# Is the Planck force a truly "Planck unit"?

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."

• Related question: What is the significance of Planck force? May 10, 2022 at 10:20
• You might prefer to call it a geometrized unit.
– J.G.
May 10, 2022 at 11:27
• Note that the Planck velocity is just $c$ :) May 10, 2022 at 15:31
• Amended the question to show that Planck units do not necessarily have to involve Planck's constant.
– jim
May 11, 2022 at 14:17

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: $$F=\frac{GM^2}{R^2}=\frac{c^4}{4G}$$

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 $$$$p(r,\alpha_{c})=\frac{4}{\kappa}\cdot \frac{1}{r^2}-\frac{4}{3}~ \rho(0,\alpha_{c}) \cdot c^2+\mathcal{O}(r^2),$$$$ with $$\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}$$, thus is equal for all solutions. This force is inversely proportional to Einstein's gravitational constant and equal twice the Planck force $$$$F(0,\alpha_{c})\equiv\lim_{r \to~0} p~4\pi\cdot r^{2}=\frac{16\pi}{\kappa}\equiv\frac{2c^{4}}{G}~.$$$$ The appearance of such a universal force upholds the idea of the maximal tension principle conjectured independently by Gibbons [1] and Schiller [2], which is also related to the dimension of the $$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.

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:

$$$$F_{P} = \frac{c^3 M_{P}^2}{\hbar}$$$$