I was reading this post where it states that

"gravity does have a coupling with negative dimension and it's the Newton constant"

However, I don't understand at all why Newton's constant leads to gravity coupling with negative dimension. Could it be possible for someone to show why gravity couples with negative dimensions?

  • 5
    $\begingroup$ Have you learned about "natural units"? The "negative dimension" of $G$ is $[M]^{-2}$ in the sense of dimensional analysis using natural units where $\hbar=c=1$. The quote isn't talking about spacetime dimensions. $\endgroup$
    – Ghoster
    Commented Aug 24, 2022 at 4:22

2 Answers 2


To expand on @Ghoster's comment, $c=\hbar=1$ implies $1/m=\hbar/(mc)$ is a length, i.e. length and time each have mass dimension $-1$, and $G$ has dimension $\mathsf{L}^3\mathsf{M}^{-1}\mathsf{T}^{-2}$ so has mass dimension $-3-1+2=-2$. By contrast, comparing the GPE $-Gm_1m_2/r$ to its electrostatic counterpart $kq_1q_2/r$ shows $k$, like charge, has mass dimension $0$.

  • $\begingroup$ Thanks! But I just want to ask, the gravitational constant has dimensions $L^3 M^{-1} T^{-2}$ but why would that then lead to $-3-1+2$? $\endgroup$ Commented Aug 24, 2022 at 9:14
  • $\begingroup$ @JungwoonSong As I said, length & time each have mass dimension $-1$. Alternatively, note $[G/(c\hbar)]=\mathsf{M}^{-2}$. $\endgroup$
    – J.G.
    Commented Aug 24, 2022 at 9:20
  • $\begingroup$ Thanks so much! $\endgroup$ Commented Aug 24, 2022 at 9:42
  1. Let's for simplicity work in units where $\hbar=1=c$. We will then classify physical quantities $Q$ according to their mass dimension $[Q]$, cf. Ref. 1.

  2. Recall that Newton's law of gravitation in $d$ spacetime dimensions is $$F~=~G\frac{m_1m_2}{r^{d-2}},\tag{1}$$ up to possible dimensionless constants. The fact that $$ [F]~=~2, \qquad [m_1]~=~1~=~[m_2], \qquad [r]~=~-1,\tag{2}$$ implies that the gravitational constant $G$ has mass dimension $$[G]~=~2-d,\tag{3}$$ which is negative for $d>2$, cf. OP's title question. See also e.g. this Phys.SE post.

  3. Now, the underlying reason OP asks their question is apparently because this fact leads to the non-renormalizability of perturbative quantum gravity, cf. their linked post. This is e.g. shown in my Phys.SE answer here.


  1. M. Srednicki, QFT, 2007; Chapter 12. A prepublication draft PDF file is available here.

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