I did some algebra...

In Planck unit, if make $\mu_0 = 4\pi$ and $\epsilon_0 = \frac{1}{4\pi}$ you get:

$$\mu_0 = 4\pi \cdot \frac{m_p l_p}{t_p^2 I_p^2} = 1.2566368452237765 \cdot 10^{-6} N \cdot A^{-2} $$

(where $4\pi$ is the supposed value of $\mu_0$, $m_p$ is Planck mass, $l_p$ is Planck length, $t_p$ is Planck time and $I_p$ is Planck current).

Which is very near to the CODATA value in SI and probably is the correct value.

CODATA: $1.25663706212(19) \cdot 10^{-6} N A^{-2}$

Similar for ε of vacuum:

$$\epsilon_0 = \frac{1}{4\pi} \cdot \frac{t_p^4 I_p^2}{m l_p^3} = 8.85419142073371 \cdot 10^{-12} F/m$$

CODATA value: $8.854 187 8128(13) x 10^{-12} F m^{-1}$

It is clear to me that the measurement are approximations of this perfect mathematical values... $4\pi$ and $\frac{1}{4\pi}$, so that $\mu_0\epsilon_0c^2=1$, and $c^2 = \frac{1}{\mu_0\epsilon_0}$ and $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, in facts:

$$\mu_0\cdot\epsilon_0 = 4\pi \cdot \frac{m_p l_p}{t_p^2 I_p^2} \cdot \frac{1}{4\pi} \cdot \frac{t_p^4 I_p^2}{m l_p^3} = \frac{t_p^2}{l_p^2} = \frac{1}{c^2}$$ in Planck units.

Couloumb constant $k_C$, at this point, is:

$$k_C = \frac{1}{4\pi\epsilon_0} = \frac{c^2\mu_0}{4\pi} = c^2 \cdot 10^{-7} H m^{-1} = 8987548129.98536 N m^2 C^{-2}$$

So, we have correct and exact values for $\epsilon_0, \mu_0, c, k_C$ in Planck units, that is, respectively: $\epsilon_0 = \frac{1}{4\pi}, \mu_0 = 4\pi, c=1, k_C=1$, and by multiplying for their dimensions expressed in Planck Units we obtain the correct, exact, values in SI.


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  • $\begingroup$ The numbers in parentheses at the end of the cited values are the digits in which there is uncertainty. As you can see, their values do not agree with 4pi or 1/4pi even in digits that we have measured precisely. In short, they are not approximations. $\endgroup$ – gabe Jun 12 at 18:34
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    $\begingroup$ First, you should format using MathJax. Second, I am not sure what is going on here. What calculus did you do? What are $m$, $l$, $t$, and $I$? These values are chosen so that $c=1$. They aren't measured. In fact we actually define what $c$ is and we adjust our units accordingly. $\endgroup$ – Aaron Stevens Jun 12 at 18:36
  • $\begingroup$ How do you adjust l and t given c? $\endgroup$ – M. Manfredi Jun 12 at 18:39
  • $\begingroup$ Did you mean “I did some algebra”? $\endgroup$ – G. Smith Jun 12 at 19:34
  • $\begingroup$ Is there some reason that you are putting the subscript 0 on the $\epsilon$ but not on the $\mu$? $\endgroup$ – G. Smith Jun 12 at 19:38

They do indeed have these values in Planck units. This is because in Planck units $1/4\pi\epsilon_0$ and $c$ are set to $1$, and


Planck units are not in everyday usage because your height would be about $10^{35}$ and most people don’t even know what that means.

  • $\begingroup$ If I'm not wrong, the meter has recently been redefined in Planck lengths... I'd have rather prefer to use a multiple of Planck length, giving it a new name, and different from the meter (like foot or knots are). So, for instance, 6.18714249917247e+34 is a meter in Planck length, you can take e33 lp to define "the new centimeter", and 100 times that gives the "new meter".... so that 1.616255 is the ratio between the old meter and the new meter, and is a limited and precise number, still a multiple of lp. $\endgroup$ – M. Manfredi Jun 12 at 20:27
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    $\begingroup$ You’re wrong. The meter was not redefined in terms of the Planck length. The Planck length is not known to high accuracy because $G$ has not been measured to high accuracy. So Planck lengths are completely unsuitable as measurement units. $\endgroup$ – G. Smith Jun 12 at 20:34
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    $\begingroup$ According to Wikipedia the Planck length is known to only 5 or 6 significant digits. To metrologists, this is pathetic. $\endgroup$ – G. Smith Jun 12 at 20:40
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    $\begingroup$ Yep, sorry. It is the Kg that has been redefined on the base of h :-) $\endgroup$ – M. Manfredi Jun 12 at 20:49
  • $\begingroup$ Following this definition, the 2 constants have the values of 4π and 1/4π. Why the CODATA values differ? $\endgroup$ – M. Manfredi Jun 13 at 4:07

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