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I am looking for a way to translate formulas written in natural units into either HLU units or SI units. Seeing the Planck constant and the speed of light would help me understand what is going on. The problem is that the advice to use dimensional analysis suggested fiddling with multiplying or dividing until it works. It was not a clearly defined procedure. Where can I find clear instructions how to rewrite a formula in natural units so that the constants will become visible?

Thomas Fritsch
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Avi
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    "It was not a clearly defined procedure." Well, you haven't seen a clear write-up yet, but the procedure is well defined. For each of the necessary outcome units there is only one combination of $\hbar$ and $c$ that will work. – dmckee --- ex-moderator kitten Sep 05 '19 at 23:02
  • I have to say, though, that something few people every bother to mention is that there is an ambiguity in the dimensionality of an equation (because you can multiply the whole thing by any number of constant factors with affecting it's equality), so you must pick a symbol somewhere and say "This symbol will be adjusted to have these units and everything else must agree." As nuclear and particle physicists tend to write everything in multiples of $\mathrm{eV}$ I always elect to write energies as energy for my starting point. – dmckee --- ex-moderator kitten Sep 05 '19 at 23:11

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What you are looking for is called dimensional analysis.

You multiply every but one term of the equation with unknown powers of $c^\alpha$ and $\hbar^\beta$, where $\alpha$ and $\beta$ are variables. Then you identify the powers of mass, length, time etc. on both sides, which should give you equations in $\alpha$ and $\beta$. After solving them, you know exactly how many $c$ and how many $\hbar$s you have to add.

This works because we are not changing the equation. In natural units, $c=1$ and $\hbar=1$ (and maybe some other constants as well) and therefore multiplying by $1^\alpha$ does not change anything.

ersbygre1
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  • You need to go term-by-term - the equation $(-\frac{1}{2m}\nabla^2 + V)\psi = E\psi$ needs an $\hbar^2$ in the kinetic term but not in the potential term. – J. Murray Sep 05 '19 at 23:05
  • That‘s absolutely true, I was just imagining a more simple equation in my head. Will edit accordingly. – ersbygre1 Sep 05 '19 at 23:06