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About E=cB

  1. Feb 17, 2010 #1
    I know for spherical or plane EM wave, there's relation E=cB, and we can prove it by the explicit expression of these two kinds of wave. But does E=cB hold for all EM waves, e.g. all possible wavelike solutions of maxwell's equation?
  2. jcsd
  3. Feb 17, 2010 #2

    electromagnetic waves have always vector character and a common relation (yield from Maxwell's equations) between the electric field [tex]\vec E(\vec r,t)[/tex] and the magnetic field [tex]\vec B(\vec r,t)[/tex] is

    [tex]\vec B(\vec r,t) = \frac{1}{\omega} \, \vec k \times \vec E(\vec r,t)[/tex]​

    As you stated waves can have a more complex form! The most general form can be represented by the fourier transformation and is a consequence of the linearity of Maxwell's equations. A general expression for the electric field in this form is

    [tex]\vec E(\vec r, t) = \frac{1}{(2\pi)^{3/2}} ~ \int ~ \mathrm{d}^3k ~ \tilde{ \vec E}(\vec k) \, e^{i(\vec k \vec r - c |\vec k|t)}[/tex]​

    where [tex]\tilde{\vec E}(\vec k)[/tex] is the amplitude of a plane wave that belongs to the wavevector [tex]\vec k[/tex]. This is often called a wave packet.

    So we are talking about the amplitudes of the electric field belonging to the plane wave with wavevector [tex]\vec k[/tex]. But what is about the amplitudes of the magnetic field?
    For the amplitudes the first equation is always true, because [tex]\tilde{\vec B}, \tilde{\vec E}, \vec k[/tex] are always perpendicular. So the amplitudes of the magnetic field are

    [tex]\tilde{\vec B}(\vec k) = \frac{1}{\omega} \vec k \times \tilde{\vec E}(\vec k)[/tex]​

    So [tex]\vec B(\vec r,t)[/tex] is analogous to [tex]\vec E(\vec r,t)[/tex]

    [tex]\vec B(\vec r, t) = \frac{1}{(2\pi)^{3/2}} ~ \int ~ \mathrm{d}^3k ~ \tilde{\vec B}(\vec k) \, e^{i(\vec k \vec r - c |\vec k|t)} = \frac{1}{(2\pi)^{3/2}} ~ \int ~ \mathrm{d}^3k ~ \, \frac{1}{\omega} \, \vec k \times \tilde{\vec E}(\vec k) ~ e^{i(\vec k \vec r - c |\vec k|t)}[/tex]​

    So, the above equation is really the general solution of Maxwell's equations! Like you have seen, nothing is as easy as it seems (remember plane waves aren't physical but the infinite sum of them are as long as they vanish in infinty).

    Hope i could help...
  4. Feb 17, 2010 #3
    And where does
    \vec B(\vec r,t) = \frac{1}{\omega} \, \vec k \times \vec E(\vec r,t)
    come from?
  5. Feb 17, 2010 #4
    I assume, that we already know that the plane waves

    [tex]\vec E(\vec r,t) = \vec E_0 ~ e^{i(\vec k \vec r - \omega t)}[/tex]
    [tex]\vec B(\vec r,t) = \vec B_0 ~ e^{i(\vec k \vec r - \omega t)}[/tex]​

    are solutions of Maxwell's equation. Let's calculate

    [tex]\mathrm{rot} ~ \vec E(\vec r,t) = - \frac{\partial \vec B(\vec r,t)}{\partial t}[/tex]

    with these solutions.

    [tex]\mathrm{rot} ~ \vec E(\vec r,t)[/tex] yields

    [tex]\mathrm{rot} ~ \vec E(\vec r,t) = \vec \nabla \times \Bigl( \vec E_0 ~ e^{i(\vec k \vec r - \omega t)} \Bigl) = i \vec k \times \vec E_0 ~ e^{i(\vec k \vec r - \omega t)}[/tex]

    [tex]\partial \vec B(\vec r,t) / \partial t[/tex] yields

    [tex]\frac{\partial \vec B(\vec r,t)}{\partial t} = - i \omega \vec B_0 ~ e^{i(\vec k \vec r - \omega t)[/tex]

    With these results we got the expression

    [tex]\vec k \times \vec E_0 = \omega ~ \vec B_0[/tex]​
  6. Feb 17, 2010 #5


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    Staff: Mentor

  7. Feb 18, 2010 #6
    Thanks a lot!
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