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I have been reading about Susskind's $Complexity = Volume$ conjecture from a set of lecture notes. In these, he gives an equation for the growth of the interior of an AdS-Schwarzschild black hole:

$$ \frac{dV(t)}{dt} \sim l_{ads}AT$$

where V is the volume of the BH interior, $l_{AdS}$ is the AdS length scale, A is the horizon area and T is the BH temperature. He does not derive this equation but says that it can be easily calculated from the AdS metric or deduced by dimensional analysis. I tried the former method but I'm not sure how to integrate the metric such that only the interior volume is considered. Could someone perhaps give me a hint as to how I should proceed?

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I suspect that Susskind arrived at the growth rate heuristically by considering the horizon as a codimension two hypersurface and then integrating over the radius of AdS space. Let's assume that the volume growth rate is constant with respect to Euclidean time to make the following argument. The SAdS$_4$ metric, and its volume form are given by $$\text{d}s^2 =g^E_{\mu\nu}\text{d} x^\mu\text{d} x^\nu= f(r)\text{d}t_{E}^2 + f(r)^{-1}\text{dr}^2 + r^2 \text{d}\theta^2+r^2\sin(\theta)^2\text{d}\phi^2,\\ \sqrt{\det{g}^E}\varepsilon = r^2\sin(\theta)\text{d}t_E\wedge\text{d}r\wedge\text{d}\theta\wedge\text{d}\phi.$$ Now consider the codimension two surface normal to the $t_E-r$ disk evaluated on the horizon given by $$\text{d}s_\partial^2 =h_{ab}\text{d}x^a\text{d}x^b= r_h^2\text{d}\theta^2 + r_h^2\sin(\theta)^2\text{d}\phi^2,\\ \sqrt{\det{h}}\varepsilon_\partial = r_h^2\sin(\theta)\text{d}\theta\wedge\text{d}\phi.$$ If we integrate the codimension two volume form, we get the surface area of the SAdS$_4$ black hole, i.e., $$A_h = \int\sqrt{\det{h}} \varepsilon_\partial = r_h^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\phi \sin(\theta) = 4\pi r_h^2.$$ Now, the trick is to consider a differential of the black hole volume as it grows in an interval $\text{d}t_E$. We have $$dV = A_hdr \implies A_h = \frac{\text{d}V}{\text{d}r} = \frac{\text{d}V}{\text{d}t_E}\frac{\text{d}t_E}{\text{d}r} \iff \int A_h \text{d}r = \frac{\text{d}V}{\text{d}t_E}\int\frac{\text{d}t_E}{\text{d}r}\text{d}r = \frac{\text{d}V}{\text{d}t_E}\int\text{d}t_E $$ where I have used that the growth rate of volume $(\text{d}V/\text{d}t_E)$ is constant. Now, integrating over the Euclidean periodicity and over AdS space, I have $$A_h \ell_{AdS} = \frac{\text{d}V}{\text{d}t_E}\int_0^\beta\text{d}t_E = \beta \frac{\text{d}V}{\text{d}t_E} \implies \frac{\text{d}V}{\text{d}t_E} = \beta^{-1}A_h\ell_{AdS} = TA_h\ell_{AdS}.$$ That's my best bet anyway, you may have to dig deeper into literature to find out if those assumptions are the one's Susskind used but this at least gives you a start.

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  • $\begingroup$ I'm sorry for the late comment; just now got the time to look at this. This was quite helpful. I think you mean to say "$\textit{volume}$ growth rate is constant" (of course your statement can be made by assuming the C=V conjecture, but I was looking for a pure GR answer to go towards the conjecture). Also, the assumption of volume growth being constant is fair (arxiv.org/abs/1411.2854). $\endgroup$ Commented Oct 12, 2020 at 5:55
  • $\begingroup$ I have another naive question about this calculation: would the equation being about growth in $\textit{Euclidean time}$ make a difference? It would give an extra factor of $\textit{i}$ making the growth rate imaginary, right? $\endgroup$ Commented Oct 12, 2020 at 6:24
  • $\begingroup$ @thunderbolt, thanks for pointing that out. I've updated the solution to include volume growth rate. The time signature surprisingly doesn't make a difference but I'll leave this as homework for you! (Here's a hint, use the chain rule; ${dV}/{dr} =( {dV}/{dt_E})({dt_E}/{dr}) = ({dV}/{dt_M})({dt_M}/{dr})$. You'll find that the i cancels everywhere once you integrate appropriatly.) $\endgroup$
    – Dhuality
    Commented Oct 12, 2020 at 13:45

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