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Hamiltons Principle Symmetries Keywords and References

Hamiltons principle and Symmetries

Sourendu Gupta

TIFR, Mumbai, India

Classical Mechanics 2012

August 13, 2012

Sourendu Gupta Classical Mechanics 2012: Lecture 3

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Action

For a mechanical system with generalized coordinates q1, q2, in

motion between times t1 and t2, the action is defined as theintegral

S[t1, t2] =

t2t1

dtL(q1, q2, , q1, q2, , t)

The value of the action depends on the world line of the particle{q1(t), q2(t), }: it is a functional of the world line.

world linetrajectory t

q1

q2

Hamiltons principle

The particles actual trajectory is the one that minimizes the action

subject to the boundary conditions imposed on it.Sourendu Gupta Classical Mechanics 2012: Lecture 3

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Functionals: easy as

An example of a functional of a curve: the area under a curve!

A[y] =

xf

xi

dxy(x).

x

y

y(x)

x

y

y(x)

x

y

y(x)

The area is a functional of the curve. Use Riemann sums to

approximate the area in terms of N variables y(x1), y(x2), etc.,

A[y] = hN

Ni=1

y(xi), where hN =xf xi

N.

Finding variations of the area with the curve is just calculus of

many variables.Sourendu Gupta Classical Mechanics 2012: Lecture 3

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Calculus of variations

Finding an extremum of a function involves setting its derivatives

to zero and then solving the resulting equations. Checking whetheran extremum is a maximum or minimum involves checking the signof the second derivative. The minimum of a functional is similar.Make small variations qk(t) around a trajectory specified by qk(t)and qk(t). Since the boundary conditions are fixed, soqk(t1) = qk(t2) = 0. The variation of L under such a variationof the trajectory is

L =

kqk

L

qk+

d

dtqk

L

qk qk

d

dtL

qkNow we put this inside the integral to evaluate S. Sincet2

t1

dtd

dt

qk

L

qk

= qk

L

qk

t2

t1

= 0,

we recover the Euler-Lagrange equations.Sourendu Gupta Classical Mechanics 2012: Lecture 3

H il P i i l S i K d d R f

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Two examples

Quartic oscillator Physical pendulum

L 12 mq2 14 V0q

4 12 m

22 mg(1 cos )

Q V0q3 mg sin

p mq m2EoM mq+ V0q

3 = 0 m2 + mg sin = 0

q+ 2q3 = 0 + 2 sin = 0

where 2 = V0m

where 2 = gLt t + sin = 0

q q/ q+ q3 = 0


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Numerical solutions

Minimization of the action can be carried out numerically very

simply using a lattice discretization. Work with D degrees offreedom, q1, q2, qD. Divide the time interval into N equalpieces. At times tj = t0 +jhN use the notation q

jk = qk(tj), so that

the action is a function of D(N 1) variables

S = mk

2h2N

D,Nk=1,j=1

qjk q

j1k

2

Nj=1

V(qj1, qj2, , q

jD)

= N2N

j=1

D

k=1

qjkq

j1k V(q

j1, q

j2, , q

jD) .

In the second line we have assumed that units have been chosen sothat mk = 1 and tN t0 = 1. Since V depends only the qk at onefixed time, we have defined V = V N2

k(qjk)

2. The only

connection between different times is the hopping term q

j

kq

j1

k .Sourendu Gupta Classical Mechanics 2012: Lecture 3


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An idiotically simple algorithm for minimization

There are many ways to minimize a function of many variables. Amindlessly simple algorithm is:

1 Choose a stopping criterion .

2 Start with a trajectory {qjk}. The corresponding value of theaction is S.

3 Select a random set of values {qjk} and compute thecorresponding value of the action, S.

4 |S S| < then stop. The trajectory is now {qjk}.

5 If S < S set qjk = qjk. Repeat from step 3.

Minimization of a function of many variables

Can you think of other ways of minimizing the action which do notuse the Euler-Lagrange equations?



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Hamilton s Principle Symmetries Keywords and References

Two Problems

Problem 6: Simple harmonic motionThe Lagrangian for a simple harmonic oscillator becomes

L =1

22

1

22, (0) =

1

2, (1) =

1

2.

Find the trajectory by numerical minimization.

Problem 7: The physical pendulum

The Lagrangian for the physical pendulum is

L =1

22 (1 cos ), (0) =

1

2, (1) =

1

2.

Find the trajectory by numerical minimization.



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Integrals of motion

Some constraints on the solutions of the equations of motion can

be found without solving the full problem, i.e., without giving qsas functions of time. One example is of a free particle in thiscase the integral of motion is the conserved value of themomentum. Similarly for a freely rotating body the conserved

angular momentum is an integral of motion.First integrals of motion

In general a first integral of motion is some relation

f(q1, q2, , q1, q2, , t) = constant.

They are called first integrals because the equations of motioninvolve qk, i.e., differential equations of the second order.However, these conditions integrate the equations once, so thatthey provide differential equations of first order.



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Symmetries of a system

For a free particle, any point in space looks the same as any other,because every point is free of forces. As a result, the Lagrangian isindependent of the coordinates, i.e., V = constant.For a freely rotating body, any orientation is the same as anyother, since there is no moment that arises as it rotates. As aresult, the Lagrangian is independent of the angle the body makeswith respect to a fixed frame.

Symmetries of mechanical systems

These notions of symmetry permit immediate generalization. Wesay that a system posseses a symmetry (or symmetries) if one (ormore) of the generalized coordinates do not appear in theLagrangian.



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Symmetries and integrals of motion

Write the Euler-Lagrange equations in the form

pk =L

qk.

If a Lagrangian possesses a symmetry with respect to one of thegeneralized coordinates, qk, then the right hand side vanishes. Thesymmetry leads to a first integral of motion, the conservation ofthe corresponding momentum.

Noethers Theorem

Every symmetry leads to a conservation law for the correspondingmomentum. Generalized coordinates associated with such integralsof motion are called cyclic coordinates.



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p y y

Motion of a charged particle

Problem 8: Motion of a charged particle

What is the canonical momentum of a charged particle in a EMfield? If the external fields are time-independent, then L does notdepend explicitly on the time. In this case check whether T + V isconserved. system?

Problem 9: A charged particle in a constant magnetic field

Are there any conserved quantities when a charged particle movesin a constant magnetic field and zero electric field? What doesyour analysis show about the general character of the trajectories?

Problem 10: A charged particle around a magnetic monopole

A magnetic monopole is any source which produces a radialmagnetic field. For a charged particle moving in the field of a fixedmagnetic monopole, what are the conserved quantities?



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Scaling symmetries

Under a scaling L L the equations of motion and its solutions

remain unchanged.Assume that under a scaling of coordinates qk qk, thepotential scales as V V. Scale the time as t t; thekinetic energy scales as T 2(1)T. For the Lagrangian toscale uniformly, one must have

2(1 ) = , i.e. = 1

2.

As a result, two trajectories related by scaling must obey the law

t

t=

l

l

(1/2).

Then Keplers third law that the cube of the distance of a planetfrom the sun is proportional to the square of its period of

revolution implies = 1, i.e., Newtons law of gravity.Sourendu Gupta Classical Mechanics 2012: Lecture 3


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Keywords and References

Keywords

action, world line, functional, lattice discretization, hopping term,algorithm, stopping criterion, first integral of motion, symmetry,

conservation law, cyclic coordinates, Noethers theorem, scaling,Keplers third law, Newtons law of gravity

References

Goldstein, Chapter 2Landau, Chapter II


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