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In Mechanical Wave v = w/k. EM wave w/k = c. How Equated ?

  1. May 13, 2015 #1

    morrobay

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    With ω/k = 2π/T / 2π/λ = velocity for both transverse mechanical waves and EM waves.
    I can understand velocity as distance over time in mechanical wave. But how is the ratio Em/Bm = ω/k = c.
    That is the maximum amplitudes of the E and B fields in the y and z planes corresponding to c in x direction ?
    Is it correct to say that the " leading edges" of the E and B fields in y and z planes are at c ?

    http://www.santarosa.edu/~lwillia2/42/WaveEquationDerivation.pdf
     
    Last edited: May 13, 2015
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  3. May 14, 2015 #2

    morrobay

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    Too late edit : With dE/dx and t constant, dB/dt and x constant --> ∂E/∂x = - ∂B/∂t .... then to ω/k = Em/Bm = c. Does answer the change in x/change in t question. Again still not sure how this is related to amplitudes ?
     
  4. May 14, 2015 #3

    SteamKing

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    The velocity of propagation for a transverse wave is not related to the amplitude of the wave at all.
     
  5. May 14, 2015 #4

    morrobay

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    Well thats my question , with ω/k = Emax/Bmax = c. (from Halliday-Resnick) The propagation velocity is related to amplitude. Unless I am misinterpreting Em/Bm
     
  6. May 14, 2015 #5

    morrobay

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    This is from a page in the text : kEm cos (kx-ωt) = ωBm cos (kx-ωt).
    ω/k = Em/Bm = c
    Thus the speed of wave c is the ratio of the amplitude of the electric and magnetic components of the wave.
    So can someone show how Em/Bm = ω/k ?
     
  7. May 15, 2015 #6

    jtbell

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    Consider the following plane wave as an example: $$\vec E = \hat x E_m \cos (kz - \omega t) \\ \vec B = \hat y B_m \cos (kz - \omega t) $$ where ##\hat x## and ##\hat y## are unit vectors in the x and y directions. That is, ##\vec E## and ##\vec B## are in the x and y directions respectively, and the wave propagates in the z direction. Substitute these into the third Maxwell equation in free space: $$\vec \nabla \times \vec E = - \frac {\partial \vec B}{\partial t}$$ and you will get the desired result.

    [added: I originally had ##E_m## instead of ##B_m## in my equation for ##\vec B## above. I've fixed this.]
     
    Last edited: May 17, 2015
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