Why are the kilogram, ampere, kelvin, and mole still considered base units and not derived? Given their current definitions, they're purely defined in terms of other units and cannot be independently measured, right?
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Replying to @sgrif
It's no different to how the metre is defined by using the speed of light - it's taking a fixed value for a physical constant and then deriving the value of the base unit.
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Replying to @pixeltrix
Right, but the Planck constant isn't something you can independently measure. The kg is now defined by giving it an exact value
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Replying to @sgrif
But that argument also applies to the speed of light doesn't it? Can't measure how fast something is without a distance. I think the phrase 'base' at this point mainly means they're the units to use when doing dimensional analysis and not some kind of fundamental value.
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Replying to @pixeltrix
> But that argument also applies to the speed of light doesn't it? Can't measure how fast something is without a distance. That's sorta the point. Distance is determined by how fast light travels. You need a defined second and measure how far light travels in a fixed time.
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Replying to @sgrif @pixeltrix
> not some kind of fundamental value. This just makes the base/derived split weird though. The name implies (and it used to mean AFAIK) a unit that was fundamentally required to describe the universe. If its definition is in entirely terms of other units, that's not the case
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Replying to @sgrif @pixeltrix
Granted I also don't fully understand how the modern definition of the kg even works since it seems like it defines it in terms of the mass of a photon but photons have no mass so ¯\_(ツ)_/¯
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Replying to @sgrif
It has energy so it's just E = mc2 I guess? Just glad it's above my pay grade to think about these things
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Replying to @pixeltrix
Right, that's how you get to J = kg*m^2*s^-1, but photons have no mass so wtf I should have gone to college
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All I'm saying is that we stopped using the mass of a litre of water because it was difficult to use in practice, and I'm not sure how determining the effects of a fixed numerical value for the Planck constant is somehow easier
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Replying to @sgrif
I guess no-one has to take special care of a lump of metal anymore is the main benefit
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