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@ -12,7 +12,7 @@ cd ~/projects/meanas
rm -rf _doc_mathimg
pdoc --pdf --force --template-dir pdoc_templates -o doc . > doc.md
pandoc --metadata=title:"meanas" --from=markdown+abbreviations --to=html --output=doc.htex --gladtex -s --css pdoc_templates/pdoc.css doc.md
gladtex -a -n -d _doc_mathimg -c white doc.htex
gladtex -a -n -d _doc_mathimg -c white -b black doc.htex
# Approach 3: html with gladtex
#pdoc3 --html --force --template-dir pdoc_templates -o doc .

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@ -1,4 +1,4 @@
"""
r"""
Tools for finite difference frequency-domain (FDFD) simulations and calculations.
These mostly involve picking a single frequency, then setting up and solving a
@ -19,71 +19,71 @@ Submodules:
From the "Frequency domain" section of `meanas.fdmath`, we have
$$
\\begin{aligned}
\\tilde{E}_{l, \\vec{r}} &= \\tilde{E}_{\\vec{r}} e^{-\\imath \\omega l \\Delta_t} \\\\
\\tilde{H}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} &= \\tilde{H}_{\\vec{r} + \\frac{1}{2}} e^{-\\imath \\omega (l - \\frac{1}{2}) \\Delta_t} \\\\
\\tilde{J}_{l, \\vec{r}} &= \\tilde{J}_{\\vec{r}} e^{-\\imath \\omega (l - \\frac{1}{2}) \\Delta_t} \\\\
\\tilde{M}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} &= \\tilde{M}_{\\vec{r} + \\frac{1}{2}} e^{-\\imath \\omega l \\Delta_t} \\\\
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}})
-\\Omega^2 \\epsilon_{\\vec{r}} \\cdot \\tilde{E}_{\\vec{r}} &= -\\imath \\Omega \\tilde{J}_{\\vec{r}} e^{\\imath \\omega \\Delta_t / 2} \\\\
\\Omega &= 2 \\sin(\\omega \\Delta_t / 2) / \\Delta_t
\\end{aligned}
\begin{aligned}
\tilde{E}_{l, \vec{r}} &= \tilde{E}_{\vec{r}} e^{-\imath \omega l \Delta_t} \\
\tilde{H}_{l - \frac{1}{2}, \vec{r} + \frac{1}{2}} &= \tilde{H}_{\vec{r} + \frac{1}{2}} e^{-\imath \omega (l - \frac{1}{2}) \Delta_t} \\
\tilde{J}_{l, \vec{r}} &= \tilde{J}_{\vec{r}} e^{-\imath \omega (l - \frac{1}{2}) \Delta_t} \\
\tilde{M}_{l - \frac{1}{2}, \vec{r} + \frac{1}{2}} &= \tilde{M}_{\vec{r} + \frac{1}{2}} e^{-\imath \omega l \Delta_t} \\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{\vec{r}})
-\Omega^2 \epsilon_{\vec{r}} \cdot \tilde{E}_{\vec{r}} &= -\imath \Omega \tilde{J}_{\vec{r}} e^{\imath \omega \Delta_t / 2} \\
\Omega &= 2 \sin(\omega \Delta_t / 2) / \Delta_t
\end{aligned}
$$
resulting in
$$
\\begin{aligned}
\\tilde{\\partial}_t &\\Rightarrow -\\imath \\Omega e^{-\\imath \\omega \\Delta_t / 2}\\\\
\\hat{\\partial}_t &\\Rightarrow -\\imath \\Omega e^{ \\imath \\omega \\Delta_t / 2}\\\\
\\end{aligned}
\begin{aligned}
\tilde{\partial}_t &\Rightarrow -\imath \Omega e^{-\imath \omega \Delta_t / 2}\\
\hat{\partial}_t &\Rightarrow -\imath \Omega e^{ \imath \omega \Delta_t / 2}\\
\end{aligned}
$$
Maxwell's equations are then
$$
\\begin{aligned}
\\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}} &=
\\imath \\Omega e^{-\\imath \\omega \\Delta_t / 2} \\hat{B}_{\\vec{r} + \\frac{1}{2}}
- \\hat{M}_{\\vec{r} + \\frac{1}{2}} \\\\
\\hat{\\nabla} \\times \\hat{H}_{\\vec{r} + \\frac{1}{2}} &=
-\\imath \\Omega e^{ \\imath \\omega \\Delta_t / 2} \\tilde{D}_{\\vec{r}}
+ \\tilde{J}_{\\vec{r}} \\\\
\\tilde{\\nabla} \\cdot \\hat{B}_{\\vec{r} + \\frac{1}{2}} &= 0 \\\\
\\hat{\\nabla} \\cdot \\tilde{D}_{\\vec{r}} &= \\rho_{\\vec{r}}
\\end{aligned}
\begin{aligned}
\tilde{\nabla} \times \tilde{E}_{\vec{r}} &=
\imath \Omega e^{-\imath \omega \Delta_t / 2} \hat{B}_{\vec{r} + \frac{1}{2}}
- \hat{M}_{\vec{r} + \frac{1}{2}} \\
\hat{\nabla} \times \hat{H}_{\vec{r} + \frac{1}{2}} &=
-\imath \Omega e^{ \imath \omega \Delta_t / 2} \tilde{D}_{\vec{r}}
+ \tilde{J}_{\vec{r}} \\
\tilde{\nabla} \cdot \hat{B}_{\vec{r} + \frac{1}{2}} &= 0 \\
\hat{\nabla} \cdot \tilde{D}_{\vec{r}} &= \rho_{\vec{r}}
\end{aligned}
$$
With $\\Delta_t \\to 0$, this simplifies to
With $\Delta_t \to 0$, this simplifies to
$$
\\begin{aligned}
\\tilde{E}_{l, \\vec{r}} &\\to \\tilde{E}_{\\vec{r}} \\\\
\\tilde{H}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} &\\to \\tilde{H}_{\\vec{r} + \\frac{1}{2}} \\\\
\\tilde{J}_{l, \\vec{r}} &\\to \\tilde{J}_{\\vec{r}} \\\\
\\tilde{M}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} &\\to \\tilde{M}_{\\vec{r} + \\frac{1}{2}} \\\\
\\Omega &\\to \\omega \\\\
\\tilde{\\partial}_t &\\to -\\imath \\omega \\\\
\\hat{\\partial}_t &\\to -\\imath \\omega \\\\
\\end{aligned}
\begin{aligned}
\tilde{E}_{l, \vec{r}} &\to \tilde{E}_{\vec{r}} \\
\tilde{H}_{l - \frac{1}{2}, \vec{r} + \frac{1}{2}} &\to \tilde{H}_{\vec{r} + \frac{1}{2}} \\
\tilde{J}_{l, \vec{r}} &\to \tilde{J}_{\vec{r}} \\
\tilde{M}_{l - \frac{1}{2}, \vec{r} + \frac{1}{2}} &\to \tilde{M}_{\vec{r} + \frac{1}{2}} \\
\Omega &\to \omega \\
\tilde{\partial}_t &\to -\imath \omega \\
\hat{\partial}_t &\to -\imath \omega \\
\end{aligned}
$$
and then
$$
\\begin{aligned}
\\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}} &=
\\imath \\omega \\hat{B}_{\\vec{r} + \\frac{1}{2}}
- \\hat{M}_{\\vec{r} + \\frac{1}{2}} \\\\
\\hat{\\nabla} \\times \\hat{H}_{\\vec{r} + \\frac{1}{2}} &=
-\\imath \\omega \\tilde{D}_{\\vec{r}}
+ \\tilde{J}_{\\vec{r}} \\\\
\\end{aligned}
\begin{aligned}
\tilde{\nabla} \times \tilde{E}_{\vec{r}} &=
\imath \omega \hat{B}_{\vec{r} + \frac{1}{2}}
- \hat{M}_{\vec{r} + \frac{1}{2}} \\
\hat{\nabla} \times \hat{H}_{\vec{r} + \frac{1}{2}} &=
-\imath \omega \tilde{D}_{\vec{r}}
+ \tilde{J}_{\vec{r}} \\
\end{aligned}
$$
$$
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}})
-\\omega^2 \\epsilon_{\\vec{r}} \\cdot \\tilde{E}_{\\vec{r}} = -\\imath \\omega \\tilde{J}_{\\vec{r}} \\\\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{\vec{r}})
-\omega^2 \epsilon_{\vec{r}} \cdot \tilde{E}_{\vec{r}} = -\imath \omega \tilde{J}_{\vec{r}} \\
$$
# TODO FDFD?

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@ -189,10 +189,10 @@ def e_tfsf_source(
def poynting_e_cross_h(dxes: dx_lists_t) -> Callable[[cfdfield_t, cfdfield_t], cfdfield_t]:
"""
r"""
Generates a function that takes the single-frequency `E` and `H` fields
and calculates the cross product `E` x `H` = $E \\times H$ as required
for the Poynting vector, $S = E \\times H$
and calculates the cross product `E` x `H` = $E \times H$ as required
for the Poynting vector, $S = E \times H$
Note:
This function also shifts the input `E` field by one cell as required

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@ -45,14 +45,14 @@ def e_full(
pec: vfdfield_t | None = None,
pmc: vfdfield_t | None = None,
) -> sparse.spmatrix:
"""
r"""
Wave operator
$$ \\nabla \\times (\\frac{1}{\\mu} \\nabla \\times) - \\Omega^2 \\epsilon $$
$$ \nabla \times (\frac{1}{\mu} \nabla \times) - \Omega^2 \epsilon $$
del x (1/mu * del x) - omega**2 * epsilon
for use with the E-field, with wave equation
$$ (\\nabla \\times (\\frac{1}{\\mu} \\nabla \\times) - \\Omega^2 \\epsilon) E = -\\imath \\omega J $$
$$ (\nabla \times (\frac{1}{\mu} \nabla \times) - \Omega^2 \epsilon) E = -\imath \omega J $$
(del x (1/mu * del x) - omega**2 * epsilon) E = -i * omega * J
@ -131,14 +131,14 @@ def h_full(
pec: vfdfield_t | None = None,
pmc: vfdfield_t | None = None,
) -> sparse.spmatrix:
"""
r"""
Wave operator
$$ \\nabla \\times (\\frac{1}{\\epsilon} \\nabla \\times) - \\omega^2 \\mu $$
$$ \nabla \times (\frac{1}{\epsilon} \nabla \times) - \omega^2 \mu $$
del x (1/epsilon * del x) - omega**2 * mu
for use with the H-field, with wave equation
$$ (\\nabla \\times (\\frac{1}{\\epsilon} \\nabla \\times) - \\omega^2 \\mu) E = \\imath \\omega M $$
$$ (\nabla \times (\frac{1}{\epsilon} \nabla \times) - \omega^2 \mu) E = \imath \omega M $$
(del x (1/epsilon * del x) - omega**2 * mu) E = i * omega * M
@ -188,28 +188,28 @@ def eh_full(
pec: vfdfield_t | None = None,
pmc: vfdfield_t | None = None,
) -> sparse.spmatrix:
"""
r"""
Wave operator for `[E, H]` field representation. This operator implements Maxwell's
equations without cancelling out either E or H. The operator is
$$ \\begin{bmatrix}
-\\imath \\omega \\epsilon & \\nabla \\times \\\\
\\nabla \\times & \\imath \\omega \\mu
\\end{bmatrix} $$
$$ \begin{bmatrix}
-\imath \omega \epsilon & \nabla \times \\
\nabla \times & \imath \omega \mu
\end{bmatrix} $$
[[-i * omega * epsilon, del x ],
[del x, i * omega * mu]]
for use with a field vector of the form `cat(vec(E), vec(H))`:
$$ \\begin{bmatrix}
-\\imath \\omega \\epsilon & \\nabla \\times \\\\
\\nabla \\times & \\imath \\omega \\mu
\\end{bmatrix}
\\begin{bmatrix} E \\\\
$$ \begin{bmatrix}
-\imath \omega \epsilon & \nabla \times \\
\nabla \times & \imath \omega \mu
\end{bmatrix}
\begin{bmatrix} E \\
H
\\end{bmatrix}
= \\begin{bmatrix} J \\\\
\end{bmatrix}
= \begin{bmatrix} J \\
-M
\\end{bmatrix} $$
\end{bmatrix} $$
Args:
omega: Angular frequency of the simulation

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@ -1,4 +1,4 @@
"""
r"""
Operators and helper functions for waveguides with unchanging cross-section.
The propagation direction is chosen to be along the z axis, and all fields
@ -12,166 +12,166 @@ As the z-dependence is known, all the functions in this file assume a 2D grid
Consider Maxwell's equations in continuous space, in the frequency domain. Assuming
a structure with some (x, y) cross-section extending uniformly into the z dimension,
with a diagonal $\\epsilon$ tensor, we have
with a diagonal $\epsilon$ tensor, we have
$$
\\begin{aligned}
\\nabla \\times \\vec{E}(x, y, z) &= -\\imath \\omega \\mu \\vec{H} \\\\
\\nabla \\times \\vec{H}(x, y, z) &= \\imath \\omega \\epsilon \\vec{E} \\\\
\\vec{E}(x,y,z) = (\\vec{E}_t(x, y) + E_z(x, y)\\vec{z}) e^{-\\gamma z} \\\\
\\vec{H}(x,y,z) = (\\vec{H}_t(x, y) + H_z(x, y)\\vec{z}) e^{-\\gamma z} \\\\
\\end{aligned}
\begin{aligned}
\nabla \times \vec{E}(x, y, z) &= -\imath \omega \mu \vec{H} \\
\nabla \times \vec{H}(x, y, z) &= \imath \omega \epsilon \vec{E} \\
\vec{E}(x,y,z) &= (\vec{E}_t(x, y) + E_z(x, y)\vec{z}) e^{-\gamma z} \\
\vec{H}(x,y,z) &= (\vec{H}_t(x, y) + H_z(x, y)\vec{z}) e^{-\gamma z} \\
\end{aligned}
$$
Expanding the first two equations into vector components, we get
$$
\\begin{aligned}
-\\imath \\omega \\mu_{xx} H_x &= \\partial_y E_z - \\partial_z E_y \\\\
-\\imath \\omega \\mu_{yy} H_y &= \\partial_z E_x - \\partial_x E_z \\\\
-\\imath \\omega \\mu_{zz} H_z &= \\partial_x E_y - \\partial_y E_x \\\\
\\imath \\omega \\epsilon_{xx} E_x &= \\partial_y H_z - \\partial_z H_y \\\\
\\imath \\omega \\epsilon_{yy} E_y &= \\partial_z H_x - \\partial_x H_z \\\\
\\imath \\omega \\epsilon_{zz} E_z &= \\partial_x H_y - \\partial_y H_x \\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{xx} H_x &= \partial_y E_z - \partial_z E_y \\
-\imath \omega \mu_{yy} H_y &= \partial_z E_x - \partial_x E_z \\
-\imath \omega \mu_{zz} H_z &= \partial_x E_y - \partial_y E_x \\
\imath \omega \epsilon_{xx} E_x &= \partial_y H_z - \partial_z H_y \\
\imath \omega \epsilon_{yy} E_y &= \partial_z H_x - \partial_x H_z \\
\imath \omega \epsilon_{zz} E_z &= \partial_x H_y - \partial_y H_x \\
\end{aligned}
$$
Substituting in our expressions for $\\vec{E}$, $\\vec{H}$ and discretizing:
Substituting in our expressions for $\vec{E}$, $\vec{H}$ and discretizing:
$$
\\begin{aligned}
-\\imath \\omega \\mu_{xx} H_x &= \\tilde{\\partial}_y E_z + \\gamma E_y \\\\
-\\imath \\omega \\mu_{yy} H_y &= -\\gamma E_x - \\tilde{\\partial}_x E_z \\\\
-\\imath \\omega \\mu_{zz} H_z &= \\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x \\\\
\\imath \\omega \\epsilon_{xx} E_x &= \\hat{\\partial}_y H_z + \\gamma H_y \\\\
\\imath \\omega \\epsilon_{yy} E_y &= -\\gamma H_x - \\hat{\\partial}_x H_z \\\\
\\imath \\omega \\epsilon_{zz} E_z &= \\hat{\\partial}_x H_y - \\hat{\\partial}_y H_x \\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{xx} H_x &= \tilde{\partial}_y E_z + \gamma E_y \\
-\imath \omega \mu_{yy} H_y &= -\gamma E_x - \tilde{\partial}_x E_z \\
-\imath \omega \mu_{zz} H_z &= \tilde{\partial}_x E_y - \tilde{\partial}_y E_x \\
\imath \omega \epsilon_{xx} E_x &= \hat{\partial}_y H_z + \gamma H_y \\
\imath \omega \epsilon_{yy} E_y &= -\gamma H_x - \hat{\partial}_x H_z \\
\imath \omega \epsilon_{zz} E_z &= \hat{\partial}_x H_y - \hat{\partial}_y H_x \\
\end{aligned}
$$
Rewrite the last three equations as
$$
\\begin{aligned}
\\gamma H_y &= \\imath \\omega \\epsilon_{xx} E_x - \\hat{\\partial}_y H_z \\\\
\\gamma H_x &= -\\imath \\omega \\epsilon_{yy} E_y - \\hat{\\partial}_x H_z \\\\
\\imath \\omega E_z &= \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x H_y - \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y H_x \\\\
\\end{aligned}
\begin{aligned}
\gamma H_y &= \imath \omega \epsilon_{xx} E_x - \hat{\partial}_y H_z \\
\gamma H_x &= -\imath \omega \epsilon_{yy} E_y - \hat{\partial}_x H_z \\
\imath \omega E_z &= \frac{1}{\epsilon_{zz}} \hat{\partial}_x H_y - \frac{1}{\epsilon_{zz}} \hat{\partial}_y H_x \\
\end{aligned}
$$
Now apply $\\gamma \\tilde{\\partial}_x$ to the last equation,
then substitute in for $\\gamma H_x$ and $\\gamma H_y$:
Now apply $\gamma \tilde{\partial}_x$ to the last equation,
then substitute in for $\gamma H_x$ and $\gamma H_y$:
$$
\\begin{aligned}
\\gamma \\tilde{\\partial}_x \\imath \\omega E_z &= \\gamma \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x H_y
- \\gamma \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y H_x \\\\
&= \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x ( \\imath \\omega \\epsilon_{xx} E_x - \\hat{\\partial}_y H_z)
- \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (-\\imath \\omega \\epsilon_{yy} E_y - \\hat{\\partial}_x H_z) \\\\
&= \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x ( \\imath \\omega \\epsilon_{xx} E_x)
- \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (-\\imath \\omega \\epsilon_{yy} E_y) \\\\
\\gamma \\tilde{\\partial}_x E_z &= \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\tilde{\\partial}_x \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y) \\\\
\\end{aligned}
\begin{aligned}
\gamma \tilde{\partial}_x \imath \omega E_z &= \gamma \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_x H_y
- \gamma \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_y H_x \\
&= \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_x ( \imath \omega \epsilon_{xx} E_x - \hat{\partial}_y H_z)
- \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_y (-\imath \omega \epsilon_{yy} E_y - \hat{\partial}_x H_z) \\
&= \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_x ( \imath \omega \epsilon_{xx} E_x)
- \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_y (-\imath \omega \epsilon_{yy} E_y) \\
\gamma \tilde{\partial}_x E_z &= \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \tilde{\partial}_x \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y) \\
\end{aligned}
$$
With a similar approach (but using $\\gamma \\tilde{\\partial}_y$ instead), we can get
With a similar approach (but using $\gamma \tilde{\partial}_y$ instead), we can get
$$
\\begin{aligned}
\\gamma \\tilde{\\partial}_y E_z &= \\tilde{\\partial}_y \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\tilde{\\partial}_y \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y) \\\\
\\end{aligned}
\begin{aligned}
\gamma \tilde{\partial}_y E_z &= \tilde{\partial}_y \frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \tilde{\partial}_y \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y) \\
\end{aligned}
$$
We can combine this equation for $\\gamma \\tilde{\\partial}_y E_z$ with
the unused $\\imath \\omega \\mu_{xx} H_x$ and $\\imath \\omega \\mu_{yy} H_y$ equations to get
We can combine this equation for $\gamma \tilde{\partial}_y E_z$ with
the unused $\imath \omega \mu_{xx} H_x$ and $\imath \omega \mu_{yy} H_y$ equations to get
$$
\\begin{aligned}
-\\imath \\omega \\mu_{xx} \\gamma H_x &= \\gamma^2 E_y + \\gamma \\tilde{\\partial}_y E_z \\\\
-\\imath \\omega \\mu_{xx} \\gamma H_x &= \\gamma^2 E_y + \\tilde{\\partial}_y (
\\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y)
)\\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{xx} \gamma H_x &= \gamma^2 E_y + \gamma \tilde{\partial}_y E_z \\
-\imath \omega \mu_{xx} \gamma H_x &= \gamma^2 E_y + \tilde{\partial}_y (
\frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y)
)\\
\end{aligned}
$$
and
$$
\\begin{aligned}
-\\imath \\omega \\mu_{yy} \\gamma H_y &= -\\gamma^2 E_x - \\gamma \\tilde{\\partial}_x E_z \\\\
-\\imath \\omega \\mu_{yy} \\gamma H_y &= -\\gamma^2 E_x - \\tilde{\\partial}_x (
\\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y)
)\\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{yy} \gamma H_y &= -\gamma^2 E_x - \gamma \tilde{\partial}_x E_z \\
-\imath \omega \mu_{yy} \gamma H_y &= -\gamma^2 E_x - \tilde{\partial}_x (
\frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y)
)\\
\end{aligned}
$$
However, based on our rewritten equation for $\\gamma H_x$ and the so-far unused
equation for $\\imath \\omega \\mu_{zz} H_z$ we can also write
However, based on our rewritten equation for $\gamma H_x$ and the so-far unused
equation for $\imath \omega \mu_{zz} H_z$ we can also write
$$
\\begin{aligned}
-\\imath \\omega \\mu_{xx} (\\gamma H_x) &= -\\imath \\omega \\mu_{xx} (-\\imath \\omega \\epsilon_{yy} E_y - \\hat{\\partial}_x H_z) \\\\
&= -\\omega^2 \\mu_{xx} \\epsilon_{yy} E_y
+\\imath \\omega \\mu_{xx} \\hat{\\partial}_x (
\\frac{1}{-\\imath \\omega \\mu_{zz}} (\\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x)) \\\\
&= -\\omega^2 \\mu_{xx} \\epsilon_{yy} E_y
-\\mu_{xx} \\hat{\\partial}_x \\frac{1}{\\mu_{zz}} (\\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x) \\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{xx} (\gamma H_x) &= -\imath \omega \mu_{xx} (-\imath \omega \epsilon_{yy} E_y - \hat{\partial}_x H_z) \\
&= -\omega^2 \mu_{xx} \epsilon_{yy} E_y
+\imath \omega \mu_{xx} \hat{\partial}_x (
\frac{1}{-\imath \omega \mu_{zz}} (\tilde{\partial}_x E_y - \tilde{\partial}_y E_x)) \\
&= -\omega^2 \mu_{xx} \epsilon_{yy} E_y
-\mu_{xx} \hat{\partial}_x \frac{1}{\mu_{zz}} (\tilde{\partial}_x E_y - \tilde{\partial}_y E_x) \\
\end{aligned}
$$
and, similarly,
$$
\\begin{aligned}
-\\imath \\omega \\mu_{yy} (\\gamma H_y) &= \\omega^2 \\mu_{yy} \\epsilon_{xx} E_x
+\\mu_{yy} \\hat{\\partial}_y \\frac{1}{\\mu_{zz}} (\\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x) \\\\
\\end{aligned}
\begin{aligned}
-\imath \omega \mu_{yy} (\gamma H_y) &= \omega^2 \mu_{yy} \epsilon_{xx} E_x
+\mu_{yy} \hat{\partial}_y \frac{1}{\mu_{zz}} (\tilde{\partial}_x E_y - \tilde{\partial}_y E_x) \\
\end{aligned}
$$
By combining both pairs of expressions, we get
$$
\\begin{aligned}
-\\gamma^2 E_x - \\tilde{\\partial}_x (
\\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y)
) &= \\omega^2 \\mu_{yy} \\epsilon_{xx} E_x
+\\mu_{yy} \\hat{\\partial}_y \\frac{1}{\\mu_{zz}} (\\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x) \\\\
\\gamma^2 E_y + \\tilde{\\partial}_y (
\\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_x (\\epsilon_{xx} E_x)
+ \\frac{1}{\\epsilon_{zz}} \\hat{\\partial}_y (\\epsilon_{yy} E_y)
) &= -\\omega^2 \\mu_{xx} \\epsilon_{yy} E_y
-\\mu_{xx} \\hat{\\partial}_x \\frac{1}{\\mu_{zz}} (\\tilde{\\partial}_x E_y - \\tilde{\\partial}_y E_x) \\\\
\\end{aligned}
\begin{aligned}
-\gamma^2 E_x - \tilde{\partial}_x (
\frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y)
) &= \omega^2 \mu_{yy} \epsilon_{xx} E_x
+\mu_{yy} \hat{\partial}_y \frac{1}{\mu_{zz}} (\tilde{\partial}_x E_y - \tilde{\partial}_y E_x) \\
\gamma^2 E_y + \tilde{\partial}_y (
\frac{1}{\epsilon_{zz}} \hat{\partial}_x (\epsilon_{xx} E_x)
+ \frac{1}{\epsilon_{zz}} \hat{\partial}_y (\epsilon_{yy} E_y)
) &= -\omega^2 \mu_{xx} \epsilon_{yy} E_y
-\mu_{xx} \hat{\partial}_x \frac{1}{\mu_{zz}} (\tilde{\partial}_x E_y - \tilde{\partial}_y E_x) \\
\end{aligned}
$$
Using these, we can construct the eigenvalue problem
$$
\\beta^2 \\begin{bmatrix} E_x \\\\
E_y \\end{bmatrix} =
(\\omega^2 \\begin{bmatrix} \\mu_{yy} \\epsilon_{xx} & 0 \\\\
0 & \\mu_{xx} \\epsilon_{yy} \\end{bmatrix} +
\\begin{bmatrix} -\\mu_{yy} \\hat{\\partial}_y \\\\
\\mu_{xx} \\hat{\\partial}_x \\end{bmatrix} \\mu_{zz}^{-1}
\\begin{bmatrix} -\\tilde{\\partial}_y & \\tilde{\\partial}_x \\end{bmatrix} +
\\begin{bmatrix} \\tilde{\\partial}_x \\\\
\\tilde{\\partial}_y \\end{bmatrix} \\epsilon_{zz}^{-1}
\\begin{bmatrix} \\hat{\\partial}_x \\epsilon_{xx} & \\hat{\\partial}_y \\epsilon_{yy} \\end{bmatrix})
\\begin{bmatrix} E_x \\\\
E_y \\end{bmatrix}
\beta^2 \begin{bmatrix} E_x \\
E_y \end{bmatrix} =
(\omega^2 \begin{bmatrix} \mu_{yy} \epsilon_{xx} & 0 \\
0 & \mu_{xx} \epsilon_{yy} \end{bmatrix} +
\begin{bmatrix} -\mu_{yy} \hat{\partial}_y \\
\mu_{xx} \hat{\partial}_x \end{bmatrix} \mu_{zz}^{-1}
\begin{bmatrix} -\tilde{\partial}_y & \tilde{\partial}_x \end{bmatrix} +
\begin{bmatrix} \tilde{\partial}_x \\
\tilde{\partial}_y \end{bmatrix} \epsilon_{zz}^{-1}
\begin{bmatrix} \hat{\partial}_x \epsilon_{xx} & \hat{\partial}_y \epsilon_{yy} \end{bmatrix})
\begin{bmatrix} E_x \\
E_y \end{bmatrix}
$$
where $\\gamma = \\imath\\beta$. In the literature, $\\beta$ is usually used to denote
where $\gamma = \imath\beta$. In the literature, $\beta$ is usually used to denote
the lossless/real part of the propagation constant, but in `meanas` it is allowed to
be complex.
An equivalent eigenvalue problem can be formed using the $H_x$ and $H_y$ fields, if those are more convenient.
Note that $E_z$ was never discretized, so $\\gamma$ and $\\beta$ will need adjustment
Note that $E_z$ was never discretized, so $\gamma$ and $\beta$ will need adjustment
to account for numerical dispersion if the result is introduced into a space with a discretized z-axis.
@ -198,7 +198,7 @@ def operator_e(
epsilon: vfdfield_t,
mu: vfdfield_t | None = None,
) -> sparse.spmatrix:
"""
r"""
Waveguide operator of the form
omega**2 * mu * epsilon +
@ -210,18 +210,18 @@ def operator_e(
More precisely, the operator is
$$
\\omega^2 \\begin{bmatrix} \\mu_{yy} \\epsilon_{xx} & 0 \\\\
0 & \\mu_{xx} \\epsilon_{yy} \\end{bmatrix} +
\\begin{bmatrix} -\\mu_{yy} \\hat{\\partial}_y \\\\
\\mu_{xx} \\hat{\\partial}_x \\end{bmatrix} \\mu_{zz}^{-1}
\\begin{bmatrix} -\\tilde{\\partial}_y & \\tilde{\\partial}_x \\end{bmatrix} +
\\begin{bmatrix} \\tilde{\\partial}_x \\\\
\\tilde{\\partial}_y \\end{bmatrix} \\epsilon_{zz}^{-1}
\\begin{bmatrix} \\hat{\\partial}_x \\epsilon_{xx} & \\hat{\\partial}_y \\epsilon_{yy} \\end{bmatrix}
\omega^2 \begin{bmatrix} \mu_{yy} \epsilon_{xx} & 0 \\
0 & \mu_{xx} \epsilon_{yy} \end{bmatrix} +
\begin{bmatrix} -\mu_{yy} \hat{\partial}_y \\
\mu_{xx} \hat{\partial}_x \end{bmatrix} \mu_{zz}^{-1}
\begin{bmatrix} -\tilde{\partial}_y & \tilde{\partial}_x \end{bmatrix} +
\begin{bmatrix} \tilde{\partial}_x \\
\tilde{\partial}_y \end{bmatrix} \epsilon_{zz}^{-1}
\begin{bmatrix} \hat{\partial}_x \epsilon_{xx} & \hat{\partial}_y \epsilon_{yy} \end{bmatrix}
$$
$\\tilde{\\partial}_x$ and $\\hat{\\partial}_x$ are the forward and backward derivatives along x,
and each $\\epsilon_{xx}$, $\\mu_{yy}$, etc. is a diagonal matrix containing the vectorized material
$\tilde{\partial}_x$ and $\hat{\partial}_x$ are the forward and backward derivatives along x,
and each $\epsilon_{xx}$, $\mu_{yy}$, etc. is a diagonal matrix containing the vectorized material
property distribution.
This operator can be used to form an eigenvalue problem of the form
@ -253,9 +253,10 @@ def operator_e(
mu_yx = sparse.diags(numpy.hstack((mu_parts[1], mu_parts[0])))
mu_z_inv = sparse.diags(1 / mu_parts[2])
op = omega * omega * mu_yx @ eps_xy + \
mu_yx @ sparse.vstack((-Dby, Dbx)) @ mu_z_inv @ sparse.hstack((-Dfy, Dfx)) + \
sparse.vstack((Dfx, Dfy)) @ eps_z_inv @ sparse.hstack((Dbx, Dby)) @ eps_xy
op = (omega * omega * mu_yx @ eps_xy
+ mu_yx @ sparse.vstack((-Dby, Dbx)) @ mu_z_inv @ sparse.hstack((-Dfy, Dfx))
+ sparse.vstack((Dfx, Dfy)) @ eps_z_inv @ sparse.hstack((Dbx, Dby)) @ eps_xy
)
return op
@ -265,7 +266,7 @@ def operator_h(
epsilon: vfdfield_t,
mu: vfdfield_t | None = None,
) -> sparse.spmatrix:
"""
r"""
Waveguide operator of the form
omega**2 * epsilon * mu +
@ -277,18 +278,18 @@ def operator_h(
More precisely, the operator is
$$
\\omega^2 \\begin{bmatrix} \\epsilon_{yy} \\mu_{xx} & 0 \\\\
0 & \\epsilon_{xx} \\mu_{yy} \\end{bmatrix} +
\\begin{bmatrix} -\\epsilon_{yy} \\tilde{\\partial}_y \\\\
\\epsilon_{xx} \\tilde{\\partial}_x \\end{bmatrix} \\epsilon_{zz}^{-1}
\\begin{bmatrix} -\\hat{\\partial}_y & \\hat{\\partial}_x \\end{bmatrix} +
\\begin{bmatrix} \\hat{\\partial}_x \\\\
\\hat{\\partial}_y \\end{bmatrix} \\mu_{zz}^{-1}
\\begin{bmatrix} \\tilde{\\partial}_x \\mu_{xx} & \\tilde{\\partial}_y \\mu_{yy} \\end{bmatrix}
\omega^2 \begin{bmatrix} \epsilon_{yy} \mu_{xx} & 0 \\
0 & \epsilon_{xx} \mu_{yy} \end{bmatrix} +
\begin{bmatrix} -\epsilon_{yy} \tilde{\partial}_y \\
\epsilon_{xx} \tilde{\partial}_x \end{bmatrix} \epsilon_{zz}^{-1}
\begin{bmatrix} -\hat{\partial}_y & \hat{\partial}_x \end{bmatrix} +
\begin{bmatrix} \hat{\partial}_x \\
\hat{\partial}_y \end{bmatrix} \mu_{zz}^{-1}
\begin{bmatrix} \tilde{\partial}_x \mu_{xx} & \tilde{\partial}_y \mu_{yy} \end{bmatrix}
$$
$\\tilde{\\partial}_x$ and $\\hat{\\partial}_x$ are the forward and backward derivatives along x,
and each $\\epsilon_{xx}$, $\\mu_{yy}$, etc. is a diagonal matrix containing the vectorized material
$\tilde{\partial}_x$ and $\hat{\partial}_x$ are the forward and backward derivatives along x,
and each $\epsilon_{xx}$, $\mu_{yy}$, etc. is a diagonal matrix containing the vectorized material
property distribution.
This operator can be used to form an eigenvalue problem of the form
@ -320,10 +321,10 @@ def operator_h(
mu_xy = sparse.diags(numpy.hstack((mu_parts[0], mu_parts[1])))
mu_z_inv = sparse.diags(1 / mu_parts[2])
op = omega * omega * eps_yx @ mu_xy + \
eps_yx @ sparse.vstack((-Dfy, Dfx)) @ eps_z_inv @ sparse.hstack((-Dby, Dbx)) + \
sparse.vstack((Dbx, Dby)) @ mu_z_inv @ sparse.hstack((Dfx, Dfy)) @ mu_xy
op = (omega * omega * eps_yx @ mu_xy
+ eps_yx @ sparse.vstack((-Dfy, Dfx)) @ eps_z_inv @ sparse.hstack((-Dby, Dbx))
+ sparse.vstack((Dbx, Dby)) @ mu_z_inv @ sparse.hstack((Dfx, Dfy)) @ mu_xy
)
return op

View File

@ -73,8 +73,10 @@ def cylindrical_operator(
omega2 = omega * omega
op = (omega2 * diag((Tx, Ty)) + pa) @ diag((a0, a1)) + \
op = (
(omega2 * diag((Tx, Ty)) + pa) @ diag((a0, a1))
- (sparse.bmat(((None, Ty), (Tx, None))) + pb / omega2) @ diag((b0, b1))
)
return op

View File

@ -1,4 +1,4 @@
"""
r"""
Basic discrete calculus for finite difference (fd) simulations.
@ -43,11 +43,11 @@ Scalar derivatives and cell shifts
----------------------------------
Define the discrete forward derivative as
$$ [\\tilde{\\partial}_x f]_{m + \\frac{1}{2}} = \\frac{1}{\\Delta_{x, m}} (f_{m + 1} - f_m) $$
$$ [\tilde{\partial}_x f]_{m + \frac{1}{2}} = \frac{1}{\Delta_{x, m}} (f_{m + 1} - f_m) $$
where $f$ is a function defined at discrete locations on the x-axis (labeled using $m$).
The value at $m$ occupies a length $\\Delta_{x, m}$ along the x-axis. Note that $m$
The value at $m$ occupies a length $\Delta_{x, m}$ along the x-axis. Note that $m$
is an index along the x-axis, _not_ necessarily an x-coordinate, since each length
$\\Delta_{x, m}, \\Delta_{x, m+1}, ...$ is independently chosen.
$\Delta_{x, m}, \Delta_{x, m+1}, ...$ is independently chosen.
If we treat `f` as a 1D array of values, with the `i`-th value `f[i]` taking up a length `dx[i]`
along the x-axis, the forward derivative is
@ -56,13 +56,13 @@ along the x-axis, the forward derivative is
Likewise, discrete reverse derivative is
$$ [\\hat{\\partial}_x f ]_{m - \\frac{1}{2}} = \\frac{1}{\\Delta_{x, m}} (f_{m} - f_{m - 1}) $$
$$ [\hat{\partial}_x f ]_{m - \frac{1}{2}} = \frac{1}{\Delta_{x, m}} (f_{m} - f_{m - 1}) $$
or
deriv_back(f)[i] = (f[i] - f[i - 1]) / dx[i]
The derivatives' values are shifted by a half-cell relative to the original function, and
will have different cell widths if all the `dx[i]` ( $\\Delta_{x, m}$ ) are not
will have different cell widths if all the `dx[i]` ( $\Delta_{x, m}$ ) are not
identical:
[figure: derivatives and cell sizes]
@ -88,19 +88,19 @@ identical:
In the above figure,
`f0 =` $f_0$, `f1 =` $f_1$
`Df0 =` $[\\tilde{\\partial}f]_{0 + \\frac{1}{2}}$
`Df1 =` $[\\tilde{\\partial}f]_{1 + \\frac{1}{2}}$
`df0 =` $[\\hat{\\partial}f]_{0 - \\frac{1}{2}}$
`Df0 =` $[\tilde{\partial}f]_{0 + \frac{1}{2}}$
`Df1 =` $[\tilde{\partial}f]_{1 + \frac{1}{2}}$
`df0 =` $[\hat{\partial}f]_{0 - \frac{1}{2}}$
etc.
The fractional subscript $m + \\frac{1}{2}$ is used to indicate values defined
The fractional subscript $m + \frac{1}{2}$ is used to indicate values defined
at shifted locations relative to the original $m$, with corresponding lengths
$$ \\Delta_{x, m + \\frac{1}{2}} = \\frac{1}{2} * (\\Delta_{x, m} + \\Delta_{x, m + 1}) $$
$$ \Delta_{x, m + \frac{1}{2}} = \frac{1}{2} * (\Delta_{x, m} + \Delta_{x, m + 1}) $$
Just as $m$ is not itself an x-coordinate, neither is $m + \\frac{1}{2}$;
Just as $m$ is not itself an x-coordinate, neither is $m + \frac{1}{2}$;
carefully note the positions of the various cells in the above figure vs their labels.
If the positions labeled with $m$ are considered the "base" or "original" grid,
the positions labeled with $m + \\frac{1}{2}$ are said to lie on a "dual" or
the positions labeled with $m + \frac{1}{2}$ are said to lie on a "dual" or
"derived" grid.
For the remainder of the `Discrete calculus` section, all figures will show
@ -113,12 +113,12 @@ Gradients and fore-vectors
--------------------------
Expanding to three dimensions, we can define two gradients
$$ [\\tilde{\\nabla} f]_{m,n,p} = \\vec{x} [\\tilde{\\partial}_x f]_{m + \\frac{1}{2},n,p} +
\\vec{y} [\\tilde{\\partial}_y f]_{m,n + \\frac{1}{2},p} +
\\vec{z} [\\tilde{\\partial}_z f]_{m,n,p + \\frac{1}{2}} $$
$$ [\\hat{\\nabla} f]_{m,n,p} = \\vec{x} [\\hat{\\partial}_x f]_{m + \\frac{1}{2},n,p} +
\\vec{y} [\\hat{\\partial}_y f]_{m,n + \\frac{1}{2},p} +
\\vec{z} [\\hat{\\partial}_z f]_{m,n,p + \\frac{1}{2}} $$
$$ [\tilde{\nabla} f]_{m,n,p} = \vec{x} [\tilde{\partial}_x f]_{m + \frac{1}{2},n,p} +
\vec{y} [\tilde{\partial}_y f]_{m,n + \frac{1}{2},p} +
\vec{z} [\tilde{\partial}_z f]_{m,n,p + \frac{1}{2}} $$
$$ [\hat{\nabla} f]_{m,n,p} = \vec{x} [\hat{\partial}_x f]_{m + \frac{1}{2},n,p} +
\vec{y} [\hat{\partial}_y f]_{m,n + \frac{1}{2},p} +
\vec{z} [\hat{\partial}_z f]_{m,n,p + \frac{1}{2}} $$
or
@ -144,12 +144,12 @@ y in y, and z in z.
We call the resulting object a "fore-vector" or "back-vector", depending
on the direction of the shift. We write it as
$$ \\tilde{g}_{m,n,p} = \\vec{x} g^x_{m + \\frac{1}{2},n,p} +
\\vec{y} g^y_{m,n + \\frac{1}{2},p} +
\\vec{z} g^z_{m,n,p + \\frac{1}{2}} $$
$$ \\hat{g}_{m,n,p} = \\vec{x} g^x_{m - \\frac{1}{2},n,p} +
\\vec{y} g^y_{m,n - \\frac{1}{2},p} +
\\vec{z} g^z_{m,n,p - \\frac{1}{2}} $$
$$ \tilde{g}_{m,n,p} = \vec{x} g^x_{m + \frac{1}{2},n,p} +
\vec{y} g^y_{m,n + \frac{1}{2},p} +
\vec{z} g^z_{m,n,p + \frac{1}{2}} $$
$$ \hat{g}_{m,n,p} = \vec{x} g^x_{m - \frac{1}{2},n,p} +
\vec{y} g^y_{m,n - \frac{1}{2},p} +
\vec{z} g^z_{m,n,p - \frac{1}{2}} $$
[figure: gradient / fore-vector]
@ -172,15 +172,15 @@ Divergences
There are also two divergences,
$$ d_{n,m,p} = [\\tilde{\\nabla} \\cdot \\hat{g}]_{n,m,p}
= [\\tilde{\\partial}_x g^x]_{m,n,p} +
[\\tilde{\\partial}_y g^y]_{m,n,p} +
[\\tilde{\\partial}_z g^z]_{m,n,p} $$
$$ d_{n,m,p} = [\tilde{\nabla} \cdot \hat{g}]_{n,m,p}
= [\tilde{\partial}_x g^x]_{m,n,p} +
[\tilde{\partial}_y g^y]_{m,n,p} +
[\tilde{\partial}_z g^z]_{m,n,p} $$
$$ d_{n,m,p} = [\\hat{\\nabla} \\cdot \\tilde{g}]_{n,m,p}
= [\\hat{\\partial}_x g^x]_{m,n,p} +
[\\hat{\\partial}_y g^y]_{m,n,p} +
[\\hat{\\partial}_z g^z]_{m,n,p} $$
$$ d_{n,m,p} = [\hat{\nabla} \cdot \tilde{g}]_{n,m,p}
= [\hat{\partial}_x g^x]_{m,n,p} +
[\hat{\partial}_y g^y]_{m,n,p} +
[\hat{\partial}_z g^z]_{m,n,p} $$
or
@ -203,7 +203,7 @@ where `g = [gx, gy, gz]` is a fore- or back-vector field.
Since we applied the forward divergence to the back-vector (and vice-versa), the resulting scalar value
is defined at the back-vector's (fore-vector's) location $(m,n,p)$ and not at the locations of its components
$(m \\pm \\frac{1}{2},n,p)$ etc.
$(m \pm \frac{1}{2},n,p)$ etc.
[figure: divergence]
^^
@ -227,23 +227,23 @@ Curls
The two curls are then
$$ \\begin{aligned}
\\hat{h}_{m + \\frac{1}{2}, n + \\frac{1}{2}, p + \\frac{1}{2}} &= \\\\
[\\tilde{\\nabla} \\times \\tilde{g}]_{m + \\frac{1}{2}, n + \\frac{1}{2}, p + \\frac{1}{2}} &=
\\vec{x} (\\tilde{\\partial}_y g^z_{m,n,p + \\frac{1}{2}} - \\tilde{\\partial}_z g^y_{m,n + \\frac{1}{2},p}) \\\\
&+ \\vec{y} (\\tilde{\\partial}_z g^x_{m + \\frac{1}{2},n,p} - \\tilde{\\partial}_x g^z_{m,n,p + \\frac{1}{2}}) \\\\
&+ \\vec{z} (\\tilde{\\partial}_x g^y_{m,n + \\frac{1}{2},p} - \\tilde{\\partial}_y g^z_{m + \\frac{1}{2},n,p})
\\end{aligned} $$
$$ \begin{aligned}
\hat{h}_{m + \frac{1}{2}, n + \frac{1}{2}, p + \frac{1}{2}} &= \\
[\tilde{\nabla} \times \tilde{g}]_{m + \frac{1}{2}, n + \frac{1}{2}, p + \frac{1}{2}} &=
\vec{x} (\tilde{\partial}_y g^z_{m,n,p + \frac{1}{2}} - \tilde{\partial}_z g^y_{m,n + \frac{1}{2},p}) \\
&+ \vec{y} (\tilde{\partial}_z g^x_{m + \frac{1}{2},n,p} - \tilde{\partial}_x g^z_{m,n,p + \frac{1}{2}}) \\
&+ \vec{z} (\tilde{\partial}_x g^y_{m,n + \frac{1}{2},p} - \tilde{\partial}_y g^z_{m + \frac{1}{2},n,p})
\end{aligned} $$
and
$$ \\tilde{h}_{m - \\frac{1}{2}, n - \\frac{1}{2}, p - \\frac{1}{2}} =
[\\hat{\\nabla} \\times \\hat{g}]_{m - \\frac{1}{2}, n - \\frac{1}{2}, p - \\frac{1}{2}} $$
$$ \tilde{h}_{m - \frac{1}{2}, n - \frac{1}{2}, p - \frac{1}{2}} =
[\hat{\nabla} \times \hat{g}]_{m - \frac{1}{2}, n - \frac{1}{2}, p - \frac{1}{2}} $$
where $\\hat{g}$ and $\\tilde{g}$ are located at $(m,n,p)$
with components at $(m \\pm \\frac{1}{2},n,p)$ etc.,
while $\\hat{h}$ and $\\tilde{h}$ are located at $(m \\pm \\frac{1}{2}, n \\pm \\frac{1}{2}, p \\pm \\frac{1}{2})$
with components at $(m, n \\pm \\frac{1}{2}, p \\pm \\frac{1}{2})$ etc.
where $\hat{g}$ and $\tilde{g}$ are located at $(m,n,p)$
with components at $(m \pm \frac{1}{2},n,p)$ etc.,
while $\hat{h}$ and $\tilde{h}$ are located at $(m \pm \frac{1}{2}, n \pm \frac{1}{2}, p \pm \frac{1}{2})$
with components at $(m, n \pm \frac{1}{2}, p \pm \frac{1}{2})$ etc.
[code: curls]
@ -287,27 +287,27 @@ Maxwell's Equations
If we discretize both space (m,n,p) and time (l), Maxwell's equations become
$$ \\begin{aligned}
\\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}} &= -\\tilde{\\partial}_t \\hat{B}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}}
- \\hat{M}_{l, \\vec{r} + \\frac{1}{2}} \\\\
\\hat{\\nabla} \\times \\hat{H}_{l-\\frac{1}{2},\\vec{r} + \\frac{1}{2}} &= \\hat{\\partial}_t \\tilde{D}_{l, \\vec{r}}
+ \\tilde{J}_{l-\\frac{1}{2},\\vec{r}} \\\\
\\tilde{\\nabla} \\cdot \\hat{B}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}} &= 0 \\\\
\\hat{\\nabla} \\cdot \\tilde{D}_{l,\\vec{r}} &= \\rho_{l,\\vec{r}}
\\end{aligned} $$
$$ \begin{aligned}
\tilde{\nabla} \times \tilde{E}_{l,\vec{r}} &= -\tilde{\partial}_t \hat{B}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}}
- \hat{M}_{l, \vec{r} + \frac{1}{2}} \\
\hat{\nabla} \times \hat{H}_{l-\frac{1}{2},\vec{r} + \frac{1}{2}} &= \hat{\partial}_t \tilde{D}_{l, \vec{r}}
+ \tilde{J}_{l-\frac{1}{2},\vec{r}} \\
\tilde{\nabla} \cdot \hat{B}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}} &= 0 \\
\hat{\nabla} \cdot \tilde{D}_{l,\vec{r}} &= \rho_{l,\vec{r}}
\end{aligned} $$
with
$$ \\begin{aligned}
\\hat{B}_{\\vec{r}} &= \\mu_{\\vec{r} + \\frac{1}{2}} \\cdot \\hat{H}_{\\vec{r} + \\frac{1}{2}} \\\\
\\tilde{D}_{\\vec{r}} &= \\epsilon_{\\vec{r}} \\cdot \\tilde{E}_{\\vec{r}}
\\end{aligned} $$
$$ \begin{aligned}
\hat{B}_{\vec{r}} &= \mu_{\vec{r} + \frac{1}{2}} \cdot \hat{H}_{\vec{r} + \frac{1}{2}} \\
\tilde{D}_{\vec{r}} &= \epsilon_{\vec{r}} \cdot \tilde{E}_{\vec{r}}
\end{aligned} $$
where the spatial subscripts are abbreviated as $\\vec{r} = (m, n, p)$ and
$\\vec{r} + \\frac{1}{2} = (m + \\frac{1}{2}, n + \\frac{1}{2}, p + \\frac{1}{2})$,
$\\tilde{E}$ and $\\hat{H}$ are the electric and magnetic fields,
$\\tilde{J}$ and $\\hat{M}$ are the electric and magnetic current distributions,
and $\\epsilon$ and $\\mu$ are the dielectric permittivity and magnetic permeability.
where the spatial subscripts are abbreviated as $\vec{r} = (m, n, p)$ and
$\vec{r} + \frac{1}{2} = (m + \frac{1}{2}, n + \frac{1}{2}, p + \frac{1}{2})$,
$\tilde{E}$ and $\hat{H}$ are the electric and magnetic fields,
$\tilde{J}$ and $\hat{M}$ are the electric and magnetic current distributions,
and $\epsilon$ and $\mu$ are the dielectric permittivity and magnetic permeability.
The above is Yee's algorithm, written in a form analogous to Maxwell's equations.
The time derivatives can be expanded to form the update equations:
@ -369,34 +369,34 @@ Each component forms its own grid, offset from the others:
The divergence equations can be derived by taking the divergence of the curl equations
and combining them with charge continuity,
$$ \\hat{\\nabla} \\cdot \\tilde{J} + \\hat{\\partial}_t \\rho = 0 $$
$$ \hat{\nabla} \cdot \tilde{J} + \hat{\partial}_t \rho = 0 $$
implying that the discrete Maxwell's equations do not produce spurious charges.
Wave equation
-------------
Taking the backward curl of the $\\tilde{\\nabla} \\times \\tilde{E}$ equation and
replacing the resulting $\\hat{\\nabla} \\times \\hat{H}$ term using its respective equation,
and setting $\\hat{M}$ to zero, we can form the discrete wave equation:
Taking the backward curl of the $\tilde{\nabla} \times \tilde{E}$ equation and
replacing the resulting $\hat{\nabla} \times \hat{H}$ term using its respective equation,
and setting $\hat{M}$ to zero, we can form the discrete wave equation:
$$
\\begin{aligned}
\\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}} &=
-\\tilde{\\partial}_t \\hat{B}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}}
- \\hat{M}_{l-1, \\vec{r} + \\frac{1}{2}} \\\\
\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}} &=
-\\tilde{\\partial}_t \\hat{H}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}} \\\\
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}}) &=
\\hat{\\nabla} \\times (-\\tilde{\\partial}_t \\hat{H}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}}) \\\\
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}}) &=
-\\tilde{\\partial}_t \\hat{\\nabla} \\times \\hat{H}_{l-\\frac{1}{2}, \\vec{r} + \\frac{1}{2}} \\\\
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}}) &=
-\\tilde{\\partial}_t \\hat{\\partial}_t \\epsilon_{\\vec{r}} \\tilde{E}_{l, \\vec{r}} + \\hat{\\partial}_t \\tilde{J}_{l-\\frac{1}{2},\\vec{r}} \\\\
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l,\\vec{r}})
+ \\tilde{\\partial}_t \\hat{\\partial}_t \\epsilon_{\\vec{r}} \\cdot \\tilde{E}_{l, \\vec{r}}
&= \\tilde{\\partial}_t \\tilde{J}_{l - \\frac{1}{2}, \\vec{r}}
\\end{aligned}
\begin{aligned}
\tilde{\nabla} \times \tilde{E}_{l,\vec{r}} &=
-\tilde{\partial}_t \hat{B}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}}
- \hat{M}_{l-1, \vec{r} + \frac{1}{2}} \\
\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l,\vec{r}} &=
-\tilde{\partial}_t \hat{H}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}} \\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l,\vec{r}}) &=
\hat{\nabla} \times (-\tilde{\partial}_t \hat{H}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}}) \\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l,\vec{r}}) &=
-\tilde{\partial}_t \hat{\nabla} \times \hat{H}_{l-\frac{1}{2}, \vec{r} + \frac{1}{2}} \\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l,\vec{r}}) &=
-\tilde{\partial}_t \hat{\partial}_t \epsilon_{\vec{r}} \tilde{E}_{l, \vec{r}} + \hat{\partial}_t \tilde{J}_{l-\frac{1}{2},\vec{r}} \\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l,\vec{r}})
+ \tilde{\partial}_t \hat{\partial}_t \epsilon_{\vec{r}} \cdot \tilde{E}_{l, \vec{r}}
&= \tilde{\partial}_t \tilde{J}_{l - \frac{1}{2}, \vec{r}}
\end{aligned}
$$
@ -406,27 +406,27 @@ Frequency domain
We can substitute in a time-harmonic fields
$$
\\begin{aligned}
\\tilde{E}_{l, \\vec{r}} &= \\tilde{E}_{\\vec{r}} e^{-\\imath \\omega l \\Delta_t} \\\\
\\tilde{J}_{l, \\vec{r}} &= \\tilde{J}_{\\vec{r}} e^{-\\imath \\omega (l - \\frac{1}{2}) \\Delta_t}
\\end{aligned}
\begin{aligned}
\tilde{E}_{l, \vec{r}} &= \tilde{E}_{\vec{r}} e^{-\imath \omega l \Delta_t} \\
\tilde{J}_{l, \vec{r}} &= \tilde{J}_{\vec{r}} e^{-\imath \omega (l - \frac{1}{2}) \Delta_t}
\end{aligned}
$$
resulting in
$$
\\begin{aligned}
\\tilde{\\partial}_t &\\Rightarrow (e^{ \\imath \\omega \\Delta_t} - 1) / \\Delta_t = \\frac{-2 \\imath}{\\Delta_t} \\sin(\\omega \\Delta_t / 2) e^{-\\imath \\omega \\Delta_t / 2} = -\\imath \\Omega e^{-\\imath \\omega \\Delta_t / 2}\\\\
\\hat{\\partial}_t &\\Rightarrow (1 - e^{-\\imath \\omega \\Delta_t}) / \\Delta_t = \\frac{-2 \\imath}{\\Delta_t} \\sin(\\omega \\Delta_t / 2) e^{ \\imath \\omega \\Delta_t / 2} = -\\imath \\Omega e^{ \\imath \\omega \\Delta_t / 2}\\\\
\\Omega &= 2 \\sin(\\omega \\Delta_t / 2) / \\Delta_t
\\end{aligned}
\begin{aligned}
\tilde{\partial}_t &\Rightarrow (e^{ \imath \omega \Delta_t} - 1) / \Delta_t = \frac{-2 \imath}{\Delta_t} \sin(\omega \Delta_t / 2) e^{-\imath \omega \Delta_t / 2} = -\imath \Omega e^{-\imath \omega \Delta_t / 2}\\
\hat{\partial}_t &\Rightarrow (1 - e^{-\imath \omega \Delta_t}) / \Delta_t = \frac{-2 \imath}{\Delta_t} \sin(\omega \Delta_t / 2) e^{ \imath \omega \Delta_t / 2} = -\imath \Omega e^{ \imath \omega \Delta_t / 2}\\
\Omega &= 2 \sin(\omega \Delta_t / 2) / \Delta_t
\end{aligned}
$$
This gives the frequency-domain wave equation,
$$
\\hat{\\nabla} \\times (\\mu^{-1}_{\\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}})
-\\Omega^2 \\epsilon_{\\vec{r}} \\cdot \\tilde{E}_{\\vec{r}} = -\\imath \\Omega \\tilde{J}_{\\vec{r}} e^{\\imath \\omega \\Delta_t / 2} \\\\
\hat{\nabla} \times (\mu^{-1}_{\vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{\vec{r}})
-\Omega^2 \epsilon_{\vec{r}} \cdot \tilde{E}_{\vec{r}} = -\imath \Omega \tilde{J}_{\vec{r}} e^{\imath \omega \Delta_t / 2} \\
$$
@ -436,48 +436,48 @@ Plane waves and Dispersion relation
With uniform material distribution and no sources
$$
\\begin{aligned}
\\mu_{\\vec{r} + \\frac{1}{2}} &= \\mu \\\\
\\epsilon_{\\vec{r}} &= \\epsilon \\\\
\\tilde{J}_{\\vec{r}} &= 0 \\\\
\\end{aligned}
\begin{aligned}
\mu_{\vec{r} + \frac{1}{2}} &= \mu \\
\epsilon_{\vec{r}} &= \epsilon \\
\tilde{J}_{\vec{r}} &= 0 \\
\end{aligned}
$$
the frequency domain wave equation simplifies to
$$ \\hat{\\nabla} \\times \\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}} - \\Omega^2 \\epsilon \\mu \\tilde{E}_{\\vec{r}} = 0 $$
$$ \hat{\nabla} \times \tilde{\nabla} \times \tilde{E}_{\vec{r}} - \Omega^2 \epsilon \mu \tilde{E}_{\vec{r}} = 0 $$
Since $\\hat{\\nabla} \\cdot \\tilde{E}_{\\vec{r}} = 0$, we can simplify
Since $\hat{\nabla} \cdot \tilde{E}_{\vec{r}} = 0$, we can simplify
$$
\\begin{aligned}
\\hat{\\nabla} \\times \\tilde{\\nabla} \\times \\tilde{E}_{\\vec{r}}
&= \\tilde{\\nabla}(\\hat{\\nabla} \\cdot \\tilde{E}_{\\vec{r}}) - \\hat{\\nabla} \\cdot \\tilde{\\nabla} \\tilde{E}_{\\vec{r}} \\\\
&= - \\hat{\\nabla} \\cdot \\tilde{\\nabla} \\tilde{E}_{\\vec{r}} \\\\
&= - \\tilde{\\nabla}^2 \\tilde{E}_{\\vec{r}}
\\end{aligned}
\begin{aligned}
\hat{\nabla} \times \tilde{\nabla} \times \tilde{E}_{\vec{r}}
&= \tilde{\nabla}(\hat{\nabla} \cdot \tilde{E}_{\vec{r}}) - \hat{\nabla} \cdot \tilde{\nabla} \tilde{E}_{\vec{r}} \\
&= - \hat{\nabla} \cdot \tilde{\nabla} \tilde{E}_{\vec{r}} \\
&= - \tilde{\nabla}^2 \tilde{E}_{\vec{r}}
\end{aligned}
$$
and we get
$$ \\tilde{\\nabla}^2 \\tilde{E}_{\\vec{r}} + \\Omega^2 \\epsilon \\mu \\tilde{E}_{\\vec{r}} = 0 $$
$$ \tilde{\nabla}^2 \tilde{E}_{\vec{r}} + \Omega^2 \epsilon \mu \tilde{E}_{\vec{r}} = 0 $$
We can convert this to three scalar-wave equations of the form
$$ (\\tilde{\\nabla}^2 + K^2) \\phi_{\\vec{r}} = 0 $$
$$ (\tilde{\nabla}^2 + K^2) \phi_{\vec{r}} = 0 $$
with $K^2 = \\Omega^2 \\mu \\epsilon$. Now we let
with $K^2 = \Omega^2 \mu \epsilon$. Now we let
$$ \\phi_{\\vec{r}} = A e^{\\imath (k_x m \\Delta_x + k_y n \\Delta_y + k_z p \\Delta_z)} $$
$$ \phi_{\vec{r}} = A e^{\imath (k_x m \Delta_x + k_y n \Delta_y + k_z p \Delta_z)} $$
resulting in
$$
\\begin{aligned}
\\tilde{\\partial}_x &\\Rightarrow (e^{ \\imath k_x \\Delta_x} - 1) / \\Delta_t = \\frac{-2 \\imath}{\\Delta_x} \\sin(k_x \\Delta_x / 2) e^{ \\imath k_x \\Delta_x / 2} = \\imath K_x e^{ \\imath k_x \\Delta_x / 2}\\\\
\\hat{\\partial}_x &\\Rightarrow (1 - e^{-\\imath k_x \\Delta_x}) / \\Delta_t = \\frac{-2 \\imath}{\\Delta_x} \\sin(k_x \\Delta_x / 2) e^{-\\imath k_x \\Delta_x / 2} = \\imath K_x e^{-\\imath k_x \\Delta_x / 2}\\\\
K_x &= 2 \\sin(k_x \\Delta_x / 2) / \\Delta_x \\\\
\\end{aligned}
\begin{aligned}
\tilde{\partial}_x &\Rightarrow (e^{ \imath k_x \Delta_x} - 1) / \Delta_t = \frac{-2 \imath}{\Delta_x} \sin(k_x \Delta_x / 2) e^{ \imath k_x \Delta_x / 2} = \imath K_x e^{ \imath k_x \Delta_x / 2}\\
\hat{\partial}_x &\Rightarrow (1 - e^{-\imath k_x \Delta_x}) / \Delta_t = \frac{-2 \imath}{\Delta_x} \sin(k_x \Delta_x / 2) e^{-\imath k_x \Delta_x / 2} = \imath K_x e^{-\imath k_x \Delta_x / 2}\\
K_x &= 2 \sin(k_x \Delta_x / 2) / \Delta_x \\
\end{aligned}
$$
with similar expressions for the y and z dimnsions (and $K_y, K_z$).
@ -485,20 +485,20 @@ with similar expressions for the y and z dimnsions (and $K_y, K_z$).
This implies
$$
\\tilde{\\nabla}^2 = -(K_x^2 + K_y^2 + K_z^2) \\phi_{\\vec{r}} \\\\
K_x^2 + K_y^2 + K_z^2 = \\Omega^2 \\mu \\epsilon = \\Omega^2 / c^2
\tilde{\nabla}^2 = -(K_x^2 + K_y^2 + K_z^2) \phi_{\vec{r}} \\
K_x^2 + K_y^2 + K_z^2 = \Omega^2 \mu \epsilon = \Omega^2 / c^2
$$
where $c = \\sqrt{\\mu \\epsilon}$.
where $c = \sqrt{\mu \epsilon}$.
Assuming real $(k_x, k_y, k_z), \\omega$ will be real only if
Assuming real $(k_x, k_y, k_z), \omega$ will be real only if
$$ c^2 \\Delta_t^2 = \\frac{\\Delta_t^2}{\\mu \\epsilon} < 1/(\\frac{1}{\\Delta_x^2} + \\frac{1}{\\Delta_y^2} + \\frac{1}{\\Delta_z^2}) $$
$$ c^2 \Delta_t^2 = \frac{\Delta_t^2}{\mu \epsilon} < 1/(\frac{1}{\Delta_x^2} + \frac{1}{\Delta_y^2} + \frac{1}{\Delta_z^2}) $$
If $\\Delta_x = \\Delta_y = \\Delta_z$, this simplifies to $c \\Delta_t < \\Delta_x / \\sqrt{3}$.
If $\Delta_x = \Delta_y = \Delta_z$, this simplifies to $c \Delta_t < \Delta_x / \sqrt{3}$.
This last form can be interpreted as enforcing causality; the distance that light
travels in one timestep (i.e., $c \\Delta_t$) must be less than the diagonal
of the smallest cell ( $\\Delta_x / \\sqrt{3}$ when on a uniform cubic grid).
travels in one timestep (i.e., $c \Delta_t$) must be less than the diagonal
of the smallest cell ( $\Delta_x / \sqrt{3}$ when on a uniform cubic grid).
Grid description
@ -515,8 +515,8 @@ to make the illustration simpler; we need at least two cells in the x dimension
demonstrate how nonuniform `dx` affects the various components.
Place the E fore-vectors at integer indices $r = (m, n, p)$ and the H back-vectors
at fractional indices $r + \\frac{1}{2} = (m + \\frac{1}{2}, n + \\frac{1}{2},
p + \\frac{1}{2})$. Remember that these are indices and not coordinates; they can
at fractional indices $r + \frac{1}{2} = (m + \frac{1}{2}, n + \frac{1}{2},
p + \frac{1}{2})$. Remember that these are indices and not coordinates; they can
correspond to arbitrary (monotonically increasing) coordinates depending on the cell widths.
Draw lines to denote the planes on which the H components and back-vectors are defined.
@ -718,14 +718,14 @@ composed of the three diagonal tensor components:
or
$$
\\epsilon = \\begin{bmatrix} \\epsilon_{xx} & 0 & 0 \\\\
0 & \\epsilon_{yy} & 0 \\\\
0 & 0 & \\epsilon_{zz} \\end{bmatrix}
\epsilon = \begin{bmatrix} \epsilon_{xx} & 0 & 0 \\
0 & \epsilon_{yy} & 0 \\
0 & 0 & \epsilon_{zz} \end{bmatrix}
$$
$$
\\mu = \\begin{bmatrix} \\mu_{xx} & 0 & 0 \\\\
0 & \\mu_{yy} & 0 \\\\
0 & 0 & \\mu_{zz} \\end{bmatrix}
\mu = \begin{bmatrix} \mu_{xx} & 0 & 0 \\
0 & \mu_{yy} & 0 \\
0 & 0 & \mu_{zz} \end{bmatrix}
$$
where the off-diagonal terms (e.g. `epsilon_xy`) are assumed to be zero.

View File

@ -62,7 +62,7 @@ def deriv_back(
def curl_forward(
dx_e: Sequence[NDArray[numpy.float_]] | None = None,
) -> fdfield_updater_t:
"""
r"""
Curl operator for use with the E field.
Args:
@ -71,7 +71,7 @@ def curl_forward(
Returns:
Function `f` for taking the discrete forward curl of a field,
`f(E)` -> curlE $= \\nabla_f \\times E$
`f(E)` -> curlE $= \nabla_f \times E$
"""
Dx, Dy, Dz = deriv_forward(dx_e)
@ -91,7 +91,7 @@ def curl_forward(
def curl_back(
dx_h: Sequence[NDArray[numpy.float_]] | None = None,
) -> fdfield_updater_t:
"""
r"""
Create a function which takes the backward curl of a field.
Args:
@ -100,7 +100,7 @@ def curl_back(
Returns:
Function `f` for taking the discrete backward curl of a field,
`f(H)` -> curlH $= \\nabla_b \\times H$
`f(H)` -> curlH $= \nabla_b \times H$
"""
Dx, Dy, Dz = deriv_back(dx_h)

View File

@ -1,4 +1,4 @@
"""
r"""
Utilities for running finite-difference time-domain (FDTD) simulations
See the discussion of `Maxwell's Equations` in `meanas.fdmath` for basic
@ -11,9 +11,9 @@ Timestep
From the discussion of "Plane waves and the Dispersion relation" in `meanas.fdmath`,
we have
$$ c^2 \\Delta_t^2 = \\frac{\\Delta_t^2}{\\mu \\epsilon} < 1/(\\frac{1}{\\Delta_x^2} + \\frac{1}{\\Delta_y^2} + \\frac{1}{\\Delta_z^2}) $$
$$ c^2 \Delta_t^2 = \frac{\Delta_t^2}{\mu \epsilon} < 1/(\frac{1}{\Delta_x^2} + \frac{1}{\Delta_y^2} + \frac{1}{\Delta_z^2}) $$
or, if $\\Delta_x = \\Delta_y = \\Delta_z$, then $c \\Delta_t < \\frac{\\Delta_x}{\\sqrt{3}}$.
or, if $\Delta_x = \Delta_y = \Delta_z$, then $c \Delta_t < \frac{\Delta_x}{\sqrt{3}}$.
Based on this, we can set
@ -27,81 +27,81 @@ Poynting Vector and Energy Conservation
Let
$$ \\begin{aligned}
\\tilde{S}_{l, l', \\vec{r}} &=& &\\tilde{E}_{l, \\vec{r}} \\otimes \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\\\
&=& &\\vec{x} (\\tilde{E}^y_{l,m+1,n,p} \\hat{H}^z_{l',\\vec{r} + \\frac{1}{2}} - \\tilde{E}^z_{l,m+1,n,p} \\hat{H}^y_{l', \\vec{r} + \\frac{1}{2}}) \\\\
& &+ &\\vec{y} (\\tilde{E}^z_{l,m,n+1,p} \\hat{H}^x_{l',\\vec{r} + \\frac{1}{2}} - \\tilde{E}^x_{l,m,n+1,p} \\hat{H}^z_{l', \\vec{r} + \\frac{1}{2}}) \\\\
& &+ &\\vec{z} (\\tilde{E}^x_{l,m,n,p+1} \\hat{H}^y_{l',\\vec{r} + \\frac{1}{2}} - \\tilde{E}^y_{l,m,n,p+1} \\hat{H}^z_{l', \\vec{r} + \\frac{1}{2}})
\\end{aligned}
$$ \begin{aligned}
\tilde{S}_{l, l', \vec{r}} &=& &\tilde{E}_{l, \vec{r}} \otimes \hat{H}_{l', \vec{r} + \frac{1}{2}} \\
&=& &\vec{x} (\tilde{E}^y_{l,m+1,n,p} \hat{H}^z_{l',\vec{r} + \frac{1}{2}} - \tilde{E}^z_{l,m+1,n,p} \hat{H}^y_{l', \vec{r} + \frac{1}{2}}) \\
& &+ &\vec{y} (\tilde{E}^z_{l,m,n+1,p} \hat{H}^x_{l',\vec{r} + \frac{1}{2}} - \tilde{E}^x_{l,m,n+1,p} \hat{H}^z_{l', \vec{r} + \frac{1}{2}}) \\
& &+ &\vec{z} (\tilde{E}^x_{l,m,n,p+1} \hat{H}^y_{l',\vec{r} + \frac{1}{2}} - \tilde{E}^y_{l,m,n,p+1} \hat{H}^z_{l', \vec{r} + \frac{1}{2}})
\end{aligned}
$$
where $\\vec{r} = (m, n, p)$ and $\\otimes$ is a modified cross product
in which the $\\tilde{E}$ terms are shifted as indicated.
where $\vec{r} = (m, n, p)$ and $\otimes$ is a modified cross product
in which the $\tilde{E}$ terms are shifted as indicated.
By taking the divergence and rearranging terms, we can show that
$$
\\begin{aligned}
\\hat{\\nabla} \\cdot \\tilde{S}_{l, l', \\vec{r}}
&= \\hat{\\nabla} \\cdot (\\tilde{E}_{l, \\vec{r}} \\otimes \\hat{H}_{l', \\vec{r} + \\frac{1}{2}}) \\\\
&= \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l, \\vec{r}} -
\\tilde{E}_{l, \\vec{r}} \\cdot \\hat{\\nabla} \\times \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\\\
&= \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\cdot
(-\\tilde{\\partial}_t \\mu_{\\vec{r} + \\frac{1}{2}} \\hat{H}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} -
\\hat{M}_{l, \\vec{r} + \\frac{1}{2}}) -
\\tilde{E}_{l, \\vec{r}} \\cdot (\\hat{\\partial}_t \\tilde{\\epsilon}_{\\vec{r}} \\tilde{E}_{l'+\\frac{1}{2}, \\vec{r}} +
\\tilde{J}_{l', \\vec{r}}) \\\\
&= \\hat{H}_{l'} \\cdot (-\\mu / \\Delta_t)(\\hat{H}_{l + \\frac{1}{2}} - \\hat{H}_{l - \\frac{1}{2}}) -
\\tilde{E}_l \\cdot (\\epsilon / \\Delta_t )(\\tilde{E}_{l'+\\frac{1}{2}} - \\tilde{E}_{l'-\\frac{1}{2}})
- \\hat{H}_{l'} \\cdot \\hat{M}_{l} - \\tilde{E}_l \\cdot \\tilde{J}_{l'} \\\\
\\end{aligned}
\begin{aligned}
\hat{\nabla} \cdot \tilde{S}_{l, l', \vec{r}}
&= \hat{\nabla} \cdot (\tilde{E}_{l, \vec{r}} \otimes \hat{H}_{l', \vec{r} + \frac{1}{2}}) \\
&= \hat{H}_{l', \vec{r} + \frac{1}{2}} \cdot \tilde{\nabla} \times \tilde{E}_{l, \vec{r}} -
\tilde{E}_{l, \vec{r}} \cdot \hat{\nabla} \times \hat{H}_{l', \vec{r} + \frac{1}{2}} \\
&= \hat{H}_{l', \vec{r} + \frac{1}{2}} \cdot
(-\tilde{\partial}_t \mu_{\vec{r} + \frac{1}{2}} \hat{H}_{l - \frac{1}{2}, \vec{r} + \frac{1}{2}} -
\hat{M}_{l, \vec{r} + \frac{1}{2}}) -
\tilde{E}_{l, \vec{r}} \cdot (\hat{\partial}_t \tilde{\epsilon}_{\vec{r}} \tilde{E}_{l'+\frac{1}{2}, \vec{r}} +
\tilde{J}_{l', \vec{r}}) \\
&= \hat{H}_{l'} \cdot (-\mu / \Delta_t)(\hat{H}_{l + \frac{1}{2}} - \hat{H}_{l - \frac{1}{2}}) -
\tilde{E}_l \cdot (\epsilon / \Delta_t )(\tilde{E}_{l'+\frac{1}{2}} - \tilde{E}_{l'-\frac{1}{2}})
- \hat{H}_{l'} \cdot \hat{M}_{l} - \tilde{E}_l \cdot \tilde{J}_{l'} \\
\end{aligned}
$$
where in the last line the spatial subscripts have been dropped to emphasize
the time subscripts $l, l'$, i.e.
$$
\\begin{aligned}
\\tilde{E}_l &= \\tilde{E}_{l, \\vec{r}} \\\\
\\hat{H}_l &= \\tilde{H}_{l, \\vec{r} + \\frac{1}{2}} \\\\
\\tilde{\\epsilon} &= \\tilde{\\epsilon}_{\\vec{r}} \\\\
\\end{aligned}
\begin{aligned}
\tilde{E}_l &= \tilde{E}_{l, \vec{r}} \\
\hat{H}_l &= \tilde{H}_{l, \vec{r} + \frac{1}{2}} \\
\tilde{\epsilon} &= \tilde{\epsilon}_{\vec{r}} \\
\end{aligned}
$$
etc.
For $l' = l + \\frac{1}{2}$ we get
For $l' = l + \frac{1}{2}$ we get
$$
\\begin{aligned}
\\hat{\\nabla} \\cdot \\tilde{S}_{l, l + \\frac{1}{2}}
&= \\hat{H}_{l + \\frac{1}{2}} \\cdot
(-\\mu / \\Delta_t)(\\hat{H}_{l + \\frac{1}{2}} - \\hat{H}_{l - \\frac{1}{2}}) -
\\tilde{E}_l \\cdot (\\epsilon / \\Delta_t)(\\tilde{E}_{l+1} - \\tilde{E}_l)
- \\hat{H}_{l'} \\cdot \\hat{M}_l - \\tilde{E}_l \\cdot \\tilde{J}_{l + \\frac{1}{2}} \\\\
&= (-\\mu / \\Delta_t)(\\hat{H}^2_{l + \\frac{1}{2}} - \\hat{H}_{l + \\frac{1}{2}} \\cdot \\hat{H}_{l - \\frac{1}{2}}) -
(\\epsilon / \\Delta_t)(\\tilde{E}_{l+1} \\cdot \\tilde{E}_l - \\tilde{E}^2_l)
- \\hat{H}_{l'} \\cdot \\hat{M}_l - \\tilde{E}_l \\cdot \\tilde{J}_{l + \\frac{1}{2}} \\\\
&= -(\\mu \\hat{H}^2_{l + \\frac{1}{2}}
+\\epsilon \\tilde{E}_{l+1} \\cdot \\tilde{E}_l) / \\Delta_t \\ \\
+(\\mu \\hat{H}_{l + \\frac{1}{2}} \\cdot \\hat{H}_{l - \\frac{1}{2}}
+\\epsilon \\tilde{E}^2_l) / \\Delta_t \\ \\
- \\hat{H}_{l+\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l+\\frac{1}{2}} \\\\
\\end{aligned}
\begin{aligned}
\hat{\nabla} \cdot \tilde{S}_{l, l + \frac{1}{2}}
&= \hat{H}_{l + \frac{1}{2}} \cdot
(-\mu / \Delta_t)(\hat{H}_{l + \frac{1}{2}} - \hat{H}_{l - \frac{1}{2}}) -
\tilde{E}_l \cdot (\epsilon / \Delta_t)(\tilde{E}_{l+1} - \tilde{E}_l)
- \hat{H}_{l'} \cdot \hat{M}_l - \tilde{E}_l \cdot \tilde{J}_{l + \frac{1}{2}} \\
&= (-\mu / \Delta_t)(\hat{H}^2_{l + \frac{1}{2}} - \hat{H}_{l + \frac{1}{2}} \cdot \hat{H}_{l - \frac{1}{2}}) -
(\epsilon / \Delta_t)(\tilde{E}_{l+1} \cdot \tilde{E}_l - \tilde{E}^2_l)
- \hat{H}_{l'} \cdot \hat{M}_l - \tilde{E}_l \cdot \tilde{J}_{l + \frac{1}{2}} \\
&= -(\mu \hat{H}^2_{l + \frac{1}{2}}
+\epsilon \tilde{E}_{l+1} \cdot \tilde{E}_l) / \Delta_t \\
+(\mu \hat{H}_{l + \frac{1}{2}} \cdot \hat{H}_{l - \frac{1}{2}}
+\epsilon \tilde{E}^2_l) / \Delta_t \\
- \hat{H}_{l+\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l+\frac{1}{2}} \\
\end{aligned}
$$
and for $l' = l - \\frac{1}{2}$,
and for $l' = l - \frac{1}{2}$,
$$
\\begin{aligned}
\\hat{\\nabla} \\cdot \\tilde{S}_{l, l - \\frac{1}{2}}
&= (\\mu \\hat{H}^2_{l - \\frac{1}{2}}
+\\epsilon \\tilde{E}_{l-1} \\cdot \\tilde{E}_l) / \\Delta_t \\ \\
-(\\mu \\hat{H}_{l + \\frac{1}{2}} \\cdot \\hat{H}_{l - \\frac{1}{2}}
+\\epsilon \\tilde{E}^2_l) / \\Delta_t \\ \\
- \\hat{H}_{l-\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l-\\frac{1}{2}} \\\\
\\end{aligned}
\begin{aligned}
\hat{\nabla} \cdot \tilde{S}_{l, l - \frac{1}{2}}
&= (\mu \hat{H}^2_{l - \frac{1}{2}}
+\epsilon \tilde{E}_{l-1} \cdot \tilde{E}_l) / \Delta_t \\
-(\mu \hat{H}_{l + \frac{1}{2}} \cdot \hat{H}_{l - \frac{1}{2}}
+\epsilon \tilde{E}^2_l) / \Delta_t \\
- \hat{H}_{l-\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l-\frac{1}{2}} \\
\end{aligned}
$$
These two results form the discrete time-domain analogue to Poynting's theorem.
@ -109,25 +109,25 @@ They hint at the expressions for the energy, which can be calculated at the same
time-index as either the E or H field:
$$
\\begin{aligned}
U_l &= \\epsilon \\tilde{E}^2_l + \\mu \\hat{H}_{l + \\frac{1}{2}} \\cdot \\hat{H}_{l - \\frac{1}{2}} \\\\
U_{l + \\frac{1}{2}} &= \\epsilon \\tilde{E}_l \\cdot \\tilde{E}_{l + 1} + \\mu \\hat{H}^2_{l + \\frac{1}{2}} \\\\
\\end{aligned}
\begin{aligned}
U_l &= \epsilon \tilde{E}^2_l + \mu \hat{H}_{l + \frac{1}{2}} \cdot \hat{H}_{l - \frac{1}{2}} \\
U_{l + \frac{1}{2}} &= \epsilon \tilde{E}_l \cdot \tilde{E}_{l + 1} + \mu \hat{H}^2_{l + \frac{1}{2}} \\
\end{aligned}
$$
Rewriting the Poynting theorem in terms of the energy expressions,
$$
\\begin{aligned}
(U_{l+\\frac{1}{2}} - U_l) / \\Delta_t
&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l + \\frac{1}{2}} \\ \\
- \\hat{H}_{l+\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l+\\frac{1}{2}} \\\\
(U_l - U_{l-\\frac{1}{2}}) / \\Delta_t
&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l - \\frac{1}{2}} \\ \\
- \\hat{H}_{l-\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l-\\frac{1}{2}} \\\\
\\end{aligned}
\begin{aligned}
(U_{l+\frac{1}{2}} - U_l) / \Delta_t
&= -\hat{\nabla} \cdot \tilde{S}_{l, l + \frac{1}{2}} \\
- \hat{H}_{l+\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l+\frac{1}{2}} \\
(U_l - U_{l-\frac{1}{2}}) / \Delta_t
&= -\hat{\nabla} \cdot \tilde{S}_{l, l - \frac{1}{2}} \\
- \hat{H}_{l-\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l-\frac{1}{2}} \\
\end{aligned}
$$
This result is exact and should practically hold to within numerical precision. No time-
@ -147,10 +147,10 @@ of the time-domain fields.
The Ricker wavelet (normalized second derivative of a Gaussian) is commonly used for the pulse
shape. It can be written
$$ f_r(t) = (1 - \\frac{1}{2} (\\omega (t - \\tau))^2) e^{-(\\frac{\\omega (t - \\tau)}{2})^2} $$
$$ f_r(t) = (1 - \frac{1}{2} (\omega (t - \tau))^2) e^{-(\frac{\omega (t - \tau)}{2})^2} $$
with $\\tau > \\frac{2 * \\pi}{\\omega}$ as a minimum delay to avoid a discontinuity at
t=0 (assuming the source is off for t<0 this gives $\\sim 10^{-3}$ error at t=0).
with $\tau > \frac{2 * \pi}{\omega}$ as a minimum delay to avoid a discontinuity at
t=0 (assuming the source is off for t<0 this gives $\sim 10^{-3}$ error at t=0).

View File

@ -12,7 +12,7 @@ def poynting(
h: fdfield_t,
dxes: dx_lists_t | None = None,
) -> fdfield_t:
"""
r"""
Calculate the poynting vector `S` ($S$).
This is the energy transfer rate (amount of energy `U` per `dt` transferred
@ -44,16 +44,16 @@ def poynting(
The full relationship is
$$
\\begin{aligned}
(U_{l+\\frac{1}{2}} - U_l) / \\Delta_t
&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l + \\frac{1}{2}} \\ \\
- \\hat{H}_{l+\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l+\\frac{1}{2}} \\\\
(U_l - U_{l-\\frac{1}{2}}) / \\Delta_t
&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l - \\frac{1}{2}} \\ \\
- \\hat{H}_{l-\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
- \\tilde{E}_l \\cdot \\tilde{J}_{l-\\frac{1}{2}} \\\\
\\end{aligned}
\begin{aligned}
(U_{l+\frac{1}{2}} - U_l) / \Delta_t
&= -\hat{\nabla} \cdot \tilde{S}_{l, l + \frac{1}{2}} \\
- \hat{H}_{l+\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l+\frac{1}{2}} \\
(U_l - U_{l-\frac{1}{2}}) / \Delta_t
&= -\hat{\nabla} \cdot \tilde{S}_{l, l - \frac{1}{2}} \\
- \hat{H}_{l-\frac{1}{2}} \cdot \hat{M}_l \\
- \tilde{E}_l \cdot \tilde{J}_{l-\frac{1}{2}} \\
\end{aligned}
$$
These equalities are exact and should practically hold to within numerical precision.

View File

@ -321,7 +321,6 @@ class _ToMarkdown:
"""Wrap URLs in Python-Markdown-compatible <angle brackets>."""
return re.sub(r'(?<![<"\'])(\s*)((?:http|ftp)s?://[^>)\s]+)(\s*)', r'\1<\2>\3', text)
import subprocess
class _MathPattern(InlineProcessor):
NAME = 'pdoc-math'