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Die Hamiltonschen Gleichungen
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=====Teilchen in Zylinderkoordinaten ganz ohne Zwnagsbedingungen===== # q1=3, q2=Phi, q3 = z # :<math>\begin{align} & x=r\cos \phi ,\dot{x}=\dot{r}\cos \phi -r\dot{\phi }\sin \phi \\ & y=r\sin \phi ,\dot{y}=\dot{r}\sin \phi +r\dot{\phi }\cos \phi \\ & z=z \\ \end{align}</math> # :<math>\begin{align} & T=\frac{1}{2}m\left( {{{\dot{x}}}^{2}}+{{{\dot{y}}}^{2}}+{{{\dot{z}}}^{2}} \right)=\frac{1}{2}m\left( {{{\dot{r}}}^{2}}+{{r}^{2}}{{{\dot{\phi }}}^{2}}+{{{\dot{z}}}^{2}} \right) \\ & V=V(r,\phi ,z) \\ & L=L(r,\phi ,z,\dot{r},\dot{\phi },\dot{z})=\frac{1}{2}m\left( {{{\dot{r}}}^{2}}+{{r}^{2}}{{{\dot{\phi }}}^{2}}+{{{\dot{z}}}^{2}} \right)-V \\ \end{align}</math> # Generalisierte Impulse: :<math>\begin{align} & {{p}_{k}}=\frac{\partial L}{\partial {{{\dot{q}}}_{k}}} \\ & {{p}_{r}}=m\dot{r} \\ & {{p}_{\phi }}=m{{r}^{2}}\dot{\phi } \\ & {{p}_{z}}=m\dot{z} \\ \end{align}</math> Radialimpuls, z-Komponente des Drehimpulses und z-Komponente des Impulses # Aufstellung der Legendretrafo: :<math>\begin{align} & H=m{{{\dot{r}}}^{2}}+m{{r}^{2}}{{{\dot{\phi }}}^{2}}+m{{{\dot{z}}}^{2}}=\frac{1}{2}m\left( {{{\dot{r}}}^{2}}+{{r}^{2}}{{{\dot{\phi }}}^{2}}+{{{\dot{z}}}^{2}} \right)+V \\ & H=\frac{1}{2m}\left( {{p}_{r}}^{2}+\frac{{{p}_{\phi }}^{2}}{{{r}^{2}}}+{{p}_{z}}^{2} \right)+V(r,\phi ,z) \\ \end{align}</math> # Kanonische Gleichungen: :<math>\begin{align} & {{{\dot{p}}}_{k}}=-\frac{\partial H}{\partial {{q}_{k}}} \\ & \frac{\partial H}{\partial {{p}_{k}}}={{{\dot{q}}}_{k}}\quad k=1,...,f \\ & \dot{r}=\frac{\partial H}{\partial {{p}_{r}}}=\frac{{{p}_{r}}}{m},\dot{\phi }=\frac{\partial H}{\partial {{p}_{\phi }}}=\frac{{{p}_{\phi }}}{m{{r}^{2}}},\dot{z}=\frac{\partial H}{\partial {{p}_{z}}}=\frac{{{p}_{z}}}{m} \\ & {{{\dot{p}}}_{r}}=-\frac{\partial H}{\partial r}=\frac{{{p}_{\phi }}^{2}}{m{{r}^{3}}}-\frac{\partial V}{\partial r},{{{\dot{p}}}_{\phi }}=-\frac{\partial H}{\partial \phi }=-\frac{\partial V}{\partial \phi },{{{\dot{p}}}_{z}}=-\frac{\partial H}{\partial z}=-\frac{\partial V}{\partial z} \\ \end{align}</math> Interessant ist das Ergebnis der Zentrifugalkraft (Scheinkraft): F(Zentrifugal)= :<math>\frac{{{p}_{\phi }}^{2}}{m{{r}^{3}}}</math>, die den radialen Impuls ändert. Bekannt aus dem Keplerproblem ist uns bereits der Fall V®, ein Zentralpotenzial bei ebener Bewegung: :<math>{{\dot{p}}_{r}}=-\frac{\partial H}{\partial r}=\frac{{{p}_{\phi }}^{2}}{m{{r}^{3}}}-\frac{\partial V}{\partial r},{{\dot{p}}_{\phi }}=0,{{\dot{p}}_{z}}=0</math> Somit sind Drehimpuls in der Ebene und z-Impuls des Systems erhalten. :<math>z,\phi </math> sind zyklische Variablen :<math>\begin{align} & {{p}_{z}}=const.=o.B.d.A.=0 \\ & {{p}_{\phi }}=const. \\ \end{align}</math> oBdA: ebene Bewegung, Drehimpulserhaltung in der Ebene ======Beispiel: eindimensionaler harmonischer Oszi:====== Das System ist skleronom wegen :<math>\frac{\partial L}{\partial t}=0</math>, also folgt Energieerhaltung: E=H=T+V :<math>\frac{1}{2}m\left( {{{\dot{q}}}^{2}}+{{\omega }_{o}}^{2}{{q}^{2}} \right)=E=\frac{1}{2}m\left( \frac{{{p}^{2}}}{{{m}^{2}}}+{{\omega }_{o}}^{2}{{q}^{2}} \right)\Rightarrow \frac{{{p}^{2}}}{2mE}+\frac{{{q}^{2}}}{\left( \frac{2E}{m{{\omega }_{o}}^{2}} \right)}=1</math> Also ist die Lösung der Phasenraumkurve eine Ellipse. Die Ellipsengröße variiert je nach Energie: Die Halbachsen sind: :<math>a=\sqrt{2mE},b=\sqrt{\frac{2E}{m{{\omega }_{o}}^{2}}}</math> (bestimmt durch 1. Integral). Als kanonische Gleichungen ergibt sich: :<math>\begin{align} & {{{\dot{p}}}_{{}}}=-\frac{\partial H}{\partial q}=-m{{\omega }_{o}}^{2}q \\ & \dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m} \\ \end{align}</math> Daraus folgt dann gerade die Bewegungsgleichung :<math>\begin{align} & \ddot{q}=\frac{d}{dt}\frac{\partial H}{\partial q}=\frac{{\dot{p}}}{\acute{\ }m}=-{{\omega }_{o}}^{2}q \\ & \ddot{q}+{{\omega }_{o}}^{2}q=0 \\ \end{align}</math> Diese definiert ein Richtungsfeld im Phasenraum
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