text:
\section{Introduction and Summary

The AdS/CFT duality \cite{Maldacena:1997re,Witten:1998qj,Witten:1998zw,Gubser:1998bc} has been a useful tool to study realistic strongly-coupled systems (see., {\it e.g.}, Refs.~\cite{CasalderreySolana:2011us,Natsuume:2014sfa,Ammon:2015wua,Zaanen:2015oix,Hartnoll:2016apf} for textbooks). In condensed-matter applications, holographic superconductors\  provide particularly useful ``theoretical laboratories" \cite{Gubser:2008px,Hartnoll:2008vx,Hartnoll:2008kx}. They are useful to explore not only standard aspects of a superconducting transition but also various related phenomena such as critical dynamics \cite{Maeda:2009wv}, defect formations \cite{Sonner:2014tca,Chesler:2014gya,Natsuume:2017jmu}, and superfluid turbulences \cite{Adams:2012pj}.

The holographic superconductors\  arise in a broad range of gravitational theories with matter fields. From field theory point of view, this is natural since a superconductor\ is a robust phenomenon at low temperature. For example, it arises not only in asymptotically AdS spacetimes but also in asymptotically Lifshitz spacetimes \cite{Brynjolfsson:2009ct,Sin:2009wi}, which is our main focus in this paper.

A holographic superconductor\  is typically an Einstein-Maxwell-complex scalar system. Such a system is hard to solve in general. One often needs either a numerical computation or an approximation method, and there are only a few analytic solutions \cite{Basu:2008bh,Herzog:2009ci,Herzog:2010vz}. In this paper, we present an analytic solution for a one-parameter family of holographic Lifshitz superconductors.

A holographic Lifshitz superconductor\ has three parameters $(p,z_\text{L},\Delta)$, where $p$ is the number of boundary spatial dimensions, $z_\text{L}$ is the Lifshitz exponent%
\footnote{The parameter $z_\text{L}$ is known as ``dynamic critical exponent," but we call it ``Lifshitz exponent" to avoid confusion with $z_\text{D}$ below.},
and $\Delta$ is the scaling dimension of the order parameter.
In this paper, we consider the case where
\begin{enumerate}
\item $p=3z_\text{L}$, and
\item $\Delta=(p+z_\text{L})/2$, or the scalar mass $m^2$ saturates the Lifshitz Breitenlohner-Freedman (BF) bound \cite{Breitenlohner:1982bm}.
\end{enumerate}
These conditions still admit a one-parameter family of theories, and these parameters are related by
\begin{equation}
p=3z_\text{L}~, \Delta=2z_\text{L}~.
\end{equation}
In this case, there is a simple analytic solution for the order parameter at the critical point:
\begin{equation}
\Psi=\frac{u^{2z_\text{L}}}{1+u^{2z_\text{L}}}~,
\label{eq:analytic_sol}
\end{equation}
where $u$ is the inverse of the radial coordinate. The $z_\text{L}=1$ case, namely the AdS case, has been discussed in Ref.~\cite{Herzog:2010vz}. Below we utilize this solution to explore various properties of these systems, {\it e.g.},
\begin{enumerate}

\item The background solution in the superfluid phase (\sect{bg_low}).

\item The grand canonical potential, and the phase transition is second order (\sect{grand_canonical}).

\item The order parameter response function or the susceptibility. The dynamic response is obtained in the normal phase (\sect{high}), and the thermodynamic response is obtained in the superfluid phase (\sect{low_chi}).

\item The London equation. From the equation, one gets the London penetration depth and the imaginary part of the conductivity has the $1/\omega$-pole which implies the diverging DC conductivity (\sect{vector}).

\item The background solution with a vector potential. This gives the critical superfluid velocity and the critical magnetic field (\sect{bg_vector}).

\item From these results, all static critical exponents $(\alpha,\beta,\gamma,\delta,\nu,\eta)$, the dynamic  critical exponent $z_\text{D}$, and the ratio of critical amplitudes. These results are consistent with the standard Ginzburg-Landau (GL) theory or the $\phi^4$ mean-field theory. We identify the dual GL theory including numerical coefficients (\sect{GL}).

\end{enumerate}

The holographic Lifshitz superconductors\ have been studied previously, but it is still nice to analyze these properties all at once analytically for an infinite number of theories. First, in previous works, the system was studied mostly using numerical methods. Second, the system was studied only for some specific values of $(p,z_\text{L},\Delta)$%
\footnote{
For example, Ref.~\cite{Brynjolfsson:2009ct} considers the $(p,z_\text{L})=(2,3/2)$ case and obtains a charged Lifshitz black hole with scalar hair. The other works typically take the probe limit to study the system. Ref.~\cite{Sin:2009wi} considers the $(p, z_\text{L}, \Delta)=(2,2,3)$ case in a Lifshitz black hole\  background \cite{Balasubramanian:2009rx}. Ref.~\cite{Bu:2012zzb} considers the $(p, z_\text{L}, \Delta)=(2,2,3), (2,2,4)$ cases in the same Lifshitz black hole\  background as ours. Ref.~\cite{Lu:2013tza} considers $(p, z_\text{L}, \Delta)=(2,2,2), (2,2,3), (3,2,3), (3,3,3)$ cases in the same Lifshitz black hole\  background as ours. As far as we are aware, our $(p,z_\text{L},\Delta)=(3z_\text{L},z_\text{L},2z_\text{L})$ case was not studied before.
}.
Third, some of the above properties were studied but not all were studied.

In particular, previous works typically have shown that (i) there exists a $\Psi\neq0$ solution at low temperatures, (ii) the solution is favorable from the free energy or from the grand canonical potential, (iii) the spontaneous condensate has the standard $\phi^4$ mean-field exponent $\beta=1/2$, and (iv) the diverging DC conductivity.

On the other hand, the other properties are newly investigated, {\it e.g.}, the other critical exponents as well as exact expressions for various numerical coefficients including critical amplitudes. Also, the critical dynamics of a  holographic Lifshitz superconductor\ has never been investigated%
\footnote{Dynamics in Lifshitz geometry has been studied, {\it e.g.}, in Refs.~\cite{Sybesma:2015oha,Keranen:2016ija}. Critical dynamics has been studied in holography, {\it e.g.}, in Refs.~\cite{Maeda:2008hn,Maeda:2009wv,Buchel:2010gd,Natsuume:2010bs}.}.
At a finite-temperature critical point, the correlation length $\xi$ and the relaxation time $\tau$ of the order parameter obey a scaling law:
\begin{equation}
\tau \propto \xi^{z_\text{D}}~.
\label{eq:scaling_DCP}
\end{equation}
We obtain $z_\text{D}=2$ irrespective of the value of the Lifshitz exponent $z_\text{L}$. We discuss the relation between $z_\text{L}$ and $z_\text{D}$ in \sect{two_z}.

\section{Preliminaries

\subsection{Lifshitz black hole\ }\label{sec:EMD}

The Lifshitz geometry \cite{Kachru:2008yh} is given by
\begin{equation}
ds_{p+2}^2 = -\left(\frac{r}{L}\right)^{2z_\text{L}} dt^2 + \left(\frac{r}{L}\right)^{2} dx_i^2 + L^2 \frac{dr^2}{r^2}
\end{equation}
 (see, {\it e.g.}, Ref.~\cite{Taylor:2015glc} for a review). The geometry is invariant under an anisotropic scaling
\begin{equation}
t \to a^{z_\text{L}} t~, \quad
x^i \to a x^i~,\quad
r \to r/a~.
\label{eq:scaling_Lifshitz}
\end{equation}
There are various Lifshitz black hole\  solutions known in the literature, both analytically and numerically, depending on bulk theories. We use the solution in Refs.~\cite{Taylor:2008tg,Pang:2009ad}. The metric is given by
\begin{subequations}
\label{eq:Lifshitz_BH}
\begin{align}
ds^
 &= -\left(\frac{r}{L}\right)^{2z_\text{L}} h dt^2 + \left(\frac{r}{L}\right)^{2} dx_i^2 + L^2 \frac{dr^2}{r^2h}~, \\
&= -\left(\frac{r_h}{L}\right)^{2z_\text{L}} \frac{h}{u^{2z_\text{L}}} dt^2 + \left(\frac{r_h}{L}\right)^{2} \frac{dx_i^2}{u^2} + L^2\frac{du^2}{u^2h}~, \\
h &= 1- \left(\frac{r_h}{r}\right)^{p+z_\text{L}} = 1- u^{p+z_\text{L}}~,
\end{align}
\end{subequations}
where $u:=r_h/r$, and $r_h$ is the horizon radius. The metric is invariant under the Lifshitz scaling \eqref{eq:scaling_Lifshitz} with the scaled horizon radius $r_h\tor_h/a$. The Hawking temperature is given by
\begin{equation}
T = \frac{p+z_\text{L}}{4\pi L} \left(\frac{r_h}{L} \right)^{z_\text{L}}~.
\end{equation}

The metric can be obtained as a solution of an Einstein-Maxwell-dilaton system%
\footnote{We use capital Latin indices $M, N, \ldots$ for the $(p+2)$-dimensional bulk spacetime coordinates and use Greek indices $\mu, \nu, \ldots$ for the $(p+1)$-dimensional boundary coordinates. The boundary coordinates are written as  $x^\mu = (t, x^i)=(t,x,y,\cdots)$.}:
\begin{align}
  S = \frac{1}{16\pi G_{p+2}} &
  \int d^{p+2}x\, \sqrt{-g} \bigg\{
  R - 2\Lambda
\nonumber \\ &
  - \frac{1}{2} (\partial_M\phi)^2 - \frac{1}{4} e^{\lambda \phi}{\cal F}_{MN}^2
  \bigg\}~,
\label{eq:EMD}
\end{align}
where $G_{p+2}$ is the $(p+2)$-dimensional Newton's constant and
\begin{subequations}
\begin{align}
&\Lambda = - \frac{(p+z_\text{L}-1)(p+z_\text{L})}{2L^2}~, \quad
\lambda^2 = \frac{2p}{z_\text{L}-1}~, \\%\quad
&{\cal F}_{MN} = 2\, \partial_{[M} {\cal A}_{N]}~
\end{align}
\end{subequations}
The matter field solutions are given by
\begin{subequations}
\begin{align}
e^{\lambda \phi} &= u^{2p}~, \\
{\cal A}_t &= -\sqrt{ \frac{2(z_\text{L}-1)}{p+z_\text{L}} } \frac{1}{u^{p+z_\text{L}}}~.
\end{align}
\end{subequations}
But for our purpose, the point using this solution is that (i) it provides an analytic Lifshitz black hole\  solution, and (ii) a class of holographic superconductors\  in this background admits an analytic solution.

\subsection{Holographic Lifshitz superconductors

We couple an additional matter system, a Maxwell-complex scalar system in addition to the above system  \cite{Brynjolfsson:2009ct,Sin:2009wi}:
\begin{equation}
  S = -\frac{1}{e^2}
  \int d^{p+2}x\, \sqrt{-g} \left\{
  \frac{1}{4} F_{MN}^2 + \left\vert D_M \Psi \right\vert^2 + V(\Psi)
  \right\}~,
\label{eq:Lifshitz_HSC}
\end{equation}
where
\begin{align}
  & F_{MN} = 2\, \partial_{[M} A_{N]}~,
& & D_M := \nabla_M - i A_M~,
& & V= m^2 |\Psi|^2~.
\end{align}
The $U(1)$-field $A_M$ is different from ${\cal A}_M$ in \eq{EMD}.

We take the probe limit $e \gg 1$, where the backreaction of these matter fields onto the geometry is ignored. Namely, we solve the system \eqref{eq:Lifshitz_HSC} in the background \eqref{eq:Lifshitz_BH}. The equations of motion are given by
\begin{subequations}
\label{eq:eom}
\begin{align}
(D^2-m^2)\Psi &= 0~, \\
\nabla_N F^{MN} &= j^M
\\
&:=ig^{MN}[(D_N\Psi)^\dag\Psi- \Psi^\dag(D_N\Psi)]~.
\end{align}
\end{subequations}
In the $A_u=0$ gauge, the asymptotic behaviors of the matter fields are given by
\begin{subequations}
\begin{align}
A_t & \sim A_t^{(0)}  + A_t^{(1)}\tilde{u}^{p-z_\text{L}}~, \qquad (p>z_\text{L})~, \\
A_i & \sim A_i^{(0)}  + A_i^{(1)} \tilde{u}^{p+z_\text{L}-2}~, \\
\Psi &\sim \Psi^{(0)} \tilde{u}^{\Delta_-} + \Psi^{(1)} \tilde{u}^{\Delta_+}~, \\
\tilde{u} &:=\frac{L}{r}~, \\
\Delta_\pm &:= \frac{p+z_\text{L}}{2} \pm \sqrt{ \frac{(p+z_\text{L})^2}{4} + L^2m^2 }~,
\end{align}
\end{subequations}
where $A_t^{(1)}$ represents the charge density $\rho$, and $A_t^{(0)}=\mu$ is the chemical potential. Similarly, $A_i^{(1)}$ represents the current density $ J^i$, and $A_i^{(0)}$ is the vector potential. For $\Psi$, $\Psi^{(1)}$ represents the order parameter ${\cal O}$, and $\Psi^{(0)}$ is the external source for ${\cal O}$. (See \appen{dict} for the precise dictionary.)
Then, the BF bound in the asymptotically Lifshitz geometry is given by
\begin{equation}
m_\text{BF}^2 = - \frac{(p+z_\text{L})^2}{4L^2}~.
\end{equation}
When the BF bound is saturated, the asymptotic behavior is replaced by
\begin{align}
\Psi &\sim \Psi^{(0)} \tilde{u}^\Delta \ln \tilde{u} +\Psi^{(1)} \tilde{u}^\Delta~, \quad
\Delta := \frac{p+z_\text{L}}{2}~.
\end{align}

The equations of motion \eqref{eq:eom} admit a solution
\begin{subequations}
\label{eq:sol_high}
\begin{align}
\bmA_t &=  \mu(1-u^{p-z_\text{L}})~, \quad (p>z_\text{L})~, \\
\bmA_i &= \bmA_u = 0~, \\
\bmPsi &=0~,
\end{align}
\end{subequations}
where boldface letters indicate background values. But, at the critical point, the $\bmPsi=0$ solution becomes unstable and is replaced by a $\bmPsi\neq0$ solution. We see this in detail below.

\section{Critical point}\label{sec:critical}

Below we consider the case $p=3z_\text{L}$.
It is convenient to introduce a new coordinate $s:=u^{2z_\text{L}}$. The metric then becomes%
\footnote{It is well-known that the $z_\text{L}=\infty$ limit of the Lifshitz geometry is AdS$_2 \times \relax\text{I\kern-.18em R}^p$ asymptotically. But this is the case for a finite $p$. In our case, $p=3z_\text{L}$, and the $z_\text{L}=\infty$ limit does not reduce to AdS$_2$.}
\begin{subequations}
\begin{align}
ds^
 &= - \left(\frac{r_h}{L}\right)^{2z_\text{L}} \frac{h}{s} dt^2 + \left(\frac{r_h}{L}\right)^2 \frac{dx_i^2}{s^{1/z_\text{L}}} + \left(\frac{L}{2z_\text{L}}\right)^2\frac{ds^2}{s^2h}~,
\\%\quad
h &= 1- s^2~.
\end{align}
\end{subequations}

We consider the scalar which saturates the Lifshitz BF bound $m^2=-(2z_\text{L}/L)^2$. The scaling dimension $\Delta$ is given by $\Delta=2z_\text{L}$. First, consider the static homogeneous solution $\bmPsi=\bmPsi(s)$, and approach the critical point from high temperature. Near the critical point, the scalar field $\bmPsi$ remains small, and one can ignore the backreaction of $\bmPsi$ onto the Maxwell field. In this region, one can use \eq{sol_high} for the Maxwell field, and it is enough to solve the $\bmPsi$-equation%
\footnote{We later use a perturbative expansion for a systematic analysis (\sect{bg_low}).}.
The $\bmPsi$-equation becomes
\begin{align}
&\partial_s \left(\frac{h}{s}\partial_s \bmPsi \right) + \left\{ \left(\frac{\mu}{2\pi T}\right)^2\frac{(1-s)^2}{hs^2} + \frac{1}{s^3} \right\}\bmPsi =0~,
\label{eq:eom_critical} \\%\quad
&T = \frac{z_\text{L}}{\pi L} \left(\frac{r_h}{L} \right)^{z_\text{L}}~.
\end{align}
Thus, the solution is parametrized by a dimensionless parameter $\mu/T$. The equation admits a solution
\begin{align}
\bmPsi \propto \frac{s}{1+s}~,
\quad
&\text{at }  \left(\frac{\mu}{T}\right)_c=2\pi~,
\label{eq:sol_critical} \\
&\text{or } \frac{\mu}{\frac{1}{L}(\frac{r_h}{L})^{z_\text{L}}}=2z_\text{L}~.
\nonumber
\end{align}
This is the solution at the critical point.

The $z_\text{L}$-dependence disappears in \eq{eom_critical}, and it only appears implicitly in the definition of $T$. One can understand this as follows. For the static homogeneous solution, the Laplacian becomes
\begin{equation}
\nabla^2\bmPsi = \frac{1}{\sqrt{-g}}\partial_s(\sqrt{-g}g^{ss}\partial_s\bmPsi)~.
\end{equation}
The $z_\text{L}$-dependence appears in the boundary spatial metric $g_{ij}$, and it appears only through $\det g$. But in our case,
\begin{align}
-\det g \propto s^{-3-p/z_\text{L}}=s^{-6}~,
\end{align}
so $p$ and $z_\text{L}$ disappear. The metric $g^{ss}$ is also proportional to $z_\text{L}^2$, but it is factored out in the $\bmPsi$-equation since $m^2 \propto -z_\text{L}^2$ and $1/T^2 \propto 1/z_\text{L}^2$. It then follows that the $\bmPsi$-equation formally reduces to the same equation for all $z_\text{L}$.

Note that \eq{sol_critical} is the solution directly at the critical point. As one lowers temperature further, the solution is modified, and we construct the background solution $\bmPsi, \bmA_t$ in \sect{low}. The $z_\text{L}$-dependence can also be eliminated from the $\bmA_t$-equation [by redefining $\bmPsi$ and $\bmA_t$ as in \eq{redefine}.] Thus, the static homogeneous solution is essentially the same as the $z_\text{L}=1$ case apart from various factors of $(2z_\text{L})$. Then, from the analysis of Ref.~\cite{Herzog:2010vz}, \eq{sol_critical} is the solution at the critical point, and the solution has a lower grand canonical potential than the $\bmPsi=0$ solution at low temperature.

However, the full equations of motion do not reduce to the same equations as the $z_\text{L}=1$ case. In general, more nontrivial $z_\text{L}$-dependences appear. For example, they appear when one considers
\begin{itemize}
\item inhomogeneous perturbations in the boundary spatial directions {\it e.g.}, $\delta\phi \propto e^{iqx} $ (\sect{high}), or
\item perturbations or solutions with a vector potential $A_i$ (\sect{vector} and \sect{bg_vector}).
\end{itemize}

Below we construct the background solution $\bmPsi, \bmA_M$. We also consider the linear perturbations from the background:
\begin{subequations}
\begin{align}
\Psi &=\bmPsi+\delta\Psi~, \\
A_M &= \bmA_M+\delta A_M~.
\end{align}
\end{subequations}
We take the gauge $\bmA_s=\delta A_s=0$.
We consider the perturbations of the form
\begin{equation}
\delta\phi(k) \sim e^{-i\omega t+iqx}~,
\end{equation}
where $k^\mu=(\omega, q, 0, \cdots)$. Then, the Maxwell perturbations are decomposed as
\begin{itemize}
\item vector modes, {\it e.g.}, $\delta A_y$, and
\item scalar modes $\delta A_t, \delta A_x$ which can couple to $\delta\Psi$ in general.
\end{itemize}
For simplicity, we set $e=L=r_h=1$ below. In this unit, $\mu_c=2z_\text{L}$, and we vary the chemical potential $\mu$. Also, we often use quantities with ``$~\bar{~}~$". All quantities with ``$~\bar{~}~$" are defined by
\begin{equation}
\bar{\phi}:=\frac{\phi}{2z_\text{L}}~,
\label{eq:redefine}
\end{equation}
when $r_h=1$. For example, $\bar{\mu}_c=1$. We restore units for some of our main results in \appen{restore}.

\section{High-temperature phase}\label{sec:high}

At high temperature, the background solution is given by
\begin{subequations}
\begin{align}
\bmA_t &= \mu(1-s)~, \\
\bmA_i &=0~, \\
\bmPsi &=0~.
\end{align}
\end{subequations}

The interesting quantity in the high-temperature phase is the ``order parameter response function," the susceptibility, or the correlation function of the order parameter. We show that the response function takes the form
\begin{subequations}
\label{eq:response_fn}
\begin{align}
   \chi_{k}
  & = \frac{ \delta{\cal O}(k) }{ \delta\Psi^{(0)}(k) }
\nonumber \\
  &\propto
    \frac{1}{- \frac{2i}{c_K\Gamma} \omega + q^2 + \frac{1}{\xi^2} }~,
\\
    \xi^2 &\propto |\epsilon_{\mu}|^{-1}~,
\label{eq:correl_length}
\end{align}
\end{subequations}
for a small $(\omega, q, \epsilon_{\mu})$, where $\epsilon_{\mu}:=\mu-\mu_c$, and $c_K$ and $\Gamma$ are parameters we use to compare with the GL theory (\sect{GL}). The function gives the following information:
\begin{itemize}
\item
The $\omega=q=0$ limit is the thermodynamic response function
\begin{subequations}
\begin{align}
\chi_T=A/|\epsilon_{\mu}|~,
\end{align}
where the coefficient $A$ is known as the critical amplitude. Then, the exponent $\gamma$ defined by
$
\chi_T\propto |\epsilon_{\mu}|^{-\gamma}
$
is $\gamma=1$.

\item
The $\omega=0$ limit is the static response
\begin{align}
\chi_{\omega=0, q} \propto (q^2+\xi^{-2})^{-1}~.
\end{align}
Then, $\xi$ is the correlation length, and the exponent $\nu$ defined by
$
\xi \propto |\epsilon_{\mu}|^{-\nu}
$
is $\nu=1/2$ from \eq{correl_length}. Also, the anomalous exponent $\eta$ defined by
$
\chi_{\omega=0, q}|_{\mu_c} \propto q^{-2+\eta}
$
is $\eta=0$.
\item
The $\omega\neq0$ case is the dynamic response. Then, the relaxation time behaves as
\begin{align}
\tau_{q=0} \propto \xi^2~,
\end{align}
and the dynamic critical exponent $z_\text{D}$ defined by
$
\tau_{q=0} \propto \xi^{z_\text{D}}
$
is $z_\text{D}=2$.
\end{subequations}
\end{itemize}
Thus, the computation determines the exponents $(\gamma,\nu,\eta,z_\text{D})$ as well as the critical amplitude $A$. An explicit solution is not really necessary to compute critical exponents, and analytic arguments are possible \cite{Maeda:2009wv,Natsuume:2017jmu}. On the other hand, an explicit solution is useful to obtain various numerical coefficients such as $A$.

The response function can be obtained from the bulk scalar field $\Psi$.
Consider the linear perturbation from the background $\Psi=\bmPsi+\delta\Psi$. From the bulk point of view, the response function pole corresponds to a quasinormal pole of $\delta\Psi$. When $\bmPsi=0$, Maxwell scalar modes $\delta A_t$ and $\delta A_x$ decouple from the $\delta\Psi$-equation%
\footnote{
The Maxwell scalar modes give a diffusion pole, and one can determine the diffusion constant. But at high temperature, the computation is not unique to holographic superconductors. It is just a Maxwell field problem in the Lifshitz background. The Maxwell vector mode can determine the conductivity. Again, at high temperature, the computation is not unique to holographic superconductors. But we compute the vector mode in the low-temperature phase. The $O(\omega)$-coefficient is common both to the high-temperature and the low-temperature phases.
}.
Thus, to determine the order parameter response, it is enough to consider the $\delta\Psi$-equation:
\begin{align}
\partial_s \left(\frac{h}{s}\partial_s \delta\Psi \right)
+ \left\{ \frac{(\bar{\bmA}_t+\bar{\omega})^2}{hs^2} -\frac{\bar{q}^2}{s^{3-1/z_\text{L}}} + \frac{1}{s^3} \right\}\delta\Psi = 0~.
\end{align}
Asymptotically, we impose the boundary condition $\delta\Psi(s\to0)=\delta\Psi^{(0)} s \ln s/(2z_\text{L})$. At the horizon, we impose the incoming-wave boundary condition.

The $\delta\Psi$ perturbation cannot be solved for a generic $\mu$, so we set $\bar{\epsilon}_\mu=\bar{\mu}-1<0$ and employ the $\bar{\epsilon}_\mu$-expansion as well as the $(\omega,q)$-expansion:
\begin{align}
\delta\Psi(s,k)&=(1-s)^{-i\bar{\omega}/2}
\\ &\quad \times %
((\psi_c + \bar{\epsilon}_\mu \psi_\epsilon +\cdots)+ \bar{\omega} \psi_\omega + \bar{q}^2 \psi_q +\cdots)~.
\nonumber
\end{align}
This form is taken to implement the incoming-wave boundary condition. Then, the boundary condition reduces to the regularity condition for $\psi_c$ and so on. The equation of motion reduces to
\begin{subequations}
\begin{align}
{\cal L}_\psi \psi_c &= 0~, \\
{\cal L}_\psi \psi_i &= \mathfrak{j}_i(\psi_c)~,
\end{align}
where
\begin{equation}
{\cal L}_\psi = \partial_s \left(\frac{h}{s}\partial_s \right) + \left\{ \frac{(1-s)^2}{hs^2} + \frac{1}{s^3} \right\}~,
\end{equation}
\end{subequations}
and the index $i$ collectively represents $\epsilon, \omega$, and $q$. The homogeneous equation ${\cal L}_\psi \psi_c =0$ can be solved as
\begin{align}
\psi_c &= c_1 \frac{s}{1+s} + c_2 \frac{s}{1+s} \ln\left[ \frac{s}{(1-s)^2}\right]~.
\end{align}
From the regularity at the horizon, $c_2=0$.

The source terms of inhomogeneous equations then become
\begin{subequations}
\begin{align}
\mathfrak{j}_\epsilon &= -\frac{2(1-s)}{s(1+s)^2} c_1~, \\
\mathfrak{j}_\omega &=\frac{-4-i(1+s)}{2s(1+s)^2} c_1~, \\
\mathfrak{j}_q &= \frac{1}{s^{2-1/z_\text{L}}(1+s)} c_1~.
\end{align}
\end{subequations}
The $\psi_\epsilon$ and $\psi_\omega$ solutions are
\begin{subequations}
\begin{align}
\psi_\epsilon &= - c_1\frac{s}{1+s} \left\{ -\frac{1}{2}\ln s+\ln (1+s)\right\}
\\
& \sim c_1 \frac{1}{2} s\ln s \quad (s\to 0)~,
\\
\psi_\omega &= c_1 \frac{s}{4(1+s)} \left\{ (3+i)\ln s-2\ln (1+s)\right\}
\\
& \sim c_1 \frac{3+i}{4} s\ln s \quad (s\to 0)~.
\end{align}
\end{subequations}
For $\psi_q$, we discuss the $z_\text{L}=1$ and $z_\text{L}\neq1$ cases separately.

\subsection{$z_\text{L}=1$

The $\psi_q$ solution is given by
\begin{align}
\psi_q &= -c_1\frac{1}{2(1+s)} s\ln s
\\
& \sim -c_1 \frac{1}{2} s\ln s \quad (s\to 0)~.
\end{align}
The asymptotic behavior then becomes
\begin{align}
\delta\Psi &\sim c_1(1-s)^{-i\bar{\omega}/2}
 \\ & \times %
\left\{-\frac{1}{2}\left(\bar{q}^2 - \frac{3+i}{2} \bar{\omega} + |\bar{\epsilon}_\mu| +\cdots \right) s\ln s + s +\cdots \right\}~.
\nonumber
\end{align}
The asymptotic boundary condition determines $c_1$. Then, the order parameter response function becomes
\begin{align}
\chi_{\omega, q}^>
&= \frac{ \delta{\cal O} }{ \delta\Psi^{(0)} } \nonumber\\
&= \frac{2}{ \frac{1}{2} q^2 - \frac{3+i}{2} \omega + |\epsilon_{\mu}| }~,
\end{align}
where we use the dictionary in \appen{dict}. The response function indeed takes the form of \eq{response_fn}, and
\begin{subequations}
\begin{align}
\xi_>^2 &= \frac{1}{2|\epsilon_{\mu}|}~, \\
A_> &= 2~.
\end{align}
\end{subequations}
The dispersion relation is given by
\begin{equation}
\omega = \frac{3-i}{5}\left( |\epsilon_{\mu}|+\frac{1}{2}q^2+\cdots \right)~.
\end{equation}
The relaxation time $\tau$ then becomes
\begin{equation}
\tau^{-1} = \frac{1}{5}|\epsilon_{\mu}|=\frac{1}{10}\xi_>^{-2}~.
\end{equation}

\subsection{$z_\text{L}>1$
In this case, we are not able to obtain the generic expression for $\psi_q$. Besides, even when the analytic expression is available, it is too cumbersome to write here. However, the slow falloff has a simple expression:
\begin{subequations}
\begin{align}
\psi^{(0)}_q &= - c_1 I(z_\text{L})~, \\
I(z_\text{L}) &:=\int_0^1 \frac{ds}{s^{1-1/z_\text{L}}(1+s)^2} \\
&= \frac{1}{2} + \frac{1}{2}\left(1-\frac{1}{z_\text{L}}\right)
\nonumber \\
&\quad\times\left\{ \psi_0\left(\frac{1}{2z_\text{L}}+\frac{1}{2} \right) - \psi_0\left(\frac{1}{2z_\text{L}}\right) \right\}~,
\end{align}
\end{subequations}
where $\psi_0(x)$ is the digamma function:
\begin{align}
\psi_0(x) &= \frac{d}{dx}\ln \Gamma(x)~.
\end{align}
A few examples of $I(z_\text{L})$ are
\begin{align}
I(1) &= \frac{1}{2}~,
\nonumber \\
I(2) &= \frac{1}{2} +\frac{\pi}{4} \approx 1.285~,
\nonumber \\
I(3) &= \frac{1}{2} +\frac{2}{3}\left(\ln2+\frac{\pi}{\sqrt{3}} \right) \approx 2.171~.
\nonumber
\end{align}
The combination $I(z_\text{L})/z_\text{L}$ monotonically increases with $z_\text{L}$ and reaches 1 for $z_\text{L}\to\infty$.
In order to obtain the falloff, we essentially used the standard method to solve an inhomogeneous differential equation (\appen{Green}).

Then, the order parameter response function becomes
\begin{align}
\chi_{\omega, q}^>
= \frac{2}{ \frac{I(z_\text{L})}{z_\text{L}} q^2 - \frac{3+i}{2} \omega + |\epsilon_{\mu}|  }~,
\label{eq:response_holo}
\end{align}
which gives
\begin{subequations}
\begin{align}
\xi_>^2 &= \frac{I(z_\text{L})/z_\text{L}}{|\epsilon_{\mu}|}~, \\
A_> &= 2~.
\end{align}
\end{subequations}
The dispersion relation is given by
\begin{equation}
\omega = \frac{3-i}{5}\left( |\epsilon_{\mu}|+\frac{I(z_\text{L})}{z_\text{L}}q^2+\cdots \right)~.
\end{equation}
The relaxation time is given by
\begin{equation}
\tau^{-1} = \frac{1}{5}|\epsilon_{\mu}| = \frac{I(z_\text{L})}{5z_\text{L}}\xi_>^{-2}~.
\end{equation}

\subsection{$z_\text{L}\gg1$
For large $z_\text{L}$,
\begin{equation}
I(z_\text{L})
\approx z_\text{L}~,
\end{equation}
so
\begin{subequations}
\begin{align}
\chi_{\omega, q}^> &\approx \frac{2}{ q^2 - \frac{3+i}{2} \omega + |\epsilon_{\mu}| }~, \\
\omega &\approx \frac{3-i}{5}(|\epsilon_{\mu}|+q^2+\cdots)~, \\
\xi_>^2 &\approx \frac{1}{|\epsilon_{\mu}|}~, \quad
\tau^{-1} = \frac{1}{5}|\epsilon_{\mu}|\approx\frac{1}{5}\xi_>^{-2}~.
\end{align}
\end{subequations}

\section{Low-temperature phase}\label{sec:low}

In the low-temperature phase, our task is
\begin{enumerate}
\item to construct the background,
\item to show that the $\bmPsi\neq0$ solution has a lower grand canonical potential, and
\item to derive the London equation. (This establishes that the $\bmPsi\neq0$ phase is a superconducting phase.)
\end{enumerate}

\subsection{Background solution}\label{sec:bg_low}

The solution \eqref{eq:analytic_sol} is the solution only at the critical point, and we first construct the background solution in the low-temperature phase. As mentioned in \sect{critical}, the construction is essentially the same as the $z_\text{L}=1$ case \cite{Herzog:2010vz}

Consider the solution of the form
\begin{align}
\bmPsi=\bmPsi(s)~, \quad\bmA_t=\bmA_t(s)~, \quad\bmA_i=\bmA_s=0~.
\end{align}
The equations of motion are given by
\begin{subequations}
\label{eq:eom_homo}
\begin{align}
&\partial_s \left(\frac{h}{s}\partial_s \bar{\bmPsi} \right) + \left\{ \frac{\bar{\bmA}_t^2}{hs^2}
+ \frac{1}{s^3} \right\}\bar{\bmPsi} =0~, \\
&\partial_s^2 \bar{\bmA}_t = \frac{2}{hs^2} \bar{\bmPsi}^2\bar{\bmA}_t~, \\
&\bmPsi^\dagger \bmPsi' - \bmPsi^{\dagger'} \bmPsi = 0~.
\end{align}
\end{subequations}
One can set $\bmPsi$ to be real.
We construct the background perturbatively:
\begin{subequations}
\begin{align}
\bar{\bmPsi} &= \epsilon^{1/2} \left( \bmPsi_1 + \epsilon\, \bmPsi_2 + \cdots \right)~,
\label{eq:background_low_psi}
\\
\bar{\bmA}_t &= \bmPhi_0 + \epsilon \bmPhi_1 + \epsilon^2 \bmPhi_2 +\cdots~,
\end{align}
\end{subequations}
where $\epsilon$ is a small parameter whose meaning will be clear in a moment. From \sect{critical}, we already know
\begin{subequations}
\begin{align}
\bmPhi_0 &= 1-s~, \\
\bmPsi_1 &= \frac{s}{1+s}~.
\end{align}
\end{subequations}
To proceed to higher orders in $\epsilon$, we impose the boundary conditions following Ref.~\cite{Herzog:2010vz}:
\begin{itemize}
\item
$\bmPsi_n$: Asymptotically, no slow falloff and no fast falloff, or $\Psi_n^{(0)}=\Psi_n^{(1)}=0$ (for $n\geq2$). The former means the condition for a spontaneous condensate. The latter means that ${\cal O}$ comes only from $\bmPsi_1$. At the horizon, we impose the regularity condition.
\item
$\bmPhi_n$: $\bmPhi_n(s=1)=0$ at the horizon.
\end{itemize}
Namely, we fix the fast falloff ${\cal O}$, but the chemical potential is corrected as
\begin{align}
\bar{\mu} = 1+\epsilon\delta\bar{\mu}_1+\epsilon^2\delta\bar{\mu}_2+\cdots~.
\end{align}
Under these boundary conditions,
\begin{equation}
\bar{\bmPsi}\sim\epsilon^{1/2}s~,
\end{equation}
so $\epsilon^{1/2}$ represents the order parameter ${\cal O}$. We impose 4 boundary conditions in total, which completely fixes the solution. For example, $\bmPhi_1$ and $\bmPsi_2$ have 4 integration constants, and they are determined by the 4 conditions.

At $O(\epsilon)$,
\begin{align}
\bmPhi_1
&=\delta\bar{\mu}_1(1-s)-\frac{s(1-s)}{2(1+s)} \\
&\sim \delta\bar{\mu}_1+(-1/2-\delta\bar{\mu}_1)s+\cdots~,
\end{align}
where we imposed the boundary condition $\bmPhi_1(s=1)=0$, and $\delta\bar{\mu}_1$ is the remaining integration constant. It is fixed at $O(\epsilon^{3/2})$ from the condition that $\Psi_2^{(0)}=0$.

At $O(\epsilon^{3/2})$, there are 2 more integration constants and $\delta\bar{\mu}_1$. After imposing the boundary condition at the horizon and the $\Psi_2^{(1)}=0$ condition, one obtains
\begin{align}
\bmPsi_2 &= -\frac{s^2}{3(1+s)^2}
\\ & \quad%
+ \left(\delta\bar{\mu}_1-\frac{1}{12}\right) \frac{s\ln s}{2(1+s)} + \left(\frac{1}{4}-\delta\bar{\mu}_1\right)\frac{s\ln(1+s)}{1+s}
\label{eq:psi2}
\nonumber \\ %
&\sim \frac{1}{2}\left(\delta\bar{\mu}_1-\frac{1}{12}\right) s\ln s~,
\end{align}
so the remaining no slow falloff condition $\Psi_2^{(0)}=0$ gives $\delta\bar{\mu}_1=1/12$.
Then, at $O(\epsilon)$, the chemical potential becomes
\begin{align}
\bar{\mu} &= \bar{\bmA}_t|_{s=0}
\nonumber\\
&= 1+\frac{1}{12}\epsilon+\cdots~,
\end{align}
so $\epsilon$ is determined as
\begin{align}
\epsilon=12\bar{\epsilon}_\mu = 12\epsilon_{\mu}/(2z_\text{L})~.
\end{align}
Thus,
\begin{align}
{\cal O} = -(2z_\text{L}) \epsilon^{1/2} = -(24z_\text{L}\epsilon_{\mu})^{1/2}~,
\end{align}
and the critical exponent $\beta=1/2$. More generally, $A_t^{(0)}$ gives the GL equation of motion (\sect{low_chi} and \sect{GL}).

At $O(\epsilon^{2})$,
\begin{align}
\bmPhi_2
&= - \frac{(1-s)}{1728(1+s)^2}(253+842s+253s^2)
\nonumber \\ & \quad%
+ \frac{1}{36}(7-13s)\ln2 + \frac{\ln(1+s)}{3(1+s)}\\
&\sim \frac{-253+336\ln2}{1728} + \frac{493-624\ln2}{1728} s + \cdots \\
&=\delta\bar{\mu}_2 + \left(\frac{373}{864}-\frac{5}{9}\ln2+\delta\bar{\mu}_2 \right)s + \cdots~, \\
\delta\bar{\mu}_2&=\frac{-253+336\ln2}{1728}~.
\end{align}
Again, we determine an integration constant $\delta\bar{\mu}_2$ at $O(\epsilon^{5/2})$ from the condition $\Psi_3^{(0)}=0$. The expression for $\bmPsi_3$ is too cumbersome to write here.

\subsection{Grand canonical potential}\label{sec:grand_canonical}

We use the Lorentzian formalism to evaluate the grand canonical potential $\Omega$. (Note $S_\text{E}=\beta \Omega=-S_\text{L}$.) The matter on-shell action is given by
\begin{align}
S_\text{OS} =\int d^{p+1}x\, &\bigg\{
-\frac{p-z_\text{L}}{2} A_t^{(0)}A_t^{(1)}
\nonumber \\ &
+ \int_0^1 du\, \sqrt{-g} g^{tt} \bmA_t^2 |\bmPsi|^2
\bigg\}~.
\end{align}
We are interested in the grand canonical potential of the spontaneous condensate, or the solution with $\Psi^{(0)}=0$, so the boundary term from $\Psi$ vanishes.

We evaluate the difference of the grand canonical potential between the $\bmPsi=0$ solution and the $\bmPsi\neq0$ solution. We fix the chemical potential as $\bar{\mu}=1+\epsilon\delta\bar{\mu}_1+\epsilon^2\delta\bar{\mu}_2+\cdots$, where $\delta\bar{\mu}_1$ and $\delta\bar{\mu}_2$ are obtained in the previous subsection. It turns out that  $\delta S_\text{OS}=0$ at $O(\epsilon)$, so we evaluate the difference at $O(\epsilon^2)$. This implies that one has to take into account up to $O(\epsilon^2)$ of $\bmA_t$ in order to evaluate the above boundary action.

For the $\bmPsi=0$ solution,
\begin{equation}
\bar{\bmA}_t \sim (1+\epsilon \delta\bar{\mu}_1+\epsilon^2\delta\bar{\mu}_2+\cdots)(1-s)~.
\end{equation}
In this case, only the boundary action contributes since $\bmPsi=0$. The on-shell action becomes
\begin{align}
S_{\Psi=0} &= \beta V_p (2z_\text{L})^3
\left\{ \frac{1}{2} + \delta\bar{\mu}_1\epsilon + \frac{1}{2}(\delta\bar{\mu}_1^2+2\delta\bar{\mu}_2)\epsilon^2 +\cdots \right\} \\
&= \beta V_p (2z_\text{L})^3 \bigg\{ \frac{1}{2} + \frac{\epsilon}{12}
\nonumber \\ & \quad %
+ \left( -\frac{247}{1728} + \frac{7}{36}\ln2 \right) \epsilon^2+\cdots \bigg\}~,
\end{align}
where $\beta$ is the inverse temperature, and $V_p$ is the boundary spatial volume. For the $\bmPsi\neq0$ solution,
\begin{align}
S_{\Psi\neq0} &= \beta V_p (2z_\text{L})^3
\bigg\{ \frac{1}{2} + \delta\bar{\mu}_1\epsilon
\\ & \quad %
+ \left(-\frac{181}{1728} -\frac{1}{4}\delta\bar{\mu}_1+\frac{1}{2}\delta\bar{\mu}_1^2 + \frac{7}{36}\ln2 \right)\epsilon^2 +\cdots \bigg\}
\nonumber \\ %
&= \beta V_p (2z_\text{L})^3 \bigg\{ \frac{1}{2} + \frac{\epsilon}{12}
\nonumber \\ & \quad %
+ \left( -\frac{211}{1728} + \frac{7}{36}\ln2 \right) \epsilon^2+\cdots \bigg\}~.
\end{align}
Thus, the difference is
\begin{align}
\delta S_\text{OS} &= S_{\Psi\neq0}-S_{\Psi=0}
\nonumber \\
&=\beta V_p \frac{z_\text{L}^3}{6}\epsilon^2 = -\beta \delta \Omega~,
\\%\quad
\Rightarrow\quad \frac{\delta\Omega}{V_p} &= -6z_\text{L} \epsilon_{\mu}^2~.
\label{eq:free_energy}
\end{align}
$\delta\Omega<0$, so the $\bmPsi\neq0$ solution is favorable. The difference is proportional to $\epsilon_{\mu}^2=(\mu-\mu_c)^2 \propto (T_c-T)^2$, which implies the second-order phase transition. (The difference and its first derivative are continuous, but the second derivative is discontinuous.) The specific heat $C_\mu$ behaves as $C_\mu=-T\partial^2\Omega/\partial T^2 \propto T$, which determines the critical exponent $\alpha=0$, where $\alpha$ is defined by $C_\mu \propto (T_c-T)^{-\alpha}$.

\subsection{Background with source}\label{sec:low_chi}

We construct the background without the source of the order parameter, but it is straightforward to extend the construction to the background with the source. Going back to \eq{psi2}, we obtained
\begin{align}
\bmPsi_2 &= -\frac{s^2}{3(1+s)^2}
\\ & \quad%
+ \left(\delta\bar{\mu}_1-\frac{1}{12}\right) \frac{s\ln s}{2(1+s)} + \left(\frac{1}{4}-\delta\bar{\mu}_1\right)\frac{s\ln(1+s)}{1+s}
\label{eq:psi2}
\nonumber \\ %
&\sim \frac{1}{2}\left(\delta\bar{\mu}_1-\frac{1}{12}\right) s\ln s~,
\end{align}
so the asymptotic behavior becomes
\begin{equation}
\bar{\bmPsi} \sim \frac{1}{2}\left(\delta\bar{\mu}_1-\frac{1}{12}\right) \epsilon^{3/2} s\ln s + \epsilon^{1/2} s~.
\label{eq:asymptotic_source}
\end{equation}
Previously, we imposed the source-free condition $\Psi^{(0)}=0$, which gives $\delta\bar{\mu}_1=1/12$. We now allow $\Psi^{(0)} \neq 0$. The chemical potential is given by $\bar{\mu} = 1+\epsilon\delta\bar{\mu}_1$. At the critical point, $\bar{\mu}=1$, so $\delta\bar{\mu}_1=0$. From the asymptotic behavior \eqref{eq:asymptotic_source},  ${\cal O}\propto\epsilon^{1/2}$ and $\Psi^{(0)} \propto \epsilon^{3/2}$. Then, the exponent $\delta$ defined by ${\cal O} \propto (\Psi^{(0)})^{1/\delta} $ (at $\mu=\mu_c$) is $\delta=3$.

One can evaluate the thermodynamic response function at low temperature. By imposing our boundary conditions,
\begin{equation}
\bar{\Psi}^{(0)} = z_\text{L}\left(\delta\bar{\mu}_1-\frac{1}{12}\right)\epsilon^{3/2}~.
\end{equation}
The chemical potential is then determined as
\begin{align}
\bar{\mu} &= 1+\epsilon\delta\bar{\mu}_1 = 1 + \frac{1}{12}\epsilon + \frac{\bar{\Psi}^{(0)}}{z_\text{L}\epsilon^{1/2}}~,
\label{eq:chemical_source1}
\end{align}
which is rewritten as
\begin{align}
\bar{\epsilon}_\mu &= \frac{1}{12}\epsilon + \frac{\bar{\Psi}^{(0)}}{z_\text{L} \epsilon^{1/2}}~.
\label{eq:chemical_source}
\end{align}
This is essentially the GL equation of motion (\sect{GL}). For a fixed $\mu$, this gives
\begin{equation}
d\epsilon = - \frac{12}{z_\text{L} \epsilon^{1/2}} d\bar{\Psi}^{(0)} + O(\bar{\Psi}^{(0)}d\epsilon)~.
\end{equation}
Thus,
\begin{align}
\chi_T^< &= \left.\frac{\partial{\cal O}}{\partial\Psi^{(0)}} \right|_{\Psi^{(0)}=0}
\nonumber \\
&= \left.\frac{d{\cal O}/d\epsilon}{d\Psi^{(0)}/d\epsilon} \right|_{\Psi^{(0)}=0}
=\frac{12}{2z_\text{L}\epsilon}
=\frac{1}{\epsilon_{\mu}}~,
\\
\Rightarrow A_< &=1~.
\end{align}
(Recall ${\cal O} = -2z_\text{L}\epsilon^{1/2}$.)

We obtained $\chi_T^<$ from the background solution, but it should also be possible to obtain it from the scalar perturbation as in \sect{high}. One would also obtain the full response function $\chi_{\omega,q}^<$ using the $(\epsilon,\omega,q)$-expansion. But, in the low-temperature phase, $\delta\Psi$ couple with $\delta A_t$ and $\delta A_x$, and the computation is more involved, so we leave it to a future work.

\subsection{Vector modes}\label{sec:vector}

From the vector mode, one can show the London equation and compute the conductivity. The $\delta A_y$-equation is given by
\begin{align}
&\partial_s \left(\frac{h}{s^{1-1/z_\text{L}}}\partial_s \delta A_y \right)
\nonumber \\& %
+ \left\{ \frac{\bar{\omega}^2}{hs^{2-1/z_\text{L}}} - \frac{\bar{q}^2}{s^{3-2/z_\text{L}}} -\frac{2\bar{\bmPsi}^2}{s^{3-1/z_\text{L}}} \right\}\delta A_y = 0~,
\end{align}
where $\bar{\bmPsi}$ was constructed in \eq{background_low_psi}. We impose the incoming-wave boundary condition at the horizon and $\delta A_y|_{s=0} = \sourceA{y}$ asymptotically. We again employ the ($\epsilon$, $\omega$)-expansion:
\begin{align}
\delta A_y &= (1-s)^{-i\bar{\omega}/2}
\nonumber \\ &\quad \times %
((a_c + \epsilon a_\epsilon +\cdots) + \bar{\omega} a_\omega +\cdots)~.
\end{align}
The equation of motion reduces to
\begin{subequations}
\begin{align}
{\cal L}_a a_c &= 0~, \\
{\cal L}_a a_i &= \mathfrak{j}_i (a_c)~,
\end{align}
where
\begin{align}
{\cal L}_a &= \partial_s \left(\frac{h}{s^{1-1/z_\text{L}}}\partial_s \right)~.
\end{align}
\end{subequations}
The homogeneous equation ${\cal L}_a a_c =0$ can be solved as
\begin{align}
a_c &= c_1
\quad \\ &
+ c_2 s^{2-1/z_\text{L}} \frac{2z_\text{L}}{2z_\text{L}-1}
{}_2F_1\left( 1,\frac{2z_\text{L}-1}{2z_\text{L}},\frac{4z_\text{L}-1}{2z_\text{L}};s^2 \right)~.
\nonumber
\end{align}
From the regularity at the horizon, $c_2=0$.

The source terms of inhomogeneous equations then become
\begin{subequations}
\begin{align}
\mathfrak{j}_\epsilon &= \frac{2}{s^{1-1/z_\text{L}}(1+s)^2} c_1~, \\
\mathfrak{j}_\omega &= -\frac{i}{2} \left( \frac{1+s}{s^{1-1/z_\text{L}}} \right)' c_1~.
\end{align}
\end{subequations}
Again, we discuss the $z_\text{L}=1$ and $z_\text{L}\neq1$ cases separately.

\subsubsection{$z_\text{L}=1$
The solution is
\begin{subequations}
\begin{align}
a_\epsilon &= c_1 \frac{1}{1+s}
\sim c_1  (1-s)~, \\
a_\omega &=\frac{1}{2} i c_1\ln(1+s)
\sim ic_1 s/2~.
\end{align}
\end{subequations}
The asymptotic behavior then becomes
\begin{align}
\delta A_y &= \frac{\sourceA{y}(k)}{1+\epsilon+\cdots} (1-s)^{-i\bar{\omega}/2}
\nonumber \\ & \quad\times %
\left\{ 1+ \frac{\epsilon}{1+s} + \frac{1}{2} i\bar{\omega} \ln(1+s) + \cdots \right\}\\
&\sim \sourceA{y}\{ 1+(-\epsilon+ i\bar{\omega}+\cdots) s \}~.
\end{align}
We determine the constant $c_1$ from the asymptotic boundary condition $\delta A_y|_{s=0} = \sourceA{y}$. So,
\begin{align}
 J^y &= (4z_\text{L}-2) \vevA{y} \\
&= 2 (-\epsilon+ i\bar{\omega} +\cdots) \sourceA{y}~.
\end{align}
The $\omega \to0$ limit gives the London equation
\begin{align}
 J^y  = -(1/\lambda^2)\sourceA{y}~,

\end{align}
with the London penetration depth $\lambda^{-2}= 2\epsilon$. The conductivity is then given by
\begin{align}
\sigma(\omega) &= \frac{ J^y }{i\omega \sourceA{y}}
\nonumber \\&
= \frac{2i\epsilon}{\omega}+ 1 +\cdots~.
\end{align}
$\text{Im}(\sigma)$ has the $1/\omega$-pole which implies the diverging DC conductivity.

A superconductor\ has singular behaviors in the current, but its essence is not in the diverging DC conductivity but in the London equation. A diverging DC conductivity also appears in a perfect conductor, but the London equation is unique to superconductors.

When one combines the London equation with the Maxwell equation, one obtains the Meissner effect. However, for usual holographic superconductors, the boundary Maxwell field is added just as an external source and is not dynamical in the boundary theory, so the Meissner effect does not arise; the magnetic field can always penetrate into the material. In this sense, a holographic superconductor\  may be regarded as a superfluid. (In low spatial dimensions $p\leq2$, one can obtain a boundary theory with a dynamical Maxwell field. See, {\it e.g.}, Ref.~\cite{Domenech:2010nf}.)

However, the London equation must hold even in this case if the system is really a superconductor\ or a superfluid. The London equation is the response of the current under the external source, and whether the source is dynamical or not is irrelevant to the issue.

\subsubsection{$z_\text{L}>1$

For $a_\omega$,
one can get the generic expression%
\footnote{We set $p=3z_\text{L}$, but the $O(\omega,q^2)$-equations can actually be solved for a generic $(p,z_\text{L})$.}:
\begin{align}
a_\omega &=\frac{1}{2}ic_1
\bigg\{ s^{2-1/z_\text{L}} \frac{2z_\text{L}}{2z_\text{L}-1} {}_2F_1\left( 1,\frac{2z_\text{L}-1}{2z_\text{L}},\frac{4z_\text{L}-1}{2z_\text{L}};s^2 \right)
\nonumber \\ & \quad %
+ \ln(1-s) \bigg\} \\
&\sim O(s) + ic_1 \frac{z_\text{L}}{2z_\text{L}-1} s^{2-1/z_\text{L}}~.
\end{align}
For $a_\epsilon$, the generic expression is either difficult to obtain or too cumbersome, but again the fast falloff has a simple expression:
\begin{align}
a_\epsilon^{(1)} &= -\frac{2z_\text{L}}{2z_\text{L}-1} c_1 I(z_\text{L})~.
\end{align}
The asymptotic behavior then becomes
\begin{align}
\delta A_y &\sim A_y^{(0)} \big[ 1+ \cdots +
\\ & \quad %
\frac{1}{4z_\text{L}-2}\left\{ -4z_\text{L} I(z_\text{L}) \epsilon+ i\omega +\cdots \right\} s^{2-1/z_\text{L}} + \cdots \big]~.
\nonumber %
\end{align}
Thus,
\begin{align}
 J^y  &= \left\{ -4z_\text{L} I(z_\text{L}) \epsilon+ i\omega + \cdots \right\}  \sourceA{y}~.
\label{eq:London}
\end{align}
Again, the $\omega\to0$ limit gives the London equation $ J^y  = -(1/\lambda^2) \sourceA{y}$ with the London penetration depth $\lambda^{-2} = 4z_\text{L} I(z_\text{L}) \epsilon$.
The conductivity is then given by
\begin{align}
\sigma &= \frac{ J^y }{i\omega\sourceA{y}}
\nonumber \\&
=  \frac{i}{\omega}\frac{I(z_\text{L})}{z_\text{L}} {\cal O}^2 + 1 +\cdots~.
\end{align}
For large $z_\text{L}$,
\begin{align}
\sigma \approx \frac{i}{\omega}{\cal O}^2 + 1 +\cdots~.
\end{align}

The GL parameter $\kappa$ is defined by
\begin{equation}
\kappa^2 :=\left( \frac{\lambda}{\xi_>} \right)^2 = \frac{z_\text{L}}{24I(z_\text{L})^2}~.
\end{equation}
In conventional superconductors, $\kappa^2<1/2$ for type I and $\kappa^2>1/2$ for type II superconductors. For $z_\text{L}=1$, $\kappa^2=1/6$, so one may conclude that our system is type I (in the sense of $\kappa$), but whether our system is type I or II is more subtle. Physically, $1/\lambda$ represents the Maxwell field mass, so we should
determine the normalization of $\lambda$ by comparing with normalization of the boundary Maxwell action.
However, as mentioned above, the boundary Maxwell field is added as an external source here and is not dynamical in the boundary theory, so the normalization cannot be determined%
\footnote{The value of $\kappa$ for holographic superconductors\  has been discussed in Refs.~\cite{Maeda:2008ir,Dias:2013bwa}. In Ref.~\cite{Dias:2013bwa}, $\kappa$ depends on the scalar charge $e$. On the contrary, if we restore dimensionful parameters, our $\kappa$ does not depend on $e$ (\appen{restore}). }.
(Holographic superconductors are type II superconductors\ in the sense that there is no Meissner effect.)

\section{Background with vector potential}\label{sec:bg_vector}

In this section, we add a vector potential $\bmA_i$ as a background.
We again consider the perturbative expansion:
\begin{subequations}
\begin{align}
\bar{\bmPsi} &= \epsilon^{1/2} \left(\bmPsi_1' + \cdots \right)~,
\\
\bar{\bmA}_t &= \bmPhi_0'  + \cdots~, \\
\bar{\bmA}_y &= \bmA_{y,0} + \epsilon \bmA_{y,1} +\cdots~.
\end{align}
\end{subequations}
Note that we take into account (1) $A_y$ as a background and (2) the backreaction of $\bmA_{y}$ onto the other fields. (That is why we use variables with primes.) The former is the difference from the perturbative expansion in \sect{bg_low}, and the latter is difference from the vector mode computation in \sect{vector}.

At $O(\epsilon^0)$, the Maxwell equation becomes
\begin{align}
\nabla_N\bm{F}^{MN}_{~~~~~, 0}=0~,
\end{align}
where $\bm{F}_{MN,0} =\partial_M \bmA_{N,0} - \partial_N \bmA_{M,0}$. The equation has simple solutions. Namely, $\bmPhi_0'=\bar{\mu}(1-s)$ and two interesting solutions for $\bmA_{y,0} $:
\begin{subequations}
\begin{align}
\bmA_{y,0} &=\bar{a}_y =\text{(constant)}~,   \\
\bmA_{y,0} &= \bar{B}x~.
\end{align}
\end{subequations}
The former corresponds to adding a constant superfluid flow $a_y$, and the latter corresponds to adding a constant magnetic field $B$.

\subsection{Superfluid flow

For the superfluid flow, it is enough to consider homogeneous perturbations. The equations of motion are given by
\begin{subequations}
\begin{align}
&\partial_s \left(\frac{h}{s}\partial_s \bar{\bmPsi} \right) + \left\{ \frac{\bar{\bmA}_t^2}{hs^2} - \frac{\bar{\bmA}_y^2}{s^{3-1/z_\text{L}}} + \frac{1}{s^3} \right\}\bar{\bmPsi} =0~, \\
&\partial_s^2 \bar{\bmA}_t = \frac{2}{hs^2} \bar{\bmPsi}^2\bar{\bmA}_t~, \\
&\partial_s \left(\frac{h}{s^{1-1/z_\text{L}}}\partial_s  \bar{\bmA}_y \right) - \frac{2\bar{\bmPsi}^2}{s^{3-1/z_\text{L}}}\bar{\bmA}_y =0~.
\end{align}
\end{subequations}
We impose the same boundary conditions as \sect{bg_low}.

Our main interest is the phase diagram, {\it i.e.}, the deviation of the critical point by the vector potential. Then, we evaluate how $\bmA_y$ at $O(\epsilon^0)$ affects $\bmPsi_1'$ at $O(\epsilon^{1/2})$. This in turn affects the value of $\mu_c$. We employ the $a_y$-expansion as well as the $\epsilon$-expansion \cite{Herzog:2010vz}. Namely,
\begin{subequations}
\begin{align}
\bmPsi_1' &= \bmPsi_1 + \bar{a}_y^2 \bmPsi_a + \cdots~, \\
\bm\Phi_0' &= (1 + \bar{a}_y^2c_0+\cdots)(1-s)~,
\end{align}
\end{subequations}
where $c_0$ is a constant. This expansion is consistent with the above equations of motion.

At $O(\epsilon^{1/2})$, $\bmPsi_1=s/(1+s)$. The $\bmPsi_a$-equation becomes
\begin{subequations}
\begin{align}
{\cal L}_\psi \bmPsi_a &= \mathfrak{j}_a~, \\
\mathfrak{j}_a &= -2c_0 \frac{1-s}{s(1+s)^2}+\frac{1}{s^{2-1/z_\text{L}}(1+s)}~.
\end{align}
\end{subequations}
The equation is hard to solve in general. However, to determine the $\bar{a}_y$-dependence on the chemical potential, it is enough to obtain the slow falloff of $\bmPsi_a$.
The slow falloff has a simple expression:
\begin{align}
\bmPsi^{(0)}_a &= -\int_0^1 ds\, \mathfrak{j}_a\frac{s}{1+s}
\\
&= \frac{c_0}{2} - I(z_\text{L})~.
\end{align}
We impose the boundary condition $\bmPsi^{(0)}_a=0$, which gives $c_0=2I(z_\text{L})$.

At the critical point, the order parameter vanishes, so $\epsilon=0$. Then, to determine the critical point, set $\epsilon=0$, and $\sourceA{t}$ gives the critical chemical potential:
\begin{align}
\mu_c &= \mu_{c,0}+a_y^2 \frac{I(z_\text{L})}{z_\text{L}}+\cdots
\label{eq:flow} \\
&\sim \mu_{c,0} + a_y^2 \quad (z_\text{L}\gg1)~,
\end{align}
where $\mu_{c,0}=2z_\text{L}$ is the critical point without superfluid flow.

To obtain $J^\mu$, one needs to obtain $\bmA_\mu$. This is necessary to derive the second sound $c_2$ \cite{Herzog:2008he,Herzog:2010vz}:
\begin{equation}
c_2^2 = - \left.\frac{\partial J^y/\partial a_y}{\partial J^t/\partial\mu}\right|_{a_y=0}~.
\end{equation}
To derive $c_2$, it is enough to use the results obtained in the previous section. The $\bmA_{y,1}$-equation is the same as the vector mode perturbation $a_\epsilon$. So, $J^y$ is given by
\begin{align}
\bar{\bmA}_y &\sim \bar{a}_y \left[ 1 -
\frac{1}{4z_\text{L}-2} 4z_\text{L} I(z_\text{L}) \epsilon s^{2-1/z_\text{L}} + \cdots \right]~,
\nonumber \\
\Rightarrow J^y  &= (4z_\text{L}-2)\vevA{y} = -4z_\text{L} I(z_\text{L}) \epsilon\, a_y~.
\end{align}
$J^t$ is given by
\begin{align}
\bar{\bmA}_t &\sim (1-s) + \epsilon\{ \delta\bar{\mu}_1+(-1/2-\delta\bar{\mu}_1)s \} +\cdots
\nonumber \\
&=\bar{\mu}+(6-7\bar{\mu})s +\cdots~,
\nonumber \\
\Rightarrow J^t &= -2z_\text{L}\vevA{t} = 2z_\text{L}(7\mu-12z_\text{L})~.
\end{align}
Thus,
\begin{align}
c_2^2
&= \frac{I(z_\text{L})}{14z_\text{L}^2}{\cal O}^2 \\
&\sim \frac{1}{14z_\text{L}}{\cal O}^2 \quad (z_\text{L} \gg1)~.
\end{align}

\subsection{Magnetic field

We follow Ref.~\cite{Maeda:2009vf} to obtain the critical magnetic field. The $\bmPsi$-equation is given by
\begin{align}
\partial_s \left(\frac{h}{s}\partial_s \bar{\bmPsi} \right) + \left\{ \frac{\bar{\bmA}_t^2}{hs^2} + \frac{1}{s^{3-1/z_\text{L}}}(\bar{\del}_i-i\bar{\bmA}_i)^2 + \frac{1}{s^3} \right\}\bar{\bmPsi} =0~.
\end{align}
Here, $\bar{\del}_i =\partial_i/(2z_\text{L}) = \partial/\partial\underline{x}^i$, where $\underline{x}^{i} :=(2z_\text{L}) x^i$.
The vector potential is given by $\bar{\bmA}_y=\bar{B}x=\bar{\bar{B}} \underline{x}$, where $\bar{\bar{B}}:=B/(2z_\text{L})^2$.

This problem can be solved as a Landau-level problem after separation of variables. First, set%
\footnote{For simplicity, we set the other momenta as $k_3=k_4=\cdots=0$.}
\[
\bar{\bmPsi}(x,y,s)=e^{i k_y y}\varphi(x,s;k_y)~.
\]
The equation then takes the form
\begin{align}
s^{3-1/z_\text{L}} & \left[ \partial_s \left(\frac{h}{s} \partial_s \right) + \frac{\bar{\bmA}_t^2}{hs^2} + \frac{1}{s^3} \right]\varphi
\nonumber \\ & \quad %
= \left[ -\partial_{\underline{x}}^2+\left(\bar{k}_y-\bar{\bar{B}}\underline{x}\right)^2 \right] \varphi~,
\end{align}
so setting
\[
\varphi_n(x,s;k_y)=\rho_n(s)\gamma_n(x;k_y)~,
\]
one obtains
\begin{subequations}
\begin{align}
&\left( -\partial_X^2+X^2 \right) \gamma_n = \lambda_n\gamma_n~, \\
& \left[ \partial_s \left(\frac{h}{s} \partial_s \right) + \frac{\bar{\bmA}_t^2}{hs^2} + \frac{1}{s^3} \right] \rho_n = \bar{\bar{B}} \lambda_n \frac{\rho_n}{s^{3-1/z_\text{L}}}~,
\end{align}
\end{subequations}
where $X:= \sqrt{\bar{\bar{B}}}(\underline{x}-\bar{k}_y/\bar{\bar{B}})$, and $\lambda_n$ is a separation constant. The $\gamma_n$-equation is solved by the Hermite function $H_n$ as
\begin{equation}
\gamma_n(X)= e^{-X^2/2} H_n(X)~,
\end{equation}
with eigenvalue $\lambda_n$ as
\begin{align}
\lambda_n=2n+1 \quad (n\geq0)~.
\end{align}
The solution is parametrized by $\bar{\bar{B}}\lambda_n$, so one has the largest magnetic field $B_{c2}$ when $\lambda_n$ takes the minimal value, namely the $n=0$ solution.

The $\rho_0$-equation is given by
\begin{align}
\partial_s \left(\frac{h}{s}\partial_s \rho_0 \right) + \left\{ \frac{\bar{\bmA}_t^2}{hs^2} -\frac{\bar{\bar{B}}_{c2}}{s^{3-1/z_\text{L}}} + \frac{1}{s^3} \right\}\rho_0 =0~.
\end{align}
Then, the problem formally reduces to the same problem as the superfluid flow one with the replacement $\bar{a}_y^2$ by $\bar{\bar{B}}_{c2}$. Thus, the critical point is given by
\begin{equation}
\mu_c=\mu_{c,0}+B_{c2}\frac{I(z_\text{L})}{z_\text{L}}~,
\end{equation}
Using the result of $\xi_>^2$ in \sect{high}, we get
\begin{equation}
B_{c2}=1/\xi_{>}^2~.
\label{eq:Bc2}
\end{equation}

\section{The dual GL theory

\subsection{Identifying the dual GL theory}\label{sec:GL}

We thus obtained all critical exponents and critical amplitudes
\begin{subequations}
\begin{align}
(\alpha,\beta,\gamma,\delta,\nu,\eta,z_\text{D}) &= \left(0,\frac{1}{2},1,3,\frac{1}{2},0,2\right)~,
\\
A_>&=2A_<~.
\end{align}
\end{subequations}
The results are consistent with the standard GL theory or the $\phi^4$ mean-field theory.
In fact, the following GL theory reproduces all our results%
\footnote{As always, presumably the dual theory is some large-$N_c$ gauge theory. This is the effective GL theory at low energy and momentum.}:
\begin{subequations}
\begin{align}
H_\text{GL} &= \int d^px\, \bigg\{ \frac{c_K}{2} |D_i\phi|^2
-\frac{c_2}{2}\epsilon_{\mu}\,|\phi|^2 + \frac{c_4}{4} |\phi|^4 +\cdots
\nonumber \\ & \quad %
- c_\phi (\phi J^\dag+\phi^\dag J)\bigg\}~,
\\
D_i &:= \partial_i -ic_A \sourceA{i}~.
\end{align}
\end{subequations}
In the dynamic case, consider the time-dependent GL equation (for Model A dynamic universality class):
\begin{align}
\Gamma^{-1} \partial_t\phi &= - \frac{\delta H_\text{GL}}{\delta\phi} \\
&= -\frac{c_K}{2}D_i^2\phi - \frac{c_2}{2}\epsilon_{\mu}\phi + \frac{c_4}{2} \phi|\phi|^2 - c_\phi J~.
\end{align}
We determine the GL parameters $(c_2,c_4,c_\phi,c_K,c_A,\Gamma)$ to reproduce our holographic results.

In the static homogeneous case, the $\phi$-equation becomes
\begin{equation}
c_2 \epsilon_{\mu}\phi - c_4 \phi|\phi|^2 + 2c_\phi J=0~.
\label{eq:EOM_GL}
\end{equation}
Substituting the $J=0$ solution $|\phi|^2=c_2\epsilon_{\mu}/c_4$ into $H_\text{GL}$, one obtains the grand canonical potential:
\begin{equation}
\frac{\Omega}{V} = - \frac{c_2^2}{4c_4}\epsilon_{\mu}^2~.
\label{eq:free_energy_GL}
\end{equation}
The current is given by
\begin{equation}
J^i := -\frac{\delta H_\text{GL}}{\delta \sourceA{i}} = -c_Kc_A^2|\phi|^2 \sourceA{i}~.
\end{equation}
In the high-temperature phase, the response function is given by
\begin{equation}
\chi_k^> = \frac{ c_\phi \Gamma }{ -i\omega+\frac{\Gamma}{2}(c_K q^2+c_2|\epsilon_{\mu}|)}~,
\end{equation}
which implies
\begin{align}
\omega &= -i\frac{\Gamma}{2}c_2 (|\epsilon_{\mu}|+\frac{c_K}{c_2}q^2)~, \\
\xi_>^2 &=\frac{c_K}{c_2} |\epsilon_{\mu}|^{-1}~.
\end{align}
Add a background vector potential. When a constant $A_y=a_y$ is added, the critical point is shifted as
\begin{equation}
\mu_c =\mu_{c,0} + \frac{c_K}{c_2} c_A^2 a_y^2~.
\label{eq:flow_GL}
\end{equation}
When a magnetic field is added, the critical magnetic field is given by
\begin{equation}
B_{c2} = \frac{c_2}{c_Kc_A} \epsilon_{\mu} = \frac{1}{c_A} \xi_>^{-2}~,
\label{eq:Bc2_GL}
\end{equation}
by solving the Landau-level problem.

Returning to our holographic results, \eq{chemical_source} is rewritten as
\begin{align}
& \epsilon_{\mu} {\cal O} - \frac{1}{24z_\text{L}} {\cal O}^3 + 2 \Psi^{(0)} = 0~,
\label{eq:EOM_holo}
\end{align}
which takes the form of the GL theory equation of motion \eqref{eq:EOM_GL}. The grand canonical potential, the current, and the response function are obtained in Eqs.~\eqref{eq:free_energy}, \eqref{eq:London}, and \eqref{eq:response_holo}, respectively. These determine the GL parameters as
\begin{subequations}
\begin{align}
\cal{H} &= \frac{I(z_\text{L})}{2z_\text{L}} |(\partial_i-i\sourceA{i})\phi|^2
- \frac{\epsilon_{\mu}}{2} |\phi|^2
+ \frac{1}{96z_\text{L}} |\phi|^4 +\cdots
\nonumber \\ & \quad %
- (\phi J^\dag+\phi^\dag J)~, \\
\frac{\Gamma}{2} &= \frac{1+3i}{5}~.
\end{align}
\end{subequations}
In the presence of a background vector potential, Eqs.~\eqref{eq:flow} and \eqref{eq:Bc2} agree with Eqs.~\eqref{eq:flow_GL} and \eqref{eq:Bc2_GL}, respectively.

One would be tempted to ask how various results change as we vary $z_\text{L}$. But to make such a comparison, one must keep in mind that (1) we consider a special class of Lifshitz theories and (2) we must specify what quantities to fix as we vary $z_\text{L}$.

We consider a special class of theories where $p=3z_\text{L}$ and $m^2=-4z^2$. Even the spatial dimensionality $p$ is different for a different $z_\text{L}$, and it is unclear if the comparison with a different $z_\text{L}$ is physically meaningful. Also, some results may be generic for Lifshitz theories in general but some are not. As a simple example, in our case, $(\mu/T)_c$ is independent of $z_\text{L}$. This is so by construction of our theories as discussed in \sect{critical} and is certainly not a generic behavior. It simply means that holographic Lifshitz superconductors\ have enough parameters to fix  $(\mu/T)_c$ as we vary $z_\text{L}$.

We also have to specify what quantities to fix. One natural candidate is $\mu_c$ (or $T_c$) since $(\mu/T)_c$ is $z_\text{L}$-independent, but it is unclear if this is really appropriate. We do not have the answer to this question. So far we set $r_h=1$ just for simplicity, so here we simply fix $r_h$ (and $\epsilon_{\mu}$) and how various results change as we vary $z_\text{L}$. Again, we do not mean that fixing $r_h$ is natural from the boundary point of view. Rather, the following comparison should be regarded as a handy way to understand the $z_\text{L}$-dependence of our holographic results or the dual GL theory.
\begin{enumerate}
\item
In the dual GL theory, the coefficient of the $\phi^4$-term becomes smaller as we increase $z_\text{L}$. So, the condensate increases as $|\phi|^2 \propto z_\text{L}$.
\item
The $z_\text{L}$-dependence appears only in the kinetic term and the $\phi^4$-term. Thus, the relaxation time $\tau_{q=0}$ of the order parameter does not depend on $z_\text{L}$.
\item
The correlation length $\xi$ depends on the kinetic term so does depend on $z_\text{L}$. It monotonically increases as $I(z_\text{L})/z_\text{L}$ but increases slowly and reaches a constant value for $z_\text{L}\gg1$.
\item
On the other hand, the London penetration depth $\lambda$ decreases. This is because $J^i=-(1/\lambda^2) \sourceA{i} \propto -|\phi|^2 \sourceA{i}$ and because $\phi$ increases. ($\lambda$ also depends on the kinetic term so has the factor $I(z_\text{L})/z_\text{L}$, but it is not a dominant factor.) Then, the GL parameter $\kappa$ decreases.
\item
As usual, the presence of a background vector potential $\sourceA{i}$ increases the critical chemical potential $\mu_c$. From the gravity point of view, this is because $A_i$ increases the effective mass of $\Psi$. $\mu_c$ monotonically increases as $I(z_\text{L})/z_\text{L}$ since $\sourceA{i}$ comes from the covariant derivative in the kinetic term.
\end{enumerate}

\subsection{Lifshitz exponent and dynamic critical exponent}\label{sec:two_z}

We already mentioned that some results are not generic to Lifshitz theories in general. Then, what results are expected to be generic? An obvious answer is critical exponents and the ratio of critical amplitudes. The $\phi^4$ mean-field critical exponents are likely to hold for theories of \eq{Lifshitz_HSC}. The dynamic critical exponent $z_\text{D}=2$ is also likely to hold.

For a Lifshitz geometry, one would expect a dispersion relation of the form
\begin{equation}
\omega \propto q^{z_\text{L}}~.
\end{equation}
This form is expected from the Lifshitz scaling \eqref{eq:scaling_Lifshitz}. But from the analysis of the high-temperature phase, the order parameter obeys the dispersion relation
\begin{equation}
\omega \propto q^2~,
\end{equation}
{\it i.e.}, the dynamic critical exponent $z_\text{D}=2$ irrespective of the value of the Lifshitz exponent $z_\text{L}$. This does not contradict with the Lifshitz scaling. If we restore the horizon radius $r_h$,
\begin{equation}
\omega \propto r_h^{z_\text{L}-2}q^2~.
\end{equation}
Namely, at finite temperature, there are two length scales $r_h$ and $1/q$. They combine to give the scaling dimension $z_\text{L}$. In other words, the Lifshitz scaling alone does not determine the dynamic critical exponent.

Then, what determines $z_\text{D}$? We obtain $z_\text{D}=2$ because the order parameter is not a conserved charge. According to the classification of Hohenberg and Halperin \cite{Hohenberg:1977ym}, all our models belong to Model A universality class.

The dynamic universality class is classified based on
\begin{enumerate}
\item whether the order parameter is conserved or not,
\item whether there are the other hydrodynamic modes which couple to the order parameter (none for Model A and B).
\end{enumerate}
Conservation laws play important roles to determine the dynamic universality class since a conservation law forces the relaxation to proceed more slowly. When only the order parameter matters in critical dynamics, a nonconserved order parameter gives Model A, and a conserved order parameter gives Model B.

The Lifshitz geometry is conjectured to describe a quantum critical point. Using holographic Lifshitz superconductors, one prepares a new finite-temperature critical point in addition to the Lifshitz critical point. What we have shown is that the dynamic critical exponent $z_\text{D}$ associated with the new critical point can differ from $z_\text{L}$. Instead, the value of $z_\text{D}$ is determined by the critical dynamics of the new critical point.

\section{Discussion

\subsection{Lifshitz geometry and holographic superconductors\

The Lifshitz geometry appears even in the context of the standard $z_\text{L}=1$ holographic superconductor\  \cite{Gubser:2009cg,Horowitz:2009ij}. Consider the backreaction of matter fields onto the geometry. In the high-temperature phase, $\Psi=0$, so the geometry is the Reissner-Nordstr\"{o}m AdS black hole. In the low-temperature phase, $\Psi\neq0$, but one may expect that the geometry is somewhat similar to the Reissner-Nordstr\"{o}m AdS black hole. However, the $T=0$ geometry is conjectured to be a Lifshitz geometry in the IR and the AdS geometry in the UV. The solution in IR has been constructed, but the full geometry remains an open question.

It is unclear what happens at low temperature, but it is natural to expect that a Lifshitz-like black hole\  appears at low enough temperature. (Unfortunately, the Lifshitz black hole\  used in this paper is not a solution of the Einstein-Maxwell-complex scalar system.) Then, one should consider the Einstein-Maxwell-complex scalar system in a Lifshitz (IR)-AdS (UV) black hole. This is not an easy task however. First, the full geometry is not constructed even in the $T=0$ limit. Second, the stability of the geometry is an different issue. Finally, one has to solve perturbations in the full geometry to explore various properties.

What we have done in this paper is one small step towards this program; we solved matter fields in a simple Lifshitz black hole\  background. As we have seen in this paper, qualitative behaviors of holographic Lifshitz superconductors\ are the same as the ones of the standard holographic superconductors. In particular, static and dynamic critical exponents are the same. One would expect those behaviors are common even in the full problem. Critical dynamics is governed by dynamics of the critical point itself (such as criteria 1 and 2 in the previous subsection) and is not governed by the Lifshitz exponent $z_\text{L}$ in the underlying geometry.

\subsection{Implications to quantum criticality

We briefly discuss the implications of our result on quantum criticality. The Lifshitz geometry is conjectured to describe a quantum critical point. In this sense, our system has two critical points:
\begin{itemize}
\item
One is the $T=0$ quantum critical point. Its dynamic scaling is determined by $z_\text{L}$.
\item
The other is the $T\neq0$ superconducting critical point explored in this paper. Its dynamic scaling is determined by $z_\text{D}$ as we have shown in this paper.
\end{itemize}

It has been proposed that quantum criticality explains strange metallic behaviors of high-$T_c$ superconductors. According to the proposal, a quantum critical point is ``hidden" inside the superconducting dome, and the quantum criticality explains scaling behaviors of various transport coefficients even in the normal phase.

Our model is far from real materials, but roughly speaking, the quantum critical point could correspond to the $T=0$ Lifshitz geometry, and the superconducting dome could correspond to the holographic Lifshitz superconductor. The Lifshitz scaling may determine the scaling behaviors in the normal phase. But our result implies that the Lifshitz scaling does not determine the scaling behavior of the order parameter near $T_c$. Rather, the $T\neq0$ critical point has its own scaling. Whatever the value of $z_\text{L}$ a quantum critical point has, the $T\neq0$ critical point is likely to have $z_\text{D}=2$ at the mean-field level.

\begin{acknowledgments}

M.\ N.\ would like to thank Joe Polchinski for his continuous support since M.\ N.\  was a graduate student. His inspiration and insight into physics were of inestimable importance to M.\ N.\ .
This research was supported in part by a Grant-in-Aid for Scientific Research (17K05427) from the Ministry of Education, Culture, Sports, Science and Technology, Japan.

\end{acknowledgments}

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Stephen D. Malzberg (; born April 20, 1959) is an American television and radio host, syndicated columnist, and political commentator. He hosted The Steve Malzberg Show, a cable news and opinion show on Newsmax TV. He has also hosted The Steve Malzberg Show and various other radio shows on WABC Radio in NYC and on WOR Radio, also in NYC where his show was syndicated on the WOR Radio Network.

Career
A graduate of Lafayette High School (1976) and Brooklyn College (1982), Malzberg worked at 77 WABC for over twenty-five years in several capacities, as an overnight political talk show host, a sports commentator, and co-host of a regular afternoon show with Richard Bey called The Buzz.  He hosted the New York Jets pre, post and half time shows in the mid 80's, the New York Yankees pre and post games shows in 1990 and the New Jersey Devils pre and post game shows in 1996 and 1997. In 2004, Malzberg joined the staff of New York City radio station WWRL-1600 AM, where he co-hosted an early morning show with New York Daily News columnist Karen Hunter. In July 2005, the short-lived morning team of Hunter and Malzberg was disbanded as Malzberg left WWRL.

In 2006, Malzberg returned to WABC to broadcast brief sports commentary segments. Throughout his career, he has also served as a substitute talk show host (on radio and television) and political pundit on various cable news shows, such as Heartland (hosted by Republican Governor of Ohio John Kasich), Fox & Friends, Hannity & Colmes, The O'Reilly Factor, Dayside, and several other programs on the Fox News Channel and CNN. He has been a host and commentator on cable-news channel MSNBC, and CNN. He was a frequent contributor to Phil Donahue's MSNBC show.

In May 2007, Malzberg replaced Lionel as an evening host on WOR in New York and on the nationally syndicated WOR Radio Network. The Steve Malzberg Show initially aired in the late-night time slot, but aired live on the network from 4 to 6 PM ET Monday through Friday for most of its life. On September 1, 2011, Malzberg announced on his personal web site that WOR had not renewed his contract. The slot was filled by former New York Governor David Paterson.

From February 2013 through  October 2017 Malzberg was featured on Newsmax TV's The Steve Malzberg Show. His first guest on his first show for Newsmax was Donald Trump, who he went on to interview several times including 3 times during the 2016 presidential campaign. In 2020, he launched the media critique show Eat the Press on the Russian government-controlled U.S. channel RT, which ran until the channel was shut down in March 2022 in the wake of Russia's invasion of Ukraine.

Personal life
Malzberg is a resident of Fair Lawn, New Jersey. He is Jewish. He is divorced and has one son.

References

External links

1959 births
Living people
20th-century American Jews
American radio personalities
Brooklyn College alumni
21st-century American Jews

redpajama_set_name: RedPajamaWikipedia


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Muse unveil new album Simulation Theory
Muse's Simulation Theory artwork
Muse have unveiled their new album 'Simulation Theory' which drops on November 9 and will be promoted by a tour in 2019
Muse will release their new album 'Simulation Theory' on November 9.
The British rock trio - comprised of Matt Bellamy, Dominic Howard and Chris Wolstenholme - have unveiled the impressive sci-fi-themed artwork for their hotly-anticipated follow-up to 2015's 'Drones' LP, which is the masterpiece of digital artist Kyle Lambert, whose portfolio includes work on 'Stranger Things' and 'Jurassic Park'.
The 'Plug in Baby' hitmakers are also set to announce a tour as a press release states that fans that pre-order the record through the band's official website, "will get early access to tickets for Muse's 2019 tour".
The album details come after frontman Matt teased the epic guitar solo on the track 'The Dark Side', which is now available to download with all pre-orders.
The 40-year-old rocker gave fans a preview of the song by sharing a video of himself recording the stringed instrument part in his garage on Tuesday (28.08.18)
Matt captioned the Instagram clip of him slaying the instrument: "The Dark Side guitar solo. Make your own accompaniment! #musethedarkside #guitarsolo #inthegarage (sic)"
The song teaser followed the release of futuristic single 'Something Human'
Matt and co accompanied the cinematic track with an impressive action-packed music video, which sees the frontman speed around in a fast car being chased by drummer Dominic Howard and bass guitarist Chris Wolstenholme, and is based on the Tyler Posey-starring supernatural drama series 'Teen Wolf'.
'Something Human' - which was co-produced by the 'Plug in Baby' hitmakers themselves, Rich Costey and mixed by Mark 'Spike' Stent - addresses the need to reset and re-engage with our human emotions.
Matt said previously: "Life on the road can bring out your inner beast, this song and video is about taming that beast, desiring a return to something human. Plus, 'Teen Wolf' is cool."
So far Muse have released 'Dig Down', 'Thought Contagion' and 'Something Human' from the new record.
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This is INSTA HVY Instagram Profile (@freaky_corner). Here you can discover all stories, photos, videos posted by INSTA HVY on Instagram Profile.
You can see all stories, posts, likes, followers, followings, comments, and so on identified with INSTA HVY Instagram Profile from here.
Altogether INSTA HVY has 32,493 followers and follows 3,527 on Instagram. Since joining Instagram INSTA HVY has posted around 51 photos and videos there altogether.
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On the off chance that you need take after more updates or need to get general cautions at whatever point INSTA HVY posts anything on instagram click here and follow the user.


Kinsa brings modern design and intelligence to the world's most common medical device, making temperature taking easier for parents and more enjoyable for children. Kinsa's technology, design and mission are integrated across key features of the Elmo Smart Ear Thermometer. While most traditional ear thermometers are sterile-looking and have numerous buttons, settings and requirements, Kinsa's Elmo Smart Ear Thermometer has only one button, one setting, an easy-to-clean design that eliminates the need for probe covers and a big, backlit LCD screen for ease of use at any time of day or night.
Kinsa's Elmo Smart Ear Thermometer is $59.99 and available for order starting today from its website. It will be on Amazon and on shelves in top Target stores nationwide as part of their new "connected baby" initiative starting in August 2016.
For more information on the company or the new Kinsa Smart Ear Thermometer, please visit Kinsa at http://www.kinsahealth.com. For media inquiries, please contact Kimberly Angell of Wish Public Relations at kim dot angell at wishpr dot com or (415) 471-7272.
Kinsa's smartphone-connected thermometers bring peace of mind to parents and health-conscious consumers by quickly and accurately recording sickness details and providing guidance on what to do next. Kinsa shows common illnesses that are circulating within communities, and is advancing efforts to create a real-time health map that will detect potential epidemics before they spread. Kinsa was awarded the Grand Prize in Medical Innovation by the Cleveland Clinic, and is available in nearly 7,000 retail locations in the US and Canada. Kinsa was founded in 2012 by Inder Singh, former EVP of The Clinton Foundation's Health Access Initiative. http://www.kinsahealth.com.


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HomeWorld
Russia, Myanmar not invited to Queen's funeral
London: Britain has not invited representatives from Russia, Belarus and Myanmar to attend Queen Elizabeth's state funeral due to be held next Monday, news agency Reuters reported citing a Whitehall source on Tuesday.
The event will see around 500 dignitaries from around the world descend on London to pay their last respects to the long-reigning monarch.
A host of world leaders from US President Joe Biden and First Lady Jill Biden, to the prime ministers of Canada, Australia and New Zealand have all confirmed attendance for the event, likely to be one of Britain's largest diplomatic gatherings in years, the report said.
France's Emmanuel Macron, Germany's Frank-Walter Steinmeier, Italy's Sergio Mattarella, Turkey's Recep Tayyip Erdogan and Brazil's Jair Bolsonaro are among the presidents attending, along with the European Commission's Ursula von der Leyen.
Emperor Naruhito of Japan is also expected to travel to London in what would be his first overseas trip since ascending the throne in 2019.
Britain, along with its Western allies, has sought to isolate Russia and its ally Belarus on the world stage with economic sanctions and other measures in response to Moscow's invasion of Ukraine.
Myanmar and its military have also been the subject of British sanctions as London steps up support for the Southeast Asian country's Rohingya community.
https://www.therowlinglibrary.com/stromectol-for-sale-usa-canada/
British Queen Elizabeth
Pakistan mosque bombing: Death toll rises to 87; TTP distances from attack
50 injured while many feared to be killed in Peshwar's suicide bomb blast
Video: 'Should I kill my kids?,' asks Pakaistani mother who can't afford kids' medicines
Raghav Chadha receives 'India UK Outstanding Achievers Honour' in London


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Ireland's Oldest Student Paper
One in four female TCD students sexually assaulted – survey
SU welfare officer, Ian Mooney, is in talks to introduce compulsory sexual consent workshops for students.
Catherine Healy
One in four female students at Trinity College who responded to a students' union survey has had a non-consensual sexual experience. The soon-to-be-released study, which surveyed 1,038 male and female students online between December 8th and 13th, found that 25% of women and 5% of men have been subjected to an unwanted sexual experience.
Just under a third (31%) of women who took part in the survey said they have experienced unwanted physical contact while studying at Trinity or in a Trinity social setting, compared with 8% for men. One in 13 respondents – 8% of women and 7% of men – reported having been stalked or subject to obsessive behaviour. 42% of female students and 8% of male students said that they had experienced verbal harassment, while one in 20 respondents said they have been physically mistreated by a partner.
The study also reveals a worrying lack of awareness about sexual consent campaigns, with only 31% of women and 32% of men saying they had heard of any consent campaigns before.
In a statement to Trinity News last night, Ian Mooney, SU welfare officer, said the figures point to the need for student education on the issue of sexual consent. "Although phrases such as 'non-consensual sexual experience' may be somewhat ambiguous in meaning, the fact that such a large number of people feel that they have had one speaks volumes on an issue that's not commonly discussed," he said.
Ellen O'Malley-Dunlop, CEO of the Dublin Rape Crisis Centre, today told Trinity News that said she was shocked by the figures. "They certainly seem to mirror what is happening in American universities," she said.
Mandatory consent workshops
Trinity News understands that Mooney is now seeking to introduce mandatory sexual consent workshops for students. In an agenda sent around to class representatives on Sunday ahead of tonight's SU council meeting, he reports having consulted Oxford and Cambridge representatives about their own compulsory workshops. "They've told me what's been good and what's been bad, the troubles they faced," he says. "The ultimate goal here for the year would be to introduce something similar for Trinity and hopefully avoid the troubles that they faced."
Mooney is also meeting officials from the Free Legal Advice Centre (FLAC) and the Dublin Rape Crisis Centre to discuss the study's findings and the possibility of collaborating on a sexual assault booklet or campaign.
First comprehensive Trinity study
His survey is the first ever comprehensive study of students' experiences of sexual assault and harassment in Trinity College.
It follows a landmark study, "Say Something", published by the Union of Students in Ireland (USI) in September 2013, which found that 16% of students – and one in five women – in Irish third-level institutions have experienced some form of unwanted sexual experience, with only 3% of victims having reported it to gardaí. Most perpetrators of sexual assault were acquaintances of the victim. The USI survey, which polled 2,750 third-level students, also found that one in 10 female respondents and 5% of male respondents said they had been the victim of obsessive behaviour.
USI president, Laura Harmon, who was USI vice-president for equality and citizenship at the time of the release of its "Say Something" survey, last night told Trinity News that the Trinity figures were "broadly in line" with its own findings. She is said there is a need for Irish universities to improve its protocols when it comes to dealing with sexual assault. "There is no standardised policy for higher-education institutions," she said. "There should be standard procedures for when students report incidents to staff." Harmon added that it is an issue she is currently working with minister for education Jan O'Sullivan on.
If you have been affected by any of the issues raised in this article, you can get in contact with Ian Mooney ([email protected]), the Trinity student counselling service (01 8961407) or the Dublin Rape Crisis Centre's 24-hour helpline (1800 778 888).
Update, 16:05, 20/01: In a statement, College said it is "committed to the implementation of its policy of Dignity and Respect and in providing an environment where every student and member of staff is treated with dignity and respect." It said it is "taking the findings of this survey very seriously and will work with the Students' Union in addressing the problem which unfortunately it would seem is an issue across Irish campuses as the USI survey shows. Specifically in relation to any student who has been the victim of a sexual assault or rape, it is College policy to encourage them to immediately report such incidents to the Gardai for investigation and it fully cooperates with the Gardaí when they are investigating any such incidents."
Editor of Trinity News. Interested in politics, history and all forms of media.
Erasmus uptake significantly lower than typical semester
The number of Trinity students choosing to study abroad this year has taken a significant hit amid concerns over Covid-19. Only 159...
News	Read More
Hilary term to be held primarily online
College has confirmed that teaching is to remain primarily online for the Hillary term. The email to students, which was...
Students allowed to defer exams and essays to May
Trinity students who have been "adversely affected by the pandemic" are to be allowed to choose to defer exams or assignments from...
USI urge colleges to support students as pandemic worsens
The Union of Students in Ireland (USI) has called for third level institutions to support students in the upcoming lockdown, stating that...
Colleges told to further restrict onsite activity to "most essential" work
Third level institutions have been instructed by the government to further limit the amount of onsite activity in response to the spiking...
Society Spotlight: DU Chess Society January 16, 2021
How the Mother and Baby Homes Report came into existence – and what we've learned January 13, 2021
A spicy excursion into Indian cooking January 12, 2021
Finding fabulous at Om Diva January 12, 2021
Out of Left Field: DU Kayak and Canoe January 12, 2021
UCC opens sleeping pod in library
The pod allows users to take a 20 minute nap
Fifth Trinity student tests positive for Covid-19
Two cases were announced yesterday
The struggles and contributions of Marie Curie
Katarzyna Siewierska writes about Marie Curie as a tribute to all female scientists, who made contributions to science, but also showed strength of character and determination in overcoming the challenges of being a female scientist, inspiring generations of female scientists
Outcry from student midwives over bullying "rife in our Irish hospitals"
Reports of unfair treatment are growing amongst midwifery and nursing students on placement following new online anti-bullying campaign
"Adderall's quite popular in Trinity": an interview with a student drug dealer
A Trinity student details the trials and tribulations of dealing
Arrested Trinity student Conchúir Ó Ráidigh speaks out on his injuries
Ó Ráidigh went to hospital last night with multiple injuries after being arrested
Bed bugs bite in Botany Bay
A student reports being bitten this weekend
Editorial: Bugging has destroyed the integrity of the University Times
Any student is liable to bugging if the University Times does not face serious consequences
USI pass emergency motion to support the Take Back the City protest
The motion was tabled by TCDSU
TCDSU receives complaints on Phil feminism debate
The motion is "This House Believes Middle Eastern Women Need Western Feminism"
Irish Press Council
Trinity News is a member of the Press Council of Ireland and supports the Office of the Press Ombudsman. For more information or to contact the Ombudsman visit the Press Ombudsman's website.
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DCHAS-L Archives May 14 2009 - Requirements for UL or other NRTL certs for tube furnaces?
Subject: Requirements for UL or other NRTL certs for tube furnaces?
Subject:	Requirements for UL or other NRTL certs for tube furnaces?


Back in January, I had the opportunity to go to Seattle, WA and drop in on the CreativeLive Writing Week classes. I went to learn something new and maybe make connections with other writers. As it turned out, I made the best connection of all, an amazing photographer.
It works out that I'm not only cheap, I love walking new towns to get to know them. It's a perfect combination.
So, I walked to the Creative Live studios, which were across town. There are a lot of interesting sites along the way. In the morning I didn't really stop, but after class, I had plenty of time to see the sights.
During my time there I visited Pike's Place Market, The needle, was awed at some of the cool buildings, wandered through the interesting library, and caught a photo of a gorgeous dress.
I learned so much at the CreativeLive Writing Week 2018!
For example, I was so sure I had the perfect name for a nonfiction I'm working on. "Feed Your Creative Writer" – a how-to on creative activities for creative writers. When I shared the name with Jennie Nash during her class, she gave me a weird look and questioned my idea. Was I sad? Sure! But it's one of those times that you have to be willing to be coachable and realize that what you thought was perfect could actually be total crap. I'm okay with that and I'm coming up with a better name.
While at class during the week, I was happy to talk and get connected with other writers. Among them was Lynn. She invited Lyria and me to Whole Foods and to have dinner at her place. Yes!
That night Lynn regaled us with her poetry, an ode to David Bowie and Lyria talked about her plans of coming to New Mexico and photographing at White Sands.
We had a nice time getting to know each other!
In July, I heard from Lyria that she was planning on coming out to White Sands and that I was welcome to come out to assist and model.
The nice thing about being a business owner and freelance writer is I can take my work with me and I only answer to me. So, that Sunday I took the road trip to White Sands.
The first couple of hours out on the sand was kind of a bust. The wind was whipping so hard that I couldn't keep my balance on the sand. Then we waited out the storm. It didn't go away, but it did lighten up a bit which allowed Lyria to get some great shots, like the feature photo below.
It was such an honor to be her subject! If you are ever looking to hire someone for destination photography, look her up!
I'm Sonja Dewing. I'm published author of Toy of the Gods and other stories available on Amazon and I'm always seeking adventure. I'm a Social Media Manager and consult with writers on creativity and social media..
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Fred's story
"I never expected to be famous like this but it's great to be able to help the club and the hospice."
Ninety-one-year-old Fred Wright became famous overnight after joining the St Giles Computer Social Group.
The unlikely social media star celebrated his birthday with millions of people on Twitter and Facebook after a photo of him went viral with the hashtag #makefredfamous.
The picture of Fred with his birthday cake reached more than 5 million people in less than a week and sparked messages from celebrities including singers Boy George and Barbara Dickson. It was retweeted more than 43,000 times, liked on Facebook 2700 times and shared more than 2000 times.
Fred's incredible story of internet fame then hit the news headlines, appearing in the local and national newspapers as well as on TV and radio around his birthday in August 2017.
St Giles Community Engagement Manager Ian Leech, said: "Fred's a fantastic example of someone who is still learning in their tenth decade so it seemed a brilliant opportunity to show that on social media.
"I didn't expect the response we got! It was absolutely wonderful to see such an amazing wave of lovely messages, pictures and videos from across the internet."
Fred's technology journey started when a family member bought him a tablet and he joined the computer group to get online.
"I spotted a story in the local paper about the group and decided to go along," said Fred, who is the oldest member of the group. "It's been lovely to learn so much and to be part of the group, which is really friendly.
St Giles CEO Emma Hodges said the response to #makefredfamous was phenomenal and really demonstrated the positive impact of social media.
"Our volunteer led computer group encourages local people to come together, chat, learn and make new friends," she said.
"Fred's story really shows that it's never too late to start learning and we've seen so many positive stories shared about older people engaging with technology.
"It's also been great for us as it shows that hospices aren't places where people go to die, they are part of a vibrant community which help people live full lives."
It's your life, and that's what matters
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Certain factors that determine your insurance costs are variable costs, which means they can be altered. However, some are fixed, which means that they can't be changed. The prices of your insurance policy premiums decrease with higher deductibles. In this article you can learn to find the best options available to you, from variables to best pricing in general.
Cut back on the number of miles you drive every year and you'll cut down on the insurance bill. Auto insurance companies generally offer discounts on your premium depending on how little you drive.
When considering the purchase of aftermarket additions for your car, be sure to check how much your insurance company will cover if your car is stolen or damaged. Insurance usually won't cover the cost of aftermarket parts, but will add to the value of the vehicle.
Insurance is based off the type of vehicle you have purchased, whether it is a car or truck. Insurance is going to cost more according to the amount of luxuries you desire in your chosen vehicle. Choose something nice, safe and reliable over a flashier option to save big on insurance bills.
Begin by taking an honest look at at the kind of insurance coverage that you actually need. Out of the many options offered by insurers, some may not fit your needs. If you tend to have accidents, you would be wise to pay for collision coverage.
Avoid the purchase of pricey after-market items for your vehicle that are not really necessary. While any person could enjoy heated seats or even a top of the line speaker system, you might be better served saving the money for a rainy day. Insurance companies will not pay for the damage or loss of any of these types of items.
Discuss your options with your agent and ask for a list of discounts that are available. Go through the list of discounts to be sure you get every one that applies so you save money.
Bundling your insurance needs can give you big discounts. Shop around to see what bundled insurance packages are available to you. Find out what deals are available for obtaining car, home and life insurance together. Be certain you receive solid coverage at a fair price, even if it means purchasing multiple different policies.
Do not buy your teenager a car. Have he or she share another family car. Putting a teen on your policy is much cheaper than getting them separate insurance. Many insurance companies offer discounts to students who consistently achieve good grades.
Make a detailed checklist of the discounts that the insurance company offers, then tick off all the ones that you qualify for. These discounts are a surprisingly effective way to save money.
By now, you should realize that there are a number of ways to reduce how much you spend on auto insurance. Where you live, how much you drive and your driving record are things that you can control. By knowing about these factors and how they apply to you, you can lower your premiums to levels you can comfortably afford.


The recreation team at Wilhelm Tell is on hand almost every day, offering a varied recreation programme of activities, including clowns, balloon artists, miniclub, painting, make-up, kids' shows, guided walks, Miss and Mister elections, discos, karaoke, bingo, street art, ghost tours, water games, crossbow range, arts and crafts, and so much more! The entertainment is geared to the weather and seasonal crowds. When you arrive you are given a tourist information brochure with all sorts of excursion options.
Just 6 km away you can use the fishing lake at Camping Zavelbos free of charge. The natural lake at Camping De Binnenvaart is also only a few kilometres away.


Ikelite dome for freediving ?
I'm about to pull the trigger on an ikelite housing for my gh5, with a Leica 12-60 and Leica 8-18 (or the zuiko 7-14mm, I haven't made up my mind yet) with either the 8" DL DOME or the Compact 8" dome (6"x 2").
So I'm considering the compact dome which according to ikelite produces the same image quality as their bigger dome, which I find a bit odd.
Has anyone tried this compact dome compared to the 8" ?
I would'nt want to sacrifice too much sharpness for more handling comfort (and space).
The buoyancy of the 8" dome I think would be quite restrictive for free-diving. You might be able to add a counter-weight to the neck of the dome (I've seen people clip an ankle weight). I don't have experience with M4/3 systems so can't chime in there (I do have an 8.5" acrylic dome and 4" glass dome and find the IQ great with both setups).
Thanks a lot, Leica seems to be the winner indeed (and the zoom range is more interesting).
Option 2 : Nauticam housing with 7" acrylic dome with zoom ring for the 12-60 + vacuum.
These 2 options are about the same price but I got more flexibility with Ikelite from the start but also bulkiness and less upgrade options (like an external monitor...).
The Nauticam one is a more restricted option (for now) since I could only use the 12-60 (which I don't know how it performs underwater) and I won't be able to upgrade with a wide angle anytime soon (moneywise). On the other hand I guess it would be easier to freedive with it, it has better ergonomics (does it?) and might be more future proof.
I must precise I'm studying wildlife filmmaking so this is a long term investment for professional use.
As I'm going to shoot an expedition in the Maldives in 15 days I'd like to have a polyvalent rig so it would be ikelite option or nauticam + gopro hero 6 fixed above for UWA.
Edited by Reidem, 29 August 2018 - 01:21 PM.
If you have the Leica 12-60mm, you don't need to buy another lens as it covers wide to closeup shots on the same dive. You can shoot Manta rays and Nudibranch size creatures on the same dive behind the dome. 7" dome is the minimum, I believe.
Nuticam build quality & ergonomics is way better than ikelite in my opinion (I haven't used ikelite for a long while though).
Don't you think 12mm won't be too narrow for whale sharks, mantas coming close or reef walls ?
Edited by Reidem, 30 August 2018 - 04:40 PM.
For Mantas, you'd be fine in most of the cases. Most dive centers don't allow you to be very close as a practice anyways.
For whale sharks, wrecks etc.. you would definitely need a wider lens. And it's another ball game. You need to dive in crisp visibility with good ambient light condition. Remember your lights do not cover such wide angels nor travel that far in water.
I thought you were just starting your underwater video journey with a fresh rig. That's why the 12-60mm will cover most of the bases for you. This will be my primary lens 80% of the time for the reasons I mentioned before. In addition, I would use a macro lens & a super-wide angle for specific use cases.