# Why $µ_0$ and $ε_0$ are not considered to be $4π$ and $1/4π$? [closed]

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.

## closed as unclear what you're asking by Aaron Stevens, Buzz, Kyle Kanos, GiorgioP, Jon CusterJun 13 at 13:50

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• 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. – gabe Jun 12 at 18:34
• 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. – Aaron Stevens Jun 12 at 18:36
• How do you adjust l and t given c? – M. Manfredi Jun 12 at 18:39
• Did you mean “I did some algebra”? – G. Smith Jun 12 at 19:34
• Is there some reason that you are putting the subscript 0 on the $\epsilon$ but not on the $\mu$? – 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
$$c=\frac{1}{\sqrt{\mu_0\epsilon_0}}.$$
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.
• 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. – G. Smith Jun 12 at 20:34