In Birrell and Davies, the author says in the Introduction that

If the gravitational field is treated as a small perturbation, and attempts are made to quantize it along the lines of quantum electrodynamics (Q.E.D.), then the square of the Planck length appears in the role of coupling constant.

But why does the square of the Planck length comes in as a coupling constant here? How does one determine what the coupling constant should be?

  • $\begingroup$ As for the coupling constant determination problem, we usually start with a Lagranigan or Hamiltonian, write down the energies for the materials and the force carrying bosons, and then the physics of the interaction between them determines what the coupling constant should be. That is, we tend to know them from prior physics understanding. We might also guess from analogy. $\endgroup$ Commented May 8, 2023 at 11:10

1 Answer 1


Let's for simplicity work in units where $\hbar=1=c$. If we try to quantize perturbatively the Einstein-Hilbert (EH) action $$ S~=~ \frac{1}{16\pi G} \int \! d^dx \sqrt{-g} (R-2\Lambda)\tag{1} $$ in $d$ spacetime dimensions, we see that the coupling constant $G$ has dimension $$[G] ~=~ L^{d-2},\tag{2}$$ since $$[S]~=~ L^0, \quad [x^{\mu}]~=~L^1, \quad [g_{\mu\nu}]~=~L^0, \quad [R]~=~L^{-2}~=~[\Lambda].\tag{3}$$

In fact the Planck length $\ell_P$ is defined as the combination of the 3 fundamental physics constants $c$, $\hbar$ and $G$ that has dimension of length, so it is not surprising that $G$ is proportional to $\ell_P^{d-2}$, cf. the statement in Birrell & Davies.

For more information, see e.g. my related Phys.SE answers here and here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.