On Thu, 30 Nov 2006, [EMAIL PROTECTED] wrote:
Is it *necessarily* true that all the bands move to lower frequency values
as the effective index of the photonic crystal is increased ? I thought
this was, in general, true. But now I have some calculations that show
otherwise, and am a little confused.
I don't know what you mean by "effective index" in this context. If you
mean "average" index, with the usual definition of "average", then this
statement is not correct in general.
There are a few statements that you can prove rigorously.
First, if you multiply the index everywhere by a constant C (i.e. for
*all* the materials by the *same* constant), then the eigenfrequencies
decrease by the same factor of C.
Second, if you change all of the dielectric constants everywhere by
\Delta\epsilon << 1, where \Delta\epsilon can be any arbitrary function of
position as long as it is small everywhere, then the frequencies decrease
(or increase) if the integral of \Delta\epsilon |E|^2 is positive (or
negative). That is, the frequencies decrease if you slightly increase the
*average* dielectric constant *weighted* by |E|^2. (This is proved from
perturbation theory.)
Third, as a consequence of the second theorem, if you change the
refractive index *everywhere* by a non-negative amount, the frequencies
must decrease (or at least cannot increase).
What does *not* follow is that increasing the index "on average", where
the average is not weighted by n|E|^2, will necessarily decrease the
frequencies. In particular, if you increase the index in some regions and
decrease it in other regions, the frequency could either increase or
decrease depending on where the electric field is concentrated.
Steven
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