservative. And much similar to this, the power of a EM wave in a loss-less medium is uniform versus time.

    The value of the intensity of the field can be estimated by 0.5*absolute(E)^2 which the E-field is a complex number, the imaginary part (represent for the oscillation phase of the wave) must not be ignored.
     
  8. Jul 17, 2015 #7
    That's a really superb analogy. Many thanks.
     
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