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@ -13,7 +13,7 @@ we have
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$$ c^2 \\Delta_t^2 = \\frac{\\Delta_t^2}{\\mu \\epsilon} < 1/(\\frac{1}{\\Delta_x^2} + \\frac{1}{\\Delta_y^2} + \\frac{1}{\\Delta_z^2}) $$
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or, if \\( \\Delta_x = \\Delta_y = \\Delta_z \\), then \\( c \\Delta_t < \\frac{\\Delta_x}{\\sqrt{3}} \\).
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or, if $\\Delta_x = \\Delta_y = \\Delta_z$, then $c \\Delta_t < \\frac{\\Delta_x}{\\sqrt{3}}$.
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Based on this, we can set
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@ -27,21 +27,21 @@ Poynting Vector and Energy Conservation
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Let
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$$ \\begin{align*}
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$$ \\begin{aligned}
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\\tilde{S}_{l, l', \\vec{r}} &=& &\\tilde{E}_{l, \\vec{r}} \\otimes \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\\\
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&=& &\\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}}) \\\\
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& &+ &\\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}}) \\\\
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& &+ &\\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}})
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\\end{align*}
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\\end{aligned}
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$$
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where \\( \\vec{r} = (m, n, p) \\) and \\( \\otimes \\) is a modified cross product
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in which the \\( \\tilde{E} \\) terms are shifted as indicated.
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where $\\vec{r} = (m, n, p)$ and $\\otimes$ is a modified cross product
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in which the $\\tilde{E}$ terms are shifted as indicated.
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By taking the divergence and rearranging terms, we can show that
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$$
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\\begin{align*}
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\\begin{aligned}
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\\hat{\\nabla} \\cdot \\tilde{S}_{l, l', \\vec{r}}
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&= \\hat{\\nabla} \\cdot (\\tilde{E}_{l, \\vec{r}} \\otimes \\hat{H}_{l', \\vec{r} + \\frac{1}{2}}) \\\\
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&= \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\cdot \\tilde{\\nabla} \\times \\tilde{E}_{l, \\vec{r}} -
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@ -49,30 +49,30 @@ $$
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&= \\hat{H}_{l', \\vec{r} + \\frac{1}{2}} \\cdot
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(-\\tilde{\\partial}_t \\mu_{\\vec{r} + \\frac{1}{2}} \\hat{H}_{l - \\frac{1}{2}, \\vec{r} + \\frac{1}{2}} -
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\\hat{M}_{l-1, \\vec{r} + \\frac{1}{2}}) -
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\\tilde{E}_{l, \\vec{r}} \\cdot (\\hat{\\partial}_t \\tilde{\\epsilon}_\\vec{r} \\tilde{E}_{l'+\\frac{1}{2}, \\vec{r}} +
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\\tilde{E}_{l, \\vec{r}} \\cdot (\\hat{\\partial}_t \\tilde{\\epsilon}_{\\vec{r}} \\tilde{E}_{l'+\\frac{1}{2}, \\vec{r}} +
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\\tilde{J}_{l', \\vec{r}}) \\\\
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&= \\hat{H}_{l'} \\cdot (-\\mu / \\Delta_t)(\\hat{H}_{l + \\frac{1}{2}} - \\hat{H}_{l - \\frac{1}{2}}) -
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\\tilde{E}_l \\cdot (\\epsilon / \\Delta_t )(\\tilde{E}_{l'+\\frac{1}{2}} - \\tilde{E}_{l'-\\frac{1}{2}})
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- \\hat{H}_{l'} \\cdot \\hat{M}_{l-1} - \\tilde{E}_l \\cdot \\tilde{J}_{l'} \\\\
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\\end{align*}
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\\end{aligned}
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$$
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where in the last line the spatial subscripts have been dropped to emphasize
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the time subscripts \\( l, l' \\), i.e.
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the time subscripts $l, l'$, i.e.
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$$
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\\begin{align*}
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\\begin{aligned}
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\\tilde{E}_l &= \\tilde{E}_{l, \\vec{r}} \\\\
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\\hat{H}_l &= \\tilde{H}_{l, \\vec{r} + \\frac{1}{2}} \\\\
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\\tilde{\\epsilon} &= \\tilde{\\epsilon}_\\vec{r} \\\\
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\\end{align*}
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\\tilde{\\epsilon} &= \\tilde{\\epsilon}_{\\vec{r}} \\\\
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\\end{aligned}
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$$
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etc.
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For \\( l' = l + \\frac{1}{2} \\) we get
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For $l' = l + \\frac{1}{2}$ we get
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$$
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\\begin{align*}
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\\begin{aligned}
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\\hat{\\nabla} \\cdot \\tilde{S}_{l, l + \\frac{1}{2}}
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&= \\hat{H}_{l + \\frac{1}{2}} \\cdot
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(-\\mu / \\Delta_t)(\\hat{H}_{l + \\frac{1}{2}} - \\hat{H}_{l - \\frac{1}{2}}) -
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@ -87,13 +87,13 @@ $$
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+\\epsilon \\tilde{E}^2_l) / \\Delta_t \\ \\
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- \\hat{H}_{l+\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
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- \\tilde{E}_l \\cdot \\tilde{J}_{l+\\frac{1}{2}} \\\\
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\\end{align*}
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\\end{aligned}
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$$
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and for \\( l' = l - \\frac{1}{2} \\),
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and for $l' = l - \\frac{1}{2}$,
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$$
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\\begin{align*}
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\\begin{aligned}
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\\hat{\\nabla} \\cdot \\tilde{S}_{l, l - \\frac{1}{2}}
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&= (\\mu \\hat{H}^2_{l - \\frac{1}{2}}
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+\\epsilon \\tilde{E}_{l-1} \\cdot \\tilde{E}_l) / \\Delta_t \\ \\
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@ -101,7 +101,7 @@ $$
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+\\epsilon \\tilde{E}^2_l) / \\Delta_t \\ \\
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- \\hat{H}_{l-\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
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- \\tilde{E}_l \\cdot \\tilde{J}_{l-\\frac{1}{2}} \\\\
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\\end{align*}
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\\end{aligned}
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$$
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These two results form the discrete time-domain analogue to Poynting's theorem.
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@ -109,16 +109,16 @@ They hint at the expressions for the energy, which can be calculated at the same
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time-index as either the E or H field:
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$$
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\\begin{align*}
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\\begin{aligned}
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U_l &= \\epsilon \\tilde{E}^2_l + \\mu \\hat{H}_{l + \\frac{1}{2}} \\cdot \\hat{H}_{l - \\frac{1}{2}} \\\\
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U_{l + \\frac{1}{2}} &= \\epsilon \\tilde{E}_l \\cdot \\tilde{E}_{l + 1} + \\mu \\hat{H}^2_{l + \\frac{1}{2}} \\\\
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\\end{align*}
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\\end{aligned}
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$$
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Rewriting the Poynting theorem in terms of the energy expressions,
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$$
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\\begin{align*}
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\\begin{aligned}
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(U_{l+\\frac{1}{2}} - U_l) / \\Delta_t
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&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l + \\frac{1}{2}} \\ \\
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- \\hat{H}_{l+\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
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@ -127,14 +127,14 @@ $$
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&= -\\hat{\\nabla} \\cdot \\tilde{S}_{l, l - \\frac{1}{2}} \\ \\
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- \\hat{H}_{l-\\frac{1}{2}} \\cdot \\hat{M}_l \\ \\
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- \\tilde{E}_l \\cdot \\tilde{J}_{l-\\frac{1}{2}} \\\\
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\\end{align*}
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\\end{aligned}
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$$
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This result is exact an should practically hold to within numerical precision. No time-
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or spatial-averaging is necessary.
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Note that each value of \\( J \\) contributes to the energy twice (i.e. once per field update)
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despite only causing the value of \\( E \\) to change once (same for \\( M \\) and \\( H \\)).
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Note that each value of $J$ contributes to the energy twice (i.e. once per field update)
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despite only causing the value of $E$ to change once (same for $M$ and $H$).
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Sources
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@ -149,8 +149,8 @@ shape. It can be written
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$$ f_r(t) = (1 - \\frac{1}{2} (\\omega (t - \\tau))^2) e^{-(\\frac{\\omega (t - \\tau)}{2})^2} $$
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with \\( \\tau > \\frac{2 * \\pi}{\\omega} \\) as a minimum delay to avoid a discontinuity at
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t=0 (assuming the source is off for t<0 this gives \\( \\sim 10^{-3} \\) error at t=0).
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with $\\tau > \\frac{2 * \\pi}{\\omega}$ as a minimum delay to avoid a discontinuity at
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t=0 (assuming the source is off for t<0 this gives $\\sim 10^{-3}$ error at t=0).
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