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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.

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marked as duplicate by John Rennie, Kyle Kanos, Brandon Enright, V. Moretti, BebopButUnsteady Apr 18 at 14:29

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1  
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. –  WetSavannaAnimal aka Rod Vance Apr 16 at 10:39
    
Note that in the proposed redefinition of SI base units (in this case, ampere), permitivity (and permeability) of vacuum will be subject to experimental error. –  Mormegil Apr 18 at 10:18
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2 Answers 2

up vote 8 down vote accepted

Although 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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I'm upvoting, but see my comment to the OP: it's been a while since I've thought about all this so I'm not 100% sure I'm not overlooking something. Your argument is simply: fix $c$, thus fixing metre, fix $\mu_0$, thus fixing ampère leaving no DOF for $\epsilon_0$, right? –  WetSavannaAnimal aka Rod Vance Apr 16 at 10:44
    
Thank you for the response and the comment. What makes me then wonder is: Why did we choose that value for $\mu_0$ ? What does this value mean? Shouldn't this choice have some physical meaning with uncertainty? –  The Quantum Physicist Apr 16 at 10:48
    
@WetSavannaAnimalakaRodVance that is exactly the argument. I simply hadn't noticed your comment when I posted the answer. –  Danu Apr 16 at 11:11
    
@WetSavannaAnimalakaRodVance I do still think it is important to point out explicitly the unity of electromagnetism (which OP seemed to be dissatisfied with), so I won't delete my answer. –  Danu Apr 16 at 11:27
    
@Danu No, don't delete it, as far as I can see it's right. –  WetSavannaAnimal aka Rod Vance Apr 16 at 11:52
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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.
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