If we consider a canonical transformation from $(q,p)$ to $(Q,P)$, it is stated in several sources that by Jacobian rules, $$ \frac{\partial(Q,P)}{\partial(q,p)} = \frac{\partial(Q,P)/\partial(q,P)}{\partial(q,p)/\partial(q,P)}. \tag{1} $$

By taking books such as Riley's Mathematical Methods for physics and engineering, I could confirm that this is indeed true (section 6.4.4). However, I have tried this myself and it seems that I am missing something. For instance the left hand side of the above equation expands in:

$$ \frac{\partial(Q,P)}{\partial(q,p)}=\begin{vmatrix} \frac{\partial Q}{\partial q} & \frac{\partial Q}{\partial p}\\ \frac{\partial P}{\partial q} & \frac{\partial P}{\partial p}\\ \end{vmatrix} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} $$

Now, the numerator of the right hand side is

$$ \frac{\partial(Q,P)}{\partial(q,P)}=\begin{vmatrix} \frac{\partial Q}{\partial q} & \frac{\partial Q}{\partial P}\\ \frac{\partial P}{\partial q} & \frac{\partial P}{\partial P}\\ \end{vmatrix} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial P} - \frac{\partial Q}{\partial P}\frac{\partial P}{\partial q}=\frac{\partial Q}{\partial q} $$

Then, the denominator is

$$ \frac{\partial(q,p)}{\partial(q,P)}=\begin{vmatrix} \frac{\partial q}{\partial q} & \frac{\partial q}{\partial P}\\ \frac{\partial p}{\partial q} & \frac{\partial p}{\partial P}\\ \end{vmatrix} = \frac{\partial q}{\partial q}\frac{\partial p}{\partial P} - \frac{\partial q}{\partial P}\frac{\partial p}{\partial q}=\frac{\partial p}{\partial P} $$

yielding the right hand side of the first equation above:

$$ \frac{\partial(Q,P)/\partial(q,P)}{\partial(q,p)/\partial(q,P)} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} $$

which corresponds to the first term of the second equation above. This means that the second term of the second equation above has to be zero. However, I fail to see why this is true. Assuming that $Q=Q(p,q)$ and $P=P(p,q)$, that term should have a value. What am I missing?

  • $\begingroup$ Are you writing the determinant of the Jacobian? THe notation was not clear to me $\endgroup$
    – innisfree
    May 23, 2020 at 6:05
  • $\begingroup$ @innisfree Yes, I meant the determinant of the Jacobian with the above notation. I am sorry it was not clear, I just took the notation found in more than one book (Riley, Hamill, etc.) and assumed it was common... $\endgroup$
    – Rapha
    May 23, 2020 at 14:28

2 Answers 2


OP's eq. (1) is true for any invertible coordinate transformation -- they don't need to be canonical coordinates. The trick is to keep track of what is kept constant during the partial differentations. In 2D eq. (1) reads:

$$\begin{align} {\rm LHS} ~=~&\left(\frac{\partial Q}{\partial q}\right)_p \left(\frac{\partial P}{\partial p}\right)_q - \left(\frac{\partial P}{\partial q}\right)_p \left(\frac{\partial Q}{\partial p}\right)_q\cr ~=~& \left[\left(\frac{\partial q}{\partial q}\right)_p \left(\frac{\partial Q}{\partial q}\right)_P + \left(\frac{\partial P}{\partial q}\right)_p \left(\frac{\partial Q}{\partial P}\right)_q\right] \left(\frac{\partial P}{\partial p}\right)_q \cr &- \left(\frac{\partial P}{\partial q}\right)_p \left[\left(\frac{\partial q}{\partial p}\right)_q \left(\frac{\partial Q}{\partial q}\right)_P +\left(\frac{\partial P}{\partial p}\right)_q \left(\frac{\partial Q}{\partial P}\right)_q \right]\cr ~=~& \left[\left(\frac{\partial Q}{\partial q}\right)_P + \left(\frac{\partial P}{\partial q}\right)_p \left(\frac{\partial Q}{\partial P}\right)_q\right] \left(\frac{\partial P}{\partial p}\right)_q - \left(\frac{\partial P}{\partial q}\right)_p \left(\frac{\partial P}{\partial p}\right)_q \left(\frac{\partial Q}{\partial P}\right)_q \cr ~=~& \left(\frac{\partial Q}{\partial q}\right)_P\left(\frac{\partial P}{\partial p}\right)_q ~=~\left(\frac{\partial Q}{\partial q}\right)_P / \left(\frac{\partial p}{\partial P}\right)_q ~=~{\rm RHS},\end{align}$$ where we used the multi-variable chain rule twice.

  • $\begingroup$ I think I finally got it. However, if we are initially calculating the Jacobian determinant $\displaystyle \frac{\partial(Q,P)}{\partial(q,p)}$, is it right to simply state that $Q=Q(q,P)$ in $\displaystyle \left( \frac{\partial Q}{\partial q}\right)_p$ and $\displaystyle \left( \frac{\partial Q}{\partial p}\right)_q$. Why is it obvious? $\endgroup$
    – Rapha
    May 26, 2020 at 3:32
  • $\begingroup$ Well, what I mean is that from your derivation (chain rule), $\displaystyle \left(\frac{\partial Q}{\partial q}\right)_p = \left(\frac{\partial Q}{\partial q}\right)_P \left(\frac{\partial q}{\partial q}\right)_p + \left(\frac{\partial Q}{\partial P}\right)_q \left(\frac{\partial P}{\partial q}\right)_p$ in the first term (and something similar for the second term as well), which is why I am assuming that $Q=Q(q,P)$ $\endgroup$
    – Rapha
    May 26, 2020 at 15:41
  • 1
    $\begingroup$ In eq. (1) it is implicitly assumed that $(q,p)\mapsto (q,P)$ and $(q,P)\mapsto (Q,P)$ are well-defined bijective maps. $\endgroup$
    – Qmechanic
    May 26, 2020 at 15:49

The expression $\frac{\partial(Q,P)/\partial(q,P)}{\partial(q,p)/\partial(q,P)}$ is ratio of Jacobians.

Evaluate each expression independently by eliminating the repeated variables - then divide them.

In your case, $\frac{\partial(Q,P)/\partial(q,P)}{\partial(q,p)/\partial(q,P)}=\frac{\partial Q/\partial q}{\partial p/\partial P}=J$.

The transformation is canonical since $\partial Q/\partial q=1$ and $\partial p/\partial P=1$ which implies $J=1$ where $J$ is the Jacobian.

  • $\begingroup$ From my understanding, a transformation is canonical if the new variables satisfy Hamilton's equation. The case where $\partial Q/\partial q =1$ and $\partial p/\partial P = 1$ is a special case for the "identity transformation" when the generating function is of the type $F_2 = q_i P_i$ (see Goldstein's Classical Mechanics section 9.1, table 9.1). $\endgroup$
    – Rapha
    May 25, 2020 at 19:15
  • $\begingroup$ The problem statement was "consider a canonical transformation from $(q,p)$ to $(Q,P)$" which is where $\partial Q/\partial q=1$ and $\partial p/\partial P = 1$ are derived - which implies the $J=1$, i.e, the volume of phase space doesn't change under the given canonical transformation. $\endgroup$ May 25, 2020 at 19:36

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