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A canonical transformation is a transformation from one set of coordinates $q,p$ to a new one $Q(q,p), P(q,p)$. For a function $f(Q,P)$ using the chain rule and using summation notation

$$\frac{\partial f}{\partial q_{i}}=\frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial q_{i}}+\frac{\partial f}{\partial P_{j}}\frac{\partial P_{j}}{\partial q_{i}}$$

$$\frac{\partial f}{\partial p_{i}}=\frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial p_{i}}+\frac{\partial f}{\partial P_{j}}\frac{\partial P_{j}}{\partial p_{i}}$$

The Poisson bracket: \begin{eqnarray} \left\{f,g\right\}&=&\frac{\partial f}{\partial q_{i}}\frac{\partial g}{\partial p_{i}}-\frac{\partial f}{\partial p_{i}}\frac{\partial g}{\partial q_{i}}\\ &=& \left(\frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial q_{i}}+\frac{\partial f}{\partial P_{j}}\frac{\partial P_{j}}{\partial q_{i}}\right)\left(\frac{\partial g}{\partial Q_{k}}\frac{\partial Q_{k}}{\partial p_{i}}+\frac{\partial g}{\partial P_{k}}\frac{\partial P_{k}}{\partial p_{i}}\right) - \left(\frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial p_{i}}+\frac{\partial f}{\partial P_{j}}\frac{\partial P_{j}}{\partial p_{i}}\right)\left(\frac{\partial g}{\partial Q_{k}}\frac{\partial Q_{k}}{\partial q_{i}}+\frac{\partial g}{\partial P_{k}}\frac{\partial P_{k}}{\partial q_{i}}\right)\\ &=& \end{eqnarray} enter image description here

For me, it is not very clear how to obtain the relations which are written in the attached image (the first two lines). What do these relations mean? What is it about? Is it about the chain rule, it is just simple multiply? How can I obtain the first two lines from image?

Thanks!

UPDATE $\displaystyle \frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial q_{i}}$ - it is about the derivative

In the calculation of Poisson bracket we have:

$\displaystyle \frac{\partial f}{\partial Q_{j}}\frac{\partial Q_{j}}{\partial q_{i}} \cdot \frac{\partial g}{\partial P_{k}}\frac{\partial P_{k}}{\partial p_{i}} $ - it is about the composition and about product which is marked with $\cdot$

when we try to make a arrangement it about what....

enter image description here $\displaystyle \frac{\partial f}{\partial Q_{j}} \frac{\partial g}{\partial P_{k}} \left(\frac{\partial Q_{j}}{\partial q_{i}} \frac{\partial P_{k}}{\partial p_{i}} \right)$ here it about what?

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It's just a rearrangement. Multiply out the products before and after the step and check that you get the same terms in each case.

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  • $\begingroup$ thanks for the answer. Yes, I did it, but I had thought it is not about multiply. Where it is: $\frac{\partial f}{\partial Q_j}\frac{\partial Q_j}{\partial q_i}$ here it is not product, no? it is the composition of the functions. $\frac{\partial f}{\partial Q_j}\frac{\partial Q_j}{\partial q_i} \cdot \frac{\partial g}{\partial P_k}\frac{\partial P_k}{\partial p_i}$ is same with $\frac{\partial f}{\partial Q_j}\frac{\partial g}{\partial P_k}\left(\frac{\partial Q_j}{\partial q_i}\frac{\partial P_k}{\partial p_i}+\ldots\right)$ Thanks $\endgroup$
    – Iuli
    Commented Sep 11, 2015 at 7:52
  • $\begingroup$ @Iuli: No, it's all just products, no composition. $\endgroup$
    – joriki
    Commented Sep 11, 2015 at 7:56
  • $\begingroup$ I want to ask you something, please. Have some papers, or a link where I can find some explanation about the derivative of the functions, something which can help me to avoid the mistakes like in my question. Thanks! $\endgroup$
    – Iuli
    Commented Sep 11, 2015 at 8:07
  • $\begingroup$ @Iuli: What do you mean by "the derivative of the functions"? Which derivative of which functions? $\endgroup$
    – joriki
    Commented Sep 11, 2015 at 8:07
  • $\begingroup$ I did a modification in the question. I hope this make sense about what I'm asking. thanks :) $\endgroup$
    – Iuli
    Commented Sep 11, 2015 at 8:24

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