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JustinLevy
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Electromagnetic waves
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.
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]
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.
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.
Last edited: