Mass — Question 2

Question 2

Realizing a kilogram still requires measurements

The SI kilogram is defined through the exact Planck constant $h=6.62607015\times10^{-34}\ \mathrm{J\,s}$, together with the metre and second. A definition fixes a unit; an apparatus must still realize that unit.

For an idealized Kibble balance, two separate operating modes yield quantities related by $$P=mgv,\qquad P=UI.$$ Here $U$ and $I$ are measured electrical quantities, $v$ is a measured speed, and $g$ is the local gravitational field strength. $P$ is their electrical product in watts; the equation combines separate modes, not simultaneous conversion of electrical power into lifting power. Take the model as supplied.

The central data are $$P_0=0.009810\ \mathrm{W},\quad g_0=9.8100\ \mathrm{m\,s^{-2}}, \quad v=0.001000\ \mathrm{m\,s^{-1}}.$$ For this exercise $v$ is exact and all remaining errors are included in the hard bounds $$P\in[0.009809,0.009811]\ \mathrm{W},\qquad g\in[9.8099,9.8101]\ \mathrm{m\,s^{-2}}.$$ Every pair of values in these bounds is allowed.

1. Infer the central mass and check its units using $1\ \mathrm{W}=1\ \mathrm{kg\,m^2\,s^{-3}}$.

2. Find the smallest and largest allowed mass. State exact endpoint expressions and approximate deviations from the central mass in milligrams. Use $1\ \mathrm{kg}=10^6\ \mathrm{mg}$.

3. A laboratory labels the result “exactly one kilogram because $h$ is exact.” Explain the error. Does a change in the apparatus’s measured result change the SI kilogram?

4. Why must this apparatus use the local $g$, even though an ideal equal-arm mass comparison cancels a common $g$?

Question 2 - Solution

Realizing a kilogram still requires measurements

(a) Solve the measurement equation for mass.

$$[m]=\frac{[P]}{[g][v]}=\frac{\mathrm{kg\,m^2\,s^{-3}}}{(\mathrm{m\,s^{-2}})(\mathrm{m\,s^{-1}})}=\mathrm{kg}.$$

$$m_0=\frac{P_0}{g_0v}=\frac{0.009810}{9.8100(0.001000)}\ \mathrm{kg} =\boxed{1.000000\ \mathrm{kg}}.$$ The units are $\mathrm{(kg\,m^2\,s^{-3})/[(m\,s^{-2})(m\,s^{-1})]}=\mathrm{kg}$.

(b) Opposite endpoints give the extreme ratios.

All values are positive. Increasing $P$ increases $m$; increasing $g$ decreases $m$. Therefore the full model interval is $$\boxed{m\in\left[ \frac{0.009809}{9.8101(0.001000)}, \frac{0.009811}{9.8099(0.001000)} \right]\ \mathrm{kg}}.$$ Its endpoints are approximately 0.9998878707 kg and 1.0001121316 kg. Relative to 1 kg, these are approximately $-112.1293\ \mathrm{mg}$ and $+112.1316\ \mathrm{mg}$. The exact fractions define the bound; the displayed decimals are approximations. No root-sum-square combination is justified by these hard bounds.

(c) Exact definitions do not remove realization error.

Fixing $h$ supplies an invariant definition of the unit. It does not make the measured electrical product or local gravitational field exact. Their allowed variations remain in the inferred mass. A changed result may reflect the realization, the test object, or conditions; it does not redefine the kilogram.

(d) There is no second gravitational mass term to cancel.

This apparatus compares a gravitational quantity involving $mg$ with electrical and speed measurements. Solving for $m$ therefore requires $g$. In an ideal equal-arm comparison, both sides contain the same factor $g$, so it cancels. The two procedures connect the object to a mass value in different ways.