# Holographic entanglement entropy (Thermal case)

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/pdf/1609.01287.pdf (page 68) but they don't have the explicit calculation. The result for this case has to be:

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

where $$\epsilon$$ is a cutoff.

My problem is that I have those 2 equations:

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

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

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

$$$$r_{\ast}=r_{+}\textrm{coth}\left(ar_{+}\right).$$$$

Thank you everyone for reading me and I would be very grateful if anyone can help me!!!!

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.
• 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. – Ulquiorra15 Aug 22 '19 at 23:17
• 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 :) – GMNafisi Aug 23 '19 at 13:30