What determines the duration of inflation? In standard cosmology, the answer usually depends on the dynamics of the inflaton field. But there is a different possibility: perhaps inflation is constrained not by field dynamics alone, but by geometry . Here is the postulate: Inflation generates macroscopic quantum modes and freezes them onto the causal boundary (the Hubble horizon). Inflation must end when the total number of accumulated modes exhausts the geometric capacity of that boundary. If that idea is correct, then the duration of inflation follows from a simple bulk-to-boundary counting argument, and the result is $$N_e = 6\pi^2 \approx 59.2.$$ 1. The Geometry of the Bulk and the Boundary Consider an observer in de Sitter space with constant Hubble parameter $H$. The causal boundary is the Hubble horizon, with radius $$R_H = \frac{1}{H}.$$ This defines two fundamental geometric quantities. The physical volume inside the horizon is $$V_H = \frac{4\pi}{3} R_...
In the static patch of de Sitter spacetime , the cosmological horizon behaves thermodynamically in ways closely analogous to a physical interface. One can assign it entropy, temperature, and even an effective surface tension . Remarkably, the familiar Young–Laplace pressure relation from surface physics appears naturally at the horizon. Scope. Everything below is formulated in the static patch of de Sitter spacetime — the causally accessible region for a single inertial observer, covered by static coordinates in which the metric $$ds^2 = -\left(1-\frac{r^2}{L^2}\right)c^2,dt^2 + \left(1-\frac{r^2}{L^2}\right)^{-1}dr^2 + r^2 d\Omega^2$$ is manifestly time-independent. The static patch admits a timelike Killing vector $\partial_t$, and it is this Killing vector that defines the notions of energy, temperature, and thermodynamic equilibrium used throughout. Global de Sitter spacetime has no timelike Killing vector; the thermodynamic framework does not extend be...