Lemma (1) Let P have a rest mass of $m$ and be in an excited state characterized by $\widetilde{\rho}$ the mean radius of its phase-components. Remember that the density of P is defined by

$\begin{align} \varrho \equiv \frac{ m c^{2} }{ \widetilde{\rho} } \end{align}$

So the condition for being a Newtonian particle also implies that the mean radius is constrained like

(1)
\begin{align} \widetilde{\rho} \ll mc^{2} / \, k_{\sf{F}} \end{align}

Lemma (2) Recall that $\rho^{\mathcal{S}}$ and $\rho^{\mathcal{A}}$ are the norms of the radius vectors of the phase-components of P. And these numbers are non-negative, they cannot cancel each other in the sum that defines the mean radius

$\widetilde{\rho} \equiv \left( \rho^{\mathcal{S}} + \rho^{\mathcal{A}} \right) \, / \, 2$

And so the Newtonian condition (1) given above necessarily implies that $\rho^{\mathcal{S}}$ also has an upper bound set by the mass

(2)
\begin{align} \rho^{\mathcal{S}} \ll mc^{2} / \, k_{\sf{F}} \end{align}

## Enthalpy of a Newtonian Particle

For Newtonian particles, the energy of the rest mass is almost the same as the absolute value of the enthalpy $H$ because the definition of mass

$m \equiv c^{-2} \sqrt{ \: H^{2} - W^{2} \ \vphantom{ {\Sigma^{2}}^{2} } }$

can be rearranged to give

$m^{2} c^{4} + W^{2} = H^{2}$

And recall that for an excited particle

$W = 2 W ^{\mathcal{S}} =2 k_{\sf{F}} \, \rho^{\mathcal{S}}$

where $\rho^{\mathcal{S}}$ is the radius of a symmetric component. Then

$m^{2} c^{4} + 4 \left( k_{\sf{F}} \, {\rho^{\mathcal{S}}} \right)^{2} = H^{2}$

But as shown in Lemma (2) above

$k_{\sf{F}} \, \rho^{\mathcal{S}} \ll mc^{2}$

So the negligible term can be dropped, and for Newtonian particles

$m^{2} c^{4} \simeq H^{2}$

Taking the square root gives