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MP350 Classical Mechanics

Solutions Problem set 3

1. (a) The canonical momentum is

p =L

q = mq cos2 q = q =

p

m cos2

q . (1.1)

(b) i. The hamiltonian is

H = pq L = mq cos2 q q m2

q 2 cos2 q + k sin(q t)

=12

mq 2 cos2 q + k sin(q t) .(1.2)

ii. Using (1.1), (1.2) we nd

H =m2

pm cos2 q

2

cos2 q + k sin(q t)

=p2

2m cos2 q + k sin(q t) .

(1.3)

(c) H is not conserved since it (and L) depends explicitly on time.

2. (a) The canonical momenta are

p =L

= mR 2 , p =L

= mR 2 sin2 . (2.1)

Since the Lagrangian does not depend explicitly on , we have

p =ddt

L

=L

= 0 , (2.2)

ie p is conserved.(b) i. The hamiltonian is given by

H = p + p L

= mR 2 2 + mR 2 2 sin2 12

mR 2 2 12

mR 2 2 sin2 +12

kR 2 sin2

=1

2mR 2 2 +

1

2mR 2 2 sin2 +

1

2kR 2 sin2 .

(2.3)

Alternatively, one may note that there are no time-dependent coordinatetransformations and no velocity-dependent potential energy here, so onemay just write down H = T + V .

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ii. From (2.1) we nd

=p

mR 2, =

pmR 2 sin2

. (2.4)

Substituting this into (2.3) we get

H =p2

2mR 2+

p22mR 2 sin2

+12

kR 2 sin2 . (2.5)

(c) There is no explicit time-dependence in the lagrangian (or hamiltonian) inthis case, so the hamiltonian is conserved.

3. (a) The EulerLagrange equation is

ddt

L x

=ddt

mxet = m xet + xet

=Lx

= m2 xet(3.1)

= x + x + 2 x = 0 . (3.2)

If > 0, this is the equation for a damped harmonic oscillator.(b) The canonical momentum is

px =L x

= mxet (3.3)

The hamiltonian is

H = px x L =12

m 2x2 x2 + 2 x2 et =12

m x2 + 2 x2 )et (3.4)

=p2x

2me t +

m2

2x2 et . (3.5)

The hamiltonian is not conserved, since it (and the lagrangian) depends ex-plicitly on time. Physically, the damping of the oscillator dissipates energy.

(c) With q = xet/ 2 we have x = qe t/ 2 and

q = xet/ 2 + x 2

et/ 2 = x = qe t/ 2 x 2

= qe t/ 2 2

qe t/ 2 (3.6)

x2 = q 2 qq + 2

4e t (3.7)

L =1

2m(x2 2 x2 )et =

1

2m q 2 qq (2 2 / 4)q 2 (3.8)

(d) The canonical momentum is

p =Lq

= mq m 2

q . (3.9)

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We can use this to eliminate q :

q =1m

p +m 2

q . (3.10)

The hamiltonian is, as a function of p and q ,

H = pq L

=p

m p +

m

2q

1

2m

1

m2 p +

m

2q

2

m p +

m

2q q 2

2

4q 2

=p2

2m+

pq 2

m2

2 q 2 .

(3.11)

If we instead write H in terms of q and q we get

H = pq L = mq 2 m 2

q q 12

mq 2 +m 2

q q +12

m(2 2 / 4)q 2

=12

mq 2 +12

m(2 2 / 4)q 2 .(3.12)

There is no explicit time dependence in this hamiltonian, so H is conserved.However, H is not the total energy, since the potential energy contains avelocity-dependent term 12 m qq . Alternatively, we may note that the coor-dinate q is introduced in a time-dependent way, ie there is a time-dependenttransformation equation between the original coordinate x and the new co-ordinate q . From this alone, we cannot conclude anything about whether thetotal energy is conserved.It is arguable that the usual arguments regarding the relation between L, H and energy do not hold in this case, since L is not really the difference betweenkinetic and potential energy of a physical system; it is rather constructed insuch a way as to obtain the correct equations of motion for a system withfrictional (damping) forces. The correct way to decide whether total energyis conserved is then to look at the physical system (damped oscillator) andrecall that the damping forces dissipate energy, and hence total energy is notconserved.