EP 222: Classical Mechanics - Lecture 28 Dipan K. Ghosh Indian Institute of Technology Bombay dipan.ghosh@gmail.com Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 28 October 4, 2014
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 ( Q i ) and the new momenta ( P i ) are functions of the old coordinates ( q i ) and old momenta ( p i ) Q i = Q i ( { q i } , { p j } , t ) P i = P i ( { q i } , { p j } , 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 Q i and P i are canonically conjugate pairs. Thus there must exist some Hamiltonian K ( Q , P , t )with respect to which these variables satisfy Hamilton;s equations Q i = ∂ K ˙ ∂ P i P i = − ∂ K ˙ ∂ Q i 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 � t 2 S = t 1 ( p ˙ q − H ) dt be minimised Variational principle requires that δ S = 0 for all paths with fixed end points δ q ( t 1 ) = δ q ( t 2 ) = 0 leads to Euler Lagrange equations. In Hamiltonian formalism we treat q and p on equal footing and vary both independently. � t 2 �� � � � � q i − ∂ H p i − ∂ H dt + [ p i δ q i ] t 2 δ S = ˙ δ p i + − ˙ δ q i t 1 ∂ p i ∂ q i t 1 For all variations of δ p i and δ q i , 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 δ S = 0 for all p i and q i gives the Hamilton’s equations Note that for the term [ p i δ q i ] t 2 t 1 to vanish, we only require δ q i ( t 1 ) = δ q i ( t 2 ) = 0 while δ p i can be free at the end points. q i and p i are not symmetric in this treatment. We could, of course put in an additional requirement that δ p i ( t 1 ) = δ p i ( t 2 ) = 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 � t 2 � � q i − H ( q , p ) + dF ( q , p ) S = p i ˙ dt dt t 1 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 � δ S = δ ( p ˙ q − H ) dt = 0 with no variation at the end points. With respect to the new variables � δ S ′ = 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 i P i ˙ A Scale Transformation � Q i − K = λ � i ( p i ˙ q i − H ) Q i − K + dF i P i ˙ A canonical transformation � dt = � i ( p i ˙ q i − H ) One could combine both to have an Extended Canonical Transformation Q i − K + dF � P i ˙ � dt = λ ( p i ˙ q i − H ) i i Necessary and sufficient condition for a transformation to be canonical is that the difference between two differential forms should be an exact differential � � ( p i dq i − H dt ) − ( P i dQ i − Kdt ) = dF i i 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 P i = ν p i ; Q i = µ q i P and Q are obtainable from a new Hamiltonian K P i = − ∂ K p i = − 1 ∂ K p i = − 1 ∂ K ˙ = ⇒ ν ˙ = ⇒ ˙ ∂ Q i µ ∂ q i µν ∂ q i p i = ∂ H Comparing with ˙ , we have K = µν H ∂ q i Thus � P i ˙ � Q i − K ( P , Q , t ) = µν ( p i q i − H ) i i Compare with � P i ˙ � Q i − K ( P , Q , t ) = λ ( p i q i − H ) i i we have λ = µν Dipan Ghosh (I.I.T. Bombay) Class. Mech. -28 October 9, 2014 7 / 13
Generating Function Q i − K + dF i P i ˙ � dt = � i ( p i ˙ q i − H ) : F is a function of p i , q i , P i , Q i and t . F is called the Generating Function of the canonical transformation. Identity Transformation: Let F = q i P i − Q i P i Q i − K + dF P i ˙ dt = P i ˙ q i P i + q i ˙ P i − ˙ Q i P i − Q i ˙ Q i − K + ˙ P i = − K + ( q i − Q i ) ˙ P i + P i ˙ q i ≡ p i ˙ q i − H The equation is satisfied if P i = p i , Q i = q i and K = H . Dipan Ghosh (I.I.T. Bombay) Class. Mech. -28 October 9, 2014 8 / 13
Generating Function- Identity Let F = f i ( q 1 , q 2 , . . . , q n ; t ) P i − Q i P i , implied summation on repeated index. � ∂ f i Q i − K + dF q j P i + ∂ f i � P i ˙ dt = P i ˙ + f i ˙ P i − ˙ Q i P i − Q i ˙ Q i − K + ˙ ∂ t P i P ∂ q j P i + ∂ f i q j P i + ∂ f i = − K + ( f i − Q i ) ˙ ˙ ∂ t P i ∂ q j ≡ p i ˙ q i − H The equation is satisfied if Q i = f i and K = H + ∂ f i ∂ t P i and requiring that q i = ∂ f i q j P i = ∂ f j ∂ f j � p i ˙ ˙ q i P j = ˙ ⇒ q i = P j ∂ q j ∂ q i ∂ q i j Dipan Ghosh (I.I.T. Bombay) Class. Mech. -28 October 9, 2014 9 / 13
Generating Function- How to find it Q i − K + dF Let K ( Q , P , t ) = H ( q , p , t ) Using P i ˙ dt = p i ˙ q i − H dF � q i − P i ˙ dt = ( p i ˙ Q i ) i Simplest way to satisfy it is F = F ( q , Q ) as � ∂ F q i + ∂ F � dF � ˙ � q i − P i ˙ dt = ˙ Q i = ( p i ˙ Q i ) ∂ q i ∂ Q i i i This gives ∂ F = p i and ∂ F = − P i ∂ q i ∂ Q i This can be satisfied by F ( { q i } , { Q i } ) = � i q i Q i so that ∂ F = Q i = p i ∂ q i ∂ F = q i = − P i ∂ Q i 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 F 1 ( q , Q , t ) , F 2 ( q , P , t ) , F 3 ( p , Q , t ) and F 4 ( p , P , t ) Type - 1 Generator dF 1 dt = ∂ F 1 q i + ∂ F 1 Q i + ∂ F 1 ˙ ˙ ∂ q i ∂ Q i ∂ t P i ˙ � � ≡ p i ˙ q i − Q i + K − H i i Comparing, p i = ∂ F 1 ∂ q i P i = − ∂ F 1 ∂ Q i K = H + ∂ F 1 ∂ t Dipan Ghosh (I.I.T. Bombay) Class. Mech. -28 October 9, 2014 11 / 13
Type 1 Generator Write q i − P i ˙ dF 1 = p i ˙ Q i + ( 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 q i − P i ˙ dt = p i ˙ Q i For instantaneous transformation dF = p i δ q i − P i δ Q i Since the left hand side is an exact differential, the condition for a transformation to be canonical ∂ p i = − ∂ P i ∂ Q i ∂ q i 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 F 1 = 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 = Q 2 2 m + 1 2 kP 2 Hamilton’s equations: Q = ∂ K ˙ ∂ P = kP P = − ∂ K ∂ Q = − Q ˙ m P = − Q Equation of motion: ¨ Q = k ˙ m ˙ Q k = − Q 0 ω Solution : Q = Q 0 cos( ω t + δ ), and P = sin( ω t + δ ) k Equate P to − q , we get q = q 0 cos( ω t + δ ′ ) with q 0 = Q 0 ω/ k and δ ′ = δ + π/ 2 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -28 October 9, 2014 13 / 13
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