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How to define a kilogram

Kilogram.jpg
The official kilogram. Credit: BIPM

By Margaret Harris

Pick the correct definition of a kilogram:

a) the mass of a body with a de Broglie wavelength of 6.626069311 x 10^-34 m at a velocity of 1 m/s

b) a mass of a body at rest such that Planck’s constant h is 6.626069311 x 10^-34 Js

c) a mass of exactly 5.0184512725 x 10^25 unbound carbon-12 atoms at rest in their ground state

d) the mass of a lump of platinum-iridium sitting under three vacuum jars in a French laboratory

Readers with an interest in metrology will know that the answer is d) — and anyone who didn’t know it could probably have guessed from the photo. But why is the kilogram, alone of all SI units, defined by something so un-fundamental as a lump of metal?

The difficulty, as Bryan Kibble explained this afternoon in a talk at the QuAMP conference in Leeds, is that several of the alternatives have problems of their own. Options a) and b) both rely on pinning down a value for Planck’s constant, and thus might seem like the best way to go; indeed, one of them may actually become the new SI definition, perhaps as early as 2011. However, Kibble argued, both options are somewhat circular, swapping uncertainty in the kilogram for uncertainty in other Planck-derived units, and there’s not really any new science involved in them.

A definition in terms of carbon-12 atoms — or indeed, any kind of atoms — would be more satisfying, Kibble says, but as efforts like the Avogadro project at the UK’s National Physical Laboratory have shown, counting atoms isn’t a trivial task.

Nobody offered any solutions during the question period after the talk, but we did manage to pin down one thing: any fluctuations in fundamental constants (like the fine structure constant, for example) will not affect the kilogram problem — at least not for around 1000 years. So that’s all right then.

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3 comments

  1. Conor Coyle

    Maybe we should go back to the imperial system and that will solve the problem of defining a kilogram…lol

  2. Anton Szautner

    What’s the problem? Why should counting exactly 5.0184512725 × 10^25 of Carbon-12 atoms equaling “One Kilogram” represent a pain in the neck for anybody, including those who so tenderly protect their precious reference “lump of platinum-iridium sitting under three vacuum jars in a French laboratory”?
    Chill the THEORETICAL idea of it to near absolutute zero, by all means, if one insists, but never mind trying to actually put exactly that number of C-12 atoms together in a single lump (good luck) let alone preserve it anywhere, an approach to standardization which SHOULD by now be considered pure folly.
    If anybody is going to do any research that requires extremely fine resolution or even a change in any constant, they had better prepare a proper in situ reference that would indicate such a change without regard to some arbitrary vaguely definable lump, no matter how protectively seated it is in its protective chambers.
    Are physicists to continue to defer to such a outmoded tradition as a standard, that is “weighed” on a spot on the surface of the Earth, upon which now measurable tidal influences and even deep mantle movements change the gravitational acceleration at that spot far more than any expected change in constants, or is there something unalterably special about France?
    I do not think it would be terribly earth-shattering to make a shift to an entirely theoretical value based on the addition of exactly 5.0184512725 × 10^25 independant Carbon-12 atoms, or that such a reference could possibly be any worse than a vague number of platinum-iridium atoms.
    Of course, experimentalists COULD go right ahead and write papers that spurn any mention of the “standard” in favor of a reference that they supply themselves.

  3. Its not necessary to count atoms.
    The Australian National Mesurement under the project leader Walter Giardini have made a sphere of pure Silicon 28 that will be contrasted against the Paris cylinder and replaced it.
    Knowing the mass of an atom of silicon 28 you can deduce the number of atoms and take that number as an Universal Constant.
    Henceforth you will make replicas by volume. This is more exact than to weigh the Paris cylinder.
    Ludovicus

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