university/S2/AnaMech/VL/AnMeVL12.typ

// Main VL template
#import "../preamble.typ": *

// Fix theorems to be shown the right way in this document
#import "@preview/ctheorems:1.1.3": *
#show: thmrules

// Main settings call
#show: conf.with(
	// May add more flags here in the future
	num: 12,
	type: 0, // 0 normal, 1 exercise
	date: datetime.today().display(),
	//date: datetime(
	//	year: 2025,
	//	month: 5,
	//	day: 1,
	//).display(),
)

E: 2.6.25

= Uebersicht

+ Ein Beispiel fuer Lagrange II
+ Erhaltungsgroessen in L I und L II
+ Symetrien u. Erhaltungsgroessen

= Lagrange Funktion

Es gilt fuer die Lagrange Funktion
$
	L (q_(k) dot(q)_(k), t) =  T (q_(k), dot(q)_(k) , t) -  V (q_(k) , t) , space k =  1, ..., f.
$
Lagrange II
$
	dif / (dif t) (partial L) / (partial dot(q)_(k) ) - (partial L) / (partial q_(k) ) =  0 , space k =  1, ..., f.
$

#example[
	MP im Kegel.

	Es wirkt nur die Gravitationskraft.

	Hier gilt fuer die Koordinaten
	$
		R = 1 => f = 2 \
		rho = z tan alpha.
	$

	Es gilt fuer eine Z.B.
	$
		g (x, y, z) =  x ^2 +  y ^2  - z ^2 tan  ^2 alpha = 0 \
		x =  r sin alpha cos phi , space theta = alpha \
		y = r sin alpha sin phi \
		z =  r cos alpha.
	$
	Dann waehle fuer die generalisierten Koordinaten
	$
		q =  {r, phi}.
	$
	Das ergibt fuer die Lagrange-Funktion
	$
		L (x,y,z,dot(x),dot(y), dot(z)) =  m/2 (dot(x)^2 + dot(y)^2 + dot(z)^2 ) - m g z \
		=> L =  m/2 (dot(r)^2 +  r ^2 dot(phi)^2 sin ^2 alpha) - underbrace(m g cos alpha r, V (r)).
	$
	Hier gibt es eine zyklische Koordinate
	$
		(partial L) / (partial phi) =  0 => p_(phi) =  (partial L) / (partial dot(phi)) =  m r ^2 sin ^2 alpha dot(phi) prop  L_(z) 
	$
	#highlight[TODO: berechne Drehimpuls in Kugelkoordinaten]
]

= Erhaltungsgroessen in L I und L II

Es gilt
$
	m_(n) dot.double(x)_(n)  =  F_(n) sum_(alpha = 1)^(R) alpha_(alpha)  (partial g_(alpha) (arrow(x), t)) / (partial x_(n) ) , space n = 1, ..., 3N \
	g_(alpha)  (arrow(x), t) = 0 , space alpha = 1, ..., R \
	=> (x_(n) , lambda_(alpha) ).
$

== Energieerhaltung

=== Lagrange I

Wir betrachten Zwangskraefte
$
	g_(alpha) (arrow(x), t) = 0 \
	arrow(Z)_(alpha)  * d arrow(r) =  0.
$

Wir wollen Erhaltung von mit $V =  V (x_(n) )$
$
	E =  T +  V \
	(dif T) / (dif t) =  dif / (dif t)  sum_(n) 1/2 m_(n) dot(x)_(n) ^2  = sum _(n)  m_(n) dot(x)_(n) dot.double(x)_(n)  \
	(dif V) / (dif t)  = sum _(n)  (partial_(n) V)dot(x)_(n) =  - sum _(n) F_(n) dot(x)_(n) , space "Konservativ" =>  partial_(n) V =  - F_(n).
$ 
Nebenrechnung
$
	(dif g_(alpha) ) / (dif t) = sum _(n) (partial g_(alpha) ) / (partial x_(n) ) dot(x)_(n) +  (partial g_(alpha) ) / (partial t) = 0.
$
Das ergibt dann
$
	dif / (dif t)  (T +  V) =  sum _(n) sum _(alpha) lambda_(alpha)  (partial g_(alpha) ) / (partial x_(n) ) dot(x)_(n) =  sum _(alpha)  lambda_(alpha) (- (partial g_(alpha) ) / (partial t) ) .
$
Wir erwarten die Energieerhaltung nur fuer abg. und kons. Systeme.

Falls die Zwangskraefte sich also nicht mit der Zeit aendern, dann gilt die Energieerhaltung
$
	(partial g_(alpha) ) / (partial t) = 0 space forall alpha \
	=>  E =  T +  V =  "const".
$

=== Lagrange II

Allgemeine Erhaltungsgroessen.

Erinnerung fuer mechanische Groessen
$
	Q =  Q (q, dot(q), t) \
	(dif Q) / (dif t)  =  0 <=> Q "erhalten".
$

Wie bekommt man im Lagrangeformalismus erhaltungsgroessen
+ Zyklische Koordinaten
	$
		underbrace((partial L) / (partial q_(k) ) =  0, q_(k) "zyklisch") => dif / (dif t) (partial L) / (partial dot(q)_(k) ) =  0 => p_(k) "erhalten" , space p_(k) :=  "gen. Impuls".
	$
	Die Frage nach zyklischen Koordinaten darf nur gestellt werden, wenn man schon $f$ unabhaengige Koordinaten hat.

Nun 
$
	(partial g_(alpha) ) / (partial t) != 0 <=>^(!)  (partial x_(n) ) / (partial t) !=  0 and (partial L) / (partial t) = 0  \
	(dif L) / (dif t) = sum _(k) (partial L) / (partial q_(k) ) dot(q)_(k) +  sum _(k) (partial L) / (partial dot(q)_(k) )  dot.double(q)_(k) +  (partial L) / (partial t) \
	=> 	 dif / (dif t) (sum _(k) (partial L) / (partial dot(q)_(k) ) - L) =  - (partial L) / (partial t) 
$.

Betrachte als Nebenrechnung
$
	dif / (dif t) sum_(i=1)^(t) (partial L) / (partial dot(q)_(i) ) dot(q)_(i) = sum _(k) dot(q)_(k) dif / (dif t) (partial L) / (partial dot(q)_(k) ) =  sum _(k) dot.double(q)_(k) (partial L) / (partial dot(q)_(k) ) =^("L II")   sum _(k) (dot(q)_(k) (partial L) / (partial q_(k) ) +  dot.double(q)_(k) (partial L) / (partial dot(q)_(k) )   ).
$

Es gilt also
$
	(partial L) / (partial t) = 0 <=> sum _(k = 1) ^(f) p_(k) dot(q)_(k) - L "erhalten"
$ <erh>

Jetzt der Fall, dass
$
	(partial g_(alpha) ) / (partial t) &=  0 , space (partial L) / (partial t) = 0 \
	&=> x_(n)  =  x_(n) (q) \
	&=> T =  sum _(i, i) m_(i k)  (q) dot(q)_(i) dot(q)_(k)  \
	&=> sum_(i=1)^(f) (partial L) / (partial dot(q)_(i) ) = sum _(k) (partial T) / (partial dot(q)_(k) ) dot(q)_(k) = 2 T (q, dot(q)) \
	(partial L) / (partial t) = 0 &=> V (q, t) =  V (q) \
	sum_(k = 1)^(f) (partial L) / (partial dot(q)_(k) ) dot(q)_(k) - L &=  2 T -  (T -  V) =  T +  V =  E.
$

Nun der naechste Fall, dass
$
	(partial L) / (partial t) != 0 => "keine Erhaltungsgroesse".
$
Zuletzt
$
	(partial L) / (partial t) =  0 , space (partial g_(alpha) ) / (partial t) != 0 =>  (partial x_(n) ) / (partial t) != 0 \
	=> "Erhaltungsgroesse durch" #[@erh].
$

= Beispiele

#example[
	MP im Kegel.

	Hier gilt fuer die Zwangsbedingung
	$
		(partial g_(alpha) ) / (partial t) = 0 => (partial x_(n) ) / (partial t) = 0 \
		=> (partial L) / (partial t) = 0 => E =  T +  V.
	$
]

#example[
	Perle auf rotierendem Draht.

	Der Draht rotiert in der Ebene mit
	$
		phi =  omega t , space omega =  "const." \
		g (x, g, t) =  tan ^(-1) (y/x) - omega t =  phi -  omega t =  0 \
		=> (partial g_(alpha) ) / (partial t) !=  0 .. (alpha = 1).
	$
	Es gilt fuer die Trafo
	$
		x =  r cos (omega t) \
		y =  r sin (omega t) \
		=> x_(n)  = x_(n)  (q, t) \
		=> L =  m/2 (dot(r)^2 +  omega ^2 r ^2 ) , space  V = 0 \
		(partial L) / (partial t) = 0 => underbrace(sum _(k) p _(k)  dot(q)_(k)  - L, = O) "erhalten" \
		O = m dot(r) dot(r) -  m/2 dot(r)^2 - m/2 omega^2 r^2 =  m/2 dot(r)^2 - m/2 omega^2 r^2 
	$
]