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The Horizon Quantum of Area

  Several independent semiclassical arguments converge on $$\boxed{\Delta A=8\pi l_{\mathrm{Pl}}^2}, \qquad l_{\mathrm{Pl}}^2=\frac{G\hbar}{c^3}$$ I got interested in this reading the 2018 paper  Bekenstein, I, and the quantum of black-hole surface area . Even though I disagree with Hod, its a great paper.  Honourable mentions also go to:   The Physical Interpretation of the Spectrum of Black Hole Quasinormal Modes , 2008  The Off-Shell Black Hole , 1994     On Black Hole Spectroscopy via Adiabatic Invariance , 2012 An Analytical Computation of Asymptotic Schwarzschild Quasinormal Frequencies , 2003  Asymptotic Black Hole Quasinormal Frequencies , 2003  1. Horizon Action Near any nonextremal horizon, imaginary time turns the normal geometry into a plane: $$ds_E^2\simeq d\rho^2+\rho^2d\Theta^2+\cdots.$$ Smoothness requires one horizon cycle: $$\oint d\Theta=2\pi.$$ The gravitational action identifies the momentum conjugate to $\The...
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A First Quantum Horizon and Cosmic Information

 

Consider a spherical Schwarzschild horizon of radius $R$. Its energy, temperature, and entropy are

$$E=\frac{c^{4}R}{2G},\qquad k_BT_H=\frac{\hbar c}{4\pi R},\qquad S=\frac{k_BA}{4L_P^2},$$

where $A=4\pi R^2$ and $L_P^2=\hbar G/c^3$.

1. Quantized Horizon

The horizon's thermal period defines an angular frequency

$$\omega_H=\frac{2\pi k_BT_H}{\hbar}=\frac{c}{2R}.$$

Treat the Euclidean horizon cycle as a periodic quantum motion and define its action by $dI=dE/\omega_H$. Since $dE=c^4dR/2G$, we obtain $dI=c^3RdR/G$, and therefore

$$I=\frac{c^3R^2}{2G} =\frac{\hbar R^2}{2L_P^2} =\frac{\hbar A}{8\pi L_P^2}.$$

Equivalently, using $dE=T_HdS$,

$$I=\frac{\hbar S}{2\pi k_B}.$$

Quantizing this periodic motion gives

$$I_n=\left(n+\frac12\right)\hbar, \qquad n=0,1,2,\ldots$$

where the half-unit is the ground-state contribution associated with the regular centre of the Euclidean horizon geometry. Hence

$$A_n=4\pi(2n+1)L_P^2, \qquad R_n=\sqrt{2n+1},L_P.$$

Thus the first nonzero horizon has

$$\boxed{I_0=\frac{\hbar}{2}} \qquad \boxed{R_0=L_P} \qquad \boxed{A_0=4\pi L_P^2}$$

and the area spacing is $\boxed{\Delta A=8\pi L_P^2}$.

2. Horizon Degree Count

Write the Schwarzschild energy as thermal equipartition: $E=\frac12N_Hk_BT_H$. Substitution gives

$$N_H=\frac{2E}{k_BT_H} =\frac{A}{L_P^2} =\frac{4S}{k_B} =8\pi\frac{I}{\hbar}.$$

The first horizon therefore has

$$\boxed{N_{H,0}=4\pi}.$$

This is an equipartition count, not a count of entropy units or microscopic states.

3. Cosmic Mode Count

For one physical field, the number of Fourier modes in comoving volume $V$ is $dN=V,d^3k/(2\pi)^3$. At Hubble crossing, $k=aH/c$, and the corresponding comoving Hubble volume is $V_H=4\pi/3k^3$. Using $d^3k=k^3,d\ln k,d\Omega$ and integrating over all directions gives

$$dN_c=\frac{2}{3\pi},d\ln(aH).$$

Therefore, over a monotonic finite crossing interval,

$$\boxed{ N_c=\frac{2}{3\pi} \left|\Delta\ln(aH)\right| }.$$

4. Matching Principle

The new physical proposal is: the complete first cosmic mode record saturates the capacity of the first nonzero horizon. Thus $N_c=N_{H,0}=4\pi$. It follows that

$$\frac{2}{3\pi} \left|\Delta\ln(aH)\right| =4\pi,$$

so

$$\boxed{ \left|\Delta\ln(aH)\right|=6\pi^2 }.$$

Equivalently,

$$ \frac{(aH){\max}}{(aH){\min}} =e^{6\pi^2} \approx5.22\times10^{25} $$

For nearly constant $H$, this corresponds to approximately $59.2$ expansion e-folds.

Result

$$ I_0=\frac{\hbar}{2}, \qquad R_0=L_P,\qquad N_c=N_{H,0}=4\pi,\qquad \left|\Delta\ln(aH)\right|=6\pi^2 $$

The derivation rests on two proposed extensions: quantising the periodic horizon action with a half-unit ground state, and matching the complete cosmic mode record to the capacity of that first horizon.

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