Spacetime Evolution and Equipartition in Lanczos-Lovelock Gravity

Abstract

In the case of general relativity it is possible to interpret the Noether charge in any bulk region of spacetime as the heat content of its boundary surface. Further, the time evolution of spacetime metric in Einstein’s theory arises due to the difference between suitably defined surface and bulk degrees of freedom. In this chapter we demonstrate that this thermodynamic interpretation can be generalized in a natural fashion to all Lanczos-Lovelock theories of gravity. The Noether charge, related to time evolution vector field, in Lanczos-Lovelock gravity evaluated in a bulk region of space is equal to the heat content of the boundary surface. Here the temperature is defined using local Rindler observers and the entropy identified as the Wald entropy. Defining the surface degrees of freedom using Wald entropy and Komar energy density to define the bulk degrees of freedom, we can also show that the time evolution of the geometry is sourced by their difference. If the spacetime admits a timelike Killing vector field the holographic equipartition for Lanczos-Lovelock gravity follows.

Appendices

Appendix A

In this appendix, we shall present the supplementary material for this chapter.

A.1 Derivation of Noether Current from Differential Identities in Lanczos-Lovelock Gravity

In this section the Noether current for Lanczos-Lovelock gravity will be derived starting from identities in differential geometry without using any diffeomorphism invariance of action principles. The conceptual importance of this approach has already been emphasized in [1], in the context of Einstein gravity, and we shall generalize the result for Lanczos-Lovelock models. We start with the fact that the covariant derivative of any vector field can be decomposed into a symmetric and an antisymmetric part. From the antisymmetric part we can define another antisymmetric tensor field as,

$$\begin{aligned} 16 \pi J^{aj}=2P^{ajki}\nabla _{k}v_{i}=P^{ajki}\left( \nabla _{k}v_{i}-\nabla _{i}v_{k}\right) \end{aligned}$$

(5.44)

It is evident from the antisymmetry of $P^{abcd}$ that a conserved current exists such that, $J^{a}=\nabla _{j}J^{aj}$. We recall the identities:

$$\begin{aligned} \left( \nabla _{j}\nabla _{k}-\nabla _{k}\nabla _{j} \right) v^{i}=R^{i}_{~cjk}v^{c} \end{aligned}$$

(5.45)

and,

$$\begin{aligned} \mathcal {L}_{v}\Gamma ^{i}_{jk}=\nabla _{j}\nabla _{k}v^{i}-R^{i}_{~kjm}v^{m} \end{aligned}$$

(5.46)

and use them in the definition in Eq. (2.36) to get:

$$\begin{aligned} \mathcal {R}^{ab}v_{b}= & {} P^{aijk}R^{b}_{~ijk}v_{b} =-P^{aijk}\left( \nabla _{j}\nabla _{k}-\nabla _{k}\nabla _{j}\right) v_{i} \nonumber \\= & {} P^{aijk}\nabla _{k}\nabla _{j}v_{i}+\left( P^{akij}+P^{ajki}\right) \nabla _{j}\nabla _{k}v_{i} \nonumber \\= & {} P^{aijk}\nabla _{k}\nabla _{j}v_{i}+P^{akij}\nabla _{j}\nabla _{k}v_{i} +\nabla _{j}\left( P^{ajki}\nabla _{k}v_{i} \right) \end{aligned}$$

(5.47)

where in the second line we have used the identity, $P^{a(bcd)}=0$. Then from Eq. (5.44) we obtain:

$$\begin{aligned} 16 \pi J^{a}= & {} 2\mathcal {R}^{ab}v_{b}-2P^{aijk}\nabla _{k}\nabla _{j}v_{i}-2P^{akij}\nabla _{j}\nabla _{k}v_{i} \nonumber \\= & {} 2\mathcal {R}^{ab}v_{b}+2P_{i}^{~ajk}\nabla _{k}\nabla _{j}v^{i}-2P_{i}^{~jak}\nabla _{j}\nabla _{k}v^{i} \nonumber \\= & {} 2\mathcal {R}^{ab}v_{b}+2P_{i}^{~ajk}\left( \mathcal {L}_{v}\Gamma ^{i}_{kj}+R^{i}_{~jkm}v^{m} \right) -2P_{i}^{~jak}\left( \mathcal {L}_{v}\Gamma ^{i}_{jk}+R^{i}_{~kjm}v^{m}\right) \nonumber \\= & {} 2\mathcal {R}^{ab}v_{b}+2P_{i}^{~jka}\mathcal {L}_{v}\Gamma ^{i}_{jk} \end{aligned}$$

(5.48)

while arriving at the third line we have used Eq. (5.46) and for the last line we have used the fact that, $P^{ijak}R_{ikjm}=P^{akij}R_{ikjm}=-P^{kaij}R_{ikjm}=P^{kaij}R_{kijm}$. Thus Eq. (2.47) can be derived without any reference to the diffeomorphism invariance of the gravitational action, using only the identities in differential geometry and various symmetry properties.

A.2 Projection of Noether Current Along Acceleration and Newtonian Limit

The analysis in [1] uses the fact that $u_aJ^a(\xi )$ is a 3-divergence, so that the spatial volume integral of $u_aJ^a(\xi )$ can be converted to a surface integral. In this case, it is natural to interpret $u_aJ^a(\xi )$ as a spatial density, viz., charge per unit volume of space. It turns out that similar results can be obtained even for the component of $J^a(\xi )$ in the direction of the normal to the equipotential surface along the following lines. It can be easily shown that

$$\begin{aligned} \hat{a}_{p}J^{p}(\xi )=-\left( g^{ij}-\hat{a}^{i}\hat{a}^{j}\right) \nabla _{i}\left( 2Nau_{j}\right) =-g^{ij}_{\perp }\nabla _{i}\left( 2Nau_{j}\right) \end{aligned}$$

(5.49)

where the tensor $g^{ij}_{\perp }$ acts as a projection tensor transverse to the unit vector $\hat{a}^i$. However in order to define a surface covariant derivative we need $\hat{a}^{i}$ to foliate the spacetime, which in turn implies $u^{i}\nabla _{i}N=0$. In this case Eq. (5.49) can be written as $\hat{a}_{p}J^{p}(\xi )=-\mathcal {D}_{i}(2Nau^{i})$, where $\mathcal {D}_{i}$ is the covariant derivative operator corresponding to the induced metric $g^{ij}_{\perp }$ on the $N=\text {constant}$ surfaces with normal $\hat{a}_{i}$. (When $u^{i}\nabla _{i}N=0$, we have $a_{j}=\nabla _{j}\ln N$.) To obtain an integral version of this result, let us transform from the original $(t,x^{\alpha })$ coordinates to a new coordinate system $(t,N,x^{A})$ using N itself as a “radial” coordinate. In this coordinate we have $a_{i}\propto \delta _{i}^{N}$ and thus $u^{N}=0$, thanks to the relation $u^{i}a_{i}=0$. Thus $\mathcal {D}_{i}(2Nau^{i})$ will transform into $\mathcal {D}_{\bar{\alpha }}(2Nau^{\bar{\alpha }})$, where $\bar{\alpha }$ stands for the set of coordinates ($t,x^{A}$) on the N = constant surface. Integrating both sides $\hat{a}_{p}J^{p}(\xi )=-\mathcal {D}_{\bar{\alpha }}(2Nau^{\bar{\alpha }})$, over the $N=\text {constant}$ surface will now lead to the result (with restoration of $1/16\pi$ factor):

$$\begin{aligned} \int d^{2}xdt\sqrt{-g_{\perp }}\,\hat{a}_{p}J^{p}(\xi )=-\int d^{2}x\sqrt{q}N\left( \frac{Na}{8\pi }\right) u^{t} =\int d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) \Bigg |^{t_1}_{t_2} \end{aligned}$$

(5.50)

where we have used the standard result $\sqrt{-g_{\perp }}=N\sqrt{q}$, with q being the determinant of the two-dimensional hypersurface. The right hand side can be thought of as the difference in the heat content $Q(t_2)-Q(t_1)$ between the two surfaces $t=t_2$ and $t=t_1$ where:

$$\begin{aligned} Q(t)\equiv \int d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) =\int d^{2}x (Ts) \end{aligned}$$

(5.51)

This looks very similar to the result we obtained in the case of the integral over $u_i J^i$ earlier (see Eq. (5.1) with $a^\alpha =r^\alpha$ on the $N=$ constant surface), but there is a difference in the interpretation of the left hand side. While $u_i J^i$ can be thought of as the charge density per unit spatial volume, the quantity $\hat{a}_{p}J^{p}(\xi )$ represents the flux of Noether current through a time-like surface; therefore, $\hat{a}_{p}J^{p}(\xi )$ should be thought of as a current per unit area per unit time. We will see later that the flux of Noether current through null surfaces leads to a very similar result.

We conclude this section with a discussion of the Newtonian limit of general relativity using the Noether current which has some amusing features. The Newtonian limit is obtained by setting $N^{2}=1+2\phi$, $g_{0\alpha }=0$ and $g_{\alpha \beta }=\delta _{\alpha \beta }$, where $\phi$ is the Newtonian potential [3]. Then the acceleration of the fundamental observers turn out to be $a_{\alpha }=\partial _{\alpha }\phi$. Since the spatial section of the spacetime is flat, the extrinsic curvature identically vanishes and so does the Lie variation term. Also $2\bar{T}_{ab}u^{a}u^{b}=\rho _\mathrm{Komar}=\rho$, which immediately leads to (with G inserted, see Eq. (5.2)):

$$\begin{aligned} \nabla ^{2}\phi =4\pi G\rho \end{aligned}$$

(5.52)

the correct Newtonian limit. The same can also be obtained using the four velocity $u_{a}$. The Noether charge associated with $u_{a}$ turns out to have the following expression [1]

$$\begin{aligned} D_{\alpha }a^{\alpha }=16\pi u_{a}J^{a}(u)=16\pi \bar{T}_{ab}u^{a}u^{b}+u_{a}g^{bc}\pounds _{u}N^{a}_{bc} \end{aligned}$$

(5.53)

In the Newtonian limit the following results hold $2\bar{T}_{ab}u^{a}u^{b}=\rho$ and $u_{a}g^{bc}\pounds _{u}N^{a}_{bc}=-D_{\alpha }a^{\alpha }$ (which follows from the Newtonian limit of the result $u_{a}g^{bc}\pounds _{\xi }N^{a}_{bc}=ND_{\alpha }a^{\alpha }+2a^{\alpha }D_{\alpha }N-Nu_{a}g^{ij}\pounds _{u}N^{a}_{ij}$ and the fact that in spacetime with flat spatial section the term $u_{a}g^{bc}\pounds _{\xi }N^{a}_{bc}$ identically vanishes). This immediately leads to Eq. (5.52).

We also see that the Noether charge is positive as long as $\rho >0$ in the Newtonian limit. In fact, the Noether charge contained inside any equipotential surface is always a positive definite quantity as long as $r^\alpha$ and $a^\alpha$ point in the same direction (which happens when $\bar{T}_{ab}u^{a}u^{b}>0$). To prove this we can integrate the Noether charge over a small region on a $t=\text {constant}$ hypersurface to obtain,

$$\begin{aligned} \int _{t=\mathrm {constant}}d^{3}x\sqrt{h}u_{a}J^{a}(u)&=\frac{1}{8\pi }\int _{N,t=\text {constant}}d^{2}x\sqrt{q}2Na^{\alpha }r_{\alpha } \nonumber \\&=\int _{N,t=\mathrm {constant}}d^{2}x\left( \frac{Na}{2\pi }\right) \left( \frac{\sqrt{q}}{4}\right) \end{aligned}$$

(5.54)

Since $\rho$ is positive definite in this case the fundamental observers are accelerating outwards and thus $r_{\alpha }a^{\alpha }=a$. The temperature as measured by these fundamental observers is a positive definite quantity and so is the entropy density and hence the positivity of Noether charge follows.

A.3 Identities Regarding Noether Current in Lanczos-Lovelock Action

The Noether potential $J^{ab}$ is antisymmetric in (a, b) and from its expression given by Eq. (2.46) it is evident that $J^{ab}(q)$ would identically vanish for $q_{a}=\nabla _{a}\phi$. We will use the above fact in order to obtain a relation between the Noether current for two vector fields $q_{a}$ and $v_{a}$ connected by $v_{a}=f(x)q_{a}$. This result, in the case of general relativity is detailed in [1]. While in the case of Lanczos-Lovelock gravity the Noether current for a vector field $v_{a}=f(x)q_{a}$ can be decomposed as:

$$\begin{aligned} 16 \pi J^{ab}(v)= & {} 2P^{abcd}\nabla _{c}\left( fq_{d}\right) \nonumber \\= & {} 2P^{abcd}q_{d}\nabla _{c}f+2fP^{abcd}\nabla _{c}q_{d} \end{aligned}$$

(5.55)

Then the corresponding Noether current has the following expression:

$$\begin{aligned} 16 \pi J^{a}(v)= & {} 2P^{abcd}\nabla _{b}\left( q_{d}\nabla _{c}f\right) +2P^{abcd}\nabla _{b}\left( f\nabla _{c}q_{d}\right) \nonumber \\= & {} 2P^{abcd}q_{d}\nabla _{b}\nabla _{c}f+2P^{abcd}\nabla _{c}f\nabla _{b}q_{d} \nonumber \\+ & {} 2P^{abcd}\nabla _{b}f\nabla _{c}q_{d}+2fP^{abcd}\nabla _{b}\nabla _{c}q_{d} \end{aligned}$$

(5.56)

From the above equation we readily arrive at:

$$\begin{aligned} 16 \pi \left\{ J^{a}(v)-fJ^{a}(q)\right\}= & {} 2P^{abcd}q_{d}\nabla _{b}\nabla _{c}f+2P^{abcd}\nabla _{c}f\nabla _{b}q_{d} +2P^{abcd}\nabla _{b}f\nabla _{c}q_{d} \nonumber \\= & {} P^{abcd}\nabla _{b}A_{cd}+16 \pi J^{ab}(q)\nabla _{b}f \end{aligned}$$

(5.57)

where we have defined the antisymmetric tensor $A_{cd}$ as $A_{cd}=q_{d}\nabla _{c}f-q_{c}\nabla _{d}f$. Now consider the following result: $q_{a}\nabla _{b}A_{cd}=\nabla _{b}\left( q_{a}A_{cd}\right) -A_{cd}\nabla _{b}q_{a}$ which leads to:

$$\begin{aligned} P^{abcd}q_{a}\nabla _{b}A_{cd}= & {} \nabla _{b}\left( P^{abcd}q_{a}A_{cd}\right) -2P^{abcd}q_{d}\nabla _{c}f\nabla _{b}q_{a} \nonumber \\= & {} \nabla _{b}\left( P^{abcd}q_{a}A_{cd}\right) -16 \pi q_{a}J^{ab}\left( q\right) \nabla _{b}f \end{aligned}$$

(5.58)

Then Eq. (5.57) can be rewritten in the following manner:

$$\begin{aligned} 16 \pi \left\{ q_{a}J^{a}(fq)-fq_{a}J^{a}(q)\right\}= & {} 16 \pi J^{ab}(q)\nabla _{b}fq_{a} +\nabla _{b}\left( P^{abcd}q_{a}A_{cd}\right) -16 \pi q_{a}J^{ab}(q)\nabla _{b}f \nonumber \\= & {} \nabla _{b}\left( 2P^{abcd}q_{a}q_{d}\nabla _{c}f\right) \end{aligned}$$

(5.59)

It can be easily verified that in the Einstein-Hilbert limit $P^{abcd}=Q^{abcd}=(1/2)\left( g^{ac}g^{bd}-g^{ad}g^{bc}\right)$, under which the above equation reduces to the respective one in general relativity.

Applying the above equation to $u_{a}=-N\nabla _{a}t$ with $q_{a}=\nabla _{a}t=-u_{a}/N$ and $f=-N$ we arrive at:

$$\begin{aligned} 16 \pi u_{a}J^{a}\left( u\right) =2N\nabla _{b}\left( P^{abcd}u_{a}u_{d}\frac{\nabla _{c}N}{N^{2}}\right) \end{aligned}$$

(5.60)

In order to proceed we define a new vector field such that:

$$\begin{aligned} \chi ^{a}= & {} -2P^{abcd}u_{b}u_{d}\frac{\nabla _{c}N}{N} \nonumber \\= & {} -2P^{abcd}u_{b}u_{d}\left( a_{c}-\frac{1}{N}u_{c}u^{j}\nabla _{j}N\right) \nonumber \\= & {} -2P^{abcd}u_{b}a_{c}u_{d} \end{aligned}$$

(5.61)

Note that in the Einstein-Hilbert limit this vector reduces to the acceleration four vector as follows:

$$\begin{aligned} \chi ^{a}=-2P^{abcd}u_{b}a_{c}u_{d}=-\left( g^{ac}g^{bd}-g^{ad}g^{bc}\right) u_{b}a_{c}u_{d} =-u^{b}u_{b}a^{a}+u^{b}a_{b}u^{a}=a^{a} \end{aligned}$$

(5.62)

Also just as in the case of acceleration for the vector $\chi ^{a}$ as well we have:

$$\begin{aligned} u_{a}\chi ^{a}= & {} -2aP^{ab\beta d}u_{a}u_{b}r_{\beta }u_{d}=0 \end{aligned}$$

(5.63)

where antisymmetry of $P^{abcd}$ in the first two components has been used. We can also have the following relation for the vector field $\chi ^{a}$:

$$\begin{aligned} Na_{b}\chi ^{b}=\chi ^{b}\nabla _{b}N +\chi ^{b}u_{b}u^{j}\nabla _{j}N=\chi ^{b}\nabla _{b}N \end{aligned}$$

(5.64)

where we have used the relation $u_{a}\chi ^{a}=0$ from Eq. (5.63). Thus Eq. (5.60) can be written in terms of the newly defined vector field $\chi ^{a}$ in the following way:

$$\begin{aligned} 16 \pi u_{a}J^{a}\left( u\right)= & {} N\nabla _{b}\left( \frac{\chi ^{b}}{N}\right) \nonumber \\= & {} \nabla _{b}\chi ^{b}-\frac{\nabla _{b}N}{N}\chi ^{b} \nonumber \\= & {} D_{\alpha }\chi ^{\alpha } \end{aligned}$$

(5.65)

The last relation follows from the fact that:

$$\begin{aligned} D_{\alpha }\chi ^{\alpha }= D_{b}\chi ^{b}=\nabla _{b}\chi ^{b}-a_{b}\chi ^{b}=\nabla _{b}\chi ^{b}-\frac{\nabla _{b}N}{N}\chi ^{b} \end{aligned}$$

(5.66)

Then it is straightforward to get the Noether current for $\xi ^{a}$ by using $q_{a}=u_{a}$ and $f=N$ in Eq. (5.59) with Eq. (5.65) as:

$$\begin{aligned} 16 \pi u^{a}J_{a}\left( \xi \right)= & {} 16 \pi Nu_{a}J^{a}(u)+\nabla _{b}\left( N\chi ^{b}\right) \nonumber \\= & {} ND_{\alpha }\chi ^{\alpha }+\nabla _{b}\left( N\chi ^{b}\right) \nonumber \\= & {} D_{\alpha }\left( 2N\chi ^{\alpha }\right) \end{aligned}$$

(5.67)

Here also we have used the following identity:

$$\begin{aligned} D_{\alpha }\left( N\chi ^{\alpha }\right)= & {} \left( g^{ij}+u^{i}u^{j}\right) \nabla _{i}\left( N\chi _{j}\right) \nonumber \\= & {} \nabla _{i}\left( N\chi ^{i}\right) +u^{i}u^{j}\nabla _{i}\left( N\chi _{j}\right) \nonumber \\= & {} N\nabla _{i}\chi ^{i}+N\chi ^{i}a_{i}-N\chi ^{j}\left( u^{i}\nabla _{i}u_{j}\right) \nonumber \\= & {} N\nabla _{i}\chi ^{i} \end{aligned}$$

(5.68)

Thus we have derived the desired relation for the Noether current of the vector field $\xi _{a}$ and it turns out to have identical structure as that of Einstein-Hilbert action with $\chi ^{a}$ playing the role of four acceleration.