标签：order uniq omega ima represent tom effective can mes
This article summarizes the famous Maxwell equations in different forms, namely, time-dependent formulation, time-harmonic formulation and their corresponding \(\vect{A}-\varphi\) formulations.
Maxwell equations are usually presented in the following time-dependent form with both partial differential equations and constitutive laws.
Partial differential equations:
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\vect{H} = \vect{J} = \vect{J}_c + \frac{\partial \vect{D}}{\partial t} \label{eq:ampere-time} \cr
& \text{Faraday‘s law: } \nabla\times\vect{E} = -\frac{\partial \vect{B}}{\partial t} \cr
& \text{Gauss’s law for $\vect{D}$: } \nabla \cdot \vect{D} = \rho \cr
& \text{Gauss’s law for $\vect{B}$: } \nabla \cdot \vect{B} = 0
\end{align}
\]
Constitutive equations:
\[
\begin{align}
\vect{D} &= \varepsilon \vect{E} \cr
\vect{B} &= \mu \vect{H}
\end{align}
\]
It should be noted that the four partial differential equations are not independent of each other. And their inter-relationships are clarified below.
(a) Apply divergence to Ampère‘s law, the continuity of total current can be obtained.
\[
\begin{equation}
\nabla \cdot (\nabla \times \vect{H}) = \nabla \cdot \vect{J} = \nabla \cdot \left( \vect{J}_c + \frac{\partial\vect{D}}{\partial t} \right) = 0.
\label{eq:total-current-continuity}
\end{equation}
\]
Then, if the charge conservation law is prescribed as
\[
\begin{equation}
\frac{\partial \rho}{\partial t} + \nabla \cdot \vect{J}_c = 0,
\end{equation}
\]
we have \(\frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t} (\nabla \cdot \vect{D})\). If we further enforce the initial condition \(\rho \vert_{t=0} = \nabla \cdot \vect{D}\vert_{t=0}\), Gauss‘s law for \(\vect{D}\) is obtained.
On the other hand, if Gauss‘s law is given first, by substituting \(\nabla \cdot \vect{D} = \rho\) into \eqref{eq:total-current-continuity}, the charge conservation law can be derived.
For short, Ampère‘s law implies the continuity of total current. The continuity of total current and charge conservation law as well as Gauss‘s law for \(\vect{D}\) which is only prescribed at \(t=0\) lead to the Gauss‘s law for \(\vect{D}\) at any time. Meanwhile, the charge conservation law can be derived from the continuity of total current and Gauss‘s law for \(\vect{D}\) at any time.
(b) Apply divergence to Faraday‘s law, we have \(\nabla \cdot (\nabla \times \vect{E}) = -\frac{\partial}{\partial t}(\nabla \cdot \vect{B}) = 0\). If the initial condition \(\nabla \cdot \vect{B} \vert_{t=0} = 0\) is prescribed, i.e. there is no magnetic charge in the universe from the origin of time, then Gauss‘s law for \(\vect{B}\) can be obtained.
(c) If an additional impressed current \(\vect{J}_i\) is involved, the charge conservation law becomes
\[
\begin{equation}
\frac{\partial \rho}{\partial t} + \nabla \cdot (\vect{J}_c + \vect{J}_i) = \frac{\partial \rho}{\partial t} + \nabla \cdot (\sigma\vect{E}) + \nabla \cdot \vect{J}_i = 0.
\end{equation}
\]
In non-conducting domain \(\Omega_I\), the conductivity \(\sigma\) is zero and there should be no time-variation of free charge density. Therefore, we have \(\nabla \cdot \vect{J}_i = 0\), i.e. the impressed current should be divergence-free.
Partial differential equations:
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\phsvect{H} = \phsvect{J} = \phsvect{J}_c + \rmi\omega\phsvect{D} = (\sigma + \rmi\omega\varepsilon) \phsvect{E} \cr
& \text{Faraday‘s law: } \nabla\times\phsvect{E} = -\rmi\omega\phsvect{B} = -\rmi\omega\mu\phsvect{H} \cr
& \text{Gauss’s law for $\vect{D}$: } \nabla \cdot \phsvect{D} = \phs{\rho} \cr
& \text{Gauss’s law for $\vect{B}$: } \nabla \cdot \phsvect{B} = 0
\end{align}
\]
Constitutive equations:
\[
\begin{align}
\phsvect{D} &= \varepsilon \phsvect{E} \cr
\phsvect{B} &= \mu \phsvect{H}
\end{align}
\]
(a) Substitute Ohm‘s law into Ampère‘s law and consider an additional impressed current \(\phsvect{J}_i\), we have
\[
\begin{equation}
\nabla \times \phsvect{H} = \sigma\phsvect{E} + \rmi\omega\varepsilon\phsvect{E} + \phsvect{J}_i = (\sigma + \rmi\omega\varepsilon)\phsvect{E} + \phsvect{J}_i.
\label{eq:ampere-harmonic-full}
\end{equation}
\]
Notice that the complex-valued conductivity \(\sigma + \rmi\omega\varepsilon\) appears.
(b) Apply divergence to Ampère‘s law as in Equation \eqref{eq:ampere-harmonic-full}, we have
\[
\begin{equation}
\nabla \cdot \left[ (\sigma + \rmi\omega\varepsilon) \phsvect{E} \right] + \nabla \cdot \phsvect{J}_i = 0.
\end{equation}
\]
Because the impressed current \(\phsvect{J}_i\) is divergence-free, the continuity of total current, i.e. conduction current plus displacement current, is obtained.
\[
\begin{equation}
\nabla \cdot \left[ (\sigma + \rmi\omega\varepsilon) \phsvect{E} \right] = 0.
\end{equation}
\]
Specifically, in non-conducting domain \(\Omega_I\), the conduction current vanishes and we have the Laplace equation
\[
\begin{equation}
\nabla \cdot \left( \varepsilon \phsvect{E} \right) = 0.
\end{equation}
\]
(c) Apply divergence to Faraday‘s law, there is then Gauss‘s law for \(\vect{B}\),
\[
\begin{equation}
\nabla \cdot (\nabla \times \phsvect{H}) = \nabla \cdot (-\rmi\omega\phsvect{B}) = \nabla \cdot \phsvect{B} = 0.
\end{equation}
\]
(d) Summarize the above, we know that in conducting domain \(\Omega_c\), Ampère‘s law leads to the continuity of total current, while in non-conducting domain \(\Omega_I\), it leads to Gauss‘s law for \(\vect{D}\). On the other hand, Faraday‘s law implies Gauss‘s law for \(\vect{B}\).
Neglecting the displacement current \(\rmi\omega\varepsilon\phsvect{E}\), we obtain the eddy current approximation of time-harmonic Maxwell equations as follows.
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\phsvect{H} = \phsvect{J} = \sigma \phsvect{E} \cr
& \text{Faraday‘s law: } \nabla\times\phsvect{E} = -\rmi\omega\phsvect{B} = -\rmi\omega\mu\phsvect{H} \cr
& \text{Gauss’s law for $\vect{D}$: } \nabla \cdot \phsvect{D} = \phs{\rho} \cr
& \text{Gauss’s law for $\vect{B}$: } \nabla \cdot \phsvect{B} = 0
\end{align}
\]
(a) Because Ampère‘s law does not include the displacement current term, Gauss‘s law for \(\vect{D}\) cannot be definitely derived from Ampère‘s law any more.
(b) In conducting domain \(\Omega_c\), \(\sigma \neq 0\) and apply divergence to Ampère‘s law. Hence
\[
\begin{equation}
\nabla \cdot (\nabla \times \phsvect{H}) = \nabla \cdot (\sigma \phsvect{E}) = 0.
\end{equation}
\]
This is the continuity of conduction current due to the absence of displacement current. If the impressed current \(\phsvect{J}_i\) appears in Ampère‘s law, because it is divergence-free, the continuity of conduction current still holds.
(c) In non-conducting domain \(\Omega_I\), \(\sigma = 0\) and Ampère‘s law becomes \(\nabla \times \phsvect{H} = 0\) or \(\nabla \times \phsvect{H} = \phsvect{J}_i\), if the impressed current is present. Then apply divergence to it will only produce \(0 = 0\), which is degenerate. Therefore, Gauss‘s law for \(\vect{D}\) cannot be derived from Ampère‘s law anymore and must be explicitly enforced in \(\Omega_I\), i.e. \(\nabla \cdot \phsvect{D} = \nabla \cdot \left( \varepsilon \phsvect{E} \right) = 0\).
(d) Summarize the above, the eddy current approximation of time-harmonic Maxwell equations is as follows.
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\phsvect{H} = \phsvect{J} = \sigma \phsvect{E} \cr
& \text{Faraday‘s law: } \nabla\times\phsvect{E} = -\rmi\omega\mu\phsvect{H} \cr
& \text{Gauss’s law for $\vect{D}$: } \nabla \cdot \phsvect{D} = 0 \; \text{in $\Omega_I$}
\end{align}
\]
Let \(\vect{B} = \nabla \times \vect{A}\) and \(\vect{E} = -\nabla \varphi - \frac{\partial \vect{A}}{\partial t}\).
(a) Substitute them into Ampère‘s law in Equation \eqref{eq:ampere-time}, hence
\[
\nabla \times \left( \frac{1}{\mu} \nabla \times \vect{A} \right) = \vect{J}_c + \frac{\partial}{\partial t} \left[ \varepsilon (-\nabla\varphi - \frac{\partial \vect{A}}{\partial t}) \right].
\]
Reorder the terms, Ampère‘s law becomes
\[
\begin{equation}
\nabla \times \left( \frac{1}{\mu} \nabla \times \vect{A} \right) + \varepsilon \nabla \frac{\partial \varphi}{\partial t} + \varepsilon \frac{\partial^2 \vect{A}}{\partial t^2} = \vect{J}_c.
\label{eq:ampere-law-aphi}
\end{equation}
\]
(b) For Faraday‘s law, due to the adopted \(\vect{A}-\varphi\) potentials,
\[
\nabla \times \vect{E} = \nabla \times \left( -\nabla\varphi - \frac{\partial \vect{A}}{\partial t} \right) = -\nabla\times(\nabla\varphi) - \frac{\partial}{\partial t} \left( \nabla\times\vect{A} \right) = -\frac{\partial \vect{B}}{\partial t}.
\]
Therefore, Faraday‘s law is automatically satisfied through the selection of \(\vect{A}-\varphi\).
(c) Gauss‘s law for \(\vect{D}\):
\[
\begin{align}
\nabla \cdot \left[ \varepsilon \left( -\nabla\varphi - \frac{\partial \vect{A}}{\partial t} \right) \right] &= \rho \cr
-\nabla \cdot \left( \varepsilon \nabla \varphi \right) - \frac{\partial}{\partial t} \nabla\cdot\left( \varepsilon\vect{A} \right) &= \rho \label{eq:gauss-law-for-d-aphi}.
\end{align}
\]
(d) Gauss‘s law for \(\vect{B}\): \(\nabla\cdot\vect{B}=0\) is automatically satisfied by the selection of \(\vect{A}\) such that \(\vect{B}=\nabla\times\vect{A}\).
(e) It should be noted that the above set of effective Maxwell equations, namely, Ampère‘s law and Gauss‘s law for \(\vect{D}\), does not uniquely solve \(\vect{A}-\varphi\). This can be further explained below.
If we assume \((\vect{A},\varphi)\) is a solution to the equation system, then for any scalar field \(\psi\), we can define
\[
\vect{A}‘=\vect{A}+\nabla\psi, \;\varphi‘=\varphi-\frac{\partial\psi}{\partial t}.
\]
Substitute \((\vect{A}‘,\varphi‘)\) into Ampère‘s law as in Equation \eqref{eq:ampere-law-aphi},
\[
\begin{aligned}
\nabla\times\left(\frac{1}{\mu}\nabla\times\vect{A}‘\right) + \varepsilon\nabla\frac{\partial \varphi‘}{\partial t} + \varepsilon\frac{\partial^2 \vect{A}‘}{\partial t^2} &= \vect{J}_c \cr
\nabla\times\left[\frac{1}{\mu}\nabla\times\left(\vect{A}+\nabla\psi\right)\right] + \varepsilon\nabla\left[\frac{\partial}{\partial t}\left(\varphi-\frac{\partial \psi}{\partial t}\right)\right] + \varepsilon\frac{\partial^2}{\partial t^2}(\vect{A}+\nabla\psi) &= \vect{J}_c \cr
\nabla\times\left(\frac{1}{\mu}\nabla\times\vect{A}\right) + \varepsilon\nabla\frac{\partial \varphi}{\partial t} - \varepsilon\nabla\frac{\partial^2 \psi}{\partial t^2} + \varepsilon\frac{\partial^2 \vect{A}}{\partial t^2} + \varepsilon\frac{\partial^2}{\partial t^2}(\nabla\psi) &= \vect{J}_c \cr
\nabla\times\left(\frac{1}{\mu}\nabla\times\vect{A}\right) + \varepsilon\nabla\frac{\partial \varphi}{\partial t} + \varepsilon\frac{\partial^2\vect{A}}{\partial t^2} &= \vect{J}_c
\end{aligned}.
\]
It can be seen that Ampère‘s law filled with \((\vect{A},\varphi)\) is recovered.
Substitute \((\vect{A}‘,\varphi‘)\) into Gauss‘s law for \(\vect{D}\) as in Equation \eqref{eq:gauss-law-for-d-aphi},
\[
\begin{aligned}
-\nabla\cdot(\varepsilon\nabla\varphi‘) - \frac{\partial}{\partial t}\nabla\cdot(\varepsilon\vect{A}‘) &= \rho \cr
-\nabla\cdot\left[\varepsilon\nabla\left( \varphi - \frac{\partial\psi}{\partial t} \right)\right] - \frac{\partial}{\partial t}\nabla\cdot\left[\varepsilon(\vect{A}+\nabla\psi)\right] &= \rho \cr
-\nabla\cdot(\varepsilon\nabla\varphi) + \nabla\cdot\left[\varepsilon\nabla\left(\frac{\partial\psi}{\partial t}\right)\right] - \frac{\partial}{\partial t}\nabla\cdot(\varepsilon\vect{A}) - \frac{\partial}{\partial t}\nabla\cdot(\varepsilon\nabla\psi) &= \rho \cr
-\nabla\cdot(\varepsilon\nabla\varphi) - \frac{\partial}{\partial t}\nabla\cdot(\varepsilon\vect{A}) &= \rho
\end{aligned}.
\]
Hence, the original Gauss‘s law for \(\vect{D}\) filled with \((\vect{A},\varphi)\) is recovered. Summarize the above, \((\vect{A}‘,\varphi‘)\) is also a solution of the Maxwell equations.
To uniquely determine the solution \((\vect{A},\varphi)\), Coulomb gauge should be appended to the system, which defines the divergence of \(\vect{A}\), i.e. \(\nabla\cdot(\varepsilon\vect{A})=0\). With Coulomb gauge adopted, Equation \eqref{eq:gauss-law-for-d-aphi} becomes
\[
\begin{equation}
-\nabla\cdot(\varepsilon\nabla\varphi) = \rho,
\end{equation}
\]
which is the classical Poisson equation for electric scalar potential.
(f) Finally, \(\vect{A}-\varphi\) formulation for the time-dependent Maxwell equations is given below.
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\left(\frac{1}{\mu}\nabla\times\vect{A}\right) + \varepsilon\nabla\frac{\partial\varphi}{\partial t} + \varepsilon\frac{\partial^2\vect{A}}{\partial t^2} &= \vect{J}_c \label{eq:ampere-law-aphi-final} \cr
& \text{Gauss‘s law for $\vect{D}$: } -\nabla\cdot(\varepsilon\nabla\varphi) = \rho \label{eq:gauss-law-for-d-aphi-final} \cr
& \text{Coulomb gauge: } \nabla\cdot(\varepsilon\vect{A}) = 0. \label{eq:coulomb-gauge-aphi-final}
\end{align}
\]
The system of equations can be directly derived from Equation \eqref{eq:ampere-law-aphi-final}, \eqref{eq:gauss-law-for-d-aphi-final} and \eqref{eq:coulomb-gauge-aphi-final}.
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\left(\frac{1}{\mu}\nabla\times\phsvect{A}\right) + \rmi\omega\varepsilon\nabla\phs{\varphi} - \varepsilon\omega^2\phsvect{A} &= \phsvect{J}_c \label{eq:ampere-law-aphi-harmo} \cr
& \text{Gauss‘s law for $\vect{D}$: } -\nabla\cdot(\varepsilon\nabla\phs{\varphi}) = \phs{\rho} \label{eq:gauss-law-for-d-aphi-harmo} \cr
& \text{Coulomb gauge: } \nabla\cdot(\varepsilon\phsvect{A}) = 0. \label{eq:coulomb-gauge-aphi-harmo}
\end{align}
\]
(a) Similar to the discussion in section 1(a), here we also have the fact that Ampère‘s law and Gauss‘s law lead to the charge conservation law with only the difference that Coulomb gauge needs to be substituted into the system. Apply divergence to Ampère‘s law, we have
\[
\nabla\cdot\left[ \nabla\times\left(\frac{1}{\mu}\nabla\times\phsvect{A}\right) \right] + \rmi\omega\nabla\cdot(\varepsilon\nabla\phs{\varphi}) - \omega^2\nabla\cdot(\varepsilon\phsvect{A}) = \nabla\cdot\phsvect{J}_c,
\]
where the first and third terms on the left hand side are zero. Then charge conservation law is obtained:
\[
\begin{equation}
\text{Charge conservation law: } \rmi\omega\phs{\rho} + \nabla\cdot\phsvect{J}_c = 0.
\label{eq:charge-conservation-harmo}
\end{equation}
\]
This suggests us that the charge density can be actually represented by the conduction current density \(\phsvect{J}_c\) as long as \(\omega>0\), whereas \(\phsvect{J}_c\) itself can be further represented by electric field \(\phsvect{E}\) due to Ohm‘s law as below.
\[
\begin{equation}
\phsvect{J}_c = \sigma\phsvect{E} = \sigma(-\nabla\phs{\varphi}-\rmi\omega\phsvect{A}) = -\sigma\nabla\phs{\varphi} - \rmi\sigma\omega\phsvect{A}.
\label{eq:current-density-aphi}
\end{equation}
\]
Hence, \(\phs{\rho}\) can be eliminated from the system with its final representation in \(\phsvect{A}-\phs{\varphi}\),
\[
\begin{equation}
\phs{\rho} = \frac{\rmi}{\omega}\nabla\cdot(-\sigma\nabla\phs{\varphi}-\rmi\sigma\omega\phsvect{A}).
\label{eq:charge-density-aphi}
\end{equation}
\]
Substitute Equation \eqref{eq:current-density-aphi} into Equation \eqref{eq:ampere-law-aphi-harmo} and Equation \eqref{eq:charge-density-aphi} into Equation \eqref{eq:gauss-law-for-d-aphi-harmo}, we have
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\left(\frac{1}{\mu}\nabla\times\phsvect{A}\right) + (\sigma+\rmi\omega\varepsilon)\nabla\phs{\varphi} + \rmi\omega(\sigma+\rmi\omega\varepsilon)\phsvect{A} = 0 \label{eq:ampere-law-aphi-harmo-final} \cr
& \text{Gauss‘s law for $\vect{D}$: } \rmi\omega\nabla\cdot(\varepsilon\nabla\phs{\varphi}) = -\nabla\cdot(\sigma\nabla\phs{\varphi} + \rmi\sigma\omega\phsvect{A}) \label{eq:gauss-law-for-d-aphi-harmo-final} \cr
& \text{Coulomb gauge: } \nabla\cdot(\varepsilon\phsvect{A}) = 0. \label{eq:coulomb-gauge-aphi-harmo-final}
\end{align}
\]
(b) According to section 2(b), for the time-harmonic case, Ampère‘s law implies the continuity of total current in conducting domain \(\Omega_c\) and Gauss‘s law for \(\vect{D}\) in non-conducting domain \(\Omega_I\). Therefore, this property may also be applicable here for the time-harmonic \(\vect{A}-\varphi\) formulation. Apply divergence to Equation \eqref{eq:ampere-law-aphi-harmo-final} and also use the Coulomb gauge in Equation \eqref{eq:coulomb-gauge-aphi-harmo-final}, we have
\[
\begin{aligned}
& \nabla\cdot\left[ (\sigma+\rmi\omega\varepsilon) \nabla\phs{\varphi} \right] + \rmi\omega\nabla\cdot\left[ (\sigma+\rmi\omega\varepsilon)\phsvect{A} \right] = 0 \cr
& \rmi\omega\nabla\cdot(\varepsilon\nabla\phs{\varphi}) = -\nabla\cdot(\sigma\nabla\phs{\varphi}) - \nabla\cdot(\rmi\sigma\omega\phsvect{A}) + \omega^2\nabla\cdot(\varepsilon\phsvect{A}) \cr
& \rmi\omega\nabla\cdot(\varepsilon\nabla\phs{\varphi}) = -\nabla\cdot(\sigma\nabla\phs{\varphi} + \rmi\sigma\omega\phsvect{A}),
\end{aligned}
\]
which is just Equation \eqref{eq:gauss-law-for-d-aphi-harmo-final}. Therefore, Ampère‘s law implies Gauss‘s law for \(\vect{D}\) under the precondition \(\omega>0\). When \(\omega=0\), Gauss‘s law for \(\vect{D}\) should be explicitly enforced.
(c) Summarize the above, the \(\vect{A}-\varphi\) formulation for time-harmonic Maxwell equations is given as below.
When \(\omega>0\),
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\left(\frac{1}{\mu}\nabla\times\phsvect{A}\right) + (\sigma+\rmi\omega\varepsilon)\nabla\phs{\varphi} + \rmi\omega(\sigma+\rmi\omega\varepsilon)\phsvect{A} = 0 \cr
& \text{Coulomb gauge: } \nabla\cdot(\varepsilon\phsvect{A}) = 0.
\end{align}
\]
When \(\omega=0\),
\[
\begin{align}
& \text{Ampère‘s law: } \nabla\times\left(\frac{1}{\mu}\nabla\times\phsvect{A}\right) + \sigma\nabla\phs{\varphi} = 0 \cr
& \text{Gauss‘s law in $\Omega_I$ with $\phs{\rho}=0$: } -\nabla\cdot(\varepsilon\nabla\phs{\varphi}) = 0 \cr
& \text{Gauss‘s law in $\Omega_c$: } -\nabla\cdot(\sigma\nabla\phs{\varphi}) = 0 \cr
& \text{Coulomb gauge: } \nabla\cdot(\varepsilon\phsvect{A}) = 0.
\end{align}
\]
Various formulations of Maxwell equations
标签：order uniq omega ima represent tom effective can mes
原文地址：https://www.cnblogs.com/peabody/p/11620044.html
