Effect of Roughness on Cavity Resonator Spectrum

[Swamp Thing]

If we start with a light mode in a perfect cavity, in a state such as $1/2\hbar\omega_m + N\hbar\omega_m$ , what happens if we introduce a small amount of roughness (something like 0.5% to 5% of $\lambda_m$) ?

Would it create a cluster of similar but non-degenerate discrete modes around the original $\omega$, or would it be a continuum spread around $\omega$ ? And how would we estimate the total energy associated with that spread out mode? Or, would the roughness have no effect at all on the spectrum?

Would the total energy increase (due to more degrees of freedom), or would it remain constant and somehow re-distribute over the spread out band?

Edit : Although the walls are rough, let's assume that they are 100% reflective, so no light can leave the cavity.

 

[Paul Colby]

Roughness would add terms to the Hamiltonian which could be modeled as a coupling between ideal modes. Like all problems, write it out and grind. Degenerate states would split and shift.

 

[tech99]

would the roughness have no effect at all on the spectrum?

Would the total energy increase (due to more degrees of freedom), or would it remain constant and somehow re-distribute over the spread out band?

I think that for single frequency excitation, a cavity resonator, as with an LC circuit, will only contain the frequency with which it is driven.If the modes are changed by the roughness, we are just altering the field distributions within the cavity but for the same frequency.
On the other hand if the cavity is driven by a continuous noise spectrum, such as white light, then we can see various frequencies being accentuated.