EP 222: Classical Mechanics - Lecture 28 Dipan K. Ghosh Indian - - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 28

Dipan K. Ghosh

Indian Institute of Technology Bombay dipan.ghosh@gmail.com

October 4, 2014

Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 28

Canonical Transformation

Classical mechanics does not provide a unique way of choosing the generalized coordinate and the corresponding momenta. Any set which spans the entire phase space is an acceptable alternative. Replacing one set of coordinates by another is called Point Transformation We need point transformations in phase space where the new coordinates (Qi) and the new momenta (Pi) are functions of the old coordinates (qi) and old momenta (pi) Qi = Qi({qi}, {pj}, t) Pi = Pi({qi}, {pj}, t)

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 1 / 13

Canonical Transformation

In Hamiltonian mechanics, only those transformations are of interest for which Qi and Pi are canonically conjugate pairs. Thus there must exist some Hamiltonian K(Q, P, t)with respect to which these variables satisfy Hamilton;s equations ˙ Qi = ∂K ∂Pi ˙ Pi = − ∂K ∂Qi The choice of Q and P are not specific to a particular mechanical system but are common to all systems having the same degree of freedom. Transformations which preserve the form of Hamilton’s equations are called canonical transformation

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 2 / 13

Principle of Least Action and Hamiltonian formalism

Principle of least action can be written in the form S = t2

t1 (p ˙

q − H)dt be minimised Variational principle requires that δS = 0 for all paths with fixed end points δq(t1) = δq(t2) = 0 leads to Euler Lagrange equations. In Hamiltonian formalism we treat q and p on equal footing and vary both independently. δS = t2

t1

• ˙

qi − ∂H ∂pi

• δpi +
• −˙

pi − ∂H ∂qi

• δqi
• dt + [piδqi]t2

t1

For all variations of δpi and δqi, the Hamilton’s equations are valid.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 3 / 13

Principle of Least Action and Hamiltonian formalism

δS = 0 for all pi and qi gives the Hamilton’s equations Note that for the term [piδqi]t2

t1 to vanish, we only require

δqi(t1) = δqi(t2) = 0 while δpi can be free at the end points. qi and pi are not symmetric in this treatment. We could, of course put in an additional requirement that δpi(t1) = δpi(t2) = 0, which would still make the above derivation valid but this is more restrictive than required. This means that we have an additional freedom to define the action integral as S = t2

t1

• pi ˙

qi − H(q, p) + dF(q, p) dt

• dt

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 4 / 13

Canonical Transformation

The variational principle for the variables (p, q) gives δS = δ

• (p ˙

q − H)dt = 0 with no variation at the end points. With respect to the new variables δS′ = δ

• (P ˙

Q − K)dt = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 5 / 13

Canonical Transformation

The two forms S and S′ are equivalent if the two integrands either differ by a scale factor or by the total time derivative of a function F

A Scale Transformation

i Pi ˙

Qi − K = λ

i(pi ˙

qi − H) A canonical transformation

i Pi ˙

Qi − K + dF dt =

i(pi ˙

qi − H)

One could combine both to have an Extended Canonical Transformation

• i

Pi ˙ Qi − K + dF dt = λ

• i

(pi ˙ qi − H) Necessary and sufficient condition for a transformation to be canonical is that the difference between two differential forms should be an exact differential (

• i

pidqi − Hdt) − (

• i

PidQi − Kdt) = dF

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 6 / 13

Scale Transformation

All old coordinates get multiplied by the same scale factor and all momenta get multiplied by a common scale factor Pi = νpi; Qi = µqi P and Q are obtainable from a new Hamiltonian K ˙ Pi = − ∂K ∂Qi = ⇒ ν ˙ pi = − 1 µ ∂K ∂qi = ⇒ ˙ pi = − 1 µν ∂K ∂qi Comparing with ˙ pi = ∂H ∂qi , we have K = µνH Thus

• i

Pi ˙ Qi − K(P, Q, t) = µν(

• i

piqi − H) Compare with

• i

Pi ˙ Qi − K(P, Q, t) = λ(

• i

piqi − H) we have λ = µν

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 7 / 13

Generating Function

• i Pi ˙

Qi − K + dF dt =

i(pi ˙

qi − H) : F is a function of pi, qi, Pi, Qi and t. F is called the Generating Function of the canonical transformation. Identity Transformation: Let F = qiPi − QiPi Pi ˙ Qi − K + dF dt = Pi ˙ Qi − K + ˙ qiPi + qi ˙ Pi − ˙ QiPi − Qi ˙ Pi = −K + (qi − Qi) ˙ Pi + Pi ˙ qi ≡ pi ˙ qi − H The equation is satisfied if Pi = pi, Qi = qi and K = H.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 8 / 13

Generating Function- Identity

Let F = fi(q1, q2, . . . , qn; t)Pi − QiPi, implied summation on repeated index. Pi ˙ Qi − K + dF dt = Pi ˙ Qi − K + ∂fi ∂qj ˙ qjPi + ∂fi ∂t Pi

• + fi ˙

Pi − ˙ QiPi − Qi ˙ P = −K + (fi − Qi) ˙ Pi + ∂fi ∂qj ˙ qjPi + ∂fi ∂t Pi ≡ pi ˙ qi − H The equation is satisfied if Qi = fi and K = H + ∂fi ∂t Pi and requiring that pi ˙ qi = ∂fi ∂qj ˙ qjPi = ∂fj ∂qi ˙ qiPj = ⇒ qi =

• j

∂fj ∂qi Pj

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 9 / 13

Generating Function- How to find it

Let K(Q, P, t) = H(q, p, t) Using Pi ˙ Qi − K + dF dt = pi ˙ qi − H dF dt =

• i

(pi ˙ qi − Pi ˙ Qi) Simplest way to satisfy it is F = F(q, Q) as dF dt =

• i

∂F ∂qi ˙ qi + ∂F ∂Qi ˙ Qi

• =
• i

(pi ˙ qi − Pi ˙ Qi) This gives ∂F ∂qi = pi and ∂F ∂Qi = −Pi This can be satisfied by F({qi}, {Qi}) =

i qiQi so that

∂F ∂qi = Qi = pi ∂F ∂Qi = qi = −Pi Simply exchange coordinates and momenta in the Hamiltonian.

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 10 / 13

Four Types of Generating Functions

F1(q, Q, t), F2(q, P, t), F3(p, Q, t) and F4(p, P, t) Type - 1 Generator dF1 dt = ∂F1 ∂qi ˙ qi + ∂F1 ∂Qi ˙ Qi + ∂F1 ∂t ≡

• i

pi ˙ qi −

• i

Pi ˙ Qi + K − H Comparing, pi = ∂F1 ∂qi Pi = −∂F1 ∂Qi K = H + ∂F1 ∂t

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 11 / 13

Type 1 Generator

Write dF1 = pi ˙ qi − Pi ˙ Qi + (K − H)dt If the Hamiltonian has no explicit time dependence, the generating function will also not have explicit time dependence and K = H dF dt = pi ˙ qi − Pi ˙ Qi For instantaneous transformation dF = piδqi − PiδQi Since the left hand side is an exact differential, the condition for a transformation to be canonical ∂pi ∂Qi = −∂Pi ∂qi

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 12 / 13

Type 1 Generator- Example

Consider the harmonic oscillator. Try a trivial type 1 generator F1 = qQ, so that p = Q, P = −q, i.e. swap coordinates and momenta with a sign adjustment. Hamiltonian has no explicit time dependence, K = H K = Q2 2m + 1 2kP2 Hamilton’s equations: ˙ Q = ∂K ∂P = kP ˙ P = −∂K ∂Q = −Q m Equation of motion: ¨ Q = k ˙ P = −Q m Solution : Q = Q0 cos(ωt + δ), and P = ˙ Q k = −Q0ω k sin(ωt + δ) Equate P to −q, we get q = q0 cos(ωt + δ′) with q0 = Q0ω/k and δ′ = δ + π/2

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -28

October 9, 2014 13 / 13