How Do Gaussian Profiles Affect Solutions to Maxwell's Equations in a Vacuum?

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JustinLevy
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Electromagnetic waves

Homework Statement


Find the solution of Maxwell's equations in vacuum for a continuous beam of light of frequency [itex]\omega[/itex] traveling in the z direction with a gaussian profile in the x and y directions.

Homework Equations


Maxwell's equations in vaccuum.
[tex]\nabla \cdot \vec{E} = 0[/tex]
[tex]\nabla \cdot \vec{B} = 0[/tex]
[tex]\nabla \times \vec{E} = - \frac{\partial}{\partial t} \vec{B}[/tex]
[tex]\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \vec{E}[/tex]

These of course can be combined to give the wave equation:
[tex]\nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E} = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{E}[/tex]



The Attempt at a Solution



I already know what the plane wave solutions look like in the z direction:
[tex] \vec{E}(x,y,z) = \vec{E}_0 \cos(kz - \omega t) [/tex]
where [tex]k = \omega/c[/tex]

No polarization was specified, so let's choose linear polarization in the x direction. Our solution should then be similar for polarization in the y direction, and we can get the general solution of any polarization by adding these with arbitrary amplitudes and phases.

So, for linear polarization in the x direction and generalizing the above with dependence in the x-y direction it would be something like:
[tex] \vec{E}(x,y,z) = \hat{x} \ E_0 \ g(x,y) \ \cos(kz - \omega t) [/tex]

Now seeing the constraint on g(x,y) we find:
[tex]0 = \nabla \cdot \vec{E} = E_{0} \ \cos(kz - \omega t) \frac{\partial}{\partial x}g(x,y)[/tex]

Which seems to say there can't be any dependence on x! What!?

What am I doing wrong?
I've tried searching around the net and haven't found any good hints on this problem either. Please help.
 
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