Proving Poisson Brackets Homework Statement

Nusc
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Homework Statement



[tex] <br /> f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...<br /> [/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

 
on Phys.org
Nusc said:

Homework Statement



[tex] <br /> f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...<br /> [/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]
 
What's the theorem that says mixed partials are commutative? Not Claurait's...
 
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]
 
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 <i>explicit</i> 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][/tex][/tex]
 
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?
 
[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]
 
Okay, now use the product rule...

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

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