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Electromagnetic wave propagation question

  1. Jan 17, 2010 #1
    From reviewing previous posts on the subject, I understand that the E field and B field propagate in-phase through space.

    From this information, does it mean that at a specific space and time the E and B fields will be zero? If we had an instrument to determine the E and B field at specific times and locations, will there be measurements where both E and B-fields are equal to zero? To take it one step further, does this mean our measurement equipment will be sensitive to aliasing, and if we sample only when the E and B fields are zero there will be no EM wave detected (when the sampling rate is equal to the frequency of the wave being sampled)?

    Additionally, does the E and B-field amplitude matter when dealing with an interaction with a material? As a thought experiment, if we have a piece of matter that we send an em wave to, will there be a different interaction at the surface atoms if the interacting em wave has its E/B field at its max or at zero?

    I am trying to get a better understanding of electromagnetic waves, and thanks in advance for your responses.

    The following posts discuss how the E and F fields are in phase.
    https://www.physicsforums.com/archive/index.php/t-15160.html
    https://www.physicsforums.com/archive/index.php/t-36797.html

    PS (added at 09:16 AM): I just thought of another related question. I am not sure if this warrants a separate post. From the previous links, the electromagnetic wave is described as a self-propagating wave with its B-field and E-fields in phase. Does this mean that the E-field propagates the B-field that is 90 degrees further along the wave, which then also propagates the E-field 90 degrees along the wave? My logic being that if the B-field equals the differential of the E-field, when the slope of the E-field is max, the amplitude of the B-field is max at 90 degrees further along the wave.
     
    Last edited: Jan 17, 2010
  2. jcsd
  3. Jan 17, 2010 #2
    Sure, as long as the time that your sensor is activated is extremely short - as in maybe 1/50 of the period of the wave -and you only measure in multiples of the period.
     
  4. Jan 17, 2010 #3
    Look at the equations that describe the electromagnetic field that is produced by a source whose current density varies sinusoidally with time:

    [tex]\vec{E} = \frac{\eta J_s}{2} cos(\omega t \mp \beta z) \hat{a_x}[/tex]

    [tex]\vec{H} = \pm \frac{J_s}{2} cos(\omega t \mp \beta z) \hat{a_y}[/tex]

    where [tex]J_s[/tex] is the current source amplitude, [tex]\omega[/tex] is the source's angular frequency, and [tex]\beta[/tex] is related to the (medium dependent) wavelength -- all three are constants here.


    Under what conditions are both E and H zero?
     
    Last edited: Jan 17, 2010
  5. Jan 17, 2010 #4
    Thanks for your response. Gnurf asked, "under what conditions are both E and H zero?"

    If my understanding is correct, then it would be when
    [tex]cos(\omega t \mp \beta z)[/tex] equals zero and when [tex]\omega t \mp \beta z[/tex] equals pi/2 + pi*n where n=0,1,2,3... This can be seen at the zeros of the following diagram.

    http://www.mtholyoke.edu/~mlyount/MySites/Pictures/e_mag.JPG [Broken]

    A few questions come up when thinking about this. I am not sure if I am misunderstanding something trivial.

    1.) Does that mean there are different surface interactions depending on whether the em wave interacts with a material when the B/E fields are equal to zero and the B/E fields are at its max/min? For example if I have two setups with each an em wave aimed at a target surface. In the first setup if a coherent em wave is at its max B/E fields when it reaches the surface, then the surface will feel an immediate large B/E field. In the second setup if a different but coherent em wave is at zero when it reaches the surface, then the surface will feel a small/zero B/E field that builds up to its max. It seems like there would be a difference, but I am not sure if I am confused and missing some theory that would say otherwise.

    2.) If I was to do an experiment to probe the B/E fields at specific times, is there a possibility that I not detect a em wave or miscalculate the frequency of the em wave due to sample aliasing (say I sample at the em frequency or 2x the em frequency)? I would suspect so, but I am wondering if there is any possible counterinutitive quantum effects I am missing.

    3.) From my impression the E-field propagates the B-field and visa versa. In a previous post, user Chi Meson mentioned that "the reason this shift does not apply to light propagation is that the collapsing B-field at any point is space is not creating the E-field at the same point; rather, the collapsing B-field at one point in space is inducing the B-field(s) at subsequent points in space." I was wondering what he means by "subsequent points in space." Is it 90 degrees subsequent in space, which equals the wavelength/4? So the collapsing E-field is generating a B-field a wavelength/4 further along space and visa versa. My logic being that if the B-field equals the differential of the E-field, when the slope of the E-field is max, the amplitude of the B-field is max at 90 degrees further along the wave.

    Thanks for your response. I am not sure if I am misunderstanding something trivial.

    Edited to add more descriptions to Q1 and Q3

     
    Last edited by a moderator: May 4, 2017
  6. Jan 18, 2010 #5
    In a (self-) propagating, non-attenuating electromagnetic wave, E and H are 90-degrees out of phase. This can be seen directly in the two curl equations in Maxwell's equations (with no conduction):

    Curl E = - μ0H/∂t and
    Curl H = +ε0E/∂t

    The partial time derivative produces a 90-degree phase shift, just like in the telegrapher's equation (for transmission lines), like in resonant LC circuits, and even like in T and V in the pendulum equation. In a propagating electromagnetic wave, the energy is conserved by oscillating between an electric field (e.g., voltage) and a magnetic field (e.g., current). E and H have an in-phase component only if the conductivity σ is non-zero.

    Bob S
     
  7. Apr 16, 2010 #6
    I thought this was true too...but then when I was brushing up on this, the sources said that [for light] the E and B fields are in phase.

    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwavecon.html#c1


    I have found other post-ers [in other threads] speak of light in the same way Bob does...that the E- and B-fields are out of phase. Can someone shed some...err...light on this?
     
  8. Apr 16, 2010 #7
    E and H fields are in phase for a freely propagating wave. (Sorry, I work in MKS units.)

    They are not 90 degrees out of phase in time, but they are 90 degrees out of phase in space. That's a really bad way of saying they are perpendicular.


    To the OP, there may or may not be a zero in the fields. For linearly polarized waves, the E and H fileds are zero at the same point in space, but that point moves forward at the speed of light.

    In the case of circular polarization the E and H fields never become zero, they simply rotate like a corkscrew through space while remaining 90 degrees perpendicular. Neither type of polarization is fundamental, you can have left or right CP photons just as easily as you have vertial or horizontal ones.
     
  9. Apr 16, 2010 #8

    Andy Resnick

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    That's almost correct- linear polarized plane waves are the same as circularly polarized, in terms of the relative phase between E and B. Other wavefronts, such as radially polarized, tangentially polarized, Bessel beams, and the near field do have zeros in E and B simultaneous within some stable region of space.

    http://docs.google.com/viewer?a=v&q...h3S6K3&sig=AHIEtbRNQZKpeDdFXrm1C8ee6cshcCfGwQ
     
  10. Apr 16, 2010 #9
    Indeed. The proper way to say it is far-field radiation but I didn't want to cloud the issue. All the other wavefronts you mentioned do not sustain themselves as such if one if far enough away from the source. (There's also elliptical polarization which is a mixture of linear and circular but it doesn't change the essentials of the discussion so I left it out.)
     
  11. Apr 16, 2010 #10

    Dale

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    Hi Antiphon, I know that you know better than this. It is not just a really bad way of saying it, it is a wrong way of saying it. Phase and direction refer to entirely different properties of the fields. Two vector fields could, in principle, be out of phase but pointed in the same direction, or in phase and pointed in the same direction, etc.
     
    Last edited: Apr 16, 2010
  12. Apr 16, 2010 #11
    Tnx Dalespam. Sometimes I cut corners too much in the terminology hoping to make a clearer point. It backfired this time.
     
  13. Apr 16, 2010 #12

    Andy Resnick

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  14. Apr 17, 2010 #13
  15. Apr 17, 2010 #14

    Andy Resnick

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    I'm not sure what your point is, as a plane wave is a limiting case at best only approximately satisfied by real beams.
     
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