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Superfluid, the Vacuum, and MOND

      We previously  discussed Verlinde's connection of the MOND acceleration scale to the entropy of de Sitter space. A different route appears in superfluid dark matter (a 2026 review is here ), where baryons interact with a phonon field whose nonlinear dynamics generate a MOND-like force. The 2026 review (Section 7.1) introduces a scale $\Lambda_{SF}$ and notes it needs to be in the order of meV to account for the MOND scale. So, we take this idea one step further and directly identify the characteristic phonon scale with the vacuum-energy scale! This converts the cosmological constant into a galactic acceleration scale and reproduces the baryonic Tully–Fisher relation. Throughout, set $c=\hbar=1$ and $M_{\rm Pl}^{-2}=8\pi G$, where $M_{\rm Pl}$ is the reduced Planck mass. Vacuum Scale Let $\Lambda_{\rm cos}$ denote the cosmological constant. Its vacuum-energy density is $\rho_\Lambda=M_{\rm Pl}^2\Lambda_{\rm cos}$. For a pure de Sitter universe, $H_\Lambda^2=\...
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Superfluid, the Vacuum, and MOND

 

 

 

We previously discussed Verlinde's connection of the MOND acceleration scale to the entropy of de Sitter space. A different route appears in superfluid dark matter (a 2026 review is here), where baryons interact with a phonon field whose nonlinear dynamics generate a MOND-like force.

The 2026 review (Section 7.1) introduces a scale $\Lambda_{SF}$ and notes it needs to be in the order of meV to account for the MOND scale. So, we take this idea one step further and directly identify the characteristic phonon scale with the vacuum-energy scale! This converts the cosmological constant into a galactic acceleration scale and reproduces the baryonic Tully–Fisher relation.

Throughout, set $c=\hbar=1$ and $M_{\rm Pl}^{-2}=8\pi G$, where $M_{\rm Pl}$ is the reduced Planck mass.

Vacuum Scale

Let $\Lambda_{\rm cos}$ denote the cosmological constant. Its vacuum-energy density is $\rho_\Lambda=M_{\rm Pl}^2\Lambda_{\rm cos}$. For a pure de Sitter universe, $H_\Lambda^2=\Lambda_{\rm cos}/3$, so $\rho_\Lambda=3M_{\rm Pl}^2H_\Lambda^2$. This is the Friedmann equation for a universe containing only vacuum energy.

Define the vacuum-energy scale $M_\Lambda$ by $M_\Lambda^4=\rho_\Lambda$. Then

$$\boxed{ M_\Lambda = 3^{1/4}\sqrt{M_{\rm Pl}H_\Lambda} = \left(M_{\rm Pl}^2\Lambda_{\rm cos}\right)^{1/4} }.$$

Using the unreduced Planck mass $M_p=\sqrt{8\pi},M_{\rm Pl}$ and the horizon energy scale $m_s=H_\Lambda=L^{-1}$, this becomes the seesaw relation

$$\boxed{ M_\Lambda = \left(\frac{3}{8\pi}\right)^{1/4} \sqrt{M_pm_s} }.$$

Numerically, $M_\Lambda \simeq 2.3\times10^{-12}\ {\rm GeV} \simeq 2.3\ {\rm meV}$. Thus the small vacuum-energy scale is the geometric mean of the very large Planck scale and the very small de Sitter horizon scale.

Superfluid Identification

In superfluid dark matter, baryons can excite a phonon field $\pi$. In the static, large-gradient regime, take

$$\mathcal L_\pi = -\frac{2\Lambda_{\rm SF}}{3}\left|\boldsymbol\nabla\pi\right|^3 -\alpha\frac{\Lambda_{\rm SF}}{M_{\rm Pl}}\pi\rho_B,$$

where $\Lambda_{\rm SF}$ is the phonon energy scale, $\alpha$ is a dimensionless coupling, and $\rho_B$ is the baryonic density. The central identification is

$$\boxed{ \Lambda_{\rm SF}^4=\rho_\Lambda }, \qquad \boxed{ \Lambda_{\rm SF}=M_\Lambda }.$$

It follows that $\Lambda_{\rm SF}^2 = M_{\rm Pl}\sqrt{\Lambda_{\rm cos}} = \sqrt{3},M_{\rm Pl}H_\Lambda$. The phonon scale is therefore fixed directly by the vacuum-energy density.

Phonon Acceleration Scale

Define the physical potential felt by baryons as $\Phi_\pi = \alpha\frac{\Lambda_{\rm SF}}{M_{\rm Pl}}\pi$. In terms of this potential, the phonon action becomes

$$\mathcal L_\pi = -\frac{\left|\boldsymbol\nabla\Phi_\pi\right|^3}{12\pi G a_{\rm SF}} -\rho_B\Phi_\pi,$$

where

$$\boxed{ a_{\rm SF} = \frac{\alpha^3\Lambda_{\rm SF}^2}{M_{\rm Pl}} }.$$

Using the vacuum identification gives $a_{\rm SF} = \alpha^3\sqrt{\Lambda_{\rm cos}} = \sqrt{3},\alpha^3H_\Lambda$ in natural units. Restoring $c$,

$$\boxed{ a_{\rm SF} = \alpha^3c^2\sqrt{\Lambda_{\rm cos}} = \sqrt{3},\alpha^3cH_\Lambda }.$$

The Planck mass cancels. Vacuum curvature therefore fixes the acceleration scale, apart from the dimensionless coupling $\alpha$.

MOND Force Law

Varying the phonon action gives

$$\boldsymbol\nabla\cdot \left( \frac{\left|\boldsymbol\nabla\Phi_\pi\right|}{a_{\rm SF}} \boldsymbol\nabla\Phi_\pi \right) = 4\pi G\rho_B.$$

For spherical symmetry, define $g_\pi=|\Phi_\pi'|$ and $M_B(r) = 4\pi\int_0^r\rho_B(r')r'^2,dr'$. Integrating the field equation gives $r^2g_\pi^2 = Ga_{\rm SF}M_B(r)$. The Newtonian baryonic acceleration is $g_B=GM_B(r)/r^2$. Therefore,

$$\boxed{ g_\pi^2=a_{\rm SF}g_B }.$$

The cubic phonon action converts the vacuum-set acceleration into the deep-MOND square-root force law. Outside an isolated mass $M$,

$$\boxed{ g_\pi = \frac{\sqrt{GMa_{\rm SF}}}{r} }.$$

For a circular orbit dominated by the phonon force, $v_f^2/r=g_\pi$, so

$$\boxed{ v_f^4=GMa_{\rm SF} }.$$

This is the baryonic Tully–Fisher relation.

Numerics

The observed MOND acceleration determines the coupling:

$$\alpha = \left( \frac{a_0^{\rm obs}}{c^2\sqrt{\Lambda_{\rm cos}}} \right)^{1/3}.$$

Using $a_0^{\rm obs} \simeq 1.2\times10^{-10}\ {\rm m,s^{-2}}$ and $\Lambda_{\rm cos} \simeq 1.09\times10^{-52}\ {\rm m^{-2}}$, gives $\alpha\simeq0.504$, close to $\alpha=1/2$. If this value is exact, then

$$\boxed{ a_0 = \frac{c^2}{8}\sqrt{\Lambda_{\rm cos}} = \frac{\sqrt{3}}{8}cH_\Lambda \simeq 1.17\times10^{-10}\ {\rm m,s^{-2}} }.$$

The corresponding vacuum-energy relation, in natural units, is

$$\boxed{ \frac{a_0^2}{8\pi G} = \frac{\rho_\Lambda}{64} }.$$

The observed MOND scale is therefore reproduced by a simple dimensionless coupling of order unity.

Physics

The complete chain is

$$\boxed{ \Lambda_{\rm cos} \longrightarrow \rho_\Lambda \longrightarrow M_\Lambda=\Lambda_{\rm SF} \longrightarrow a_0 \longrightarrow g_\pi^2=a_0g_B \longrightarrow v_f^4=GMa_0 }.$$

Clifford spacetime algebra makes the geometry explicit: the superfluid selects a time direction, leaving three-dimensional space in which scale invariance selects the cubic phonon action. This action produces the MOND force law, while vacuum curvature supplies its acceleration scale.

The two additional assumptions are

$$\boxed{ \Lambda_{\rm SF}^4=\rho_\Lambda, \qquad \alpha=\frac{1}{2} }.$$

Together, these relations reproduce the observed MOND acceleration with no remaining dimensional parameter.

A complete theory must explain why the phonon scale equals the vacuum-energy scale, why the baryonic coupling equals $1/2$, and how the required superfluid phase forms and remains stable inside galaxies. It must also recover acceptable cosmological behaviour outside the galactic superfluid regime.

The result is therefore a sketch of an idea: Clifford spacetime algebra explains why three-dimensional geometry selects the force law, while vacuum curvature fixes its acceleration scale.

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