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I have a quantity that I need to convert to different units.

First, I need to note that the value is in the natural units system. The quantity is in cm^-1 (or, to be accurate, $cm^{-1}\hbar c /k$, see https://physics.stackexchange.com/a/187749/65616). I need to convert this quantity to inverse nanoseconds (converting to inverse seconds will answer my question perfectly fine). How can I do this? The concept of natural units is confusing me because I know that $\hbar=c=k=1$ but when we do calculations with them, we use the SI equivalents. So how do I do the conversion?

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  • $\begingroup$ @Acid Jazz That link doesn't work, and why not format it as a link? I did flag your comment. $\endgroup$ – mistermarko Aug 8 '15 at 2:41
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Calculating energies when given inverse wavelengths

Frequencies are related to energies by $E = 2\pi\hbar~f$ and frequencies to wavelengths by $f = c / \lambda$, so generally if you have something which is an inverse wavelength you'd multiply by $2\pi\hbar~c$ to get its corresponding energy, while converting to a frequency like inverse nanoseconds is done by multiplying by $c$.

Natural units

It is best for these to identify some natural units with certain dimensions. One way I would do this in college involved writing down a "powers list" for my units of mass, length, time, and charge, in that order. So I'd write the natural unit of electric potential as the powers list $\{1~2~\bar 2~\bar 1\}$ which stood for the unit $(\text{cm}^{-1} \text{ mass})^1 (\text{cm length})^2 (\text{cm time})^{-2} (\text{cm}^0 \text{ charge})^{-1},$ rather than a nebulous $\text{cm}^{-1}$ term which might mean anything. The conversion factors are therefore not strictly $1$ but have units associated with them:$$\hbar = 1~\{1~2~\bar 1~0\}\\c = 1~\{0~1~\bar 1~0\}\\k = 1~\{1~3~\bar 2~\bar 2\}$$Note that the over-bars are used for minus signs, and these represent powers of the appropriate natural-unit dimensions in a very specific order; $\hbar$ is one "natural mass * squared natural length per natural time," is what the $\{1 2 \bar 1 0\}$ means.

In SI units you can then use these constants to "work out" the equivalent sizes of things by converting everything to the units that you know in SI (in this case, length in centimeters). So one "natural mass" can be written as $$\{1~0~0~0\} = \{0~\bar1~0~0\} \cdot\frac{\hbar}{c} = 1\,\text{cm}^{-1} \cdot\frac{\hbar}{c}\approx 3.518 \times 10^{-41} \text{kg}$$when, say, your length unit is known to be $1~\text{cm}.$ If we're abusing notation we'd call this mass to be "one cm-1 of mass" in the natural units.

So in your case you have units which are $\text{cm}^{-1} \cdot \hbar c / k$ which I'd write as $\{0~\bar 1~0~2\}.$

Now the question you need to ask yourself is, "what am I trying to calculate, here?" You've said that "more accurately" your parameters are in the dimensions of $\text{charge}^2/\text{length}$. I don't know what physical quantities you're thinking about which would have those dimensions: but once you figure out what your squared charge per length $\mathcal Q$ is doing, you might very well have an expression that there is some sort of energy $k~\mathcal Q$ associated with it, which then needs to be related to a photon with frequency $f = k~\mathcal Q / h$. If that's true then in these units where $\hbar$ is being set to 1, it means you need to take your expression in terms of $\text{cm}^{-1}$, then multiply that unit by $c/2 \pi$, and this will give the frequency you're looking for.

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  • $\begingroup$ is this valid in the natural units system? $\endgroup$ – TanMath Aug 5 '15 at 17:36
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    $\begingroup$ It is valid regardless of unit system, for any wave which is quantized into particles by $E = h f$ and which travels at speed $c$. It happens to be the case that in many natural units $c = \hbar = k = 1$. Probably the best way to deal with these is to start aggressively including dimension, speaking of "cm distance", "cm time", "cm^-1 energy", "cm^-1 mass", "cm^0 charge" as different units with for example $\hbar = 1 (\text{cm}^{-1}\text{ energy})*(\text{cm time})$. $\endgroup$ – CR Drost Aug 5 '15 at 19:31
  • $\begingroup$ Can you include an example? $\endgroup$ – TanMath Aug 5 '15 at 19:43
  • $\begingroup$ @TanMath: I've updated the above answer to include a bit more discussion of what you can do with natural units and an example of how to convert your expression to inverse seconds. $\endgroup$ – CR Drost Aug 5 '15 at 20:28
  • $\begingroup$ I am so sorry, but I don't understand tue power list concept, and please see the edits to my question.. I am sorry for asking so much from you.. $\endgroup$ – TanMath Aug 5 '15 at 21:13

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