I'm trying to calculate the entanglement entropy in CFT2/AdS3 in the thermal case for a finite interval (-a,a). I'm reading the paper of Takayanagi and Rangamani (2016): https://arxiv.org/abs/1609.01287 (page 68) but they don't have the explicit calculation. The result for this case has to be:

\begin{equation}S_{A}= \frac{c}{3}log\left(\frac{\beta}{\pi\epsilon}\sinh\left(\frac{2\pi a}{\beta}\right)\right). \end{equation}

where $\epsilon$ is a cutoff.

My problem is that I have those 2 equations:

The metric of BTZ BH: \begin{equation} ds^{2}=-\frac{(r^{2}-r_{+}^{2})}{l^{2}_{AdS}}dt^{2}+ \frac{dr^{2}}{(r^{2}-r_{+}^{2})}+ \frac{r^{2}}{l^{2}_{AdS}}dx^{2} \end{equation}

and from geodesics equations: \begin{equation} \frac{dr}{dx}= \frac{r}{l^{2}_{AdS}}\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r^{2}_{\ast}}-1\right)}. \end{equation}

I've tried the same procedure that I used with the normal case (without temperature): \begin{align} \textrm{Length}& =\int \textrm{d}x\sqrt{r^{2}+r^{2}\left(r^{2}-r_{+}^{2}\right)\left(\frac{r^{2}}{r_{\ast}^{2}}-1\right) \frac{1}{r^{2}-r_{+}^{2}}}= \int\textrm{d}x\sqrt{r^{2}+\frac{r^{2}r^{2}}{r_{\ast}^{2}}-r^{2}}\\ &= \int \textrm{d}x\frac{r^{2}}{r_{\ast}}=\int_{r_{\ast}}^{\infty} \frac{\textrm{d}r}{r}\frac{1}{\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r_{\ast}^{2}}-1\right)}}\frac{r^{2}}{r^{\ast}}= 2\ln\left( \sqrt{r^{2}-r_{+}^{2}}+\sqrt{r^{2}-r_{\ast}^{2}}\right)\Bigg|_{r_{\ast}}^{\infty} \end{align}

Now I don't know how to get the result that they obtained... because I don't know how to continue. The only thing they especified in their article is that

\begin{equation} r_{\ast}=r_{+}\textrm{coth}\left(ar_{+}\right). \end{equation}


1 Answer 1


First, introduce a UV cutoff $r_b\equiv (1/\epsilon)\to\infty$. Then perform the length integral on $[r_*,r_b]$ which you'll get a linear combination of natural logarithms in terms of $r_+, r_*$ and $r_b$ (let's denote the result by $\mathcal{L}\,$). Now by integrating the following equation $$\frac{dr}{dx}= r\,\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r^{2}_{\ast}}-1\right)}$$ on the half interval for $x$, you'll get an expression for $r_*$ in terms of $a$, $r_+$ and $r_b\,$ and you can approximate it in the limit $r_b\to \infty$ which will give you $$r_*=r_+\coth\,(a\, r_+)$$ Substituting this result in $\mathcal{L}$ and again approximating the obtained result in $r_b\to \infty$ limit yields $$2\ln\,\left(\frac{2\,r_b}{r_+}\,\sinh(a\,r_+) \right)$$ Finally, by substituting $r_b=1/\epsilon\,$, $r_+=2\pi/\beta\,$ and $c=\frac{3}{2G_N}$ together with $S=\frac{\mathcal{L}}{4G_N}\,$, you'll will get the desired result. Please note that for convenience, I set the $\small AdS$ radius to unity. I hope this helps you.

  • $\begingroup$ I'm really grateful with your answer GMNafisi. I was about 2 months to solve this calculus because I didnt know what to do with the coth. I did the whole calculus and I got the result, but I had a bit problem with the factor of 2 that you put inside the ln. $2r_{b}$. Where does this 2 appear? Thank you for all your help ;)! It's my first time in this forum, if I can rate you possitively please, tell me how. $\endgroup$ Aug 22, 2019 at 23:17
  • $\begingroup$ You're welcome! The factor 2 inside $\log$ comes after when you substitute $r_*$ in $\mathcal{L}$ and series expand it around $r_b\to \infty$ up to $\mathcal{O}(r_b)$ (I did the whole thing with Mathematica) which will get you $2\ln(r_b)+\ln\left(4\sinh(a\,r_+)^2/r_+^2 \right)\,$. Now you can use $\log(x^d)=d\,\log(x)$ property and get the result. Btw, I'm a new guy as well, but I think clicking the upper triangle icon would be nice :) $\endgroup$
    – GMNafisi
    Aug 23, 2019 at 13:30
  • $\begingroup$ @GMNafisi I have spent about 3 weeks to perform this derivation, thank you so much! $\endgroup$ Apr 18, 2021 at 13:21
  • $\begingroup$ @ArtemAlexandrov Glad to be of help! $\endgroup$
    – GMNafisi
    Apr 19, 2021 at 16:53

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