${\displaystyle {\begin{aligned}{\frac {dp_{x}}{du_{y}}}&={\frac {d}{du_{y}}}{\frac {d\epsilon }{d{\dot {q}}_{x}}}={\frac {d}{du_{y}}}\sum _{r=1}^{r=n}{\frac {d\epsilon }{du_{r}}}{\frac {du_{r}}{d{\dot {q}}_{x}}}\\&=\sum _{r=1}^{r=n}{\bigg (}{\frac {d^{2}\epsilon }{du_{y}\,du_{r}}}{\frac {du_{r}}{d{\dot {q}}_{x}}}{\bigg )}={\frac {d}{d{\dot {q}}_{x}}}{\frac {d\epsilon }{du_{y}}}={\frac {du_{y}}{d{\dot {q}}_{x}}}\end{aligned}}}$

(137)

Therefore

${\displaystyle {\frac {d(p,\ldots p_{n})}{d(u,\ldots u_{n})}}={\frac {d(u,\ldots u_{n})}{d({\dot {q}},\ldots {\dot {q}}_{n})}}}$

(138)

and

${\displaystyle {\bigg (}{\frac {d(p,\ldots p_{n})}{d(u,\ldots u_{n})}}{\bigg )}^{2}={\bigg (}{\frac {d(u,\ldots u_{n})}{d({\dot {q}},\ldots {\dot {q}}_{n})}}{\bigg )}^{2}={\frac {d(p,\ldots p_{n})}{d({\dot {q}},\ldots {\dot {q}}_{n})}}}$

(139)

These determinants are all functions of the ${\displaystyle q}$'s alone.^[1] The last is evidently the Hessian or determinant formed of the second differential coefficients of the kinetic energy with respect to ${\displaystyle {\dot {q}}_{1},\ldots {\dot {q}}_{n}}$. We shall denote it by ${\displaystyle \Delta _{\dot {q}}}$. The reciprocal determinant

${\displaystyle {\frac {d({\dot {q}}_{1}\ldots {\dot {q}}_{n})}{d(p_{1}\ldots p_{n})}},}$

which is the Hessian of the kinetic energy regarded as function of the ${\displaystyle p}$'s, we shall denote by ${\displaystyle \Delta _{p}}$.

If we set

${\displaystyle {\begin{aligned}e^{-{\frac {\psi _{p}}{\Theta }}}&=\int \limits _{-\infty }^{+\infty }\ldots \int \limits _{-\infty }^{+\infty }e^{-{\frac {\epsilon _{p}}{\Theta }}}\,\Delta _{p}{}^{\frac {1}{2}}\,dp,\ldots dp_{n}\\&=\int \limits _{-\infty }^{+\infty }\ldots \int \limits _{-\infty }^{+\infty }e^{\frac {-u_{1}{}^{2}\ldots -u_{n}{}^{2}}{2\Theta }}du_{1}\ldots du_{n}=(2\pi \Theta )^{\frac {n}{2}},\end{aligned}}}$

(140)

and

${\displaystyle \psi _{q}=\psi -\psi _{p},}$

(141)

1. ↑ It will be observed that the proof of (137) depends on the linear relation between the ${\displaystyle u}$'s and ${\displaystyle {\dot {q}}}$'s, which makes ${\displaystyle {\frac {du_{r}}{d{\dot {q}}_{x}}}}$ constant with respect to the differentiations here considered. Compare note on p. 12.