I found this interesting numerical coincidence between Newtonian constant of gravitation $G$ and other fundamental physical constants (use numerical values in SI-units and forget units):

$$G = {1 \over c} \sqrt[21]{\alpha^2\mu_0c\hbar} = (6.674\ 067\ 595 \pm0.000\ 000\ 009)\times10^{-11}.$$

According to CODATA 2014, experimental value of $$G = (6.674\ 08 \pm0.000\ 31)\times10^{-11}.$$

I have left out units intentionally because formula above makes no sense if you consider units: it is relatively easy to show (based on dimensions only) that $G$ cannot be constructed from combination of other fundamental constants $\hbar$, $c$, $\mu_0$ and $\epsilon_0$, thus this is just a numerical coincidence.

I wonder when experimental accuracy in measuring $G$ is by factor of 10 or 100 better in order to see how many digits are actually correct in this equation. Someone may already argue there are too many correct digits this not being just a coincidence. But what kind of bizarre theory leads in to such a weird equation like this?


  • $\begingroup$ Related: physics.stackexchange.com/q/44017/2451 $\endgroup$ – Qmechanic Oct 12 '16 at 14:02
  • 1
    $\begingroup$ This question is a bit out there, sure, but I don't think it's worth all the downvotes. I mean, you have to learn it sometime. I think that @Jim had an appropriate response. $\endgroup$ – InertialObserver Jan 8 '17 at 9:14

As you said, this is a numerical coincidence. You rightfully pointed out that when units are included, it becomes nonsensical. What does that mean? It means this result is itself meaningless. You have thrown together a collection of numbers in a certain order to result in another number. Give me any two numbers and I can find a way to combine them to make any third number. You used slightly more than two numbers, so obviously you can find some combination that leads to almost the exact same number as $G$.

However, this means nothing if the units don't agree. The units are what turn a number into something. $4>1$ but $4s$ and $1m$ are incomparable. $6=6$ but $6m\ne6km$. Essentially, this boils down to the following statement "Hey look, these constants combine to form a number that is not at all equal to this other number".

To prove the meaninglessness of this to yourself, try doing the exact same calculation with each constant in different units. Use cm instead of m, hr instead of s, and g instead of kg. The numbers will be different.

This is a meaningless result


I think you are just falling into the realm of numerology. Numerical coincidences are funny and intriguing (see http://xkcd.com/1047/) but that is precisely just what they are, coincidences.

  • $\begingroup$ Numerology can fun and exciting or dangerous even. This is a good example of latter: relatively simple and nice looking equation with apparent high accuracy, but physically meaningless. It is a good dimensional analysis exercise for students to show such a link is not possible and therefore it would be waste of time searching such equation. I would be glad to see other similar nonsense equations. $\endgroup$ – HaanKa Oct 13 '16 at 5:23
  • $\begingroup$ In some cases, they are not just coincidences of course, as in the case of Monstrous moonshine and Umbral moonshine. $\endgroup$ – JamalS Dec 30 '16 at 12:47

Here is my contribution to 'coincidences' regarding the constant of gravitation: $$G = {8c^3\hbar}{\left(\alpha^2(2-\alpha)\over \mu_0\;w_u\right)^2}=6.6729195334\times10^{-11} \;\mathrm m^3 \mathrm K \mathrm g^{-1} \mathrm s^{-2}$$ Where ${w_u=16\pi^4}{Q_u^2\over\ t_u}=1558.545 \ldots \mathrm C^2 \mathrm s^{-1} $ is the ratio of unitary charge squared over unitary time. The resulting number for ${G}$ is a few standard deviations below the Codata value but is in line with the experimental data on atom interferometry found by Rosi et al. in 2014.
Admittedly, ${w_u}$ could be seen just as a trick to get the units right and a sustainable reasoning would be necessary to justify the equation.
To link ${G}$ with other constants is a step towards quantum gravity. The above 'coincidence', together with others, can be found in a table that is likely to generate more questions than answers.

  • 1
    $\begingroup$ Welcome on Physics SE :) I think your post would benefit from quoting sources and explaining how/why this should have something to do with $G$ ... $\endgroup$ – Sanya Dec 30 '16 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.