Suppose I am working with a system of units where $c = G = \hbar = 1$.

I can then write e.g. a distance in units of kg by converting with a factor of

$$ \frac{c^2}{G} $$

Now if I have an energy in joules, how would I then convert this given that $c = G = \hbar = 1$?



Suppose, the energy you have is $χ\, \text{J}$ with $x∈ℝ^+$. Then you have to solve the equation:

$$ χ\, \text{J} = χ\, \frac{\text{kg}\, \text{m}^2}{\text{s}^2} = ξ\, c^x G^y \hbar^z,$$

where $x,y,z ∈ ℕ$, and $ξ ∈ ℝ$ is your desired value in the respective natural units. By inserting the SI values for $c$, $G$, and $\hbar$ and going to the logarithmic domain, you will find that this is equivalent to the linear equation system:

$$ \begin{pmatrix} \log_{10}(χ/ξ)\\ \hphantom{-}1\\ \hphantom{-}2\\ -2 \end{pmatrix} = \begin{pmatrix} 8.5 & -10.2 & -34 \\ \hphantom{-}0 & -1 & \hphantom{-}1 \\ \hphantom{-}1 & \hphantom{-}3 & \hphantom{-}2 \\ -1 & -2 & -1 \\ \end{pmatrix} · \begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}, $$

where the first column corresponds to the decadic logarithm of the numerical values, and the other columns are the powers of $\text{kg}$, $\text{m}$, and $\text{s}$. The solution to this is $x=\frac{5}{2}$, $y=-\frac{1}{2}$ and $z=\frac{1}{2}$ and thus $\log_{10}(χ/ξ)=9.29$.

For further reading, I wrote a didactical paper (preprint) on this.

  • $\begingroup$ Isn't this just scaling by the Planck energy: $E_p = \sqrt{\frac{\hbar c^5}{G}}$? $\endgroup$ – JEB Feb 4 '18 at 17:13
  • $\begingroup$ @JEB: Sure, but being the solution to this (specific) problem is the definition or point, respectively, of the Planck energy. You do not have something like this available in every dimension for every natural unit system. $\endgroup$ – Wrzlprmft Feb 4 '18 at 17:30

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