I see, this makes sense, thank you. So this is not a proper method of determining the resonance outgoing-coupling constant D, as it is typically denoted in coupled mode theories? (as in s_out = C * s_in + D*a)
On Fri, Mar 8, 2019 at 11:42 AM Steven G. Johnson <stevenj....@gmail.com> wrote: > > On Mar 7, 2019, at 11:19 PM, Alexander Cerjan <alexcer...@gmail.com> > wrote: > Naively, I would have expected that if I do a time-domain simulation and > place narrow-in-frequency sources inside of the structure (central > frequency on the resonance band of the 2d phc slab, very small df), that > upon collecting the outgoing flux sufficiently far away from the slab that > I would see more outgoing flux for excitations at low-Q resonances than at > high-Q resonances. Instead, I am seeing somewhat the opposite, larger > outgoing flux magnitudes when I excite high-Q resonances than low-Q ones. > > > No, it sounds like what you are seeing is exactly correct. You are > measuring the LDOS (power expended by a dipole source at a given frequency > and position) and are seeing Purcell enhancement: a current source gets a > Q/V enhancement of its emitted power (for a fixed current J), so the same > current source in a higher Q cavity emits more power (if the coupling to > the mode is the same). See also the review in section 4.4 of our textbook > chapter: https://arxiv.org/abs/1301.5366 > > (On the other hand, if you were to put in a dipole source with a fixed > *voltage*, instead of a fixed *current*, you would get less power out > with higher Q. In the antenna community, the Purcell enhancement factor > Q/V is proportional to something called "radiation resistance" R. If you > fix current I, then power I²R increases with resistance, as in your > simulation, whereas if you fix voltage V then the power V²/R decreases with > resistance.) > > Steven >
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