On fluctuations of the gravitational potential?

Question

In the book "Gravitation", written by Charles W. Misner, Kip S. Thorne, and John A. Wheeler, in § 43.4, p.1192 "Fluctuations in geometry", and in § 44.2, p.1200, by analogy with electrodynamics, the following formula (43.29) is obtained for the fluctuations of the gravitational potential:

$$\Delta g\sim\frac{\ell_P}{L}\,\,\,\,\,\,\,\,\,(1)$$ Here $g$ is the gravitational potential, $\ell_P=1.6\times 10^{-33} cm$ is the so-called Planck length, $L$ is the region of dimension.

However, detailed analysis shows (see T. Regge, Nuovo Cim. 7, 215 (1958). ''Gravitational fields and quantum mechanics'') that the formula for the gravitational potential fluctuations should have the form: $$\Delta g\sim\frac{\ell^{\,2}_P}{L^2}\,\,\,\,\,\,\,\,\,(2)$$

This formula also follows from the Bohr-Rosenfeld uncertainty relation (see here https://arxiv.org/abs/gr-qc/9403008v2, chapter 5, p.12]): $$\Delta g\Delta L^2\ge\ell^{\,2}_P$$ Which of the formulas is correct: (1) or (2)?

Answer 1

The answer can be the following. The gravitational field performs zero-point oscillations, and the geometry associated with it also oscillates. The ratio of the circumference to the radius varies near the Euclidean value . The smaller the scale, the greater the deviations from the Euclidean geometry. Let us estimate the order of the wavelength of zero gravitational oscillations, at which the geometry becomes completely unlike the Euclidean geometry. The degree of deviation $\zeta$ of geometry from Euclidean geometry in the gravitational field is determined by the ratio of the gravitational potential $\varphi$ and the square of the speed of light $c$: $\zeta=\varphi/c^2$. When $\zeta\ll 1$, the geometry is close to Euclidean geometry; for $\zeta\sim 1$, all similarities disappear.The energy of the oscillation of scale $L$ is equal to $E=\hbar\nu\sim \hbar c/L$ (where $c/L$ is the order of the oscillation frequency). The gravitational potential created by the mass $m$, at this length is $\varphi=Gm/L$, where $G$ is the constant of universal gravitation. Instead of $m$, we must substitute a mass, which, according to Einstein's formula, corresponds to the energy $E$ (where $m=E/c^2$). We get $\varphi=GE/L\,c^2=G\hbar/L^2c$. Dividing this expression by $c^2$, we obtain the value of the deviation $\zeta=G\hbar/c^3L^2=\ell^2_P/L^2$. Equating $\zeta=1$, we find the length at which the Euclidean geometry is completely distorted. It is equal to Planck length $\ell_P=\sqrt{G\hbar/c^3}\approx 10^{-35}$m. Therefore deviations from Euclidean geometry ($g_{00}=1$) on the Planck scale are equal to: $g_{00}=1-\zeta\approx 1-\ell^2_P/L^2$. It follows that $\zeta=\Delta g\sim\ell^2_P/L^2$.