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E-M waves detail

  1. Nov 17, 2005 #1

    quasar987

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    Basically, my book (Modern Optics by Robert Guenther) presents the "proof" of the interdependancy of the plane waves E and B as follow:

    Suppose [itex]\vec{E}[/itex] is an electric plane wave:

    [tex]\vec{E} = \vec{E_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)}[/tex].

    Then we find that

    [tex]\frac{\partial \vec{E}}{\partial t}=i\omega \vec{E}[/tex].

    And if [itex]\vec{B}[/itex] is a plane wave in-phase with [itex]\vec{E}[/itex], such as

    [tex]\vec{B} = \vec{B_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)}[/tex],

    then

    [tex]\vec{\nabla}\times \vec{B} = -i\vec{k}\times \vec{B}[/tex].

    And thus, given [itex]\vec{E}[/itex] a plane wave, [itex]\vec{B}[/itex] a plane in-phase satify the Maxwell equation

    [tex]\vec{\nabla}\times \vec{B} = \mu\epsilon \frac{\partial \vec{E}}{\partial t}[/tex]

    under the simple condition that [itex]E_0 = cB_0[/itex] but what tells me that given [itex]\vec{E}[/itex] a plane wave, this the only solution? It's this little detail that bugs me.
     
    Last edited: Nov 17, 2005
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  3. Nov 18, 2005 #2

    Galileo

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    If I understand your problem correctly. The author has shown the given solution for B satisfied Maxwell's equations. What you'd rather want is to show that, given E, Maxwell's equations imply that B must be of that form correct?
    That is indeed the way I'd prefer it too. Since you are given [itex]\vec E(\vec r,t)[/itex], Maxwell tells you that, in vacuum:
    [tex]\vec \nabla \cdot E =0[/tex]
    [tex]\vec \nabla \cdot \vec B=0[/tex]
    [tex]\vec \nabla \times \vec B=\frac{1}{c^2}\frac{\partial \vec E}{\partial t}[/tex]
    [tex]\vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t}[/tex]

    Just use these to see how the plane wave looks like. It'll give you a set of interdependent equations. The plane wave satisfies Maxwell's equations only under certain conditions. The first for example (divergence of E vanishes) tells you that k is perpendicular to E. Using the others you can show that B and E are in phase and mutually perpendicular. Give it a shot.

    Hint: Not neccessary, but for simplicity, choose your axes so that E is point in the x direction and k in the z direction. No loss of generality there after you've shown that k and E are perpendicular.
     
  4. Nov 18, 2005 #3

    Physics Monkey

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    On a related note, one can demonstrate fairly generally that the electric and magnetic fields of a spatially confined system of charges and currents satisfy [tex] \vec{B} = \hat{n}\times \vec{E} [/tex] in the so called "radiation zone" far away from the charges. One way to do so is to use the retarded potentials in the Lorentz gauge and calculate the leading contribution to the fields at large distances (this is the [tex] 1/r [/tex] radiation field). After a little playing around, you can find the above relation without too much trouble. You have to use conservation of charge at one point in the calculation, so you do need to be careful with the retarded time in your calculation.
     
    Last edited: Nov 18, 2005
  5. Nov 18, 2005 #4

    quasar987

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    Nevermind, I guess my question doesn't make sense.
    E plane and B plane are acceptable solutions of the wave equations provided they meet the criterions

    i) |E|/|B| = c/n
    ii) E, B and k are mutually perpendicular
    iii) E and B are in phase

    That's all there is to it
     
    Last edited: Nov 18, 2005
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