%  le6.tex     LOESUNGEN zur Klausur  Lehrer II 99/00

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\def\folg{\curvearrowright}
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\def\pu#1{\\[-5mm]\makebox[1mm][r]{}\hfill \ffall #1}
\def\blatt{\rl{\vbox{
  \hbox{\bf Theoretische Physik II f\"ur Lehrer} 
  \hbox{Universit\"at Hannover, WS 1999/2000}} 
  \hspace*{-.4cm}} \vspace*{1.5cm} \pn} 
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\begin{document}
\blatt

\vskip -16mm
{\LARGE\bf L\"OSUNGEN } \ {\large zur Klausur \ am  28. Januar 2000 }

\vskip 6mm \hrule \vskip 4mm

\no{1}{
$\dis t_1={a\0c} + {x_1\0c}$ \ , \ \ $\ma{a + x_1 \cr x_1 \cr}
= \ma{\g & \b \g \cr \b\g & \g \cr} \ma{3a \cr 0 \cr}\;$, \
d.h. \ $\matrix{\hfill a+x_1 = \phantom{\b}3\g a \cr \hfill x_1= 3\b\g a}$ 
\ $\;\folg$ \\ 
$\dis 3(1-\b)={1\0\g} = \wu{(1-\b)(1+\b)}\;$, 
$\dis \;9(1-\b)=1+\b\;$ : \ $\dis \;\b = {4\05}\,$, $\dis \g = {5\03}$ 
und $\dis x_1 = 4a\;$. }
 
\no{2}{
\vskip -4mm
$ \vc E^\prime = \ma{0 \cr 0 \cr \g (E+\b b) \cr}$ , \
$c\vc B^\prime = \ma{ 0 \cr \g (b + \b E) \cr 0 \cr} $
\quad \parbox[t]{6.4cm}{\vskip -2mm {\ft
Na! Ich wei\ss\ doch,  da\ss\ dies 
eine Invariante ist. Aber per  Rechnung folgt es  
 ebenfalls~:}} \\[1mm] {\ft $[ \;\uparrow$ \under{Rel 20}, (65) $]$}
 \hfill
$\dis \vc E^{\prime\,2} - c^2 \vc B^{\prime\,2} 
= \g^2 \lk (E+\b b)^2 - (\b E + b)^2 \rk = E^2-b^2\,$ .  }

\vskip 3mm
\hspace*{-3cm} $- - - - - - - - - - - -$

\no{3}{
$\dis \lk  - \( \6_\rho^2 + {1\0\rho} \6_\rho + {1\0\rho^2} \6_\ph^2 
  + \6_z^2 \) + \a \rho^2 \rk \,  e^{im\ph}\,e^{ikz} \ph (\rho ) 
= E\;  e^{im\ph}\,e^{ikz} \ph (\rho ) $  \\[-3mm] 
 $\phantom{a} $ \hfill $\folg$ \ \ \ 
$\dis \big[  \quad ? \quad \big] = - \6_\rho^2 - {1\0\rho} \6_\rho
 + \a \rho^2 + {m^2\0 \rho^2} + k^2$ . }

\no{4}{
Links und rechts geradlinig , \ innen~: \ 
$\dis \lk - \6_x^2 - \kappa^2 \rk \ph = 0$ , \
$\ph_{\rm innen} = A \cos (\kappa x )$ , \ 
$\ph_{\rm rechts} = B (b-x) $ , \ 
$\dis \matrix{\, \ \ \ A\cos(\kappa a) 
  = B (b-a) \cr -A\kappa \sin(\kappa a)  = - B \hfill \cr}$ , \ 
$\dis \hbox{ obere} \0 \, -\hbox{untere }$ \ \ 
$\dis \folg \ \ \  b=a + {1\0\kappa}\; {\rm cot}(\kappa a)$ .}

\no{5}{
$\schl V = \int\! d^3r \; e^{...} V$ : \ $\big[\, \schl V \,\big] =
Vol \cdot E = {Vol \cdot \hbar \0 Zeit}$ , \ 
$\lk \s \rk = \({m \0 \hbar}\, {\schl V \0 \hbar}  \)^2 =
\({m \0 m \ell  v}\, \ell^2 v \)^2 = $ {\sl Fl\"ache} . }

\no{6}{
$\dis \schl V = \a \int_0^\infty \! dr \; r^2 \, \d (r-R) \, 2\pi
\int_{-1}^1 \! du \; e^{ikru} = 4\pi \a R \; {\sin (kR)\0k}$ . \ 
\\ Dies wird bei $k \to 0$ zu $4\pi \a R^2$ \ \  \ --- \ \  
und $\int \! d^3r \; \a\,\d (r-R) =4\pi \a R^2\;$ ebenfalls. }

\no{7}{\vskip -3mm
$\dis \ma{ {7\09} & {4\09} + i {4\09} \cr
  {4\09} - i {4\09} & -{7\09} \cr} \ma{2+2i \cr 1 \cr}
 = \,\ldots\, =  \ma{2+2i \cr 1 \cr}$ ; \hfill $\dis \a 
 = {1\03}$ , \hfill
$\dis \vc \chi (t) =  e^{-i\o t \s_3}\;\cdot \quad $ \\[1mm]
$\dis \cdot\lk \ma{ {2\03} (1+i) \cr 0 \cr} 
             + \ma{ 0 \cr {1\03}\cr}  \rk 
 = \ma{ {2\03} (1+i) e^{-i\o t} \cr  {1\03} e^{i\o t} \cr}$ , \
$\dis P_+ = \left| \ma{1\cr 0 \cr } \vc \chi (t) \right|^2 
 = {8\09}$ , \ $\dis P_- = {1\09}\;$. }

\no{8}{
$H=-{\hbar^2\02\theta}\,\6_\ph^2$ , \ 
\unitlength .8mm \begin{picture}(8,5)
 \put(0,-3){\Large\bf --}  \put(0,-1){\Large\bf --}
 \put(5,-1){\Large\bf --}   \put(0,5){\Large\bf --}
 \put(5,5){\Large\bf --}   \end{picture} \ 
\unitlength .8mm \begin{picture}(8,5)
 \put(0,-3){\Large\bf --}  \put(0,-1){\Large\bf --}
 \put(5,-1){\Large\bf --}  \put(0,5){\Large\bf --}
 \put(5,5){\Large\bf --}   \end{picture}  \ , \ 
$\dis \eta = 4$ , \ 
$\dis \e_F = {\hbar^2 \0 2\theta} \cdot 4$ , \
$\dis E_0 = 0 + 4 \cdot {\hbar^2 \0 2\theta} 
  + {\hbar^2 \0 2\theta} \cdot 4 = 4\,{\hbar^2 \0 \theta}$ . }

\vskip 3mm
\hspace*{-3cm} $- - - - - - - - - - - -$

\no{9}{
$\dis p_1={1\02}$ , \ $\dis p_2=p_4=p_6={1\06}$ , \
$\dis S= {1\02} \,\ln(2) + {1\02} \, \ln(6)$ , \
$\dis \lw\, \nu \, \rw ={1\02} \cdot 1 + {1\06}\,(2+4+6) = 2.5$ . }

\no{10}{
$\dis Z = \sum_{m=0}^\infty \eta_m\,
 e^{-\b {\hbar^2 m^2 \0 2\theta}}\,$ mit 
 $\,\eta_0=2\,$ und $\,\eta_{m \geqslant 1}=4\;$ , \ 
$\dis Z = 2 + 4\, e^{-\b \hbar^2/(2\theta )}\;$ ; \\ 
$\dis Z \to 4 \int_0^\infty \!\! dm \; 
  e^{-\b {\hbar^2 m^2 \02\theta}}  = {1\0 \wu \b } 
  \cdot \const_T$ , \ 
$\dis E = -\6_\b {1\02} \, \ln \({1\0\b}\) 
  = \,{1\02}\,T $ \quad $\matrix{ 
 \lower 5pt\hbox{\scriptsize IST ja auch ein transla--} \cr
 \hbox{\scriptsize torischer Freiheitsgrad~.}\cr}$  }

\no{11}{
$\dis E = - \6_\b \,3N\,\ln(1/\b) = 3NT$ , \ 
$\dis F = - T\,N\,\Big(\,\ln(V) + \const_V \Big)$ , \ 
$\dis p = - \6_V F = NT/V$ , \ $pV = NT = {1\03}E$ .
\quad $\dis \xi={1\03}\;${\scriptsize (statt ${2\03}$ f\"ur 
\glqq klassisch--ideales Gas\grqq ) ist typisch f\"ur 
ultrarelativistische Materie.} }

\vskip 6mm \hrule

\end{document}