Skip to main content

If Lambda Is Equilibrium; Viscosity Is the Friction of Approach to It

The idea that bulk viscosity 

could be an alternative to dark energy for a cosmological effective theory has been around for a while. For example, , 2011 Dark goo: bulk viscosity as an alternative to dark energy, or Hu, 2024 Viscous universe with cosmological constant,  or Khan, 2025 Spatial Phonons: A Phenomenological Viscous Dark Energy Model for DESI. Paul, 2025 Origin of bulk viscosity in cosmology and its thermodynamic implications, uses FLRW expansion gradients with apparent-horizon thermodynamics.
 
The real issue however, is: Is viscosity relaxation toward de Sitter, or dark energy itself? 
This is equivalent to saying:  Is $\Lambda$ geometry itself (equilibrium curvature, the standard view), or an emergent attractor from dissipation (NESS fixed point).    

 

If viscosity replaces $\Lambda$ (NESS fixed point)

Let

$$\varepsilon_\Lambda = \frac{3H^2c^2}{8\pi G}$$

be the de Sitter energy density.

If viscosity itself mimics dark energy, impose

$$p_{\rm vis}=-3\zeta H=-\varepsilon_\Lambda.$$

Then

$$\boxed{\zeta_{\Lambda} = \frac{Hc^2}{8\pi G}.}$$

This is the viscous-dark-energy value.

It gives exact de Sitter at the background level:

$$H=\text{constant}, \qquad p_{\rm vis}=-\varepsilon_\Lambda.$$

So this idea says:

$$\boxed{\text{viscosity replaces }\Lambda.}$$

But then exact de Sitter is not strict equilibrium. It is a steady dissipative state.

Mean-free-path derivation

Paul's kinetic picture makes the scale transparent.

Take layers separated by a mean free path $$\lambda_m$$ The Hubble flow is

$$v(d)=Hd.$$

Momentum transport gives

$$p_{\rm vis} = -\frac{2\varepsilon\lambda_m H}{c}.$$

Comparing with

$$p_{\rm vis}=-3\zeta H$$

gives

$$\boxed{\zeta = \frac{2\varepsilon\lambda_m}{3c}.}$$

Now choose a horizon-scale mean free path,

$$\lambda_m = \frac{c}{2H}.$$

Using

$$\varepsilon_\Lambda = \frac{3H^2c^2}{8\pi G},$$

we get

$$\zeta = \frac{2}{3c}\frac{3H^2c^2}{8\pi G}\frac{c}{2H} = \boxed{\frac{Hc^2}{8\pi G}.}$$

So full horizon-scale transport gives the dark-energy viscosity scale.

Viscosity as quasi-de Sitter relaxation

Now impose instead:

$$\boxed{\text{exact de Sitter is equilibrium.}}$$

Then

$$\dot H=0, \qquad R_A=\frac{c}{H}=\text{constant}, \qquad \dot S_A=0.$$

Define

$$\epsilon_H = -\frac{\dot H}{H^2}.$$

The apparent-horizon entropy is

$$S_A = \frac{\pi k_Bc^5}{G\hbar H^2}.$$

Since $$S_A\propto H^{-2}$$,

$$\dot S_A = 2\epsilon_HHS_A.$$

Now use the bulk-viscous entropy production form

$$\dot S_i = \frac{\zeta_{\rm qdS}\theta^2V_A}{T_A},$$

where

$$\theta=3H, \qquad V_A=\frac{4\pi c^3}{3H^3}, \qquad T_A=\frac{\hbar H}{2\pi k_B}.$$

Demand

$$\dot S_i=\dot S_A.$$

Solving gives

$$\boxed{\zeta_{\rm qdS} = \frac{\epsilon_HHc^2}{12\pi G}.}$$

Equivalently,

$$\boxed{\zeta_{\rm qdS} = -\frac{c^2}{12\pi G}\frac{\dot H}{H} = \frac{c^2}{12\pi G}\frac{\dot R_A}{R_A}.}$$

So exact de Sitter gives

$$\epsilon_H=0 \quad\Rightarrow\quad \zeta_{\rm qdS}=0.$$

This idea (the standard) says:

$$\boxed{\Lambda \text{ is equilibrium curvature; viscosity is relaxation.}}$$

Mean-free-path version of quasi-dS

The general kinetic result was

$$\zeta = \frac{2\varepsilon\lambda}{3c}.$$

To reproduce

$$\zeta_{\rm qdS} = \frac{\epsilon_HHc^2}{12\pi G},$$

with

$$\varepsilon = \frac{3H^2c^2}{8\pi G},$$

we need

$$\boxed{\lambda_{\rm tr} = \frac{\epsilon_Hc}{3H}.}$$

So the two closures differ by the effective transport length:

$$\boxed{\lambda_\Lambda = \frac{c}{2H}}$$

for viscosity-as-dark-energy, but

$$\boxed{\lambda_{\rm qdS} = \frac{\epsilon_Hc}{3H}}$$

for quasi-de Sitter relaxation.

Thus the horizon still sets the scale, but only the non-equilibrium fraction participates in irreversible transport.

Relation between the two viscosities

The two results are related by

$$\boxed{\zeta_{\rm qdS} = \frac{2}{3}\epsilon_H\zeta_\Lambda.}$$

So:

$$\zeta_\Lambda = \frac{Hc^2}{8\pi G}$$

is full horizon-scale transport.

Whereas:

$$\zeta_{\rm qdS} = \frac{\epsilon_HHc^2}{12\pi G}$$

is suppressed by departure from de Sitter.

Pressure caveat

For the dark-energy closure, one intentionally sets

$$p_{\rm vis} = -3\zeta_\Lambda H = -\varepsilon_\Lambda.$$

That is the model.

For the quasi-dS closure, however, $$\zeta_{\rm qdS}$$ is best read as a horizon-fluid entropy-production coefficient, not automatically as the full FLRW pressure viscosity.

If one writes

$$\Pi=-3\zeta_{\rm qdS}H,$$

then

$$\Pi = -\frac{2}{3}\epsilon_H\varepsilon.$$

Adding this to

$$p_{\rm eq}=-\varepsilon$$

would give

$$p_{\rm eff} = -\varepsilon-\frac{2}{3}\epsilon_H\varepsilon,$$

which is phantom-like for $$\epsilon_H>0$$.

But ordinary non-phantom quasi-de Sitter requires

$$w_{\rm eff} = -1+\frac{2}{3}\epsilon_H.$$

So:

$$\boxed{\zeta_{\rm qdS} \text{ is geometric/horizon-fluid viscosity, not automatically the total cosmic pressure.}}$$

Wrap

Paul supplies:

$$\boxed{\text{FLRW expansion } \Rightarrow \text{ viscosity } \Rightarrow \text{ horizon entropy transfer.}}$$

There are then two ideas.

Viscosity replaces dark energy

$$\boxed{\zeta_\Lambda = \frac{Hc^2}{8\pi G}, \qquad \lambda=\frac{c}{2H}.}$$

This makes de Sitter a steady dissipative state.

Viscosity is quasi-dS relaxation (more likely) 

$$\boxed{\zeta_{\rm qdS} = \frac{\epsilon_HHc^2}{12\pi G}, \qquad \lambda_{\rm tr} = \frac{\epsilon_Hc}{3H}.}$$

This makes exact de Sitter equilibrium. So:

$$\boxed{\text{full horizon mean-free path } \Rightarrow \text{viscosity mimics }\Lambda.}$$

$$\boxed{\text{non-equilibrium horizon fraction } \Rightarrow \text{viscosity measures relaxation to de Sitter.}}$$

Or, in slogan form:

$$\boxed{\Lambda \text{ is equilibrium curvature; viscosity is the friction of approach to it.}}$$

Comments

Post a Comment

Popular posts from this blog

Blurring the horizon - the quantum width of the cosmic event horizon

A  paper by Zurek applied a random walk argument to a black hole horizon. Credit, Zurek, 2021 Zurek  ( Snowmass 2021 White Paper: Observational Signatures of Quantum Gravity )  called this a blurring of the horizon — a fuzzy, or uncertain horizon — and went through derivations supporting the idea that this length scale is the quantum uncertainty in the position of the black hole horizon: a dynamic quantum width of an event horizon. This is a concept which fundamentally applies to the Universe's own Cosmic Event Horizon (CEH). The Bekenstein-Hawking entropy gives the number of quantum degrees of freedom that can fluctuate. Below, we step out our own cosmic de Sitter derivation of the random walk argument. To do this, let $l_{\Lambda}$ represent the generalised de Sitter horizon scale. Due to the holographic UV/IR correspondence, this scale manifests dually: at the fundamental microscopic limit as $l_{UV} = 2L_p$ (the gravitational/casual limit, aka the Schwarzschild radi...
Post a Comment
Read more

The Cosmic Strange Metal

      Strange metals, quantum spin liquids, and SYK-like systems share a striking transport pattern: no quasiparticles, strong collective dynamics, Planckian relaxation, near-minimal viscosity, and maximal chaos. Their characteristic data are $$\frac{\eta}{s}=\frac{\hbar}{4\pi k_B}, \qquad \lambda_L=\frac{2\pi k_BT}{\hbar}, \qquad \tau_P=\frac{\hbar}{k_BT}.$$ The claim is not that the three-dimensional de Sitter bulk is literally a strange metal. The sharper claim is: $$\boxed{\text{The de Sitter stretched horizon belongs to the same transport universality class as a Planckian strange metal.}}$$ The correspondence applies to the horizon membrane, not to bulk spacetime. Membrane Paradigm and the KSS Value In the membrane paradigm, an event horizon behaves for exterior observers as a stretched viscous membrane with transport coefficients fixed by Einstein gravity. This does not require an assumed AdS/CFT dual. For de Sitter, $$\ell_\Lambda=\frac{c}{H}, \qquad T_{dS...
Post a Comment
Read more

Our cosmic event horizon on a string

A Cosmic Stringy Adventure ! As we previously discussed, our spacetime characterised by a positive cosmological constant $\Lambda$. The natural bounds are then a minimal  ultraviolet (UV) length $l_{UV} = 2L_P$ and an infrared (IR) cosmological horizon $l_{\Lambda}$.  This dual-boundary spacetime enforces a fundamental Compton–gravitational duality . Every geometric scale $r$ carries two natural mass definitions: $$m_C(r) = \frac{\hbar}{rc}, \qquad m_G(r) = \frac{c^2}{4G}\ r$$ The product of these masses, $m_C \ m_G = M_P^2/4$, is scale-independent. They intersect exclusively at the UV boundary $r = l_{UV}$, defining a maximal local force in GR: $F_{max} = c^4 / 4G$. At the opposite extreme, the Compton mass evaluated at the IR horizon yields the fundamental  spectral gap  (not a particle) of the universe: $m_s = \hbar / (l_{\Lambda} c)$.  In this post, to explore how energy propagates through this dual-scale geometry, we model the mass gap $m_s$ as a null-ener...
Post a Comment
Read more