\underbrace{\tfrac{1}{2}\,m_\text{eff}\,v_\text{rot,inner}^2}_{\substack{\text{the vortex's full spin energy}\\[1pt] \text{at the inner speed}}} \;=\; E \;=\; \underbrace{m_e\,c^2}_{\substack{\text{Einstein's rest energy}\\[1pt] \text{at the rim speed}}}
\alpha_{mf} = \tan^2\theta_W = 0.3008
the visible mass fraction
m_e = \alpha_{mf}\,m_\text{eff} \qquad\qquad {v_\text{rot,inner} = \sqrt{2\alpha_{mf}}\;c = 0.776\,c}
\tfrac{1}{2}\,m_\text{eff}\big(2\alpha_{mf}\,c^2\big) \;=\; \big(\alpha_{mf}\,m_\text{eff}\big)\,c^2 \;=\; \boxed{m_e\,c^2}
Q = -\frac{\hbar^2}{2m}\,\frac{\nabla^2 R}{R}\,, \qquad R = \sqrt{\rho}
\nabla \cdot \mathbf{F}_{ns} \;\propto\; \nabla^2\!\left(\frac{\nabla^2 R}{R}\right)
HVBK mutual friction in steady state — integrating back gives Q as the reaction force
\text{\footnotesize vortex diffusivity}\quad D = \frac{\kappa_q}{4\pi\alpha_{mf}} \quad \text{\footnotesize set equal to} \quad \frac{\hbar}{2m}\,:
\hbar = 2mD \;\;\Longrightarrow\;\; \mathbf{m_\text{eff}\,\alpha_{mf} = m}
v_\text{ebb} = \sqrt{\frac{2GM}{r}} \;\;\Rightarrow\;\; ds^2 = -c^2 dt^2 + \big(dr - v_\text{ebb}\,dt\big)^2 + r^2 d\Omega^2
Schwarzschild, exactly — space as a river - the rare leak through the stack of boundary layers
f_\text{cross} = \frac{4\pi G}{v_L} \approx 10^{-15}
weight is the ebbing leak, the weak nature of gravity, one boundary crossing per quadrillion
a_0 = c\sqrt{G\rho_\text{DM}} = 1.16\times10^{-10}\ \text{m/s}^2
the MOND acceleration scale from the substrate density
\sqrt{\rho_\Lambda/\rho_\text{Pl}} \;\approx\; \big(\ell_\text{Pl}/\xi\big)^2 \;\approx\; 10^{-61.5}
the cosmological-constant number and weak gravity are the same geometric ratio
{\color{#184d85} c} = \frac{\hbar}{m_1\,\xi}, \qquad \xi = \frac{\hbar}{m_1\,{c}}
Volovik’s quasiparticle speed, equivalently \xi is the vacuum particle’s Compton wavelength
\xi = \left(\frac{\hbar}{\rho_\text{DM}\,c}\right)^{\!1/4} f^{\,1/4} \approx 97\ \mu\text{m}, \qquad f = \frac{4\pi}{K\sqrt{2}} = 0.5666
Planck’s dark-matter density using the vortex to lattice cell occupancy fraction finds the lattice cell size
m_1 c^2 = \hbar c/\xi \approx 2.0\ \text{meV}
the cell size as the vacuum particle’s Compton wavelength shows its visible mass
\nu = m_\text{eff}/m_1 = 8.35\times10^8