> On Jun 25, 2017, at 8:45 PM, mayawei <maya...@stu.xjtu.edu.cn> wrote:
> First, if eq. 4.22 in book chapter: http://arxiv.org/abs/1301.5366
> <http://arxiv.org/abs/1301.5366>, does not
> apply to lossy metals which epsilon is complex (Lorentz-Drude model), now can
> (
> dft-ldos f 0 1) apply to? According to the web http://ab-initio.mit.edu/wiki/
> <http://ab-initio.mit.edu/wiki/>
> index.php/Meep_Tutorial/Local_density_of_states, LDOS can be calculated
> directly from the power radiated by a dipole and Fourier-transforming the
> result of a pulse using the dft-ldos command, which is eq. 4.21 in book
> chapter:
> http://arxiv.org/abs/1301.5366 <http://arxiv.org/abs/1301.5366>. Then,if
> dft-ldos command can calculate LDOS of
> lossy metals. My model is a metal disc.
If you put a point source *inside* a lossy material (e.g. a Drude metal), then
the power expended by a dipole diverges as you increase the resolution. (In
the limit of infinite resolution, infinite power is absorbed). That is why
LDOS is not well defined for points inside of a lossy material. It's not a
question of computational technique.
However, LDOS is perfectly well defined for points *outside* the lossy
material. e.g. you can put a point outside your metal disc and calculate the
LDOS, and it will converge to a finite value as you increase the resolution.
(It is possible to define a "radiative" LDOS that is only the power radiated to
infinity by a point source, and not radiated+absorbed. This would be finite
even for points inside a lossy material. However, this definition is
problematic in other ways — it can't be used for many of the things that LDOS
is usually used for, e.g. spontaneous emission rates. It is also not what
Meep's dft-ldos function computes. You could compute it by computing the power
flux out through the boundaries of the computational cell if there is no lossy
material there, however.)
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