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EM wave orthoganality

  1. May 13, 2009 #1
    So im trying to work through the proof why why the direction of proporgation, the E field and B field are all orthogonal to one another.

    What i have is...

    [tex]E=E_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]
    [tex]B=B_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]

    [tex]\nabla \times E= -\frac{dB}{dt} \Rightarrow k \times E_{0}= \omega B_{0}[/tex]

    [tex]\nabla \times B= \mu_{0}\epsilon_{0}\frac{dE}{dt} \Rightarrow k \times B_{0}= \mu_{0}\epsilon_{0}\omega E_{0}[/tex]

    and i can see from this why k, B and E must be orthogonal. What im having difficulty with is how to get from the left to the right...

    Any ideas?
     
    Last edited: May 13, 2009
  2. jcsd
  3. May 13, 2009 #2
    It is interesting and very useful to think about two signals passing one another in a coax cable (e.g., RG-8) without interference. I show cartesian coordinates below, but in a coax, E is radial, and H is azimuthal. Also, for RG-8, E/H = 50 ohms

    Ex = E0 exp[j(k z - w t)]
    Hy = H0 exp[j(k z - w t)]

    and
    Ex = E0 exp[j(-k z - w t)]
    Hy = -H0 exp[j(-k z - w t)]

    So one pulse goes from left to right, and the other from right to left, without a "collision". There are instruments (directional couplers) that can measure the amplitude AND direction of each pulse simultaneously.

    Edit: changed k x to k z four places
     
    Last edited: May 13, 2009
  4. May 13, 2009 #3

    jtbell

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    You mean how to work out the derivatives?
     
  5. May 13, 2009 #4

    Andy Resnick

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    Have you tried taking a dot product?

    Edit- I just noticed the form you wrote for E and B- spherical wavefronts. That can complicate the maths, so it's easier to use plane waves.
     
    Last edited: May 13, 2009
  6. May 13, 2009 #5
    For a plane wave, we can use the vector identity (ignoring any currents J)
    div(E x H) = H curl E - E curl H
    = -E dD/dt - H dB/dt = -1/2 (d/dt)(e0E2 + u0 H2). (= energy loss per meter). So
    S= E x H is a vector (called the Poynting vector), whose direction is the direction, and magnitude is the average power flow, in watts per square meter. Both E and H are orthogonal to direction of power flow.
     
  7. May 13, 2009 #6

    gabbagabbahey

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    Actually, the OP is using plane waves (and just not specifying the direction of propagation in Cartesian coords, explicitly). Spherical waves have a much more complicated form.
     
  8. May 14, 2009 #7

    gabbagabbahey

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    If you are having trouble working out the derivatives, it will help you to look at my responses to question 2. in this thread since you are essentially both trying to work out the same curls....Are you also having trouble with the time derivatives?
     
  9. May 14, 2009 #8
    That thread has proved helpful - i have cracked the derivatives. Thanks.

    Another question though, k is the wavenumber right? Is the wavenumber equivalent to the direction of proporgation?

    I mean, from the solution k, E and B must be orthogonal, but how does this relate to the direction of proporgtaion?
     
  10. May 14, 2009 #9

    gabbagabbahey

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    You're welcome!:smile:

    The scalar [itex]k[/itex] is the wavenumber, the vector [itex]\vec{k}[/itex] which appears in your plane waves (it's a good idea to get into the habit of explicitly writing vectors with some sort of marking to distinguish them from scalars!) , is defined as the product of the wavenumber with the unit vector in the direction of propagation. In other words, [itex]\vec{k}[/itex] points in the direction of propagation and its magnitude is the wavenumber.
     
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