Editing Das ideale Fermigas (section)

====Definition: Fermi- Dirac- Integral der Ordnung s:====

:<math>\begin{align}

& {{F}_{s}}\left( \eta  \right):=\frac{1}{\Gamma \left( s+1 \right)}\int_{0}^{\infty }{{}}dy\frac{{{y}^{s}}}{\left( {{e}^{y-\eta }}+1 \right)} \\

& s>0 \\

\end{align}</math>

<u>'''Entwicklung für'''</u>

:<math>\eta >>1\Rightarrow \xi >>1</math>, also Entartung:

:<math>\begin{align}

& \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right):=\int_{0}^{\infty }{{}}dy\frac{{{y}^{s}}}{\left( {{e}^{y-\eta }}+1 \right)}=\frac{1}{s+1}\int_{0}^{\infty }{{}}dy\frac{d}{dy}\left( {{y}^{s+1}} \right)\frac{1}{\left( {{e}^{y-\eta }}+1 \right)} \\

& =\frac{1}{s+1}\left. \left[ \left( {{y}^{s+1}} \right)\frac{1}{\left( {{e}^{y-\eta }}+1 \right)} \right] \right|_{0}^{\infty }+\frac{1}{s+1}\int_{0}^{\infty }{{}}dy{{y}^{s+1}}\frac{{{e}^{y-\eta }}}{{{\left( {{e}^{y-\eta }}+1 \right)}^{2}}} \\

& \frac{1}{s+1}\left. \left[ \left( {{y}^{s+1}} \right)\frac{1}{\left( {{e}^{y-\eta }}+1 \right)} \right] \right|_{0}^{\infty }=0 \\

\end{align}</math>

weitere Substitution:

:<math>\begin{align}

& x=y-\eta  \\

& \Rightarrow \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right)=\frac{1}{s+1}\int_{0}^{\infty }{{}}dy{{y}^{s+1}}\frac{{{e}^{y-\eta }}}{{{\left( {{e}^{y-\eta }}+1 \right)}^{2}}}=\frac{1}{s+1}\int_{-\eta }^{\infty }{{}}dx{{\left( x+\eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}} \\

& \eta >>1 \\

\end{align}</math>

Somit kann man die Grenzen erweitern, da <math>\eta >>1</math>

:

:<math>\begin{align}

& x=y-\eta  \\

& \Rightarrow \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right)=\frac{1}{s+1}\int_{-\eta }^{\infty }{{}}dx{{\left( x+\eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}\approx \frac{1}{s+1}\int_{-\infty }^{\infty }{{}}dx{{\left( x+\eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+O\left( {{e}^{-\eta }} \right) \\

& O\left( {{e}^{-\eta }} \right)<<1 \\

\end{align}</math>

Dies kann man durch Entwicklung von

:<math>{{\left( x+\eta  \right)}^{s+1}}</math>

lösen:

:<math>{{\left( x+\eta  \right)}^{s+1}}\approx {{\left( \eta  \right)}^{s+1}}+\left( s+1 \right){{\left( \eta  \right)}^{s}}x+\frac{s\left( s+1 \right)}{2}{{\left( \eta  \right)}^{s-1}}{{x}^{2}}+....</math>

Somit:

:<math>\begin{align}

& \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right)=\frac{1}{s+1}\int_{-\eta }^{\infty }{{}}dx{{\left( x+\eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}\approx \frac{1}{s+1}\int_{-\infty }^{\infty }{{}}dx{{\left( x+\eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+O\left( {{e}^{-\eta }} \right) \\

& \approx \frac{1}{s+1}\int_{-\infty }^{\infty }{{}}dx{{\left( \eta  \right)}^{s+1}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+\int_{-\infty }^{\infty }{{}}dx{{\left( \eta  \right)}^{s}}x\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+\frac{s}{2}\int_{-\infty }^{\infty }{{}}dx{{\left( \eta  \right)}^{s-1}}{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}} \\

& =\frac{{{\left( \eta  \right)}^{s+1}}}{s+1}\int_{-\infty }^{\infty }{{}}dx\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+{{\left( \eta  \right)}^{s}}\int_{-\infty }^{\infty }{{}}dx\frac{x{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}+\frac{s}{2}{{\left( \eta  \right)}^{s-1}}\int_{-\infty }^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}} \\

\end{align}</math>

Für die Terme gilt im Einzelnen:

:<math>\begin{align}

& \int_{-\infty }^{\infty }{{}}dx\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=\left[ \frac{-1}{\left( {{e}^{x}}+1 \right)} \right]_{-\infty }^{\infty }=1 \\

& \int_{-\infty }^{\infty }{{}}dx\frac{x{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=0\quad da\ Integrand\ ungerade \\

& \int_{-\infty }^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}:=I \\

\end{align}</math>

Bleibt Integral I zu lösen:

:<math>\begin{align}

& I=\int_{-\infty }^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=2\int_{0}^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=-2\left[ {{x}^{2}}\frac{1}{\left( {{e}^{x}}+1 \right)} \right]_{0}^{\infty }+4\int_{0}^{\infty }{{}}dx\frac{x}{\left( {{e}^{x}}+1 \right)} \\

& \left[ {{x}^{2}}\frac{1}{\left( {{e}^{x}}+1 \right)} \right]_{0}^{\infty }=0 \\

& \int_{0}^{\infty }{{}}dx\frac{x}{\left( {{e}^{x}}+1 \right)}=\frac{{{\pi }^{2}}}{12} \\

& \Rightarrow I=\frac{{{\pi }^{2}}}{3} \\

\end{align}</math>

Somit ergibt sich das Fermi- Dirac- Integral gemäß

:<math>\begin{align}

& \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right)\approx \frac{{{\left( \eta  \right)}^{s+1}}}{s+1}+\frac{s}{2}{{\left( \eta  \right)}^{s-1}}\frac{{{\pi }^{2}}}{3} \\

& \Gamma \left( s+1 \right){{F}_{s}}\left( \eta  \right)=\frac{{{\left( \eta  \right)}^{s+1}}}{s+1}+\frac{s}{2}{{\left( \eta  \right)}^{s-1}}\frac{{{\pi }^{2}}}{3}+O\left( {{\left( \eta  \right)}^{s-3}} \right) \\

& \Rightarrow {{F}_{s}}\left( \eta  \right)=\frac{1}{\Gamma \left( s+1 \right)}\left[ \frac{{{\left( \eta  \right)}^{s+1}}}{s+1}+\frac{s{{\pi }^{2}}}{6}{{\left( \eta  \right)}^{s-1}}+O\left( {{\left( \eta  \right)}^{s-3}} \right) \right] \\

\end{align}</math>

'''Speziell:'''

:<math>\begin{align}

& {{F}_{\frac{1}{2}}}\left( \eta  \right)\approx \frac{2}{\sqrt{\pi }}\left[ \frac{{{\left( \eta  \right)}^{\frac{3}{2}}}}{\frac{3}{2}}+\frac{{{\pi }^{2}}}{12}{{\left( \eta  \right)}^{-\frac{1}{2}}} \right] \\

& {{F}_{\frac{3}{2}}}\left( \eta  \right)\approx \frac{4}{3\sqrt{\pi }}\left[ \frac{{{\left( \eta  \right)}^{\frac{5}{2}}}}{\frac{5}{2}}+\frac{{{\pi }^{2}}}{4}{{\left( \eta  \right)}^{\frac{1}{2}}} \right] \\

\end{align}</math>

Also:

:<math>\begin{align}

& \bar{N}=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2mkT \right)}^{\frac{3}{2}}}\int_{0}^{\infty }{{}}dy\frac{{{y}^{\frac{1}{2}}}}{\left( {{e}^{y-\eta }}+1 \right)}=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2mkT \right)}^{\frac{3}{2}}}\left[ \frac{2}{3}{{\left( \frac{\mu }{kT} \right)}^{\frac{3}{2}}}+\frac{{{\pi }^{2}}}{12}{{\left( \frac{\mu }{kT} \right)}^{-\frac{1}{2}}} \right] \\

& \Rightarrow \bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m\mu  \right)}^{\frac{3}{2}}}\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right] \\

\end{align}</math>

<u>Definition: Fermi- Energie:</u>

:<math>{{E}_{F}}:=\mu \left( T=0,\bar{N},V \right)</math>

Bei T= 0 Kelvin sind die Zustände mit <math>E<{{E}_{F}}</math>

voll besetzt, die anderen leer!

Wir können dann <math>\mu \left( T=0,\bar{N},V \right)</math>

durch <math>{{E}_{F}}</math>

und <math>\bar{N}</math>

eliminieren:

<u>'''T→0'''</u>

:<math>\begin{align}

& \bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m\mu  \right)}^{\frac{3}{2}}}\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right]=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m{{E}_{F}} \right)}^{\frac{3}{2}}} \\

&  \\

\end{align}</math>

Für größere Temperaturen T>0 wird nun

:<math>\begin{align}

& \bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m{{E}_{F}} \right)}^{\frac{3}{2}}} \\

&  \\

\end{align}</math>

in niedrigster Ordnung in <math>\frac{kT}{{{E}_{F}}}</math>

entwickelt und diese Entwicklung dann eingesetzt in die Formel

:<math>\begin{align}

& \bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m\mu  \right)}^{\frac{3}{2}}}\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right]=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m{{E}_{F}} \right)}^{\frac{3}{2}}} \\

& \Rightarrow {{\left( \mu  \right)}^{\frac{3}{2}}}\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right]\approx {{\left( {{E}_{F}} \right)}^{\frac{3}{2}}} \\

& \Rightarrow \mu \approx {{E}_{F}}{{\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right]}^{-\frac{2}{3}}} \\

\end{align}</math>

'''Jetzt wird '''in niedrigster Ordnung in <math>\frac{kT}{{{E}_{F}}}</math>

entwickelt:

Das heißt, für kT=1 zeigt µ über Ef etwa folgenden verlauf:

'''die Kurve wird für höhere Temperaturen immer weiter auseinandergedehnt!'''

<u>'''Innere Energie'''</u>

:<math>\begin{align}

& U=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2mkT \right)}^{\frac{3}{2}}}kT\int_{0}^{\infty }{{}}dy\frac{{{y}^{\frac{3}{2}}}}{\left( {{e}^{y-\eta }}+1 \right)} \\

& {{F}_{\frac{3}{2}}}\left( \eta  \right)\approx \frac{4}{3\sqrt{\pi }}\left[ \frac{{{\left( \eta  \right)}^{\frac{5}{2}}}}{\frac{5}{2}}+\frac{{{\pi }^{2}}}{4}{{\left( \eta  \right)}^{\frac{1}{2}}} \right] \\

\end{align}</math>

Also:

:<math>\begin{align}

& U=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m \right)}^{\frac{3}{2}}}{{\left( kT \right)}^{\frac{5}{2}}}\left[ \frac{2}{5}{{\left( \frac{\mu }{kT} \right)}^{\frac{5}{2}}}+\frac{{{\pi }^{2}}}{4}{{\left( \frac{\mu }{kT} \right)}^{\frac{1}{2}}} \right] \\

& =\frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( \mu  \right)}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{\mu } \right)}^{2}} \right] \\

\end{align}</math>

Verwende:

So dass:

:<math>\begin{align}

& U=\frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( \mu  \right)}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{\mu } \right)}^{2}} \right] \\

& \approx \frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( {{E}_{F}} \right)}^{\frac{5}{2}}}{{\left[ 1-\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right] \\

\end{align}</math>

Mit

:<math>\bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m{{E}_{F}} \right)}^{\frac{3}{2}}}</math>

folgt:

:<math>\begin{align}

& U\approx \frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( {{E}_{F}} \right)}^{\frac{5}{2}}}{{\left[ 1-\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right] \\

& \frac{2}{5}\left( \frac{V}{{{h}^{3}}} \right)4\pi \frac{\left( 2s+1 \right)}{2}{{\left( 2m \right)}^{\frac{3}{2}}}{{\left( {{E}_{F}} \right)}^{\frac{5}{2}}}\approx \frac{3}{5}\bar{N}{{E}_{F}} \\

& {{\left[ 1-\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]}^{\frac{5}{2}}}\left[ 1+\frac{5}{2}\frac{{{\pi }^{2}}}{4}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]\approx 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \\

& \Rightarrow U\approx \frac{3}{5}\bar{N}{{E}_{F}}\left[ 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right] \\

\end{align}</math>

Somit haben wir die '''kalorische Zustandsgleichung'''

:<math>U\approx \frac{3}{5}\bar{N}{{E}_{F}}\left[ 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]</math>

und die '''thermische Zustandsgleichung'''

:<math>pV=\frac{2}{3}U\approx \frac{2}{5}\bar{N}{{E}_{F}}\left[ 1+5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}} \right]</math>

Das bedeutet:

Der Druck des fermigases ist um einen Faktor  <math>\frac{{{E}_{F}}}{kT}</math>

größer als in klassischen idealen Gasen

Beispiel:

:<math>{{E}_{F}}\approx 1eV\Rightarrow T\tilde{\ }{{10}^{4}}K</math>

1 eV entspricht 10.000 K!!

'''Grund ''' ist das Pauli- Prinzip!!

Also eine effektive Abstoßung der Teilchen! Dies bewirkt für niedrige Temperaturen den enormen Faktor

:<math>\frac{{{E}_{F}}}{kT}</math>
,
 mit dem der Druck gegenüber dem idealen Gas zu multiplizieren ist.

Für sehr hohe Temperaturen überwiegt dann der hintere teil, und es gilt:

Der Fermidruck ist etwa

:<math>pV=\frac{2}{3}U\approx \frac{2}{5}\bar{N}{{E}_{F}}5\frac{{{\pi }^{2}}}{12}{{\left( \frac{kT}{{{E}_{F}}} \right)}^{2}}=\frac{{{\pi }^{2}}}{6}\bar{N}kT\left( \frac{kT}{{{E}_{F}}} \right)</math>

Also auch größer als beim klassischen idealen Gas, nämlich um den Faktor <math>\left( \frac{kT}{{{E}_{F}}} \right)</math>

!