Solve Poisson Brackets: (g,h) = 1, (g^n,h) = ng^{n-1}

[sundriedtomato]

[SOLVED] Poisson brackets.

Homework Statement

Show that, if Poisson brackets (g,h) = 1, then (g^{n},h) = ng^{n-1}
where g = g(p,q) and h = h(p,q)

p and q are canonical coordinates

The Attempt at a Solution

I suppose that this is purely mathematical, but I am still searching for a detailed example in literature.
I also would like to ask - what book/author can you recommend, where alike problem is discussed.

Thank You!
P.S. I tried search function, but found nothing similar.

Solutions:

(g,h) \equiv \frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q} = 1

so, from here we have

(g^{n},h) \equiv \frac{\delta g^{n}}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g^{n}}{\delta p}\frac{\delta h}{\delta q}

=> ng^{n-1}\frac{\delta g^{n}}{\delta q}*\frac{\delta h}{\delta p} - ng^{n-1}\frac{\delta g}{\delta p}*\frac{\delta h}{\delta q} =>

ng^{n-1}(\frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q})
and the part in brackets is = 1 as we know from given Poisson bracket =>

(g^{n},h) \equiv ng^{n-1}

Yep! Or, using the fact that \{.,h\} acts like a derivative, via the Leibniz rule:

\{ab,h\} = a\{b,h\} + \{a,h\}b

So,

\{g^n,h\} = g^{n-1}\{g,h\} + \{g^{n-1},h\}g

Giving you a recursive relation, that should be solvable.

 

[nrqed]

Homework Statement

Show that, if Poisson brackets (g,h) = 1, then (g^{n},h) = ng^{n-1}
where g = g(p,q) and h = h(p,q)

p and q are canonical coordinates

The Attempt at a Solution

I suppose that this is purely mathematical, but I am still searching for a detailed example in literature.
I also would like to ask - what book/author can you recommend, where alike problem is discussed.

Thank You!
P.S. I tried search function, but found nothing similar.

What have you already seen on Poisson brackets? I could give you the answer but it will make more sense to you if you can get the answer from what you have learned.

First of all, have you seen the definition of a PB?

 

[sundriedtomato]

Yes, I have seen the definition of PB, and brief explanation of its properties.
Should I involve partial integration by time?
I will try to work out what (g^2,h) will give.

 

[nrqed]

Yes, I have seen the definition of PB, and brief explanation of its properties.
Should I involve partial integration by time?
I will try to work out what (g^2,h) will give.

There are no integrations involved, just partial derivatives.

Actually, I take back what I wrote. If you know the definition, just plug in g^n and you will directly get the final answer!

 

[genneth]

The important thing is that the Poisson bracket is like a derivative operator. Use the Leibnez rule.

 

[sundriedtomato]

Well, here is what I have:

(g,h) \equiv \frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q} = 1

so, from here we have

(g^{n},h) \equiv \frac{\delta g^{n}}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g^{n}}{\delta p}\frac{\delta h}{\delta q}

=> ng^{n-1}*\frac{\delta h}{\delta p} - ng^{n-1}*\frac{\delta h}{\delta q} (??)





I am actually ashamed of being unable to do such an easy task, which is basically depends on elementary calculus.

 

[nrqed]

Well, here is what I have:

(g,h) \equiv \frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q} = 1

so, from here we have

(g^{n},h) \equiv \frac{\delta g^{n}}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g^{n}}{\delta p}\frac{\delta h}{\delta q}

=> ng^{n-1}*\frac{\delta h}{\delta p} - ng^{n-1}*\frac{\delta h}{\delta q} (??)





I am actually ashamed of being unable to do such an easy task, which is basically depends on elementary calculus.

You are applying the chain rule, so you should have written, for example,
\frac{\delta g^n}{\delta q} = n g^{n-1} \frac{\delta g}{\delta q} (you forgot the delta g/delta q)
and the same for the derivative with respect to p.

Then you will see that it works out.

Patrick

 

[sundriedtomato]

(g,h) \equiv \frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q} = 1

so, from here we have

(g^{n},h) \equiv \frac{\delta g^{n}}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g^{n}}{\delta p}\frac{\delta h}{\delta q}

=> ng^{n-1}\frac{\delta g^{n}}{\delta q}*\frac{\delta h}{\delta p} - ng^{n-1}\frac{\delta g}{\delta p}*\frac{\delta h}{\delta q} =>

ng^{n-1}(\frac{\delta g}{\delta q}\frac{\delta h}{\delta p} - \frac{\delta g}{\delta p}\frac{\delta h}{\delta q})
and the part in brackets is = 1 as we know from given Poisson bracket =>

(g^{n},h) \equiv ng^{n-1}

?

Simon

 

[genneth]

Yep! Or, using the fact that \{.,h\} acts like a derivative, via the Leibniz rule:

\{ab,h\} = a\{b,h\} + \{a,h\}b

So,

\{g^n,h\} = g^{n-1}\{g,h\} + \{g^{n-1},h\}g

Giving you a recursive relation, that should be solvable.

 

[sundriedtomato]

That is beautiful. Thank You everybody who took part in this!