- #1
dEdt
- 288
- 2
Hey guys,
I'm trying to understand the properties of normal modes of the electromagnetic field inside an arbitrary cavity, but I'm having some trouble.
By definition, for a normal mode we have [itex]\mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 (\mathbf{x}) e^{i \omega_1 t}[/itex] and [itex]\mathbf{B}(\mathbf{x},t) = \mathbf{B}_0 (\mathbf{x}) e^{i \omega_2t}[/itex]. It's easy to show that [itex]\omega_1 = \omega_2[/itex]. Substituting these two equations into the wave equation gives the standard equation for the normal modes of a system, namely
[tex]\nabla^2 \mathbf{E}_0 + \frac{\omega^2}{c^2}\mathbf{E}_0 = 0[/tex]
and
[tex]\nabla^2 \mathbf{B}_0 + \frac{\omega^2}{c^2}\mathbf{B}_0 = 0.[/tex]
For any allowed value of [itex]\omega[/itex], there will be three linearly independent solutions to each of these equations -- unless we have degeneracy, of course. The conditions
[tex]\nabla \cdot \mathbf{E}_0 = 0[/tex]
and
[tex]\nabla \cdot \mathbf{B}_0 = 0[/tex]
reduce this number to two.
In the case of a rectangular cavity, it's easy to show that
1) the electric and magnetic fields are perpendicular; and
2) the two linearly independent solutions mentioned above correspond to the two possible polarization states.
My intuition tells me that these properties extend to the case of an arbitrarily shaped cavity, but I can't prove that they do. In particular, I'm not sure how polarization is defined if our electromagnetic field solution doesn't have a well-defined direction of propagation.
Any ideas?
I'm trying to understand the properties of normal modes of the electromagnetic field inside an arbitrary cavity, but I'm having some trouble.
By definition, for a normal mode we have [itex]\mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 (\mathbf{x}) e^{i \omega_1 t}[/itex] and [itex]\mathbf{B}(\mathbf{x},t) = \mathbf{B}_0 (\mathbf{x}) e^{i \omega_2t}[/itex]. It's easy to show that [itex]\omega_1 = \omega_2[/itex]. Substituting these two equations into the wave equation gives the standard equation for the normal modes of a system, namely
[tex]\nabla^2 \mathbf{E}_0 + \frac{\omega^2}{c^2}\mathbf{E}_0 = 0[/tex]
and
[tex]\nabla^2 \mathbf{B}_0 + \frac{\omega^2}{c^2}\mathbf{B}_0 = 0.[/tex]
For any allowed value of [itex]\omega[/itex], there will be three linearly independent solutions to each of these equations -- unless we have degeneracy, of course. The conditions
[tex]\nabla \cdot \mathbf{E}_0 = 0[/tex]
and
[tex]\nabla \cdot \mathbf{B}_0 = 0[/tex]
reduce this number to two.
In the case of a rectangular cavity, it's easy to show that
1) the electric and magnetic fields are perpendicular; and
2) the two linearly independent solutions mentioned above correspond to the two possible polarization states.
My intuition tells me that these properties extend to the case of an arbitrarily shaped cavity, but I can't prove that they do. In particular, I'm not sure how polarization is defined if our electromagnetic field solution doesn't have a well-defined direction of propagation.
Any ideas?