*2.2. Horizon Entropy Force*

The entropy of the black hole horizon is the (dimensionless) Bekenstein–Hawking entropy:

$$S\_{BH} = \frac{A\_{BH}}{4\lambda\_P^2} = \frac{\pi R\_S^2 c^3}{G\hbar} \tag{5}$$

with *λ<sup>P</sup>* the Planck length, and its corresponding black-body radiation temperature is the Hawking temperature:

$$T^{BH} = k\_B T\_H = \hbar c^3 / 8\pi GM \tag{6}$$

here given in energy units. This yields trivially the Bekenstein–Hawking energy *EBH* = *TBHSBH* = 1 <sup>2</sup> *Mc*<sup>2</sup> of the horizon, just half the black hole energy. For a classical black hole, the horizon has no width, suggesting an infinite gradient when crossing it. In order to obtain the force, the relevant distance

*Entropy* **2019**, *21*, 716

taken for the gradient is the diameter 2*RS* of the black hole, yielding for the modulus of the outward directed entropy force:

$$F\_S^{BH} \sim \frac{T^{BH} S\_{BH}}{2R\_S} = \frac{c^4}{8G} = \frac{\mathcal{F}\_S}{4} \approx 1.5 \times 10^{43} \text{ N} \tag{7}$$

a quarter of the Schwarzschild constant. A more precise calculation would correct for the numerical factor, which, however, is of order *O*(1). Formally, the Schwarzschild constant, and thus also the entropy force at the horizon, is in fact a very strong force. Its presence, if real, at the horizon is rather surprising. It suggests that the physics of what happens inside the black hole is not really well enough understood.

Forming the ratio of the gravitational force *FBH <sup>G</sup>* <sup>=</sup> <sup>−</sup>*GmM*/*R*<sup>2</sup> *<sup>S</sup>* any particle of mass *m* experiences when approaching and touching the black hole horizon to this horizon entropy force, one obtains:

$$\left| F\_G^{BH} / F\_S^{BH} \right| \sim m / M \tag{8}$$

For any massive black hole and any normal mass particle, this is a small number. A light particle *m M* will barely overcome this repulsion when hitting the horizon. To overcome it, it requires the collision of two black holes of nearly equal mass *M*<sup>1</sup> ∼ *M*2, which would make the ratio *m*/*M* → *M*1/*M*<sup>2</sup> ≈ 1. Collisions of such nearly-equal mass black holes have only recently been detected by the spectacular observation of gravitational waves.

At its horizon, the entropy force of a massive black hole *M m* compensates by far for the black hole's gravitational attraction on *m*. This is a consequence of the enormous sharpness of the entropy gradient at the horizon where the entropy is restricted to the surface of the horizon only, whose width is not precisely given, but as generally assumed, is of the order of a few Planck lengths *λ<sup>P</sup>* only. This force is remarkable only at the horizon itself when the mass *m* gets into contact with the horizon. It will not be susceptible at some larger finite distance.

This follows from the fact that there is no known classical entropy field that would allow the entropy force to extend a distance ahead of the black hole into the surrounding space and is conjectured from the complete absence of any radial dependence of the force outside the horizon. In this picture, the entropy gradient is felt only locally across the horizon of the black hole when the particle touches it, an instant that is never seen or experienced by an external observer for whom the time the particle approaches the horizon stretches out to infinity. The particle, however, does in fact experience the presence of the black hole and, assuming that it remains intact having survived the enormous attraction during its inward spiraling motion, in its proper frame at proper time, really touches the horizon and wants to cross it. Shortly before this instant, however, the horizon entropy comes into play and stops the particle.

The gravitational force on the particle of mass *m* (assuming it retains its mass till reaching the horizon, which is certainly not the case) would overcome the entropic force still only at a small fraction of the radius given by:

$$\frac{\Delta r}{R\_S} \approx \left(\frac{M\_\odot}{M}\right)^{\frac{1}{2}} \left(\frac{m}{M\_\odot}\right)^{\frac{1}{2}} \approx 3 \times 10^{-28} \left(\frac{M\_\odot}{M}\right)^{\frac{1}{2}}\tag{9}$$

where on the right, we assumed a proton. For instance, this distance for a proton and a *<sup>M</sup>* <sup>=</sup> 108*M* massive black hole is of the order of only <sup>Δ</sup>*<sup>r</sup>* <sup>∼</sup> <sup>10</sup>−<sup>23</sup> m, deep inside the submicroscopic domain, though ten orders of magnitude larger than the Planck length. If it applies, then it would cause accumulation of matter in a film of roughly this width only.

The classical picture does not inform about the microscopic physics going on when this happens. Elucidating the real physics requires a quantum electrodynamic calculation for instance along the paths drawn by Hawking when calculating the black-body black hole radiation. Referring to Hawking's results implies that the horizon will be surrounded by a dilute and thin radial dust film of newly- and

continuously-created virtual particles, which sustain and support the weak Hawking radiation when tunneling into reality. The radial extension of this dust film implies a softening of the radial entropy gradient corresponding to a finite radial extension of the action region of the entropy force. It would be this radial domain where the light particle in its inward spiraling motion becomes trapped and retarded and is ultimately stopped and prevented from entering the interior of the black hole.
