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Lecture 1: Lagrange equations for N particles
Geometrical derivation
of Lagrangian Mechanics
(James Casey. Am. J. Phys. 62 (9), 1994)
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Newton’s second law in configuration space
1,2,3; 1,2,i n N
1( ) ,F n 2( ) ,F n3( )F n
( ) ( ) ( )i iF n M n x n
1 2 3
1 2 3
1 2 3
(1), (1), (1),
(2), (2), (2),
.............................
( ), ( ), ( )
x x x
x x x
x N x N x N
1 2 3 3 2 3 1 3, , , , , ,N N NP x x x x x x
1x
2x
3x
Physical Space (3-D) Cartesian Configuration Space (3N-D)
N Particles
P
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(1), (1), (1), , ( ), ( ), ( )M M M M N M N M N
1 2 3 3 2 3 1 3, , , , ,N N Nm m m m m m Cartesian mass components
1 2 3 1 2 3(1), (1), (1), , ( ), ( ), ( )F F F F N F N F N
1 2 3 3 2 3 1 3, , , , ,N N Nf f f f f f Cartesian force components
Components of Newton’s equation in configuration space
1, 2, , 3
,k
k k
k N
f m x
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Correspondence:
Physical space Configuration space
3 3
3 3
3 2 3 1 3
( 1, 2,3; 1, , )( ) ,
( ) ,
( ) ,
n i
i
i n i
n n n
i n Nx n x
F n f
M n m m m
2 2 2
1 2 3
1
1( ) ( ) ( ) ( )
2
N
i
T M i x i x i x i
Kinetic energy:
32
1
1( )
2
Nk
k
k
m x
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Configuration vectorial space
• New Cartesian coordinates of P: ,k k kmx x
m
where
1
( ),N
i
m M i
• The square of the distance from the origin to the position occupied by P is
defined as:
3 32 2 2
1 1
1( ) ( ) ,
N Nk k
OP k
k k
d x m xm
• We may represent P by its position vector in configuration space, and define
an inner product We choose a fixed orthonormal basis ,
and express as:
r2 .OPr r d 1 2 3, , , Ne e e
r
3
1
,N
k
k
k
r x e
1
2
3
1, 0, 0,0, ,0 ,
0,1, 0,0, ,0 ,
,
0, 0, 0,0, ,1 ,N
e
e
e
( 1, 2, , 3 ; 1, 2, , 3 )
1, ;
0, ;j k jk
j N k N
j ke e
j k
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• Besides the orthonormal basis it is useful to have a fixed pair of
reciprocal basis and , defined by ke ke
/ ,k k ke e m m / ,k
k ke e m m
( 1, 2, , 3 ; 1, 2, , 3 ),k k
j j j N k Ne e
3
1
Nk
k
k
r x e
3
1
,N
k
k
k
drv x e
dt
• Next, we construct a force vector in configuration space by setting f
3
1
,N
k
k
k
f f e
• We may write the position vector , and the velocity of P as: ( )r ( )v
3
1
,N
k
k
k
x e
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• The vectorial expression of Newton’s second law in configuration space is
• We observe that :
3 3
1 1
1 1,
2 2
N Nk j
k j
k j
mv v m x x e e T
,dv
f mdt
1 1 1( )2 2 2
dT d dv dv dvmv v m v mv m v f v
dt dt dt dt dt
(square of length of a line element ) 2 2 22,ds dr dr v v dt Tdt
m
,
1,2, ,3 ,
k k
dvf e m e
dt
k N
( ) ,
1,2, ,3 ,
kk
k k k k
dxf m e e m x
dt
k N
/km m
1, 2, , 3
,k
k k
k N
f m x
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Configuration manifold (M ) and its geometry
( 1, 2, , 3 ),( , ) ,j
j j Nx q t
Suppose we introduce a system of curvilinear coordinates,
called generalized coordinates, 1, 2, , 3 ,N ,q
1 2 3, , ,( , ), ,
Nq q q qr r q t
, 1, 2, , 3 ,r
a Nq
These 3N vector define an 3N-dimensional tangent space to M. denoted by
TPM. . Any vector ` TPM. can be resolved on the basis as w ,a
.w w a
P
Tangent to the
coordinates lines
• The velocity of P can be expressed as ( , )
,dr r q t
v q adt t
• We denote as generalized velocity. Notice that TPM.
q q a
M
,a
M
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2
21 12 1 02 2
( , ),
r q tT mv m q a T T T
t
3 3
2
1 1
1( ) 0,
2
N N
T m a a q q
• The kinetic energy of P can also be expressed as:
3
1
1
( ) ,N r
T m a qt
2
0
1.
2
rT m
t
3 3
2
1 1
( ) 0,N N
t cteds dr dr a a dq dq
• M is a manifold which moves through configuration space and whose
geometry is called Riemannian.
where
2220
Tdt
m
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Lagrange equations
,mv a f a
• Now, using the identities :
• Take a scalar product of both sides of Newton’s equation with each of the basis
vectors in the tangent space a
• Let (generalized force components) ,Q f a
; ,r v da v
aq q dt q
( ) , 1, ,3 .d T T
Q Ndt q q
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Geometrical constraints (holonomic)
• Suppose the system of N particles is subject to M
geometrical constraints (holonomic) of type:
( 1, 2, , 3 )( , ) 0 ,j j M Nr t
• Each of these constraints defines a hypersurface of
dimension 3N-1 . The intersection of the M hypersurfaces is
a subset M of dimension n=3N-M . The particle P must
remains on M and n (degrees of freedom) is the minimum
number of coordinates to locate P at time t on M . We may
describe M by n Gaussian coordinates called
generalized coordinates:
q
1 2, , , .( , ),
nq q q qr r q t
These n vectors define an n-dimensional tangent space to M denoted by
TPM. . Any vector TPM. can be resolved on the basis as
, 1, 2, ,r
a nq
w ,a .w w a
P
Subset M of
dimension n
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• The velocity of P can be expressed as
( , ),
r q tv q a
t
• We denote as generalized velocity. Notice that TPM. . q
2 1 0,T T T T
2
1 1
1( ) 0,
2
n n
T m a a q q
aq a
• The kinetic energy of P can also be expressed as:
1
1
( ) ,n r
T m a qt
2
0
1.
2
rT m
t
2 22
1 1
2( ) 0,
n n Tds a a dq dq dt
m
• M is a manifold which moves through configuration space and whose geometry
is called Riemannian.
where
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Lagrange equations
• Take a scalar product of both sides of Newton’s equation with each of the basis
vectors in the tangent space
,f a mv a
• Let (generalized force components).
• Now, using the identities :
a
,Q f a
;r v
aq q
,da v
dt q
( ) , 1, , .d T T
Q ndt q q
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Ideal constraints (ideal holonomic)
• The total force on the particle P in configuration space can be
written as
• We say a constraint is ideal if the corresponding constraint
force is orthogonal to the constraint :
where
* ,CHf f f
1, , ; 1, , 30, ( 0, )CH
j j M n N Mf a a
Given forces
Constraints forces 1
,M
CH
j
j
f
1, , ,,CH
j j j Mf 3
1
.N
j j k
j kk
er x
1
,M
CH
j j
j
f
P
CHf
Subset M of
dimension n
Thus and as a consequence
and j Subset M of dimension n
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Lagrange equations with ideal holonomic
constraints
• The ideal constraints forces do not appear in Lagrange equations
* CHf a f a f a * ,Q
*( ) , 1, , .d T T
Q ndt q q
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Lagrange equations with ideal non-holonomic constraints
* ,CH CNf f f f
CN
r r
r
f a B
1
( , ) ( , ) 0, 1, , ;n
r rB q t q B q t r n
,CN CN
r r r
r r
f f b
1
, ;n
r rb B a a a a
TPM.
*
1
( ) , 1, , ,
( , ) ( , ) 0, 1, , ;
r r
r
n
r r
d T TQ B n
dt q q
B q t q B q t r n
• In addition, let suppose we have ideal non-holonomic constrains of type:
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Lagrange function:
• In addition, let suppose that the given forces can be derived from the
generalized potential as
* ,U d U
fr dt v
( , , )L q q t
( , , )U r v t
• Let . Then, 1( , , , )nr r q q t
* * ,a
U d UQ f a
q dt q
• We define the Lagrange function as .
Then, the Lagrange equations can be finally written as:
( , , ) ( , , ) ( , , )L q q t T q q t U q q t
3 3
1 1
, ,N N
k k
k kj j
U U U Ue e
r x v x
1
( ) , 1, , ,
( , ) ( , ) 0, 1, , ;
r r
r
n
r r
d L LB n
dt q q
B q t q B q t r n
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Definición de sistema lagrangiano:
U sistema de partículas sin ligaduras (o con ligaduras holónomas
ideales) bajo la acción de unas fuerzas que derivan de un potencial
generalizado se dice que es un sistema lagrangiano.
Un sistema físico cuya evolución en el tiempo ( ) se
determina a partir de las ecuaciones
para una cierta función , se dice que es un sistema
lagrangiano.
0, 1,2,3,j j
d L Lj
dt q q
( ), 1,2,3,jq t j
( , , )L q q t
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Introducción a las leyes de conservación o
constantes del movimiento.
Definición:
Las ecuaciones de Euler-Lagrange constituyen un sistema de
ecuaciones diferenciales ordinarias (EDO) de segundo orden en las
coordenadas generalizadas . Cualquier que
permanece constante durante el movimiento del sistema se llama
constante del movimiento o integral primera :
1 2, ,q q q ( , , )q q t
donde son las trayectorias o soluciones de las ecuaciones del
movimiento.
( )q t
Si es una integral primera deberá verificar : ( , , )q q t
( )
0,j j
j j q q t
dq q
dt t q q
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Casos “triviales” de leyes de conservación Energía
1)
Supongamos un sistema con la lagrangiana, , independiente del
tiempo, y sometido a dos, una o ninguna ligadura no holónoma ideal sin
término independiente, es decir . Bajo estas
condiciones la función
( , ) ,i
i i
LE q q q L
q
( , )L q q
es una constante del movimiento o integral primera.
( , ) 0ri i
i
B q t q
Demostración:
El sistema de ecuaciones que determina el movimiento es
,r ri
ri i
d L LB
dt q q
( , ) 0,ri i
i
B q t q y derivando E se obtiene
i i i i
i i i ii i i i
dE L d L L L Lq q q q
dt q dt q q q t
0,i r ri i
i r ii i
d L Lq B q
dt q q
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Caso particular de 1)
Demostración:
( )jk j
jk k
L Tc q q
q q
• Si , con (función solo de las q) y una
función cuadrática homogénea de las velocidades generalizadas, es
decir, , entonces .
L T U ( )U U q ( , )T q q
12
,
( )ij i j
i j
T c q q q ( , )E q q T U
( , ) ( ) 2 ,i ij i j
i i ji
LE q q q T U c q q q T U T T U T U
q
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Demostración:
a)
i i
d d
dt q q dt
b) Es consecuencia directa de a) y 1)
i i
d d d
dt q dt q dt
j j
j ji j i j
dq q
dt q t q q t q
2 2
j j
j ji j i i j
dq q
dt q q q t q q
2 2
j
ji i i j
dq
dt q q t q q
2 2 2 2
0, ( ).j j i
j ji j i i i j
q q q tq t q q q t q q
2)
Supongamos un sistema lagrangiano con lagrangiana , tal que
admite la descomposición
a) Las soluciones de las ecuaciones de Lagrange con lagrangiana
son idénticas a las soluciones con lagrangiana .
b) Si no depende de t explícitamente, , el sistema tiene la
ley de conservación: .
( , , )L q q t
( , , ) ( , , ) ( , ).d
L q q t L q q t q tdt
LL
( , ) i
i i
LE q q q L
q
L ( , )L q q
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Definición: En un sistema lagrangiano se denomina coordenada
cíclica a aquella coordenada generalizada que no aparece
explícitamente en la expresión de la lagrangiana.
Definición de momento canónico conjugado a una
coordenada .
Supongamos un sistema lagrangiano con lagrangiana . El momento
canónico, , conjugado a la coordenada es la función de ,
definida por:
, ,q q t
( , , ) .j
j
Lp q q t
q
jqjpL
jq
Ley de conservación: El momento canónico conjugado a una
coordenada cíclica es una constante del movimiento.
Demostración: sea la coordenada cíclica, De la
ecuación de Lagrange correspondiente a esa coordenada se tiene:
kq 0.k
L
q
0, ( , , ) .k
k k
d L Lp q q t const
dt q q
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Ejemplo 1 (ejercicio nº2 de apuntes).
Dos partículas de masas M1 y M2 están unidas a través de un hilo ideal que
pasa por un agujero taladrado en el plano horizontal (sin rozamiento) de la figura.
Determinar: Variedad de configuración, fuerzas generalizadas, ecuaciones de
Lagrange, etc.
x
y
z
2
1
g
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x
y
z
2
1
g
hT
Partícula 1: Partícula 2: ( , )x y z
Espacio de config. Cartesiano:
1 2 3, , , ,x x x x y z
1 2 3 1 1 2 1 2, , , , ; ;m m m M M M m M M
Espacio vect. de config: 1 2 31,0,0 , 0,1,0 , 0,0,1 ,e e e
1 1 2
1 2 1 2 1 21 1 2 2 3 3; ; ;M M M
M M M M M Me e e e e e
1 2 1 2 1 2
1 1 2
1 2 3
1 2 3; ; ;M M M M M M
M M Me e e e e e
2M g
1 2 3;r xe ye ze
hT
3 2 312 1 2
2 2 2 2( ) ;CH
h h h
x yf M ge f T e T e T M g e
x y x y
Ecuación de ligadura: 2 2 2 2
1 0,x y z x y z
2 2 2 311
2 2 2 2,
x ye e x y e
x y x y
2 311
2 2 2 2;CH h
h
h
x y Tf T e e e
Tx y x y
Ligadura ideal:
1 1 1;CHf ;h hT T
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Coordenadas generalizadas: (no usamos la ligadura en la parametrización de la
variedad de configuración)
( , , )q x y z
x
y
z
2
1
g
2 2 2 21 1 11 22 2 2( )T mv M x y M z
1 1 2 22 2 2 2
3 3 2
; ;
;
h h
h
x yQ f e T Q f e T
x y x y
Q f e T M g
1
2
3
,
,
,
d T TQ
dt x x
d T TQ
dt y y
d T TQ
dt z z
2M g
hT
hT
2 2
1 0,x y z
1 2 3;dr
v xe ye zedt
12 2
12 2
2 2
,
,
,
h
h
h
xM x T
x y
yM y T
x y
M z T M g
1 1 2 2 3 3, , ,r
a e a e a ex
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Coordenadas generalizadas: (usamos la ligadura en la parametrización de la
variedad de configuración!!!!)
( , )q
x
y
z
2
1
g
2 312 1 2
1 2 3 2
cos sin ( ) ( sin cos ) 0,
(cos sin ) ,
h h hQ f a T e T e T M g e e e
Q f a f e e e M g
,
,
d T TQ
dt
d T TQ
dt
2M g
hT
hT
,dr
v a adt
1 2 3( , ) (cos sin ) ( ) ;r e e e
cos , sin , ,x y z
1 2 3 1 2cos sin , ( sin cos ),r r
a e e e a e e
2 2 2 2 2 2 2 21 1 1 1 11 2 1 2 22 2 2 2 2( ) ( ) ,T mv M M M M M
2
2
2
1 2 2 2
( )0,
( ) ,
dM
dt
M M M M g
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Obtened la lagrangiana usando
y 2 leyes de conservación.
( , )q
x
y
z
2
1
g
2M g
hT
hT
0, ( , ) .L L L
E q q L T U constt
2 2 21 11 2 2 22 2
( ) ( )L T U M M M M g
2 2 ( ),U M g z M g
2
20, .L L
p M const
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30
Determinación de fuerzas de ligadura holónomas ideales
* ,CH CNf f f f
1
,CN
r r
r
f b
• Supongamos que el sistema (N partículas: espacio de configuración 3N) posee M
ligaduras holónomas y ligaduras no holónomas (ambas ideales):
( , ) 0,
1,2, , .
j r t
j M
3
1
( , ) ( , ) 0,
1, , 3 .
N
r rA r t x A r t
r n N M
1
,M
CH
j j
j
f
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1
( ) ( ) , 1, , 3 1 ,
( , ) ( , ) 0, 1, , ;
( , ) 0,
M M r r
r
n
r r
M
d T TQ a B n N M
dt q q
B q t q B q t r n
q t
31
Determinación de fuerzas de ligadura holónomas ideales
1 2, , , ; 3 1 .nq q q q n N M
• Las ecuaciones de Lagrange se pueden plantear con coordenadas generalizadas
sobre una variedad de configuración de dimension con ,
dependiendo del nº de ligaduras holónomas que tomemos en la parametrización. n 3 3N M n N
• Las fuerzas de ligadura holónomas (los ) no apareceran si se escoge la
variedad de configuración de dimensión mínima posible: . 3n n N M
• Las fuerzas de ligadura no holónomas (los ) siempre apareceran cualquiera
que sea la dimensión de la variedad de configuración. r
j
• Supongamos que queremos determinar la fuerza de ligadura hónoma ejercida por
la ligadura, por ejemplo, : j M CH
M M Mf
( , ) 0,
1,2, , 1.
j r t
j M
( , ), , 1,2, ,r q t a n
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32
Potencial de fuezas giroscópicas
0Q q
• Def: Se denominan fuerzas giroscópicas a aquellas cuya potencia es nula:
1 2, ,q q q
( )d d
Q q q qq dt q q dt
; ; ;q q qq q q q
• Ejemplo: El potencial generalizado es giroscópico. ( , ) ( )U q q q q
Q q
0.q q