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Planck's constant, as derived to explain the curve of black-body radiation, makes intuitive sense to me. What I do not understand is the connection between this constant and why it should be used as a limit for the smallest possible length in which "length" retains a consistent meaning. Is there an intuitive thought process that can be followed to understand why Planck's constant from black-body spectral analysis should be used to also define the smallest unit of "length" and "time"?

Einstein's explanation of the photoelectric effect was no help to me. I just want to understand a deeper meaning than a simple definitional usage.

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Planck's constant is a different concept from Planck length. Planck's constant is exactly as you describe. Planck's length is derived by considering a combination of G (the gravitational constant), h (Planck's constant), and c (the speed of light) which gives units of length. Since these are the constants that determine the scale on which a theory of quantum gravity becomes important, it is this length that is the smallest length that we can make sense of without a theory of quantum gravity.

Edit: any theory that combines quantum mechanics and special relativity must have planck's constant and the speed of light in it. Planck's constant sets the scale at which quantum effects become important, and the speed of light sets the universal speed limit. After we have these two constants, the only fundamental constant that would have a bearing on a quantum theory of gravity is the gravitational constant, which determines how strong the gravitational force is. This is why it is these three constants that are used to derive the Planck length.

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  • $\begingroup$ Could you elaborate more on the process of why Planck's constant is used to derive the length, as opposed to some other arbitrary number in its place though? I guess I just don't understand why anyone would be like, "Hey, this equation could use a little bit of Planck's constant!" $\endgroup$
    – aiwyn
    Commented Mar 6, 2017 at 22:54
  • $\begingroup$ aiwyn, is the new edit helpful? $\endgroup$
    – Travis
    Commented Mar 7, 2017 at 1:28
  • $\begingroup$ It provides more information. I think all I can do at this point is to continue studying the chronology of thought processes leading to the need for Planck's constant to come up with the quantized length. Marked as answer. $\endgroup$
    – aiwyn
    Commented Mar 7, 2017 at 3:45
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    $\begingroup$ @aiwyn The Planck length may not have anything to do with quantization of space. There is no evidence at the moment that space is quantized in the first place. The Planck length is just a quantity with the dimensions of length that we can create from the fundamental constants; you could multiply it by any dimensionless constant (like $\alpha$, for example, or any power of $\alpha$) and it would still have the same units. It's impossible to know what dimensionless factors we're missing in front of the quantity, so its value may have nothing to do with space quantization, even if it exists. $\endgroup$ Commented Jun 15, 2018 at 13:09
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John P. Ralston might have an answer for you, he proposes (https://arxiv.org/abs/1203.5557) "a modern approach where Planck’s constant is absent: it is unobservable except as a constant of human convention... In the new approach Planck’s constant is tied to macroscopic conventions of Newtonian origin, which are dispensable."

Case in point, the quantization condition: $$ [x, p] = [x, -i\hbar\partial/\partial x] = i\hbar. $$ "Introducing $\hbar$ made the first time in history where multiplying a math identity by the same constant on both sides was reported to make a new physical principle". It comes from $$ [x, -i\partial/\partial x] = i, $$ which is the trivial identity it appears to be.

Exhibit 2, the path integrand of massless Dirac spinor: $$ e^{\frac{i}{\hbar}S_{Dirac}} = e^{\frac{i}{\hbar}\int i\hbar \bar{\psi}\not{\partial}\psi} = e^{i\int i\bar{\psi}\not{\partial}\psi}. $$ If the two $\hbar$s net out, why do we bother to introduce $\hbar$ in the first place? As for the mass term in Dirac action, $\hbar$ can be simply absorbed into the redefinition of mass $m$.

Exhibit 3, the fine-structure constant: $$ \alpha = \frac{e^2}{\hbar c}. $$ Measurements of $\alpha$, $c$, and $e$ seem to get you $\hbar$. The catch is that the whole schema hinges on the convention of unit of electron charge $e$ and measurement thereof. If you do proper rescaling of gauge field $A$ in the QED path integrand, only the fine-structure constant $\alpha$ remains. Electron charge $e$ drops out completely and you don't need $e$ anywhere in the Lagrangian. We don't sustain any loss of information if we abandon the notion of $e$ and only invoke $\alpha$ in theory and in experiment (thus foregoing $\hbar$ as well ). Planck constant $\hbar$ is only an arbitrary intermediate step which is subject to human convention.

An added note. When you do rescaling of certain field and then a parameter changes size or pops up at a different Lagrangian term, it’s a change of physics unit. However, if two parameters in a theory collapse into one parameter after rescaling ($\hbar$, $e$ -> $\alpha$), you might suspect there must be something fishy and redundant, which is nothing but the Planck’s constant $\hbar$. Given the historical role $\hbar$ played, physicists have a certain emotional attachment to it. And the widely circulated folklore surrounding $\hbar$ lends it a mystical aura of importance. It’s not surprising that the view expressed here in this thread would receive plenty of down votes. But if you pause for moment and think twice about it, you will be rewarded handsomely by the insight gained.

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1203.5557 $\endgroup$
    – Qmechanic
    Commented Jun 14, 2018 at 19:58
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Commented Jun 15, 2018 at 15:36

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