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Question about the Poynting vector

  1. Feb 19, 2010 #1
    Suppose I have a monochromatic electromagnetic plane wave with the E-field linearly polarized in the x-direction (and the B-field linearly polarized in the y-direction). Then the Poynting vector should be pointing in the z direction with a magnitude equal to the product of the B and E-field magnitudes divided by the magnetic constant. But because they are complex, the magnitudes of the B and E field don't depend on time or position, which doesn't make sense, as the Poynting vector shouldn't be constant over all time. Of course, this can be solved by taking only the real parts of both solutions and multiplying them, but this would require breaking up the complex exponentials into cosines and sines. Is there anyway to do this without having to break up the complex exponentials?
     
  2. jcsd
  3. Feb 20, 2010 #2

    Born2bwire

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    The Poynting vector is

    [tex]\mathbf{S} = \mathbf{E}\times\mathbf{H} [/tex]

    not

    [tex]\mathbf{S} \neq |\mathbf{E}|\times|\mathbf{H}| [/tex]

    If the fields vary in time, so does the Poynting vector.
     
  4. Feb 20, 2010 #3
    Well that's true, but if I have an E-field like

    [tex]E_0 \hat{x} e^{-i(kz -\omega t)}[/tex]

    And consequently, a B-field like

    [tex] B_0 \hat{y} e^{-i(kz - \omega t)} [/tex]

    Then shouldn't the magnitude of the Poynting vector just be [tex]|\mathbf{S}| = |\mathbf{E}||\mathbf{H}| = \frac{E_0 B_0}{\mu_0}[/tex]?
     
    Last edited: Feb 20, 2010
  5. Feb 20, 2010 #4

    Born2bwire

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    In this case, yes.
     
  6. Feb 20, 2010 #5
    Hi,

    A small question concerning the calculation above.
    When we take the expressions for E and B and evalute the cross product, the obtained result is a vector or a vector field ?

    In other words, given a EM wave field it correct to say that the Poyinting Vector is also a vector field ?

    Thank you in advance,

    Best Regards,

    DaTario
     
  7. Feb 21, 2010 #6
    I think it is a vector field, as every z position has a different E and B field and thus, a different Poynting vector.

    I'm not sure if the calculation I did was correct; a plane wave should not have a constant Poynting vector, as there are points where the E and B field are 0, and the magnitude of the Poynting vector should also be 0. I think there should be a sinusoidal dependence on z and t, but whenever I use complex exponentials, the dependence vanishes...
     
  8. Feb 22, 2010 #7

    Born2bwire

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    A plane wave has constant magnitude and the example you gave maintains a constant phase relationship between the two fields, so of course its Poynting vector also has constant magnitude.
     
  9. Feb 22, 2010 #8
    The magnitude of the wave is only constant across planes perpendicular to the wave's motion. The magnitude of the E and B fields still vary sinusoidally with z (parallel planes with a different z component have a different magnitude), so shouldn't the magnitude of the Poynting vector change with z?
     
  10. Feb 22, 2010 #9

    turin

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    Maybe "amplitude" is a better word. Anyway, you are probably restricting your consideration to a particular instant in time. Think about what happens as time passes. There will be a particular maximum value for E and B at a given point, and, since you are talking about a plane wave, this value is the same for every point. The phase is an arbitrary choice that you make, and is no more physical than the location of z=0.
     
  11. Feb 22, 2010 #10

    turin

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    When using the complex amplitudes, I believe that the expression should be

    [tex]\mathbf{S}=\frac{1}{2}\mathbf{E}\times\mathbf{H}^*[/tex]

    This is the complex Poynting vector. The exponential factors that make the E and B fields a plane wave cancel out, due to the complex conjugation.
     
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