Unit restoration and scale checks for the two-rotor KAM statement: the scaling identities at $\mathcal{I}=E_0=1$, the physical angular frequencies, the exclusion of rational frequency labels, and the equal-rescaling invariance

Dimensions: the physical Hamiltonian has $[{\mathrm{\Pi }}_i^2\mathrm{/}\mathcal{I}]=\mathrm{k}{\mathrm{g}}^2{\mathrm{m}}^4{\mathrm{s}}^{-2}\mathrm{/}(\mathrm{k}\mathrm{g}{\mathrm{m}}^2)=\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-2}=\mathrm{J}$ and $[E_0\epsilon ]=1\cdot \mathrm{J}$ with $\epsilon$ dimensionless; ${\omega }_0=\sqrt{E_0\mathrm{/}\mathcal{I}}$ has $[{\omega }_0]=\sqrt{\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-2}\mathrm{/}(\mathrm{k}\mathrm{g}{\mathrm{m}}^2)}={\mathrm{s}}^{-1}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). With the declared values, $J_0=1\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, $t_0=1\mathrm{s}$, ${\omega }_0=1{\mathrm{s}}^{-1}$.

Scaling of the equations, first equation: with ${\mathrm{\Pi }}_i=J_0{I}_i$ and $t=t_{0}s$ the chain rule gives ${\dot{\theta }}_i=t_0^{-1}{\theta }_i^{\mathrm{\prime }}$ and ${\dot{\mathrm{\Pi }}}_i=t_0^{-1}{J}_0{I}_i^{\mathrm{\prime }}$; substituting into ${\dot{\theta }}_i={\mathrm{\Pi }}_i\mathrm{/}\mathcal{I}$ and using $J_0{t}_0=\mathcal{I}$ yields $t_0^{-1}{\theta }_i^{\mathrm{\prime }}=J_0{I}_i\mathrm{/}\mathcal{I}=I_i\mathrm{/}{t}_0$, that is the dimensionless equation ${\theta }_i^{\mathrm{\prime }}=I_i$.

Rational labels are non-Diophantine: let $\nu =(p_1\mathrm{/}q,p_2\mathrm{/}q)\in C^{\ast }$ with integers $p_1,p_2$ and $q\ge 1$. If $p_1=p_2=0$, choose $k=(1,0)$; otherwise set $k:=(p_2,-p_1)\mathrm{\ne }0$; in either case $k\cdot \nu =0<\alpha \mid k{\mid }_1^{-\tau }$, so $\nu \mathrm{\notin }{\mathrm{\Omega }}_{\alpha ,\tau }$ by [L2]. Hence the Diophantine set of the theorem contains no rational point of the cube; the positive measure it retains is carried by irrational labels.

Scaling of the equations, second equation: substituting into ${\dot{\mathrm{\Pi }}}_i=E_0\epsilon \sin {\theta }_i\cos {\theta }_{3-i}$ gives $t_0^{-1}{J}_0{I}_i^{\mathrm{\prime }}=E_0\epsilon \sin {\theta }_i\cos {\theta }_{3-i}$; since $J_0\mathrm{/}{t}_0=J_0{E}_0\mathrm{/}{J}_0=E_0$, this is equivalent to $I_i^{\mathrm{\prime }}=\epsilon \sin {\theta }_i\cos {\theta }_{3-i}$, the dimensionless Hamilton equation of the potential $\epsilon \cos {\vartheta }_1\cos {\vartheta }_2$. With the declared values $J_0\mathrm{/}{t}_0=1\mathrm{J}$ and $E_0=1\mathrm{J}$ the substitution is an identity.

Physical frequencies: by [L1] a surviving torus with label $\nu$ carries $\phi (t)={\phi }_0+{\omega }_0\nu t$ in its torus parameter; the physical angle is its embedded angular component; with ${\omega }_0=1{\mathrm{s}}^{-1}$ the components ${\omega }_0{\nu }_i={\nu }_i{\mathrm{s}}^{-1}$ are the physical angular frequencies, of dimension ${\mathrm{s}}^{-1}$ as in step 1.1.

Equal rescaling: replace $(\mathcal{I},E_0)$ by $(\lambda \mathcal{I},\lambda E_0)$, $\lambda >0$; then ${\omega }_0=\sqrt{E_0\mathrm{/}\mathcal{I}}$ is unchanged, $J_0=\sqrt{\mathcal{I}{E}_0}$ is multiplied by $\lambda$ and $t_0=J_0\mathrm{/}{E}_0$ is unchanged, so the dimensionless Hamiltonian, the frequency cube, the Diophantine set and the smallness condition are all unchanged, while the physical momenta ${\mathrm{\Pi }}_i=J_0{I}_i$ are rescaled by $\lambda$. The constant $c$ of the theorem is not numerically evaluated in this check; only the exact scaling behaviour is asserted.