text:
\section{Introduction}\label{Sec-1}

The unusual properties of non-trivial scale invariance of matter in the IR regime was first studied by Banks and Zaks \cite{Banks:1981nn}. Georgi suggested \cite{Georgi:2007ek, Georgi:2007si} a coupling between the standard model (SM) fields and this scale invariant sector of a high energy physics described by the Banks-Zaks (BZ) fields. The interactions between the standard model fields and the BZ fields are mediated by the exchange of highly massive particles $M_U$, obeying the scale-supressed non-renormalizable Lagrangian density
\begin{equation}
\mathcal{L}=\frac{1}{M_U^k}\mathcal{O}_{SM}\mathcal{O}_{BZ},
\end{equation}
where $\mathcal{O}$ stands for the field operators of dimension $d_{SM}$ and $d_{BZ}$, while $k=d_{SM}+d_{BZ}-4$ in order to guarantee a dimensionless action. It was argued by Georgi \cite{Georgi:2007si} that below some scale energy limit ($\Lambda_U$) the interactions between the standard model and the hidden sector become stronger. As the scale-invariance is manifested, massive particles are not allowed and this sector was dubbed ``unparticles'', since the notion of mass in this sector is meaningless.

Since the proposal of the unparticle physics several papers were published in particle physics \cite{Aliev:2017bme, Bagheri:2017roa, VanSoa:2018pkb}, condensed matter analogue models \cite{Limtragool:2016ghz, LeBlanc:2014xpa}, extra dimensions \cite{Soa:2019whl} among others. A very interesting work by Gaete and Spallucci \cite{Gaete:2008wg} study the unparticle dynamics in a standard quantum field theory scenario, and they construct effective actions for scalar, gravitation and vector gauge unparticles. Also in cosmological and astrophysical scenarios, several were carried out including studies with black holes \cite{Gaete:2010sp, Mureika:2010je, Lee:2011wc, Alencar:2018vvb} and white dwarfs \cite{Moussa:2021fsb, deSouza:2012ap}. One of these black holes works draws the attention to the fact that the exterior event horizon presents a surface with fractal dimension equals to $d_U$ \cite{Gaete:2010sp}. Recently, some of the present authors generalized the work of \cite{Gaete:2010sp} to rotating black holes and found that the the fractalization is also present\cite{Alencar:2018vvb}. It is also noticeable that a recent paper on Casimir effect with unparticles shows a fractalization in the parallel plates dimension \cite{Frassino:2013lya}. The dimensional reduction seems to be a natural behavior in high energy  and its comprehension in fundamental at this regime scales\cite{Stojkovic:2013xcj,Carlip:2017eud}.
 However, as far as we know, no studies were carried out in order to investigate the influence of the unparticle physics in cosmic strings and black strings.

Objects such as cosmic strings may have been formed due to phase transitions in the early universe \cite{Vilenkin}. The spacetime geometry related to an infinitely long and straight cosmic string is characterized by a planar angle deficit on the two-surface orthogonal to the string. In addition to that, the spacetime is flat, locally, except on the top of the string. Cosmic strings can also be described in the realm of a classical field theory when the energy-momentum tensor related to the vortex configuration of the Higgs-Maxwell system \cite{Nielsen:1973cs} couples to the Einstein equations.

On the other hand, black strings are higher dimensional solutions of the Einstein equations in D-dimensional spacetime \cite{Duff:1987cs}. These vacuum objects can be understood as symmetric generalizations of black holes under translations being the event horizon topologically equivalent to $S_2 \times R$ (or $S_2 \times S_1$ in the case of black rings) and spacetime is asymptotically $M_{D-1} \times S_1$ for a zero cosmological constant or $AdS_{D-1} \times S_1$ for a negative one \cite{Emparan:2001wn, Bellucci:2010gb}. A black string is a particular case of a black p-brane solution (with $p = 1$), where $p$ is the number of extra spatial dimensions to ordinary space. Hence a p-brane gives rise to a (p + 1)-dimensional world-volume in spacetime. In string theory, a black string is described by a D1-brane surrounded by a horizon. It is widely known that since an open string propagates through spacetime, it is necessary that its endpoints to lie on a D2-brane on which it satisfies Dirichlet boundary condition, and the dynamics on the D-brane worldvolume is described by a gauge field theory \cite{Polchinski:1995mt, Polchinski:1994fq}. In some models our universe is understood as a higher dimensional world-brane, thus the gravitational collapse of galactic matter would produce black holes lying on the brane. Black string solutions are nevertheless unstable to long wavelength perturbations (at least in asymptotically flat spaces) since the localized black hole is entropically preferred to a long segment of string. The string's horizon therefore has a tendency to form a line of black holes. However, in AdS the space acts like a confining box which prevents fluctuations of long wavelengths from developing. For instance, in a Randall Sundrum domain wall in a five dimensional AdS spacetime the black string is stable at least far from the AdS horizon. Near the horizon, the string is likely to pinch off to become a stable shorter cigar-like singularity \cite{Chamblin:1999by}.

In this paper we study the unparticle corrections for cosmic and black strings. This paper is organized as follows: In the next section we present a brief review of the unparticle theory. In section III we study the unparticle corrections for cosmic strings and in section IV we obtain the black string solution in the unparticle scenario. In section V we study the thermodynamic properties of black strings in the unparticle physics domain and in section VI we draw our conclusions.

\section{Unparticle Theory}\label{Sec-2}
The un-gravity action is \cite{Gaete:2010sp}
\begin{eqnarray}\label{1}
    S_u = \frac{1}{2\kappa^2} \int d^4x \sqrt{-g} \left[1 + \frac{A_{d_U}}{(2d_U -1)\sin{(\pi d_U)}}
    \frac{\kappa_{*}^{2}}{\kappa^{2}}\left(-\frac{D^2}{\Lambda^2_U}\right)^{1-d_U}\right]^{-1} \mathcal{R},
\end{eqnarray}
where $\kappa^2 = 8\pi G$, $\kappa_{*}^2$ is represents the un-gravitational Newton constant, and $ A_{d_U} = 16\pi^{5/2} \Gamma(d_U + 1/2)\left[(2\pi)^{2d_U} \Gamma(2d_U) \Gamma(d_U-1)\right]^{-1} $  \cite{Gaete:2010sp}, and $D^2$ is the generally covariant D'alambertian. The Einstein Field Equations (EFE) is given by
\begin{eqnarray}\label{2}
    R^\mu_\nu - \frac{1}{2} \delta^\mu_\nu \mathcal{R} &=& \kappa^2 T^\mu_\nu+ \frac{\kappa_*^2A_{d_U}}{\sin{(\pi d_U)}}\frac{\Lambda_U^{2 - 2d_U}}{(2d_U -1)}\left(-\frac{1}{D^2}\right)^{1-d_U} T^\mu_\nu \nonumber \\
    &=& \kappa^2 T^\mu_\nu + \frac{\kappa_*^2A_{d_U}}{\sin{(\pi d_U)}} {T_U}^\mu_\nu
\end{eqnarray}
The second term of right side of EFE is the unparticle correction of energy-momentum tensor. In Ref. \cite{Mureika:2010je} it was found an exact solution of EFE for vector particles, and this solution is Reissner-Nordstrom like. In this paper, we'll consider an axially symmetric \textit{ansatz} for the metric, $(g_{\mu\nu})=\text{diag}\left(g_{tt}(r),g_{rr}(r),g_{\phi\phi}(r),g_{zz}(r)\right)$, since we are interested in studying solutions of Cosmic String and Black String with unparticle corrections. Moreover, we adopt the metric's signature $(-1,1,1,1)$, so  we consider the energy-momentum tensor as $(T^\mu_\nu) = \text{diag}(-\rho,p,p_\phi,p_z)$, which can be represented by
\begin{eqnarray}\label{energy-momentum}
    T^\mu_\nu = (\rho + p_\phi) u^\mu u_\nu + (p-p_\phi)w^\mu w_\nu+ (p_z-p_\phi)Z^\mu Z_\nu + p_\phi~\delta^\mu_\nu,
\end{eqnarray}
where $(u_\mu) = \left(\sqrt{-g_{tt}},0,0,0\right)$ is the four-velocity field, $(w_\mu) = \left(0,\sqrt{g_{rr}},0,0\right)$ and $(Z_\mu) = \left(0,0,0,\sqrt{g_{zz}}\right)$ are  units spacelike vectors.

Let us consider a static string which has a density of energy $\rho(r) =\mu(2\pi r)^{-1} \delta(r)= \mu \delta(x)\delta(y)$, where $\mu$ is the linear density of mass. We can compute the contribution of unparticle in equation (\ref{2}) by calculating the term $T_U$. We achieve that by switching to isotropic (free-falling), Cartesian-like coordinate for computational convenience,
\begin{eqnarray}\label{6}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 &=& - \mu \left(-\frac{1}{\nabla_{\vec{X}}^2}\right)^{1-d_U} \delta \left(\vec{X}\right) \nonumber\\
    &=& - \mu \left(-\frac{1}{\nabla_{\vec{X}}^2}\right)^{1-d_U} \frac{1}{(2\pi)^2}\int d^2k e^{i \vec{k}\cdot\vec{X}} \nonumber \\
    &=&- \mu \frac{1}{(2\pi)^2}\int d^2k \left(\frac{1}{k^2}\right)^{1-d_U} e^{i \vec{k}\cdot\vec{X}},
\end{eqnarray}
where $\vec{X}$ is a vector in a direction perpendicular to the string. Hence, by applying the Schwinger representation for $(1/k^2)^{1-d_U}$, we obtain
\begin{eqnarray}\label{7}
    \left(\frac{1}{k^2}\right)^{1-d_U} = \frac{1}{\Gamma(1-d_U)}\int_0^\infty ds s^{-d_U} e^{-sk^2}.
\end{eqnarray}
Yielding for the equation (\ref{6})
\begin{eqnarray}\label{8}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 = - \mu \frac{1}{(2\pi)^2}\int d^2k \left[\frac{1}{\Gamma(1-d_U)}\int_0^\infty ds s^{-d_U} e^{-sk^2}\right] e^{i \vec{k}\cdot\vec{X}}.
\end{eqnarray}
Finally, when we compute this integration, we find
\begin{eqnarray}\label{9}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 = - \mu \frac{2^{2d_U - 2}}{\pi} \frac{\Gamma(d_U)}{\Gamma(1-d_U)} \left(\frac{1}{\vec{X}^2}\right)^{d_U},
\end{eqnarray}
which allow us to write a proper expression for the energy density of a static string with unparticle as
\begin{eqnarray}\label{density0}
    \rho(r) = \frac{\mu}{2\pi}\frac{\delta(r)}{r} + \mu\left(\frac{\kappa^{2}_{*}}{\kappa^2}\right) \frac{ A_{d_U}}{\sin{\pi d_U}}\frac{\Lambda^{2-2d_U}_U}{(2d_U-1)} \frac{2^{2d_U-2}}{\pi}\frac{\Gamma(d_U)}{\Gamma(1-d_U)} \left(\frac{1}{r}\right)^{2d_U}.
\end{eqnarray}
When we replace $ A_{d_U} = 16\pi^{5/2} \Gamma(d_U + 1/2)\left[(2\pi)^{2d_U} \Gamma(2d_U) \Gamma(d_U-1)\right]^{-1}$ in equation (\ref{density0}), we find
\begin{eqnarray}\label{density}
    \rho(r) = \frac{\mu}{2\pi}\left[\frac{\delta(r)}{r}+\frac{R^{2d_U-2}}{r^{2d_U}}\right],
\end{eqnarray}
where
\begin{eqnarray}\label{unradiu}
    R = \left[\left(\frac{\kappa_*^2}{\kappa^2}\right)\frac{8\Lambda_U^{2 - 2d_U}}{\sin{(\pi d_U)}(2d_U -1)\pi^{2d_U-5/2}}\frac{ \Gamma(d_U)\Gamma(d_U + 1/2)}{\Gamma(1-d_U) \Gamma(2d_U) \Gamma(d_U-1)}\right]^\frac{1}{2d_U-2}.
\end{eqnarray}

Also it is important to highlight that the $R$ parameter sets a proper unparticle length scale so that, similarly to the authors in \cite{Gaete:2010sp, Alencar:2018vvb}, it is possible to distinguish between two regimes. The gravity dominated (GD) regime is defined by $R<<r$ and the ungravity dominated (UGD) regime by $R>>r$. Note that in the GD regime the second term of (\ref{density}) vanishes and we recover the usual cosmic string energy density.

\begin{figure}[h]
    \centering
    \includegraphics[width=10cm]{unparticleR-plot.pdf}
    \caption{Behavior of $R$ for different $\Lambda_U$ as function of $d_U$, with $(\kappa_{*}/\kappa)^2=0.5$}
    \label{fig-beta-constant}
\end{figure}

In the next section, we will present the solution of static cosmic string with  unparticle corrections.

\section{Cosmic String with unparticle corrections}
 In this section, we will study the spacetime solution which is static and has cylindrical symmetry. Consider the \textit{ansatz} of the line-element as follows
 \begin{eqnarray}\label{3}
     ds^2 = -dt^2 + dr^2 + b(r)^2 d\phi^2 + dz^2.
 \end{eqnarray}
 The components of the Ricci tensor are $R_{\mu\nu}=R^\alpha_{\mu\alpha\nu} = \partial_{[\alpha}\Gamma^{\alpha}_{\nu]\mu}+ \Gamma^{\alpha}_{\beta[\alpha}\Gamma^{\beta}_{\nu]\mu}$, where $V_{[\mu\nu]}= V_{\mu\nu}-V_{\nu\mu}$. The Einstein Tensor is
 \begin{eqnarray}\label{4}
     G^\mu_\nu = \frac{1}{b(r)}\frac{d^2b(r)}{dr^2} ~\text{diag}\left(1,0,0,1\right).
 \end{eqnarray}
We know that the $T^0_0 = - \rho(r)$, so we can see by EFE $G^\mu_\nu = \kappa^2 T^\mu_\nu$ that $T^\mu_\nu = -\rho(r) \text{diag}(1,0,0,1)$. Then the EFE resumes to
\begin{eqnarray}\label{5}
    \frac{1}{b(r)}\frac{d^2b(r)}{dr^2} = - \kappa^2 \rho(r).
\end{eqnarray}
This equation can be rewritten as
\begin{eqnarray}\label{12}
    \frac{d^2b(r)}{dr^2} +\frac{\mu\kappa^2}{2\pi}\left[ \frac{\delta(r)}{r} + R^{2d_U-2} r^{-2d_U}\right] b(r) = 0.
\end{eqnarray}

Now, we need to properly solve the equation (\ref{12}). Let us perform the following variable transformation $b(r) = \sqrt{r}f(r)$, such that the equation (\ref{12}) becomes
\begin{eqnarray}\label{13}
    \left\{\frac{d^2f}{dr^2}+ \frac{1}{r}\frac{df}{dr}+\left[\frac{\mu\kappa^2}{2\pi} \frac{\delta(r)}{r} + \frac{\mu\kappa^2}{2\pi}R^{2d_U-2}~ r^{-2d_U}- \frac{1}{4r^2}\right]f\right\} \sqrt{r} = 0.
\end{eqnarray}
First, we need to solve this equation for $r>0$. After we impose that when $R/r \rightarrow 0$ we must find cosmic string spacetime. When we consider $r>0$ the equation (\ref{13}) simplifies to
\begin{eqnarray}\label{14}
    \frac{d^2f}{dr^2}+ \frac{1}{r}\frac{df}{dr}+\left[\frac{\mu\kappa^2}{2\pi}R^{2d_U-2} r^{-2d_U}- \frac{1}{4r^2}\right]f = 0.
\end{eqnarray}
Now, we can perform a transformation of variable $ x = \sqrt{\frac{\mu\kappa^2}{2\pi}}\frac{R^{d_U-1}}{\left(d_U-1\right)} r^{1-d_U}$, allowing the equation (\ref{14}) to be written as
\begin{eqnarray}\label{15}
    \frac{d^2f}{dx^2}+ \frac{1}{x}\frac{df}{dx}+\left[1- \frac{\left(\frac{1}{2du-2}\right)^2}{x^2}\right]f = 0.
\end{eqnarray}
This is the Bessel differential equation. The solution of equation (\ref{15}) depends on $\left(\frac{1}{2du-2}\right)$. When we define $\xi \equiv 1/(2du-2)$ as a positive real number the equation (\ref{15}) becomes
\begin{eqnarray}\label{16}
    \frac{d^2f}{dx^2}+ \frac{1}{x}\frac{df}{dx}+\left[1- \frac{\xi^2}{x^2}\right]f = 0,
\end{eqnarray}
whose solution is
\begin{eqnarray}\label{17}
    f(x) = C_1 J_\xi (x) + C_2 N_\xi (x).
\end{eqnarray}
Here $J_\xi(x)$ and $N_\xi(x)$ are the first and second kind Bessel function of order $\xi$, respectively.
Let us present some considerations regarding the boundary conditions. Initially, we want that when $R/r \rightarrow 0$, $b(r) = (1-8G\mu)^{1/2}r$, that is, the spacetime becomes the cosmic string spacetime. Since $x$ is proportional to $(R/r)^{d_U-1}$ and $b(r)=\sqrt{r}f(x(r))$, we can study the equation (\ref{17}) when $x \ll 1$. In regime $x \ll 1$ and $\xi > 0$, equation (\ref{17}) can be written as
\begin{eqnarray}\label{18}
    f(x)= C_1\frac{1}{\Gamma{(\xi+1)}}\left(\frac{x}{2}\right)^\xi - C_2 \frac{\Gamma(\xi)}{\pi}\left(\frac{x}{2}\right)^{-\xi}.
\end{eqnarray}
Replacing $ x = \sqrt{\frac{\mu\kappa^2}{2\pi}}\frac{R^{d_U-1}}{\left(d_U-1\right)} r^{1-d_U}$ in equation (\ref{18}), when we impose the limit $R/r \ll 1$ we shall find $f(r) = (1-8G\mu)^{1/2}\sqrt{r}$. Hence $C_1 = 0$ and
\begin{eqnarray*}
C_2 = -(1-8G\mu)^{1/2}\frac{\pi}{\Gamma{(\xi)}}\left(\xi \sqrt{\frac{\kappa^2\mu}{2\pi}}\right)^\xi R^{1/2}.
\end{eqnarray*}
Consequently, the component $g_{\phi\phi}$ of the metric is
\begin{eqnarray}\label{21}
    b^2(r) = (1-8G\mu)r^2\left[\frac{\pi^2}{\Gamma^2{(\xi)}}\xi^{2\xi}\left( 4G\mu\right)^{\xi} \frac{R}{r}\right]  {N_\xi}^2 \left(4\xi\sqrt{G\mu}\left(\frac{R}{r}\right)^{\frac{1}{2\xi}}\right).
\end{eqnarray}
The metric component $g_{\phi\phi} = b^2(r)$, the behavior of this component is given at Figure \ref{metric-component-phi}.
\begin{figure}[h]
    \centering
    \includegraphics[width=10cm]{gPhiPhi-plot.pdf}
    \caption{Metric component $g_{\phi\phi}$ of the unparticle static cosmic string as a function of the radial coordinate $r$ for some values of $d_U$ , with $\Lambda_U = 10$, $\mu = 0.7$, $\kappa_* = 10^3$}
    \label{metric-component-phi}
\end{figure}

In order to understand the structure of this spacetime, that is, how the unparticles modify the cosmic string spacetime, We must analyze the curvature scalars. Ricci's scalar for the ansatz $ds^2 = -dt^2 + dr^2 + b^2(r)d\phi^2 + dz^2$ is $\mathcal{R} = - \frac{2}{b}\frac{d^2b}{dr^2}$. Then Ricci's scalar for our solution is given by
\begin{eqnarray}\label{22}
    \mathcal{R}=\frac{\mu\kappa^2}{\pi}\left[ \frac{\delta(r)}{r} + \frac{1}{r^2}\left(\frac{R}{r}\right)^{2d_U-2}\right],
\end{eqnarray}
where we used equation (\ref{13}).

\section{Black string with unparticle corrections}

Let us consider the following line element for the black string
\begin{eqnarray}\label{23}
    ds^2=-f(r)dt^2+\frac{1}{f(r)}dr^2+r^2d\phi^2+\alpha^2r^2dz^2,
\end{eqnarray}
where $t\in(-\infty, \infty)$, the radial coordinate $r\in[0,\infty)$, the angular coordinate $\phi\in[0,2\pi)$ and the axial coordinate $z\in(-\infty,\infty)$. The $\alpha$ parameter is considered as $\alpha^2=-\Lambda/3$.

For the black string in the un-gravity scenario, the Einstein-Hilbert effective action is composed by (\ref{1}) added by the cosmological constant contribution, i.e.,
\begin{eqnarray}\label{24}
    S_u = \frac{1}{2\kappa^2} \int d^4x \sqrt{-g} \left\{\left[1 + \frac{A_{d_U}}{(2d_U -1)\sin{(\pi d_U)}}
    \frac{\kappa_{*}^{2}}{\kappa^{2}}\left(-\frac{D^2}{\Lambda^2_U}\right)^{1-d_U}\right]^{-1} \mathcal{R}-2\Lambda\right\}.
\end{eqnarray}
The Einstein equation in (\ref{2}) remains valid, however the left-hand-side will be modified due to the different ansatz for the metric in (\ref{23}). The EFE for this ansatz are
\begin{eqnarray}
    G^t_t -3\alpha^2 &=& \frac{1}{r} \frac{df(r)}{d r}  + \frac{f(r)}{r^2} - 3\alpha^2, \label{25}\\
    G^r_r -3\alpha^2 &=& \frac{1}{r} \frac{df(r)}{d r}  + \frac{f(r)}{r^2} - 3\alpha^2, \label{26}\\
    G^\phi_\phi -3\alpha^2 &=& \frac{1}{2}\frac{d^{2}f(r)}{d r^{2}}+ \frac{1}{r} \frac{df(r)}{d r} - 3\alpha^2, \label{27}\\
    G^z_z -3\alpha^2 &=&\frac{1}{2}\frac{d^{2}f(r)}{d r^{2}}+ \frac{1}{r} \frac{df(r)}{d r} - 3\alpha^2.\label{28}
\end{eqnarray}
We can see that the energy-momentum tensor for the ansatz of equation (\ref{23}) is $T^\mu_\nu = -\rho(r)\text{ diag}(1,1,0,0) + p_l(r)\text{ diag}(0,0,1,1)$, where $p_l = p_\phi = p_z$. This way we can find $f(r)$ by solving $G^t_t -3\alpha^2 = -\kappa^2 \rho(r)$, we find

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{f_r_1.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{f_r_2.eps}
  \caption{}

\end{subfigure}%
    \caption{Plot of $f(r)$ (scalar case (a) and vector case (b)) for some values of $d_U$ with $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$}
    \label{f(r)plot}
\end{figure*}

\begin{eqnarray}\label{29}
    f(r) = \alpha^2r^2 - \frac{4\mu}{\alpha r} \pm \frac{\kappa^2\mu}{2\pi(2d_U-3)}\left(\frac{R}{r}\right)^{2d_U-2},
\end{eqnarray}
with $R$ given by equation (\ref{unradiu}). Note that the unparticle black string solution is divergent for $d_U=3/2$. The plus signal is taken for the scalar unparticle case, while the minus one is for the vector case.

The above solution for the black string in the ungravity scenario recovers the usual black string solution \cite{Lemos:1994xp} in the GD regime, i.e.,
\begin{eqnarray}\label{31}
    f(r)=\left(\alpha^2r^2-\frac{4\mu}{\alpha r}\right).
\end{eqnarray}

Also, it is important to highlight here that the unparticle black string may exhibit naked singularity, violating therefore the cosmic censorship, for the scalar case when $3/2<d_U<2$.

\section{Thermodynamic properties of the Black String with unparticle corrections}
To make easy the calculations, we can rewrite the equation (\ref{29}) as $f(r) = h(r) + \mu s(r)$, where
\begin{eqnarray}
    h(r) &=& \alpha^2 r^2, \\
    s(r) &=& \frac{4}{\alpha r}\left[-1 \pm \frac{\alpha r \kappa^2}{8\pi (2d_U-3)}\left(\frac{R}{r}\right)^{2d_U -2}\right].
\end{eqnarray}

Our black string solution in the un-gravity scenario has the horizon curves defined by $f(r_h)=0$, so $\mu = - \frac{h(r_h)}{s(r_h)}$, which gives us
\begin{eqnarray}\label{32}
    \mu= -\frac{\alpha^3r_h^3}{4}\left[-1\pm\frac{\alpha r_h \kappa^2}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]^{-1}.
\end{eqnarray}

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{mu_scalar.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{mu_vectorial.eps}
  \caption{}

\end{subfigure}%
    \caption{Linear mass parameter $\mu$ as a function of $d_U$ for three values of $r$. In (a) we are considering the scalar unparticle case while in (b) we have the vectorial unparticle case. We have considered $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$.}
    \label{mu}
\end{figure*}

In the GD regime, the mass parameter $\mu$ reduces to $\mu=\frac{\alpha^3}{4}r_h^3$, which is the usual mass parameter for the black-string in General Relativity. In \cite{Alencar:2018vvb} the authors have found that for an unparticle black hole the mass parameter becomes negative for the vector unparticle case. For the unparticle black string, as we can see in fig.(\ref{mu}), both vector and scalar cases present the possibility of negative mass parameter. For the scalar unparticle case we have a negative mass parameter for values of $3/2<d_U<2$. For the vector unparticle case the mass parameter becomes negative for $1<d_U<3/2$.

The authors in \cite{Mureika:2010je} does not take into consideration the regions where the mass parameter is negative. However it is known that a negative mass parameter in a black hole solution suggests that it has a nontrivial topology \cite{Mann:1997jb}. Therefore we can expect that these negative mass regions are also related to nontrivial topologies of the the black string.

In possession of the solution for the static black string in the ungravity scenario given by (\ref{29}), we are able to study the thermodynamics of the black string by computing the Hawking's temperature by means of $T_H=\frac{1}{4\pi}f'(r_h)$, where $r_h$ is the radius of the horizon. Note that in terms of $h(r)$ and $s(r)$ we find
\begin{eqnarray}\label{temp}
    T_H &=& \frac{1}{4\pi}\left[h'(r_h)+ \mu ~s'(r_h)\right] \nonumber \\
     &=& \frac{s(r_h)}{4\pi}\frac{d}{dr_h}\left(\frac{h(r_h)}{s(r_h)}\right),
\end{eqnarray}
where we used $\mu = -\frac{h(r_h)}{s(r_h)}$. When we substitute the expressions for $h(r_h)$ and $s(r_h)$, we find
\begin{eqnarray}\label{33}
    T_H = \frac{3\alpha^2 r_h}{4\pi}\frac{\left[-1\pm\frac{d_U\alpha r_h\kappa^2}{12\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]}{\left[-1\pm\frac{\alpha r_h\kappa^2}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]}.
\end{eqnarray}
The usual black string Hawking temperature, i.e. $T_H=\frac{3 \alpha ^2 }{4 \pi }r_h$, is recovered in the GD regime and in UGD regime, we find
\begin{eqnarray}\label{33-1}
    T_H = \frac{\alpha^2 2d_U}{4\pi}r_h,
\end{eqnarray}
for both GD and UGD regimes. The Hawking temperature for the unparticle black string presents a similar behaviour for both scalar and vector unparticle cases. The behaviour of the Hawking temperature as a function of $r$ for the scalar unparticle case for some values of $d_U$ is presented in figure (\ref{HawkingTemperature}). We can see from figure (\ref{HawkingTemperature}) that when $1<d_U<3/2$ the temperature increases slowly with the radius in comparison with the usual black string temperature. When $3/2<d_U<2$ the unparticle correction promotes an increasing in the temperature in comparison with the black string usual case.

\begin{figure}[h!]
    \centering
    \includegraphics[scale=0.8]{T__H_s.eps}
    \caption{Hawking temperature as a function of $r$ for the scalar unparticle case for three values of $d_U$. We have considered $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$.}
    \label{HawkingTemperature}
\end{figure}

Note that the temperature is aproximally proportional to $D-1$, where $D$ is the dimension of the spacetime. So, when we confront the equation (\ref{33-1}) with this fact, we can see the fractalization of the event horizon with
\begin{eqnarray}\label{33.2}
 d_H=2d_U-1 ,
\end{eqnarray}
where $d_H = D - 2$. To reinforce the fractalization of the event horizon, we can compute the density of entropy by length $\mathcal{S}$ althrough  $d\mathcal{S} = d\mu/T_H$, which in terms of $h(r)$ and $s(r)$,
\begin{eqnarray}
    d\mathcal{S} = - \frac{4\pi}{s(r_h)} dr_h,
\end{eqnarray}
where we used equation (\ref{temp}) and again $\mu = - \frac{h(r_h)}{s(r_h)}$. So we find
\begin{eqnarray}\label{differentialOfEntropy}
    d\mathcal{S} = \frac{\pi\alpha r_h ~dr_h}{\left[-1 \pm \frac{\alpha R \kappa^2}{8\pi (2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U -3}\right]}.
\end{eqnarray}
Now, we want to find analytically the entropy by length of the black string with unparticle. It can be done by integration of equation (\ref{differentialOfEntropy}), but it is necessary to analyze the regions due to the possible singularity of this equation. Up to now we have discussed the scalar and vector cases simultaneously, but now, for the sake of simplicity we will consider them separately.

\subsection{scalar case:}

Let us begin by considering the case when $d_U<3/2$. In this case we can rewrite the equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{differentialOfEntropyDSmallerThan3/2}
    d\mathcal{S} = \frac{\pi\alpha r_h ~dr_h}{\left[1 + \frac{\alpha R \kappa^2}{8\pi (3-2d_U )}\left(\frac{R}{r_h}\right)^{2d_U -3}\right]}.
\end{eqnarray}
So, in case $d_U<3/2$ there is not singularity. By using the identity
\begin{eqnarray}\label{identity}
\int \frac{x dx}{1\pm Ax^\gamma} = \frac{x^2}{2}\hyper{1}{\frac{2}{\gamma}}{1+\frac{2}{\gamma}}{\mp Ax^\gamma} + \text{const.},
\end{eqnarray}
where $\hyper{a}{b}{c}{x}$ is the hypergeometric function and $\gamma>0$, the integration of equation (\ref{differentialOfEntropyDSmallerThan3/2}) can be written as
\begin{eqnarray}\label{entropy1}
    S = \frac{2\pi\alpha {r_h}^2}{4}\hyper{1}{\frac{2}{3-d_U}}{1+\frac{2}{3-d_U}}{- \frac{\alpha R \kappa^2}{8\pi (3-2d_U )}\left(\frac{R}{r_h}\right)^{2d_U -3}}.
\end{eqnarray}
Note that, in GD regime ($R\rightarrow 0$) the argument of hypergeometric funcion is proportional to $R^{2d_U-2}$, which approaches to $0$ when $R \rightarrow 0$. It is, the entropy in GD recovers  $S = \frac{2\pi\alpha {r_H}^2}{4}$ as shall be.

The heat capacity and the free energy of the system are defined by
\begin{eqnarray}
    \label{heat}C_V&=&\frac{d\mu}{dT_H}\\
    \label{free}F&=&\mu-T_H S.
\end{eqnarray}
The heat capacity can be obtained directly from the equations (\ref{32}) and (\ref{33}), which yields
\begin{eqnarray}
    C_V=\frac{8\pi^2(-3+2d_U)\alpha(r_h R)^2\left[12(-3+2d_U)\pi R^2-d_U\kappa^2\alpha R^3\left(\frac{R}{r_h}\right)^{2d_U-3}\right]}{96(3-2d_U)^2\pi^2R^4+8\pi(-9+27d_U-20d_U^2+4d_U^3)\kappa^2r_h^3R^2\alpha\left(\frac{R}{r_h}\right)^{2d_U}+d_U\alpha^2\kappa^4r_h^6\left(\frac{R}{r_h}\right)^{4d_U}}.
\end{eqnarray}
The free energy can also be obtained from (\ref{32}), (\ref{33}) and (\ref{entropy1}), yielding
\begin{eqnarray}
    \nonumber F&=&-\frac{\alpha^3 r_h^3}{4 \alpha  \kappa^2 r_h^{3-2 d_U} R^{2 d_U}+32 \pi  (3-2 d_U) R^2} \left[8 \pi  (2 d_U-3) R^2\right.\\
    &&+\left.\left(\alpha d_U \kappa^2 r_h^{3} \left(\frac{R}{r_h}\right)^{2 d_U}+12 \pi  (3-2 d_U) R^2\right) \, _2F_1\left(1,\frac{2}{3-2 d_U};1+\frac{2}{3-2 d_U};-\frac{\alpha  \kappa^2 r_h^{3-2 d_U} R^{2 d_U-2}}{24 \pi -16 \pi  d_U}\right)\right].
\end{eqnarray}

The behaviour of the Hawking temperature, entropy, heat capacity and free energy for $d_U<3/2$ is depicted in figure (\ref{therm_quant_1}).

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{therm_quant_1.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{therm_quant_3.eps}
  \caption{}

\end{subfigure}%
    \caption{Hawking temperature, entropy, heat capacity and free energy for $d_U=1.4$ (a) and for $d_U=1.6$ (b)}
    \label{therm_quant_1}
\end{figure*}

Now, we need to investigate the case $d_U>3/2$. In this case, there is a singularity in equation (\ref{differentialOfEntropy}) when
\begin{eqnarray}\label{singularityForr_H}
    r_0 = \left[\frac{\alpha R \kappa^2}{8\pi (2d_U-3)}\right]^{\frac{1}{2d_U -3}}R,
\end{eqnarray}
It is easy to see that we have two disjunte regions and when $r_h >r_0$ $\left(r_h <r_0\right)$  the differential of entropy by length, equation (\ref{differentialOfEntropy}), approaches infinity (minus infinity). First, let us compute the entropy for $r_h >r_0$.We can rewrite equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{differentialOfentropyOfDLargerThan3/2}
    d\mathcal{S} = \frac{\alpha \pi r_hdr_h}{\left[1 -\left(\frac{r_0}{r_h}\right)^{2d_U-3}\right]}.
\end{eqnarray}
We can make a changing of variable $z = (r_o/r_h)^{2d_U-3}$, equation (\ref{differentialOfentropyOfDLargerThan3/2}) becomes
\begin{eqnarray}\label{differentialOfentropyOfDLargerThan3/2-withz}
    d\mathcal{S} = -\frac{\alpha \pi{r_0}^2}{(2d_U-3)}~\frac{ z^{-\left(\frac{2d_U-1}{2d_U-3}\right)}dz}{1 -z}.
\end{eqnarray}
We can use the identity
\begin{eqnarray}\label{identity2}
    \int \frac{x^{\pm c}dx}{1-x} =  \frac{x^{1\pm c}}{1\pm c}~\hyper{1}{1\pm c}{1+(1\pm c)}{x} + \text{const.},
\end{eqnarray}
where $c>0$. So, when we integrate the equation (\ref{differentialOfentropyOfDLargerThan3/2}), we find
\begin{eqnarray}\label{entropyOfDLargerThan3/2-1}
\mathcal{S} = \frac{2\pi \alpha r_h^2}{4}~\hyper{1}{\frac{2}{3-2d_U}}{1+\frac{2}{3-2d_U}}{\left(\frac{r_0}{r_h}\right)^{2d_U-3}}.
\end{eqnarray}
Note that equation (\ref{entropyOfDLargerThan3/2-1}) recovers $\mathcal{S} = \frac{2\pi \alpha r_h^2}{4}$ in GD regime. The ultimate case is when $r_h <r_0$. We can rewrite equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{changingOfVariable2}
    d\mathcal{S} =-\alpha \pi \left(\frac{r_h}{r_0}\right)^{2d_U-3}\frac{ r_h dr_h}{\left[1-\left(\frac{r_h}{r_0}\right)^{2d_U-3}\right]}.
\end{eqnarray}
When we make a changing of variable $z = (r_h/r_0)^{2d_U-3}$, equation (\ref{changingOfVariable2}) becomes
\begin{eqnarray}\label{changingOfVariable2-3}
    d\mathcal{S} =-\frac{\alpha \pi {r_0}^2}{(2d_U-3)}~ \frac{ z^{\frac{2}{2d_U-3}} dz}{1-z}.
\end{eqnarray}

So, when we integrate the equation (\ref{changingOfVariable2-3}) by using equation (\ref{identity2}), we find
\begin{eqnarray}\label{entropyOfDLargerThan3/2-2}
    \mathcal{S} = - \frac{\pi \alpha {r_0}^2}{(2d_U-1)}\left(\frac{r_h}{r_0}\right)^{2d_U-1} \hyper{1}{\frac{2d_U-1}{2d_U-3}}{\frac{4(d_U-1)}{2d_U-3}}{\left(\frac{r_h}{r_0}\right)^{2d_U-3}}
\end{eqnarray}
Note that in UGD regime the depencence of $r_h$ in equation (\ref{entropyOfDLargerThan3/2-2}) becomes $r_h^{d_h} = r_h^{2d_U-1}$. Then $d_h = 2d_U -1 $, which reinforces the fractalization and is in agreement with equation (\ref{33-1}).

The heat capacity can be obtained straightforwardly from the equations (\ref{32}) and (\ref{33}), yielding
\begin{eqnarray}
    C_v=\frac{8 \pi ^2 \alpha  (2 d_U-3) R^2 r_h^{2 d_U+2} \left(12 \pi  (2 d_U-3) R^2 r_h^{2 d_U}-\alpha  d \kappa^2 r_h^3 R^{2 d_U}\right)}{\alpha ^2 d \kappa^4 r_h^6 R^{4 d_U}+8 \pi  \alpha  (d_U-3) (2 d_U-3) (2 d_U-1) \kappa^2 r_h^3 R^2 (r_h R)^{2 d_U}+96 \pi ^2 (3-2 d_U)^2 R^4 r_h^{4 d_U}}.
\end{eqnarray}

The free energy can also be obtained from (\ref{32}), (\ref{33}) and (\ref{entropy1}), from which follows that

\begin{eqnarray}
    F=\frac{\alpha ^3 r_h \left(r_0^2 B_{\left(\frac{r}{r_0}\right)^{2 d_U-3}}\left(1+\frac{2}{2 d_U-3},0\right) \left(\alpha  d \kappa^2 r_h^3 R^{2 d_U}-12 \pi  (2 d_U-3) R^2 r_h^{2 d_U}\right)+4 \pi  (3-2 d_U)^2 R^2 r_h^{2 d_U+2}\right)}{2 (2 d_U-3) \left(8 \pi  (2 d_U-3) R^2 r_h^{2 d}-\alpha  \kappa^2 r_h^3 R^{2 d_U}\right)},
\end{eqnarray}
where $B_{z}\left(a,b\right)$ is the incomplete Euler Beta function.

As it is widely known, the thermodynamical stability of black holes (black strings for our case) is directly related to the sign of the heat capacity. A positive heat capacity indicates that the system is thermodynamically stable, while its negativity imply a thermodynamical instability. As we can see in figure (\ref{therm_quant_1}), the value of the $d_U$ controls the black string stability, being the black string stable in the ungravity regime for $d_U<3/2$ and unstable for $d_U>3/2$.

\subsection{vector case:}

Let us consider the case when $d_U>3/2$, which for the vector case avoid the singularity, so that the equation (\ref{differentialOfEntropy}) can be written as
\begin{eqnarray}\label{dsvector}
    d\mathcal{S}=\frac{\pi\alpha r_hdr_h}{\left[1+\frac{\alpha\kappa^2 R}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-3}\right]}.
\end{eqnarray}
We can clearly see that, in virtue of the similarity between the expressions (\ref{dsvector}) and (\ref{differentialOfEntropyDSmallerThan3/2}), the same analysis performed for the scalar case when $d_U<3/2$ can be done for the vector case when $d_U>3/2$. Also the results for the scalar case when $d_U>3/2$ are identical to those of the vector case when $d_U<3/2$.

Therefore, the thermodynamic quantities depicted in figure (\ref{therm_quant_1}a) for the scalar case when $d_U<3/2$ also describe the thermodynamic quantities for the vector case when $d_U>3/2$. Likewise, the thermodynamic quantities depicted in figure (\ref{therm_quant_1}b) for the scalar case when $d_U>3/2$ are also describing the thermodynamic quantities for the vector case when $d_U<3/2$.

\section{Conclusion}\label{Sec-6}

In this paper we study the unparticle corrections for cosmic and black strings. Initially we have discussed some general features about the unparticle physics from which we obtained a general expression for the energy density for a string. We could identify a proper length scale $R$ associated with the unparticle scenario. such length scale define two regimes, namely, a gravity dominated regime and an ungravity dominated regime. We find that the unparticle corrects the a string source by
\begin{eqnarray}\label{source}
    \rho(r) = \frac{\mu}{2\pi}\left[\frac{\delta(r)}{r}+\frac{R^{2d_U-2}}{r^{2d_U}}\right],
\end{eqnarray}

With the above source, we first have considered an static cosmic string ansatz from which we obtain the solution in terms of first and second kind Bessel functions. The structure of the cosmic string spacetime in the unparticle scenario was studied by means of the Ricci scalar.

 However, in order to study  fractalization, we must consider the black hole analogous with cylindrical symmetry. In order to get this, first we obtained the solution for black string in the unparticle scenario directly from the Einstein field equations. The solutions were studied for both scalar and vector unparticle case, and in both cases the solution is divergent when $d_U=3/2$. In both cases the unparticle black string solution recovers the usual black string solution in the gravity dominated regime, as expected. Also, it is important to highlight that, in the black string, all the cases considered  may exhibit a naked singularity depending on the range of $d_U$. For the unparticle black hole this was a feature present only for the vector case\cite{Alencar:2018vvb}.

The thermodynamic properties of the black string with the unparticle corrections were also addressed. We could verify that either in scalar unparticle case or in vector case, there are regions with negative mass parameter. For the scalar case, the mass becomes negative when $3/2<d_U<2$, while for the vector case the mass becomes negative when $1<d_U<3/2$. As pointed out in \cite{Mann:1997jb}, black holes (black strings, for our case) with negative mass can exist and present non-trivial topology. By investigating the Hawking temperature we could see that both scalar and vector cases present a very similar behaviour. When $1<d_U<3/2$ the temperature increases slowly with the radius in comparison with the usual black string temperature. When $3/2<d_U<2$ the unparticle correction promotes an increasing in the temperature in comparison with the black string usual case.

With the above solution we analyze if, similarly to the unparticle black hole cases, we can have a fractalization. We could identify that for the black string in the UGD regime we have an effective dimension given by $d_H=2d_U-1$. We should point that, for the static and rotating black hole, the fractal dimension is given by $d_H=2d_U$. The fractalization can also be obtained from the entropy, which was analytically obtained for the scalar and vector cases. For both cases it is given by the Hypergeometric confluent functions. In the UGD regime we find again a fractal dimension given by  $d_H=2d_U-1$.  Entropy and other thermodynamical quantities also give important information about the stability of the system. We perform our analysis separately for $d_U>3/2$ and $d_U<3/2$ in order to avoid singularities. We compute the specific heat and free energy for the scalar case and plots are depicted in figure (\ref{therm_quant_1}) for a particular choice of parameters. We could see that the value of the $d_U$ controls the black string stability in the ungravity scenario, being the black string stable for $d_U<3/2$ and unstable for $d_U>3/2$.

Finally we point that the source \ref{source} will give modified light deflection and could be found in the present era of high precision astrophysical measurements. It would also be interesting to study if the rotating case give the same fractal dimension, as in the black hole case. In fact, the study of fractal dimension in unparticle can be done to many other objects. A more ambitious project would be to show that horizon fractalization is always present in unparticle gravity.

\vfill
\section*{Acknowledgement}
The authors would like to thanks Alexandra Elbakyan and sci-hub, for removing all barriers
in the way of science. We acknowledge the financial support provided by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior or (CAPES), the Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq) and Fundaçao Cearense de Apoio ao Desenvolvimento Científico e
Tecnológico (FUNCAP) through PRONEM PNE0112- 00085.01.00/16.

\newpage

\section{Introduction}\label{Sec-1}

The unusual properties of non-trivial scale invariance of matter in the IR regime was first studied by Banks and Zaks \cite{Banks:1981nn}. Georgi suggested \cite{Georgi:2007ek, Georgi:2007si} a coupling between the standard model (SM) fields and this scale invariant sector of a high energy physics described by the Banks-Zaks (BZ) fields. The interactions between the standard model fields and the BZ fields are mediated by the exchange of highly massive particles $M_U$, obeying the scale-supressed non-renormalizable Lagrangian density
\begin{equation}
\mathcal{L}=\frac{1}{M_U^k}\mathcal{O}_{SM}\mathcal{O}_{BZ},
\end{equation}
where $\mathcal{O}$ stands for the field operators of dimension $d_{SM}$ and $d_{BZ}$, while $k=d_{SM}+d_{BZ}-4$ in order to guarantee a dimensionless action. It was argued by Georgi \cite{Georgi:2007si} that below some scale energy limit ($\Lambda_U$) the interactions between the standard model and the hidden sector become stronger. As the scale-invariance is manifested, massive particles are not allowed and this sector was dubbed ``unparticles'', since the notion of mass in this sector is meaningless.

Since the proposal of the unparticle physics several papers were published in particle physics \cite{Aliev:2017bme, Bagheri:2017roa, VanSoa:2018pkb}, condensed matter analogue models \cite{Limtragool:2016ghz, LeBlanc:2014xpa}, extra dimensions \cite{Soa:2019whl} among others. A very interesting work by Gaete and Spallucci \cite{Gaete:2008wg} study the unparticle dynamics in a standard quantum field theory scenario, and they construct effective actions for scalar, gravitation and vector gauge unparticles. Also in cosmological and astrophysical scenarios, several were carried out including studies with black holes \cite{Gaete:2010sp, Mureika:2010je, Lee:2011wc, Alencar:2018vvb} and white dwarfs \cite{Moussa:2021fsb, deSouza:2012ap}. One of these black holes works draws the attention to the fact that the exterior event horizon presents a surface with fractal dimension equals to $d_U$ \cite{Gaete:2010sp}. Recently, some of the present authors generalized the work of \cite{Gaete:2010sp} to rotating black holes and found that the the fractalization is also present\cite{Alencar:2018vvb}. It is also noticeable that a recent paper on Casimir effect with unparticles shows a fractalization in the parallel plates dimension \cite{Frassino:2013lya}. The dimensional reduction seems to be a natural behavior in high energy  and its comprehension in fundamental at this regime scales\cite{Stojkovic:2013xcj,Carlip:2017eud}.
 However, as far as we know, no studies were carried out in order to investigate the influence of the unparticle physics in cosmic strings and black strings.

Objects such as cosmic strings may have been formed due to phase transitions in the early universe \cite{Vilenkin}. The spacetime geometry related to an infinitely long and straight cosmic string is characterized by a planar angle deficit on the two-surface orthogonal to the string. In addition to that, the spacetime is flat, locally, except on the top of the string. Cosmic strings can also be described in the realm of a classical field theory when the energy-momentum tensor related to the vortex configuration of the Higgs-Maxwell system \cite{Nielsen:1973cs} couples to the Einstein equations.

On the other hand, black strings are higher dimensional solutions of the Einstein equations in D-dimensional spacetime \cite{Duff:1987cs}. These vacuum objects can be understood as symmetric generalizations of black holes under translations being the event horizon topologically equivalent to $S_2 \times R$ (or $S_2 \times S_1$ in the case of black rings) and spacetime is asymptotically $M_{D-1} \times S_1$ for a zero cosmological constant or $AdS_{D-1} \times S_1$ for a negative one \cite{Emparan:2001wn, Bellucci:2010gb}. A black string is a particular case of a black p-brane solution (with $p = 1$), where $p$ is the number of extra spatial dimensions to ordinary space. Hence a p-brane gives rise to a (p + 1)-dimensional world-volume in spacetime. In string theory, a black string is described by a D1-brane surrounded by a horizon. It is widely known that since an open string propagates through spacetime, it is necessary that its endpoints to lie on a D2-brane on which it satisfies Dirichlet boundary condition, and the dynamics on the D-brane worldvolume is described by a gauge field theory \cite{Polchinski:1995mt, Polchinski:1994fq}. In some models our universe is understood as a higher dimensional world-brane, thus the gravitational collapse of galactic matter would produce black holes lying on the brane. Black string solutions are nevertheless unstable to long wavelength perturbations (at least in asymptotically flat spaces) since the localized black hole is entropically preferred to a long segment of string. The string's horizon therefore has a tendency to form a line of black holes. However, in AdS the space acts like a confining box which prevents fluctuations of long wavelengths from developing. For instance, in a Randall Sundrum domain wall in a five dimensional AdS spacetime the black string is stable at least far from the AdS horizon. Near the horizon, the string is likely to pinch off to become a stable shorter cigar-like singularity \cite{Chamblin:1999by}.

In this paper we study the unparticle corrections for cosmic and black strings. This paper is organized as follows: In the next section we present a brief review of the unparticle theory. In section III we study the unparticle corrections for cosmic strings and in section IV we obtain the black string solution in the unparticle scenario. In section V we study the thermodynamic properties of black strings in the unparticle physics domain and in section VI we draw our conclusions.

\section{Unparticle Theory}\label{Sec-2}
The un-gravity action is \cite{Gaete:2010sp}
\begin{eqnarray}\label{1}
    S_u = \frac{1}{2\kappa^2} \int d^4x \sqrt{-g} \left[1 + \frac{A_{d_U}}{(2d_U -1)\sin{(\pi d_U)}}
    \frac{\kappa_{*}^{2}}{\kappa^{2}}\left(-\frac{D^2}{\Lambda^2_U}\right)^{1-d_U}\right]^{-1} \mathcal{R},
\end{eqnarray}
where $\kappa^2 = 8\pi G$, $\kappa_{*}^2$ is represents the un-gravitational Newton constant, and $ A_{d_U} = 16\pi^{5/2} \Gamma(d_U + 1/2)\left[(2\pi)^{2d_U} \Gamma(2d_U) \Gamma(d_U-1)\right]^{-1} $  \cite{Gaete:2010sp}, and $D^2$ is the generally covariant D'alambertian. The Einstein Field Equations (EFE) is given by
\begin{eqnarray}\label{2}
    R^\mu_\nu - \frac{1}{2} \delta^\mu_\nu \mathcal{R} &=& \kappa^2 T^\mu_\nu+ \frac{\kappa_*^2A_{d_U}}{\sin{(\pi d_U)}}\frac{\Lambda_U^{2 - 2d_U}}{(2d_U -1)}\left(-\frac{1}{D^2}\right)^{1-d_U} T^\mu_\nu \nonumber \\
    &=& \kappa^2 T^\mu_\nu + \frac{\kappa_*^2A_{d_U}}{\sin{(\pi d_U)}} {T_U}^\mu_\nu
\end{eqnarray}
The second term of right side of EFE is the unparticle correction of energy-momentum tensor. In Ref. \cite{Mureika:2010je} it was found an exact solution of EFE for vector particles, and this solution is Reissner-Nordstrom like. In this paper, we'll consider an axially symmetric \textit{ansatz} for the metric, $(g_{\mu\nu})=\text{diag}\left(g_{tt}(r),g_{rr}(r),g_{\phi\phi}(r),g_{zz}(r)\right)$, since we are interested in studying solutions of Cosmic String and Black String with unparticle corrections. Moreover, we adopt the metric's signature $(-1,1,1,1)$, so  we consider the energy-momentum tensor as $(T^\mu_\nu) = \text{diag}(-\rho,p,p_\phi,p_z)$, which can be represented by
\begin{eqnarray}\label{energy-momentum}
    T^\mu_\nu = (\rho + p_\phi) u^\mu u_\nu + (p-p_\phi)w^\mu w_\nu+ (p_z-p_\phi)Z^\mu Z_\nu + p_\phi~\delta^\mu_\nu,
\end{eqnarray}
where $(u_\mu) = \left(\sqrt{-g_{tt}},0,0,0\right)$ is the four-velocity field, $(w_\mu) = \left(0,\sqrt{g_{rr}},0,0\right)$ and $(Z_\mu) = \left(0,0,0,\sqrt{g_{zz}}\right)$ are  units spacelike vectors.

Let us consider a static string which has a density of energy $\rho(r) =\mu(2\pi r)^{-1} \delta(r)= \mu \delta(x)\delta(y)$, where $\mu$ is the linear density of mass. We can compute the contribution of unparticle in equation (\ref{2}) by calculating the term $T_U$. We achieve that by switching to isotropic (free-falling), Cartesian-like coordinate for computational convenience,
\begin{eqnarray}\label{6}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 &=& - \mu \left(-\frac{1}{\nabla_{\vec{X}}^2}\right)^{1-d_U} \delta \left(\vec{X}\right) \nonumber\\
    &=& - \mu \left(-\frac{1}{\nabla_{\vec{X}}^2}\right)^{1-d_U} \frac{1}{(2\pi)^2}\int d^2k e^{i \vec{k}\cdot\vec{X}} \nonumber \\
    &=&- \mu \frac{1}{(2\pi)^2}\int d^2k \left(\frac{1}{k^2}\right)^{1-d_U} e^{i \vec{k}\cdot\vec{X}},
\end{eqnarray}
where $\vec{X}$ is a vector in a direction perpendicular to the string. Hence, by applying the Schwinger representation for $(1/k^2)^{1-d_U}$, we obtain
\begin{eqnarray}\label{7}
    \left(\frac{1}{k^2}\right)^{1-d_U} = \frac{1}{\Gamma(1-d_U)}\int_0^\infty ds s^{-d_U} e^{-sk^2}.
\end{eqnarray}
Yielding for the equation (\ref{6})
\begin{eqnarray}\label{8}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 = - \mu \frac{1}{(2\pi)^2}\int d^2k \left[\frac{1}{\Gamma(1-d_U)}\int_0^\infty ds s^{-d_U} e^{-sk^2}\right] e^{i \vec{k}\cdot\vec{X}}.
\end{eqnarray}
Finally, when we compute this integration, we find
\begin{eqnarray}\label{9}
    \left(-\frac{1}{D^2}\right)^{1-d_U} T^0_0 = - \mu \frac{2^{2d_U - 2}}{\pi} \frac{\Gamma(d_U)}{\Gamma(1-d_U)} \left(\frac{1}{\vec{X}^2}\right)^{d_U},
\end{eqnarray}
which allow us to write a proper expression for the energy density of a static string with unparticle as
\begin{eqnarray}\label{density0}
    \rho(r) = \frac{\mu}{2\pi}\frac{\delta(r)}{r} + \mu\left(\frac{\kappa^{2}_{*}}{\kappa^2}\right) \frac{ A_{d_U}}{\sin{\pi d_U}}\frac{\Lambda^{2-2d_U}_U}{(2d_U-1)} \frac{2^{2d_U-2}}{\pi}\frac{\Gamma(d_U)}{\Gamma(1-d_U)} \left(\frac{1}{r}\right)^{2d_U}.
\end{eqnarray}
When we replace $ A_{d_U} = 16\pi^{5/2} \Gamma(d_U + 1/2)\left[(2\pi)^{2d_U} \Gamma(2d_U) \Gamma(d_U-1)\right]^{-1}$ in equation (\ref{density0}), we find
\begin{eqnarray}\label{density}
    \rho(r) = \frac{\mu}{2\pi}\left[\frac{\delta(r)}{r}+\frac{R^{2d_U-2}}{r^{2d_U}}\right],
\end{eqnarray}
where
\begin{eqnarray}\label{unradiu}
    R = \left[\left(\frac{\kappa_*^2}{\kappa^2}\right)\frac{8\Lambda_U^{2 - 2d_U}}{\sin{(\pi d_U)}(2d_U -1)\pi^{2d_U-5/2}}\frac{ \Gamma(d_U)\Gamma(d_U + 1/2)}{\Gamma(1-d_U) \Gamma(2d_U) \Gamma(d_U-1)}\right]^\frac{1}{2d_U-2}.
\end{eqnarray}

Also it is important to highlight that the $R$ parameter sets a proper unparticle length scale so that, similarly to the authors in \cite{Gaete:2010sp, Alencar:2018vvb}, it is possible to distinguish between two regimes. The gravity dominated (GD) regime is defined by $R<<r$ and the ungravity dominated (UGD) regime by $R>>r$. Note that in the GD regime the second term of (\ref{density}) vanishes and we recover the usual cosmic string energy density.

\begin{figure}[h]
    \centering
    \includegraphics[width=10cm]{unparticleR-plot.pdf}
    \caption{Behavior of $R$ for different $\Lambda_U$ as function of $d_U$, with $(\kappa_{*}/\kappa)^2=0.5$}
    \label{fig-beta-constant}
\end{figure}

In the next section, we will present the solution of static cosmic string with  unparticle corrections.

\section{Cosmic String with unparticle corrections}
 In this section, we will study the spacetime solution which is static and has cylindrical symmetry. Consider the \textit{ansatz} of the line-element as follows
 \begin{eqnarray}\label{3}
     ds^2 = -dt^2 + dr^2 + b(r)^2 d\phi^2 + dz^2.
 \end{eqnarray}
 The components of the Ricci tensor are $R_{\mu\nu}=R^\alpha_{\mu\alpha\nu} = \partial_{[\alpha}\Gamma^{\alpha}_{\nu]\mu}+ \Gamma^{\alpha}_{\beta[\alpha}\Gamma^{\beta}_{\nu]\mu}$, where $V_{[\mu\nu]}= V_{\mu\nu}-V_{\nu\mu}$. The Einstein Tensor is
 \begin{eqnarray}\label{4}
     G^\mu_\nu = \frac{1}{b(r)}\frac{d^2b(r)}{dr^2} ~\text{diag}\left(1,0,0,1\right).
 \end{eqnarray}
We know that the $T^0_0 = - \rho(r)$, so we can see by EFE $G^\mu_\nu = \kappa^2 T^\mu_\nu$ that $T^\mu_\nu = -\rho(r) \text{diag}(1,0,0,1)$. Then the EFE resumes to
\begin{eqnarray}\label{5}
    \frac{1}{b(r)}\frac{d^2b(r)}{dr^2} = - \kappa^2 \rho(r).
\end{eqnarray}
This equation can be rewritten as
\begin{eqnarray}\label{12}
    \frac{d^2b(r)}{dr^2} +\frac{\mu\kappa^2}{2\pi}\left[ \frac{\delta(r)}{r} + R^{2d_U-2} r^{-2d_U}\right] b(r) = 0.
\end{eqnarray}

Now, we need to properly solve the equation (\ref{12}). Let us perform the following variable transformation $b(r) = \sqrt{r}f(r)$, such that the equation (\ref{12}) becomes
\begin{eqnarray}\label{13}
    \left\{\frac{d^2f}{dr^2}+ \frac{1}{r}\frac{df}{dr}+\left[\frac{\mu\kappa^2}{2\pi} \frac{\delta(r)}{r} + \frac{\mu\kappa^2}{2\pi}R^{2d_U-2}~ r^{-2d_U}- \frac{1}{4r^2}\right]f\right\} \sqrt{r} = 0.
\end{eqnarray}
First, we need to solve this equation for $r>0$. After we impose that when $R/r \rightarrow 0$ we must find cosmic string spacetime. When we consider $r>0$ the equation (\ref{13}) simplifies to
\begin{eqnarray}\label{14}
    \frac{d^2f}{dr^2}+ \frac{1}{r}\frac{df}{dr}+\left[\frac{\mu\kappa^2}{2\pi}R^{2d_U-2} r^{-2d_U}- \frac{1}{4r^2}\right]f = 0.
\end{eqnarray}
Now, we can perform a transformation of variable $ x = \sqrt{\frac{\mu\kappa^2}{2\pi}}\frac{R^{d_U-1}}{\left(d_U-1\right)} r^{1-d_U}$, allowing the equation (\ref{14}) to be written as
\begin{eqnarray}\label{15}
    \frac{d^2f}{dx^2}+ \frac{1}{x}\frac{df}{dx}+\left[1- \frac{\left(\frac{1}{2du-2}\right)^2}{x^2}\right]f = 0.
\end{eqnarray}
This is the Bessel differential equation. The solution of equation (\ref{15}) depends on $\left(\frac{1}{2du-2}\right)$. When we define $\xi \equiv 1/(2du-2)$ as a positive real number the equation (\ref{15}) becomes
\begin{eqnarray}\label{16}
    \frac{d^2f}{dx^2}+ \frac{1}{x}\frac{df}{dx}+\left[1- \frac{\xi^2}{x^2}\right]f = 0,
\end{eqnarray}
whose solution is
\begin{eqnarray}\label{17}
    f(x) = C_1 J_\xi (x) + C_2 N_\xi (x).
\end{eqnarray}
Here $J_\xi(x)$ and $N_\xi(x)$ are the first and second kind Bessel function of order $\xi$, respectively.
Let us present some considerations regarding the boundary conditions. Initially, we want that when $R/r \rightarrow 0$, $b(r) = (1-8G\mu)^{1/2}r$, that is, the spacetime becomes the cosmic string spacetime. Since $x$ is proportional to $(R/r)^{d_U-1}$ and $b(r)=\sqrt{r}f(x(r))$, we can study the equation (\ref{17}) when $x \ll 1$. In regime $x \ll 1$ and $\xi > 0$, equation (\ref{17}) can be written as
\begin{eqnarray}\label{18}
    f(x)= C_1\frac{1}{\Gamma{(\xi+1)}}\left(\frac{x}{2}\right)^\xi - C_2 \frac{\Gamma(\xi)}{\pi}\left(\frac{x}{2}\right)^{-\xi}.
\end{eqnarray}
Replacing $ x = \sqrt{\frac{\mu\kappa^2}{2\pi}}\frac{R^{d_U-1}}{\left(d_U-1\right)} r^{1-d_U}$ in equation (\ref{18}), when we impose the limit $R/r \ll 1$ we shall find $f(r) = (1-8G\mu)^{1/2}\sqrt{r}$. Hence $C_1 = 0$ and
\begin{eqnarray*}
C_2 = -(1-8G\mu)^{1/2}\frac{\pi}{\Gamma{(\xi)}}\left(\xi \sqrt{\frac{\kappa^2\mu}{2\pi}}\right)^\xi R^{1/2}.
\end{eqnarray*}
Consequently, the component $g_{\phi\phi}$ of the metric is
\begin{eqnarray}\label{21}
    b^2(r) = (1-8G\mu)r^2\left[\frac{\pi^2}{\Gamma^2{(\xi)}}\xi^{2\xi}\left( 4G\mu\right)^{\xi} \frac{R}{r}\right]  {N_\xi}^2 \left(4\xi\sqrt{G\mu}\left(\frac{R}{r}\right)^{\frac{1}{2\xi}}\right).
\end{eqnarray}
The metric component $g_{\phi\phi} = b^2(r)$, the behavior of this component is given at Figure \ref{metric-component-phi}.
\begin{figure}[h]
    \centering
    \includegraphics[width=10cm]{gPhiPhi-plot.pdf}
    \caption{Metric component $g_{\phi\phi}$ of the unparticle static cosmic string as a function of the radial coordinate $r$ for some values of $d_U$ , with $\Lambda_U = 10$, $\mu = 0.7$, $\kappa_* = 10^3$}
    \label{metric-component-phi}
\end{figure}

In order to understand the structure of this spacetime, that is, how the unparticles modify the cosmic string spacetime, We must analyze the curvature scalars. Ricci's scalar for the ansatz $ds^2 = -dt^2 + dr^2 + b^2(r)d\phi^2 + dz^2$ is $\mathcal{R} = - \frac{2}{b}\frac{d^2b}{dr^2}$. Then Ricci's scalar for our solution is given by
\begin{eqnarray}\label{22}
    \mathcal{R}=\frac{\mu\kappa^2}{\pi}\left[ \frac{\delta(r)}{r} + \frac{1}{r^2}\left(\frac{R}{r}\right)^{2d_U-2}\right],
\end{eqnarray}
where we used equation (\ref{13}).

\section{Black string with unparticle corrections}

Let us consider the following line element for the black string
\begin{eqnarray}\label{23}
    ds^2=-f(r)dt^2+\frac{1}{f(r)}dr^2+r^2d\phi^2+\alpha^2r^2dz^2,
\end{eqnarray}
where $t\in(-\infty, \infty)$, the radial coordinate $r\in[0,\infty)$, the angular coordinate $\phi\in[0,2\pi)$ and the axial coordinate $z\in(-\infty,\infty)$. The $\alpha$ parameter is considered as $\alpha^2=-\Lambda/3$.

For the black string in the un-gravity scenario, the Einstein-Hilbert effective action is composed by (\ref{1}) added by the cosmological constant contribution, i.e.,
\begin{eqnarray}\label{24}
    S_u = \frac{1}{2\kappa^2} \int d^4x \sqrt{-g} \left\{\left[1 + \frac{A_{d_U}}{(2d_U -1)\sin{(\pi d_U)}}
    \frac{\kappa_{*}^{2}}{\kappa^{2}}\left(-\frac{D^2}{\Lambda^2_U}\right)^{1-d_U}\right]^{-1} \mathcal{R}-2\Lambda\right\}.
\end{eqnarray}
The Einstein equation in (\ref{2}) remains valid, however the left-hand-side will be modified due to the different ansatz for the metric in (\ref{23}). The EFE for this ansatz are
\begin{eqnarray}
    G^t_t -3\alpha^2 &=& \frac{1}{r} \frac{df(r)}{d r}  + \frac{f(r)}{r^2} - 3\alpha^2, \label{25}\\
    G^r_r -3\alpha^2 &=& \frac{1}{r} \frac{df(r)}{d r}  + \frac{f(r)}{r^2} - 3\alpha^2, \label{26}\\
    G^\phi_\phi -3\alpha^2 &=& \frac{1}{2}\frac{d^{2}f(r)}{d r^{2}}+ \frac{1}{r} \frac{df(r)}{d r} - 3\alpha^2, \label{27}\\
    G^z_z -3\alpha^2 &=&\frac{1}{2}\frac{d^{2}f(r)}{d r^{2}}+ \frac{1}{r} \frac{df(r)}{d r} - 3\alpha^2.\label{28}
\end{eqnarray}
We can see that the energy-momentum tensor for the ansatz of equation (\ref{23}) is $T^\mu_\nu = -\rho(r)\text{ diag}(1,1,0,0) + p_l(r)\text{ diag}(0,0,1,1)$, where $p_l = p_\phi = p_z$. This way we can find $f(r)$ by solving $G^t_t -3\alpha^2 = -\kappa^2 \rho(r)$, we find

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{f_r_1.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{f_r_2.eps}
  \caption{}

\end{subfigure}%
    \caption{Plot of $f(r)$ (scalar case (a) and vector case (b)) for some values of $d_U$ with $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$}
    \label{f(r)plot}
\end{figure*}

\begin{eqnarray}\label{29}
    f(r) = \alpha^2r^2 - \frac{4\mu}{\alpha r} \pm \frac{\kappa^2\mu}{2\pi(2d_U-3)}\left(\frac{R}{r}\right)^{2d_U-2},
\end{eqnarray}
with $R$ given by equation (\ref{unradiu}). Note that the unparticle black string solution is divergent for $d_U=3/2$. The plus signal is taken for the scalar unparticle case, while the minus one is for the vector case.

The above solution for the black string in the ungravity scenario recovers the usual black string solution \cite{Lemos:1994xp} in the GD regime, i.e.,
\begin{eqnarray}\label{31}
    f(r)=\left(\alpha^2r^2-\frac{4\mu}{\alpha r}\right).
\end{eqnarray}

Also, it is important to highlight here that the unparticle black string may exhibit naked singularity, violating therefore the cosmic censorship, for the scalar case when $3/2<d_U<2$.

\section{Thermodynamic properties of the Black String with unparticle corrections}
To make easy the calculations, we can rewrite the equation (\ref{29}) as $f(r) = h(r) + \mu s(r)$, where
\begin{eqnarray}
    h(r) &=& \alpha^2 r^2, \\
    s(r) &=& \frac{4}{\alpha r}\left[-1 \pm \frac{\alpha r \kappa^2}{8\pi (2d_U-3)}\left(\frac{R}{r}\right)^{2d_U -2}\right].
\end{eqnarray}

Our black string solution in the un-gravity scenario has the horizon curves defined by $f(r_h)=0$, so $\mu = - \frac{h(r_h)}{s(r_h)}$, which gives us
\begin{eqnarray}\label{32}
    \mu= -\frac{\alpha^3r_h^3}{4}\left[-1\pm\frac{\alpha r_h \kappa^2}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]^{-1}.
\end{eqnarray}

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{mu_scalar.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{mu_vectorial.eps}
  \caption{}

\end{subfigure}%
    \caption{Linear mass parameter $\mu$ as a function of $d_U$ for three values of $r$. In (a) we are considering the scalar unparticle case while in (b) we have the vectorial unparticle case. We have considered $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$.}
    \label{mu}
\end{figure*}

In the GD regime, the mass parameter $\mu$ reduces to $\mu=\frac{\alpha^3}{4}r_h^3$, which is the usual mass parameter for the black-string in General Relativity. In \cite{Alencar:2018vvb} the authors have found that for an unparticle black hole the mass parameter becomes negative for the vector unparticle case. For the unparticle black string, as we can see in fig.(\ref{mu}), both vector and scalar cases present the possibility of negative mass parameter. For the scalar unparticle case we have a negative mass parameter for values of $3/2<d_U<2$. For the vector unparticle case the mass parameter becomes negative for $1<d_U<3/2$.

The authors in \cite{Mureika:2010je} does not take into consideration the regions where the mass parameter is negative. However it is known that a negative mass parameter in a black hole solution suggests that it has a nontrivial topology \cite{Mann:1997jb}. Therefore we can expect that these negative mass regions are also related to nontrivial topologies of the the black string.

In possession of the solution for the static black string in the ungravity scenario given by (\ref{29}), we are able to study the thermodynamics of the black string by computing the Hawking's temperature by means of $T_H=\frac{1}{4\pi}f'(r_h)$, where $r_h$ is the radius of the horizon. Note that in terms of $h(r)$ and $s(r)$ we find
\begin{eqnarray}\label{temp}
    T_H &=& \frac{1}{4\pi}\left[h'(r_h)+ \mu ~s'(r_h)\right] \nonumber \\
     &=& \frac{s(r_h)}{4\pi}\frac{d}{dr_h}\left(\frac{h(r_h)}{s(r_h)}\right),
\end{eqnarray}
where we used $\mu = -\frac{h(r_h)}{s(r_h)}$. When we substitute the expressions for $h(r_h)$ and $s(r_h)$, we find
\begin{eqnarray}\label{33}
    T_H = \frac{3\alpha^2 r_h}{4\pi}\frac{\left[-1\pm\frac{d_U\alpha r_h\kappa^2}{12\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]}{\left[-1\pm\frac{\alpha r_h\kappa^2}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-2}\right]}.
\end{eqnarray}
The usual black string Hawking temperature, i.e. $T_H=\frac{3 \alpha ^2 }{4 \pi }r_h$, is recovered in the GD regime and in UGD regime, we find
\begin{eqnarray}\label{33-1}
    T_H = \frac{\alpha^2 2d_U}{4\pi}r_h,
\end{eqnarray}
for both GD and UGD regimes. The Hawking temperature for the unparticle black string presents a similar behaviour for both scalar and vector unparticle cases. The behaviour of the Hawking temperature as a function of $r$ for the scalar unparticle case for some values of $d_U$ is presented in figure (\ref{HawkingTemperature}). We can see from figure (\ref{HawkingTemperature}) that when $1<d_U<3/2$ the temperature increases slowly with the radius in comparison with the usual black string temperature. When $3/2<d_U<2$ the unparticle correction promotes an increasing in the temperature in comparison with the black string usual case.

\begin{figure}[h!]
    \centering
    \includegraphics[scale=0.8]{T__H_s.eps}
    \caption{Hawking temperature as a function of $r$ for the scalar unparticle case for three values of $d_U$. We have considered $\Lambda_{U}=1$, $\alpha=0.5$, $\mu=0.7$, $\kappa=1$ and $\kappa_{*}=10^2$.}
    \label{HawkingTemperature}
\end{figure}

Note that the temperature is aproximally proportional to $D-1$, where $D$ is the dimension of the spacetime. So, when we confront the equation (\ref{33-1}) with this fact, we can see the fractalization of the event horizon with
\begin{eqnarray}\label{33.2}
 d_H=2d_U-1 ,
\end{eqnarray}
where $d_H = D - 2$. To reinforce the fractalization of the event horizon, we can compute the density of entropy by length $\mathcal{S}$ althrough  $d\mathcal{S} = d\mu/T_H$, which in terms of $h(r)$ and $s(r)$,
\begin{eqnarray}
    d\mathcal{S} = - \frac{4\pi}{s(r_h)} dr_h,
\end{eqnarray}
where we used equation (\ref{temp}) and again $\mu = - \frac{h(r_h)}{s(r_h)}$. So we find
\begin{eqnarray}\label{differentialOfEntropy}
    d\mathcal{S} = \frac{\pi\alpha r_h ~dr_h}{\left[-1 \pm \frac{\alpha R \kappa^2}{8\pi (2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U -3}\right]}.
\end{eqnarray}
Now, we want to find analytically the entropy by length of the black string with unparticle. It can be done by integration of equation (\ref{differentialOfEntropy}), but it is necessary to analyze the regions due to the possible singularity of this equation. Up to now we have discussed the scalar and vector cases simultaneously, but now, for the sake of simplicity we will consider them separately.

\subsection{scalar case:}

Let us begin by considering the case when $d_U<3/2$. In this case we can rewrite the equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{differentialOfEntropyDSmallerThan3/2}
    d\mathcal{S} = \frac{\pi\alpha r_h ~dr_h}{\left[1 + \frac{\alpha R \kappa^2}{8\pi (3-2d_U )}\left(\frac{R}{r_h}\right)^{2d_U -3}\right]}.
\end{eqnarray}
So, in case $d_U<3/2$ there is not singularity. By using the identity
\begin{eqnarray}\label{identity}
\int \frac{x dx}{1\pm Ax^\gamma} = \frac{x^2}{2}\hyper{1}{\frac{2}{\gamma}}{1+\frac{2}{\gamma}}{\mp Ax^\gamma} + \text{const.},
\end{eqnarray}
where $\hyper{a}{b}{c}{x}$ is the hypergeometric function and $\gamma>0$, the integration of equation (\ref{differentialOfEntropyDSmallerThan3/2}) can be written as
\begin{eqnarray}\label{entropy1}
    S = \frac{2\pi\alpha {r_h}^2}{4}\hyper{1}{\frac{2}{3-d_U}}{1+\frac{2}{3-d_U}}{- \frac{\alpha R \kappa^2}{8\pi (3-2d_U )}\left(\frac{R}{r_h}\right)^{2d_U -3}}.
\end{eqnarray}
Note that, in GD regime ($R\rightarrow 0$) the argument of hypergeometric funcion is proportional to $R^{2d_U-2}$, which approaches to $0$ when $R \rightarrow 0$. It is, the entropy in GD recovers  $S = \frac{2\pi\alpha {r_H}^2}{4}$ as shall be.

The heat capacity and the free energy of the system are defined by
\begin{eqnarray}
    \label{heat}C_V&=&\frac{d\mu}{dT_H}\\
    \label{free}F&=&\mu-T_H S.
\end{eqnarray}
The heat capacity can be obtained directly from the equations (\ref{32}) and (\ref{33}), which yields
\begin{eqnarray}
    C_V=\frac{8\pi^2(-3+2d_U)\alpha(r_h R)^2\left[12(-3+2d_U)\pi R^2-d_U\kappa^2\alpha R^3\left(\frac{R}{r_h}\right)^{2d_U-3}\right]}{96(3-2d_U)^2\pi^2R^4+8\pi(-9+27d_U-20d_U^2+4d_U^3)\kappa^2r_h^3R^2\alpha\left(\frac{R}{r_h}\right)^{2d_U}+d_U\alpha^2\kappa^4r_h^6\left(\frac{R}{r_h}\right)^{4d_U}}.
\end{eqnarray}
The free energy can also be obtained from (\ref{32}), (\ref{33}) and (\ref{entropy1}), yielding
\begin{eqnarray}
    \nonumber F&=&-\frac{\alpha^3 r_h^3}{4 \alpha  \kappa^2 r_h^{3-2 d_U} R^{2 d_U}+32 \pi  (3-2 d_U) R^2} \left[8 \pi  (2 d_U-3) R^2\right.\\
    &&+\left.\left(\alpha d_U \kappa^2 r_h^{3} \left(\frac{R}{r_h}\right)^{2 d_U}+12 \pi  (3-2 d_U) R^2\right) \, _2F_1\left(1,\frac{2}{3-2 d_U};1+\frac{2}{3-2 d_U};-\frac{\alpha  \kappa^2 r_h^{3-2 d_U} R^{2 d_U-2}}{24 \pi -16 \pi  d_U}\right)\right].
\end{eqnarray}

The behaviour of the Hawking temperature, entropy, heat capacity and free energy for $d_U<3/2$ is depicted in figure (\ref{therm_quant_1}).

\begin{figure*}[h!]
    \centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{therm_quant_1.eps}
  \caption{}

\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[scale=0.8]{therm_quant_3.eps}
  \caption{}

\end{subfigure}%
    \caption{Hawking temperature, entropy, heat capacity and free energy for $d_U=1.4$ (a) and for $d_U=1.6$ (b)}
    \label{therm_quant_1}
\end{figure*}

Now, we need to investigate the case $d_U>3/2$. In this case, there is a singularity in equation (\ref{differentialOfEntropy}) when
\begin{eqnarray}\label{singularityForr_H}
    r_0 = \left[\frac{\alpha R \kappa^2}{8\pi (2d_U-3)}\right]^{\frac{1}{2d_U -3}}R,
\end{eqnarray}
It is easy to see that we have two disjunte regions and when $r_h >r_0$ $\left(r_h <r_0\right)$  the differential of entropy by length, equation (\ref{differentialOfEntropy}), approaches infinity (minus infinity). First, let us compute the entropy for $r_h >r_0$.We can rewrite equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{differentialOfentropyOfDLargerThan3/2}
    d\mathcal{S} = \frac{\alpha \pi r_hdr_h}{\left[1 -\left(\frac{r_0}{r_h}\right)^{2d_U-3}\right]}.
\end{eqnarray}
We can make a changing of variable $z = (r_o/r_h)^{2d_U-3}$, equation (\ref{differentialOfentropyOfDLargerThan3/2}) becomes
\begin{eqnarray}\label{differentialOfentropyOfDLargerThan3/2-withz}
    d\mathcal{S} = -\frac{\alpha \pi{r_0}^2}{(2d_U-3)}~\frac{ z^{-\left(\frac{2d_U-1}{2d_U-3}\right)}dz}{1 -z}.
\end{eqnarray}
We can use the identity
\begin{eqnarray}\label{identity2}
    \int \frac{x^{\pm c}dx}{1-x} =  \frac{x^{1\pm c}}{1\pm c}~\hyper{1}{1\pm c}{1+(1\pm c)}{x} + \text{const.},
\end{eqnarray}
where $c>0$. So, when we integrate the equation (\ref{differentialOfentropyOfDLargerThan3/2}), we find
\begin{eqnarray}\label{entropyOfDLargerThan3/2-1}
\mathcal{S} = \frac{2\pi \alpha r_h^2}{4}~\hyper{1}{\frac{2}{3-2d_U}}{1+\frac{2}{3-2d_U}}{\left(\frac{r_0}{r_h}\right)^{2d_U-3}}.
\end{eqnarray}
Note that equation (\ref{entropyOfDLargerThan3/2-1}) recovers $\mathcal{S} = \frac{2\pi \alpha r_h^2}{4}$ in GD regime. The ultimate case is when $r_h <r_0$. We can rewrite equation (\ref{differentialOfEntropy}) as
\begin{eqnarray}\label{changingOfVariable2}
    d\mathcal{S} =-\alpha \pi \left(\frac{r_h}{r_0}\right)^{2d_U-3}\frac{ r_h dr_h}{\left[1-\left(\frac{r_h}{r_0}\right)^{2d_U-3}\right]}.
\end{eqnarray}
When we make a changing of variable $z = (r_h/r_0)^{2d_U-3}$, equation (\ref{changingOfVariable2}) becomes
\begin{eqnarray}\label{changingOfVariable2-3}
    d\mathcal{S} =-\frac{\alpha \pi {r_0}^2}{(2d_U-3)}~ \frac{ z^{\frac{2}{2d_U-3}} dz}{1-z}.
\end{eqnarray}

So, when we integrate the equation (\ref{changingOfVariable2-3}) by using equation (\ref{identity2}), we find
\begin{eqnarray}\label{entropyOfDLargerThan3/2-2}
    \mathcal{S} = - \frac{\pi \alpha {r_0}^2}{(2d_U-1)}\left(\frac{r_h}{r_0}\right)^{2d_U-1} \hyper{1}{\frac{2d_U-1}{2d_U-3}}{\frac{4(d_U-1)}{2d_U-3}}{\left(\frac{r_h}{r_0}\right)^{2d_U-3}}
\end{eqnarray}
Note that in UGD regime the depencence of $r_h$ in equation (\ref{entropyOfDLargerThan3/2-2}) becomes $r_h^{d_h} = r_h^{2d_U-1}$. Then $d_h = 2d_U -1 $, which reinforces the fractalization and is in agreement with equation (\ref{33-1}).

The heat capacity can be obtained straightforwardly from the equations (\ref{32}) and (\ref{33}), yielding
\begin{eqnarray}
    C_v=\frac{8 \pi ^2 \alpha  (2 d_U-3) R^2 r_h^{2 d_U+2} \left(12 \pi  (2 d_U-3) R^2 r_h^{2 d_U}-\alpha  d \kappa^2 r_h^3 R^{2 d_U}\right)}{\alpha ^2 d \kappa^4 r_h^6 R^{4 d_U}+8 \pi  \alpha  (d_U-3) (2 d_U-3) (2 d_U-1) \kappa^2 r_h^3 R^2 (r_h R)^{2 d_U}+96 \pi ^2 (3-2 d_U)^2 R^4 r_h^{4 d_U}}.
\end{eqnarray}

The free energy can also be obtained from (\ref{32}), (\ref{33}) and (\ref{entropy1}), from which follows that

\begin{eqnarray}
    F=\frac{\alpha ^3 r_h \left(r_0^2 B_{\left(\frac{r}{r_0}\right)^{2 d_U-3}}\left(1+\frac{2}{2 d_U-3},0\right) \left(\alpha  d \kappa^2 r_h^3 R^{2 d_U}-12 \pi  (2 d_U-3) R^2 r_h^{2 d_U}\right)+4 \pi  (3-2 d_U)^2 R^2 r_h^{2 d_U+2}\right)}{2 (2 d_U-3) \left(8 \pi  (2 d_U-3) R^2 r_h^{2 d}-\alpha  \kappa^2 r_h^3 R^{2 d_U}\right)},
\end{eqnarray}
where $B_{z}\left(a,b\right)$ is the incomplete Euler Beta function.

As it is widely known, the thermodynamical stability of black holes (black strings for our case) is directly related to the sign of the heat capacity. A positive heat capacity indicates that the system is thermodynamically stable, while its negativity imply a thermodynamical instability. As we can see in figure (\ref{therm_quant_1}), the value of the $d_U$ controls the black string stability, being the black string stable in the ungravity regime for $d_U<3/2$ and unstable for $d_U>3/2$.

\subsection{vector case:}

Let us consider the case when $d_U>3/2$, which for the vector case avoid the singularity, so that the equation (\ref{differentialOfEntropy}) can be written as
\begin{eqnarray}\label{dsvector}
    d\mathcal{S}=\frac{\pi\alpha r_hdr_h}{\left[1+\frac{\alpha\kappa^2 R}{8\pi(2d_U-3)}\left(\frac{R}{r_h}\right)^{2d_U-3}\right]}.
\end{eqnarray}
We can clearly see that, in virtue of the similarity between the expressions (\ref{dsvector}) and (\ref{differentialOfEntropyDSmallerThan3/2}), the same analysis performed for the scalar case when $d_U<3/2$ can be done for the vector case when $d_U>3/2$. Also the results for the scalar case when $d_U>3/2$ are identical to those of the vector case when $d_U<3/2$.

Therefore, the thermodynamic quantities depicted in figure (\ref{therm_quant_1}a) for the scalar case when $d_U<3/2$ also describe the thermodynamic quantities for the vector case when $d_U>3/2$. Likewise, the thermodynamic quantities depicted in figure (\ref{therm_quant_1}b) for the scalar case when $d_U>3/2$ are also describing the thermodynamic quantities for the vector case when $d_U<3/2$.

\section{Conclusion}\label{Sec-6}

In this paper we study the unparticle corrections for cosmic and black strings. Initially we have discussed some general features about the unparticle physics from which we obtained a general expression for the energy density for a string. We could identify a proper length scale $R$ associated with the unparticle scenario. such length scale define two regimes, namely, a gravity dominated regime and an ungravity dominated regime. We find that the unparticle corrects the a string source by
\begin{eqnarray}\label{source}
    \rho(r) = \frac{\mu}{2\pi}\left[\frac{\delta(r)}{r}+\frac{R^{2d_U-2}}{r^{2d_U}}\right],
\end{eqnarray}

With the above source, we first have considered an static cosmic string ansatz from which we obtain the solution in terms of first and second kind Bessel functions. The structure of the cosmic string spacetime in the unparticle scenario was studied by means of the Ricci scalar.

 However, in order to study  fractalization, we must consider the black hole analogous with cylindrical symmetry. In order to get this, first we obtained the solution for black string in the unparticle scenario directly from the Einstein field equations. The solutions were studied for both scalar and vector unparticle case, and in both cases the solution is divergent when $d_U=3/2$. In both cases the unparticle black string solution recovers the usual black string solution in the gravity dominated regime, as expected. Also, it is important to highlight that, in the black string, all the cases considered  may exhibit a naked singularity depending on the range of $d_U$. For the unparticle black hole this was a feature present only for the vector case\cite{Alencar:2018vvb}.

The thermodynamic properties of the black string with the unparticle corrections were also addressed. We could verify that either in scalar unparticle case or in vector case, there are regions with negative mass parameter. For the scalar case, the mass becomes negative when $3/2<d_U<2$, while for the vector case the mass becomes negative when $1<d_U<3/2$. As pointed out in \cite{Mann:1997jb}, black holes (black strings, for our case) with negative mass can exist and present non-trivial topology. By investigating the Hawking temperature we could see that both scalar and vector cases present a very similar behaviour. When $1<d_U<3/2$ the temperature increases slowly with the radius in comparison with the usual black string temperature. When $3/2<d_U<2$ the unparticle correction promotes an increasing in the temperature in comparison with the black string usual case.

With the above solution we analyze if, similarly to the unparticle black hole cases, we can have a fractalization. We could identify that for the black string in the UGD regime we have an effective dimension given by $d_H=2d_U-1$. We should point that, for the static and rotating black hole, the fractal dimension is given by $d_H=2d_U$. The fractalization can also be obtained from the entropy, which was analytically obtained for the scalar and vector cases. For both cases it is given by the Hypergeometric confluent functions. In the UGD regime we find again a fractal dimension given by  $d_H=2d_U-1$.  Entropy and other thermodynamical quantities also give important information about the stability of the system. We perform our analysis separately for $d_U>3/2$ and $d_U<3/2$ in order to avoid singularities. We compute the specific heat and free energy for the scalar case and plots are depicted in figure (\ref{therm_quant_1}) for a particular choice of parameters. We could see that the value of the $d_U$ controls the black string stability in the ungravity scenario, being the black string stable for $d_U<3/2$ and unstable for $d_U>3/2$.

Finally we point that the source \ref{source} will give modified light deflection and could be found in the present era of high precision astrophysical measurements. It would also be interesting to study if the rotating case give the same fractal dimension, as in the black hole case. In fact, the study of fractal dimension in unparticle can be done to many other objects. A more ambitious project would be to show that horizon fractalization is always present in unparticle gravity.

\vfill
\section*{Acknowledgement}
The authors would like to thanks Alexandra Elbakyan and sci-hub, for removing all barriers
in the way of science. We acknowledge the financial support provided by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior or (CAPES), the Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq) and Fundaçao Cearense de Apoio ao Desenvolvimento Científico e
Tecnológico (FUNCAP) through PRONEM PNE0112- 00085.01.00/16.

\newpage

redpajama_set_name: RedPajamaArXiv


[S52] x NSW BDM Indexes, 1845268 48 .
[S48] Ancestry.co.uk, Ancestry.com. Australia, Births and Baptisms, 1792-1981 [database on-line]. Provo, UT, USA: Ancestry.com Operations, Inc., 2014.
[S2792] Richmond and Windsor Gazette, Friday 6 July 1923.
capacity by the late Arthur Dight, Messrs.
[S52] x NSW BDM Indexes, 14813/1923 .
[S52] x NSW BDM Indexes, Reg. Richmond 1868 3270.


Naruto funny pics!!!!. heres some funny punda pics of Naruto so enjoy! :). Wallpaper and background images in the Naruto club tagged: naruto anime funny.
Wkwk.... That's why... When i first read it, I was thinking,like, OMG, Itach's supposed 2 say this 2 Sasuke!
When I saw the second row I think Itachi said this to Kisame or he is practicing with Kisame to say that.LOL!
I thought Itachi was talking to Kisame when I first got 2 the 2nd row and why is Itachi wearing sunglasses ?


text:
William P. Connery Sr. (September 1855 – November 1928) was a Massachusetts politician who served as the 34th mayor of Lynn, Massachusetts.  He was the father of U.S. Congressional Representatives Lawrence J. Connery and William P. Connery, Jr.

References

Notes

1855 births
1928 deaths
Mayors of Lynn, Massachusetts

redpajama_set_name: RedPajamaWikipedia


text:
Q: Rigorous proof of shifting of center during integration In statistical mechanics, we see a lot of high-dimensional integrals. When using only a pairwise potential, we tend to "integrate away" one of the dimension by shifting it to the origin.
For example, in the energy equation,
$$U^{ex} = \int \int u(|\mathbf{r}_1-\mathbf{r}_2|)g(|\mathbf{r}_1-\mathbf{r}_2|)d\mathbf{r}_1 d\mathbf{r}_2$$
Now they say, shifting $\mathbf{r}_1$ to be the origin, they say they can integrate out $\mathbf{r}_1$ and say that
$$U^{ex} = V\int u(r)g(r)d\mathbf{r}$$
I want to know the step-by-step procedure of how something like this is allowed. I know the standard basics of integration by parts and substitution, but I have not been able to come up with ways to drag this high 3N (N=2) dimensional integral down to a one dimensional, isotropic integral.
Any advice you have would be appreciated.

A: Define $\vec{r}=\vec{r}_1-\vec{r}_2$, treating $\vec{r}_2$ as a constant in the inner integral. Then,
$$
d^3r_1 = d^3u,
$$
because the Jacobian of the transformation clearly has determinant one. Assuming we are integrating over "all space", the limits are the same, and under the transformation,
$$
|\vec{r}_1-\vec{r}_2|=|\vec{r}|=r.
$$
Then, the integrand no longer depends on $\vec{r}_2$, and so we just integrate a constant over "all space", which brings out a factor of the volume $V$ of "all space".  (Note of course that we are doing some standard physics "tricks", which is partly to assume that we are integrating over a region large enough so that the shift in the coordinates doesn't make a difference.  So, asking for "rigorous" in physics isn't quite right, but perhaps the transformation above is what you were looking for.)

redpajama_set_name: RedPajamaStackExchange


I had my old west bend for 51 years, and it has been well used, and it had it's problems. The stainless looks wonderful, is easy to clean, and cooks beautifully, and is a larger capacity. I hope i get as many years of use as the old one.
I love the versatility and the option of being able to use the pot on the stove and in the oven. It makes browning meat, etc. The size is also a major plus. Great for storing, but not lacking in space.
I had a very old version of this slow cooker and loved it. The base broke from dropping, but still worked with duck tape, and the pot lost its no-stick surface, so i decided it was time to replace it. This newer version is no longer made in the usa, and the quality is not quite as good, but it still works like i want it too. After my first time to use this new pot it became warped and was not able to sit flat upon the cooking surface. I was unable to remove spots within the pot from searing my meat before cooking. I complained about the pot and immediately received a brand new pot (and a new glass cover, which i really didn't need) at no charge. So i am very grateful to west bend for their excellent customer service. Despite the replacement and the excellent customer service, i must report that the new pot also warped. The company desperately needs to do something to replace these poorly made pots.
, Perfect for me !
This is the third cooker like this that i have purchased and i have loved everyone of them. The only thing that i don't like about this one is the handles are hot to touch. It would have been nice if they were manufactured so that a person could pick the pot up without getting burned. Other wise i would have given this cooker five stars.
This cooker was a disappointment in the fact, the actual cooking vessel did not sit flat on the cooking base. It had a left to right wobble. I would imagine that is why some people said it had hot spots (center was higher than the sides) i was upset about this fact and my better half took it to the work shop and i have no idea what he did but it is now flat to the cooking surface. Be aware of this and you will not be disappointed.
I've wandered into those kitchen shops where you drool over everything and can't afford anything. And i lusted after one of their upscale brand slow cookers that move from stovetop burner to oven to fridge with ease (for a few hundred dollars). And it made me wonder whether there wasn't a cheaper way to have the utility (if not the cachet) of those things. Enter the west bend versatility slow cooker. Because the pot is really just a stainless steel dutch oven (with a teflon base to prevent sticking), you can do whatever you want with it–including putting it on the burner unit that comes with it. But you can also use it to brown your meat on the stove in the same pot you use for slow cooking (saving you the trouble of putting a second pot in the dishwasher). You can also dispense with the base and put the thing right in your oven, if you choose. And it's light enough that you won't dislocate a shoulder doing so. But the mystique of the crock pot, with the stoneware container emitting radiant heat all around your roast or stew–is gone. What you get here is–just a pot and a spare burner.
I highly recommend the west bend 6-quart versatility slow cooker in stainless. I had to replace my 35+ year old butterscotch-color west bend slo-cooker plus. The glass cover was chipped on the edge. The no-stick interior had worn spots, but no peeling. It continued to cook well, so i was lazy about doing the research to replace it. My husband bought a crockpot slow cooker for me in december, 2014. All it did was boil no matter what setting it was on. The cooking elements in that type of ceramic crockpot go around the bottom and up the sides. We returned it, thanks to a very liberal warehouse return policy.
My sister told me she didn't have a slow cooker and asked to borrow mine. So i told her she could have my 4 qt west bend and i could get this new 6 qt one. I love versatility, so i knew i wanted another west bend. I like making pots of chili, soups and casseroles. The stainless steel pot looks so nice is easy to clean. My daughter likes to make scrambled eggs and pancakes on the grill and has no problem with cleaning it. The glass cover makes it nice to see inside and the big handle keeps my daughter from burning her knuckles when she lends a hand to stir the pot. This slow cooker has my vote and my daughters.
I would recommend this product as it cooks the food good, and i love the grill part where you can cook a grill sandwich. The only thing l was concern about was that the inside part coating has scraped off some. But the looks of it is very appealing and l love the high and low temps.
This holds a lot, can go on top of the stove and even in the oven.
Way nicer to use than the typical crock pot. One bought for myself and one for my sister-in-law.
This is the only slow cooker i enjoy. With seperate pot and heater unit you can brown food first, or get it boiling, then place on heater to slow cook. I have notticed lowest setting is too cool for serving temperature.
I had one like this years ago, and i loved it. It's taken me a long time to find this one. It's much easier do deal with than the crockpots with the stoneware insert – those weigh a ton.This one is light and durable. It cleans up well – goes right in the dishwasher. The pot is totally separate from the base.
As long as i live, this is the only slow cooker i will own. It doesn't stick, it doesn't burn, it cooks evenly and i can leave it on all day long even if i'm away from the house. I've owned only this slow cooker for many years.
This is my second versatility, the first one being the basic black version. I ordered the second one because the nonstick coating on mine was coming off after quite a lot of use. The only problem with this one is that the bottom of the pan tends to warp when used on the stovetop. Also, i don't need the grill feature. Otherwise i'd recommend this line of slow cookers if you want to brown meat straight in the pan, something i prefer. There is another make of versatile cooker out there, but it costs in the order of $250 so this is a good low cost alternative.
I grew up knowing the west bend slow cooker as the go-to cooking appliance in our home. From appetizers to desserts, my grandma & mom cooked it all. So of course i received one when i moved out and used it for many years until the internal coating became too scratched up. It was difficult to find a replacement as this particular slow cooker is not found in stores. Once i located them on amazon, i purchased the 6 qt. As i now feed a growing family. I am happy with the cooker's performance and results (tender roasts, perfectly cooked one dish meals, soups, using the griddle in camper trailer, having a temp control dial not just low-high, etc. ) but the construction is disappointing. At purchase, i realized it had a different lid than its predecessor ( glass lid could be used as a serving tray) and it just doesn't provide the same fit\seal as older version. The pot is also a lighter material and it was 'warped -wobbly' from first use.
I have purchased at least 5 of these cookers in the last 10 years. My only issue with them is that the teflon coating is thin and even using wooden or plastic spoons does not prevent wearing away of the surface. What i love is the fact that the pot can be used on top of the stove thereby making starting dishes on a burner possible( i. Browning hamburger to put in spaghetti sauce, sautéing onions). Also the temperature setting is unlimited- not just low, medium or high. I use mine all of the time and love it.
As i said we still use the old one to keep things warm. Trusting this will last 10 years plus. Service at amazon was again wonderful. Package came in a few days very well packed ******** we will buy from amazon again and again.
It was time to retire my 26-year-old west bend slow cooker. Not because it quit working, but because the non-stick surface was beginning to fade a little. As i shopped at a lot of local stores, the only slow cookers i could find were the ones that had temperature settings of just low and high. Not good enough for me, as my west bend had a sweep-able thermostat that went from low, to medium, to high. What really surprised me was fact that i could not find the west bend brand anywhere. I figured they must have gone out of business. So maybe retirement wasn't going to happen after all. And i was not going to compromise. Not being able to find anything locally, i turned to amazon. And as you have found out as well, west bend is alive and well here. After reading some of the reviews, i settled for this model. Now first off, i'm not a slow cooker fanatic. I use it for spaghetti sauce, beef stew, chili, roasts, and soups. Yes its got a sweep-able thermostat. That in itself was enough for me. Some folks dislike the fact that this cooker is probably not one to leave unattended all day. Which is true, to some degree.
I love this versatility slow cooker. It is the 3rd one i have purchased. One for my main residence and one for my winter home. The first one was damaged in shipping from one place to the other that is the reason for purchasing the next 2.
Nice slow cooker , like the element being at the bottom.
I liked this slow cooker when i got it and used it until the bottom warped and had to be replaced. I went back to the crockpot. I feel i have better control over the cooking in the crockpot.
It is a replacement for one i've had for 25 years & i love it. Works great & it's bigger than the one i had.


text:
More on the SEC investigation
The industry-wide probe of videogames companies by the Securities and Exchange Commission is now known to extent to four publishers - but key analysts believe that the investigation will exonerate the industry's accounting practices.
News by Rob Fahey Contributing Editor
It has been confirmed that an industry-wide probe of videogames companies by the Securities and Exchange Commission now covers four publishers - but key analysts believe that the investigation will exonerate the industry's accounting practices.
Yesterday we reported that Activision, Acclaim and THQ had all confirmed that they had been approached for information by the SEC [read more]; today it has emerged that Midway has also been asked to furnish accounting details. In ways, this means that the probe actually covers five publishers - Take Two was already under official investigation by the SEC, and it's thought that that investigation may have been the catalyst for this wider probe of the industry.
It's not clear which other publishers may have been approached for information, but it's hard to believe that any broadly-focused probe of the games industry would leave out key companies such as Electronic Arts.
Stocks in publishers have come under pressure as a result of the probe, as markets reacted nervously to what is an unusually broad investigation by the SEC. However, Wedbrush Morgan analyst Michael Pachter believes that the investigation is likely to exonerate the accounting practices of the leading publishers who have been approached for information.
He believes that the SEC is seeking to uncover any possibility of game companies smoothing over earnings figures by maintaining large reserves, a practice which is carried out by a number of leading companies in the industry. "We think that the SEC investigation is unlikely to lead to criminal or disciplinary action for any of the companies... It is our belief that these companies properly account for sales of their products, and that each is appropriately conservative in maintaining its reserves," according to Pachter.
UBS Investment Research analyst Michael Wallace, meanwhile, believes that the investigation may be focused on the practice of "stuffing the channel" prior to quarterly results being posted - a way of manipulating reserves by shipping unwanted products to retailers in time for quarterly figures, and then taking it back via excess reserves in the next quarter.
It's this kind of practice which led to the investigation of Take Two, which was forced to restate nearly two years worth of quarterly results last year due to the way it recognised revenues in its accounting.
"We believe that there is some potential that the SEC will require or recommend that the companies shift to 'sell-through' recognition of revenues, which could cause increased volatility of earnings going forward," explains Pachter. He sees the drop in share prices caused by the investigation as an opportunity for investors, however, commenting that due to the low probability of any adverse action resulting from the probe, this is a chance for investors to "buy on the dip".
Rob Fahey
Rob Fahey is a former editor of GamesIndustry.biz who spent several years living in Japan and probably still has a mint condition Dreamcast Samba de Amigo set.

redpajama_set_name: RedPajamaCommonCrawl


Perfect for events and exhibitions our Pop Up Display Units attract attention and add impact for your brand.
We print the PVC panels that are incorporated into this clever apparatus that can be set up and disassembled in minutes.
Supplied in a hard case and lights come as standard.


After deciding I wanted an older PUG for my next "product machine" I called Richard at Route 6x6. These old workhorses are not easy to find but as usual Richard came through and located a worn out early 1970's unit that had been used on a farm. A few of these machines were manufactured with hydraulic dump beds (vs. personnel carrier beds) which this one had making it an even better find.
Once back at the shop I cleaned the layers of "cement" hard manure and grime from the PUG and found it to be rust free under the protection of all that crust. The fiberglass body was broken up and missing sections, but repairable. The first thing I noticed after beginning the tear down was how well the PUG was built compared to some of the other machines of this era. The 3 speed transaxle and rear end are massive as are the frame sections, Pugs were defiantly designed with hard work in mind. The transaxle came apart easy enough and looked new once inside. After hot tanking all the parts it was reassembled using all new bearings and seals as a precaution. Most all the drive train parts are standard ag. components which were easy to locate. After learning that the brakes are from a 1964 Ford Fairlane a trip to NAPA provided me with all the necessary parts for a complete rebuild. The original engine was either a 10 or 16 HP Briggs. After some thought whether to rebuild the engine or replace it I decided to go the latter route. The modern air cooled engines of today are just too good. The 12 HP Kawasaki is incredibly smooth and provides plenty of power, especially in the ultra low 1st gear.
One issue that sometimes has to be dealt with when replacing an older engine with a modern one is the fact that the new engines idle faster. This is because the older engines were splash lubricated vs. the pressure lubricated modern engines. The faster idle supplies the needed oil but due to belt drag sometimes spins the transmission just enough to grind gears when shifting from neutral into gear. Shifting from a gear to neutral if the transmission is still spinning is not a problem. I addressed this by installing a small motorcycle brake disc on the output shaft of the transmission. A go-kart caliper on the disc and bicycle cable/lever attached to the shift lever actuate the brake for shifting and provide an additional machine brake that actually stops the PUG instantly.
The PUG is a fun machine to drive even though the speed sensitive driven clutch (vs. a torque sensitive clutch) and articulated steering take some getting used to. Besides the articulated steering both body halves also pivot independently on a front to rear horizontal axis. This feature allows the PUG to crawl over very difficult terrain while keeping the tires in contact with the ground. Even with open diffs it gets amazing traction.
With regular maintenance and inside storage this machine should easily last another 35 years.
-All steering parts replaced or rebuilt.
-Combination roll cage, bumper and winch mount fabricated.
-Installed new seats that fold up for engine/trans access.