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by mananaysiempre 1681 days ago
... Seriously, though, if you’re actually need this type of calculation regularly and didn’t just pick a random example, atomic-scale calculations are absolutely miserable to do in SI (and this is not a problem, it’s a human-scale, engineering system, after all; and its metrological aspects, which were the actual advance originally, are completely unimportant here).

If I had to do this in my head or with a desk calculator, I’d just do it in high-energy units (c = ℏ = 1, mass and energy in eV, length and time in eV^-1). So,

  4 amu = 4 × 0.93 GeV (a proton weighs 939 MeV, an amu is slightly smaller due do binding energy, rounding to 1 GeV is good enough for most purposes) ≈ 4 GeV,

  (1000 nm / s)^2 = (1e4 Å / s)^2 = (1e4 / 1.97 keV^-1 s^-1)^2 (an angstrom is a typical atomic size, a keV is a typical [large] atomic energy, a fermi aka femtometer is a typical nuclear size, a MeV is a typical [not so large] nuclear energy, remember any of 197 MeV fm = 1.97 keV Å = 1, though again 200 is almost always good enough) ≈ (1e4 / 2 keV^-1 s^-1)^2 = 25e6 keV^-2 s^-2,

  4 GeV × 25e6 keV^-2 s^-2 = 4e6 keV × 100e6/4 × keV^-2 s^-2 = 1e14 keV^-1 s^-2.
This is slightly inconvenient, we wanted energy in eV, but the seconds don’t seem to want to go away. I don’t remember Planck’s constant in eV s, but I do remember 2 keV Å ≈ 1 and 300e3 km/s = 3e8 m/s = 1, so let’s sprinkle it with those,

  1e14 keV^-1 s^-2 ≈ 1e14 keV^-1 s^-2 × (2 keV Å)^2 / (3e8 m/s)^2 = 4/9 × 1e14 × 1e-16 keV Å^2 m^-2 = 0.44 × 1e14 × 1e-16 keV × (1e-10)^2 ≈ 0.44e-22 keV ≈ 0.44e-19 eV.
The hardest part is pretending to be a normal person: you have to remember what an electronvolt actually is in normal units. Good thing this is numerically the same as remembering the charge of an electron in coulombs (1 eV = 1.6e-19 J),

  0.44e-19 eV = 0.44e-19 eV × 1.6e-19 J / eV (turns out converting to a decimal fraction wasn’t a good idea after all, powers of two FTW) ≈ 4/9 × 16 × 1e-1 × 1e-19 × 1e-19 J = 64/9 × 1e-39 J ≈ 63/9 × 1e-39 J = 7e-39 J.
Good enough to a couple percent.

OK, I won’t pretend that this is easy or that I did it flawlessly the first time just now, but I do think this looks like a skill you could plausibly learn, unlike the textbook “SI all the things” calculation. The good news is that you’ve just seen essentially all the relevant constants you’re going to have to remember, except maybe Avogadro’s number if you’re going to have moles somewhere.

(One place where this doesn’t help is first-principles chemistry, things like electrolysis, because you need to subtract large binding energies to get a change that’s hundreds to thousands times smaller. Calculating things to a couple percent just isn’t good enough.)

1 comments

Yes, I am familiar with this system. If anything, being a physicist is all the better reason to want a computer to deal with the units though...

My example was entirely contrived of course, a less contrived one would be estimating how long a gas cylinder will last. The tank name plate might say it has 200 cubic feet (sigh) and you need to flow at 10mL/min. How many months does the tank last? I'm talking about quick engineering tasks, not theory.

BTW, the answer is about 13 months, whatever that is in eV^{-1}:

https://www.google.com/search?hl=en&q=200%20cubic%20feet%20%...

Which took me about 15 seconds to type. Just different use cases.

units(1) can handle this but by default it gives you the answer in seconds.

    You have: 200 ft**3 / (10 mL/min)
    You want: months
     * 12.921493