I have been curious to find what could be the significance of Planck force? It is calculated by the formula $c^4/G = 1.21031359\times 10^{44} \, \mathrm{N}$, where $c$ is the speed of light and $G$ is the gravitational constant.
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5 Answers
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There is no particular physical significance; it's just a unit. Of course, in any system where such a large force is exerted, our current theories should not be accurate, and a quantum theory of gravity or some as-yet-unknown theory would be needed to accurately describe its behavior.
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3$\begingroup$ I wish people made it clearer there is no physical significance to any Planck unit. $\endgroup$ Commented Jun 1, 2016 at 8:44
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$\begingroup$ Quantum gravity isn't needed to explain a constant that has no dependence on $\hbar$. @OrangeDog, it's not true that there is no physical significance to any Planck unit. The Planck velocity, $c$, has a clear meaning in special relativity, and the Planck angular momentum, $\hbar$, has a clear meaning in QM. $c^4/G$ has an interpretation as a maximum gravitational force in classical GR, as Johannes and Alex Meiburg's answers say. It's only the units depending on both $G$ and $\hbar$ whose interpretation is unclear. $\endgroup$– benrgCommented Sep 26, 2022 at 21:49
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When describing the formation of a black hole (or the merger of two equally-sized black holes) in a Newtonian gravity framework, Planck-scale gravitational forces of the order $c^4/G$ enter into the description. This is independent of the mass of the black hole. Such can easily be seen by modeling an infalling spherical shell of dust with mass $M$ under the influence of a Newtonian gravitational force.
Of course, we know that black hole formation and black hole mergers require a general relativistic description, and studying these phenomena in a Newtonian framework means stretching Newtonian gravity beyond its range of applicability. So treat the statement "Planck-scale forces are the forces that occur in the formation of black holes" as nothing more than a rough intuitive scaling argument.
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Building on Johannes's answer, if you have two equal-sized spherical black holes of Schwarzschild radius $r$, at distance $\alpha r$ from each other, then the force is $F_{\textrm{planck}}/\alpha^2$.
This is only accurate when $\alpha \gg 1$, as otherwise spacetime is curved in more complicated ways, and even defining 'force' is not immediately straightforward. But you can very loosely think of a Planck force as: the force two black holes would have on each other if their centers were placed on the event horizon's of each other. Less loosely, it's the force they exert, when distance is measured in terms of their size.
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Planck's constant is a force associated with each cycle of a photon. For instance a photon with the 500 nm wavelength actually oscillates at a frequency more than 600 trillion times per second. Each time it oscillates, Planck's Constant is applied.
answered Apr 7, 2016 at 17:37
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2$\begingroup$ What do you mean when you say "Planck's constant is applied?" How do you "apply" a constant? $\endgroup$– AsherCommented Apr 7, 2016 at 21:33
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I think, the physical signifcance of the Planck force comes fromthe formula: looks like it's the force needed to accelerate the Planck mass to the speed of light in the Planck time right?
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$\begingroup$ That is the definition of plank force, not the significance thereof. $\endgroup$– GonencCommented Jul 9, 2015 at 11:36
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5$\begingroup$ There does not exist a force capable of accelerating the Planck mass to the speed of light in any finite amount of time $\endgroup$– JimCommented Jul 9, 2015 at 13:14
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$\begingroup$ You might be interested in the help center page on merging multiple accounts: physics.stackexchange.com/help/merging-accounts. $\endgroup$ Commented Jul 9, 2015 at 14:59