# Natural units

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Definitions of speed of light, reduced Planck's constant, and electronvolt in terms of kilograms, meters, seconds:

$c=299792458\ {m \over s}$ $\hbar =1.054571817\times 10^{-34}\ kg{m^{2} \over s}$ $eV=1.602176463\times 10^{-19}\ kg{m^{2} \over s^{2}}$ Then solve for second:

${\hbar \over eV}=6.582120268\times 10^{-16}s$ $s=1.519267287\times 10^{15}{\hbar \over eV}$ Solve for kilogram:

$c^{2}=8.987551787\times 10^{16}\ {m^{2} \over s^{2}}$ ${eV \over c^{2}}=1.782661731\times 10^{-36}\ kg$ $kg=5.609589202\times 10^{35}\ {eV \over c^{2}}$ Solve for meter:

${\hbar \over c}=3.51767294\times 10^{-43}\ kg\cdot m$ ${\hbar \over c}=1.973270014\times 10^{-7}\ {eV \over c^{2}}m$ $5067730.179\ {\hbar \over c}{c^{2} \over eV}=m$ $m=5067730.179\ {\hbar c \over eV}$ Summarize results:

$kg=5.609589202\times 10^{35}\ {eV \over c^{2}}$ $m=5067730.179\ {\hbar c \over eV}$ $s=1.519267287\times 10^{15}{\hbar \over eV}$ Dimensional conversion: $kg^{x}m^{y}s^{z}\sim c^{y-2x}\hbar ^{y+z}eV^{x-y-z}$ Now omit c and ℏ:

$kg=5.609589202\times 10^{35}\ eV$ $m=5,067,730.179\ {1 \over eV}$ $s=1.519267287\times 10^{15}{1 \over eV}$ Dimensional conversion: $kg^{x}m^{y}s^{z}\sim eV^{x-y-z}$ Based on: http://www.physics.gla.ac.uk/~dmiller/lectures/RQM_2008.pdf, page 12