Proving Poisson Brackets Homework Statement

In summary, the problem at hand is to prove the equality f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+... and we can do so by expanding f(q,p) as a Taylor series, and using the Poisson bracket to show that if f(q,p) is a constant of motion, then \frac{df(q,p)}{dt}=\{f,H\}. The theorem that states mixed partial derivatives are commutative is still unknown, but it is known that if f(q,p) is a constant of motion, then \frac{df
  • #1
Nusc
760
2

Homework Statement



[tex]

f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...

[/tex]
Prove the above equality. p & q are just coords and momenta

How do we do this if we don't know what H is?
Where do we start?

Homework Equations


The Attempt at a Solution

 
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  • #2
Nusc said:

Homework Statement



[tex]

f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...

[/tex]
Prove the above equality. p & q are just coords and momenta

How do we do this if we don't know what H is?
Where do we start?

Homework Equations





The Attempt at a Solution


Looks like you may not need to know what the Hamiltonian is, just how the Hamiltonian works in the Poisson bracket (assume [itex]f[/itex] is not a constant of motion):

[tex]
\frac{df(q,p)}{dt}&=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{dq}{dt}+\frac{\partial f}{\partial p}\frac{dp}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}=\frac{\partial f}{\partial t}+\{f,H\}
[/tex]

So, using this, it looks like you would expand [itex]f(q,p)[/itex] as a Taylor series. Note that if [itex]f(q,p)[/itex] is a constant of motion, [itex]f_{,t}=0[/itex] so that you have

[tex]
\frac{df(q,p)}{dt}=\{f,H\}
[/tex]
 
  • #3
What's the theorem that says mixed partials are commutative? Not Claurait's...
 
  • #4
Also

I know

[tex]
\frac{\partial }{\partial t}\{H,f\} = \{\frac{\partial H}{\partial t},f\} + \{H,\frac{\partial f}{\partial t}\}
[/tex]

What about for ordinary derivatives?
[tex]
\frac{d}{d t}\{H,f\} =?
[/tex]
 
  • #5
jdwood983 said:
Note that if [itex]f(q,p)[/itex] is a constant of motion, [itex]f_{,t}=0[/itex] so that you have

[tex]
\frac{df(q,p)}{dt}=\{f,H\}
[/tex]

No, if [itex]f(q,p)[/itex] is a constant of motion, then [tex]\frac{df}{dt}=0[/itex]. The fact that [itex]f[/itex] has no explicit time dependence (it is given as a function of [itex]q[/itex] and [itex]p[/itex] only), tells you that [tex]\frac{\partial f}{\partial t}=0[/itex]
 
  • #6
Nusc said:
What about for ordinary derivatives?
[tex]
\frac{d}{d t}\{H,f\} =?
[/tex]

You tell us...expand the Poisson bracket and calculate the derivatives...what do you get?
 
  • #7
[tex]
\frac{d}{dt}\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{d}{dt}\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}
[/tex]
 
  • #8
Okay, now use the product rule...

[tex]\frac{d}{dt}\left(\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}\right)=[/itex]

?
 

What is a Poisson Bracket?

A Poisson Bracket is a mathematical tool used to calculate the time evolution of a dynamical system. It is represented by two functions and is used to find the derivative of one function with respect to another.

How do you prove Poisson Brackets?

To prove Poisson Brackets, you must first define the two functions and then use the properties of the bracket to manipulate the equations. This involves using the Leibniz rule, the Jacobi identity, and the linearity property of the bracket.

What are the properties of Poisson Brackets?

The properties of Poisson Brackets include the Leibniz rule, the Jacobi identity, and the linearity property. The Leibniz rule states that the bracket of two functions multiplied by a third function is equal to the bracket of the first two functions multiplied by the third function plus the first two functions multiplied by the bracket of the third function. The Jacobi identity states that the bracket of three functions is equal to the sum of the cyclic permutations of the bracket of the three functions. The linearity property states that the bracket is a linear operator.

How are Poisson Brackets used in physics?

Poisson Brackets are used in physics to study the dynamics of systems with continuously varying quantities, such as in classical mechanics and electromagnetism. They can help determine the equations of motion and reveal symmetries and conservation laws of a system. They are also used in Hamiltonian mechanics to find canonical transformations.

Can Poisson Brackets be generalized?

Yes, Poisson Brackets can be generalized to work in non-commutative and quantum systems. This is known as the Moyal Bracket, which is used in quantum mechanics to study the dynamics of systems with discrete energy levels.

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