PHYS 705: Classical Mechanics Hamilton-Jacobi Equation 2
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PHYS 705: Classical Mechanics Hamilton-Jacobi Equation 2
Hamilton-Jacobi Equation
There is also a very elegant relation between the Hamiltonian Formulation of Mechanics and Quantum Mechanics.
To do that, we need to derive the Hamilton-Jacobi equation.
The Hamilton-Jacobi equation also represents a very general method in solving mechanical problems.
To start…
Let say we are able to find a canonical transformation taking our 2n phase q, p space variables i i directly to 2n constants of motion, i , i i.e.,
Qi i Pi i 3
Hamilton-Jacobi Equation
One sufficient condition to ensure that our new variables are constant in time is that the transformed Hamiltonian K shall be identically zero.
If that is the case, the equations of motion will be, K Qi 0 Pi K Pi 0 Qi Recall that we have the new and old Hamiltonian, K and H, relating through the generating function F by:
F K H t 4
Hamilton-Jacobi Equation
With K identically equals to zero, we can write: F H qpt, , 0 (*) t For convenience in later calculations, we shall take F to be of Type 2
so that its independent variables are the old coordinates q i and the
new momenta P i . In the same notations as previously, we can write
FqQPt ,,, FqPt2 (,,) QPi i
Now, in order to write H in terms of this chosen set of variables qPt ,, ,
we will replace all the p i in H using the transformation equation:
F2 qPt,, pi qi 5
Hamilton-Jacobi Equation F F Then, one can formally rewrite Eq. (*) as: note that 2 since t t
FqPt ,, FqPt2 (,,) QPi i F2 F 2 F 2 Hqq1, ,n ; , , ; t 0 q1 qn t Fq ,, t p 2 This PDE is known as the Hamilton-Jacobi Equation. i qi
Notes: -Since P i i are constants, the HJ equation constitutes a partial differential equation of (n+1) independent variables: q1,,, qtn - We used the fact that the new P’s and Q’s are constants but we have not specified in how to determine them yet.
- It is customary to denote the solution F 2 by S and called it the Hamilton’s Principal Function. 6
Hamilton-Jacobi Equation
Suppose we are able to find a solution to this 1st order partial differential equation in (n+1) variables…
FSSqq2 1,,,n 1,, n ; t
where we have explicitly write out the constant new momenta: Pi i
Recall that for a Type 2 generating function, we have the following partial derivatives describing the canonical transformation: Sq ,, t S pi pi ( T 1) qi qi Sq,, t S Q Qi i ( T 2) i Pi i 7
Hamilton-Jacobi Equation
After explicitly taking the partial derivative on the RHS of Eq. (T1) Sq ,, t pi qi
and evaluating them at the initial time t 0 , we will have n equations that
one can invert to solve for the n unknown constants i in terms of the
initial conditions qp 0, 0 , t 0 , i.e.,
i i qp0, 0, t 0
Similarly, by explicitly evaluating the partial derivatives on the RHS of Eq.
(T2) at time t 0 , we obtain the other n constants of motion i Sq ,, t Qi i i qq0, tt 0 8
Hamilton-Jacobi Equation
With all 2n constants of motion i , i solved, we can now use Eq.
(T2) again to solve for q i in terms of the i , i at a later time t.
Sq ,, t qqi i ,, t i i p Then, with ii , , and q i known, we can use Eq. (T1) again to evaluate i
in terms of i , i at a later time t. Sq ,,,, t t p pp,, t i q i i i
The two boxed equations constitute the desired complete solutions of the Hamilton equations of motion. 9
Hamilton Principal Function and Action
Now, let consider the total time derivative of the Hamilton Principal function S, dSq ,, t SS qi dt qi t Using Eq. (T1) and the Hamilton-Jacobi equation itself, we can
replace the two partial derivatives on the right by p i and H, S S dS pi H 0 t pqi i H L qi dt So, the Hamilton Principal Function is differed at most from the action by a constant, i.e., S Ldt constant 10
Example: Solving Harmonic Oscillator with HJ
m Recall, we have:
1 2 0 x H pmqE2 2 2 2 k m kq2 2m f( q ) kq U( q ) 2 The Hamilton-Jacobi equation gives:
2 1 SS SS m2 2 q 2 0 Hq, ; t 0 q t 2mq t
Recall that under this scheme, we have:
K 0 Q and P 11
Example: Solving Harmonic Oscillator with HJ
m 1 H pmqE2 2 2 2 2 k m 2m 0 x kq2 f( q ) kq U( q ) When H does not explicitly depend on time, 2 we can generally try this trial solution for S:
SqtWq ,,, t (q and t are separable)
With this trial solution, we have S t and the HJ equation becomes: 2 2 1 SS 2 2 2 1 W 2 2 2 recall: LHS H m q 0 m q 2mq t 2m q so H 12
Example: Solving Harmonic Oscillator with HJ
m The previous equation can be immediately integrated:
2 2 1 W W 2 2 2 m2 2 q 2 mq 2 m q 2m q 2 W 2 2 2 2m m q q
2 2 2 W 2 m m2 2 q 2 dq and S 2 m m q dq t 13
Example: Solving Harmonic Oscillator with HJ
m We don’t need to integrate yet since we only need SS to use , in the transformation eqs: q
Sq ,, t S 2 2 2 #1 eq: i Q 2 m m q dq t i 1 dq m dq 2m t t 2 2m m2 2 q 2 2 1 m 2 q 2 2 m Letting x q , we can rewrite the integral as: 2 1 dx 1 1 m t tarcsin x arcsin q 2 1 x 2 14
Example: Solving Harmonic Oscillator with HJ
m 1 1 m tarcsin x arcsin q 2
Inverting the arcsin, we have
2 qt sin t ' with ' m 2
Sq ,,,, t t nd p The 2 transformation equation gives, i q i S #2 eq: p2 m m2 2 q 2 dq t 2m m2 2 q 2 q q 15
Example: Solving Harmonic Oscillator with HJ
Substituting , we have, m q t
m 2 2 pt2 m 1 sin2 t ' 2 2 m
pt2 m 1sin 2 t ' with '
pt2 m cos2 t '
pt 2 m cos t ' 16
Example: Solving Harmonic Oscillator with HJ
2 m qt sin t ' m 2 pt 2 m cos t '
Completing the calculations, we still need to link the two constants of
motion , to initial conditions: @ t 0, q q0 , p p 0
Recall that the constant of motion is H=E of the system so that we can calculate this by squaring q and p at time t 0 :
1 2 2 2 2 p0 m q 0 2m q 1 And, we can calculate at time t 0 by dividing the 2 eqs: 0 tan ' p0 m 17
Another example in using the Hamilton-Jacobi Method (t-dependent H) (G 10.8) Suppose we are given a Hamiltonian to a system as,
p2 H mAtq where A is a constant 2m Our task is to solve for the equation of motion by means of the Hamilton’s principle function S with the initial conditions
t0, q0 0, pmv 0 0
The Hamilton-Jacobi equation for S is:
SS Hq; ; t 0 q t 18
Example in using the Hamilton-Jacobi Method
Recall that the Hamilton’s principle function S is a Type 2 generating function with independent variables qPt,, S SqPt ,, with the condition that the canonically transformed variables are constants, i.e., Qand P
For a Type 2 generating function, we have the following partial derivatives for S:
Sq ,, t Sq ,, t p ( T 1) Q ( T 2) q 19
Example in using the Hamilton-Jacobi Method
Writing H out explicitly in the Hamilton-Jacobi equation, we have 2 1 S S mAtq 0 (*) 2mq t
S Hq; ; t q
Here, we assume a solution to be separatable of the form: Sq ,;, t f tqgt
Substituting this into the above Eq (*), we have,
1 2 f, t mAtq f ' , tqgt ' 0 ( ' is time 2m derivative here) 20
Example in using the Hamilton-Jacobi Method
1 2 f, t mAtq f ',t qg' t 0 2m Concentrating on the q dependent terms, they have to be independently add up to zero… So, we have, mAt 2 f', t mAt ft, f 2 0 And, requiring the remaining two terms adding up to zero also, we have 1 gt' ft2 2m Substituting f (t) from above, we have 22 2 2 4 1mAt 1 mAt 2 2 gt' f0 mAtff 0 0 2m 2 2 m 4 21
Example in using the Hamilton-Jacobi Method
Integrating wrt time on both sides, we then have, mA2 t 5 Af f 2 g t 0 t3 0 t g 40 6 2m 0 Putting both f (t) and g(t) back into the Hamilton’s principle function, we have,
2 2 5 2 mAt mAt Af0 3 f 0 S q, ; t f0 q t t g 0 2 40 6 2m
Since the Hamilton-Jacobi Equation only involves partial derivatives of S, g 0 can be taken to be zero without affect the dynamics and for simplicity, we
will take the integration constant f 0 to be simply , i.e., f0 2 2 5 2 mAt mAt A3 Sqt, ; q tt 2 40 6 2m 22
Example in using the Hamilton-Jacobi Method mAt2 mA2t 5 A 2 We can then solve for using Eq. T2, S q, ; t q q tt3 2 40 6 2m Sq ,, t A Q qtt3 6 m Solving for q, we then have, A qt t3 t 6 m
From the initial conditions, we have t0, q0 0, pmv 0 0
This implies that q0 0 And, Sq ,, t mAt2 p0 mv q 2 0 t0 t0 23
Example in using the Hamilton-Jacobi Method
Putting these two constants , back into our equation for q(t), we finally arrive at an explicit equation of motion for the system:
A qt t3 vt 6 o
Using Eq. T2 again, we have ,
Sq ,, t mAt2 p t q 2
We just found that mv 0 , mAt2 pt mv 2 0 24
Connection to the Schrödinger equation in QM (This connection was first derived by David Bohm.)
We first start by writing a quantum wavefunction r , t in phase- amplitude form, r,,t A r te iSr, t
where A r , t and S r , t are real functions of the position r and t. Let say this quantum particle is moving under the influence of a conservative potential U and its time evolution is then given by the Schrodinger equation: 2 i 2 U t2 m 25
Connection to the Schrödinger equation in QM
Now, we substitute the polar form of the wavefuntion into the
Schrodinger equation term by term: r,,t A r te iSr, t
iS A 1 S i : i ei A color code: Re Im t t t t
2 iS i : e A A S 2 iiS i iS i e S A A S e A A S i1 2 ii iS iS 2 2 eSA 2 ASe A ASAS
iS 2 A 2 i 2 color code: Re Im eAS 2 2 A S A S 26
Connection to the Schrödinger equation in QM
Substituting these terms into the Schrodinger equation, dividing out the common factor e iS and separating them into real/imaginary parts,
IM (black): i t 2
2 A 1 2 2 ASAS t2 m A 1 Simplifying 2 A2 2 A ASAS A 2 and 2 A : t2 m
2 AS2 Regrouping: A t m 27
Connection to the Schrödinger equation in QM
Now, for the real part of the equation, we have,
i t 2 U RE (blue): 2 S 2 A 2 A A2 SUA t2 m
2 2 S A 1 2 Simplifying: S U t2 mA 2 m
2 2 1 2 S A Rearranging: S U 2m tmA 2 28
Connection to the Schrödinger equation in QM
Taking the limit 0 , we have, 1 2 S S U 0 2m t
This is the Hamilton Jacobi equation if we identify the quantum phase S of with the Hamilton Principal Function or the classical Action.
To see that explicitly, recall the Hamilton Jacobi equation is,
SSS Hqq1, ,n ; , , ; t 0 q1 qn t 29
Connection to the Schrödinger equation in QM
Now, for a particle with mass m moving under the influence of a conservative potential U, its Hamiltonian H is given by its total energy:
p2 H U 2m Using the “inverse” canonical transformation, Eq. (T1), we can write H as,
1SS 1 2 S H USU p 2mqq 2 m q Substituting this into the Hamilton-Jacobi Equation, we have,
1 2 S S U 0 (This is the same equation for the phase of 2m t the wavefunction from Schrödinger Eq) 30
Connection to the Schrödinger equation in QM
Notes: 2 1. The neglected term proportional to is called the Bohm’s quantum potential, 2 2 A Q 2m A Note that this potential is nonlocal due to its spatial diffusive term and it can be interpreted as the source of nonlocality in QM. 2. The real part of the Schrödinger equation can be interpreted as the continuity equation for the conserved probability density A 2 * with the velocity field given by v Sm pm / : 2 AS2 A v t m t 31
Hamilton’s Characteristic Function
Let consider the case when the Hamiltonian is constant in time but we don’t a priori know what it is,
H qi, p i 1
Now, let also consider a canonical transformation under which the new momenta are all constants of the motion, (the transformed Q i are not restricted a priori. ) Pi i
AND we choose the new canonical momentum 1 to be the constant H itself.
Then, we seek to determine a time-independent generating function
Wq i, P i (Type-2) producing the desired CT. 32
Hamilton’s Characteristic Function
Similar to the development of the Hamilton’s Principal Function, since WqP , is Type-2, the corresponding equations of transformation are W q, p ( T 1) i (Note: the indices inside W(q, P) are qi W q, being suppressed.) Q ( T 2) i i W q, Now, since WqP , is time-independent, 0 and we have t
W W H qi , K 1 qi t 33
Hamilton’s Characteristic Function
WqP , is called the Hamilton’s Characteristic Function and
W H qi , 1 0 qi
is the partial differential equation (Hamilton-Jacobi Equation) for W.
Here, we have n independent constants i (with 1 H ) in determining this partial diff. eq.
And, through Eqs T1 and T2, W generates the desired canonical transformation in which all transformed momenta are constants ! 34
Hamilton’s Characteristic Function
In the transformed variables , , and the EOM is given by Qi, P i K 1 the Hamilton’s Equations, K (as required) Pi 0 or Pi i Qi K 1,i 1 Qi i 0,i 1 By integrating, this immediately gives,
Q1 t 1 where i are some integration
Qi i , i 1 constants determined by ICs.
Note that Q 1 is basically time and its conjugate momenta P 1 1 K is the Hamiltonian. 35
Hamilton’s Characteristic Function
By solving the H-J Eq, we can obtain W q,
Then, to get a solution for q i , p i , we use the transformation equations (T1 & T2),
W qi, i t1, i 1 W qi, i Qi pi i i ,i 1 qi
pi can be directly evaluated qi can be solved by taking the by taking the derivatives of W derivatives of W on the RHR and inverting the equation
The set of 2n constants i , i are fixed through the 2n initial
conditions qi0 , p i 0 36
Hamilton’s Characteristic Function
When the Hamiltonian does not dependent on time explicitly, one can use either
- The Hamilton Principal Function S qPt ,, or - The Hamilton Characteristic Function W qP, to solve a particular mechanics problem using the H-J equation and they are related by:
SqPt ,,, WqP 1 t
Note: As we have seen earlier, the Hamilton Principal Function can be used when H dep on time explicitly but no the HCF. 37
Action-Angle Variables in 1dof
- Often time, for a system which oscillates in time, we might not be interested in the details about the EOM but we just want information about the frequencies of its oscillations.
- The H-J procedure in terms of the Hamilton Characteristic Function can be a powerful method in doing that.
- To get a sense on the power of the technique, we will examine the simple case when we have only one degree of freedom.
- We continue to assume a conservative system with H 1 being a constant 38
Action-Angle Variables in 1dof
Let say we know that the dynamic of a system is periodic so that qtT qt
Recall from our discussion for a pendulum on dynamical systems, we have two possible periodic states:
libration rotation
p p
39
Action-Angle Variables in 1dof
- Now, we introduce a new variable
J p dq called the Action Variable, where the path integral is taken over one full cycle of the periodic motion.
- Now, since the Hamiltonian is a constant, we have
H q, p 1
- Then, by inverting the above equation to solve for p, we have
p p q,1 40
Action-Angle Variables in 1dof
- Then, we “integrate out” the q dependence in J p q, dq 1
One can then write that J is a function of 1 alone or vice versa,
1 H HJ
Since J is a function of the constant 1 alone, it is itself a constant.
- Now, instead of requiring our new momenta P to be 1 , we requires
P J (another constant instead of 1 ) 41
Action-Angle Variables in 1dof - Then, our Hamilton Characteristic Function can be written as W WqJ ,
- From the transformation equations, the generalized coordinate Q corresponding to P is given by Eq. T2 W Q P - Enforcing our new momenta P to be J and calling its conjugate coordinate w , we have W w J
- Here, in this context, w is called the Angle Variable. 42
Action-Angle Variables in 1dof
- Note, since the generating function W(q, P) is time independent, the Hamiltonian in the transformed coordinate K = H so that
1 H HJ KJ
- Correspondingly, using the Hamilton Equations, the EOM for w is, K J w vJ J
- Now, since K is a constant, v J will also be a constant function of J . 43
Action-Angle Variables in 1dof
w vJ The above diff. eq. can be immediately integrated to get w vt
where is an integration constant depending on IC.
- Thus, this Angle Variable w is a linear function of time ! 44
Action-Angle Variables in 1dof
Now, let integrate w over one period T of the periodic motion. w w dw dq T q T W Using the transformation equation, w , we then have, J 2W w dq q J T 45
Action-Angle Variables in 1dof
Since J is a constant with no q dependence, we can move the derivative wrt J outside of the q integral,
2W dW w dq dq J q dJ q T T W Then, using the other transformation equation, p , we can write, q d dJ w p dq 1 dJT dJ
(recall J p dq ) 46
Action-Angle Variables in 1dof
w 1 This result means that w changes by unity as q goes through a complete period T. Then, since we have w vt , we can write
wwT( ) w (0) vT 1
-vJ 1 T gives the frequency of the periodic oscillation associated with q ! - and it can be directly evaluated thru K J v J (*) J without finding the complete EOM 47
Action-Angle Variables in 1dof: Example
Let consider a linear harmonic oscillator given by the Hamiltonian,
2 Motion is periodic in q p m 2 2 H q is the const total E 2m 2 k m is the natural freq Solving for p, we have,
p2 m m2 2 q 2
Then, we can substitute this into the integral for the Action Variable,
J 2 m m2 2 q 2 dq T 48
Action-Angle Variables in 1dof: Example
Using the following coordinate change, 2 q sin m2
The integrand can be simplified,
2 22 2 2 2 2 2 2m m q dq 2m m 2 sin 2 cosd m m
2 2 2m 1 sin 2 cosd m 2 2 2m cos2 cos d cos 2 d m2 49
Action-Angle Variables in 1dof: Example
Putting this back into our integral for the Action Variable, we have,
2 2 J cos2 d 0 This gives, 2 J
Inverting to solve for in terms of J, we have,
J H K 2 50
Action-Angle Variables in 1dof: Example
Finally, applying our result for v using Eq. (*), we have
K J v J J 2 2 Thus,
1 k v 2 2 m
Here, we have derived the frequency of the linear harmonic oscillator by only calculating the Action Variable without explicitly solving for the EOM !