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J.C.A. Wevers } \newfont{\sfd}{cmssdc10 scaled\magstep0} \newfont{\cmu}{cmu10 scaled\magstep0} \def\jwline#1#2#3#4{% \put(#1,#2){\special{em:moveto}}% \put(#3,#4){\special{em:lineto}}} \def\newpic#1{} \def\npar{\par\medskip} %\def\vec#1{\mbox{\boldmath$#1$\unboldmath}} % Uncomment this when you don't like the arrows \def\vvec#1{\vec{#1}\,} \def\rr{\vec{r}} \def\vv{\vec{v}} \def\aaa{\vec{a}} \def\e#1{\vec{e}_{\rm #1}} \def\ee#1{\vec{e}_{#1}} \def\Q#1#2{\frac{\partial #1}{\partial #2}} \def\QQ#1#2{\frac{\partial^2 #1}{\partial #2^2}} \def\Qc#1#2#3{\left(\frac{\partial #1}{\partial #2}\right)_{#3}} \def\LL{{\cal L}} \def\RR{I\hspace{-1mm}R} \def\NN{I\hspace{-1mm}N} \def\TT{\mbox{\sfd T}} \def\DD{\mbox{\sfd D}} \def\half{\mbox{$\frac{1}{2}$}} \def\kwart{\mbox{$\frac{1}{4}$}} \def\av#1{\left\langle #1 \right\rangle} \def\oiint{\int\hspace{-2ex}\int\hspace{-3ex}\bigcirc~} \def\iint{\int\hspace{-1.5ex}\int} \def\iiint{\int\hspace{-1.5ex}\int\hspace{-1.5ex}\int} \def\dd{d\hspace{-1ex}\rule[1.25ex]{2mm}{0.4pt}} \def\lrarrow{~\lower.2ex\hbox{$\rightarrow$}\kern-2.4ex\raise.7ex\hbox{$\leftarrow$}~} \def\rlarrow{~\lower.2ex\hbox{$\leftarrow$}\kern-2.3ex\raise.7ex\hbox{$\rightarrow$}~} \def\ne{n_{\rm e}} \def\ni{n_{\rm i}} \def\no{n_{\rm 0}} \def\me{m_{\rm e}} \def\mi{m_{\rm i}} \def\Te{T_{\rm e}} \def\Ti{T_{\rm i}} \def\labelenumii{\Roman{enumii}.} \def\theenumii{\Roman{enumii}} \def\p@enumii{\theenumi} \pagestyle{fancyplain} \lhead[\fancyplain{}{\thepage}]{\fancyplain{}{\cmu Physics Formulary by ir. J.C.A. Wevers}} \rhead[\fancyplain{}{\cmu Physics Formulary by ir. J.C.A. Wevers}]{\fancyplain{}{\thepage}} \cfoot{\fancyplain{\thepage}{}} \thispagestyle{empty} \setcounter{page}{0} \hrule \rule{.4pt}{22.85cm}\hspace*{154mm}\rule{.4pt}{22.85cm} \vspace*{-18cm} \begin{center} \Huge \fbox{\fbox{Physics Formulary}} \end{center} \vspace{2cm} \centerline{\Large\bf By ir. J.C.A. Wevers} \vspace*{2cm} %\epsfysize 10cm %\centerline{\epsffile{euterpe.ps}~~~~~~} \vfill \hrule \newpage \thispagestyle{empty} \copyright~1995, 2012~~J.C.A. Wevers\hfill Version: July 8, 2012 \npar \hrule \par \bigskip Dear reader, \npar This document contains a 108 page \LaTeX\ file which contains a lot equations in physics. It is written at advanced undergraduate/postgraduate level. It is intended to be a short reference for anyone who works with physics and often needs to look up equations. \npar This, and a Dutch version of this file, can be obtained from the author, Johan Wevers\linebreak ({\tt johanw@xs4all.nl}). \npar It can also be obtained on the WWW. See {\tt http://www.xs4all.nl/\~{}johanw/index.html}, where also Postscript and PDF versions are available. \npar If you find any errors or have any comments, please let me know. I am always open for suggestions and possible corrections to the physics formulary. \npar The Physics Formulary is made with te\TeX\ and \LaTeX\ version 2.09. It can be possible that your \LaTeX\ version has problems compiling the file. The most probable source of problems would be the use of large bezier curves and/or em\TeX\ specials in pictures. If you prefer the notation in which vectors are typefaced in boldface, uncomment the redefinition of the {\tt $\backslash$vec} command in the \TeX\ file and recompile the file. \npar This work is licenced under the Creative Commons Attribution 3.0 Netherlands License. To view a copy of this licence, visit {\tt http://creativecommons.org/licenses/by/3.0/nl/} or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California 94105, USA. \npar Johan Wevers \vfill \hrule \newpage \pagenumbering{Roman} \addcontentsline{toc}{chapter}{Contents} \typeout{Contents} \tableofcontents \newpage \pagenumbering{arabic} \chapter*{\center Physical Constants} \typeout{Physical Constants} \addcontentsline{toc}{chapter}{Physical Constants} \begin{center} \begin{tabular}{||l|lll||} \hline {\bf Name}&{\bf Symbol}&{\bf Value}&{\bf Unit}\\ \hline \hline Number $\pi$ &$\pi$&3.14159265358979323846&\\ Number e &e &2.71828182845904523536&\\ Euler's constant &\multicolumn{3}{|l||}{$\gamma=\lim\limits_{n\rightarrow\infty}\left(\sum\limits_{k=1}^n 1/k-\ln(n)\right)=0.5772156649$}\\ \hline Elementary charge &$e$&$1.60217733\cdot10^{-19}$&C\rule{0pt}{13pt}\\ Gravitational constant &$G,\kappa$&$6.67259\cdot10^{-11}$&m$^3$kg$^{-1}$s$^{-2}$\\ Fine-structure constant &$\alpha=e^2/2hc\varepsilon_0$&$\approx1/137$&\\ Speed of light in vacuum &$c$&$2.99792458\cdot10^8$&m/s (def)\\ Permittivity of the vacuum &$\varepsilon_0$&$8.854187\cdot10^{-12}$&F/m\\ Permeability of the vacuum &$\mu_0$&$4\pi\cdot10^{-7}$&H/m\\ $(4\pi\varepsilon_0)^{-1}$ &&$8.9876\cdot10^9$&Nm$^2$C$^{-2}$\\ \hline Planck's constant &$h$&$6.6260755\cdot10^{-34}$&Js\rule{0pt}{13pt}\\ Dirac's constant &$\hbar=h/2\pi$&$1.0545727\cdot10^{-34}$&Js\\ Bohr magneton &$\mu_{\rm B}=e\hbar/2m_{\rm e}$&$9.2741\cdot10^{-24}$&Am$^2$\\ Bohr radius &$a_0$&$0.52918$&\AA\\ Rydberg's constant &$Ry$&13.595&eV\\ Electron Compton wavelength &$\lambda_{\rm Ce}=h/\me c$&$2.2463\cdot10^{-12}$&m\\ Proton Compton wavelength &$\lambda_{\rm Cp}=h/m_{\rm p}c$&$1.3214\cdot10^{-15}$&m\\ Reduced mass of the H-atom &$\mu_{\rm H}$&$9.1045755\cdot10^{-31}$&kg\\ \hline Stefan-Boltzmann's constant &$\sigma$&$5.67032\cdot10^{-8}$&Wm$^{-2}$K$^{-4}$\rule{0pt}{13pt}\\ Wien's constant &$k_{\rm W}$&$2.8978\cdot10^{-3}$&mK\\ \hline Molar gasconstant &$R$&8.31441&J$\cdot$mol$^{-1}\cdot$K$^{-1}$\\ Avogadro's constant &$N_{\rm A}$&$6.0221367\cdot10^{23}$&mol$^{-1}$\\ Boltzmann's constant &$k=R/N_{\rm A}$&$1.380658\cdot10^{-23}$&J/K\\ \hline Electron mass &$m_{\rm e}$&$9.1093897\cdot10^{-31}$&kg\rule{0pt}{13pt}\\ Proton mass &$m_{\rm p}$&$1.6726231\cdot10^{-27}$&kg\\ Neutron mass &$m_{\rm n}$&$1.674954\cdot10^{-27}$&kg\\ Elementary mass unit &$m_{\rm u}=\frac{1}{12}m(^{12}_{~6}$C)&$1.6605656\cdot10^{-27}$&kg\\ Nuclear magneton &$\mu_{\rm N}$&$5.0508\cdot10^{-27}$&J/T\\ \hline Diameter of the Sun &$D_\odot$&$1392\cdot10^6$&m\rule{0pt}{13pt}\\ Mass of the Sun &$M_\odot$&$1.989\cdot10^{30}$&kg\\ Rotational period of the Sun &$T_\odot$&25.38&days\\ Radius of Earth &$R_{\rm A}$&$6.378\cdot10^6$&m\\ Mass of Earth &$M_{\rm A}$&$5.976\cdot10^{24}$&kg\\ Rotational period of Earth &$T_{\rm A}$&23.96&hours\\ Earth orbital period &Tropical year&365.24219879&days\\ Astronomical unit &AU&$1.4959787066\cdot10^{11}$&m\\ Light year &lj&$9.4605\cdot10^{15}$&m\\ Parsec &pc&$3.0857\cdot10^{16}$&m\\ Hubble constant &$H$&$\approx(75\pm25)$&km$\cdot$s$^{-1}\cdot$Mpc$^{-1}$\\ \hline \end{tabular} \end{center} \renewcommand{\chaptermark}[1]{\markboth{#1}{#1}} \lhead[\fancyplain{}{\thepage}]{\fancyplain{}{\cmu Chapter \thechapter: \leftmark}} \rhead[\fancyplain{}{\cmu Physics Formulary by ir. J.C.A. Wevers}]{\fancyplain{}{\thepage}} \chapter{Mechanics} \typeout{Mechanics} \section{Point-kinetics in a fixed coordinate system} \subsection{Definitions} The position $\rr$, the velocity $\vv$ and the acceleration $\aaa$ are defined by: $\rr=(x,y,z)$, $\vv=(\dot{x},\dot{y},\dot{z})$, $\aaa=(\ddot{x},\ddot{y},\ddot{z})$. The following holds: \[ s(t)=s_0+\int|\vv(t)|dt~;~~~\rr(t)=\rr_0+\int\vv(t)dt~;~~~\vv(t)=\vv_0+\int\aaa(t)dt \] When the acceleration is constant this gives: $v(t)=v_0+at$ and $s(t)=s_0+v_0t+\half at^2$. For the unit vectors in a direction $\perp$ to the orbit $\e{t}$ and parallel to it $\e{n}$ holds: \[ \e{t}=\frac{\vv}{|\vv|}=\frac{d\rr}{ds}~~~\dot{\e{t}}= \frac{v}{\rho}\e{n}~;~~~\e{n}=\frac{\dot{\e{t}}}{|\dot{\e{t}}|} \] For the {\it curvature} $k$ and the {\it radius of curvature} $\rho$ holds: \[ \vec{k}=\frac{d\e{t}}{ds}=\frac{d^2\rr}{ds^2}=\left|\frac{d\varphi}{ds}\right| ~;~~~\rho=\frac{1}{|k|} \] \subsection{Polar coordinates} Polar coordinates are defined by: $x=r\cos(\theta)$, $y=r\sin(\theta)$. So, for the unit coordinate vectors holds: $\dot{\ee{r}}=\dot{\theta}\ee{\theta}$, $\dot{\ee{\theta}}=-\dot{\theta}\ee{r}$ \npar The velocity and the acceleration are derived from: $\rr=r\ee{r}$, $\vv=\dot{r}\ee{r}+r\dot{\theta}\ee{\theta}$, $\aaa=(\ddot{r}-r\dot{\theta}^2)\ee{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\ee{\theta}$. \section{Relative motion} For the motion of a point D w.r.t. a point Q holds: $\displaystyle\rr_{\rm D}=\rr_{\rm Q}+\frac{\vec{\omega}\times\vv_{\rm Q}}{\omega^2}$ with $\vec{\rm QD}=\rr_{\rm D}-\rr_{\rm Q}$ and $\omega=\dot{\theta}$. \npar Further holds: $\alpha=\ddot{\theta}$. $'$ means that the quantity is defined in a moving system of coordinates. In a moving system holds:\\ $\vv=\vv_{\rm Q}+\vv\,'+\vec{\omega}\times\rr\,'$ and $\aaa=\aaa_{\rm Q}+\aaa\,'+\vec{\alpha}\times\rr\,'+2\vec{\omega}\times\vv\,'+\vec{\omega}\times(\vec{\omega}\times\rr\,')$\\ with $\vec{\omega}\times(\vec{\omega}\times\rr\,')=-\omega^2\rr\,'_n$ \section{Point-dynamics in a fixed coordinate system} \subsection{Force, (angular)momentum and energy} Newton's 2nd law connects the force on an object and the resulting acceleration of the object where the {\it momentum} is given by $\vec{p}=m\vv$: \[ \vec{F}(\rr,\vv,t)=\frac{d\vec{p}}{dt}=\frac{d(m\vv\,)}{dt}=m\frac{d\vv}{dt}+ \vv\,\frac{dm}{dt}\mathop{=}\limits^{m={\rm const}}m\aaa \] Newton's 3rd law is given by: $\vec{F}_{\rm action}=-\vec{F}_{\rm reaction}$. \npar For the power $P$ holds: $P=\dot{W}=\vec{F}\cdot\vv$. For the total energy $W$, the kinetic energy $T$ and the potential energy $U$ holds: $W=T+U~;~~~\dot{T}=-\dot{U}$ with $T=\half mv^2$. \npar The {\it kick} $\vec{S}$ is given by: $\displaystyle\vec{S}=\Delta\vec{p}=\int\vec{F}dt$ \npar The work $A$, delivered by a force, is $\displaystyle A=\int\limits_1^2\vec{F}\cdot d\vec{s}=\int\limits_1^2F\cos(\alpha)ds$ \npar The torque $\vec{\tau}$ is related to the angular momentum $\vec{L}$:~~$\vec{\tau}=\dot{\vec{L}}=\rr\times\vec{F}$; and\\ $\vec{L}=\rr\times\vec{p}=m\vv\times\rr$, $|\vec{L}|=mr^2\omega$. The following equation is valid: \[ \tau=-\Q{U}{\theta} \] Hence, the conditions for a mechanical equilibrium are: $\sum\vec{F}_i=0$ and $\sum\vec{\tau}_i=0$. \npar The {\it force of friction} is usually proportional to the force perpendicular to the surface, except when the motion starts, when a threshold has to be overcome: $F_{\rm fric}=f\cdot F_{\rm norm}\cdot\e{t}$. \subsection{Conservative force fields} A conservative force can be written as the gradient of a potential: $\vec{F}_{\rm cons}=-\vec{\nabla}U$. From this follows that $\nabla\times\vec{F}=\vec{0}$. For such a force field also holds: \[ \oint\vec{F}\cdot d\vec{s}=0~\Rightarrow~U=U_0-\int\limits_{r_0}^{r_1}\vec{F}\cdot d\vec{s} \] So the work delivered by a conservative force field depends not on the trajectory covered but only on the starting and ending points of the motion. \subsection{Gravitation} The Newtonian law of gravitation is (in GRT one also uses $\kappa$ instead of $G$): \[ \vec{F}_{\rm g}=-G\frac{m_1 m_2}{r^2}\ee{r} \] The gravitational potential is then given by $V=-Gm/r$. From Gauss law it then follows: $\nabla^2 V=4\pi G\varrho$. \subsection{Orbital equations} If $V=V(r)$ one can derive from the equations of Lagrange for $\phi$ the conservation of angular momentum: \[ \Q{\LL}{\phi}=\Q{V}{\phi}=0\Rightarrow\frac{d}{dt}(mr^2\phi)=0\Rightarrow L_z=mr^2\phi=\mbox{constant} \] For the radial position as a function of time can be found that: \[ \left(\frac{dr}{dt}\right)^2=\frac{2(W-V)}{m}-\frac{L^2}{m^2r^2} \] The angular equation is then: \[ \phi-\phi_0=\int\limits_0^r\left[\frac{mr^2}{L}\sqrt{\frac{2(W-V)}{m}-\frac{L^2}{m^2r^2}}~\right]^{-1}dr \stackrel{r^{-2}{\rm field}}{=} \arccos\left(1+\frac{\frac{1}{r}-\frac{1}{r_0}}{\frac{1}{r_0}+km/L_z^2}\right) \] If $F=F(r)$: $L=$constant, if $F$ is conservative: $W=$constant, if $\vec{F}\perp\vv$ then $\Delta T=0$ and $U=0$. \subsubsection{Kepler's orbital equations} In a force field $F=kr^{-2}$, the orbits are conic sections with the origin of the force in one of the foci (Kepler's 1st law). The equation of the orbit is: \[ r(\theta)=\frac{\ell}{1+\varepsilon\cos(\theta-\theta_0)}~,~~\mbox{or:~~} x^2+y^2=(\ell-\varepsilon x)^2 \] with \[ \ell=\frac{L^2}{G\mu^2M_{\rm tot}}~;~~~\varepsilon^2=1+\frac{2WL^2}{G^2\mu^3M^2_{\rm tot}}=1-\frac{\ell}{a} ~;~~~a=\frac{\ell}{1-\varepsilon^2}=\frac{k}{2W} \] $a$ is half the length of the long axis of the elliptical orbit in case the orbit is closed. Half the length of the short axis is $b=\sqrt{a\ell}$. $\varepsilon$ is the {\it excentricity} of the orbit. Orbits with an equal $\varepsilon$ are of equal shape. Now, 5 types of orbits are possible: \begin{enumerate} \item $k<0$ and $\varepsilon=0$: a circle. \item $k<0$ and $0<\varepsilon<1$: an ellipse. \item $k<0$ and $\varepsilon=1$: a parabole. \item $k<0$ and $\varepsilon>1$: a hyperbole, curved towards the centre of force. \item $k>0$ and $\varepsilon>1$: a hyperbole, curved away from the centre of force. \end{enumerate} Other combinations are not possible: the total energy in a repulsive force field is always positive so $\varepsilon>1$. \npar If the surface between the orbit covered between $t_1$ and $t_2$ and the focus C around which the planet moves is $A(t_1,t_2)$, Kepler's 2nd law is \[ A(t_1,t_2)=\frac{L_{\rm C}}{2m}(t_2-t_1) \] Kepler's 3rd law is, with $T$ the period and $M_{\rm tot}$ the total mass of the system: \[ \frac{T^2}{a^3}=\frac{4\pi^2}{GM_{\rm tot}} \] \subsection{The virial theorem} The virial theorem for one particle is: \[ \av{m\vv\cdot\rr}=0\Rightarrow\av{T}=-\half\av{\vec{F}\cdot\rr}=\half\av{r\frac{dU}{dr}}= \half n\av{U}\mbox{ if } U=-\frac{k}{r^n} \] The virial theorem for a collection of particles is: \[ \av{T}=-\half\av{\sum\limits_{\rm particles}\vec{F}_i\cdot\rr_i+ \sum\limits_{\rm pairs}\vec{F}_{ij}\cdot\rr_{ij}} \] These propositions can also be written as: $2E_{\rm kin}+E_{\rm pot}=0$. \section{Point dynamics in a moving coordinate system} \subsection{Apparent forces} The total force in a moving coordinate system can be found by subtracting the apparent forces from the forces working in the reference frame: $\vec{F}\,'=\vec{F}-\vec{F}_{\rm app}$. The different apparent forces are given by: \begin{enumerate} \item Transformation of the origin: $F_{\rm or}=-m\aaa_a$ \item Rotation: $\vec{F}_{\alpha}=-m\vec{\alpha}\times\rr\,'$ \item Coriolis force: $F_{\rm cor}=-2m\vec{\omega}\times\vv$ \item Centrifugal force: $\vec{F}_{\rm cf}=m\omega^2\rr_n\,'=-\vec{F}_{\rm cp}$ ;~~$\displaystyle\vec{F}_{\rm cp}=-\frac{mv^2}{r}\ee{r}$ \end{enumerate} \subsection{Tensor notation} Transformation of the Newtonian equations of motion to $x^\alpha=x^\alpha(x)$ gives: \[ \frac{dx^\alpha}{dt}=\Q{x^\alpha}{\bar{x}^\beta}\frac{d\bar{x}^\beta}{dt}; \] The chain rule gives: \[ \frac{d}{dt}\frac{dx^\alpha}{dt}=\frac{d^2x^\alpha}{dt^2}=\frac{d}{dt} \left(\Q{x^\alpha}{\bar{x}^\beta}\frac{d\bar{x}^\beta}{dt}\right)= \Q{x^\alpha}{\bar{x}^\beta}\frac{d^2\bar{x}^\beta}{dt^2}+ \frac{d\bar{x}^\beta}{dt}\frac{d}{dt}\left(\Q{x^\alpha}{\bar{x}^\beta}\right) \] so: \[ \frac{d}{dt}\Q{x^\alpha}{\bar{x}^\beta}=\Q{}{\bar{x}^\gamma}\Q{x^\alpha}{\bar{x}^\beta}\frac{d\bar{x}^\gamma}{dt}= \frac{\partial^2x^\alpha}{\partial\bar{x}^\beta\partial\bar{x}^\gamma}\frac{d\bar{x}^\gamma}{dt} \] This leads to: \[ \frac{d^2x^\alpha}{dt^2}=\Q{x^\alpha}{\bar{x}^\beta}\frac{d^2\bar{x}^\beta}{dt^2}+ \frac{\partial^2x^\alpha}{\partial\bar{x}^\beta\partial\bar{x}^\gamma}\frac{d\bar{x}^\gamma}{dt} \left(\frac{d\bar{x}^\beta}{dt}\right) \] Hence the Newtonian equation of motion \[ m\frac{d^2x^\alpha}{dt^2}=F^\alpha \] will be transformed into: \[ m\left\{\frac{d^2x^\alpha}{dt^2}+\Gamma_{\beta\gamma}^\alpha \frac{dx^\beta}{dt}\frac{dx^\gamma}{dt}\right\}=F^\alpha \] The apparent forces are taken from he origin to the effect side in the way $\displaystyle\Gamma_{\beta\gamma}^\alpha\frac{dx^\beta}{dt}\frac{dx^\gamma}{dt}$. \section{Dynamics of masspoint collections} \subsection{The centre of mass} The velocity w.r.t. the centre of mass $\vec{R}$ is given by $\vv-\dot{\vec{R}}$. The coordinates of the centre of mass are given by: \[ \rr_{\rm m}=\frac{\sum m_i\rr_i}{\sum m_i} \] In a 2-particle system, the coordinates of the centre of mass are given by: \[ \vec{R}=\frac{m_1\rr_1+m_2\rr_2}{m_1+m_2} \] With $\rr=\rr_1-\rr_2$, the kinetic energy becomes: $T=\half M_{\rm tot}\dot{R}^2+\half\mu\dot{r}^2$, with the {\it reduced mass} $\mu$ given by: $\displaystyle\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$\\ The motion within and outside the centre of mass can be separated: \[ \dot{\vec{L}}_{\rm outside}=\vec{\tau}_{\rm outside}~;~~~ \dot{\vec{L}}_{\rm inside}=\vec{\tau}_{\rm inside} \] \[ \vec{p}=m\vv_{\rm m}~;~~~\vec{F}_{\rm ext}=m\aaa_{\rm m}~;~~~\vec{F}_{12}=\mu\vec{u} \] \subsection{Collisions} With collisions, where B are the coordinates of the collision and C an arbitrary other position, holds: $\vec{p}=m\vv_{\rm m}$ is constant, and $T=\half m\vv_{\rm m}^{\,2}$ is constant. The changes in the {\it relative velocities} can be derived from: $\vec{S}=\Delta\vec{p}=\mu(\vv_{\rm aft}-\vv_{\rm before})$. Further holds $\Delta\vec{L}_{\rm C}=\vec{\rm CB}\times\vec{S}$, $\vec{p}~\parallel~\vec{S}=$constant and $\vec{L}$ w.r.t. B is constant. \section{Dynamics of rigid bodies} \subsection{Moment of Inertia} The angular momentum in a moving coordinate system is given by: \[ \vec{L}'=I\vec{\omega}+\vec{L}_n' \] where $I$ is the {\it moment of inertia} with respect to a central axis, which is given by: \[ I=\sum\limits_i m_i\rr_i~^2~;~~~T'=W_{\rm rot}=\half\omega I_{ij}\vec{e}_i\vec{e}_j=\half I\omega^2 \] or, in the continuous case: \[ I=\frac{m}{V}\int{r'}^2_ndV=\int{r'}^2_ndm \] Further holds: \[ L_i=I^{ij}\omega_j~;~~~I_{ii}=I_i~;~~~I_{ij}=I_{ji}=-\sum\limits_km_kx_i'x_j' \] Steiner's theorem is: $I_{\rm w.r.t. D}=I_{\rm w.r.t. C}+m(DM)^2$ if axis C $\parallel$ axis D. \begin{center} \begin{tabular}{||l|l||l|l||} \hline {\bf Object}&$I$&{\bf Object}&$I$\\ \hline \hline \rule{0pt}{15pt}Cavern cylinder &$I=mR^2$ &Massive cylinder &$I=\half mR^2$\\ \rule{0pt}{15pt}Disc, axis in plane disc through m&$I=\kwart mR^2$ &Halter &$I=\half\mu R^2$\\ \rule{0pt}{15pt}Cavern sphere &$I=\frac{2}{3}mR^2$ &Massive sphere &$I=\frac{2}{5}mR^2$\\ \rule{0pt}{15pt}Bar, axis $\perp$ through c.o.m. &$I=\frac{1}{12}ml^2$ &Bar, axis $\perp$ through end &$I=\frac{1}{3}ml^2$\\ \rule[-5pt]{0pt}{20pt}Rectangle, axis $\perp$ plane thr. c.o.m.&$I=\frac{1}{12}m(a^2+b^2)$&Rectangle, axis $\parallel b$ thr. m &$I=ma^2$\\ \hline \end{tabular} \end{center} \subsection{Principal axes} Each rigid body has (at least) 3 principal axes which stand $\perp$ to each other. For a principal axis holds: \[ \Q{I}{\omega_x}=\Q{I}{\omega_y}=\Q{I}{\omega_z}=0~~\mbox{so}~~L'_n=0 \] The following holds: $\dot{\omega}_k=-a_{ijk}\omega_i\omega_j$ with $\displaystyle a_{ijk}=\frac{I_i-I_j}{I_k}$ if $I_1\leq I_2\leq I_3$. \subsection{Time dependence} For torque of force $\vec{\tau}$ holds: \[ \vec{\tau}\,'=I\ddot{\theta}~;~~~\frac{d''\vec{L}'}{dt}=\vec{\tau}\,'-\vec{\omega}\times\vec{L}' \] The {\it torque} $\vec{T}$ is defined by: $\vec{T}=\vec{F}\times\vec{d}$. \section{Variational Calculus, Hamilton and Lagrange mechanics} \subsection{Variational Calculus} Starting with: \[ \delta\int\limits_a^b\LL(q,\dot{q},t)dt=0\mbox{~~with~~}\delta(a)=\delta(b)=0\mbox{~~and~~} \delta\left(\frac{du}{dx}\right)=\frac{d}{dx}(\delta u) \] the equations of Lagrange can be derived: \[ \frac{d}{dt}\Q{\LL}{\dot{q}_i}=\Q{\LL}{q_i} \] When there are additional conditions applying to the variational problem $\delta J(u)=0$ of the type $K(u)=$constant, the new problem becomes: $\delta J(u)-\lambda\delta K(u)=0$. \subsection{Hamilton mechanics} The {\it Lagrangian} is given by: $\LL=\sum T(\dot{q}_i)-V(q_i)$. The {\it Hamiltonian} is given by: $H=\sum\dot{q}_ip_i-\LL$. In 2 dimensions holds: $\LL=T-U=\half m(\dot{r}^2+r^2\dot{\phi}^2)-U(r,\phi)$. \npar If the used coordinates are {\it canonical} the Hamilton equations are the equations of motion for the system: \[ \frac{dq_i}{dt}=\Q{H}{p_i}~;~~~\frac{dp_i}{dt}=-\Q{H}{q_i} \] Coordinates are canonical if the following holds: $\{q_i,q_j\}=0,~\{p_i,p_j\}=0,~\{q_i,p_j\}=\delta_{ij}$ where $\{,\}$ is the {\it Poisson bracket}: \[ \{A,B\}=\sum\limits_i\left[\Q{A}{q_i}\Q{B}{p_i}-\Q{A}{p_i}\Q{B}{q_i}\right] \] The Hamiltonian of a Harmonic oscillator is given by $H(x,p)=p^2/2m+\half m\omega^2 x^2$. With new coordinates $(\theta,I)$, obtained by the canonical transformation $x=\sqrt{2I/m\omega}\cos(\theta)$ and $p=-\sqrt{2Im\omega}\sin(\theta)$, with inverse $\theta=\arctan(-p/m\omega x)$ and $I=p^2/2m\omega+\half m\omega x^2$ it follows: $H(\theta,I)=\omega I$. \npar The Hamiltonian of a charged particle with charge $q$ in an external electromagnetic field is given by: \[ H=\frac{1}{2m}\left(\vec{p}-q\vvec{A}\right)^2+qV \] This Hamiltonian can be derived from the Hamiltonian of a free particle $H=p^2/2m$ with the transformations $\vec{p}\rightarrow\vec{p}-q\vec{A}$ and $H\rightarrow H-qV$. This is elegant from a relativistic point of view: this is equivalent to the transformation of the momentum 4-vector $p^\alpha\rightarrow p^\alpha-qA^\alpha$. A gauge transformation on the potentials $A^\alpha$ corresponds with a canonical transformation, which make the Hamilton equations the equations of motion for the system. \subsection{Motion around an equilibrium, linearization} For natural systems around equilibrium the following equations are valid: \[ \left(\Q{V}{q_i}\right)_0=0~;~~~V(q)=V(0)+V_{ik}q_iq_k\mbox{~~with~~} V_{ik}=\left(\frac{\partial^2V}{\partial q_i\partial q_k}\right)_0 \] With $T=\half(M_{ik}\dot{q}_i\dot{q}_k)$ one receives the set of equations $M\ddot{q}+Vq=0$. If $q_i(t)=a_i\exp(i\omega t)$ is substituted, this set of equations has solutions if ${\rm det}(V-\omega^2 M)=0$. This leads to the eigenfrequencies of the problem: $\displaystyle\omega^2_k=\frac{a_k^{\rm T}Va_k}{a_k^{\rm T}Ma_k}$. If the equilibrium is stable holds: $\forall k$ that $\omega^2_k>0$. The general solution is a superposition if eigenvibrations. \subsection{Phase space, Liouville's equation} In phase space holds: \[ \nabla=\left(\sum_i\Q{}{q_i},\sum_i\Q{}{p_i}\right)\mbox{~~so~~} \nabla\cdot\vv=\sum_i\left(\Q{}{q_i}\Q{H}{p_i}-\Q{}{p_i}\Q{H}{q_i}\right) \] If the equation of continuity, $\partial_t\varrho+\nabla\cdot(\varrho\vv\,)=0$ holds, this can be written as: \[ \{\varrho,H\}+\Q{\varrho}{t}=0 \] For an arbitrary quantity $A$ holds: \[ \frac{dA}{dt}=\{A,H\}+\Q{A}{t} \] Liouville's theorem can than be written as: \[ \frac{d\varrho}{dt}=0~;~~~\mbox{or:~}\int pdq=\mbox{constant} \] \subsection{Generating functions} Starting with the coordinate transformation: \[ \left\{\begin{array}{l} Q_i=Q_i(q_i,p_i,t)\\ P_i=P_i(q_i,p_i,t) \end{array}\right. \] one can derive the following Hamilton equations with the new Hamiltonian $K$: \[ \frac{dQ_i}{dt}=\Q{K}{P_i}~;~~~\frac{dP_i}{dt}=-\Q{K}{Q_i} \] Now, a distinction between 4 cases can be made: \begin{enumerate} \item If $\displaystyle p_i\dot{q}_i-H=P_iQ_i-K(P_i,Q_i,t)-\frac{dF_1(q_i,Q_i,t)}{dt}$, the coordinates follow from: \[ p_i=\Q{F_1}{q_i}~;~~~P_i=-\Q{F_1}{Q_i}~;~~~K=H+\Q{F_1}{t} \] \item If $\displaystyle p_i\dot{q}_i-H=-\dot{P}_iQ_i-K(P_i,Q_i,t)+\frac{dF_2(q_i,P_i,t)}{dt}$, the coordinates follow from: \[ p_i=\Q{F_2}{q_i}~;~~~Q_i=\Q{F_2}{P_i}~;~~~K=H+\Q{F_2}{t} \] \item If $\displaystyle-\dot{p}_iq_i-H=P_i\dot{Q}_i-K(P_i,Q_i,t)+\frac{dF_3(p_i,Q_i,t)}{dt}$, the coordinates follow from: \[ q_i=-\Q{F_3}{p_i}~;~~~P_i=-\Q{F_3}{Q_i}~;~~~K=H+\Q{F_3}{t} \] \item If $\displaystyle-\dot{p}_iq_i-H=-P_iQ_i-K(P_i,Q_i,t)+\frac{dF_4(p_i,P_i,t)}{dt}$, the coordinates follow from: \[ q_i=-\Q{F_4}{p_i}~;~~~Q_i=\Q{F_4}{P_i}~;~~~K=H+\Q{F_4}{t} \] \end{enumerate} The functions $F_1$, $F_2$, $F_3$ and $F_4$ are called {\it generating functions}. \chapter{Electricity \& Magnetism} \typeout{Electricity & Magnetism} \section{The Maxwell equations} The classical electromagnetic field can be described by the {\it Maxwell equations}. Those can be written both as differential and integral equations: \[ \begin{array}{ll} \displaystyle\oiint(\vec{D}\cdot\vec{n}\,)d^2A=Q_{\rm free,included}~~~~~~~~~~~~~ &\displaystyle\nabla\cdot\vec{D}=\rho_{\rm free}\\ \displaystyle\oiint(\vec{B}\cdot\vec{n}\,)d^2A=0 &\displaystyle\nabla\cdot\vec{B}=0\\ \displaystyle\oint\vec{E}\cdot d\vec{s}=-\frac{d\Phi}{dt} &\displaystyle\nabla\times\vec{E}=-\Q{\vec{B}}{t}\\ \displaystyle\oint\vec{H}\cdot d\vec{s}=I_{\rm free,included}+\frac{d\Psi}{dt} &\displaystyle\nabla\times\vec{H}=\vec{J}_{\rm free}+\Q{\vec{D}}{t} \end{array} \] For the fluxes holds: $\displaystyle \Psi=\iint(\vec{D}\cdot\vec{n}\,)d^2A$, $\displaystyle\Phi=\iint(\vec{B}\cdot\vec{n}\,)d^2A$. \npar The electric displacement $\vec{D}$, polarization $\vec{P}$ and electric field strength $\vec{E}$ depend on each other according to: \npar $\vec{D}=\varepsilon_0\vec{E}+\vec{P}=\varepsilon_0\varepsilon_{\rm r}\vec{E}$,~~ $\vec{P}=\sum\vec{p}_0/{\rm Vol}$, $\varepsilon_{\rm r}=1+\chi_{\rm e}$, with $\displaystyle\chi_{\rm e}=\frac{np_0^2}{3\varepsilon_0kT}$ \npar The magnetic field strength $\vec{H}$, the magnetization $\vec{M}$ and the magnetic flux density $\vec{B}$ depend on each other according to: \npar $\vec{B}=\mu_0(\vec{H}+\vec{M})=\mu_0\mu_{\rm r}\vec{H}$,~~ $\vec{M}=\sum\vec{m}/{\rm Vol}$, $\mu_{\rm r}=1+\chi_{\rm m}$, with $\displaystyle\chi_{\rm m}=\frac{\mu_0nm_0^2}{3kT}$ \section{Force and potential} The force and the electric field between 2 point charges are given by: \[ \vec{F}_{12}=\frac{Q_1Q_2}{4\pi\varepsilon_0\varepsilon_{\rm r}r^2}\ee{r} ~;~~~\vec{E}=\frac{\vec{F}}{Q} \] The Lorentzforce is the force which is felt by a charged particle that moves through a magnetic field. The origin of this force is a relativistic transformation of the Coulomb force: $\vec{F}_{\rm L}=Q(\vv\times\vvec{B})=l(\vec{I}\times\vvec{B})$. \npar The magnetic field in point $P$ which results from an electric current is given by the {\it law of Biot-Savart}, also known as the law of Laplace. In here, $d\vec{l}\parallel\vec{I}$ and $\rr$ points from $d\vec{l}$ to $P$: \[ d\vec{B}_P=\frac{\mu_0I}{4\pi r^2}d\vec{l}\times\ee{r} \] If the current is time-dependent one has to take {\it retardation} into account: the substitution $I(t)\rightarrow I(t-r/c)$ has to be applied. \npar The potentials are given by: $\displaystyle V_{12}=-\int\limits_1^2\vec{E}\cdot d\vec{s}$ and $\vec{A}=\half\vec{B}\times\rr$. \npar Here, the freedom remains to apply a {\it gauge transformation}. The fields can be derived from the potentials as follows: \[ \vec{E}=-\nabla V-\Q{\vec{A}}{t}~,~~~\vec{B}=\nabla\times\vec{A} \] Further holds the relation: $c^2\vec{B}=\vv\times\vec{E}$. \section{Gauge transformations} The potentials of the electromagnetic fields transform as follows when a gauge transformation is applied: \[ \left\{\begin{array}{l} \vec{A}'=\vec{A}-\nabla f\\ \displaystyle V'=V+\Q{f}{t} \end{array}\right. \] so the fields $\vec{E}$ and $\vec{B}$ do not change. This results in a canonical transformation of the Hamiltonian. Further, the freedom remains to apply a limiting condition. Two common choices are: \begin{enumerate} \item Lorentz-gauge: $\displaystyle\nabla\cdot\vec{A}+\frac{1}{c^2}\Q{V}{t}=0$. This separates the differential equations for $\vec{A}$ and $V$: $\displaystyle\Box V=-\frac{\rho}{\varepsilon_0}$, $\Box\vec{A}=-\mu_0\vec{J}$. \item Coulomb gauge: $\nabla\cdot\vec{A}=0$. If $\rho=0$ and $\vec{J}=0$ holds $V=0$ and follows $\vec{A}$ from $\Box\vec{A}=0$. \end{enumerate} \section{Energy of the electromagnetic field} The energy density of the electromagnetic field is: \[ \frac{dW}{d{\rm Vol}}=w=\int HdB+\int EdD \] The energy density can be expressed in the potentials and currents as follows: \[ w_{\rm mag}=\half\int\vec{J}\cdot\vec{A}\,d^3x~~,~~w_{\rm el}=\half\int\rho Vd^3x \] \section{Electromagnetic waves} \subsection{Electromagnetic waves in vacuum} The wave equation $\Box\Psi(\rr,t)=-f(\rr,t)$ has the general solution, with $c=(\varepsilon_0\mu_0)^{-1/2}$: \[ \Psi(\rr,t)=\int\frac{f(\rr,t-|\rr-\rr\,'|/c)}{4\pi|\rr-\rr\,'|}d^3r' \] If this is written as: $\vec{J}(\rr,t)=\vec{J}(\rr\,)\exp(-i\omega t)$ and $\vec{A}(\rr,t)=\vec{A}(\rr\,)\exp(-i\omega t)$ with: \[ \vec{A}(\rr\,)=\frac{\mu}{4\pi}\int\vec{J}(\rr\,')\frac{\exp(ik|\rr-\rr\,'|)}{|\rr-\rr\,'|}d^3\rr\,'~~,~~~ V(\rr\,)=\frac{1}{4\pi\varepsilon}\int\rho(\rr\,')\frac{\exp(ik|\rr-\rr\,'|)}{|\rr-\rr\,'|}d^3\rr\,' \] A derivation via multipole expansion will show that for the radiated energy holds, if $d,\lambda\gg r$: \[ \frac{dP}{d\Omega}=\frac{k^2}{32\pi^2\varepsilon_0c}\left|\int J_\perp(\rr\,'){\rm e}^{i\vec{k}\cdot\rr}d^3r'\right|^2 \] The energy density of the electromagnetic wave of a vibrating dipole at a large distance is: \[ w=\varepsilon_0E^2=\frac{p^2_0\sin^2(\theta)\omega^4}{16\pi^2\varepsilon_0r^2c^4}\sin^2(kr-\omega t)~,~~~ \av{w}_t=\frac{p^2_0\sin^2(\theta)\omega^4}{32\pi^2\varepsilon_0r^2c^4}~,~~ P=\frac{ck^4|\vvec{p}|^2}{12\pi\varepsilon_0} \] The radiated energy can be derived from the {\it Poynting vector} $\vec{S}$: $\vec{S}=\vec{E}\times\vec{H}=cW\vec{e}_v$. The {\it irradiance} is the time-averaged of the Poynting vector: $I=\langle|\vvec{S}|\rangle_t$. The radiation pressure $p_{\rm s}$ is given by $p_{\rm s}=(1+R)|\vvec{S}|/c$, where $R$ is the coefficient of reflection. \subsection{Electromagnetic waves in matter} The wave equations in matter, with $c_{\rm mat}=(\varepsilon\mu)^{-1/2}$ the lightspeed in matter, are: \[ \left(\nabla^2-\varepsilon\mu\QQ{}{t}-\frac{\mu}{\rho}\Q{}{t}\right)\vec{E}=0~,~~ \left(\nabla^2-\varepsilon\mu\QQ{}{t}-\frac{\mu}{\rho}\Q{}{t}\right)\vec{B}=0 \] give, after substitution of monochromatic plane waves: $\vec{E}=E\exp(i(\vec{k}\cdot\rr-\omega t))$ and $\vec{B}=B\exp(i(\vec{k}\cdot\rr-\omega t))$ the dispersion relation: \[ k^2=\varepsilon\mu\omega^2+\frac{i\mu\omega}{\rho} \] The first term arises from the displacement current, the second from the conductance current. If $k$ is written in the form $k:=k'+ik''$ it follows that: \[ k'=\omega\sqrt{\half\varepsilon\mu}\sqrt{1+\sqrt{1+\frac{1}{(\rho\varepsilon\omega)^2}}}~~~\mbox{and}~~~ k''=\omega\sqrt{\half\varepsilon\mu}\sqrt{-1+\sqrt{1+\frac{1}{(\rho\varepsilon\omega)^2}}} \] This results in a damped wave: $\vec{E}=E\exp(-k''\vec{n}\cdot\rr\,)\exp(i(k'\vec{n}\cdot\rr-\omega t))$. If the material is a good conductor, the wave vanishes after approximately one wavelength, $\displaystyle k=(1+i)\sqrt{\frac{\mu\omega}{2\rho}}$. \section{Multipoles} Because $\displaystyle \frac{1}{|\rr-\rr\,'|}=\frac{1}{r}\sum_0^\infty\left(\frac{r'}{r}\right)^lP_l(\cos\theta)$ the potential can be written as: $\displaystyle V=\frac{Q}{4\pi\varepsilon}\sum_n\frac{k_n}{r^n}$ \npar For the lowest-order terms this results in: \begin{itemize} \item Monopole: $l=0$, $k_0=\int\rho dV$ \item Dipole: $l=1$, $k_1=\int r\cos(\theta)\rho dV$ \item Quadrupole: $l=2$, $k_2=\half\sum\limits_i(3z^2_i-r^2_i)$ \end{itemize} \begin{enumerate} \item The electric dipole: dipole moment: $\vec{p}=Ql\e{}$, where $\e{}$ goes from $\oplus$ to $\ominus$, and $\vec{F}=(\vec{p}\cdot\nabla)\vec{E}_{\rm ext}$, and $W=-\vec{p}\cdot\vec{E}_{\rm out}$.\\ Electric field: $\displaystyle \vec{E}\approx\frac{Q}{4\pi\varepsilon r^3}\left(\frac{3\vec{p}\cdot\rr}{r^2}-\vec{p}\right)$.~~ The torque is: $\vec{\tau}=\vec{p}\times\vec{E}_{\rm out}$ \item The magnetic dipole: dipole moment: if $r\gg\sqrt{A}$: $\vec{\mu}=\vec{I}\times(A\ee{\perp})$, $\vec{F}=(\vec{\mu}\cdot\nabla)\vec{B}_{\rm out}$\\ $\displaystyle|\mu|=\frac{mv^2_\perp}{2B}$, $W=-\vec{\mu}\times\vec{B}_{\rm out}$\\ Magnetic field: $\displaystyle\vec{B}=\frac{-\mu}{4\pi r^3}\left(\frac{3\mu\cdot\rr}{r^2}-\vec{\mu}\right)$.~~ The moment is: $\vec{\tau}=\vec{\mu}\times\vec{B}_{\rm out}$ \end{enumerate} \section{Electric currents} The continuity equation for charge is: $\displaystyle\Q{\rho}{t}+\nabla\cdot\vec{J}=0$. The {\it electric current} is given by: \[ I=\frac{dQ}{dt}=\iint(\vec{J}\cdot\vvec{n})d^2A \] For most conductors holds: $\vec{J}=\vec{E}/\rho$, where $\rho$ is the {\it resistivity}. \npar If the flux enclosed by a conductor changes this results in an {\it induced voltage} $\displaystyle V_{\rm ind}=-N\frac{d\Phi}{dt}$. If the current flowing through a conductor changes, this results in a self-inductance which opposes the original change: $\displaystyle V_{\rm selfind}=-L\frac{dI}{dt}$. If a conductor encloses a flux $\Phi$ holds: $\Phi=LI$. \npar The magnetic induction within a coil is approximated by: $\displaystyle B=\frac{\mu NI}{\sqrt{l^2+4R^2}}$ where $l$ is the length, $R$ the radius and $N$ the number of coils. The energy contained within a coil is given by $W=\half LI^2$ and $L=\mu N^2A/l$. \npar The {\it capacity} is defined by: $C=Q/V$. For a capacitor holds: $C=\varepsilon_0\varepsilon_{\rm r}A/d$ where $d$ is the distance between the plates and $A$ the surface of one plate. The electric field strength between the plates is $E=\sigma/\varepsilon_0=Q/\varepsilon_0A$ where $\sigma$ is the surface charge. The accumulated energy is given by $W=\half CV^2$. The current through a capacity is given by $\displaystyle I=-C\frac{dV}{dt}$. \npar For most PTC resistors holds approximately: $R=R_0(1+\alpha T)$, where $R_0=\rho l/A$. For a NTC holds: $R(T)=C\exp(-B/T)$ where $B$ and $C$ depend only on the material. \npar If a current flows through two different, connecting conductors $x$ and $y$, the contact area will heat up or cool down, depending on the direction of the current: the {\it Peltier effect}. The generated or removed heat is given by: $W=\Pi_{xy}It$. This effect can be amplified with semiconductors. \npar The {\it thermic voltage} between 2 metals is given by: $V=\gamma(T-T_0)$. For a Cu-Konstantane connection holds: $\gamma\approx0.2-0.7$ mV/K. \npar In an electrical net with only stationary currents, {\it Kirchhoff's} equations apply: for a knot holds: $\sum I_n=0$, along a closed path holds: $\sum V_n=\sum I_nR_n=0$. \section{Depolarizing field} If a dielectric material is placed in an electric or magnetic field, the field strength within and outside the material will change because the material will be polarized or magnetized. If the medium has an ellipsoidal shape and one of the principal axes is parallel with the external field $\vec{E}_0$ or $\vec{B}_0$ then the depolarizing is field homogeneous. \begin{eqnarray*} \vec{E}_{\rm dep}=\vec{E}_{\rm mat}-\vec{E}_0=-\frac{{\cal N}\vec{P}}{\varepsilon_0}\\ \vec{H}_{\rm dep}=\vec{H}_{\rm mat}-\vec{H}_0=-{\cal N}\vec{M} \end{eqnarray*} $\cal N$ is a constant depending only on the shape of the object placed in the field, with $0\leq{\cal N}\leq1$. For a few limiting cases of an ellipsoid holds: a thin plane: ${\cal N}=1$, a long, thin bar: ${\cal N}=0$, a sphere: ${\cal N}=\frac{1}{3}$. \section{Mixtures of materials} The average electric displacement in a material which is inhomogenious on a mesoscopic scale is given by: $\av{D}=\av{\varepsilon E}=\varepsilon^*\av{E}$ where $\displaystyle \varepsilon^*=\varepsilon_1\left(1-\frac{\phi_2(1-x)}{\Phi(\varepsilon^*/\varepsilon_2)}\right)^{-1}$ where $x=\varepsilon_1/\varepsilon_2$. For a sphere holds: $\Phi=\frac{1}{3}+\frac{2}{3}x$. Further holds: \[ \left(\sum_i \frac{\phi_i}{\varepsilon_i}\right)^{-1}\leq\varepsilon^*\leq\sum_i \phi_i\varepsilon_i \] \chapter{Relativity} \typeout{Relativity} \section{Special relativity} \subsection{The Lorentz transformation} The Lorentz transformation $(\vvec{x}',t')=(\vvec{x}'(\vec{x},t),t'(\vec{x},t))$ leaves the wave equation invariant if $c$ is invariant: \[ \QQ{}{x}+\QQ{}{y}+\QQ{}{z}-\frac{1}{c^2}\QQ{}{t}= \QQ{}{x'}+\QQ{}{y'}+\QQ{}{z'}-\frac{1}{c^2}\QQ{}{t'} \] This transformation can also be found when $ds^2=ds'^2$ is demanded. The general form of the Lorentz transformation is given by: \[ \vvec{x}'=\vec{x}+\frac{(\gamma-1)(\vec{x}\cdot\vv\,)\vv}{|v|^2}-\gamma\vv t~~,~~ t'=\gamma\left(t-\frac{\vec{x}\cdot\vv}{c^2}\right) \] where \[ \gamma=\frac{1}{\sqrt{1-\frac{\displaystyle v^2}{\displaystyle c^2}}} \] The velocity difference $\vv\,'$ between two observers transforms according to: \[ \vv\,'=\left(\gamma\left(1-\frac{\vv_1\cdot\vv_2}{c^2}\right)\right)^{-1} \left(\vv_2+(\gamma-1)\frac{\vv_1\cdot\vv_2}{v_1^2}\vv_1-\gamma\vv_1\right) \] If the velocity is parallel to the $x$-axis, this becomes $y'=y$, $z'=z$ and: \begin{eqnarray*} &&x'=\gamma(x-vt)~,~~~x=\gamma(x'+vt')\\ &&t'=\gamma\left(t-\displaystyle\frac{xv}{c^2}\right)~,~~~t=\gamma\left(t'+\displaystyle\frac{x'v}{c^2}\right)~,~~~ v'=\frac{\displaystyle v_2-v_1}{\displaystyle 1-\frac{v_1v_2}{c^2}} \end{eqnarray*} If $\vv=v\vec{e}_x$ holds: \[ p'_x=\gamma\left(p_x-\frac{\beta W}{c}\right)~,~~~W'=\gamma(W-vp_x) \] With $\beta=v/c$ the electric field of a moving charge is given by: \[ \vec{E}=\frac{Q}{4\pi\varepsilon_0r^2}\frac{(1-\beta^2)\ee{r}}{(1-\beta^2\sin^2(\theta))^{3/2}} \] The electromagnetic field transforms according to: \[ \vec{E}'=\gamma(\vec{E}+\vv\times\vvec{B})~~,~~~ \vec{B}'=\gamma\left(\vec{B}-\frac{\vv\times\vec{E}}{c^2}\right) \] Length, mass and time transform according to: $\Delta t_{\rm r}=\gamma\Delta t_{\rm 0}$, $m_{\rm r}=\gamma m_0$, $l_{\rm r}=l_0/\gamma$, with $_{\rm 0}$ the quantities in a co-moving reference frame and $_{\rm r}$ the quantities in a frame moving with velocity $v$ w.r.t.\ it. The proper time $\tau$ is defined as: $d\tau^2=ds^2/c^2$, so $\Delta\tau=\Delta t/\gamma$. For energy and momentum holds: $W=m_{\rm r}c^2=\gamma W_0$, $W^2=m_0^2c^4+p^2c^2$. $p=m_{\rm r}v=\gamma m_0v=Wv/c^2$, and $pc=W\beta$ where $\beta=v/c$. The {\it force} is {\it defined} by $\vec{F}=d\vec{p}/dt$. \npar 4-vectors have the property that their modulus is independent of the observer: their components {\it can} change after a coordinate transformation but not their modulus. The difference of two 4-vectors transforms also as a 4-vector. The 4-vector for the velocity is given by $\displaystyle U^\alpha=\frac{dx^\alpha}{d\tau}$. The relation with the ``common'' velocity $u^i:=dx^i/dt$ is: $U^\alpha=(\gamma u^i,ic\gamma)$. For particles with nonzero restmass holds: $U^\alpha U_\alpha=-c^2$, for particles with zero restmass (so with $v=c$) holds: $U^\alpha U_\alpha=0$. The 4-vector for energy and momentum is given by: $p^\alpha=m_0U^\alpha=(\gamma p^i,iW/c)$. So: $p_\alpha p^\alpha=-m_0^2c^2=p^2-W^2/c^2$. \subsection{Red and blue shift} There are three causes of red and blue shifts: \begin{enumerate} \item Motion: with $\vec{e}_v\cdot\vec{e}_r=\cos(\varphi)$ follows: $\displaystyle \frac{f'}{f}=\gamma\left(1-\frac{v\cos(\varphi)}{c}\right)$.\\ This can give both red- and blueshift, also $\perp$ to the direction of motion. \item Gravitational redshift: $\displaystyle\frac{\Delta f}{f}=\frac{\kappa M}{rc^2}$. \item Redshift because the universe expands, resulting in e.g.~the cosmic background radiation:\\ $\displaystyle\frac{\lambda_0}{\lambda_1}=\frac{R_0}{R_1}$. \end{enumerate} \subsection{The stress-energy tensor and the field tensor} The stress-energy tensor is given by: \[ T_{\mu\nu}=(\varrho c^2+p)u_\mu u_\nu+pg_{\mu\nu}+\frac{1}{c^2} \left(F_{\mu\alpha}F^\alpha_\nu+\kwart g_{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}\right) \] The conservation laws can than be written as: $\nabla_\nu T^{\mu\nu}=0$. The electromagnetic field tensor is given by: \[ \mbox{\fbox{$\displaystyle F_{\alpha\beta}=\Q{A_\beta}{x^\alpha}-\Q{A_\alpha}{x^\beta}$}} \] with $A_\mu:=(\vec{A},iV/c)$ and $J_\mu:=(\vec{J},ic\rho)$. The Maxwell equations can than be written as: \[ \partial_\nu F^{\mu\nu}=\mu_0J^\mu~,~~ \partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0 \] The equations of motion for a charged particle in an EM field become with the field tensor: \[ \mbox{\fbox{$\displaystyle\frac{dp_\alpha}{d\tau}=qF_{\alpha\beta}u^\beta$}} \] \section{General relativity} \subsection{Riemannian geometry, the Einstein tensor} The basic principles of general relativity are: \begin{enumerate} \item The geodesic postulate: free falling particles move along geodesics of space-time with the proper time $\tau$ or arc length $s$ as parameter. For particles with zero rest mass (photons), the use of a free parameter is required because for them holds $ds=0$. From $\delta\int ds=0$ the equations of motion can be derived: \[ \frac{d^2x^\alpha}{ds^2}+\Gamma_{\beta\gamma}^{\alpha}\frac{dx^\beta}{ds}\frac{dx^\gamma}{ds}=0 \] \item The {\it principle of equivalence}: inertial mass $\equiv$ gravitational mass $\Rightarrow$ gravitation is equivalent with a curved space-time were particles move along geodesics. \item By a proper choice of the coordinate system it is possible to make the metric locally flat in each point $x_i$: $g_{\alpha\beta}(x_i)=\eta_{\alpha\beta}:=$diag$(-1,1,1,1)$. \end{enumerate} \npar The {\it Riemann tensor} is defined as: $R^\mu_{\nu\alpha\beta}T^\nu:=\nabla_\alpha\nabla_\beta T^\mu-\nabla_\beta\nabla_\alpha T^\mu$, where the covariant derivative is given by $\nabla_j a^i=\partial_ja^i+\Gamma_{jk}^ia^k$ and $\nabla_j a_i=\partial_ja_i-\Gamma_{ij}^ka_k$. Here, \[ \Gamma_{jk}^i=\frac{g^{il}}{2}\left(\frac{\partial g_{lj}}{\partial x^k}+ \frac{\partial g_{lk}}{\partial x^j}-\frac{\partial g_{_jk}}{\partial x^l}\right), \mbox{ for Euclidean spaces this reduces to: } \Gamma_{jk}^i=\frac{\partial^2\bar{x}^l}{\partial x^j\partial x^k}\Q{x^i}{\bar{x}^l}, \] are the {\it Christoffel symbols}. For a second-order tensor holds: $[\nabla_\alpha,\nabla_\beta]T_\nu^\mu=R_{\sigma\alpha\beta}^\mu T^\sigma_\nu+R^\sigma_{\nu\alpha\beta}T^\mu_\sigma$, $\nabla_k a^i_j=\partial_ka^i_j-\Gamma_{kj}^la_l^i+\Gamma_{kl}^ia_j^l$, $\nabla_k a_{ij}=\partial_ka_{ij}-\Gamma_{ki}^la_{lj}-\Gamma_{kj}^la_{jl}$ and $\nabla_k a^{ij}=\partial_ka^{ij}+\Gamma_{kl}^ia^{lj}+\Gamma_{kl}^ja^{il}$. The following holds: $R_{\beta\mu\nu}^\alpha=\partial_\mu\Gamma_{\beta\nu}^\alpha-\partial_\nu\Gamma_{\beta\mu}^\alpha+ \Gamma_{\sigma\mu}^\alpha\Gamma_{\beta\nu}^\sigma-\Gamma_{\sigma\nu}^\alpha\Gamma_{\beta\mu}^\sigma$. \npar The {\it Ricci tensor} is a contraction of the Riemann tensor: $R_{\alpha\beta}:=R^\mu_{\alpha\mu\beta}$, which is symmetric: $R_{\alpha\beta}=R_{\beta\alpha}$. The {\it Bianchi identities} are: $\nabla_\lambda R_{\alpha\beta\mu\nu}+\nabla_\nu R_{\alpha\beta\lambda\mu}+ \nabla_\mu R_{\alpha\beta\nu\lambda}=0$. \npar The {\it Einstein tensor} is given by: $G^{\alpha\beta}:=R^{\alpha\beta}-\half g^{\alpha\beta}R$, where $R:=R_\alpha^\alpha$ is the {\it Ricci scalar}, for which holds: $\nabla_\beta G_{\alpha\beta}=0$. With the variational principle $\delta\int(\LL(g_{\mu\nu})-Rc^2/16\pi\kappa)\sqrt{|g|}d^4x=0$ for variations $g_{\mu\nu}\rightarrow g_{\mu\nu}+\delta g_{\mu\nu}$ the {\it Einstein field equations} can be derived: \[ \mbox{\fbox{$\displaystyle G_{\alpha\beta}=\frac{8\pi\kappa}{c^2}T_{\alpha\beta} $}} ~~~\mbox{, which can also be written as}~~~ R_{\alpha\beta}=\frac{8\pi\kappa}{c^2}(T_{\alpha\beta}-\half g_{\alpha\beta}T^{\mu}_{\mu}) \] For empty space this is equivalent to $R_{\alpha\beta}=0$. The equation $R_{\alpha\beta\mu\nu}=0$ has as only solution a flat space. \npar The Einstein equations are 10 independent equations, which are of second order in $g_{\mu\nu}$. From this, the Laplace equation from Newtonian gravitation can be derived by stating: $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, where $|h|\ll1$. In the stationary case, this results in $\nabla^2 h_{00}=8\pi\kappa\varrho/c^2$. \npar The most general form of the field equations is: $\displaystyle R_{\alpha\beta}-\half g_{\alpha\beta}R+\Lambda g_{\alpha\beta}=\frac{8\pi\kappa}{c^2}T_{\alpha\beta}$ \npar where $\Lambda$ is the {\it cosmological constant}. This constant plays a role in inflatory models of the universe. \subsection{The line element} The {\it metric tensor} in an Euclidean space is given by: $\displaystyle g_{ij}=\sum_k\Q{\bar{x}^k}{x^i}\Q{\bar{x}^k}{x^j}$. \npar In general holds: $ds^2=g_{\mu\nu}dx^\mu dx^\nu$. In special relativity this becomes $ds^2=-c^2dt^2+dx^2+dy^2+dz^2$. This metric, $\eta_{\mu\nu}:=$diag$(-1,1,1,1)$, is called the {\it Minkowski metric}. \npar The {\it external Schwarzschild metric} applies in vacuum outside a spherical mass distribution, and is given by: \[ ds^2=\left(-1+\frac{2m}{r}\right)c^2dt^2+\left(1-\frac{2m}{r}\right)^{-1}dr^2+r^2d\Omega^2 \] Here, $m:=M\kappa/c^2$ is the {\it geometrical mass} of an object with mass $M$, and $d\Omega^2=d\theta^2+\sin^2\theta d\varphi^2$. This metric is singular for $r=2m=2\kappa M/c^2$. If an object is smaller than its event horizon $2m$, that implies that its escape velocity is $>c$, it is called a {\it black hole}. The Newtonian limit of this metric is given by: \[ ds^2=-(1+2V)c^2dt^2+(1-2V)(dx^2+dy^2+dz^2) \] where $V=-\kappa M/r$ is the Newtonian gravitation potential. In general relativity, the components of $g_{\mu\nu}$ are associated with the potentials and the derivatives of $g_{\mu\nu}$ with the field strength. \npar The Kruskal-Szekeres coordinates are used to solve certain problems with the Schwarzschild metric near $r=2m$. They are defined by: \begin{itemize} \item $r>2m$: \[ \left\{\begin{array}{ccl} u&=&\displaystyle{\sqrt{\frac{r}{2m}-1}\exp\left(\frac{r}{4m}\right)\cosh\left(\frac{t}{4m}\right)}\\ &&\\ v&=&\displaystyle{\sqrt{\frac{r}{2m}-1}\exp\left(\frac{r}{4m}\right)\sinh\left(\frac{t}{4m}\right)} \end{array}\right. \] \item $r<2m$: \[ \left\{\begin{array}{ccl} u&=&\displaystyle{\sqrt{1-\frac{r}{2m}}\exp\left(\frac{r}{4m}\right)\sinh\left(\frac{t}{4m}\right)}\\ &&\\ v&=&\displaystyle{\sqrt{1-\frac{r}{2m}}\exp\left(\frac{r}{4m}\right)\cosh\left(\frac{t}{4m}\right)} \end{array}\right. \] \item $r=2m$: here, the Kruskal coordinates are singular, which is necessary to eliminate the coordinate singularity there. \end{itemize} The line element in these coordinates is given by: \[ ds^2=-\frac{32m^3}{r}{\rm e}^{-r/2m}(dv^2-du^2)+r^2d\Omega^2 \] The line $r=2m$ corresponds to $u=v=0$, the limit $x^0\rightarrow\infty$ with $u=v$ and $x^0\rightarrow-\infty$ with $u=-v$. The Kruskal coordinates are only singular on the hyperbole $v^2-u^2=1$, this corresponds with $r=0$. On the line $dv=\pm du$ holds $d\theta=d\varphi=ds=0$. \npar For the metric outside a rotating, charged spherical mass the Newman metric applies: \begin{eqnarray*} ds^2&=&\left(1-\frac{2mr-e^2}{r^2+a^2\cos^2\theta}\right)c^2dt^2- \left(\frac{r^2+a^2\cos^2\theta}{r^2-2mr+a^2-e^2}\right)dr^2- (r^2+a^2\cos^2\theta)d\theta^2-\\ &&\left(r^2+a^2+\frac{(2mr-e^2)a^2\sin^2\theta}{r^2+a^2\cos^2\theta}\right)\sin^2\theta d\varphi^2+ \left(\frac{2a(2mr-e^2)}{r^2+a^2\cos^2\theta}\right)\sin^2\theta(d\varphi)(cdt) \end{eqnarray*} where $m=\kappa M/c^2$, $a=L/Mc$ and $e=\kappa Q/\varepsilon_0c^2$.\\ A rotating charged black hole has an event horizon with $R_{\rm S}=m+\sqrt{m^2-a^2-e^2}$. \npar Near rotating black holes frame dragging occurs because $g_{t\varphi}\neq0$. For the Kerr metric ($e=0$, $a\neq0$) then follows that within the surface $R_{\rm E}=m+\sqrt{m^2-a^2\cos^2\theta}$ (de ergosphere) no particle can be at rest. \subsection{Planetary orbits and the perihelium shift} To find a planetary orbit, the variational problem $\delta\int ds=0$ has to be solved. This is equivalent to the problem $\delta\int ds^2=\delta\int g_{ij}dx^idx^j=0$. Substituting the external Schwarzschild metric yields for a planetary orbit: \[ \frac{du}{d\varphi}\left(\frac{d^2u}{d\varphi^2}+u\right)=\frac{du}{d\varphi}\left(3mu+\frac{m}{h^2}\right) \] where $u:=1/r$ and $h=r^2\dot{\varphi}=$constant. The term $3mu$ is not present in the classical solution. This term can in the classical case also be found from a potential $\displaystyle V(r)=-\frac{\kappa M}{r}\left(1+\frac{h^2}{r^2}\right)$. \npar The orbital equation gives $r=$constant as solution, or can, after dividing by $du/d\varphi$, be solved with perturbation theory. In zeroth order, this results in an elliptical orbit: $u_0(\varphi)=A+B\cos(\varphi)$ with $A=m/h^2$ and $B$ an arbitrary constant. In first order, this becomes: \[ u_1(\varphi)=A+B\cos(\varphi-\varepsilon\varphi)+\varepsilon \left(A+\frac{B^2}{2A}-\frac{B^2}{6A}\cos(2\varphi)\right) \] where $\varepsilon=3m^2/h^2$ is small. The perihelion of a planet is the point for which $r$ is minimal, or $u$ maximal. This is the case if $\cos(\varphi-\varepsilon\varphi)=0\Rightarrow\varphi\approx2\pi n(1+\varepsilon)$. For the perihelion shift then follows: $\Delta\varphi=2\pi\varepsilon=6\pi m^2/h^2$ per orbit. \subsection{The trajectory of a photon} For the trajectory of a photon (and for each particle with zero restmass) holds $ds^2=0$. Substituting the external Schwarzschild metric results in the following orbital equation: \[ \frac{du}{d\varphi}\left(\frac{d^2u}{d\varphi^2}+u-3mu\right)=0 \] \subsection{Gravitational waves} Starting with the approximation $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ for weak gravitational fields and the definition $h'_{\mu\nu}=h_{\mu\nu}-\half\eta_{\mu\nu}h^{\alpha}_{\alpha}$ it follows that $\Box h'_{\mu\nu}=0$ if the gauge condition $\partial h'_{\mu\nu}/\partial x^\nu=0$ is satisfied. From this, it follows that the loss of energy of a mechanical system, if the occurring velocities are $\ll c$ and for wavelengths $\gg$ the size of the system, is given by: \[ \frac{dE}{dt}=-\frac{G}{5c^5}\sum_{i,j}\left(\frac{d^3Q_{ij}}{dt^3}\right)^2 \] with $Q_{ij}=\int\varrho(x_ix_j-\frac{1}{3}\delta_{ij}r^2)d^3x$ the mass quadrupole moment. \subsection{Cosmology} If for the universe as a whole is assumed: \begin{enumerate} \item There exists a global time coordinate which acts as $x^0$ of a Gaussian coordinate system, \item The 3-dimensional spaces are isotrope for a certain value of $x^0$, \item Each point is equivalent to each other point for a fixed $x^0$. \end{enumerate} then the {\it Robertson-Walker metric} can be derived for the line element: \[ ds^2=-c^2dt^2+\frac{R^2(t)}{r_0^2\left(1-\displaystyle\frac{kr^2}{4r_0^2}\right)}(dr^2+r^2d\Omega^2) \] For the {\it scalefactor} $R(t)$ the following equations can be derived: \[ \frac{2\ddot{R}}{R}+\frac{\dot{R}^2+kc^2}{R^2}=-\frac{8\pi\kappa p}{c^2}+\Lambda ~~~\mbox{and}~~~ \frac{\dot{R}^2+kc^2}{R^2}=\frac{8\pi\kappa\varrho}{3}+\frac{\Lambda}{3} \] where $p$ is the pressure and $\varrho$ the density of the universe. If $\Lambda=0$ can be derived for the {\it deceleration parameter} $q$: \[ q=-\frac{\ddot{R}R}{\dot{R}^2}=\frac{4\pi\kappa\varrho}{3H^2} \] where $H=\dot{R}/R$ is {\it Hubble's constant}. This is a measure of the velocity with which galaxies far away are moving away from each other, and has the value $\approx(75\pm25)$ km$\cdot$s$^{-1}\cdot$Mpc$^{-1}$. This gives 3 possible conditions for the universe (here, $W$ is the total amount of energy in the universe): \begin{enumerate} \item {\bf Parabolical universe}: $k=0$, $W=0$, $q=\half$. The expansion velocity of the universe $\rightarrow0$ if $t\rightarrow\infty$. The hereto related {\it critical density} is $\varrho_{\rm c}=3H^2/8\pi\kappa$. \item {\bf Hyperbolical universe}: $k=-1$, $W<0$, $q<\half$. The expansion velocity of the universe remains positive forever. \item {\bf Elliptical universe}: $k=1$, $W>0$, $q>\half$. The expansion velocity of the universe becomes negative after some time: the universe starts collapsing. \end{enumerate} \chapter{Oscillations} \typeout{Oscillations} \section{Harmonic oscillations} The general form of a harmonic oscillation is: $\Psi(t)=\hat{\Psi}{\rm e}^{i(\omega t\pm\varphi)}\equiv\hat{\Psi}\cos(\omega t\pm\varphi)$, \npar where $\hat{\Psi}$ is the {\it amplitude}. A superposition of several harmonic oscillations {\it with the same frequency} results in another harmonic oscillation: \[ \sum_i \hat{\Psi}_i\cos(\alpha_i\pm\omega t)=\hat{\Phi}\cos(\beta\pm\omega t) \] with: \[ \tan(\beta)=\frac{\sum\limits_i\hat{\Psi}_i\sin(\alpha_i)}{\sum\limits_i\hat{\Psi}_i\cos(\alpha_i)}~~~\mbox{and}~~~ \hat{\Phi}^2=\sum_i\hat{\Psi}^2_i+2\sum_{j>i}\sum_i\hat{\Psi}_i\hat{\Psi}_j\cos(\alpha_i-\alpha_j) \] For harmonic oscillations holds: $\displaystyle\int x(t)dt=\frac{x(t)}{i\omega}$ and $\displaystyle\frac{d^nx(t)}{dt^n}=(i\omega)^n x(t)$. \section{Mechanic oscillations} For a construction with a spring with constant $C$ parallel to a damping $k$ which is connected to a mass $M$, to which a periodic force $F(t)=\hat{F}\cos(\omega t)$ is applied holds the equation of motion $m\ddot{x}=F(t)-k\dot{x}-Cx$. With complex amplitudes, this becomes $-m\omega^2 x=F-Cx-ik\omega x$. With $\omega_0^2=C/m$ follows: \[ x=\frac{F}{m(\omega_0^2-\omega^2)+ik\omega}~~,\mbox{and for the velocity holds:}~~ \dot{x}=\frac{F}{i\sqrt{Cm}\delta+k} \] where $\displaystyle\delta=\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega}$. The quantity $Z=F/\dot{x}$ is called the {\it impedance} of the system. The {\it quality} of the system is given by $\displaystyle Q=\frac{\sqrt{Cm}}{k}$. \npar The frequency with minimal $|Z|$ is called {\it velocity resonance frequency}. This is equal to $\omega_0$. In the {\it resonance curve} $|Z|/\sqrt{Cm}$ is plotted against $\omega/\omega_0$. The width of this curve is characterized by the points where $|Z(\omega)|=|Z(\omega_0)|\sqrt{2}$. In these points holds: $R=X$ and $\delta=\pm Q^{-1}$, and the width is $2\Delta\omega_{\rm B}=\omega_0/Q$. \npar The {\it stiffness} of an oscillating system is given by $F/x$. The {\it amplitude resonance frequency} $\omega_{\rm A}$ is the frequency where $i\omega Z$ is minimal. This is the case for $\omega_{\rm A}=\omega_0\sqrt{1-\half Q^2}$. \npar The {\it damping frequency} $\omega_{\rm D}$ is a measure for the time in which an oscillating system comes to rest. It is given by $\displaystyle\omega_{\rm D}=\omega_0\sqrt{1-\frac{1}{4Q^2}}$. A weak damped oscillation $(k^2<4mC)$ dies out after $T_{\rm D}=2\pi/\omega_{\rm D}$. For a {\it critical damped} oscillation $(k^2=4mC)$ holds $\omega_{\rm D}=0$. A strong damped oscillation $(k^2>4mC)$ drops like (if $k^2\gg 4mC$) $x(t)\approx x_0\exp(-t/\tau)$. \section{Electric oscillations} The {\it impedance} is given by: $Z=R+iX$. The phase angle is $\varphi:=\arctan(X/R)$. The impedance of a resistor is $R$, of a capacitor $1/i\omega C$ and of a self inductor $i\omega L$. The quality of a coil is $Q=\omega L/R$. The total impedance in case several elements are positioned is given by: \begin{enumerate} \item Series connection: $V=IZ$, \[ Z_{\rm tot}=\sum_i Z_i~,~~L_{\rm tot}=\sum_i L_i~,~~ \frac{1}{C_{\rm tot}}=\sum_i\frac{1}{C_i}~,~~Q=\frac{Z_0}{R}~,~~ Z=R(1+iQ\delta) \] \item parallel connection: $V=IZ$, \[ \frac{1}{Z_{\rm tot}}=\sum_i\frac{1}{Z_i}~,~~ \frac{1}{L_{\rm tot}}=\sum_i\frac{1}{L_i}~,~~ C_{\rm tot}=\sum_i C_i~,~~Q=\frac{R}{Z_0}~,~~ Z=\frac{R}{1+iQ\delta} \] Here, $\displaystyle Z_0=\sqrt{\frac{L}{C}}$ and $\displaystyle\omega_0=\frac{1}{\sqrt{LC}}$. \end{enumerate} The power given by a source is given by $P(t)=V(t)\cdot I(t)$, so $\av{P}_t=\hat{V}_{\rm eff}\hat{I}_{\rm eff}\cos(\Delta\phi)$\\ $=\half\hat{V}\hat{I}\cos(\phi_v-\phi_i)=\half\hat{I}^2{\rm Re}(Z)= \half\hat{V}^2{\rm Re}(1/Z)$, where $\cos(\Delta\phi)$ is the work factor. \section{Waves in long conductors} These cables are in use for signal transfer, e.g.\ coax cable. For them holds: $\displaystyle Z_0=\sqrt{\frac{dL}{dx}\frac{dx}{dC}}$.\\ The transmission velocity is given by $\displaystyle v=\sqrt{\frac{dx}{dL}\frac{dx}{dC}}$. \section{Coupled conductors and transformers} For two coils enclosing each others flux holds: if $\Phi_{12}$ is the part of the flux originating from $I_2$ through coil 2 which is enclosed by coil 1, than holds $\Phi_{12}=M_{12}I_2$, $\Phi_{21}=M_{21}I_1$. For the coefficients of mutual induction $M_{ij}$ holds: \[ M_{12}=M_{21}:=M=k\sqrt{L_1L_2}=\frac{N_1\Phi_1}{I_2}=\frac{N_2\Phi_2}{I_1}\sim N_1N_2 \] where $0\leq k\leq1$ is the {\it coupling factor}. For a transformer is $k\approx1$. At full load holds: \[ \frac{V_1}{V_2}=\frac{I_2}{I_1}=-\frac{i\omega M}{i\omega L_2+R_{\rm load}}\approx-\sqrt{\frac{L_1}{L_2}}=-\frac{N_1}{N_2} \] \section{Pendulums} The oscillation time $T=1/f$, and for different types of pendulums is given by: \begin{itemize} \item Oscillating spring: $T=2\pi\sqrt{m/C}$ if the spring force is given by $F=C\cdot\Delta l$. \item Physical pendulum: $T=2\pi\sqrt{I/\tau}$ with $\tau$ the moment of force and $I$ the moment of inertia. \item Torsion pendulum: $T=2\pi\sqrt{I/\kappa}$ with $\displaystyle\kappa=\frac{2lm}{\pi r^4\Delta\varphi}$ the constant of torsion and $I$ the moment of inertia. \item Mathematical pendulum: $T=2\pi\sqrt{l/g}$ with $g$ the acceleration of gravity and $l$ the length of the pendulum. \end{itemize} \chapter{Waves} \typeout{Waves} \section{The wave equation} The general form of the wave equation is: $\Box u=0$, or: \[ \nabla^2u-\frac{1}{v^2}\QQ{u}{t}=\QQ{u}{x}+\QQ{u}{y}+\QQ{u}{z}-\frac{1}{v^2}\QQ{u}{t}=0 \] where $u$ is the disturbance and $v$ the {\it propagation velocity}. In general holds: $v=f\lambda$. By definition holds: $k\lambda=2\pi$ and $\omega=2\pi f$. \npar In principle, there are two types of waves: \begin{enumerate} \item Longitudinal waves: for these holds $\vec{k}\parallel\vv\parallel\vec{u}$. \item Transversal waves: for these holds $\vec{k}\parallel\vv\perp\vec{u}$. \end{enumerate} The {\it phase velocity} is given by $v_{\rm ph}=\omega/k$. The {\it group velocity} is given by: \[ v_{\rm g}=\frac{d\omega}{dk}=v_{\rm ph}+k\frac{dv_{\rm ph}}{dk}= v_{\rm ph}\left(1-\frac{k}{n}\frac{dn}{dk}\right) \] where $n$ is the refractive index of the medium. If $v_{\rm ph}$ does not depend on $\omega$ holds: $v_{\rm ph}=v_{\rm g}$. In a dispersive medium it is possible that $v_{\rm g}>v_{\rm ph}$ or $v_{\rm g}vt \end{array}\right. \end{eqnarray*} Further holds the relation: $\displaystyle Q(x,x',t)=\Q{P(x,x',t)}{t}$ \section{Waveguides and resonating cavities} The boundary conditions for a perfect conductor can be derived from the Maxwell equations. If $\vec{n}$ is a unit vector $\perp$ the surface, pointed from 1 to 2, and $\vec{K}$ is a surface current density, than holds: \[ \begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array} \] In a waveguide holds because of the cylindrical symmetry: $\vec{E}(\vec{x},t)=\vec{{\cal E}}(x,y){\rm e}^{i(kz-\omega t)}$ and $\vec{B}(\vec{x},t)=\vec{{\cal B}}(x,y){\rm e}^{i(kz-\omega t)}$. From this one can now deduce that, if ${\cal B}_z$ and ${\cal E}_z$ are not $\equiv0$: \[ \begin{array}{ll} \displaystyle {\cal B}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\Q{{\cal B}_z}{x}-\varepsilon\mu\omega\Q{{\cal E}_z}{y}\right)~~&~~ \displaystyle {\cal B}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\Q{{\cal B}_z}{y}+\varepsilon\mu\omega\Q{{\cal E}_z}{x}\right)\\ \displaystyle {\cal E}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\Q{{\cal E}_z}{x}+\varepsilon\mu\omega\Q{{\cal B}_z}{y}\right)~~&~~ \displaystyle {\cal E}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\Q{{\cal E}_z}{y}-\varepsilon\mu\omega\Q{{\cal B}_z}{x}\right) \end{array} \] Now one can distinguish between three cases: \begin{enumerate} \item $B_z\equiv0$: the Transversal Magnetic modes (TM). Boundary condition: ${\cal E}_z|_{\rm surf}=0$. \item $E_z\equiv0$: the Transversal Electric modes (TE). Boundary condition: $\displaystyle\left.\Q{{\cal B}_z}{n}\right|_{\rm surf}=0$. \npar For the TE and TM modes this gives an eigenvalue problem for ${\cal E}_z$ resp.\ ${\cal B}_z$ with boundary conditions: \[ \left(\QQ{}{x}+\QQ{}{y}\right)\psi=-\gamma^2\psi~~\mbox{with eigenvalues}~~ \gamma^2:=\varepsilon\mu\omega^2-k^2 \] This gives a discrete solution $\psi_\ell$ with eigenvalue $\gamma_\ell^2$: $k=\sqrt{\varepsilon\mu\omega^2-\gamma_\ell^2}$. For $\omega<\omega_\ell$, $k$ is imaginary and the wave is damped. Therefore, $\omega_\ell$ is called the {\it cut-off frequency}. In rectangular conductors the following expression can be found for the cut-off frequency for modes TE$_{m,n}$ of TM$_{m,n}$: \[ \lambda_\ell=\frac{2}{\sqrt{(m/a)^2+(n/b)^2}} \] \item $E_z$ and $B_z$ are zero everywhere: the Transversal electromagnetic mode (TEM). Than holds: $k=\pm\omega\sqrt{\varepsilon\mu}$ and $v_{\rm f}=v_{\rm g}$, just as if here were no waveguide. Further $k\in\RR$, so there exists no cut-off frequency. \end{enumerate} In a rectangular, 3 dimensional resonating cavity with edges $a$, $b$ and $c$ the possible wave numbers are given by: $\displaystyle k_x=\frac{n_1\pi}{a}~,~~k_y=\frac{n_2\pi}{b}~,~~k_z=\frac{n_3\pi}{c}$ This results in the possible frequencies $f=vk/2\pi$ in the cavity: \[ f=\frac{v}{2}\sqrt{\frac{n_x^2}{a^2}+\frac{n_y^2}{b^2}+\frac{n_z^2}{c^2}} \] For a cubic cavity, with $a=b=c$, the possible number of oscillating modes $N_{\rm L}$ for longitudinal waves is given by: \[ N_{\rm L}=\frac{4\pi a^3f^3}{3v^3} \] Because transversal waves have two possible polarizations holds for them: $N_{\rm T}=2N_{\rm L}$. \section{Non-linear wave equations} The {\it Van der Pol} equation is given by: \[ \frac{d^2x}{dt^2}-\varepsilon\omega_0(1-\beta x^2)\frac{dx}{dt}+\omega_0^2x=0 \] $\beta x^2$ can be ignored for very small values of the amplitude. Substitution of $x\sim{\rm e}^{i\omega t}$ gives: $\omega=\half\omega_0(i\varepsilon\pm2\sqrt{1-\frac{1}{2}\varepsilon^2})$. The lowest-order instabilities grow as $\half\varepsilon\omega_0$. While $x$ is growing, the 2nd term becomes larger and diminishes the growth. Oscillations on a time scale $\sim\omega_0^{-1}$ can exist. If $x$ is expanded as $x=x^{(0)}+\varepsilon x^{(1)}+\varepsilon^2x^{(2)}+\cdots$ and this is substituted one obtains, besides periodic, {\it secular terms} $\sim\varepsilon t$. If it is assumed that there exist timescales $\tau_n$, $0\leq\tau\leq N$ with $\partial\tau_n/\partial t=\varepsilon^n$ and if the secular terms are put 0 one obtains: \[ \frac{d}{dt}\left\{\frac{1}{2}\left(\frac{dx}{dt}\right)^2+\half\omega_0^2x^2\right\}= \varepsilon\omega_0(1-\beta x^2)\left(\frac{dx}{dt}\right)^2 \] This is an energy equation. Energy is conserved if the left-hand side is 0. If $x^2>1/\beta$, the right-hand side changes sign and an increase in energy changes into a decrease of energy. This mechanism limits the growth of oscillations. \npar The {\it Korteweg-De Vries} equation is given by: \[ \Q{u}{t}+\Q{u}{x}-\underbrace{au\Q{u}{x}}_{\rm non-lin}+ \underbrace{b^2\Q{^3u}{x^3}}_{\rm dispersive}=0 \] This equation is for example a model for ion-acoustic waves in a plasma. For this equation, soliton solutions of the following form exist: \[ u(x-ct)=\frac{-d}{\cosh^2(e(x-ct))} \] with $c=1+\frac{1}{3}ad$ and $e^2=ad/(12b^2)$. \chapter{Optics} \typeout{Optics} \section{The bending of light} For the refraction at a surface holds: $n_i\sin(\theta_i)=n_t\sin(\theta_t)$ where $n$ is the {\it refractive index} of the material. Snell's law is: \[ \frac{n_2}{n_1}=\frac{\lambda_1}{\lambda_2}=\frac{v_1}{v_2} \] If $\Delta n\leq1$, the change in phase of the light is $\Delta\varphi=0$, if $\Delta n>1$ holds: $\Delta\varphi=\pi$. The refraction of light in a material is caused by scattering from atoms. This is described by: \[ n^2=1+\frac{n_{\rm e}e^2}{\varepsilon_0m}\sum_j\frac{f_j}{\omega_{0,j}^2-\omega^2-i\delta\omega} \] where $n_{\rm e}$ is the electron density and $f_j$ the {\it oscillator strength}, for which holds: $\sum\limits_j f_j=1$. From this follows that $v_{\rm g}=c/(1+(n_{\rm e}e^2/2\varepsilon_0m\omega^2))$. From this the equation of Cauchy can be derived: $n=a_0+a_1/\lambda^2$. More general, it is possible to expand $n$ as: $\displaystyle n=\sum_{k=0}^n\frac{a_k}{\lambda^{2k}}$. \npar For an electromagnetic wave in general holds: $n=\sqrt{\varepsilon_{\rm r}\mu_{\rm r}}$. \npar The path, followed by a light ray in material can be found from {\it Fermat's principle}: \[ \delta\int\limits_1^2 dt=\delta\int\limits_1^2\frac{n(s)}{c}ds=0\Rightarrow \delta\int\limits_1^2 n(s)ds=0 \] \section{Paraxial geometrical optics} \subsection{Lenses} The Gaussian lens formula can be deduced from Fermat's principle with the approximations $\cos\varphi=1$ and $\sin\varphi=\varphi$. For the refraction at a spherical surface with radius $R$ holds: \[ \frac{n_1}{v}-\frac{n_2}{b}=\frac{n_1-n_2}{R} \] where $|v|$ is the distance of the object and $|b|$ the distance of the image. Applying this twice results in: \[ \frac{1}{f}=(n_{\rm l}-1)\left(\frac{1}{R_2}-\frac{1}{R_1}\right) \] where $n_{\rm l}$ is the refractive index of the lens, $f$ is the focal length and $R_1$ and $R_2$ are the curvature radii of both surfaces. For a double concave lens holds $R_1<0$, $R_2>0$, for a double convex lens holds $R_1>0$ and $R_2<0$. Further holds: \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{b} \] $D:=1/f$ is called the dioptric power of a lens. For a lens with thickness $d$ and diameter $D$ holds to a good approximation: $1/f=8(n-1)d/D^2$. For two lenses placed on a line with distance $d$ holds: \[ \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2} \] In these equations the following signs are being used for refraction at a spherical surface, as is seen by an incoming light ray: \begin{center} \begin{tabular}{||c|l|l||} \hline {\bf Quantity}&\multicolumn{1}{c|}{\boldmath$+$}&\multicolumn{1}{c||}{\boldmath$-$}\\ \hline \hline $R$&Concave surface&Convex surface\\ $f$&Converging lens&Diverging lens\\ $v$&Real object&Virtual object\\ $b$&Virtual image&Real image\\ \hline \end{tabular} \end{center} \subsection{Mirrors} For images of mirrors holds: \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{b}=\frac{2}{R}+\frac{h^2}{2}\left(\frac{1}{R}-\frac{1}{v}\right)^2 \] where $h$ is the perpendicular distance from the point the light ray hits the mirror to the optical axis. Spherical aberration can be reduced by not using spherical mirrors. A parabolical mirror has no spherical aberration for light rays parallel with the optical axis and is therefore often used for telescopes. The used signs are: \begin{center} \begin{tabular}{||c|l|l||} \hline {\bf Quantity}&\multicolumn{1}{c|}{\boldmath$+$}&\multicolumn{1}{c||}{\boldmath$-$}\\ \hline \hline $R$&Concave mirror&Convex mirror\\ $f$&Concave mirror&Convex mirror\\ $v$&Real object&Virtual object\\ $b$&Real image&Virtual image\\ \hline \end{tabular} \end{center} \subsection{Principal planes} \parbox[t]{115mm}{ The {\it nodal points} N of a lens are defined by the figure on the right. If the lens is surrounded by the same medium on both sides, the nodal points are the same as the principal points H. The plane $\perp$ the optical axis through the principal points is called the {\it principal plane}. If the lens is described by a matrix $m_{ij}$ than for the distances $h_1$ and $h_2$ to the boundary of the lens holds: \[ h_1=n\frac{m_{11}-1}{m_{12}}~~,~~~h_2=n\frac{m_{22}-1}{m_{12}} \] }\hfill \parbox[b]{35mm}{ \begin{picture}(35,0)(0,25) \put(10,15){\line(1,0){20}} \bezier{120}(15,5)(10,15)(15,25) \bezier{120}(25,5)(30,15)(25,25) \put(20,15){\circle*{1}} \jwline{10}{1}{16}{15} \jwline{13.4}{8.82}{26.5}{21.23} \jwline{24}{15}{29.65}{29} \put(24,15){\circle*{1}} \put(16,15){\circle*{1}} \put(16,18){\makebox(0,0)[cc]{N$_1$}} \put(24,11){\makebox(0,0)[cc]{N$_2$}} \put(20,11){\makebox(0,0)[cc]{O}} \end{picture} } \subsection{Magnification} The {\it linear magnification} is defined by: $\displaystyle N=-\frac{b}{v}$ \npar The {\it angular magnification} is defined by: $\displaystyle N_{\alpha}=-\frac{\alpha_{\rm syst}}{\alpha_{\rm none}}$ \npar where $\alpha_{\rm sys}$ is the size of the retinal image with the optical system and $\alpha_{\rm none}$ the size of the retinal image without the system. Further holds: $N\cdot N_{\alpha}=1$. For a telescope holds: $N=f_{\rm objective}/f_{\rm ocular}$. The {\it f-number} is defined by $f/D_{\rm objective}$. \section{Matrix methods} A light ray can be described by a vector $(n\alpha,y)$ with $\alpha$ the angle with the optical axis and $y$ the distance to the optical axis. The change of a light ray interacting with an optical system can be obtained using a matrix multiplication: \[ \left(\begin{array}{c}n_2\alpha_2\\y_2\end{array}\right)=M \left(\begin{array}{c}n_1\alpha_1\\y_1\end{array}\right) \] where ${\rm Tr}(M)=1$. $M$ is a product of elementary matrices. These are: \begin{enumerate} \item Transfer along length $l$: $\displaystyle M_{\rm R}= \left(\begin{array}{cc}1&0\\l/n&1\end{array}\right)$ \item Refraction at a surface with dioptric power $D$: $\displaystyle M_{\rm T}=\left(\begin{array}{cc}1&-D\\0&1\end{array}\right)$ \end{enumerate} \section{Aberrations} Lenses usually do not give a perfect image. Some causes are: \begin{enumerate} \item {\bf Chromatic aberration} is caused by the fact that $n=n(\lambda)$. This can be partially corrected with a lens which is composed of more lenses with different functions $n_i(\lambda)$. Using $N$ lenses makes it possible to obtain the same $f$ for $N$ wavelengths. \item {\bf Spherical aberration} is caused by second-order effects which are usually ignored; a spherical surface does not make a perfect lens. Incomming rays far from the optical axis will more bent. \item {\bf Coma} is caused by the fact that the principal planes of a lens are only flat near the principal axis. Further away of the optical axis they are curved. This curvature can be both positive or negative. \item {\bf Astigmatism}: from each point of an object not on the optical axis the image is an ellipse because the thickness of the lens is not the same everywhere. \item {\bf Field curvature} can be corrected by the human eye. \item {\bf Distorsion} gives abberations near the edges of the image. This can be corrected with a combination of positive and negative lenses. \end{enumerate} \section{Reflection and transmission} If an electromagnetic wave hits a transparent medium part of the wave will reflect at the same angle as the incident angle, and a part will be refracted at an angle according to Snell's law. It makes a difference whether the $\vec{E}$ field of the wave is $\perp$ or $\parallel$ w.r.t. the surface. When the coefficients of reflection $r$ and transmission $t$ are defined as: \[ r_\parallel\equiv\left(\frac{E_{0r}}{E_{0i}}\right)_\parallel~,~~ r_\perp\equiv\left(\frac{E_{0r}}{E_{0i}}\right)_\perp~,~~ t_\parallel\equiv\left(\frac{E_{0t}}{E_{0i}}\right)_\parallel~,~~ t_\perp\equiv\left(\frac{E_{0t}}{E_{0i}}\right)_\perp \] where $E_{0r}$ is the reflected amplitude and $E_{0t}$ the transmitted amplitude. Then the Fresnel equations are: \[ r_\parallel=\frac{\tan(\theta_i-\theta_t)}{\tan(\theta_i+\theta_t)}~~~,~~~ r_\perp =\frac{\sin(\theta_t-\theta_i)}{\sin(\theta_t+\theta_i)}\\ \] \[ t_\parallel=\frac{2\sin(\theta_t)\cos(\theta_i)}{\sin(\theta_t+\theta_i)\cos(\theta_t-\theta_i)}~~~,~~~ t_\perp =\frac{2\sin(\theta_t)\cos(\theta_i)}{\sin(\theta_t+\theta_i)} \] The following holds: $t_\perp-r_\perp=1$ and $t_\parallel+r_\parallel=1$. If the coefficient of reflection $R$ and transmission $T$ are defined as (with $\theta_i=\theta_r$): \[ R\equiv\frac{I_r}{I_i}~~~\mbox{and}~~~T\equiv\frac{I_t\cos(\theta_t)}{I_i\cos(\theta_i)} \] with $I=\langle|\vec{S}|\rangle$ it follows: $R+T=1$. A special case is $r_\parallel=0$. This happens if the angle between the reflected and transmitted rays is $90^\circ$. From Snell's law it then follows: $\tan(\theta_i)=n$. This angle is called {\it Brewster's angle}. The situation with $r_\perp=0$ is not possible. \section{Polarization} The polarization is defined as: $\displaystyle P=\frac{I_{\rm p}}{I_{\rm p}+I_{\rm u}}=\frac{I_{\rm max}-I_{\rm min}}{I_{\rm max}+I_{\rm min}} $ \npar where the intensity of the polarized light is given by $I_{\rm p}$ and the intensity of the unpolarized light is given by $I_{\rm u}$. $I_{\rm max}$ and $I_{\rm min}$ are the maximum and minimum intensities when the light passes a polarizer. If polarized light passes through a polarizer {\it Malus law} applies: $I(\theta)=I(0)\cos^2(\theta)$ where $\theta$ is the angle of the polarizer. \npar The state of a light ray can be described by the {\it Stokes-parameters}: start with 4 filters which each transmits half the intensity. The first is independent of the polarization, the second and third are linear polarizers with the transmission axes horizontal and at $+45^\circ$, while the fourth is a circular polarizer which is opaque for $L$-states. Then holds $S_1=2I_1$, $S_2=2I_2-2I_1$, $S_3=2I_3-2I_1$ and $S_4=2I_4-2I_1$. \npar The state of a {\it polarized} light ray can also be described by the {\it Jones vector}: \[ \vec{E}=\left(\begin{array}{c} E_{0x}{\rm e}^{i\varphi_x}\\ E_{0y}{\rm e}^{i\varphi_y} \end{array}\right) \] For the horizontal $P$-state holds: $\vec{E}=(1,0)$, for the vertical $P$-state $\vec{E}=(0,1)$, the $R$-state is given by $\vec{E}=\half\sqrt{2}(1,-i)$ and the $L$-state by $\vec{E}=\half\sqrt{2}(1,i)$. The change in state of a light beam after passage of optical equipment can be described as $\vec{E}_2=M\cdot\vec{E}_1$. For some types of optical equipment the Jones matrix $M$ is given by: \npar \begin{tabular}{lr} Horizontal linear polarizer: &$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$\\[4mm] Vertical linear polarizer: &$\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$\\[4mm] Linear polarizer at $+45^\circ$&$\half\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$\\[4mm] Lineair polarizer at $-45^\circ$&$\half\left(\begin{array}{cc}1&-1\\-1&1\end{array}\right)$\\[4mm] \kwart-$\lambda$ plate, fast axis vertical &${\rm e}^{i\pi/4}\left(\begin{array}{cc}1&0\\0&-i\end{array}\right)$\\[4mm] \kwart-$\lambda$ plate, fast axis horizontal&${\rm e}^{i\pi/4}\left(\begin{array}{cc}1&0\\0&i\end{array}\right)$\\[4mm] Homogene circular polarizor right&$\half\left(\begin{array}{cc}1&i\\-i&1\end{array}\right)$\\[4mm] Homogene circular polarizer left &$\half\left(\begin{array}{cc}1&-i\\i&1\end{array}\right)$ \end{tabular} \section{Prisms and dispersion} A light ray passing through a prism is refracted twice and aquires a deviation from its original direction $\delta=\theta_i+\theta_{i'}+\alpha$ w.r.t. the incident direction, where $\alpha$ is the apex angle, $\theta_i$ is the angle between the incident angle and a line perpendicular to the surface and $\theta_{i'}$ is the angle between the ray leaving the prism and a line perpendicular to the surface. When $\theta_i$ varies there is an angle for which $\delta$ becomes minimal. For the refractive index of the prism now holds: \[ n=\frac{\sin(\half(\delta_{\rm min}+\alpha))}{\sin(\half\alpha)} \] The dispersion of a prism is defined by: \[ D=\frac{d\delta}{d\lambda}=\frac{d\delta}{dn}\frac{dn}{d\lambda} \] where the first factor depends on the shape and the second on the composition of the prism. For the first factor follows: \[ \frac{d\delta}{dn}=\frac{2\sin(\half\alpha)}{\cos(\half(\delta_{\rm min}+\alpha))} \] For visible light usually holds $dn/d\lambda<0$: shorter wavelengths are stronger bent than longer. The refractive index in this area can usually be approximated by Cauchy's formula. \section{Diffraction} Fraunhofer diffraction occurs far away from the source(s). The Fraunhofer diffraction of light passing through multiple slits is described by: \[ \frac{I(\theta)}{I_0}=\left(\frac{\sin(u)}{u}\right)^2\cdot \left(\frac{\sin(Nv)}{\sin(v)}\right)^2 \] where $u=\pi b\sin(\theta)/\lambda$, $v=\pi d\sin(\theta)/\lambda$. $N$ is the number of slits, $b$ the width of a slit and $d$ the distance between the slits. The maxima in intensity are given by $d\sin(\theta)=k\lambda$. \npar The diffraction through a spherical aperture with radius $a$ is described by: \[ \frac{I(\theta)}{I_0}=\left(\frac{J_1(ka\sin(\theta))}{ka\sin(\theta)}\right)^2 \] The diffraction pattern of a rectangular aperture at distance $R$ with length $a$ in the $x$-direction and $b$ in the $y$-direction is described by: \[ \frac{I(x,y)}{I_0}=\left(\frac{\sin(\alpha')}{\alpha'}\right)^2\left(\frac{\sin(\beta')}{\beta'}\right)^2 \] where $\alpha'=kax/2R$ and $\beta'=kby/2R$. \npar When X rays are diffracted at a crystal holds for the position of the maxima in intensity {\it Bragg's relation}: $2d\sin(\theta)=n\lambda$ where $d$ is the distance between the crystal layers. \npar Close at the source the Fraunhofermodel is invalid because it ignores the angle-dependence of the reflected waves. This is described by the {\it obliquity} or {\it inclination factor}, which describes the directionality of the secondary emissions: $E(\theta)=\frac{1}{2}E_0(1+\cos(\theta))$ where $\theta$ is the angle w.r.t. the optical axis. \npar Diffraction limits the {\it resolution} of a system. This is the minimum angle $\Delta\theta_{\rm min}$ between two incident rays coming from points far away for which their refraction patterns can be detected separately. For a circular slit holds: $\Delta\theta_{\rm min}=1.22\lambda/D$ where $D$ is the diameter of the slit. \npar For a grating holds: $\Delta\theta_{\rm min}=2\lambda/(Na\cos(\theta_m))$ where $a$ is the distance between two peaks and $N$ the number of peaks. The minimum difference between two wavelengths that gives a separated diffraction pattern in a multiple slit geometry is given by $\Delta\lambda/\lambda=nN$ where $N$ is the number of lines and $n$ the order of the pattern. \section{Special optical effects} \begin{itemize} \item {\bf Birefringe and dichroism}. $\vec{D}$ is not parallel with $\vec{E}$ if the polarizability $\vec{P}$ of a material is not equal in all directions. There are at least 3 directions, the {\it principal axes}, in which they are parallel. This results in 3 refractive indices $n_i$ which can be used to construct Fresnel's ellipsoid. In case $n_2=n_3\neq n_1$, which happens e.g. at trigonal, hexagonal and tetragonal crystals there is one optical axis in the direction of $n_1$. Incident light rays can now be split up in two parts: the {\it ordinary wave} is linear polarized $\perp$ the plane through the transmission direction and the optical axis. The {\it extraordinary wave} is linear polarized in the plane through the transmission direction and the optical axis. {\it Dichroism} is caused by a different absorption of the ordinary and extraordinary wave in some materials. {\it Double images} occur when the incident ray makes an angle with the optical axis: the extraordinary wave will refract, the ordinary will not. \item {\bf Retarders: waveplates and compensators}. Incident light will have a phase shift of $\Delta\varphi=2\pi d(|n_0-n_{\rm e}|)/\lambda_0$ if an uniaxial crystal is cut in such a way that the optical axis is parallel with the front and back plane. Here, $\lambda_0$ is the wavelength in vacuum and $n_0$ and $n_{\rm e}$ the refractive indices for the ordinary and extraordinary wave. For a quarter-wave plate holds: $\Delta\varphi=\pi/2$. \item {\bf The Kerr-effect}: isotropic, transparent materials can become birefringent when placed in an electric field. In that case, the optical axis is parallel to $\vec{E}$. The difference in refractive index in the two directions is given by: $\Delta n=\lambda_0KE^2$, where $K$ is the {\it Kerr constant} of the material. If the electrodes have an effective length $\ell$ and are separated by a distance $d$, the retardation is given by: $\Delta\varphi=2\pi K\ell V^2/d^2$, where $V$ is the applied voltage. \item {\bf The Pockels} or linear electro-optical effect can occur in 20 (from a total of 32) crystal symmetry classes, namely those without a centre of symmetry. These crystals are also {\it piezoelectric}: their polarization changes when a pressure is applied and vice versa: $\vec{P}=pd+\varepsilon_0\chi\vec{E}$. The retardation in a Pockels cell is $\Delta\varphi=2\pi n_0^3 r_{63}V/\lambda_0$ where $r_{63}$ is the 6-3 element of the electro-optic tensor. \item {\bf The Faraday effect}: the polarization of light passing through material with length $d$ and to which a magnetic field is applied in the propagation direction is rotated by an angle $\beta={\cal V}Bd$ where $\cal V$ is the {\it Verdet constant}. \item {\bf \u{C}erenkov radiation} arises when a charged particle with $v_q>v_{\rm f}$ arrives. The radiation is emitted within a cone with an apex angle $\alpha$ with $\sin(\alpha)=c/c_{\rm medium}=c/nv_q$. \end{itemize} \section{The Fabry-Perot interferometer} \parbox[b]{59mm}{ For a Fabry-Perot interferometer holds in general: $T+R+A=1$ where $T$ is the transmission factor, $R$ the reflection factor and $A$ the absorption factor. If $F$ is given by $F=4R/(1-R)^2$ it follows for the intensity distribution: \[ \frac{I_t}{I_i}=\left[1-\frac{A}{1-R}\right]^2\frac{1}{1+F\sin^2(\theta)} \] The term $[1+F\sin^2(\theta)]^{-1}:={\cal A}(\theta)$ is called the {\it Airy function}. }\hfill \parbox[b]{9cm}{ \begin{picture}(90,50) \put(10,30){\line(1,0){80}} \put(10,10){\framebox(2.13,39.99)[cc]{}} \bezier{250}(25,10)(20,30)(25,50) \bezier{250}(25,10)(30,30)(25,50) \special{em:linewidth 0.5pt} \jwline{35.05}{10}{39.94}{50} \jwline{39.94}{50}{42}{50} \jwline{42}{50}{42}{10} \jwline{42}{10}{35.05}{10} \jwline{49.94}{10}{56.96}{10} \jwline{56.96}{10}{52.07}{50} \jwline{52.07}{50}{49.94}{50} \jwline{49.94}{50}{49.94}{10} \bezier{250}(70,10)(65,30)(70,50) \bezier{250}(70,10)(75,30)(70,50) \jwline{70}{50}{75}{50} \jwline{75}{50}{75}{10} \jwline{75}{10}{70}{10} \put(88,10){\framebox(1.91,40)[cc]{}} \put(42,15){\rule{0.85\unitlength}{30\unitlength}} \put(49,15){\rule{0.85\unitlength}{30\unitlength}} \special{em:linewidth 0.25pt} \jwline{42}{10}{42}{5.10} \jwline{49.94}{10}{49.94}{5.10} \put(45,7){\vector(-1,0){2.98}} \put(44,7){\vector(1,0){5.96}} \put(10,5){\makebox(0,0)[cc]{Source}} \put(25,5){\makebox(0,0)[cc]{Lens}} \put(45,4){\makebox(0,0)[cc]{$d$}} \put(70,2){\makebox(0,0)[cc]{Focussing lens}} \put(79,6){\makebox(0,0)[lc]{Screen}} \put(12.08,17.02){\vector(3,-1){12.97}} \jwline{25.06}{13}{49}{18.93} \jwline{49}{18.93}{67.59}{23.61} \jwline{67.59}{23.61}{75.04}{25.10} \jwline{75.04}{25.10}{88.01}{39.99} \jwline{49}{18.93}{42.92}{20.42} \jwline{42.92}{20.42}{49}{21.91} \jwline{49}{21.91}{43.14}{23.40} \jwline{43.14}{23.40}{48.88}{24.88} \jwline{48.88}{24.88}{43.14}{26.37} \jwline{43.14}{26.37}{48.88}{27.86} \jwline{48.88}{27.86}{43.14}{29.35} \jwline{43.14}{29.35}{49}{30.84} \jwline{49}{30.84}{42.92}{32.33} \jwline{42.92}{32.33}{49}{33.82} \jwline{49}{33.82}{42.92}{35.31} \jwline{42.92}{35.31}{49}{36.80} \jwline{49}{36.80}{42.92}{38.28} \jwline{42.92}{38.28}{49}{39.77} \jwline{49.03}{21.95}{75}{32.29} \jwline{75}{32.29}{88.03}{40.04} \jwline{49.03}{24.89}{75}{36.52} \jwline{75}{36.52}{88.03}{40.04} \jwline{49.03}{27.82}{75}{41.22} \jwline{75}{41.22}{88.03}{40.04} \jwline{49.03}{30.88}{75}{44.04} \jwline{75}{44.04}{88.03}{40.04} \jwline{49.03}{33.82}{75}{46.97} \jwline{75}{46.97}{88.03}{40.04} \jwline{49.03}{36.75}{75}{48.97} \jwline{75}{48.97}{88.03}{40.04} \special{em:linewidth 0.75pt} \jwline{42.11}{12.01}{49.87}{12.01} \jwline{49.87}{11.09}{42.11}{11.09} \jwline{42.11}{48.97}{49.87}{48.97} \jwline{49.87}{48.05}{42.11}{48.05} \end{picture} } \npar The width of the peaks at half height is given by $\gamma=4/\sqrt{F}$. The {\it finesse} $\cal F$ is defined as ${\cal F}=\half\pi\sqrt{F}$. The maximum resolution is then given by $\Delta f_{\rm min}=c/2nd{\cal F}$. \chapter{Statistical physics} \typeout{Statistical physics} \section{Degrees of freedom} A molecule consisting of $n$ atoms has $s=3n$ degrees of freedom. There are 3 translational degrees of freedom, a linear molecule has $s=3n-5$ vibrational degrees of freedom and a non-linear molecule $s=3n-6$. A linear molecule has 2 rotational degrees of freedom and a non-linear molecule 3. \npar Because vibrational degrees of freedom account for both kinetic and potential energy they count double. So, for linear molecules this results in a total of $s=6n-5$. For non-linear molecules this gives $s=6n-6$. The average energy of a molecule in thermodynamic equilibrium is $\av{E_{\rm tot}}=\frac{1}{2}skT$. Each degree of freedom of a molecule has in principle the same energy: the {\it principle of equipartition}. \npar The rotational and vibrational energy of a molecule are: \[ W_{\rm rot}=\frac{\hbar^2}{2I}l(l+1)=Bl(l+1)~,~~W_{\rm vib}=(v+\half)\hbar\omega_0 \] The vibrational levels are excited if $kT\approx\hbar\omega$, the rotational levels of a hetronuclear molecule are excited if $kT\approx2B$. For homonuclear molecules additional selection rules apply so the rotational levels are well coupled if $kT\approx6B$. \section{The energy distribution function} The general form of the equilibrium velocity distribution function is\\ $P(v_x,v_y,v_z)dv_xdv_ydv_z=P(v_x)dv_x\cdot P(v_y)dv_y\cdot P(v_z)dv_z$ with \[ P(v_i)dv_i=\frac{1}{\alpha\sqrt{\pi}}\exp\left(-\frac{v_i^2}{\alpha^2}\right)dv_i \] where $\alpha=\sqrt{2kT/m}$ is the {\it most probable velocity} of a particle. The average velocity is given by $\av{v}=2\alpha/\sqrt{\pi}$, and $\av{v^2}=\frac{3}{2}\alpha^2$. The distribution as a function of the absolute value of the velocity is given by: \[ \frac{dN}{dv}=\frac{4N}{\alpha^3\sqrt{\pi}}~v^2\exp\left(-\frac{mv^2}{2kT}\right) \] The general form of the energy distribution function then becomes: \[ P(E)dE=\frac{c(s)}{kT}\left(\frac{E}{kT}\right)^{\frac{1}{2}s-1}\exp\left(-\frac{E}{kT}\right)dE \] where $c(s)$ is a normalization constant, given by: \begin{enumerate} \item Even $s$: $s=2l$: $\displaystyle c(s)=\frac{1}{(l-1)!}$ \item Odd $s$: $s=2l+1$: $\displaystyle c(s)=\frac{2^l}{\sqrt{\pi}(2l-1)!!}$ \end{enumerate} \section{Pressure on a wall} The number of molecules that collides with a wall with surface $A$ within a time $\tau$ is given by: \[ \iiint d^3N=\int\limits_0^\infty \int\limits_0^\pi \int\limits_0^{2\pi} nAv\tau\cos(\theta)P(v,\theta,\varphi)dvd\theta d\varphi \] From this follows for the particle flux on the wall: $\Phi=\kwart n\av{v}$. For the pressure on the wall then follows: \[ d^3p=\frac{2mv\cos(\theta)d^3N}{A\tau}~,~~\mbox{so}~~p=\frac{2}{3}n\av{E} \] \section{The equation of state} If intermolecular forces and the volume of the molecules can be neglected then for gases from $p=\frac{2}{3}n\av{E}$ and $\av{E}=\frac{3}{2}kT$ can be derived: \[ pV=n_sRT=\frac{1}{3}Nm\av{v^2} \] Here, $n_s$ is the number of {\it moles} particles and $N$ is the total number of particles within volume $V$. If the own volume and the intermolecular forces cannot be neglected the {\it Van der Waals} equation can be derived: \[ \left(p+\frac{an_s^2}{V^2}\right)(V-bn_s)=n_sRT \] There is an isotherme with a horizontal point of inflection. In the Van der Waals equation this corresponds with the {\it critical temperature, pressure} and {\it volume} of the gas. This is the upper limit of the area of coexistence between liquid and vapor. From $dp/dV=0$ and $d^2p/dV^2=0$ follows: \[ T_{\rm cr}=\frac{8a}{27bR}~,~~p_{\rm cr}=\frac{a}{27b^2}~,~~V_{\rm cr}=3bn_s \] For the critical point holds: $p_{\rm cr}V_{m,\rm cr}/RT_{\rm cr}=\frac{3}{8}$, which differs from the value of 1 which follows from the general gas law. \npar Scaled on the critical quantities, with $p^*:=p/p_{\rm cr}$, $T^*=T/T_{\rm cr}$ and $V_m^*=V_m/V_{m,\rm cr}$ with $V_m:=V/n_s$ holds: \[ \left(p^*+\frac{3}{(V_m^*)^2}\right)\left(V_m^*-\mbox{$\frac{1}{3}$}\right)= \mbox{$\frac{8}{3}$}T^* \] Gases behave the same for equal values of the reduced quantities: the {\it law of the corresponding states}. A {\it virial expansion} is used for even more accurate views: \[ p(T,V_m)=RT\left(\frac{1}{V_m}+\frac{B(T)}{V_m^2}+\frac{C(T)}{V_m^3}+\cdots\right) \] The {\it Boyle temperature} $T_{\rm B}$ is the temperature for which the 2nd virial coefficient is 0. In a Van der Waals gas, this happens at $T_{\rm B}=a/Rb$. The {\it inversion temperature} $T_{\rm i}=2T_{\rm B}$. \npar The equation of state for solids and liquids is given by: \[ \frac{V}{V_0}=1+\gamma_p\Delta T-\kappa_T\Delta p= 1+\frac{1}{V}\Qc{V}{T}{p}\Delta T+\frac{1}{V}\Qc{V}{p}{T}\Delta p \] \section{Collisions between molecules} The collision probability of a particle in a gas that is translated over a distance $dx$ is given by $n\sigma dx$, where $\sigma$ is the {\it cross section}. The mean free path is given by $\displaystyle\ell=\frac{v_1}{nu\sigma}$ with $u=\sqrt{v_1^2+v_2^2}$ the relative velocity between the particles. If $m_1\ll m_2$ holds: $\displaystyle\frac{u}{v_1}=\sqrt{1+\frac{m_1}{m_2}}$, so $\displaystyle\ell=\frac{1}{n\sigma}$. If $m_1=m_2$ holds: $\displaystyle\ell=\frac{1}{n\sigma\sqrt{2}}$. This means that the average time between two collisions is given by $\displaystyle\tau=\frac{1}{n\sigma v}$. If the molecules are approximated by hard spheres the cross section is: $\sigma=\kwart\pi(D_1^2+D_2^2)$. The average distance between two molecules is $0.55n^{-1/3}$. Collisions between molecules and small particles in a solution result in the {\it Brownian motion}. For the average motion of a particle with radius $R$ can be derived: $\av{x_i^2}=\frac{1}{3}\av{r^2}=kTt/3\pi\eta R$. \npar A gas is called a {\it Knudsen gas} if $\ell\gg$ the dimensions of the gas, something that can easily occur at low pressures. The equilibrium condition for a vessel which has a hole with surface $A$ in it for which holds that $\ell\gg\sqrt{A/\pi}$ is: $n_1\sqrt{T_1}=n_2\sqrt{T_2}$. Together with the general gas law follows: $p_1/\sqrt{T_1}=p_2/\sqrt{T_2}$. \npar If two plates move along each other at a distance $d$ with velocity $w_x$ the {\it viscosity} $\eta$ is given by: $\displaystyle F_x=\eta\frac{Aw_x}{d}$. The velocity profile between the plates is in that case given by $w(z)=zw_x/d$. It can be derived that $\eta=\frac{1}{3}\varrho\ell\av{v}$ where $v$ is the {\it thermal velocity}. \npar The heat conductance in a non-moving gas is described by: $\displaystyle\frac{dQ}{dt}=\kappa A\left(\frac{T_2-T_1}{d}\right)$, which results in a temperature profile $T(z)=T_1+z(T_2-T_1)/d$. It can be derived that $\kappa=\frac{1}{3}C_{mV}n\ell\av{v}/N_{\rm A}$. Also holds: $\kappa=C_V\eta$. A better expression for $\kappa$ can be obtained with the {\it Eucken correction}: $\kappa=(1+9R/4c_{mV})C_V\cdot\eta$ with an error $<$5\%. \section{Interaction between molecules} For dipole interaction between molecules can be derived that $U\sim-1/r^6$. If the distance between two molecules approaches the molecular diameter $D$ a repulsing force between the electron clouds appears. This force can be described by $U_{\rm rep}\sim\exp(-\gamma r)$ or $V_{\rm rep}=+C_s/r^s$ with $12\leq s\leq20$. This results in the {\it Lennard-Jones} potential for intermolecular forces: \[ U_{\rm LJ}=4\epsilon\left[\left(\frac{D}{r}\right)^{12}-\left(\frac{D}{r}\right)^6\right] \] with a minimum $\epsilon$ at $r=r_{\rm m}$. The following holds: $D\approx0.89r_{\rm m}$. For the Van der Waals coefficients $a$ and $b$ and the critical quantities holds: $a=5.275 N_{\rm A}^2D^3\epsilon$, $b=1.3N_{\rm A}D^3$, $kT_{\rm kr}=1.2\epsilon$ and $V_{\rm m,kr}=3.9N_{\rm A}D^3$. \npar A more simple model for intermolecular forces assumes a potential $U(r)=\infty$ for $r1$. \npar {\bf Isobaric processes}\\[2mm] Here holds: $H_2-H_1=\int_1^2 C_pdT$. For a reversible isobaric process holds: $H_2-H_1=Q_{\rm rev}$. \npar {\bf The throttle process}\\[2mm] This is also called the {\it Joule-Kelvin} effect and is an adiabatic expansion of a gas through a porous material or a small opening. Here $H$ is a conserved quantity, and $dS>0$. In general this is accompanied with a change in temperature. The quantity which is important here is the {\it throttle coefficient}: \[ \alpha_H=\Qc{T}{p}{H}=\frac{1}{C_p}\left[T\Qc{V}{T}{p}-V\right] \] The {\it inversion temperature} is the temperature where an adiabatically expanding gas keeps the same temperature. If $T>T_{\rm i}$ the gas heats up, if $T2300$: $L_{\rm e}/2R\approx50$ \end{enumerate} For gas transport at low pressures (Knudsen-gas) holds: $\displaystyle\Phi_V=\frac{4R^3\alpha\sqrt{\pi}}{3}\frac{dp}{dx}$ \npar For flows at a small Re holds: $\nabla p=\eta\nabla^2\vv$ and div$\vv=0$. For the total force on a sphere with radius $R$ in a flow then holds: $F=6\pi\eta Rv$. For large Re holds for the force on a surface $A$: $F=\half C_WA\varrho v^2$. \section{Potential theory} The {\it circulation} $\Gamma$ is defined as: $\displaystyle \Gamma=\oint(\vv\cdot\e{t})ds=\iint({\rm rot}\vv\,)\cdot\vec{n}d^2A=\iint(\vec{\omega}\cdot\vvec{n})d^2A $ \npar For non viscous media, if $p=p(\varrho)$ and all forces are conservative, Kelvin's theorem can be derived: \[ \frac{d\Gamma}{dt}=0 \] For rotationless flows a velocity potential $\vv={\rm grad}\phi$ can be introduced. In the incompressible case follows from conservation of mass $\nabla^2\phi=0$. For a 2-dimensional flow a flow function $\psi(x,y)$ can be defined: with $\Phi_{AB}$ the amount of liquid flowing through a curve $s$ between the points A and B: \[ \Phi_{AB}=\int\limits_A^B (\vv\cdot\vvec{n})ds=\int\limits_A^B(v_xdy-v_ydx) \] and the definitions $v_x=\partial\psi/\partial y$, $v_y=-\partial\psi/\partial x$ holds: $\Phi_{AB}=\psi(B)-\psi(A)$. In general holds: \[ \QQ{\psi}{x}+\QQ{\psi}{y}=-\omega_z \] In polar coordinates holds: \[ v_r=\frac{1}{r}\Q{\psi}{\theta}=\Q{\phi}{r}~~,~~ v_\theta=-\Q{\psi}{r}=\frac{1}{r}\Q{\phi}{\theta} \] For source flows with power $Q$ in $(x,y)=(0,0)$ holds: $\displaystyle\phi=\frac{Q}{2\pi}\ln(r)$ so that $v_r=Q/2\pi r$, $v_\theta=0$. \npar For a dipole of strength $Q$ in $x=a$ and strength $-Q$ in $x=-a$ follows from superposition: $\displaystyle\phi=-Qax/2\pi r^2$ where $Qa$ is the dipole strength. For a vortex holds: $\phi=\Gamma\theta/2\pi$. \npar If an object is surrounded by an uniform main flow with $\vv=v\vec{e}_x$ and such a large Re that viscous effects are limited to the boundary layer holds: $F_x=0$ and $F_y=-\varrho\Gamma v$. The statement that $F_x=0$ is d'Alembert's paradox and originates from the neglection of viscous effects. The lift $F_y$ is also created by $\eta$ because $\Gamma\neq0$ due to viscous effects. Henxe rotating bodies also create a force perpendicular to their direction of motion: the {\it Magnus effect}. \section{Boundary layers} \subsection{Flow boundary layers} If for the thickness of the boundary layer holds: $\delta\ll L$ holds: $\delta\approx L/\sqrt{\rm Re}$. With $v_\infty$ the velocity of the main flow it follows for the velocity $v_y$ $\perp$ the surface: $v_yL\approx\delta v_\infty$. Blasius' equation for the boundary layer is, with $v_y/v_\infty=f(y/\delta)$: $2f'''+ff''=0$ with boundary conditions $f(0)=f'(0)=0$, $f'(\infty)=1$. From this follows: $C_W=0.664~{\rm Re}_x^{-1/2}$. \npar The momentum theorem of Von Karman for the boundary layer is: $\displaystyle \frac{d}{dx}(\vartheta v^2)+\delta^* v\frac{dv}{dx}=\frac{\tau_0}{\varrho} $ \npar where the displacement thickness $\delta^*v$ and the momentum thickness $\vartheta v^2$ are given by: \[ \vartheta v^2=\int\limits_0^\infty (v-v_x)v_xdy~~,~~~ \delta^*v=\int\limits_0^\infty (v-v_x)dy~~\mbox{and}~~ \tau_0=-\eta\left.\Q{v_x}{y}\right|_{y=0} \] The boundary layer is released from the surface if $\displaystyle\Qc{v_x}{y}{y=0}=0$. This is equivalent with $\displaystyle\frac{dp}{dx}=\frac{12\eta v_\infty}{\delta^2}$. \subsection{Temperature boundary layers} \begin{tabbing} If the thickness of the temperature boundary layer $\delta_T\ll L$ holds:~~~~\= 1. If ${\rm Pr}\leq1$: $\delta/\delta_T\approx\sqrt{\rm Pr}$.\\ \>2. If ${\rm Pr}\gg1$: $\delta/\delta_T\approx\sqrt[3]{\rm Pr}$. \end{tabbing} \section{Heat conductance} For non-stationairy heat conductance in one dimension without flow holds: \[ \Q{T}{t}=\frac{\kappa}{\varrho c}\QQ{T}{x}+\Phi \] where $\Phi$ is a source term. If $\Phi=0$ the solutions for harmonic oscillations at $x=0$ are: \[ \frac{T-T_\infty}{T_{\rm max}-T_\infty}=\exp\left(-\frac{x}{D}\right)\cos\left(\omega t-\frac{x}{D}\right) \] with $D=\sqrt{2\kappa/\omega\varrho c}$. At $x=\pi D$ the temperature variation is in anti-phase with the surface. The one-dimensional solution at $\Phi=0$ is \[ T(x,t)=\frac{1}{2\sqrt{\pi at}}\exp\left(-\frac{x^2}{4at}\right) \] This is mathematical equivalent to the diffusion problem: \[ \Q{n}{t}=D\nabla^2n+P-A \] where $P$ is the production of and $A$ the discharge of particles. The flow density $J=-D\nabla n$. \section{Turbulence} The time scale of turbulent velocity variations $\tau_{\rm t}$ is of the order of: $\tau_{\rm t}=\tau\sqrt{\rm Re}/{\rm Ma^2}$ with $\tau$ the molecular time scale. For the velocity of the particles holds: $v(t)=\av{v}+v'(t)$ with $\av{v'(t)}=0$. The Navier-Stokes equation now becomes: \[ \Q{\av{\vv\,}}{t}+(\av{\vv\,}\cdot\nabla)\av{\vv\,}=-\frac{\nabla\av{p}}{\varrho}+ \nu\nabla^2\av{\vv\,}+\frac{{\rm div}\mbox{\sfd S}_R}{\varrho} \] where ${\mbox{\sfd S}_R}_{ij}=-\varrho\av{v_i v_j}$ is the turbulent stress tensor. Boussinesq's assumption is: $\tau_{ij}=-\varrho\av{v_i'v_j'}$. It is stated that, analogous to Newtonian media: {\sfd S}$_R=2\varrho\nu_t\av{\DD}$. Near a boundary holds: $\nu_t=0$, far away of a boundary holds: $\nu_t\approx\nu{\rm Re}$. \section{Self organization} For a (semi) two-dimensional flow holds: $\displaystyle\frac{d\omega}{dt}=\Q{\omega}{t}+J(\omega,\psi)=\nu\nabla^2\omega$ \npar With $J(\omega,\psi)$ the Jacobian. So if $\nu=0$, $\omega$ is conserved. Further, the kinetic energy$/mA$ and the enstrofy $V$ are conserved: with $\vv=\nabla\times(\vec{k}\psi)$ \[ E\sim(\nabla\psi)^2\sim\int\limits_0^\infty {\cal E}(k,t)dk=\mbox{constant}~~,~~ V\sim(\nabla^2\psi)^2\sim\int\limits_0^\infty k^2{\cal E}(k,t)dk=\mbox{constant} \] From this follows that in a two-dimensional flow the energy flux goes towards large values of $k$: larger structures become larger at the expanse of smaller ones. In three-dimensional flows the situation is just the opposite. \chapter{Quantum physics} \typeout{Quantum physics} \section[~~Introduction to quantum physics]{Introduction to quantum physics} \subsection{Black body radiation} Planck's law for the energy distribution for the radiation of a black body is: \[ w(f)=\frac{8\pi hf^3}{c^3}\frac{1}{{\rm e}^{hf/kT}-1}~~,~~~ w(\lambda)=\frac{8\pi hc}{\lambda^5}\frac{1}{{\rm e}^{hc/\lambda kT}-1} \] Stefan-Boltzmann's law for the total power density can be derived from this: $P=A\sigma T^4$. Wien's law for the maximum can also be derived from this: $T\lambda_{\rm max}=k_{\rm W}$. \subsection{The Compton effect} For the wavelength of scattered light, if light is considered to exist of particles, can be derived: \[ \lambda'=\lambda+\frac{h}{mc}(1-\cos\theta)=\lambda+\lambda_{\rm C}(1-\cos\theta) \] \subsection{Electron diffraction} Diffraction of electrons at a crystal can be explained by assuming that particles have a wave character with wavelength $\lambda=h/p$. This wavelength is called the Broglie-wavelength. \section[~~Wave functions]{Wave functions} The wave character of particles is described by a wavefunction $\psi$. This wavefunction can be described in normal or momentum space. Both definitions are each others Fourier transform: \[ \Phi(k,t)=\frac{1}{\sqrt{h}}\int\Psi(x,t){\rm e}^{-ikx}dx~~~\mbox{and}~~~ \Psi(x,t)=\frac{1}{\sqrt{h}}\int\Phi(k,t){\rm e}^{ikx}dk \] These waves define a particle with group velocity $v_{\rm g}=p/m$ and energy $E=\hbar\omega$. \npar The wavefunction can be interpreted as a measure for the probability $P$ to find a particle somewhere (Born): $dP=|\psi|^2d^3V$. The expectation value $\av{f}$ of a quantity $f$ of a system is given by: \[ \av{f(t)}=\iiint\Psi^* f\Psi d^3V~~,~~\av{f_p(t)}=\iiint\Phi^*f\Phi d^3V_p \] This is also written as $\av{f(t)}=\av{\Phi|f|\Phi}$. The normalizing condition for wavefunctions follows from this: $\av{\Phi|\Phi}=\av{\Psi|\Psi}=1$. \section[~~Operators in quantum physics]{Operators in quantum physics} In quantum mechanics, classical quantities are translated into operators. These operators are hermitian because their eigenvalues must be real: \[ \int\psi_1^*A\psi_2d^3V=\int\psi_2(A\psi_1)^*d^3V \] When $u_n$ is the eigenfunction of the eigenvalue equation $A\Psi=a\Psi$ for eigenvalue $a_n$, $\Psi$ can be expanded into a basis of eigenfunctions: $\Psi=\sum\limits_nc_nu_n$. If this basis is taken orthonormal, then follows for the coefficients: $c_n=\av{u_n|\Psi}$. If the system is in a state described by $\Psi$, the chance to find eigenvalue $a_n$ when measuring $A$ is given by $|c_n|^2$ in the discrete part of the spectrum and $|c_n|^2da$ in the continuous part of the spectrum between $a$ and $a+da$. The {\it matrix element} $A_{ij}$ is given by: $A_{ij}=\av{u_i|A|u_j}$. Because $(AB)_{ij}=\av{u_i|AB|u_j}=\langle u_i|A\sum\limits_n|u_n\rangle\av{u_n|B|u_j}$ holds: $\sum\limits_n|u_n\rangle\langle u_n|=1$. \npar The time-dependence of an operator is given by (Heisenberg): \[ \frac{dA}{dt}=\Q{A}{t}+\frac{[A,H]}{i\hbar} \] with $[A,B]\equiv AB-BA$ the {\it commutator} of $A$ and $B$. For hermitian operators the commutator is always complex. If $[A,B]=0$, the operators $A$ and $B$ have a common set of eigenfunctions. By applying this to $p_x$ and $x$ follows (Ehrenfest): $md^2\av{x}_t/dt^2=-\av{dU(x)/dx}$. \npar The first order approximation $\av{F(x)}_t\approx F(\av{x})$, with $F=-dU/dx$ represents the classical equation. \npar Before the addition of quantummechanical operators which are a product of other operators, they should be made symmetrical: a classical product $AB$ becomes $\half(AB+BA)$. \section[~~The uncertainty principle]{The uncertainty principle} If the uncertainty $\Delta A$ in $A$ is defined as: $(\Delta A)^2=\av{\psi|A_{\rm op}-\av{A}|^2\psi}=\av{A^2}-\av{A}^2$ it follows: \[ \Delta A\cdot\Delta B\geq\half|\av{\psi|[A,B]|\psi}| \] From this follows: $\Delta E\cdot\Delta t\geq\half\hbar$, and because $[x,p_x]=i\hbar$ holds: $\Delta p_x\cdot\Delta x\geq\half\hbar$, and $\Delta L_x\cdot\Delta L_y\geq\half\hbar L_z$. \section[~~The Schr\"odinger equation]{The Schr\"odinger equation} The momentum operator is given by: $p_{\rm op}=-i\hbar\nabla$. The position operator is: $x_{\rm op}=i\hbar\nabla_p$. The energy operator is given by: $E_{\rm op}=i\hbar\partial/\partial t$. The Hamiltonian of a particle with mass $m$, potential energy $U$ and total energy $E$ is given by: $H=p^2/2m+U$. From $H\psi=E\psi$ then follows the {\it Schr\"odinger equation}: \begin{center} \fbox{\mbox{$\displaystyle -\frac{\hbar^2}{2m}\nabla^2\psi+U\psi=E\psi=i\hbar\Q{\psi}{t} $}} \end{center} The linear combination of the solutions of this equation give the general solution. In one dimension it is: \[ \psi(x,t)=\left(\sum+\int dE\right)c(E)u_E(x)\exp\left(-\frac{iEt}{\hbar}\right) \] The current density $J$ is given by: $\displaystyle J=\frac{\hbar}{2im}(\psi^*\nabla\psi-\psi\nabla\psi^*)$ \npar The following conservation law holds: $\displaystyle\Q{P(x,t)}{t}=-\nabla J(x,t)$ \section[~~Parity]{Parity} The parity operator in one dimension is given by ${\cal P}\psi(x)=\psi(-x)$. If the wavefunction is split in even and odd functions, it can be expanded into eigenfunctions of $\cal P$: \[ \psi(x)= \underbrace{\half(\psi(x)+\psi(-x))}_{\rm even:~\hbox{$\psi^+$}}+ \underbrace{\half(\psi(x)-\psi(-x))}_{\rm odd:~\hbox{$\psi^-$}} \] $[{\cal P},H]=0$. The functions $\psi^+=\half(1+{\cal P})\psi(x,t)$ and $\psi^-=\half(1-{\cal P})\psi(x,t)$ both satisfy the Schr\"odinger equation. Hence, parity is a conserved quantity. \section[~~The tunnel effect]{The tunnel effect} The wavefunction of a particle in an $\infty$ high potential step from $x=0$ to $x=a$ is given by $\psi(x)=a^{-1/2}\sin(kx)$. The energylevels are given by $E_n=n^2h^2/8a^2m$. \npar If the wavefunction with energy $W$ meets a potential well of $W_0>W$ the wavefunction will, unlike the classical case, be non-zero within the potential well. If 1, 2 and 3 are the areas in front, within and behind the potential well, holds: \[ \psi_1=A{\rm e}^{ikx}+B{\rm e}^{-ikx}~~,~~~ \psi_2=C{\rm e}^{ik'x}+D{\rm e}^{-ik'x}~~,~~~\psi_3=A'{\rm e}^{ikx} \] with $k'^2=2m(W-W_0)/\hbar^2$ and $k^2=2mW$. Using the boundary conditions requiring continuity: $\psi=$continuous and $\partial\psi/\partial x=$continuous at $x=0$ and $x=a$ gives $B$, $C$ and $D$ and $A'$ expressed in $A$. The amplitude $T$ of the transmitted wave is defined by $T=|A'|^2/|A|^2$. If $W>W_0$ and $2a=n\lambda'=2\pi n/k'$ holds: $T=1$. \section[~~The harmonic oscillator]{The harmonic oscillator} For a harmonic oscillator holds: $U=\half bx^2$ and $\omega_0^2=b/m$. The Hamiltonian $H$ is then given by: \[ H=\frac{p^2}{2m}+\half m\omega^2 x^2=\half\hbar\omega+\omega A^\dagger A \] with \[ A=\sqrt{\half m\omega}x+\frac{ip}{\sqrt{2m\omega}}~~\mbox{and}~~ A^\dagger=\sqrt{\half m\omega}x-\frac{ip}{\sqrt{2m\omega}} \] $A\neq A^\dagger$ is non hermitian. $[A,A^\dagger]=\hbar$ and $[A,H]=\hbar\omega A$. $A$ is a so called {\it raising ladder operator}, $A^\dagger$ a {\it lowering ladder operator}. $HAu_E=(E-\hbar\omega)Au_E$. There is an eigenfunction $u_0$ for which holds: $Au_0=0$. The energy in this ground state is $\half\hbar\omega$: the zero point energy. For the normalized eigenfunctions follows: \[ u_n=\frac{1}{\sqrt{n!}}\left(\frac{A^\dagger}{\sqrt{\hbar}}\right)^nu_0~~\mbox{with}~~ u_0=\sqrt[4]{\frac{m\omega}{\pi\hbar}}\exp\left(-\frac{m\omega x^2}{2\hbar}\right) \] with $E_n=(\half+n)\hbar\omega$. \section[~~Angular momentum]{Angular momentum} For the angular momentum operators $L$ holds: $[L_z,L^2]=[L_z,H]=[L^2,H]=0$. However, cyclically holds: $[L_x,L_y]=i\hbar L_z$. Not all components of $L$ can be known at the same time with arbitrary accuracy. For $L_z$ holds: \[ L_z=-i\hbar\Q{}{\varphi}=-i\hbar\left(x\Q{}{y}-y\Q{}{x}\right) \] The ladder operators $L_\pm$ are defined by: $L_\pm=L_x\pm iL_y$. Now holds: $L^2=L_+L_-+L_z^2-\hbar L_z$. Further, \[ L_\pm=\hbar{\rm e}^{\pm i\varphi}\left(\pm\Q{}{\theta}+i\cot(\theta)\Q{}{\varphi}\right) \] From $[L_+,L_z]=-\hbar L_+$ follows: $L_z(L_+Y_{lm})=(m+1)\hbar(L_+Y_{lm})$. \npar From $[L_-,L_z]=\hbar L_-$ follows: $L_z(L_-Y_{lm})=(m-1)\hbar(L_-Y_{lm})$. \npar From $[L^2,L_\pm]=0$ follows: $L^2(L_\pm Y_{lm})=l(l+1)\hbar^2(L_\pm Y_{lm})$. \npar Because $L_x$ and $L_y$ are hermitian (this implies $L_\pm^\dagger=L_\mp$) and $|L_\pm Y_{lm}|^2>0$ follows: $l(l+1)-m^2-m\geq0\Rightarrow-l\leq m\leq l$. Further follows that $l$ has to be integral or half-integral. Half-odd integral values give no unique solution $\psi$ and are therefore dismissed. \section[~~Spin]{Spin} For the spin operators are defined by their commutation relations: $[S_x,S_y]=i\hbar S_z$. Because the spin operators do not act in the physical space $(x,y,z)$ the uniqueness of the wavefunction is not a criterium here: also half odd-integer values are allowed for the spin. Because $[L,S]=0$ spin and angular momentum operators do not have a common set of eigenfunctions. The spin operators are given by $\vec{\vec{S}}=\half\hbar\vec{\vec{\sigma}}$, with \[ \vec{\vec{\sigma}}_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)~~,~~ \vec{\vec{\sigma}}_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)~~,~~ \vec{\vec{\sigma}}_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right) \] The eigenstates of $S_z$ are called {\it spinors}: $\chi=\alpha_+\chi_++\alpha_-\chi_-$, where $\chi_+=(1,0)$ represents the state with spin up ($S_z=\half\hbar$) and $\chi_-=(0,1)$ represents the state with spin down ($S_z=-\half\hbar$). Then the probability to find spin up after a measurement is given by $|\alpha_+|^2$ and the chance to find spin down is given by $|\alpha_-|^2$. Of course holds $|\alpha_+|^2+|\alpha_-|^2=1$. \npar The electron will have an intrinsic magnetic dipole moment $\vec{M}$ due to its spin, given by $\vec{M}=-eg_S\vec{S}/2m$, with $g_S=2(1+\alpha/2\pi+\cdots)$ the gyromagnetic ratio. In the presence of an external magnetic field this gives a potential energy $U=-\vec{M}\cdot\vec{B}$. The Schr\"odinger equation then becomes (because $\partial\chi/\partial x_i\equiv0$): \[ i\hbar\Q{\chi(t)}{t}=\frac{eg_S\hbar}{4m}\vec{\sigma}\cdot\vec{B}\chi(t) \] with $\vec{\sigma}=(\vec{\vec{\sigma}}_x,\vec{\vec{\sigma}}_y,\vec{\vec{\sigma}}_z)$. If $\vec{B}=B\vec{e}_z$ there are two eigenvalues for this problem: $\chi_\pm$ for $E=\pm eg_S\hbar B/4m=\pm\hbar\omega$. So the general solution is given by $\chi=(a{\rm e}^{-i\omega t},b{\rm e}^{i\omega t})$. From this can be derived: $\av{S_x}=\half\hbar\cos(2\omega t)$ and $\av{S_y}=\half\hbar\sin(2\omega t)$. Thus the spin precesses about the $z$-axis with frequency $2\omega$. This causes the normal Zeeman splitting of spectral lines. \npar The potential operator for two particles with spin $\pm\half\hbar$ is given by: \[ V(r)=V_1(r)+\frac{1}{\hbar^2}(\vec{S}_1\cdot\vec{S_2})V_2(r)= V_1(r)+\half V_2(r)[S(S+1)-\mbox{$\frac{3}{2}$}] \] This makes it possible for two states to exist: $S=1$ (triplet) or $S=0$ (Singlet). \section[~~The Dirac formalism]{The Dirac formalism} If the operators for $p$ and $E$ are substituted in the relativistic equation $E^2=m_0^2c^4+p^2c^2$, the {\it Klein-Gordon} equation is found: \[ \mbox{\fbox{$\displaystyle \left(\nabla^2-\frac{1}{c^2}\QQ{}{t}-\frac{m_0^2c^2}{\hbar^2}\right)\psi(\vec{x},t)=0 $}} \] The operator $\Box-m_0^2c^2/\hbar^2$ can be separated: \[ \nabla^2-\frac{1}{c^2}\QQ{}{t}-\frac{m_0^2c^2}{\hbar^2}= \left\{\gamma_\lambda\Q{}{x_\lambda}-\frac{m_0c}{\hbar}\right\} \left\{\gamma_\mu\Q{}{x_\mu}+\frac{m_0c}{\hbar}\right\} \] where the Dirac matrices $\gamma$ are given by: $\{\gamma_\lambda,\gamma_\mu\}= \gamma_\lambda\gamma_\mu+\gamma_\mu\gamma_\lambda=2\delta_{\lambda\mu}$ (In general relativity this becomes $2g_{\lambda\mu}$). From this it can be derived that the $\gamma$ are hermitian $4\times4$ matrices given by: \[ \gamma_k=\left(\begin{array}{cc}0&-i\sigma_k\\i\sigma_k&0\end{array}\right)~~,~~ \gamma_4=\left(\begin{array}{cc}I&0\\0&-I\end{array}\right) \] With this, the Dirac equation becomes: \[ \mbox{\fbox{$\displaystyle \left(\gamma_\lambda\Q{}{x_\lambda}+\frac{m_0c}{\hbar}\right)\psi(\vec{x},t)=0 $}} \] where $\psi(x)=(\psi_1(x),\psi_2(x),\psi_3(x),\psi_4(x))$ is a spinor. \section[~~Atomic physics]{Atomic physics} \subsection[~~Solutions]{Solutions} The solutions of the Schr\"odinger equation in spherical coordinates if the potential energy is a function of $r$ alone can be written as: $\psi(r,\theta,\varphi)=R_{nl}(r)Y_{l,m_l}(\theta,\varphi)\chi_{m_s}$, with \[ Y_{lm}=\frac{C_{lm}}{\sqrt{2\pi}}P_l^m(\cos\theta){\rm e}^{im\varphi} \] For an atom or ion with one electron holds: $\displaystyle R_{lm}(\rho)=C_{lm}{\rm e}^{-\rho/2}\rho^l L_{n-l-1}^{2l+1}(\rho)$ \npar with $\rho=2rZ/na_0$ with $a_0=\varepsilon_0 h^2/\pi\me e^2$. The $L_i^j$ are the associated Laguere functions and the $P_l^m$ are the associated Legendre polynomials: \[ P_l^{|m|}(x)=(1-x^2)^{m/2}\frac{d^{|m|}}{dx^{|m|}}\left[(x^2-1)^l\right]~~,~~ L_n^m(x)=\frac{(-1)^mn!}{(n-m)!}{\rm e}^{-x}x^{-m}\frac{d^{n-m}}{dx^{n-m}}({\rm e}^{-x}x^n) \] The parity of these solutions is $(-1)^l$. The functions are $2\sum\limits_{l=0}^{n-1}(2l+1)=2n^2$-folded degenerated. \subsection[~~Eigenvalue equations]{Eigenvalue equations} The eigenvalue equations for an atom or ion with with one electron are: \begin{center} \begin{tabular}{||l|l|l||} \hline {\bf Equation}&{\bf Eigenvalue}&{\bf Range}\\ \hline \hline $H_{\rm op}\psi=E\psi$ &$E_n=\mu e^4Z^2/8\varepsilon_0^2h^2n^2$&$n\geq1$\rule{0pt}{14pt}\\ $L_{z\rm op}Y_{lm}=L_zY_{lm}$ &$L_z=m_l\hbar$ &$-l\leq m_l\leq l$\rule{0pt}{14pt}\\ $L^2_{\rm op}Y_{lm}=L^2Y_{lm}$&$L^2=l(l+1)\hbar^2$&$l$E_n=E_n^0+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\cdots$ \end{tabbing} and this is put into the Schr\"odinger equation the result is: $E_n^{(1)}=\av{\phi_n|H_1|\phi_n}$ and\\ $\displaystyle c_{nm}^{(1)}=\frac{\av{\phi_m|H_1|\phi_n}}{E_n^0-E_m^0}$ if $m\neq n$. The second-order correction of the energy is then given by:\\ $\displaystyle E^{(2)}_n=\sum_{k\neq n}\frac{|\av{\phi_k|H_1|\phi_n}|^2}{E^0_n-E^0_k}$. So to first order holds: $\displaystyle\psi_n=\phi_n+\sum_{k\neq n} \frac{\av{\phi_k|\lambda H_1|\phi_n}}{E_n^0-E_k^0}~\phi_k$. \npar In case the levels are degenerated the above does not hold. In that case an orthonormal set eigenfunctions $\phi_{ni}$ is chosen for each level $n$, so that $\av{\phi_{mi}|\phi_{nj}}=\delta_{mn}\delta_{ij}$. Now $\psi$ is expanded as: \[ \psi_n=N(\lambda)\left\{\sum_i\alpha_i\phi_{ni}+\lambda\sum_{k\neq n} c_{nk}^{(1)}\sum_i\beta_i\phi_{ki}+\cdots\right\} \] $E_{ni}=E_{ni}^0+\lambda E_{ni}^{(1)}$ is approximated by $E_{ni}^0:=E_n^0$. Substitution in the Schr\"odinger equation and taking dot product with $\phi_{ni}$ gives: $\sum\limits_i\alpha_i\av{\phi_{nj}|H_1|\phi_{ni}}=E_n^{(1)}\alpha_j$. Normalization requires that $\sum\limits_i|\alpha_i|^2=1$. \subsection[~~Time-dependent perturbation theory]{Time-dependent perturbation theory} From the Schr\"odinger equation $\displaystyle i\hbar\Q{\psi(t)}{t}=(H_0+\lambda V(t))\psi(t)$ \npar and the expansion $\displaystyle\psi(t)=\sum_nc_n(t)\exp\left(\frac{-iE_n^0t}{\hbar}\right)\phi_n$ with $c_n(t)=\delta_{nk}+\lambda c_n^{(1)}(t)+\cdots$ \npar follows: $\displaystyle c_n^{(1)}(t)=\frac{\lambda}{i\hbar}\int\limits_0^t\av{\phi_n|V(t')|\phi_k} \exp\left(\frac{i(E_n^0-E_k^0)t'}{\hbar}\right)dt'$ \section[~~N-particle systems]{N-particle systems} \subsection[~~General]{General} Identical particles are indistinguishable. For the total wavefunction of a system of identical indistinguishable particles holds: \begin{enumerate} \item Particles with a half-odd integer spin (Fermions): $\psi_{\rm total}$ must be antisymmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles. The Pauli principle results from this: two Fermions cannot exist in an identical state because then $\psi_{\rm total}=0$. \item Particles with an integer spin (Bosons): $\psi_{\rm total}$ must be symmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles. \end{enumerate} For a system of two electrons there are 2 possibilities for the spatial wavefunction. When $a$ and $b$ are the quantum numbers of electron 1 and 2 holds: \[ \psi_{\rm S}(1,2)=\psi_a(1)\psi_b(2)+\psi_a(2)\psi_b(1)~~,~~ \psi_{\rm A}(1,2)=\psi_a(1)\psi_b(2)-\psi_a(2)\psi_b(1) \] Because the particles do not approach each other closely the repulsion energy at $\psi_{\rm A}$ in this state is smaller. The following spin wavefunctions are possible: \[ \chi_{\rm A}= \begin{array}{ll} \half\sqrt{2}[\chi_+(1)\chi_-(2)-\chi_+(2)\chi_-(1)]&m_s=0 \end{array} \] \[ \chi_{\rm S}= \left\{\begin{array}{ll} \chi_+(1)\chi_+(2)&m_s=+1\\ \half\sqrt{2}[\chi_+(1)\chi_-(2)+\chi_+(2)\chi_-(1)]&m_s=0\\ \chi_-(1)\chi_-(2)&m_s=-1 \end{array}\right. \] Because the total wavefunction must be antisymmetric it follows: $\psi_{\rm total}=\psi_{\rm S}\chi_{\rm A}$ or $\psi_{\rm total}=\psi_{\rm A}\chi_{\rm S}$. \npar For $N$ particles the symmetric spatial function is given by: \[ \psi_{\rm S}(1,...,N)=\sum\psi(\mbox{all permutations of }1..N) \] The antisymmetric wavefunction is given by the determinant $\displaystyle\psi_{\rm A}(1,...,N)=\frac{1}{\sqrt{N!}}|u_{E_i}(j)|$ \subsection[~~Molecules]{Molecules} The wavefunctions of atom $a$ and $b$ are $\phi_a$ and $\phi_b$. If the 2 atoms approach each other there are two possibilities: the total wavefunction approaches the bonding function with lower total energy $\psi_{\rm B}=\half\sqrt{2}(\phi_a+\phi_b)$ or approaches the anti-bonding function with higher energy $\psi_{\rm AB}=\half\sqrt{2}(\phi_a-\phi_b)$. If a molecular-orbital is symmetric w.r.t. the connecting axis, like a combination of two s-orbitals it is called a $\sigma$-orbital, otherwise a $\pi$-orbital, like the combination of two p-orbitals along two axes. \npar The energy of a system is: $\displaystyle E=\frac{\av{\psi|H|\psi}}{\av{\psi|\psi}}$. \npar The energy calculated with this method is always {\it higher} than the real energy if $\psi$ is only an approximation for the solutions of $H\psi=E\psi$. Also, if there are more functions to be chosen, the function which gives the lowest energy is the best approximation. Applying this to the function $\psi=\sum c_i\phi_i$ one finds: $(H_{ij}-ES_{ij})c_i=0$. This equation has only solutions if the {\it secular determinant} $|H_{ij}-ES_{ij}|=0$. Here, $H_{ij}=\av{\phi_i|H|\phi_j}$ and $S_{ij}=\av{\phi_i|\phi_j}$. $\alpha_i:=H_{ii}$ is the Coulomb integral and $\beta_{ij}:=H_{ij}$ the exchange integral. $S_{ii}=1$ and $S_{ij}$ is the overlap integral. \npar The first approximation in the molecular-orbital theory is to place both electrons of a chemical bond in the bonding orbital: $\psi(1,2)=\psi_{\rm B}(1)\psi_{\rm B}(2)$. This results in a large electron density between the nuclei and therefore a repulsion. A better approximation is: $\psi(1,2)=C_1\psi_{\rm B}(1)\psi_{\rm B}(2)+C_2\psi_{\rm AB}(1)\psi_{\rm AB}(2)$, with $C_1=1$ and $C_2\approx0.6$. \npar In some atoms, such as C, it is energetical more suitable to form orbitals which are a linear combination of the s, p and d states. There are three ways of hybridization in C: \begin{enumerate} \item SP-hybridization: $\psi_{\rm sp}=\half\sqrt{2}(\psi_{\rm 2s}\pm\psi_{{\rm 2p}_z})$. There are 2 hybrid orbitals which are placed on one line under $180^\circ$. Further the 2p$_x$ and 2p$_y$ orbitals remain. \item SP$^2$ hybridization: $\psi_{\rm sp^2}=\psi_{\rm 2s}/\sqrt{3}+c_1\psi_{{\rm 2p}_z}+c_2\psi_{{\rm 2p}_y}$, where $(c_1,c_2)\in\{(\sqrt{2/3},0),(-1/\sqrt{6},1/\sqrt{2})$ $,(-1/\sqrt{6},-1/\sqrt{2})\}$. The 3 SP$^2$ orbitals lay in one plane, with symmetry axes which are at an angle of $120^\circ$. \item SP$^3$ hybridization: $\psi_{\rm sp^3}=\half(\psi_{\rm 2s}\pm\psi_{{\rm 2p}_z}\pm\psi_{{\rm 2p}_y}\pm\psi_{{\rm 2p}_x})$. The 4 SP$^3$ orbitals form a tetraheder with the symmetry axes at an angle of $109^\circ 28'$. \end{enumerate} \section[~~Quantum statistics]{Quantum statistics} If a system exists in a state in which one has not the disposal of the maximal amount of information about the system, it can be described by a {\it density matrix} $\rho$. If the probability that the system is in state $\psi_i$ is given by $a_i$, one can write for the expectation value $a$ of $A$: $\av{a}=\sum\limits_ir_i\langle\psi_i|A|\psi_i\rangle$. \npar If $\psi$ is expanded into an orthonormal basis $\{\phi_k\}$ as: $\psi^{(i)}=\sum\limits_k c_k^{(i)}\phi_k$, holds: \[ \av{A}=\sum_k(A\rho)_{kk}={\rm Tr}(A\rho) \] where $\rho_{lk}=c_k^*c_l$. $\rho$ is hermitian, with Tr$(\rho)=1$. Further holds $\rho=\sum r_i|\psi_i\rangle\langle\psi_i|$. The probability to find eigenvalue $a_n$ when measuring $A$ is given by $\rho_{nn}$ if one uses a basis of eigenvectors of $A$ for $\{\phi_k\}$. For the time-dependence holds (in the Schr\"odinger image operators are not explicitly time-dependent): \[ i\hbar\frac{d\rho}{dt}=[H,\rho] \] For a macroscopic system in equilibrium holds $[H,\rho]=0$. If all quantumstates with the same energy are equally probable: $P_i=P(E_i)$, one can obtain the distribution: \[ P_n(E)=\rho_{nn}=\frac{{\rm e}^{-E_n/kT}}{Z}~~\mbox{with the state sum}~~ Z=\sum_n{\rm e}^{-E_n/kT} \] The thermodynamic quantities are related to these definitions as follows: $F=-kT\ln(Z)$, $U=\av{H}=\sum\limits_np_nE_n=\displaystyle-\Q{}{kT}\ln(Z)$, $S=-k\sum\limits_nP_n\ln(P_n)$. For a mixed state of $M$ orthonormal quantum states with probability $1/M$ follows: $S=k\ln(M)$. \npar The distribution function for the internal states for a system in thermal equilibrium is the most probable function. This function can be found by taking the maximum of the function which gives the number of states with Stirling's equation: $\ln(n!)\approx n\ln(n)-n$, and the conditions $\sum\limits_kn_k=N$ and $\sum\limits_kn_kW_k=W$. For identical, indistinguishable particles which obey the Pauli exclusion principle the possible number of states is given by: \[ P=\prod_k\frac{g_k!}{n_k!(g_k-n_k)!} \] This results in the {\it Fermi-Dirac statistics}. For indistinguishable particles which {\it do not} obey the exclusion principle the possible number of states is given by: \[ P=N!\prod_k\frac{g_k^{n_k}}{n_k!} \] This results in the {\it Bose-Einstein statistics}. So the distribution functions which explain how particles are distributed over the different one-particle states $k$ which are each $g_k$-fold degenerate depend on the spin of the particles. They are given by: \begin{enumerate} \item Fermi-Dirac statistics: integer spin. $n_k\in\{0,1\}$, $\displaystyle n_k=\frac{N}{Z_g}\frac{g_k}{\exp((E_k-\mu)/kT)+1}$\\ with $\ln(Z_{\rm g})=\sum g_k\ln[1+\exp((E_i-\mu)/kT)]$. \item Bose-Einstein statistics: half odd-integer spin. $n_k\in\NN$, $\displaystyle n_k=\frac{N}{Z_g}\frac{g_k}{\exp((E_k-\mu)/kT)-1}$\\ with $\ln(Z_{\rm g})=-\sum g_k\ln[1-\exp((E_i-\mu)/kT)]$. \end{enumerate} Here, $Z_{\rm g}$ is the large-canonical state sum and $\mu$ the chemical potential. It is found by demanding $\sum n_k=N$, and for it holds: $\displaystyle\lim\limits_{T\rightarrow0}\mu=E_{\rm F}$, the Fermi-energy. $N$ is the total number of particles. The Maxwell-Boltzmann distribution can be derived from this in the limit $E_k-\mu\gg kT$: \[ n_k=\frac{N}{Z}\exp\left(-\frac{E_k}{kT}\right)~~\mbox{with}~~ Z=\sum_kg_k\exp\left(-\frac{E_k}{kT}\right) \] With the Fermi-energy, the Fermi-Dirac and Bose-Einstein statistics can be written as: \begin{enumerate} \item Fermi-Dirac statistics: $\displaystyle n_k=\frac{g_k}{\exp((E_k-E_{\rm F})/kT)+1}$. \item Bose-Einstein statistics: $\displaystyle n_k=\frac{g_k}{\exp((E_k-E_{\rm F})/kT)-1}$. \end{enumerate} \chapter{Plasma physics} \typeout{Plasma physics} \section{Introduction} The {\it degree of ionization} $\alpha$ of a plasma is defined by: $\displaystyle\alpha=\frac{\ne}{\ne+\no}$ \npar where $\ne$ is the electron density and $\no$ the density of the neutrals. If a plasma contains also negative charged ions $\alpha$ is not well defined. \npar The probability that a test particle collides with another is given by $dP=n\sigma dx$ where $\sigma$ is the {\it cross section}. The collision frequency $\nu_{\rm c}=1/\tau_{\rm c}=n\sigma v$. The {\it mean free path} is given by $\lambda_{\rm v}=1/n\sigma$. The {\it rate coefficient} $K$ is defined by $K=\av{\sigma v}$. The number of collisions per unit of time and volume between particles of kind 1 and 2 is given by $n_1n_2\av{\sigma v}=Kn_1n_2$. \npar The potential of an electron is given by: \[ V(r)=\frac{-e}{4\pi\varepsilon_0r}\exp\left(-\frac{r}{\lambda_{\rm D}}\right) ~~~\mbox{with}~~~ \lambda_{\rm D}=\sqrt{\frac{\varepsilon_0k\Te\Ti}{e^2(\ne\Ti+\ni\Te)}}\approx \sqrt{\frac{\varepsilon_0k\Te}{\ne e^2}} \] because charge is shielded in a plasma. Here, $\lambda_{\rm D}$ is the {\it Debye length}. For distances $<\lambda_{\rm D}$ the plasma cannot be assumed to be quasi-neutral. Deviations of charge neutrality by thermic motion are compensated by oscillations with frequency \[ \omega_{\rm pe}=\sqrt{\frac{\ne e^2}{m_{\rm e}\varepsilon_0}} \] The distance of closest approximation when two equal charged particles collide for a deviation of $\pi/2$ is $2b_0=e^2/(4\pi\varepsilon_0\half mv^2)$. A ``neat'' plasma is defined as a plasma for which holds: $b_0<\ne^{-1/3}\ll\lambda_{\rm D}\ll L_{\rm p}$. Here $L_{\rm p}:=|\ne/\nabla\ne|$ is the gradient length of the plasma. \section{Transport} Relaxation times are defined as $\tau=1/\nu_{\rm c}$. Starting with $\sigma_{\rm m}=4\pi b_0^2\ln(\Lambda_{\rm C})$ and with $\half mv^2=kT$ it can be found that: \[ \tau_{\rm m}=\frac{4\pi\varepsilon_0^2m^2v^3}{ne^4\ln(\Lambda_{\rm C})}= \frac{8\sqrt{2}\pi\varepsilon_0^2\sqrt{m}(kT)^{3/2}}{ne^4\ln(\Lambda_{\rm C})} \] For momentum transfer between electrons and ions holds for a Maxwellian velocity distribution: \[ \tau_{\rm ee}=\frac{6\pi\sqrt{3}\varepsilon_0^2\sqrt{\me}(k\Te)^{3/2}}{\ne e^4\ln(\Lambda_{\rm C})}\approx\tau_{\rm ei}~~,~~ \tau_{\rm ii}=\frac{6\pi\sqrt{3}\varepsilon_0^2\sqrt{\mi}(k\Ti)^{3/2}}{\ni e^4\ln(\Lambda_{\rm C})} \] The energy relaxation times for identical particles are equal to the momentum relaxation times. Because for e-i collisions the energy transfer is only $\sim2\me/\mi$ this is a slow process. Approximately holds: $\tau_{\rm ee}:\tau_{\rm ei}:\tau_{\rm ie}:\tau_{\rm ie}^E=1:1:\sqrt{\mi/\me}:\mi/\me$. \npar The relaxation for e-o interaction is much more complicated. For $T>10$ eV holds approximately: $\sigma_{\rm eo}=10^{-17}v_{\rm e}^{-2/5}$, for lower energies this can be a factor 10 lower. \npar The resistivity $\eta=E/J$ of a plasma is given by: \[ \eta=\frac{\ne e^2}{\me\nu_{\rm ei}}=\frac{e^2\sqrt{\me}\ln(\Lambda_{\rm C})}{6\pi\sqrt{3}\varepsilon_0^2(k\Te)^{3/2}} \] The diffusion coefficient $D$ is defined by means of the flux $\Gamma$ by $\vec{\Gamma}=n\vv_{\rm diff}=-D\nabla n$. The equation of continuity is $\partial_tn+\nabla(nv_{\rm diff})=0\Rightarrow\partial_tn=D\nabla^2n$. One finds that $D=\frac{1}{3}\lambda_{\rm v}v$. A rough estimate gives $\tau_{\rm D}=L_{\rm p}/D=L_{\rm p}^2\tau_{\rm c}/\lambda_{\rm v}^2$. For magnetized plasma's $\lambda_{\rm v}$ must be replaced with the cyclotron radius. In electrical fields also holds $\vec{J}=ne\mu\vec{E}=e(\ne\mu_{\rm e}+\ni\mu_{\rm i})\vec{E}$ with $\mu=e/m\nu_{\rm c}$ the mobility of the particles. The Einstein ratio is: \[ \frac{D}{\mu}=\frac{kT}{e} \] Because a plasma is electrically neutral electrons and ions are strongly coupled and they don't diffuse independent. The {\it coefficient of ambipolar diffusion} $D_{\rm amb}$ is defined by $\vec{\Gamma}=\vec{\Gamma}_{\rm i}=\vec{\Gamma}_{\rm e}=-D_{\rm amb}\nabla n_{\rm e,i}$. From this follows that \[ D_{\rm amb}=\frac{k\Te/e-k\Ti/e}{1/\mu_{\rm e}-1/\mu_{\rm i}}\approx\frac{k\Te\mu_{\rm i}}{e} \] In an external magnetic field $B_0$ particles will move in spiral orbits with {\it cyclotron radius} $\rho=mv/eB_0$ and with cyclotron frequency $\Omega=B_0e/m$. The helical orbit is perturbed by collisions. A plasma is called {\it magnetized} if $\lambda_{\rm v}>\rho_{\rm e,i}$. So the electrons are magnetized if \[ \frac{\rho_{\rm e}}{\lambda_{\rm ee}}=\frac{\sqrt{\me}e^3\ne\ln(\Lambda_{\rm C})}{6\pi\sqrt{3}\varepsilon_0^2(k\Te)^{3/2}B_0}<1 \] Magnetization of only the electrons is sufficient to confine the plasma reasonable because they are coupled to the ions by charge neutrality. In case of magnetic confinement holds: $\nabla p=\vec{J}\times\vec{B}$. Combined with the two stationary Maxwell equations for the $B$-field these form the ideal magneto-hydrodynamic equations. For a uniform $B$-field holds: $p=nkT=B^2/2\mu_0$. \npar If both magnetic and electric fields are present electrons and ions will move in the same direction. If $\vec{E}=E_r\vec{e}_r+E_z\vec{e}_z$ and $\vec{B}=B_z\vec{e}_z$ the $\vec{E}\times\vec{B}$ drift results in a velocity $\vec{u}=(\vec{E}\times\vvec{B})/B^2$ and the velocity in the $r,\varphi$ plane is $\dot{r}(r,\varphi,t)=\vec{u}+\dot{\vec{\rho}}(t)$. \section{Elastic collisions} \subsection{General} \parbox[t]{85mm}{ The scattering angle of a particle in interaction with another particle, as shown in the figure at the right is: \[ \chi=\pi-2b\int\limits_{r_a}^\infty \frac{dr}{r^2\sqrt{\displaystyle 1-\frac{b^2}{r^2}-\frac{W(r)}{E_0}}} \] Particles with an impact parameter between $b$ and $b+db$, moving through a ring with $d\sigma=2\pi bdb$ leave the scattering area at a solid angle $d\Omega=2\pi\sin(\chi)d\chi$. The {\it differential cross section} is then defined as: \[ I(\Omega)=\left|\frac{d\sigma}{d\Omega}\right|=\frac{b}{\sin(\chi)}\Q{b}{\chi} \] }\hfill \parbox[t]{6cm}{ \special{em:linewidth 0.4pt} \begin{picture}(60,35) \jwline{5}{10}{50}{10} \jwline{25}{0}{60}{30} \bezier{310}(5,15)(35,15)(55,35) \put(7,13){\vector(0,1){2}} \put(7,13){\vector(0,-1){3}} \put(56,27){\vector(-1,1){4.67}} \put(56,27){\vector(1,-1){0}} \jwline{37}{10}{31.33}{19.67} \jwline{37}{10}{20.33}{16.33} \jwline{31}{19}{30}{18.5} \jwline{31}{19}{31.5}{18} \put(45,13){\makebox(0,0)[cc]{$\chi$}} \put(37,6){\makebox(0,0)[cc]{M}} \put(4,13){\makebox(0,0)[cc]{$b$}} \put(56,32){\makebox(0,0)[cc]{$b$}} \put(37,17){\makebox(0,0)[cc]{$r_a$}} \put(23,13){\makebox(0,0)[cc]{$\varphi$}} \end{picture} } \npar For a potential energy $W(r)=kr^{-n}$ follows: $I(\Omega,v)\sim v^{-4/n}$. \npar For low energies, $\cal O$(1 eV), $\sigma$ has a {\it Ramsauer minimum}. It arises from the interference of matter waves behind the object. $I(\Omega)$ for angles $0<\chi<\lambda/4$ is larger than the classical value. \subsection{The Coulomb interaction} For the Coulomb interaction holds: $2b_0=q_1q_2/2\pi\varepsilon_0mv_0^2$, so $W(r)=2b_0/r$. This gives $b=b_0\cot(\half\chi)$ and \[ I(\Omega=\frac{b}{\sin(\chi)}\Q{b}{\chi}=\frac{b_0^2}{4\sin^2(\half\chi)} \] Because the influence of a particle vanishes at $r=\lambda_{\rm D}$ holds: $\sigma=\pi(\lambda_{\rm D}^2-b_0^2)$. Because $dp=d(mv)=mv_0(1-\cos\chi)$ a cross section related to momentum transfer $\sigma_{\rm m}$ is given by: \[ \sigma_{\rm m}=\int(1-\cos\chi)I(\Omega)d\Omega=4\pi b_0^2\ln\left(\frac{1}{\sin(\half\chi_{\rm min})}\right)= 4\pi b_0^2\ln\left(\frac{\lambda_{\rm D}}{b_0}\right):=4\pi b_0^2\ln(\Lambda_{\rm C}) \sim\frac{\ln(v^4)}{v^4} \] where $\ln(\Lambda_{\rm C})$ is the {\it Coulomb-logarithm}. For this quantity holds: $\Lambda_{\rm C}=\lambda_{\rm D}/b_0=9n(\lambda_{\rm D})$. \subsection{The induced dipole interaction} The induced dipole interaction, with $\vec{p}=\alpha\vec{E}$, gives a potential $V$ and an energy $W$ in a dipole field given by: \[ V(r)=\frac{\vec{p}\cdot\vec{e}_r}{4\pi\varepsilon_0r^2}~~,~~ W(r)=-\frac{|e|p}{8\pi\varepsilon_0 r^2}=-\frac{\alpha e^2}{2(4\pi\varepsilon_0)^2r^4} \] with $\displaystyle b_a=\sqrt[4]{\frac{2e^2\alpha}{(4\pi\varepsilon_0)^2\half mv_0^2}}$ holds: $\displaystyle\chi=\pi-2b\int\limits_{r_a}^\infty \frac{dr}{r^2\sqrt{\displaystyle 1-\frac{b^2}{r^2}+\frac{b_a^4}{4r^4}}}$ \npar If $b\geq b_a$ the charge would hit the atom. Repulsing nuclear forces prevent this to happen. If the scattering angle is a lot times $2\pi$ it is called capture. The cross section for capture $\sigma_{\rm orb}=\pi b_a^2$ is called the Langevin limit, and is a lowest estimate for the total cross section. \subsection{The centre of mass system} If collisions of two particles with masses $m_1$ and $m_2$ which scatter in the centre of mass system by an angle $\chi$ are compared with the scattering under an angle $\theta$ in the laboratory system holds: \[ \tan(\theta)=\frac{m_2\sin(\chi)}{m_1+m_2\cos(\chi)} \] The energy loss $\Delta E$ of the incoming particle is given by: \[ \frac{\Delta E}{E}=\frac{\half m_2v_2^2}{\half m_1v_1^2}=\frac{2m_1m_2}{(m_1+m_2)^2}(1-\cos(\chi)) \] \subsection{Scattering of light} Scattering of light by free electrons is called Thomson scattering. The scattering is free from collective effects if $k\lambda_{\rm D}\ll1$. The cross section $\sigma=6.65\cdot10^{-29}$m$^2$ and \[ \frac{\Delta f}{f}=\frac{2v}{c}\sin(\half\chi) \] This gives for the scattered energy $E_{\rm scat}\sim n\lambda_0^4/(\lambda^2-\lambda_0^2)^2$ with $n$ the density. If $\lambda\gg\lambda_0$ it is called Rayleigh scattering. Thomson sccattering is a limit of Compton scattering, which is given by $\lambda'-\lambda=\lambda_{\rm C}(1-\cos\chi)$ with $\lambda_{\rm C}=h/mc$ and cannot be used any more if relativistic effects become important. \section{Thermodynamic equilibrium and reversibility} Planck's radiation law and the Maxwellian velocity distribution hold for a plasma in equilibrium: \[ \rho(\nu,T)d\nu=\frac{8\pi h\nu^3}{c^3}\frac{1}{\exp(h\nu/kT)-1}d\nu~~,~~ N(E,T)dE=\frac{2\pi n}{(\pi kT)^{3/2}}\sqrt{E}\exp\left(-\frac{E}{kT}\right)dE \] ``Detailed balancing'' means that the number of reactions in one direction equals the number of reactions in the opposite direction because both processes have equal probability if one corrects for the used phase space. For the reaction \[ \sum_{\rm forward}{\rm X}_{\rm forward}\rlarrow\sum_{\rm back}{\rm X}_{\rm back} \] holds in a plasma in equilibrium {\it microscopic} reversibility: \[ \prod_{\rm forward}\hat{\eta}_{\rm forward}=\prod_{\rm back}\hat{\eta}_{\rm back} \] If the velocity distribution is Maxwellian, this gives: \[ \hat{\eta}_{x}=\frac{n_x}{g_x}\frac{h^3}{(2\pi m_xkT)^{3/2}}{\rm e}^{-E_{\rm kin}/kT} \] where $g$ is the statistical weight of the state and $n/g:=\eta$. For electrons holds $g=2$, for excited states usually holds $g=2j+1=2n^2$. \npar With this one finds for the Boltzmann balance, ${\rm X}_p+{\rm e}^-\rlarrow{\rm X}_1+{\rm e}^-+(E_{1p})$: \[ \frac{n_p^{\rm B}}{n_1}=\frac{g_p}{g_1}\exp\left(\frac{E_p-E_1}{k\Te}\right) \] And for the Saha balance, ${\rm X}_p+{\rm e}^-+(E_{pi})\rlarrow{\rm X}_1^++2{\rm e}^-$: \[ \frac{n_p^{\rm S}}{g_p}=\frac{n_1^+}{g_1^+}\frac{\ne}{g_{\rm e}} \frac{h^3}{(2\pi\me k\Te)^{3/2}}\exp\left(\frac{E_{pi}}{k\Te}\right) \] Because the number of particles on the left-hand side and right-hand side of the equation is different, a factor $g/V_{\rm e}$ remains. This factor causes the {\it Saha-jump}. \npar From microscopic reversibility one can derive that for the rate coefficients $K(p,q,T):=\av{\sigma v}_{pq}$ holds: \[ K(q,p,T)=\frac{g_p}{g_q}K(p,q,T)\exp\left(\frac{\Delta E_{pq}}{kT}\right) \] \section{Inelastic collisions} \subsection{Types of collisions} The kinetic energy can be split in a part {\it of} and a part {\it in} the centre of mass system. The energy {\it in} the centre of mass system is available for reactions. This energy is given by \[ E=\frac{m_1m_2(v_1-v_2)^2}{2(m_1+m_2)} \] Some types of inelastic collisions important for plasma physics are: \begin{enumerate} \item Excitation: ${\rm A}_p+{\rm e}^-\rlarrow{\rm A}_q+{\rm e}^-$ \item Decay: ${\rm A}_q\rlarrow{\rm A}_p+hf$ \item Ionisation and 3-particles recombination: ${\rm A}_p+{\rm e}^-\rlarrow{\rm A}^++2{\rm e}^-$ \item radiative recombination: ${\rm A}^++{\rm e}^-\rlarrow{\rm A}_p+hf$ \item Stimulated emission: ${\rm A}_q+hf\rightarrow{\rm A}_p+2hf$ \item Associative ionisation: $\rm A^{**}+B\rlarrow AB^++e^-$ \item Penning ionisation: b.v. $\rm Ne^*+Ar\rlarrow Ar^++Ne+e^-$ \item Charge transfer: $\rm A^++B\rlarrow A+ B^+$ \item Resonant charge transfer: $\rm A^++A\rlarrow A+A^+$ \end{enumerate} \subsection{Cross sections} Collisions between an electron and an atom can be approximated by a collision between an electron and one of the electrons of that atom. This results in \[ \frac{d\sigma}{d(\Delta E)}=\frac{\pi Z^2 e^4}{(4\pi\varepsilon_0)^2E(\Delta E)^2} \] Then follows for the transition $p\rightarrow q$: $\displaystyle \sigma_{pq}(E)=\frac{\pi Z^2e^4\Delta E_{q,q+1}}{(4\pi\varepsilon_0)^2E(\Delta E)_{pq}^2} $ \npar For ionization from state $p$ holds to a good approximation: $\displaystyle \sigma_p=4\pi a_0^2 Ry\left(\frac{1}{E_p}-\frac{1}{E}\right)\ln\left(\frac{1.25\beta E}{E_p}\right) $ \npar For resonant charge transfer holds: $\displaystyle\sigma_{\rm ex}=\frac{A[1-B\ln(E)]^2}{1+CE^{3.3}}$ \section{Radiation} In equilibrium holds for radiation processes: \[ \underbrace{n_pA_{pq}}_{\rm emission}+ \underbrace{n_pB_{pq}\rho(\nu,T)}_{\rm stimulated~emission}= \underbrace{n_qB_{qp}\rho(\nu,T)}_{\rm absorption} \] Here, $A_{pq}$ is the matrix element of the transition $p\rightarrow q$, and is given by: \[ A_{pq}=\frac{8\pi^2e^2\nu^3|r_{pq}|^2}{3\hbar\varepsilon_0c^3}~~\mbox{with}~~ r_{pq}=\langle\psi_p|\rr\,|\psi_q\rangle \] For hydrogenic atoms holds: $A_p=1.58\cdot10^8Z^4p^{-4.5}$, with $A_p=1/\tau_p=\sum\limits_qA_{pq}$. The intensity $I$ of a line is given by $I_{pq}=hfA_{pq}n_p/4\pi$. The Einstein coefficients $B$ are given by: \[ B_{pq}=\frac{c^3A_{pq}}{8\pi h\nu^3}~~\mbox{and}~~\frac{B_{pq}}{B_{qp}}=\frac{g_q}{g_p} \] A spectral line is broadened by several mechanisms: \begin{enumerate} \item Because the states have a finite life time. The natural life time of a state $p$ is given by $\tau_p=1/\sum\limits_q A_{pq}$. From the uncertainty relation then follows: $\Delta(h\nu)\cdot\tau_p=\half\hbar$, this gives \[ \Delta\nu=\frac{1}{4\pi\tau_p}=\frac{\sum\limits_qA_{pq}}{4\pi} \] The natural line width is usually $\ll$ than the broadening due to the following two mechanisms: \item The Doppler broadening is caused by the thermal motion of the particles: \[ \frac{\Delta\lambda}{\lambda}=\frac{2}{c}\sqrt{\frac{2\ln(2)k\Ti}{\mi}} \] This broadening results in a Gaussian line profile:\\ $k_\nu=k_0\exp(-[2\sqrt{\ln 2}(\nu-\nu_0)/\Delta\nu_D]^2)$, with $k$ the coefficient of absorption or emission. \item The Stark broadening is caused by the electric field of the electrons: \[ \Delta\lambda_{1/2}=\left[\frac{\ne}{C(\ne,\Te)}\right]^{2/3} \] with for the H-$\beta$ line: $C(\ne,\Te)\approx3\cdot10^{14}$\AA$^{-3/2}$cm$^{-3}$. \end{enumerate} The natural broadening and the Stark broadening result in a Lorentz profile of a spectral line: $k_\nu=\half k_0\Delta\nu_L/[(\half\Delta\nu_L)^2+(\nu-\nu_0)^2]$. The total line shape is a convolution of the Gauss- and Lorentz profile and is called a {\it Voigt profile}. \npar The number of transitions $p\rightarrow q$ is given by $n_pB_{pq}\rho$ and by $n_pn_{hf}\av{\sigma_{\rm a} c}=n_p(\rho d\nu/h\nu)\sigma_{\rm a}c$ where $d\nu$ is the line width. Then follows for the cross section of absorption processes: $\sigma_{\rm a}=B_{pq}h\nu/cd\nu$. \npar The background radiation in a plasma originates from two processes: \begin{enumerate} \item Free-Bound radiation, originating from radiative recombination. The emission is given by: \[ \varepsilon_{fb}=\frac{C_1}{\lambda^2}\frac{z_{\rm i}\ni\ne}{\sqrt{k\Te}}\left[1-\exp\left(-\frac{hc}{\lambda k\Te}\right)\right]\xi_{fb}(\lambda,\Te) \] with $C_1=1.63\cdot10^{-43}$ Wm$^4$K$^{1/2}$sr$^{-1}$ and $\xi$ the {\it Biberman factor}. \item Free-free radiation, originating from the acceleration of particles in the EM-field of other particles: \[ \varepsilon_{ff}=\frac{C_1}{\lambda^2}\frac{z_{\rm i}\ni\ne}{\sqrt{k\Te}}\exp\left(-\frac{hc}{\lambda k\Te}\right)\xi_{ff}(\lambda,\Te) \] \end{enumerate} \section{The Boltzmann transport equation} It is assumed that there exists a distribution function $F$ for the plasma so that \[ F(\rr,\vv,t)=F_r(\rr,t)\cdot F_v(\vv,t)=F_1(x,t)F_2(y,t)F_3(z,t)F_4(v_x,t)F_5(v_y,t)F_6(v_z,t) \] Then the BTE is: $\displaystyle \frac{dF}{dt}=\Q{F}{t}+\nabla_r\cdot(F\vv\,)+\nabla_v\cdot(F\aaa\,)=\Qc{F}{t}{\rm coll-rad} $ \npar Assuming that $v$ does not depend on $r$ and $a_i$ does not depend on $v_i$, holds $\nabla_r\cdot(F\vv\,)=\vv\cdot\nabla F$ and $\nabla_v\cdot(F\aaa\,)=\aaa\cdot\nabla_vF$. This is also true in magnetic fields because $\partial a_i/\partial x_i=0$. The velocity is separated in a thermal velocity $\vv_{\rm t}$ and a drift velocity $\vec{w}$. The total density is given by $n=\int Fd\vv$ and $\int\vv Fd\vv=n\vec{w}$. \npar The balance equations can be derived by means of the moment method: \begin{enumerate} \item Mass balance: $\displaystyle\int({\rm BTE})d\vv\Rightarrow\Q{n}{t}+\nabla\cdot(n\vec{w})=\Qc{n}{t}{\rm cr}$ \item Momentum balance: $\displaystyle\int({\rm BTE})m\vv d\vv\Rightarrow mn\frac{d\vec{w}}{dt}+\nabla\TT'+\nabla p=mn\av{\aaa~}+\vec{R}$ \item Energy balance: $\displaystyle\int({\rm BTE})mv^2d\vv\Rightarrow\frac{3}{2}\frac{dp}{dt}+\frac{5}{2}p\nabla\cdot\vec{w}+\nabla\cdot\vec{q}=Q$ \end{enumerate} Here, $\av{\aaa~}=e/m(\vec{E}+\vec{w}\times\vvec{B})$ is the average acceleration, $\vec{q}=\half nm\av{\vv_{\rm t}^{~2}\vv_{\rm t}}$ the heat flow,\\ $\displaystyle Q=\int\frac{mv_{\rm t}^2}{r}\Qc{F}{t}{\rm cr}d\vv$ the source term for energy production, $\vec{R}$ is a friction term and $p=nkT$ the pressure. \npar A thermodynamic derivation gives for the total pressure: $\displaystyle p=nkT=\sum_ip_i-\frac{e^2(\ne+z_{\rm i}\ni)}{24\pi\varepsilon_0\lambda_{\rm D}}$ \npar For the electrical conductance in a plasma follows from the momentum balance, if $w_{\rm e}\gg w_{\rm i}$: \[ \eta\vec{J}=\vec{E}-\frac{\vec{J}\times\vec{B}+\nabla p_{\rm e}}{e\ne} \] In a plasma where only elastic e-a collisions are important the equilibrium energy distribution function is the {\it Druyvesteyn distribution}: \[ N(E)dE=C\ne\left(\frac{E}{E_0}\right)^{3/2}\exp\left[-\frac{3\me}{m_0}\left(\frac{E}{E_0}\right)^2\right]dE \] with $E_0=eE\lambda_{\rm v}=eE/n\sigma$. \section{Collision-radiative models} These models are first-moment equations for excited states. One assumes the Quasi-steady-state solution is valid, where $\forall_{p>1}[(\partial n_p/\partial t=0)\wedge(\nabla\cdot(n_p\vec{w}_p)=0)]$. This results in: \[ \Qc{n_{p>1}}{t}{\rm cr}=0~~,~~ \Q{n_1}{t}+\nabla\cdot(n_1\vec{w}_1)=\Qc{n_1}{t}{\rm cr}~~,~~ \Q{\ni}{t}+\nabla\cdot(\ni\vec{w}_{\rm i})=\Qc{n_{\rm i}}{t}{\rm cr} \] with solutions $n_p=r_p^0n_p^{\rm S}+r_p^1n_p^{\rm B}=b_pn_p^{\rm S}$. Further holds for all collision-dominated levels that $\delta b_p:=b_p-1=b_0p_{\rm eff}^{-x}$ with $p_{\rm eff}=\sqrt{Ry/E_{p{\rm i}}}$ and $5\leq x\leq6$. For systems in ESP, where only collisional (de)excitation between levels $p$ and $p\pm1$ is taken into account holds $x=6$. Even in plasma's far from equilibrium the excited levels will eventually reach ESP, so from a certain level up the level densities can be calculated. \npar To find the population densities of the lower levels in the stationary case one has to start with a macroscopic equilibrium: \[ \mbox{Number of populating processes of level}~p~=~ \mbox{Number of depopulating processes of level}~p~, \] When this is expanded it becomes: \[ % population \underbrace{\ne\sum_{qp}n_qK_{qp}}_{\rm coll.~deexcit.}+ \underbrace{\sum_{q>p}n_qA_{qp}}_{\rm rad.~deex.~to}+ \underbrace{\ne^2\ni K_{+p}}_{\rm coll.~recomb.}+ \underbrace{\ne\ni\alpha_{\rm rad}}_{\rm rad.~recomb}= \] \[ % depopulation \underbrace{\ne n_p\sum_{qp}K_{pq}}_{\rm coll.~excit.}+ \underbrace{n_p\sum_{q0$: $S_i$ and $S_j$ become parallel: the material is a ferromagnet. \item $J<0$: $S_i$ and $S_j$ become antiparallel: the material is an antiferromagnet. \end{enumerate} Heisenberg's theory predicts quantized spin waves: magnons. Starting from a model with only nearest neighbouring atoms interacting one can write: \[ U=-2J\vec{S}_p\cdot(\vec{S}_{p-1}+\vec{S}_{p+1})\approx \vec{\mu}_p\cdot\vec{B}_p~~~\mbox{with}~~~ \vec{B}_p=\frac{-2J}{g\mu_{\rm B}}(\vec{S}_{p-1}+\vec{S}_{p+1}) \] The equation of motion for the magnons becomes: $\displaystyle\frac{d\vec{S}}{dt}=\frac{2J}{\hbar}\vec{S}_p\times(\vec{S}_{p-1}+\vec{S}_{p+1})$ \npar From here the treatment is analogous to phonons: postulate traveling waves of the type $\vec{S}_p=\vec{u}\exp(i(pka-\omega t))$. This results in a system of linear equations with solution: \[ \ho=4JS(1-\cos(ka)) \] \section{Free electron Fermi gas} \subsection{Thermal heat capacity} The solution with period $L$ of the one-dimensional Schr\"odinger equation is: $\psi_n(x)=A\sin(2\pi x/\lambda n)$ with $n\lambda_n=2L$. From this follows \[ E=\frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2 \] In a linear lattice the only important quantum numbers are $n$ and $m_s$. The {\em Fermi level} is the uppermost filled level in the ground state, which has the {\em Fermi-energy} $E_{\rm F}$. If $n_{\rm F}$ is the quantum number of the Fermi level, it can be expressed as: $2n_{\rm F}=N$ so $E_{\rm F}=\hbar^2\pi^2 N^2/8mL$. In 3 dimensions holds: \[ k_{\rm F}=\left(\frac{3\pi^2N}{V}\right)^{1/3}~~\mbox{and}~~ E_{\rm F}=\frac{\hbar^2}{2m}\left(\frac{3\pi^2N}{V}\right)^{2/3} \] The number of states with energy $\leq E$ is then: $\displaystyle N=\frac{V}{3\pi^2}\left(\frac{2mE}{\hbar^2}\right)^{3/2}$. \npar and the density of states becomes: $\displaystyle D(E)=\frac{dN}{dE}=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{E}=\frac{3N}{2E}$. \npar The heat capacity of the electrons is approximately 0.01 times the classical expected value $\frac{3}{2}Nk$. This is caused by the Pauli exclusion principle and the Fermi-Dirac distribution: only electrons within an energy range $\sim kT$ of the Fermi level are excited thermally. There is a fraction $\approx T/T_{\rm F}$ excited thermally. The internal energy then becomes: \[ U\approx NkT\frac{T}{T_{\rm F}}~~\mbox{and}~~C=\Q{U}{T}\approx Nk\frac{T}{T_{\rm F}} \] A more accurate analysis gives: $C_{\rm electrons}=\frac{1}{2}\pi^2NkT/T_{\rm F}\sim T$. Together with the $T^3$ dependence of the thermal heat capacity of the phonons the total thermal heat capacity of metals is described by: $C=\gamma T+AT^3$. \subsection{Electric conductance} The equation of motion for the charge carriers is: $\vec{F}=md\vv/dt=\hbar d\vec{k}/dt$. The variation of $\vec{k}$ is given by $\delta\vec{k}=\vec{k}(t)-\vec{k}(0)=-e\vec{E}t/\hbar$. If $\tau$ is the characteristic collision time of the electrons, $\delta\vec{k}$ remains stable if $t=\tau$. Then holds: $\av{\vv\,}=\mu\vec{E}$, with $\mu=e\tau/m$ the {\it mobility} of the electrons. \npar The current in a conductor is given by: $\vec{J}=nq\vv=\sigma\vec{E}=\vec{E}/\rho=ne\mu\vec{E}$. Because for the collision time holds: $1/\tau=1/\tau_L+1/\tau_i$, where $\tau_L$ is the collision time with the lattice phonons and $\tau_i$ the collision time with the impurities follows for the resistivity $\rho=\rho_L+\rho_i$, with $\lim\limits_{T\rightarrow 0}\rho_L=0$. \subsection{The Hall-effect} If a magnetic field is applied $\perp$ to the direction of the current the charge carriers will be pushed aside by the Lorentz force. This results in a magnetic field $\perp$ to the flow direction of the current. If $\vec{J}=J\vec{e}_x$ and $\vec{B}=B\vec{e}_z$ than $E_y/E_x=\mu B$. The Hall coefficient is defined by: $R_{\rm H}=E_y/J_x B$, and $R_{\rm H}=-1/ne$ if $J_x=ne\mu E_x$. The Hall voltage is given by: $V_{\rm H}=Bvb=IB/neh$ where $b$ is the width of the material and $h$ de height. \subsection{Thermal heat conductivity} With $\ell =v_F\tau$ the mean free path of the electrons follows from $\kappa=\frac{1}{3}C\av{v}\ell$: $\kappa_{\rm electrons}=\pi^2nk^2T\tau/3m$. From this follows for the {\it Wiedemann-Franz ratio}: $\kappa/\sigma=\frac{1}{3}(\pi k/e)^2T$. \section{Energy bands} In the {\it tight-bond} approximation it is assumed that $\psi={\rm e}^{ikna}\phi(x-na)$. From this follows for the energy: $\av{E}=\av{\psi|H|\psi}=E_{\rm at}-\alpha-2\beta\cos(ka)$. So this gives a cosine superimposed on the atomic energy, which can often be approximated by a harmonic oscillator. If it is assumed that the electron is nearly free one can postulate: $\psi=\exp(i\vec{k}\cdot\rr\,)$. This is a traveling wave. This wave can be decomposed into two standing waves: \begin{eqnarray*} \psi(+)&=&\exp(i\pi x/a)+\exp(-i\pi x/a)=2\cos(\pi x/a)\\ \psi(-)&=&\exp(i\pi x/a)-\exp(-i\pi x/a)=2i\sin(\pi x/a) \end{eqnarray*} The probability density $|\psi(+)|^2$ is high near the atoms of the lattice and low in between. The probability density $|\psi(-)|^2$ is low near the atoms of the lattice and high in between. Hence the energy of $\psi(+)$ is also lower than the energy of $\psi)(-)$. Suppose that $U(x)=U\cos(2\pi x/a)$, than the bandgap is given by: \[ E_{\rm gap}=\int\limits_0^1 U(x)\left[ |\psi(+)|^2-|\psi(-)|^2\right] dx=U \] \section{Semiconductors} The band structures and the transitions between them of direct and indirect semiconductors are shown in the figures below. Here it is assumed that the momentum of the absorbed photon can be neglected. For an indirect semiconductor a transition from the valence- to the conduction band is also possible if the energy of the absorbed photon is smaller than the band gap: then, also a phonon is absorbed. \begin{center} \begin{picture}(140,45)(-70,0) \put(-60,0){\framebox(50,45)[b]{\em Direct transition\rule[-5pt]{0pt}{10pt}}} \put(-37,8.5){\vector(0,1){32}} \put(-37,33){\vector(0,1){0}} \put(-37,20){\oval(3,3)[l]} \put(-37,26){\oval(3,3)[l]} \put(-37,23){\oval(3,3)[r]} \put(-37,29){\oval(3,3)[r]} \put(-36,41.5){$E$} \put(-27,41.7){conduction} \put(-24,38.5){band} \put(-34,25){$\omega_{\rm g}$} \put(-37.9,16.5){$\circ$} \bezier{250}(-47,44)(-37,23)(-27,44) \bezier{250}(-57,10.5)(-37,24)(-17,10.5) \put(-37.9,33){$\bullet$} \put(10,0){\framebox(50,45)[b]{\em Indirect transition\rule[-5pt]{0pt}{10pt}}} \put(33,8.5){\vector(0,1){32}} % photon \put(33,30.5){\vector(0,1){0}} \put(33,20){\oval(3,3)[l]} \put(33,26){\oval(3,3)[l]} \put(33,23){\oval(3,3)[r]} \put(32,41.5){$E$} \put(32.1,16.5){$\circ$} % conduction band \bezier{180}(10,38)(15,23)(20,38) \bezier{180}(46,38)(51,23)(56,38) \bezier{100}(26,38)(22,43)(20,38) \bezier{100}(46,38)(44,43)(40,38) \bezier{200}(26,38)(33,30)(40,38) \bezier{250}(13,10.5)(33,24)(53,10.5) \put(14.3,30){$\bullet$} % phonon \put(33,30.5){\line(-1,0){7}} \put(22.3,30.5){\vector(-1,0){7}} \jwline{22.3}{30.5}{23.225}{32.8} \jwline{23.225}{32.8}{25.075}{28.2} \jwline{25.075}{28.2}{26}{30.5} \put(36,25){$\omega$} \put(23,24){$\Omega$} \end{picture} \end{center} This difference can also be observed in the absorption spectra: \begin{center} \begin{picture}(140,45)(-70,-15) \put(-60,-15){\framebox(50,45)[b]{\em Direct semiconductor\rule[-5pt]{0pt}{10pt}}} \put(-58,-3.5){\vector(0,1){28}} \put(-58,-3.5){\vector(1,0){40}} \put(-59,26.5){absorption} \put(-17,-4){$E$} \put(-43,-7){$\ho_{\rm g}$} \multiput(-42,-3.5)(0,4){5}{\rule{0.4pt}{2.5mm}} \bezier{300}(-42,-3.5)(-34,22)(-12,26.5) \put(10,-15){\framebox(50,45)[b]{\em Indirect semiconductor\rule[-5pt]{0pt}{10pt}}} \put(12,-3.5){\vector(0,1){28}} \put(12,-3.5){\vector(1,0){40}} \put(11,26.5){absorption} \put(54,-4){$E$} \put(14,-7){$E_{\rm g}+\hbar\Omega$} \put(17,-3.5){\line(2,3){7}} \multiput(23.67,6.7)(0.5,0.75){10}{.} \multiput(23.33,6.7)(-0.15,-1){11}{.} \bezier{300}(23.67,6.7)(25,24)(53,26.5) \end{picture} \end{center} So indirect semiconductors, like Si and Ge, cannot emit any light and are therefore not usable to fabricate lasers. When light is absorbed holds: $\vec{k}_{\rm h}=-\vec{k}_{\rm e}$, $E_{\rm h}(\vec{k}_{\rm h})=-E_{\rm e}(\vec{k}_{\rm e})$, $\vv_{\rm h}=\vv_{\rm e}$ and $m_{\rm h}=-m_{\rm e}^*$ if the conduction band and the valence band have the same structure. \npar Instead of the normal electron mass one has to use the {\it effective mass} within a lattice. It is defined by: \[ m^*=\frac{F}{a}=\frac{dp/dt}{dv_{\rm g}/dt}=\hbar\frac{dK}{dv_{\rm g}}= \hbar^2\left(\frac{d^2E}{dk^2}\right)^{-1} \] with $E=\ho$ and $v_{\rm g}=d\omega/dk$ and $p=\hbar k$. \npar With the distribution function $f_{\rm e}(E)\approx\exp((\mu-E)/kT)$ for the electrons and $f_{\rm h}(E)=1-f_{\rm e}(E)$ for the holes the density of states is given by: \[ D(E)=\frac{1}{2\pi^2}\left(\frac{2m^*}{\hbar^2}\right)^{3/2}\sqrt{E-E_c} \] with $E_c$ the energy at the edge of the conductance band. From this follows for the concentrations of the holes $p$ and the electrons $n$: \[ n=\int\limits_{E_c}^{\infty}D_{\rm e}(E)f_{\rm e}(E)dE=2\left(\frac{m^*kT}{2\pi\hbar^2} \right)^{3/2}\exp\left(\frac{\mu-E_c}{kT}\right) \] For the product $np$ follows: $\displaystyle np=4\left(\frac{kT}{2\pi\hbar^2}\right)^3 \sqrt{m^*_{\rm e}m_{\rm h}}\exp\left(-\frac{E_{\rm g}}{kT}\right)$ \npar For an intrinsic (no impurities) semiconductor holds: $n_{\rm i}=p_{\rm i}$, for a $n-type$ holds: $n>p$ and in a $p-type$ holds: $n0}\psi_l~~\mbox{and}~~\psi_0=\frac{\sin(kr)}{kr} \] If the potential is approximately rectangular holds: $\displaystyle\psi_0'=C\frac{\sin(kr+\delta_0)}{kr}$ \npar The cross section is then $\displaystyle \sigma(\theta)=\frac{\sin^2(\delta_0)}{k^2}~~\mbox{so}~~ \sigma=\int\sigma(\theta)d\Omega=\frac{4\pi\sin^2(\delta_0)}{k^2}$ \npar At very low energies holds: $\displaystyle\sin^2(\delta_0)=\frac{\hbar^2k^2/2m}{W_0+W}$ \npar with $W_0$ the depth of the potential well. At higher energies holds: $\displaystyle\sigma=\frac{4\pi}{k^2}\sum_l\sin^2(\delta_l)$ \subsection{Conservation of energy and momentum in nuclear reactions} If a particle $P_1$ collides with a particle $P_2$ which is in rest w.r.t. the laboratory system and other particles are created, so \[ P_1+P_2\rightarrow\sum_{k>2} P_k \] the total energy $Q$ gained or required is given by $Q=(m_1+m_2-\sum\limits_{k>2}m_k)c^2$. \npar The minimal required kinetic energy $T$ of $P_1$ in the laboratory system to initialize the reaction is \[ T=-Q\frac{m_1+m_2+\sum m_k}{2m_2} \] If $Q<0$ there is a threshold energy. \section{Radiation dosimetry} {\it Radiometric quantities} determine the strength of the radiation source(s). {\it Dosimetric quantities} are related to the energy transfer from radiation to matter. Parameters describing a relation between those are called {\it interaction parameters}. The intensity of a beam of particles in matter decreases according to $I(s)=I_0\exp(-\mu s)$. The deceleration of a {\it heavy} particle is described by the {\it Bethe-Bloch equation}: \[ \frac{dE}{ds}\sim\frac{q^2}{v^2} \] \npar The {\it fluention} is given by $\Phi=dN/dA$. The {\it flux} is given by $\phi=d\Phi/dt$. The energy loss is defined by $\Psi=dW/dA$, and the energy flux density $\psi=d\Psi/dt$. The {\it absorption coefficient} is given by $\mu=(dN/N)/dx$. The {\it mass absorption coefficient} is given by $\mu/\varrho$. \npar The {\it radiation dose} $X$ is the amount of charge produced by the radiation per unit of mass, with unit C/kg. An old unit is the R\"ontgen: 1Ro$=2.58\cdot10^{-4}$ C/kg. With the energy-absorption coefficient $\mu_E$ follows: \[ X=\frac{dQ}{dm}=\frac{e\mu_E}{W\varrho}\Psi \] where $W$ is the energy required to disjoin an elementary charge. \npar The {\it absorbed dose} $D$ is given by $D=dE_{\rm abs}/dm$, with unit Gy=J/kg. An old unit is the rad: 1 rad=0.01 Gy. The {\it dose tempo} is defined as $\dot{D}$. It can be derived that \[ D=\frac{\mu_E}{\varrho}\Psi \] The {\it Kerma} $K$ is the amount of kinetic energy of secundary produced particles which is produced per mass unit of the radiated object. \npar The {\it equivalent dose} $H$ is a weight average of the absorbed dose per type of radiation, where for each type radiation the effects on biological material is used for the weight factor. These weight factors are called the quality factors. Their unit is Sv. $H=QD$. If the absorption is not equally distributed also weight factors $w$ per organ need to be used: $H=\sum w_k H_k$. For some types of radiation holds: \begin{center} \begin{tabular}{||l|r||} \hline {\bf Radiation type}&\boldmath$Q$\unboldmath\\ \hline \hline R\"ontgen, gamma radiation&1\\ $\beta$, electrons, mesons&1\\ Thermic neutrons&3 to 5\\ Fast neutrons&10 to 20\\ protons&10\\ $\alpha$, fission products&20\\ \hline \end{tabular} \end{center} \chapter{Quantum field theory \& Particle physics} \typeout{Quantum field theory & Particle physics} \section[~~Creation and annihilation operators]{Creation and annihilation operators} A state with more particles can be described by a collection occupation numbers $|n_1n_2n_3\cdots\rangle$. Hence the vacuum state is given by $|000\cdots\rangle$. This is a complete description because the particles are indistinguishable. The states are orthonormal: \[ \av{n_1n_2n_3\cdots|n_1'n_2'n_3'\cdots}=\prod_{i=1}^\infty\delta_{n_in_i'} \] The time-dependent state vector is given by \[ \Psi(t)=\sum_{n_1n_2\cdots}c_{n_1n_2\cdots}(t)|n_1n_2\cdots\rangle \] The coefficients $c$ can be interpreted as follows: $|c_{n_1n_2\cdots}|^2$ is the probability to find $n_1$ particles with momentum $\vec{k}_1$, $n_2$ particles with momentum $\vec{k}_2$, etc., and $\av{\Psi(t)|\Psi(t)}=\sum|c_{n_i}(t)|^2=1$. The expansion of the states in time is described by the Schr\"odinger equation \[ i\frac{d}{dt}|\Psi(t)\rangle=H|\Psi(t)\rangle \] where $H=H_0+H_{\rm int}$. $H_0$ is the Hamiltonian for free particles and keeps $|c_{n_i}(t)|^2$ constant, $H_{\rm int}$ is the interaction Hamiltonian and can increase or decrease a $c^2$ at the cost of others. \npar All operators which can change occupation numbers can be expanded in the $a$ and $a^\dagger$ operators. $a$ is the {\it annihilation operator} and $a^\dagger$ the {\it creation operator}, and: \begin{eqnarray*} a(\vec{k}_i)|n_1n_2\cdots n_i\cdots\rangle&=&\sqrt{n_i}~|n_1n_2\cdots n_i-1\cdots\rangle\\ a^\dagger(\vec{k}_i)|n_1n_2\cdots n_i\cdots\rangle&=&\sqrt{n_i+1}~|n_1n_2\cdots n_i+1\cdots\rangle \end{eqnarray*} Because the states are normalized holds $a|0\rangle=0$ and $a(\vec{k}_i)a^\dagger(\vec{k}_i)|n_i\rangle=n_i|n_i\rangle$. So $aa^\dagger$ is an occupation number operator. The following commutation rules can be derived: \[ [a(\vec{k}_i),a(\vec{k}_j)]=0~~,~~~ [a^\dagger(\vec{k}_i),a^\dagger(\vec{k}_j)]=0~~,~~~ [a(\vec{k}_i),a^\dagger(\vec{k}_j)]=\delta_{ij} \] Hence for free spin-0 particles holds: $H_0=\sum\limits_i a^\dagger(\vec{k}_i)a(\vec{k}_i)\hbar\omega_{k_i}$ \section[~~Classical and quantum fields]{Classical and quantum fields} Starting with a real field $\Phi^\alpha(x)$ (complex fields can be split in a real and an imaginary part), the {\it Lagrange density} $\LL$ is a function of the position $x=(\vec{x},ict)$ through the fields: $\LL=\LL(\Phi^\alpha(x),\partial_\nu\Phi^\alpha(x))$. The Lagrangian is given by $L=\int\LL(x)d^3x$. Using the variational principle $\delta I(\Omega)=0$ and with the action-integral $I(\Omega)=\int\LL(\Phi^\alpha,\partial_\nu\Phi^\alpha)d^4x$ the field equation can be derived: \[ \Q{\LL}{\Phi^\alpha}-\Q{}{x_\nu}\Q{\LL}{(\partial_\nu\Phi^\alpha)}=0 \] The {\it conjugated field} is, analogous to momentum in classical mechanics, defined as: \[ \Pi^\alpha(x)=\Q{\LL}{\dot{\Phi}^\alpha} \] With this, the Hamilton density becomes ${\cal H}(x)=\Pi^\alpha\dot{\Phi}^\alpha-\LL(x)$. \npar Quantization of a classical field is analogous to quantization in point mass mechanics: the field functions are considered as operators obeying certain commutation rules: \[ [\Phi^\alpha(\vec{x}),\Phi^\beta(\vvec{x}')]=0~~,~~~ [\Pi^\alpha(\vec{x}),\Pi^\beta(\vvec{x}')]=0~~,~~~ [\Phi^\alpha(\vec{x}),\Pi^\beta(\vvec{x}')]=i\delta_{\alpha\beta}(\vec{x}-\vvec{x}') \] \section[~~The interaction picture]{The interaction picture} Some equivalent formulations of quantum mechanics are possible: \begin{enumerate} \item Schr\"odinger picture: time-dependent states, time-independent operators. \item Heisenberg picture: time-independent states, time-dependent operators. \item Interaction picture: time-dependent states, time-dependent operators. \end{enumerate} The interaction picture can be obtained from the Schr\"odinger picture by an unitary transformation: \[ |\Phi(t)\rangle={\rm e}^{iH_0^{\rm S}}|\Phi^{\rm S}(t)\rangle~~\mbox{and}~~ O(t)={\rm e}^{iH_0^{\rm S}}O^{\rm S}{\rm e}^{-iH_0^{\rm S}} \] The index $^{\rm S}$ denotes the Schr\"odinger picture. From this follows: \[ i\frac{d}{dt}|\Phi(t)\rangle=H_{\rm int}(t)|\Phi(t)\rangle~~\mbox{and}~~ i\frac{d}{dt}O(t)=[O(t),H_0] \] \section[~~Real scalar field in the interaction picture]{Real scalar field in the interaction picture} It is easy to find that, with $M:=m_0^2c^2/\hbar^2$, holds: \[ \Q{}{t}\Phi(x)=\Pi(x)~~\mbox{and}~~\Q{}{t}\Pi(x)=(\nabla^2-M^2)\Phi(x) \] From this follows that $\Phi$ obeys the Klein-Gordon equation $(\Box-M^2)\Phi=0$. With the definition $k_0^2=\vec{k}^2+M^2:=\omega_k^2$ and the notation $\vec{k}\cdot\vec{x}-ik_0t:=kx$ the general solution of this equation is: \[ \Phi(x)=\frac{1}{\sqrt{V}}\sum_{\vec{k}}\frac{1}{\sqrt{2\omega_k}} \left(a(\vvec{k}){\rm e}^{ikx}+a^\dagger(\vvec{k}){\rm e}^{-ikx}\right)~,~~ \Pi(x)=\frac{i}{\sqrt{V}}\sum_{\vec{k}}\sqrt{\half\omega_k} \left(-a(\vvec{k}){\rm e}^{ikx}+a^\dagger(\vvec{k}){\rm e}^{-ikx}\right) \] The field operators contain a volume $V$, which is used as normalization factor. Usually one can take the limit $V\rightarrow\infty$. \npar In general it holds that the term with ${\rm e}^{-ikx}$, the positive frequency part, is the creation part, and the negative frequency part is the annihilation part. \npar the coefficients have to be each others hermitian conjugate because $\Phi$ is hermitian. Because $\Phi$ has only one component this can be interpreted as a field describing a particle with spin zero. From this follows that the commutation rules are given by $[\Phi(x),\Phi(x')]=i\Delta(x-x')$ with \[ \Delta(y)=\frac{1}{(2\pi)^3}\int\frac{\sin(ky)}{\omega_k}d^3k \] $\Delta(y)$ is an odd function which is invariant for proper Lorentz transformations (no mirroring). This is consistent with the previously found result $[\Phi(\vec{x},t,\Phi(\vvec{x}',t)]=0$. In general holds that $\Delta(y)=0$ outside the light cone. So the equations obey the locality postulate. \npar The Lagrange density is given by: $\LL(\Phi,\partial_\nu\Phi)=-\half(\partial_\nu\Phi\partial_\nu\Phi+m^2\Phi^2)$. The energy operator is given by: \[ H=\int{\cal H}(x)d^3x=\sum_{\vec{k}}\hbar\omega_ka^\dagger(\vvec{k})a(\vvec{k}) \] \section[~~Charged spin-0 particles, conservation of charge]{Charged spin-0 particles, conservation of charge} The Lagrange density of charged spin-0 particles is given by: $\LL=-(\partial_\nu\Phi\partial_\nu\Phi^*+M^2\Phi\Phi^*)$. \npar Noether's theorem connects a continuous symmetry of $\LL$ and an additive conservation law. Suppose that $\LL\left((\Phi^\alpha)',\partial_\nu(\Phi^\alpha)'\right)= \LL\left(\Phi^\alpha,\partial_\nu\Phi^\alpha\right)$ and there exists a continuous transformation between $\Phi^\alpha$ and ${\Phi^\alpha}'$ such as ${\Phi^\alpha}'=\Phi^\alpha+\epsilon f^\alpha(\Phi)$. Then holds \[ \Q{}{x_\nu}\left(\Q{\LL}{(\partial_\nu\Phi^\alpha)}f^\alpha\right)=0 \] This is a continuity equation $\Rightarrow$ conservation law. Which quantity is conserved depends on the symmetry. The above Lagrange density is invariant for a change in phase $\Phi\rightarrow\Phi{\rm e}^{i\theta}$: a global gauge transformation. The conserved quantity is the current density $J_\mu(x)=-ie(\Phi\partial_\mu\Phi^*-\Phi^*\partial_\mu\Phi)$. Because this quantity is 0 for real fields a complex field is needed to describe charged particles. When this field is quantized the field operators are given by \[ \Phi(x)=\frac{1}{\sqrt{V}}\sum_{\vec{k}}\frac{1}{\sqrt{2\omega_k}} \left(a(\vvec{k}){\rm e}^{ikx}+b^\dagger(\vvec{k}){\rm e}^{-ikx}\right) ~,~~ \Phi^\dagger(x)=\frac{1}{\sqrt{V}}\sum_{\vec{k}}\frac{1}{\sqrt{2\omega_k}} \left(a^\dagger(\vvec{k}){\rm e}^{ikx}+b(\vvec{k}){\rm e}^{-ikx}\right) \] Hence the energy operator is given by: \[ H=\sum_{\vec{k}}\hbar\omega_k\left(a^\dagger(\vvec{k})a(\vvec{k})+b^\dagger(\vvec{k})b(\vvec{k})\right) \] and the charge operator is given by: \[ Q(t)=-i\int J_4(x)d^3x\Rightarrow Q=\sum_{\vec{k}}e\left(a^\dagger(\vvec{k})a(\vvec{k})-b^\dagger(\vvec{k})b(\vvec{k})\right) \] From this follows that $a^\dagger a:=N_+(\vvec{k})$ is an occupation number operator for particles with a positive charge and $b^\dagger b:=N_-(\vvec{k})$ is an occupation number operator for particles with a negative charge. \section[~~Field functions for spin-$\frac{1}{2}$ particles]{Field functions for spin-$\frac{1}{2}$ particles} Spin is defined by the behaviour of the solutions $\psi$ of the Dirac equation. A {\it scalar} field $\Phi$ has the property that, if it obeys the Klein-Gordon equation, the rotated field $\tilde{\Phi}(x):=\Phi(\Lambda^{-1}x)$ also obeys it. $\Lambda$ denotes 4-dimensional rotations: the proper Lorentz transformations. These can be written as: \[ \tilde{\Phi}(x)=\Phi(x){\rm e}^{-i\vec{n}\cdot\vec{L}}~~\mbox{with}~~ L_{\mu\nu}=-i\hbar\left(x_\mu\Q{}{x_\nu}-x_\nu\Q{}{x_\mu}\right) \] For $\mu\leq3,\nu\leq3$ these are rotations, for $\nu=4,\mu\neq4$ these are Lorentz transformations. \npar A rotated field $\tilde{\psi}$ obeys the Dirac equation if the following condition holds: $\tilde{\psi}(x)=D(\Lambda)\psi(\Lambda^{-1}x)$. This results in the condition $D^{-1}\gamma_\lambda D=\Lambda_{\lambda\mu}\gamma_\mu$. One finds: $D={\rm e}^{i\vec{n}\cdot\vec{S}}$ with $S_{\mu\nu}=-i\half\hbar\gamma_\mu\gamma_\nu$. Hence: \[ \tilde{\psi}(x)={\rm e}^{-i(S+L)}\psi(x)={\rm e}^{-iJ}\psi(x) \] Then the solutions of the Dirac equation are given by: \[ \psi(x)=u_\pm^r(\vvec{p}){\rm e}^{-i(\vec{p}\cdot\vec{x}\pm Et)} \] Here, $r$ is an indication for the direction of the spin, and $\pm$ is the sign of the energy. With the notation $v^r(\vvec{p})=u^r_-(-\vvec{p})$ and $u^r(\vvec{p})=u^r_+(\vvec{p})$ one can write for the dot products of these spinors: \[ u_+^r(\vvec{p})u_+^{r'}(\vvec{p})=\frac{E}{M}\delta_{rr'}~~,~~ u_-^r(\vvec{p})u_-^{r'}(\vvec{p})=\frac{E}{M}\delta_{rr'}~~,~~ u_+^r(\vvec{p})u_-^{r'}(\vvec{p})=0 \] Because of the factor $E/M$ this is not relativistic invariant. A Lorentz-invariant dot product is defined by $\overline{a}b:=a^\dagger\gamma_4b$, where $\overline{a}:=a^\dagger\gamma_4$ is a row spinor. From this follows: \[ \overline{u^r(\vvec{p})}u^{r'}(\vvec{p})=\delta_{rr'}~~,~~ \overline{v^r(\vvec{p})}v^{r'}(\vvec{p})=-\delta_{rr'}~~,~~ \overline{u^r(\vvec{p})}v^{r'}(\vvec{p})=0 \] Combinations of the type $a\overline{a}$ give a $4\times4$ matrix: \[ \sum_{r=1}^2 u^r(\vvec{p})\overline{u^r(\vvec{p})}=\frac{-i\gamma_\lambda p_\lambda+M}{2M}~~,~~ \sum_{r=1}^2 v^r(\vvec{p})\overline{v^r(\vvec{p})}=\frac{-i\gamma_\lambda p_\lambda-M}{2M} \] The Lagrange density which results in the Dirac equation and having the correct energy normalization is: \[ \LL(x)=-\overline{\psi(x)}\left(\gamma_\mu\Q{}{x_\mu}+M\right)\psi(x) \] and the current density is $J_\mu(x)=-ie\overline{\psi}\gamma_\mu\psi$. \section[~~Quantization of spin-$\frac{1}{2}$ fields]{Quantization of spin-$\frac{1}{2}$ fields} The general solution for the fieldoperators is in this case: \[ \psi(x)=\sqrt{\frac{M}{V}}\sum_{\vec{p}}\frac{1}{\sqrt{E}}\sum_r \left(c_r(\vvec{p})u^r(\vvec{p}){\rm e}^{ipx}+ d^\dagger_r(\vvec{p})v^r(\vvec{p}){\rm e}^{-ipx}\right) \] and \[ \overline{\psi(x)}=\sqrt{\frac{M}{V}}\sum_{\vec{p}}\frac{1}{\sqrt{E}}\sum_r \left(c^\dagger_r(\vvec{p})\overline{u^r}(\vvec{p}){\rm e}^{-ipx}+ d_r(\vvec{p})\overline{v^r}(\vvec{p}){\rm e}^{ipx}\right) \] Here, $c^\dagger$ and $c$ are the creation respectively annihilation operators for an electron and $d^\dagger$ and $d$ the creation respectively annihilation operators for a positron. The energy operator is given by \[ H=\sum_{\vec{p}}E_{\vec{p}}\sum_{r=1}^2\left(c_r^\dagger(\vvec{p})c_r(\vvec{p})-d_r(\vvec{p})d^\dagger_r(\vvec{p})\right) \] To prevent that the energy of positrons is negative the operators must obey anti commutation rules in stead of commutation rules: \[ [c_r(\vvec{p}),c_{r'}^\dagger(\vvec{p})]_+=[d_r(\vvec{p}),d_{r'}^\dagger(\vvec{p})]_+=\delta_{rr'}\delta_{pp'}~~,~~ \mbox{all other anti commutators are 0.} \] The field operators obey \[ [\psi_\alpha(x),\psi_\beta(x')]=0~~,~~ [\overline{\psi_\alpha(x)},\overline{\psi_\beta(x')}]=0~~,~~ [\psi_\alpha(x),\overline{\psi_\beta(x')}]_+=-iS_{\alpha\beta}(x-x') \] \[ \mbox{with}~~~S(x)=\left(\gamma_\lambda\Q{}{x_\lambda}-M\right)\Delta(x) \] The anti commutation rules give besides the positive-definite energy also the Pauli exclusion principle and the Fermi-Dirac statistics: because $c_r^\dagger(\vvec{p})c_r^\dagger(\vvec{p})=-c_r^\dagger(\vvec{p})c_r^\dagger(\vvec{p})$ holds: $\{c_r^\dagger(p)\}^2=0$. It appears to be impossible to create two electrons with the same momentum and spin. This is the exclusion principle. Another way to see this is the fact that $\{N_r^+(\vvec{p})\}^2=N_r^+(\vvec{p})$: the occupation operators have only eigenvalues 0 and 1. \npar To avoid infinite vacuum contributions to the energy and charge the {\it normal product} is introduced. The expression for the current density now becomes $J_\mu=-ieN(\overline{\psi}\gamma_\mu\psi)$. This product is obtained by: \begin{itemize} \item Expand all fields into creation and annihilation operators, \item Keep all terms which have no annihilation operators, or in which they are on the right of the creation operators, \item In all other terms interchange the factors so that the annihilation operators go to the right. By an interchange of two fermion operators add a minus sign, by interchange of two boson operators not. Assume hereby that all commutators are zero. \end{itemize} \section[~~Quantization of the electromagnetic field]{Quantization of the electromagnetic field} Starting with the Lagrange density $\displaystyle\LL=-\half\Q{A_\nu}{x_\mu}\Q{A_\nu}{x_\mu}$ \npar it follows for the field operators $A(x)$: \[ A(x)=\frac{1}{\sqrt{V}}\sum_{\vec{k}}\frac{1}{\sqrt{2\omega_k}}\sum_{m=1}^4 \left(a_m(\vvec{k})\epsilon^m(\vvec{k}){\rm e}^{ikx}+ a^\dagger(\vvec{k})\epsilon^m(\vvec{k})^*{\rm e}^{-ikx}\right) \] The operators obey $[a_m(\vvec{k}),a_{m'}^\dagger(\vvec{k})]=\delta_{mm'}\delta_{kk'}$. All other commutators are 0. $m$ gives the polarization direction of the photon: $m=1,2$ gives transversal polarized, $m=3$ longitudinal polarized and $m=4$ timelike polarized photons. Further holds: \[ [A_\mu(x),A_\nu(x')]=i\delta_{\mu\nu}D(x-x')~~\mbox{with}~~D(y)=\Delta(y)|_{m=0} \] In spite of the fact that $A_4=iV$ is imaginary in the classical case, $A_4$ is still defined to be hermitian because otherwise the sign of the energy becomes incorrect. By changing the definition of the inner product in configuration space the expectation values for $A_{1,2,3}(x)\in\RR$ and for $A_4(x)$ become imaginary. \npar If the potentials satisfy the Lorentz gauge condition $\partial_\mu A_\mu=0$ the $E$ and $B$ operators derived from these potentials will satisfy the Maxwell equations. However, this gives problems with the commutation rules. It is now demanded that only those states are permitted for which holds \[ \Q{A^+_\mu}{x_\mu}|\Phi\rangle=0 \] This results in:\hspace{4.4cm}$\displaystyle\av{\Q{A_\mu}{x_\mu}}=0$. \npar From this follows that $(a_3(\vvec{k})-a_4(\vvec{k}))|\Phi\rangle=0$. With a local gauge transformation one obtains $N_3(\vvec{k})=0$ and $N_4(\vvec{k})=0$. However, this only applies to free EM-fields: in intermediary states in interactions there can exist longitudinal and timelike photons. These photons are also responsible for the stationary Coulomb potential. \section[~~Interacting fields and the S-matrix]{Interacting fields and the S-matrix} The $S$(scattering)-matrix gives a relation between the initial and final states of an interaction: $|\Phi(\infty)\rangle=S|\Phi(-\infty)\rangle$. If the Schr\"odinger equation is integrated: \[ |\Phi(t)\rangle=|\Phi(-\infty)\rangle-i\int\limits_{-\infty}^t H_{\rm int}(t_1)|\Phi(t_1)\rangle dt_1 \] and perturbation theory is applied one finds that: \[ S=\sum_{n=0}^\infty\frac{(-i)^n}{n!}\int\cdots\int T\left\{{\cal H}_{\rm int}(x_1)\cdots{\cal H}_{\rm int}(x_n)\right\}d^4x_1\cdots d^4x_n \equiv\sum_{n=0}^\infty S^{(n)} \] Here, the $T$-operator means a {\it time-ordered product}: the terms in such a product must be ordered in increasing time order from the right to the left so that the earliest terms act first. The $S$-matrix is then given by: $S_{ij}=\av{\Phi_i|S|\Phi_j}=\av{\Phi_i|\Phi(\infty)}$. \npar The interaction Hamilton density for the interaction between the electromagnetic and the electron-positron field is: ${\cal H}_{\rm int}(x)=-J_\mu(x)A_\mu(x)=ieN(\overline{\psi}\gamma_\mu\psi A_\mu)$ \npar When this is expanded as: ${\cal H}_{\rm int}=ieN\left((\overline{\psi^+}+\overline{\psi^-})\gamma_\mu(\psi^++\psi^-)(A_\mu^++A_\mu^-)\right)$ \npar eight terms appear. Each term corresponds with a possible process. The term $ie\overline{\psi^+}\gamma_\mu\psi^+A_\mu^-$ acting on $|\Phi\rangle$ gives transitions where $A_\mu^-$ creates a photon, $\psi^+$ annihilates an electron and $\overline{\psi^+}$ annihilates a positron. Only terms with the correct number of particles in the initial and final state contribute to a matrix element $\av{\Phi_i|S|\Phi_j}$. Further the factors in ${\cal H}_{\rm int}$ can create and thereafter annihilate particles: the {\it virtual particles}. \npar The expressions for $S^{(n)}$ contain time-ordered products of normal products. This can be written as a sum of normal products. The appearing operators describe the minimal changes necessary to change the initial state into the final state. The effects of the virtual particles are described by the (anti)commutator functions. Some time-ordened products are: \begin{eqnarray*} T\left\{\Phi(x)\Phi(y)\right\}&=&N\left\{\Phi(x)\Phi(y)\right\}+\half\Delta^{\rm F}(x-y)\\ T\left\{\psi_\alpha(x)\overline{\psi_\beta(y)}\right\}&=&N\left\{\psi_\alpha(x)\overline{\psi_\beta(y)}\right\}-\half S_{\alpha\beta}^{\rm F}(x-y)\\ T\left\{A_\mu(x)A_\nu(y)\right\}&=&N\left\{A_\mu(x)A_\nu(y)\right\}+\half\delta_{\mu\nu}D_{\mu\nu}^{\rm F}(x-y) \end{eqnarray*} Here, $S^{\rm F}(x)=(\gamma_\mu\partial_\mu-M)\Delta^{\rm F}(x)$, $D^{\rm F}(x)=\Delta^{\rm F}(x)|_{m=0}$ and \[ \Delta^{\rm F}(x)=\left\{\begin{array}{ll} \displaystyle\frac{1}{(2\pi)^3}\int\frac{{\rm e}^{ikx}}{\omega_{\vec{k}}}d^3k&~\mbox{if}~x_0>0\\ \\ \displaystyle\frac{1}{(2\pi)^3}\int\frac{{\rm e}^{-ikx}}{\omega_{\vec{k}}}d^3k&~\mbox{if}~x_0<0 \end{array}\right. \] The term $\half\Delta^{\rm F}(x-y)$ is called the contraction of $\Phi(x)$ and $\Phi(y)$, and is the expectation value of the time-ordered product in the vacuum state. Wick's theorem gives an expression for the time-ordened product of an arbitrary number of field operators. The graphical representation of these processes are called {\it Feynman diagrams}. In the $x$-representation each diagram describes a number of processes. The contraction functions can also be written as: \[ \Delta^{\rm F}(x)=\lim_{\epsilon\rightarrow0}\frac{-2i}{(2\pi)^4} \int\frac{{\rm e}^{ikx}}{k^2+m^2-i\epsilon}d^4k ~~~\mbox{and}~~~ S^{\rm F}(x)=\lim_{\epsilon\rightarrow0}\frac{-2i}{(2\pi)^4} \int{\rm e}^{ipx}\frac{i\gamma_\mu p_\mu-M}{p^2+M^2-i\epsilon}d^4p \] In the expressions for $S^{(2)}$ this gives rise to terms $\delta(p+k-p'-k')$. This means that energy and momentum is conserved. However, virtual particles do not obey the relation between energy and momentum. \section[~~Divergences and renormalization]{Divergences and renormalization} It turns out that higher orders contribute infinite terms because only the sum $p+k$ of the four-momentum of the virtual particles is fixed. An integration over one of them becomes $\infty$. In the $x$-representation this can be understood because the product of two functions containing $\delta$-like singularities is not well defined. This is solved by discounting all divergent diagrams in a renormalization of $e$ and $M$. It is assumed that an electron, if there would not be an electromagnetical field, would have a mass $M_0$ and a charge $e_0$ unequal to the observed mass $M$ and charge $e$. In the Hamilton and Lagrange density of the free electron-positron field appears $M_0$. So this gives, with $M=M_0+\Delta M$: \[ \LL_{\rm e-p}(x)=-\overline{\psi(x)}(\gamma_\mu\partial_\mu+M_0)\psi(x)= -\overline{\psi(x)}(\gamma_\mu\partial_\mu+M)\psi(x)+\Delta M\overline{\psi(x)}\psi(x) \] and ${\cal H}_{\rm int}=ieN(\overline{\psi}\gamma_\mu\psi A_\mu)-i\Delta eN(\overline{\psi}\gamma_\mu\psi A_\mu)$. \section[~~Classification of elementary particles]{Classification of elementary particles} Elementary particles can be categorized as follows: \begin{enumerate} \setlength{\itemsep}{0mm} \item {\bf Hadrons:} these exist of quarks and can be categorized in: \begin{enumerate} \setlength{\itemsep}{0mm} \item {\bf Baryons:} these exist of 3 quarks or 3 antiquarks. \item {\bf Mesons:} these exist of one quark and one antiquark. \end{enumerate} \item {\bf Leptons:} e$^\pm$, $\mu^\pm$, $\tau^\pm$, $\nu_{\rm e}$, $\nu_\mu$, $\nu_\tau$, $\overline{\nu}_{\rm e}$, $\overline{\nu}_\mu$, $\overline{\nu}_\tau$. \item {\bf Field quanta:} $\gamma$, W$^\pm$, Z$^0$, gluons, gravitons (?). \item {\bf Higgs particle}: $\phi$. \end{enumerate} An overview of particles and antiparticles is given in the following table: \begin{center} \begin{tabular}{||c|c@{}cccccccrr|c||} \hline Particle&spin ($\hbar$)&B&L&T&T$_3$&S&C&B$^*$&charge ($e$)&$m_0$ (MeV)&antipart.\\ \hline \hline u&1/2&1/3&0&1/2&1/2 &0 &0 &0 &$+2/3$& 5&$\overline{\rm u}$\\ d&1/2&1/3&0&1/2&$-1/2$&0 &0 &0 &$-1/3$& 9&$\overline{\rm d}$\\ s&1/2&1/3&0&0 &0 &$-1$&0 &0 &$-1/3$& 175&$\overline{\rm s}$\\ c&1/2&1/3&0&0 &0 &0 &1 &0 &$+2/3$& 1350&$\overline{\rm c}$\\ b&1/2&1/3&0&0 &0 &0 &0 &$-1$&$-1/3$& 4500&$\overline{\rm b}$\\ t&1/2&1/3&0&0 &0 &0 &0 &0 &$+2/3$&173000&$\overline{\rm t}$\\ \hline e$^-$ &1/2&0&1&0&0&0&0&0&$-1$& 0.511&e$^+$\\ $\mu^-$ &1/2&0&1&0&0&0&0&0&$-1$& 105.658&$\mu^+$\\ $\tau^-$ &1/2&0&1&0&0&0&0&0&$-1$&1777.1 &$\tau^+$\\ $\nu_{\rm e}$&1/2&0&1&0&0&0&0&0& 0 & 0(?)&$\overline{\nu}_{\rm e}$\\ $\nu_\mu$ &1/2&0&1&0&0&0&0&0& 0 & 0(?)&$\overline{\nu}_{\mu}$\\ $\nu_\tau$ &1/2&0&1&0&0&0&0&0& 0 & 0(?)&$\overline{\nu}_{\tau}$\\ \hline $\gamma$ &1&0&0&0&0&0&0&0&$0$&0&$\gamma$\\ gluon &1&0&0&0&0&0&0&0&$0$&0&$\overline{\rm gluon}$\\ $\rm W^+$&1&0&0&0&0&0&0&0&$+1$&80220&$\rm W^-$\\ Z &1&0&0&0&0&0&0&0&$0$&91187&Z\\ graviton &2&0&0&0&0&0&0&0&$0$&0&graviton\\ \hline Higgs &0&0&0&0&0&0&0&0&$0$&125600&Higgs\\ \hline \end{tabular} \end{center} Here B is the baryon number and L the lepton number. It is found that there are three different lepton numbers, one for e, $\mu$ and $\tau$, which are separately conserved. T is the isospin, with $T_3$ the projection of the isospin on the third axis, C the charmness, S the strangeness and B$^*$ the bottomness. The anti particles have quantum numbers with the opposite sign except for the total isospin T. The composition of (anti)quarks of the hadrons is given in the following table, together with their mass in MeV in their ground state: \begin{center} \begin{tabular}{||l|cr@{.}l||l|cr@{.}l||l|cr@{.}l||} \hline $\pi^0$&$\half\sqrt{2}$(u$\overline{\rm u}$+d$\overline{\rm d}$)&134&9764&J/$\Psi$&c$\overline{\rm c}$&3096&8&$\overline{\Sigma^+}$&$\overline{\rm d}~\overline{\rm d}~\overline{\rm s}$&1197&436\rule{0pt}{15pt}\\ $\pi^+$&u$\overline{\rm d}$&139&56995&$\Upsilon$&b$\overline{\rm b}$&9460&37 &$\Xi^0$&u~s~s&1314&9\\ \cline{5-8} $\pi^-$&d$\overline{\rm u}$&139&56995&p$^+$&u~u~d&938&27231&$\overline{\Xi}^0$&$\overline{\rm u}~\overline{\rm s}~\overline{\rm s}$&1314&9\\ K$^0$&s$\overline{\rm d}$&497&672&p$^-$&$\overline{\rm u}~\overline{\rm u}~\overline{\rm d}$&938&27231&$\Xi^-$&d~s~s&1321&32\\ K$^0$&d$\overline{\rm s}$&497&672&n$^0$&u~d~d&939&56563&$\Xi^+$&$\overline{\rm d}~\overline{\rm s}~\overline{\rm s}$&1321&32\\ K$^+$&u$\overline{\rm s}$&493&677&$\overline{\rm n}^0$&$\overline{\rm u}~\overline{\rm d}~\overline{\rm d}$&939&56563&$\Omega^-$&s~s~s&1672&45\\ K$^-$&s$\overline{\rm u}$&493&677&$\Lambda$&u~d~s&1115&684&$\Omega^+$&$\overline{\rm s}~\overline{\rm s}~\overline{\rm s}$&1672&45\\ D$^+$&c$\overline{\rm d}$&1869&4&$\overline{\Lambda}$&$\overline{\rm u}~\overline{\rm d}~\overline{\rm s}$&1115&684&$\Lambda_c^+$&u~d~c&2285&1\\ D$^-$&d$\overline{\rm c}$&1869&4&$\Sigma^+$&u~u~s&1189&37&$\Delta^{2-}$&$\overline{\rm u}~\overline{\rm u}~\overline{\rm u}$&1232&0\\ D$^0$&c$\overline{\rm u}$&1864&6&$\overline{\Sigma^-}$&$\overline{\rm u}~\overline{\rm u}~\overline{\rm s}$&1189&37&$\Delta^{2+}$&u~u~u&1232&0\\ D$^0$&u$\overline{\rm c}$&1864&6&$\Sigma^0$&u~d~s&1192&55&$\Delta^+$&u~u~d&1232&0\\ F$^+$&c$\overline{\rm s}$&1969&0&$\Sigma^0$&$\overline{\rm u}~\overline{\rm d}~\overline{\rm s}$&1192&55&$\Delta^0$&u~d~d&1232&0\\ F$^-$&s$\overline{\rm c}$&1969&0&$\Sigma^-$&d~d~s&1197&436&$\Delta^-$&d~d~d&1232&0\\ \hline \end{tabular} \end{center} Each quark can exist in two spin states. So mesons are bosons with spin 0 or 1 in their ground state, while baryons are fermions with spin $\half$ or $\frac{3}{2}$. There exist excited states with higher internal $L$. Neutrino's have a helicity of $-\half$ while antineutrino's have only $+\half$ as possible value. \npar The quantum numbers are subject to conservation laws. These can be derived from symmetries in the Lagrange density: continuous symmetries give rise to additive conservation laws, discrete symmetries result in multiplicative conservation laws. \npar {\it Geometrical conservation laws} are invariant under Lorentz transformations and the CPT-operation. These are: \begin{enumerate} \setlength{\itemsep}{0mm} \item Mass/energy because the laws of nature are invariant for translations in time. \item Momentum because the laws of nature are invariant for translations in space. \item Angular momentum because the laws of nature are invariant for rotations. \end{enumerate} {\it Dynamical conservation laws} are invariant under the CPT-operation. These are: \begin{enumerate} \setlength{\itemsep}{0mm} \item Electrical charge because the Maxwell equations are invariant under gauge transformations. \item Colour charge is conserved. \item Isospin because QCD is invariant for rotations in T-space. \item Baryon number and lepton number are conserved but not under a possible SU(5) symmetry of the laws of nature. \item Quarks type is only conserved under the colour interaction. \item Parity is conserved except for weak interactions. \end{enumerate} The elementary particles can be classified into three families: \begin{center} \begin{tabular}{||l|c|c|c|c||} \hline &leptons&quarks&antileptons& antiquarks\\ \hline 1st generation&e$^-$ &d&e$^+$&$\overline{\rm d}$\rule{0pt}{12pt}\\ &$\nu_{\rm e}$&u&$\overline{\nu}_{\rm e}$&$\overline{\rm u}$\\ \hline 2nd generation&$\mu^-$ &s&$\mu^+$&$\overline{\rm s}$\rule{0pt}{12pt}\\ &$\nu_\mu$&c&$\overline{\nu}_\mu$&$\overline{\rm c}$\\ \hline 3rd generation&$\tau^-$ &b&$\tau^+$&$\overline{\rm b}$\rule{0pt}{12pt}\\ &$\nu_\tau$&t&$\overline{\nu}_\tau$&$\overline{\rm t}$\\ \hline \end{tabular} \end{center} Quarks exist in three colours but because they are {\it confined} these colours cannot be seen directly. The color force does {\it not} decrease with distance. The potential energy will become high enough to create a quark-antiquark pair when it is tried to disjoin an (anti)quark from a hadron. This will result in two hadrons and not in free quarks. \section[~~P and CP-violation]{P and CP-violation} It is found that the weak interaction violates P-symmetry, and even CP-symmetry is not conserved. Some processes which violate P symmetry but conserve the combination CP are: \begin{enumerate} \item $\mu$-decay: $\mu^-\rightarrow {\rm e}^-+\nu_\mu+\overline{\nu}_{\rm e}$. Left-handed electrons appear more than $1000\times$ as much as right-handed ones. \item $\beta$-decay of spin-polarized $^{60}$Co: $\rm ^{60}Co\rightarrow^{60}Ni+e^-+\overline{\nu}_e$. More electrons with a spin parallel to the Co than with a spin antiparallel are created: (parallel$-$antiparallel)/(total)=20\%. \item There is no connection with the neutrino: the decay of the $\Lambda$ particle through: $\Lambda\rightarrow{\rm p}^++\pi^-$ and $\Lambda\rightarrow{\rm n}^0+\pi^0$ has also these properties. \end{enumerate} The CP-symmetry was found to be violated by the decay of neutral Kaons. These are the lowest possible states with a s-quark so they can decay only weakly. The following holds: $\rm C|K^0\rangle=\eta|\overline{K^0}\rangle$ where $\eta$ is a phase factor. Further holds $\rm P|K^0\rangle=-|K^0\rangle$ because $\rm K^0$ and $\rm\overline{K^0}$ have an intrinsic parity of $-1$. From this follows that $\rm K^0$ and $\rm\overline{K^0}$ are not eigenvalues of CP: $\rm CP|K^0\rangle=|\overline{K^0}\rangle$. The linear combinations \[ \rm|K_1^0\rangle:=\half\sqrt{2}(|K^0\rangle+|\overline{K^0}\rangle)~~\mbox{and}~~ |K_2^0\rangle:=\half\sqrt{2}(|K^0\rangle-|\overline{K^0}\rangle) \] are eigenstates of CP: $\rm CP|K_1^0\rangle=+|K_1^0\rangle$ and $\rm CP|K_2^0\rangle=-|K_2^0\rangle$. A base of $\rm K_1^0$ and $\rm K_2^0$ is practical while describing weak interactions. For colour interactions a base of $\rm K^0$ and $\rm\overline{K^0}$ is practical because then the number u$-$number$~\overline{\rm u}$ is constant. The expansion postulate must be used for weak decays: \[ \rm|K^0\rangle=\half(\langle K_1^0|K^0\rangle+\langle K_2^0|K^0\rangle) \] The probability to find a final state with CP$=-1$ is $\rm\half|\av{K_2^0|K^0}|^2$, the probability of CP=+1 decay is $\rm\half|\av{K_1^0|K^0}|^2$. \npar The relation between the mass eigenvalues of the quarks (unaccented) and the fields arising in the weak currents (accented) is $(u',c',t')=(u,c,t)$, and: \begin{eqnarray*} \left(\begin{array}{c}d'\\s'\\b'\end{array}\right)&=& \left(\begin{array}{ccc} 1&0&0\\ 0&\cos\theta_2&\sin\theta_2\\ 0&-\sin\theta_2&\cos\theta_2 \end{array}\right) \left(\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&{\rm e}^{i\delta} \end{array}\right) \left(\begin{array}{ccc} \cos\theta_1&\sin\theta_1&0\\ -\sin\theta_1&\cos\theta_1&0\\ 0&0&1\\ \end{array}\right)\\ &&\left(\begin{array}{ccc} 1&0&0\\ 0&\cos\theta_3&\sin\theta_3\\ 0&-\sin\theta_3&\cos\theta_3\\ \end{array}\right) \left(\begin{array}{c}d\\s\\b\end{array}\right) \end{eqnarray*} $\theta_1\equiv\theta_{\rm C}$ is the {\it Cabibbo angle}: $\sin(\theta_{\rm C})\approx0.23\pm0.01$. \section[~~The standard model]{The standard model} When one wants to make the Lagrange density which describes a field invariant for local gauge transformations from a certain group, one has to perform the transformation \[ \Q{}{x_\mu}\rightarrow\frac{D}{Dx_\mu}=\Q{}{x_\mu}-i\frac{g}{\hbar}L_kA_\mu^k \] Here the $L_k$ are the generators of the gauge group (the ``charges'') and the $A_\mu^k$ are the gauge fields. $g$ is the matching coupling constant. The Lagrange density for a scalar field becomes: \[ \LL=-\half(D_\mu\Phi^*D^\mu\Phi+M^2\Phi^*\Phi)-\kwart F_{\mu\nu}^aF_a^{\mu\nu} \] and the field tensors are given by: $F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gc^a_{lm}A_\mu^lA_\nu^m$. \subsection[~~The electroweak theory]{The electroweak theory} The electroweak interaction arises from the necessity to keep the Lagrange density invariant for local gauge transformations of the group SU(2)$\otimes$U(1). Right- and left-handed spin states are treated different because the weak interaction does not conserve parity. If a fifth Dirac matrix is defined by: \[ \gamma_5:=\gamma_1\gamma_2\gamma_3\gamma_4=- \left(\begin{array}{cccc}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{array}\right) \] the left- and right- handed solutions of the Dirac equation for neutrino's are given by: \[ \psi_{\rm L}=\half(1+\gamma_5)\psi~~~\mbox{and}~~~\psi_{\rm R}=\half(1-\gamma_5)\psi \] It appears that neutrino's are always left-handed while antineutrino's are always right-handed. The {\it hypercharge} $Y$, for quarks given by $Y={\rm B+S+C+B^*+T'}$, is defined by: \[ Q=\half Y+T_3 \] so $[Y,T_k]=0$. The group U(1)$_Y\otimes$SU(2)$_T$ is taken as symmetry group for the electroweak interaction because the generators of this group commute. The multiplets are classified as follows: \begin{center} \begin{tabular}{||l|c|c|c|c|c||} \hline &e$^-_{\rm R}$&$\nu_{\rm eL}$~~e$^-_{\rm L}$&u$_{\rm L}$~~d$'_{\rm L}$&u$_{\rm R}$&d$_{\rm R}$\\ \hline \rule{0pt}{15pt}$T$&0&\half&\half&0&0\\ \rule{0pt}{15pt}$T_3$&0&$\half~~-\half$&$\half~~-\half$&0&0\\ \rule[-5pt]{0pt}{20pt}$Y$&$-2$&$-1$&$\frac{1}{3}$&$\frac{4}{3}$&$-\frac{2}{3}$\\ \hline \end{tabular} \end{center} Now, 1 field $B_\mu(x)$ is connected with gauge group U(1) and 3 gauge fields $\vec{A}_\mu(x)$ are connected with SU(2). The total Lagrange density (minus the fieldterms) for the electron-fermion field now becomes: \begin{eqnarray*} \LL_{0,\rm EW}&=&-(\overline{\psi_{\rm\nu e,L}},\overline{\psi_{\rm eL}})\gamma^\mu \left(\partial_\mu-i\frac{g}{\hbar}\vec{A}_\mu\cdot(\half\vec{\sigma})-\half i\frac{g'}{\hbar}B_\mu\cdot(-1)\right) \left(\begin{array}{c}\psi_{\rm\nu e,L}\\ \psi_{\rm eL}\end{array}\right)-\\ &&\overline{\psi_{\rm eR}}\gamma^\mu\left(\partial_\mu-\half i\frac{g'}{\hbar}(-2)B_\mu\right)\psi_{\rm eR} \end{eqnarray*} Here, $\half\vec{\sigma}$ are the generators of $T$ and $-1$ and $-2$ the generators of $Y$. \subsection[~~Spontaneous symmetry breaking: the Higgs mechanism]{Spontaneous symmetry breaking: the Higgs mechanism} All leptons are massless in the equations above. Their mass is probably generated by {\it spontaneous symmetry breaking}. This means that the dynamic equations which describe the system have a symmetry which the ground state does not have. It is assumed that there exists an isospin-doublet of scalar fields $\Phi$ with electrical charges +1 and 0 and potential $V(\Phi)=-\mu^2\Phi^*\Phi+\lambda(\Phi^*\Phi)^2$. Their antiparticles have charges $-1$ and 0. The extra terms in $\LL$ arising from these fields, $\LL_H=(D_{L\mu}\Phi)^*(D_L^\mu\Phi)-V(\Phi)$, are globally U(1)$\otimes$SU(2) symmetric. Hence the state with the lowest energy corresponds with the state $\Phi^*(x)\Phi(x)=v=\mu^2/2\lambda=$constant. The field can be written (were $\omega^\pm$ and $z$ are Nambu-Goldstone bosons which can be transformed away, and $m_\phi=\mu\sqrt{2}$) as: \[ \Phi=\left(\begin{array}{c}\Phi^+\\ \Phi^0\end{array}\right)= \left(\begin{array}{c}i\omega^+\\ (v+\phi-iz)/\sqrt{2}\end{array}\right)\mbox{~and~} \av{0|\Phi|0}=\left(\begin{array}{c}0\\ v/\sqrt{2}\end{array}\right) \] Because this expectation value $\neq0$ the SU(2) symmetry is broken but the U(1) symmetry is not. When the gauge fields in the resulting Lagrange density are separated one obtains: \begin{eqnarray*} W_\mu^-&=&\half\sqrt{2}(A_\mu^1+iA_\mu^2)~~,~~W_\mu^+~=~\half\sqrt{2}(A_\mu^1-iA_\mu^2)\\ Z_\mu&=&\frac{gA_\mu^3-g'B_\mu}{\sqrt{g^2+g'^2}}\equiv A_\mu^3\cos(\theta_{\rm W})-B_\mu\sin(\theta_{\rm W})\\ A_\mu&=&\frac{g'A_\mu^3+gB_\mu}{\sqrt{g^2+g'^2}}\equiv A_\mu^3\sin(\theta_{\rm W})+B_\mu\cos(\theta_{\rm W}) \end{eqnarray*} where $\theta_{\rm W}$ is called the {\it Weinberg angle}. For this angle holds: $\sin^2(\theta_{\rm W})=0.255\pm0.010$. Relations for the masses of the field quanta can be obtained from the remaining terms: $M_W=\half vg$ and $M_Z=\half v\sqrt{g^2+g'^2}$, and for the elementary charge holds: $\displaystyle e=\frac{gg'}{\sqrt{g^2+g'^2}}=g'\cos(\theta_{\rm W})=g\sin(\theta_{\rm W})$ \par Experimentally it is found that $M_W=80.022\pm0.26$ GeV/c$^2$ and $M_Z=91.187\pm0.007$ GeV/c$^2$. According to the weak theory this should be: $M_W=83.0\pm0.24$ GeV/c$^2$ and $M_Z=93.8\pm2.0$ GeV/c$^2$. \subsection[~~Quantumchromodynamics]{Quantumchromodynamics} Coloured particles interact because the Lagrange density is invariant for the transformations of the group SU(3) of the colour interaction. A distinction can be made between two types of particles: \begin{enumerate} \item ``White'' particles: they have no colour charge, the generator $\vec{T}=0$. \item ``Coloured'' particles: the generators $\vec{T}$ are 8 $3\times3$ matrices. There exist three colours and three anticolours. \end{enumerate} The Lagrange density for coloured particles is given by \[ \LL_{\rm QCD}=i\sum_k\overline{\Psi_k}\gamma^\mu D_\mu\Psi_k+ \sum_{k,l}\overline{\Psi_k}M_{kl}\Psi_l-\kwart F^a_{\mu\nu}F_a^{\mu\nu} \] The gluons remain massless because this Lagrange density does not contain spinless particles. Because left- and right- handed quarks now belong to the same multiplet a mass term can be introduced. This term can be brought in the form $M_{kl}=m_k\delta_{kl}$. \section[~~Path integrals]{Path integrals} The development in time of a quantum mechanical system can, besides with Schr\"odingers equation, also be described by a {\it path integral} (Feynman): \[ \psi(x',t')=\int F(x',t',x,t)\psi(x,t)dx \] in which $F(x',t',x,t)$ is the amplitude of probability to find a system on time $t'$ in $x'$ if it was in $x$ on time $t$. Then, \[ F(x',t',x,t)=\int\exp\left(\frac{iS[x]}{\hbar}\right)d[x] \] where $S[x]$ is an action-integral: $S[x]=\int L(x,\dot{x},t)dt$. The notation $d[x]$ means that the integral has to be taken over all possible paths $[x]$: \[ \int d[x]:=\lim_{n\rightarrow\infty}\frac{1}{N}\prod_n\left\{\int\limits_{-\infty}^\infty dx(t_n)\right\} \] in which $N$ is a normalization constant. To each path is assigned a probability amplitude $\exp(iS/\hbar)$. The classical limit can be found by taking $\delta S=0$: the average of the exponent vanishes, except where it is stationary. In quantum fieldtheory, the probability of the transition of a fieldoperator $\Phi(\vec{x},-\infty)$ to $\Phi'(\vec{x},\infty)$ is given by \[ F(\Phi'(\vec{x},\infty),\Phi(\vec{x},-\infty))=\int\exp\left(\frac{iS[\Phi]}{\hbar}\right)d[\Phi] \] with the action-integral \[ S[\Phi]=\int\limits_\Omega \LL(\Phi,\partial_\nu\Phi)d^4x \] \section[~~Unification and quantum gravity]{Unification and quantum gravity} The strength of the forces varies with energy and the reciprocal coupling constants approach each other with increasing energy. The SU(5) model predicts complete unification of the electromagnetical, weak and colour forces at $10^{15}$GeV. It also predicts 12 extra X bosons which couple leptons and quarks and are i.g.\ responsible for proton decay, with dominant channel ${\rm p}^+\rightarrow\pi^0+{\rm e}^+$, with an average lifetime of the proton of $10^{31}$ year. This model has been experimentally falsified. \npar Supersymmetric models assume a symmetry between bosons and fermions and predict partners for the currently known particles with a spin which differs \half. The supersymmetric SU(5) model predicts unification at $10^{16}$GeV and an average lifetime of the proton of $10^{33}$ year. The dominant decay channels in this theory are ${\rm p}^+\rightarrow{\rm K}^++\overline{\nu}_\mu$ and ${\rm p}^+\rightarrow{\rm K}^0+\mu^+$. \npar Quantum gravity plays only a role in particle interactions at the Planck dimensions, where $\lambda_{\rm C}\approx R_{\rm S}$: $m_{\rm Pl}=\sqrt{hc/G}=3\cdot10^{19}$ GeV, $t_{\rm Pl}=h/m_{\rm Pl}c^2=\sqrt{hG/c^5}=10^{-43}$ sec and $r_{\rm Pl}=ct_{\rm Pl}\approx 10^{-35}$ m. \chapter{Astrophysics} \typeout{Astrophysics} \section{Determination of distances} The {\it parallax} is mostly used to determine distances in nearby space. The parallax is the angular difference between two measurements of the position of the object from different view-points. If the annual parallax is given by $p$, the distance $R$ of the object is given by $R=a/\sin(p)$, in which $a$ is the radius of the Earth's orbit. The {\it clusterparallax} is used to determine the distance of a group of stars by using their motion w.r.t. a fixed background. The tangential velocity $v_{\rm t}$ and the radial velocity $v_{\rm r}$ of the stars along the sky are given by \npar \parbox{11cm}{ \[ v_{\rm r}=V\cos(\theta)~~~,~~~v_{\rm t}=V\sin(\theta)=\omega R \] where $\theta$ is the angle between the star and the {\it point of convergence} and $\hat{R}$ the distance in pc. This results, with $v_{\rm t}=v_{\rm r}\tan(\theta)$, in: \[ R=\frac{v_{\rm r}\tan(\theta)}{\omega}~\Rightarrow~\hat{R}=\frac{1''}{p} \] where $p$ is the parallax in arc seconds. The parallax is then given by \[ p=\frac{4.74\mu}{v_{\rm r}\tan(\theta)} \] }\hfill \parbox{35mm}{ \begin{picture}(35,35) \put(5,5){\framebox(30,30)[cc]{}} \jwline{5}{7.52}{15}{9} \jwline{15}{10}{27.97}{23} \jwline{15}{18}{35}{35.04} \put(16.01,7.97){\makebox(0,0)[lc]{\small RR-Lyrae}} \put(23,16){\makebox(0,0)[lc]{\small Type 2}} \put(23,27){\makebox(0,0)[rc]{\small Type 1}} \put(5,2){\makebox(0,0)[cc]{\small 0,1}} \put(10,2){\makebox(0,0)[cc]{\small 0,3}} \put(15,2){\makebox(0,0)[cc]{\small 1}} \put(20,2){\makebox(0,0)[cc]{\small 3}} \put(25,2){\makebox(0,0)[cc]{\small 10}} \put(30,2){\makebox(0,0)[cc]{\small 30}} \put(35,2){\makebox(0,0)[cc]{\small 100}} \put(2,5){\makebox(0,0)[rc]{\small1}} \put(2,10){\makebox(0,0)[rc]{\small0}} \put(2,15){\makebox(0,0)[rc]{-\small1}} \put(2,20){\makebox(0,0)[rc]{-\small2}} \put(2,25){\makebox(0,0)[rc]{-\small3}} \put(2,30){\makebox(0,0)[rc]{-\small4}} \put(2,35){\makebox(0,0)[rc]{-\small5}} \multiput(10,5)(5,0){5}{\line(0,1){1}} \multiput(10,35.1)(5,0){5}{\line(0,-1){1}} \multiput(5,10)(0,5){5}{\line(1,0){1}} \multiput(35,10)(0,5){5}{\line(-1,0){1}} \put(10,-3){$P$ (days) $\rightarrow$} \put(-8,23){$\av{M}$} \end{picture} } \npar with $\mu$ de proper motion of the star in $''$/yr. A method to determine the distance of objects which are somewhat further away, like galaxies and star clusters, uses the period-Brightness relation for Cepheids. This relation is shown in the above figure for different types of stars. \section{Brightness and magnitudes} The {\it brightness} is the total radiated energy per unit of time. Earth receives $s_0=1.374$ kW/m$^2$ from the Sun. Hence, the brightness of the Sun is given by $L_\odot=4\pi r^2s_0=3.82\cdot10^{26}$ W. It is also given by: \[ L_\odot=4\pi R_\odot^2\int\limits_0^\infty \pi F_\nu d\nu \] where $\pi F_\nu$ is the monochromatic radiation flux. At the position of an observer this is $\pi f_\nu$, with $f_\nu=(R/r)^2F_\nu$ if absorption is ignored. If $A_\nu$ is the fraction of the flux which reaches Earth's surface, the transmission factor is given by $R_\nu$ and the surface of the detector is given by $\pi a^2$, then the apparent brightness $b$ is given by: \[ b=\pi a^2\int\limits_0^\infty f_\nu A_\nu R_\nu d\nu \] The {\it magnitude} $m$ is defined by: \[ \frac{b_1}{b_2}=(100)^{\frac{1}{5}(m_2-m_1)}=(2.512)^{m_2-m_1} \] because the human eye perceives lightintensities logaritmical. From this follows that $m_2-m_1=2.5\cdot^{10}\log(b_1/b_2)$, or: $m=-2.5\cdot^{10}\log(b)+C$. The apparent brightness of a star if this star would be at a distance of 10 pc is called the {\it absolute brightness} $B$: $B/b=(\hat{r}/10)^2$. The absolute magnitude is then given by $M=-2.5\cdot^{10}\log(B)+C$, or: $M=5+m-5\cdot^{10}\log(\hat{r})$. When an interstellar absorption of $10^{-4}$/pc is taken into account one finds: \[ M=(m-4\cdot10^{-4}\hat{r})+5-5\cdot^{10}\log(\hat{r}) \] If a detector detects all radiation emitted by a source one would measure the {\it absolute bolometric magnitude}. If the {\it bolometric correction} $BC$ is given by \[ BC=2.5\cdot^{10}\log\left(\frac{\mbox{Energy flux received}}{\mbox{Energy flux detected}}\right)= 2.5\cdot^{10}\log\left(\frac{\int f_\nu d\nu}{\int f_\nu A_\nu R_\nu d\nu}\right) \] holds: $M_b=M_V-BC$ where $M_V$ is the visual magnitude. Further holds \[ M_b=-2.5\cdot^{10}\log\left(\frac{L}{L_\odot}\right)+4.72 \] \section{Radiation and stellar atmospheres} The radiation energy passing through a surface $dA$ is $dE=I_\nu(\theta,\varphi)\cos(\theta)d\nu d\Omega dAdt$, where $I_\mu$ is the {\it monochromatical intensity} [Wm$^{-2}$sr$^{-1}$Hz$^{-1}$]. When there is no absorption the quantity $I_\nu$ is independent of the distance to the source. Planck's law holds for a black body: \[ I_\nu(T)\equiv B_\nu(T)=\frac{c}{4\pi}w_\nu(T)=\frac{2h\nu^3}{c^2}\frac{1}{\exp(h\nu/kT)-1} \] The radiation transport through a layer can then be written as: \[ \frac{dI_\nu}{ds}=-I_\nu\kappa_\nu+j_\nu \] Here, $j_\nu$ is the {\it coefficient of emission} and $\kappa_\nu$ the {\it coefficient of absorption}. $\int ds$ is the thickness of the layer. The {\it optical thickness} $\tau_\nu$ of the layer is given by $\tau_\nu=\int\kappa_\nu ds$. The layer is optically thin if $\tau_\nu\ll1$, the layer is optically thick if $\tau_\nu\gg1$. For a stellar atmosphere in LTE holds: $j_\nu=\kappa_\nu B_\nu(T)$. Then also holds: \[ I_\nu(s)=I_\nu(0){\rm e}^{-\tau_\nu}+B_\nu(T)(1-{\rm e}^{-\tau_\nu}) \] \section{Composition and evolution of stars} The structure of a star is described by the following equations: \begin{eqnarray*} \frac{dM(r)}{dr}&=&4\pi\varrho(r)r^2\\ \frac{dp(r)}{dr}&=&-\frac{GM(r)\varrho(r)}{r^2}\\ \frac{L(r)}{dr}&=&4\pi\varrho(r)\varepsilon(r)r^2\\ \left(\frac{dT(r)}{dr}\right)_{\rm rad}&=&-\frac{3}{4}\frac{L(r)}{4\pi r^2}\frac{\kappa(r)}{4\sigma T^3(r)}~,~~\mbox{(Eddington), or}\\ \left(\frac{dT(r)}{dr}\right)_{\rm conv}&=&\frac{T(r)}{p(r)}\frac{\gamma-1}{\gamma}\frac{dp(r)}{dr}~,~~\mbox{(convective energy transport)} \end{eqnarray*} Further, for stars of the solar type, the composing plasma can be described as an ideal gas: \[ p(r)=\frac{\varrho(r)kT(r)}{\mu m_{\rm H}} \] where $\mu$ is the average molecular mass, usually well approximated by: \[ \mu=\frac{\varrho}{nm_{\rm H}}=\frac{1}{2X+\frac{3}{4}Y+\half Z} \] where $X$ is the mass fraction of H, $Y$ the mass fraction of He and $Z$ the mass fraction of the other elements. Further holds: \[ \kappa(r)=f(\varrho(r),T(r),\mbox{composition})~~\mbox{and}~~ \varepsilon(r)=g(\varrho(r),T(r),\mbox{composition}) \] Convection will occur when the star meets the Schwartzschild criterium: \[ \left(\frac{dT}{dr}\right)_{\rm conv}<\left(\frac{dT}{dr}\right)_{\rm rad} \] Otherwise the energy transfer takes place by radiation. For stars in quasi-hydrostatic equilibrium hold the approximations $r=\half R$, $M(r)=\half M$, $dM/dr=M/R$, $\kappa\sim\varrho$ and $\varepsilon\sim\varrho T^\mu$ (this last assumption is only valid for stars on the main sequence). For pp-chains holds $\mu\approx5$ and for the CNO chains holds $\mu=12$ tot 18. It can be derived that $L\sim M^3$: the {\it mass-brightness relation}. Further holds: $L\sim R^4\sim T^8_{\rm eff}$. This results in the equation for the main sequence in the Hertzsprung-Russel diagram: \[ ^{10}\log(L)=8\cdot^{10}\log(T_{\rm eff})+\mbox{constant} \] \section{Energy production in stars} The net reaction from which most stars gain their energy is: $\rm 4{}^1H\rightarrow{}^4He+2e^++2\nu_{\rm e}+\gamma$.\\ This reaction produces 26.72 MeV. Two reaction chains are responsible for this reaction. The slowest, speed-limiting reaction is shown in boldface. The energy between brackets is the energy carried away by the neutrino. \begin{enumerate} \item The proton-proton chain can be divided into two subchains:\\ \boldmath$\bf ^1H+p^+\rightarrow{}^2D+e^++\nu_{\bf e}$,\unboldmath ~and then $\rm ^2D+p\rightarrow{}^3He+\gamma$. \begin{enumerate} \item pp1: $\rm ^3He+^3He\rightarrow2p^++{}^4He$. There is 26.21 + (0.51) MeV released. \item pp2: $\rm ^3He+\alpha\rightarrow{}^7Be+\gamma$ \begin{enumerate} \item $\rm ^7Be+e^-\rightarrow{^7Li}+\nu$, then $\rm ^7Li+p^+\rightarrow2 ^4He+\gamma$. 25.92 + (0.80) MeV. \item $\rm ^7Be+p^+\rightarrow{^8B}+\gamma$, then $\rm ^8B+e^+\rightarrow2^4He+\overline{\nu}$. 19.5 + (7.2) MeV. \end{enumerate} Both $^7$Be chains become more important with raising $T$. \end{enumerate} \item The CNO cycle. The first chain releases 25.03 + (1.69) MeV, the second 24.74 + (1.98) MeV. The reactions are shown below. \npar \begin{tabular}{ccccc} &&$\longrightarrow$&$\searrow$&\\ $\nearrow$&$\rightarrow$&$\rm~^{15}N+p^+\rightarrow\alpha+^{12}C$&&$\rm^{15}N+p^+\rightarrow~^{16}O+\gamma$\\ &&$\downarrow$&&$\downarrow$\\ $\rm^{15}O+e^+\rightarrow~^{15}N+\overline{\nu}$&&$\rm^{12}C+p^+\rightarrow{}^{13}N+\gamma$&&$\rm^{16}O+p^+\rightarrow~^{17}F+\gamma$\\ $\uparrow$&&$\downarrow$&&$\downarrow$\\ \boldmath$\bf ^{14}N+p^+\rightarrow~^{15}O+\gamma$\unboldmath&&$\rm^{13}N\rightarrow~^{13}C+e^++\nu$&&$\rm^{17}F\rightarrow~^{17}O+e^++\nu$\\ &&$\downarrow$&&$\downarrow$\\ $\nwarrow$&$\leftarrow$&$\rm ^{13}C+p^+\rightarrow{}^{14}N+\gamma$&&$\rm^{17}O+p^+\rightarrow\alpha+{}^{14}N$\\ &&$\longleftarrow$&$\swarrow$& \end{tabular} \end{enumerate} \newpage \lhead[\fancyplain{}{\thepage}]{\fancyplain{}{\cmu The $\nabla$ operator}} \rhead[\fancyplain{}{\cmu The $\nabla$ operator}]{\fancyplain{}{\thepage}} \section*{The $\nabla$-operator} \typeout{The nabla-operator} \addcontentsline{toc}{chapter}{The $\nabla$-operator} In cartesian coordinates $(x,y,z)$ holds: \[ \vec{\nabla}=\Q{}{x}\ee{x}+\Q{}{y}\ee{y}+\Q{}{z}\ee{z}~~,~~ {\rm grad}f=\vec{\nabla}f=\Q{f}{x}\ee{x}+\Q{f}{y}\ee{y}+\Q{f}{z}\ee{z} \] \[ {\rm div}~\aaa=\vec{\nabla}\cdot\aaa=\Q{a_x}{x}+\Q{a_y}{y}+\Q{a_z}{z}~~,~~ \nabla^2 f=\QQ{f}{x}+\QQ{f}{y}+\QQ{f}{z} \] \[ {\rm rot}~\aaa=\vec{\nabla}\times\aaa= \left(\Q{a_z}{y}-\Q{a_y}{z}\right)\ee{x}+ \left(\Q{a_x}{z}-\Q{a_z}{x}\right)\ee{y}+ \left(\Q{a_y}{x}-\Q{a_x}{y}\right)\ee{z} \] In cylinder coordinates $(r,\varphi,z)$ holds: \[ \vec{\nabla}=\Q{}{r}\ee{r}+\frac{1}{r}\Q{}{\varphi}\ee{\varphi}+\Q{}{z}\ee{z}~~,~~ {\rm grad}f=\Q{f}{r}\ee{r}+\frac{1}{r}\Q{f}{\varphi}\ee{\varphi}+\Q{f}{z}\ee{z} \] \[ {\rm div}~\aaa=\Q{a_r}{r}+\frac{a_r}{r}+\frac{1}{r}\Q{a_\varphi}{\varphi}+\Q{a_z}{z}~~,~~ \nabla^2 f=\QQ{f}{r}+\frac{1}{r}\Q{f}{r}+\frac{1}{r^2}\QQ{f}{\varphi}+\QQ{f}{z} \] \[ {\rm rot}~\aaa=\left(\frac{1}{r}\Q{a_z}{\varphi}-\Q{a_\varphi}{z}\right)\ee{r}+ \left(\Q{a_r}{z}-\Q{a_z}{r}\right)\ee{\varphi}+ \left(\Q{a_\varphi}{r}+\frac{a_\varphi}{r}-\frac{1}{r}\Q{a_r}{\varphi}\right)\ee{z}\\ \] In spherical coordinates $(r,\theta,\varphi)$ holds: \begin{eqnarray*} \vec{\nabla} &=&\Q{}{r}\ee{r}+\frac{1}{r}\Q{}{\theta}\ee{\theta}+\frac{1}{r\sin\theta}\Q{}{\varphi}\ee{\varphi}\\ {\rm grad}f &=&\Q{f}{r}\ee{r}+\frac{1}{r}\Q{f}{\theta}\ee{\theta}+\frac{1}{r\sin\theta}\Q{f}{\varphi}\ee{\varphi}\\ {\rm div}~\aaa&=&\Q{a_r}{r}+\frac{2a_r}{r}+\frac{1}{r}\Q{a_\theta}{\theta}+\frac{a_\theta}{r\tan\theta}+\frac{1}{r\sin\theta}\Q{a_\varphi}{\varphi}\\ {\rm rot}~\aaa&=&\left(\frac{1}{r}\Q{a_\varphi}{\theta}+\frac{a_\theta}{r\tan\theta}-\frac{1}{r\sin\theta}\Q{a_\theta}{\varphi}\right)\ee{r}+ \left(\frac{1}{r\sin\theta}\Q{a_r}{\varphi}-\Q{a_\varphi}{r}-\frac{a_\varphi}{r}\right)\ee{\theta}+\\ &&\left(\Q{a_\theta}{r}+\frac{a_\theta}{r}-\frac{1}{r}\Q{a_r}{\theta}\right)\ee{\varphi}\\ \nabla^2 f &=&\QQ{f}{r}+\frac{2}{r}\Q{f}{r}+\frac{1}{r^2}\QQ{f}{\theta}+\frac{1}{r^2\tan\theta}\Q{f}{\theta}+\frac{1}{r^2\sin^2\theta}\QQ{f}{\varphi} \end{eqnarray*} General orthonormal curvelinear coordinates $(u,v,w)$ can be obtained from cartesian coordinates by the transformation $\vec{x}=\vec{x}(u,v,w)$. The unit vectors are then given by: \[ \ee{u}=\frac{1}{h_1}\Q{\vec{x}}{u}~,~~\ee{v}=\frac{1}{h_2}\Q{\vec{x}}{v}~,~~ \ee{w}=\frac{1}{h_3}\Q{\vec{x}}{w} \] where the factors $h_i$ set the norm to 1. Then holds: \begin{eqnarray*} {\rm grad}f &=&\frac{1}{h_1}\Q{f}{u}\ee{u}+\frac{1}{h_2}\Q{f}{v}\ee{v}+\frac{1}{h_3}\Q{f}{w}\ee{w}\\ {\rm div}~\aaa&=&\frac{1}{h_1h_2h_3}\left(\Q{}{u}(h_2h_3a_u)+\Q{}{v}(h_3h_1a_v)+\Q{}{w}(h_1h_2a_w)\right)\\ {\rm rot}~\aaa&=&\frac{1}{h_2h_3}\left(\Q{(h_3a_w)}{v}-\Q{(h_2a_v)}{w}\right)\ee{u}+ \frac{1}{h_3h_1}\left(\Q{(h_1a_u)}{w}-\Q{(h_3a_w)}{u}\right)\ee{v}+\\ &&\frac{1}{h_1h_2}\left(\Q{(h_2a_v)}{u}-\Q{(h_1a_u)}{v}\right)\ee{w}\\ \nabla^2 f &=&\frac{1}{h_1h_2h_3}\left[\Q{}{u}\left(\frac{h_2h_3}{h_1}\Q{f}{u}\right)+ \Q{}{v}\left(\frac{h_3h_1}{h_2}\Q{f}{v}\right)+ \Q{}{w}\left(\frac{h_1h_2}{h_3}\Q{f}{w}\right)\right] \end{eqnarray*} \newpage \lhead[\fancyplain{}{\thepage}]{\fancyplain{}{\cmu The SI units}} \rhead[\fancyplain{}{\cmu The SI units}]{\fancyplain{}{\thepage}} \section*{\center The SI units} \typeout{The SI units} \addcontentsline{toc}{chapter}{The SI units} \begin{tabular}[t]{||l|ll||} \multicolumn{3}{@{}l}{\large\bf Basic units}\\[1mm] \hline \bf Quantity &\bf Unit&\bf Sym.\\ \hline \hline Length &metre &m\\ Mass &kilogram&kg\\ Time &second &s\\ Therm. temp. &kelvin &K\\ Electr. current &ere &A\\ Luminous intens. &candela &cd\\ Amount of subst. &mol &mol\\ \hline \multicolumn{3}{l}{}\\ \multicolumn{3}{@{}l}{\large\bf Extra units}\\[1mm] \hline Plane angle &radian &rad\\ solid angle &sterradian&sr\\ \hline \end{tabular} \hfill \begin{tabular}[t]{||l|lll||} \multicolumn{4}{@{}l}{\large\bf Derived units with special names}\\[1mm] \hline \bf Quantity &\bf Unit&\bf Sym.&\bf Derivation\\ \hline \hline Frequency &hertz &Hz &$\rm s^{-1}$\rule{0pt}{11pt}\\ Force &newton &N &$\rm kg\cdot m\cdot s^{-2}$\\ Pressure &pascal &Pa &$\rm N\cdot m^{-2}$\\ Energy &joule &J &$\rm N\cdot m$\\ Power &watt &W &$\rm J\cdot s^{-1}$\\ Charge &coulomb &C &$\rm A\cdot s$\\ El.\ Potential &volt &V &$\rm W\cdot A^{-1}$\\ El.\ Capacitance &farad &F &$\rm C\cdot V^{-1}$\\ El.\ Resistance &ohm &$\Omega$&$\rm V\cdot A^{-1}$\\ El.\ Conductance &siemens &S &$\rm A\cdot V^{-1}$\\ Mag.\ flux &weber &Wb &$\rm V\cdot s$\\ Mag.\ flux density &tesla &T &$\rm Wb\cdot m^{-2}$\\ Inductance &henry &H &$\rm Wb\cdot A^{-1}$\\ Luminous flux &lumen &lm &$\rm cd\cdot sr$\\ Illuminance &lux &lx &$\rm lm\cdot m^{-2}$\\ Activity &bequerel&Bq &$\rm s^{-1}$\\ Absorbed dose &gray &Gy &$\rm J\cdot kg^{-1}$\\ Dose equivalent &sievert &Sv &$\rm J\cdot kg^{-1}$\\ \hline \end{tabular} \section*{\center Prefixes} \begin{center} \begin{tabular}{||lll|lll|lll|lll|lll||} \hline yotta&Y&$10^{24}$&giga &G&$10^9$&deci &d &$10^{-1}$ &pico &p&$10^{-12}$\rule{0pt}{11pt}\\ zetta&Z&$10^{21}$&mega &M&$10^6$¢i&c &$10^{-2}$ &femto&f&$10^{-15}$\\ exa &E&$10^{18}$&kilo &k&$10^3$&milli&m &$10^{-3}$ &atto &a&$10^{-18}$\\ peta &P&$10^{15}$&hecto&h&$10^2$µ&$\mu$&$10^{-6}$&zepto&z&$10^{-21}$\\ tera &T&$10^{12}$&deca&da&$10 $&nano &n&$10^{-9}$ &yocto&y&$10^{-24}$\\ \hline \end{tabular} \end{center} \end{document}