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S2/AnaMech/VL/AnMeVL11.typ
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243
S2/AnaMech/VL/AnMeVL11.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 11,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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E: 26.05.2025
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= Wiederholung
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Die Hoersaaluebung am 30.5. findet statt.
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Naechste Vorlesung VL12 werden die Erhaltungssaetze in Lagrange I und II diskutiert.
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Die letzten HA beinhalten nicht alle Themen der Klausur.
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Es wird mehr Bonuspunkt geben.
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In Lagrange I wird mehr geloest als benoetigt wird.
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= Mechanische Systeme mit Zwangsbedingungen
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Einmal mit N MP in 3D. Hier gibt es dann R ZB der Form
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$
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g_(alpha) = (arrow(x), t) = 0 , space alpha = 1, ..., R , space arrow(x) in RR^(3 N) .
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$
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Die BWGL in Lagrange I lauten dann
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$
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m_(n) dot.double(x)_(n) = F_(n) + sum_(alpha = 1)^(R) lambda_(alpha) arrow(nabla) g_(alpha) (arrow(x), t).
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$
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Hier sind dann noch 3N Gleichungen zu bestimmen und die R $lambda$ mit dem $arrow(x)$.
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Die Zahl der unabhaengigen Koordinaten ist
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$
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f = 3 N - R.
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$
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Dieses Problem mit Zwangsbedingungen wird als ein Problem mit Nebenbedingungen bezeichnet.
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= Generalisierte Koordinaten
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Die Generaliserten Koordinaten haben die Form und Eigenschaft
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$
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q_(k), space k = 1,...f , space x_(n) = x_(n) (q, t), \
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q = {q_1, q_2, ..., q_(f) }.
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$
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Es gilt dann fuer alle Werte von $q_(k) $ und fuer alle Zeiten
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$
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g_(alpha) (x(q_1, ...,q_(f) ), t) = 0.
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$
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#example[
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Fuer einen MP auf einer Kugeloberflaeche gilt
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$
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x_(1) ^2 + y_(1) ^2 + z_(1) ^2 - R^2 = 0.
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$
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Hier gibt es dann die generaliserten Koordinaten
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$
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theta "und" phi.
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$
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]
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#example[
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Das Doppelpendel.
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Hier ist eine Skizze des Pendels in kartesischen Koordinaten.
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Wir sind in der Ebene $==>$ Es gibt 4 kartesischen Koordinaten..
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Die Zwangsbedingungen sind dann
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$
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x_(1) ^2 + y_(1) ^2- l^2 = 0 , space abs(arrow(r_(1) ))= l_(1) \
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(x_(1) - x_(2) )^2 + (y_(1) - y_(2) )^2 - l_(2) ^2 = 0 , space abs(arrow(r)_(1) - arrow(r)_(2) ) = l_(2).
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$
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Die Generalisierten Koordinaten sind dann die Winkel
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$
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phi_(1) "und" phi_(2)
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$
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mit den Trafos
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$
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x_(1) = l_(1) sin phi_(1) \
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y_(1) = - l_(1) cos phi_(1) \
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x_(2) = x_(1) + l_(2) sin phi_(2) \
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y_(2) = - l_(1) cos phi_(1) - l_(2) cos phi_(2).
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$
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]
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= Elimieren der Zwangskraefte
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Es gibt hier $f$ BWGL fuer $q_(k) $.
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Lagrange II folgt
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$
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dif / (dif t) (partial L) / (partial dot(q)_(k) ) - (partial L) / (partial q_(k) ) = 0 , space L = T - V.
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$
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Als Ansatz gilt dass die $g_(alpha) $ nicht varrieren bei Varriation von $q_k $!
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Es gilt so
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$
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forall q_(k): (partial g_(alpha) ) / (partial q_(k) ) = 0 =>^("Kettenregel") sum_(n = 1)^(3 N) (partial g_(alpha) ) / (partial x_(n) ) (partial x_(n) ) / (partial q_(k) ) = 0 space forall k = 1, ... ,f.
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$
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Fixiere $q_(k) $ dann multipliziere alle 3N Gleichungen mit $(partial x_(n) ) / (partial q_(k) ) $ und dann bilde die Summe
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ueber alle 3N Lagrange I Gleichungen.
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Es gibt hier $f$ Moeglichkeiten $==>$ $f$ Gleichungen.
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Dadurch folgt
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$
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sum_(n) m_(n) dot.double(x)_(n) (partial x_(n) ) / (partial q_(k) ) = sum _(n) F_(n) (partial x_(n) ) / (partial q_(k) ) + underbrace( sum _(alpha) lambda_(alpha) sum _(n) (partial g_(alpha) ) / (partial x_(n) ) (partial x_(n) ) / (partial q_(k) ), = 0).
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$
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So entstehen $f$ Gleichungen.
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Beispiel fuer die Notation
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$
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h (q) &= g (r (q)) \
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&= g (f (q_1 ), ..., f_(R) (q_(1) ) ).
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$
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= Generaliserte Geschwindigkeiten
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Es gilt fuer die Geschwindigkeiten
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$
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dot(q)_(k) = (dif q_(k) ) / (dif t) , space k = 1, ..., f \
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x_(n) = x_(n) (q,t) => ^(?) dot(x)_(n) = dot(x)_(n) (q, dot(q), t) \
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dot(x)_(n) = dif / (dif t) x_(n) (q,t) = sum_(k = 1)^(f) (partial x_(n) ) / (partial q_(k) ) dot(q)_(k) + (partial x_(n) ) / (partial t)
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=> (partial dot(x)_(n) ) / (partial dot(q)_(k) ) = (partial x_(n) ) / (partial q_(k) ).
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$
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Erinnerung 1MP
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$
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T = sum_(i=1)^(3) m/2 dot(x)_(i) ^2 = sum_(j, k = 1)^(3) m/2 g_(i k) dot(q)_(j) dot(q)_(k) , space g_(j k) = arrow(g)_(j) * arrow(g)_(k).
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$
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Hier folgt dann
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$
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T &= sum_(i=1)^(3 N) m_(n) /2 dot(x)_(n) ^2 = sum_(i=1)^(3 N) m_(n) /2 (sum_(i=1)^(f) (partial x_(n) ) / (partial q_(k) ) + (partial x_(n) ) / (partial t) ) (sum_(i=1)^(f) (partial x_(n) ) / (partial q_(j) ) dot(q)_(j) + (partial x_(n) ) / (partial t) ) \
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&= sum_(k, j = 1)^(f) m_(k j) dot(q)_(k) dot(q)_(j) + underbrace(sum_(k)^(f) b_(k) (q,t) dot(q)_(k) + c (q,t), "nur wenn" (partial x_(n) ) / (partial t) != 0).
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$
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Hier steht dann insgesamt
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$
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sum_(k, j) sum_(n) (m_(n) /2 (partial x_(n) ) / (partial q_(k) ) (partial x_(n) ) / (partial q_(j) ) )dot(q)_(k) dot(q)_(j).
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$
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Ferner gilt
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$
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g_(alpha) (arrow(x), t)= 0 \
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=> x_(n) = x_(n) (q,t).
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$
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= Partielle Ableitungen von der kinetischen Energie
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Schreibe
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$
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T = sum _(n) m_(n) /2 dot(x)_(n) ^2 = T (q, dot(q), t) \
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(partial T) / (partial q_(k) ) = sum _(n) m_(n) dot(x)_(n) (partial dot(x)_(n) ) / (partial q_(k) ).
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$
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Betrachte nun
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$
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(partial T) / (partial dot(q)_(k) ) = sum _(n) m_(n) dot(x)_(n) (partial dot(x)_(n) ) / (partial dot(q)_(k) ) = sum _(n) m_(n) dot(x)_(n) (partial x_(n) ) / (partial q_(k) ) \
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$
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Nun die totale Zeitableitung
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$
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dif / (dif t) (partial T) / (partial dot(q)_(k) ) = sum_n m_(n) dot.double(x)_(n) (partial x_(n) ) / (partial q_(k) ) + sum _(n) m_(n) dot(x)_(n) underbrace(dif / (dif t)(partial x_(n) ) / (partial q_(k) ), = (partial dot(x)_(n) ) / (partial q_(k) ) )
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$
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Der Faktor $1/2$ verschwindet hier durch die Kettenregel.
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Zusammen ergibt das dann
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$
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dif / (dif t) (partial T) / (partial dot(q)_(k) ) - (partial T) / (partial q_(k) ) = sum _(n) m_(n) dot.double(x)_(n) + underbrace(sum _(n) m_(n) dot(x)_(n) (partial dot(x)_(n) ) / (partial q_(k) ) - sum _(n) dot(x)_(n) (partial dot(x)_(n) ) / (partial q_(k) ), = 0) \
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= sum _(n) F_(n) (partial x_(n) ) / (partial q_(k) ) , space k = 1, ..., f.
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$
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#definition[
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Generalisierte Kraefte sind gegeben durch
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$
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Q_(k) = sum_(i=1)^(3 N) F_(n) (partial x_(n) ) / (partial q_(k) ) \
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=> dif / (dif t) (partial T) / (partial dot(q)_(k) ) - (partial T) / (partial q_(k) ) = Q_(k) , space k = 1,...,f.
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$
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]
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Es gilt fuer konservative Kraefte mit $L = L (q, dot(q), t)$
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$
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F_(n) = - (partial V) / (partial x_(n) ) => Q_(k) = sum _(n) F_(n) (partial x_(n) ) / (partial q_(k) ) = - sum _(n) (partial V) / (partial x_(n) ) (partial x_(n) ) / (partial q_(k) ) = - (partial V (q, t)) / (partial q_(k) ) \
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dif / (dif t) (partial (T-V)) / (partial dot(q)_(k) ) - (partial (T-V)) / (partial q_(k) ) = 0 \
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=> dif / (dif t) (partial L) / (partial dot(q)_(k) ) - (partial L) / (partial q_(k) ) = 0 , space 1, ..., f.
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$
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Die Grundaufgabe ist herrauszufinden welche Aussagen ueber Lagrangefunktionen gemacht werden koenne.
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#example[
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MP auf einer rotierenden Stange.
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Wir geben vor
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$
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arrow(omega) = omega arrow(e)_(z) \
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=> phi = omega t \
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f = 2 -1 = 1 \
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$
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Waehle generalisierte Koordinaten
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$
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r = r (t) \
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x = x (r, t) = r cos (omega t) \
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y = y (r, t) = r sin (omega t)
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$
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Fuer die Lagrangefunktion ergib sich
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$
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V &= 0 => L = T (r, dot(r), t) \
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T &= m/2 (dot(x)^2 + dot(y)^2 + dot(z)^2 ) = m/2 (dot(r)^2 + omega^2 r^2 ) \
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&= m/2 (dot(r)^2 + dot(phi)^2 r^2 ).
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$
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Dann bilde die Ableitungen
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$
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(partial L) / (partial dot(r)) = m dot(r) \
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(partial L) / (partial r) = omega ^2 m r \
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=> m dot.double(r) - omega^2 m r = 0.
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$
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Dadurch folgt fuer die Loesung
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$
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r (t) = a e ^(omega t) + b e ^(- omega t).
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$
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]
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