Tuesday, August 22, 2017

electromagnetism - Uncertainty of permittivity of vacuum


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.



Answer



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SI units that were in use at the time of writing; these are due to change in 2019, however, making this answer slightly outdated.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.


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