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Electric and magnetic waves orthogonal to each other?

  1. May 5, 2007 #1
    In my intro to E&M course, in the section on electromagnetic waves, my textbook just says that electric and magnetic waves are orthogonal to each other, but it doesn't say why. How do we know this? Is it from solving the wave partial differential equation? If so, given that I've tooken a course on intro to DEs that slightly covered PDEs, is it possible for me to solve the wave equation and find out the the E and M waves are orthogonal to each other?
  2. jcsd
  3. May 6, 2007 #2


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    Staff: Mentor

    You get this not from the differential wave equations for [itex]\vec E[/itex] and [itex]\vec B[/itex], but from the third Maxwell equation, [itex]\nabla \times \vec E = - \partial \vec B / \partial t[/itex]. See


    in particular the section beginning with equation 448.
  4. May 7, 2007 #3
    I agree that light is transverse wave, i.e. E and B are all normal to the propagation direction and normal to each other.

    However, I saw some definitions about TE, TM and TEM stating that TE is transverse wave where you have only E component normal to the propagation, and so on for TM, TEM. This is meaningless since you would never have E or B NOT normal to the light direction, so all light is TEM.

    Am I lost somewhere?
  5. May 7, 2007 #4
    You are not talking about waves in free space, but about waves in a waveguide (i.e. guided waves). Propagation of TE and TM modes can be seen as two plane waves reflecting in zig-zag against the walls of the guide. E and B field are orthogonal, but one of them is not orthogonal to the direction of propagation. Try to find a drawing of the shape of electric and magnetic fields in a waveguide.
  6. May 12, 2007 #5
    That link doesn't explain why it is k vector dot r vector. My textbooks say that its just k*r, where k and r are scalars
  7. May 14, 2007 #6
    Ah yes, after digging a chapter for the waveguides, I got it now.

    Thank you, lpfr.
  8. May 14, 2007 #7
    Proton, look at eq. 451:


    This implies that k and E are orthogonal.
    If they were not, the result of the LHS would be another vector that's not parallel with E.

    Draw the product of v=kxE, and then kxv, you will see that.

    Or, another way is using axbxc=b(ac)-c(ab) where ab is scalar product.
    Last edited: May 14, 2007
  9. May 14, 2007 #8
    ok I found another textbook today that derived that E and B are orthogonal in a similar way to the link jtbell provided. I just found it strange that k and r were vectors, but I managed to figure it out.
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