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I am running a cosmological simulation and am having some trouble putting things into code units. The physical distance units in my simulation are in terms of $\text{Mpc/h}$, where $h$ is the dimensionless Hubble parameter. This makes enough sense because, as noted elsewhere, simulations are often scale-free so it makes sense to factor out the $h$ dependence and make it explicit. This unit convention is causing me some confusion however. In one calculation I have to do during the simulation, I essentially (ignoring context which I can provide later) have to multiply the speed of light $c$ by an inverse distance $1/x_0$ which is given in units of $\text{Mpc/h}$.

In order to properly have the units cancel, I first put $c$ in units of $\text{Mpc/s}$ to get $$ 9.716 \times10^{-15} \text{Mpc/s} $$ However, should I know factor out the $h$ dependence? These seems strange to me because in my mind, the value of the speed of light should not depend on the underlying cosmology I have simulated. On the other hand, I feel that I should not cancel units of $\text{Mpc}$ with units of $\text{Mpc/h}$. To make things concrete, let's assume I have a value of $h=.7$. Should I then take the quantity above and multiply it by $.7$ to yield $$ 6.802 \times 10^{-15} \text{(Mpc/h)/s} $$

and use that result in my calculations? I think this situation confuses me because it doesn't involve measurements, where it is clear how $h$ can enter, and it involves a constant of nature, which should be independent of the assumed cosmology.

Let me know if more info is needed.

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  • $\begingroup$ If the result of your computation is in units of $hs^{-1}$ then that can be ok. Depending on the value of $h$ the process is longer or shorter in physical time. $\endgroup$
    – user26287
    Commented May 14, 2020 at 12:18

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Partial answer; although the question is 17 days old, I hope that this guide to the dependence on h in cosmology can give you a good insight into this concept.

Quick guide to h dependence in cosmology

which links to Damn You, Little h! (Or, Real-World Applications of the Hubble Constant Using Observed and Simulated Data). The abstract says:

The Hubble constant, H0, or its dimensionless equivalent, "little h", is a fundamental cosmological property that is now known to an accuracy better than a few per cent. Despite its cosmological nature, little h commonly appears in the measured properties of individual galaxies. This can pose unique challenges for users of such data, particularly with survey data. In this paper we show how little h arises in the measurement of galaxies, how to compare like-properties from different datasets that have assumed different little h cosmologies, and how to fairly compare theoretical data with observed data, where little h can manifest in vastly different ways. This last point is particularly important when observations are used to calibrate galaxy formation models, as calibrating with the wrong (or no) little h can lead to disastrous results when the model is later converted to the correct h cosmology. We argue that in this modern age little h is an anachronism, being one of least uncertain parameters in astrophysics, and we propose that observers and theorists instead treat this uncertainty like any other. We conclude with a `cheat sheet' of nine points that should be followed when dealing with little h in data analysis.

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As argued in the paper you cite, you shouldn't think of $h$ as a unit, but rather as a part of the value. Although it's commonly seen, and although numerically it's not wrong, I therefore think the term "$10h^{-1}\,\mathrm{Mpc}$" is preferred to "$10\,\mathrm{Mpc}/h$".

There's no need to express $c$ in terms of $h$, because our lack of knowledge about the value of $h$ doesn't affect the value of $c$. If your distance is, say, $x_0 = 10h^{-1}\,\mathrm{Mpc}$, so that your inverse distance is $1/x_0 = 0.1h\,\mathrm{Mpc}^{-1}$, your calculation becomes $$ \begin{array}{rcl} c\frac{1}{x_0} & = & \left[9.716\times10^{-15}\,\mathrm{Mpc}\,\mathrm{s}^{-1}\right] \times \left[0.1h\,\mathrm{Mpc}^{-1}\right] \\ & = & 9.716\times10^{-16}h\,\mathrm{s}^{-1} \\ \big( & \simeq & 0.7\times10^{-15}\,\mathrm{s}^{-1}.\big) \end{array} $$

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