This seems like a simple question, but I cannot wrap my head around it. If $\hbar = c = 1$ then length is time, and mass is inverse length or inverse time. Hence $G$ should have dimensions of length squared or time squared or inverse mass squared. But what would be its numerical value (since that is not set to 1) in these natural units ?
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In natural units, we also set $G=1$. – Prahar Nov 08 '23 at 18:29
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@Prahar how can that be possible? $c=1$ implies $L=T$. From mass-energy equivalence, we also have $E=M$. $\hbar=1$ implies $E=T^{-1}$. From $G=1$ we have $EL=M^2$, but $EL$ is unitless, so $M$ is unitless. But, $E=M$, so energy is unitless too, along with everything else considered so far. So do we take everything to be unitless in this scenario? – MattHusz Jan 25 '24 at 14:49
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1The dimensions for these constants are $[c] = LT^{-1}$, $[\hbar] = M L^2 T^{-1}$ and $[G] = M^{-1} L^3 T^{-2}$. Setting $c=1$ fixes $L=T$. Next, setting $\hbar=1$ implies $M=L^{-1}$ so in total we have $M^{-1} = L = T$. Finally, setting $G=1$ gets rid of all the dimensions and makes everything unitless. – Prahar Jan 25 '24 at 16:15
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If $\hbar = c = 1$ then length is time, and mass is inverse length or inverse time.
This is correct. In this way, since $[G]=[M^{-1}L^3T^{-2}]$ in SI units, its dimensionality reduces to $$ [G] =[M^{-1}L^3T^{-2}] =[M^{-1}M^{-3}M^{2}] =[M^{-2}] . $$ This tells you that there is a product of powers of $\hbar$ and $c$ such that $G$ becomes an inverse-squared mass when you divide by it, and this product turns out to be $\hbar c$. This then gives us the correct expression: $$ G = \hbar c \times 6.7\times 10^{-57} (\mathrm{eV}/c^2)^{-2}, $$ as a valid identity in SI units, which then reduces to $$ G = 6.7 \times 10^{-57} \mathrm{eV}^{-2} $$ if you set $\hbar = c = 1$.
Emilio Pisanty
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