- #1
sundriedtomato
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[SOLVED] Poisson brackets.
Show that, if Poisson brackets (g,h) = 1, then (g[tex]^{n}[/tex],h) = ng[tex]^{n-1}[/tex]
where g = g(p,q) and h = h(p,q)
p and q are canonical coordinates
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) [tex]\equiv[/tex] [tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex] = 1
so, from here we have
(g[tex]^{n}[/tex],h) [tex]\equiv[/tex] [tex]\frac{\delta g^{n}}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g^{n}}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex]
=> ng[tex]^{n-1}[/tex][tex]\frac{\delta g^{n}}{\delta q}[/tex]*[tex]\frac{\delta h}{\delta p}[/tex] - ng[tex]^{n-1}[/tex][tex]\frac{\delta g}{\delta p}[/tex]*[tex]\frac{\delta h}{\delta q}[/tex] =>
ng[tex]^{n-1}[/tex]([tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex])
and the part in brackets is = 1 as we know from given Poisson bracket =>
(g[tex]^{n}[/tex],h) [tex]\equiv[/tex] ng[tex]^{n-1}[/tex]
Homework Statement
Show that, if Poisson brackets (g,h) = 1, then (g[tex]^{n}[/tex],h) = ng[tex]^{n-1}[/tex]
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) [tex]\equiv[/tex] [tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex] = 1
so, from here we have
(g[tex]^{n}[/tex],h) [tex]\equiv[/tex] [tex]\frac{\delta g^{n}}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g^{n}}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex]
=> ng[tex]^{n-1}[/tex][tex]\frac{\delta g^{n}}{\delta q}[/tex]*[tex]\frac{\delta h}{\delta p}[/tex] - ng[tex]^{n-1}[/tex][tex]\frac{\delta g}{\delta p}[/tex]*[tex]\frac{\delta h}{\delta q}[/tex] =>
ng[tex]^{n-1}[/tex]([tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex])
and the part in brackets is = 1 as we know from given Poisson bracket =>
(g[tex]^{n}[/tex],h) [tex]\equiv[/tex] ng[tex]^{n-1}[/tex]
genneth said:Yep! Or, using the fact that [tex]\{.,h\}[/tex] acts like a derivative, via the Leibniz rule:
[tex]\{ab,h\} = a\{b,h\} + \{a,h\}b[/tex]
So,
[tex]\{g^n,h\} = g^{n-1}\{g,h\} + \{g^{n-1},h\}g[/tex]
Giving you a recursive relation, that should be solvable.
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