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PHYS 705: Classical Mechanics Canonical Transformation 2

Canonical Variables and Hamiltonian Formalism

As we have seen, in the Hamiltonian Formulation of Mechanics,

 qj, p j are independent variables in phase space on equal footing

 The Hamilton’s Equation for q j , p j are “symmetric” (symplectic, later) H  H qj and p  j  pj  q j  This elegant formal structure of mechanics affords us the freedom in selecting other appropriate canonical variables as our phase space “coordinates” and “momenta”

- As long as the new variables formally satisfy this abstract structure (the form of the Hamilton’s Equations. 3

Canonical Transformation

Recall (from hw) that the Euler-Lagrange Equation is invariant for a point transformation: Qj Qqt j ( ,) L d  L  i.e., if we have,   0,   qj dt  q j  L d  L  then,   0,   Qj dt  Q j  Now, the idea is to find a generalized (canonical) transformation in phase space (not config. space) such that the Hamilton’s Equations are invariant !

Qj Qqpt j (, ,) (In general, we look for transformations which Pj Pqpt j (, ,) are invertible.) 4

Invariance of EL equation for Point Transformation First look at the situation in config. space first: dL   L Given:   0, and a point transformation: Q Qqt( ,)   j j dt q j   q j dL   L   0  Need to show:   dt Q j   Q j

L  Lqi  L  q i Formally, calculate:    (chain rule) Qji  qQ ij i  qQ ij L L q  L q i   i Q j i  qiQ j i  q i Q j

From the inverse point transformation equation q i  qQt i ( ,) , we have,

qi q q q  q  0 and i  i qi Q  i Q i k  j Q j Qj k Qk  t 5

Invariance of EL equation for Point Transformation dL   L Forming the LHS of EL equation with Q : LHS   j   dt Q j   Q j   d Lqi   L  q i  L  q i LHS       dt iqQij  i  qQ ij i  qQ  ij       Ldqi dL   q i  Lqi  L  q i        q dt  Q  dt q Q qQ  qQ i i j    ij   iiij ij

    d Lqi  L qi L d qi L  q i          dt qQ q  Q q dt  Q  q Q i  i j i j i j  i j  6

Invariance of EL equation for Point Transformation (exchange order of diff)

       d L  Lqi L d qi  q i  LHS            dt q q Q q dt  Q  Q i i i   j i  jj 

 dq q i   i  0 Qj dt  Q j (Since that’s what given !) q  q i   i  0 Qj  Q j dL   L LHS  0   0   dt Q j   Q j 7

Canonical Transformation

Now, back to phase space, we need to find the appropriate (canonical) transformation Qjj Qqpt(,,)and P jj  Pqpt (,,)

such that there exist a transformed Hamiltonian KQPt( , ,)

with which the Hamilton’s Equations are satisfied:

K  K Q j and P j  Pj  Q j

(The form of the EOM must be invariant in the new coordinates.)

** It is important to further stated that the transformation considered must also be problem-independent meaning that ( Q , P ) must be canonical coordinates for all system with the same number of dofs. 8

Canonical Transformation

To see what this condition might say about our canonical transformation, we need to go back to the Hamilton’s Principle:

Hamilton’s Principle: The motion of the system in configuration space is such that the action I has a stationary value for the actual path, .i.e.,

t2 I   Ldt  0 t1 Now, we need to extend this to the 2n-dimensional phase space

1. The integrant in the action integral must now be a function of the

independent conjugate variable q j , p j and their derivatives qj, p  j 2. We will consider variations in all 2n phase space coordinates 9

Hamilton’s Principle in Phase Space

1. To rewrite the integrant in terms of q j ,,, p jj q  p  j , we will utilize the definition for the Hamiltonian (or the inverse Legendre Transform):

H pqjj  L Lpq jj   Hqpt(, ,) (Einstein’s sum rule) Substituting this into our variation equation, we have

t2 t 2 I Ldt   p q  H( q , p , t )  dt  0   j j  t1 t 1

2. The variations are now for nqj ' s and n pj ' s : (all q’s and p’s are independent)

The rewritten integrant is formally a (,,)qqp  pqj  j Hqpt (,,) q,,, p q p p function of j j  j  j but in fact it does not depend on  j , i.e.  p j  0 This fact will proved to be useful later on.

again, we will required the variations for the q j to be zero at ends 10

q q  Hamilton’s Principle in Phase Space j0 j j pj p0 j  j

Affecting the variations on all 2n variables q j , p j , we have,

t2     I  qj   q j  q' s d   d   d  j   q   q    j j j  t1   pj   p j   p' s d  ddt    0 j p  p   j j j 

As in previous discussion, the second term in the sum for  q j ' s can be rewritten using integration by parts:

t t2 2 t 2     q j   q j  q j d  dt     dt  q  q    dt  q   jj t    j  t 1 1 t1 11

Hamilton’s Principle in Phase Space

 Previously, we have required the variations for the q j to be zero at end pts

t t2 2 t 2    q j  q j   q j q j d  so that,  0  dt     dt  q  q    dt  q  t t, t  j  j t   j  1 2 t 1 1 t1

So, the first sum with  q j ' s can be written as:

t2  q  q d      jd j   d   dt  q   dt  q    j j j   t1 t2  d     q    q dtwhere  q  j d    q dt q  j j   j j j   t1 12

Hamilton’s Principle in Phase Space

Now, perform the same integration by parts to the corresponding term for  pj ' s t t2 2 t 2    we have,  p j   p j  p j d  dt     dt  p  p    dt  p   jj t    j  t 1 1 t1 t2  p  Note, since , j without enforcing the p j  0  0 p j  t1 variations for p j to be zero at end points.

nd This gives the result for the 2 sum in the variation equation for  p j ' s :

t2  d     p    p dtwhere  p  j d    p dt p  j j   j j j   t1 13

Hamilton’s Principle in Phase Space

Putting both terms back together, we have:

t2 I    d    d      d    qj     p j  dt  0    q dt q   p dt  p  jjj  j jj      t1 1 2

Since both variations are independent, 1 and2 must vanish independently !

d    (,,)qqp  pq Hqpt (,,)   0 j  j 1   dt q j   q j   H p j and  q j  q jj q H  p   0  j   H q j  (one of the Hamilton’s   p j  q j equations) 14

Hamilton’s Principle in Phase Space

t2 I    d    d      d    qj     p j  dt  0    q dt q   p dt  p  jjj  j jj      t1 1 2

d    (,,)qqp  pq Hqpt (,,)   0 j  j 2   dt p j   p j   H 0 and q j p j  p j p j H  0q j   0  H nd p j  (2 Hamilton’s   q j  p j equations) 15

Hamilton’s Principle in Phase Space

So, we have just shown that applying the Hamilton’s Principle in Phase Space, the resulting dynamical equation is the Hamilton’s Equations.

H p j  q j

H q j  p j 16

Hamilton’s Principle in Phase Space

Notice that a full time derivative of an arbitrary function F of (, q pt ,) can be put into the integrand of the action integral without affecting the

variations: t2  dF   pj q j  H(, q p ,) t  dt  dt  t 1 t t 2 2  dF p q H(, q p ,) t  dt   dt  j j   dt t1 t1

t2 dF  F t2  const  t1 t1

Thus, when variation is taken, this constant term will not contribute ! 17

Canonical Transformation

Now , we come back to the question: When is a transformation to Q , P canonical?

 We need Hamilton’s Equations to hold in both systems

This means that we need to have the following variational conditions:

 p q H( q , p , t )  dt  0 AND  P Q K( Q , P , t )  dt  0  j j   j j 

 For this to be true simultaneously, the integrands must equal

 And, from our previous slide, this is also true if they are differed by a full time derivative of a function of any of the phase space variables involved + time: dF pq Hqpt(,,)(,,),,,,  PQ KQPt   qpQPt  jj jj dt 18

Canonical Transformation dF pq Hqpt(,,)  PQ KQPt (,,)   qpQPt ,,,,  (1)(9.11) G jj jj dt F is called the Generating Function for the canonical transformation: Qj Qqpt j (, ,) (,)qpjj  QP jj ,:   Pj Pqpt j (, ,)

 As the name implies, different choice of F give us the ability to

generate different Canonical Transformation to get to different Qj, P j 

 F is useful in specifying the exact form of the transformation if it contains half of the old variables and half of the new variables. It, then, acts as a bridge between the two sets of canonical variables. 19

Canonical Transformation dF pq Hqpt( , , )  PQ KQPt ( , , )   qpQPt , , , ,  (*) ( G 9.11) jj jj dt F is called the Generating Function for the canonical transformation: Qj Qqpt j (, ,) (,)qpjj  QP jj ,:   Pj Pqpt j (, ,)

 Depending on the form of the generating functions (which pair of canonical variables being considered as the independent variables for the Generating Function), we can classify canonical transformations into four basic types. 20

Canonical Transformation: 4 Types dF p q H(,,)(,,),, q p t  P Q K Q P t   old new t  jj jj dt

Type 1: F1 F1 F1 pj   qQt,,  Pj   qQt,,  K H  F  F1(,q Q ,) t q j Qj t

Type 2: F2 F2 F2 p  qPt,,  Qj   qPt,,  K H  FF(,q P ,) t  QP j 2 i i q j Pj t

Type 3: F F F q 3  pQt,,  P 3  pQt,,  K H  3 FF(,p Q ,) t  q p j j t 3 i i p j Qj

Type 4: F F F q 4 pPt,, Q 4  pPt,,  K H  4 FF(,p P ,) tqp  Q P j   j 4 ii ii p j Pj t 21

Canonical Transformation: Type 1 (old) (new)

Type 1:F FqQt1(, ,) | F is a function of q and Q + time

Writing out the full time derivative for F, Eq (1) becomes: F F F p qHPQ  K  1  1 q  1 Q  (again E’s jjj j  j j sum rule) t q j Qj

Or, we can write the equation in differential form: dq dQ (write out q  j , and Q   j and multiply the equation by dt )  jdt j dt

F F  F  p  1 dqP  1 dQ  K H 1 dt  0 j jj  j   q j Qj  t  22

Canonical Transformation: Type 1

Since all the q j , Q j are independent, their coefficients must vanish independently. This gives the following set of equations:

F1 F1 F1 pj   qQt,,  (1) C Pj   qQt, ,  ( C 2) KH  ( C 3) q j Qj t

For a given specific expression for F 1  qQt ,,  , e.g. FqQt1  ,,   qQj j

 Eq.( C 1) are n relations defining p j in terms of q j ,, Q j t and they can be inverted to get the 1st set of the canonical transformation:

In the specific example, we have:

F1 pj q,, Q t   Q jjj Q p q j 23

Canonical Transformation: Type 1

 Eq. ( C 2) are n relations defining P j in terms of q j ,, Q j t . Together with nd our results for the Q j , the 2 set of the canonical transformation can be obtained. Again, in the specific example, we have:

F1 Pj q,, Q t   q jjj  P  q Qj  Eq.( C 3) gives the connection between K and H: F K H  1 K  H t

(note:KQPt( , ,) is a function of the new variables so that the RHS needs

to be re-express in terms of Q j , P j using the canonical transformation.) 24

Canonical Transformation: Type 1 In summary, for the specific example of a Type 1 generating function:

FqQt1  ,,   qQj j We have the following: Qj p j Canonical Transformation Pj q j

and K H Transformed Hamiltonian

Note: this example results in basically swapping the generalized coordinates with their conjugate momenta in their dynamical role and this exercise demonstrates that swapping them basically results in the same situation !

 Emphasizing the equal role for q and p in Hamiltonian Formalism ! 25

Canonical Transformation: Type 2 (old) (new) Type 2: , where F is a function of q and P + time F Fq2 (,P ,) t  QPj j 2

(One can think of F2 as the Legendre transform of FqQt(, ,) in exchanging the variables Q and P.)

Substituting into our defining equation for canonical transformation, Eq. (1): dFFF  F pq H  PQ K2 2 q  2 PPQ   PQ pqjj  H PQ jj jj  K  j jjj j j dttqj  P j

Again, writing the equation in differential form:

F   F    F  p2 dq  Q 2 dP  K H 2 dt  0 j  jj  j   qj   P j    t  26

Canonical Transformation: Type 2

Since all the q j , P j are independent, their coefficients must vanish independently. This gives the following set of equations:

F2 F2 F2 pj   qPt,,  Qj   qPt,,  K H  q j Pj t

For a given specific expression for F 2  qPt ,,  , e.g. F2  qPt,,   qPj j

F2 F2 pj qPt,,   P j Qj q,, P t   q j q j Pj K H Pj  p j Qj  q j

Thus, the identity transformation is also a canonical transformation ! 27

Canonical Transformation: Type 2

Let consider a slightly more general example for type 2: F FqPt2  ,,   QPj j

with FqPt2 ,,,,,,,,  fq 1 qtPn j  gq 1  qt n  where f and g are function of q’s only + time Going through the same procedure, we will get:

F2 F p j  2 Qj  q j f  g Pj KH P  tj  t f  g Q  fq,, qt pj  P j  j1  n  qj  q j

Notice that the Q equation is the general point transformation in the configuration space. In order for this to be canonical, the P and H transformations must be handled carefully (not necessary simple functions). 28

Canonical Transformation: Summary

The remaining two basic types are Legendre transformation of the remaining two variables:

F F3 (,p Q ,) t  pjq j q  p

F F4 (,,)p P t  pjj q Q jj P q  p & Q P (Results are summarized in Table 9.1 on p. 373 in Goldstein .)

Canonical Transformations form a group with the following properties:

1. The identity transformation is canonical (type 2 example) 2. If a transformation is canonical, so is its inverse 3. Two successive canonical transformations (“product”) is canonical 4. The product operation is associative 29

Qj Qqpt j (, ,) P Pqpt(, ,) Canonical Transformation: 4 Types j j dF p q H(,,)(,,),, q p t  P Q K Q P t   old new t  jj jj dt dep var Type 1: F1 F1 F1 pj  q,, Q t  Pj  q,, Q t  K H  F  F1(,q Q ,) t q j Qj t ind var Type 2: F2 F2 F2 p q,, P t  Qj  q,, P t  K H  FF(,q P ,) t  QP j 2 i i q j Pj t

Type 3: F F F q 3 p,, Q t  P 3 p,, Q t  K H  3 FF(,p Q ,) t  q p j j t 3 i i p j Qj

Type 4: F F F q 4 p,, P t Q 4 p,, P t  K H  4 FF(,p P ,) tqp  Q P j   j 4 ii ii p j Pj t 30

Canonical Transformation (more)

If we are given a canonical transformation Q Qqpt(, ,) j j (*) Pj Pqpt j (, ,) How can we find the appropriate generating function F ?

st - Let say, we wish to find a generating function of the 1 type, i.e., F FqQt1(, ,) (Note: generating function of the other types can be obtain through an appropriate Legendre transformation.)

- Since our chosen generating function (1st type) depends on q, Q, and t explicitly, we will rewrite our p and P in terms of q and Q using Eq. (*):

pj pqQt j (, ,) Pj PqQt j (, ,) 31

Canonical Transformation (more) Now, from the pair of equations for the Generating Function Derivatives (Table 9.1), we form the following diff eqs,

F1  qQt,,  pj pqQt j (, ,)  q j

F1  qQt,,  Pj PqQt j (, ,)  Qj can then be obtained by directly integrating the above equations F1(, qQt ,) and combining the resulting expressions.

Note: Since dF1 is an exact differential wrt q and Q, then

2 2 Pj FqQtPj  ,,,,F    F  FqQt p   p (We will give the full 11   1 1  i  i    qqiiqi  Q  j qQ ij    Qq Qq jj ji  Q j   Q j list of relations later.) 32

Canonical Transformation (more)

Example (G8.2): We are given the following canonical transformation for a system with 1 dof:

Q Q( q , p )  q cos  p sin  (HW: showing this P P(,) q p  q sin  p cos  trans. is canonical) (Q and P is rotated in phase space from q and p by an angle )

st We seek a generating function of the 1 kind: F1(, qQ )

First, notice that the cross-second derivatives for F1 are equal as required for a canonical transformation:

F1    Q  1   q cot    Qq    Q sin  sin

F1   q  1   Q cot    q Q   q sin  sin 33

Canonical Transformation (more)

Rewrite the transformation in terms of q and Q (indep. vars of F1 ): F F 1 p  pqQ(, ) 1 P PqQ(, ) q Q Q Q cos    q cot qsin   q  cos  sin sin sin   q   Q cot sin Now, integrating the two partial differential equations: Qq q2 qQ Q2 F cot  hQ  F1  cot  g q  1 sin 2 sin 2

Comparing these two expression, one possible solution for F 1 is,

Qq 1 2 2 FqQ1 ,    qQ   cot sin 2 34

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, we can directly use the fact that F2 is the

Legendre transform of F1,

F1 FqPt 2 (, ,)  QPj j

FqPt2(,,) FqQt 1 (,,)  QP

Qq 1 2 2 FqP2 ,    qQ   cot  QP sin 2 Now, from the CT, we can write Q by q and P: q Q qcos  p sin  Q  P tan P qsin  p cos  cos 35

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, we can directly use the fact that F2 is the

Legendre transform of F1,

F FqPt2 (, ,)  QPj j

FqPt2(,,) FqQt 1 (,,)  QP

Qq 1 2 2 FqP2 ,    qQ   cot  QP sin 2 This then gives:

2 q q 1  q   FqPP,  P tan  q2 P tan  cot 2       sin cos  2 cos    36

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, we can directly use the fact that F2 is the

Legendre transform of F1,

F FqPt2 (, ,)  QPj j 2 q q 1  q   FqPP,  P tan  q2 P tan  cot 2       sin cos  2 cos   

2qP q2   P2 tan cos sin  cos

2 12q 2 qP 2   qcot  P tan    2 cos sin cos    37

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, we can directly use the fact that F2 is the

Legendre transform of F1,

F FqPt2 (, ,)  QPj j 2 q q 1  q   FqPP,  P tan  q2 P tan  cot 2       sin cos  2 cos   

2qP q2   P2 tan cos sin  cos 1 q2 2qP   q2 cot    P2 tan      2 cos sin  cos   38

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, we can directly use the fact that F2 is the

Legendre transform of F1,

F FqPt2 (, ,)  QPj j 2 q q 1  q   FqPP,  P tan  q2 P tan  cot 2       sin cos  2 cos   

Finally, qP 1 2 2 FqP2 ,    qP   tan cos 2 39

Canonical Transformation (more)

nd Now, let say we wish to seek a generating function of the 2 type: F2 (, qP )

As we have discussed previously, alternatively, we can substitute the term into Q P Eq. (*), results in replacing the PQ j  j term by j  j in our condition for a canonical transformation,

F FqPt2 (, ,)  QPj j dF pq H  QP  K jj jj dt

Recall, this gives us the two partial derivatives relations for F2:

F qPt,,  F2  qPt,,  p  2 Q  [Or, use the Table] j j P q j j 40

Canonical Transformation (more) Q Q( q , p )  q cos  p sin  P P( q , p )  q sin  p cos  To solve for in our example, again, we rewrite our given canonical F2 (, qPt ,) transformation in q and P explicitly.

F F 2 p  pqP(, ) 2 Q  QqP(, ) q P P P sin    q tan qcos   q  sin  cos cos   cos q   P tan cos Integrating and combining give,

qP 1 2 2 FqP2 ,    qP   tan cos 2 41

Q Q( q , p )  q cos  p sin  Canonical Transformation (more) P P( q , p )  q sin  p cos  Notice that when   0 , sin  0

so that our coordinate transformation is just the identity transformation: Q  q and P p

p, P CANNOT be written explicitly in terms of q and Q !

so that our assumption for using the type 1 generating function (with q and Q as indp var) cannot be fulfilled.

Consequently, blow up and cannot be used to derive the F1  qQ,  canonical transformation:

Qq 1 2 2 FqQ1 ,    qQ   cot  as   0 sin 2 But, using a Type 2 generating function will work. 42

Canonical Transformation (more) Q Q( q , p )  q cos  p sin  P P( q , p )  q sin  p cos   Similarly, we can see that when   , cos  0 2 our coordinate transformation is a coordinate switch Q  p , P q

p, Q CANNOT be written explicitly in terms of q and P !

so that the assumption for using the type 2 generating function (with q and P as indp var) cannot be fulfilled.

Consequently, blow up and cannot be used to derive the F2  qP,  canonical transformation:

qP 1 2 2 FqP2 ,    qP  tan  as   0 cos 2

But, using a Type 1 generating function will work in this case. 43

Canonical Transformation (more)

- A suitable generating function doesn’t have to conform to only one of the four types for all the degrees of freedom in a given problem ! - There can also be more than one solution for a given CT - First , we need to choose a suitable set of independent variables for the generating function. For a generating function to be useful, it should depends on half of the old and half of the new variables As we have done in the previous example, the procedure in solving for F involves integrating the partial derivative relations resulted from “consistence” considerations using the main condition for a canonical transformation, i.e., dF pq Hqpt( , , )  PQ KQPt ( , , )  ( G 9.11) jj jj dt 44

Canonical Transformation (more)

For these partial derivative relations to be solvable, one must be able to feed-in 2n independent coordinate relations (from the given CT) in terms of a chosen set of ½ new + ½ old variables.

- In general, one can use ANY one of the four types of generating functions for the canonical transformation as long as the RHS of the transformation can be written in terms of the associated pairs of phase space coordinates: (q, Q, t), (q, P, t), (q, Q, t), or (p, P, t). T1 T 2 T3 T 4 - On the other hand, if the transformation is such that the RHS cannot written in term of a particular pair: (q, Q, t), (q, P, t), (q, Q, t), or (p, P, t), then that associated type of generating functions cannot be used. 45

Canonical Transformation: an example with two dofs

- To see in practice how this might work… Let say, we have the following

transformation involving 2 dofs: qpqp1,,,,,, 1 2 2   QPQP 1 1 2 2 

Q1 q 1 (1 a ) Q2 p 2 (2 a )

P1 p 1 (1 b ) P2 q 2 (2 b )

- As we will see, this will involve a mixture of two different basic types. 46

Canonical Transformation: an example with two dofs

Q1 q 1 (1 a ) Q2 p 2 (2 a )

P1 p 1 (1 b ) P2 q 2 (2 b )

- First, let see if we can use the simplest type (type 1) for both dofs, i.e., F will depend only on the q-Q’s:

FqqQQt(,1 2 , 1 , 2 ,)

Notice that Eq (1a) is a relation linking q 1 , Q 1 only , they CANNOT both be independent variables  Type 1 (only) WON’T work !

- As an alternative, we can try to use the set p 1 ,,, q 2 Q 1 Q 2 as our independent variables. This will give an F which is a mixture of Type 3 and 1.

(In Goldstein (p. 377), another alternative was using qq 1 ,,, 2 PQ 1 2 resulted in a different generating function which is a mixture of Type 2 and 1.) 47

Canonical Transformation: an example with two dofs

From our CT, we can write down the following relations: Dependent variables Independent variables

q1,,, p 2 PP 1 2 pq1,,, 2 QQ 1 2

q1  Q1 p  Q 2 2 (*) P1  p1

P2 q2

Now, with this set of ½ new + ½ old independent variable chosen, we need to

derive the set of partial derivative conditions by substituting FpqQQt(,,1 2 1 , 2 ,) into Eq. 9.11 (or look them up from the Table). 48

Canonical Transformation: an example with two dofs

The explicit independent variables (those appear in the differentials) in Eq. 9.11 are the q-Q’s. To do the conversion:

q1,,,q2 QQ 1 2 (Eq. 9.11’s explicit ind vars) 

p1,,,q2Q 1 Q 2 (our preferred ind vars) we will use the following Legendre transformation: F F'( p1 , q 2 , Q 1 , Q 2 ,) t  q 1 p 1

Substituting this into Eq. 9.11, we have: dF p q pq  H PQ  PQ  K 1 1 2 2 1 1 2 2 dt F'  F '  F '  F ' F ' P1Q 1  PQ 2 2  K  p1 q 2  Q1 Q  2  q 1 p  1 p 1q  1  p1 q 2 Q 1  Q 2 t 49

Canonical Transformation: an example with two dofs FFFFF'''''F ' F ' F '  F ' ppqHPQPQK222 q 2HPQ1 1122 1PQ2  2 K p  1  q 2  Q 1  QQqp 2 211 q1 p 1  ppqQQ1 q2 Q1 Q2  t Comparing terms, we have the following conditions:

F ' F ' q  P1  1 p Q F ' 1 1 K H  F ' F ' t p2  P2  q2 Q2

As advertised, this is a mixture of Type 3 and 1 of the basic CT. 50

Canonical Transformation: an example with two dofs q1  Q1 p  Q Substituting our coordinates transformation [Eq. (*)] into the partial 2 2 P1  p1 derivative relations, we have : P2 q2 F ' q1  Q 1 F' Qp1 1  fqQQ (,,) 2 1 2 p1 F ' P1  p 1 F' pQ1 1  gpqQ (,,) 1 2 2 Q1 F'  pQ1 1  qQ 2 2 F ' F' qQ  hpqQ (,,) P2  q 2  2 2 1 2 1 Q2 F ' F' Qq2 2  kpQQ (,,) 1 1 2 p2  Q 2 q2

(Note: Choosing q 1 ,,, q 2 P 1 Q 2 instead, Goldstein has F ''  qP 1 1  qQ 2 2 . Both of these are valid generating functions.) 51

Qj Qqpt j (, ,) P Pqpt(, ,) Canonical Transformation: Review j j dF p q H(,,)(,,),, q p t  P Q K Q P t   old new t  jj jj dt

Type 1: F1 F1 F1 pj   qQt,,  Pj   qQt,,  K H  F FqQt1(, ,) q j Qj t

Type 2: F2 F2 F2 p  qPt,,  Qj   qPt,,  K H  F FqPt(, ,)  QP j 2 i i q j Pj t

Type 3: F F F q 3  pQt,,  P 3  pQt,,  K H  3 F F(, p Q ,) t  q p j j t 3 i i p j Qj

Type 4: F F F q 4 pPt,, Q 4  pPt,,  K H  4 F FpPt(, ,)  qp QP j   j 4 ii ii p j Pj t 52

Canonical Transformation: Review

- Generating function is useful as a bridge to link half of the original set of coordinates (either q or p) to another half of the new set (either Q or P).

- In general, one can use ANY one of the four types of generating functions for the canonical transformation as long as the transformation can be written in terms of the associated pairs of phase space coordinates: (q, Q, t), (q, P, t), (q, Q, t), or (p, P, t).

- On the other hand, if the transformation is such that it cannot be written in term of a particular pair: (q, Q, t), (q, P, t), (q, Q, t), or (p, P, t), then that associated type of generating functions cannot be used.

- the procedure in solving for F involves integrating the resulting partial derivative relations from the CT condition