11.7: Jacobian proof of Liouville’s Theorem

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After this rather long detour into Jacobian theory, recall we are trying to establish that the volume of a region in phase space is unaffected by a canonical transformation, we need to prove that

\[\int d Q_{1} \ldots d Q_{s} d P_{1} \ldots d P_{s}=\int d q_{1} \ldots d q_{s} d p_{1} \ldots d p_{s}\]

and that means we need to show that the Jacobian

\[D=\frac{\partial\left(Q_{1}, \ldots, Q_{s}, P_{1}, \ldots, P_{s}\right)}{\partial\left(q_{1}, \ldots, q_{s}, p_{1}, \ldots, p_{s}\right)}=1\]

Using the theorems above about the inverse of a Jacobian and the chain rule product,

\[D=\frac{\partial\left(Q_{1}, \ldots, Q_{s}, P_{1}, \ldots, P_{s}\right)}{\partial\left(q_{1}, \ldots, q_{s}, P_{1}, \ldots, P_{s}\right)} / \frac{\partial\left(q_{1}, \ldots, q_{s}, p_{1}, \ldots, p_{s}\right)}{\partial\left(q_{1}, \ldots, q_{s}, P_{1}, \ldots, P_{s}\right)}\]

Now invoking the rule that if the same variables appear in both numerator and denominator, they can be cancelled,

\[D=\left\{\frac{\partial\left(Q_{1}, \ldots, Q_{s}\right)}{\partial\left(q_{1}, \ldots, q_{s}\right)}\right\}_{P=\text { constant }} /\left\{\frac{\partial\left(p_{1}, \ldots, p_{s}\right)}{\partial\left(P_{1}, \ldots, P_{s}\right)}\right\}_{q=\text { constant }}\]

Up to this point, the equations are valid for any nonsingular transformation—but to prove the numerator and denominator are equal in this expression requires that the equation be canonical, that is, be given by a generating function, as explained earlier.

Recall now the properties of the generating function \(\Phi(q, P, t)\)

\[d \Phi(q, P, t)=d\left(F+\sum P_{i} Q_{i}\right)=\sum p_{i} d q_{i}+\sum Q_{i} d P_{i}+\left(H^{\prime}-H\right) d t\]

from which

\[p_{i}=\partial \Phi(q, P, t) / \partial q_{i}, \quad Q_{i}=\partial \Phi(q, P, t) / \partial P_{i}, \quad H^{\prime}=H+\partial \Phi(q, P, t) / \partial t\]

In the expression for the Jacobian \(D\), the \(i\),\(k\) element of the numerator is \(\partial Q_{i} / \partial q_{k}\).

In terms of the generating function \(\Phi(q, P) \text { this element is } \partial^{2} \Phi / \partial q_{k} \partial P_{i}\).

Exactly the same procedure for the denominator gives the \(i\),\(k\) element to be \(\partial P_{i} / \partial p_{k}=\partial^{2} \Phi / \partial q_{i} \partial P_{k}\)

In other words, the two determinants are the same (rows and columns are switched, but that doesn’t affect the value of a determinant). This means D=1, and Liouville’s theorem is proved.