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Plane EM wave Euler's identity

  1. Oct 24, 2014 #1
    For EM wave, magnetic and electrical components are in phase, meaning when E = 0, then B = 0.

    Thus, I understand if it is written:

    f(x,t) = A(cos(kx - wt) + icos(kx - wt))

    Then why plane wave is always described:

    f(x,t) = Aei(kx-wt) = A(cos(kx-wt) + isin(kx - wt))

    Implying that Real and Imaginary components are not in phase, which is not the case?

    Thank you.
     
  2. jcsd
  3. Oct 24, 2014 #2

    jtbell

    User Avatar

    Staff: Mentor

    Electric versus magnetic fields have nothing to do with real versus imaginary numbers. If you have a plane EM wave traveling in the z-direction, with the electric field parallel to the x-axis and the magnetic field parallel to the y-axis, then we have $$\vec E = E_0 \cos (kz - \omega t + \phi_0) \hat x \\
    \vec B = B_0 \cos (kz - \omega t + \phi_0) \hat y$$ where the unit vectors ##\hat x## and ##\hat y## give the directions of the fields.

    It makes some mathematical manipulations easier if we write the oscillation using a complex exponential instead of an ordinary trig function: $$\vec E = E_0 e^{i(kz - \omega t + \phi_0)} \hat x = E_0 \left[ \cos(kz - \omega t + \phi_0) + i \sin (kz - \omega t + \phi_0) \right] \hat x \\ \vec B = B_0 e^{i(kz - \omega t + \phi_0)} \hat y = B_0 \left[ \cos(kz - \omega t + \phi_0) + i \sin (kz - \omega t + \phi_0) \right] \hat y$$ but in this case it is always understood that in the end we use only the real part of the final result of our calculations.
     
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