The International System of Units — Question 4

Question 4

Defining a kilogram without a defining lump of metal

The SI fixes the following numerical values exactly: $$\begin{aligned} \Delta\nu_{\mathrm{Cs}}&=9\,192\,631\,770\ \mathrm{s^{-1}},\\ c&=299\,792\,458\ \mathrm{m\,s^{-1}},\\ h&=6.626\,070\,15\times10^{-34}\ \mathrm{J\,s}. \end{aligned}$$ Here $\Delta\nu_{\mathrm{Cs}}$ is the specified unperturbed caesium-133 transition frequency, $c$ is the speed of light in vacuum, and $h$ is the Planck constant. You may use $1\ \mathrm{J}=1\ \mathrm{kg\,m^2\,s^{-2}}$.

1. Explain the order second $\rightarrow$ metre $\rightarrow$ kilogram shown above. Rewrite the unit of $h$ using base units and rearrange its exact relation to express $1\ \mathrm{kg}$ in terms of $h$, $1\ \mathrm{m}$, and $1\ \mathrm{s}$.

2. Find exponents $a,b,d$ so that $h^a c^b(\Delta\nu_{\mathrm{Cs}})^d$ has the dimension of mass. Does having the dimension of mass make its value exactly one kilogram?

3. A laboratory improves its apparatus for realizing the kilogram. Does that improvement redefine the kilogram or make every measured mass exact? Explain.

Question 4 - Solution

Defining a kilogram without a defining lump of metal

(a) The constants constrain linked unit sizes.

The fixed frequency sets the second. With that second, the fixed value of $c$ sets the metre. Since $\mathrm{J\,s}=\mathrm{kg\,m^2\,s^{-1}}$, fixing $h$ then sets the kilogram once metre and second are specified. With $H=6.626\,070\,15\times10^{-34}$, a pure number, $$h=H(1\ \mathrm{kg})(1\ \mathrm{m})^2(1\ \mathrm{s})^{-1},\qquad \boxed{1\ \mathrm{kg}=\frac{h}{H}\frac{1\ \mathrm{s}}{(1\ \mathrm{m})^2}}.$$

(b) Solve the dimension constraints.

$$[h^a c^b\nu^d]=\mathrm{kg}^{a}\mathrm m^{2a+b}\mathrm s^{-a-b-d};\quad a=1,\ 2a+b=0,\ -a-b-d=0.$$

Write dimensions as $\mathsf{M,L,T}$. The product has dimension $\mathsf{M}^{a}\mathsf{L}^{2a+b}\mathsf{T}^{-a-b-d}$. Mass alone requires $a=1$, $2a+b=0$, and $-a-b-d=0$; thus $$\boxed{a=1,\ b=-2,\ d=1},\qquad \frac{h\Delta\nu_{\mathrm{Cs}}}{c^2}\text{ has mass dimension.}$$ Its value is about $6.78\times10^{-41}\ \mathrm{kg}$, not $1\ \mathrm{kg}$. Dimensions determine the kind of unit, not a numerical normalization.

(c) Definition versus realization.

A realization implements the definition experimentally. Better apparatus can reduce realization uncertainty without changing the fixed value of $h$ or the unit definition. Alignment, environmental corrections, and other experimental effects can still leave measured masses uncertain. A material object can serve as a calibrated reference without defining the kilogram.