Post on 08-Jan-2016
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Variational Principles
and Lagrange’s Equations
Definitions
• Lagrangian density:
• Lagrangian:
• Action:
• How to find the special value for action corresponding to observable ?
nin
nmi
xx
xη,
)(L tzyxx
i
n ,,,
?...3,2,1,0
dxdydzL L
t
dt
trdL
im
i
,)(
dtt
dt
trdLI
im
i
,)(
)(tr Joseph Louis
Lagrange/Giuseppe Luigi
Lagrangia (1736 – 1813)
zyxm rrrr ,,
Variational principle
• Maupertuis: Least Action Principle
• Hamilton: Hamilton’s Variational Principle
• Feynman: Quantum-Mechanical Path Integral Approach
Pierre-Louis Moreau de Maupertuis (1698 – 1759)
Sir William Rowan Hamilton
(1805 – 1865)
Richard Phillips Feynman
(1918 – 1988)
Functionals
• Functional: given any function f(x), produces a number S
• Action is a functional:
• Examples of finding special values of functionals using variational approach:
shortest distance between two points on a plane;the brachistochrone problem;minimum surface of revolution; etc.
)]([ xfS
2
1
,)(
)]([t
ti
mi
dttdt
trdLtrI
Shortest distance between two points on a plane
• An element of length on a plane is
• Total length of any curve going between points 1 and 2 is
• The condition that the curve is the shortest path is that the functional I takes its minimum value
22 dydxds
2
1
dsI
2
1
2
1 dxdx
dy
The brachistochrone problem
• Find a curve joining two points, along which a particle falling from rest under the influence of gravity travels from the highest to the lowest point in the least time
• Brachistochrone solution: the value of the functional t [y(x)] takes its minimum value
dtdsv /
2
1
12 v
dst
2
1
2
2
/1dx
gy
dxdy
22 dydxds
mgymv
2
2
gyv 2
Calculus of variations
• Consider a functional of the following type
• What function y(x) yields a stationary value (minimum, maximum, or saddle) of J ?
2
1
),...,,,,()]([x
x
dxxyyyyfxyJ dx
dyy
x
y
0
),( 22 yx
),( 11 yx
Calculus of variations
• Assume that function y0(x) yields a stationary value and consider all possible functions in the form:
x
y
0
),( 22 yx
),( 11 yx
...)()()(),( 12
0 xxxyxy
0)()( 21 xx
Calculus of variations
• In this case our functional becomes a function of α:
• Stationary value condition:
)()(),( 0 xxyxy 0)()( 21 xx
)(),()],([2
1
JdxxfxyJx
x
0)()(
0)(),( 0
d
dJ
d
dJ
xyxy
Stationary value
2
1
),...,,,()(
x
x
dxxyyyfd
d
d
dJ
2
1
),...,,,(x
x
dxd
xyyydf
2
1
...x
x
dxy
y
fy
y
fy
y
f
1
2
3
2
1
.1x
x
dxy
y
f
2
1
x
x
dxy
f
)()(),( 0 xxyxy
Stationary value
2
1
),...,,,()(
x
x
dxxyyyfd
d
d
dJ
2
1
),...,,,(x
x
dxd
xyyydf
2
1
...x
x
dxy
y
fy
y
fy
y
f
1
2
3
2
1
.2x
x
dxy
y
f
2
1
2x
x
dxx
y
y
f
u
dv
2
1
x
x
y
y
f
u
v
2
1
x
x
dxy
f
dx
dy
v
du
)()(),( 0 xxyxy
2
1
x
xy
f
2
1
x
x
dxy
f
dx
d
0)(
0)(
2
1
x
x
Stationary value
2
1
),...,,,()(
x
x
dxxyyyfd
d
d
dJ
2
1
),...,,,(x
x
dxd
xyyydf
2
1
...x
x
dxy
y
fy
y
fy
y
f
1
2
3
2
1
.3x
x
dxy
y
f
2
1
x
x
y
y
f
2
1
x
x
dxy
f
dx
dy
2
1
x
xy
f
2
1
x
xy
f
dx
d
2
1
2
2x
x
dxy
f
dx
d
Stationary value
2
1
),...,,,()(
x
x
dxxyyyfd
d
d
dJ
2
1
),...,,,(x
x
dxd
xyyydf
2
1
...x
x
dxy
y
fy
y
fy
y
f
1
2
3
2
1
x
xy
f
2
1
2
2x
x
dxy
f
dx
d
2
1
x
x
dxy
f
2
1
x
x
dxy
f
dx
d
2
1
...2
2x
x
dxy
f
dx
d
y
f
dx
d
y
f
2
1
x
xy
f
...
...
Stationary value
... ...2
1
2
1
2
2
x
x
x
x y
fdx
y
f
dx
d
y
f
dx
d
y
f
d
dJ
),( xyff
2
1
x
x
dxy
f
d
dJ
0)(
0
d
dJ0
2
1 0 )(),(
x
x xyxy
dxy
f
arbitrary
0y
fTrivial …
Stationary value
... ...2
1
2
1
2
2
x
x
x
x y
fdx
y
f
dx
d
y
f
dx
d
y
f
d
dJ
),,( xyyff
2
1
x
x
dxy
f
dx
d
y
f
d
dJ
0)(
0
d
dJ0
2
1 0
x
x
dxy
f
dx
d
y
f
arbitrary0
y
f
dx
d
y
f Nontrivial !!!
Shortest distance between two points on a plane
2
1
2
1 dxdx
dyI 21 yf
0
y
f
dx
d
y
f
01
02
y
y
dx
d
c
y
y
21
21 c
cy
bx
c
cy
21
Straight line!
The brachistochrone problem
2
1
2
122
/1dx
gy
dxdyt
gy
yf
2
1 2
0
y
f
dx
d
y
f
0
1222
123
2
ygy
y
dx
d
gy
y
Scary!
Recipe
• 1. Bring together structure and fields
• 2. Relate this togetherness to the entire system
• 3. Make them fit best when the fields have observable dependencies:
Structure
FieldsFields
Structure
Physical Laws
Best F
it
mη mη
Back to trajectories and Lagrangians
• How to find the special values for action corresponding to observable trajectories ?
• We look for a stationary action using variational principle
2
1
,)(
)]([t
ti
mi
dttdt
trdLtrI
)()(),( 0 ttrtr mmm
0)()( 21 tt mm 0)(
0
d
dI
2
1
,),(
)],([)(t
ti
mi
m dttdt
trdLtrII
Recipe
• 1. Bring together structure and fields
• 2. Relate this togetherness to the entire system
• 3. Make them fit best when the fields have observable dependencies:
Structure
FieldsFields
Structure
Physical Laws
Best F
it
mη mη
Back to trajectories and Lagrangians
• For open systems, we cannot apply variational principle in a consistent way, since integration in not well defined for them
• We look for a stationary action using variational principle for closed systems:
2
1
,),(
)],([)(t
ti
mi
m dttdt
trdLtrII
0)(
0
d
dI
dxdydzL L dxdydzdtI L
Stationary value
... ...2
1
2
1
2
2
x
x
x
x y
fdx
y
f
dx
d
y
f
dx
d
y
f
d
dJ
),,( xyyff
2
1
x
x
dxy
f
dx
d
y
f
d
dJ
0)(
0
d
dJ0
2
1 0
x
x
dxy
f
dx
d
y
f
0
y
f
dx
d
y
f Nontrivial !!!
Simplest non-trivial case
• Let’s start with the simplest non-trivial result of the variational calculus and see if it can yield observable trajectories
2
1
,),(
)(t
ti
mi
dttdt
trdLI
0)(
0
d
dI
2
1
,),(
),,(t
t
mm dtt
dt
tdrtrL
zyxm
i
,,
1,0
2
1
,,t
t
mm dttrrL
Stationary value
... ...2
1
2
1
2
2
x
x
x
x y
fdx
y
f
dx
d
y
f
dx
d
y
f
d
dJ
),,( xyyff
2
1
x
x
dxy
f
dx
d
y
f
d
dJ
0)(
0
d
dJ0
2
1 0
x
x
dxy
f
dx
d
y
f
0
y
f
dx
d
y
f Nontrivial !!!
Euler- Lagrange equations
• These equations are called the Euler- Lagrange equations
0)(
0
d
dI 2
1
,,t
t
mm dttrrLI
0
y
f
dx
d
y
f
0
mm r
L
dt
d
r
L
Joseph Louis Lagrange
(1736 – 1813)
Leonhard Euler (1707 – 1783)
Recipe
• 1. Bring together structure and fields
• 2. Relate this togetherness to the entire system
• 3. Make them fit best when the fields have observable dependencies:
Structure
FieldsFields
Structure
Physical Laws
Best F
it
mη mη
How to construct Lagrangians?
• Let us recall some kindergarten stuff
• On our – classical-mechanical – level, we know several types of fundamental interactions:
• Gravitational• Electromagnetic• That’s it
0
mm r
L
dt
d
r
L
Gravitation
• For a particle in a gravitational field, the trajectory is described via 2nd Newton’s Law:
• This system can be approximated as closed
• The structure (symmetry) of the system is described by the gravitational potential
gUdt
vmd
)(
gm
),,,( tzyxgg
Sir Isaac Newton(1643 – 1727)
Electromagnetic field
• For a charged particle in an electromagnetic field, the trajectory is described via 2nd Newton’s Law:
• This system can be approximated as closed
• The structure (symmetry) of the system is described by the scalar and vector potentials
),,,(
),,,(
tzyx
tzyxAA
)( AvqqAqvmdt
d Really???
Electromagnetic field
)( AvqqAqvmdt
d
)()(
Avqqdt
Adq
dt
vmd
dt
Ad
),,,( tzyxAA
zz
Ay
y
Ax
x
A
t
A
Avt
A
)(
Avqt
AqAvqq
dt
vmd
)()()(
AvAvqt
Aqq
dt
vmd
)()()(
Electromagnetic field
AvAvqt
Aqq
dt
vmd
)()()(
FGFGGFGF
)()()(
GF
vAvAAvAv
)()()(
Av
Avqt
Aqq
dt
vmd
)(
Electromagnetic field
• Lorentz force!
Avqt
Aq
dt
vmd
)(
t
AE
AB
BvEqdt
vmd
)(
Hendrik Lorentz(1853-1928)
Kindergarten
• Thereby:
• In component form
0)( Aqvmdt
dAvqq
0)()(
dt
rmd
r
m j
j
g
0)(
dt
vmdm g
0)())((
dt
qArmd
r
Arqq jj
j
How to construct Lagrangians?
• Kindergarten stuff:
• The “kindergarten equations” look very similar to the Euler-Lagrange equations! We may be on the
right track!
0
jj r
L
dt
d
r
L
0)()(
dt
rmd
r
m j
j
g
0)())((
dt
qArmd
r
Arqq jj
j
Gravitation
0
jj r
L
dt
d
r
L 0
)()(
dt
rmd
r
m j
j
g
j
g
j r
m
r
L
)(
j
j
r
L
dt
d
dt
rmd
)(
),,,( trrrTmL zyxg C
r
Lrm
jj
),,,( trrrS zyx),,,( trrr zyxgg
2
)( 222zyx rrrm
L
Gravitation
gzyx mtrrrTL ),,,(
),,,(2
)( 222
trrrSrrrm
L zyxzyx
gzyx mrrrm
L
2
)( 222
Electromagnetism
0
jj r
L
dt
d
r
L
0)())((
dt
qArmd
r
Arqq jj
j
)(2
)( 222
Arqqrrrm
L zyx
Bottom line
• We successfully demonstrated applicability of our recipe
• This approach works not just in classical mechanics only, but in all other fields of physics
Structure
Physical LawsBes
t Fit
Some philosophy
• de Maupertuis on the principle of least action (“Essai de cosmologie”, 1750): “In all the changes that take place in the universe, the sum of the products of each body multiplied by the distance it moves and by the speed with which it moves is the least that is possible.”
• How does an object know in advance what trajectory corresponds to a stationary action???
• Answer: quantum-mechanical pathintegral approach
Pierre-Louis Moreau de Maupertuis (1698 – 1759)
Some philosophy
• Feynman: “Is it true that the particle doesn't just "take the right path" but that it looks at all the other possible trajectories? ... The miracle of it all is, of course, that it does just that. ... It isn't that a particle takes the path of least action but that it smells all the paths in the neighborhood and chooses the one that has the least action ...”
Richard Phillips Feynman
(1918 – 1988)
Some philosophy
• Dyson: “In 1949, Dick Feynman told me about his "sum over histories" version of quantum mechanics. "The electron does anything it likes," he said. "It just goes in any direction at any speed, forward or backward in time, however it likes, and then you add up the amplitudes and it gives you the wave-function." I said to him, "You're crazy." But he wasn't.”
Freeman John Dyson (born 1923)
Some philosophy
• Philosophical meaning of the Lagrangian formalism: structure of a system determines its observable behavior
• So, that's it?
• Why do we need all this?
• In addition to the deep philosophical meaning, Lagrangian formalism offers great many advantages compared to the Newtonian approach
Lagrangian approach: extra goodies
• It is scalar (Newtonian – vectorial)
• Allows introduction of configuration space and efficient description of systems with constrains
• Becomes relatively simpler as the mechanical system becomes more complex
• Applicable outside Newtonian mechanics
• Relates conservation laws with symmetries
• Scale invariance applications
• Gauge invariance applications
Simple example
• Projectile motion gzyx mrrrm
L
2
)( 222
zzyx mgrrrrm
L
2
)( 222
0
yy r
L
dt
d
r
L
0
xx r
L
dt
d
r
L
0
zz r
L
dt
d
r
L
0xrmdt
d
0yrmdt
d
constrm x
constrm y
zg gr
mgrmdt
dz constgtrz
Another example
• Another Lagrangian
• What is going on?!
xy
zx mgrrm
rrmL 2
2
0
yy r
L
dt
d
r
L
0
xx r
L
dt
d
r
L
0
zz r
L
dt
d
r
L
0 zrmdt
d
0yrmdt
d
constrm x
constrm y
constgtrz mg
0 0 xrmdt
d
Gauge invariance
• For the Lagrangians of the type
• And functions of the type
• Let’s introduce a transformation (gauge transformation):
trrL ii ,,
dt
trdFtrrLtrrL i
iiii
,,,,,'
trF i ,
Gauge invariance
dt
dFLL ' j
j j
rr
F
t
F
dt
dF
trFF i ,
dt
dF
rr
L
r
L
iii
'
jj jii
rr
F
t
F
rr
L
jj jiii
rrr
F
tr
F
r
L
22
Gauge invariance
dt
dFLL ' j
j j
rr
F
t
F
dt
dF
trFF i ,
dt
dF
rr
L
r
L
iii '
jj jii
rr
F
t
F
rr
L
ii r
F
r
L
iii r
F
dt
d
r
L
dt
d
r
L
dt
d
'
ii r
F
tr
L
dt
d
jj ij
rr
F
r
Gauge invariance
jj ijiii
rr
F
rr
F
tr
L
dt
d
r
L
dt
d
'
jj ijii
rrr
F
rt
F
r
L
dt
d
22
jj jiiii
rrr
F
tr
F
r
L
r
L
22'
ii r
L
r
L
dt
d
'' ii r
L
r
L
dt
d
0
Back to the question: How to construct Lagrangians?
• Ambiguity: different Lagrangians result in the same equations of motion
• How to select a Lagrangian appropriately?
• It is a matter of taste and art
• It is a question of symmetries of the physical system one wishes to describe
• Conventionally, and for expediency, for most applications in classical mechanics:
VTL
Cylindrically symmetric potential
• Motion in a potential that depends only on the distance to the z axis
• It is convenient to work in cylindrical coordinates
• Then
22
222
2
)(yx
zyx rrVrrrm
L
zrrrrr zyx ;sin ;cos
zr
rrr
rrr
z
y
x
cossin
sincos
Cylindrically symmetric potential
• How to rewrite the equations of motion in cylindrical coordinates?
22
222
2
)(yx
zyx rrVrrrm
L
22222
sincos2
rrVzm
2
)cossin( 2 rrm
2
)sincos( 2 rrm
)(2
)( 2222
rVzrrm
0
jj r
L
dt
d
r
L
Generalized coordinates
• Instead of re-deriving the Euler-Lagrange equations explicitly for each problem (e.g. cylindrical coordinates), we introduce a concept of generalized coordinates
• Let us consider a set of coordinates
• Assume that the Euler-Lagrange equations hold for these variables
• Consider a new set of (generalized) coordinates
),...,,(: 21 Ni rrrr
0
ii r
L
dt
d
r
L
),,...,,( 21 trrrqq Njj
Generalized coordinates
• We can, in theory, invert these equations:
• Let us do some calculations:
N
i m
i
im q
r
r
L
q
L
1
),,...,,( 21 trrrqq Nmm
),,...,,( 21 tqqqrr Mii
M
mm
m
iii q
q
r
t
rr
1
N
i m
i
im q
r
r
L
dt
d
q
L
dt
d
1
N
i m
i
i q
r
r
L
dt
d
1
m
iN
i i
N
i m
i
i q
r
dt
d
r
L
q
r
r
L
dt
d
11
m
i
m
i
q
r
q
r
Generalized coordinates
• The Euler-Lagrange equations are the same in generalized coordinates!!!
M
kk
m
i
k
i
m
i
m
i qq
r
q
r
q
r
tq
r
dt
d
1
m
iN
i i
N
i m
i
im q
r
dt
d
r
L
q
r
r
L
dt
d
q
L
dt
d
11
M
kk
k
ii
m
r
t
r
q 1
M
mm
m
iii q
q
r
t
rr
1
m
i
q
r
mq
L
dt
d m
iN
i i q
r
r
L
1 mq
L
N
i m
i
i q
r
r
L
1
Generalized coordinates
• If the Euler-Lagrange equations are true for one set of coordinates, then they are also true for the other set
ii r
L
r
L
dt
d
),,...,,( 21 trrrqq Nmm
mm q
L
q
L
dt
d
Cylindrically symmetric potential
• Radial force causes a change in radial momentum and a centripetal acceleration
)(2
)( 2222
rVzrrm
L
0
jj q
L
dt
d
q
L
),,(: zrqi 0
r
L
dt
d
r
L
r
rV
)(
0)(
dt
rmd
dt
rmdmr
r
rV )()( 2
2mr
Cylindrically symmetric potential
• Angular momentum relative to the z axis is a constant
)(2
)( 2222
rVzrrm
L
0
jj q
L
dt
d
q
L
),,(: zrqi 0
L
dt
dL
0 0)( 2
dt
mrd
constmrrmr )(2
0)( 2
dt
mrd
Cylindrically symmetric potential
• Axial component of velocity does not change
)(2
)( 2222
rVzrrm
L
0
jj q
L
dt
d
q
L
),,(: zrqi 0
z
L
dt
d
z
L
0 0)(
dt
zmd
constzm
0)(
dt
zmd
Symmetries and conservation laws
• The most beautiful and useful illustration of the “structure vs observed behavior” philosophy is the link between symmetries and conservation laws
• Conjugate momentum for coordinate :
• If Lagrangian does not depend on a certain coordinate, this coordinate is called cyclic (ignorable)
• For cyclic coordinates, conjugate momenta are conserved
)( iqfL
mq
L
mq
0
jj q
L
dt
d
q
L
0
jq
L
dt
d
Symmetries and conservation laws
• For cyclic coordinates, conjugate momenta are conserved
p =
cons
t p ≠ const
Cylindrically symmetric potential
• Cyclic coordinates:
• Rotational symmetry Translational symmetry
• Conjugate momenta:
)(2
)( 2222
rVzrrm
L
0
L
dt
dL
constmr 2
0
z
L
dt
d
z
L
constzm
z
Electromagnetism
• Conjugate momenta:
)(2
)( 222
Arqqrrrm
L zyx
jr
L
jrm jqA jrm
Noether’s theorem
• Relationship between Lagrangian symmetries and conserved quantities was formalized only in 1915 by Emmy Noether:
• “For each symmetry of the Lagrangian, there is a conserved quantity”
• Let the Lagrangian be invariant under the change of coordinates:
• α is a small parameter. This invariance
has to hold to the first order in α
),,...,,(~21 tqqqqq Niii
Emmy Noether/Amalie Nöther(1882 – 1935)
Noether’s theorem
• Invariance of the Lagrangian:
• Using the Euler-Lagrange equations
0ddL
N
i
i
i
i
i
q
q
Lq
q
L
d
dL
1
~
~
~
~
N
ii
i
ii q
L
q
L
1~~
N
ii
i
i
i q
L
q
L
dt
d
1~~
N
ii
iq
L
dt
d
1~ 0
constq
LN
ii
i
1
),,...,,(~21 tqqqqq Niii
Example
• Motion in an x-y plane of a mass on a spring (zero equilibrium length):
• The Lagrangian is invariant (to the first order in α) under the following change of coordinates:
• Then, from Noether’s theorem it follows that
2
)(
2
)( 2222yxyx rrkrrm
L
xyyyxx rrrrrr ~ ;~
constr
L
r
Ly
yx
x
yxrrm xyrrm const
Example
• In polar coordinates:
• The conserved quantity:
• Angular momentum in the x-y plane is conserved
constrrmrrm xyyx
sin
cos
rr
rr
y
x
cossin
sincos
rrr
rrr
y
x
xyyx rrmrrm sin)sincos( rrrm
cos)cossin( rrrm 2mr const
Example
• For the same problem, we can start with a Lagrangian expressed in polar coordinates:
• The Lagrangian is invariant (to any order in α) under the following change of coordinates:
• The conserved quantity from Noether’s theorem:
constmr 12
2
)(
2
)( 2222yxyx rrkrrm
L
22
)( 2222 krrrm
1~
constL
Back to trajectories and Lagrangians
• How to find the special values for action corresponding to observable trajectories ?
• We look for a stationary action using variational principle
2
1
,)(
)]([t
ti
mi
dttdt
trdLtrI
)()(),( 0 ttrtr mmm
0)()( 21 tt mm 0)(
0
d
dI
2
1
,),(
)],([)(t
ti
mi
m dttdt
trdLtrII
),,...,,(~21 tqqqqq Nmmm
Stationary value
2
1
),...,,,()(
x
x
dxxyyyfd
d
d
dJ
2
1
),...,,,(x
x
dxd
xyyydf
2
1
...x
x
dxy
y
fy
y
fy
y
f
1
2
3
2
1
.2x
x
dxy
y
f
2
1
2x
x
dxx
y
y
f
u
dv
2
1
x
x
y
y
f
u
v
2
1
x
x
dxy
f
dx
dy
v
du
)()(),( 0 xxyxy
2
1
x
xy
f
2
1
x
x
dxy
f
dx
d
0)(
0)(
2
1
x
x
constq
LN
ii
i
1
More on symmetries
• Full time derivative of a Lagrangian:
• From the Euler-Lagrange equations:
• If
dt
dL
M
m
M
mm
mm
m
Lq
q
L
t
L
1 1
M
m
M
mm
mm
m
Lq
q
L
dt
d
t
L
1 1
M
mm
m
L
dt
d
t
L
1
Lqq
L
dt
d
t
L M
mm
m1
dt
dH
0t
L constLqq
LH
M
mm
m
1
What is H?
• Let us expand the Lagrangian in powers of :
• Form calculus, for a homogeneous function f of
degree n (Euler’s theorem) :
......),,...,,(
),,...,,(),,...,,(
3210,
212
211210
LLLLqqtqqql
qtqqqltqqqLL
jji
iMij
iiMiM
fnx
fx
i ii
iq
...210
ii
iii
iii
iii
i
Lq
q
Lq
q
Lq
q
L
...320 321 LLL
What is H?
• If the Lagrangian has a form:
• Then
• For electromagnetism:
Lqq
LH
M
mm
m
1
...32 321 LLL
...)( 3210 LLLL ...2 320 LLL
210 LLLL
02 LLH
)(2/2 ArqqrmL
2L 0L 1L
02 LLH qrm 2/2 EVT
Conservation of energy
• In the field formalism, the conservation of H is a part of Noether’s theorem
210 LLLL
EH
constEt
L
0
The brachistochrone problem
• Similarly to the “H-trick”:
2
1
12 dxft gy
yf
2
1 2 0
x
f
0
1222
123
2
ygy
y
dx
d
gy
y
Scary!
constfy
fy
H
gy
y
ygy
yy
2
1
12
2
2
constygy
212
1
21/ yyC
!!!
The brachistochrone problem
• Change of variables:
• Parametric solution (cycloid)
21/ yyC
2sinCy
dx
dy
y
Cy 1 dy
yC
ydx
)sin(sin
sin 22
2
Cd
CC
Cdx
dC sin2 2
BdCx sin2 2
2sin
)2/)2(sin(
Cy
CBx
)2/)2(sin( CB
Scale invariance
• For Lagrangians of the following form:
• And homogeneous L0 of degree k
• Introducing scale and time transformations
• Then
jji
iijM qqlqqqLLLL ,
221020 ),...,,( constl ij 2
tt
qq ii
'
'
),...,,(),...,,(' 2102100 Mk
M qqqLqqqLL
jji
iij qql
,2
2
ii qq
' jji
iij qqlL ''',
22
Scale invariance
• Therefore, after transformations
• If
• Then
• The Euler-Lagrange equations after transformations
• The same!
2
2
0' LLL k
k
2
LL k'
0''
jj q
L
dt
d
q
L
0)()(
j
k
j
k
q
L
dt
d
q
L
0
jj q
L
dt
d
q
L
Scale invariance
• So, the Euler-Lagrange equations after transformations are the same if
• Free fall
• Let us recall
k
2
2/1 k2/1
2/1 ''k
i
ik
q
q
t
t
mgzzm
L 2
21k
2/1''
k
z
z
t
t
2/1'
z
z
2/12z
g
zt
Scale invariance
• So, the Euler-Lagrange equations after transformations are the same if
• Mass on a spring
• Let us recall
k
2
2/1 k2/1
2/1 ''k
i
ik
q
q
t
t
22
22 KzzmL
2k
2/1''
k
z
z
t
t
0'
z
z
02 zK
mT
Scale invariance
• So, the Euler-Lagrange equations after transformations are the same if
• Kepler’s problem
• Let us recall 3rd Kepler’s law
k
2
2/1 k2/1
2/1 ''k
i
ik
q
q
t
t
2
2
2 r
MmG
rmL
1k
2/1''
k
z
z
t
t
2/3'
z
z
2/3RT
Johannes Kepler(1571-1630)
How about open systems?
• For some systems we can neglect their interaction with the outside world and formulate their behavior in terms of Lagrangian formalism
• For some systems we can not do it
• Approach: to describe the system without “leaks” and “feeds” and then add them to the description of the system
How about open systems?
• For open systems, we first describe the system without “leaks” and “feeds”
• After that we add “leaks” and “feeds” to the description of the system
• Q: Non-conservative generalized forces
jj q
L
q
L
dt
d
jQ
Generalized forces
• Forces
• 1: Conservative (Potential)
• 2: Non-conservative
j
jj
T
q
T
dt
d
UU
jjjjj
Qqqdt
d
q
T
q
T
dt
d
UU
...),,...,,( 21 tqqqV NU
12
1
UTL
Generalized forces
• In principle, there is no need to introduce generalized forces for a closed system fully described by a Lagrangian
• Feynman: “…The principle of least actiononly works for conservative systems — where all forces can be gotten from a potential function. … On a microscopic level — on the deepest level of physics — there are no non-conservative forces. Non-conservative forces, like friction, appear only because we neglect microscopic complications — there are just too many particles to analyze.”
• So, introduction of non-conservative forces is a result of the open-system approach
Richard Phillips Feynman
(1918 – 1988)
Degrees of freedom
• The number of degrees of freedom is the number of independent coordinates that must be specified in order to define uniquely the state of the system
• For a system of N free particle there are 3N degrees of freedom (3N coordinates)
N) ..., ,2 ,1(
ˆˆˆ
Ni
rkrjrir ziyixii
Constraints
• We can imposed k constraints on the system
• The number of degrees of freedom is reduced to 3N – k = s
• It is convenient to think of the remaining s independent coordinates as the coordinates of a single point in an s-dimensional space: configuration space N
), ..., , ,(
...
), ..., , ,(
321
32111
tqqqrr
tqqqrr
kNNN
kN
k
Types of constraints
• Holonomic (integrable) constraints can be expressed in the form:
• Nonholonomic constraints cannot be expressed in this form
• Rheonomous constraints – contain time dependence explicitly
• Scleronomous constraints – do not contain time dependence explicitly
kj
tqqqf nj
,...,2,1
0), ..., , ,( 21
Analysis of systems with holonomic constraints
• Elimination of variables using constraints equations
• Use of independent generalized coordinates
• Lagrange’s multiplier method
Double 2D pendulum
• An example of a holonomic scleronomous constraint
• The trajectories of the system are very complex
• Lagrangian approach produces equation of motion
• We need 2 independent generalized coordinates (N = 2, k = 2 + 2, s = 3 N – k = 2)
0)( 21
21 lr 0)( 2
22
21 lrr
1 2
Double 2D pendulum
• Relative to the pivot, the Cartesian coordinates
• Taking the time derivative, and then squaring
• Lagrangian in Cartesian coordinates:
11,1 sinlr x
11,1 coslr z 2211,2 sinsin llr x
2211,2 coscos llr z
21
21
21 lr
)sinsincos(cos2 212121212
22
22
12
12
2 llllr
)(2 ,22,11
222
211
zz rmrmgrmrm
L
Double 2D pendulum
• Lagrangian in new coordinates:
• The equations of motion:
)coscos(cos
2
)cos(2
2
22112111
2121212
22
22
12
122
12
11
llgmglm
llllmlmL
222212
1212
212
12122
22
22
1121212
2212
212
22122
12
121
sin)sin(
)cos(0
sin)()sin(
)cos()(0
glmllm
llmlm
glmmllm
llmlmm
Double 2D pendulum
• Special case
• The equations of motion:
• More fun at:
http://www.mathstat.dal.ca/~selinger/lagrange/doublependulum.html
21 mm 0, 21 lll 21
221
121
0
220
gl
gl
Lagrange’s multiplier method
• Used when constraint reactions are the object of interest
• Instead of considering 3N - k variables and equations, this method deals with 3N + k variables
• As a results, we obtain 3N trajectories and k constraint reactions
• Lagrange’s multiplier method can be applied to some nonholonomic constraints
Lagrange’s multiplier method
• Let us explicitly incorporate constraints into the structure of our system
• For observable trajectories
• So
kjtqqqf nj ,...,2,1 ;0), ..., , ,( 21
k
jnjj tqqqftLL
121 ), ..., , ,()('
0), ..., , ,( 21 tqqqf nj
k
jjj fLL
1
' L
ii q
L
q
L
dt
d
''
k
j i
jj
ii q
f
q
L
q
L
dt
d
1
0
0
Lagrange’s multiplier method
• - constraint reactions
• Now we have 3N + k equations for and
kjtqqqf nj ,...,2,1 ;0), ..., , ,( 11
k
j i
jj
ii q
f
q
L
q
L
dt
d
1
iQ
iQ
Niq
f
q
L
q
L
dt
d k
j i
jj
ii
3,...,2,1 ;1
iq j
Application to a nonholonomic case
• A particle on a smooth hemisphere
• One nonholonomic constraint:
• While the particle remains on the sphere, the constraint is holonomic
• And the reaction from the surface is not zero
02222 arrr zyx
02222 arrr zyx
Application to a nonholonomic case
• Constraint equation in cylindrical coordinates:
• New Lagrangian in cylindrical coordinates:
• Equations of motion
0 ar
)(cos2
)(' 1
2222
armgrzrrm
L
r
f
r
L
dt
d
r
L
1
1
01cos 12 mgmrrm
Application to a nonholonomic case
• Constraint equation in cylindrical coordinates:
• New Lagrangian in cylindrical coordinates:
• Equations of motion
0 ar
)(cos2
)(' 1
2222
armgrzrrm
L
0
L
dt
dL
0sin2 mgrmr
Application to a nonholonomic case
• Constraint equation in cylindrical coordinates:
• New Lagrangian in cylindrical coordinates:
• Equations of motion
• Trivial
0 ar
)(cos2
)(' 1
2222
armgrzrrm
L
0
z
L
dt
d
z
L
0zm
Application to a nonholonomic case
• Constraint reaction:
ar
0cos 12 mgmrrm
0sin2 mgrmr
a
mg /cos 12
sina
g sin22
a
g
cos22
a
g
dt
d
dt
d
0
0
)cos1(
22 a
g
)2cos3(1 mg
)2cos3(111
1 mgr
f
Application to a nonholonomic case
• Constraint reaction:
• Reaction disappears when
• The particle becomes airborne
)2cos3(1 mg
2cos3
3
2cos 1
ar