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Setting $h=c=1$ the familiar Wikipedia result is that the Gravitational constant, $G$ goes like

$$ G = \frac{1}{m_{Planck}^2}$$ However, I suspect that this is only true in 4 or (3+1) dimensions? Or just 3 spatial dimensions. But how would this change if I work in d-dimensions? I guess I would have

$$ G_d = \frac{1}{m_{Planck}^{f(d)}}$$ where $f(d)$ is a dimensional dependent number. But then what is $f(d)$?

Qmechanic
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Johan Hansen
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  • Do you want $d$ to denote the dimension of space viz. @Cham's answer, or of spacetime? – J.G. Jun 24 '21 at 14:08
  • This question is written kind of backwards. The first equation is the definition of the Planck mass in the units that you're using, not a result about the Gravitational constant. – Brick Jun 24 '21 at 17:53

1 Answers1

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This is a fascinating subject!

I'll use "natural" units so that $\hbar = 1$ and $c = 1$. Poisson's equation in $d$ space dimensions is $$\tag{1} \nabla^2 \phi = 4 \pi G \rho, $$ where the mass density has units $\rho \sim \mathrm{L}^{-d - 1}$, since mass has units $m \sim \mathrm{L}^{-1}$ (in natural units). The left part of equation (1) has units $\nabla^2 \phi \sim \mathrm{L}^{-2}$, since $\phi \sim \mathrm{L}^0$ (no units! This comes from the potential energy or the force equation... and also from general relativity since $\phi$ is just a part of the spacetime metric). Thus it's easy to find the units of $G$: $$\tag{2} G \sim \mathrm{L}^{d - 1}. $$ For $d = 1$ (two spacetime dimensions), $G$ doesn't have any unit! For $d = 3$ space dimensions, we have $G \sim \mathrm{L}^{2}$ (the squared Planck lenght).

Notice that the electromagnetic fine-structure $\alpha \equiv \frac{k e^2}{\hbar c} \approx \frac{1}{137}$ have units for $d \ne 3$. The same exercice (using Maxwell equations) gives $$\tag{3} \alpha \sim \mathrm{L}^{d-3}. $$ It's interesting to note that the ratio $G/\alpha$ have units that don't depend on the number of dimensions: $$\tag{4} \frac{G}{\alpha} \sim \mathrm{L}^{2}. $$ Like gravitation (when $d \ne 1$), electromagnetism imposes a favored scale in universes of dimensions $d \ne 3$.

Cham
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  • The factor of $4\pi$ doesn't work for arbitrary $d$. – J.G. Jun 24 '21 at 14:10
  • @J.G. It depends. You can absorb the $4 \pi$ factor (or its generalisation) into your definition of $G$ and $\alpha$. The $\Omega_d$ solid angle doesn’t have any units, wathever the space you consider. – Cham Jun 24 '21 at 14:18
  • True, but a general-$d$ treatment would usually absorb so that the coefficient is $1$ or the solid angle, not that it's $4\pi$ specificially. Anyway, as you've said, it doesn't affect the dimensional analysis. – J.G. Jun 24 '21 at 14:20
  • You write "$d=1$ ( two spacetime dimensions)" and then write "$d=3$ spacetime dimensions" which seems inconsistent to me, would that not be labeled as 4 spacetime dimensions as the usual prescription is $n+1$ to account for time? – Triatticus Jun 24 '21 at 17:30
  • @Triatticus, please re-read what I wrote. I didn’t said d = 3 spacetime dimensions. – Cham Jun 24 '21 at 17:42