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@@ -88,6 +88,8 @@ \section{Vector Analysis Recap} +\subsection{Partial derivatives} + \begin{definition}[Partial derivative] A vector valued function \(f: \mathbb{R}^m\to\mathbb{R}\), with \(\vec{v}\in\mathbb{R}^m\), has a partial derivative with respect to \(v_i\) @@ -117,6 +119,8 @@ To illustrate the previous theorem, in a simpler case with \(f(x,y)\), we get Beware that this is valid only if \(x\) and \(y\) are indipendent. If there is a relation \(x(y)\) or \(y(x)\) the above does not hold. +\subsection{Vector derivatives} + \begin{definition}[Gradient vector] The \emph{gradient} of a function \(f(\vec{x}), \vec{x}\in\mathbb{R}^m\) is a column vector containing the partial derivatives @@ -228,11 +232,13 @@ If there is a relation \(x(y)\) or \(y(x)\) the above does not hold. \begin{theorem}[Stokes' theorem] \[ - \int_\mathcal{S} \curl \vec{F} \dotp d\vec{s} - = \oint_{\partial\mathcal{S}} \vec{F} \dotp d\vec{r} + \int_S \curl \vec{F} \dotp d\vec{s} + = \oint_{\partial S} \vec{F} \dotp d\vec{r} \] \end{theorem} +\subsection{Second vector derivatives} + \begin{definition}[Laplacian operator] A second vector derivative is so important that it has a special name. For a scalar function \(f: \mathbb{R}^m \to \mathbb{R}\) the divergence of the @@ -330,15 +336,15 @@ Maxwell's equations in matter in their integral form are Where \(\vec{J}\) and \(\rho\) are the \emph{free current density} and \emph{free charge density} respectively. -\subsection{Isotropic linear materials and boundary conditions} +\subsection{Linear materials and boundary conditions} -Inside of so called isotropic linear materials the fields and flux (or current) -densities are proportional, i.e. +Inside of so called isotropic linear materials fluxes and current +densities are proportional and parallel to the fields, i.e. \begin{align*} - \vec{D} &= \varepsilon \vec{E}, & \vec{J} &= \sigma \vec{E}, & \vec{B} &= \mu \vec{H}. + \vec{D} &= \epsilon \vec{E}, & \vec{J} &= \sigma \vec{E}, & \vec{B} &= \mu \vec{H}. \end{align*} -Between two materials (1) and (2) the following boundary conditions must be satisfied: +Where two materials meet the following boundary conditions must be satisfied: \begin{align*} &\uvec{n} \dotp \vec{D}_1 = \uvec{n} \dotp \vec{D}_2 + \rho_s & &\uvec{n} \crossp \vec{E}_1 = \uvec{n} \crossp \vec{E}_2 \\ @@ -348,9 +354,57 @@ Between two materials (1) and (2) the following boundary conditions must be sati &\uvec{n} \crossp \vec{M}_1 = \uvec{n} \crossp \vec{M}_2 + \vec{J}_{s,m} \end{align*} -\subsection{Magnetic vector potential} +\subsection{Potentials} + +Because \(\vec{E}\) is often conservative (\(\curl \vec{E} = \vec{0}\)), and +\(\div \vec{B}\) is always zero, it is often useful to use \emph{potentials} to +describe these quantities instead. The electric scalar potential and magnetic +vector potentials are in their integral form: +\begin{align*} + \varphi &= \int_\mathsf{A}^\mathsf{B} \vec{E} \dotp d\vec{l}, & + \vec{A} &= \frac{\mu_0}{4\pi} \int_V \frac{\vec{J} dv}{R} +\end{align*} +With differential operators: +\begin{align*} + \vec{E} &= - \grad \varphi, & + \mu_0 \vec{J} &= - \vlaplacian \vec{A}. +\end{align*} +By taking the divergence on both sides of the equation with the electric field +we get \(\rho/\epsilon = - \laplacian \varphi\), which also contains the +Laplacian operator. We will study equations with of form in \S \ref{sec:poisson}. + +% \subsection{Energy density} + +\section{Laplace and Poisson's equations} \label{sec:poisson} + +The so called \emph{Poisson's equation} has the form +\[ + \laplacian \varphi = - \frac{\rho}{\epsilon}. +\] +When the right side of the equation is zero, it is also known as \emph{Laplace's +equation}. + +\subsection{Easy solutions of Laplace and Poisson's equations} -\section{Laplace and Poisson's equation} +\subsubsection{Geometry with zenithal and azimuthal symmetries (\"Ubung 2)} +Suppose we have a geometry where, using spherical coordinates, there is a +symmetry such that the solution does not depend on \(\phi\) or \(\theta\). +Then Laplace's equation reduces down to +\[ + \laplacian \varphi = \frac{1}{r^2} \partial_r ( r^2 \partial_r \varphi) = 0, +\] +which has solutions of the form +\[ + \varphi(r) = \frac{C_1}{r} + C_2. +\] + +\subsection{Geometry with azimuthal and translational symmetry (\"Ubung 3)} + +Suppose that when using cylindrical coordinates, the solution does not depend +on \(\phi\) or \(z\). Then Laplace's equation becomes +\[ + \laplacian A_z = \frac{1}{r} \partial_r (r \partial_r A_z) = 0. +\] \end{document} |