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量について

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$\newcommand{\X}{\mathrm{X}}\newcommand{\kg}{\mathrm{kg}}\newcommand{\m}{\mathrm{m}}\newcommand{\s}{\mathrm{s}}\newcommand{\K}{\mathrm{K}}\newcommand{\J}{\mathrm{J}}\newcommand{\N}{\mathrm{N}}\newcommand{\A}{\mathrm{A}}\newcommand{\b}{\mathrm{b}}\newcommand{\B}{\mathrm{B}}\newcommand{\Ki}{\mathrm{Ki}}\newcommand{\Mi}{\mathrm{Mi}}\newcommand{\Gi}{\mathrm{Gi}}$

次元の除去

物理量や情報量を単位抜きで扱いたい.数についての対数表示を利用すれば十分実用的だと思われるし,古典物理学や量子力学を具体的な数値として量的に把握することができるような気がする.

$4 \pi G = c = \varepsilon_0 = \mu_0 = k = h/2 \pi i = 1$とする.

普通は$\hbar = h/2\pi = 1$とするが,それでは次元が退化しすぎる感じだし,
$pq - qp = h/2\pi i$だけ見ても,$h = 2\pi i$とみなしたくなる.

これらの量を慣用の単位系で表して
$c = 299792458\ \m\ \s^{-1} = 1$
$h = 6.626070040 \times 10^{-34}\ \kg\ \m^2 \s^{-1} = 2 \pi i$
$G = 6.67408 \times 10^{-11}\ \kg^{-1} \m^3 \s^{-2} = 1/4\pi$
とすれば,$\mathbb{C}$で解いて
$\kg = 18.908488.125\ \X = \exp(18.908488 + 2\pi i \times 0.125)$
$\m = 78.844871.125\ \X$
$\s = 98.363472.125\ \X$
となる.ついでに
$k = 1.3806488 \times 10^{-23} \J\ \K^{-1} = 1$
$\mu_0 = 4\pi \times 10^{-7}\ \N\ \A^{-2} = 1$
とすれば,
$\K = -72.765617.125\ \X$
$\A = -56.280327 \ \X$
となる.また,
$\b = \log 2 = 0.693147 = -0.366513\ \X$
$\B = \log 2^8 = 8 \log 2 = 1.712929\ \X$
$\Ki\B = 2^{10}\ \B = 8.644400\ \X$
$\Mi\B = 2^{20}\ \B = 15.575872\ \X$
$\Gi\B = 2^{30}\ \B = 22.507344\ \X$
などとなる.

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