EDIT: I am now convinced that the sign of the logarithmic terms in the equations after 3.29 and 3.30 are wrong (unless I have missed something else). These identites come from looking at singularitites of the propagator given by their equation (3.19) as $$ K_\pm \sim \lim_{\varepsilon \rightarrow 0^+} \frac{1}{\left(-2 r Y_B\cdot Y_b - 2 \sqrt{r^2-1} \cosh(s-s_B \pm i\varepsilon) \right)^\Delta} $$ For example, we may look for singularitites in $s_B$ given $s$. Then we are looking to solve $$ \frac{r Y_B \cdot Y_b}{\sqrt{r^2-1}} + \cosh(s-s_B) = 0 $$ This is solved by $s_B = s - \text{arccosh}\left(\frac{-r Y_B \dot Y_b}{\sqrt{r^2-1}}\right)$. Defining the argument of the arccosh to be $\alpha$, we use that $$ \text{arccosh}(\alpha)= \ln(\alpha \pm \sqrt{\alpha^2-1}) $$ (in priniciple we might want to be careful about a branch cut in the logarithm. This should not be an issue since we will be exponentiating all appearances of t_B as we continue) and it follows that the singularities are at $$ s_B = s - \ln(\alpha \pm \sqrt{\alpha^2-1}) $$ which differs from the expression given by faulkner et al by exactly the sign I needed below. For $\alpha \leq -1$ the argument of the log is always negative so the cosh can reasonably cancel the positive term it is supposed to cancel.

I have been working through a paper by Faulkner et al. (1705.03026) titled "nonlinear gravity from entanglement in conformal field theories". I have been having serious trouble reproducing their equations 3.46 and 3.47. I will refer to equation numbers in the reference for all definitions. I will try to formulate my problem in a way such that the rest of the paper does not need to be understood. Undeclared variables are unimportant in the sense that they will just appear in the final answer, needing no further specificaiton in the present context.

We are considering the limit as $l^-_B \rightarrow 0$ of an integral. In this limit we are interested in expanding $$ e^{s_{*}^{+}} $$ in powers of l_B^-. In terms of a parameter $\alpha$, $s_{*}^{+}$ is given after equation 3.30 in the paper by $s_{*}^{+} = s_B - \ln(\alpha - \sqrt{\alpha^2-1})$. Therefore, I conclude that $$ e^{s_{*}^{+}} = e^{s_B} (\alpha - \sqrt{\alpha^2-1})^{-1} $$ Now, this is supposed to be expressed in terms of the lightcone coordinates $l^+_B = \sqrt{r^2-1}e^{s_B}$ and $l_B^- = \sqrt{r^2-1}e^{-s_B}$ (defined in equation (3.35) ). The quantity $\alpha$ is given by $$ \alpha= -\frac{r Y_B\cdot Y_b}{\sqrt{r^2-1}} \text{ (after equation (3.29))} $$ We now want some derived identities. For example we may note that $$ \frac{l^+_B}{l^-_B} = e^{2 s_B} \Rightarrow e^{s_B} = \sqrt{\frac{l^+_B}{l^-_B}} $$ Next we may note that \begin{equation} \begin{aligned} \alpha &= - \left(1 + l_B^+ l_B^-\right)^{1/2} \frac{Y_B \cdot Y_b}{\sqrt{l^+_B l^-_B}}\\ & = -\frac{Y_B\cdot Y_b}{\sqrt{l^+_B l^-_B}}\left(1+\frac{l^+_B l^-_B}{2} \right) + \mathcal{O}(l^-_B)^{3/2} %-\frac{(l^+_B l^-_B)^2}{8} \end{aligned} \end{equation} since $\sqrt{l^+_B l^-_B} = \sqrt{r^2-1}$ and $r = \sqrt{l^+_B l^-_B +1}$. Similarily we may write \begin{equation} \begin{aligned} \sqrt{\alpha^2 - 1} &= \left[\left( 1+ \frac{1}{l^+_B l^-_B} \right)(Y_B\cdot Y_b)^2 - 1 \right]^{\tfrac{1}{2}} \\ &= \frac{Y_B \cdot Y_b}{\sqrt{l_B^+ l_B^-}}\left[1 +\left( 1 - \frac{1}{(Y_B \cdot Y_b)^2} \right)l^+_B l^-_B\right]^{\tfrac{1}{2}}\\ &= \frac{Y_B \cdot Y_b}{\sqrt{l_B^+ l_B^-}} \left[1 +\frac{1}{2}\left( 1 - \frac{1}{(Y_B \cdot Y_b)^2} \right)l^+_B l^-_B \right] + \mathcal{O}(l^-_B)^{3/2}%- \frac{1}{8}\left( 1 - \frac{1}{(Y_B \cdot Y_b)^2} \right)^2l^+_B{}^2 l^-_B{}^2 \end{aligned} \end{equation} We can then plug all of this into our expression for $e^{s_{*}^{+}}$ \begin{equation} \begin{aligned} e^{s_*^+} &= e^{t_B}\left(\alpha - \sqrt{\alpha^2-1} \right)^{-1} \\ &= \frac{l^+_B}{Y_B\cdot Y_b} \left[ -\left(1+\frac{l^+_B l^-_B}{2} \right) - 1 -\frac{1}{2}\left( 1 - \frac{1}{(Y_B \cdot Y_b)^2} \right)l^+_B l^-_B \right]^{-1}\\ &= \frac{l^+_B}{2 Y_B\cdot Y_b} \left[ -1 -\left( \frac{1}{2} - \frac{1}{(2 Y_B \cdot Y_b)^2} \right)l^+_B l^-_B \right]^{-1} \\ &= -\frac{l^+_B}{2 Y_B\cdot Y_b} \left[ 1 - \left( \frac{1}{2} - \frac{1}{(2 Y_B \cdot Y_b)^2} \right)l^+_B l^-_B \right] \end{aligned} \end{equation} where I have suppressed the $\mathcal{O}$'s (maybe a little sloppy). The result given in equation (3.46) of the reference is $$ e^{s^+_*} = \frac{-2Y_B \cdot Y_b}{l^-_B} + \left( \frac{l^+_B}{2Y_B\cdot Y_b} - l^+_B Y_B \cdot Y_B \right) $$ which can be rewritten more similarlily to my form as $$ e^{s^+_*} = \frac{-2 Y_b \cdot Y_B}{l^-_B}\left[1 + \left(\frac{1}{2} - \frac{1}{(2Y_B \cdot Y_b)^2}\right)l^+_Bl^-_B \right] $$ The mismatch can be solved if I just inver the $(\alpha - \sqrt{\alpha^2-1})^{-1}$ term, but I feel that doing so would be unphysical/wrong and it would be ignoring the definitions of my reference. I can not for the life of me figure out what is wrong, and this problem has been simmering in the back of my head for months, but i've had other stuff to write. I would be incredibly grateful if someone managed to find where this goes wrong.

  • $\begingroup$ Just a general comment: if you are still have unresolved questions about this, you could probably just email the authors of that paper, since this is a fairly detailed and technical question. I'm guessing they would be able to give you a response or confirm whether there is a typo/error. $\endgroup$ – asperanz Apr 26 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.