# Prove a change of coordinates is symplectic

I have to find the value $$\alpha\in\mathbb{R}$$ such that the following change of coordinates is symplectic:

$$\varphi(p,q)\rightarrow (P,Q)$$

where

$$Q = q^2 + \alpha\sqrt{q^2+p}$$

$$P = q + \sqrt{q^2+p}$$

My first approach to solve it was to directly verify for which $$\alpha$$ it holds $$dp\wedge dq = dP \wedge dQ$$ and if comes out $$\alpha=2$$, which should be right.

But on some notes, I found that to prove that this holds it could be even imposed that the determinant of the Jacobian of the transformation $$\varphi$$ is equal 1.

Is this related to Liouville volume preserving theorem? I really don't get why we should have that this last verification implies that $$\varphi$$ is a symplectomorphism.

I one sentence: You are in 2D, the symplectic form is the volume form.

In more words: Jacobian determinant measures (local) volume stretching, so is being 1 at every point means volume is preserved, but in 2D this is the same as being a symplectomorphism. You can see it very explicitly: one one hand, Jacobian determinant is $$\frac{\partial{P}}{\partial{p}} \frac{\partial{Q}}{\partial{q}} - \frac{\partial{Q}}{\partial{p}} \frac{\partial{P}}{\partial{q}}$$; on the other hand $$dP= \frac{\partial{P}}{\partial{p}} dp+ \frac{\partial{P}}{\partial{q}} dq$$ and $$dQ= \frac{\partial{Q}}{\partial{p}} dp+ \frac{\partial{Q}}{\partial{q}} dq$$ so $$dP\wedge dQ= (\frac{\partial{P}}{\partial{p}} \frac{\partial{Q}}{\partial{q}} - \frac{\partial{Q}}{\partial{p}} \frac{\partial{P}}{\partial{q}}) dp \wedge dq$$. Hence $$dP\wedge dQ=dp\wedge dq$$ is the same as Jacobian 1.

In higher dimensions, the volume form is $$\omega^n$$. Preserving $$\omega$$ (symplectomorphism) implies preserving volume $$\omega^n$$. This is Liouville theorem. However, when $$n\geq 2$$ (i.e. in 4D and up) the converse is not true.

To begin just a definition: A symplectic linear structure on $$\mathbb{R}^{2n}$$ is a non-degenerate bilinear skew symmetric 2-form. This form is called the skew-scalar product such that $$[a,b]=-[b,a]$$.

Now a transformation $$S:\mathbb{R}^{2n} \rightarrow \mathbb{R}^{2n}$$ of the symplectic space is symplectic iff it preserves the differential 2-Form. In your case where $$n=1$$ this means that $$dp \wedge dq$$ is preserved and maps to $$dP \wedge dQ$$. Which is equivelent to $$[Sa,Sb]=[a,b]$$

Moreover a transformation is symplectic iff the Matrix Representation of S satisfies $$S^T\Omega S=\Omega$$ where $$\Omega$$ is the block matrix

$$\begin{bmatrix} 0 & -I \\ I & 0 \end{bmatrix}$$

and I is the identity matrix; becasue: $$[\Omega Sa,Sb]=[\Omega a,b]=[S^T \Omega Sa,Sb]$$.

Now then you could find the matrix representation of the transformation $$\phi$$ and then compute $$\phi ^T \Omega \phi$$, let it equal to $$\Omega$$ and solve for $$\alpha$$.

Furthermore I recommend the chapter on Symplectic Manifolds from V.I Arnold's book on Classical Mechanics.