I have a short question: what does it mean when one says that General relativity is not renormalizable?
I guess unlike the Quantum Field Theory, but what is the idea behind this statement?
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When we say that GR is non-renormalizable, we generally mean that the tools we used in other field theories to get rid of infinities are not working in this case. For example if we treat gravity as a quantum field theory, we can expand around flat spacetime. $$ g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu} $$
If we perturb around the field $h_{\mu\nu}$, the action becomes$$ \mathcal{S}_{gravity}\sim \frac{1}{16\pi G}\int{d^4x} \;(\partial h)^2(1+h^2+...) $$ $$ \sim\int{d^4x}(dh)^2\left[{\frac{1}{16\pi G}}+\left({\frac{1}{16\pi G}}\right)^2h^2+..\right] $$
Now, the couplings of each $(1+h^2+..)$ will have higher and higher powers. The couplings in the action (namely the terms which contain the gravitational constant $G$) have mass dimension -2. This means our couplings have mass dimensions $[\lambda] < 0$. In general for a theory to be renormalizable we need $[\lambda] \ge 0$. In other words, there are more divergencies in our theory than there are counter-terms.
answered Jan 28, 2020 at 22:39
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3$\begingroup$ Thanks for your quick answer. Just a little detail : you do a Taylor development, right ? How do you get : $\left(\dfrac{1}{16\pi G}\right)^{2}$ into $$\sim\int{d^4x}(dh)^2\left[{\frac{1}{16\pi G}}+\left({\frac{1}{16\pi G}}\right)^2h^2+..\right]$$ $\endgroup$– rob ♦Commented Jan 28, 2020 at 23:43
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$\begingroup$ @redhood : just a reminder if you have the time : how do you make appear the terms $\left(\dfrac{1}{16\pi G}\right)^{2}$ and others superior orders ? Regards $\endgroup$– user87745Commented Jul 24, 2020 at 13:14
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$\begingroup$ We expand $\sqrt{-g}$ which is a power series in $h$ and then by dimensional analysis we have increasing powers of $\frac{1}{16\pi G}$. If you are wondering how to expand $\sqrt{-g}$ check this question(physics.stackexchange.com/questions/3873/…) $\endgroup$ Commented Jul 26, 2020 at 13:53