This maximum free molecule size can easily be found from Ideal gas law.

$22.414 ~{\rm dm^3/mol}$, at $0 ~^\circ{\rm C}$ and 1 atmosphere.
This means $2.68678\times 10^{25}~{\rm mol/m^3}$ which means that a 1 m row of molecules in these conditions is $\sqrt[3]{2.68678\times 10^{25}}= 299509596.1 \textrm{ molecule/m}$

This alouds to calculate the diameter of a single molecule; $$ d=\frac{1\textrm{ molecule}}{299509596.1 \textrm{ molecule/m}}=3.3 \times 10^{-9} ~{\rm m}$$

When this diameter of a sphere, is used to calculate the area momentum of inertia, which is just a geometrical property. The equation for circular shape in $z$-direction is

$$I_z=\frac{\pi}{2}r^4$$ The meaning of "$z$-direction" can be understood as perpendicular to circular-plane;Second area moment of circle

Calculating this gives $$I_z=\frac{\pi}{2}r^4=\frac{\pi}{2}\frac{d^4}{2^4}==\frac{\pi\cdot (3.3 \times 10^{-9})^4}{32}=1.164 \times 10^{-35}\textrm{ m}^4$$

And as this area momentum of inertia is used to calculate Deflections, which have a pure exponential shape. (The own mass of structure is neglected, with own mass the form is Catenary.) The inverse of such an exponential shape is natural logarithm. So multiplying this result with $e^4$ gives;

$$h=e^4\frac{\pi}{2}r^4=6.356\times 10^{-34}\textrm{ m}^4$$

Question; Can Planck's constant be derived from the maximum free Molecule size? Means; Is this logic valid? (Compared to ie. Numerology)

  • $\begingroup$ Related; physics.stackexchange.com/questions/284717/… and the full background of my thoughts; researchgate.net/publication/… $\endgroup$
    – Jokela
    Oct 12, 2016 at 7:47
  • $\begingroup$ I'm completely mystified as to how a question this poor got upvoted. $\endgroup$ Oct 12, 2016 at 13:17
  • $\begingroup$ As the good answer of Pisanty reall points out the importancy of units. I want to make a clear note, how the value "molecule/m" is actually the speed of light, but using it's unit m/s instead of the molecule/m would give a wrong result. So I divided this part of the path out of this question. I am also aware that the unit of Planck is ML^2/T, but I do have conceptual underpinnings to claim that M=L^3 and T=L which produces L^5/L = L^4 as a unit. But It's ofcourse possible that this whole idea is wrong. Yet already Minkowski unified length and time to 4-D spacetime. $\endgroup$
    – Jokela
    Oct 12, 2016 at 14:30

1 Answer 1


Can Planck's constant be derived from [insert quantity with dimensions other than action or $\text{action}^\text{power}$]?


Period. Can't do it.

You can only have an invariant relationship (i.e. one where the numbers are equal in every unit system) between quantities of the same dimensionality. If you start with a dimensionful quantity $X,$ then the only way to get a quantity with different dimensions is to take its powers, $X^n$, but that's about it. If none of the powers of $X$ matches the dimensions of action, there is no invariant operation you can perform on $X$ that will make it equal to $\hbar$.

In other words, your "solution" completely ignores units. This is something that should cause an automatic fail in physics tests from middle school onwards.

In particular, for example:

  • The size of [insert whatever system you want] is a length, and it knows nothing about time; it would be exactly the same if our clocks ran faster or slower. This makes any length incapable of addressing the Planck constant, whose dimensionality of $[ML^2/T]$ includes a factor of time, and would therefore change in that scenario.
  • Similarly, to take, say, the speed of light, it knows nothing of mass (it would have the same value if our reference masses were lighter or heavier), so again it cannot be used on its own to say anything about the value of Planck's constant.

For your purposes, then, your two options are (1) using a single constant whose dimensions are action or a power of it (which are in short supply), or (2) using multiple constants.

However, I give you essentially zero chances of actually coming up with something meaningful this way. As an example, the orbital angular momentum $L$ of the Earth about the Sun is $$L=2.661\times 10^{40}\:\mathrm{J\:s}\approx 2.523\times 10^{74}\:\hbar,$$ or in other words $$ L \approx \frac{\text{Milky Way galactic radius}}{\text{radius of a caesium atom}} \frac{\text{frequency of Lyman alpha line}}{\text{typical human respiration rate}} \: \hbar $$ Coincidence? I mean, what are the chances? ... well, about 100%.

Similarly, unless you actually have conceptual underpinnings for whatever numerical coincidence you want to highlight, you're essentially being this guy. (Note however, that the guy in that comic is actually much better off than your attempts here and in your previous question - his units actually check out.)

  • $\begingroup$ Thanks. -Units-, Yes.... I suddenly remember whilst doing my engineer studies some 22 years ago, I ended up fighting with my physics teacher, because I calculated the results with some equations he didn't recognize. And thus he reduced my score, though the result was correct. -As I had just calculated it straight forward with units. So I asked if I would get full score by only giving him the answer without showing my way to get there. He said "YES, but then you take a risk of getting zero points if the answer is not correct." I answered only "That's not your problem,,," But thanks.-Really! $\endgroup$
    – Jokela
    Oct 12, 2016 at 13:21
  • $\begingroup$ To issue; -The invariant relationship.- If we reduce the c=1 then this would make any powers of c to remain "1". Ofcourse. but what it comes to geometrical properties, the unit doesnt matter. you can get same pi in inches or meters. or what ever. So if we say the "c=1" then "h=5.3601" And this is still only math which comes from geometry, as "pi" comes from geometry only. $\hbar$ would be 0.853 $\endgroup$
    – Jokela
    Oct 12, 2016 at 13:26
  • 1
    $\begingroup$ I'm not sure what your point is with that anecdote, but it seems I have failed to impress upon you just how basic a mistake you've made, and just how off-topic further questions along this line would be. $\endgroup$ Oct 12, 2016 at 13:28
  • $\begingroup$ Regarding your second comment, that just shows a complete lack of understanding of how dimensionless systems works. You will get nothing more than a tautology that way, and if you can't see why then you should stop, or take it elsewhere. $\endgroup$ Oct 12, 2016 at 13:32
  • $\begingroup$ No. You hit perfectly to point. THE UNITS are of great importance. It has been the hardest work for me to sort out the relationships between Mass time and lenght. It has been my way to think physics. through basic units; Length, Time and Mass. Now I am in a middle of communication problem, and I don't really know what to say. But your answer; it's great. Honestly. It's like Robert Millikan's critic. -very valuable indeed. "underpinnings" -yes, $\endgroup$
    – Jokela
    Oct 12, 2016 at 13:40

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