I understand the conversion factors. But if you look at the tables, they take an SI unit, say 1 kg, convert it into geometrized units, say $$1$$ m, and then reconvert it into SI units, which is $$1.3466\times 16^{27}$$ kg. I refer specifically to the table on pg. 4.

1. When we reconvert an SI unit back into an SI unit, shouldn't we get back 1 kg?

2. On pg. 2, they say $$E$$ can be any unit of energy. How is that? Clearly assuming $$E=1$$ J and $$E= 1$$ GeV gives us different answers.

• Welcome to physics.SE. The usual expectation on SE is that questions should be self-contained. Otherwise when the external link evaporates, the question becomes useless. Could you edit the question to summarize the relevant material? BTW, I assume $16^{27}$ was meant to be $10^{27}$, unless you're really into hexadecimal. – Ben Crowell Feb 4 at 22:46

For reference, here is table 2 from page 4 in reference [$$1$$]: 1. When we reconvert an SI unit back into an SI unit, shouldn't we get back 1 kg?

Yes. The source of confusion here seems to be a misunderstanding of what the first line in table 2 means. It means that in a system of units in which the speed of light ($$c$$) and Newton's constant ($$G$$) are both equal to $$1$$, the SI unit "$$1$$ meter" can also be expressed as a number of kilograms, specifically $$1.3466\times 10^{27}$$ kg. The table is not starting with $$1$$ kg and then ending up with $$1.3466\times 10^{27}$$ kg. Instead, it is showing how to express "$$1$$ meter" in kilograms. Here's the explicit calculation, using $$c=G=1$$, keeping only two significant digits for simplicity: \begin{align} 1\text{ meter} &= 1\text{ m }\times\frac{c^2}{G} \\ &\approx 1\text{ m }\times\frac{ (3.0\times 10^8\text{ m/s})^2 }{ 6.7\times 10^{-11}\text{ m}^3/(\text{kg}\cdot\text{s}^2) } \\ &\approx 1\text{ m }\times\frac{1.3\times 10^{27}\text{ kg}}{1\text{ m}} \\ &\approx 1.3\times 10^{27}\text{ kg}. \end{align} The first step simply multiplies by $$1$$, expressed as $$c^2/G$$. Since we're using units in which $$c=G=1$$, we could also multiply by $$1$$ expressed as, say, $$c^{42}G^{7/3}$$, if we wanted to, because that's also equal to $$1$$ in these units, and the result would still be legitimate. However, that would give an awkward combination of the SI units "meter" and "kilogram" on the right-hand side. The reason for multiplying by $$c^2/G$$ is that all of the meters cancel, leaving only kilograms, so we can use this to express a given number of meters as some number of kilograms, or conversely. For example, the mass $$M$$ of the sun is $$2.0\times 10^{30}$$ kg, which can be expressed in meters like this: \begin{align} M\approx 2.0\times 10^{30}\text{ kg} &\approx 2.0\times 10^{30}\text{ kg}\times \frac{1\text{ m }}{1.3\times 10^{27}\text{ kg}} \\ &\approx 1.5\times 10^3\text{ m}. \end{align} The Schwarzschild radius of the sun is $$R=2GM/c^2$$, which can be written simply as $$R=2M$$ in units where $$c=G=1$$. Either way, it comes out to be $$R\approx 3$$ km.

1. On pg. 2, they say $$E$$ can be any unit of energy. How is that? Clearly assuming $$E=1$$ J and $$E=1$$ GeV gives us different answers.

Of course, $$1$$ Joule and $$1$$ GeV are two entirely different amounts of energy. They are not equivalent. On the contrary, $$1$$ GeV is equivalent to $$\approx 1.6\times 10^{-10}$$ Joules, according to page 126 in reference  and also acknowledged explicitly on page 2 in reference [$$1$$]. Page 2 in reference [$$1$$] is saying that if we use units in which $$c$$ and Planck's constant $$\hbar$$ are both equal to $$1$$ (that is, $$c=\hbar=1$$), then we can express kilograms, meters, and seconds all in units of energy. Once we have a quantity in units of energy, we can express it either in GeV or in Joules — with different numeric values, of course, because $$1$$ GeV $$\approx 1.6\times 10^{-10}$$ Joules. That's all reference [$$1$$] means by "where $$E$$ is an arbitrarily chosen energy unit" below equation (2); nothing novel, just the usual freedom to express a given amount of energy using either GeV, or Joules, or ergs ($$1$$ erg $$=10^{-7}$$ J), or kilowatt-hours, or whatever happens to be convenient.

References:

[$$1$$] Myers, "NATURAL SYSTEM OF UNITS IN GENERAL RELATIVITY," https://www.seas.upenn.edu/~amyers/NaturalUnits.pdf

 "The International System of Units (SI), 8th edition," International Bureau of Weights and Measures (BIPM), http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf