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Maxwell's Equations in Vacuum: Constraints on Wave

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Condensed/simplified problem statement

    [itex]\vec{E} = f_{y}(x-ct)\hat{y} + f_{z}(x-ct)\hat{z} \\
    \vec{B} = g_{y}(x-ct)\hat{y} + g_{z}(x-ct)\hat{z} \\[/itex]

    All the f and g functions go to zero as their parameters go to ±∞.

    Show that gy = fz and gz = -fy



    2. Relevant equations
    [itex]\nabla \cdot \vec{E} = 0 \\
    \nabla \cdot \vec{B} = 0 [/itex]
    [itex]\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} [/itex]
    [itex]\nabla \times \vec{B} = \frac{1}{c^{2}}\frac{\partial \vec{E}}{\partial t}
    [/itex]



    3. The attempt at a solution
    [itex]\nabla \times \vec{E} = -\frac{\partial f_{z}(x-ct)}{\partial x}\hat{y} + \frac{\partial f_{y}(x-ct)}{\partial x}\hat{z} = -\frac{\partial \vec{B}}{\partial t} = \\-(g^{'}_{y}(x-ct)(-c)\hat{y} + g^{'}_{z}(x-ct)(-c)\hat{z}) = c(g^{'}_{y}(x-ct)\hat{y} + g^{'}_{z}(x-ct)\hat{z}) [/itex]

    The -c factors come from the chain rule when differentiating B with respect to t. So then I have the following.

    [itex]-\frac{\partial f_{z}(x-ct)}{\partial x} = -f^{'}_{z}(x-ct) = cg^{'}_{y}(x-ct) \\
    \frac{\partial f_{y}(x-ct)}{\partial x} = f^{'}_{y}(x-ct) = cg^{'}_{z}(x-ct)[/itex]




    The problem is to relate the functions directly and not by their derivatives. We are given that the functions all go to zero as the parameter goes to +/- infinity. So maybe it would be okay to just integrate without concern for a constant added on. But supposedly the c factor isn't there. The answer is supposed to have[itex]g_{y} = -f_{z}[/itex]. How do I get there?
     
    Last edited: Feb 6, 2013
  2. jcsd
  3. Feb 6, 2013 #2

    fzero

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    B and E do not have the same units, so f and g must be related with an appropriate factor of c. There's probably a typo in the problem. Your argument about the integration constants makes sense.
     
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