Question: The value of permittivity of vacuum, $\epsilon_0$, is given with absolutely no uncertainty in NIST

Why is this the case?

More details:

The permeability of vacuum can be given by

$$\mu_0=\frac{1}{\epsilon_0 c^2}$$

which comes from the definition of a magnetic field in special relativity, where we solve the problem of a wire with electrons flowing in, and we calculate the force exerted on an external charge moving with some velocity (for details, refer to the book Electricity and Magnetism, E. M. Purcell), and we define a new field called "magnetic field" with the form of Lorentz force, where the magnetic field is

$$B=\frac{I}{2\pi \epsilon_0 c^2 r}$$

where $I$ is the current due to the flow of electrons in the wire, and $r$ is the distance of the external charge from the wire. And there we get the definition of $\mu_0$ that makes $B$: $$B=\frac{\mu_0 I}{2\pi r}$$

and there starts the concept "magnetism".

Why am I giving this detailed example? Because I wouldn't like to get the answer that $\mu_0$ has no error, and that's why $\epsilon_0$ has no error, and then we fall into circular logic. So I expect a reason which is independent of $\mu_0$.

Thank you in advance.

  • 1
    $\begingroup$ But can't you fix $\mu_0$, thus defining current? Fixing $c$ defines the metre, then you fix $\mu_0$ to define current and you've got no DOF left for $\epsilon_0$. It has been a while since I thought about all this stuff, though and you've likely got a counter to my argument: if so, I suggest putting it in the question, because if I'm overlooking something, it's just vaguely possible that someone else may be too. $\endgroup$ Commented Apr 16, 2014 at 10:39

3 Answers 3


The answer below is based on the definitions of SI units that were in use at the time of writing; these are due to change in 2019, however, making this answer slightly outdated. Emilio Pisanty has posted an up-to-date answer.

lthough you might not like to hear it, the answer really DOES lie in the definition of $\mu_0$ (and $c$). $\mu_0$ is defined to be exactly $4\pi *10^{-7}\ \text{H m}^{-1}$. Similarly, $c$ is defined as exactly $299792458\ \text{ms}^{-1}$. It immediately follows from the relation $$\epsilon_0=\frac{1}{\mu_0 c^2}$$ that $\epsilon_0$ also has no uncertainty.

Maybe you don't like this because it makes explicit reference to a concept from magnetism, and you would like to see a formulation of electric effects that is separated from magnetic effects. Such a thing is simply not possible, since a simple change of reference frame can turn an electric effect into a magnetic one or vice versa. Electromagnetism really is a single unified framework. There is also no circularity in this argument, as far as I can tell.

  • $\begingroup$ Note that this has changed with the new SI system to be adopted from May 2019. $\endgroup$
    – ProfRob
    Commented Nov 18, 2018 at 17:44
  • $\begingroup$ @RobJeffries Next time, please feel free to update the answer with an edit in the future (I have done so myself now)! $\endgroup$
    – Danu
    Commented Nov 18, 2018 at 19:21
  • $\begingroup$ The note was not a rebuke. $\endgroup$
    – ProfRob
    Commented Nov 18, 2018 at 19:52
  • $\begingroup$ @RobJeffries I didn't take it that way (sorry if my comment came across that way..). It was more of a "I agree, and you should feel free to add "This answer will be outdated as of the SI update in 2019" or something like that to the answer" :-) $\endgroup$
    – Danu
    Commented Nov 18, 2018 at 19:57
  • $\begingroup$ You two would be in violent agreement if you weren't so polite about it. $\endgroup$ Commented Nov 18, 2018 at 23:19

The existing answers are correct as of 2018, but it's important to note that in May 2019 the SI is getting an overhaul which changes the answer.

  • In the current SI, the ampere is defined using the force between two wires, which sets an exact value for $\mu_0$, and the relativistically-invariant nature of electromagnetism then demands an equally exactly-set value for $\varepsilon_0 = 1/c^2\mu_0$.

  • In the new SI, the definition of the ampere is altered so that the value of the elementary charge $e$ is fixed (together with the values of $c$ and $h$). That therefore means that the Coulomb constant must be measured experimentally as the force between two elementary charges separated by a unit distance (in principle); a more feasible and more accurate determination is via the identity $$\varepsilon_0 = \frac{e^2}{2hc \, \alpha}$$ in terms of the experimentally-determined fine-structure constant $\alpha$. This means that the relative uncertainty in $\mu_0$ will be directly inherited from (and equal to) that of $\alpha$.

    As a neat fact, this unit is so important that it gets its own specific mention alongside $M_\mathrm{IPK}$, $m({^{12}\mathrm{C}})$ and $T_\mathrm{TPW}$ in Appendix 2 of the Resolution that implements the revision (which will presumably be named as Resolution 1 of the CGPM 2018):

    the vacuum magnetic permeability $\mu_0$ is equal to $4\pi\times 10^{-7}\:\rm H \: m^{-1}$ within a relative standard uncertainty equal to that of the recommended value of the fine-structure constant $\alpha$ at the time this Resolution was adopted, namely $2.3\times 10^{-10}$ and that in the future its value will be determined experimentally

A full table of the effect of the change on the status of various important physical constants is available in this section of the Wikipedia page. For more on the change on this site see e.g. What are the proposed realizations in the New SI for the kilogram, ampere, kelvin and mole? or What will be the uncertainty in $\mu_0$ under the new SI scheme?, together with their Linked questions.


Just to add to Danu's Answer, which I believe to be right. The relative scalings of the "electro" and "magnetism" parts of the unified electromagnetism whole are somewhat arbitrary; we're only required to ensure that $c=\frac{1}{\sqrt{\mu_0\,\epsilon_0}}$ to achieve a valid set of Maxwell equations.

As we change these relative scalings, we change the numerical values of the strength of the field's sources. Thus:

  1. We fix $c$ to a defined value; given that we define a second as a fixed number of Caesium atom transitions, the fixing of $c$ defines the SI metre (more generally, defines our unit system's unity length in terms of its unity time interval);
  2. We can then fix either $\epsilon_0$ or $\mu_0$, the other is then automatically set by our choice. At this step we fix what the numerical value of electric charge / current is. In SI, we fix $\mu_0$ to be $4\,\pi\times10^{-7}$, thus fixing the ampère to be the current needed to flow in two infinite, parallel conductors 1 metre apart so as to achieve a force (attractive if the current is in the same direction) of $2\times10^{-7}$ newton per metre.
  • $\begingroup$ Note Emilio's answer. This all changes with the new SI, when neither the permittivity or permeability is a fixed number. $\endgroup$
    – ProfRob
    Commented Nov 18, 2018 at 17:45

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