documentation: add example of non-orthogonal lattice

[sp94]

We should add examples of non-orthogonal lattices to the documentation, with some pictures

Interface

Assuming both lattice vectors a1 and a2 should have length == 1

In this case it is only necessary to specify the angles of the lattice vectors:
geometry = Geometry(epf, muf, th1, th2, d1, d2)
where th1 and th2 are scalars

More generally

If the lattice vector lengths should not be equal to 1, then use
geometry = Geometry(epf, muf, a1, a2, d1, d2) where a1 and a2 are 2-vectors

Example

Triangular lattice example:
geometry = Geometry(epf, muf, 0, 60, d1, d2)
or equivalently
geometry = Geometry(epf, muf, [cosd(0),sind(0)], [cosd(60),sind(60)], d1, d2)

To do

Add an example from Joannopoulos' book, such as the TE and TM modes of triangular array of air columns in dielectric substrate

[kabume]

Hi @sp94 , I have built a model of triangle lattice by changing the function wu_ep(...). But the bandgap diagram is not exact for the K point is not degeneracy. The codes are attached:

using Peacock
a1 = [1, 0]; a2 = [1/2, sqrt(3)/2]

function epf(x, y)
#    angles=[60n for n in 0:5]
    angles = [30 210]
    ep_bg = 1
    ep_cyl = 15
    Rs = 1/3
    d1s = d2s = 2Rs/3
    if length(Rs) == 1
        Rs = fill(Rs[1], length(angles))
    end
    if length(d1s) == 1
        d1s = fill(d1s[1], length(angles))
    end
    if length(d2s) == 1
        d2s = fill(d2s[1], length(angles))
    end
    @assert length(Rs) == length(d1s) == length(d2s) == length(angles)
    # Triangular lattice sites
    lx, ly = 1, 1*sqrt(3)
    sites = [
        (0, 0),
        (-lx, 0),
        (+lx, 0),
        (-lx/2, -ly/2),
        (+lx/2, -ly/2),
        (+lx/2, +ly/2),
        (-lx/2, +ly/2),
    ]
    # Hexagonal ring of cylinders at each site
    for (x0,y0) in sites
        for (R,d1,d2,th) in zip(Rs,d1s,d2s,angles)
            x_, y_ = [cosd(th) sind(th); -sind(th) cosd(th)] * [(x-x0); (y-y0)]
            x_ = (x_-R) / (d1/2)
            y_ = (y_-0) / (d2/2)
            if x_^2 + y_^2 <= 1
                return ep_cyl
            end
        end
    end
    return ep_bg
end

muf(x, y) = 1

#G = BrillouinZoneCoordinate(  0,   0, "Γ")
#M = BrillouinZoneCoordinate(  0, 1/2, "M")
#K = BrillouinZoneCoordinate(1/3, 1/3, "K")
G = [0,0]; M = [0,2*pi/sqrt(3)]; K = 2*pi*[1/3,1/sqrt(3)]
ks = [G, M, K, G]

d1, d2 = 0.01, 0.01

geometry = Geometry(epf, muf, a1, a2, d1, d2)
plot(geometry)

solver_PWE =  Solver(geometry, 7)
figure(figsize=(4,3))
plot_band_diagram(solver_PWE, ks, TM, color="blue"; bands=1:2, dk=0.1, frequency_scale=1/2pi)

[sp94]

Hi @kabume, for a simple triangular lattice you only need one cylinder per unit cell so it is quite simple

I think you were trying to reproduce the figure from Joannopoulos book where it is air cylinders in dielectric background so I also swapped those

using Peacock, PyPlot

th1 = 0;
th2 = 120;

epf(x,y) = x^2 + y^2 < 1/3^2 ? 1 : 15
muf(x,y) = 1
d1, d2 = 0.01, 0.01

geometry = Geometry(epf, muf, th1, th2, d1, d2)
plot(geometry)

G = BrillouinZoneCoordinate(  0,   0, "Γ")
M = BrillouinZoneCoordinate(  0, 1/2, "M")
K = BrillouinZoneCoordinate(1/3, 1/3, "K")
ks = [G, M, K, G]

solver_PWE =  Solver(geometry, 7)
figure(figsize=(4,3))
plot_band_diagram(solver_PWE, ks, TM, color="blue"; bands=1:2, dk=0.1, frequency_scale=1/2pi)