- #1
simonstar76
- 1
- 0
- TL;DR
- The textbook Classical Mechanics by Deriglazov presents a proof of the fact that Poisson brackets are at most function of time in a canonical transformation. While the proof seems convincing for a 2D phase space, it appears to the OP at least incomplete for greater dimensions
The book Classical Mechanics by Alexei Deriglazov defines as canonical a transformation Z=Z(z,t) that preserves the Hamiltonian form of the equation of motion for any H. After taking the divergence of the vector equation relating the components of the time derivative of Z in the two coordinate systems, he writes (p. 116, I modify slightly the notation) the following equation (it is intended that repeated indices sum):
$$
\frac{\partial{}}{\partial{Z_k}}\left(\left\{Z_k,Z_l\right\}_z|_{z(Z,t)}\right)\frac{\partial{H(z(Z,t))}}{\partial{Z_l}}+\frac{\partial{}}{\partial{Z_k}}\left(\frac{\partial{Z(z,t)}}{\partial{t}}\Big{|}_{z(Z,t)}\right)=0.
$$
He asserts that, since H is arbitrary, the partial derivatives of each Poisson bracket
$$
\left\{Z_k,Z_l\right\}_z\Big{|}_{z(Z,t)}
$$
with respect to the Z coordinates must be 0 and so we can at most write (the substitution z=z(Z,t) can be omitted):
$$
\left\{Z_k,Z_l\right\}_z=c_{kl}\left(t\right)
$$
The derivation seems confirmed by an explicit calculation that the author does in the following chapter (p. 138 and 142), where he considers the phase space of dimension 2d=2. For the case e.g. of a time-independent transformation, he takes the derivatives of the two equations for the time derivatives of Q and P and adds them, according to the discussion above, obtaining (without explicitly showing the variable dependency):
$$
\frac{\partial{}}{\partial{Q}}\left(\left\{Q,P\right\}\right)\frac{\partial{H}}{\partial{P}}-\frac{\partial{}}{\partial{P}}\left(\left\{Q,P\right\}\right)\frac{\partial{H}}{\partial{Q}}=0
$$
In this case, the arbitrariness of H forces us to conclude that both partial derivatives of {Q,P} with respect to Q and P are 0, thus verifying the assertion for this case (there's no time dependence here so we can conclude {Q,P}=c=constant).
Turning back to the general assertion for 2d>2, how can such an approach lead to the desired conclusion (i.e. that all partial derivatives of the {Z
i,Zj} with respect to Zk are 0)
if we have only one equation in the 2d arbitrary partial derivatives of H which potentially multiply 2d
2(d-1)/2 partial derivatives of the Poisson brackets ? The d terms would collect many different partial Poisson bracket derivatives, leaving us with no possibility to conclude that each of the four partial derivative, for every Poisson bracket, is zero.
$$
\frac{\partial{}}{\partial{Z_k}}\left(\left\{Z_k,Z_l\right\}_z|_{z(Z,t)}\right)\frac{\partial{H(z(Z,t))}}{\partial{Z_l}}+\frac{\partial{}}{\partial{Z_k}}\left(\frac{\partial{Z(z,t)}}{\partial{t}}\Big{|}_{z(Z,t)}\right)=0.
$$
He asserts that, since H is arbitrary, the partial derivatives of each Poisson bracket
$$
\left\{Z_k,Z_l\right\}_z\Big{|}_{z(Z,t)}
$$
with respect to the Z coordinates must be 0 and so we can at most write (the substitution z=z(Z,t) can be omitted):
$$
\left\{Z_k,Z_l\right\}_z=c_{kl}\left(t\right)
$$
The derivation seems confirmed by an explicit calculation that the author does in the following chapter (p. 138 and 142), where he considers the phase space of dimension 2d=2. For the case e.g. of a time-independent transformation, he takes the derivatives of the two equations for the time derivatives of Q and P and adds them, according to the discussion above, obtaining (without explicitly showing the variable dependency):
$$
\frac{\partial{}}{\partial{Q}}\left(\left\{Q,P\right\}\right)\frac{\partial{H}}{\partial{P}}-\frac{\partial{}}{\partial{P}}\left(\left\{Q,P\right\}\right)\frac{\partial{H}}{\partial{Q}}=0
$$
In this case, the arbitrariness of H forces us to conclude that both partial derivatives of {Q,P} with respect to Q and P are 0, thus verifying the assertion for this case (there's no time dependence here so we can conclude {Q,P}=c=constant).
Turning back to the general assertion for 2d>2, how can such an approach lead to the desired conclusion (i.e. that all partial derivatives of the {Z
i,Zj} with respect to Zk are 0)
if we have only one equation in the 2d arbitrary partial derivatives of H which potentially multiply 2d
2(d-1)/2 partial derivatives of the Poisson brackets ? The d terms would collect many different partial Poisson bracket derivatives, leaving us with no possibility to conclude that each of the four partial derivative, for every Poisson bracket, is zero.