text:
Rowan Beaird
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Ohio, Summer
"The boy sits on the concrete lot that juts out onto the water, his feet dangling over the nook where water meets sand. There is no one on the shore but an old man, oddly dressed in long sleeves, trousers, and socks. The television forecast said steady rain but it has held off, clouds knitting together in the distance. The lake is dark as ash, uneven and unsteady before the storm. He imagines the crests, these wide-mouthed gasps of water, in his bathtub, swallowing him whole, splashing out onto the tile, soaking the soiled towels and bath mat. He looks for fish, but today he has seen only shadows, the receding tide revealing an empty bottle, shells, a spindly rope of algae. There are planks of wood in the wet sand, still as fossils."
"Ohio, Summer" was published in The American Literary Review. It's about a missing limb and a missing father. To read the full story, visit their site.


text:
300 Canal St
300 Canal St,
Bonneau, SC 29431
Est. Refinance $392/mo
Est. Value* : $96,910
300 Canal St is a single family home in Bonneau, SC. This home includes 1,236 square feet of living space. Additional features include a parking garage, and fireplace. In 11/1/2017, this property sold for $75,000. Today, the estimated home value is around $96,910. Frequently asked questions about 300 Canal St and the surrounding community.
How is the neighborhood? 300 Canal St is located in a community that is home to 5,147 residents, which has decreased by -6.1% since 2000. 80.2% of the residents are families with children. The largest age group is between 45 to 54 (15.88%), while the second-largest age group is between 35 to 44 (13.06%).
What is the cost of living? The average family living near 300 Canal St has a median household income of $45,886 and spends about $40,613 annually.
What is the crime rate? The neighborhood surrounding 300 Canal St is relatively unsafe since the personal crime risk (including both violent and non-violent crimes) is 17.4%, which is higher than the state average of 13.6%. Property crime risk is 17.4%, which is higher than the state average of 11.1%.
How are the schools? The schools closest to 300 Canal St include Berkeley County Middle College High, Henry E. Bonner Elementary, and Macedonia Middle. The average SAT score for this area is about 956 (477 for Math and 479 for English), which is higher than the state average SAT score of 945.
How's the job market? About 58.9% of the residents near 300 Canal St have blue-collar jobs, while the remaining 41.1% have white-collar jobs. The unemployment rate is about 4.3%, which is higher than the state unemployment rate of 4.0%.
What is the weather like year-round? The average temperature for 300 Canal St is about 46.0 degrees in January and 80.6 degrees in July. The average rainfall is around 49.1 inches per year. The average snowfall is around 1.0 inches per year.
Is there a risk of a natural disaster? 300 Canal St resides in an area with low flood risk, high tornado risk, and high earthquake risk.
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11/1/2017 $75,000 Sold
1113 Ole South Court,
Moncks Corner, SC 29461
Melissa Cook - Coldwell Banker Realty - 21000708
1229 Sheridan Court,
2 Bd | 1 bath | 972 sqft
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458 Four Seasons Blvd,
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107 Bluff Isle Court,
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162 Calm Water Way,
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211 Sabal Palmetto Court,
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429 Whispering Breeze Lane,
Andrea Bell - Coldwell Banker Realty - 19027181
414 Ambergate Lane,
Stephanie Lukac - Coldwell Banker Realty - 21001113
2018 $2,001 $0 $0 $85,700
2017 $418 $0 $0 $109,700
2014 $322 $0 $0 $4,160
Berkeley County Middle College High 11-12 98 4 24.5 12.11 mi
Henry E. Bonner Elementary PK-5 715 41 17.4 7.5 mi
Macedonia Middle 6-8 373 26 14.3 7.3 mi

redpajama_set_name: RedPajamaCommonCrawl


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text:
\section{Introduction}

The theory of the unified weak and electromagnetic interaction
between elementary particles including vector particles (Standard
Model) is renormalized field theory \cite{Hooft}. The crucial role
plays the spontaneous breaking symmetry when massless vector
fields acquire masses due to the Higgs mechanism. Nowadays, it is
of great importance for searching the scalar Higgs bosons at Large
Hadronic Collider. Anyway, the old problem of describing massive
and massless vector particles is of theoretical interest.  It is
well known that the Proca equations for vector particles can be
cast into the first-order (the matrix form) Duffin$-$Kemmer$-$Petiau
(DKP) relativistic wave equation (RWE) \cite{Petiau},
\cite{Duffin}, \cite{Kemmer}, \cite{Umezawa}. One can
find early references on DKP equations in \cite{Krajcik}. The
matrix form of RWE is also convenient for the formulation of
higher derivative field equations \cite{Kruglov}, \cite{Kruglov(a)}, \cite{Kruglov(b)},
\cite{Kruglov(c)}, fields with
multi-spin \cite{Kruglov1}, \cite{Kruglov1(a)}, \cite{Kruglov1(b)},
\cite{Kruglov1(c)}, \cite{Kruglov1(d)}, \cite{Kruglov1'}, Einstein gravity
equations \cite{Fedorov1}, fields in curved space-time
\cite{Pimentel}, \cite{Pimentel1}, \cite{Bogush} and quantum chromodynamics
\cite{Gribov}. There is a vast number of papers devoted to DKP
equations, and, therefore, we mention only some part of them.

The goal of this paper is to give the systematic description of
massive and massless vector fields in the DKP formalism, to find
solutions in the form of projection matrix-dyads, to obtain the
quantum-mechanical Hamiltonian, to investigate the dilatation
symmetry, and to perform the canonical quantization.

The paper is organized as follows. In Sec.2, we consider massive
and massless vector fields in the form of RWE with two mass
parameters. Solutions of the wave equation for a free massive
field are obtained in the form of matrix-dyads. The
quantum-mechanical Hamiltonians are found for massive and massless
vector fields in Sec.3. We obtain the canonical and the
symmetrical Belinfante energy momentum tensors and the dilatation
current in Sec.4. It is demonstrated that the dilatation symmetry
is broken for massive fields but in the massless case the modified
dilatation current is conserved. In Sec.5, the canonical
quantization of massive fields is performed and the propagator of
fields is obtained in the DKP formalism. We draw a conclusion in
Sec.6. In Appendix, we demonstrate that the system of equations with
two mass parameter introduced can describe massive or massless fields.

The Euclidean metric is used and the system of units $\hbar =c=1$
is explored.

\section{Vector particles in the first-order formalism}

To consider the massive and massless vector fields, let us introduce the system of
equations with two mass parameters
\[
\partial _\nu \psi _{[\mu \nu ]}+m_1 \psi _\mu=0,
\]
\vspace{-8mm}
\begin{equation}
\label{1}
\end{equation}
\vspace{-8mm}
\[
\partial _\nu \psi _\mu -\partial _\mu \psi _\nu +m_2\psi _{[\mu \nu]}=0.
\]
System of equations (1) is the generalization of Proca equations, and describes massive or massless particles
depending on parameters $m_1$, $m_2$ chosen (see Appendix). At $m_1=m_2$, after the renormalization of fields, one arrives at the Proca equations. At $m_1=0$, we have the massless case corresponding to
the Maxwell equations. Excluding the antisymmetrical tensor $\psi
_{[\mu \nu ]}$ in Eq.(1) (see (A1) in Appendix), in the general case, we find the wave
equation for the field $\psi _\mu$ possessing the mass
$m=\sqrt{m_1m_2}$. Thus, Eq.(1) gives the convenient parametrization for description of
massive or massless vector fields. The fields $\psi_A$ ($A=\mu$,
$[\mu\nu]$) have the same dimension due to the presence of the
mass parameter $m_2$.

Introducing the wave function
\begin{equation}
\Psi (x)=\left\{ \psi _A(x)\right\} =\left(
\begin{array}{c}
\psi_\mu (x)\\
\psi_{[\mu\nu]}(x)\\
\end{array}
\right) \hspace{0.5in}(A=\mu , [\mu\nu]),\label{2}
\end{equation}
and the elements of the entire matrix algebra $\varepsilon
^{A,B}$, with properties
\begin{equation}
\left( \varepsilon ^{M,N}\right) _{AB}=\delta _{MA}\delta _{NB},
\hspace{0.5in}\varepsilon ^{M,A}\varepsilon ^{B,N}=\delta
_{AB}\varepsilon ^{M,N}, \label{3}
\end{equation}
where $A,B,M,N=\mu,[\mu\nu]$, the system of equations (1) can be
represented in the matrix form
\begin{equation}
\left[\partial _\mu \left(\varepsilon ^{\nu,[\nu\mu]}+ \varepsilon
^{[\nu\mu],\nu}\right) + m_1\varepsilon ^{\mu,\mu}+
m_2\frac{1}{2}\varepsilon ^{[\nu\mu],[\nu\mu]}\right] _{AB}\Psi
_B(x)=0 . \label{4}
\end{equation}
We imply a summation over all repeated indices. Defining the
$10\times 10$ matrices
\begin{equation}
\beta_\mu=\varepsilon ^{\nu,[\nu\mu]}+ \varepsilon
^{[\nu\mu],\nu},~~~\overline{P}=\varepsilon ^{\mu,\mu},~~~
P=\frac{1}{2}\varepsilon ^{[\nu\mu],[\nu\mu]}, \label{5}
\end{equation}
Eq.(4) takes the form of the first-order RWE
\begin{equation}
\left( \beta _\mu \partial _\mu + m_1\overline{P}+ m_2P\right)
\Psi (x)=0 . \label{6}
\end{equation}
The matrices $\beta_\mu$ are Hermitian matrices,
$\beta_\mu^+=\beta_\mu$. The projection operator $\overline{P}=
\overline{P}^+$ extracts the four-dimensional vector subspace
($\psi_\mu$) of the wave function $\Psi$, and the projection
operator $P=P^+$ extracts the six-dimensional tensor subspace
corresponding to the $\psi_{[\mu\nu]}$. The matrices $\beta _\mu$
obey the DKP algebra
\begin{equation}
\beta _\mu \beta _\nu \beta _\alpha +\beta _\alpha \beta _\nu
\beta _\mu =\delta _{\mu \nu }\beta _\alpha+\delta _{\alpha \nu
}\beta _\mu , \label{7}
\end{equation}
and matrices $\overline{P}$, $P$ are projection matrices
\[
\overline{P}^2=\overline{P},~~P^2=P,~~\overline{P}+P=1,~~
\overline{P}P=P\overline{P}=0,
\]
\vspace{-8mm}
\begin{equation}
\label{8}
\end{equation}
\vspace{-8mm}
\[
\beta_\mu\overline{P}+\overline{P}\beta_\mu=\beta_\mu,~~ \beta_\mu
P+P\beta_\mu=\beta_\mu.
\]
At $m_1=m_2=m$, from Eq.(6), we arrive at the DKP equation for
massive vector fields
\begin{equation}
\left( \beta _\mu \partial _\mu + m\right) \Psi (x)=0 . \label{9}
\end{equation}
For massless DKP equation, corresponding to the Maxwell equations,
we put $m_1=0$ in Eq.(6), and arrive at
\begin{equation}
\left( \beta _\mu \partial _\mu + m_2P\right) \Psi (x)=0 .
\label{10}
\end{equation}
In \cite{Harish}, \cite{Moroz} the case $m_1=0$, Eq.(10), was
considered. It should be noted that massless DKP model describes electromagnetic fields and is
invariant under a local U(1) gauge symmetry and being a fundamental
requirement for their description. The gauge transformations in DKP formalism are given by
\[
\Psi'=\Psi+\overline{P}\Phi.
\]
Then with the help of Eq.(8),(10), we obtain
\[
\left( \beta _\mu \partial _\mu + m_2P\right) \Psi'=(1-\overline{P})\beta _\mu \partial _\mu\Phi .
\]
Thus Eq.(10) is invariant under gauge transformations if the function $\Phi$ obeys the equation as follows (see also\cite{Harish}): $P\beta_\mu\partial_\mu\Phi=0$. It is easy to verify that this equation is valid for the function
\[
\Phi(x)=\left(
\begin{array}{c}
\partial_\mu \Lambda(x)\\
0
\end{array}\right),
\]
which leads to the gauge transformations: $\psi'_\mu(x)= \psi_\mu (x)+\partial_\mu \Lambda (x)$.
The gauge invariance of massless DKP equation was also discussed
in \cite{Lunardi}.

Now, we investigate the general case, Eq.(6), including two mass parameters, $m_1$ and $m_2$.
The Lorentz group generators in the $10$-dimension representation space are  given by
\begin{equation}
J_{\mu\nu}=\beta_\mu\beta_\nu-\beta_\nu\beta_\mu=
\varepsilon^{[\lambda\mu],[\lambda\nu]}+\varepsilon^{\mu,\nu}-
\varepsilon^{[\lambda\nu],[\lambda\mu]}-\varepsilon^{\nu,\mu},
 \label{11}
\end{equation}
and obey the commutation relations
\[
\left[J_{\rho\sigma},J_{\mu\nu}\right]=\delta_{\sigma\mu}J_{\rho\nu}+
\delta_{\rho\nu}J_{\sigma\mu}-\delta_{\rho\mu}J_{\sigma\nu}-\delta_{\sigma\nu}J_{\rho\mu},
\]
\vspace{-8mm}
\begin{equation}
\label{12}
\end{equation}
\vspace{-8mm}
\[
\left[\beta_{\lambda},J_{\mu\nu}\right]=\delta_{\lambda\mu}\beta_\nu
-\delta_{\lambda\nu}\beta_\mu,~~\left[\overline{P},J_{\mu\nu}\right]=
0,~~\left[P,J_{\mu\nu}\right]= 0 .
\]
Eq.(6) is form-invariant under the Lorentz transformations because
of Eq.(12). The Lorentz-invariant is $\overline{\Psi }\Psi =\Psi
^{+}\eta \Psi$, where $\Psi ^{+}$ is the Hermitian-conjugated wave
function, and the Hermitianizing matrix, $\eta $, is given by
\begin{equation}
\eta =\varepsilon ^{m,m}-\varepsilon ^{4,4} +\varepsilon
^{[m4],[m4]}-\frac{1}{2}\varepsilon ^{[mn],[mn]}. \label{13}
\end{equation}
The $\eta $ is the Hermitian matrix, $\eta^+ =\eta $, and obeys
the relations: $\eta \beta _m=-\beta _m\eta$ (m=1,2,3), $\eta
\beta _4=\beta _4\eta$. With the help of these relations, one
finds the "conjugated" equation
\begin{equation}
\overline{\Psi}(x)\left( \beta _\mu \overleftarrow{\partial}_\mu -
m_1\overline{P}- m_2P\right) =0 . \label{14}
\end{equation}
The Lagrangian leading to Eq.(6),(14) is
\[
\mathcal{L}=-\frac{1}{2}\overline{\Psi }(x)\left( \beta _\mu
\partial _\mu + m_1\overline{P} + m_2 P\right)\Psi (x)
\]
\vspace{-8mm}
\begin{equation}
\label{15}
\end{equation}
\vspace{-8mm}
\[
+ \frac{1}{2}\overline{\Psi }(x)\left(\beta _\mu
\overleftarrow{\partial _\mu} -m_1\overline{P}- m_2 P\right) \Psi
(x).
\]
In terms of fields $\psi_A$, the Lagrangian (15) reads
\begin{equation}
{\cal L}=\frac{1}{2}\left(\psi_{[\rho\mu]}^*\partial
_\mu\psi_\rho-\psi_\rho^*\partial _\mu\psi_{[\rho\mu]} - m_1
\psi_\mu^*\psi_\mu +
\frac{m_2}{2}\psi_{[\rho\mu]}^*\psi_{[\rho\mu]}\right)+ c.c.
,\label{16}
\end{equation}
where the $c.c.$ means the complex conjugated expression; complex
conjugation $*$ does not act on the metric imaginary unit $i$ of
fourth components in Eq.(16), $\psi_\mu^*=(\psi_m^*,i\psi_0^*)$,
and so on, and we used $\overline{\Psi
}=(\psi_\mu^*,-\psi_{[\mu\nu]}^*)$. It is easy to verify that
Euler-Lagrange equations $\partial{\cal
L}/\partial\psi_A-\partial_\mu\left(\partial{\cal L}/
\partial\partial_\mu\psi_A\right)=0$ lead to the equations of motion
 Eq.(1). The Lagrangian (15) (and (16)) vanishes for fields $\psi_A$ obeying Eq.(1) (or
Eq.(6),(14))).

\section{Solutions to the matrix equation}

Eq.(6), in the momentum space, for the positive ($+p$) and
negative ($-p$) energies, reads
\begin{equation}
\left( \pm i\widehat{p} + m_1\overline{P}+ m_2 P\right) \Psi (\pm
p)=0 , \label{17}
\end{equation}
where $\widehat{p}=\beta _\mu p _\mu$, and the four-momentum being
$p_\mu=(\textbf{p},ip_0)$ ($p^2= \textbf{p}^2-p_0^2$). One can
verify that the matrix of equation (17)
\begin{equation}
\Lambda_\pm = \pm i\widehat{p} + m_1\overline{P}+ m_2 P,
\label{18}
\end{equation}
obeys the ``minimal" matrix equation:
\begin{equation}
\left(\Lambda_\pm-m_1\right)\left(\Lambda_\pm-m_2 \right)
\left[\left(\Lambda_\pm-m_1\right)\left(\Lambda_\pm-m_2
\right)+p^2 \right]=0. \label{19}
\end{equation}
The non-trivial solutions to Eq.(17) exist if $\det \Lambda_\pm
=0$. Therefore, the eigenvalue of the matrix $\Lambda_\pm$ should
be zero, and this requirement results to the dispersion relation
\begin{equation}
p^2+m_1 m_2=0 . \label{20}
\end{equation}
Other eigenvalues of the matrix $\Lambda_\pm$ are $m_1$ and $m_2$,
that follows from Eq.(19). On-shell, $p^2=-m_1 m_2$, when $m_1\neq
m_2$, Eq.(19) becomes
\begin{equation}
\Lambda_\pm\left(\Lambda_\pm-m_1-m_2 \right)
\left(\Lambda_\pm-m_1\right)\left(\Lambda_\pm-m_2 \right)=0.
\label{21}
\end{equation}
Solutions to Eq.(17) in the form of the projection matrix
\cite{Fedorov} follow from Eq.(21):
\[
\Pi_\pm=N\left(\Lambda_\pm-m_1-m_2 \right)
\left(\Lambda_\pm-m_1\right)\left(\Lambda_\pm-m_2 \right)
\]
\vspace{-8mm}
\begin{equation}
\label{22}
\end{equation}
\vspace{-8mm}
\[
=\mp Ni\widehat{p}\left(\pm im_1\widehat{p}P\pm
im_2\widehat{p}\overline{P} -m_1 m_2\right),
\]
so that $\Lambda_\pm\Pi_\pm=0$, where $N$ is the normalization
constant. Every column of the matrix $\Pi_\pm$ is the solution to
Eq.(17). The projection matrix obeys the equation
\begin{equation}
\Pi_\pm^2=\Pi_\pm, \label{23}
\end{equation}
that leads to the normalization constant
\begin{equation}
N=-\frac{1}{m_1 m_2(m_1+m_2)} . \label{24}
\end{equation}
This can be verified with the help of Eq.(21). We notice that
Eq.(21) is the minimal polynomial only for the non-degenerate case
$m_1\neq m_2$. For the special case, $m_1=m_2$, we have to use the
minimal matrix equation on-shell:
\begin{equation}
\Lambda_\pm\left(\Lambda_\pm-m\right)\left(\Lambda_\pm-2m\right)=0.
\label{25}
\end{equation}
We obtain the projection operator from Eq.(25)
\begin{equation}
\Pi^{DKP}_\pm=\frac{1}{2m^2}\left(\Lambda_\pm^2-3m
\Lambda_\pm+2m^2\right) =\frac{\pm i\widehat{p}\left(\pm
i\widehat{p}-m\right)}{2m^2}. \label{26}
\end{equation}
Eq.(22) at $m_1=m_2$ becomes Eq.(26) .

We use the spin projection operators \cite{Fedorov}:
\begin{equation}
S_{(\pm 1)}=\frac 12\sigma _p\left( \sigma _p\pm 1\right)
,\hspace{0.5in}S_{(0)}=1-\sigma _p^2 .\label{27}
\end{equation}
where
\begin{equation}
\sigma _p=-\frac i{2\mid \mathbf{p}\mid }\epsilon
_{abc}p_aJ_{bc}=-\frac i{\mid \mathbf{p}\mid }\epsilon
_{abc}p_a\beta _b\beta _c. \label{28}
\end{equation}
One can check the relations $[\Lambda_\pm,\sigma _p]=0$, $S_{(\pm
1)}^2=S_{(\pm 1)}$, $S _{(\pm 1)}S_{(0)}=0$, $S_{(0)}^2=S_{(0)}$,
$[\Lambda_\pm,S_{(\pm 1)}]=0$, $[\Lambda_\pm,S_{(0)}]=0$. The
projection operators extracting solutions to Eq.(17) with spin
projections $\pm 1$, $0$ in the form of matrix-dyads are given by
\begin{equation}
\Pi_\pm S_{(\pm 1)}=\Psi_{\pm 1}\cdot \overline{\Psi}_{\pm 1}
,~~~~\Pi_\pm S_{(0)}=\Psi_{0}\cdot \overline{\Psi}_{0}. \label{29}
\end{equation}
The matrix-dyad has the matrix elements $(\Psi\cdot
\overline{\Psi})_{AB}= \Psi_A \overline{\Psi}_B $. Eq.(29) allow
us to make calculations of different electrodynamics processes
with vector particles in the covariant form \cite{Fedorov}.

For the case of massless vector particles (photons), the parameter
$m_1 =0$ and $p^2=0$. Then the matrix of equation (10) is
\begin{equation}
\Lambda^{(0)}_\pm= \pm i\widehat{p} + m_2 P, \label{30}
\end{equation}
and satisfies the minimal matrix equation
\begin{equation}
\Lambda^{(0)2}_\pm\left(\Lambda^{(0)2}_\pm-m_2\right)^2 =0.
\label{31}
\end{equation}
In this case zero eigenvalues of the operator $\Lambda^{(0)}_\pm $
are degenerated and it is impossible to obtain solutions in the
form of projection matrix-dyads \cite{Fedorov}, \cite{Moroz}. In
the case of generalized Maxwell equations such difficulty is
absent \cite{Kruglov2}, \cite{Kruglov2(a)}.

\section{The Hamiltonian form of equations}
\subsection{Massive fields}

It is very useful to obtain the quantum mechanical Hamiltonian
corresponding to equations (1) (or (6)) because the non-dynamical
components will be absent. There was a suggestion in \cite{Ghose}
to couple the electromagnetic field in the DKP equation only at
the level of the Hamiltonian form. Some aspects of Hamiltonian
form of DKP equations were considered in \cite{Nowakowski}, \cite{Nowakowski(a)},
\cite{Nowakowski(b)}. To
exclude the non-dynamical components, we rewrite Eq.(1) in the
form of two systems
\begin{equation}
m_2\psi _{[4 m]}=\partial _4 \psi _m -\partial _m \psi _4 , ~~~~
\partial _4 \psi _{[m 4 ]}+\partial _n \psi _{[m n]}+m_1 \psi _m =
0,
\label{32}
\end{equation}
\begin{equation}
m_2\psi _{[mn]}=\partial _m \psi _n -\partial _n \psi _m , ~~~~
\partial _n \psi _{[4n]}+m_1 \psi _4 = 0. \label{33}
\end{equation}
Equations (32) contain derivatives on the time of the dynamical
components $\psi_m$, $\psi_{[m4]}$, and Eq.(33) possess only
spatial derivatives on auxiliary (non-dynamical) components
$\psi_4$, $\psi_{[mn]}$. Replacing non-dynamical components from
Eq.(33) into Eq.(32), we arrive at the equations for the dynamical
components
\[
i\partial _t \psi _m=-m_2\psi _{[4 m]} +\frac{1}{m_1}\partial
_m\partial_n \psi _{[4 n]},
\]
\vspace{-8mm}
\begin{equation}
\label{34}
\end{equation}
\vspace{-8mm}
\[
i\partial _t \psi _{[m 4 ]}=\frac{1}{m_2}\left(\partial_m\partial
_n \psi _n-\partial_n^2\psi_m\right)+m_1 \psi _m .
\]
We can represent Eq.(34) in the matrix form introducing the
six-component wave function
\begin{equation}
\Phi (x)=\left(
\begin{array}{c}
\psi_n (x)\\
\psi_{[m 4]}(x)\\
\end{array}
\right).\label{35}
\end{equation}
With the help of the elements of the matrix algebra Eq.(3), we
rewrite Eq.(34) in the Schr\"{o}dinger-like form
\[
i\partial_t\Phi
(x)=\biggl[m_1\varepsilon^{[m4],m}-m_2\varepsilon^{n,[4n]}+\frac{1}{m_1}\varepsilon^{n,[4m]}\partial_n\partial_m
\]
\vspace{-7mm}
\begin{equation}
\label{36}
\end{equation}
\vspace{-7mm}
\[
+ \frac{1}{m_2}\biggl(\varepsilon^{[m4],n}\partial_m\partial_n\
-\varepsilon^{[m4],m}\partial_n^2\biggr)\biggr]\Phi (x),
\]
where the Hamiltonian is given by
\[
{\cal H}=m_1\varepsilon^{[m4],m}-m_2\varepsilon^{n,[4n]}
\]
\vspace{-8mm}
\begin{equation}
\label{37}
\end{equation}
\vspace{-8mm}
\[
+\frac{1}{m_1}\varepsilon^{n,[4m]}\partial_n\partial_m
+ \frac{1}{m_2}\biggl(\varepsilon^{[m4],n}\partial_m\partial_n\
-\varepsilon^{[m4],m}\partial_n^2\biggr).
\]
We have implied that $m_1\neq 0$, $m_2\neq 0$. The Hamiltonian
(37) is simplified for the choice $m_1=m_2\equiv m$:
\begin{equation}
{\cal H}=m\left(\varepsilon^{[m4],m}-\varepsilon^{n,[4n]}\right)+\frac{1}{m}\left[
\biggl(\varepsilon^{n,[4m]}+\varepsilon^{[m4],n}\biggr)\partial_m\partial_n\
-\varepsilon^{[m4],m}\partial_n^2\right]. \label{38}
\end{equation}
The six-component wave function (35) corresponds to three spin
states with positive (for particles) and negative (for
antiparticles) energies and does not contain auxiliary components.
We can introduce the minimal interaction of vector particles with
electromagnetic fields by the replacement $\partial_\mu\rightarrow
\partial_\mu-ieA_\mu$, where $A_\mu$ is the vector-potential of electromagnetic
fields. One may verify using the properties (3) that the matrix
Hamiltonian (37) in the momentum space obeys the minimal equation
\begin{equation}
\left({\cal H}^2- \textbf{p}^2-m_1m_2\right)\left({\cal H}^2+
\textbf{p}^2-m_1m_2\right)=0.\label{39}
\end{equation}
Equation (39) is an operator identity because in the momentum space, $\partial_\mu\rightarrow ip_\mu$,
the Hamiltonian is the matrix operator. If $m_1>0$ and $m_2>0$, the physical eigenvalue of the Hamiltonian
squared follows from Eq.(39): $p^2_0=\textbf{p}^2+m_1m_2$. The
projection operator extracting states with positive ($p_0$) and
negative ($-p_0$) energies is given by
\begin{equation}
\Sigma_{\pm}=\pm\frac{1}{4p_0\textbf{p}^2}\left({\cal H}\pm
p_0\right)\left({\cal H}^2+ \textbf{p}^2-m_1m_2\right),\label{40}
\end{equation}
so that $\Sigma_{\pm}^2=\Sigma_{\pm}$, $\Sigma_{\pm} {\cal
H}={\cal H}\Sigma_{\pm}=\pm p_0\Sigma_{\pm}$, where
$p_0=\sqrt{\textbf{p}^2+m_1m_2}$. The projection operator (40)
allows us to get solutions to the Hamiltonian equation (36).

\subsection{Massless fields}

In the case of massless vector fields (photons), we put $m_1=0$ in
Eq.(32),(33) and arrive at
\begin{equation}
m_2\psi _{[4 m]}=\partial _4 \psi _m -\partial _m \psi _4 , ~~~~
\partial _4 \psi _{[m 4 ]}+\partial _n \psi _{[m n]} =
0, \label{41}
\end{equation}
\begin{equation}
m_2\psi _{[mn]}=\partial _m \psi _n -\partial _n \psi _m , ~~~~
\partial _n \psi _{[4n]} = 0. \label{42}
\end{equation}
It is impossible to exclude the component $\psi_4$ from
Eq.(41),(42). Therefore, we need to consider the $\psi_4$ as a
dynamical component for the massless fields. To have the evolution
of the $\psi_4$ in time, we add to Eq.(41),(42) the Lorentz
condition $\partial_m\psi_m+\partial_4\psi_4=0$. After the
exclusion the non-dynamical component $\psi_{[mn]}$ from Eq.(41),
one finds the equations as follows:
\[
i\partial _t \psi _m=-m_2\psi _{[4 m]} -\partial_m \psi _4 ,
\]
\begin{equation}
i\partial _t \psi _4=\partial_n \psi _n , \label{43}
\end{equation}
\[
i\partial _t \psi _{[m 4 ]}=\frac{1}{m_2}\left(\partial_m\partial
_n \psi _n-\partial_n^2\psi_m\right) .
\]
Introducing the seven-component wave function
\begin{equation}
\Phi_0 (x)=\left(
\begin{array}{c}
\psi_\mu (x)\\
\psi_{[m 4]}(x)\\
\end{array}
\right),\label{44}
\end{equation}
and using the elements of the entire matrix algebra Eq.(3), we
represent Eq.(43) in the Hamiltonian form
\[
i\partial_t\Phi_0 (x)={\cal H}_0\Phi_0 (x),
\]
\vspace{-8mm}
\begin{equation}
\label{45}
\end{equation}
\vspace{-8mm}
\[
{\cal
H}_0=-m_2\varepsilon^{n,[4n]}+\left(\varepsilon^{4,m}-\varepsilon^{m,4}\right)\partial_m
+ \frac{1}{m_2}\biggl(\varepsilon^{[m4],n}\partial_m\partial_n\
-\varepsilon^{[m4],m}\partial_n^2\biggr).
\]
With the help of Eq.(3) one can check that the Hamiltonian (45) in
the momentum space satisfies the minimal equation
\begin{equation}
{\cal H}_0^2\left({\cal H}_0^2- \textbf{p}^2\right)=0.\label{46}
\end{equation}
The projection operator extracting states with positive
($p_0=|\textbf{p}|$) and negative ($p_0=-|\textbf{p}|$) energies
is
\begin{equation}
\Sigma^0_{\pm}=\pm\frac{1}{2|\textbf{p}|^3}\left({\cal H}_0\pm
|\textbf{p}|\right){\cal H}_0^2,\label{47}
\end{equation}
and $\left(\Sigma_{\pm}^0\right)^2=\Sigma^0_{\pm}$,
$\Sigma^0_{\pm} {\cal H}_0={\cal H}_0\Sigma^0_{\pm}=\pm|\textbf{p}|
\Sigma^0_{\pm}$, where $|\textbf{p}|=\sqrt{p_1^2+p_2^2+p_3^2}$.

\section{The energy-momentum tensor and dilatation current}

Now, we construct the energy-momentum tensor for a given Lagrangian (15).
With the help of the general equation \cite{Ahieser}
\begin{equation}
T^c_{\mu\nu}=\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu
\Psi (x)\right)}\partial_\nu \Psi (x)+\partial_\nu \overline{\Psi
}(x) \frac{\partial\mathcal{L}}{\partial\left(\partial_\mu
\overline{\Psi} (x)\right)}-\delta_{\mu\nu} \mathcal{L}
,\label{48}
\end{equation}
we obtain the canonical energy-momentum tensor
\begin{equation}
T^c_{\mu\nu}=\frac{1}{2}\left(\partial_\nu \overline{\Psi}
(x)\right)\beta_\mu \Psi (x)-\frac{1}{2} \overline{\Psi}
(x)\beta_\mu \partial_\nu\Psi (x), \label{49}
\end{equation}
where we took into consideration that the Lagrangian (15) vanishes
for fields obeying the equations of motion. With the aid of
Eq.(2),(3),(5) the canonical energy-momentum tensor in the
component form becomes
\begin{equation}
T^c_{\mu\nu}=\frac{1}{2}\left(\psi_{[\rho\mu]}^\ast\partial_\nu
\psi_\rho-\psi_\rho^\ast\partial_\nu \psi_{[\rho\mu]}\right)+c.c.
. \label{50}
\end{equation}
It is easy to verify, using Eq.(1), that the energy-momentum
tensor (50) (and (49)) is conserved tensor, $\partial_\mu
T^c_{\mu\nu}=0$. The canonical energy-momentum tensor is not the
symmetric tensor, $T^c_{\mu\nu}\neq T^c_{\nu\mu}$, and its trace
is
\begin{equation}
T_{\mu\mu}^{c}=m_1\psi_\mu^\ast\psi_\mu-\frac{1}{2}m_2
\psi_{[\mu\nu]}^\ast\psi_{[\mu\nu]}. \label{51}
\end{equation}
One expects a nonzero trace of the energy-momentum
tensor and a broken dilatation current in a massive theory. But even for massless fields,
$m_1=0$, $m_2\neq 0$, the trace of the energy-momentum tensor in (51) is non-zero.
Therefore, it is of interest to investigate the scale invariance in the general case.
Thus, we consider the canonical dilatation current \cite{Coleman}
\begin{equation}
D_\mu^c=x_\alpha T_{\mu\alpha}^{c}+\Pi_\mu \textbf{d}\Psi,
\label{52}
\end{equation}
where
\begin{equation}
\Pi_\mu=\frac{{\partial\cal
L}}{\partial\left(\partial_\mu\Psi\right)} =-\overline{\Psi
}\beta_\mu, \label{53}
\end{equation}
and the matrix $\textbf{d}$ defines the field dimension. For the
Bose fields the $\textbf{d}$ is the unit matrix. From
Eq.(52),(53), we obtain
\begin{equation}
\partial_\mu D_\mu^c=T_{\mu\mu}^{c}, \label{54}
\end{equation}
where the conservation of the current was used
\begin{equation}
\partial_\mu j_\mu=\partial_\mu(i\overline{\Psi}\beta_\mu\Psi)=0. \label{55}
\end{equation}
The analogous relation follows from \cite{Coleman}
\[
\partial_\mu D_\mu^c=\Pi_\mu\left(\textbf{d}+1\right)\partial_\mu\Psi
 +\frac{\partial{\cal L}}{\partial\Psi}\textbf{d}\Psi
 +\overline{\Psi}\textbf{d}\frac{\partial{\cal L}}{\partial\overline{\Psi}}-4{\cal L}
\]
\vspace{-8mm}
\begin{equation}
\label{56}
\end{equation}
\vspace{-8mm}
\[
=\overline{\Psi
}\left(m_1\overline{P}+m_2P\right)\Psi=m_1\psi_\mu^\ast\psi_\mu-\frac{1}{2}m_2
\psi_{[\mu\nu]}^\ast\psi_{[\mu\nu]}.
\]
In Eq.(56), we took into account that for the charged particles
the $\Psi$ and $\overline{\Psi}$ are the independent wave
functions. The dilatation symmetry is broken because of massive
parameters $m_1$ and $m_2$. In the massless case, $m_1=0$, the
dilatation current $D_\mu^c$ is also not conserved, but later we
will introduce the modified conserved current.

To obtain the symmetrical the energy-momentum, we use the
expression for the Belinfante tensor \cite{Coleman}
 \begin{equation}
T_{\mu\alpha}^{B}=\frac{1}{2}\left(T_{\mu\alpha}^{c}+
\partial_\beta X_{\beta\mu\alpha}\right), \label{57}
\end{equation}
where
\begin{equation}
X_{\beta\mu\alpha}=\frac{1}{2}\left[\Pi_\beta J_{\mu\alpha}\Psi-
\Pi_\mu J_{\beta\alpha}\Psi-\Pi_\alpha
J_{\beta\mu}\Psi\right]+c.c.. \label{58}
\end{equation}
The additional complex conjugated term in Eq.(58) and the factor
$1/2$ in Eq.(57) are specific for our first-order formulation of
charged fields. From Eq.(2),(5),(53), we obtain the tensor
$X_{\beta\mu\alpha}$:
\begin{equation}
X_{\beta\mu\alpha}=\delta_{\alpha\beta}\psi_\lambda^\ast\psi_{[\lambda\mu]}-
\delta_{\alpha\mu}\psi_\lambda^\ast\psi_{[\lambda\beta]}
-\psi_\alpha^\ast\psi_{[\beta\mu]}+\psi_{[\mu\beta]}^\ast\psi_\alpha+c.c..
\label{59}
\end{equation}
With the help of expressions (59),(57), and equations of motion,
we obtain the Belinfante symmetric energy-momentum tensor
\[
T_{\mu\alpha}^{B}=m_2\left(\psi_{[\lambda\mu]}\psi_{[\alpha\lambda]}^\ast
+\psi_{[\lambda\mu]}^\ast\psi_{[\alpha\lambda]}\right)-m_1\left(\psi_\mu^
\ast\psi_\alpha +\psi_\alpha^\ast\psi_\mu\right)
\]
\vspace{-8mm}
\begin{equation}
\label{60}
\end{equation}
\vspace{-8mm}
\[
-\frac{1}{2}\delta_{\alpha\mu}\partial_\beta\left(\psi_\lambda^\ast\psi_{[\lambda\beta]}
+\psi_{[\lambda\beta]}^\ast\psi_\lambda\right).
\]
From Eq.(60), one finds the trace of the energy-momentum tensor
\begin{equation}
T_{\mu\mu}^{B}=2m_1\psi_\lambda^\ast\psi_\lambda .\label{61}
\end{equation}
In the case of massless fields, $m_1=0$, the trace of the
Belinfante symmetric energy-momentum tensor vanishes. We note that
the trace of the canonical energy-momentum tensor (51) does not
equal zero for massless fields. We evaluate a modified Belinfante
dilatation current \cite{Coleman}
 \begin{equation}
D_\mu^B=x_\alpha T_{\mu\alpha}^{B}+V_\mu, \label{62}
\end{equation}
where the field-virial $V_\mu$ is given by
 \begin{equation}
V_\mu=\Pi_\mu \Psi-\Pi_\alpha
J_{\alpha\mu}\Psi=-\overline{\Psi}\beta_\mu\Psi+\overline{\Psi
}\beta_\alpha J_{\alpha\mu}\Psi. \label{63}
\end{equation}
Using (2),(5),(11), one finds
\begin{equation}
\partial_\mu V_\mu=-m_1\psi_\lambda^\ast\psi_\lambda-\frac{1}{2}m_2
\psi_{[\mu\nu]}^\ast\psi_{[\mu\nu]}. \label{64}
\end{equation}
As a result of Eq.(61),(62),(64), we obtain
\begin{equation}
\partial_\mu D_\mu^B=T_{\mu\mu}^{B}+\partial_\mu V_\mu=m_1\psi_\lambda^\ast\psi_\lambda-\frac{1}{2}m_2
\psi_{[\mu\nu]}^\ast\psi_{[\mu\nu]}=\partial_\mu D_\mu^c.
\label{65}
\end{equation}
i.e. the same result (see Eq.(56)). For the massive fields,
$m_1\neq 0$, the dilatation symmetry is broken as it should be. For massless
fields ($m_1=0$), the currents $D_\mu^c$, $D_\mu^B$ also are not
conserved, but one can introduce the new conserved
current
\begin{equation}
D_\mu=D_\mu^B-V_\mu=x_\alpha T_{\mu\alpha}^{B}, \label{66}
\end{equation}
so that $\partial_\mu D_\mu=0$. Thus, massless fields (charged and
neutral electromagnetic fields) possess the dilatation symmetry
with the new dilatation current (66). Conformal symmetry of massless DKP equation also was
investigated in another formalism in \cite{Casana}.

\section{The canonical quantization of massive fields}

Let us consider the massive case implying that $m_1\neq 0$, $m_2\neq 0$.
The normalized solutions to Eq.(6) with definite energy-momentum
in the form of plane waves can be written as follows:
\begin{equation}
\Psi_s^{(\pm)}(x)=\sqrt{\frac{m_1+m_2}{2p_0 V}}\Psi_s(\pm
p)\exp(\pm ipx) , \label{67}
\end{equation}
where $p^2= \textbf{p}^2-p_0^2=-m_1m_2$, $V$ is the normalization
volume, and $s=\pm 1,0$ is the spin index. The function
$\Psi_s(\pm p)$ obeys Eq.(17), and we explore the normalization
conditions on the charge
\begin{equation}
\int_V \overline{\Psi}^{(\pm)}_{s}(x)\beta_4 \Psi^{(\pm)}_{s'
}(x)d^3 x=\pm\delta_{ss'} ,~~~~\int_V
\overline{\Psi}^{(\pm)}_{s}(x)\beta_4 \Psi^{(\mp)}_{s' }(x)d^3 x=0
, \label{68}
\end{equation}
where $\overline{\Psi}^{(\pm)}_{s}(x)=\left(\Psi^{(\pm)}_{s
}(x)\right)^+ \eta$. The field operators in the second quantized
theory are given by
\[
\Psi(x)=\sum_{p,s}\left[a_{p,s}\Psi^{(+)}_{s}(x) +
b^+_{p,s}\Psi^{(-)}_{s}(x)\right] ,
\]
\vspace{-7mm}
\begin{equation} \label{69}
\end{equation}
\vspace{-7mm}
\[
\overline{\Psi}(x)=\sum_{p,s}\left[a^+_{p,s}
\overline{\Psi}^{(+)}_{s}(x)+ b_{p,s}
\overline{\Psi}^{(-)}_{s}(x)\right] ,
\]
where the positive and negative parts of the wave function
$\Psi_s^{(\pm)}(x)$ are defined by Eq.(67). The creation and
annihilation operators of particles, $a^+_{p,s}$, $a_{p,s}$ and
antiparticles $b^+_{p,s}$, $b_{p,s}$ obey the usual commutation
relations
\[
[a_{p,s},a^+_{p',s'}]=\delta_{ss'} \delta_{pp'} ,~~~
[a_{p,s},a_{p',s'}]=[a^+_{p,s},a^+_{p',s'}]=0,
\]
\begin{equation}
[b_{p,s},b^+_{p',s'}]=\delta_{ss'} \delta_{pp'} ,~~~
[b_{p,s},b_{p',s'}]=[b^+_{p,s},b^+_{p',s'}]=0, \label{70}
\end{equation}
\[
[a_{p,s},b_{p',s'}]=[a_{p,s},b^+_{p',s'}]=
[a^+_{p,s},b_{p',s'}]=[a^+_{p,s},b^+_{p',s'}]=0.
\]
It follows from commutation relations (70) that the metric is positive-definite.
Thus, we construct vector space which is really a Hilbert space.
For quantization of the massless fields, one needs to introduce indefinite metric (see Ref. \cite{Kruglov2(a)}).
From Eq.(68)-(70), one can find the commutation relations for different times
\[
\left[\Psi_M(x),\Psi_N(x')]=[\overline{\Psi}_{ M}(x),
\overline{\Psi}_{ N}(x')\right] =0, ~~\left[\Psi_{
M}(x),\overline{\Psi}_{N}(x')\right]=N_{ MN}(x,x'),
\]
\[
N_{MN}(x,x')=N^+_{MN}(x,x')-N^-_{ MN}(x,x'),
\]
\vspace{-7mm}
\begin{equation} \label{71}
\end{equation}
\vspace{-7mm}
\[
N^+_{MN}(x,x')=\sum_{p,s}\left(\Psi^{(+)}_{s}(x)\right)_M
\left(\overline{\Psi}^{(+)}_{s}(x')\right)_N ,
\]
\[
N^-_{MN}(x,x')=\sum_{p,s}\left(\Psi^{(-)}_{s}(x)\right)_M
\left(\overline{\Psi}^{(-)}_{s}(x')\right)_N .
\]
We obtain from Eq.(67),(71)
\begin{equation}
N^\pm_{MN}(x,x')=\sum_{p,s}\frac{m_1+m_2}{2p_0
V}\left(\Psi_{s}(\pm p)\right)_M\left(\overline{\Psi}_{s}(\pm
p)\right)_N\exp [\pm ip(x-x')] .
 \label{72}
\end{equation}
From Eq.(27), one finds that the equation
\begin{equation}
S_{(+ 1)}+ S_{(- 1)}+S_{(0)}=1,\label{73}
\end{equation}
is valid. Then, from Eq.(22),(29), we obtain
\begin{equation}
\Pi_{\pm}=\sum_s\left(\Psi_{s}(\pm
p)\right)\cdot\left(\overline{\Psi}_{s}(\pm p)\right) = \frac{\pm
i\widehat{p}\left(\pm im_1\widehat{p}P\pm
im_2\widehat{p}\overline{P}
-m_1m_2\right)}{m_1m_2\left(m_1+m_2\right)}. \label{74}
\end{equation}
and Eq.(72) reads
\[
N^\pm_{MN}(x,x')=\sum_{p} \frac{1}{2p_0 V} \left(\frac{\pm
i\widehat{p}\left(\pm im_1\widehat{p}P\pm
im_2\widehat{p}\overline{P} -m_1 m_2\right)}{m_1 m_2}\right)_{MN}
\exp [\pm ip(x-x')]
\]
\vspace{-6mm}
\begin{equation} \label{75}
\end{equation}
\vspace{-6mm}
\[
= \left( \frac{\beta_\mu\partial_\mu\left(m_1\beta_\mu
P\partial_\mu + m_1\beta_\mu \overline{P}\partial_\mu -m_1
m_2\right)}{m_1 m_2} \right)_{MN} \Delta_{\pm}(x-x'),
\]
where we introduce the singular functions \cite{Ahieser}
\begin{equation}
\Delta_+(x)=\sum_{p}\frac{1}{2p_0V}\exp
(ipx),~~~~\Delta_-(x)=\sum_{p}\frac{1}{2p_0V}\exp (-ipx).
\label{76}
\end{equation}
With the aid of the function \cite{Ahieser}
\begin{equation}
\Delta_0 (x)=i\left(\Delta_+(x)-\Delta_-(x)\right), \label{77}
\end{equation}
and using Eq.(71),(75)-(77), one finds
\begin{equation}
N_{MN}(x,x')=-i\left(\frac{\beta_\mu\partial_\mu\left(m_1\beta_\mu
P\partial_\mu + m_2\beta_\mu \overline{P}\partial_\mu -m_1
m_2\right)}{m_1 m_2}\right)_{MN} \Delta_{0}(x-x').
 \label{78}
\end{equation}
The function $\Delta_{0}(x)$ vanishes when $x^2=
\textbf{x}^2-t^2>0$ \cite{Ahieser}. One can check with the help of
Eq.(7), that the relation
\[
 \left(\beta_\mu\partial_\mu+ m_1
\overline{P}+m_2 P
\right)\frac{\beta_\mu\partial_\mu\left(m_1\beta_\mu P\partial_\mu
+ m_2\beta_\mu \overline{P}\partial_\mu -m_1 m_2\right)}{m_1 m_2}
\]
\vspace{-7mm}
\begin{equation} \label{79}
\end{equation}
\vspace{-7mm}
\[
= \frac{\left( m_1 \overline{P}+m_2 P\right)\beta_\mu\partial_\mu
}{m_1 m_2}\left(\partial_\mu^2-m_1 m_2\right).
 \]
holds. Using the equation \cite{Ahieser} $\left(\partial_\mu^2-m_1
m_2\right)\Delta_\pm (x)=0$, from Eq.(75),(79), we arrive at
\begin{equation}
\left(\beta_\mu\partial_\mu + m_1 \overline{P}+m_2
P\right)N^\pm(x,x')=0 .
 \label{80}
\end{equation}
The vacuum expectation of chronological pairing of operators (the
propagator) is defined as \cite{Ahieser}
\[
\langle T\Psi_{M}(x)\overline{\Psi}_{ N}(y)\rangle_0=N^c_{
MN}(x-y)
\]
\vspace{-8mm}
\begin{equation} \label{81}
\end{equation}
\vspace{-8mm}
\[
=\theta\left(x_0 -y_0\right)N^+_{MN}(x-y)+\theta\left(y_0
-x_0\right)N^-_{MN}(x-y) ,
\]
where $\theta(x)$ is the theta-function. Then, one obtains from
Eq.(81) the propagator:
\begin{equation}
\langle T\Psi(x)\cdot\overline{\Psi}(y)\rangle_0
=\frac{\beta_\mu\partial_\mu\left(m_1\beta_\mu P\partial_\mu +
m_2\beta_\mu \overline{P}\partial_\mu -m_1 m_2\right)}{m_1
m_2}\Delta_c (x-y),
 \label{82}
\end{equation}
where
\begin{equation}
\Delta_c (x-y)=\theta\left(x_0
-y_0\right)\Delta_+(x-y)+\theta\left(y_0 -x_0\right)\Delta_-(x-y)
. \label{83}
\end{equation}
With the help of the equation \cite{Ahieser}
$\left(\partial_\mu^2-m_1 m_2\right)\Delta_c (x)=i\delta(x)$, and
from Eq.(79),(82),(83), we find
\begin{equation}
\left(\beta_\mu\partial_\mu+ m_1 \overline{P}+m_2 P \right)\langle
T\Psi(x)\cdot\overline{\Psi}(y)\rangle_0 = i\frac{ m_1
\overline{P}+m_2 P }{m_1 m_2}\beta_\mu\partial_\mu\delta (x-y).
 \label{84}
\end{equation}
At the chose $m_1=m_2=m$, and taking into consideration Eq.(8),
$P+\overline{P}=1$, the propagator (82) and Eq.(84) are
simplified. The above equations are not valid for the massless
case, $m_1=0$, because of singularities in Eq.(78),(79),(82),84). For the massive case, the
propagator (82) can be used for calculations of quantum processes
with vector particles in the first-order formalism.
The difficulty to quantize massless fields in the 10-component form of equations is connected with
the fact that eigenvalues of the operator $\Lambda^{(0)}_\pm $, Eq.(31), are
degenerated. To avoid this difficulty, one needs to introduce a general gauge and
to use the 11-component RWE. Quantization of massless fields (in
the general gage) in the first-order formalism was performed in Ref. \cite{Kruglov2(a)}.

\section{Conclusion}

We have considered the massive and massless vector fields in the
DKP formalism. Solutions in the form of matrix-dyads obtained
allow us to make quantum-electrodynamics calculations of processes
with vector particles in the covariant form. After the exclusion
of the non-dynamical components the Hamiltonian forms of equations
for massive and massless fields are obtained. One may consider
particles in external electromagnetic fields at the level of the
Hamiltonian form. The canonical and symmetrical Belinfante
energy-momentum tensors found possess their nonzero traces. We
investigate the dilatation symmetry in the first-order formalism.
It was demonstrated that the dilatation symmetry is broken in the
massive case but in the massless case the modified dilatation
current is conserved. The canonical quantization is considered in
the DKP form and the propagator of the massive fields is obtained.
Quantization of fields and the propagator found make it possible
to use the perturbation theory for different quantum calculations
in a simple manner.

\vspace{5mm}

\textsc{\textbf{Appendix}}

\vspace{5mm}

Now, we clarify the physical meaning of parameters $m_1$, $m_2$ introduced in Eq.(1).
Eliminating the antisymmetric tensor in Eq.(1), one obtains an equation as follows:
\[
\left(\delta_{\mu\nu}\partial_\alpha^2-\partial_\mu\partial_\nu\right)\psi_\nu-m_1m_2\psi_\mu=0,~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A1)
\]
Let us consider the momentum space, $\partial_\mu\rightarrow ip_\mu$. Then Eq.(A1)
becomes the matrix equation
\[
M_{\mu\nu}\psi_\nu=0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A2)
\]
where the matrix of the equation is given by
\[
M_{\mu\nu}=\delta_{\mu\nu}\left(p^2+m_1m_2\right)-p_\mu p_\nu,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A3)
\]
where $p^2=p_\mu^2=\textbf{p}^2-p_0^2$. One can verify that the matrix $M=\{M_{\mu\nu}\}$ obeys the ``minimal" polynomial equation
\[
\left(M-p^2-m_1m_2\right)\left(M-m_1m_2\right)=0.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A4)
\]
It follows from Eq.(A4) that the matrix $M$ has the eigenvalues $\lambda_1=p^2+m_1m_2$ and $\lambda_1=m_1m_2$.
The algebraic matrix equation (A2) possesses non-trivial solutions if $\det M=0$. This leads to $\lambda_1=0$ that is equivalent to the dispersion equation
\[
p_0^2=\textbf{p}^2+m_1m_2,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A5)
\]
or $\lambda_2=0$. Eq.(A5) tells us that the mass of a particle is $m=\sqrt{m_1m_2}$. The equation $\lambda_2=0$ is satisfied if $m_1=0$ (we do not discuss the trivial case $m_2=0$). Equation $m_1=0$ corresponds to massless case because it leads to $p_0^2=\textbf{p}^2$. It should be noted that Eq.(A1) (or system (1)) does not describe simultaneously two particles (massive and massless). Eq.(A1) describes massive particles if $m_1\neq 0$, $m_2\neq 0$ or massless particles if $m_1= 0$, $m_2\neq 0$. Thus, we have the convenient parametrization which allows us to consider both cases: massive and massless. Applying the 4-divergence to Eq.(A1), one finds that the equation $\partial_\mu\psi_\mu=0$ holds automatically only for massive case $m_1\neq 0$, $m_2\neq 0$. The equation $\partial_\mu\psi_\mu=0$ leaves only three physical degrees from 4-vector $\psi_\mu$ corresponding to three spin projections of vector particles. For massless fields ($m_1=0$) the Lorentz condition $\partial_\mu\psi_\mu=0$ does not follow from the equation of motion (A1) which becomes the wave equation for the four-potential \cite{Itzykson}.

redpajama_set_name: RedPajamaArXiv


I'm the chief evangelist of object-oriented UX, a methodology I started popularizing in 2013. I teach my OOUX methodologies at conferences, within companies, and through 1-on-1 coaching. I love mentoring— I lead the Atlanta chapter of Ladies that UX and I created the UX Hustle Summit Conference and I host the UX Hustle Podcast.
Before starting Rewired, I led UX efforts for clients such as AT&T, Blue Cross Blue Shield, Athena Healthcare, Coca-Cola, the IRS, and the Australian Tax Office. As a UX designer at CNN.com, I designed the responsive 2012 election night experience viewed by over 200 million. In 2016, I returned (as a consultant) to do it again.
Today, I focus on bringing the magic of OOUX to companies and teams around the world.
Interested in having me at your conference or meetup? Let's talk!
Email me at Sophia@RewiredUX.com or set up a call.
I will be running a full day OOUX workshop and delivering my UX for Lizard Brains talk about what UX designers can learn from the psychology of perception.
A two-day OOUX workshop in beautiful Sofia, Bulgaria.
A zero-fluff quarterly-ish email on object-oriented UX.
I can keep telling you more about me. Let's talk ikigai—the Japanese word for "reason to get up in the morning." Mine is simple: create systems, structures, and frameworks that help people live happier, healthier, and more intentional lives.
When I was seven, guess what I was doing? Basically doing just that—imagining theme parks, summer camps, and layouts for playgrounds (see above). I was designing environments where people can thrive. And thankfully, I still am.
"Do not ask yourself what the world needs.


Welcome to the podcast and "How Much Water Do You Need?"
Insulin Resistance is at a crisis point in the modern world. With our stressful lives, poor diets and excessive consumption of caffeine, sugar and alcohol, it is no wonder that health statistics are as scary as they are. If you want to understand, and more importantly avoid this destructive condition, check out the video.
The way we practice medicine in the 21st Century is changing…. And that is a good thing! In this video I explain the difference between your Genes and your Epigenetic system that keeps your genes from expressing any disease you may have inherited from your family.
How Much Water Does Your Body Actually Need in a Day?
There are a lot of myths and truths about how much water your body needs every day. In this video I take the myths apart, share the optimum amount of water to drink, when to drink it and how to make sure you are getting enough.
Is it time to turn things around?
Why Should You Do a Cleanse?


text:
If I ruled the world: Susan Greenfield
Re-introduce Classics
by Susan Greenfield	/ November 17, 2016 / Leave a comment
©Colin McPherson/Corbis via Getty Images
If I ruled the world, I would re-introduce the Classics into all schools. I get very angry when people label Greek and Latin as elitist, or dismiss them as simply dead languages. It's my firm belief that everyone would benefit from studying them, irrespective of the path they choose in life afterwards.
Unfortunately, nowadays Classics is a minority subject taught mainly in independent schools. It has also been eclipsed, mistakenly I think, by science subjects—as though we had to choose. Personally, I came late to science: I am a neuroscientist now, but I did Classics to entry level at Oxford so I have a foot in both camps.
Neuroscientists have been able to see parallels in the kind of questions the Greeks were asking. For example, in Euripides' The Bacchae, Pentheus tries to suppress women who worship the wine god Dionysus because they're getting out of hand with orgies and so on. But the prophet Tiresias cautions him against it, saying "no, you need a balance in life, there's a wine force and a bread force, and you need both those things." Science tells us that we have something called the prefrontal cortex, the frontal part of the brain, which, when it's under-functioning, will lead to a sensational "living for the moment" mentality where you abnegate the sense of self—similar to what the Dionysus worshippers go through—as opposed to most of the time when you're aware of a past, present and future, and of your own identity. As a neuroscientist I find it very interesting that one can see parallels like that.
Classical literature touches on large themes such as the great depth of human nature and, for want of a better term, the meaning of life. Even young children could learn a lot from reading this literature. At primary school level, one could read Homer in translation. The Iliad and The Odyssey are wonderful stories: not only are they rich in characters and adventure, but they also bring into play ideas of magic, the supernatural and of destiny, and of the choices you make in life, topics of great relevance at a formative age. Game of Thrones has nothing on Homer—so why not go for the original rather than a knock-off version? It trains your attention span and encourages…
9529210285d317db77ae768.49916381
What classics can teach us about the Brexit labyrinth
Charlotte Higgins / May 8, 2019
The original myth is no less weird and insane than the UK's current political...
Do Hermes couriers know about their classical links to the underworld?
Charlotte Higgins / June 10, 2019
Hermes, Trojan, Rubicon—for the Classically-minded, brand names can be a major source...
Paola Trimarco
As a linguist, I agree with a lot of these points. But why not study Homer in translation and learn other languages, such as French, from a young age? French involves conjugations and gender and number agreement which can also help young minds to think in terms of sturctures etc. And the advantage would be having a living language to use as well.
Susan Greenfield
Susan Greenfield is a scientist and writer. Her latest book is "A Day in the Life of the Brain" (Allen Lane), which will be discussed at the Prospect Book Club.


text:
The American spirit didn't die on Nov. 8
Posted: Friday 11.18.2016 5:43 am The Arab American News In Our Opinion
Eeman Abbasi speaks during a protest on the University of Connecticut campus against the election of Donald Trump. Photograph: Pat Eaton-Robb/AP
Arab and Muslim Americans have many good reasons to be fearful
of the Trump presidency. Donald Trump made fear and
xenophobia a central part of his campaign. This was a valid reason for Muslim and Arab Americans
to be apprehensive of his victory.
Even though a minute number of Muslim and Arab Americans voted
for Trump on Nov. 8, hoping that his offensive language and radical proposed
policies would be obliterated once he was victorious, a majority of the
community is fearing the worst– mainly for two reasons.
First, it is a gross violation of the U.S. Constitution to grant
or prevent entry to the country based on religion, race, sexual orientation or
color. Second, Muslims, like the rest of those who were insulted and threatened
by Trump's offensive remarks, were hoping that after his victory speech, he
would embark on a mission to unite the country and erase the divisive overtones
of his campaign.
Muslims also know very well that the U.S. system of government
is built on check and balances. The power of the presidency is important, but
is not exclusive.
More than half of American voters cast their votes against the president-elect.
It is safe to assume that most American Muslims are among the majority.
Even those who voted for Donald Trump didn't necessarily agree
with all of his policies and rhetoric. The goodness of the American spirit
didn't surrender or die on November 8. In fact, it is exemplified in the
outpouring of protests against his actions. It has
spurred advocates across the nation to pledge to fight for the dignity of all
Americans.
in Lansing told their Muslim neighbors they reject Trump's bigotry. Dozens of female students at a high school in Minnesota
stood in solidarity with a Muslim student who was feeling outcasted by showing
up wearing a hijab. Arab Americans and Muslim aren't
alone and can't sit on the sideline of this fight. They must be at the center
and the forefront of the battle to reclaim and protect our rights.
It is a sad fact that our nation is divided. It is also true
that the division is deeper and more dangerous than we have ever seen since the
American civil war era.
The demonstrations that swept the country after Trump's victory
wasn't orchestrated by political operatives, like he insinuated. It was a
popular movement in reaction to systematic corruption and policies that placed
Americans in a predicament to select "the lesser of two evils" as
their president.
And for many, Trump has become the face of a distorted nation.
Yes, Trump is the president-elect; and Americans, including
Arabs and Muslims, must bow to democracy and to the will of the voters and
abide by the result.
But, that doesn't mean that we should surrender to bigotry, xenophobia and fear.
We must not succumb to an administration that intends to strip our rights and
drive us into hiding and humiliation. We owe it to America to stand up and join
the fight.
Trump's initial pick to his administration rapidly eradicated
any hope that the campaign talk was just a rhetoric. It is already sparking fears that we are heading into uncharted
and dangerous territory or –perhaps – setting us back 100 years.
A Trump advisor is pushing for a Muslim registry. Another
surrogate claims that the forced internment of Japanese Americans during world
War II sets a precedent for the Trump administration to create a registry for
Muslims living in the U.S.
These are challenging times. We the people know that the Constitution
protects all of us.
Arab Americans and Muslims have a greater responsibility today.
They should
redouble their efforts in building coalitions with other minority and ethnic
groups, locally and across the nation. It is imperative that we search for
common grounds with African Americans, Jewish Americans, Latinos, Asians and
community must strengthen its relationship with civil rights advocates and
civil rights organizations such as the American Civil Liberties Union (ACLU), the NAACP and the Southern Poverty Law
Center (SPLC) among others. Also, Arab Americans must shoulder the
responsibility for supporting their own civil rights organizations while
holding them accountable to protect their rights in the difficult times ahead.
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text:
Leaps of Faith: The resentful optimism of 'Catastrophe'
Leaps of Faith
Not your typical TV romance, Catastrophe looks at the messy middle of long term relationships with honesty, bleak humour and resentful optimism.
By Lauren Carroll Harris
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Allison Gallagher
Conversation with Tatiana Salem Levy
Gerard Elson
Posthumous and Personal: Remembering David Rakoff
Stephanie Van Schilt
Sharon Horgan and Rob Delaney in Catastrophe. Image: Channel 4.
An Irish school teacher and an American advertising executive on a business trip meet in London. Their six-night fling results in a pregnancy; they decide to go through with it, and to marry. The series is called Catastrophe. 'A terrible thing has happened,' says Rob to Sharon. 'Let's make the best of it.'
It's that sense of irony, derived from the optimism required to work through a terrifying false step, that drives Sharon Horgan and Rob Delaney's brilliant, hard-edged British sitcom, inspired by Horgan's own pregnancy six months into a relationship. Far beyond the usual rom-com arc – the halo period of falling in love – or the ending of a slow, painful breakup, Catastrophe throws us into the muddy middle – staying in love, and in like, while surviving a long term relationship. There's no fate, no destiny, and no pat conclusions. There aren't even any declarations of love. Here, a marriage is a pragmatic, yet deeply felt and impactful task shared between two perennial strangers.
There's no fate, no destiny, and no pat conclusions. There aren't even any declarations of love.
The first season shows Rob and Sharon getting to know each other through the project of pregnancy and parenthood. The second season begins in the middle of an argument, and traces the couple as they reveal their nastier selves – Rob a name-calling alcoholic, Sharon a selfish, bridge-burning hedonist – while the third descends into the realm of two people making irreparable life errors. Darker in tone, season three starts moments after the end of the second, as Rob confronts Sharon about a secret birth control purchase. She's not the kind of person who can hide anything, and soon confesses a drunken infidelity.
In the next episode, he laments: 'You had the female response, which is to go nuclear. I had the male response, which is to bury the pain, and go jerk off in the basement.'
'Well, I can't keep saying sorry. If I could cut off the hand that…' she trails off.
'Well, I would endorse that,' scowls Rob, turning away passive aggressively, 'if you having less hands wouldn't fuck up my life and make things more difficult.' As with many scenes, it ends abruptly: Horgan and Delaney know that the bleak truth is funnier than any throwaway gag executed in the traditional sitcom rhythm.
Horgan and Delaney know that the bleak truth is funnier than any throwaway gag executed in the traditional sitcom rhythm.
Following the grand tradition of bleak British comedy, Catastrophe – comprising tightly-written six-episode narrative arcs rather than an endless spool of half-hour episodes – is loaded with these kinds of frustrated, guilt-laden exchanges of marital intimacy (many of the arguments unfold while they're in bed). But it's realism, not nihilism, that governs the series. Sharon is a wonderful creation – often infuriating, a brazenly unskilled liar making adolescent mistakes even as she laughs them off. And yet Catastrophe is a show not about difficult women, but difficult relationships – as all relationships are. Any form of togetherness, coupledom, is fraught. Why do some run from it, and others cling to it? Why do some crave it even when they are apparently incapable of expressing it?
The sitcom form has always been anchored around the indestructible nuclear family: the infuriating, wacky characters (sexist patriarchs, eye-rolling mothers, zany children) of Everybody Loves Raymond and The Simpsons, or even more grown-up family comedy-dramas like Transparent and Six Feet Under, drive each other to the brink of madness, only to ultimately confirm their attachments to each other. Catastrophe's radicalness is that it refuses to provide a TV model of marriage. Rather, it looks face-on at a relationship, constantly on the brink, within the chaos of an accidental family unit. That dramatic tension – that Sharon and Rob could feasibly break up at any moment – gives even the quietest, snidest remarks a bitter bite.
Throughout the series, at the centre of Rob and Sharon's marriage is their genuine friendship. They get along, they're goofy together, they're allies who like one another. When Sharon calls a friend a cunt behind their back, Rob just laughs. When she calls herself a simple person – 'I am. I'm Irish, we're simple people' – he responds with true endearment: 'An Irish milkmaid is a simple person. You're a cosmopolitan clothes fiend who consciously left Ireland to come here and shop.' So, late in season three, after the sting of her unfaithfulness has turned to resentment, it hurts to watch when she puts her head on Rob's shoulder and asks, 'when are you going to start liking me again?'
By this stage, Rob has sunk deep into old habits. An incurable emotional eater, he stinks constantly of cheese and onion chips, slinks off to Burger King after an especially vicious fight, and wraps hot dogs in ham before dipping them in hummus late at night while giving Sharon the cold shoulder.
Catastrophe's radicalness is that it refuses to provide a TV model of marriage.
In Australia, the series hasn't yet found the devoted audience, nor critical adoration, as it has in the UK and the US. Perhaps that's to do with the undignified time-slot allocated by the ABC (season three's finale aired at 11.30pm on a Sunday night) and the absence of a strong marketing push by either its broadcaster or DVD distributor. Many of the finest comedies of the present moment swim around a go-nowhere theme – Love, Search Party, Atlanta, Enlightened and Please Like Me have all plunged us into the worlds of characters who are at a dead end and yet don't seem to realise it.
But Catastrophe is not about people in their thirties making the same blunders as they did in their twenties. Even at the end of season three, Rob and Sharon's children Frankie and Muireann are rarely seen on-screen, though their presence raises the dramatic stakes massively. It's a sly ongoing gag that keeps the focus suffocatingly close on Sharon and Rob: the show isn't about the parent-child relationship, nor how families work, but how having a family complicates a couple's bond forever.
Catastrophe never blinks. It remains deeply committed, in a teeth-grittingly optimistic manner, to the project of relationships. It's cynical in the sunniest way possible. Rob and Sharon have as good a chance as anyone at lasting the distance; all relationships are an irrational leap of faith. And that, perhaps, is the series' wryest, cagiest, most ironic joke of all.
Lauren Carroll Harris is a writer and curator. She has contributed cultural criticism and arts journalism to Cineaste, The Baffler, The Toast, The Saturday Paper, 7am podcast, Radio National's The Screen Show, Guardian Australia, Senses of Cinema and The Monthly's website, among others. In 2019, she founded Prototype, an experimental video art unit and moving image platform.
More from Lauren Carroll Harris
Fauna by Donna Mazza: An Extract
Donna Mazza
My Inner Fangirl
Emma Maguire
Stranger in the House
Elke Wakefield
A Sky Full of Pride
Omar Sakr
Twitter, Tears and Strategic Powerlessness
Ruby Hamad
Turning and Returning
Lauren Carroll Harris


Frederick Fennell (July 2, 1914 – December 7, 2004) was an internationally recognized conductor, and one of the primary figures in promoting the wind ensemble as a performing group. He was also influential as a band pedagogue, and greatly affected the field of music education in the USA and abroad. In Fennell's The New York Times obituary, colleague Jerry F. Junkin was quoted, saying "He was arguably the most famous band conductor since John Philip Sousa."


I claim this nest in the name of all things feathery!
European Starlings tend to get a undeserved bad reputation. People consider them messy and annoying, and in part it's true: They tend to be aggressive in defending their territory, in large groups they leave a lot of droppings and a flock of starlings can pick a berry fruit tree dry in an instant.
When our beloved pet starling passed away, we decided that it would be a nice memorial to have some starling visitors in the yard for the summer. We do this every year in January to allow for time for the neighborhood starlings to find, claim and use their newfound nest. In order to do this, we had to figure out the best way to make a nest that would be of interest to these little darlings. At the same time, we're constantly saddened by the number of websites dedicated to the entrapment (and subsequent extermination) of European Starlings. We simply decided to take things half-way: Built the nest, minus the trap.
Why not build many bird houses at once?
You'll be able to build your nest from a single board. While we use cedar to make it more "weather-proof", chances are you'll need to replace the nest each year (starling nests tend to be messy). If you decide to paint your nest (for enhanced weatherization), you'll be disappointed for two reasons. First, the nest will need to be disposed of each year (see below). Second, if you do this to make it weather-proof, any seal you make will actually increase the rot and mildew that can accumulate from the inside (via bird droppings). If you must paint the nest, make certain to only paint the outside, making sure that the entrance hole will look dark from the outside (very good for attracting the birds). Divide your board up using the following height measurements and cut using your instrument of choice. Once done, you'll have six sections with one left-over piece.
One adjustment that your starling visitors will love is making the bottom panel just as long as the top (14″) instead of 11.75″; this way, they'll have something to perch on that's a bit sturdier than a round dowel/peg, and you'll get to see the starlings sit outside of their nest more often.
The measurements listed above as 7 3/8″ are the variable ones; if the width of your board is something strange like 7.75″, make sure to adjust those numbers to match. Otherwise, you'll have that odd gap mentioned above. After you've cut out the sections of the board, drill a 2″ hole on the front panel. The center of the hole should be 2″ from the top. Don't use anything smaller, otherwise starlings won't be able to enter the nest. Drill three holes on the right side of the front panel and prep three screws. You'll need to make sure to place the right side panel underneath in order to drill it as well.
Once you're done, you'll have one side of the nest in the shape of an "L". This part is intended to be removable, so we won't be drilling any more holes into it whatsoever. Now, do the same to the left and back panel, making the opposite "L" shape. Since the left/back panel pair is not removable, go ahead and secure it to the bottom board, pre-drilling the holes to avoid splitting the wood. You should now have two functional parts of the nest.
Add and secure the top panel, making sure that the extra length extends past the front of the nest. Use the front/right panel to help steady the roof as you attach it, making sure not to secure this bit as well. When you're done, you should have two fully functional parts of the nest at your disposal. The front/right panel will be held in place by the pressure of the the top/bottom of the nest. It will actually fit nice and snug.
Prep the nest for attaching to its final resting location. From what we've seen, starlings seem to love interior corners of buildings. In our case, we'll need to secure the nest using the back and the top. Had we thought this out differently, we could have made the left side removable, however this will suffice.
We purposefully made the removable part of the nest inaccessible to us to prevent undue prying eyes lest temptation get the better of us. Also, as starlings tend to stuff their nest full of anything they can get their beaks on, making a removable side panel can be temperamental and easily dislodged (permanently freaking out the mother bird and making her abandon her eggs should the side wall fall off; we've seen it happen). Attaching it in-place with a screw from the bottom is probably a good idea.
Sometimes, the nest is so packed full that it spills out of the edges!
Placing the nest in a corner near a structure is your best bet to have your nest attract lots of attention. Starlings like nooks and corners, and this is appealing to them. It may very well have to do with their nature to prefer human habitations for survival over non-natural ones; indeed, they do tend to thrive quite well by themselves. Starlings will investigate, claim ownership of and occasionally hang out near or on the nest during most of the year. The only time they'll actively use it is during breeding season. Here in the Pacific NorthWest, this tends to mean that our nest must be in place by February, with chicks hatching in May/June.
You've found a starling sitting near or on your nest, flapping his wings and proclaiming proudly to the world. Congratulations! A male starling has staked out his territory and is attempting to attract a female. He may have placed small twigs, flowers or other assorted items in (or often hanging out of) the nest as if to say, "this is some place I've got here!" When a female joins him, she'll promptly remove all items from the nest and redecorate it (go figure, my wife says). Once the nest is ready, she'll start laying between 5-6 eggs at the rate of one egg per day and sit on them for the next 12-14 days until they hatch. Be careful not to be too curious as investigation of the interior of the nest will likely make the mother bird abandon the nest (along with any eggs or chicks inside).
Shoo! Go away! Nothing to see here!
Once hatched, the mother bird will start flying out from the nest on a regular basis ever 5-15 minutes to hunt for food. As she leaves, she'll grab the feces left over from the baby birds which are, at this age, conveniently still contained in a self-enclosed sack. As she flies off, she'll drop the "droppings" along her path. Take note that this can affect your decision as to where to place the nest! Her job also involves staying with them to provide warmth and to feed them on a regular basis. Where is the husband in all of this? He's around, although he tends to stay elsewhere during the nights. Once the starlings hatch and leave the nest, you'll need to remove the nest and dispose of it. Chances are that it'll be covered in bugs and other assorted critters. It's the price of starling appreciation, I'm afraid to say. We end up making at least two each year and placing them in various areas of the yard where the squirrels are less likely to go.
That's a good question. We have noticed that certain nests are not occupied when facing a certain direction, yet when we rotate them, they're occupied the next day. This might be relative to the location (assessment of danger, etc) or a specific geographic preference. Most of our nests face west or east; the few north facing ones we've used have never been occupied (but could easily be a fluke).


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Audrey Rich Not being nice to her older brother, stomping around the house, arguing with her parents.
Lucas Rich Not being nice to his sister...and those stupid braids.
C.J. Griggs Wearing all those funky crazy socks that don't match.
Emily K $500 to see a Harry Potter Play!?
Kathy Kelly For starting Krampus in AP!
The Naughty List is now Closed. Please come back next year for another chance to add a name.
Disclaimer: Names added to the Naughty List post immediately online and will be read out loud during Krampus festivities. By submitting the form above, you agree that you are not sharing sensitive information that will be harmful or hateful towards others. You agree that the information you are sharing is appropriate and child friendly.
Please submit as many times as you like to add more people to the list!


The Dandux Poly Elevated Trucks feature an ergonomically correct raised platform height that helps reduce fatigue associated with bending or reaching to retrieve loads. Perfect for under counter storage and anywhere space is at a premium. Extra-thick 0.225″ polyethylene body is easily cleaned and offers superior durability. Body is secured to a rugged powder coated steel frame for extra strength.
Roto-molded from high quality polyethylene resin, Dandux Elevated Plastic Box Trucks are rugged, tough and virtually indestructible. The smooth seamless body comes in an array of colors to match décor or color code identification and features drain holes for easy cleaning.
Four 3" rubber swivel casters offer easy maneuverability.
Optional Spring Lift Platform is available for the 6 bushel truck to make loading and unloading faster, easier and safer. Glosstex covered powder coated steel frame with large "S" hooks and bungee cords allows for easy installation and removal.
*Measurements are nominal and can vary slightly.


A river, about 415 km (260 mi) long, of western Bulgaria and northeast Greece flowing southward to an inlet of the Aegean Sea.
2. Botany A cushionlike swelling at the base of a moss capsule.
stru·mat′ic (-măt′ĭk), stru′mose′ (-mōs′), stru′mous (-məs) adj.
2. a cushionlike swelling on a plant organ, as that at one side of the base of the capsule in many mosses.
a river in S Europe, flowing SE through SW Bulgaria and NE Greece into the Aegean. 225 mi. (362 km) long.
n. L. estruma, engrosamiento de la tiroides; pop. bocio.
Up to moderate southern wind in Eastern Bulgaria and along the Struma valley, light wind in the Danube plain from east-northeast.
During World War II, the Struma vessel, carrying Jewish refugees who were fleeing Nazi persecutions sank after it was torpedoed on 24 February 1942 in the international waters of the Black Sea.
With only money to spare for a boat ticket, Jacob arranged for David to board a small ship called the Struma, docked at Constanta, Romania and bound for Istanbul, Turkey--a neutral country in World War II.
In addition, lot 1 of the Struma motorway and 8km of the Hemus highway were also completed in August 2013.
When a ship called the Struma was dragged to the port of Old ystanbul, Alaton was a 15-year-old witness to the agony onboard.
Bona fide cases of struma ovarii that eventually act in a clinically aggressive fashion (biologically malignant) are exceedingly rare, with literature composed mainly, until recently, of case reports.
Struma and Krupnik fault zones are the most prominent tectonic structures in Southwestern Bulgaria.
Struma ovarii is a unique variant of the monodermal teratomas of the ovary, which is entirely composed of thyroid tissue.
The Struma, which sank with 768 refugees aboard after the crippled boat was forced back into the Black Sea by the Turkish authorities.


text:
Meghan Markle Accuses Buckingham Palace Of "Perpetuating Falsehoods"
By Michael Bamidele
04 March 2021 | 12:53 pm
Meghan Markle says Buckingham Palace is "perpetuating falsehoods" about her and her husband Prince Harry, according to part of an interview she has done with Oprah Winfrey. The explosive claim is the latest salvo in an increasingly heated public relations battle between the British royal family and the US-based couple. The clip, released Wednesday, came…
Meghan Markle and Prince Harry on set with Oprah Winfrey | Image: AP
Meghan Markle says Buckingham Palace is "perpetuating falsehoods" about her and her husband Prince Harry, according to part of an interview she has done with Oprah Winfrey.
The explosive claim is the latest salvo in an increasingly heated public relations battle between the British royal family and the US-based couple.
The clip, released Wednesday, came hours after Buckingham Palace said it was investigating claims Markle had bullied household staff in the months after her fairytale wedding to the prince.
"I don't know how they could expect that after all of this time we would still just be silent if there's an active role that The Firm is playing in perpetuating falsehoods about us," Markle told Winfrey.
"The Firm" is a sometimes-derisive name for the British royal family.
"There's a lot that's been lost already."
CBS Presents Oprah with Meghan and Harry: A Primetime Special this Sunday at 8/7c. #OprahMeghanHarry pic.twitter.com/iKFjiVP3IH
— CBS (@CBS) March 4, 2021
Harry and Meghan announced last year they were stepping back from royal duties, in what became an increasingly acrimonious split from the family. Their permanent removal from duties was confirmed earlier this year, with the royal household and the couple jockeying to control the narrative of the separation — and public sympathies.
The interview with Winfrey is due to air in the US on Sunday, and in Britain on Monday.
In a clip released by US broadcaster CBS earlier this week, Harry said he was worried about "history repeating itself", in an apparent reference to the death of his mother Diana, who died in a high-speed car crash after being relentlessly pursued by photographers through Paris.
Prince Harry, Meghan Markle and Baby Archie | Image: Toby Melville/Pool-Getty
On Wednesday, Buckingham Palace said it was looking into newspaper reports of bullying by Markle against members of household staff. "We are clearly very concerned about allegations in The Times following claims made by former staff of The Duke and Duchess of Sussex… our HR team will look into the circumstances," it said.
The Times said the allegations dated back to October 2018, a few months after the couple's spectacular wedding.
"Members of staff involved at the time, including those who have left the Household, will be invited to participate to see if lessons can be learned," said the palace.
"The Royal Household… does not and will not tolerate bullying or harassment in the workplace."
Markle earlier released a statement directly addressing the report.
"The duchess is saddened by this latest attack on her character, particularly as someone who has been the target of bullying herself," her spokesman said.
"She is determined to continue her work building compassion around the world and will keep striving to set an example for doing what is right."
The couple said The Times was being used to "peddle a wholly false narrative" ahead of their television interview.
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text:
Q: Flask: token-based authorization I am building a basic app with Flask: it has a unique route that requests just a token authentication, i.e. if the token provided in the headers is correct, then the request is satisfied. To do so, I installed both Flask and Flask-Security. This is a snippet of my app:
from flask import Flask
from flask.ext.security import auth_token_required
from flask.ext.security import Security

app = Flask(__name__)
app.config['SECURITY_TOKEN_AUTHENTICATION_KEY'] = 'mytoken'
security = Security(app)

@app.route('/myurl')
@auth_token_required
def myrequest():
    return 'OK'

if __name__ == '__main__':
    app.run(debug=True)

I test it by running:
$ curl -H 'Authorization: Token token="mytoken"' localhost:5000/myurl
or even:
$ curl localhost:5000/myurl
However, I get the following error:
Traceback (most recent call last):
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1477, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1381, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1475, in full_dispatch_request
    rv = self.dispatch_request()
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask/app.py", line 1461, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask_security/decorators.py", line 113, in decorated
    if _check_token():
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask_security/decorators.py", line 50, in _check_token
    header_key = _security.token_authentication_header
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/werkzeug/local.py", line 338, in __getattr__
    return getattr(self._get_current_object(), name)
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/werkzeug/local.py", line 297, in _get_current_object
    return self.__local()
  File "/Users/username/.virtualenvs/my_env/lib/python2.7/site-packages/flask_security/decorators.py", line 24, in <lambda>
    _security = LocalProxy(lambda: current_app.extensions['security'])
KeyError: 'security'

Do you know where the error lies? Is it in the app or in some conflicts among the Flask libraries?
EDIT: added Security(app) initialization

A:     user_datastore = SQLAlchemyUserDatastore(db, User, Role)

security = Security(app, user_datastore)

redpajama_set_name: RedPajamaStackExchange


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These computer-generated images chart the development of severe weather patterns on the highly eccentric exoplanet HD 80606b during the days after its closest approach to its parent star.
The images were produced by computer simulations that modeled NASA's Spitzer Space Telescope's measurements of heat radiating from the planet. This animation shows a computer simulation of the planet HD 80606b from an observer located at a point in space lying between the Earth and the HD 80606 system. On Thursday, The New York Times officially released the details of its long-awaited paywall digital subscription offerings. The newest Times digital subscription offerings come nearly four years after the Grey Lady threw caution to the wind and opened up full access to the website and its archives. The Times will still be offering readers free access to front page stories, as well as the ability to view 20 articles per month. Back in 2007, the Times reported that it had 227,000 paying subscribers for its TimesSelect offering.
On Sunday 29 May 2016 Choose from Our List of 11830 Online Newspapers & ePapers to Get Your Daily Newspaper Fix!
Energy supplier SSE is fined A?10.5m by Ofgem for "prolonged and extensive" mis-selling, the regulator's biggest ever fine for a supplier. A mother and father convicted of killing six of their children in a house fire were "good" parents, a court hears.
Author Iain Banks posts a message on his official website saying he has late stage cancer and is unlikely to live for more than a year. PROFILE: Perhaps the most famous of Britain's quality newspapers, The Times is a daily national newspaper published in London since 1785, when it was known as The Daily Universal Register. About Us: The Paperboy helps you find thousands of online newspapers, ePapers and front pages from the US and around the world.
The blue glow of the crescent is starlight that has been scattered and reflected by the planet.
The new access policies — which go into effect March 17 in Canada and March 28 in the United States — will require frequent readers to pay at least $15 per month for unlimited access to the New York Times website or mobile app. Additionally, links from Twitter, Facebook or other blogs will continue to work, even if the article limit has been reached. It will be interesting to see how many subscribers the newspaper will be able to retain under the new plan. Let us know by taking this poll and feel free to expand on your thoughts in the comments below.
Powered by its own proprietary technology, Mashable is the go-to source for tech, digital culture and entertainment content for its dedicated and influential audience around the globe.
The Times and its sister paper The Sunday Times are published by Times Newspapers Limited, a subsidiary of News International, owned by the News Corporation group, headed by Rupert Murdoch.
The blue areas are reflected starlight (the blue color arises mainly from absorption by sodium and potassium in the planetary atmosphere). Print subscribers will continue to have access to the website and all other digital apps at no additional charge. Though traditionally a moderately centre-right newspaper and a supporter of the Conservatives, it supported the Labour Party in the 2001 and 2005 general elections. In other words, not having a digital subscription won't mean that users can't still access Times content online.
For the 2010 general election, however, the newspaper declared its support for the Tories once again. The Times was printed in broadsheet format for 219 years, but switched to tabloid size in 2004 partly in an attempt to appeal to younger readers and partly to appeal to commuters using public transport.
Profile extracted from Wikipedia and used under the Creative Commons Attribution-ShareAlike License.
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