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Der Satz von Liouville
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====Beweis (integrale Form):==== Gegeben sei eine Menge von Anfangskonfigurationen (to), die das Phasenraumgebiet Uto mit dem Volumen Vto ausfüllen: :<math>{{V}_{to}}=\int\limits_{{{U}_{to}}}^{{}}{{{d}^{2f}}{{x}_{0}}}</math> Bei t: :<math>{{V}_{t}}=\int\limits_{{{U}_{t}}}^{{}}{{{d}^{2f}}{{x}_{0}}}=\int\limits_{{{U}_{{{t}_{0}}}}}^{{}}{{{d}^{2f}}{{x}_{0}}}\det \left( \frac{\partial x}{\partial {{x}_{0}}} \right)=\int\limits_{{{U}_{{{t}_{0}}}}}^{{}}{{{d}^{2f}}{{x}_{0}}}\det \left( D{{\Phi }_{t,{{t}_{0}}}}({{{\bar{x}}}_{0}}) \right)</math> Mit der Jacobi- Matrix: :<math>{{\left( D{{\Phi }_{t,{{t}_{0}}}}({{{\bar{x}}}_{0}}) \right)}_{ik}}:=\frac{\partial {{\Phi }^{i}}_{t,{{t}_{0}}}({{{\bar{x}}}_{0}})}{\partial {{x}_{0}}^{k}}=\frac{\partial {{x}^{i}}}{\partial {{x}_{0}}^{k}}</math> Dies kann für Zeiten nahe t0 reihenentwickelt werden: :<math>\begin{align} & {{\Phi }_{t,{{t}_{0}}}}({{{\bar{x}}}_{0}})={{{\bar{x}}}_{0}}+\bar{F}({{{\bar{x}}}_{0}},t)(t-{{t}_{0}})+O({{(t-{{t}_{0}})}^{2}}) \\ & \bar{F}({{{\bar{x}}}_{0}},t)=J{{{\bar{H}}}_{,x}}=\left( \begin{matrix} \frac{\partial H}{\partial p} \\ -\frac{\partial H}{\partial q} \\ \end{matrix} \right) \\ \end{align}</math> Somit folgt: :<math>\frac{\partial {{\Phi }^{i}}_{t,{{t}_{0}}}({{{\bar{x}}}_{0}})}{\partial {{x}_{0}}^{k}}={{\delta }_{ik}}+\frac{\partial {{{\bar{F}}}^{i}}({{{\bar{x}}}_{0}},t)}{\partial {{x}_{0}}^{k}}(t-{{t}_{0}})+O({{(t-{{t}_{0}})}^{2}})</math> Mit Hilfe :<math>\det \left( 1+B\varepsilon \right)=1+\varepsilon tr(B)+O({{\varepsilon }^{2}})</math> folgt: :<math>\begin{align} & \det \left( D{{\Phi }_{t,{{t}_{0}}}} \right)=\left| \frac{\partial {{\Phi }^{i}}_{t,{{t}_{0}}}({{{\bar{x}}}_{0}})}{\partial {{x}_{0}}^{k}} \right|=1+(t-{{t}_{0}})\sum\limits_{i=1}^{2f}{{}}\frac{\partial {{{\bar{F}}}^{i}}({{{\bar{x}}}_{0}},t)}{\partial {{x}_{0}}^{i}}(t-{{t}_{0}})+O({{(t-{{t}_{0}})}^{2}}) \\ & \sum\limits_{i=1}^{2f}{{}}\frac{\partial {{{\bar{F}}}^{i}}({{{\bar{x}}}_{0}},t)}{\partial {{x}_{0}}^{i}}=div\bar{F}=\frac{\partial }{\partial q}\frac{\partial H}{\partial p}-\frac{\partial }{\partial p}\frac{\partial H}{\partial q}=0 \\ \end{align}</math> Der Fluß im Phasenraum ist also divergenzfrei. Dann folgt jedoch für die Jacobideterminante: :<math>\det \left( D{{\Phi }_{t,{{t}_{0}}}} \right)=\left| \frac{\partial {{\Phi }^{i}}_{t,{{t}_{0}}}({{{\bar{x}}}_{0}})}{\partial {{x}_{0}}^{k}} \right|=1+O({{(t-{{t}_{0}})}^{2}})\cong 1</math> :<math>\Rightarrow {{V}_{t}}=\int\limits_{{{U}_{{{t}_{0}}}}}^{{}}{{{d}^{2f}}{{x}_{0}}}\det \left( D{{\Phi }_{t,{{t}_{0}}}}({{{\bar{x}}}_{0}}) \right)=\int\limits_{{{U}_{{{t}_{0}}}}}^{{}}{{{d}^{2f}}{{x}_{0}}}\left( 1+O{{(t-{{t}_{0}})}^{2}} \right)\cong {{V}_{t0}}</math> Nebenbemerkung: Der Satz von Liouville kann auch in der LOKALEN Form formuliert werden: Für den Fluß :<math>{{\Phi }_{t,{{t}_{0}}}}</math> zu <math>\dot{\bar{x}}:=J{{\bar{H}}_{,x}}</math> ist <math>D{{\Phi }_{t,{{t}_{0}}}}</math> eine symplektische Matrix, das heißt :<math>\det \left( D{{\Phi }_{t,{{t}_{0}}}} \right)=1</math>. Das bedeutet, das Volumenelement :<math>d{{x}^{1}}...d{{x}^{2f}}</math> im Phasenraum ist unter dem Fluß invariant: :<math>d{{x}^{1}}...d{{x}^{2f}}=Det(D\Phi )d{{x}_{0}}^{1}....d{{x}_{0}}^{2f}=d{{x}_{0}}^{1}....d{{x}_{0}}^{2f}</math> Whoa, things just got a whole lot eiaser.
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