text:
\section{Introduction}

As far as we know today, the known high energy physics is well explained
by the Standard Model (SM).
SM is an SU(3)$\times$SU(2)$\times$U(1) nonabelian gauge theory  \cite{cl}.
Hence, the practical calculations in electroweak and strong
interactions are calculations in a gauge theory
requiring a choice of gauge. The two choices of gauges most
frequently employed are the Lorentz-type
and the axial-type gauges. (The latter include the light-cone gauges (LCG)
and planar gauges, while the former include $R_\xi$-gauges in spontaneously
broken gauge theories (SBGT).) The Lorentz-type gauges have been
popular on account of their Lorentz covariance, simplicity of
Feynman rules and availability of a gauge parameter to ensure
gauge independence of physical results. The disadvantage of
Lorentz-type gauges is however the presence of Faddeev-Popov
ghosts and relatively large number of Feynman diagrams
needed for evaluation of Green's functions.
The axial-type gauges, on the other hand, have the advantage
of formal decoupling of ghosts \cite{diff}. This leads to a much  smaller
number of diagrams to be  evaluated. These gauges are particularly useful
in perturbative QCD calculations \cite{bass1}. In fact the first
QCD calculations  were done in these gauges \cite{gw}.

The main disadvantage of axial-type gauges arises from the lack of Lorentz
covariance and especially from the appearance of $1/(\eta\cdot k)^p$-type
spurious singularities in propagators.  Much literature
has been devoted  to the question of how these singularities should
be treated \cite{diff,bass2}. Prescriptions
have been proposed to deal with this issue:
two important ones of these are the ``principal-value prescription "
(PVP)\cite{pvp} and the Mandelstam-Leibbrandt (ML) prescriptions
\cite{lm}. These, however, have lead in many cases to difficulties. The PVP
procedure fails for LCG already at the on loop level and yields the wrong
answer for the Wilson loop to order $g^4$\cite{diff}.

Moreover, there are instances where
the ghosts need to be taken into account \cite{diff}. In canonical
quantization, the treatment of $1/\eta\cdot k$-type singularities has
been given for axial gauges of the from $A_1+\lambda A_3=0$\cite{landsh}
(This does not include LCG).

In this work, we advocate an ab-initio and fresh
approach to the question of the axial gauge propagator
based on earlier works \cite{jm,jb,talk,BRSpap1,BRS2lett}. The approach
here utilizes a finite field-dependent BRS transformation established earlier
between the Lorentz-type and the axial-type gauge\cite{jm,jb,talk}. This
transformation has been used to write down a compact expression that interpolates
between Green's functions from the
axial-type gauges to the Lorentz-type gauges \cite{BRSpap1}.
We apply the results of \cite{BRSpap1} to the relation between the
axial and the Lorentz gauge propagator. The procedure we adopt is detailed
below.

We know how the $1/k^2$ singularities of the propagator are handled in the
Lorentz gauges. These are in effect,
treated by replacing $k^2$ by $k^2+i\epsilon$. Where the poles have physical
interpretation, this amount to propagation of positive frequency waves into
future and negative frequency waves into
the past as the Feynman propagator for a physical field shows. This
is taken into account in a Lorentz covariant
manner by introduction of a term $\epsilon\int d^4x(A^2/2-{\bar c}c)$ to
the action. Introduction of such a term also has
natural interpretation in the Minkowskian formulation of Lorentz gauge theories.
This is elaborated in Section 3. We start from this well-established
procedure in Lorentz gauges. We then perform a finite field-dependent
BRS transformation \cite{jm,jb} (a nonlocal field transformation) that
converts the Lorentz to the axial gauges. This procedure,
following the work \cite{BRS2lett} leads us to an expression
for the tree $\epsilon$-dependent
propagator for the gauge field in the axial gauges. We suggest
that this expression, should, in principle be used for
the axial  gauges propagator. We analyze
the singularity structure of the propagator and find that for
real $\eta\cdot k$, there are no spurious
poles. The propagator reduces to the usual
propagator for $|\eta\cdot k|>>\epsilon$;
however, it show a complex structure in a small region near
$\eta\cdot k =0$. We show that this propagator can equally
well be replaced by an effective simpler expression.

Our prescription, by its very construction, has the
desirable propertry that it preserves the value of the Wilson
loop \cite{CT}. This is so because the field transformation of \cite{jm} and
\cite{BRSpap1} was explicitly constructed to preserve the expectation values of
gauge-invariant observables as you go from gauge to gauge. This is
unlike the other prescriptions
where the property has to be imposed as a check on the prescription. For
more comments, please see reference \cite{inprepar}.

We now briefly state the plan of the paper. In Section 2,
we review the results needed in this work. In particular
we introduce the FFRBS transformations and the results of
references \cite{BRSpap1} and \cite{BRS2lett}. In Section 3, we obtain
the effect
of $\epsilon\int d^4x(A^2/2-{\bar c}c)$ term in the Lorentz
gauge generating functional. In Section 4, we
work out the $\epsilon$-dependent axial gauge propagator in detail.
In Section 5, we examine the singularity structure of
the propagator. In Section 6, we show how the propagator can effectively be
replaced by a much simpler expression.
Section 7 has the conclusion.

\section{Summary of Results on FFBRS Transformation  between Lorentz-
and Axial-type Gauges}

\subsection{Notations and Conventions}

We start with the Faddeev-Popov effective action (FPEA) in linear
Lorentz-type gauges:
\begin{equation}
\label{eq:FPEAL}
S^L_{\rm eff}[A, c, {\bar c}]=\int d^4x\biggl(-{1\over 4} F^\alpha_{\mu\nu}
F^{\alpha, \mu\nu}\biggr)+S_{\rm gf}+S_{\rm gh},
\end{equation}
where the gauge-fixing
action $S_{\rm gf}$ is given by:
\begin{equation}
\label{eq:gfL}
S^L_{\rm gf}=-{1\over{2\lambda}}
\int d^4x\sum_{\alpha}(\partial\cdot A^\alpha)^2\equiv-{1\over{2\lambda}}
\int d^4x\sum_{\alpha}(f^\alpha_L[A])^2,
\end{equation}
and the ghost action $S_{\rm gh}$ is given by:
\begin{equation}
\label{eq:ghL}
S^L_{\rm gh}=-\int d^4x {\bar c}^\alpha M^{\alpha\beta}c^\beta,
\end{equation}
where
\begin{equation}
\label{eq:MLdef}
M^{\alpha\beta}[A(x)] \equiv \partial^\mu D^{\alpha\beta}_\mu(A,x).
\end{equation}
The covariant derivative is defined by:
\begin{equation}
\label{eq:covDdef}
{\rm D}^{\alpha\beta}_\mu\equiv\delta^{\alpha\beta}\partial_\mu
+gf^{\alpha\beta\gamma}A_\mu^\gamma.
\end{equation}

In a similar manner, the FPEA in axial-type gauges, is given by:
\begin{equation}
\label{eq:FPEAA}
S^A_{\rm gf}\equiv-{1\over{2\lambda}}\int d^4x\sum_{\alpha}
(\eta\cdot A^\alpha)^2\equiv-{1\over{2\lambda}}\sum_{\alpha}\int d^4x
(f_A^\alpha[A])^2.
\end{equation}
We require $\eta_\mu$ to be real, but otherwise unrestricted.
and
\begin{equation}
\label{eq:ghA}
S^A_{\rm gh}=-\int d^4x {\bar c}^\alpha\tilde{M}^{\alpha\beta}c^\beta,
\end{equation}
with
\begin{equation}
\label{eq:tildeMdef}
\tilde{M}^{\alpha\beta}=\eta^\mu D^{\alpha\beta}_\mu.
\end{equation}
In the $\lambda\rightarrow 0$,
\begin{equation}
\label{eq:lambdazero}
e^{iS^A_{\rm gf}}\sim \prod_{\alpha,x}\delta
\biggl(\eta\cdot A^\alpha(x)\biggr).
\end{equation}
Thus, in the presence of the delta function, the $A$-dependent term in
$\tilde{M}$ can be dropped leading to the formally ghost-free matrix.
As is well known, $S^L_{\rm eff}$ and $S^A_{\rm eff}$
are invariant under the BRS transformations:
\begin{eqnarray}
\label{eq:BRS1}
& & \delta
 A^\alpha_\mu(x)=D_\mu^{\alpha\beta}c^\beta(x)\delta\Lambda
\nonumber\\
& & \delta c^\alpha(x)=-{g\over 2}f^{\alpha\beta\gamma}c^\beta(x)
c^\gamma(x)\delta \Lambda
\nonumber\\
& & \delta{\bar c}^\alpha(x)={{f^\alpha[A]}\over{\lambda}}\delta\Lambda,
\end{eqnarray}
where $f^\alpha[A]=\partial\cdot A^\alpha$ or $\eta\cdot A^\alpha$,
depending  on whether one  has action in the Lorentz or the axial-type gauges.
We also need the interpolating mixed gauge with
\begin{equation}
\label{eq:Mgf}
S^M_{\rm g.f.}=-{1\over{2\lambda}}\int d^4x
[\partial\cdot A^\alpha(1-\kappa)+\kappa\eta\cdot A^\alpha]^2
\end{equation}
and the associated ghost term
\begin{equation}
\label{eq:Mgh}
S^M_{\rm gh}=-\int d^4x{\bar c}\biggl[(1-\kappa)M+\kappa\tilde M\biggr]c.
\end{equation}
The net effective action $S^M_{\rm eff}$
has a BRS symmetry under transformations
(\ref{eq:BRS1}) with
$f^\alpha[A]\to (1-\kappa)\partial\cdot A^\alpha+\kappa\eta\cdot A^\alpha$.

We write in general, the BRS transformations in this case as:
\begin{equation}
\label{eq:del12def}
\delta_{\rm BRS}\phi_i=(\tilde\delta_{1i}[\phi]
+\kappa\tilde\delta_{2i}[\phi])\delta\Lambda.
\end{equation}

\subsection{FFBRS Transformations}

As observed by Joglekar and Mandal \cite {jm}, in (\ref{eq:BRS1}),
$\delta\Lambda$ need not be infinitesimal
 nor need it be field-independent as long as it  does not  depend on
$x$ explicitly for (\ref{eq:BRS1}) to by a symmetry of FPEA
In fact, the following finite field-dependent BRS (FFBRS)
transformations were introduced:
\begin{eqnarray}
\label{eq:BRS2}
& &
{{A^\prime}}\ ^\alpha_\mu = A^\alpha_\mu
+ D_\mu^{\alpha\beta}c^\beta(x)\Theta[\phi]
\nonumber\\
& & {{c^\prime}}\ ^\alpha = c^\alpha-{g\over 2}f^{\alpha\beta\gamma}c^\beta(x)
c^\gamma(x)\Theta[\phi]
\nonumber\\
& & {{\bar c^\prime}}\ ^\alpha = {\bar c}^\alpha
+ {{f^\alpha[A]}\over{\lambda}}\Theta[\phi],
\end{eqnarray}
or generically
\begin{equation}
\label{eq:BRS3}
\phi^\prime_i(x)=\phi_i(x)+\delta_{\rm BRS}\phi_i(x)\Theta[\phi],
\end{equation}
where $\Theta[\phi]$ is an $x$-independent functional of $A,\ c,\ {\bar c}$ (generically denoted
by $\phi_i$) and these were also the symmetry of the FPEA.
The transformations of the  form ({\ref{eq:BRS2}) were used to
connect actions of different kinds for Yang-Mills theory in \cite {jm} and
\cite {jb}. The FPEA is invariant under (\ref{eq:BRS2}), but the
functional measure is not invariant under the
(nonlocal) transformations (\ref{eq:BRS2}). The Jacobian for the
FFBRS transformations can be expressed (in special
cases dealt with in \cite{jm,jb}) effectively as
$exp(iS_1)$ and this $S_1$ explains the difference between the two
effective actions. Such FFBRS transformations were constructed in
\cite{jm},\cite{jb} by
integration of an infinitesimal field-dependent BRS
(IFBRS) transformation:
\begin{equation}
\label{eq:BRS4}
{d\phi_i(x,\kappa)\over{d\kappa}}
=\delta_{\rm BRS}[\phi(x,\kappa)]\Theta^\prime[\phi(x,\kappa)]
\end{equation}
The integration of (\ref{eq:BRS4}) from $\kappa=0$ to 1, leads to the FFBRS transformation
of (\ref{eq:BRS3})
with $\phi(\kappa=1)\equiv \phi^\prime$ and $\phi(\kappa=0)=\phi$.
Further $\Theta$ in (\ref{eq:BRS3}) was related to $\Theta^\prime$ by:
\begin{equation}
\label{eq:BRS5}
\Theta[\phi]=\Theta^\prime[\phi]{{exp[f[\phi]]-1}\over{f[\phi]}},
\end{equation}
where
\begin{equation}
\label{eq:BRS6}
f[\phi]=\sum_i\int d^4x{\delta\Theta^\prime\over{\delta\phi_i(x)}}\delta_{\rm BRS}\phi_i(x)
\end{equation}
FFBRS transformations of the type (\ref{eq:BRS3}) were used
to connect the FPEA in Lorentz-type gauges with  gauge parameter $\lambda$
to (i) the most general BRS/anti-BRS symmetric action in linear gauges,
(ii)FPEA in quadratic gauges, (iii) the FPEA in Lorentz-type gauges with another gauge parameter
$\lambda^\prime$ in \cite{jm}. It was also used to connect the former to FPEA
in axial-type gauges in \cite{jb}. We shall now summarize the results of \cite{jb}
in {\bf 2.3}.

\subsection{ FFBRS Transformation for Lorentz to Axial Gauge $S_{\rm eff}$}

We give the results for the FFBRS transformation that connects
the Lorentz-type gauges (See (\ref{eq:FPEAL})) with  gauges parameter $\lambda$ to
axial gauges (See (\ref{eq:FPEAA})) with same gauge parameter $\lambda$.
[The same calculation can be used to connect it to axial gauges
with another  gauge
parameter $\lambda^\prime$: one simply rescales $\eta$ suitably.]
They are obtained by integrating:
\begin{equation}
\label{eq:BRS7}
{d\phi_i(\kappa)\over{d\kappa}}=\delta_{\rm BRS}[\phi]\Theta^\prime[\phi],
\end{equation}
with
\begin{equation}
\label{eq:Thetaprimedef}
\Theta^\prime=i\int d^4x{\bar c}^\alpha
(\partial\cdot A^\alpha-\eta\cdot A^\alpha).
\end{equation}
the consequent $\Theta[\phi]$ is given by (\ref{eq:BRS5}) with
\begin{equation}
\label{eq:fdef}
f[\phi]=i\int d^4x\Biggl[{\partial\cdot A^\alpha\over\lambda}
(\partial\cdot A^\alpha-\eta\cdot A^\alpha)
+{\bar c}(\partial\cdot{\rm D}-\eta\cdot{\rm D})c^\alpha\Biggr].
\end{equation}

The meaning of these field transformations is as follows.
Suppose we begin with vacuum expectation  value of  a gauge invariant functional
$G[\phi]$ in the Lorentz-type gauges:
\begin{equation}
\label{eq:vev}
\langle\langle G[\phi]\rangle\rangle\equiv\int{\cal D}
\phi G[\phi]e^{iS^L_{\rm eff}[\phi]}.
\end{equation}
Now, we perform the transformation $\phi\rightarrow\phi^\prime$ given by (\ref{eq:BRS3}).
Then we have [with $G[\phi^\prime]=G[\phi]$ by gauge invariance]
\begin{equation}
\label{eq:vevFFBRS}
\langle\langle G[\phi]
\rangle\rangle\equiv\langle\langle G[\phi^\prime]\rangle\rangle
=\int {\cal D}\phi^\prime J[\phi^\prime]G[\phi^\prime]e^{iS_{\rm eff}^L[\phi^\prime]}
\end{equation}
on account of the BRS invariance of $S_{\rm eff}^L$.
Here $J[\phi^\prime]$ is the Jacobian
\begin{equation}
\label{eq:Jdef}
{\cal D}\phi={\cal D}\phi^\prime J[\phi^\prime].
\end{equation}
As was shown in \cite{jm}, for the special case $G[\phi]\equiv{\bf 1}$,  the
Jacobian $J[\phi^\prime]$ in (\ref{eq:Jdef}), can be replaced by
$e^{iS_1[\phi^\prime]}$ where
\begin{equation}
\label{eq:S_eff^Adef}
S^L_{\rm eff}[\phi^\prime]+S_1[\phi^\prime]=S^A_{\rm eff}[\phi^\prime].
\end{equation}
As shown in Section III of \cite{BRSpap1}, this
replacement is valid for  any gauge invariant $G[\phi]$
functional of  $A$. If one were to live  with vacuum expectation values of
gauge invariant observables,  the FFBRS in  \cite{jm} would be
sufficient. But as  seen in \cite{BRSpap1},  general Green's functions need
a modified treatment.

\subsection{Relations between Green's  functions in Axial-
and Lorentz-type gauges}

The FFBRS in {\bf 2.3} was used to correlate arbitrary Green's
functions in  the Lorentz-type and Axial-type gauges
\cite{BRSpap1} Let $O[\phi]$ represent any field operator (local or
multi-local). Then the relation between the Green's functions  in the two
gauges is given by:
\begin{eqnarray}
\label{eq:relAL}
& & \langle\langle O[\phi]\rangle\rangle_A\equiv\int{\cal D}\phi^\prime
O[\phi^\prime]e^{iS^A_{\rm eff}[\phi^\prime]}\nonumber\\
& & =\int {\cal D}\phi\biggl(O[\phi]+\sum_i\delta_i\phi[\phi]
{\delta O\over{\delta\phi_i}}\biggr)e^{iS^L_{\rm eff}[\phi]}\nonumber\\
& & \equiv\langle\langle O[\phi]+\sum_i\delta_i\phi[\phi]
{\delta O\over{\delta\phi_i}}\rangle\rangle_L,
\end{eqnarray}
where
\begin{eqnarray}
\label{eq:result1}
& & \phi^\prime=\phi+\biggl(\tilde\delta_1[\phi]\Theta_1[\phi]
+\tilde\delta_2[\phi]\Theta_2[\phi]\biggr)\Theta^\prime[\phi]
\nonumber\\
& & \equiv\phi+\delta\phi[\phi]
\end{eqnarray}
is an FFBRS \cite{jm} with
\begin{equation}
\label{eq:Theta12def}
\Theta_{1,2}[\phi]\equiv\int_0^1 d\kappa (1,\kappa)exp\biggl(\kappa
f_1[\phi]
+{\kappa^2\over 2}f_2[\phi]\biggr);
\end{equation}
\begin{eqnarray}
\label{eq:f12def2}
& & f_1[\phi]\equiv i\int d^4x\biggl[{\partial\cdot A^\alpha\over\lambda}
(\partial\cdot A^\alpha-\eta\cdot A^\alpha)+{\bar c}
(\partial\cdot{\rm D}-\eta\cdot{\rm D})c\biggr]
\nonumber\\
& & f_2[\phi]\equiv -{i\over\lambda}
\int d^4x (\partial\cdot A^\alpha-\eta\cdot A^\alpha)^2,
\end{eqnarray}and
\begin{equation}
\label{Thprimedef}
\Theta^\prime\equiv i\int d^4x {\bar c}^\alpha
(\partial\cdot A^\alpha-\eta\cdot A^\alpha).
\end{equation}
The relation (\ref{eq:relAL}) can be used to related the ordinary
Green's functions, operator Green's functions,
etc. in the two set of gauges depending on
the choice of $O[\phi]$.

A much more convenient and tractable result was
also derived in \cite{BRS2lett}
\begin{equation}
\label{eq:AtoL2}
\langle {\cal O}\rangle_A=\langle{\cal O}\rangle_L+\int_0^1 d\kappa\int
D\phi
\sum_i\biggl(\tilde\delta_{1,i}[\phi]+\kappa\tilde\delta_{2,i}[\phi]
\biggr)\Theta^\prime[\phi]
{\delta{\cal O}\over{\delta\phi_i}} e^{iS^M_{\rm eff}},
\end{equation}
where $S_{\rm eff}^M$ is the
FPEA  for the
mixed gauge function with the gauge fixing term
defined in (\ref{eq:Mgf}) and $\tilde\delta_{1,i}$ and
$\tilde\delta_{2,i}$ have been defined in
(\ref{eq:del12def}).
Of course, $\tilde\delta_{2,i}$ is non-vanishing only for ${\bar c}$ field.

\section{The General Procedure for Generating Prescription}

In this section, we shall outline the general procedure for
generating the correct treatment for ${1\over{\eta\cdot k}}$
singularities in the axial-type gauges
starting from the Lorentz-type gauges. In the Lorentz-type
gauges, also there is a
singularity in the propagator at $k^2=0$ in both the gauge and
the ghost propagators. This, in
analogy with the scalar particle, is dealt with by adding
an $i\epsilon$ term, viz $k^2\rightarrow k^2+i\epsilon$
($\epsilon$ is small positive) in
the denominators. As is well known, this
prescription allows the propagation of positive
energy solutions into future and the negative energy
solutions into the past. The role of this prescription
in the Lorentz-type
gauges can also be understood clearly
in the Minkowskian formulation of quantum field theory.
The above prescription is implemented
by an addition of the term $-i{\epsilon\over 2}A_\mu A^\mu$
to the gauge field action.
In the context of a scalar theory, the $i\epsilon\phi^2$ term provides a damping
in the path integral:
\begin{equation}
\label{eq:scleps}
W=\int{\cal D}\phi e^{i(S+i\epsilon\phi^2)}
\end{equation}
for large $\phi$. In the context of gauge theories, we expect the
$\epsilon$-dependent term to be determined by similar damping
provided in the transverse degrees of freedom. Then the form of the term in
the context of a covariant formulation viz $-i\epsilon A_\mu A^\mu/2$ is
determined by covariance.
Thus, this treatment of the ${1\over k^2}$-type singularity is well
understood in the Lorentz-type gauges. Further,
there are WT identities for Green's
functions that have terms that involve both the
ghost and the gauge propagators.
Their exact preservation requires that a similar
modification be made in
the ghost propagator poles $1/k^2\rightarrow 1/{k^2+i\epsilon}$.
Thus, the path integral for $\langle 0|0\rangle$ in the Lorentz-
type gauges we normally start with, is given in Minkowski
space, by:
\begin{eqnarray}
\label{eq:presp2}
& & W^L=\int {\cal D}\phi e^{i[S^L_{\rm eff}-i{\epsilon\over 2}
A_\mu A^\mu+i\epsilon {\bar c}c]}\nonumber\\
& & \equiv\int {\cal D}\phi
e^{iS^L_{\rm eff}[\phi]+iO_1[A,c,{\bar c},\epsilon]}.
\end{eqnarray}
We now expect that if we start with this $W^L$ that has no pole
prescription ambiguities and make suitable field transformation
(as outlined in Section 2) to the axial gauges, we
should obtain an ambiguity-free treatment of the axial gauge
propagator. We thus imagine making the field transformation
of (\ref{eq:BRS3}) viz:
\begin{equation}
\label{eq:presp3}
\phi^\prime_i(x)=\phi_i(x)+\delta_{\rm BRS}[\phi_i]\Theta[\phi]
\end{equation}
with $\Theta[\phi]$ given explicitly by (\ref{eq:BRS5}), (\ref{eq:Thetaprimedef})
and (\ref{eq:fdef}).

As shown in \cite{BRSpap1}, the effect of this field transformation can be
evaluated via the formula (\ref{eq:relAL}), and is given by:
\begin{eqnarray}
\label{eq:presp4}
& & W^L=\int{\cal D}\phi e^{iS^L_{\rm eff}[\phi]+i\epsilon O_1[\phi]}
\nonumber\\
& & =\int {\cal D}\phi^\prime e^{iS^A_{\rm eff}[\phi^\prime]+i\epsilon
O_1^\prime[\phi^\prime]}
\end{eqnarray}
with
\begin{equation}
\label{eq:presp5}
O_1^\prime=O_1+\sum_i\delta\phi_i[\phi]{\delta O\over{\delta\phi_i}}.
\end{equation}
We regard the net exponent, including the $new$ ${\cal O}(\epsilon)$
terms, viz.
\begin{equation}
\label{eq:presp6}
{S^A_{\rm eff}}\ ^\prime\equiv S^A_{\rm eff}+\epsilon O_1^\prime
\end{equation}
as given correct the treatment of the axial gauge poles. We can now,
in principle, evaluate the effect of the $O_1^\prime$ term by looking
at the new effective quadratic form in (\ref{eq:presp6}).
This turns out to be a more cumbersome procedure. We proceed
along an alternate route as below.

Consider the effect of the net $\epsilon$-term in (\ref{eq:presp6}) on an
axial gauge Green's function:
\begin{equation}
\label{eq:presp7}
\langle O\rangle_A=\int{\cal D}\phi^\prime O[\phi^\prime]
e^{iS^A_{\rm eff}[\phi^\prime]}.
\end{equation}
This is given by the modification $S^A_{\rm eff}\rightarrow {S^A_{\rm eff}}\ ^\prime$ in
(\ref{eq:presp7}),viz:
\begin{equation}
\label{eq:presp8}
\langle O\rangle_A=\int{\cal D}\phi^\prime O[\phi^\prime]
e^{iS^A_{\rm eff}+i\epsilon O_1^\prime}.
\end{equation}
We now proceed to relate
(\ref{eq:presp8}) to the corresponding Green's functions
in Lorentz gauges as done in \cite{BRSpap1}. We reexpress $\langle O\rangle_A$ as:
\begin{eqnarray}
\label{eq:presp9}
& & \langle O\rangle_A={1\over i}
{\delta\over {\delta N}}\int {\cal D}\phi^\prime e^{iS^A_{\rm eff}[\phi^\prime]
+i\epsilon O_1^\prime[\phi^\prime]+iN
O[\phi^\prime]}|_{N=0}\nonumber\\
& & \equiv{1\over i}{\delta\over{\delta N}}
W^A[N]|_{N=0}.
\end{eqnarray}
Now, we consider the quantity:
\begin{eqnarray}
\label{eq:presp10}
& & W^A[N]
\equiv\int {\cal D}\phi^\prime e^{iS^A_{\rm eff}[\phi^\prime]}
.\{e^{i\epsilon O_1^\prime[\phi^\prime]+iN O[\phi^\prime]}\}\nonumber\\
& & \equiv
\int {\cal D}\phi^\prime e^{iS^A_{\rm eff}[\phi^\prime]}f[\phi^\prime]
\end{eqnarray}
We now apply the procedure of \cite{BRSpap1} (following equation (34) of
that work) to the above expression
where $O[\phi^\prime]$ there is replaced by the
curly bracket above. Then, using identity (53) of \cite{BRSpap1}, we
obtain that
\begin{equation}
\label{eq:presp11}
W^A[N]=\int{\cal D}\phi e^{iS^L_{\rm eff}[\phi]}\{f[\phi]+\sum_i\delta\phi_i
[\phi]{\delta f\over{\delta\phi_i}}\}.
\end{equation}
Now,
\begin{eqnarray}
\label{eq:presp12}
& & f[\phi]+\sum_i\delta\phi_i[\phi]{\delta f\over{\delta\phi_i}}
=exp\biggl[i\biggl(\epsilon(O_1^\prime+\sum\delta\phi_i[\phi]
{\delta O_1^\prime\over{\delta\phi_i}})\nonumber\\
& &  +N(O+\sum_i\delta\phi_i[\phi]{\delta O\over{\delta\phi_i}})\biggr)\biggr].
\end{eqnarray}
In writing
(\ref{eq:presp12}), we have used the nilpotency of $\Theta^\prime$ contained
in each of $\delta\phi_i[\phi]$ [See equation (\ref{eq:result1})]. Hence,
\begin{equation}
\label{eq:presp13}
W^A[N]=\int{\cal D}\phi e^{iS^L_{\rm eff}[\phi]}
e^{i(\epsilon[O_1^\prime+\sum\delta\phi_i[\phi]
{\delta O_1^\prime\over{\delta\phi_i}}]
+N[O+\sum_i\delta\phi_i[\phi]{\delta O\over{\delta\phi_i}}])}.
\end{equation}
Now at $N=0$, the above must coincide with $W^L$ of (\ref{eq:presp2}). Hence,
\begin{equation}
\label{eq:presp14}
\epsilon\biggl[O_1^\prime
+\sum_i\delta\phi_i[\phi]{\delta O_1^\prime\over{\delta\phi_i}}
\biggr]=\epsilon O_1[\phi]=-i\int d^4x\epsilon(A^2/2-{\bar c}c)
\end{equation}
Now, following the transition from (53) to
(54) of \cite{BRSpap1}, we can make a transition in (\ref{eq:presp10}) above.
This amounts to substitution of $\delta\phi_i$
via (\ref{eq:result1}) and (\ref{eq:Thetaprimedef}). (Here we note that
this is possible because the
${\cal O}(\epsilon)$ terms in (\ref{eq:presp10}) are independent
of $\kappa$) We then have
\begin{eqnarray}
\label{eq:presp15}
& & \langle O\rangle_A = \langle O\rangle_L\nonumber\\
& & +\int_0^1 d\kappa\int{\cal D}\phi e^{i[S^M_{\rm eff}[\phi,\kappa]
-i\epsilon\int d^4x(A^2/2-{\bar c}c)]}\sum_i\biggl(\delta_{1,i}[\phi]
+\kappa\delta_{2,i}[\phi]\biggr)\Theta^\prime[\phi]
{\delta O\over{\delta\phi_i}}.\nonumber\\
& &
\end{eqnarray}
Thus, while the ${\cal O}(\epsilon)$ terms needed to be calculated
in the Green's function calculation for (\ref{eq:presp7})
(viz $\epsilon O_1^\prime$) are very complicated),when $\langle O\rangle_A$
is reexpressed as an integral over $\kappa$, the effect
of $\epsilon$ terms in this integral is simply to modify
$S_M[\phi,\kappa]\rightarrow S_M[\phi,\kappa]+i\epsilon
\int d^4x(-A^2/2+{\bar c}c)$. Thus, the form (\ref{eq:presp15}) facilitates the
evaluation of the effect of ${\cal O}(\epsilon)$ terms on
the axial gauge Green's
functions, as the modification there is
$\kappa$-independent.

We shall now use (\ref{eq:presp15})  to write down the expression for the axial
gauge propagator. So we consider:
\begin{equation}
\label{eq:AA}
O[\phi]\equiv A^\alpha_\mu(x)A^\beta_\nu(y).
\end{equation}
Then, with obvious notations
\begin{eqnarray}
\label{eq:exres}
& & iG^{A\ \alpha\beta}_{\mu\nu}(x-y)=iG^{L\ \alpha\beta}_{\mu\nu}(x-y)
+i\int_0^1 d\kappa\int{\cal D}\phi e^{iS_{\rm eff}^M[\phi,\kappa]-i
\epsilon\int(A^2/2-{\bar c}c)d^4x}\nonumber\\
& & \times
\biggl(({\rm D}_\mu c)^\alpha(x)A^\beta_\nu(y)
+A^\alpha_\mu(x)
({\rm D}_\nu c)^\beta(y)\biggr)
\int d^4z{\bar c}(z)(\partial\cdot A^\gamma-\eta\cdot A^\gamma)(z)
\nonumber\\
& &
\end{eqnarray}
The above is an exact result for  the relation between propagators
valid  to all orders.  For
obtaining the correct treatment for the $1/(\eta\cdot q)$-singularity,
we  are however interested in the tree propagator. For this, we collect the
$O(g^0)$ terms on the right-hand side. Noting
\begin{eqnarray}
\label{eq:ghprop}
& & \int {\cal D}\phi c^\alpha(x){\bar c}^\beta(y)
e^{i[S^M_{\rm eff}-i\epsilon\int(A^2/2-{\bar c}c)d^4x]}
=i\delta^{\alpha\beta}\tilde G^{0 M}(x-y)\nonumber\\
& & =i\delta^{\alpha\beta}\int d^4q
{e^{-iq\cdot(x-y)}\over{[(\kappa-1)q^2-i\kappa q\cdot\eta -i\epsilon]}}.
\end{eqnarray}
We can write, for  the tree propagator
 $G^{0}_{\mu\nu}$:
\begin{eqnarray}
\label{eq:exres1}
&  & G^{0 A\ \alpha\beta}_{\mu\nu}(x-y)=G^{0 L\ \alpha\beta}_{\mu\nu}(x-y)
\nonumber\\
& & +i\int_0^1 d\kappa\biggl[-i\partial_\mu^x
\tilde G^{0M}(x-y)(\partial_z^\sigma-\eta^\sigma)
\tilde G^{0M\ \alpha\beta}_{\sigma\nu}
+(\mu,x,\alpha)\leftrightarrow(\nu,y,\beta)\biggr].
\nonumber\\
& &
\end{eqnarray}

In the next section, we shall use the result (\ref{eq:exres1}) to obtain
the correct $\epsilon$-dependent propagator in the axial gauges.

\section{Evaluation of the Axial Gauge Propagator}

In this section , we shall evaluate the $\epsilon$-dependent axial gauge
propagator using (\ref{eq:exres1}). We shall show
that over most of the real $\eta\cdot k$ axial, we recover the
naive axial gauge propagator. However, we find a non-trivial
complex structure in a small region near $\eta\cdot k=0$.
In this section, we shall content
ourselves with the algebraic study of $G^{0 A}_{\mu\nu}$.
In the next section, we shall study, in detail, the analytic structure
of $G^{0 A}_{\mu\nu}$ over the complex $\eta\cdot k$ plane.

We express (\ref{eq:exres1}), in momentum space as:
\begin{eqnarray}
\label{eq:exres2}
&  & G^{0 A}_{\mu\nu}(k)=G^{0 L}_{\mu\nu}(k)
\nonumber\\
& & +i\int_0^1 d\kappa\biggl[k_\mu^x
\tilde G^{0M}(k,\kappa)(-ik^\sigma-\eta^\sigma)
\tilde G^{0M}_{\sigma\nu}(k,\kappa)
+(\mu,k)\leftrightarrow(\nu,-k)\biggr].
\nonumber\\
& &
\end{eqnarray}
Here
\begin{equation}
\label{eq:eps5}
\tilde G^{0 M}(k,\kappa)
={1\over{(\kappa-1)k^2-i\kappa k\cdot\eta-i\epsilon}}
\end{equation}
and
\begin{equation}
\label{eq:exres3}
\tilde G^{0 M}_{\mu\rho}(k,\kappa)=Z^{-1}_{\mu\rho}
\end{equation}
with
\begin{equation}
\label{eq:exres4}
Z_{\mu\nu}\equiv
-(k^2+i\epsilon)\biggl[g_{\mu\rho}+\biggl({1\over\lambda}(1-\kappa)^2-1\biggr)
{k_\mu k_\rho\over{k^2+i\epsilon}}
-i{{\kappa(1-\kappa)}\over\lambda}{k_{[\mu}\eta_{\rho]}\over{k^2+i\epsilon}}
+{\kappa^2\over\lambda}{\eta_\mu\eta_\rho\over{k^2+i\epsilon}}\biggr],
\end{equation}
is the quadratic form in momentum space arising from the
$\epsilon$-dependent
action $S^M_{\rm eff}-i\epsilon\int (A^2/2-{\bar c}c)$
in (\ref{eq:exres}).
[It should be emphasized $\tilde G^{0M}_{\mu\nu}$ and
$\tilde G^{0M}$ are only
$intermediate$ objects occurring in calculations
and are $not$ the actual   ghost   and gauge
propagators  in the mixed gauges. For
example, while $\tilde G^{0M}_{\mu\nu}(k,\kappa=0)$ is equal to
$\tilde G^{0 L}_{\mu\nu}$,
$\tilde G^{0M}_{\mu\nu}(k,\kappa=1)\neq
G^{0A}_{\mu\nu}(k)$ because the latter has to be
evaluated with  the exact $\epsilon O_1^\prime$ terms in the
exponent and not $\epsilon(A^2/2-{\bar c}c)$ as occurring in
(\ref{eq:exres}). The $actual$ tree propagator in
mixed gauges would similarly be evaluated with
an appropriate term $\epsilon O_1[\phi,\kappa]$
and not from (\ref{eq:exres}); this is not required in our
evaluation.]
We express $\tilde G^{0M}_{\mu\rho}$ as:
\begin{eqnarray}
\label{eq:pres6}
& & \tilde G^{0 M}_{\mu\rho}(k)
=-{1\over{k^2+i\epsilon}}\Biggl[g_{\mu\rho}+\nonumber\\
& & {{\biggl(\biggl[[(1-\kappa)^2-\lambda]-{\eta^2\kappa^2\over{k^2+i\epsilon}}
\biggr]k_\mu k_\rho
-i\kappa(1-\kappa)k_{[\mu}\eta_{\rho]}
+{\kappa^2\eta\cdot k\over{k^2+i\epsilon}}k_{[\mu}\eta_{\rho]_+}
+i{\kappa^2\epsilon\over{k^2+i\epsilon}}\eta_\mu\eta_\rho\biggr)}
\over{\biggl(-{\kappa^2\over{(k^2+i\epsilon)}}
\biggl[(\eta\cdot k)^2-\eta^2k^2+(k^2+\eta^2)(k^2+i\epsilon)\biggr]
+2k^2\kappa
-i\epsilon\lambda-k^2\biggr)}}\Biggr].
\nonumber\\
& &
\end{eqnarray}
\begin{eqnarray}
\label{eq:Result1}
& & \tilde G^{0 A}_{\mu\nu}-\tilde G^{0 L}_{\mu\nu}
={-i\over{(k^2+i\epsilon)^2(1-i\xi_1-i\xi_2)(1-i\xi_2+\xi_1^2+i\xi_2\xi_3)}}
\nonumber\\
& & \times\int_0^1 d\kappa{\Biggl[k_\mu k_\nu\biggl(\kappa
+\biggl[{i\lambda-\xi_1(1-\lambda)\over{\xi_1+i\xi_3}}\biggr]
\biggr)
(\xi_1+i\xi_3)+\eta_\mu k_\nu
\biggl(\kappa+\biggl[{1-i\xi_2(1-\lambda)\over{-1-i\xi_1+i\xi_2}}\biggr]\biggr)
(-1-i\xi_1+i\xi_2)\Biggr]
\over{(\kappa-a_1)(\kappa^2-2\gamma\kappa+\beta)}}\nonumber\\
& & +(k\rightarrow-k,\ \mu\leftrightarrow\nu)
\end{eqnarray}
with
\begin{eqnarray}
\label{eq:defins}
& & \xi_1\equiv{\eta\cdot k\over{k^2+i\epsilon}};\nonumber\\
& & \xi_2\equiv{\epsilon\over{k^2+i\epsilon}};\nonumber\\
& & \xi_3\equiv{\eta^2\over{k^2+i\epsilon}};\nonumber\\
& & a_1\equiv{1\over{1-i\xi_1-i\xi_2}};\nonumber\\
& & \gamma\equiv{(1-i\xi_2)\over{1-i\xi_2+\xi_1^2+i\xi_2\xi_3}}\equiv{1-i\xi_2\over D};\nonumber\\
& & \beta\equiv{1+i\xi_2(\lambda-1)\over{1-i\xi_2+\xi_1^2+i\xi_2\xi_3}}
=\gamma +{i\xi_2\lambda\over D}
\end{eqnarray}
The quadratic in the denominator
can be rewritten as $(\kappa-\kappa_1)(\kappa-\kappa_2)$ with
\begin{eqnarray}
\label{eq:kappa12def}
& & \kappa_{1,2}=\gamma\pm\sqrt{\gamma^2-\beta}
={{1-i\xi_2}\pm
\sqrt{(1-i\xi_2)^2
-[1+i\xi_2(\lambda-1)](1-i\xi_2+\xi_1^2+i\xi_2\xi_3)}\over D}\nonumber\\
& & \equiv{1-i\xi_2\pm\sqrt{Y}\over D}
\end{eqnarray}
We  note that of  the three zeros of the denominators,
two are equal are $\epsilon=0$, since
\begin{eqnarray}
\label{eq:a1kappa1}
& & \kappa_1|_{\epsilon=0}
={1\over{1+\xi_1^2}}+\sqrt{\biggl(
{1\over{1+\xi_1^2}}\biggr)^2-{1\over{1+\xi_1^2}}}\nonumber\\
& & ={1+i\xi_1\over{1+\xi_1^2}}=a_1|_{\epsilon=0}.
\end{eqnarray}
We shall now state an important convention in defining the square roots in
(\ref{eq:kappa12def}). The square root $\sqrt{Y}$ has branch points
at $\pm\sqrt{{-i\xi_2[(1-i\xi_2)\xi_3+\lambda(1-i\xi_2+i\xi_2\xi_3)]
\over{1-i\xi_2(1-\lambda)}}}$ and these lie a distance O$(\sqrt{\epsilon})$
away from the origin [For LCG, in the $k^2=0$ subspace,
however, $\sqrt{Y}=i\sqrt{\lambda}\xi_1$
has no branch cut in $\xi_1$-plane]. We choose
the branch cut joining these. To obtain the value of
$+\sqrt{Y}$ at any point $\xi^\prime$
not on the branch cut, we consider $\sqrt{Y}$
for $\xi_1=M\xi^\prime$ as $M\to+\infty$.
Then we can ignore $\epsilon$ terms in this
case and $\sqrt{Y}=\sqrt{-\xi_1^2}$. This, we define to be
$i\xi_1$. We then define $\sqrt{Y}$ for $\xi_1=\xi^\prime$ by requiring that
the phase of $\sqrt{Y}$ is a continuous function of $M$ for $1\leq M<\infty$.
>From this and from the fact that $Y\equiv Y(\xi_1^2)$, we learn that
$\sqrt{Y}(-\xi_1)=-\sqrt{Y}(\xi_1)$. Hence, $\kappa_2(-\xi_1)=\kappa_1(\xi_1)$.
We further note that this prescription defines
uniquely $\sqrt{Y}$ for  real $\eta\cdot k\neq 0$ since  the branch cut  cuts
the real $\eta\cdot k$ axis only at the
origin $\eta\cdot k=0$.

We further note that both the $k_\mu k_\nu$
and the $\eta_\mu k_\nu$ terms involve an integral of the same form
\begin{equation}
\label{eq:kappaint}
\int_0^1 d\kappa
{(\kappa+\alpha)\over{(\kappa-a_1)(\kappa-\kappa_1)(\kappa-\kappa_2)}},
\end{equation}
the constant $\alpha$ being different for the $k_\mu k_\nu$ and
$\eta_\mu k_\nu$ terms.
This can be evaluated and reorganized as:
\begin{eqnarray}
\label{eq:pres10}
& & {(a_1+\alpha)\over{(a_1-\kappa_1)(a_1-\kappa_2)}}
ln\biggl[{{1-a_1}\over{-a_1}}\biggr]\nonumber\\
& & +{(\kappa_1+\alpha)\over{(\kappa_1-a_1)(\kappa_1-\kappa_2)}}
ln\biggl[{{1-\kappa_1}\over{-\kappa_1}}\biggr]
+{(\kappa_2+\alpha)\over{(\kappa_2-a_1)
(\kappa_2-\kappa_1)}}ln\biggl[{{1-\kappa_2}\over{-\kappa_2}}\biggr]
\nonumber\\
& & \equiv {1\over{a_1-\kappa_2}}\biggl({(\alpha+a_1)\over{(a_1-\kappa_1)}}
ln\biggl[{{\kappa_1-\kappa_1a_1}\over{a_1-a_1\kappa_1}}\biggr]
-{(\kappa_2+\alpha)\over{(\kappa_1-\kappa_2)}}
ln\biggl[{{\kappa_2-\kappa_1\kappa_2}
\over{\kappa_1-\kappa_1\kappa_2}}\biggr]\biggr).
\end{eqnarray}
It is shown in Appendix A that the contribution of the second term
vanishes in the limit $\epsilon\to 0$.
Hence the propagator (\ref{eq:Result1}) is given in terms of the first term
in (\ref{eq:pres10}):
\begin{equation}
\label{eq:mainln}
{a_1+\alpha\over{(a_1-\kappa_2)(a_1-\kappa_1)}}
ln\biggl[{{\kappa_1-\kappa_1a_1}\over{a_1-a_1\kappa_1}}\biggr]
\end{equation}
substituted for (\ref{eq:kappaint}) in (\ref{eq:Result1}). Hence, we shall study the
structure of (\ref{eq:mainln}) in detail.
The singularity structure of (\ref{eq:mainln}) is dependent on the
denominators and the logarithm. The equation
(\ref{eq:mainln}), in general reads:
\begin{equation}
\label{eq:lnP}
{(a_1+\alpha)D(1-i\xi_1-i\xi_2)^2
\over{i\xi_2(1-\lambda)P(\xi_1)}}ln\biggl[{-i(\xi_1+\xi_2)
\over{{-i\xi_2\lambda\over{1-i\xi_2(1-\lambda)}}
-\sqrt{[{1-i\xi_2\over{1-i\xi_2(1-\lambda)}}]^2-{D\over{1-i\xi_2(1-\lambda)}}}}}
\biggr]
\end{equation}
with
\begin{equation}
\label{eq:defP}
P(\xi_1)\equiv
\xi_1^2+2i\xi_1(1-i\xi_2)+{\lambda+i\xi_2(1-2\lambda)
+\xi_2^2(1-\lambda)+\xi_3\over{1-\lambda}}.
\end{equation}
The apparent complexity of (\ref{eq:lnP})
actually exists only in the small region of the $\eta\cdot k$ complex
plane near the  origin. We note  that for
$|a_1-\kappa_1|<|a_1(1-\kappa_1)|$, the
expression
(\ref{eq:mainln}) can be expressed as
\begin{equation}
\label{eq:expln}
{1\over{a_1-\kappa_1}}
ln\biggl[{\kappa_1-a_1\kappa_1\over{a_1-a_1\kappa_1}}\biggr]
=-{1\over{a_1-a_1\kappa_1}}
+O(a_1-\kappa_1).
\end{equation}
The condition $|a_1-\kappa_1|<|a_1(1-\kappa_1)|$ implies
\begin{equation}
\label{eq:condition}
Im\biggl({\sqrt{-(\eta\cdot k)^2-i\epsilon\eta^2}\over
{\sqrt{{k^2\over{k^2+i\epsilon}}}(\eta\cdot k+\epsilon)}}\biggr)>{1\over 2}
\end{equation}
and this covers all of real $\eta\cdot k$ axis save the region
$(-\epsilon,0)$ for $\eta^2\neq0$ and $(-\epsilon,\epsilon)$ for LCG. Thus,
(\ref{eq:expln}) reads neglecting O$(\epsilon)$ terms
\begin{equation}
\label{eq:expln2}
-{1\over{a_1(1-\kappa_1)(a_1-\kappa_2)}}.
\end{equation}
For $|\eta\cdot k|>>\epsilon$, this is easily seen to be
\begin{equation}
\label{eq:smalln.k}
-{(1-i\xi_1)^2(1+\xi_1^2)\over{2\xi_1^2}}
\end{equation}
and leads
 to the usual behavior of the axial propagator
when substituted into
(\ref{eq:Result1}), which then reads (See Appendix C):
\begin{equation}
\label{eq:eps=0}
\tilde G^{0 A}_{\mu\nu}-\tilde G^{0 L}_{\mu\nu}=
-{1\over k^2}k_\mu k_\nu
\biggl({(\lambda k^2+\eta^2)\over{(\eta\cdot k)^2}}+{(1-\lambda)\over k^2}
\biggr) + {k_{[\mu}\eta_{\nu]_+}\over{k^2\eta\cdot k}}.
\end{equation}

We finally summarize our results. We find:
\begin{equation}
\label{eq:freslt1}
\tilde G^{0 A}_{\mu\nu}=\tilde G^{0 L}_{\mu\nu}\
+\biggl[\biggl(k_\mu k_\nu\Sigma_1+\eta_\mu k_\nu\Sigma_2\biggr)ln\Sigma_3
+(k\to -k;\mu\leftrightarrow\nu)\biggr]
\end{equation}
where
\begin{eqnarray}
\label{eq:freslt2}
& & \Sigma_1\equiv {-(k^2-i\eta\cdot k)\biggl({\eta\cdot k+i\eta^2
\over{k^2-i\eta\cdot k}}+i\lambda
-{(1-\lambda)\eta\cdot k\over{k^2+i\epsilon}}\biggr)
\over{\epsilon\Sigma}}\nonumber\\
& & \Sigma_2\equiv {-(k^2-i\eta\cdot k)\biggl(
-\biggl[{k^2+i\eta\cdot k\over{k^2-i\eta\cdot k}}
\biggr]+1-{i\epsilon(1-\lambda)\over{k^2+i\epsilon}}\biggr)
\over{\epsilon\Sigma}}\nonumber\\
\nonumber\\
& & \Sigma_3\equiv{-i(\eta\cdot k+\epsilon)(k^2+i\epsilon\lambda)
\over{(k^2+i\epsilon)\biggl(-i\epsilon\lambda-\sqrt{k^4-(k^2+i\epsilon\lambda)
\biggl[k^2+{(\eta\cdot k)^2+i\epsilon\eta^2\over{k^2+i\epsilon}}\biggr]}\biggr)}},\nonumber\\
& & {\rm and}\nonumber\\
& & \Sigma\equiv\biggl[(1-\lambda)[(\eta\cdot k)^2+2ik^2\eta\cdot k]
+i\epsilon k^2(1-2\lambda)+\lambda(k^2+i\epsilon)^2+\eta^2(k^2+i\epsilon)\biggr].
\nonumber\\
& &
\end{eqnarray}

\section{Singularity  Structure of the Propagator}

In  this section, we shall
study the singularity structure of the
propagator both on the real  $\eta\cdot k$ axis
as well as the $\eta\cdot k$ complex plane in general.
The singularity structure on  the  real axis is important from
the point of vie w of  the well-defined  nature  of the
propagator for real $k_\mu$ while the singularity
structure in the complex $\eta\cdot k$ plane is  relevant the
question of Wick rotation.

As shown in Section 4,  the quantity (\ref{eq:lnP}) is relevant
to both the propagator terms of  $k_\mu k_\nu$ and
$(k_\mu\eta_\nu+k_\nu\eta_\mu)$ kind. We shall first analyze its
structure. The singularities of (\ref{eq:lnP}) arise from
those of $P(\xi_1)$ and from those of $ln$ term. We shall
first analyze the singularities of  $P(\xi_1)$.

$P(\xi_1)$ is  a   quadratic polynomial in
$\xi_1={\eta\cdot k\over{k^2+i\epsilon}}$.
It has two zeros; they are:
\begin{equation}
\label{eq:zerosP1}
 \xi_1=-i(1-i\xi_2)\pm i\sqrt{{1-i\xi_2+\xi_3\over{1-\lambda}}},
\end{equation}
i.e. at
\begin{equation}
\label{eq:zerosP2}
\eta\cdot k=-ik^2\pm i\sqrt{{(k^2+\eta^2)(k^2+i\epsilon)\over{1-\lambda}}}.
\end{equation}
We note that the above roots vanish only when $k^2$ satisfies:
\begin{equation}
\label{eq:zerosP3}
\lambda k^4+k^2(\eta^2+i\epsilon)+i\epsilon\eta^2=0.
\end{equation}
The above equation for $k^2=0$ has physical (i.e. real $k^2$)
root(s) only in $\eta^2=0$ (except for $\lambda=1$). For $\eta^2=0$,
it is easily seen that both the roots in (\ref{eq:zerosP2})
vanish at $k^2=0$. We thus note that in the
light cone gauge ($\eta^2=0$), the point $k^2=0$ needs
to be treated with care. We shall first discuss the case $\eta^2\neq0$.

\underline{Case I: $\eta^2>0$}

(i) $k^2>0$: Hence, $(k^2+\eta^2)k^2>k^4>0$. We may set $\epsilon=0$. The roots
are:
\begin{equation}
\label{eq:zerosP4}
\eta\cdot k=-ik^2\biggl[1\mp\sqrt{{1\over{1-\lambda}}\biggl(1+{\eta^2\over k^2}\biggr)}\biggr]
\end{equation}
and lie on imaginary axis on wither side of the real axis at a finite
distance even as $\epsilon\to 0$.

(ii) $k^2=0$: The roots are at
\begin{equation}
\label{eq:zerosP5}
\eta\cdot k=\pm i\sqrt{{i\eta^2\epsilon\over{1-\lambda}}}=\pm e^{{3i\pi\over4}}
\sqrt{{\eta^2\epsilon\over{1-\lambda}}}.
\end{equation}
The roots  again lie on either side of the real axis at an
infinitesimal distance on the line $\theta={3\pi\over 4}$.

(iii) $0>k^2>-\eta^2$: Here $(k^2+\eta^2)k^2<0$. We can therefore
write the roots as:
\begin{equation}
\label{eq:zerosP6}
\eta\cdot k=-ik^2\pm\sqrt{{|(k^2+\eta^2)k^2|-i\epsilon|k^2+\eta^2|
\over{1-\lambda}}}.
\end{equation}
Again, we may set $\epsilon=0$ here.
Then, the roots
\begin{equation}
\label{eq:zerosP7}
\eta\cdot k=-ik^2\pm\sqrt{{|(k^2+\eta^2)k^2|\over{1-\lambda}}}
\end{equation}
are on the same side of the real axis, i.e. in the the upper half plane (UHP)
for $k^2<0$.

(iv) $k^2<-\eta^2$:
Here $0<k^2(k^2+\eta^2)<k^4$. Then (setting $\epsilon=0$), the roots
are at:
\begin{equation}
\label{eq:zerosP8}
\eta\cdot k=i|k^2|\biggl[1\pm\sqrt{{1\over{\lambda-1}}\biggl(1
-{\eta^2\over{|k^2|}}}\biggr)\biggr],
\end{equation}
and both lie in the UHP.

\underline{Case II: $\eta^2<0$}

We express (\ref {eq:zerosP2}) as
\begin{equation}
\label{eq:zerosP9}
\eta\cdot k=i(-k^2)\pm i\sqrt{{(-k^2-\eta^2)(-k^2-i\epsilon)\over{1-\lambda}}}.
\end{equation}
We note that the above expression is analogous to the RHS
 of (\ref{eq:zerosP2}) with
$-\eta^2>0$ and $k^2\to -k^2,\ \epsilon\to-\epsilon$. Hence,
a discussion parallel to that given above applies. We find:

(i) $k^2<0$: The poles are at
\begin{equation}
\label{eq:zerosP10}
\eta\cdot k=i|k^2|\biggl[1\pm\sqrt{{1\over{1-\lambda}}\biggl(1+{|\eta^2|
\over |k^2|}\biggr)}\biggr]
\end{equation}
and lie on imaginary axis on wither side of the real axis at a finite
distance even as $\epsilon\to 0$.

(ii) $k^2=0$
\begin{equation}
\label{eq:zerosP11}
\eta\cdot k=\pm i\sqrt{{-i|\eta^2|\epsilon\over{1-\lambda}}}=\pm e^{{i\pi\over4}}
\sqrt{{|\eta^2|\epsilon\over{1-\lambda}}}.
\end{equation}
The roots  again lie on either side of the real axis at an
infinitesimal distance on the line $\theta={\pi\over 4}$.

(iii) $0<k^2<-\eta^2$: Here $(k^2+\eta^2)k^2<0$. We can therefore
write the roots as:
\begin{equation}
\label{eq:zerosP12}
\eta\cdot k=ik^2\pm\sqrt{{|(k^2+\eta^2)k^2|\over{1-\lambda}}}
\end{equation}
and  both lie in the UHP.

(iv) $k^2<-\eta^2$:
Here $0<k^2(k^2+\eta^2)<k^4$. Then (setting $\epsilon=0$), the roots
are at:
\begin{equation}
\label{eq:zerosP13}
\eta\cdot k=-ik^2\biggl[1\mp \sqrt{{1\over{\lambda-1}}\biggl(1
-{|\eta^2|\over{|k^2|}}}\biggr)\biggr],
\end{equation}
and  both lie in the LHP.

Finally, we consider the important case of the light
cone gauge (LCG). It is this case, where
 most of the difficulties associated with the
prescriptions obtained by others are located.

\underline{Case III: $\eta^2=0$}
 We note that the roots of (\ref{eq:zerosP2})
now read:
\begin{equation}
\label{eq:zerosP14}
\eta\cdot k=-ik^2\pm i\sqrt{{k^4+i\epsilon k^2\over{1-\lambda}}}.
\end{equation}
For $k\neq0$, we can write the roots as:
\begin{equation}
\label{eq:zerosP15}
\eta\cdot k=-ik^2\biggl[1\mp\sqrt{{1\over{1-\lambda}}\biggl(1+i{\epsilon
\over k^2}\biggr)}\biggr].
\end{equation}
Without loss of generality, we may assume that
$\epsilon<|k^2|$ and expand the square root. The we find:
\begin{equation}
\label{eq:zerosP16}
\eta\cdot k=-ik^2\biggl[1\mp{1\over{\sqrt{1-\lambda}}}\mp
{1\over{8\sqrt{1-\lambda}}}{\epsilon^2\over k^4}\biggr]\mp
{\epsilon\over{2\sqrt{1-\lambda}}}.
\end{equation}
We note that for $0<\lambda<1$, the roots lie on either side of the real axis.
In particular for small positive $\lambda$, we may set $\epsilon=0$ and find:
\begin{equation}
\label{eq:zerosP17}
\eta\cdot k=-2ik^2,\ {i\lambda k^2\over 2}.
\end{equation}
On the other hand, we may set $\lambda=0$ in
the beginning. Then we find it necessary to
take into account the $O(\epsilon^2)$ term in (\ref{eq:zerosP15}).
We then find:
\begin{equation}
\label{eq:zerosP18}
\eta\cdot k=-2ik^2,\ {i\epsilon\over{8k^2}}.
\end{equation}
Thus, in either case, the roots are at:
(i) $\eta\cdot k=-2ik^2$, and
(ii) on imaginary axis, in UHP for $k^2>0$
and  LHP for $k^2<0$.
We note that this discussion clearly fails for $k^2=0$
in the LCG. We then necessarily have
to obtain the treatment for the LCG by a limiting procedure $\eta^2\to0$
for this subspace of momenta. From (\ref{eq:zerosP2}),
for $k^2=0$, the roots $\eta\cdot k$ are:
\begin{equation}
\label{eq:zerosP19}
\eta\cdot k=\pm\sqrt{{-i\epsilon\eta^2\over{1-\lambda}}}.
\end{equation}
For $\eta\neq0$, (irrespective of where $\eta^2>0$ or $<0$)
these lie on the opposite sides of the
real axis at an infinitesimal distance away.

\section{Effective Treatment for the Axial Propagator}

We obtained an exact $\epsilon$-dependent expression for the axial gauge propagator
from its connection to the Lorentz gauge Green's functions.  As remarked
earlier, the propagator is effectively the same
as the usual one except in  a small region near $\eta\cdot k=0$.
It is in this region that the many
treatments of the propagator have been suggested, differ. The expression
we have obtained ab initio, however, has complicated structure in this
region. We wish to show that it can be replaced by a much simpler
expression which yields
the same (coordinate space) propagator and will facilitate the
actual axial gauge calculations rather than using the expressions of
(\ref{eq:freslt1}) and (\ref{eq:freslt2}).

We consider the tree axial gauge propagator in coordinate space:
\begin{eqnarray}
\label{eq:simpeq1}
& & \Delta_{\mu\nu}(x-y)=\int{d^4k\over{(2\pi)^4}}
e^{-ik\cdot(x-y)}G^{0A}_{\mu\nu}(k)
\nonumber\\
& & =\int{d^3k\over{(2\pi)^3}}e^{i\vec k\cdot(\vec x-\vec y)}
\int{dk^0\over{2\pi}}e^{-ik^0(x^0-y^0)}G^{0A}_{\mu\nu}(k)
\end{eqnarray}
We introduce the variable:
\begin{equation}
\label{eq:simpeq2}
\zeta\equiv\eta^0k^0-\vec\eta\cdot\vec  k.
\end{equation}
We shall deal with the case $\eta^0\neq0$. (This
is always possible by a proper choice of a Lorentz frame.) We further assume
for simplicity of treatment , $\eta^0=1$. (This can always
by arranged by rescaling $\eta_\mu$ if necessary.) Thus,
we use:
$\zeta=k^0-\vec\eta\cdot\vec k$ and $\vec k$ as
integration variables. We express
\begin{equation}
\label{eq:simpeq3}
\Delta_{\mu\nu}(x-y)=\int{d^3k\over{(2\pi)^3}}
e^{i\vec k\cdot(\vec x-\vec y)-i\vec\eta\cdot\vec k(x^0-y^0)}
\int{d\zeta\over{(2\pi)}}e^{-i\zeta(x^0-y^0)}G^{0A}_{\mu\nu}(\zeta,\vec k).
\end{equation}
We shall focus our attention on those terms in
$G^{0A}_{\mu\nu}(k)=G^{0L}_{\mu\nu}
+[G^{0A}_{\mu\nu}(k)-G^{0L}_{\mu\nu}(k)]$ which have nontrivial
structure near $\zeta=0$. We shall in effect show that
\begin{equation}
\label{eq:simpeq4}
\int_{-\infty}^\infty
{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}G^{0A}_{\mu\nu}(\zeta,\vec k)
\end{equation}
can, in the limit $\epsilon\to0$, be effectively replaced by two terms: (i) one
of which is
\begin{equation}
\label{eq:simpeq5}
\int_C{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}G^{0A}_{\mu\nu,{\rm eff}}(\zeta,\vec k)
\end{equation}
where $G^{0A}_{\mu\nu,{\rm eff}}$ has a simple
structure near $\zeta=0$ and the contour
$C$ is a contour suitably distorted near $\zeta=0$ as will
be specified soon, (ii) and a contribution having
a relatively simple form arising from the region near $\zeta=0$.

To achieve this, we replace,
\begin{equation}
\label{eq:simpeq6}
\int_{-\infty}^\infty=\int_Cd\zeta+\oint_{C_1}d\zeta
\end{equation}
where $C$ runs from $(-\infty,-\alpha\sqrt{\epsilon})$ and
$(\alpha\sqrt{\epsilon},\infty)$ and is completed by adding a
semicircle of radius $\alpha\sqrt{\epsilon}$ in the LHP. Contour $C_1$
on the other hand is a closed contour to compensate for the left hand
side (See Fig 1). Here $\alpha$ is a (large enough) arbitrary positive
number. We then show that (i) on $C$, $G^{0A}_{\mu\nu}(\zeta,\vec k)$, the
$\epsilon$-dependence can, in  fact, be ignored and be replaced by the naive
axial propagator (with $\zeta$ complex over the semicircle);
(ii) The contour $C_1$ can be shrunk so that the contribution over $C_1$ can be
replaced by that around the branch cut (from $-i\sqrt{i\epsilon\eta^2}$ to
$i\sqrt{i\epsilon\eta^2}$) which will then be evaluated (See Fig. 2).
This contribution,
in the limit $\epsilon\to0$, can be replaced by a simple expression as shown
later.

[Here, we clarify the location of the contour on the left hand side
of (\ref{eq:simpeq6}). We recall that in (\ref{eq:freslt1}), the $ln$
factor:
\begin{equation}
\label{eq:lnfac1}
ln\biggl({\zeta+\epsilon\over{\sqrt{\zeta^2+i\epsilon\eta^2}}}\biggr).
\end{equation}
is in principle, multi-valued.
However, this factor, has arisen out of
the expression (\ref{eq:mainln}) whose value is unambiguous in
in the original complex integral (\ref{eq:kappaint}). In particular,
we have already defined $\sqrt{\zeta^2+i\epsilon\eta^2}$ earlier for
$|\zeta|>>\epsilon$, the factor boils down to
\begin{equation}
\label{eq:lnfac2}
ln\biggl({\zeta\over{i\zeta}}\biggr)=-i{\pi\over2}.
\end{equation}
In the region $|\zeta|\sim\epsilon$, we
must define the phase of the logarithm  recalling that it has
arisen from the unambiguous integrals
\begin{eqnarray}
\label{eq:lnfac3}
& & \int_0^1{d\kappa\over{\kappa-a_1}}=ln\biggl({1-a_1\over{-a_1}}\biggr)
\\
& & \int_0^1{d\kappa\over{\kappa-\kappa_1}}=ln\biggl({1-\kappa_1\over{-\kappa_1}}\biggr).
\end{eqnarray}
This defines the way the contour on the left hand side of
(\ref{eq:simpeq6}) should be drawn near $\zeta=0,\ -\epsilon$.]

In this work, we shall  only deal with the case $\eta^2\neq0$ and consider the
propagator for $k^2\neq0$. We may then choose $\epsilon<<|k^2|$, in which
case we may replace $k^2/(k^2+i\epsilon)$ by 1 wherever possible. We shall also set
$\lambda=0$for simplicity. [In the case of LCG, we may need to keep
$\lambda\neq0$ till the end; we shall not deal with this here however.]

On the contour $C$, we have $|\zeta|>>\sqrt{\epsilon\eta^2}$ for sufficiently
large $\alpha$. Here we can employ the treatment in Appendices
B and C to conclude that the propagator over $C$ can be replaced
by the naive axial propagator (with complex $\zeta$):
\begin{equation}
\label{eq:naiveaxprop}
G^{0A}_{\mu\nu}(k,\epsilon=0)=-{1\over k^2}\biggl(g_{\mu\nu}
+{{(\lambda k^2+\eta^2)}\over{(\eta\cdot k)^2}}k_\mu k_\nu
-{k_{[\mu}\eta_{\nu]_+}\over{\eta\cdot k}}\biggr).
\end{equation}

We recall from the discussion of Section  4, that for $k^2\neq0,\eta^2\neq0$,
the singularities of $P(\xi_1)$ are at a finite distance away from $\zeta=0$.
Hence $C_1$ does not enclose these.
We however, have the branch cut defined earlier (see discussion following
(\ref{eq:a1kappa1})) and ones arising from the presence of the
logarithm. We shrink the contour $C_1$ so that  it in effect goes around
the branch cut in the LHP, and receives a contribution proportional to the
discontinuity in $ln(\sqrt{\zeta^2+i\epsilon\eta^2})$ across the branch
cut.

We summarize the procedure followed for evaluating the effective replacement.
We express
\begin{eqnarray}
\label{eq:simpeq7}
& &
\int{d^3k\over{(2\pi)^3}}e^{i\vec k\cdot(\vec x-\vec y)}
\oint_{C_1}
{dk^0\over{(2\pi)}}e^{-ik^0(x^0-y^0)}\biggl[k_\mu k_\nu A_1(\zeta,\vec k)
+\eta_\mu k_\nu A_2(\zeta,\vec k)+\eta_\nu k_\mu
A_3(\zeta,\vec k)\biggr]\nonumber\\
& & =-\partial^x_\mu\partial^x_\nu
\int{d^3k\over{(2\pi)^3}}e^{i\vec k\cdot(\vec x-\vec y)
-i\vec\eta\cdot\vec k(x^0-y^0)}\oint_{C_1}{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}A_1(\zeta,\vec k)\nonumber\\
& & +i\eta_\mu\partial^x_\nu
\int{d^3k\over{(2\pi)^3}}e^{i\vec k\cdot(\vec x-\vec y)
-i\vec\eta\cdot\vec k(x^0-y^0)}\oint_{C_1}{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}A_2(\zeta,\vec k)\nonumber\\
& & +i\eta_\nu\partial^x_\mu
\int{d^3k\over{(2\pi)^3}}
e^{i\vec k\cdot(\vec x-\vec y)-i\vec\eta\cdot\vec k(x^0-y^0)}
\oint_{C_1}{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}A_3(\zeta,\vec k)
\end{eqnarray}
We then evaluate
\begin{equation}
\label{eq:simpeq8}
\oint_{C_1}
{d\zeta\over{(2\pi)}}e^{-i\zeta(x^0-y^0)}A_i(\zeta,\vec k)
\end{equation}
by replacing $C_1$ by a contour that goes around the branch cut as
mentioned earlier. The contribution comes only from the discontinuity of $A_i$
across the branch cut, and is equal to
 \begin{equation}
\label{eq:simpeq9}
\int_0^{-i\sqrt{i\epsilon\eta^2}}
{d\zeta\over{(2\pi)}}e^{-i\zeta(x^0-y^0)}{\rm Disc}A_i(\zeta,\vec k)
\end{equation}
We expand Disc$A_i(\zeta,\vec k)$ around $\zeta=0$:
\begin{equation}
\label{eq:simpeq10}
{\rm Disc} A_i(\zeta,\vec k)={1\over\epsilon}\sum_n a_{i(n)}\zeta^n
\end{equation}
where we have  explicitly shown the ${1\over\epsilon}$ dependence
of the discontinuity
We have:
\begin{eqnarray}
\label{eq:simpeq11}
& &
{1\over\epsilon}\int_0^{-i\sqrt{i\epsilon\eta^2}}{d\zeta\over{(2\pi)}}
e^{-i\zeta(x^0-y^0)}\sum_{n=0}^\infty a_{i(n)}\zeta^n
\nonumber\\
& & =
{1\over\epsilon}\int_0^{-i\sqrt{i\epsilon\eta^2}}{d\zeta\over{(2\pi)}}
\biggl(a_{i(0)}[1-i\zeta(x^0-y^0)]+a_{i(1)}\zeta\biggr)+O(\sqrt{\epsilon})
\nonumber\\
& & = -a_{i(0)}{1\over{2\pi}}i\sqrt{{i\eta^2\over\epsilon}}
e^{-\sqrt{i\epsilon\eta^2}(x^0-y^0)\over 2}
-{i\eta^2\over{2(2\pi)}}a_{i(1)}+O(\sqrt{\epsilon})
\nonumber\\
& &
={e^{-\sqrt{i\epsilon\eta^2}(x^0-y^0)\over 2}\over{2\pi}}
\biggl[-i\sqrt{{i\eta^2\over\epsilon}}a_{i(0)}
-{i\eta^2\over 2}a_{i(1)}\biggr]+O(\sqrt{\epsilon})\nonumber\\
& & =\int_{-\infty}^\infty{d k^0\over{2\pi}}e^{-ik^0(x^0-y^0)}
\delta\biggl(k^0-{1\over2}\sqrt{{\epsilon\eta^2\over i}}\biggr)
\biggl[-i\sqrt{{i\eta^2\over\epsilon}}a_{i(0)}
-{i\eta^2\over 2}a_{i(1)}\biggr]+O(\sqrt{\epsilon})\nonumber\\
\end{eqnarray}
We can thus in effect replace the contribution in (\ref{eq:simpeq7}) by:
\begin{eqnarray}
\label{eq:simpeq12}
& & \int{d^4k\over{(2\pi)^4}}e^{-ik\cdot(x-y)}\biggl\{\delta\biggl(k^0
-{1\over2}\sqrt{{\epsilon\eta^2\over i}}-\vec\eta\cdot\vec k\biggr)
\biggl(k_\mu k_\nu
\biggl[-i\sqrt{{i\eta^2\over\epsilon}}a_{1(0)}
-{i\eta^2\over 2}a_{1(1)}\biggr]\nonumber\\
& & +\eta_\mu k_\nu\biggl[-i\sqrt{{i\eta^2\over\epsilon}}a_{2(0)}
-{i\eta^2\over 2}a_{2(1)}\biggr]\nonumber\\
& & +\eta_\nu k_\mu\biggl[-i\sqrt{{i\eta^2\over\epsilon}}a_{3(0)}
-{i\eta^2\over 2}a_{3(1)}\biggr]\biggr)\biggr\};
\end{eqnarray}
the curly bracket above gives the effective addition to the propagator.

We
quote the values of $a_{i(n)},\ n=0,1;\ i=1,2,3$ for completeness:
\begin{eqnarray}
\label{eq:simpeq13}
& & a_{1(0)}=-{2i\eta^2\over{\cal K}_1}ln(-1)={2\pi\eta^2\over{\cal K}_1},
\nonumber\\
& & a_{1(1)}={i\eta^2[{\cal K}_2+k\rightarrow-k]\over{\cal K}_1^2}ln(-1)
={-4i\pi\eta^2\biggl[(\vec\eta\cdot\vec k)^2-\vec k^2\biggr]\over{\cal K}_1^2};
\nonumber\\
& & a_{2(0)}=0,\nonumber\\
& & a_{2(1)}={2i\over{\cal K}_1}ln(-1)=-{2\pi\over{\cal K}_1};
\nonumber\\
& & a_{3(0)}=0,\nonumber\\
& & a_{3(1)}=-{2i\over{\cal K}_1}ln(-1)={2\pi\over{\cal K}_1},
\end{eqnarray}
where
\begin{eqnarray}
\label{eq:simpeq14}
& & {\cal K}_1\equiv\biggl((\vec\eta\cdot\vec k)^2-\vec k^2\biggr)
(\eta^2+i\epsilon);
\nonumber\\
& & {\cal K}_2\equiv 2i\biggl((\vec\eta\cdot\vec k)^2-\vec k^2\biggr)
+2\vec\eta\cdot\vec k(\eta^2+i\epsilon).
\end{eqnarray}

In the treatment we have given in this section, we have found it necessary to
keep $\eta^2$ nonzero in the intermediate stages.
We can, however, define the LCG
as a limit of the final result as $\eta^2\rightarrow 0$.
The procedure, here, is very
analogous to the one we had to adopt in Appendix A, where we found it
necessary to keep $\eta^2$ nonzero in the intermediate stages of calculations.
We need, however, take this limit keeping $\epsilon$ nonzero.
We find, by taking the limit
$\eta^2\rightarrow0$ in (\ref{eq:simpeq12})-(\ref{eq:simpeq14}),
that these extra terms
vanish for LCG. This result can also be seen ab intio by looking at the contour
integral over $C_1$.
Here we note that the width of the branch cut shrinks as $\eta^2\rightarrow0$.
This together with the fact that [for $\epsilon$ nonzero] the discontinuity
is finite for $\eta^2=0$, leads to the above result.

\section{Conclusion}

In this work, we have dealt with the question of the treatment of
$1/(\eta\cdot k)^p$-type singularities in
the axial/LCG gauges in an ab-initio manner.
We have used the known treatment in the Lorentz gauges and connection
between Lorentz and axial gauges
to achieve this. We have used the results established earlier on
the FFBRS transformation that connect Green's functions in these two
gauges. We evaluated  an $\epsilon$-dependent propagator
in axial gauges via this procedure. We have suggested
that this should give the  correct
way to deal with the $1/(\eta\cdot k)^p$-type
singularities in the axial propagator.

We find that this propagator, not surprisingly,
coincides with the usual axial propagator except in a
small region around $\eta\cdot k=0$, The propagator does not
show spurious poles
at $\eta\cdot k=0$. It is however complicated
but {\it predetermined in form} in a small region near $\eta\cdot k=0$.
We have shown however that there is a way of effectively replacing
this complicated structure of the propagator by a much simpler expression.
We believe, this is a first ab-initio treatment of axial gauge poles in
the path-integral formalism.

As mentioned in the introducton and elaborated in [15], the prescription
obtained here also preserves the value of the Wilson loop.

redpajama_set_name: RedPajamaArXiv


Season 6, Episode 12 Bravo. What could possibly go wrong? See more ideas about Mike and jessica shahs, Jessica shahs and Mike shahs of sunset. This season of , we get a front-row seat to the ups and downs of the first year of marriage for and. Look, five years together, she was 21 when I met her, so we shared a lot of memories in very pivotal years of her life. Past shahs include: Lilly Ghalichi seasons 2—3 , Asifa Mirza season 4 and Shervin Roohparvar seasons 5—6. The series debuted on March 11, 2012. However, while rumors of a potential engagement began swirling weeks later, the former couple's whirlwind romance quickly fizzled.
She graduated with a Bachelor of Arts degree in English Literature from in 2002. And alleges that Mike is having a hard time accepting their relationship is over. It is definitely a difficult process to go through. Born and raised in Los Angeles, Nema's love for his family and friends can only be matched by his love of Rose wine, his Lexus and his 2 dogs. I'm embarking on making baby shoes.
I'm trying to stay focused on my goals instead of worrying about things that don't matter. In December 2012 she told that she was engaged to Omid Kalantari, who has a recurring role in season 2. When asked about the messages, Shouhed's answer didn't please his wife. Jessica Parido enjoyed a whirlwind romance after her divorce After filing for divorce, Parido stepped out with a new man in Los Angeles. She graduated from with a double major in Psychology and Philosophy. She is from a Persian family; her parents and her brother, Arta, all appear on the show. So, where is it now? Mike Shouhed comes clean about his past mistakes with Jessica Parido during Sunday's new episode of Shahs of Sunset season five.
Mike Shouhed, Jessica Parido Relationship: Are They Divorced?
I have loved and lost. It is a violation of the for an employer to retaliate in any way against employees for their exercising their rights under the law. See more ideas about Mike and jessica shahs, Jessica shahs and Mike shahs of sunset. Looks like things may not be quite over for these two. I need to start putting my life back together.
Any fan of The Kardashians or The Real Housewives is sure to love Shahs. This show and its cast members has been on Celebrity Net Worth's radar for months now, and today we're happy to bring you the net worth of each cast member of The Shahs of Sunset! Jessica Parido is moving on from her highly publicized split from Mike Shouhed and focusing on her Glam Envy business. Jessica Parido's new boyfriend doesn't appear to be the best catch. Jessica, 26, filed for divorce from 37-year-old Mike last month after less than eight months of marriage, claiming he cheated on her with multiple women. And it's hard cause you're losing something that was very near and dear to you. Shahs of Sunset's Mike Shouhed proposed to Jessica Parido while skydiving on June See the ring! See more ideas about Mike and jessica shahs, Jessica shahs and Mike shahs of sunset. It's amazing how the mind and the heart play tug of war in situations like this where you don't know how to feel.
Shahs of Sunset follows a group of six extremely wealthy Persian-Americans who live the dream party life in Beverly Hills while trying to balance the pressures of their traditional Iranian families and culture. Shouhed also has a moving company called Royalty Moving. The two broke off their engagement the following month. In an episode he reveals his father converted to Islam to marry his mother. Sending my boy good vibes to get through this difficult time, we all love and support you! Now divorce is my only option. Her paintings were published in the art book Imagining Ourselves, published by the in San Francisco. Your heart and your mind are very complex organs and it's easy for an outsider to say, 'Oh thank god you're done and you got out safe.
He loves Jessica and always will. Meanwhile, Parido's ex-husband, Shouhed, doesn't appear to have dated anyone after their split. He's offered his heartfelt apology to her. On October 27, 2016, the show was renewed for a sixth season which premiered Sunday, July 16, 2017. He and Shouhed joked with TooFab that Saddam Hussein actually inspired the new look.
They later broke up due what she described as bad habits, notably cheating on her. Shervin is a bachelor looking for the right lady. During the Season 2 reunion, Javid admitted she is a , having been convicted of in 1994. We're in this position because of what he's done. I'm learning to get out of my comfort zone and become the best Mike I can be. No word yet on whether or not Parido will be making an appearance.


text:
High volume of farmland stock set to come to market, claims Carter Jonas
By Matthew Lane
Welcome back to Women in Property Investment (or WIPI), where we focus on inspiring women...
Last week we covered the initial works being started on a new luxury development in...
Many property pundits have been espousing the benefits of investing in the North West to...
In our latest development round-up, we zone in on projects in Staffordshire and Barking. A Worcestershire-based...
British luxury property developer Caudwell has sold another of its penthouses at Parc du Cap...
New data from property consultancy Carter Jonas has revealed that a high volume of farmland stock is set to come to market over the coming months.
While severe weather conditions in the first quarter of 2018 had a damaging impact on activity, 6,000 acres of farmland are set to enter the market over the current and subsequent quarter, mostly consisting of smaller lots.
A slight dip in average land values in England and Wales was recorded in the first three months of 2018, falling by 0.6% to £8,917. This was reflective of market fluctuations and the ongoing uncertainty surrounding Brexit negotiations – with rural businesses in particular currently reliant on support payments from the EU. While there is uncertainty over the protracted negotiations, there is also uncertainty for rural firms over their future funding.
Farmland values remain at their highest in Southern and Central England, where demand for arable land is most marked. However, land in the North West has witnessed an average rise of £500 per acre, now making it some of the most expensive in the country.
Meanwhile, values dipped in the Eastern regions, reaching an average of £8,500 per acre, while prices in Wales, the South West and the Midlands have remained constant.
Despite the wider economic factors and uncertainty at play, the research found that the rural sector as a whole continues to show resilience, with demand for good quality, correctly priced land remaining robust across the UK.
"The volume of stock that is set to come to the market is encouraging, and welcome," Andrew Fallows, head of rural agency at Carter Jonas, said. "Across England and Wales, values continue to be polarised, with neighboruing pockets of land attracting significantly different levels of interest."
He added: "While farmland still offers strong, long term investment opportunities, thanks to the counter-cyclical nature of the market, the lack of certainty surrounding the lengthening of sector support is discouraging some businesses from confidently committing to rural assets in the short term, and has created a more cautious, discerning buyer, one willing to wait for the best value land."
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redpajama_set_name: RedPajamaCommonCrawl


text:
package org.apache.hadoop.hbase.spark.example.hbasecontext

import org.apache.hadoop.hbase.spark.HBaseContext
import org.apache.spark.SparkContext
import org.apache.hadoop.hbase.{TableName, HBaseConfiguration}
import org.apache.hadoop.hbase.util.Bytes
import org.apache.hadoop.hbase.client.Scan
import org.apache.spark.SparkConf
/**
 * This is a simple example of scanning records from HBase
 * with the hbaseRDD function in Distributed fashion.
 */
object HBaseDistributedScanExample {
  def main(args: Array[String]) {
    if (args.length < 1) {
      println("HBaseDistributedScanExample {tableName} missing an argument")
      return
    }

    val tableName = args(0)

    val sparkConf = new SparkConf().setAppName("HBaseDistributedScanExample " + tableName )
    val sc = new SparkContext(sparkConf)

    try {
      val conf = HBaseConfiguration.create()

      val hbaseContext = new HBaseContext(sc, conf)

      val scan = new Scan()
      scan.setCaching(100)

      val getRdd = hbaseContext.hbaseRDD(TableName.valueOf(tableName), scan)

      getRdd.foreach(v => println(Bytes.toString(v._1.get())))

      println("Length: " + getRdd.map(r => r._1.copyBytes()).collect().length);
    } finally {
      sc.stop()
    }
  }

}

redpajama_set_name: RedPajamaGithub


Hail damage; Colorado has seen it's fair share over the past couple of months. If you live in Boulder County, chances are your home's roof may have been damaged due to the recent hail storms we've seen pass through. The thought of repairing or replacing a roof after storm damage can seem overwhelming. Having a reliable roofing contractor inspect and repair or replace your damaged roof and work with your insurance carrier throughout the claim process can make all the difference when it comes to these unfortunate events. 3R Roofing, LLC is a Boulder based roofing contractor who will do just that. They have proudly offered their services in Boulder County for over two decades, and will work with you, the homeowner, throughout the entire process to make your roof replacement or repairs stress free.
How do you know you might need your roof inspected after a hail storm? Unfortunately, most homeowners will assume that if there aren't any leaks that they have not been affected. While you may not notice obvious signs on the interior of the home, such as warping, cracked paint, or peeling wall paper caused by leaking, it does not mean that your exterior didn't experience any damage.
If you live in Boulder County and you notice these, or any other signs of damage to your roof, this could be a good indication that your home was impacted by the recent hail storms. Getting a roof inspection and making any necessary repairs can make all the difference when it comes to the extensive damage that can occur if ignored. 3R Roofing, LLC is a trusted Boulder roofing contractor with the expertise to make your roof replacement and repairs a painless process. 3R Roofing will work directly with your insurance company, are eco-friendly, and are experienced in the repairs and replacement of a variety of roofing systems. Contact us today for a free estimate.
This entry was posted in Hail Damage. Bookmark the permalink.


CLAY COUNTY,Miss. (WCBI) – Columbus' Spring Pilgrimage starts Thursday and the home tours draw visitors from all across the country.
This year's pilgrimage branches out of town a little bit to include a home that hasn't been featured for several years.
This year it's back on the Columbus Pilgrimage, and guests can expect to see a few changes coming to this historic home.
Tour guide Jimmy Denning said each year more than 300 people visit the Waverley Mansion.
"Been spending the last few weeks 'lemoning oiling' all the furniture inside and wiping some of the mildews from the walls, it it's takes a little bit usually a month I start in advance," said Denning.
Charlie and Dana Stephenson are the new owners of Waverley Manson.
The couple purchased the house three months ago.
The Stephenson's are planning to restore parts of the historic home.
Denning has worked at the Waverley for 13 years.
"If I go in in an old closet I see some of the back parts and it's neat looking at the construction of the house and recently ,of course, architects have come in through here inspecting the house and they point out new things about the architecture I didn't see before," said Denning.
Denning said he's excited about the planned renovations.
"It's going to be fascinating watching him restore the house it does, it has some issues and it does need to be fully restored it's going to look beautiful.".
The new buyers are planning to restore the roof and update the kitchen.
Since 1852, Waverley has changed owners only four times.
"It's fascinating hearing the stories of how the Young's built it and the Snow's came in and restored it and now a new page is being turned ,the Stephenson's are coming in to fully restore it back to her historic glory," said Denning.
The Waverley Mansion can be found on the Orange Tour.
Tours for Waverley Mansion are Tuesday through Saturday from 9 a.m. to 5p.m.
There are no tours on Monday.


A tragic lady transformed into a succubus through a Mystic's curse. She loved a prince from the Waltner family of Gludio, but was tricked into murdering him. Now she feeds on the energy of travellers in Gludio, delaying her death day by day.
1 A tragic lady transformed into a succubus through a Mystic's curse. She loved a prince of the Waltner family of Gludio, but was tricked into murdering him. Now, she feeds on the energy of travelers in Gludio, delaying her death day by day.


text:
Brain Injury Golf Tournament | Hosted by the Brain Injury Alliance of Washington
Meridian Valley Country Club
24830 136th Ave SE,
Kent, Washington 98042
Thursday, September 02, 2021, 10:00 AM PST
About the Tournament & the Cause
On September 2nd, the Brain Injury Alliance of Washington will host the Brain Injury Golf Tournament at the exclusive Meridian Valley Country Club in Kent. The event will benefit BIAWA's pediatric brain injury support and prevention program. Registration includes all green fees and carts, lunch, reception & dinner. Fun contests too!
WRPA members organize and administer sports leagues and/or coordinate with your sports leagues that use public facilities. This event is a great opportunity to connect, raise awareness, and show alignment in the community that we all work in—our youth.
Solo Golfer | $250.00
Participation in the Tournament + contests! Includes all green fees and cart in the cost of registration. Lunch and Dinner Reception will be served to all players. Solo players will be grouped together to make a foursome.
Foursome | $900.00
Participation in the Tournament + contests! Includes all green fees and cart in the cost of registration. Lunch & Dinner Reception will be served to all players.
Foursome + Hole Sponsor | $1,500.00
Participation in the Tournament + contests! Includes all green fees and cart in the cost of registration. Lunch and Dinner Reception will be served to all players.
BIAWA's Connection to Parks & Recreation:
There is a direct connection to the parks and recreation field and the work of the Brain Injury Alliance of Washington (BIAWA) in the community through our work with youth and sports. BIAWA fully embraces the importance of play, participation, and engagement in sports. That said, we also recognize the possibility that injuries can and do occur during play.
BIAWA is a state-wide resource for individuals with brain injury. We provide resources, classes, advocacy, and much more. Our vision is a world where brain injury is prevented, and every person already affected by its impact benefits from resources to maximize their quality of life. Our mission is to increase public awareness, support, and hope for those affected by brain injury through education, assistance, and advocacy.
As advocates, BIAWA played an integral part in the 2009 passage of the Zackery Lystedt Law, the rigorous return to play law that was the first of its kind in the nation. The Lystedt Law, along with the WIAA's recently passed Kenny Bui law, help keep kids safe playing the sports they love. The BIAWA team is here to support youth and their families in concussion management, advocating for therapeutic interventions, return to school and educational advocacy, and more.
Learn more about our work at www.biawa.org.
Contact: Brain Injury Alliance of Washington | (206) 467-4800


Yi's bouquet was a lush mix of bamboo branches, fern curls, Schwartzwalder mini callas, Amnesia roses, Eloquence spray roses, scabiosa pods, seeded eucalyptus, cymbidium orchids and scented geranium.
Unfortunately, Yi's bouquet was the only thing I had time to photograph. The flower board I created for her describes the rest of the personal flowers.


IT IS said that if you want to know the pulse of the people, listen to taxi drivers.
I was riding a calibrated taxi cab and hoping against hope it would not reach monstrous proportions like the horror stories I have heard the past weeks.
The taxi driver meekly told me as I was embarking that his meter has been calibrated as if asking me if I really wanted to take the ride, I was rushing and had grocery bags in too and so having no choice, bravely said, yes to the calibrated ride.
An honest man, I thought, so what the heck.
The cab driver said it is inevitable that all cabs will be calibrated at some point, explaining the LTFRB now calibrates at random, rather than the schedule earlier presented, so avoiding taxi plates ending in 1-2-3, does not really cut it now.
It is now common knowledge that to ride a taxi is like throwing away hard earned mullah for the continence of a private ride and the cumbersome task of walking.
If you have been living under a rock for the past months and do not know what the hullabaloo is all about, let me be the one to break the bad news; the meter of the cab will add P13.50 per kilometer with an added P2.00 per minute as you go along your way, oblivious of time and space.
The cab driver rued his predicament saying the hike in fare does not really translate into an added income for drivers, saying the boundary of each has also been upped by operators.
He surmised with the increase in fare, the riding public would now think twice before riding a cab, giving them a decrease in riders and ultimately, make life harder of drivers in the long run.
I was in a chatty mood that day and the cabbie's jovial tone made me want to talk to him, if only to take my mind off the taxi meter.
The driver said with the increase in taxi fare, a corresponding increase in salaries should be set forth, to level the field for everyone.
He said only the policemen were given a substantial increase in salaries and he suspected that taxpayers will, for a long time, be paying for the Duterte-approved salary hike for the nation's finest.
My cabbie said any hike in price should be done in a staggered pace, so that ordinary people don't get a heart attack.
He now alluded to the wisdom of North Korea, with only one leader being followed and a seeming equality for all, comparing it to life in China, which he said, he sometimes wished for.
He said he did not vote for Duterte but instead for his favorite actor's daughter, Grace Poe.
He said it is sad that Poe lost to the strongman; she could have served the country better as he believed instead of the present leader who can badmouth even the Pope.
A sensible man, my cabbie was, I thought.
The driver then launched into a speech on who was the richest in the city, pointing to the members of a local mall and grocery chain and concluding what a charmed life the led.
He said the luxury cars one family member owned before his death were a sight to behold as well as a puzzle to figure out why the BIT did not give a fuss out of it.
Then he remembered the fate that befell one of the family, a grisly murder unsolved till today, and said you can't really have it all.
The ride ended after 10 minutes, spanning 4 kilometers, there was no traffic, and the cabbie printed me a receipt and walked me through how to compute if my fare was correct.
It was accurate to the last peso.


text:
Missing Person-Request For Public Assistance
The Saskatoon Police Service is asking for the public's assistance in locating a 12 year old male. Casey Gardipy was last seen at 2:40p.m. on July 18, 2019, leaving the area of Marquis Downs…
Request For Public Assistance - Persons Of Interest In Homicide File
The Saskatoon Police are requesting the public's assistance in identifying two persons of interest in relation to the homicide of 36 year old, Daniel Morrison. A photo of the two is attached…
Update - Charges Laid - Homicide 200 Block Avenue V South
A 32 year old Moose Jaw man is facing a 2nd degree murder in the death of 52 year old Noreen Moore. The male was arrested on July 17, 2019 by members…
UPDATE: Search Warrant - 200 Blk Ave. W S
Fourteen people are facing 476 charges following a search warrant executed by Saskatoon Police last night. Officers had reason to believe that a number of people in a residence in the 200 block…
Use the Citizens Online Police Reporting System to submit a non urgent report immediately. Start Here
Online Bicycle Registration
Register your bicycle serial number with the Saskatoon Police Service.
Online Record Checks
Criminal Record Checks
Vulnerable Sector Checks
View reported crime incidents and traffic collisions in your neighbourhood or across the city. View Map
Annual Police Community BBQ
Saskatoon Police Service
July 25 - 11:00am - 2:00pm
Prairieland Park
August 6 to August 11 - 12:00pm - 12:00am
Recruitment Information Night
SPS Headquarters - Community Room
August 21 - 7:00pm
SIGMA/MMPI Testing
August 22 - 8:00am
The Vulnerable Person Registry is a voluntary service offered by the Saskatoon Police Service.
On October 17, 2018 Cannabis was legalized in Canada.
@SaskatoonPolice
Strengthening Families Saskatoon
This new federally funded initiative is a proven family-based skills training program.
211 Saskatchewan
211 Saskatchewan is a free, confidential, and searchable website of human services in Saskatchewan, with over 5,000 listings of social, community, non-clinical health, and government services across the province. The categories of listings include, but are not limited to, mental health and addictions, homelessness, income support, health care, food security and community programs.
SPS Member Finds Kidney Donor
"I've been floating on a cloud." That's how SPS civilian member Monica Goulet has been feeling since finding out a match had been found for a kidney transplant...
Crisis Negotiators
These special team members attend high-risk calls to negotiate a peaceful resolution with suspects.
Are you interested in a career in policing?
Abirawan Aninda Khan (DOB: 03/17/98) - Wanted for Aggravated Assault
Unknown Male - Wanted for Robberies
Unknown Male - Described as Indigenous, 18- 20yrs old, 6', shaggy dark hair. Wanted for Sexual Assault
Unknown Male - Described as Indigenous, late teens or early 20's,120-130lbs, brush cut, scar over left eyebrow. Wanted for Sexual Assault
Timloh "Butchang" Nkem (DOB: 11/25/88) - Described as 5'11'' 170 lbs, black hair and brown eyes. Wanted for Sexual Assault and Failing to Attend Court for sentencing.
Justin Arthur Robinson (Age 29) - Described as violent, 5'8" tall, 205 lbs, brown hair, and brown eyes. Wanted for assault with a weapon, common assault (x2), breach of recognizance (x3), and drive while disqualified.
Desmond Fischer - Last seen June 25, 2018
Happy Mary Charles - Last seen April 3, 2017 in Prince Albert, SK
Jordan Charles Walker - Last seen December 2, 2015
Kandice Marie Singbeil - Last seen May 25, 2015
Kenneth Fehr - Last seen November 27, 2012


Is your child in grades four or five? Would your child enjoy learning the Ontario curriculum through the Arts?
If so, the York Region District School Board offers an Elementary integrated Arts program at Baythorn Public School. Students currently in grades four and five may apply for the arts@baythorn program, where they will have the opportunity to explore the disciplines of Dance, Drama, Visual Arts, and Music while meeting the expectations of the Ontario Curriculum for grades five, six, seven, and eight.
To learn more about the arts@baythorn program, students and their families are invited to attend an information evening at Baythorn Public School on Thursday, January 16, 2014 @ 6:30 p.m.
Applications will be on-line via the school website and available after January 16, 2014. www.baythorn.ps.yrdsb.edu.on.ca.
Please call Jacqueline Linton, Principal or Jennifer Rosenberg, Vice-Principal at Baythorn Public School (905) 889-7992, if you require further information.


Operators of some overseas scams will use American middle-men to collect money from victims before sending it on to the scammers. Federal prosecutors have accused one Florida man of being the cash conduit for a "lottery" scam being run out of Jamaica.
We've covered Jamaican lottery scams before. They're a form of fraud that exploits our innate greed, where callers tell victims that they've won millions of dollars in a lottery that they never entered, but must wire money to cover taxes or fees. Lottery scammers return to the same victims to collect more money, stringing them along for as long as possible, promising that the payout will come.
To state the obvious, however, there is no lottery. There's just a large pool of people in Jamaica trained to work in call centers servicing the United States and Canada who figured out that scamming people was much more lucrative than collecting an hourly wage.
In an indictment [PDF], the feds describe how the Florida accomplice collected payments ranging from a wire transfer of $150 to a cashier's check sent to him through the mail for $12,500. Prosecutors say that he would keep a portion of the payments and send the rest on to his accomplices in Jamaica.
It's a tragically popular scam: In 2015, around the time that the defendant in this case was allegedly receiving payments and sending them on to Jamaica, American victims were sending $300 million in fake lottery taxes and fees to that country every year.
"These charges demonstrate our commitment to combating international lottery fraud, especially efforts to target potentially vulnerable individuals" said Acting Assistant Attorney General Chad A. Readler of the JusticeDepartment's Civil Division said in a statement. "Financial schemes like these interfere with Americans' financial security and will not be tolerated." Lottery scams target elderly Americans, who are often ashamed to tell their family members what has happened.


text:
Mustapha Hadji (ur. 16 listopada 1971 w Ifran) – marokański piłkarz.

Kariera
Urodził się w Maroku, jednak w młodym wieku, wraz z rodziną wyemigrował do Francji. Na początku mieszkali w Saint-Étienne, później w Monceau, aż wreszcie marokańska rodzina osiedliła się w Creutzwald. To właśnie tutaj młody Mustapha stawiał swoje pierwsze piłkarskie kroki. Pierwszym profesjonalnym klubem Marokańczyka było AS Nancy, w barwach którego występował w latach 1991-1996. W tym czasie dostał również propozycję gry w młodzieżowej reprezentacji Francji - odmówił jednak, wybierając występy dla swojej ojczyzny.

W 1996 Hadji przeniósł się na Półwysep Iberyjski. Przez pierwszy sezon grał w Sportingu. Nie czuł się tutaj jednak dobrze i w 1997 przeszedł do Deportivo La Coruña. Podczas, gdy w Sportingu rozegrał 27 spotkań, w Deportivo zagrał tylko 31-krotnie. Hadji przeniósł się do Coventry City, gdzie występował wraz ze swoim rodakiem, Youssefem Chippo. Z drużyny odszedł po spadku klubu do niższej klasy rozgrywkowej. W lipcu 2001 Aston Villi. W lutym 2004 stał się wolnym zawodnikiem i pozyskał go Espanyol Barcelona. W sierpniu tegoż roku przeszedł do klubu ze Zjednoczonych Emiratów Arabskich - Al Ain. W Azji pozostał jednak tylko rok. Przed sezonem 2005/06 skusił się na ofertę klubu z niemieckiej 2. Bundesligi, 1. FC Saarbrücken. Hadji przebił się do pierwszego składu. Ze swoją drużyną spadł do Regionalligi. W klubie z Kraju Saary pozostał do końca sezonu 2006/07. W sierpniu 2007 podpisał kontrakt z II-ligowym klubem luksemburskim Fola Esch. W lipcu 2010 zakończył karierę.

Z reprezentacją Maroka Hadji uczestniczył w Mistrzostwach Świata 1994 i 1998. Podczas mundialu we Francji zdobył bramkę w spotkaniu z reprezentacją Norwegii.

Od maja 2014 jest asystentem selekcjonera reprezentacji Maroka.

Przypisy

Urodzeni w 1971
Reprezentanci Maroka w piłce nożnej
Piłkarze AS Nancy
Piłkarze Sportingu CP
Piłkarze Deportivo La Coruña
Piłkarze Coventry City F.C.
Piłkarze Aston Villa F.C.
Piłkarze RCD Espanyol
Piłkarze Al-Ain FC
Piłkarze 1. FC Saarbrücken
Piłkarze Fola Esch
Uczestnicy Mistrzostw Świata w Piłce Nożnej 1994
Uczestnicy Mistrzostw Świata w Piłce Nożnej 1998
Uczestnicy Pucharu Narodów Afryki 1998
Uczestnicy Pucharu Narodów Afryki 2000

redpajama_set_name: RedPajamaWikipedia


text:
Education Faith Health Money Politics U.S. WND News Center WorldWND VIDEO
WATCH: AP bizarrely warns voters about Dems losing majority
Then GOP would 'decide what bills come to the House floor'
By WND Staff
Published November 7, 2022 at 3:09pm
Meg Kinnard of the Associated Press (Video screenshot)
We now know that during the 2020 election, Mark Zuckerberg's $400 million in handouts to elections officials almost without a doubt influenced the results.
So did the FBI's lobbying Big Tech to suppress accurate information about the Biden family's international business dealings.
This year it's AP that apparently is trying to influence the election.
With a warning from a political writer about what happens if Democrats "lose" their congressional majority.
Not about what happens of the Republicans win; about the Democrats' loss.
A commentary at Twitchy pinpointed the issue:
"So has the AP given up pretending they're not a totally and completely biased ***? That's all we can assume when we see them posting a melodramatic video from their 'National Politics Reporter,' Meg Kinnard, talking about **GASP** what happens if (when) Democrats LOSE."
Are the national news media actively electioneering in favor of Democrats?
100% (1007 Votes)
It continued, "Beyond the fact that this seems like blatant campaigning for Democrats (really, they should note this as some sort of in-kind contribution), there is something really weird going on with her face, just sayin'."
What happens if Democrats lose their majority in Congress after the midterm elections?
AP national politics reporter @megkinnardap explains. #Election2022 pic.twitter.com/B1uQsqQzDJ
— The Associated Press (@AP) November 7, 2022
The commentary continued, "Creepy, right? Like she's some weird AI talking head babbling about the HORRORS of Republicans taking over the House … and c'mon, there's a chance they'll take over the Senate as well.
"And boohoo, poor Biden won't be able to push his abortion agenda if Republicans win. What a disgusting display of partisanship hackery from the AP."
Kinnard warns GOP could have votes to decide "the next speaker of the House," and members of the party would "run every committee."
That means, they could "decide what bills come to the House floor."
Biden's abortion and gun control agendas, she warned "would be sidelined."
EDITOR'S NOTE: Leftwing billionaire George Soros bankrolls everything evil – most recently the 75 pro-criminal "Soros district attorneys" who have transformed America's most dynamic cities into degraded, crime-infested hellscapes of skyrocketing violent crime. Soros also funds drug legalization, euthanasia, open borders, globalism, Black Lives Matter, "defunding the police," devaluing America's currency and destroying her sovereignty. But why?
Whistleblower magazine reveals and explores the stunning truth: Soros, an atheist, literally says he's "some kind of god," obsessed with re-creating the world in his image. Here are a few actual quotes from Soros: "If truth be known, I carried some rather potent messianic fantasies with me from childhood…" "It is a sort of disease when you consider yourself some kind of god, the creator of everything, but I feel comfortable about it now since I began to live it out." "You know, in my personal capacity I'm not actually a selfless philanthropic person. I've very much self-centered." "I am kind of a nut who wants to have an impact." "Next to my fantasies about being God, I also have very strong fantasies of being mad. … I have a lot of madness in my family. So far I have escaped it."
Except he hasn't "escaped it." Not only is Soros totally mad, but he's imposing his insanity on America. See Soros and the leftwing world he has created and funded as never before in "THE BILLIONAIRE WHO THINKS HE'S 'GOD': Why George Soros is obsessed with destroying America."
WND Staff
WND news editors compile reports for our readers.
Court reviewing just what constitutes a 'threat' on social media
Ronna Romney McDaniel re-elected to run Republican National Committee
Report: DOJ ready to seek warrant for Biden's home


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