Hawking temperature in the eternal BTZ black hole: an example of holography in AdS spacetime

Abstract

We review the relation between AdS spacetime in 1 $+$ 2 dimensions and the BTZ black hole (BTZbh). Later we show that a ground state in AdS spacetime becomes a thermal state in the BTZbh. We show that this is true in the bulk and in the boundary of AdS spacetime. The existence of this thermal state is tantamount to say that the Unruh effect exists in AdS spacetime and becomes the Hawking effect for an eternal BTZbh. In order to make this we use the correspondence introduced in algebraic holography between algebras of quasi-local observables associated to wedges and double cones regions in the bulk of AdS spacetime and its conformal boundary respectively. Also we give the real scalar quantum field as a concrete heuristic realization of this formalism.

Appendix: Global conformal transformations in two dimensions

In Sect. 5 we found explicitly the subgroups of the global conformal group in two dimensions generated by $J_{uy}$ and $J_{vx}$. In this “Appendix” we give the others subgroups of this group.

First let us remember the elements of the Lie algebra of the AdS group. These elements are

$$\begin{aligned} \begin{array}{cc} J_{uv}=u\partial _{v}-v\partial _{u}&\quad \quad J_{xy}=x\partial _{y}-y\partial _{x} \\ J_{ux}=u\partial _{x}+x\partial _{u}&\quad J_{uy}=u\partial _{y}+y\partial _{u} \\ J_{vx}=v\partial _{x}+x\partial _{v}&\quad J_{vy}=v\partial _{y}+y\partial _{v} \end{array} \end{aligned}$$

(67)

It is well known that in a representation on the vector space $\mathbb R ^{2,2}$ these generators can be represented by the matrices

$$\begin{aligned}&J_{uv}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&-1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{array} \right)\quad J_{xy}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{array} \right) \end{aligned}$$

(68)

$$\begin{aligned}&J_{ux}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{array} \right)\quad J_{uy}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{array} \right) \end{aligned}$$

(69)

$$\begin{aligned}&J_{vx}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{array} \right)\quad J_{vy}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&1&0&0 \end{array} \right) \end{aligned}$$

(70)

For our purposes the relevant elements of the Lie algebra of the AdS group are

$$\begin{aligned}&A=J_{ux}-J_{uv}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&1&1&0 \\ -1&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{array} \right),\end{aligned}$$

(71)

$$\begin{aligned}&B=J_{xy}+J_{vy}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&-1 \\ 0&1&1&0 \end{array} \right), \end{aligned}$$

(72)

$$\begin{aligned}&C=J_{uv}+J_{ux}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&-1&1&0 \\ 1&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{array} \right), \end{aligned}$$

(73)

$$\begin{aligned}&D=J_{xy}-J_{vy}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&0&-1 \\ 0&-1&1&0 \end{array} \right) \end{aligned}$$

(74)

Let $\mathbf{a}=(a,b)$ be a two dimensional vector. Then

$$\begin{aligned} \Lambda (\mathbf{a})=e^{aA+bB}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 1&a&a&0 \\ -a&1+\frac{b^{2}-a^{2}}{2}&\frac{b^{2}-a^{2}}{2}&b \\ a&\frac{a^{2}-b^{2}}{2}&1+\frac{a^{2}-b^{2}}{2}&-b \\ 0&b&b&1 \end{array} \right). \end{aligned}$$

(75)

If we apply this transformation to $X^{T}=(u,v,x,y)$ we get

$$\begin{aligned} \left( \begin{array}{c} u^{\prime } \\ v^{\prime } \\ x^{\prime } \\ y^{\prime } \end{array} \right)=\left( \begin{array}{cccc} u+av+ax \\ -au+\left(1+\frac{b^{2}-a^{2}}{2}\right)v + \left(\frac{b^{2}-a^{2}}{2}\right)x+by \\ au+\left(\frac{a^{2}-b^{2}}{2}\right)v + \left(1+\frac{a^{2}-b^{2}}{2}\right)x-by \\ bv+vx+y \end{array} \right). \end{aligned}$$

(76)

Using (29) we finally get

$$\begin{aligned} \xi {^{\prime }}^{1}=\xi ^{1}+a\quad \xi {^{\prime }}^{2}=\xi ^{2}+b. \end{aligned}$$

(77)

Hence $\Lambda (\mathbf{a})$ generate the translation subgroup on $\xi ^{1}$ and $\xi ^{2}$.

The special conformal transformations can be obtained in similar way. Let $\mathbf{c}=(c,d)$ be a two dimensional vector. Then

$$\begin{aligned} \Lambda (\mathbf{c})=e^{cC+dD}=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} 1&-c&c&0 \\ c&1+\frac{d^{2}-c^{2}}{2}&\frac{c^{2}-d^{2}}{2}&-d \\ c&\frac{d^{2}-c^{2}}{2}&1+\frac{c^{2}-d^{2}}{2}&-d \\ 0&-d&d&1 \end{array} \right). \end{aligned}$$

(78)

Applying this transformation to $X^{T}=(u,v,x,y)$ we get

$$\begin{aligned} \left( \begin{array}{c} u^{\prime } \\ v^{\prime } \\ x^{\prime } \\ y^{\prime } \end{array} \right)=\left( \begin{array}{cccc} u-cv+cx \\ cu+\left(1+\frac{d^{2}-c^{2}}{2}\right)v + \left(\frac{c^{2}-d^{2}}{2}\right)x -dy\\ cu+\left(\frac{d^{2}-c^{2}}{2}\right)v + \left(1+\frac{c^{2}-d^{2}}{2}\right)x -dy\\ -dv+dx+y \end{array} \right). \end{aligned}$$

(79)

Using again (29) we get

$$\begin{aligned} \xi {^{\prime }}^{1}=\frac{\xi ^{1}-c\left(\xi \cdot \xi \right)}{1-2\left(\xi \cdot \mathbf{c}\right)+\left(\mathbf{c}\cdot \mathbf{c} \right)\left(\xi \cdot \xi \right)}\quad \quad \quad \xi {^{\prime }}^{2} =\frac{\xi ^{2}-d\left(\xi \cdot \xi \right)}{1-2\left(\xi \cdot \mathbf{c} \right)+\left(\mathbf{c}\cdot \mathbf{c}\right)\left(\xi \cdot \xi \right)},\qquad \end{aligned}$$

(80)

where the inner product is with respect to the metric $\mathrm{diag}=(-1,1)$. Hence $\Lambda (\mathbf{c})$ generate the special conformal subgroup on $\xi ^{1}$ and $\xi ^{2}$.

Finally we will transform the coordinates $\xi ^{1}$ and $\xi ^{2}$ to complex coordinates in $\mathbb C ^{2}$. We have

$$\begin{aligned} \xi ^{1}+\xi ^{2}=\tan \left(\frac{\lambda +\theta }{2}\right)\quad \xi ^{1}-\xi ^{2}=\tan \left(\frac{\lambda -\theta }{2}\right). \end{aligned}$$

(81)

If we define

$$\begin{aligned} z_{1}\equiv \frac{1+i(\xi ^{1}+\xi ^{2})}{1-i(\xi ^{1}+\xi ^{2})} \quad z_{2}\equiv \frac{1+i(\xi ^{1}-\xi ^{2})}{1-i(\xi ^{1}-\xi ^{2})}, \end{aligned}$$

(82)

then

$$\begin{aligned} z_{1}=e^{i(\lambda -\theta )}\quad z_{2}=e^{i(\lambda +\theta )}. \end{aligned}$$

(83)

If now we make $\tau =i\lambda$ then

$$\begin{aligned} z_{1}=e^{\tau }e^{-i\theta }\quad z_{2}=e^{\tau }e^{i\theta }. \end{aligned}$$

(84)

We can consider $z_{1}$ and $z_{2}$ as complex conjugate of each other and define on the complex plane. However following the usual approach to conformal field theory [20] they can be considered as independent complex variables and define $\mathbb C ^{2}$. At this point we could apply the standard conformal field theory to our problem by using $z_{1}$ and $z_{2}$ as our complex variables.