Poisson's Identity: Solving ((φλ)χ)+((λχ)φ)+((χφ)λ)=0

[aggarwal]

please help me to solve this identity

$$((\phi\lambda)\chi)+((\lambda\chi)\phi)+((\chi\phi)\lambda)=0$$


where () = poisson bracket

$$\phi=\phi(t,q_{i},p_{i})$$
$$\chi=\chi(t,q_{i},p_{i})$$
$$\lambda=\lambda(t,q_{i},p_{i})$$
for i=1,2,...,n

 

[dextercioby]

This is a very lengthy proof. For the beginning what is $\left [\phi,\lambda]_{PB}$ eual to ?

 

[tacman]

To solve this identity, we can use the properties of the Poisson bracket and the Jacobi identity. Firstly, we can expand each term in the identity using the definition of the Poisson bracket:

((\phi\lambda)\chi) = \phi\lambda\chi + \lambda\phi\chi

((\lambda\chi)\phi) = \lambda\chi\phi + \chi\lambda\phi

((\chi\phi)\lambda) = \chi\phi\lambda + \phi\chi\lambda

Next, we can use the Jacobi identity which states that for any three functions f,g,h:

[f,[g,h]] + [g,[h,f]] + [h,[f,g]] = 0

Applying this to our identity, we get:

((\phi\lambda)\chi) + ((\lambda\chi)\phi) + ((\chi\phi)\lambda) = [\phi,\lambda\chi] + [\lambda,\chi\phi] + [\chi,\phi\lambda]

Using the properties of the Poisson bracket, we can rewrite this as:

[\phi,\lambda]\chi + \lambda[\phi,\chi] + \chi[\phi,\lambda] + [\lambda,\chi]\phi + \chi[\lambda,\phi] + \phi[\lambda,\chi] = 0

Since the Poisson bracket is bilinear and satisfies the Leibniz rule, we can further simplify this to:

[\phi,\lambda]\chi + \lambda[\phi,\chi] + \chi[\phi,\lambda] = 0

Finally, we can use the fact that the Poisson bracket is anti-symmetric, meaning that [f,g] = -[g,f], to rewrite this as:

[\phi,\lambda]\chi + [\phi,\chi]\lambda + [\lambda,\phi]\chi = 0

which is equivalent to our original identity. Therefore, we have shown that the identity holds true for any choice of functions \phi, \chi, and \lambda.