Gravitational backreaction in AdS/CFT

Question

When we add a scalar field into AdS space time, under which limit can we ignore the gravitational back reaction? In the case of massless scalar, can we totally ignore the backreation to the background geometry?

Answer 1

Most of this comes from reading "lore" and working backwards but hopefully it's not too imprecise.

Solving for the cosmological constant in terms of the $AdS_{d + 1}$ length $L$, Einstein's equations read \begin{align} G_{\mu\nu} - \frac{d(d - 1)}{2L^2} g_{\mu\nu} = 8\pi G T_{\mu\nu}. \end{align} Based on this, we want all components of the stress tensor to be less than $\frac{1}{L^2 G}$ so that most of the curvature does not come from the matter. But an important point, which can be seen by solving the wave equation, is that particles in $AdS$ behave a lot like particles in a box of volume $L^d$. So writing the energy density as $T_{00} = E / L^d$, the criterion becomes \begin{align} E \ll \frac{L^{d - 2}}{G}. \end{align} So in a sense, massless scalar fields are the safest. If the field has too large a mass $m$, it is possible for even the lowest excitation (whose energy is just $m$) to already give significant backreaction. For values of $m$ which are small though, the associated primary operators in the CFT with dimension \begin{equation} \Delta = \frac{d}{2} \pm \frac{1}{2} \sqrt{d^2 + m^2 L^2} \end{equation} will be good candidates for having their correlation functions computed perturbatively from the gravity side. There's a quick sanity check showing that operators which are too heavy really should be thought of as backreacting to create a new geometry. Just take the mass to be $L^{d - 2} / G$ and notice that, up to some order 1 factor, this is the mass of a black hole with a horizon radius of $L$. In the 80s, Hawking and Page found that black holes growing beyond a size of $L$ undergo a phase transition which makes them no longer evaporate. They come to equilibrium with their own Hawking radiation instead. So this large black hole, being the first stable saddle point besides the AdS vacuum, is a good candidate for what the above operator is creating.

A word of warning though is that, in between the lightest operators with good perturbative dynamics and the heaviest operators which correspond to black holes, there can be several other regimes. Let's solve for $E = m$ in terms of the scaling dimension to write \begin{equation} \Delta \ll \frac{L^{d - 1}}{G} \end{equation} In the classic example relating $\mathcal{N} = 4$ SYM to type IIB, this says that $\Delta$ is much less than $N^2$ from the gauge group. Abstractly, this is the central charge which becomes the expansion parameter on the CFT side. But, as is well known from that example, holographic computations don't just require $N^2$ to be large to avoid creating black holes (or $N$ to be large to avoid creading branes). They also require a large 'tHooft coupling $\lambda$ because exciting stringy modes becomes possible for dimensionless energies of order $\lambda^{\frac{1}{4}}$. The remarkable thing about this theory is that we don't just have to wave our hands and say that some operator we might never guess has this scaling dimension. Anomalous dimensions for very simple unprotected operators like \begin{align} \mathrm{Tr}[\Phi_I \Phi^I] \end{align} have been computed and found to agree with $\lambda^{\frac{1}{4}}$. I'm getting off topic though because a bottom-up model found by adding a scalar to AdS doesn't have stringy modes yet. In that case, it should be valid to study scattering processes for descendants at levels below $\frac{L^{d - 1}}{G} - mL$. And it could also be interesting to explicitly seek out backreacting examples like boson stars where the Einstein and Klein-Gordon equations need to be solved together.