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The problem I'm supposed to solve is finding $Q$, such that $(p,q)\rightarrow(P,Q)$ is a canonical transformation. In this case $\mathcal{H}=\frac{p^{2}+q^{2}}{2}$ and the new hamiltonian $\mathcal{K}$ is $\mathcal{K}=P$.

This means $\dot{q}=p$ and $\dot{p}=-q$

Since $\mathcal{H}$ and $\mathcal{K}$ are time independent $\mathcal{H}=\mathcal{K}$ and $P=\frac{p^{2}+q^{2}}{2}$. Now I use a generating function of canonical transformations $F_{4}=F_{4}(p,P)$ so:

$\frac{\partial F_{4}}{\partial p}=-q\quad\quad\quad\mbox{and}\quad\quad\quad\frac{\partial F_{4}}{\partial P}=Q$

$P=\frac{p^{2}+q^{2}}{2}\quad\Rightarrow\quad q=\sqrt{2P-p^{2}}$

Then

\begin{equation} F_{4}=-\int\sqrt{2P-p^{2}}dp\quad\Rightarrow\quad Q=-\int \frac{\partial\sqrt{2P-p^{2}}}{\partial P}dp=-arcsin\left(\frac{p}{\sqrt{2P}}\right)=-arcsin\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right) \end{equation}

$\{Q,P\}= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=\frac{p}{p^{2}+q^{2}}p-\left(-\frac{q}{p^{2}+q^{2}}\right)q=1$.

Therefore this transformation is canonical. However I also tried to find $Q$ with the generating function $F_{1}=F_{1}(q,Q)$, where

\begin{equation} \frac{\partial F_{1}}{\partial Q}=-P\quad\quad\mbox{and}\quad\quad\frac{\partial F_{1}}{\partial q}=p \end{equation}

Then

\begin{equation} F_{1}=\int\frac{-p^{2}-q^{2}}{2}dQ\quad\Rightarrow\quad p=\int \frac{\partial\left(\frac{-p^{2}-q^{2}}{2}\right)}{\partial q}dQ=\int -qdQ=-qQ\quad\Rightarrow\quad Q=-\frac{p}{q} \end{equation}

This is very different with respect to the first $Q$ found, and $\{Q,P\}=\frac{p}{q^{2}}p+\frac{1}{q}q=\frac{p^{2}}{q^{2}}+1$ which can only be equal to 1 if $p=0$.

But if we assume this is a canonical transformation then $\dot{Q}=1$ and $\dot{P}=0$, and

\begin{equation} \dot{Q}=\frac{\partial Q}{\partial q}\dot{q}+\frac{\partial Q}{\partial p}\dot{p}=\frac{p^{2}}{q^{2}}+1=1\Rightarrow p=0 \end{equation}

I think the second result can't be possible, if $p=0$ then $Q=0$; so my question is why I could not obtain $Q$ with $F_{1}$, did I miss something?

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I am not that much familiar with Hamiltonian mechanics, but are you not supposed to write $F_1$ as a function of $q$ and $Q$ only? You need to replace $p$ in $F_1$ by a combination of $q$ and $Q$, which will obviously have a non-zero partial derivative with respect to $q$, thus changing your calculation.

I will be "cheating" since I will use the first result in the second part, but I don't know if there is a way to do it differently.

Since $Q = - \mathrm{arcsin}\left(\frac{p}{\sqrt{p^2+q^2}}\right)$, we can write that $\mathrm{sin}^2(-Q) = \frac{p^2}{p^2+q^2}$, or $p^2 = \frac{\mathrm{sin}^2(-Q)}{1 - \mathrm{sin}^2(-Q)} q^2$.

Thus:

$$F_{1}=\int\frac{-p^{2}-q^{2}}{2}dQ = \int\frac{-\frac{\mathrm{sin}^2(-Q)}{1 - \mathrm{sin}^2(-Q)}-1}{2} q^2dQ = \int -\frac{1}{2 \mathrm{cos}^2(-Q)} q^2 dQ$$

which leads to:

$p=\int \frac{\partial\left(-\frac{1}{2(1 - \mathrm{sin}^2(-Q))} q^2\right)}{\partial q}dQ = \int -\frac{q}{\mathrm{cos}^2(-Q)}dQ = \int q d(\mathrm{tan}(-Q)) = q \,\mathrm{tan}(-Q)$

or finally:

$Q = - \mathrm{arctan}\left(\frac{p}{q}\right) = - \mathrm{arcsin}\left( \frac{p}{\sqrt{p^2+q^2}}\right).$

The very last equality can be easily derived by remembering the fact that the tangent of an angle $\theta$ in a triangle can be expressed as the ratio of the opposite side length $p$ over the adjacent side length $q$, whereas the sine of the same angle is expressed as the ratio of $p$ over the hypotenuse length $\sqrt{p^2+q^2}$. But $\mathrm{arctan(tan}(\theta)) = \mathrm{arcsin(sin}(\theta)) = \theta$.

Of course, this would not be useful to derive the expression for $Q$, as in this solution I've used the expression of $Q$ from the first part of your answer to find the same expression in the end. This is merely a safety check that the equations on $F_1$ are correct. I don't know how if you can derive the same result using $F_1$ from scratch. The problem here is that you don't have a nice way to express $F_1$ using explicitely $q$ and $Q$ only.

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