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Question about intensity of EM waves

  1. Jul 29, 2015 #1


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    For electromagnetic wave if it's reflected from a perfect conductor standing wave can be form. I wonder why Poynting vector can be used to describe the intensity of standing EM wave. (see p.19 of http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide13.pdf ).

    From some textbooks the Poynting vector is used to describe the intensity of traveling wave. It is derived from wave without reflection and since Maxwell's equations are obeyed and therefore cB=E. However in standing wave since E=2Eo cos(kx)cos(ωt) and B=2Bo sin(kx)sin(ωt), the E and B do not obey cB=E at every point and the overall wave is not traveling in any direction.

    Another question is, the Poynting vector of standing wave is pointing perpendicular to the E and B field and sometime has positive and negative value. It means that energy is transmitting to either directions in different times. However for standing wave cB=E is not obeyed, does it contradict Maxwell's equation?
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  3. Jul 29, 2015 #2


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    One thing to keep in mind is that Poynting's vector is, well, a vector: ##\mathbf{S}=\mathbf{E}\times\mathbf{H} ##. You really do want to keep the vectorial nature whenever dealing with it. Also, Poynting's vector comes out of Poynting's theorem:


    This is derived for any electromagnetic field, not just waves. (It does not matter whether or not E=cB). In your case you have no Joule heating (no ##\mathbf{J \cdot E}## term). Consider the integral version in the link: it simply says that for a volume V that is enclosed in some surface S, the electromagnetic energy stored int he volume (volume integral of u) decreases when there is a net flux of power through the surface (integral of Poynting vector).

    For your case it would be interesting to compute and plot the terms in the differential version. In your case, if the waves are propagating in the ##\pm \hat{\mathbf{z}} ##, ## \nabla \cdot \mathbf{S} = \frac{\partial}{\partial z} S_z ##, where ##S_z## is the z component of the POynting vector.


    EDIT: I found the notes you referenced a little hard to skim - but in case you are not familiar with D and H in the link I gave, in free space (like you have), ##\mathbf{D}=\epsilon_0 \mathbf{E}## and ##\mathbf{B}=\mu_0 \mathbf{H}##.
    Last edited: Jul 29, 2015
  4. Jul 29, 2015 #3


    Staff: Mentor

    No. Maxwells equations do not require cB=E in all cases.
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