Is the Planck force a truly "Planck unit"?

Question

The Planck force appears to be defined as the ratio of the Planck energy to the Planck distance, $ F_P = E_P/l_P $ that can be rewritten as $$ F_P = \frac{ E_P }{ l_P} = \frac{ c^4 }{ G }. $$

Isn't it rather odd that it doesn't involve Planck's constant? Is there some other acceptable interpretation of the force?

According to Wikipedia: "In particle physics and physical cosmology, Planck units are a set of units of measurement defined exclusively in terms of four universal physical constants, in such a manner that these physical constants take on the numerical value of 1 when expressed in terms of these units. Originally proposed in 1899 by German physicist Max Planck, these units are a system of natural units because their definition is based on properties of nature, more specifically the properties of free space, rather than a choice of prototype object. They are relevant in research on unified theories such as quantum gravity."

Answer 1

The Planck force can be interpreted with general relativity only, without need for quantum mechanics.

It is roughly the force between two black holes of mass $M$, located at each other's event horizon, i.e. at a distance given by the Schwarzschild radius $R=\frac{2GM}{c^2}$. Of course Newtonian mechanics is not applicable here anymore. But we can still use it to get the order of magnitude for the gravitational force between the two black holes: $$F=\frac{GM^2}{R^2}=\frac{c^4}{4G}$$

Answer 2

Yes, there is some another interesting interpretation for $c^4/G$. In all static spherically symmetric perfect fluid solutions of Einstein field equations the pressure behavior near the central initial event horizon reads \begin{equation} p(r,\alpha_{c})=\frac{4}{\kappa}\cdot \frac{1}{r^2}-\frac{4}{3}~ \rho(0,\alpha_{c}) \cdot c^2+\mathcal{O}(r^2), \end{equation} with $\kappa\equiv 8\pi G/c^4$ and $\alpha_{c}$ the critical compactness parameter ($r_{S}/R \le 8/9$) for a given solution. Remarkable, whereas the pressure near the singularity diverges, the force generated by it is finite and does not dependent of $\alpha_{c}$, thus is equal for all solutions. This force is inversely proportional to Einstein's gravitational constant and equal twice the Planck force \begin{equation} F(0,\alpha_{c})\equiv\lim_{r \to~0} p~4\pi\cdot r^{2}=\frac{16\pi}{\kappa}\equiv\frac{2c^{4}}{G}~. \end{equation} The appearance of such a universal force upholds the idea of the maximal tension principle conjectured independently by Gibbons [1] and Schiller [2], which is also related to the dimension of the $3+1$ space-time [3] and is equivalent to the holographic principle [4]. It is the force necessary to create an initial event horizon, or causally speaking, to generate a "crack" in the spacetime continuum. The different factor in the above derivation and that in referenced papers, here 2 and there 1/4, is due to the fact the former has been achieved by applying Newton's gravitation theory.

[1] https://arxiv.org/abs/hep-th/0210109

[2] https://arxiv.org/abs/physics/0607090v1

[3] https://arxiv.org/abs/2005.06809v1

[4] https://arxiv.org/abs/1507.02839

Answer 3

Isn't it rather odd that it doesn't involve Planck's constant? Is there some other acceptable interpretation of the force?

It depends on how you write it. Here the Planck constant magically appears:

\begin{equation} F_{P} = \frac{c^3 M_{P}^2}{\hbar} \end{equation}