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Let's say I'm working in a natural unit system defined by a set of physical constants set to dimensionless numbers. How can I convert quantities between that natural unit system and a more conventional unit system like SI units?

MattHusz
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1 Answers1

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To illustrate the general procedure, consider quantities that can be written using SI units of $\mathrm{kg}$, $\mathrm{m}$, and $\mathrm{s}$ and a natural unit system defined by $c=\hbar=G=1$. I'll discuss incorporating quantities with ampere and kelvin at the end of this post with a different natural unit system.

First, in this natural unit system, all quantities in natural units are dimensionless (see e.g., this comment). The procedure below also works for natural unit systems in which some quantities are not dimensionless. See, e.g., the resource at the end of this post for the common case where all units are expressed in powers of $\mathrm{GeV}$.

To find the conversion factors between these systems, start with: $$ \begin{equation} \mathrm{kg}^a \mathrm{m}^b \mathrm{s}^c = [c]^\alpha [\hbar]^\beta [G]^\gamma. \end{equation} $$

Now, substitute in the corresponding units for the defining constants and simplify: $$ \begin{align} \mathrm{kg}^a \mathrm{m}^b \mathrm{s}^c &= \left(\mathrm{m} \mathrm{s}^{-1}\right)^\alpha \left(\mathrm{kg} \mathrm{m}^2 \mathrm{s}^{-1}\right)^\beta \left(\mathrm{kg}^{-1} \mathrm{m}^3 \mathrm{s}^{-2}\right)^\gamma,\\ &= \mathrm{kg}^{\beta-\gamma} \mathrm{m}^{\alpha+2\beta+3\gamma} \mathrm{s}^{-\alpha-\beta-2\gamma}. \end{align} $$

So, $$ \begin{align} a &= \beta - \gamma,\\ b &= \alpha + 2\beta + 3\gamma,\\ c &= -\alpha - \beta - 2\gamma. \end{align} $$

Or, rearranged to solve for Greek indices in terms of Latin ones: $$ \begin{align} \alpha &= \frac{1}{2} a - \frac{3}{2} b - \frac{5}{2} c,\\ \beta &= \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} c,\\ \gamma &= -\frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} c. \end{align} $$

Applying this to a number of different quantities, we get: $$ \begin{array}{l l l l l} \text{Quantity} & \text{SI Unit} & \text{Factor} & \text{Conversion} \\ \text{mass} & \mathrm{kg} & c^{1/2} \hbar^{1/2} G^{-1/2} & 2.1764\times 10^{-8}\,\mathrm{kg}\\ \text{length} & \mathrm{m} & c^{-3/2} \hbar^{1/2} G^{1/2} & 1.6163\times 10^{-35}\,\mathrm{m}\\ \text{time} & \mathrm{s} & c^{-5/2} \hbar^{1/2} G^{1/2} & 5.3912\times 10^{-44}\,\mathrm{s}\\ \text{energy} & \mathrm{kg}\, \mathrm{m}^2\, \mathrm{s}^{-2} & c^{5/2} \hbar^{1/2} G^{-1/2} & 1.9561\times 10^9\,\mathrm{J}\\ \text{momentum} & \mathrm{kg}\, \mathrm{m}\, \mathrm{s}^{-1} & c^{3/2} \hbar^{1/2} G^{-1/2} & 6.5248\times 10^0\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}\\ \text{velocity} & \mathrm{m}\, \mathrm{s}^{-1} & c & 2.9979\times 10^8\,\mathrm{m}\,\mathrm{s}^{-1}\\ \text{angular momentum} & \mathrm{kg}\, \mathrm{m}^2\, \mathrm{s}^{-1} & \hbar & 1.0546\times 10^{-34}\,\mathrm{J}\, \mathrm{s}\\ \text{force} & \mathrm{kg}\, \mathrm{m}\, \mathrm{s}^{-2} & c^4 G^{-1} & 1.2103\times 10^{44}\,\mathrm{N}\\ \end{array} $$

To convert from a quantity in natural units to one in SI units, we multiply by the value in the "Factor" column. Obviously, then, to go from SI units to natural units we divide by this factor. The "Conversion" column gives the value of this factor in SI units.

This same procedure can be applied to other natural unit system definitions, such as $c=\hbar=G=k_{\mathrm{B}}=1$, $c=\hbar=G=\epsilon_0=1$, or $c=\hbar=G=k_{\mathrm{B}}=\epsilon_0=1$, in which case we could also make dimensionless natural unit system quantities corresponding to SI quantities with units of kelvin, ampere, or kelvin and ampere, respectively.

It's also possible to apply this procedure to natural unit systems in which certain defining constants are set to dimensionless values other than 1 (often incorporating multiples of $\pi$). To use an artifical example, if we set $c=4\pi$, then to go from a velocity in natural units to one in SI units, we would multiply by a factor of $c/4\pi$.

To convert an equation from one unit system to another, start with the equation in the starting unit system and convert all quantities to the desired unit system. See this answer for a more in-depth discussion.

Resources

MattHusz
  • 229