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The Black-Hole 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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The Black-Hole 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:  

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 $\Theta$ as

$$P_\Theta=\frac{c^3A}{8\pi G}.$$

The closed action of one cycle is therefore

$$J_H=\oint P_\Theta d\Theta =\frac{c^3A}{4G}.$$

Semiclassical quantisation gives

$$\Delta J_H=2\pi\hbar.$$

Hence

$$\frac{c^3}{4G}\Delta A=2\pi\hbar,$$

and therefore

$$\boxed{\Delta A =8\pi\frac{G\hbar}{c^3} =8\pi l_{\mathrm{Pl}}^2}.$$

The canonical relation alone does not prove discreteness. The additional step is semiclassical quantisation of the closed horizon cycle.

2. Black-Hole Resonance Frequencies

A damped resonance has complex frequency

$$\omega=\omega_R+i\omega_I.$$

Its effective oscillator frequency is

$$\Omega=\sqrt{\omega_R^2+\omega_I^2}.$$

For highly damped Schwarzschild resonances,

$$8\pi M\omega_n = \ln3+2\pi i\left(n+\frac12\right)+O(n^{-1/2}).$$

It follows that

$$\Omega_n-\Omega_{n-1} \longrightarrow \frac{1}{4M} =\kappa.$$

The correspondence principle gives

$$\Delta M=\hbar\kappa.$$

The black-hole first law gives

$$\Delta M=\frac{\kappa}{8\pi}\Delta A.$$

Combining these equations yields

$$\boxed{\Delta A=8\pi\hbar =8\pi l_{\mathrm{Pl}}^2}.$$

The alternative result $4\ln3 l_{\mathrm{Pl}}^2$  (Hod) uses the absolute real part of the resonance frequency rather than the difference between adjacent highly excited frequencies. The real part also depends on the perturbing field, whereas the leading adjacent full-frequency spacing approaches the intrinsic horizon frequency.

3. Minimum Absorption

A particle of mass $\mu$, lowered to proper distance $b$ above the horizon, carries redshifted energy

$$E_\infty\simeq\mu\kappa b.$$

The first law gives

$$\Delta A = \frac{8\pi}{\kappa}E_\infty \simeq8\pi\mu b.$$

Quantum localization requires approximately

$$b\gtrsim\frac{\hbar}{\mu}.$$

Therefore,

$$\Delta A_{\min} \gtrsim8\pi\hbar = 8\pi l_{\mathrm{Pl}}^2.$$

This argument is heuristic because localisation and backreaction can change order-one factors, but it independently supports the same result.

4. Entropy Check

Black-hole entropy is

$$S_{\mathrm{BH}} = \frac{k_BA}{4l_{\mathrm{Pl}}^2}.$$

The area spacing therefore gives

$$\Delta S_{\mathrm{BH}} = \frac{k_B\Delta A}{4l_{\mathrm{Pl}}^2} = 2\pi k_B.$$

This agrees with the horizon-action relation

$$J_H=\frac{\hbar}{k_B}S_{\mathrm{BH}},$$

because one action quantum, $\Delta J_H=2\pi\hbar$, corresponds to $\Delta S_{\mathrm{BH}}=2\pi k_B$.

Conclusion

The independent routes give $8\pi l_{\mathrm{Pl}}^2$

  1. closed horizon action
  2.  adjacent damped-resonance spacing 
  3. minimum quantum absorption

So, the asymptotic semiclassical spectrum is

$$\boxed{ A_n = 8\pi l_{\mathrm{Pl}}^2(n+\mu)+o(n) }$$

and therefore

$$\boxed{ \lim_{n\to\infty} \left(A_{n+1}-A_n\right) = 8\pi l_{\mathrm{Pl}}^2. }$$

All this means $8\pi l_{\mathrm{Pl}}^2$ is the asymptotic semiclassical black-hole area quantum.

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