Is the Time Reversal Transformation Canonical?

In summary, the given transformation of time reversal is canonical in that it preserves the form of the Hamiltonian equations of motion. However, it does not satisfy the invariance of the fundamental Poisson bracket relations. The mismatch between the two aspects may be due to the assumption that the generating function F has no explicit time dependence, which may be related to the discrepancy in the fundamental Poisson bracket. Further investigation is needed to determine the exact cause of the mismatch.
  • #1
neelakash
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1

Homework Statement



Show that the time reversal transformation given by Q = q, P = − p and T = − t, is canonical, in the sense that the form of the Hamiltonian equations of motion is preserved. However, it does not satisfy the invariance of the fundamental Poisson Bracket relations. This is an example when the two criteria are not equal.

Homework Equations


The Attempt at a Solution



This I have done...I just ask you to check if the procedure is correct.( ' denotes d/dt )

Q'=(dQ/dT)=(dQ/dt)(dt/dT)= -(dQ/dt)= -[(∂Q/∂q)q' + (∂Q/∂p)p' + (∂Q/∂t)]= -[q']= -(∂H/∂p)

Also, (∂K/∂P)=(∂H/∂p)(∂p/∂P)= -(∂H/∂p)...[we write (∂K/∂P)=(∂H/∂p) as Kamiltonian K is a

function of Q,P,T and Hamiltonian H is a function of q,p,t]

Thus, Q'= (∂K/∂P)...1st of Hamilton's canonical equations is proved.

Similarly, P'= (dP/dT)= -(dP/dt)= -[(∂P/∂q)q' + (∂P/∂p)p' + (∂P/∂t)]= p'= -(∂H/∂q)

Then, (∂K/∂Q)=(∂H/∂q)(∂q/∂Q)=(∂H/∂q)

Thus, P'= -(∂K/∂Q)

This shows that the given transformation leads (q,p) to canonically conjugate variables(Q,P)

Evaluating the Poisson brackets it is easy to show that they do not satisfy fundamental Poisson bracket.

Can anyone suggest why there is a mismatch between the two aspects?
 
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  • #2
Are people not interested for I have not used TeX?

My attempt:

[tex]\dot {Q}= (dQ/dT) = (dQ/dt)(dt/dT) = -(dQ/dt) = -[(\partial Q / \partial q) \dot {q} + (\partial Q / \partial p) \dot {p} + (\partial Q / \partial t)] = -\dot {q} = -(\partial H / \partial p)[/tex]


[tex](\partial K / \partial P)=(\partial H / \partial p)(\partial\ p / \partial P)= - (\partial H / \partial p)[/tex]

Thus, [tex]\dot {Q}= (dQ/dT) = (\partial K / \partial P) [/tex]

Similarly,

[tex]\dot {P}= (dP/dT) = (dP/dt)(dt/dT) = -(dP/dt) = -[(\partial P / \partial q) \dot {q} + (\partial P / \partial p) \dot {p} + (\partial P / \partial t)] = \dot {p} = -(\partial H / \partial q)[/tex]


[tex](\partial K / \partial Q)=(\partial H / \partial q)(\partial\ q / \partial Q)= (\partial H / \partial q)[/tex]

Thus, [tex]\dot {P}= (dP/dT) = -(\partial K / \partial Q) [/tex]

The most sensitive part of this observation is ofcourse the identification that [tex]\ K = \ H [/tex] that I have

assumed.In general,however,there will be a partial time derivative of a generating function F so that the equation lokks like

[tex]\ K = \ H + (\partial F / \partial t)[/tex]

I have assumed that F has no explicit time dependence.So the later time derivative is zero and we have Hamilton's equations

satisfied...

My question is if my assumption is justified for this problem...Note that, the problem already ensures that the transformation does

not satisfy the fundamental Poisson brackets.My calculation gave the value of Poissson Bracket -1 instead of 1 (the correct

one). In that sense,it is not a canonical transformation...

I am not sure, but it looks to me, my assumption that F does not have explicit time dependence may be somehow related to the fact of fundamnental Poisson Bracket not

being satisfied...

Can anyone please take interest?
 
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FAQ: Is the Time Reversal Transformation Canonical?

1. What is a canonical transformation problem?

A canonical transformation problem is a mathematical concept used in classical mechanics to transform one set of coordinates to another set in order to simplify the equations of motion. It is often used to solve problems involving Hamiltonian dynamics.

2. What is the purpose of a canonical transformation?

The purpose of a canonical transformation is to simplify the equations of motion and make them easier to solve. It can also be used to describe the same physical system in different coordinate systems, providing different perspectives and insights into the system.

3. How is a canonical transformation different from a regular coordinate transformation?

A canonical transformation preserves the form of the Hamiltonian equations, while a regular coordinate transformation does not. This means that the equations of motion and the physical properties of the system remain the same after a canonical transformation, but may change after a regular coordinate transformation.

4. What are the types of canonical transformations?

There are two types of canonical transformations: point transformations and generating function transformations. Point transformations are defined by a set of equations that directly relate the old coordinates to the new coordinates. Generating function transformations are defined by a set of equations that involve a generating function, which can be used to transform the old coordinates to the new coordinates.

5. How is a canonical transformation related to the principle of least action?

The principle of least action states that the path taken by a physical system is the one that minimizes the action, which is a mathematical function defined in terms of the system's coordinates and velocities. A canonical transformation can be used to transform the action from one coordinate system to another, while still preserving the physical system's path. This makes it a powerful tool in solving problems involving the principle of least action.

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