1. Introduction
During the past two decades, the AdS/CFT correspondence has become a powerful tool for studying the real-time dynamics of strongly-coupled systems [1-4]. It provides us with a geometric approach to calculate transport coefficients of boundary systems explicitly in the AdS bulk by considering various black hole solutions that can dissipate the surrounding fluctuations outside their horizons. One of the most important discoveries of this method is that for a wide class of interacting systems, there exists a lower bound on the shear viscosity. And especially, for those having a gravity dual that can be described by Einstein gravity, the ratio of shear viscosity to entropy density η/s meets a universal value ħ/4πkB which is called the Kovtun-Son-Starinet (KSS) bound [5-7]. Remarkably, such a finding in theory also explains why the ratio of shear viscosity to entropy density of the quark-gluon-plasma (QGP) is much smaller when compared to the result from the perturbative calculations of quantum field theory [8, 9]. This bound has been tested in various experiments of different realistic systems [10-14].
Nevertheless, it was found that η/s can be corrected when the finite N effect is considered. In this case, the viscosity bound can be slightly pushed down [15, 16].3 In addition, the viscosity bound can be strongly violated when matter fields are included. Numerous specific examples have shown that this can happen when the rotational symmetry or the translational symmetry is broken (Examples of these two classes can be seen in [27-41] and [42-57], respectively.). In most of these cases, the way of breaking the symmetries is explicit. However, a recent holographic study on the p-wave superfluid model shows that the ratio η/s can be enhanced even in an anisotropic system if the rotations are broken in a spontaneous manner [58]. So far, this model appears to be the only known anisotropic case that obeys the KSS bound.
In this work, we investigate the shear elasticity and the shear viscosity of an anisotropic holographic axion model at the perturbative level. This model allows us to realize the broken translations and the broken rotations explicitly and/or spontaneously via different setups. The boundary dual of this model is supposed to be certain viscoelastic strongly-coupled solids that exhibit both elastic responses as normal crystals and viscous damping as fluids. We expect that such a holographic tool will deepen our understanding of the viscoelastic properties of these complex materials in the real world. In particular, we would like to investigate the fate of the KSS bound in the case of spontaneous breaking of the symmetries. Our result shows that in contrast to the p-wave superfluid, the KSS bound is always violated in our model.
2. Anisotropic black hole solutions
We consider a simple holographic axion model [43, 48, 52, 59-62] in 4-dimensional spacetime, of which the action is given by 2 ). Following the standard quantization, the leading mode of the axion fields sets the external source for the dual scalar operators on the boundary. The expectation values of the scalar operators correspond to the subleading mode of the axion fields. For n < 5/2, the r-independent term in (8 ) dominates over the expansion and plays the role of the external source. Then, the profile (2 ) with kx ≠ ky implies that the translations and the rotations are both broken explicitly. While, for n > 5/2, the r-independent term becomes subleading and the breaking of symmetries is spontaneous [63]. In the next section, we will investigate how the broken translations and the broken rotations affect the shear viscosity-entropy density ratio in these two scenarios.
$\begin{eqnarray}\begin{array}{rcl}S & = & \displaystyle \int {{\rm{d}}}^{4}x\,\sqrt{-g}\left(R+\frac{6}{{L}^{2}}-{\lambda }^{2}\,V(X)\right),\\ X & \equiv & \frac{1}{2}{\partial }_{\mu }{\phi }^{i}{\partial }^{\mu }{\phi }^{i},\,\,i=x,y\end{array}\end{eqnarray}$
where we have adopted the convention that 16πG ≡ 1, R is the Ricci scalar, λ2 is the effective coupling constant with the dimension of mass square,4V is a general function of X, and L is the AdS radius. To break the translations of the boundary system, the profiles of the scalars in the bulk should be chosen as $\begin{eqnarray}{\phi }^{i}={{ \mathcal M }}_{j}^{i}\,{x}^{j},\quad {{ \mathcal M }}_{j}^{i}=\left(\begin{array}{cc}{k}_{x} & 0\\ 0 & {k}_{y}\end{array}\right).\end{eqnarray}$
Then, the general AdS black hole ansatz should be taken as $\begin{eqnarray}\begin{array}{rcl}{\rm{d}}{s}^{2} & = & {g}_{tt}{\rm{d}}{t}^{2}+{g}_{rr}{\rm{d}}{r}^{2}+{g}_{xx}{\rm{d}}{x}^{2}+{g}_{yy}{\rm{d}}{y}^{2}\\ & = & -A(r){\rm{d}}{t}^{2}+\frac{{\rm{d}}{r}^{2}}{A(r)}+B(r){\rm{d}}{x}^{2}+C(r){\rm{d}}{y}^{2}\end{array}\end{eqnarray}$
of which the metric components, at the AdS boundary r → ∞, should satisfy $\begin{eqnarray}A(r\to \infty )=B(r\to \infty )=C(r\to \infty )=\frac{{r}^{2}}{{L}^{2}}.\end{eqnarray}$
The location of the horizon rh is then defined by A(r = rh) = 0. For simplicity, we will set L to be one,5 and consider a class of specific models with $\begin{eqnarray}V(X)={X}^{n}.\end{eqnarray}$
By considering ${\phi }^{i}=\bar{{\phi }^{i}}+\delta {\phi }^{i}$ and expanding the action to the second order in δΦi, we obtain $\begin{eqnarray}\frac{1}{2}V^{\prime} (\bar{X}){\partial }_{\mu }\delta {\phi }^{i}{\partial }^{\mu }\delta {\phi }^{i}+\bar{X}V^{\prime\prime} (\bar{X}){({\partial }_{i}\delta {\phi }^{i})}^{2}+\cdots \,.\end{eqnarray}$
The absence of ghost requires monotonic potentials $\begin{eqnarray}V^{\prime} (\bar{X})\gt 0,\end{eqnarray}$
which gives us n > 0 [59]. For general n > 0, the axions behave like $\begin{eqnarray}{\phi }^{i}(r,{x}^{\mu })={\phi }_{(0)}^{i}({x}^{\mu })+{\phi }_{(1)}^{i}({x}^{\mu })\,{r}^{2n-5}+\cdots \end{eqnarray}$
near the AdS boundary, where the r-independent term corresponds to the profiles of the axions (In the following, we solve the background solutions perturbatively in two cases:
One-axion case
First, let us consider the single-axion case where kx = 0, ky ≠ 0 are imposed without loss of generality. Note that the low anisotropic regime can be achieved by setting λ ≪ 1 and fixing the value of ky. Then, it is found that the background can be expressed perturbatively up to the leading order for ${ \mathcal O }({\lambda }^{2})$, 12 )-(14 ) are obtained by choosing the integral constants properly, which ensures the regularity of the background solution in the bulk and does not disrupt the asymptotic AdS geometry. For more details about the derivation, one refers to Appendix A . Then, the Hawking temperature can be read off as
$\begin{eqnarray}A(r)={r}^{2}\left(1-\frac{{r}_{h}^{3}}{{r}^{3}}\right)+{\lambda }^{2}\,{a}_{1}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
$\begin{eqnarray}B(r)={r}^{2}+{\lambda }^{2}\,{b}_{1}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
$\begin{eqnarray}C(r)={r}^{2}-{\lambda }^{2}\,{b}_{1}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
with $\begin{eqnarray}{a}_{1}^{(n)}(r)=\frac{{k}_{y}^{2n}}{{2}^{n+1}\,(2n-3)}\frac{{r}^{3-2n}-{r}_{h}^{3-2n}}{r},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{b}_{1}^{(n)}(r) & = & \displaystyle \frac{n\,{k}_{y}^{2n}\,{r}^{2}\,{r}_{h}^{-2n}}{3\cdot {2}^{n}\,(2n-3)}\left[\,\rm{log}\,\left(1-\displaystyle \frac{{r}_{h}^{3}}{{r}^{3}}\right)\right.\\ & & +\,\left.\,\rm{Re}\,\,{{ \mathcal B }}_{0}\left(\displaystyle \frac{{r}^{3}}{{r}_{h}^{3}};1-\displaystyle \frac{2n}{3},0\right)\right]+{\gamma }_{1(n)}\,{r}^{2},\end{array}\end{eqnarray}$
where Re${{ \mathcal B }}_{0}\left(x;1-\displaystyle \frac{2n}{3},0\right)$ represents the regular sector of the real part of the incomplete beta function ${ \mathcal B }\left(x;1-\frac{2n}{3},0\right)$ and the constant Υ1(n) is $\begin{eqnarray}{\gamma }_{1(n)}\equiv \left\{\begin{array}{ll}\displaystyle \frac{n\,\pi \,{k}_{y}^{2n}\,{r}_{h}^{-2n}}{3\,(2n-3)\,{2}^{n}}\,\,\rm{cot}\,\left(\displaystyle \frac{2\pi n}{3}\right),\quad & n\ne \displaystyle \frac{3}{2}{\boldsymbol{z}},\\ 0,\quad & n=\displaystyle \frac{3}{2}({\boldsymbol{z}}+1),\end{array}\right.\end{eqnarray}$
with z denoting positive integers. Here, the special case $n=\frac{3}{2}$ has been excluded. It should be mentioned that the above results ( $\begin{eqnarray}T=\frac{A^{\prime} ({r}_{h})}{4\pi }=\frac{3\,{r}_{h}}{4\pi }-{\lambda }^{2}\,\frac{{{k}_{y}}^{2n}{{r}_{h}}^{1-2n}}{{2}^{n+3}\,\,\pi }+{ \mathcal O }({\lambda }^{4}).\end{eqnarray}$
While for $n=\frac{3}{2}$, one should solve it separately and obtain the corrections in the background metric as follows
$\begin{eqnarray}{a}_{1}^{(n)}(r)=\frac{{k}_{y}^{3}}{4\sqrt{2}\,r}\,\rm{log}\,\left(\frac{{r}_{h}}{r}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{b}_{1}^{(n)}(r) & = & -\frac{{r}^{2}\,{k}_{y}^{3}}{72\sqrt{2}{r}_{h}^{3}}\left[{\pi }^{2}+27{\left(\,\rm{log}\,\frac{r}{{r}_{h}}\right)}^{2}\right.\\ & & \left.+6\,{\,\rm{Li}\,}_{2}\left(1-\frac{{r}^{3}}{{r}_{h}^{3}}\right)\right],\end{array}\end{eqnarray}$
with the polylogarithm function Li2(x). And the Hawking temperature becomes $\begin{eqnarray}T=\frac{3\,{r}_{h}}{4\pi }-{\lambda }^{2}\frac{{k}_{y}^{3}}{16\,\sqrt{2}\,\pi \,{r}_{h}^{2}}+{ \mathcal O }({\lambda }^{4}).\end{eqnarray}$
2.1. Two-axion case
Now, we consider that there are two axions along both x and y directions, which means to set kx, ky ≠ 0. For the general cases, one can assume that kx ≠ ky. In the low anisotropic regime, we still have
$\begin{eqnarray}A(r)={r}^{2}\left(1-\frac{{r}_{h}^{3}}{{r}^{3}}\right)+{\lambda }^{2}\,{a}_{2}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
$\begin{eqnarray}B(r)={r}^{2}+{\lambda }^{2}\,{b}_{2}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
$\begin{eqnarray}C(r)={r}^{2}-{\lambda }^{2}\,{b}_{2}^{(n)}(r)+{ \mathcal O }({\lambda }^{4}),\end{eqnarray}$
where $\begin{eqnarray}{a}_{2}^{(n)}(r)=\frac{{({k}_{x}^{2}+{k}_{y}^{2})}^{n}}{{2}^{n+1}(2n-3)}\frac{{r}^{3-2n}-{r}_{h}^{3-2n}}{r},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{b}_{2}^{(n)}(r) & = & \displaystyle \frac{n\,{({k}_{x}^{2}+{k}_{y}^{2})}^{n-1}\,{r}^{2}\,{r}_{h}^{-2n}}{3\cdot {2}^{n}\,(2n-3)}({k}_{y}^{2}-{k}_{x}^{2})\\ & & \times \,\left[\,\rm{log}\,\left(1-\displaystyle \frac{{r}_{h}^{3}}{{r}^{3}}\right)+\,\rm{Re}\,\,{{ \mathcal B }}_{0}\left(\displaystyle \frac{{r}^{3}}{{r}_{h}^{3}};1-\displaystyle \frac{2n}{3},0\right)\right]\\ & & +\,{\gamma }_{2(n)}\,{r}^{2},\end{array}\end{eqnarray}$
with $\begin{eqnarray}{\gamma }_{2(n)}\equiv \left\{\begin{array}{ll}\displaystyle \frac{n\,\pi \,({k}_{y}^{2}-{k}_{x}^{2})\,{({k}_{x}^{2}+{k}_{y}^{2})}^{n-1}\,{r}_{h}^{-2n}}{3\,(2n-3)\,{2}^{n}}\,\,\rm{cot}\,\left(\displaystyle \frac{2\pi n}{3}\right),\quad & n\ne \displaystyle \frac{3}{2}{\boldsymbol{z}},\\ 0,\quad & n=\displaystyle \frac{3}{2}({\boldsymbol{z}}+1),\end{array}\right.\end{eqnarray}$
for $n\ne \frac{3}{2}$. The Hawking temperature then is given by $\begin{eqnarray}T=\frac{A^{\prime} ({r}_{h})}{4\pi }=\frac{3\,{r}_{h}}{4\pi }-{\lambda }^{2}\,\frac{({{k}_{x}^{2}+{k}_{y}^{2})}^{n}{{r}_{h}}^{1-2n}}{{2}^{n+3}\,\,\pi }+{ \mathcal O }({\lambda }^{4}).\end{eqnarray}$
For $n=\frac{3}{2}$, the results turn into
$\begin{eqnarray}{a}_{2}^{(n)}(r)=\frac{{({k}_{x}^{2}+{k}_{y}^{2})}^{3/2}}{4\sqrt{2}\,r}\,\rm{log}\,\left(\frac{{r}_{h}}{r}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{b}_{2}^{(n)}(r) & = & -\frac{{r}^{2}\,({k}_{y}^{2}-{k}_{x}^{2})\,\sqrt{{k}_{x}^{2}+{k}_{y}^{2}}}{72\sqrt{2}{r}_{h}^{3}}\\ & & \times \,\left[{\pi }^{2}+27{\left(\,\rm{log}\,\frac{r}{{r}_{h}}\right)}^{2}+6\,{\,\rm{Li}\,}_{2}\left(1-\frac{{r}^{3}}{{r}_{h}^{3}}\right)\right],\end{array}\end{eqnarray}$
and $\begin{eqnarray}T=\frac{3\,{r}_{h}}{4\pi }-{\lambda }^{2}\,\frac{{\left({k}_{x}^{2}+{k}_{y}^{2}\right)}^{3/2}}{16\,\sqrt{2}\,\pi \,{r}_{h}^{2}}+{ \mathcal O }({\lambda }^{4}).\end{eqnarray}$
3. Viscoelastic response
In the hydrodynamic limit (low frequency and small momentum), the retarded Green function of the stress tensor at zero momentum can be expanded as [65]
$\begin{eqnarray}{G}_{{T}_{xy}{T}_{xy}}^{R}(\omega ,{\boldsymbol{p}}=0)={\mu }_{xy}-{\rm{i}}\,\omega \,{\eta }_{xy}+{ \mathcal O }({\omega }^{2}),\end{eqnarray}$
where the non-dissipative part ${\mu }_{xy}=\mathop{\mathrm{lim}}\limits_{\omega \to 0}\,\rm{Re}\,\,{G}_{{T}_{xy}{T}_{xy}}^{R}(\omega ,{\boldsymbol{p}}=0)$ is interpreted as the shear elastic modulus, and the dissipative coefficient ηxy in the imaginary part is associated to entropy production and should be understood as the shear viscosity.In the holographic framework, to compute ${G}_{{T}_{xy}{T}_{xy}}^{R}(\omega ,{\boldsymbol{p}}=0)$, we need to introduce time-dependent perturbations δgxy upon the background in the bulk. Note that this metric fluctuation is decoupled from other possible fluctuations for zero p because δgxy is the solo fluctuation with the tensor channel in this sector. For ${g}_{ij}={\bar{g}}_{ij}+\delta {g}_{ij}$ (From now on, we will use the bar to denote the background metric we obtained in the last section.), taking the Fourier transformation $\delta {g}_{j}^{i}(t,r,{x}^{i})\unicode{x0007E}{h}_{j}^{i}(r)\,{{\rm{e}}}^{-{\rm{i}}\omega t}$, we achieve the equations of ${h}_{y}^{x}$ and ${h}_{x}^{y}$ as follows6:
$\begin{eqnarray}\begin{array}{l}\frac{1}{\sqrt{-g}}{\partial }_{r}\left(\sqrt{-g}{Z}_{i}^{j}(r){\bar{g}}^{rr}{\partial }_{r}{h}_{j}^{i}\right)-{\omega }^{2}{\bar{g}}^{tt}{Z}_{i}^{j}(r){h}_{j}^{i}\\ \quad =\,{m}_{g}^{i}{(r)}^{2}{Z}_{i}^{j}(r){h}_{j}^{i},\,(i=x,y\,\,\,\,j=x,y).\end{array}\end{eqnarray}$
where ${Z}_{i}^{j}(r)\equiv {\bar{g}}^{jj}{\bar{g}}_{ii}$ and the square of the graviton masses ${{m}_{g}^{i}}^{2}$ match the formula proposed in [47] which is given by $\begin{eqnarray}{{m}_{g}^{i}}^{2}\equiv {g}^{ii}{T}_{ii}-\frac{\delta {T}_{ij}}{\delta {g}_{ij}},\end{eqnarray}$
where the bulk stress tensor in our model is given by $\begin{eqnarray}{T}_{MN}=-{g}_{MN}{\lambda }^{2}\,V+{\lambda }^{2}\,V^{\prime} (X)\,{\partial }_{M}{\phi }^{i}{\partial }_{N}{\phi }^{i},\end{eqnarray}$
with $V^{\prime} \equiv \frac{dV}{dX}$. Obviously, the graviton masses originate from the non-trivial profiles of the scalars in the bulk. For the isotropic case, i.e., kx = ky = k, we get ${Z}_{i}^{j}=1$ and a unique graviton mass mg. Then, the graviton behaves like a massive scalar in the bulk.In our anisotropic model, there exist two distinct graviton masses which are
$\begin{eqnarray}{{m}_{g}^{i}}^{2}={\lambda }^{2}\,{k}_{i}^{2}\,V^{\prime} (\bar{X})\,{\bar{g}}^{ii}.\end{eqnarray}$
Therefore, the linearized perturbative equations of δgxy can be expressed more explicitly as $\begin{eqnarray}\begin{array}{r}{\partial }_{r}\left(\sqrt{-g}{Z}_{i}^{j}{\bar{g}}^{rr}{\partial }_{r}{h}_{j}^{i}\right)-{\omega }^{2}\sqrt{-g}{\bar{g}}^{tt}{Z}_{i}^{j}{h}_{j}^{i}\\ \quad ={\lambda }^{2}\,{k}_{i}^{2}\sqrt{-g}{\bar{g}}^{jj}{V}^{{\prime} }{h}_{j}^{i}.\end{array}\end{eqnarray}$
Next, we calculate the shear viscoelasticity both in the single-axion case and the two-axion case.3.1. One-axion case
For kx = 0, ky ≠ 0, the metric perturbation ${h}_{y}^{x}$ satisfies the following equation Appendix A , we have shown that b1(rh) < 0, which implies that, whether spontaneous or explicit, the rotational symmetry breaking always results in a violation of the KSS bound. In addition, the shear modulus in this case is zero due to the vanishing graviton mass.
$\begin{eqnarray}{\partial }_{r}\left(\sqrt{-g}{Z}_{x}^{y}{\bar{g}}^{rr}{\partial }_{r}{h}_{y}^{x}\right)-{\omega }^{2}\sqrt{-g}{Z}_{x}^{y}{\bar{g}}^{tt}{h}_{y}^{x}=0.\end{eqnarray}$
Since the field is massless, one can apply the membrane paradigm which allows us to express the shear viscosity in terms of the horizon data [6, 17]7. For the boundary system with two spatial dimensions, there can only be one shear viscosity ηxy and one shear modulus μxy. We will from now on omit the spatial indexes for simplicity. As a result, the shear viscosity to entropy density ratio can be expressed as $\begin{eqnarray}\frac{\eta }{s}=\frac{{Z}_{x}^{y}({r}_{h})}{4\pi }=\frac{1}{4\pi }\frac{{\bar{g}}_{xx}({r}_{h})}{{\bar{g}}_{yy}({r}_{h})}=\frac{1}{4\pi }\frac{B({r}_{h})}{C({r}_{h})}.\end{eqnarray}$
For the small λ, we obtain that $\begin{eqnarray}\frac{\eta }{s}=\frac{1}{4\pi }+{\lambda }^{2}\,\frac{{b}_{1}({r}_{h})}{2\,\pi \,{{r}_{h}}^{2}}+{ \mathcal O }({\lambda }^{4}).\end{eqnarray}$
In Two-axion case
For kx, ky ≠ 0, the rotational symmetry is entirely broken. In this case, ${h}_{y}^{x}$ becomes massive which satisfies the following equation Appendix B . As the result, μ and η/s are
$\begin{eqnarray}\begin{array}{r}{\partial }_{r}\left(\sqrt{-g}{Z}_{x}^{y}{\bar{g}}^{rr}{\partial }_{r}{h}_{y}^{x}\right)-{\omega }^{2}\sqrt{-g}{Z}_{x}^{y}{\bar{g}}^{tt}{h}_{y}^{x}\\ \quad ={\lambda }^{2}\,{k}_{x}^{2}\sqrt{-g}{\bar{g}}^{yy}{V}^{{\prime} }{h}_{y}^{x}.\end{array}\end{eqnarray}$
One can show that the graviton mass introduces a non-zero real part of ${G}_{{T}_{xy}{T}_{xy}}^{R}(\omega ,{\boldsymbol{p}}=0)$. Unlike the previous case, both μ and η now depend on the full geometry. The membrane paradigm is no longer applicable. However, for small λ, one can analytically calculate them order by order. The details are shown in $\begin{eqnarray}\mu ={\lambda }^{2}\frac{n\,{k}_{x}^{2}\,{\left({k}_{x}^{2}+{k}_{y}^{2}\right)}^{n-1}}{{2}^{n-1}\,(2n-3)}{r}_{h}^{3-2n}+{ \mathcal O }({\lambda }^{4}),\quad n\gt \frac{3}{2}\end{eqnarray}$
and $\begin{eqnarray}\frac{\eta }{s}=\frac{1}{4\pi }\frac{B({r}_{h})}{C({r}_{h})}-\left\{\begin{array}{ll}{\lambda }^{2}\frac{n\,{k}_{x}^{2}\,{\left({k}_{x}^{2}+{k}_{y}^{2}\right)}^{n-1}}{3\cdot {2}^{n}\,(2n-3)\,\pi \,{r}_{h}^{2n}}{ \mathcal H }(\frac{2n-3}{3})+{ \mathcal O }({\lambda }^{4}),\quad & n\ne \frac{3}{2},\\ {\lambda }^{2}\frac{\pi \,{k}_{x}^{2}\sqrt{{k}_{x}^{2}+{k}_{y}^{2}}}{72\,\sqrt{2}\,{r}_{h}^{3}}+{ \mathcal O }({\lambda }^{4}),\quad & n=\frac{3}{2},\\ \quad \end{array}\right.\end{eqnarray}$
where ${ \mathcal H }(x)$ is harmonic number.In order to separate the effects of broken translations and anisotropy in a manifest way, one can rotate the basis spatial coordinates as in [66],
$\begin{eqnarray}{\phi }^{i}=k\left(\begin{array}{cc}\sqrt{\frac{{\epsilon }^{2}}{4}+1} & \frac{\epsilon }{2}\\ \frac{\epsilon }{2} & \sqrt{\frac{{\epsilon }^{2}}{4}+1}\\ \end{array}\right)\left(\begin{array}{c}\tilde{x}\\ \tilde{y}\\ \end{array}\right),\end{eqnarray}$
where k and ε are two dimensionless parameters characterizing the strengths of translational symmetry breaking and anisotropy, respectively. If we assume that kx > ky, then they are related to k and ε as follows, $\begin{eqnarray}{k}_{x}=k\,\left(\sqrt{\frac{{\epsilon }^{2}}{4}+1}+\frac{\epsilon }{2}\right),\end{eqnarray}$
$\begin{eqnarray}{k}_{y}=k\,\left(\sqrt{\frac{{\epsilon }^{2}}{4}+1}-\frac{\epsilon }{2}\right).\end{eqnarray}$
Finally, (39 ) and (40 ) can be re-expressed as Appendix A , we show that ${ \mathcal H }(\frac{2n-3}{3})\lt 0$ for $0\lt n\lt \frac{3}{2}$ and ${ \mathcal H }(\frac{2n-3}{3})\gt 0$ for $n\gt \frac{3}{2}$. Therefore, η/s decreases monotonically with k and ε for all the cases. Furthermore, its value is suppressed more significantly for larger values of n, which implies that the KSS bound is always violated in the axion model, regardless of whether the symmetries are broken explicitly or spontaneously. In Figures 1-3, we plot μ and η/s as functions of k and ε for different cases.
$\begin{eqnarray}\begin{array}{rcl}\mu & = & {\lambda }^{2}\displaystyle \frac{n\,{k}^{2n}\,{\left(2+{\epsilon }^{2}\right)}^{n-1}\,{\left(\sqrt{4+{\epsilon }^{2}}+\epsilon \right)}^{2}}{{2}^{n+1}\,(2n-3)\,{r}_{h}^{2n-3}}\\ & & +{ \mathcal O }({\lambda }^{4}),\,\,n\gt \displaystyle \frac{3}{2},\end{array}\end{eqnarray}$
and $\begin{eqnarray}\frac{\eta }{s}=\frac{1}{4\pi }-\left\{\begin{array}{ll}{\lambda }^{2}\,\frac{n\,{k}^{2n}{\left({\epsilon }^{2}+2\right)}^{n}\,{ \mathcal H }(\frac{2n-3}{3})}{3\cdot {2}^{n+1}\,(2n-3)\,\pi \,{r}_{h}^{2n}}+{ \mathcal O }({\lambda }^{4}),\quad & n\ne \frac{3}{2},\\ {\lambda }^{2}\,\frac{\pi \,{\left[{k}^{2}\left({\epsilon }^{2}+2\right)\right]}^{3/2}}{144\,\sqrt{2}\,{r}_{h}^{3}}+{ \mathcal O }({\lambda }^{4}),\quad & n=\frac{3}{2}.\\ \quad \end{array}\right.\end{eqnarray}$
From the above results, it is obvious to see that μ just increases monotonically with k and ε, and its value is enhanced faster for larger values of n. In 
Figure 1. The shear modulus μ is the function of the k and ε, depicted by ( |
Figure 2. The ratio η/s as the function of the k and ε, depicted by ( |
Figure 3. The ratio η/s as the function of the k and ε, depicted by ( |
4. Discussion and outlook
In this paper, the shear elasticity and the shear viscosity of an anisotropic holographic axion model with broken translations have been investigated. We find that a positive shear modulus emerges when the translational symmetry is broken spontaneously. In particular, the shear modulus is doubly enhanced in the presence of broken translations and anisotropy. Furthermore, the η/s ratio is always suppressed no matter whether the translations as well as the rotations are broken explicitly or spontaneously which is in contrast to the holographic p-wave superfluid case. In the p-wave superfluid, the spontaneous breaking of rotational symmetry caused by a vector condensate enhances the value of η/s, competing against the effects of explicit symmetry breaking. The KSS bound is hence never violated [58]. From this perspective, we can conclude that the fate of the KSS bound in holographic systems cannot be solely attributed to whether rotational symmetry is broken, or to the specific manner of that breaking (whether spontaneous or explicit). The η/s ratio also depends very much on the details of the operator that gives rise to the symmetry breaking.
In the case of the spontaneous symmetry breaking with n > 5/2, the holographic model is dual to a solid on the boundary. Then, one may naively expect that the violation of the viscosity bound cannot happen and that the shear viscosity should be divergent in a perfect crystal without topological defects, say, the dislocation flow. However, this does not contradict our results. We consider the frequency-dependent viscosity of the crystal given by the Kubo formula [67], ${\eta }^{\,\rm{crystal}\,}(\omega )\equiv \,\rm{Im}\,[{G}_{{T}_{xy}{T}_{xy}}^{R}(\omega ,{\boldsymbol{p}}=0)/\omega ]\approx \mu \,\delta (\omega )+\eta $, where the delta function is from the 1/ω divergence of the real part according to the Kramers-Kronig relation and η is the dissipative coefficient addressed in this work. In the DC limit, we have ηcrystal → ∞ due to the delta function. Therefore, although the KSS bound is violated, the crystal viscosity receives a divergent contribution from the rigidity of the translational order.
Nevertheless, our study in this work is limited to the perturbative region, i.e., the symmetry breaking is very weak. Therefore, it is not clear whether our conclusion holds or not when the breaking of the symmetries is finite. In a more general case, the anisotropic background solutions cannot be obtained in an analytic way. However, to achieve a complete answer, it is necessary to extend our study by, for instance, following the numeric methods used in [29, 30] to obtain the background solutions for finite values of λ and analyze how the viscoelastic property is affected in the strongly anisotropic region. We leave this for future work.
