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We have seen the reverse statement: Lioville's Theorem states that canonical transformations preserve volume (and orientation as well). Is the reverse true? If I demand a map from the phase space to the phase space to preserve volume, is it necessarilly a canoncial transformation? I couldn't come up with a counter example, that's why I ask.

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In dimension $2n>2$ they are not equivalent since (for time-independent transformations) canonical is equivalent to $$\sum_{k=1}^n dq^k\wedge dp_k = \sum_{k=1}^n dQ^k\wedge dP_k\tag{1}$$ whereas conservation of oriented volume means $$dq^1\wedge \cdots \wedge dq^n \wedge dp_1\cdots \wedge dp_n = dQ^1\wedge \cdots \wedge dQ^n \wedge dP_1\cdots \wedge dP_n\:.\tag{2}$$ The former is much more restrictive. The latter only requires that the Jacobian matrix has determinant $1$. Already with $4\times 4$ matrices there are easy conterexamples.

$Q^1 = aq^1$,

$Q^2= b q^2$,

$P_1 = b^{-1} p_1$,

$P_2 = a^{-1}p_2$

where coordinates are over $\mathbb R^4$ and with constants $a,b>0$ satisfying $a\neq b$. This transformation satisfies (2) but not (1).

Instead, for $2n=2$, (1) and (2) are evidently equivalent.

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  • $\begingroup$ Is your condition (1) sufficient or do you need to include all the Liouville invariants? $\endgroup$ – ZeroTheHero Dec 18 '17 at 16:11
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    $\begingroup$ Condition (1) states that the symplectic form is invariant under the coordinate transformation. For time-independent transformations this is equivalent to canonicity of the transformation. $\endgroup$ – Valter Moretti Dec 18 '17 at 16:26
  • $\begingroup$ I'm sorry if my query was unclear. Your Eq.(1) is one condition on one invariant. There are other invariants such as $\sum_{i\ne k}dQ^i\wedge dP^i\wedge dQ^k\wedge dP^k$ which ought to be preserved by canonical transformations. Is the one condition contained in your (1) necessary and sufficient, or do we also need to show that $\sum_{i\ne k}dQ^i\wedge dP^i\wedge dQ^k\wedge dP^k$ and all other higher order Liouville invariants are separately preserved? $\endgroup$ – ZeroTheHero Dec 18 '17 at 20:01
  • $\begingroup$ I do not understand: a transformation independent from time is canonical if and only if it satisfies (1). If canonical transformations preserve other forms, then, evidently, (1) is a sufficient condition for these further conservation results. $\endgroup$ – Valter Moretti Dec 18 '17 at 20:15
  • $\begingroup$ I did not know (1) was necessary and sufficient. I thought it was just necessary and that one also required the CT to preserve all the other Liouville invariants. It seems weird that a condition on only one invariant should be enough but I'll take your word for it, although I will investigate more. $\endgroup$ – ZeroTheHero Dec 18 '17 at 20:17
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  1. Counterexample: The transformation $$Q^1~=~2q^1 ,\qquad P_1~=~p_1,\qquad Q^2~=~\frac{1}{2}q^2 ,\qquad P_2~=~p_2 $$ preserves phase space volume & orientation, but is not a symplectomorphism.$^1$

  2. For 2D phase space, the canonical phase space volume form $$\Omega~=~\frac{1}{n!}\omega^{\wedge n}$$ is the symplectic 2-form $\omega$ itself, so that the orientation & volume preserving transformations are the symplectomorphisms.

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$^1$ Here we will assume that OP defines a canonical transformation (CT) as a symplectomorphism. Be aware that several non-equivalent definitions of CTs appear in the literature, cf. e.g. this Phys.SE post.

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It boils down to the structure of the Phase Space. You know that the Phase Space must have a two-differential form, which is represented by your Poisson Brackets. This two-form "keeps track" of your orientation thanks to its intrinsic skew-symmetric nature, and due to it being the external product of your canonical momentum and position variable, it defines "hypervolumes" in your Phase Space. Any space equipped with Poisson Brackets is a symplectic space and the Poisson Bracket will be a symplectic form. When you change variables, you want to keep this structure, hence you want to apply a symplectomorphism (a.k.a. a canonical transformation). It is demonstrable that the Jacobian determinant of a symplectomorphism is always $1$ so it does not deform volumes in the phase space. Due to the definition of symplectomorphism, it will conserve your orientation too, since it keeps the symplectic structure unchanged. So, is a transformation with $\left|\underline{J}\right|=1$ a symplectomorphism? In general, it isn't, since you don't know if it keeps the symplectic form on your space unchanged.

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