Alternative units in SI-unit-system

Question

I was wondering, what would be the value of Vacuum Permettivity in the case 1 meter (I will call this 1m") would be defined as the distance we nowadays see as 1,10 meters.

At first this looks easy:

$E_0 = 8.8541878128 \times 10^{-12}$ F / m with normal meters

$E_0" = E_0 \times 1,1$

$E_0" = 9.739660659408 \times 10^{-12}$ F / m" with converted meters.

However, when I look at Farads, this is defined as $$1F = \frac{s{^4} \times A{^2}}{1m{^2} \times kg}$$ so Farads should be converted too.

And Amperes are defined as:

"The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to $2 \times 10{^{-7}}$ newtons per metre of length."

which also involves meters, and Newtons, which are defined as $$1N = \frac{kg \times 1m}{s{^2}}$$

this all makes it too tough for me.

So, in short, my question is: is the calculated version of E0" correct?

Answer 1

The simplest way to approach this is via the relationship $$ c^2 = \frac{1}{\epsilon_0\mu_0} \tag{$*$} $$ between the vacuum permittivity and permeability and the speed of light.

• If the value of the meter increases by 10% (i.e. multiplied by $r=1.1$), then the value of $c$ must decrease by 10% (since there are now fewer (new) meters in the distance covered by light in one second).

• In the pre-2019 SI, the definition of the ampere effectively means that $\mu_0$ has a fixed value of $$\mu_0=4\pi\times 10^{-7} \:\rm H/m=4\pi\times 10^{-7} \:\rm kg \: m \: s^{−2} \: A^{−2}.$$ If you retain the text of the old definition but now use the updated meter, then you're fixing the numerical value of $\mu_0$, i.e., the ampere must increase by $\sqrt{r}$ (roughly, increase by 4.88%$\approx$5%), but $\mu_0$ stays constant. This implies that $\epsilon_0$ goes up by $r^2$.

• On the other hand if the ampere stays fixed (i.e. the actual current) then $\mu_0$ must go up with $r$. In this case, $\epsilon_0$ goes up as $r$.

• The 2019 redefinition of the SI units changes the ampere so that it is metrologically independent from the meter (since it now depends only on the second and the elementary charge $e$). Within this setting, it pays to use a modified version of $(*)$, derived from the definition of the fine-structure constant, $$\alpha = \frac{1}{4\pi\epsilon_0}\frac{e^2}{\hbar c},$$ rearranged into the form $$\epsilon_0 = \frac{e^2}{2\alpha h c}.\tag{$**$}$$ Here $e$, $h$ and $c$ all have explicit values fixed in the new SI, and $\alpha$ has an experimentally-fixed dimensionless value.

  • As we saw above, you probably want to change $c$, decreasing by $r$, for consistency.
  • If you want to be consistent, then you probably also want to change the numerical value of $h=6.62607015×10^{−34}\:\rm kg\:m^2\:s^{-1}$, which should decrease by $r^2$, if you want to keep the value (not the definition) of the kilogram unchanged.
  • The value of $e^2$ should probably stay untouched, which means the ampere does not change (as opposed to the pre-2019 ampere).

  In these conditions, $\epsilon_0$ should increase by $r^3$.

This is just to emphasize a point I made on an earlier thread: you cannot, cannot, change one part of a measurement system without being absolutely clear about which other parts you're keeping constant, because the impacts will depend on multiple key choices you need to make when you're making that specification.