text:
\section{Introduction}

Many observables in finite-entropy quantum systems exhibit a type of pseudorandom ``noise'' that reflects the underlying discreteness of the Hilbert space. An important problem is to identify where this noise comes from in the gravity dual \cite{Maldacena:2001kr,Dyson:2002nt,Barbon:2004ce,Barbon:2014rma,Cotler:2016fpe}. This is challenging, partly because the specific pseudorandom noise for a given system will be quite complicated. However, one can imagine a simple effective description where the pseudorandom noise is approximated by true randomness with the right statistical properties. And at least in some cases \cite{Saad:2018bqo,Saad:2019pqd,Cotler:2020ugk,Belin:2020hea,Blommaert:2020seb}, this ``true random'' theory is actually reproduced by simple gravity computations.

Wormholes are the key ingredient in the bulk description of this averaged noise. Somewhat similar wormholes have also been shown to underly the ``Page curve'' computations \cite{Penington:2019npb,Almheiri:2019psf} of the entropy of radiation emitted by an evaporating black hole, see \cite{Almheiri:2019qdq,Penington:2019kki}. To understand the relationship, we will study the ``noise'' that underlies the Page curve -- the erratic off-diagonal elements that must correct Hawking's thermal answer for the density matrix of the radiation.

Consider the state of a partially evaporated black hole $\mathsf{B}$ and its radiation $\mathsf{R}$
\begin{align}\label{entstate}
|\Psi\rangle &\propto \sum_{i}|\psi_i\rangle_\mathsf{B}|i\rangle_\mathsf{R}.
\end{align}
The $|i\rangle$ form an orthonormal basis of the radiation corresponding to definite sequences of Hawking emissions, and $|\psi_i\rangle$ is the state of the black hole that is left behind after emitting sequence $i$:
\begin{align}
|i\rangle &= \text{sequence of Hawking radiation emissions}\\
|\psi_i\rangle &= \text{leftover state of BH}.
\end{align}
The states $|\psi_i\rangle$ are not necessarily orthogonal: they are just whatever you get after emitting sequence $i$. Concretely, they can be prepared by acting with annihilation operators on the initial state of the black hole, removing one Hawking particle at a time:
\begin{equation}\label{states}
|\psi_{i_W}\rangle = W_{n}(t_{n})\dots W_2(t_2) W_1(t_1)|\psi\rangle.
\end{equation}
Of course, we can also let the $W$ operators be more general than annihilation operators. Then (\ref{entstate}) would represent the state of a black hole interacting with a bath in some other way.

Now, consider the matrix of inner products $\langle \psi_i|\psi_j\rangle$ for states constructed this way, but with different sequences of operators. This directly determines the density matrix of the radiation, as one can see by tracing over the black hole system $\mathsf{B}$,
\begin{equation}
\rho_\mathsf{R}  = \text{Tr}_{\sf{B}}|\Psi\rangle\langle\Psi|\propto \sum_{ij}\langle \psi_j|\psi_i\rangle_\mathsf{B} \,\, |i\rangle\langle j|_\mathsf{R}.
\end{equation}
What should we expect for the matrix of inner products? If the operator sequences are quite different from each other, it might be reasonable to model the $|\psi_i\rangle$ as random states in an $e^{S}$-dimensional Hilbert space. Then (if the states are normalized) one would expect to find something of the form
\begin{equation}\label{expectedR}
\langle \psi_i|\psi_j\rangle = \delta_{ij} + e^{-S/2}R_{ij}.
\end{equation}
This has a big diagonal term and a small off-diagonal term proportional to a pseudorandom matrix $R_{ij}$ of complex numbers with magnitudes of order one. This ``noise'' is small but important: without it, the density matrix of the radiation would be diagonal, with an entropy that grows forever. But with the noise, unitarity (and the Page curve) can be restored \cite{Papadodimas:2013kwa}.

As an illustration of $R_{ij}$, one can consider states of the form (\ref{states}) in the SYK model \cite{Sachdev:1992fk,KitaevTalks,Kitaev:2017awl}, where the sequence of operators is a random string of fermions, with real time evolution in between. The matrix of inner products can be plotted as a heatmap:
\begin{equation}\label{heatmaps}
\text{Re} \ \langle \psi_i|\psi_j\rangle = \includegraphics[valign = c, width = .27\textwidth]{fixed/notSubtracted.pdf},
\hspace{50pt} \text{Re} \ R_{ij} =
\includegraphics[valign = c, width = .27\textwidth]{fixed/subtracted.pdf}.
\end{equation}
Here red means positive, blue mean negative, and white means zero. At right, we removed the diagonal and rescaled the remainder, to make the $e^{-S/2}R_{ij}$ term visible.

To what extent can gravity describe $R_{ij}$? More precisely, to what extent can a low-energy semiclassical description of gravity (without branes, strings, and whatever degrees of freedom account for the microstates of the black hole) describe it? On the one hand, $R_{ij}$ is a complicated quantity, and it doesn't seem likely that a simple gravity theory will be able to produce it. But on the other hand, recent successful computations of the Page curve \cite{Penington:2019npb,Almheiri:2019psf} show that simple gravity theories must know {\it something} about this noise.

To probe this, one can compute the matrix of inner products directly, using semiclassical gravity. In a sense, this is asking an unreasonable question -- something that a simple gravity theory should not be able to answer -- and one might expect to find some breakdown in the calculation. Instead, gravity finds a creative way to sidestep the complexity of the $R_{ij}$ matrix and to give a simple answer. Schematically, it predicts
\begin{align}
\langle \psi_i|\psi_j\rangle &= \delta_{ij} \label{innerprod1}\\
|\langle \psi_i|\psi_j\rangle|^2 &= \delta_{ij} + e^{-S}.\label{innerprod2}
\end{align}
In the first line, we find the naive answer, with the noise term $R_{ij}$ set to zero. But when we compute the square of the inner product, there is an off-diagonal $e^{-S}$ term coming from a wormhole contribution. This piece is not the actual erratic term that one would have expected from the quantum noise, but it has the right average value. Following \cite{Coleman:1988cy,Giddings:1988cx,Maldacena:2004rf,ArkaniHamed:2007js} and recent work \cite{Saad:2018bqo,Saad:2019lba,Saad:2019pqd}, this was interpreted in \cite{Penington:2019kki} as indicating that gravity is secretly computing the average over some kind of ensemble of quantum theories. In this ensemble, $R_{ij}$ has mean zero and a variance of order one, so the averages are
\begin{align}
\mathbb{E}\Big\{\langle \psi_i|\psi_j\rangle\Big\} &= \delta_{ij}\\
\mathbb{E}\Big\{|\langle \psi_i|\psi_j\rangle|^2\Big\}&= \delta_{ij} + e^{-S}.
\end{align}
So the gravity answers make sense if we interpret them as implicitly computing $\mathbb{E}$ of whatever we were actually trying to compute. This wouldn't make much difference for self-averaging quantities (like the entropy), but it matters for $\langle \psi_i|\psi_j\rangle$.

We will explore the wormhole that gives the $e^{-S}$ term in (\ref{innerprod2}), in a special case where the initial state in $|\psi_{i_W}\rangle = W_{n}\dots W_1|\psi\rangle$, is the thermofield double state, and where the operators act on only one side. Then $\langle \psi_{i_V}|\psi_{i_W}\rangle$ is just a special case of a thermal correlation function:
\begin{equation}\label{pictorial}
\langle \psi_{i_V}|\psi_{i_W}\rangle \hspace{5pt} = \hspace{10pt} \includegraphics[width = .5\textwidth,valign = c]{fixed/innerProds.pdf}.
\end{equation}
Here the circular parts of the figures represent Euclidean evolution, the straight parts represent real time evolution, and the $W$s and $V$s are operator insertions. The RHS is a thermal trace of real-time evolved operators. It will be convenient to also allow a bit of Euclidean evolution between the operators, so the general thing we will study is
\begin{equation}
\text{Tr}\Big[\prod_{k} e^{-\tau_k H} \mathcal{O}_k\Big], \hspace{20pt} \tau_k \in \mathbb{C}, \hspace{20pt} \text{Re}(\tau_k) \ge 0,\label{genCorr}
\end{equation}
where the $\mathcal{O}_k$ operators run over both the $W$ and $V$ operators. The quantity $\langle \psi_i|\psi_j\rangle$ is a special case of this general correlator, with the somewhat unusual feature that (for $i\neq j$) the naive gravitational answer is zero, or at least very small, and the expected quantum answer is dominated by the pseudorandom noise.

In a perturbative bulk computation, the $\mathcal{O}_j$ are sources for propagators of corresponding bulk fields, and these propagators will link up in some way. Suppose first that we are computing the diagonal term $\langle \psi_i|\psi_i\rangle$. Then the operators come in pairs (one from the ket and one from the bra), and the propagator sourced by each ket operator can end at the corresponding bra operator. This gives a large answer for the correlation function.

Next suppose that we are computing $\langle \psi_i|\psi_j\rangle$ with $i\neq j$. Since the strings of operators in the bra and ket are now different, the propagators will not find natural pairs of operators to connect. Depending on the details, the gravitational answer can either be zero or just small compared to the diagonal inner product (exponentially smaller in the number of insertions).

Finally, consider $|\langle \psi_i|\psi_j\rangle|^2$. In the bulk theory, this means a computation with two asymptotic boundaries: one with operators inserted to compute $\langle \psi_i|\psi_j\rangle$, and one with operators inserted to compute $\overline{\langle \psi_i|\psi_j\rangle}$. In this situation, there is again a natural pairing of the operator insertions, where each insertion on boundary 1 is paired with a corresponding insertion on boundary 2. If a wormhole connects the two boundaries together, then the propagators sourced at one boundary can end at the other one. This again gives a large answer for the correlation function.

However, the gravitational action suppresses this wormhole geometry, leading to the $e^{-S}$. To some extent the suppression depends on details of the strings of operators, and on how much they change the energy of the state they act on. For this reason, the $e^{-S}$ in (\ref{innerprod2}) is a bit schematic. But it is easy to see that it is correct at the level of the topological term in the gravitational action, which weights each topology by $e^{S_0\chi}$, where $\chi$ is the Euler characteristic. For the $i = j$ term we have a large contribution from a topology of two separate disks, while for the $i \neq j$ term, we need a cylinder, connecting the two boundaries together. So
\begin{equation}\label{suppression}
\frac{|\langle \psi_i|\psi_j\rangle|^2}{\langle\psi_i|\psi_i\rangle\langle \psi_j|\psi_j\rangle} \propto \frac{e^{S_0\chi_{\text{cyl}}}}{e^{2S_0\chi_{\text{disk}}}}= \frac{e^{0S_0}}{e^{2S_0}} = e^{-2S_0}.
\end{equation}
Since we are discussing a system where the initial state is a thermofield double of two copies of the black hole, the topological contribution to the entropy is $2S_0$, so (\ref{suppression}) agrees with (\ref{innerprod2}).

In the rest of the paper, we study these wormholes in more detail. There are two main points:
\begin{enumerate}
\item The wormholes that compute $|\langle \psi_i|\psi_j\rangle|^2$ are classical solutions that depend on a small amount of backreaction from the matter fields. They resemble the wormholes that compute the Renyi entropy, but they require an ensemble interpretation in a more direct way.
\item In dilaton gravity coupled to matter fields, the contribution of the wormholes agrees quite precisely with what one would expect for an ensemble average $\mathbb{E}\{|\langle \psi_i|\psi_j\rangle|^2\}$.
\end{enumerate}

For additional recent work with a range of viewpoints on wormholes, factorization, and averaging in gravity, see \cite{Bousso:2019ykv,Pollack:2020gfa,Marolf:2020xie,McNamara:2020uza,Bousso:2020kmy,Afkhami-Jeddi:2020ezh,Maloney:2020nni,Chen:2020tes,Liu:2020jsv,MarolfSantos}.

\section{Classical calculations}
With this motivation, we will consider the wormhole that contributes to the square of  (\ref{genCorr}):
\begin{equation}
\bigg|\text{Tr}\Big[\prod_{j = 1}^n e^{-\tau_j H} \mathcal{O}_j\Big]\bigg|^2\hspace{5pt} \supset \hspace{5pt}\includegraphics[valign = c, width = .18\textwidth]{fixed/13a.pdf}\label{CORR}
\end{equation}
The sketch at right emphasizes that the operators on the left boundary connect with propagators (dashed lines) to the corresponding operators on the right boundary. Other contractions can be suppressed by taking the operators to be different, or by taking the Lorentzian time separations (imaginary parts of $\tau_j$) large. We have sketched a Euclidean configuration: the Lorentzian evolution will be important, but it is hard to draw.

We will study the wormhole in a two-dimensional dilaton gravity theory, coupled to matter, with the Euclidean action
\begin{equation}\label{Action}
I = -S_0 \chi -\frac{1}{2}\int \mathrm{d}^2x\sqrt{g}\left[\phi R + V(\phi)\right] + I_{\text{matter}} + \text{bdy terms}.
\end{equation}
The special case $V = 2\phi$ is known as JT gravity \cite{Teitelboim:1983ux,Jackiw:1984je}. The dependence on the $S_0$ term was analyzed in (\ref{suppression}) above, and we will drop it below.

Let's first discuss some qualitative aspects. Wormhole solutions sometimes do and sometimes do not exist \cite{Witten:1999xp,Maldacena:2004rf,ArkaniHamed:2007js}. In dilaton gravity, the basic challenge is that the action does not stabilize the length $\ell$ across the wormhole. Indeed, if $\ell$ is fixed by hand, and all other modes are put on shell, then the leftover action for $\ell$ is (see appendix \ref{app:stabilization})
\begin{equation}\label{clac}
I = (\beta_1+\beta_2) E(\ell)
\end{equation}
where $E$ is the ADM energy measured at either end, and $\beta_1,\beta_2$ are the renormalized lengths around the boundary circles. The energy $E$ is a decreasing function of $\ell$ so there will not be a stationary point for $\ell$ (in JT, $E \propto e^{-\ell}$). We can interpret this as saying that in a microcanonical ensemble at fixed energy, the wormhole is classically independent of $E$, so in a canonical ensemble, the Boltzmann factors cause $E$ to run away to small values \cite{Saad:2018bqo}.

Even in the canonical ensemble, the instability can sometimes be removed by the effect of matter fields. In the  Maldacena-Qi wormhole \cite{Maldacena:2018lmt}, this is accomplished by adding an interaction between the two boundaries \cite{Gao:2016bin}. In our case, the operator insertions will have a similar effect, because they weight configurations according to the propagator from one end of the wormhole to the other,
\begin{equation}\label{opin}
\mathcal{O}_j\overline{\mathcal{O}_j} \sim \exp(-m_j \ell).
\end{equation}
Such a factor can stabilize $\ell$, but unless we add a large number of operators, it will be stabilized at a large value, where $E$ is small. This means that with only a few operator insertions, the dominant contribution to (\ref{CORR}) comes from low energies. To focus on the contributions from a particular energy, one can again use a microcanonical ensemble, where $\beta$ is free to vary. For small enough $\beta$, the $(\beta_1+\beta_2) E(\ell)$ term in the action will balance against the $m_j\ell$ term.

However, this seems to create another problem: if $\beta_{1,2}$ are small, it sounds like the wormhole will have a short periodic direction, so the temperature of the bulk fields will be high, and one may worry about quantum effects. This problem will be avoided by including Lorentzian evolution. If the time between operator insertions is long (even if it goes forwards and then backwards), the geodesic distance around the wormhole will be long.

Let's now be more systematic. We will work out the classical dynamics with respect to a time coordinate that goes around the periodic direction of the wormhole. A nice simplification is that it is possible to work entirely within the phase space of the pure gravity theory. The matter fields just provide impulsive forces that modify the dynamics of this theory at locations where operators are inserted.

At a given instant of the periodic time coordinate, we have the phase space of the dilaton gravity theory on an interval stretching from one boundary of the wormhole to the other. This phase space is very simple, and in fact two-dimensional \cite{LouisMartinez:1993eh,Cavaglia:1998xj,Harlow:2018tqv}. One can represent points in phase space by thinking about spatial slices through a two-sided ``thermofield double'' black hole, or Einstein-Rosen bridge. One phase space coordinate can be taken to be the aysmptotic energy measured at either end (it is equal at the two ends), and the other coordinate is the difference in Killing time between the two endpoints of the spatial slice, $\delta t$. Different slices with the same endpoints are gauge-equivalent and correspond to the same point in the physical phase space.

These two coordinates, $\delta t$ and $E$, are conjugate to each other, so that the Poisson bracket is
 $\{\delta t,E\} = 1.$
Instead of $\delta t$, we will find it a bit more convenient to use a coordinate $\theta$ defined as the angle between the two endpoints of the spatial slice in Euclidean signature:
\begin{equation}\label{theta}
\theta \hspace{5pt}= \hspace{5pt} \includegraphics[valign = c,width = .08\textwidth]{fixed/13c.pdf} \hspace{5pt}= \hspace{5pt}\pi - \frac{2\pi \mathrm{i}}{\beta_E}\delta t,
\end{equation}
Here $\beta_E$ is the inverse temperature associated to the asymptotic energy $E$. For a particular gravity theory, it would be a concrete function of energy. For example, in JT gravity, $E = \pi^2/\beta_E^2$. From $\{\delta t,E\} = 1$ and (\ref{theta}), the Poisson bracket between $E$ and $\theta$ is
\begin{equation}\label{PB}
\{E,\theta\} = \frac{2\pi}{\beta_E}\mathrm{i}.
\end{equation}

Our goal is to understand the dynamics in this phase space as we go around the periodic direction of the wormhole, applying (complexified) time evolution and acting with operators. This type of procedure is a bit more familiar in quantum mechanics, so we will keep the following translation in mind. Suppose that a system evolves with quantum Hamiltonian $\widehat{H}$
\begin{equation}
\widehat{U}(t) = \exp(- i \widehat{H}t)
\end{equation}
so that (Heisenberg picture) operators evolve according to
\begin{equation}
\widehat{A}\rightarrow \widehat{A}(t)  = \exp(it [\widehat{H},\cdot])\widehat{A},
\end{equation}
where $[\cdot,\cdot]$ is the commutator. In a classical approximation, the operators are represented by functions on phase space, and the corresponding statement is that if we evolve the system forwards using the classical Hamiltonian $H$, then these functions evolve as
\begin{equation}
A\rightarrow A(t) = \exp(-t\{H,\cdot\})A
\end{equation}
where $\{\cdot,\cdot\}$ is the Poisson bracket.

This general rule can be applied to our case. In between operator insertions, the quantum evolution would be via the operator
\begin{equation}\label{evolution}
\widehat{U} = \exp(-\tau_j \widehat{H}_L - \overline{\tau_j}\widehat{H}_R).
\end{equation}
By complexifying $t$, and applying the above rule, this translates to the classical evolution
\begin{align}
A&\rightarrow \exp\left(-i(\tau_j + \overline{\tau_j}) \{E,\cdot\}\right)A
\end{align}
where we used that $H_L = H_R = E$. Using the Poisson bracket (\ref{PB}), one can see that this is just a translation in $\theta$:
\begin{align}
E\rightarrow E, \hspace{20pt} \theta\rightarrow \theta - \frac{2\pi}{\beta_E}(\tau_j + \overline{\tau_j}).\label{euclideEv}
\end{align}

At locations where the operators $\mathcal{O}_j$ and $\overline{\mathcal{O}_j}$ are inserted, the propagator between them gives an impulsive force that pulls the two boundaries together slightly \cite{Gao:2016bin,Maldacena:2017axo,Maldacena:2018lmt}. We can analyze the effect as follows. In a large-mass approximation for the bulk fields, the $\mathcal{O}_j\overline{\mathcal{O}_j}$ propagator is
\begin{equation}\label{applyOp}
\widehat{G}_{\mathcal{O}_j} = \exp\left(-m_j \widehat{\ell}\right)
\end{equation}
where $m_j \gg 1$ is the mass of the bulk field dual to the operator $\mathcal{O}_j$, and $\ell$ is the regularized geodesic length from one boundary to the other at the locations where the operators act. Classically, this length is some concrete function of $E$ and $\theta$ that is fixed once we decide what gravity theory we are studying. For example, in JT gravity,
\begin{equation}
\ell(E,\theta) = \log\left(\frac{1}{E}\sin^2\Big(\frac{\theta}{2}\Big)\right).
\end{equation}
So, the effect of the operator insertions is to apply the operator (\ref{applyOp}), which acts in the Hilbert space of the pure gravity theory. In a classical approximation, this operator induces the following evolution in phase space
\begin{equation}
A \rightarrow \exp\left(-im_j\{\ell(E,\theta),\cdot\}\right)A.
\end{equation}
If the mass $m_j \ll E$ is small enough then we can linearize the effect. Using the Poisson bracket (\ref{PB}), this implies
\begin{equation}
E \rightarrow E + m_j\frac{2\pi}{\beta_E} \partial_\theta \ell(E,\theta), \hspace{20pt} \theta\rightarrow \theta - m_j \frac{2\pi}{\beta_E}\partial_E\ell(E,\theta).
\end{equation}
Together with (\ref{euclideEv}), this describes the evolution around the periodic direction.

The condition to have a classical solution is that the $E$ and $\theta$ variables should be periodic. This gives two equations, and we can solve these for the two ``starting'' values $\theta_1,E_1$ at some arbitrary point along the evolution. Alternatively, in the microcanonical ensemble, we would instead fix $E_1$ in advance, and look for a solution by varying the $\theta_1$ variable and also varying the amount of Euclidean evolution in the first step $\tau_1 + \overline{\tau_1}$.\footnote{One can also consider a more fine-grained microcanonical ensemble, where energies $E_j$ at some other locations along the contour are also fixed in advance, and the corresponding $\tau_j + \overline{\tau_j}$ parameters are allowed to vary.}

\subsection{An example with two operator insertions}\label{sectionTwoInsertions}
Let's discuss a simple case in more detail. Suppose that there are just two operator insertions, separated in Lorentzian time and also separated by Euclidean time $\beta/2$:
\begin{equation}\label{quant}
\text{Tr}\left[e^{-\frac{\beta}{2}H} W(t)e^{-\frac{\beta}{2}H}V\right] \hspace{5pt}= \hspace{10pt}\includegraphics[width = .25\textwidth,valign = c]{fixed/twoOps.pdf}.
\end{equation}
The RHS is a diagram of the time contour in the boundary theory that would be used to compute the LHS using the path integral. The circular portion represents Euclidean evolution and the straight portions represent forwards and backwards real time evolution.

The square of this quantity gets a contribution from a wormhole, with two propagators connecting the boundaries, and with two patches of pure gravity evolution in between. When the masses of the bulk fields dual to $W$ and $V$ are equal, then there is a symmetry that interchanges $W\leftrightarrow V$ which implies that the energy in the two patches must be identical. This means $E$ can't change when we apply an operator pair, $\partial_\theta \ell(E,\theta) = 0$, with solution $\theta = \pi$. The other requirement is that the change in the $\theta$ variable when we pass an operator pair should cancel the change due to the evolution in between, which implies
\begin{equation}\label{expli}
m\partial_E\ell(E,\theta) + \beta = 0.
\end{equation}
For the present case $\beta_1 = \beta_2 = \beta$, this is the same equation that we would find by balancing the action (\ref{clac}) against two insertions of (\ref{opin}), considered as functions of $E$.

In JT gravity, (\ref{expli}) is explicitly
\begin{equation}\label{betamE}
\beta = \frac{m}{E}.
\end{equation}
This is quite different from the thermodynamic relationship bewteen energy and inverse temperature that we get from the disk topology in JT gravity, $
\beta = \pi/\sqrt{E}$. This means that on the cylinder topology, the relationship between $\beta$ and energy is different from the thermodynamic one, and the square of the correlator (\ref{quant}) receives most of its contributions from very low energies.\footnote{If we insert enough operators, then high energies will dominate. The main point is just that the energy that dominates is not the one associated to $\beta$ by thermodynamics; instead it depends on the operator insertions.} As mentioned above, to focus on the contributions near a given fixed energy, we can instead work in the microcanonical ensemble. To do so, we just view $E$ as an input parameter, and view $\beta$ as a parameter of the solution, determined by (\ref{betamE}).

\noindent{\bf Size of the wormhole:} An important property of the wormhole is its circumference. This can be defined by measuring the length of the a geodesic that wraps the wormhole. This geodesic will exist in the complexified wormhole geometry.

There does not seem to be an easy way to compute this using the two-dimensional phase space approach discussed above. But in appendix \ref{appJTtwoOps}, we construct the solution in JT gravity using a more complicated but explicit method. There, the cylinder is formed from a quotient of the hyperbolic space by an SL(2,R) group element $T_2(4y)$, and the length of the closed geodesic on that geometry is $b = 4y$. From (\ref{y}), one finds that in the large $t$ limit, this length is
\begin{align}
 b &= 4y \approx  \frac{2\pi}{\beta_{E}}\left(2t - 4t_*\right)
\end{align}
where we have introduced the scrambling time
\begin{equation}
t_* = \frac{\beta_{E}}{2\pi}\log(\frac{2\pi}{\beta_{E}m}).
\end{equation}
Importantly, when the time separation of the operators $t$ is large, the size of the wormhole $b$ becomes large. This retrospectively justifies the fact that we neglected nontrivial geodesics connecting the operators together that wind around the wormhole. The contribution of such geodesics will be exponentially suppressed in $t$.

\vspace{5pt}

\noindent {\bf Pattern of correlation:} The actual geometry of the wormhole is a little difficult to visualize, because (with nonzero Lorentzian evolution) it is a complex geometry. But to get a sense for the rough properties of the solution, one can ask about the distances between different points on the boundary, measured along geodesics through the bulk. We will show the basic pattern by computing the distances between the points indicated with $t_1,t_2,t_3$ here (note the direction of the arrows, which indicates what positive $t$ means for each case)
\begin{equation}
\includegraphics[valign = c, width = .5\textwidth]{fixed/simple2.pdf}
\end{equation}
For simplicity, let's assume that the points are all closer to the end with the $W$ operators than to the end with the $V$ operators, so that the periodicity of the cylinder is not important (remember that at large times, the periodic direction is very long, proportional to $t$).

The basic qualitative feature is that the points 1 and 2 are close together through the wormhole, provided that $t_1$ and $t_2$ are similar, but points 1 and 3 are far apart, because of the chaos sourced by the $W$ operator. This is the same pattern of correlation as in numerical ``replica wormhole'' solutions for the Renyi entropies in the SYK model \cite{Penington:2019kki,Chen:2020wiq}. More explicitly, in appendix \ref{appA} we find (for small $m\beta_E$) the distances $D_{ij}$
\begin{align}
e^{D_{12}} &\approx e^{2\rho}\cosh^2(\tfrac{\pi}{\beta_E}t_{12})\\
e^{D_{13}} &\approx e^{2\rho}\left[\mathrm{i}\sinh(\tfrac{\pi}{\beta_E}t_{13}) + m\tfrac{\beta_E}{\pi}\sinh(\tfrac{\pi}{\beta_E}t_1)\sinh(\tfrac{\pi}{\beta_E}t_3)\right]^2,
\end{align}
where $\rho$ is the hyperbolic radius of the regularized boundary. The distance between points 1 and 2 is the same as on opposite sides of the thermofield double, and the distance between points 1 and 3 is the same as on opposites sides of an OTOC timefold.

\subsection{An example with many operator insertions}
Generally, the evolution equations described above lead to an energy that varies with time around the wormhole. This reflects the fact that acting with a simple operator tends to raise the energy of the state, and Euclidean evolution tends to decrease the energy. We can balance these against each other at some equilibrium energy $E_0$ by inserting Euclidean time evolution $\tau$ in between the operators, with
\begin{equation}
2\tau + m\,\partial_E\ell(E,\pi)\Big|_{E = E_{0}} = 0.
\end{equation}
Then there will be a simple equilibrium solution with $E = E_{0}$ and $\theta = \pi$. Including this Euclidean evolution may seem somewhat artificial, but one can think of it as a simple modification of the operators $\mathcal{O}_j$ so that they do not raise the energy.

We can use this approach to build a configuration with a large number of operators for which the contribution of the wormhole can be evaluated easily. Consider two strings of operators
\begin{align}
\mathbb{W} &= e^{-\tau H} W_n(t_n)\dots e^{-\tau H} W_2(t_2) e^{-\tau H} W_1(t_1)\\
\mathbb{V} &= e^{-\tau H} V_{n'}(t_{n'}')\dots e^{-\tau H} V_2(t_2') e^{-\tau H} V_1(t_1')
\end{align}
where the value of $\tau$ is chosen as described above. We can form states by acting with these operators on one side of a microcanonical version of the thermofield double state at energy $E = E_{0}$, which we will refer to as $|E_{0}\rangle$:
\begin{align}
|\psi_\mathbb{W}\rangle = \mathbb{W}|E_{0}\rangle, \hspace{20pt} |\psi_\mathbb{V}\rangle = \mathbb{V}|E_{0}\rangle.
\end{align}
The state $|E_{0}\rangle$ could be prepared by doing an integral over thermofield double states
\begin{equation}
|E_0\rangle = \int_{i\mathbb{R}} \mathrm{d}\beta\, e^{\frac{\beta}{2}E_0 + \epsilon \beta^2}|\text{TFD}(\beta)\rangle.
\end{equation}
Here $\epsilon$ determines the width of the microcanonical ensemble.

Then in a classical approximation, we claim that
\begin{equation}
\frac{|\langle \psi_\mathbb{W}|\psi_{\mathbb{V}}\rangle|^2}{\langle \psi_\mathbb{W}|\psi_{\mathbb{W}}\rangle\langle \psi_\mathbb{V}|\psi_{\mathbb{V}}\rangle} \hspace{5pt} = \hspace{5pt}\frac{\includegraphics[valign =c, scale = .4]{fixed/16.pdf}}{\includegraphics[valign =c, scale = .4]{fixed/17.pdf}} \hspace{5pt}= \hspace{5pt}e^{-2S_{BH}(E_{0})},\label{indof}
\end{equation}
where $S_{BH}(E)$ is the thermodynamic entropy of a single black hole at energy $E$. Because we are considering two copies of the system, the entropy of the total system is $2S_{BH}$. This agrees with (\ref{innerprod2}) with $S$ equal to the standard thermodynamic entropy.

The ratio (\ref{indof}) is independent of the normalization of the states, but to understand why this is the answer, it is convenient to use states normalized in the natural way that the gravity path integral gives, so that $\langle \text{TFD}(\beta)|\text{TFD}(\beta)\rangle = Z(\beta)$. In this same normalization, we have (up to a subleading factor that depends on $\epsilon$)
\begin{equation}\label{inn}
\langle E|E\rangle = e^{S_{BH}(E)}.
\end{equation}
The main point underlying (\ref{indof}) is that the effect of the operator insertions cancels out between the numerator and the denominator. Part of the reason this is possible is that the same set of operators appear in both cases, and they will be connected by propagators in the same pattern. But by itself, that isn't enough, because the contribution of the operators depends on the detailed geometrical configuration, in particular, it depends on the $E,\theta$ variables at the locations where the operators are inserted. However, since we matched the energy of the state $|E_0\rangle$ to the equilibrium energy for the strings of operators $\mathbb{W}$ and $\mathbb{V}$, we will have a simple solution $E = E_0$ and $\theta = \pi$ throughout the region where the operators act. This will be true both in the numerator and the denominator, so the insertions will not be able to ``feel'' whether they are in the wormhole in the numerator or the disks in the denominator.

This means that the operator insertions cancel out, and we are left with the pure gravity contribution of a microcanonical cylinder divided by two microcanonical disks. The disks compute the square of the inner product (\ref{inn}), and the action for the cylinder is zero. (One way to see this is to note that the on-shell microcanonical cylinder with no insertions is formally a cylinder with zero length in the periodic direction, $\beta = 0$, and zero classical action (\ref{clac}).)

\section{Quantum calculations}
Let's now repeat the same discussion quantum mechanically, using the exact quantization approach that was applied to JT gravity in \cite{Bagrets:2017pwq,Kitaev:2018wpr,Yang:2018gdb,Saad:2019pqd}. We will discuss the computation for a general dilaton gravity theory, making a simplifying assumption that the Lorentzian time separations between the operator insertions is large.

We continue to omit the topological $S_0$ term from the dilaton gravity action, since it was already seen in (\ref{suppression}) to give the expected contribution.

As in the classical discussion, it is convenient to think about a time coordinate that goes around the periodic direction of the wormhole. At each instant of this time, we have the Hilbert space of dilaton gravity on an interval, which can be parametrized in terms of states $|E\rangle$ with definite asymptotic energy, or states $|\ell\rangle$ of definite regularized length. Boundary time evolution is simple in the $|E\rangle$ basis, and the operator insertions are simple in the $|\ell\rangle$ basis. To translate between the two, we need the inner product $\langle \ell|E\rangle$. It will be convenient to normalize this so that the measure for energy contains a factor of $\rho(E)$, the density of states in the disk approximation. With this convention, the completeness and orthogonality relationships are
\begin{align}\label{111}
\int_{-\infty}^\infty \mathrm{d}\ell\, \langle E|\ell\rangle\langle \ell|E'\rangle &= \frac{\delta(E-E')}{\rho(E)}\\
\int_0^\infty \mathrm{d}E\, \rho(E) \langle \ell|E\rangle\langle E|\ell'\rangle &= \delta(\ell-\ell').
\end{align}
For the special case of JT gravity, the functions $\langle \ell|E\rangle$ \cite{Bagrets:2017pwq,Kitaev:2018wpr,Yang:2018gdb,Saad:2019pqd} and $\rho(E)$ \cite{Cotler:2016fpe,Stanford:2017thb} are known:
\begin{equation}
\langle \ell|E\rangle  = 2^{3/2} K_{2i\sqrt{E}}(4 e^{-\ell/2}), \hspace{20pt} \rho(E) = \frac{1}{(2\pi)^2}\sinh(2\pi\sqrt{E}).
\end{equation}
However, to demonstrate agreement with the predictions of an ensemble, we will not need these explicit functions, so the discussion will apply to a more general dilaton gravity theory.

In between operator insertions, the boundary time evolution is given by the operator $e^{-(\tau_L + \tau_R)\widehat{E}}$, where $\tau_L$ and $\tau_E$ are the amounts of (complexified) Euclidean time evolution on each boundary. Because the energy at the two asymptotic boundaries is the same, the operator only depends on the sum of the two times. The matrix elements of this operator in the $\ell$ basis are
\begin{equation}
\langle \ell'|e^{-(\tau_L + \tau_R)\widehat E}|\ell\rangle \hspace{5pt}= \hspace{10pt} \includegraphics[valign =c, width = .12\textwidth]{fixed/5.pdf}\hspace{10pt} = \hspace{5pt} \int_0^\infty \mathrm{d}E\,\rho(E) e^{-E(\tau_L+\tau_R)}\langle \ell'|E\rangle\langle E|\ell\rangle.\label{prop}
\end{equation}
This can be understood as the exact dilaton-gravity path integral on this rectangular geometry, with two geodesic boundaries and two asymptotic boundaries \cite{Yang:2018gdb,Saad:2019pqd}. With our conventions, which differ slightly from the JT gravity computations in \cite{Yang:2018gdb,Saad:2019pqd}, this rectangle can be glued together with another rectangle by integrating over the length of their shared geodesic boundary with measure $\mathrm{d}\ell$. To see that this is the right measure, one can use (\ref{111}) to check that gluing together two rectangles produces another rectangle:
\begin{equation}
\int_{-\infty}^\infty \mathrm{d}\ell' \langle \ell|e^{-(\tau_L + \tau_R)\widehat E}|\ell'\rangle \langle \ell'|e^{-(\tau_L' + \tau_R')\widehat E}|\ell''\rangle = \langle \ell|e^{-(\tau_L + \tau_R + \tau_L' + \tau_R')\widehat E}|\ell''\rangle.
\end{equation}

Next consider the operator insertions. Because the Lorentian time evolution between operators is assumed to be large, we only need to consider the case where the bulk propagators connect each operator $\mathcal{O}_j$ to its partner $\mathcal{O}_j^\dagger$ on the other boundary. Also, as we saw in the classical calculation, the large time evolution results in a cylinder with a long periodicity, so the propagators will not ``feel'' the periodicity. Inserting the operator pair on the two boundaries can therefore be translated to an operator insertion $\widehat{G}_{\mathcal{O}}$ that acts on the Hilbert space of dilaton gravity on an interval. For example, for a sufficiently massive field, the geodesic approximation to the two point function gives
\begin{equation}
\widehat{G}_{\mathcal{O}} = \int_{-\infty}^\infty \mathrm{d}\ell  \ |\ell\rangle e^{-m\ell}\langle \ell|
\end{equation}
however, we will not need to assume this form. Below, we will use this operator in the energy basis. We will not need the explicit answer, but for the particular case of JT gravity, one has \cite{Bagrets:2017pwq,Mertens:2017mtv,Yang:2018gdb}
\begin{equation}
\langle E|\widehat{G}_{\mathcal{O}}|E'\rangle = \int_{-\infty}^\infty \mathrm{d}\ell\, \langle E|\ell\rangle e^{-\Delta \ell}\langle \ell|E'\rangle = \frac{\Gamma(\Delta\pm i\sqrt{E}\pm i\sqrt{E'})}{2^{2\Delta-1}\Gamma(2\Delta)}.
\end{equation}
Here the symbol in the numerator means a product of four Gamma functions, one for each of the four different choices of $\pm$ signs, and $\Delta$ is the conformal dimension of the operator $\mathcal{O}$.

Having understood both the time evolution and the operator insertions, one would naively expect to be able to form the cylindrical wormhole by taking a product of the time evolution operators and the $\widehat{G}_{\mathcal{O}}$ operators (which combine to make a long rectangle) and then gluing the ends together. Near this ``seam,'' the gluing looks the same as the gluing we do to join two rectangles together to form a larger rectangle. Locality of JT gravity therefore implies that the same $\mathrm{d}\ell$ measure will be correct for this gluing as well.\footnote{For JT gravity we also checked this explicitly by starting with the Weil-Petersson measure. It's a bit tricky.}\footnote{In principle, one should also weight by the partition function of the matter fields. Neglecting this is a good aproximation if the wormhole has a large circumference $b$, since fields will get projected into their ground state (at least for conformal matter \cite{Saad:2018bqo}). In the classical solution, $b$ is proportional to time, so this is a good approximation for late time. However, in the full path integral, we integrate over $b$, and in fact there is an exponential divergence as $b\to 0$ \cite{Saad:2019lba}. We are assuming that this divergence can be cured in a more complete bulk theory \cite{Maldacena:2018lmt}. But we should view the quantum computation above as correctly including all fluctuation effects near the classical solution, but not correctly computing the contribution from very small wormholes, which will depend on more details of the bulk theory. If the $b\to 0$ region is made finite, it should give a small answer at late time.}

However, there is a global problem with this naive gluing, because the geodesic connecting different points on a cylinder is non-unique. As explained in \cite{Saad:2019pqd}, by integrating freely over the geodesic length, we are including winding geodesics, and therefore counting the same cylinder multiple times. Mathematically, we are failing to account for the mapping class group $\mathbb{Z}$ for the cylinder. Fortunately, when the total Lorentzian time evolution $t$ is large, the classical solution we found above has a very large circumference, $b\sim t$. This means that winding geodesics will be much longer than non-winding geodesics. Since the lengths of the geodesics that we glue across determine the correlation functions of the operator insertions, the winding geodesics will give exponentially small contributions $e^{-\# t}$. So for large $t$ it is actually allowable to forget about winding geodesics altogether, and to glue rectangles together to form a cylinder in the naive way.

We can now write a final expression for the wormhole contribution to the square of the correlator, by taking the trace of a product of the time evolution operators and the $\widehat{G}$ insertions
\begin{align}
\bigg|\text{Tr}\Big[\prod_{j = 1}^n e^{-\tau_j H} \mathcal{O}_j\Big]\bigg|^2 &=\int_{-\infty}^\infty \mathrm{d}\ell  \ \langle \ell| \prod_j e^{-(\tau_j + \overline{\tau_j})\widehat E}\widehat{G}_{\mathcal{O}_j}|\ell\rangle\\
&=  \prod_{j = 1}^n \int_0^\infty \mathrm{d}E_j \rho(E_j)e^{-(\tau_j + \overline{\tau_j})E_j}\langle E_j|\widehat{G}_{\mathcal{O}_j}|E_{j-1}\rangle.\label{finalans}
\end{align}
We have not given answers for $\rho$ or for $\langle E| G|E'\rangle$ except in the special case of JT gravity. But even so, in the next section, we will show that it agrees with the predictions of a simple ensemble average of the answer one expects based on a statistical theory of the ``noise.'' Finally, because we will use it below, here is a special case of (\ref{finalans}) for the configuration with two operators discussed in section \ref{sectionTwoInsertions}:
\begin{equation}
\bigg|\text{Tr}\left[e^{-\frac{\beta}{2}H} W(t)e^{-\frac{\beta}{2}H}V\right]\bigg|^2 = \int_0^\infty \mathrm{d}E\mathrm{d}E' \rho(E)\rho(E') e^{-\beta(E + E')}\langle E|\widehat{G}_W|E'\rangle \langle E'|\widehat{G}_V|E\rangle.\label{twoOps}
\end{equation}

\section{Matching to an ensemble answer}
As in previous work by Saad \cite{Saad:2019pqd} and Blommaert \cite{Blommaert:2020seb} on related quantities (see section 6.2 of \cite{Saad:2019pqd}), the formula (\ref{finalans}) can be interpreted in terms of a formal ensemble average. This is formal because we do not have a precise ensemble of theories in mind. But for late-time quantities there is some degree of universality, so that different ensembles will lead to approximately the same answers. So, in this context it makes sense to talk about the predictions of an ``ensemble'' without actually specifying very much about the ensemble.

For simplicity, we will study the case with two operators arranged as in (\ref{twoOps}). To start, one can write out this quantity in terms of matrix elements in the energy basis
\begin{equation}
f(t) = \text{Tr}\left[e^{-\frac{\beta}{2}H} W(t)e^{-\frac{\beta}{2}H}V\right] = \sum_{nm}\langle n|W|m\rangle\langle m|V|n\rangle e^{-\frac{\beta}{2}(E_n + E_m)} e^{it(E_n  - E_m )}.
\end{equation}
It is important to emphasize that in this section, the quantum states are quantum states in the boundary Hilbert space, not in the Hilbert space of gravity on an interval. The states $|n\rangle,|m\rangle$ represent energy eigenstates of the boundary quantum mechanics.

Now, we would like to make a prediction for the ensemble average of this quantity, over some ensemble of boundary quantum systems. In a continuous ensemble of theories, we expect the average of $f(t)$ to converge at late time to some value, which can also be obtained by averaging over a long time window. Such time averaging will force $E_n = E_m$ in order to avoid phase cancellations, so
\begin{equation}
\lim_{t\to\infty}\mathbb{E}\Big\{f(t)\Big\} = \sum_{n}\mathbb{E}\Big\{\langle n|W|n\rangle\langle n|V|n\rangle\Big\} e^{-\beta E_n}.
\end{equation}
Repeating the same logic for the ensemble average of $|f(t)|^2$, we find
\begin{equation}
\lim_{t\to\infty} \hspace{10pt} \mathbb{E}\Big\{|f(t)|^2\Big\} - \left|\mathbb{E}\Big\{f(t)\Big\}\right|^2 = \sum_{n,m}\mathbb{E}\Big\{|\langle n|W|m\rangle|^2|\langle m|V|n\rangle|^2\Big\} e^{-\beta (E_n+E_m)}.
\end{equation}
This is the quantity that we would like to compare to the wormhole contribution. We will assume that the $W$ and $V$ matrix elements are approximately uncorrelated, so that this expectation value can be computed using the functions
\begin{align}
\mathbb{E}\Big\{|\langle n|W|m\rangle|^2|\Big\}= |W_{E_n,E_m}|^2, \hspace{20pt}
\mathbb{E}\Big\{|\langle n|V|m\rangle|^2|\Big\}= |V_{E_n,E_m}|^2\label{Wen}.
\end{align}
In terms of these quantities, we find
\begin{equation}\label{predFor}
\lim_{t\to\infty} \hspace{10pt} \mathbb{E}\Big\{|f(t)|^2\Big\} - \left|\mathbb{E}\Big\{f(t)\Big\}\right|^2 = \int \mathrm{d}E\mathrm{d}E' \uprho(E)\uprho(E') |W_{E,E'}|^2|V_{E,E'}|^2e^{-\beta(E + E')}.
\end{equation}

To finish the job and turn this into a useful prediction, we need to determine the functions $|W_{E,E'}|^2$ and $|V_{E,E'}|^2$ that were defined in (\ref{Wen}). These can be determined by consistency with thermal two point functions, which are self-averaging:
\begin{align}
\text{Tr}\Big[e^{-\beta H} W(\tau) W^\dagger(0)\Big] &=  \sum_{m,n}|\langle n|W|m\rangle|^2e^{-(\beta-\tau)E_n - \tau E_m} \\ &\approx \int \mathrm{d}E\mathrm{d}E' \rho(E)\rho(E') |W_{E,E'}|^2 e^{-(\beta-\tau)E - \tau E'}.\label{ETH2}
\end{align}
So, given the thermal two point function on the LHS, in principle one can solve this equation to determine $|W_{E,E'}|^2$, and then make a prediction for (\ref{predFor}). In fact, it is easy to carry this out in practice and show (see \cite{Bagrets:2017pwq,Mertens:2017mtv,Yang:2018gdb} and appendix \ref{apptwoPt}) that for a boundary theory dual to dilaton gravity,
\begin{equation}
|W_{E,E'}|^2 = \langle E|\widehat{G}_W|E'\rangle.
\end{equation}
Plugging into the RHS of (\ref{predFor}), we find precise agreement with the wormhole computation (\ref{twoOps}). So the wormhole computation agrees with the universal answer expected at late times in an ensemble of theories. An interesting feature is that in the bulk computation, the late-time behavior sets in quite quickly, on scrambling timescales. This provides some additional information about the ensemble.

One can generalize this to the case with more operator insertions, and again one finds agreement with (\ref{finalans}) provided that the time separations between operator insertions are large.

\section{Replica wormholes and factorization}
When spacetime wormholes connect together disconnected boundaries, it is tempting to interpret them in the context of an ensemble average \cite{Coleman:1988cy,Giddings:1988cx,Maldacena:2004rf,ArkaniHamed:2007js}. This is because the contradiction in
\begin{equation}
Z \hspace{5pt} = \hspace{5pt} \includegraphics[valign = c, scale = .3]{fixed/a2.pdf} \hspace{80pt} Z^2 \hspace{5pt} = \hspace{5pt} \includegraphics[valign = c, scale = .3]{fixed/a4.pdf}
\end{equation}
can be avoided by the interpretation
\begin{equation}
\mathbb{E}\{Z\} = \includegraphics[valign = c, scale = .3]{fixed/a2.pdf}  \hspace{80pt} \mathbb{E}\{Z^2\} = \includegraphics[valign = c, scale = .3]{fixed/a4.pdf}
\end{equation}
In these drawings, we are assuming $Z$ is some boundary observable that is computed in the holographic dual by filling in a boundary circle with 2d geometry.

Do the replica wormholes involved in the Page curve computations have to be interpreted this way? If we treat the computation as a black box, then the answer is no. For a black hole evaporating into a bath, the replica wormhole for the Renyi entropy looks schematically like
\begin{equation}\label{wh1}
\text{Tr}(\rho^2) \hspace{5pt} =  \hspace{5pt}\includegraphics[valign = c, width = .35\textwidth]{fixed/d4.pdf}
\end{equation}
The boundaries are not actually disconnected, because of the interactions between them, so there is no obvious factorization problem. Unlike $Z^2$ in the previous example, $\text{Tr}(\rho^2)$ is not supposed to be the square of any simpler quantity.

But we can look inside the box and ask how much gravity knows about $\rho$ itself, beyond just its entropy. To examine individual matrix elements, one can replace the interactions by operator insertions. Following \cite{Penington:2019kki}, the main point of this paper was that the wormhole still exists:
\begin{equation}\label{wh2}
|\rho_{i_Wi_V}|^2 \hspace{5pt} =  \hspace{5pt}\includegraphics[valign = c, width = .35\textwidth]{fixed/jj.pdf}
\end{equation}
And now there {\it is} a problem with factorization. This seems consistent with the idea that simple gravity (without branes/strings/whatever makes up the black hole microstates) is trying to describe some (formal) ensemble of boundary theories.

For a more complete bulk theory that contains microstates, one hopes that (\ref{wh1}) is still a step in the right direction. What about (\ref{wh2})? For many purposes, actual randomness is a good practical description of pseudorandomness. But is it a first step in a precise description of it?

\subsection*{Acknowledgements} We are especially grateful to Pouria Dadras and Alexei Kitaev for discussions. We also thank Juan Maldacena, Don Marolf, Geoff Penington, Phil Saad, Steve Shenker, Zhenbin Yang, and Ying Zhao. Research was supported in part by the DOD under grant 13104630. Part of the work was carried out at the Kavli Institute for Theoretical Physics, supported in part by NSF grant PHY-1748958.


The Parks department ensures that the park and recreation facilities of the City are properly developed and maintained. The Parks employees install and maintain the City parks and recreational trails.
Spirit Lake has many family-friendly city parks. Within Dickinson County are some of the most attractive State and local parks in the State of Iowa.
Located throughout the Lake Okoboji and Spirit Lake area, several Iowa parks offer travelers and residents alike places to picnic, fish and camp, and includes boat launch areas, beaches, playgrounds and other things that outdoor enthusiast would want.
This neighborhood park is the City's newest park. Playground Equipment, sand play area, benches and easy access to the bike trail.
1 acre Shelter house, trail system, playground equipment, grills, park benches, picnic tables, bicycle rack, handicap-accessible restrooms.
5 acres Shelter house, walking trail, 2 floating docks, approximately 500 feet of shoreline along East Lake Okoboji, basketball court, playground equipment, park benches, picnic tables, grills, handicap-accessible restrooms.
5 acres Approximately 1,000 feet of shoreline of Upper East Lake Okoboji, boardwalk, 2 floating docks, canoe ramp, walking trail, RV dump, picnic tables, grills, handicap-accessible restrooms, and fishing piers.
3 acres Shelter house, picnic tables, playground equipment, ball field, restrooms.
10 acres Shelter house, 3 ball diamonds, 6 tennis courts, playground equipment, picnic tables, handicap-accessible restrooms.
Walking and bicycling trail through city of Spirit Lake: 2 trail heads, 4 fishing piers, 3 restrooms along the trail system, 1 pedestrian underpass. Connects to over 100 miles of trails in Dickinson County. Urban and rural sections with beautiful natural scenery of lakes, wetlands and prairie.
140 feet of shoreline along south shore of Big Spirit Lake, largest natural lake in Iowa, sandy beach, park benches, trail system. The Spirit Lake Protective Association has provided a public dock facility for boaters to access the area.
50 feet of shoreline on Center Lake, picnic tables.


The automatic setup imports your project's URLconf in order to add the Debug Toolbar's URLs. This may trigger circular imports when the URLconf imports views that import models. If you're hitting an ImportError, follow the explicit setup instructions.
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Lady Li's pups are, thus far, doing well. They are in a great foster home! They are living in the bedroom of a teen-ager who has the most cheerful, cute bedroom I have ever seen. And she has two sisters who love to hold the puppies. Life is good thus far and I am hoping it stays good for this mom and her three pups. But she was left abandoned in a house, and then she did sit at the animal shelter for two weeks and she was horribly infested with fleas, on the other hand, we have given her two distemper/parvo shots and she appears healthy. Keep a hoping and keep a praying!
Meanwhile, here are some incredibly cute photos.
The cream and the brindle are girls – the boy is brown and white. They were born November 27, 2011.
You can make a donation to help pay for the surgery that is required to fix Lady Mama's eye.
Her eye is, in my words, deflated.
Her eye needs to be cleaned out and sewed up — an expensive procedure.
You can buy a calendar too!!!
Another mom that came to us with serious medical issues is featured in our 2012 calendar. She is Miss October and came to us with puppies AND breast cancer!
Purchase of calendars helps us pay our vet bills.


While visiting the city of Coorg, it is hard to miss the famous locality of Abyathamangala Coorg. It is a beautiful area in Coorg, Dubare Elephant Camp and Nanjarayapattna Dubare Elephant Camp are some of the major landmarks associated with Abyathamangala Coorg.
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The 4-star and 5-star hotels in Abyathamangala Coorg give you the taste of luxury. They are equipped with a lot of amenities to make your stay a comfortable one. Some of the amenities you will find in these hotels are Wifi, Restaurant, and Gym.


text:
News of
The King of
The Croods:
Scott (Pete Davidson) has been a case of arrested development ever since his firefighter father died when he was seven. He's now reached his mid-20s having achieved little, chasing a dream of becoming a tattoo artist that seems far out of reach. As his ambitious younger sister (Maude Apatow) heads off to college, Scott is still living with his exhausted ER nurse mother (Marisa Tomei) and spends his days hanging with the guys and secretly hooking up with his childhood friend Kelsey (Bel Powley). But when his mother starts dating a loudmouth firefighter named Ray (Bill Burr), it sets off a chain of events that will force Scott to grapple with his grief and take his first tentative steps toward moving forward in life.
Co-starring Steve Buscemi and Pamela Adlon, The King of Staten Island is directed by Judd Apatow from a screenplay by Apatow, Pete Davidson and Dave Sirus. The film is produced by Apatow and Barry Mendel.
Movie Date
CONFIRM RSVP

redpajama_set_name: RedPajamaCommonCrawl


text:
Community, eNews, Siesta Key News
Another widespread effort of installing mini-reef filtration systems in Siesta Key's Grand Canal will take place in October, project leader Jean Cannon reports. She hopes to install 15 to 20 more during the month, adding to the approximately 250 that currently exist on Siesta Key — 210 of them being in the Grand Canal.
"I want to double that amount by this time next year," said Cannon, who's part of the Siesta Key Association civic group.
The mini-reefs are ideal for anyone with a dock. Cannon said there are 873 homes with docks on Grand Canal alone.
The reefs cost $325 and play a large role in regenerating the area's damaged aquatic life. Visit the Siesta Key Association's website at siestakeyassociation.com and click on the "projects" tab for more information.
"We're already beginning to see a juvenile fish population," Cannon said of the project, which began late last year.
Tags: Jean Cannon, Siesta Key Association, Siesta Key Mini-Reefs
Beach walk to benefit fight against cystic fibrosis
Join some special puppies Sept. 24


text:
Remember the Exxon Valdez?
by Lanny J. Davis
Post date: 10.06.06 – From TNR Online
Ask any professional crisis manager–or any academic who teaches this discipline in business schools–to name the worst-handled crisis in the history of the universe, and Exxon's mishandling of the Valdez oil-tanker spill is inevitably at the top of the list. High-ranking Exxon managers found a way to break every rule of Crisis Management 101: They didn't immediately get all of the facts out; they waited too long to even show up at the scene; and, most importantly, they delayed far too long before telling the truth and accepting responsibility. And so the bad news got worse–the story continued to dribble out, and the damage to the environment was nearly exceeded by the damage to Exxon's reputation. In short, Exxon and Valdez became the poster children for how to mishandle a crisis.
Well, Valdez, move over. This week introduced the world to an even more inept performance: Dennis Hastert's handling of the Mark Foley matter. It is, from start to finish, a case study in how not to manage a scandal.
Last spring is when Hastert and his Republican colleagues say they first learned that Foley had sent "overfriendly" (but not sexually explicit) emails to a House page. They also say they knew even these messages were serious enough that the parents of the young man had asked that Foley stop. That should have been enough to execute the following Crisis Management 101 plan:
The speaker and his colleagues should have called in Foley, showed him the emails, and asked him to submit to a full, bipartisan investigation by either the House Page Board or the House Ethics Committee. They should have notified both the chairman and the ranking Democrat on the House Page Board. They should have told Minority Leader Nancy Pelosi and asked her to support the bipartisan investigation. And then they should have issued a joint statement. It's possible that Foley would have reacted to this plan by resigning and seeking counseling (which he has only now sought). But, in any event, everything would have been on the table.
This may sound politically counterintuitive–especially in today's hyperpartisan environment–but it would have paid off. There would definitely have been a short-term Republican political crisis, but that would have been better in the spring than now, just weeks before the midterm elections. And the political damage to the overall Republican cause would likely have been minimal, since the crisis would have focused on Foley's conduct alone–not on the appearance that, in hiding the scandal, Republicans were either negligent or motivated by politics.
Of course, people in the center of crises rarely see the value of a "tell it all/tell it early/tell it yourself" strategy. So what should Hastert do now to minimize the crisis rather than magnify it? Hint: Don't sink into a defensive crouch. The law of crisis management is: Stop the bleeding first. Don't ask for, or expect, a complete cure, at least not in the short term. Hastert announced Thursday that he took responsibility for what happened, but if the speaker asked for my help, here's the statement I would have him deliver:
I and my colleagues screwed up–we should have reacted to these warning signals about Mark Foley differently and sought a bipartisan investigation when we first learned of this information. We allowed political considerations to get in the way–that is the only explanation I can come up with as to why we didn't inform the Democratic member of the House Page Board at the time but only the Republican chair. And for that, I and my colleagues are truly sorry.
Someone asked me whether I could think of any historical examples where crises had abated through transparency and accountability. Actually, I wrote about one in a book I recently published on the scandal culture: In the closing weeks of the tight 1884 presidential campaign, the Democratic nominee, New York Governor Grover Cleveland, was told that his local newspaper was about to report that he had fathered of an out-of-wedlock child. When his panicked campaign manager asked Cleveland what should they tell the press, he is reported to have answered: "Tell the truth." And so he admitted that he had fathered the child and also that he had supported the child and his mother for years.
The Republicans, of course, huffed and puffed about immorality in the wake of the nasty newspaper stories. But then the public, apparently appreciating both Cleveland's honesty and his human weaknesses, moved on. Cleveland won the election–the first Democrat to win the White House since the Civil War. Hastert should read his history.
And what about the Democrats? Here, the strategy is obvious: Avoid making this into a partisan issue, stick to the facts, call for an investigation, and shut up. So far, the Democratic congressional leadership, for the most part, seems to be following this strategy. The one exception was the Democratic National Committee (DNC), which couldn't resist putting out "talking points" to its email list, some of them in red ink. They were pretty restrained, given the usual rhetoric that comes from the DNC. But I would advise the DNC to keep quiet and let the congressional Democratic leadership carry the message, if anything further needs to be said.
Lanny J. Davis served as President Bill Clinton's special counsel from 1996-1998 and heads the Legal Crisis Communications practice group at Orrick, Herrington & Sutcliffe. He is the author of Scandal: How 'Gotcha' Politics Is Destroying America.
17 ottobre , 2006
Le città come brand – parte terza
Safe for diversity


As much as I wanted to do a totally full and proper reveal of my very first Hermes Bag, you know the box,wrapping, ribbon and all other info pertaining to ones purchase, the situation sadly does not allow me to at this time.
Acquired at the Hermes Boutique/store at the Vancouver International Airport this past Saturday, catching a flight to my other home in PV, Mexico. With that being said, I had to pass on my Orange Box & ribbon.Wanting to keep this purchase as low key as possible with the other passengers on my flight, only made sense to me doing that way.
Extra packaging was just not going to work.
I decided to do my reveal yesterday, and as I went to get everything all together, then I noticed that I didn't even have my camera here. I own two professional ones, and knew of one being back in Vancouver, but the other should have been here. Well it wasn't, so I ended up having to take the photos using photo booth in my MacBook Pro. Talk about a work out, trying to get the right angles etc was no easy chore. So I have to apologize that my photos are not the best resoltion and clarity with some shots. The leather is much more beautiful then the photos appear.
Cont'D with more photos and thank you for letting me share.
Congrats on a gorgeous bag!
Congrats on your first H bag! You wear it very well. The Vancouver airport is where I got my first ever H item, a clic clac.
I only wish that I had one of my cameras here with me to allow a much better viewing experience, lol.
I had a choice of the "Gold" color or the same model in Etoupe.
My whole face lit up when wearing the gold versus the Etoupe. The color on me, did nothing at all, no personality, so of course after careful deliberation with myself, the gold was my choice.
Oh congrats to you too on your first purchase @VIA Hermes.
This bag was not my first purchase there, just first Bag.
congrats on your gp!! a classic..
Gorgeous bag and it looks great on you! Congrats!
congratulations on your first H bag! very exciting! GP looks great on you!
Thank you very much and of course I can.
According to its measurements it is the 36, but my invoice says TPM, is that the same? Not sure of the proper codes and sizes.
It does look kinda big in the photo though. and hopefully not to big on me.
chincac - thank you for letting me share.
katielure- Awe thanks so much. I purposely didn't dress up or color coordinate to show what it looks like. Your very sweet to say.
luckylove - Thanks so much, yes exciting for the first bag.
You know I had totally forgot that my first Hermes purchase, was a belt.
What a beautiful Goldilocks! Congratulations--you wear her well. Lovely!


Stand-up comedy group All India Bakchod (AIB) is "super excited" to announce its "most ambitious" project.
"Super excited to announce our latest project. Our most ambitious one yet. Here's a sneak peek," AIB wrote on Twitter along with the link of a YouTube teaser for the yet-untitled show.
According to the video, the show is coming soon to Hotstar, Star Plus and Star World.
AIB comprises stand-up comics Tanmay Bhat, Gursimran Khamba, Rohan Joshi and Ashish Shakya. Known for its quirky, satirical humour, the humour collective has gained a massive following online across social media platforms.
The group had collaborated with actor Irrfan Khan for a spoof video on Bollywood item numbers in August.
Earlier this year, the group ran into trouble following an infamous, controversial "roast", which also featured Bollywood A-listers.
The Roast poked fun at Bollywood actors Ranveer Singh and Arjun Kapoor, and was hosted by filmmaker Karan Johar.


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MPS.itOnly italian version
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ENGITA
Vittorio Calvanico
Operating Officer
Degree in Business Administration from Bocconi University in Milan and an MBA from Yale University.
An Executive Manager of Banca Monte dei Paschi di Siena since 2001, Mr. Calvanico currently serves as Chief Operating Officer, responsible for organisation, purchasing and logistics, real estate, security, back office services as well as the outsourcing and monitoring of strategic projects. As of 25th August 2018 Mr. Calvanico also serves as Head of Organisation and Operations.
Within the bank, he previously held the positions of Chief Program and Cost Officer, Deputy CFO - Head of Planning and Chief Data Officer, as well as Head of "Strategy, M&A and Equity Investments", "CFO Division Staff Unit" (responsible for Extraordinary Transactions), "Monetary Project", "International Activities" and "Strategic Development".
As part of his duties, he is a permanent member of the Bank's Steering Committee and Risk Management Committee; he chairs the Operational Projects Committee and takes part in the Disciplinary Affairs Committee.
He is also the Operational Manager in charge of Business Continuity & Crisis Management and chairs the Operational Session of the Business Continuity & Crisis Management Committee.
He currently sits on the Board of Directors of the subsidiary MPS Capital Services and of the investee company Nomisma. In the past, he served on the Boards of several other subsidiaries (MP Venture, Banca 121, Intermonte, Consorzio Operativo di Gruppo, MPS Immobiliare) and investee companies (Consorzio Bancomat, SI Holding, Cartasi, Iconcard, MP Venture, Antonveneta Assicurazioni, Antonveneta Vita and Consorzio Triveneto).
Before joining the Montepaschi Group, Mr. Calvanico provided strategic consulting services within McKinsey & Company's Financial Institutions sector and was responsible for training and professional development in Studio Ambrosetti.
Banca Monte dei Paschi di Siena S.p.A. P.I. 00884060526
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Banca Monte dei Paschi di Siena S.p.A. P.I. 0088406052


text:
Q: Uploading file using getElementByClassName I have the code as shown below. I am trying to upload multiple files onchamge,

document.getElementsByClassName('fileUpload').onchange = function () {
    alert("changed");
   /* var field = document.getElementsByClassName('fileUpload');
    var file = field[0].files[0];*/
    var filename = this.value;
    alert(filename);

    var a = filename.split(".");
    alert(a);

    if( a.length === 1 || ( a[0] === "" && a.length === 2 ) ) {
        return "";
    }
    var suffix = a.pop().toLowerCase();
    //if( suffix != 'jpg' && suffix != 'jpeg' && suffix != 'png' && suffix != 'pdf' && suffix != 'doc'){
    if (!(suffix in {jpg:'', jpeg:'', png:'', pdf:'', doc:''})){
        document.getElementById('fileUpload').value = "";
        alert('Please select an correct file.');
    }
};
<input type="file" name="image" id="fileUpload">
<input type="file" name="image" class="fileUpload">
<input type="file" name="image" class="fileUpload">

but since an ID is set only to one element I am trying to change the code to use getElementByClassName. Please help me alter my code to get this to work thanks.

A: You need to create a shared onchange function, then apply that to each element:

// Iterate over each element with the fileUpload class and assign the handler
[].forEach.call(document.getElementsByClassName('fileUpload'), function(element) {
  element.onchange = onFileChanged;
});

// Shared handler for the event
function onFileChanged() {
  alert("changed");
  var field = this; // 'this' is the current file element
  var file = field.files[0];
  var filename = this.value;
  alert(filename);

  var a = filename.split(".");
  alert(a);

  if( a.length === 1 || ( a[0] === "" && a.length === 2 ) ) {
    return "";
  }
  var suffix = a.pop().toLowerCase();
  //if( suffix != 'jpg' && suffix != 'jpeg' && suffix != 'png' && suffix != 'pdf' && suffix != 'doc'){
  if (!(suffix in {jpg:'', jpeg:'', png:'', pdf:'', doc:''})){
    field.value = "";
    alert('Please select an correct file.');
  }
};
<input type="file" name="image" id="fileUpload">
<input type="file" name="image" class="fileUpload">
<input type="file" name="image" class="fileUpload">

A: This
getElementsByClassName('fileUpload')

returns an Array of items and not a single one. Just make a loop through instead:
var array = getElementsByClassName('fileUpload');

for (var i=0;i<array.length;i++) {
  array[i].onchange = ...
}

A: <!DOCTYPE html>

<input type="file" id="myFile" multiple size="50" onchange="myFunction()">

Tip: Use the Control or the Shift key to select multiple files.

Here, good luck, useful me, thanks you.

redpajama_set_name: RedPajamaStackExchange


Lawruk Machine was founded and has been in the machine industry since 1966. Mike Frederick purchased the company in 1990. Frederick's Firearms has preformed numerous military and government projects. Firearm making has been in the making for 3 years for various companies. Veteren designer/operator developed the new FF-15 & FF-308. All our matched sets machined on HAAS CNC Machines.
Your choice of finish coating anodized, black cerakote or hydrographics as your choice.


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Apple TV ($149 - $199): Similar to the Amazon Fire, the Apple TV is fantastic for dedicated Mac families, allowing them to sync programing between iPhones, iPads and laptops. Apple also has one of the better interfaces for finding and organizing content, with an app simply called "TV" that's designed to function a lot like a DVR, keeping the latest episodes of your favorite shows in an easily accessible queue. Note: Netflix shows can be searched via the "TV" app, but they can't be added to its queue; users are simply redirected to the Netflix app.
We just started using PlayStation Vue. So far we really like it. I looked into Sling TV when we were getting ready to cut the cord, but then I came across PlayStation Vue and it had more channels and more of the channels we as a family wanted. My husband gets a ton of his sports channels, including the Golf channel and we still get Disney Jr. for our son. We also have Netflix and a HD antenna for local channels. PlayStation Vue was only $5 more a month so it was worth it for us. We access it through the Amazon Fire stick.
Fubo TV is a sports-centric service that also offers a number of other channels including local OTA stations (except ABC) -- and more RSNs (regional sports networks) than any other service. Especially for fans of professional baseball, basketball and hockey teams, Fubo might be the only way to watch regular-season games without cable. There's no ESPN, however, and a convoluted user interface and high price mean it's not the first service we'd choose.
Netflix is a great place for binge-watching entire seasons all at once. But unless it's a Netflix original series, you'll just have to wait until a season finishes airing to get started. But hey, no commercials! Accessing the service shouldn't be a problem either. You probably have 10 devices in your house right now that came preloaded with the Netflix app. But if you want to use Netflix on more than one device at once, you'll have to upgrade to the Standard ($10) or Premium ($12) plan.
Well, to figure out the best plan I would start with what shows I primarily watch. Make a list and add who carries them, such as CNN, ABC, NBC, FOX, etc. That should help you sort out which plan has the majority of what you want to watch. Next if you have a smart TV with internet you may not need a streaming device. You can try connecting to the internet from your smart TV without one. Then, if you experience issues you may still need to invest in a streaming device.
Steven, not sure why you're so angry. If you go back through the article, in the options discussed, yes, not all of them are completely free. Some of them have up front costs or costs for equipment when you first start. After that, however, many of them are free or monthly subscription cost free (not all of them).The main one, using an antenna and watching over-the-air television, is something you can do without a recurring monthly cost. If you don't have a TV and antenna up front, yes, you'll have to pay for those. You'll also have to pay for an over-the-air DVR if you want to record programming. But after you pay for those costs there are no monthly costs. Sorry you weren't happy with the article, but there are quite a few options in the article that you can do for free. Best of luck to you, and happy new year!
Netflix: Offer available 1/31 – 4/3/2019 and must be redeemed by 8/31/2019. Valid for 12 months of Netflix service on the Netflix Premium UHD streaming plan for $15.99/month (total value of $191.88) with subscription to eligible Verizon Fios plan. Must maintain qualifying Fios services for 31 days after installation, with no past-due balance to receive a one-time bill credit of $191.88 applicable directly to customer's Netflix account. A Netflix compatible device (manufactured and sold separately) and broadband internet connection are required. 4K Ultra HD availability subject to device capabilities and content availability. Not redeemable or refundable for cash. Value may be applied to a different Netflix streaming plan; exchanges in this manner may alter the duration of the offer. Netflix Service price plans subject to change. Not available to subscribers billed through iTunes or Google play unless subscriber begins a new subscription billed via alternate payment provider or via Netflix.com. See www.netflix.com/termsofuse
Yes, in theory, the higher the dB gain number the better. Although, overall performance is just as important. You must also consider where you live and where the broadcast towers are located. If they are over the visual horizon, a higher number is better. If not, a lower number is ok. You must watch out for high dB numbers that are marketing ploys to get you to pay more for a product you may not need.
As far as what you need to connect your projector to the computer, I'm not sure, it's hard to know exactly without knowing your exact setup. If your computer has a tuner built in you could plug in an antenna like this one to get your over the air signal, and then use some sort of wireless video streaming device to get the video source to your projector.
DIRECTV NOW was DIRECTV's way of keeping its satellite TV service available for users who don't want a dish installed or multi-year contracts. You can use the Just Right package and add HBO (Game of Thrones and Westworld for only $5 more per month instead of $15? Yeah, we're in). Consider also that you can get your favorite networks like HGTV, Sundance TV, and the Travel Channel. Plus, you can also record up to 20 hours of TV to hold onto for 30 days with the included cloud DVR.
ISP's vary by location. I'm maintaining a list of providers with affordable internet only plans you can use to cut the cord. If you can't find one on that page, try your existing provider. Now I know the "big" internet providers seem only to offer "triple play" packages bundling phone, TV, and internet. However, if you dig around on their site you should be able to find an internet service offering.
Investors also pressured media companies to take Netflix's cash. Take, for instance, Time Warner Inc., which is now owned by AT&T Inc. While Disney, CBS, and others licensed many of their old shows to Netflix, Time Warner initially held out. Starting in 2009, Time Warner and Comcast Corp. tried to rally the industry around an idea to slow Netflix by making TV episodes available online—but only to cable subscribers. The idea was called TV Everywhere.
The next time you come across the question, "can you still get local channels without cable?" you will know the answer is one big yes! From indoor TV antennas to outdoor TV antennas to mobile apps, there are quite a few options when it comes to accessing your favorite local channels. Wave goodbye to that cable bill and start watching local channels your way.
Hi, We have been considering giving up our Directv for sometime, Running across this forum has made the decision for us. I see that a Chromecast would be needed per TV that you want to use. My question is, Does a laptop or computer HAVE to be used to do any of this? We have a PS3, So would I be able to use the PS3 for Hulu, Netflix, Playon, and whatever other channels we find? And the other question is, On any of these options, can you watch the show/movie live or do you have to wait a week or so after it airs? We are more worried about our shows like Outlander, Game of Thrones, Big Bang theory, Homeland, etc??
Another approach to cord cutting is an online streaming package similar to traditional cable or satellite TV plans. Several companies now offer variations of this idea: PlayStation Vue from Sony (which works without a PlayStation console, despite the name), Sling TV from Dish Network, DirecTV NOW, Hulu Live TV, YouTube TV (not to be confused with YouTube Red), Philo, and Fubo.tv. Their services cost $16 to $40 per month, for their most basic plans, and they offer largely the same variety of channels as cable but give you a different experience than cable or satellite—one with both benefits and disadvantages.
The upper echelon of antennas starts with the 8 Element Bowtie. It boasts lots of features like a 70-mile range, and is 4K ready out of the box. But where this antenna really shines is in its design, which allows for multi-directional signal pickups. The bracket allows for both sides to be adjusted to catch multiple signals, if possible, which will only help with channel clarity and numbers.
Remember, the cable TV is cheaper than internet for me. I'm more willing to pay $17 a month for cable TV, but not $52 for the internet. There are certain places like Panera Restaurants and public libraries that have free wifi. Places I steer clear of are Borders, Barnes and Noble and Starbucks because they have an agreement with T-Mobile and charge for access.
You probably want amplification, unless you're living next door to the local broadcast tower. They don't make the signal stronger coming in the house; they make an already low signal strong enough for the TV tuner to use. Even some of the flat antennas have amplification options; but amplification ups the cost. Setup is easy, but you'll have to play with the antenna position to maximize reception—just like fiddling with rabbit ear antennas in the 1970s. Some outdoor antennas can work from inside if they're up high—say in your attic—if there isn't a lot of obstruction.
DISH also has the best DVR available. The Hopper 3 Smart DVR lets you record up to 16 shows at once, and you can record 2 ½ times more HD content (500 hours) than the Genie from DIRECTV (200 hours). However, keep in mind that the advertised package price doesn't include the DVR price. You'll have to pay an extra $10 per month for the Hopper and an additional $5–$10 per month for each added receiver.
So, I really appreciate all this information and the clear explanations you've given, as in where to get the channels I do watch and the cost. I really miss 3 local stations, but my favorite one which I've watched all my life and depended on for weather was already dropped from the cable lineup a few months ago. Luckily, I just learned of the new version of "rabbit ears" and have ordered one (2019 version) that claims to have a 120 mile range, which would bring me my lost station if it works. It's gotten a solid 5 star review from many customers on Amazon, so I'm hopeful.
And, as I so often say over on Cordcutting.com, saying goodbye to cable doesn't have to mean saying goodbye to live TV. Everything cable has to offer can be matched by cord cutting alternatives – and that's true even of the local stations that offer you local news, sporting events, and more. Here's how to watch local channels on your TCL Roku TV, no cable required.
DirecTV Now, Hulu Live, YouTube TV, and Playstation Vue all offer local channels without cable, but they are currently only available for specific areas. You will have to check with each service to see if it is available for you. These subscriptions, while usually cheaper than the cable or satellite setup, will be a little more expensive than just using an Antenna. Packages range anywhere from $30 – $80 a month depending on the channels you want to see.
We also recommend testing your internet speed at peak streaming hours (between 6 – 10 p.m. weekdays) to determine if your neighborhood struggles under the strain of heavy traffic. For instance, if you routinely get around 10Mbps downloads during the day, but that figure takes a dive to about 3Mbps around dinner time, you'll want to call your internet provider to see if anything can be done. Fortunately, this is an increasingly rare problem outside of rural areas, but better to check ahead.
If you want a power upgrade from the ClearStream 2V, there's the ClearStream 4. The mantra with this antenna is simple: bigger is better. Because of that, the range is bumped up to 70 miles and the antenna is 4K-ready out of the box. Unfortunately, that will make setup a bit harder because of the bigger size, but given its improved performance over its counterpart, that definitely makes up for it.
We like the Leaf Metro because its small profile easily tucks away, without sacrificing much functionality. Though its range is limited to approximately 25 miles, it's perfect for those living in smaller apartments or rented rooms, especially in urban environments where over-air TV signals are plentiful. To compound the versatility enabled by its tiny size, the antenna comes in either black or white, and you can also paint it to match your interior. Plus, its adhesive coating means it will stick to most any surface and can be moved to other locations with ease. An included 10-foot coaxial cable allows for fairly flexible installation.
Just because you subscribe to cable or satellite doesn't mean you have to use the provided set-top box. In an effort to become part of the cord-cutting landscape, many of the major cable and satellite providers, who double as internet service providers, let users subscribe and then access all programing via apps rather than the cable box or even the tuner in the TV.
Many local libraries have movies and television shows on DVD, and some even offer BluRay. Borrowing one is completely free as long as you are eligible for a library card, and you usually have a generous return window too. The only caveats are that your selection may be limited and other borrowers may not have been kind to the DVD when they borrowed it, so some of them may not work. But, when the cost is $0, it isn't as painful when that happens.
Streaming is hotter than ever these days, with on-demand services such as Netflix, Hulu, and a litany of others, along with multiple live TV streaming services such as Sling TV, DirecTV Now, and PlayStation Vue, all looking to capitalize on the cord-cutting phenomenon. If it's available to see with your eyes and hear with your ears, there's a good chance you can find it on the web — for a fee. Add in free HD broadcasts and there's never been a better time to kick cable to the curb.
In the sports arena, you can use fuboTV to watch NFL, NBA, MLB, NHL, NCAA basketball, NCAA football, soccer, racing, and more on channels including FOX, NBC Sports, NBA TV, NFL Network, and many others. One of the biggest draws of fuboTV is that you get tons of regional sports channels. That means you can watch your home team games without cable on fuboTV — something the other streaming services don't provide without the workaround described below.


text:
INTRA REPOSE
Pictured as: the Defender (Intracoastal Marine) Photo by: Birk Thomas
Built in 1982, by Sherman Shipyard Incorporated of Panama City, Florida (hull #21) as the Captain Red for the STC Company of Wilmington, Delaware.
In 1990, the tug was bare boat chartered to Marine Petroleum Transportation Company of Wilmington, Delaware. Where she retained her name.
In 1996, she was sold and renamed as the Merchants Hope.
In 1999, the tug was acquired by the BayGulf Trading Corporation Norfolk, Virginia. Where she was renamed as the Barret L.
In 2007, the BayGulf Trading Corporation was acquired by the K-Sea Transportation Partnership Incorporated of East Brunswick, New Jersey. Where the tug was renamed as the Davis Sea.
In 2008, the tug was outfitted with an upper wheelhouse, and a JAK-400 coupler system at Feeney's Enterprises of Kingston, New York.
In 2011, the K-Sea Transportation Partnership Incorporated was acquired by the Kirby Corporation of Channelview, Texas. Where the tug retained her name.
In 2018, she was phased out of active service, and was "laid up" at Houston, Texas.
In 2019, the tug was acquired by IntraTug (the Intracoastal Tug and Barge Company) of Maurice, Louisiana. Where she was renamed as the Intra Repose.
In 2020, she was acquired by Intracoastal Marine Incorporated of Chesapeake, Virginia. Where the tug was renamed as the Defender.
Powered by two, Cummins KTA-38 diesel engines, with Twin Disc MG 540 reduction gears. She is a twin screw tug, rated at 2,000 horsepower.
Her electrical service is provided by two, John Deere 4045 generator sets. The tug's capacities are 22,550 gallons of fuel, 330 gallons of lube oil, and 22,550 gallons of water.
Vessel Name: INTRA REPOSE
Ship Builder: SHERMAN SHIPYARD INC
Hailing Port: CHESAPEAKE, VA.
Gross Tonnage: 95
CAPTAIN RED, MERCHANTS HOPE, BARRETT L., DAVIS SEA, DAVIS SEA, INTRA REPOSE
STC Co., MBJ CORPORATION, K-SEA OPERATING PARTNERSHIP L P, K-SEA OPERATING LLC, KIRBY OFFSHORE MARINE OPERATING LLC, KIRBY INLAND MARINE LP, Intracoastal Tug and Barge Co.
Pictured as: the Davis Sea (K-Sea Transportation Partnership Incorporated) Photo by: Brian Fisher
Pictured as: the Davis Sea (K-Sea Transportation Partnership Incorporated) Photo by: Tom Turner
Pictured as: the Davis Sea (K-Sea Transportation Partnership Incorporated) Photo by: Birk Thomas
Pictured as: the Davis Sea (Kirby Corporation) Photo by: John Skelson
Pictured as: the Davis Sea (Kirby Corporation) Photo by: Birk Thomas
Pictured as: the Defender (Intracoastal Marine Incorporated) Photo by: Captain Pat Folan
Photo by: IntraTug Incorporated