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Linearized Einstein’s equation around pure BTZ from entanglement thermodynamics

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Abstract

It is known that the linearized Einstein’s equation around the pure AdS can be obtained from the constraint \( \Delta S = \Delta \langle H \rangle \), known as the first law of entanglement, on the boundary CFT. The corresponding dual state in the boundary CFT is the vacuum state around which the linear perturbation is taken. In this paper we revisit this question, in the context of \( AdS _3/ CFT _2 \), with the state of the boundary \( CFT _2\) as a thermal state. The corresponding dual geometry is a planar BTZ black hole. By considering the linearized perturbation around this black brane we show that Einstein’s equation follows from the first law of entanglement.

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Notes

  1. For related discussions one can also see, e.g, [22,23,24,25,26]

  2. For relative entropy of excited states in two dimensional CFT see [38, 39].

  3. We thank David Blanco for poiting out reference [40] to us.

  4. Our convention is that we use indices KL for the three bulk coordinates \( \lbrace z,t,x \rbrace \) and indices \( \mu ,\,\nu \) for two boundary coordinates \( \lbrace t, x \rbrace \).

  5. Linear perturbations around BTZ black brane were also considered in [43] and the first order correction to holographic entanglement entropy was calculated. They have also discussed the dynamics of the shift of holographic entanglement entropy. The fact that \( 2+1 \)-dimensional gravity has no propagating modes indeed imposes two constraints on the \( \Delta S_A \) in the 2-dimensional boundary theory. This can be seen, for example, in the equations (2.16) and (2.17) of [43]. Equation (2.16) is special for \( 2+1 \)-dimensional gravity, it determines the time evolution of \( \Delta S_A \). Whereas in the higher dimensional case one can only get one constraint, equation (2.17), for \( \Delta S_A \). One can also go the other way: from boundary field theory one can derive the time evolution equation of \( \Delta S_A \) (equations (2.16) of [43]) and then using the first law of entanglement thermodynamics one can derive the Eq. (34). This can be seen in equation (VI.11) and the discussion around that equation in [26].

  6. We are grateful to David Blanco for bringing this paper to our attention.

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Acknowledgements

We are grateful to Shamik Banerjee for suggesting this problem and guiding us throughout the project. We are also thankful to him for carefully reading the draft and for correcting some parts of it. We are thankful to Amitabh Virmani and Jyotirmay Bhattacharya for helpful discussions on related matters. We would also like to acknowledge the hospitality at Chennai Mathematical Institute during the workshop ‘Student Talks in Trending Topics in Theory \((\hbox {ST}^4)\)’ where part of this work was done. PR is grateful to Debangshu Mukherjee for his generous help and multiple discussions on this work. PR is partially supported by a grant to CMI from the Infosys Foundation.

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Appendix A: Analysis to show that \( \Delta ^{(n)}_{\mu \nu }(t_0,x_0) = 0 \) for all n

Appendix A: Analysis to show that \( \Delta ^{(n)}_{\mu \nu }(t_0,x_0) = 0 \) for all n

The jth derivative of \( I_{n,m}(R) \) (46) gives

$$\begin{aligned} \partial _R^j I_{n,m}(R) = \sum _{i=0}^{j-1}\partial _R^{j-i-1}\left( \partial _R^i f_{n,m}(x,R)|_{x=R} \right) +\int _0^R \partial _R^j f_{n,m}(x,R) dx \end{aligned}$$
(A1)

where

$$\begin{aligned} f_{n,m}(x,R) = x^{2m}(\cosh R + \cosh x)^{-\frac{n}{2}+1}(\cosh R - \cosh x)^{\frac{n}{2}+1}. \end{aligned}$$
(A2)

Thus from (48) we have

$$\begin{aligned}&\sum _{n,m}\frac{2^{n+1}}{(2m)!}\partial _x^{2m} \Delta ^{(n)}_{xx}(t_0,x_0)\left[ \sum _{i=0}^{J+2}\partial _R^{J-i+2}\left( \partial _R^i f_{n,m}(x,R)|_{x=R}\right) \right] _{R=0} \nonumber \\&\quad = 0. \end{aligned}$$
(A3)

First we will show that \(\Delta _{xx}=0\) for even integers. For odd integers, we check it numerically below. Writing 2l for the even integer n, and simplifying a bit, we get

$$\begin{aligned} \sum _{l=0}^{J+1}\sum _{m}\frac{(l+1)!\,2^{l}}{(2m)!}\partial _x^{2m} \Delta ^{(2l)}_{xx}(t_0,x_0)C^{J,l,m}(R=0) = 0, \end{aligned}$$
(A4)

where

$$\begin{aligned} C^{J,l,m}(R=0) = \left[ \partial _R^{J-(l-1)} \left( R^{2m}\cosh ^{1-l}R\, \sinh ^{1+l}R \right) \right] _{R=0}. \end{aligned}$$
(A5)

Notice that we have restricted l to range from 0 to \(J+1.\) This is because the terms with \(l>J+1\) all vanish at \(x=R\). Now, (A4) can be written as

$$\begin{aligned} \sum _{l}(l+1)!2^{l} \Delta ^{(2l)}_{xx}(t_0,x_0)C^{J,l,0}(R=0)= & {} -\sum _{l,m \ne 0}\frac{(l+1)!\,2^{l}}{(2m)!} \times \nonumber \\&\partial _x^{2m}\Delta ^{(2l)}_{xx}(t_0,x_0)C^{J,l,m}(R=0).\nonumber \\ \end{aligned}$$
(A6)

Setting \( J=0, \) we get

$$\begin{aligned} \Delta _{xx}^{(0)}(t_0,x_0) = 0. \end{aligned}$$
(A7)

\(J=2\) gives

$$\begin{aligned} \Delta _{xx}^{(2)}(t_0,x_0)&= \ -\frac{1}{2}\Delta _{xx}^{(0)}(t_0,x_0) + \frac{3}{8}\partial _x^2\Delta _{xx}^{(0)}(t_0,x_0),\nonumber \\ \Rightarrow \qquad \Delta _{xx}^{(2)}(t_0,x_0)&= \ 0. \end{aligned}$$
(A8)

\(J=4\) gives

$$\begin{aligned} \Delta _{xx}^{(4)}(t_0,x_0)&= \ -\frac{2}{9}\Delta _{xx}^{(2)}(t_0,x_0) + \frac{1}{3}\partial _x^2\Delta _{xx}^{(2)}(t_0,x_0)\nonumber \\&\quad -\frac{1}{9}\Delta _{xx}^{(0)}(t_0,x_0) + \frac{5}{18}\partial _x^2\Delta _{xx}^{(0)}(t_0,x_0)\nonumber \\&\quad + \frac{5}{144}\partial _x^4\Delta _{xx}^{(0)}(t_0,x_0)\nonumber \\ \Rightarrow \qquad \Delta _{xx}^{(4)}(t_0,x_0)&= \ 0 \end{aligned}$$
(A9)

Let’s check it for \( J=2N \). It is easy to see that for all \( l \ge N+1 \) the coefficients C are zero for all m. For \( l = N \)

$$\begin{aligned} C^{2N,N,0}(R=0)= & {} [\partial _R^{N+1}(\cosh ^{1-N}R\sinh ^{1+N}R)]_{R=0} \nonumber \\= & {} (N+1)! \end{aligned}$$
(A10)

So the last non vanishing term in the l series in (A6) with \(m=0\) is \( \Delta _{xx}^{(2N)}(t_0,x_0) \) and with \( m \ne 0 \) it will be a lower order term. Now notice that this term is a linear combination of all the lower order terms and their derivatives, which are zero. Hence \( \Delta _{xx}^{(2N)}(t_0,x_0) \) is also zero for \(N=0,1,\ldots \ .\)

Fig. 4
figure 4

\( C^4_{n,m} \) versus R for different values of n and m

Fig. 5
figure 5

\( C^6_{n,m} \) versus R for different values of n and m

Fig. 6
figure 6

\( C^8_{n,m} \) versus R for different values of n and m. These figures show for a particular j what coefficients \( C^j_{n,m} \) are non-zero as \( R\rightarrow 0 \)

Fig. 7
figure 7

\( C^4_{xx} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

Fig. 8
figure 8

\( C^4_{tx} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

Fig. 9
figure 9

\( C^4_{tt} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

Fig. 10
figure 10

\( C^5_{xx} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

Fig. 11
figure 11

\( C^5_{tx} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

Fig. 12
figure 12

\( C^5_{tt} \) versus R for different values of \(n, \ m_t \) and \(m_x\)

1.1 Numerical analysis for odd n

Consider the expansion (47):

$$\begin{aligned} \sum _{n,m,j}\frac{2^{n+1}}{(2m)!}\partial _x^{2m} \Delta ^{(n)}_{xx}(t,x_0)R^{j} C^j_{n,m}(0) = 0 , \end{aligned}$$
(A11)

where

$$\begin{aligned} C^j_{n,m}(0)=\frac{1}{j!}\left[ \partial ^j_R I_{n,m}(R) \right] _{R=0}. \end{aligned}$$

We have already seen that the first two terms in this series do not give any constraint and that the terms with odd powers of R (i.e, odd j) impose constraints on even n. Thus one should expect that terms with alternative powers of R (i.e, even j) should impose constraints on odd n. Let us check the term of order \( R^4 .\) Using numerics one can see that the coefficient \( C^4_{n,m}(0) \) is zero for all nm except for \( n=1,m=0 \). The relevant plots are shown in Fig. 4.

At order \( R^6 \), there are three non-zero coefficients, namely, \( C^6_{3,0},C^6_{1,0},C^6_{1,1} \). This implies that \( \Delta ^{(3)}_{xx}(t_0,x_0) \) can be written in terms of \( \Delta ^{(1)}_{xx}(t_0,x_0) \) and \( \partial _x^2\Delta ^{(1)}_{xx}(t_0,x_0) \), and hence \( \Delta ^{(3)}_{xx}(t_0,x_0) \) is also 0. Figure 5 shows the behavior of coefficients at \( R=0 \).

At order \( R^8 \), there are several non-zero coefficients which are shown in Fig. 6. From the plot, one can deduce that \( \Delta ^{(5)}(t_0,x_0) \) can be written as a linear combination of the lower order terms and their derivatives, and hence is 0.

From this pattern, we conclude that in general the higher order coefficients can be written in terms of lower order coefficients. This means that

$$\begin{aligned} \Delta ^{(n)}_{xx}(t_0,x_0)=0 \end{aligned}$$

for all n.

Some numerical computations for boosted black brane

Here we have plotted different coefficients of \(R^4\hbox {th}\) and \(R^5\hbox {th}\) terms in the expansion series (65) around \( R=0 \). They are denoted by \( C^4_{\mu \nu } \) and \( C^5_{\mu \nu } \) respectively.

Figures 7, 8 and 9 show that all the coefficients of \(R^4\hbox {th}\) term are zero as \( R\rightarrow 0 \) except when \(n=1\) and \( m_t=m_x=0 \) which give the first equation in (69) with \( \beta =0.5 \). Similarly for \(R^5\hbox {th}\) term the only non-zero coefficient with maximum value of n as \( R\rightarrow 0 \) is \( n=2 \) and \( m_t=m_x=0 \) (Figs. 10, 11, 12). This can be expressed as a linear combination of the terms with lower values of n which are already zero. These give the second equation in (69).

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Paul, P., Roy, P. Linearized Einstein’s equation around pure BTZ from entanglement thermodynamics. Gen Relativ Gravit 51, 155 (2019). https://doi.org/10.1007/s10714-019-2636-9

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