text:
\section{Introduction}
\label{intro}
The identification of the constituents that can promote the inflationary period and the present accelerated era of the Universe is still an object of intense investigation. Generally one admits the existence of an exotic component in the Universe, with negative pressure, which is responsible for the current cosmic acceleration \cite{Zlatev, RevModPhys.75.559, Binder, RudiRaila}. The first candidate to represent this \textit{dark energy} was the cosmological constant, but other models for dark energy were also proposed. For the inflationary period, most of the models suppose a scalar field coupled with the gravitational field \cite{RevModPhys.75.559, PhysRevLett.80.1582, PhysRevD.62.063508}. Fermionic fields were also employed as candidates for the \textit{inflaton} or dark energy \cite{Sahagengrav, PhysRevD.69.124010, PhysRevD.74.124030, Marlosprd, Marlosepl, Rudifermions, fermioniccosmologies}. Important extending works concerning the fermionic fields in cosmology can be found in the references \cite{Piccongreene, Phys.Rev.D77:123535, Class.Quant.Grav.25:165014, Phys.Rev.D80:023501, Physics Letters B 718, gr-qc 1402.5880}.

The fact that common matter cannot account for the total matter of the Universe is another old problem which is still without a solution. Several candidates for \textit{dark matter} were proposed to explain this problem. In face with the problems of the dark matter and dark energy - the so-called dark sector - works have been made with two different fields to represent the dark matter and dark energy. Some of these works also admit an interaction between these fields \cite{macorra, rudicanonical, PhysRevD.79.063518}.

The aim of the present work is to describe a spatially flat homogeneous and isotropic Universe whose constituents are a fermionic field, a baryonic matter field (dust) and a radiation field (radiation and non-relativistic neutrinos). We investigate two models: i) a fermion field non-minimally coupled with the gravitational field and ii) radiation, matter (baryonic), dark matter and dark energy fields, where the dark sector and inflaton are described by the fermionic field. The first one is interesting for the study concerning the possibility that the fermion field plays the role of the inflaton. In the second one we analyse the whole evolution of the Universe, comprising its decelerated and accelerated eras (from inflation through radiation and matter eras up to the present acceleration).

The forms of the gravitational coupling and the self-interaction potential are obtained via Noether symmetry for the generic point-like Lagrangians of the two cases. Several works have already used the Noether symmetry approach to search for the forms of the coupling and potential of quintessence models \cite{PhysRevD.42.1091, PhysRevD.44.3136, Capozzielonoether, PhysRevD.80.104030, Demianskiastro, Rudifermions, Capozzielofder, Ruditaquion, Zhangyun2010, Sharif2013}. The Noether symmetry can be seen as a first principle for choosing the unknown functions of a given Lagrangian instead of an \textit{ad hoc} procedure. Our basic field equations in each case follow from the generic point-like Lagrangian corresponding to the action of each model. The resulting Dirac equations coupled with gravity and modified Friedmann equations are then solved for the couplings and potentials found through the Noether symmetry approach.

This paper is structured as follows: in the second section we present the general action from which the field equations of the two cases will be derived. In the third section we derive the Einstein and Dirac equations from the point-like Lagrangian for the first case. The Noether symmetry imposition for this case is done in subsection {3.1} and in subsection {3.2} we obtain the respective cosmological solutions. The point-like Lagrangian for the more general case and its respective field equations are presented in section {4}. The Noether symmetry analysis and the cosmological solutions for this case are done in subsection {4.1} and {4.2}, respectively. The last section is reserved for final remarks and conclusions. In this work we will adopt the metric signature $ (+,-,-,-) $ and the natural units $ 8\pi G=c=\hbar=1 $.

\section{Action}
\label{field equations}
We are interested in investigating a Universe modeled by a mixture whose constituents are the fermion field, matter and radiation. The action for a fermion field non-minimally coupled with the gravitational field reads
\ben\no
S&=&\int d^{4}x\sqrt{-g}\left\{{F(\Psi)R+\frac{i}{2}\left[\overline{\psi}\Gamma^{\mu}D_{\mu}\psi-({\overline{D}_{\mu}\overline{\psi}})\Gamma^{\mu}\psi\right]-V(\Psi)}\right\}\\\label{1}
 &+&S_{m}+S_{r},
\een
where $ R$ is the Ricci scalar, $\psi$ and $\overline \psi=\psi^\dag \gamma^0$ are the spinor field and its adjoint, respectively. In agreement to the general covariant principle  the Pauli matrices are replaced by  $ \Gamma^{\mu}=e^{\mu}_{\nu}\gamma^{\nu} $ where $e^{\mu}_{\nu}$ are tetrad fields. The generalized Dirac matrices obey the Clifford algebra $ \lbrace\Gamma^{\mu},\Gamma^{\nu}\rbrace=2g^{\mu\nu} $. $ F(\Psi) $ is the (generic) function that describes the coupling and $ V(\Psi) $ the self-interaction potential density of the fermionic field. Papers that consider $ F $ and $ V $ as functions of the bilinear $ \Psi=\overline{\psi}\psi $ are commonly found in the literature \cite{Marlosprd, Rudifermions} and the few ones that consider $ F $ and $ V $ as functions of the pseudo-scalar $ \Psi=\overline{\psi}\gamma^{5}\psi $ do it in an \textit{ad hoc} way \cite{Sahagengrav, PhysRevD.69.124010, Marlosprd}. Here we shall consider that $ F $ and $ V $ are only functions of the pseudo-scalar $ \Psi=\overline{\psi}\gamma^{5}\psi $ which will be found by the Noether symmetry criterion. Furthermore, $ S_m $ is the action of the matter field and $ S_r $ the action of the radiation field. The covariant derivatives in (\ref{1}) read
\ben\label{1a}
D_\mu\psi= \partial_\mu\psi-\Omega_\mu\psi,\qquad
D_\mu\overline\psi=\partial_\mu\overline\psi+\overline\psi\Omega_\mu,\\\Omega_\mu=-\frac{1}{4}g_{\rho\sigma}\left[\Gamma^\rho_{\mu\delta}
-e_b^\rho\left(\partial_\mu e_\delta^b\right)\right]\Gamma^\delta\Gamma^\sigma.
\een
Here $\Omega_\mu$ denotes  the spin connection
and $\Gamma^\nu_{\sigma\lambda}$  the Christoffel symbols.

We will firstly analyse the  case where the fermion field non-minimally couples with the gravitational field in the absence of matter and radiation, i.e., $ S_m =0 $ and $ S_r = 0$.

\section{Field equations for the fermion field}

For a spatially flat Friedmann-Robertson-Walker metric $ds^2=dt^2-a(t)^2(dx^2+dy^2+dz^2)$ --
where $a(t)$ denotes the cosmological scale factor -- the  Dirac-Pauli matrices and spin connection become
\begin{equation}\label{1c}
\Gamma^0=\gamma^0,\qquad
\Gamma^i=\frac{1}{a(t)}\gamma^i, \qquad \Omega_0=0, \qquad\Omega_i=\frac{1}{2}\dot a(t)\gamma^i\gamma^0,
\end{equation}
with the dot denoting time derivative.

In this case we can obtain -- after a partial integration of the action (\ref{1}) (with $ S_m =0 $ and $ S_r = 0$) -- the point-like Lagrangian
\begin{equation}\label{2}
\mathcal{L} =6a\dot{a}^{2}F+6a^{2}\dot{a}\dot{\Psi}F^{\prime}+a^{3}\frac{i}{2}\left(\dot{\overline{\psi}}\gamma^{0}\psi-\overline{\psi}\gamma^{0}\dot{\psi}\right)+a^{3}V.
\end{equation}
Here the derivative with respect to the pseudo-scalar $ \Psi $ is represented by a prime.

The Dirac equations for the spinor field and its adjoint follow from the Euler-Lagrange equations for $ \psi $ and $ \overline{\psi} $ applied to the Lagrangian (\ref{2}), namely,
\ben\label{3}
\dot{\overline{\psi}}+\frac{3}{2}H\overline{\psi}-iV^{\prime}\overline{\psi}\gamma^{5}\gamma^{0}+6iF^{\prime}\overline{\psi}\gamma^{5}\gamma^{0}\left(\dot{H}+2H^{2}\right)=0,
\\\label{4}
\dot{\psi}+\frac{3}{2}H{\psi}+iV^{\prime}\gamma^{0}\gamma^{5}{\psi}-6iF^{\prime}\gamma^{0}\gamma^{5}{\psi}\left(\dot{H}+2H^{2}\right)=0,
\een
where $ H=\dot{a}/a $ denotes de Hubble parameter.

From the Euler-Lagrange equations for $ a $ applied to the Lagrangian (\ref{2}) we obtain the acceleration equation
\begin{equation}\label{5}
\frac{\ddot{a}}{a}=-\frac{\rho_{f}+3p_{f}}{12F},
\end{equation}
where the energy density and pressure of the fermion field, read
\ben\label{6a}
\rho_{f}=V-6HF^{\prime}\dot{\Psi},\\\lb{6b}
p_{f}=\left[V^{\prime}-6F^{\prime}(\dot{H}+2H^{2})\right]\Psi -V+2\left(\ddot{\Psi}F^{\prime}+\dot{\Psi}^{2}F^{\prime\prime}+2\dot{\Psi}F^{\prime}H\right).
\een

Finally by imposing that the energy function associated with the Lagrangian (\ref{2}) vanishes:
\ben\label{8}
E_{\mathcal{L}}=\frac{\partial \mathcal{L}}{\partial \dot{a}}\dot{a}+\dot{\overline{\psi}}\frac{\partial \mathcal{L}}{\partial \dot{\overline{\psi}}}+\frac{\partial \mathcal{L}}{\partial \dot{\psi}}\dot{\psi}-\mathcal{L}=0,
\een
we get the Friedmann equation
\begin{equation}\label{9}
H^{2}=\frac{V-6HF^{\prime}\dot{\Psi}}{6F}=\frac{\rho_{f}}{6F}.
\end{equation}

Note that we supposed that the spinors which are only functions of time are compatible with the homogeneity and isotropy of a flat Friedmann-Robertson-Walker metric. In principle, this can be achieved for the classical spinors we are considering once the corresponding energy-momentum tensor is not anisotropic, so that the description of the energy density and pressure of the fluid is consistent with this metric. However, even before a semiclassical quantization of the spinor field in the presence of gravity, the ansatz for fermions (classical) compatible with the Friedmann-Robertson-Walker background is not a trivial subject. The marriage of the fermions and homogeneous and isotropic geometry of general relativity is discussed by references \cite{Christodoulakis, Christodoulakis1} in the context of Dirac fields concerning the classical field equations and Dirac's formalism for constrained Hamiltonian systems. The corresponding results appear to be in favor of the compatibility of the ansatz $\psi(t)$ with a homogeneous and isotropic Universe. The mentioned references are on the long road of investigation on the issue of quantum fields living in the space-time of general relativity, focused on a semiclassical approach in the cosmological context. Former papers which attacked this problem can be found in references \cite{Nelson, Nelson1}, as well as an extension of these approaches can be seen in \cite{D'Eath}. It is important to point out that the effective behavior of the cosmological fluid is mathematically described by a classical spinor field, which is not necessarily saying that the fluid is fundamentally composed by fermionic particles which are likely to be detected. In principle, classical spinor fields do not have anything to do with real fermionic particles. In fact, physically speaking, fermionic particles only exist in the quantum and relativistic level, whose marriage produces the known Dirac equation. The same can be said when describing inflation with a classical scalar field. Such a field models the effective behavior of an exotic fluid, but this not necessarily mean that the fluid is composed by unknown (or known) bosonic particles. A clear discussion on cosmological applications of classical fermion fields can be also found in the work \cite{Magueijo}.

\subsection{Noether symmetry}

In terms of the components of the spinor field, $ \psi=(\psi_{1},\psi_{2},\psi_{3},\psi_{4})^{T} $ and its adjoint $ \overline{\psi}=(\psi^{\dagger}_{1},\psi^{\dagger}_{2},-\psi^{\dagger}_{3},-\psi^{\dagger}_{4}) $, the Lagrangian (\ref{2}) can be written as
\ben\no
 \mathcal{L}=6a\dot{a}^{2}F+6a^{2}\dot{a}F^{\prime}\sum_{i,j=1}^4 \left(\dot{\psi}^{\dagger}_{i}\psi_{j}+\psi^{\dagger}_{i}\dot{\psi_{j}}\right)\varepsilon_{ij}
\\\label{10}
+\frac{i}{2}a^{3}\sum_{i=1}^4 \left(\dot{\psi}^{\dagger}_{i}\psi_{i}-\psi^{\dagger}_{i}\dot{\psi_{i}}\right)+a^{3}V,
\een
which is only function of ($ a $, $ \dot{a}$, $\psi^{\dagger}_{l} $, $ \dot{\psi}^{\dagger}_{l} $, $\psi_{l} $, $\dot{\psi}$). In the Lagrangian (\ref{10}) it was introduced the symbol $ \varepsilon_{ij} $, which assumes the values
 \begin{equation}\label{11}
\left\{
  \begin{array}{lllll}
    \varepsilon_{ij}=-1  & \hbox{for} &\varepsilon_{13} &\hbox{or}  &\varepsilon_{24}, \\
    \varepsilon_{ij}=+1 & \textrm{for} & \varepsilon_{31} & \textrm{or} &\varepsilon_{42}, \\
   \varepsilon_{ij}=0 &  \textrm{otherwise.} & & &
  \end{array}
\right.
  \end{equation}

The Noether symmetry is satisfied by the condition (see Appendix)
\begin{equation}\label{12}
L_{\textbf{X}}\mathcal{L}=\textbf{X}\mathcal{L}=0,
\end{equation}
where \textbf{X} is the infinitesimal generator of symmetry defined by
\begin{equation}\label{13}
\textbf{X}=C_0\frac{\partial}{\partial a}+\dot{C_0}\frac{\partial}{\partial \dot{a}}+\sum_{l=1}^{4}\left(C_{l}\frac{\partial}{\partial \psi_{l}^{\dagger}}+D_l\frac{\partial}{\partial \psi_l}+\dot{C_l}\frac{\partial}{\partial \dot{\psi_{l}}^{\dagger}}+\dot{D_l}\frac{\partial}{\partial \dot{\psi_l}}\right),
\end{equation}
and $ L_{\textbf{X}} $ is the Lie derivative of $ \mathcal{L}$ with respect to the vector \textbf{X} which is defined in the tangent space. Furthermore, the parameters $ C_0 $, $ C_l $ and $ D_l $ are arbitrary functions of ($ a $, $ \psi_{l} $, $ \psi^{\dagger}_{l} $).

Applying the condition (\ref{12}) to the Lagrangian (\ref{10}), one has an equation whose terms depend explicitly on $ \dot{a}\dot{\psi}_l^{\dagger} $, $ \dot{a}\dot{\psi}_l $, $ \dot{a}^2 $, $ \dot{\psi}_l^{\dagger} $, $ \dot{\psi}_l $, $ \dot{\psi}_l^{\dagger}\dot{\psi}_m $, $ \dot{\psi}^{\dagger}_l\dot{\psi}^{\dagger}_m $, $ \dot{\psi}_l\dot{\psi}_m $ and $ \dot{a} $. By equating the coefficients of the referred terms to zero, we obtain the following system of coupled differential equations
\ben\no
 F^{\prime}\psi_j\varepsilon_{lj}\left(2C_0+a\frac{\partial C_{0}}{\partial a}\right)+2F\frac{\partial C_0}{\partial \psi^{\dagger}_l}+aF^{\prime\prime}\psi_j\varepsilon_{lj}\sum_{i,k=1}^{4}\left(C_i\psi_k+\psi^{\dagger}_{i}D_k\right)\varepsilon_{ik}
\\\label{14}+aF^{\prime}\left[D_{j}\varepsilon_{lj}+\sum_{i,k=1}^{4}\left(\frac{\partial C_i}{\partial \psi_{l}^{\dagger}}\psi_k+\psi^{\dagger}_{i}\frac{\partial D_k}{\partial \psi_{l}^{\dagger}}\right)\varepsilon_{ik}\right]=0,
\\\no
 F^{\prime}\psi^{\dagger}_{j}\varepsilon_{jl}\left(2C_{0}+a\frac{\partial C_{0}}{\partial a}\right)+2F\frac{\partial C_{0}}{\partial \psi_{l}}+aF^{\prime\prime}\psi^{\dagger}_{j}\varepsilon_{jl}\sum_{i,k=1}^{4}\left(C_{i}\psi_{k}+\psi^{\dagger}_{i}D_{k}\right)\varepsilon_{ik}
\\\label{15}
+aF^{\prime}\left[C_{j}\varepsilon_{jl}+\sum_{i,k=1}^{4}\left(\frac{\partial C_{i}}{\partial \psi_{l}}\psi_{k}+\psi^{\dagger}_{i}\frac{\partial D_k}{\partial \psi_l}\right)\varepsilon_{ik}\right]=0,
\\\no
C_{0}F+2aF\frac{\partial C_0}{\partial a}+aF^{\prime}\sum_{i,j=1}^{4}\left(C_{i}\psi_{j}+\psi_{i}^{\dagger}D_{j}\right)\varepsilon_{ij}
\\\label{16}
+a^{2}F^{\prime}\sum_{i,j=1}^{4}\left(\frac{\partial C_i}{\partial a}\psi_j+\psi^{\dagger}_{i}\frac{\partial D_j}{\partial a}\right)\varepsilon_{ij}=0,
\\\label{17}
3C_{0}\psi_l+aD_l+a\sum_{i=1}^{4}\left(\psi_{i}\frac{\partial C_i}{\partial \psi^{\dagger}_{l}}-\psi^{\dagger}_{i}\frac{\partial D_i}{\partial \psi^{\dagger}_{l}}\right)=0,
\\\label{18}
3C_{0}\psi^{\dagger}_{l}+aC_l-a\sum_{i=1}^{4}\left(\psi_{i}\frac{\partial C_i}{\partial \psi_{l}}-\psi^{\dagger}_{i}\frac{\partial D_i}{\partial \psi_{l}}\right)=0,
\\\label{19}
F^{\prime}\left(\psi^{\dagger}_{i}\frac{\partial C_0}{\partial \psi^{\dagger}_{j}}+\psi_{j}\frac{\partial C_0}{\partial \psi_i}\right)\varepsilon_{ij}=0,\qquad F^{\prime}\left(\psi_{j}\frac{\partial C_0}{\partial \psi^{\dagger}_{m}}+\psi_{m}\frac{\partial C_0}{\partial \psi^{\dagger}_{j}}\right)\varepsilon_{ij}=0,
\\\label{20}
F^{\prime}\left(\psi^{\dagger}_{i}\frac{\partial C_0}{\partial \psi_{l}}+\psi^{\dagger}_{l}\frac{\partial C_0}{\partial \psi_{i}}\right)\varepsilon_{ij}=0,\qquad \sum_{i=1}^{4}\left(\psi_{i}\frac{\partial C_i}{\partial a}-\psi^{\dagger}_{i}\frac{\partial D_i}{\partial a}\right)=0.
\een

There remains an equality that involves the potential density, namely,
\begin{equation}\label{21}
3C_{0}V+aV^{\prime}\sum_{i,j=1}^{4}(C_{i}\psi_{j}+\psi^{\dagger}_{i}D_j)\varepsilon_{ij}=0.
\end{equation}

The above system with 55 differential equations, (\ref{14}) through (\ref{21}), will be examined in the following. From (\ref{19}) and $ (\ref{20})_{1} $ we infer that there are two possibilities for the coupling, $ F^{\prime}=0 $ and $ F^{\prime}\neq0 $. These two possibilities will be analyzed separately.

\textbf{(1) Case} $ F'=0 $:

If $ F'=0 $ it follows that $ F= $ constant and equations (\ref{19}) and $ (\ref{20})_{1} $ are automatically satisfied. Furthermore, from equations (\ref{14}) and (\ref{15}) we have that $ C_0 = C_0 (a)$. Hence, equation (\ref{16}) determines the form of $C_0$, namely,
$C_{0}=\lambda/a^{1/2}$,
where $ \lambda $ is a constant. Equations $ (\ref{20})_{2} $ and (\ref{17})  furnish an expression for $ C_j $ and $ D_j $
\begin{equation}\label{23}
C_j=-\frac{3}{2}\frac{\lambda\psi^{\dagger}_{j}}{a^{3/2}}+\beta\psi^{\dagger}_{i}\varepsilon_{ij},\qquad D_j=-\frac{3}{2}\frac{\lambda\psi_{j}}{a^{3/2}}+\beta\psi_{i}\varepsilon_{ij},
\end{equation}
with $ \beta $ being a constant.

Finally, from (\ref{21}) we determine the potential
\begin{equation}\label{24}
V=\alpha\Psi,
\end{equation}
where $ \alpha $ is a constant.

\textbf{(2) Case} $ F'\neq 0 $:

Now we will analyse the case where $ F $ can be an arbitrary function of the pseudo-scalar $ \Psi $. For this end, we write equation (\ref{21}) as follows
\begin{equation}\label{25}
\sum_{i,j=1}^{4}(C_{i}\psi_{j}+\psi^{\dagger}_{i}D_j)\varepsilon_{ij}=-3\frac{C_0}{a}\frac{V}{V^{\prime}}.
\end{equation}
By the differentiation of the above equation with respect to $ a $, one has
\begin{equation}\label{26}
\sum_{i,j=1}^{4}\left(\frac{\partial C_{i}}{\partial a}\psi_{j}+\psi^{\dagger}_{i}\frac{\partial D_j}{\partial a}\right)\varepsilon_{ij}=\frac{3}{a}\frac{V}{V^{\prime}}\left(\frac{C_0}{a}-\frac{\partial C_0}{\partial a}\right).
\end{equation}

	Now we insert equations (\ref{25}) and (\ref{26}) into (\ref{16}) and, recalling that $ F $ and $ V $ are only functions of $ \Psi $, the corresponding result is
\begin{equation}\label{27}
\frac{a}{C_0}\frac{\partial C_0}{\partial a}=\frac{FV^{\prime}}{3F^{\prime}V-2FV^{\prime}}=k,
\end{equation}
	where $ k $ is a constant.

	By analyzing equations (\ref{19}) and $ (\ref{20})_{1} $ with $ F^{\prime}\neq0 $, we infer that $ C_0 $ does not depend on $ \psi^{\dagger}_{j} $ and $ \psi_{j} $, i.e., $ C_0=C_{0}(a) $. Then, from equation (\ref{27}) we obtain the solution $C_0=\lambda a^{k}$, where $ \lambda $ is a constant.

	By combining equations $ (\ref{20})_{2} $ with (\ref{17})  we find the expressions for $ C_j $ and $ D_j $, respectively,
\begin{equation}\label{29}
C_j=-\frac{3}{2}\lambda a^{k-1}\psi^{\dagger}_{j}+\beta\psi^{\dagger}_{i}\varepsilon_{ij}\qquad D_j=-\frac{3}{2}\lambda a^{k-1}\psi_{j}+\beta\psi_{i}\varepsilon_{ij}.
\end{equation}

	From equation (\ref{21}) it follows that the potential has the same linear form obtained in the case for $F'=0$, namely, $V=\alpha\Psi.$ From equations (\ref{24}) and (\ref{27}) we obtain a power-law function for the coupling, $ F=\omega \Psi ^{p} $, where $ \omega $ is a constant. The exponent $p$ of the power-law coupling results from equations (\ref{15}), (\ref{16}) and  (\ref{27}) and must satisfy the following relationships
 \ben\label{30}
\left\{
  \begin{array}{ll}
    3pk=1+2k, \\
   3p=2+k,
  \end{array}
\right.
\qquad\hbox{which implies}\qquad
\left\{
  \begin{array}{ll}
    (1,1), \\
    (-1,1/3),
  \end{array}
\right.
\een
from which we determined $p$ and $ k $.

	Hence, we conclude that for $ F'\neq 0$ there are two Noether symmetries, namely,

\textit{Symmetry 1} (for $ k=1 $ and $ p=1 $):

\ben
F=\omega\Psi,\qquad V=\alpha\Psi,\qquad C_{0}=\lambda a, \\
C_j=-\frac{3}{2}\lambda\psi^{\dagger}_{j}+\beta\psi^{\dagger}_{i}\varepsilon_{ij}, \qquad D_j=-\frac{3}{2}\lambda\psi_{j}+\beta\psi_{i}\varepsilon_{ij}.
\een

\textit{Symmetry 2} (for $ k=-1 $ and $ p=1/3 $):

\ben
 F=\omega\Psi^{1/3},\qquad V=\alpha\Psi,\qquad C_{0}=\frac{\lambda}{a},\\ C_j=-\frac{3}{2}\frac{\lambda\psi^{\dagger}_{j}}{a^{2}}+\beta\psi^{\dagger}_{i}\varepsilon_{ij}, \qquad D_j=-\frac{3}{2}\frac{\lambda\psi_{j}}{a^{2}}+\beta\psi_{i}\varepsilon_{ij}.
\een

\subsection{Cosmological solutions}

 From the Dirac equations of the spinor field and its adjoint coupled with the gravitational field, (\ref{3}) and (\ref{4}), we obtain a differential equation for $\Psi $,
 \begin{equation}\label{31}
\frac{d\Psi}{dt}+3\frac{\Psi}{a}\frac{da}{dt}=0,\;\;\;\;\; \textrm{so that,}\;\;\;\;\;  \Psi=\frac{\Psi_0}{a^{3}},
\end{equation}
	where $ \Psi_0 $ is a constant.

	From this result, we are able to solve the Friedmann equation and determine the time evolution of the scale factor. As we did before, the two cases $ F'=0 $ and $F'\neq0 $ will be analyzed separately.

\subsubsection{The case $F'=0$:}

The choice $F=$ constant	$=1/2$ refers to a minimal coupling of the fermionic field with the gravitational field. Hence, the Friedmann equation (\ref{9}) leads to a power-law solution for the time evolution of the scale factor, i.e.,
\begin{equation}\label{32}
a(t)=[\Omega(t-t_{0})]^{2/3},\qquad \textrm{with,}\qquad \Omega=\frac{3}{2}\sqrt{\frac{\alpha\Psi_{0}}{3}}.
\end{equation}
Such a solution describes a decelerated Universe with a matter dominated behavior.

Equations (\ref{6a}) and (\ref{6b}) furnish the energy density and pressure of the fermion field, namely,
\begin{equation}\label{33}
\rho_{f}=\frac{\alpha\Psi_0}{a^{3}},\qquad p_{f}=0.
\end{equation}
From the above expressions we conclude that the $ F'=0$ case describes a pressureless matter field.\\

\subsubsection{The case $F'\neq0$:}

Solving the Friedmann equation (\ref{9}) for the \textit{Symmetry 1}, we have the following solution for the scale factor
\begin{equation}\label{34}
a(t)=e^{\Lambda(t-t_{0})},\;\;\;\; \textrm{with,}\;\;\;\; \Lambda=\sqrt{\frac{-\alpha}{12\omega}}.
\end{equation}
The above exponential solution shows that the fermion field plays the role of an inflaton.

The corresponding energy density and pressure now read
\begin{equation}\label{35}
\rho_f=-\frac{\alpha\Psi_0}{2a^{3}},\;\;\;\;\;\;\;\;\;\;p_f=-\rho_f .
\end{equation}

	From the weak energy condition we know that the energy density is  always a positive quantity \cite{hawking}. Hence, we infer from $ (\ref{35})_1 $ that $ \alpha \Psi_{0} < 0 $. Since the coupling has to be a positive quantity, we should have $ \omega \Psi_0 > 0 $, i.e., the constants $ \alpha $ and $ \omega $ always will have opposite sings, which ensures a positive real value for $ \Lambda $. Furthermore, we can infer form equation $ (\ref{35})_2 $ that the pressure of the fermion field is always negative and proportional to the energy density.

 In this exponential scenario, we get from equation $ (\ref{31})_2 $ the following solution for the time evolution of $ \Psi $
\begin{equation}\label{36}
\Psi (t)=\Psi_{0}e^{-3\Lambda(t-t_0)}.
\end{equation}

Now one can also determine the time evolution of the energy density and pressure of the fermion field, which reads
\begin{equation}\label{37}
p_f (t)=-6\omega\Lambda^{2}\Psi_{0}e^{-3\Lambda(t-t_0)}=-\rho_f (t).
\end{equation}

Although equation (\ref{34}) indicates an eternal accelerated expansion, equations (\ref{36}) and (\ref{37}) show that the source of this expansion should come to an end, since the energy and potential densities of the fermion field tend to zero. Then another regime should take the place of the exponential phase.

For the \textit{Symmetry 2} the Friedmann equation does not have a solution.

\section{Field equations for the fermion, radiation and matter fields}

Now we will look at the action with the fermion, matter and radiation fields contributions. By integrating by parts the action (\ref{1}) for a spatially flat Friedmann-Robertson-Walker metric, we have the point-like Lagrangian
\begin{equation}\label{38}
\mathcal{L} =6a\dot{a}^{2}F+6a^{2}\dot{a}\dot{\Psi}F^{\prime}+a^{3}\frac{i}{2}\left(\dot{\overline{\psi}}\gamma^{0}\psi-\overline{\psi}\gamma^{0}\dot{\psi}\right)+a^{3}V+\rho^{0}_{m}+\frac{\rho^{0}_{r}}{a},
\end{equation}
where $ \rho^{0}_{m} $ and $ \rho^{0}_{r}$ are the energy density of the baryonic matter and radiation at a initial instant $ t=t_{0} $, respectively.

Note that the new Lagrangian (\ref{38}) does not present extra terms with respect to the spinor field in comparison to the Lagrangian (\ref{2}). Hence, the Dirac equations are the same as those given by (\ref{3}) and (\ref{4}). However, by imposing that the energy function associated with the Lagrangian (\ref{38}) vanishes, we obtain a different Friedmann equation
\begin{equation}\label{39}
H^{2}=\frac{\rho_{T}}{6F},\qquad\hbox{where}\qquad \rho_{T}=V-6HF'\dot{\Psi}+\frac{\rho^{0}_{m}}{a^{3}}+\frac{\rho^{0}_{r}}{a^{4}}
\end{equation}
 stands for the total energy density, which includes the energy densities of the matter and radiation fields. In the acceleration equation (\ref{5}) one has only to include the radiation pressure $p_r=\rho_r/3$, since the matter field is supposed to be pressureless.

\subsection{Noether symmetry}

  As we did before we rewrite the Lagrangian (\ref{38}) in terms of the components of the spinors as
\ben\no
 \mathcal{L}=6a\dot{a}^{2}F+6a^{2}\dot{a}F^{\prime}\sum_{i,j=1}^4 \left(\dot{\psi}^{\dagger}_{i}\psi_{j}+\psi^{\dagger}_{i}\dot{\psi_{j}}\right)\varepsilon_{ij}
\\\label{41}
+\frac{i}{2}a^{3}\sum_{i=1}^4 \left(\dot{\psi}^{\dagger}_{i}\psi_{i}-\psi^{\dagger}_{i}\dot{\psi_{i}}\right)+a^{3}V+\rho^{0}_{m}+\frac{\rho^{0}_{r}}{a}.
\een

 By imposing the Noether symmetry to the above Lagrangian and performing the same analysis as in previous section, we get the same system of coupled differential equations (\ref{14}) -- (\ref{20}), except for the differential equation which involves  the potential. Such an equation now reads
\begin{equation}\label{42}
C_{0}\left(3V-\frac{\rho^{0}_{r}}{a^{4}}\right)+aV^{\prime}\sum_{i,j=1}^{4}(C_{i}\psi_{j}+\psi^{\dagger}_{i}D_j)\varepsilon_{ij}=0.
\end{equation}

As in the previous section, we infer from  (\ref{19}) and $ (\ref{20})_{1} $ that there are two possibilities for the coupling, $ F^{\prime}=0 $ and $ F^{\prime}\neq0 $.

If $ F^{\prime}=0 $ it follows that $ F=$ constant and equations (\ref{19}) and $ (\ref{20})_{1} $ are identically satisfied. Equation (\ref{42}) is solved for $ C_{0}=0$ and $ V=$ constant or $ C_{i}=-f(a)\psi^{\dagger}_{i} $ and $ D_{j}=f(a)\psi_{j} $, with $ f(a) $ being an arbitrary function of $ a $. With $ C_{0}=0$ and $ F^{\prime}=0 $ equations (\ref{14}) -- (\ref{16}) are also satisfied as well as (\ref{17})  with $ C_{i}=-f(a)\psi^{\dagger}_{i} $ and $ D_{j}=f(a)\psi_{j} $. From $ (\ref{20})_{2} $ it follows that $ f(a)=\gamma $, where $ \gamma $ is a constant, and
$C_{l}=- \gamma \psi^{\dagger}_{l}$, $D_{l}= \gamma \psi_{l}$.

We note from the above results that  (\ref{42}) is also satisfied by an arbitrary potential $ V=V(\Psi) $, including $V=$ constant.
Observe that the system (\ref{14}) -- (\ref{20}) and (\ref{42}) are solved with the same symmetry for $ F^{\prime}= 0 $ and $ F^{\prime}\neq 0 $ (i.e., for an arbitrary $F$, including $F=$ constant).

Then for $C_{0}=0$, $C_l=- \gamma \psi^{\dagger}_{l}$ and $D_l=\gamma \psi_{l}$ the self-interaction potential of the fermion field is an arbitrary function of the pseudo scalar, when the
 coupling function is a constant or it is an arbitrary function of the pseudo scalar, i.e.,
\begin{equation}
\left\{
  \begin{array}{ll}
    F=\hbox{constant}\Longrightarrow V=V(\Psi), \\
    F=F(\Psi)\Longrightarrow V=V(\Psi).
  \end{array}
\right.
\end{equation}

\subsection{Cosmological solutions}
\subsubsection{The case $F'=0$:}
	The solution for the pseudo-scalar is the same as that in the previous section, namely, $ \Psi=\Psi_{0}/a^{3} $ and with $F=$ constant $=1/2 $ the fermion field is minimally coupled with the gravitational field. For the potential density we propose a linear combination of two powers of the pseudo-scalar
\begin{equation}
V(\Psi)=V_{0}\Psi ^{n} + \Lambda \Psi ^{l},
\end{equation}
where $ V_{0} $, $ \Lambda $, $ n $ and $ l $ are free parameters. Here we are working with the case where we have the contributions of matter and radiation. For a more complete model of the present status of the Universe we need the contributions of dark matter and dark energy. In this way, we may take the adequate powers as $ n=1 $ and $ l=0 $. Thus, as the energy density of the fermionic model for a minimally coupled field   according to (\ref{6a}) is $ \rho_{f}=V $, it follows that
\begin{equation}\lb{pot}
\rho_{f}=V_{0}\Psi + \Lambda =\rho_{dm}+\rho_{de}
\end{equation}
where $ \rho_{dm}=V_{0}\Psi_0/a^3 $ and $ \rho_{de}=\Lambda $ represent the energy density of the dark matter and dark energy, respectively.

The Friedmann equation for this case does not have an analytic solution for the time evolution of the scale factor. So we perform a change of variables to analyse the cosmological behavior of the proposed model. The red-shift will be used as a variable instead of time thanks to the following relationships:
\begin{equation}
\frac{1}{a}=z+1, \qquad \frac{d}{dt}=-H(1+z)\frac{d}{dz}.
\end{equation}

Afterwards, the total energy density and the total pressure can be expressed in terms of the red-shift, yielding,
\begin{equation}
\rho_{T} (z)= (V_{0}\Psi_0+\rho^{0}_{m})(z+1)^3 +\rho_{r}^{0}(z+1)^4 +\Lambda,
\quad
p_T (z) =\frac{1}{3}\rho^{0}_{r}(1+z)^{4}-\Lambda.\quad
\end{equation}

In order to compare our results with the observational data, we divide all the above equations by $ \rho_0 $ -- the critic density when $ z=0 $ -- from which one has the following expressions:
  \begin{equation}
\overline{\rho_{T}(z)}=\overline{\rho_{dm}}+\overline{\rho_{m}}+\overline{\rho_{r}}+\overline{\rho_{de}},
\qquad  \overline{p_{T}(z)}=\frac{1}{3}\overline{\rho^{0}_{r}}(z+1)^{4}-\overline{\rho_{de}},
  \end{equation}
where the bar indicates that the quantity was divided by $ \rho _{0} $ and
\begin{equation}
\overline{\rho_{m}}=\overline{\rho_{m}^{0}}(z+1)^3,\quad\overline{\rho_{r}} =\overline{\rho_{r}^{0}}(z+1)^4,
\quad
\overline{\rho_{dm}} =\overline{V_0\Psi_0}(z+1)^3,\quad \overline{\rho_{de}} =\overline{\Lambda}.\quad
\end{equation}

The plots of the density parameter were done with the initial conditions which match the astronomical data. At $z=0$ we introduce the quantities $ \overline{\rho_{m}}(0)=\rho^{0}_{m}/ \rho_{0}=\Omega^{0}_{m} $, $ \overline{\rho_{r}}(0)=\rho^{0}_{r}/ \rho_{0}=\Omega^{0}_{r} $, $ \overline{\rho_{dm}}(0)=\rho^{0}_{dm}/ \rho_{0}=\Omega^{0}_{dm} $ and $ \overline{\rho_{de}}(0)=\rho^{0}_{de}/ \rho_{0}=\Omega^{0}_{de} $, where $\Omega^{0}_{i}$ denotes the value of the density parameter of each component at present time. The values adopted here are:  $\Omega^{0}_{m}=0.0463$, $\Omega^{0}_{dm}=0.233$, $\Omega^{0}_{de}=0.721$ and, $\Omega^{0}_{r}=8.5\times 10^{-5}$ (see \textit{e.g.} \cite{fukugita2004, WMAP9}).
\begin{figure}
\begin{minipage}[b]{0.5\linewidth}
\centering
\includegraphics[width=7cm]{bigpequenoz.eps}
\caption{Density parameters for \- $ 0\leq z \leq 1 $.}
\end{minipage}
\hspace{0.3cm}
\begin{minipage}[b]{0.5\linewidth}
\centering
\includegraphics[width=7.05cm]{bigz3000.eps}
\caption{Density parameters for $ 1000\leq z \leq 4000 $.}
\end{minipage}
\end{figure}

In Fig. 1 the density parameters are plotted as functions of the red-shift for values in the range $ 0\leq z \leq 1 $. From this figure we observe that the Universe is dominated by the dark energy, here described by the constant component of the fermionic field, which decreases as we go back in time. At a red-shift $ z\approx 0.45 $ the dark matter contribution begins to dominate the Universe. At this point the radiation does not have a representative contribution to the density parameter. The plot for large values of the red-shift is given in Fig. 2. The equality of the contribution of the matter and radiation fields occurs when $ z\approx 3250 $. This result is in good agreement with the parameterized observational data, since from reference \cite{WMAP9} we have $ z_{eq}=3265^{+106}_{-105} $. It is noteworthy that here we have taken into account the degrees of freedom of the relativistic neutrinos.
\begin{figure}[htbp]
    \centering
    \includegraphics[width=0.5\textwidth]{graficoqz.eps}
    \caption{Deceleration parameter as a function of the red-shift.}
    \label{fig:last2}
\end{figure}

 The deceleration parameter as a function of the red-shift is represented in Fig. 3 and we observe from this graphic that the present value of the deceleration parameter is $ q(0)=-0.55 $, and the red-shift of the deceleration-acceleration transition is: $ z_{T}=0.67$. We have compared these values with the astronomical data in Table 1 (see \textit{e.g.} \cite{PhysRevD.86.083509}) and we infer that the obtained values of $ q_0 $ and $ z_T $  are in good agreement with the observational data.
\begin{table}
\begin{center}
 \begin{tabular}{|c|c|c|}
 \hline
    & Parametrization I & Parametrization II  \\
   $ q_0 $ & $ -0.61^{\tiny +0.06}_{\tiny -0.07} $ & $ -0.56^{\tiny +0.35}_{\tiny -0.22} $  \\
  $ z_T $ & $ 0.71^{\tiny +0.14}_{\tiny -0.17} $ &  $ 0.77^{\tiny +0.52}_{\tiny -0.57} $  \\ \hline
  &  Parametrization III & Fermionic Model \\
  $ q_0 $ &  $ -0.60\pm 0.06 $ & $ -0.55 $ \\
  $ z_T $ &  $ 0.72^{\tiny +0.27}_{\tiny -0.21} $ & $ 0.67 $ \\ \hline
    \end{tabular}
\end{center}
  \caption{Comparison of the deceleration parameter $ q_0 $ and the transition red-shift $z_T$ with the observational data (see \textit{e.g.} \cite{PhysRevD.86.083509})}
\end{table}

\subsubsection{The case $F'\neq0$:}

Let us search for a solution of the Friedmann equation (\ref{39}) when the self-interaction potential of the fermion field  and the non-minimal coupling function are given by
\ben
V=\Lambda+V_0\Psi=\Lambda+\frac{V_0\Psi_0}{a^3},\qquad F=\frac{1}{2}(1-\xi\Psi)=\frac{1}{2}\left(1-\frac{\xi\Psi_0}{a^3}\right),
\een
where $\xi$  is supposed to be a positive small coupling constant.

In this case the Friedmann equation can be rewritten as the following differential equation for the scale factor
\ben\lb{Fri}
3\left(a^3+2\xi\Psi_0\right)\dot a^2=\Lambda a^5 +\left(V_0\Psi_0+\rho_m^0\right){a^2}+{\rho_r^0}{a}.
\een

Three asymptotic cosmological solutions can be derived from the Friedmann equation (\ref{Fri}), namely,
\begin{enumerate}
  \item \emph{Primordial Universe in the presence of radiation:} in this case the terms $\Lambda$, $V_0\Psi_0$ and  $\rho_m^0$ can be neglected and (\ref{Fri}) becomes
  \ben\lb{s1}
  \left(a^2+\frac{2\xi\Psi_0}{a}\right)\dot a^2=\frac{\rho_r^0}{3}.\label{Approx}
  \een
  Two cases follow from the analysis of the above equation: (a) when the scale factor is small the approximated solution of (\ref{s1}) is
  \ben
  a(t)\simeq\frac{\rho_r^0}{24\xi\Psi_0}\left(t-t_0\right)^2,
  \een
  which describes an accelerated Universe characterized by a power-law inflationary period; (b) when the scale factor grows the non-minimal coupling dilutes and the solution of (\ref{s1}) reads
  \ben
  a(t)\simeq\sqrt{2}\left(\frac{\rho_r^0}{3}\right)^{1/4}\left(t-t_0\right)^{1/2},\label{RadEra}
  \een
  which refers to a radiation dominated period.
  \item \emph{Universe dominated by matter:} here the $\Lambda$ term can be neglected, the non-minimal coupling dilutes, the matter dominates the radiation  and (\ref{Fri}) reduces to
  \ben
  a\dot a^2=\frac{1}{3}\left(V_0\Psi_0+\rho_m^0\right),
  \een
  whose solution for the scale factor refers to the matter dominated period, namely,
  \ben
  a(t)\simeq\left({\frac{3}{2}}\right)^{2/3}\left(\frac{V_0\Psi_0+\rho_m^0}{3}\right)^{1/3}\left(t-t_0\right)^{2/3}.
  \een
  \item \emph{Present Universe:} since the radiation can be neglected in comparison to the other constituents and the non-minimal coupling is diluted, (\ref{Fri}) can be written as
  \ben
  a\dot a^2=\frac{1}{3}\left(\Lambda a^3 +V_0\Psi_0+\rho_m^0\right).
  \een
  The dilution of the non-minimal coupling implies that we are in the presence of the case analyzed in the last section where $F=1/2$, whose solution predicts the present decelerated-accelerated period, i.e., a matter dominated period goes into a de Sitter Universe in the future.

\end{enumerate}

\section{Hydrodynamical perturbations}

Let us now analyse the above solutions under small perturbations. In view of the difficulty of perturbing the spinor field through the Dirac equations, which are complicatedly coupled to the metric, we make an estimate of the perturbation dynamics through the hydrodynamic approach. To do this using the known method for the hydrodynamical perturbations, we will write the non-minimal model in a more adequate way, as follows.

By defining an effective energy-momentum tensor in the form $\widetilde{T}_{\mu\nu}={T_{\mu\nu}}/{2F}$, the non-minimal coupling model can be recast in the form of the Einstein's equations. Hence from action (\ref{1}) we have the generalized field equations which are recast in the form
\begin{equation}
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=-\frac{T_{\mu\nu}}{2F}=-\widetilde{T}_{\mu\nu}.
\end{equation}
So one has the Einstein's equations with the source of gravitational field being represented by the energy momentum-tensor $\widetilde{T}_{\mu\nu}$.

As a consequence, for a perfect fluid the canonical form of the energy-momentum tensor holds
\begin{equation}
\widetilde{T}_{\mu\nu}=\left(\widetilde{\rho}+\widetilde{p}\right)U_\mu U_\nu-\widetilde{p}g_{\mu\nu},
\end{equation}
where $U^\mu$ is the four-velocity and the effective energy density and pressure read
\begin{eqnarray}
\widetilde{\rho}=\frac{\rho}{2F}, \qquad \widetilde{p}=\frac{p}{2F}.\label{effFluid}
\end{eqnarray}
Thus the related Friedmann and acceleration equations are
\begin{eqnarray}
H^2=\frac{\widetilde{\rho}}{3}, \qquad \frac{\ddot a}{a}=-\frac{\widetilde{\rho}+3\widetilde{p}}{6},
\end{eqnarray}
and the conservation equation of the effective fluid has the usual form
\begin{equation}
\frac{d\widetilde{\rho}}{dt}+3H\left(\widetilde{\rho}+\widetilde{p}\right)=0.
\end{equation}

In this way, we can treat this hydrodynamical problem in the same way one does usually. The case for $F=1/2$ and $V=\Lambda+V_0\Psi$ produces the cosmological behavior of the models with baryonic and dark matter in the presence of a cosmological constant, whose perturbed solutions are known to be stable \cite{Mukhanov, Mukhanov1}. However, it is necessary to verify the stability against perturbations for the case with $F=(1-\xi\Psi)/2$ and $V=\Lambda+V_0\Psi$, when the non-minimal coupling presents some influence. This comprises the transition from the inflation to the radiation era whose approximated background solution was obtained from (\ref{Approx}), where the non-minimal coupling effect is present.

Since the background fluid is not anisotropic, the line element for small scalar perturbations in the longitudinal gauge \cite{Bardeen, Mukhanov, Mukhanov1} can be expressed as
\begin{equation}
ds^2=a^2\left[\left(1+2\Phi\right)d\eta^2-\left(1-2\Phi\right)\delta_{ij}dx^idx^j\right],
\end{equation}
where $\eta$ is the conformal time defined by $d\eta=dt/a$ and $\Phi$ is the gauge-invariant perturbation potential.

Once we are working in the framework of Einstein's equations with effective quantities, all the corresponding known results for the gauge-invariant scalar perturbations hold. But note that all the hydrodynamical quantities will refer to the effective ones above defined. Thus the perturbation potential equation for adiabatic perturbations is the usual one \cite{Mukhanov, Mukhanov1}
\begin{equation}
\Phi''+3\left(1+c_S^2\right)\mathcal{H}\Phi'-c_S^2\nabla^2\Phi+\left[2\mathcal{H}'+\left(1+3c_S^2\right)\mathcal{H}^2\right]\Phi=0,\label{PertPot}
\end{equation}
with the prime denoting derivative with respect to conformal time and $c_S$ is the speed of sound related to the effective fluid, $c_S^2=\partial \widetilde{p}/\partial\widetilde{\rho}$. The Hubble parameter is written in terms of the conformal time, $\mathcal{H}=a'/a^2$.

When the Universe is dominated by a fluid with barotropic state equation, $\rho=wp$, where $w$ is a positive constant, one has $c_S^2=w$ and the scale factor evolves as $a\propto \eta^{2/(1+3w)}$. For such a case, considering plane wave perturbations, $\Phi(\textbf{x}, \eta)=\Phi_{\textbf{k}}(\eta)e^{i\textbf{k}.\textbf{x}}$, equation (\ref{PertPot}) gives the solution
\begin{equation}
\Phi_{\textbf{k}}=\eta^{-n}\left[AJ_n\left(\sqrt{w}k\eta\right)+BY_n\left(\sqrt{w}k\eta\right)\right], \qquad n=\frac{1}{2}\left(\frac{5+3w}{1+3w}\right),\label{SolPert}
\end{equation}
where $A$, $B$ are constants and $J_n$, $Y_n$ are the Bessel functions of the first and second kind, respectively.

The barotropic equation of radiation is also valid in the effective form, $\widetilde{p}_r=\widetilde{\rho}_r/3$, as can be seen through (\ref{effFluid}). For $w=1/3$ one has $a\propto\eta$, which is the same approximated solution (\ref{RadEra}) in the presence of a non-minimal coupling sufficiently diluted.  The corresponding perturbation potential comes from (\ref{SolPert}) with $w=1/3$ and can be written in the following form
\begin{equation}
\Phi_{\textbf{k}}=\frac{1}{\widetilde{\eta}^2}\left[\left(\frac{A}{\widetilde{\eta}}+B\right)\sin{\widetilde{\eta}}+\left(\frac{B}{\widetilde{\eta}}-A\right)
\cos{\widetilde{\eta}}\right],\label{SolOne}
\end{equation}
where $\widetilde{\eta}=k\eta/\sqrt{3}$. Under the above considerations, this solution is a good approximation for the perturbations related to the asymptotic background solution (\ref{RadEra}).

Hence the time evolution of the energy density perturbations \cite{Mukhanov, Mukhanov1} of the effective fluid reads
\begin{eqnarray}
\frac{\delta\widetilde{\rho}_r}{\widetilde{\rho}_r}=2A\left[\left(\frac{2-\widetilde{\eta}^2}{\widetilde{\eta}^2}\right)
\left(\frac{\sin{\widetilde{\eta}}}{\widetilde{\eta}}-\cos\widetilde{\eta}\right)-\frac{\sin{\widetilde{\eta}}}{\widetilde{\eta}}\right]\nonumber\\
+4B\left[\left(\frac{1-\widetilde{\eta}^2}{\widetilde{\eta}^2}\right)
\left(\frac{\cos{\widetilde{\eta}}}{\widetilde{\eta}}+\sin\widetilde{\eta}\right)+\frac{\sin{\widetilde{\eta}}}{2}\right],\label{densPert}
\end{eqnarray}
where ${\delta\widetilde{\rho}_r}$ is a gauge-invariant perturbation. From (\ref{densPert}) we can assure that the amplitude of the effective energy density perturbations decays with time and asymptotically freezes out for large times.

Now we need to know if the amplitude of the original energy density perturbations, i.e. of $\rho_r$, also decays with time. Using our initial definition, $\widetilde{\rho}=\rho/2F$, we can write the effective density perturbations in terms of the original quantities, from which one obtains
\begin{equation}
\frac{\delta\widetilde{\rho}_r}{\widetilde{\rho}_r}=\frac{\delta{\rho}_r}{{\rho}_r}-\frac{\delta F}{F}.
\end{equation}

As the non-minimal coupling in question is $F=(1-\xi\Psi)/2$, one has
\begin{equation}
\frac{\delta\widetilde{\rho}_r}{\widetilde{\rho}_r}=\frac{\delta{\rho}_r}{{\rho}_r}+\xi\frac{\delta\Psi}{1-\xi\Psi}
\approx\frac{\delta{\rho}_r}{{\rho}_r}+\xi{\delta\Psi}.
\end{equation}
Note that this approximation holds because $F$ is sufficiently diluted and $\xi$ is very small.
From this expression we conclude that ${\delta{\rho}_r}/{{\rho}_r}$ as well as $\xi\delta\Psi$ must decay, since the corresponding sum gives a quantity that decays, i.e. ${\delta\widetilde{\rho}_r}/{\widetilde{\rho}_r}$. So the amplitude of the energy density perturbations of radiation and the $\Psi$ perturbations decay with time and asymptotically stabilize for large times. Thus, under the considered approximations, the non-minimal fermionic model in question appears to be stable against small perturbations.

\section{Final remarks and conclusions}

In this work we have considered a classical fermionic field minimally and non-minimally coupled with the gravitational field. The fermionic field is represented by a spinor field which is understood as a set of complex-valued space-time functions which transforms according to the Lorentz group. Classical spinors were discussed by Armend\'ariz-Pic\'on and Greene in the reference \cite{Piccongreene}. About such a consideration we point out that: (i) the expectation value of a spinor field in a physical state is a complex number and not a Grassmannian number, (ii) the spinor field can be treated classically if its state is close to the vacuum and (iii) we have extrapolated the validity of the classical spinor to the inflation despite a classical field theory could fail at the beginning of this regime.

\textit{Universe described by a fermion field} -- In the literature \cite{PhysRevD.69.124010, PhysRevD.74.124030, Marlosprd, Marlosepl} several forms for the potential density and coupling of the fermion fields were proposed in order to describe cosmological models with accelerated and decelerated periods. The  present work shows that, if the Noether symmetry is satisfied, the potential density and coupling have very restrictive forms.
The results for this case are: (i) the minimally coupled fermion field recovers the standard model for a Universe composed of matter and one has only a decelerated regime and (ii) in the non-minimal coupling with gravity the fermion field behaves as an inflaton.

\textit{Universe described by fermion, matter and radiation fields} -- When we impose the Noether symmetry  the minimal and non-minimal coupling cases admit a generic self interacting potential of the pseudo scalar. Here the results are: (i) the minimal coupling model reproduces a decelerated-accelerated regime, comprising the radiation era in the beginning and passing through the matter domination until the era when the dark energy dominates, which is described by the constant term of the fermionic potential and (ii) the non-minimal coupling can describe a Universe that begins with an accelerated expansion which goes into the radiation dominated era when the non-minimal coupling dilutes and after that it enters into the decelerated-accelerated period (matter-dark energy era).

In the literature \cite{rudicanonical, PhysRevD.79.063518, rudibosonfermion} several models with two different fields representing the dark energy and the dark matter were proposed in order to describe cosmological models with decelerated and accelerated periods. It is interesting that we obtained the same description with just one field representing either the dark matter and the dark energy, i.e., the fermionic field. To sum up, the Noether potentials and couplings of the model with their respective cosmological scenarios are displayed in Table 2.

\begin{table}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Cases & $V$ & $F$ & Inflation & Radiation era & Matter era & Current Accel  \\
\hline
 I & $\alpha\Psi$ & $\omega\Psi$ & $a\propto e^{\Lambda t}$ & --- & --- & --- \\
 \hline
 I & $\alpha\Psi$ & $\frac{1}{2}$ & --- & --- & Yes & --- \\
\hline
 II & $\Lambda+V_0\Psi$ & $\frac{1}{2}$ & --- & Yes & Yes & Yes \\
\hline
 II & $\Lambda+V_0\Psi$ & $\frac{1}{2}(1-\xi\Psi)$ & $a\propto t^2$ & Yes & Yes & Yes \\
\hline
\end{tabular}
\caption{Potentials and couplings and their cosmological scenarios.}
\end{table}

\section*{Acnowledgments}

G.M.K. acknowledges the financial support by CNPq and G.G. by CAPES.

\section*{Appendix. Noether symmetry condition}

Let us take a Lagrangian that does not depend explicitly on time, $\mathcal{L}=\mathcal{L}(q_i, \dot q_i)$, with the coordinates $q_i=q_i(t)$. The Euler-Lagrange equations of this Lagrangian read
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot q_i}\right)=0.
\end{equation}
Consider now the following vector field
\begin{equation}
\textbf{X}=\alpha_i\frac{\partial}{\partial q_i}+\frac{d\alpha_i}{dt}\frac{\partial}{\partial \dot q_i},
\end{equation}
where the coefficients $\alpha_i$ are functions of the coordinates $q_i$. This vector field describes point transformations of the coordinates, $q_i\longrightarrow q'_i$, which induce velocity transformations,
\begin{equation}
q_i'=\frac{\partial q'_i}{\partial q_j}q_j \quad \Longrightarrow \quad \dot q_i'=\frac{\partial q'_i}{\partial q_j}\dot q_j.
\end{equation}
One assumes that these transformations are invertible. The quantity $\textbf{X}$ is also called infinitesimal generator of symmetry.

If the Lagrangian $\mathcal{L}$ is invariant under the point transformations represented by $\textbf{X}$, the following identity must hold
 \begin{equation}
L_{\textbf{X}}\mathcal{L}=\alpha_i\frac{\partial \mathcal{L}}{\partial q_i}+\frac{d\alpha_i}{dt}\frac{\partial\mathcal{L}}{\partial \dot q_i}=0,
\end{equation}
where $ L_{\textbf{X}}\mathcal{L}$ stands for the Lie derivative of $\mathcal{L}$ with respect to the vector field $\textbf{X}$.
Let us see the meaning of this invariance. If we contract the Euler-Lagrange equations of $\mathcal{L}$ with the coefficients $\alpha_i$, one obtains
\begin{eqnarray}
\alpha_i \frac{\partial \mathcal{L}}{\partial q_i}-\alpha_i\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial\dot q_i}\right)=0\\
\alpha_i\frac{\partial\mathcal{L}}{\partial q_i}-\left[\frac{d}{dt}\left(\alpha_i\frac{\partial\mathcal{L}}{\partial\dot q_i}\right)-\frac{d\alpha_i}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_i}\right]=0\\
\frac{d}{dt}\left(\alpha_i\frac{\partial\mathcal{L}}{\partial\dot q_i}\right)=\alpha_i\frac{\partial\mathcal{L}}{\partial q_i}+\frac{d\alpha_i}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_i}\equiv L_{\textbf{X}}\mathcal{L}.
\end{eqnarray}
Thus, since $L_{\textbf{X}}\mathcal{L}=0$ holds, it follows the Noether's theorem:\\

If $L_{\textbf{X}}\mathcal{L}=0$, there is a constant of motion given by $M_0=\alpha_i \frac{\partial \mathcal{L}}{\partial \dot q_i}$.\\

In other words, if the Lagrangian $\mathcal{L}$ is invariant under point transformations represented by $\textbf{X}$, there is a Noether symmetry associated with  $\mathcal{L}$.

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'No time for us to dwell on defeat' says Big Nev
RIGHT, forget about England. It's done. Let's get focused on the next job in hand, Poland.
My biggest worry is the lads will not be able to lift themselves for what is now a 'must-win' game. But look at it this way, if we CAN win in midweek the opportunity is there to take four points off Austria next March and put ourselves in the box seat for second spot.
I feel disappointed with yesterday but I think in all fairness, all anyone really needed to know about the difference between the two sides was on the bench.
Andy Melville missed out and his experience was a big loss. But say the same had happened to England. Sol Campbell is injured, in steps John Terry.
We've got a kid, young James Collins and that's it. No amount of preparation this week, with Melville in doubt, could have given him the experience he needed.
After yesterday's result we now need a really good show against Poland in Sparky's final game in charge.
Defeat is unthinkable and would end our World Cup qualification hopes.
That would be a terrible way for Sparky to sign off because it will be the end of an era, and a good era at that.
The greatest result from Sparky's time in charge of Wales has been the incredible rise in expectations.
They have risen ten-fold in the five years he has been at the helm.
No longer do Wales go into games expecting to lose.
People have pointed to the fact that Wales haven't won a competitive fixture since March last year.
But funny as it may sound, results only tell half the story.
You have to remember where we came from when Sparky took over. I know because I was involved. The situation in which he took charge in 1999 was a depressing one.
We never seemed to win, no-one expected us to win and there were no exciting players coming through the ranks.
Can you imagine playing Azerbaijan at the Millennium Stadium in those dark days?
It would have been me, the cleaner and a programme seller. But since taking charge Sparky has transformed the side into a winning one that the people of Wales want to see.
Because of that it has brought an awful lot of money into the country and into the FAW coffers that wasn't there before.
In short he has helped to change a constantly negative perception of Welsh football into a positive.
For the past couple of years people have actually talked about the Welsh football team and how good we are.
He also brought back the organisation that had been lacking since the days of Terry Yorath. And importantly he has made the players believe that they are good enough to win.
The players believed in him because of what he achieved as a player and then as their manager and of course he showed unswerving loyalty to them.
Whoever comes into the job next will have some big boots to fill.
They will need to be patient because it will be another period of transition for the squad.
But they should not lose sight that there are some good players in the squad.
If he can make subtle changes behind the scenes while maintaining a degree of calmness on the surface he will have done well.
Like any new manager, the new guy will need and want to make changes in the near future but I truly believe there are a string of good players lining up to ensure Welsh football has a bright future.
There is hope for the future.


Although there were rumblings and rumors in Paris this year that the Indians might have trouble closing the deal on their Medium, Multi-Role Combat Aircraft program, a report in AvWeek makes it sound as though it's on the glide slope for a decision soon. Indian defense officials next will go through their formal process of deciding which of the final two competitors, the Eurofighter Typhoon and the Dassault Rafale, would end up being the cheapest. Reading between the lines, it sounds as though the Indians like the capabilities of both aircraft -- they want to up their buy to 189 jets, whichever one they pick -- and now it's all down to cost.
Even though the government says it has determined how to benchmark the two offers, industry watchers warn that there could still be hurdles. For example, the source selection has left the government with two of the most expensive aircraft in the competition, which will try the available budget—the Boeing F/A-18E/F, Lockheed Martin F-16, MiG-35 and Saab Gripen were eliminated in the April downselect.
"If the [defense ministry] has benchmarked realistically, it will almost definitely mean that the government will need more money to sign a contract for either the Typhoon or Rafale, even accounting for inflation. Is the ministry willing to go down that road?" asks a manager at Boeing.
Both Eurofighter and Dassault have publicly insisted that their bids are competitive and privately hope the government has taken several factors into consideration while arriving at their benchmark price, in particular that the aircraft are more capable and newer than some that were in the running.
There is concern, however, that if the government looks across all six competitors originally in the field, the benchmark price will be below that proffered by either Eurofighter or Dassault, which could lead to questions about how the source selection process should proceed.
Perhaps anticipating a rapid change of fortunes, the local MMRCA team offices of the four eliminated contenders remain fully staffed. Officials at all four acknowledge that while they have been given satisfactory debriefs on the downselect, they also believe that things are delicate enough in the final phase for them to remain in the country. "We are ready to respond to the government in any eventuality," says a manager for Rosoboronexport, the Russian agency fronting the MiG-35 offer. Saab officials have also suggested the competition could open again.
There's a lot of smoke here, but there's some fire -- that detail that these defense contractors are keeping their local representatives around, just in case, is telling. With America's and Europe's defense budgets no longer guaranteed sources of infinite growth, these aerospace companies really, really want this Indian deal.


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NYC Emo Pioneers Return to the Stage in Brooklyn
JORDAN MELTZER/THE OBSERVER
What is punk? Jawbreaker's concert at Brooklyn Steel proved to the New York crowd that punk is doing whatever the hell you love no matter what stands in your way … even a two-decade hiatus.
By JORDAN MELTZER
The year was 1995. Blake Schwarzenbach was playing guitar and singing for up-and-coming punk band Jawbreaker, which had just signed to a major label, at a show at Irving Plaza in Manhattan — its largest city gig yet. Suddenly, a rowdy fan jumped onstage and began moshing with Schwarzenbach. Blood was spilled.
Such was NYC punk at the time. But Jawbreaker was taking that punk movement and transforming it into something new: emo (although few people knew then just how influential the music would become).
That was the last time Jawbreaker took the stage in its beloved hometown for the next 23 years, seeing as the band broke up the following year. But after its 2017 reunion, Jawbreaker did a three-show run in 2018 at Brooklyn Steel, a relatively new venue in Greenpoint, selling it out each night. In many respects, it was not the first time the band dove headfirst into uncharted territory.
Now, in 2019, the punks have returned to that very same space for multiple nights, selling it out once again. It's doubtful that anyone hearing emo in the '90s could have predicted such a phenomenon.
Schwarzenbach, the principal songwriter for the band, opened the March 24 show with the punchy guitar chords of "West Bay Invitational" from their 1994 LP, "24 Hour Revenge Therapy." That record, like many others in the emo genre, is characterized by emotional lyrics sung to punk and hardcore music. Appropriately, the concert's opening number is a love letter to the punk scene in Oakland, Calif.
"We kissed a shot of Kentucky straight," Schwarzenbach sang about a punk party in Oakland. "I swore this life is worth the wait." The audience was not rowdy as it was at Irving that night in '95; it was more pensive, basking in appreciation of the band's genius in retrospect. The emotionality — the nostalgia — flowed through the room.
Seamlessly, the trio transitioned into "Want," a cult favorite from Jawbreaker's 1990 debut, "Unfun." Noticeably, the guitar became much louder, as Schwarzenbach seemed to have several guitar pedals that corresponded to each of the band's albums (plus a clean tone for softer moments during the songs). The attention to detail in the musicianship was apparent. The crowd ate it up, cheering loudly and dancing along to the fast-paced beat of the song.
Drummer Adam Pfahler, owner of Bay Area-based Blackball Records, was on point all night. Each hit was perfectly on time, each fill ever so tasteful and each beat exactly as it sounded on the record. Bringing studio sounds to the stage is hard work, but Pfahler made it seem effortless.
Chris Bauermeister, meanwhile, played with a bass tone so chunky Bruno Mars would be jealous. The sound of his plectrum plucking the lower register of his instrument, when mixed with his high-midrange tone, made for a powerful and in-your-face presence that was unmistakably Jawbreaker.
The punks played tracks from their entire discography, performing songs from all four official full-length releases. They even included two songs from their B-side compilation "Etc." Fans jumped in surprise at some songs' inclusions, including "In Sadding Around:" The '94 track was first performed at the legendary 924 Gilman St. Project in Berkeley, Calif., and had only been performed eight times since the reunion.
The grand finale was the mega-hit "Boxcar," an absurdly self-aware track as it is a punk song that criticizes the punk scene. When fans accused Jawbreaker of not being "punk enough" in the '90s, the band responded with a song that rhetorically asked: "One, two, three, four/Who's punk? What's the score?" As music elitism and snobbery runs rampant, the lyrics still ring true today — which would explain why the whole crowd belted the words at the top of its lungs.
On that note, Schwarzenbach and crew left us with a question: What is punk? Historically, it has been a number of things: rebellion, anti-establishmentism, disenfranchisement and whatever makes people break the status quo. More than anything, though, Jawbreaker's concert at Brooklyn Steel proved to the New York crowd that punk is doing whatever the hell you love no matter what stands in your way … even a two-decade hiatus.
Jordan Meltzer
JORDAN MELTZER, News Editor Emeritus
Jordan Meltzer, Gabelli School of Business '20, is a news editor emeritus for The Observer. A global business major concentrating in communication and media management, Jordan has written for all five sections of The Observer, but his dedication is to the news section. Topics of particular interest to Jordan are politics, music and the many goings-on at Fordham Lincoln Center. Jordan is also a Yankees fan and advocates for the universalization of the designated hitter.


Description "35 [age] ; Shoe Maker ; 885 [value of real estate owned] ; 480 [value of personal estate] ; Alabama [place of birth]."
Source Includes data from the 1860 Census Population Schedules, Paris Post Office, Lamar County, Texas, Series: M653, Roll: 1299, p. 163 (stamped), entry 330.
Artisan or Artist Hines, J. W.


Will Millennials ever know Serendipity?
Apart from loving the word, I wanted to write about Serendipity as I sometimes wonder whether it is going to be a rare thing for coming generations. For those that are not sure the definition of Serendipity is basically about coming across something pleasant by accident. Now I started thinking about this from what I read online and updates etc and it struck me that unless you work hard at it, serendipity is not what it used to be.
I work in a business where we deliver exact ads to exact cookies which is another shot in the eye for Serendipity when compared with TV advertising that still trades on the fact you buy one audience but could pick up many other 'free' audiences. Online advertising is just one part of the equation though, it is also about my Twitter feed, Linkedin and so on.
If you think about it the World Wide Web is closing in on us everyday we use it. We have Google showing us Ads related to searches and sites we visit and then content that is determined by our own behaviours and our friends social behaviour. I have 500 people I follow that I have hand picked based on their subject matter expertise whether that is personal or professional or indeed they are just friends. By definition I am presented with content of the same genre, often lots of the same content over and over. Of course you can add random people but most people forget to do that. To help make your readership less varied Twitter then suggest people like you so you keep adding more of the same.
Facebook is of course again a hand picked bunch but for some reason I see the same people adding updates. I keep being told that this is based on my interaction with them, I know for sure that is not the case so there must be some other science at work, either way I want more variety and it is killing my Facebook experience and interest. On top of that I am being shown highly targeted Ads based on interest and friends. It all adds up to a pretty repetitive and unsurprising experience.
Linkedin are usually people from the same genre and updates cross over with Twitter, Ads are targeted and so again I am seeing the same stuff. If I go shopping on Amazon they are predicting my tastes and showing me content that they think I will want whether it is books, films or electricals. This leaves very little revelation in my experience. That extends into most shopping as targeting becomes more sophisticated where very personalised content will be shown just for me across many sites. Of course we are then flocking to sites like Zite that learns fast or asks what content you like until you end up with a painfully myopic view. To be honest the examples go on and on with iTunes and other music systems proposing other songs and albums, but always based on something that you already like.
Right now we are primarily talking about ads and people but when we get into search that becomes more and more tailored we may never see an alternative view. If Google learns that I am a conservative and look at that content, that is what I will see, I will never be given alternative content. Same for any belief or interest. The technology we use is driving our experience to be more and more focused on understanding the user and making sure what they see is relevant. That might be good for Ads but I feel like we need a serendipity button.
We need to mix this up, challenge ourselves and give ourselves an opportunity to see something new and exciting before we target ourself into the most narrcow cast existence with no little surprises around the corner. Let us all press the Serendipity button. Aleks Krotoski has allowed us to do this, the image for the blog above is her Serendipity engine which actually creates it through a physical representation. It takes some getting your head around but it is fascinating – click here for more.
I discussed serendipity in a recent pitch. Maybe it's evolving, where we (as advertisers) provide opportunities of serendipity through our use of data and targeting? We can show things of relevance and interest that delight, but may not be expected. The concert you didn't know was on, the latest crime novel you didn't know about, that YouTube video we know you'll like but you weren't aware of. As long as we don't only chase down the 'known converters' (QED, retargeting) then serendipity in some form lives on?


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Mummay History, Family Crest & Coats of Arms
Mummay
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The vast movement of people that followed the Norman Conquest of England of 1066 brought the Mummay family name to the British Isles. They lived in Northumberland. The name, however, is a reference to the family's place of residence prior to the Norman Conquest of England in 1066, Montbrai, in La Manche, Normandy. However, another source claims the family claim descent from "the ancient barony of Mowbray, called by Odericus Vitalis Molbraium, [which] was identical with the village of Monbrai, in the canton or Perci, an arrondissement of St. Lo in Normandy." [1]CITATION[CLOSE]
Lowe, Mark Anthony, Patronymica Britannica, A Dictionary of Family Names of the United Kingdom. London: John Russel Smith, 1860. Print.
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Early Origins of the Mummay family
The surname Mummay was first found in Northumberland Geoffrey de Montbray (d. 1093,) bishop of Coutances was a warrior, administrator and close assistant of William the Conqueror.
After the death of William, Geoffrey settled in Bristol, (as listed in the Domesday Book) where he built a strong castle but frequently feuded with William II." A strong castle was erected [at Thirsk in the North Riding of Yorkshire] about 979 by the family of Mowbray, where Roger de Mowbray in the time of Henry II., having become a confederate of the King of Scotland, erected his standard against his lawful sovereign: upon the suppression of the revolt, this fortress, with many others, was entirely demolished by order of the king." [2]CITATION[CLOSE]
Lewis, Samuel, A Topographical Dictionary of England. Institute of Historical Research, 1848, Print. The same Roger de Mowbray also held Bambrough Castle in Northumberland. "After the Norman Conquest it was held by Robert de Mowbray, on whose insurrection against William Rufus it was besieged, and, after an obstinate defence, surrendered to that monarch, who threatened, unless it were given up, to put out the eyes of Mowbray, who had been taken prisoner." [2]CITATION[CLOSE]
Lewis, Samuel, A Topographical Dictionary of England. Institute of Historical Research, 1848, Print.
The markettown and parish of Kirby-Malzeard in the West Riding of Yorkshire was another ancient family seat. "This place was the property of the Mowbray family, afterwards dukes of Norfolk, whose castle here was demolished in the reign of Henry II., and one of whom, John de Mowbray, obtained for the inhabitants in the reign of Edward I. a charter for a weekly market and two annual fairs." [2]CITATION[CLOSE]
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Early History of the Mummay family
This web page shows only a small excerpt of our Mummay research.
Another 209 words (15 lines of text) covering the years 1297, 1366, 1399, 1377, 1396, 1444, 1476, 1475, 1225, 1314, 1365, 1399, 1397, 1385, 1405 and 1405 are included under the topic Early Mummay History in all our PDF Extended History products and printed products wherever possible.
Mummay Spelling Variations
Anglo-Norman names are characterized by a multitude of spelling variations. When the Normans became the ruling people of England in the 11th century, they introduced a new language into a society where the main languages of Old and later Middle English had no definite spelling rules. These languages were more often spoken than written, so they blended freely with one another. Contributing to this mixing of tongues was the fact that medieval scribes spelled words according to sound, ensuring that a person's name would appear differently in nearly every document in which it was recorded. The name has been spelled Mowbray, Moubray, Mowbrey, Moubrey and others.
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Early Notables of the Mummay family (pre 1700)
Outstanding amongst the family at this time was Thomas Mowbray (1365-1399), an English nobleman, created 1st Duke of Norfolk in 1397, by King Richard II of England; and Thomas de Mowbray (1385-1405), 4th Earl of Norfolk, 2nd Earl of Nottingham, 8th Baron Segrave, 7th Baron Mowbray, English nobleman and rebel, after death of father, allowed to succeed him as Earl of Norfolk and Nottingham. He received his father's...
Another 68 words (5 lines of text) are included under the topic Early Mummay Notables in all our PDF Extended History products and printed products wherever possible.
Migration of the Mummay family to the New World and Oceana
For many English families, the political and religious disarray that plagued their homeland made the frontiers of the New World an attractive prospect. Thousands migrated, aboard cramped disease-ridden ships. They arrived sick, poor, and hungry, but were welcomed in many cases with far greater opportunity than at home in England. Many of these hardy settlers went on to make important contributions to the emerging nations in which they landed. Among early immigrants bearing the name Mummay or a variant listed above were: Thomas Moubray, who came to West Indies in 1678; William and Hannah Mowbury, who arrived in North Carolina in 1680; William Moubray, who settled in Maryland in 1716.
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^ Lowe, Mark Anthony, Patronymica Britannica, A Dictionary of Family Names of the United Kingdom. London: John Russel Smith, 1860. Print.
^ Lewis, Samuel, A Topographical Dictionary of England. Institute of Historical Research, 1848, Print.
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Home Sports Olivia Ruiz: what becomes of the famous 43-year-old singer?
Olivia Ruiz: what becomes of the famous 43-year-old singer?
In 2001, singer Olivia Ruiz became known to the general public thanks to her participation in the musical program Star Academy. Despite her defeat in the semi-final against the singer Jenifer, she imposed herself on the musical sphere thanks to her albums J'aime pas l'amour, released in 2003 and La Femme chocolat, in 2005.
It was during this period that she met a certain Mathias Malzieu, leader of the Dionysos group. A relationship that will last until 2011, as she had confided to our colleagues at Gala during an interview. "If we have been separated for a year now, the bond is not broken. This separation is one of the periods of mourning to overcome that I was just telling you about. You know, seven years of life together, there is necessarily a unwavering attachment between us," she said.
After their separation, Olivia Ruiz found love again in the arms of a certain Nicolas, with whom she had a son named Nino, in November 2015. The birth of the latter made her slow down her career as an artist. , as she had declared to Paris Match. "I wanted, in 2018, to take care of my son. He had accompanied me on my last tour because he was very small. But there, I could no longer continue at such a pace", explained -she.
Very close to his fans, the 43-year-old artist continues to maintain the link with them via his social networks. Alongside her family life, she devoted herself to new professional projects that Planet invites you to discover in our slideshow.
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Happy World Penguin Day
Today is World Penguin Day— not that we need a reason to celebrate these amazing little creatures, but it's great to have a designated day on which everyone thinks about these birds and how they're faring in such a rapidly changing world.
Tawaki, or Fiordland-crested penguins, photographed in New Zealand by John Yunker.
So, how exactly are the penguins doing? According to the International Union for Conservation of Nature (IUCN) Red List of Threatened Species, of the eighteen species of penguins listed, four are stable (the Royal, Snare, Gentoo, and Little penguins), two are increasing in numbers (the Adélies and the Kings), and the status of the Emperors is classified as unknown. This means that, when it comes to the rest of the penguins of the world, their numbers are decreasing — and in some cases, they are decreasingly alarmingly fast.
King penguins on South Georgia Island.
The penguins in the most danger of becoming extinct are three species we've been very fortunate to meet: the Galápagos penguin(with an estimated 1,200 individuals left), the Yellow-eyed penguin(with fewer than 3,500 left), and New Zealand's Fiordland-crested penguin, also known by its Māori name, Tawaki, meaning crested, which the IUCN lists at between 2,500 and 9,999 individuals (yet when we visited in 2017, local researchers' estimates were capped at 3,000 individuals).
These are pretty scary numbers — and the fact is, the lifestyles of each of these species make them very hard to accurately count, which means that while there could be more than we think, it's likely that there could be far fewer than we realize.
So what can we do on World Penguin Day to help make the world a better place for them? Here are a few ideas to start.
Re-think your consumption of seafood. Overfishing is one of the biggest causes of penguin death, whether it's because humans are eating their food (such as krill) or because they get killed by fishing nets and longlines. Even "sustainable" seafood has an impact on the oceans and wildlife.
Be a thoughtful traveler and a respectful birdwatcher. If you must travel to see penguins (and it's pretty irresistible), choose places that can handle your human footprints — and always go with eco-friendly tour companies. Once there, always pay close attention to guides and naturalists who know how to keep a safe distance. If you're traveling without a group or guide, be sure to study up; learn about the birds' habitat so you can be sure to stay out of their way.
Do all that you can to combat climate change (see the Climate Reality Projectand Cowspiracyfor some good tips).
Support such conservation efforts as the Center for Ecosystem Sentinels, which monitors penguins and works on the ground to ensure protections for them.
Yellow-eyed (Tawaki) penguin, photographed in New Zealand by John Yunker.
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Like most people, I am breaking down my hike into smaller goals along the way. My immediate focus is usually on the next town stop and I typically have one larger goal in mind too. I rarely think about Katahdin - it's just too far away right now. When we started out, my larger goal had been to get to Damascus, VA. It was our home for nearly three years and I was really looking forward to seeing all of our friends. Plus it marked a new state for us on the journey.
This past week we were in Damascus and it was so much more fun and restorative than I could have hoped for. I'm actually kind of surprised I was able to tear myself away.
Backing up, let's do a rundown of how the week went. Last weekend was a trail festival called Trail Days. It's arguably the biggest trail gathering of the year and many hikers find their way to town, even if they are somewhere else on the trail. We were in Hampton, TN at the time of the festival so our friend JT came and picked us up. We spent a night at the Letter People Farm and Rooney got to play with his doggy friend Runner.
The next day we moved to a house in town to stay with our friends Brian and Christine (Trace and 1/3) who met on the trail in 2003. They're such a loving and awesome couple and it was so nice to spend time with them and their two dogs, Bear and Chief.
We wandered town and did the traditional Trail Days stuff like going to the gear booths and walking in the parade. The parade is basically just hikers walking down the main road in town and getting pelted by water balloons. Locals line the streets with balloons, super-soakers and hoses. They really don't hold back and I was soaked by the end of it all.
After the weekend, we moved to our friend Dennett's house. He has an amazing yard right on the water and the Creeper Trail and he very generously allows hikers to camp in his yard. At the high point of the weekend, he had over 30 tents set up in his yard - it was crazy. The weather was kind to us and there was a lot of lazy hanging out during the day followed by big bonfires and live music at night. The perfect scene for a bunch of hippie hikers. Thank you, Dennett, for being so patient and kind to us all. You really made everyone's Trail Days a great experience and we're lucky to have you as a friend.
After three much needed zeros we finally decided to get back on the trail for some slackpacking. Slackpacking is when you take only what you need for the day and leave the bulk of your pack contents behind. I carried a pack with just snacks for the day and an empty water bottle to drink at water sources. It couldn't have been more than 5-6 pounds and it felt glorious. We were able to do almost 45 miles in two days which let us get from Hampton to Damascus quickly. JT shuttled us (have I mentioned how fantastic JT is?!) and our friend Angie lent us her car so we were able to easily make the miles happen.
After the slackpacking we took one more day off to take advantage of bad tv and having a kitchen, then it was back to the trail and goodbye to fabulous Damascus. After some serious tears and goodbye hugs, we were back in the woods. It wasn't hard to leave town comforts, I actually missed the trail, but leaving our friends was really difficult. I know we will see them again, but we won't be living there again which is hard for me to come to terms with. I really loved my home there and will miss seeing our friends regularly. There are some really special people in Damascus and they will undoubtedly be friends for life.
Once we were back on the trail, we had some really amazing spots to enjoy. The Grayson Highlands was my first hike when I moved to the area and it is home to the most amazing trails I have experienced to date. The Highlands are beautiful and we took our time through this area, just enjoying the views. The weather was perfect and our campsite was fantastic. I couldn't have asked for more and it felt great to be back on the trail.
Yesterday we hit our 500 mile mark on the trail and it felt great! We are in Virginia and ready to take on the trail's longest state. We'll be in VA for 500 miles and many find this state to be the most challenging mentally, but I am really looking forward to it. There are so many amazing things to see in VA and when we're done we will be almost halfway through our journey. Plus we are in pretty great physical shape now so it will be nice to crank through the miles and really see what we are capable of.
But first, we are taking a day off to visit with Serial's parents. They picked us up on the trail and we're spending two nights with them in their RV. They live full-time in their fifth wheel and travel the country just exploring and having adventures. How cool. Last night at dinner the four of us were reflecting on our current journeys and how lucky we all felt.
Tomorrow it's back to the trail for us. My next "big" goal will be Harpers Ferry, WV where we will meet back up with Serial's parents AND my parents. It will be our unofficial halfway celebration and a little family reunion - lots to look forward to!
Locals vs. Hikers kickball game in Damascus.
Slinging water balloons at unsuspecting folks on the Creeper Trail.
Minutes's shoes after 425 miles.
Rooney cooling off on a warm day.
I can practically see Virginia from here! Well, that's a slight exaggeration considering we are still 70 solid miles from Damascus, but it feels like we are close.
Trail Days is a yearly festival and arguably the biggest social event of the season for hikers. Many people get off the trail and find their way to Damascus for the weekend, regardless of where they are in their thru-hike. Initially we had planned to walk into Damascus just as Trail Days started, but throw in a few unexpected zeros (it's just so fun in towns!) and we are a little off schedule. But that's the beauty of the trail... there is no schedule. Thanks to a very sweet and generous friend (we love you, JT!) we will be getting picked up in Hampton later this week and driving in with him for the weekend. All of our trail friends will be in town and we will get to see all of our Damascus friends that we have been missing so very much. Giant WIN.
After Trail Days we are going to try to do big miles (and perhaps some slack-packing to aid in said miles) to meet up with Serial's parents the following weekend. They are living full time in an RV and traveling the country so we are going to stay with them for a night or two and catch up. Basically we have some really awesome weekends in our immediate future.
But let's back up a minute and fill you in on the past few weeks. We're just shy of our 400th mile on the trail and we are still loving it. Our month-long streak of sunshine was finally broken and we got our first (and second, third, fourth...) taste of hiking in heavy rain. It's crazy how quickly you can become a soggy, muddy mess. Hiking in the rain isn't so bad, but waking up the next morning to put the wet clothes back on again... that's the real kicker. Such is life though and as long as we have dry camp clothes to sleep in, life is good.
We also had a few amazing trail magic experiences. Trail magic is anything unexpected along the trail that brings a smile to a hiker's face. It can be food or drinks waiting for us at a road crossing, donated by a kind stranger. Or it can be someone who offers to take your trash for you (when you are carrying 3-4 days worth of wrappers this is a nice surprise). The trail magic we had though... man, it was good. Prepare to be jealous.
First we got a visit from our friends The Letter People. JT, L and E (get it?) are the sweetest family you can imagine and they drove up to a fire tower with a massive amount of food and drinks to join us for a night. We had amazing views from the tower and had such a great night chatting and eating. JT & L made us all veggie burritos and guacamole for dinner and egg burritos from their farm-raised free-range chickens the next morning. They also brought in endless fresh fruit and margaritas to celebrate Cinco de Mayo. It was amazing. Our friend Piedmont put it best the next morning when he muttered "did that just really happen??" in a dazed voice. Trail magic at it's finest. Thank you, Letter People!!
Then a few days later we got another visit from a Damascus friend, Dennett! He was traveling back from work and passing by us so he decided to run the trail to meet us then hike back out with us. He ended up running about 7 miles in and then hiking 7 back out... and they were not flat miles! He's a machine. He brought a bunch of pop, beer and snacks for us and ended up sharing it with a group of hikers who were all dodging the rain that day. It was such a fun surprise and awesome to see him. Thank you, DD!
And as if those two weren't magic enough... we had our first trail magic from a complete stranger. Two former thru-hikers had set up on top of Beauty Spot, which is where had planned to camp that night. They brought curry vegetables and rice, homemade cookies and a massage chair. Yep, you heard me. A MASSAGE CHAIR. One kind stranger was studying to be a massage therapist and needed to complete his community service hours so he thought, who better to practice on than hikers. We are sore pretty much everyday so to say we were excited was a giant understatement. Thank you, Tim and Andrea!!
So now you are all caught up. As always, thanks for the love and support. We miss you all!!
View from Beauty Spot, looking at Unaka mountain.
Unaka mountain, so dense and cold up there.
Hello, friends! Let's catch up. We are doing well and still really enjoying our hike. We are 270 miles in and just out of the Smokies. The trail has been riding the NC/TN state border and we cross between the states often until we reach VA. We are getting more and more comfortable on the trail and physically the hiking is becoming easier. I look forward to hiking each day and I feel so much stronger than I did a few weeks ago. Our mileage will steadily increase and we should be moving along pretty well in the next few weeks. I don't think I have my full blown "hiker legs" but I can tell they are on their way. I can't wait for that day!
The Smoky Mountains were amazing and as beautiful as everyone said they would be. The higher elevation meant tough climbs and fantastic views, but also a wide range of weather. We had hot and sunny days followed by cold, wet, hail-filled days. It was all manageable though and made for some good laughs and stories when we huddled in shelters.
I haven't seen many animals out here. I think having Rooney with us makes it less likely. So far I've seen 4 snakes, 5 deer, 2 salamanders (they make us think of our friend Ryland!) and countless bugs. Oh, and one mouse fell from the roof of a shelter on to my neck while I was packing my food bag one morning. So gross! Ticks haven't been a big issue, thankfully.They were around early on, but we haven't seen them much in NC or TN.
My appetite is finally starting to kick in. I track everything I eat and how many hours I hike each day, just out of curiousity, and so far I've been eating about 2,000-2,500 calories a day and burning about 3,000-4,000 on the trail, depending on how many miles we do. In towns I house food and can easily eat 1,000 calories in one meal without batting an eye. I weighed myself on a scale that was pretty unreliable, but I would guess I've lost about 7-8 lbs. I'm losing fat, but gaining muscle so the scale isn't that different, but my body is definitely changing. It's going to be crazy when my appetite really kicks in...I know this is just the tip of the iceberg.
Rooney is still doing really well. We boarded him at The Standing Bear hostel while we were in the Smokies and it's nice to have him back again. I'll admit there was a part of me that liked not having the additional responsibility of caring for a dog, especially in towns, but by the end of the week I was really missing my (not so) little blue pupper. I really feel for the hikers who can't see their dogs for 5-6 months.
Socially, the trail is really interesting. There are so many different people out here from all over the world. I'm actually hiking with a few others from Chicago and one from Downers Grove...small world. Thru-hikers, section-hikers and day-hikers all add to the social dynamic, but what makes it really fun is seeing familiar faces along the way. There are about 20-25 hikers that we see regularly and its nice. Like a little hiker family out here.
Serial and I have settled in to a hiking routine that works well for us. Usually we start the day together and eventually Serial picks up his pace and we hike separately for several hours. We meet up for lunch about 2/3 of the way into our day and take a break. Then we either hike together to camp or hike slightly apart to camp, depending on the terrain and how fast I'm going. Some couples out here hike all of their miles together, and we do that some days, but our paces and styles are too different to do that consistently. This just works better for us and allows us to spend some time together and some doing our own thing.
Which reminds me...I finally got a trail name! Serial and I both keep a decent pace and we pass a handful of hikers each day and for the first week everyone was always saying, "you just missed Serial, he's 10 minutes up ahead!". Early on the trail there were a lot of hikers out there and I was hearing "10 minutes" a bunch of times every day. So...my trail name became Ten Minutes. Eventually Serial kicked it into high gear and it was more like 30 minutes ha. So my name was shortened and now it's just "Minutes".
That's it for now. Gotta get my pack situated and get ready for the trail. Much love to everyone following along and supporting us!
First sign with mileage to Katahdin...gulp.


Marsy's Law, known as the victim's bill of rights, goes into effect on July 1, and the Montana department of Justice is assisting state agencies in compliance with a new website.
DOJ spokesman Eric Sell said the website is specifically designed for county attorneys and sheriff's offices. One important feature is a mock up of what is called a 'Marsy's Card' specifically for crime victims.
"What the card has on it is the 19 constitutional rights that the voters approved back in November," Sell said. "It also has a place where victims can provide their contact information to law enforcement and vice versa, law enforcement can also provide their information."
Sell said state agencies have many concerns about complying with the new law.
"Cost is certainly one concern we're hearing time and time again," he said. "The cost of perhaps additional employees or system upgrades to implement Marsy's Law, but I think in addition to that, it's also, what do we need to do to follow the letter and spirit of the law."
HB 600 (Rep. Frank Garner), passed by the 2017 Montana Legislature, directed the attorney general's office to develop a model "Marsy's Card," and make the card available to local law enforcement and prosecutors. The card serves as an example, providing general guidance to law enforcement and prosecutors as they develop a card for their own jurisdiction.


In the high-pressure, all-or-nothing cauldron of finals football, it often takes an extra special performance from one player to get a team over the line. September footy is littered with..
We select the St George-Illawarra Dragons' all-time greatest team. The best-ever team includes three rugby league immortals & a host of Dragons legends.
For most Rugby League fans, the Grand Final is the biggest day of the year. For some players, it's a day they'd probably rather forget. Graeme Langlands, Anthony Mundine, Ben Hunt, David Williams, Neil Glover.


text:
Q: Firebase: Get child of snapshot.val() object I'm trying to get an object of the first matching child using Firebase's admin SDK. So far I was able to get an object inside a parent object:
{ -LVOBaJXoai9n1mYrs3F: { created: 1546610959716, displayEnterpriseNumber: "BE 0479.312.137", email: "anthony.magnus@gmail.com", enterpriseNumber: "479312137" } }

but I want
{ created: 1546610959716, email: "john.doe@gmail.com", token: '51234', isActive: true }

Code
In a firebase function, I have the following code:
try {
   const token = req.body.token;
   const couponSnapshot = await admin.database().ref(`coupons`).orderByChild('token').equalTo(token).limitToFirst(1).once('value');
   const coupon = couponSnapshot.val();

   console.log(coupon) //{-LVOBaJXoai9n1mYrs3F: {…}} Object in Object...

} catch(error) {
   //error handling
}

How can I get the child object from the parent object?
I tried to loop over the couponSnapshot and attach the child value to the coupon variable but I'm getting an error.
const couponSnapshot = await admin.database().ref(`coupons`).orderByChild('token').equalTo(token).limitToFirst(1).once('value');

let coupon = {};

couponSnapshot.forEach(childSnapshot => {
  coupon = childSnapshot.val();
});

console.log(coupon.email);

The error:
error TS2345: Argument of type '(childSnapshot: DataSnapshot) => void' is not assignable to parameter of type '(
a: DataSnapshot) => boolean'.
  Type 'void' is not assignable to type 'boolean'.

Property 'email' does not exist on type '{}'.

Can someone help me out?

A: You've told TypeScript that you want to strongly type coupon to be an empty object, and only an empty object.  You're later telling it to access a property on the object email that doesn't exist in an empty object.  That's why you're coming up with the error: Property 'email' does not exist on type '{}'.
You should not provide any typing hints at all.  Simply let email; with no default assignment if you want it to be of type any, which lets you access arbitrary properties on the object.  Of course, you may also need to check if it's defined before reaching into it, as that would be an error as well.

redpajama_set_name: RedPajamaStackExchange


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Elk Grove Village-based manufacturer Material Sciences Corp. said Wednesday it is taking several cost-cutting measures, including eliminating 18 salaried positions, to help reduce overhead and increase operating efficiencies.
The company expects the restructuring to produce $2 million in pretax savings on an annualized basis. Officials said Material Sciences would take an after-tax charge of $500,000 to $700,000 in the third fiscal quarter, ending Nov. 30, to account for severance and other related costs.
"This reorganization is part of an ongoing effort to improve (our) overall efficiency and increase its competitiveness," Gerald G. Nadig, chairman, president and CEO, said in a statement. "Restructuring our manufacturing organization will result in a more streamlined . . .operating structure, which will make us more effective while reducing overall costs."
The actions largely affect Material Sciences manufacturing operations, Mr. Nadig said, and not areas that offer greater growth potential, such as the marketing, selling, and product development functions in the company's electronics and acoustical/thermal areas.
Material Sciences has taken other actions this year to trim expenses. The company sold off a struggling steel plant near San Francisco in May, reducing its workforce by 225, leaving 800 workers companywide. And the company said pay increases were held to 3.5% this year, down from 4.0% in 2001 (Crain's, Sept. 9).