text:
\section{Introduction}\label{intro}

It is well-known \cite{terada,sor,epole} that the presence of a
pole in the kinetic term of the inflaton gives rise to
inflationary models collectively named $\alpha$-attractors
\cite{alinde, eno5}. Confining ourselves to poles of order one and
two, the lagrangian of the homogenous inflaton field $\sg=\sg(t)$
can be written as
\beq \label{action1} {\cal  L} = \sqrt{-\mathfrak{g}}
\left(\frac{\fkk}{2\fk^2} \dot\sg^2-
\Vhi(\sg)\right)~~\mbox{with}~~\fk=1-\sg^p,~p=1,2~~\mbox{and}~~\fkk>0.\eeq
Also, $\mathfrak{g}$ is the determinant of the background
Friedmann-Robertson-Walker metric $g^{\mu\nu}$ with signature
$(+,-,-,-)$, dot stands for derivation \emph{with respect to}
({\sf\small w.r.t}) the cosmic time and $V_{\rm I}=V_{\rm I}(\sg)$
is a polynomial function of $\sg$. Expressing the canonically
normalized field, $\se$, in terms of $\sg$ as follows
\beq \label{VJe}
\frac{d\se}{d\phi}=J=\frac{\sqrt{\fkk}}{\fk}~~\Rightarrow~~\sg=
\begin{cases}
1-e^{-\se/\sqrt{N_1}} &\mbox{for}~~p=1,\\
\tanh{\lf\frac{\se}{\sqrt{N_2}}\rg} &\mbox{for}~~p=2\,,
\end{cases}
\eeq
we easily infer that  $V_{\rm I}(\se)$ develops a plateau for
$\se\gg1$, and so it becomes suitable for driving inflation of
chaotic type. It is called \emph{E-Model Inflation} ({\sf\small
EMI}) \cite{alinde, linde21} (or $\alpha$-Starobinsky model
\cite{ellis21}) and \emph{T-Model Inflation} ({\sf\small \tmd})
\cite{tmodel, linde21} for $p=1$ and $2$ respectively.

In this talk, based on \cref{sor, epole}, we present novel
realizations of \emph{E and/or T-Model Inflation} ({\sf\small
ETI}) in the context of \emph{Supergravity} ({\sf\small SUGRA}).
Namely, in \Sref{fhim} we describe our general strategy in
establishing those models and then -- in \Sref{fhi3} -- we specify
three models (\dci, \ca, \cb) employing a gauge singlet inflaton
and three (\dhi, \ha, \hb) using a gauge non-singlet inflaton. We
end up with our numerical results in \Sref{num} and our
conclusions in \Sref{con}.

Throughout the text, the subscript $,\chi$ denotes derivation
w.r.t the field $\chi$, charge conjugation is denoted by a star
($^*$) and we use units where the reduced Planck scale $\mP =
2.44\cdot 10^{18}~\GeV$ is set equal to unity.

\section{Strategy of the SUGRA Embedding}\label{fhim}

We start our investigation presenting the basic formulation of
scalar theory within SUGRA in \Sref{sugra1} and then -- in
\Sref{sugra2} and \ref{sugra3} -- we outline our methodology in
constructing viable ETI.

\subsection{General Formalism} \label{sugra1}

The part of the SUGRA lagrangian including the (complex) scalar
fields $z^\al$ can be written as
\beq\label{Saction1} {\cal  L} = \sqrt{-\mathfrak{g}} \lf
K_{\al\bbet} D_\mu z^\al D^\mu z^{*\bbet}-V_{\rm SUGRA}\rg, \eeq
where the kinetic mixing is controlled by the K\"ahler potential
$K$ and the relevant metric defined as
\beqs \beq \label{kddef} K_{\al\bbet}={\Khi_{,z^\al
z^{*\bbet}}}>0\>\>\>\mbox{with}\>\>\>K^{\bbet\al}K_{\al\bar
\gamma}=\delta^\bbet_{\bar \gamma}.\eeq
Also, the covariant derivatives for the scalar fields $z^\al$ are
given by
\beq D_\mu z^\al=\partial_\mu z^\al+ig A^{\rm a}_\mu T^{\rm
a}_{\al\bt} z^\bt\eeq\eeqs
with $A^{\rm a}_\mu$ being the vector gauge fields, $g$ the
(unified) gauge coupling constant and $T^{\rm a}$ with ${\rm
a}=1,...,\mbox{\sf\small dim}\Ggut$ the generators of a
\emph{Grand Unified Theory} ({\sf \ftn GUT}) gauge group $\Ggut$.
Here and henceforth, the scalar components of the various
superfields are denoted by the same superfield symbol.

The SUGRA scalar potential, $V_{\rm SUGRA}$, is given in terms of
$K$, and the superpotential, $W$, by
\beq V_{\rm SUGRA}=\Ve_{\rm F}+ \Ve_{\rm D}\>\>\>\mbox{with}\>\>\>
\Ve_{\rm F}=e^{\Khi}\left(K^{\al\bbet}D_\al W D^*_\bbet
W^*-3{\vert W\vert^2}\right) \>\>\>\mbox{and}\>\>\>\Ve_{\rm D}=
g^2 \mbox{$\sum_{\rm a}$} {\rm D}_{\rm a}^2/2. \label{Vsugra} \eeq
The K\"ahler covariant derivative and the D-terms read
\beq \label{Kinv} D_\al W=W_{,z^\al}
+K_{,z^\al}W\>\>\>\mbox{and}\>\>\>{\rm D}_{\rm a}= z_\al\lf T_{\rm
a}\rg^\al_\bt K^\bt\>\>\>\mbox{with}\>\>\>
K^{\al}={\Khi_{,z^\al}}\,.\eeq
Therefore, ETI introduced in \Sref{intro} can be supersymmetrized,
if we select conveniently the functions $K$ and $W$ so that
\Eref{Saction1} reproduces \Eref{action1}.

\subsection{Inflaton's Sector}\label{sugra2}

We concentrate on ETI driven by $V_{\rm F}$ assuring that $V_{\rm
D}=0$ during it. This condition may be attained identifying the
inflaton either with (the radial part of) a gauge singlet
superfield $z^2:=\Phi$ or with  the radial part of a conjugate
pair of Higgs superfields, $z^2:=\Phi$ and $z^3:=\bar\Phi$.

To achieve a kinetic term in \Eref{Saction1} similar to that in
\Eref{action1} for $p=1$ and $2$, we need to establish suitable
$K$'s so that (during ETI)
\beq \vevi{K}= -N
\ln\fk~~~\mbox{and}~~~\vevi{K_{\al\bbet}}=N/\fk^2\eeq
with $N$ related to $N_p$. However, from the F-term contribution
to \Eref{Vsugra}, we remark that $K$ affects -- besides the
kinetic mixing -- $V_{\rm SUGRA}$, which, in turn, depends on $W$
too. Therefore, $\fk$ is generically expected to emerge also in
the denominator of $V_{\rm SUGRA}$ making difficult the
establishment of an inflationary era. This problem can be
addressed \cite{sor, epole} either by tuning the terms of $W$ so
that the pole is removed from $V_{\rm SUGRA}$ thanks to
cancellations or adopting a more structured $K$ which yields the
desired kinetic terms in \Eref{action1} but lets $V_{\rm SUGRA}$
immune.

\subsection{Stabilizer's Sector}\label{sugra3}

We reserved $\al=1$ for a gauge singlet superfield, $z^1=S$ called
stabilizer or goldstino, which assists \cite{rube} us to formulate
ETI of chaotic type in SUGRA. We assume that $S$ and $W$ are
equally charged under a global $R$ symmetry and so it appears
linearly in $W$ multiplying its other terms. Also, it generates
for $\vevi{S}=0$ the inflationary potential via the only term of
$V_{\rm SUGRA}$ which remains alive
\beq \Vhi=\vevi{V_{\rm F}}= \vevi{e^{K}K^{SS^*}|W_{,S}|^2},
\label{Vhio}\eeq
where the symbol ``$\vevi{Q}$" denotes the value of a quantity $Q$
during ETI. Such an adjustment assures the boundedness of $\Vhi$,
whereas $S$ can be stabilized at $\vevi{S}=0$, if we select
\cite{su11}
\beq \label{K2}
K_2=N_S\ln\lf1+|S|^2/N_S\rg~~\Rightarrow~~\vevi{K_2^{SS^*}}=1~~\mbox{with
$0<N_S<6$}.\eeq $K_2$ parameterizes the compact manifold
$SU(2)/U(1)$. Note that for $\vevi{S}=0$, $S$ is canonically
normalized and so we do not mention it again henceforth.

\section{Inflationary Settings}\label{fhi3}

We here explain the construction of the EMI and TMI in
\Sref{fhi33} and \ref{fhi1} respectively.

\subsection{E-Model Inflation}\label{fhi33}

This setting is realized in presence of two gauge singlet
superfields $S$ and $\Phi$. The relevant SUGRA setup is presented
in \Sref{fhi30} whereas the resulting models are derived in
\Sref{fhi31}.

\subsubsection{SUGRA Set-up}\label{fhi30}

We adopt the most general renormalizable $W$ consistent with the
$R$ symmetry mentioned in \Sref{sugra3}, i.e.,
\beq W= S\lf \lda \phc+\ldb\phc^2-M^2\rg \label{whi} \eeq
where $\lda, \ldb$ and $M$ are free parameters. As regards $K$,
this includes, besides $K_2$ in \Eref{K2}, one of the following
$K$'s:
\beq \kas=-N\ln\left(1-\phc/2-\phc^*/2\right)~~~\mbox{or}~~~
\tkas=-N\ln\frac{(1-\phc/2-\phc^*/2)}{(1-\phc)^{1/2}(1-\phc^*)^{1/2}},\label{tkas}\eeq
with $\Re(\phc)<1$ and $N>0$.  For the considered versions of EMI,
\dci\ and \ca -- \cb, the total $K$ respectively is
\beq {\sf\small (a)}~\kbas=\kb+\kas~~\mbox{and}~~{\sf\small
(b)}~\tkbas=\kb+\tkas. \label{kbas}\eeq The elimination of pole in
$\Vhi$ for \dci\ can be achieved if we set \beq
N=2~~~\mbox{and}~~~\rss=-\ldb/\lda\simeq
1+\rs\,~~\mbox{with}~~~\rs\sim0~~~\mbox{and}~~M\ll1.\label{kbas1}
\eeq
No denominator exists in $\Vhi$ for \ca\ and \cb\ and so $N$,
$\lda$, $\ldb$ and $M$ are free parameters.

\subsubsection{Inflationary Configuration}\label{fhi31}

The inflationary potential can be derived from \Eref{Vhio}
specifying the inflationary trajectory as follows
\beq \vevi{S}=0\>\>\>\mbox{and}\>\>\>\vevi{\th}:=\mbox{\sf\small
arg}\vevi{\Phi}=0. \label{inftr3}\eeq
Inserting the quantities above into \Eref{Vhio} and taking into
account \Eref{K2} and
\beq \label{eK} \vevi{e^{K}}=\begin{cases}
\fr^{-N}&\mbox{for}\>\>\>K=\kbas,\\
1& \mbox{for}\>\>\>K=\tkbas,
\end{cases}\eeq
we arrive at the following master equation
\beq\Vhi=\ld^2 \begin{cases}
{\lf  \sg-\rss\sg^2-\mma^2\rg^2}/{\fr^{N}}&\mbox{for \dci},\\
{\lf \sg-\rss\sg^2-\mma^2\rg^2}&\mbox{for \ca},\\
{\lf \sg^2-\rrs\sg-\mmb^2\rg^2}&\mbox{for
\cb},\end{cases}\label{Vmab}\eeq
where $\sg=\Re(\Phi)$, $r_{ij}=-\ld_i/\ld_j$ with $i,j=1,2$ and
$\ld$ and $M_i$ are identified as follows
\beq\ld= \begin{cases}
\lda~~\mbox{and}~~\mma=M/\sqrt{\lda}&\mbox{for~\dci\ and \ca},\\
\ldb~~\mbox{and}~~\mmb=M/\sqrt{\ldb}&\mbox{for~\cb}.\end{cases}\label{ldab}\eeq
As advertised in \Sref{fhi30}, the pole in $\fr$ is presumably
present in $\Vhi$ for \dci, but it disappears for \ca\ and \cb.
The arrangement of \Eref{kbas1}, though, renders the pole harmless
for \dci.

We introduce the canonically normalized fields, $\se$ and $\what
\th$, as follows
\beq \label{K3} \vevi{K_{\Phi\Phi^*}}|\dot \Phi|^2
\simeq\frac12\lf\dot{\what\phi}^{2}+\dot{\what
\th}^{2}\rg~~\Rightarrow~~\frac{d\se}{d\sg}=J={\sqrt{N/2}\over\fr}~~\mbox{and}~~
\widehat{\theta}\simeq
J\sg\theta~~~\mbox{with}~~~\vevi{K_{\Phi\Phi^*}}=\frac{N}{4f_1^2}.
\eeq
We see that the relation between $\sg$ and $\se$ is identical with
\Eref{VJe} for $p=1$, if we do the replacement $N_1=N/2$.

\renewcommand{\arraystretch}{1.4}
\begin{table}[!t]
\bec\begin{tabular}{cccll}\br {\sc Fields}&{\sc Eigen-}&
\multicolumn{3}{c}{\sc Masses Squared}\\\cline{3-5} &{\sc states}&
& {$K=\kbas$}&{$K=\tkbas$}\cr \hline\hline $1$ real scalar
&$\widehat \theta$ & $\widehat m^2_{\theta}$&
\multicolumn{2}{c}{$6H_{\rm I}^2$}\cr
$2$ real scalars &$\what s_1,~\what s_2$ & $\what m^2_{
s}$&\multicolumn{2}{c}{$6H_{\rm I}^2/N_S$}\cr\mr
$2$ Weyl spinors & ${(\what{\psi}_{\Phi}\pm
\what{\psi}_{S})/\sqrt{2}}$& $\what m^2_{ \psi\pm}$&
\multicolumn{2}{c}{$6n(1-\sg)^2H_{\rm I}^2/N\sg^2$}\cr\br
\end{tabular}\eec
\caption{\sl Mass spectrum of our EMI along the trajectory of
\Eref{inftr3} -- we take $n=1$ for \textsf{$\delta$\small EM} and
\textsf{\small EM2} whereas $n=2$ for \textsf{\small
EM4}.}\label{tab3}
\end{table}\renewcommand{\arraystretch}{1}

To check the stability of $V_{\rm SUGRA}$ in \Eref{Vsugra} along
the trajectory in \Eref{inftr3} w.r.t the fluctuations of
$z^\alpha$'s, we construct the mass spectrum of the theory. Our
results are summarized in \Tref{tab3}. Taking the limit
$\rs=\mma=0$ for \dci, $\rss=\mma=0$ for \ca\ and $\rrs=\mmb=0$
for \cb, we find the expressions of the masses squared $\what
m^2_{\chi^\al}$ (with $\chi^\al=\th$ and $s$) arranged in
\Tref{tab3}. We there display the masses $\what m^2_{\psi^\pm}$ of
the corresponding fermions too -- we define
$\what\psi_{\Phi}=J\psi_{\Phi}$ where $\psi_\Phi$ and $\psi_S$ are
the Weyl spinors associated with $S$ and $\Phi$ respectively. We
notice that the relevant expressions can take a unified form for
all models -- recall that we use $N=2$ in \dci\ -- and approach,
close to $\sg=\sgx\simeq1$, rather well the quite lengthy, exact
ones employed in our numerical computation. From them we can
appreciate the role of $N_S<6$ in retaining positive $\what
m^2_{s}$. Also, we confirm that $\what
m^2_{\chi^\al}\gg\Hhi^2\simeq\Vhio/3$ for $\sgf\leq\sg\leq\sgx$.

\subsection{T-Model Inflation}\label{fhi1}

Although TMI is mostly realized by a gauge-singlet inflaton
\cite{alinde}, we here propose an alternative scheme where the
inflaton is included in a conjugate pair of Higgs superfields,
$\bar\Phi$ and $\Phi$ with charges $B-L=-1$ and $1$ respectively.
These fields break the GUT symmetry $\Ggut=\Gsm\times U(1)_{B-L}$
down to the \emph{Standard Model} ({\small\sf SM}) gauge group
$\Gsm$ through their \emph{vacuum expectation values} ({\small\sf
v.e.vs}). We below outline the SUGRA setting in \Sref{fhi10} and
its inflationary outcome in \Sref{fhi11}.

\subsubsection{SUGRA Set-up}\label{fhi10}

Consistently with the imposed symmetries we adopt the following
$W$
\beq W= S\lf \frac12\ldb\bar\Phi\Phi+\ldd
(\bar\Phi\Phi)^2-\frac14M^2\rg,\label{whih} \eeq
where $\ldb, \ldd$ and $M$ are free parameters. This type of $W$
is well-known from the models of F-term hybrid inflation -- for
reviews see, e.g., \cref{fhi}. The invariance of $K$ under $\Ggut$
enforces us to introduce a pole of order two in the $\phcb-\phc$
kinetic terms. One possible option -- for another equivalent one
see \cref{sor} -- is
\beq\kaa=-\frac{N}{2}\ln\left(1-2|\phc|^2\right)\left(1-2|\phcb|^2\right)\>\>\>\mbox{or}\>\>\>
\tkaa=-\frac{N}{2}\ln\frac{\left(1-2|\phc|^2\right)\left(1-2|\phcb|^2\right)}
{(1-2\phcb\phc)^{1/2}(1-2\phcb^*\phc^*)^{1/2}},\label{tkba}\eeq
which parameterizes the manifold ${\cal
M}_{(11)^2}=SU(1,1)/U(1)_\Phi \times SU(1,1)/U(1)_{\bar\Phi}$
\cite{sor} with scalar curvature ${\cal R}_{(11)^2}=-{8}/{N}$.
From the selected above $W$ and $K$'s, we obtain the models \dhi\
and \ha\ -- \hb, which employ the following total $K$'s
correspondingly
\beq {\sf\small (a)}~\kbaa=\kb+\kaa~~\mbox{and}~~{\sf\small
(b)}~\tkbaa=\kb+\tkaa. \label{kbba} \eeq
Within \dhi, an elimination of the singular denominator appearing
in $\Vhi$ is obtained setting
\beq N=2~~~\mbox{and}~~~\tss=-\ldd/\ldb\simeq
1+\ts\,~~\mbox{with}~~~\ts\sim0~~~\mbox{and}~~~M\ll1.\label{kbba1}
\eeq
On the other hand, no singularity exists in $\Vhi$, within \ha\
and \hb\ and so $N$, $\ldb$, $\ldd$ and $M$ remain free
parameters.

\subsubsection{Inflationary Configuration}\label{fhi11}

Adopting for the relevant fields the parameterization  \beq
\Phi=\phi e^{i\theta}\cos\theta_\Phi\>\>\>
\mbox{and}\>\>\>\bar\Phi=\phi e^{i\thb}\sin\theta_\Phi\>\>\>
\mbox{with}\>\>\>0\leq\thn\leq{\pi}/{2}~~~\mbox{and}~~~S= \lf{s
+i\bar s}\rg/{\sqrt{2}},\eeq
we can easily verify that a D-flat direction is \beq
\vevi{\theta}=\vevi{\thb}=0,\>\vevi{\thn}={\pi/4}\>\>\>\mbox{and}\>\>\>\vevi{S}=0,\label{inftr}\eeq
which can be qualified as inflationary path. Regarding the
exponential prefactor of $V_{\rm F}$ in \Eref{Vsugra} we obtain
\beq \label{eKh} \vevi{e^{K}}=\begin{cases}
\frr^{-N}&\mbox{for}\>\>\>K=\kaa,\\
1& \mbox{for}\>\>\>K=\tkaa,\end{cases}\eeq
Substituting it and \eqs{K2}{whih} into \Eref{Vhio}, the
inflationary potential $\Vhi$ takes its master form
\beq\Vhi=\frac{\ld^2}{16} \begin{cases}
{\lf  \sg^2-\tss\sg^4-\mmb^2\rg^2}/{\frr^{N}}&\mbox{for \dhi},\\
{\lf \sg^2-\tss\sg^4-\mmb^2\rg^2}&\mbox{for \ha},\\
{\lf \sg^4-\tts\sg^2-\mmc^2\rg^2}&\mbox{for
\hb},\end{cases}\label{Vmabh}\eeq
where $r_{ij}=-\ld_i/\ld_j$ with $i,j=1,2$ and $\ld$ and $M_i$ are
identified as follows
\beq\ld= \begin{cases}
\ldb~~\mbox{and}~~\mmb=M/\sqrt{\ldb}&\mbox{for~\dhi\ and \ha},\\
\ldd~~\mbox{and}~~\mmc=M/\sqrt{\ldd}&\mbox{for~\hb}.\end{cases}\label{ldbd}\eeq
From \Eref{Vmabh}, we infer that  the pole in $\frr$ is presumably
present in $\Vhi$ of \dhi\ but it disappears in $\Vhi$ of \ha\ and
\hb\ and so no $N$ dependence in $\Vhi$ arises. The elimination of
the pole in the regime of \Eref{kbba1} lets open the realization
of \dhi, though.

\renewcommand{\arraystretch}{1.4}
\begin{table}[!t]
\begin{center}
\lineup\begin{tabular}{cccll}\br {\sc Fields}&{\sc Eigen-}&
\multicolumn{3}{c}{\sc Masses Squared}\\\cline{3-5} &{\sc states}&
&
\multicolumn{1}{c}{$K=\kbaa$}&\multicolumn{1}{c}{$K=\tkbaa$}\cr\mr
2 real&$\widehat\theta_{+}$&$m_{\widehat\theta+}^2$&
\multicolumn{2}{c}{$3\Hhi^2$}\cr
scalars&$\widehat \theta_\Phi$ &$\widehat m_{
\theta_\Phi}^2$&\multicolumn{2}{c}{$M^2_{BL}+6\Hhi^2(1+4/N-2/N\sg^2-2\sg^2/N)$}
\cr\mr
1 complex&$s, {\bar{s}}$ &$ \widehat m_{
s}^2$&\multicolumn{1}{c}{$6\Hhi^2(1/N_S-8(1-\sg^2)/N+N\sg^2/2$}&{$6\Hhi^2(1/N_S-4/N$}\cr
scalar&&&\multicolumn{1}{c}{$+2(1-2\sg^2)+8\sg^2/N)$}&\multicolumn{1}{c}{$+2/N\sg^2+2\sg^2/N)$}\cr\mr
1 gauge boson &{$A_{BL}$}&{$M_{BL}^2$}&
\multicolumn{2}{c}{$2Ng^2\sg^2/\frr^2$}\cr\mr
$4$ Weyl  & $\what \psi_\pm$ & $\what m^2_{ \psi\pm}$&
\multicolumn{2}{c}{${12\frr^2\Hhi^2/N^2\sg^2}$}\cr
spinors&$\ldu_{BL},
\widehat\psi_{\Phi-}$&$M_{BL}^2$&\multicolumn{2}{c}{$2Ng^2\sg^2/\frr^2$}\cr
\br
\end{tabular}\end{center}
\caption{\sl\small Mass spectrum for TMI along the inflationary
trajectory of \Eref{inftr}. }\label{tab1}
\end{table}\renewcommand{\arraystretch}{1.}

The canonically normalized (hatted) fields of the $\phcb-\phc$
system during TMI, are defined as follows
\beqs\beq \vevi{K_{\al\bbet}}\dot z^\al \dot z^{*\bbet} \simeq
\frac12\lf\dot{\widehat \sg}^2+\dot{\widehat
\th}_+^2+\dot{\widehat \th}_-^2+\dot{\widehat
\th}_\Phi^2\rg~~~\mbox{for}~~~\al=2,3. \label{kzzn}\eeq
where, via \Eref{tkba}, we find
\beq \lf \vevi{K_{\al\bbet}}\rg=
\vevi{M_{\phc\phcb}}~~\mbox{with}~~
\vevi{M_{\phc\phcb}}=\kp\,\diag(1,1)\>\>\mbox{and}~~
\kp={N}/{\frr^{2}}.\eeq\eeqs
Inserting the expressions above in \Eref{kzzn} we obtain the
hatted fields
\beq {d\se}/{d\sg}=J={\sqrt{2N}/\frr},~~\widehat{\theta}_\pm
={\sqrt{\kp}}\sg\theta_\pm\>\>\mbox{and}~~\widehat \theta_\Phi =
\sqrt{2\kp}\sg\lf\theta_\Phi-{\pi}/{4}\rg,\eeq
where $\th_{\pm}=\lf\bar\th\pm\th\rg/\sqrt{2}$. From the first
equation above we conclude that \Eref{VJe} for $p=2$ is reproduced
for $N_2=2N$.

We can also verify that the direction of \Eref{inftr} is stable
w.r.t the fluctuations of the non-inflaton fields. Approximate,
quite precise though, expressions for $\sg=\sgx\sim1$ are arranged
in \Tref{tab1}. We confine ourselves to the limits $\ts=\mmb=0$
for \dhi, $\tss=\mmb=0$ for \ha\ and $\tts=\mmc=0$ for \hb. As in
the case of the spectrum in \Tref{tab3}, $N_S<6$ plays a crucial
role in retaining positive and heavy enough $\what m^2_{s}$. Here,
however, we also display  the masses, $M_{BL}$, of the gauge boson
$A_{BL}$ (which signals the fact that $U(1)_{B-L}$ is broken
during TMI) and the masses of the corresponding fermions. It is
also evident that $A_{BL}$ becomes massive absorbing the massless
Goldstone boson associated with $\what\th_-$.

\section{Inflationary Observables}\label{num}

We here constrain the parameters of our \emd\ and \tmd\ -- in
\Sref{num1} and \ref{num2} respectively -- taking into account a
number of observational and theoretical requirements described in
\Sref{obs}.

\subsection{Constraints} \label{obs}

We impose to ETI the following observational and theoretical
requirements:

\subparagraph{\bf (a)} The number of e-foldings $\Ns$ that the
scale $\ks=0.05/{\rm Mpc}$ experiences during ETI must be enough
for the resolution of the  problems of standard Big Bang, i.e.,
\cite{plcp}
\begin{equation} \label{Nhi}  \Ns=\int_{\sef}^{\sex}
d\se\frac{\Vhi}{\Ve_{\rm I,\se}}\simeq61.3+\frac{1-3w_{\rm
rh}}{12(1+w_{\rm rh})}\ln\frac{\pi^2g_{\rm
rh}\Trh^4}{30\Vhi(\sgf)}+\frac14\ln{\Vhi(\sgx)^2\over g_{\rm
rh}^{1/3}\Vhi(\sgf)},\eeq
where $\sex$ is the value of $\se$ when $\ks$ crosses the
inflationary horizon whereas $\se_{\rm f}$ is the value of $\se$
at the end of ETI, which can be found, in the slow-roll
approximation, from the condition
\beqs\beq\mbox{\sf\small max}\{\epsilon(\sg_{\rm
f}),|\eta(\sg_{\rm f})|\}=1,\>\>\>~\mbox{where}\>\>\>\epsilon=
{1\over2}\left(\frac{\Ve_{\rm I,\se}}{\Ve_{\rm
I}}\right)^2\>\>\>\mbox{and}\>\>\>\eta= \frac{\Ve_{\rm
I,\se\se}}{\Ve_{\rm I}}\,.\label{sr}\eeq
Also we assume that ETI is followed in turn by an oscillatory
phase with mean equation-of-state parameter $w_{\rm rh}$,
radiation and matter domination. We determine it applying the
formula \cite{epole}
\beq w_{\rm rh}=2\frac{\int_{\sgn}^{\sgm} d\sg J(1-
\Vhi/\Vhi(\sgm))^{1/2}}{\int_{\sgn}^{\sgm} d\sg J(1-
\Vhi/\Vhi(\sgm))^{-1/2}}-1,\label{wrh}\eeq\eeqs
where $\sgn=\vev{\sg}$ is the v.e.v of $\sg$ after ETI and $\sgm$
is the amplitude of the $\sg$ oscillations. Motivated by
implementations \cite{univ} of non-thermal leptogenesis, which may
follow ETI, we set $\Trh\simeq10^9~\GeV$ for the reheat
temperature. Finally, $g_{\rm rh}=228.75$ is the energy-density
effective number of degrees of freedom include  corresponding to
the \emph{Minimal SUSY SM} ({\small\sf MSSM}) spectrum.

\subparagraph{\bf (b)}  The amplitude $\As$ of the power spectrum
of the curvature perturbations generated by $\sg$ at  $\ks$ has to
be consistent with data~\cite{plcp}, i.e.,
\begin{equation}  \label{Prob}
A_{\rm s}={\Ve_{\rm I}(\sex)^{3}}/{12\, \pi^2}{\Ve_{\rm
I,\se}(\sex)^2} \simeq2.105\cdot 10^{-9}\,.
\end{equation}

\subparagraph{\bf (c)} The remaining inflationary observables
($\ns$, its running $\as$ and $r$) have to be consistent with the
latest \emph{Planck release 4} ({\sf\small PR4}), \emph{Baryon
Acoustic Oscillations} ({\sf\small BAO}), CMB-lensing and
BICEP/{\it Keck} ({\sf\small BK18}) data \cite{plin,gws}, i.e.,
\begin{equation}  \label{nswmap}
\mbox{\sf
(i)}\>\>\ns=0.965\pm0.009\>\>\>~\mbox{and}\>\>\>~\mbox{\sf
(ii)}\>\>r\leq0.032,
\end{equation}
at 95$\%$ \emph{confidence level} ({\sf c.l.}) -- pertaining to
the $\Lambda$CDM$+r$ framework with $|\as|\ll0.01$. These
observables are estimated through the relations
\beq\label{ns} \mbox{\sf (i)}\>\>\ns=\: 1-6\eph_\star\ +\
2\ith_\star,\>\>\>\mbox{\sf (ii)}\>\> \as
=\frac23\left(4\ith^2-(\ns-1)^2\right)-2\what\xi_\star\>\>\>~
\mbox{and}\>\>\>~\mbox{\sf (iii)}\>\>r=16\eph_\star\,, \eeq
with $\xi={\Ve_{\rm I,\se} \Ve_{\rm I,\se\se\se}/\Ve_{\rm I}^2}$
-- the variables with subscript $\star$ are evaluated at
$\sg=\sgx$.

\subparagraph{\bf (d)}  The effective theory describing ETI has to
remain valid up to a \emph{Ultra Violet} ({\small\sf UV}) cutoff
scale $\Qef\simeq\mP$ to ensure the stability of our inflationary
solutions, i.e.,
\beq \label{uv}\mbox{\sf (i)}\>\> \Vhi(\sgx)^{1/4}\leq\Qef
\>\>\>~\mbox{and}\>\>\>~\mbox{\sf (ii)}\>\>\sgx\leq\Qef.\eeq

Finally, we quantify the tuning needed for the attaintment of
efficient ETI defining
\beq \Dex=1 - \sgx.\label{dex}\eeq
The naturalness of the achievement of ETI increases with $\Dex$.

\subsection{E-Model Inflation}\label{num1}

After imposing \eqs{Nhi}{Prob} the free parameters of
$$\mbox{\dci, ~\ca, ~\cb\ ~~are}~~~(\rs,\mma),~ (N, \rss, \mma)
~~\mbox{and}~~ (N,\rrs,\mmb), $$
respectively. Recall that we use $N=2$ exclusively for \dci.
Fixing $\mma=0.001$ for \dci,  $\mma=0.01$ and $\rss=0.001$ for
\ca\ and $\mmb=0.01$ and $\rrs=0.001$ for \cb, we obtain the
curves plotted and compared to the observational data in
\Fref{fig1}.  We observe that:

\begin{figure}[t]\vspace*{-0.6cm}
\begin{minipage}{75mm}
\includegraphics[height=9cm,angle=-90]{Figures/nsrEM}
\end{minipage}
\hfill
\begin{minipage}{95mm}
\begin{center}
\lineup{\small
\begin{tabular}{cccc}\br
{Model:} &{\dci}&{\ca}&{\cb}\cr\mr
${\rs}$ / $\rss$ / $\rrs$&$-1.7\cdot10^{-5}$&{$0.001$}&{$0.001$}\\
$N$&{$2$}&$10$&$10$\cr
$\sgx/0.1$&$9.9$&$9.53$&{$9.84$}\cr
$\Dex (\%)$&$1$&$4.7$&{$2$}\cr
$\sgf/0.1$&{$6.66$}&$3.7$&{$5.6$}\cr\mr
$\wrh$&$-0.24$&$-0.08$&{$0.26$}\cr
$\Ns$&$44.4$&$51.5$&{$55.5$}\cr\mr
$\ld/10^{-5}$&$1.2$&$2.1$&{$1.9$}\cr\mr
$\ns/0.1$&$9.65$&$9.64$&{$9.65$}\cr
$-\as/10^{-4}$&$11.4$ &$6.7$&{$6.2$}\cr
$r/10^{-2}$&$0.44$&$1.3$&{$1.1$}\cr\br
\end{tabular}}
\end{center}
\end{minipage}
\caption{\sl Allowed curves in the $\ns-r$ plane for (i)
\textsf{\small $\delta$EM}, $\mma=0.01$ and various $\rs$'s
indicated on the solid line or (ii) \textsf{\small EM2},
$\mma=0.01$ and $\rss=0.001$ or \textsf{\small EM4}, $\mmb=0.01$
and $\rrs=0.001$ and various $N$'s indicated on the dashed or
dot-dashed line respectively. The marginalized joint $68\%$
[$95\%$] c.l. regions \cite{gws} from PR4, {\sffamily\small BK18},
BAO and lensing data-sets are depicted by the dark [light] shaded
contours. The relevant field values, parameters and observables
corresponding to points shown in the plot are listed in the
Table.} \label{fig1}\end{figure}

\subparagraph{\bf (a)} For \dci\ the resulting $\ns$ and $r$
increase with $|\rs|$ -- see solid line in \Fref{fig1}. From the
considered data we collect the results
\beq \label{resm1} 0\lesssim\rs/10^{-5}\lesssim3.3,
\>\>\>3.5\lesssim {r/10^{-3}}\lesssim5.3\>\>\>
\mbox{and}\>\>\>9\cdot10^{-3}\lesssim\Dex\lesssim0.01. \eeq
In all cases we obtain $\Ns\simeq44$ consistently with \Eref{Nhi}
and the resulting $\wrh\simeq-0.237$ from \Eref{wrh}. Fixing
$\ns=0.965$, we find $\rs=-1.7\cdot 10^{-5}$ and $r=0.0044$ -- see
the leftmost column of the Table in \Fref{fig1}.

\subparagraph{\bf (b)}  For \ca\ and \cb, $\ns$ and $r$ increase
with $N$ and $\Dex$ which increases w.r.t its value in \dci.
Namely, for \ca\ -- see dashed line in \Fref{fig1} -- we obtain
\beqs\beq \label{resm2}
0.96\lesssim\ns\lesssim0.9654,\>\>\>0.1\lesssim N\lesssim
65,\>\>\>0.05\lesssim{\Dex}/{10^{-2}}\lesssim
16.7\>\>\>\mbox{and}\>\>\> 0.0025\lesssim {r}\lesssim0.039\eeq
with $\wrh\simeq-0.05$ and $\Ns\simeq50$. On the other hand, for
\cb\ -- see dot-dashed line in \Fref{fig1} -- we obtain
\beq \label{resm3} 0.963\lesssim\ns\lesssim0.965,\>\>\>0.1\lesssim
N\lesssim 55,\>\>\>0.23\lesssim{\Dex}/{10^{-2}}\lesssim
8.5\>\>\>\mbox{and}\>\>\> 0.0001\lesssim {r}\lesssim0.04\eeq\eeqs
with $\wrh\simeq(0.25-0.39)$ and $\Ns\simeq54-56$. In both
equations above the lower bound on $N$ is just artificial. For
$N=10$, specific values of parameters and observables are arranged
in the rightmost columns of the Table in \Fref{fig1}.

\subsection{T-Model Inflation} \label{num2}

After enforcing \eqs{Nhi}{Prob} -- which yield $\ld$ together with
$\sgx$ -- the free parameters of the models
$$\mbox{\dhi, ~\ha, ~\hb\ ~~are}~~~(\ts,\mmb),~ (N, \tss, \mmb)
~~\mbox{and}~~ (N,\tts,\mmc), $$
respectively. Recall that we use $N=2$ exclusively for \dhi. Also,
we determine $\mmb$ and $\mmc$ demanding that the GUT scale within
MSSM, $\Mgut\simeq2/2.433\times10^{-2}$, is identified with the
value of $M_{BL}$ -- see \Tref{tab1} -- at the vacuum value of
$\sg$, $\vev{\sg}$. We approximately obtain $M_2$ and
$M_4\leq0.001$. By varying the remaining parameters for each model
we obtain the allowed curves in the $\ns-r$ plane-- see
\Fref{fig2}. A comparison with the observational data is also
displayed there.  We observe that:

\begin{figure}[t]\vspace*{-0.6cm}
\begin{minipage}{75mm}
\includegraphics[height=9cm,angle=-90]{Figures/nsrTM}
\end{minipage}
\hfil
\begin{minipage}{95mm}
\begin{center}
\lineup{\small \begin{tabular}{cccc}\br {Model:}
&{\dhi}&{\ha}&{\hb}\cr\mr
${\ts}$ / $\tss$ /
$\tts$&$-3.6\cdot10^{-5}$&{$0.01$}&{$10^{-6}$}\cr
$N$&{$2$}&$12$&$12$\cr\mr
$\sgx/0.1$&$9.9555$&$9.75$&{$9.877$}\cr
$\Dex (\%)$&$0.445$&$2.5$&{$1.23$}\cr
$\sgf/0.1$&{$5.9$}&$3.9$&{$6.5$}\cr\mr
$\wrh$&$0.33$&$0.266$&{$0.58$}\cr
$\Ns$&$55.2$&$56.4$&{$58$}\cr\mr
$\ld/10^{-5}$&$3.6$&$8.6$&{$8.5$}\cr\mr
$\ns/0.1$&$9.65$&$9.64$&{$9.65$}\cr
$-\as/10^{-4}$&$6.6$ &$6.4$&{$5.98$}\cr
$r/10^{-2}$&$0.26$&$1.4$&{$1.3$}\cr\br
\end{tabular}}
\end{center}
\end{minipage}
\caption{\sl Allowed curves in the $\ns-r$ plane fixing
$\mbl=\mgut$ for (i) \textsf{\small $\delta$TM} and various
$\ts$'s indicated on the solid line or (ii) \textsf{\small TM4}
and $\tss=0.01$ or \textsf{\small TM8} and $\tts=10^{-6}$ and
various $N$'s indicated on the dashed and dot-dashed line
respectively. The shaded corridors are identified as in Fig.~1.
The relevant field values, parameters and observables
corresponding to points shown in the plot are listed in the
Table.} \label{fig2}\end{figure}

\subparagraph{\bf (a)} For \dhi\ -- see the solid line in
\Fref{fig2} -- we obtain results similar to those obtained for
\dci\ in \Sref{num1}. Namely, the resulting $\ns$ and $r$ increase
with $|\ts|$ with $\ns$ covering the whole allowed range in
\Eref{nswmap}. From the considered data we collect the results
\beq \label{resh1}2\lesssim-\ts/10^{-5}\lesssim5.5, \>\>\>
2\lesssim
{r/10^{-3}}\lesssim3.6\>\>\>\mbox{and}\>\>\>4\lesssim\Dex/10^{-3}\lesssim4.75.
\eeq
Also, \Eref{Nhi} yields $\Ns\simeq(54.8-55.7)$ and \Eref{wrh}
results to $\wrh\simeq0.3$. Fixing $\ns=0.965$ we find
$\ts=-3.6\cdot 10^{-5}$ and $r=0.0026$ -- see the leftmost column
of the Table in \Fref{fig2}.

\subparagraph{\bf (b)}  For \ha\ and \hb, $\ns$ and $r$ increase
with $N$ and $\Dex$ which is larger than that obtained in \dhi.
More specifically, for \ha\  -- see dashed line in \Fref{fig2} --
we obtain
\beqs\beq \label{resh2}
0.963\lesssim\ns\lesssim0.964,\>\>\>0.1\lesssim N\lesssim
36,\>\>\>0.09\lesssim{\Dex}/{10^{-2}}\lesssim
7.6\>\>\>\mbox{and}\>\>\> 0.0005\lesssim {r}\lesssim0.039\,,\eeq
with $\wrh\simeq0.3$ and $\Ns\simeq56$. On the other hand, for
\hb\ -- see dot-dashed line in \Fref{fig2} -- we obtain
\beq \label{resh3} 0.963\lesssim\ns\lesssim0.965,\>\>\>0.1\lesssim
N\lesssim40,\>\>\>0.45\lesssim{\Dex}/{10^{-2}}\lesssim
3.8\>\>\>\mbox{and}\>\>\> 0.0001\lesssim
{r}\lesssim0.039\,,\eeq\eeqs
with $\wrh\simeq(0.25-0.6)$ and $\Ns\simeq(54.6-60)$. For $N=12$,
specific values of parameters and observables are arranged in the
rightmost columns of the Table in \Fref{fig2}.

\section{Conclusions}\label{con}

We reviewed the implementation of ETI in the context of SUGRA. We
employed as inflaton a gauge singlet or non-singlet superfield for
EMI or TMI respectively. The models are relied on the
superpotentials in \eqs{whi}{whih} which respect an $R$ symmetry
and include, besides inflaton, an inflaton-accompanying field
which facilitates the establishment of ETI. In each model we
singled out three subclasses of models (\dci, \ca\ and \cb) and
(\dhi, \ha\ and \hb). The models \dci\ and \dhi\ are based on the
\Ka s in \sEref{kbas}{a} and (\ref{kbba}{\sf\small a}) whereas
(\ca, \cb) and (\ha, \hb) in those shown in \sEref{kbas}{b} and
(\ref{kbba}{\sf\small b}). Within \dci\ and \dhi\ any
observationally acceptable $\ns$ is attainable by tuning $\rs$ and
$\ts$ respectively to values of the order $10^{-5}$, whereas $r$
is kept at the level of $10^{-3}$ -- see \eqs{resm1}{resh1}. On
the other hand, \ca, \cb, \ha\ and \hb\ avoid any tuning, larger
$r$'s are achievable as $N$ increases beyond $2$, while $\ns$ lies
close to its central observational value -- see \eqs{resm2}{resm3}
for EMI and \eqs{resh2}{resh3} for TMI.

\ack Work supported by the Hellenic Foundation for Research and
Innovation (H.F.R.I.) under the ``First Call for H.F.R.I. Research
Projects to support Faculty members and Researchers and the
procurement of high-cost research equipment grant'' (Project
Number: 2251).

\section*{References}

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redpajama_set_name: RedPajamaArXiv


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redpajama_set_name: RedPajamaCommonCrawl


text:
Rutherford's gold foil experiment: can alphas be deflected by electrons vs. nucleus?
In this experiment, is it possible that some of the alpha particles are deflected by the electrons? Gold, after all, usually also has ~79 of them in each atom.
Since the alpha particles want electrons, couldn't some of them steal electrons on their way through, creating Helium+ ions (or atoms)?
Update: @pwf found some good quotes below from Rutherford's report that seem to confirm that he assumed the effects of the electrons on the alpha beam would be low, but he did not measure them nor account for them in his measurements. So this question is updated to recognize that there seems to be general agreement that the deflection by electrons would happen, but that the frequency and angle of deflection would be low. One question remains: If Rutherford didn't bother to measure this effect experimentally, has anyone?
Found this detailed "thought experiment" where he seems to lay out all the calculations very well. Perhaps he or others have conducted the actual experiment:
http://www.med.harvard.edu/JPNM/physics/didactics/physics/charged/lect.html
quantum-mechanics electrons quantum-chemistry
Brad Cooper - Purpose Nation
Brad Cooper - Purpose NationBrad Cooper - Purpose Nation
$\begingroup$ The incident alphas definitely do interact with electrons - it is the electronic stopping part of the energy loss term. However, kinematics will quickly show you that an alpha cannot back scatter off an electron - it is not physically possible. $\endgroup$ – Jon Custer Mar 7 '15 at 18:28
$\begingroup$ @Floris yes, thank you I should have kept it at "...or even Helium ions or atoms..." $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 14:37
$\begingroup$ @john custer, thank you. Vladimir below seems to think "Electrons can deflect a heavy alpha-particle backward if they are sufficiently energetic." I guess the key is how "sufficiently energetic" they would need to be. I think you are saying they could "scatter," but just not "back scatter"? Just curious if the effect of the electron scattering could be distinguishable from the nucleus scattering and, if so, if that has even been measured? thanks $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:00
$\begingroup$ Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:21
$\begingroup$ The amount of additional electrons you can add is minuscule compared to the number of electrons already there. This would be a very hard experiment. Better use a collider - shoot really high energy electrons in one direction and alpha particles in the opposite direction. I suspect it has been done. $\endgroup$ – Floris Mar 8 '15 at 16:19
Rutherford modeled the atom as an extremely compact positive nucleus surrounded by a uniform ball of negative charge the "size" of the atom. He included the effect of scattering by the electrons under the assumption that they acted like such a diffuse cloud of negative charge, and showed that such a cloud had negligible probability of scattering the alpha particles by more than a degree. That's not because it lacks mass, but because it is diffuse, or rather, composed of many small charges rather than one compact point of high charge, so the alpha particle trajectory would be composed of multiple small deflections ("deflexions" in his spelling!) from the negative charges rather than a single large one. So yes, the electrons could scatter the alpha particles, but for a diffuse electron cloud the effect is negligible. (Actually, as far as I can tell the data would also be consistent with a compact negative nucleus surrounded by a diffuse positive cloud. How Rutherford justified the conclusion that the nucleus was the positive part is mysterious to me.)
Edit: From Rutherford, Philosophical Magazine 21, 669-688 (1911), penultimate paragraph: "The deductions from the theory so far considered are independent of the sign of the central charge, and it has not so far been found possible to obtain definite evidence to determine whether it be positive or negative." He goes on to explain why a positive nucleus might make more sense in terms of $\beta$ absorption and $\alpha$ emission (e.g. why $\alpha$ particles are emitted with such high velocity).
From the same article: Rutherford explains that J. J. Thomson's model of distributed positive and negative charges involves multiple small scattering events, and predicts only small overall "deflexions" (pp 669-670).
After laying out his own model of the compact central charge surrounded by a uniformly dense negative charge, he writes (pp. 671-2), "Since R [the radius of the ball of negative charge] is supposed to be of the order of the radius of the atom, viz. $10^{-8}$ cm., it is obvious that the $\alpha$ particle before being turned back penetrates so close to the central charge, that the field due to the uniform distribution of negative electricity may be neglected. In general, a simple calculation shows that for all deflexions greater than a degree, we may without sensible error suppose the deflexion due to the field of the central charge alone. Possible single deviations due to the negative electricity, if distributed in the form of corpuscles, are not taken into account at this stage of the theory. It will be shown later that its effect is in general small compared with that due to the central field."
He gets around to that discussion near the end, which may be most relevant to your question (p. 686): "In comparing the theory outlined in this paper with the experimental results, it has been supposed that the atom consists of a central charge supposed concentrated at a point, and that the large single deflexions of the $\alpha$ and $\beta$ particles are mainly due to their passage through the strong central field. The effect of the equal and opposite compensating charge supposed distributed uniformly throughout a sphere has been neglected. Some of the evidence in support of these assumptions will now be briefly considered. For concreteness, consider the passage of a high speed $\alpha$ particle through an atom having a positive central charge Ne, and surrounded by a compensating charge of N electrons. Remembering that the mass, momentum, and kinetic energy of the $\alpha$ particle are very large compared with the corresponding values for an electron in rapid motion, it does not seem possible from dynamic considerations that an $\alpha$ particle can be deflected through a large angle by a close approach to an electron, even if the latter be in rapid motion and constrained by strong electrical forces. It seems reasonable to suppose that the chance of single deflexions through a large angle due to this cause, if not zero, must be exceedingly small compared with that due to the central charge."
BTW, I highly recommend reading the original article. Once you decipher the somewhat archaic notation, it's extremely readable.
pwfpwf
$\begingroup$ By the time of Rutherford's experiment the cathode rays had been identified as negatively charged and much lower in mass than atoms. Thus the expectation that the positive charge would be associated with the largest part of the atoms mass which was also present in the "plum pudding" model of the atom that Rutherford used in devising his expectations about how the experiment would come out. And the mass of the electrons does matter if they have low momentum (implied in the models of the time and largely true). $\endgroup$ – dmckee --- ex-moderator kitten Mar 10 '15 at 0:16
$\begingroup$ Yes, I agree there was enough evidence to suspect that electrons were low-mass, given e/m measurements and existing estimates of e. But I think Rutherford's analysis and results were consistent with the low-mass, negative charge being the clumped, highly concentrated charge within the atom and the positive (though massive) charge to be the distributed part. The question was only how the charge was distributed in the atom; in terms of inertia, the atom was treated as a unit. (That makes sense assuming the positive and negative charges are reasonably tightly bound together.) (cont'd) $\endgroup$ – pwf Mar 10 '15 at 3:57
$\begingroup$ (cont'd) Besides, I don't think it was entirely accepted that the charges in cathode rays (assuming they were discrete particles) were the same as the negative charges in atoms, was it? In any case, I don't think there's any mention of the negative charges' mass in Rutherford's paper, independent of the atom's mass. $\endgroup$ – pwf Mar 10 '15 at 4:03
$\begingroup$ @pwf: thank you. Do you have a source for this: "He...showed that such a cloud had negligible probability of scattering the alpha particles by more than a degree."? This seems to be the closest so far in answering the question. It seems he realized the electrons could deflect, but I wonder if the "more than a degree" was a guess, or did he somehow isolate and verify this assumption through testing? $\endgroup$ – Brad Cooper - Purpose Nation Mar 10 '15 at 13:58
$\begingroup$ I'll add some quotes from Rutherford (1911) showing how far he considered this question to my answer. $\endgroup$ – pwf Mar 10 '15 at 16:40
The key difference is that the alpha particle is several thousand times heavier than the electron. It would be like you rolled a bowling ball at a marble and it bounced backward. As Rutherford said:
"It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge."
zeldredgezeldredge
$\begingroup$ Thank you. But even if you rolled a bowling ball and it hit a marble, the marble could still deflect the bowling ball off a straight course. And if the marble stuck to the bowling ball as it rolled, it would put it into an erratic roll as well. It seems like the alpha wants to grab electrons and seems like a hydrogen atom would act differently as it traveled than an alpha particle? $\endgroup$ – Brad Cooper - Purpose Nation Mar 7 '15 at 18:23
$\begingroup$ Well, maybe, but remember that the detector probably doesn't give a click or a dot or whatever from a collision with neutral hydrogen. $\endgroup$ – zeldredge Mar 7 '15 at 18:42
$\begingroup$ A standard surface barrier detector will register energetic neutrals. $\endgroup$ – Jon Custer Mar 7 '15 at 18:46
$\begingroup$ @PurposeNation The small angle deflections due to electrons can not be disentangled from those due to the nuclei. More over, in the original experiment the beam was only roughly columnated, so neither could be easily distinguished from the natural angular spread of the beam. $\endgroup$ – dmckee --- ex-moderator kitten Mar 7 '15 at 22:39
$\begingroup$ @dmckee Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:21
Electrons can deflect a heavy alpha-particle backward if they are sufficiently energetic. There are such electrons indeed, but their concentration is very small, I guess.
Vladimir KalitvianskiVladimir Kalitvianski
$\begingroup$ "Their concentration is very small"? No - there are lots of electrons. The point is that the alpha particle will not be very bothered by the interaction with the electrons - that's just a consequence of their relative mass. This means that it is hard to measure this; but it's quite possible that some alpha particles will come out of the gold foil with an electron (or even two) attached. $\endgroup$ – Floris Mar 7 '15 at 23:19
$\begingroup$ @Floris I think he means that electrons with large Fermi momentum back along the beam (which could in fact scatter an incident alpha pretty substantially) are rare, which is a vast understatement when we're talking about momenta comparable to that of the beam alphas. $\endgroup$ – dmckee --- ex-moderator kitten Mar 8 '15 at 2:15
$\begingroup$ @dmckee re-reading the answer I see your point - thanks $\endgroup$ – Floris Mar 8 '15 at 3:21
$\begingroup$ If this is true, I wonder if Rutherford (or anyone doing this experiment since) attempted to measure and account for this type of electron deflection vs. deflection by the gold nucleus? has this effect ever been measured? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 14:56
$\begingroup$ @Vladimir: Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:23
Note that there is an effect on the electrons that is much larger than caused by direct interaction with the alpha particle. If the alpha partcle collides with the nucleus then the nucleus will change its momentum. To a good approximation this is an instanteneous change and then the so-called "sudden approximation" applies to the electron state. So, the electrons find themselves in a new moving potential well relative to the old potential well. Because this change happens almost instantenously, the wavefunction stays the same. But the old wavefunction is not an energy eigenstate, this leads to a finite probability that the electron configuration will not be the ground state configuration.
Count IblisCount Iblis
Exept that alpha particles are much heavier than electrons and it wouldn't make sense for them to change the direction when in contact with electrons, the point of the experiment was that, because of the deflected alpha particles ONLY in a very tiny, central area of the atom, most of the atom space is EMPTY. This conclusion wouldn't been made if alpha particles and electrons would somehow interact.
Ndrina LimaniNdrina Limani
$\begingroup$ Thanks, Ndrina. However, as confirmed by @pwf, Rutherford himself acknowledged that some deflection by electrons would occur (see the references). $\endgroup$ – Brad Cooper - Purpose Nation Mar 15 '15 at 15:27
Backscattering requires a repulsive force while an electron would attract the alpha particle. Furthermore, the electron is lighter than the alpha particle : in a collision between a truck and a mosquito, the truck doesn't rebound at all.
EDIT : unidimensionnal motion (ie. the particle moving back and forth on the same line) requires a repulsive force, but dmckee argued that an attractive force could produce a result that would be experimentally undistinguishable. To rule out the possibility of scattering by an electron, only the difference in mass in necessary : for a negative charge to backscatter an alpha particle, it would need to be much heavier than an electron. Now, the mass of the electron was known to be very small at the time of Rutherford's experiment. Even if ALL the electrons in the gold atom would gatter in a small nucleus, it would not be enough : by Newton's third law, an equal force would be exerted on the alpha particle and on the negative charge, therefore the latter would have 80 times more acceleration. The negative charge would be scattered, not the alpha particle.
Teacher77Teacher77
$\begingroup$ Notice that the sun's gravity (a strictly attractive force) can scatter comets back toward their original direction of approach... $\endgroup$ – dmckee --- ex-moderator kitten Mar 7 '15 at 22:36
$\begingroup$ actually no, it cannot. The only way something is scattered at 180 degrees is with a repulsive force. $\endgroup$ – Teacher77 Mar 9 '15 at 0:38
$\begingroup$ Consider an object on a parabolic orbit ($E = 0$) at time long before and long after the encounter. It's final momentum is essentially anti-parallel to it's initial momentum. Allow "long time" to grow without bound and it is exactly anti-parallel. $\endgroup$ – dmckee --- ex-moderator kitten Mar 9 '15 at 0:46
$\begingroup$ In that limit, the incident direction and the returning direction would be two parallel directions with an infinite distance between the two. NOT the same direction. Perfect back scattering (ie. one dimensional motion where the particle simply inverts its motion) can only be acheived with a repulsive force. P.S. A parabolic orbit has e = 1, not 0... $\endgroup$ – Teacher77 Mar 9 '15 at 3:58
$\begingroup$ Compute the angle of each track. They approach the same value because the in-and-out coordinate grows faster than the width coordinate. And in particle physics the "solar system" is less than a nanometer across, so the detector a few cm away is a good approximation to infinite remove. $\endgroup$ – dmckee --- ex-moderator kitten Mar 9 '15 at 4:04
Not the answer you're looking for? Browse other questions tagged quantum-mechanics electrons quantum-chemistry or ask your own question.
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Electrons in Rutherford's gold foil experiment
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Electron interaction with the field vs. wave
In the Rutherford scattering experiment, does increasing the mass (with constant charge) of the nucleus affect the deflection?


text:
using System;
namespace Rhino.ServiceBus.FileMessaging
{
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    {
        Allow = 1,
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        Revoke = 4,
    }
}

redpajama_set_name: RedPajamaGithub


Superstar Salman Khan, who unveiled Academy-award winning musician A R Rahman's new album 'Raunaq' here, today gave full marks to the lyrics penned by Union Law Minister Kapil Sibal for the musical composition.
'Raunaq' is a unique collaboration between the legendary musician and the Central minister, who wrote all the songs for the album.
"It is a great pleasure to be here for this launch. The songs are nice. I think Kapil sir is amazing and Rahman is average," Salman said with a chuckle. The album has more than seven songs and are sung by best of artists in the industry including Lata Mangeshkar and Shreya Ghoshal among others.
Speaking at the launch, Salman said there is one thing common between him and Rahman, which is 'Jai Ho', the title of his latest release. 'Jai Ho' is also the most popular track from Rahman's Oscar-winning film 'Slumdog Millionaire'.
On working with the Bollywood superstar, Rahman said, "We did work together on 'Yuvvraj'. He has to do films that I like. There has to be an extraordinary film for us to come together." To which Salman said, "When you (A R Rahman) will work for us... Hope we work together."


At 34, Lasith Malinga can no more bowl at 90 mph for a longer duration, especially after returning from an 18-month knee injury layoff.
Till a few years back Lasith Malinga was the synonym of destruction in the cricketing world. His toe crushing yorkers were almost impossible to deal with making him one of the finest to play the sport, especially in the shorter format. However, as injuries took a toll, the Sri Lankan lost the spark, often going for more runs than desired. At 34, he can no more bowl at 90 mph for a longer duration, especially after returning from an 18-month knee injury layoff. This has forced Malinga to do some serious thinking and he has turned towards former Indian fast bowler Zaheer Khan for help, according to a report by The Indian Express.
Realising that he has lost pace, Malinga now wants to focus on movement in the air and who can be a better person to teach that than his old Mumbai Indians mate Zak. In fact, Zaheer himself faced a similar situation. He started as one of the fastest bowlers in the world in 2000 but continuously evolved to trouble the batsmen with his swing. Now, Malinga who is standing on the cusp of his 300th ODI wicket was seen seeking advice from the Indian great before the start of the third one-dayer in Pallekele.
"Zaheer is more of a seam and swing bowler, and I want to learn more of that kind of thing over the next few years. That's what I discussed with him," the Sri Lankan quick said about his interaction with Zaheer. "He (Zaheer) is a legend of the Indian bowling attack in previous years. We always talk with so many bowlers around the world. When I meet them, I want to get their experiences also. Because they are in the commentary box, they look at every single one of my body movements and my bowling action, and how I swing the ball. They have a good idea of how I'm going now. Then, every time I have an opportunity, we speak about my bowling action, and how I have to improve," said the veteran ahead of the fourth ODI against India, as per IE.
Interestingly, Malinga isn't the first opponent to seek help from Zak. Earlier, England fast bowler Jimmy Anderson had revealed that he learned the intricacies of reverse swing from Zaheer Khan during India's tour of England in 2007.
Malinga who has had a below par series so far as Sri Lanka lost all the three ODIs so far faces yet another challenge in Colombo on Thursday. After Chamara Kapugedera got injured in the 3rd ODI, the slinging pacer was named Sri Lanka's stand-in captain.


text:
Netflix Orders First Netflix Original Russian Series 'Better Than People'
Rachel Beede January 13th, 2019 - 12:39 PM
In their push for originals and movement to European markets, Netflix has ordered Better Than People, the first Russian series for "Netflix Originals." The series, titled Luchshe, chem lyudi in Russian, is set to be comprised of 16 episodes and will be translated into 25 languages. It is produced by Yellow, Black and White, a Russian production company. It will be available in the US and internationally, except in China, and, ironically, Russia. This move may jumpstart an otherwise tedious market.
It has been noted that the Russian frontier will not be the easiest for international studios to navigate. While Hollywood blockbusters have done extraordinarily well in Russia this past year, Russia has placed restrictions on when and how such films are released. For one, an additional tax has been imposed on foreign companies selling digital content in Russia in order to keep local markets safe. Additionally, the government has also limited international content in order to protect Russian films. Culture minister Vladimir Medinsky shared his concerns with RIA Novosti, a state-run news company: "Government protectionism in the area of cinema should be stepped up. Otherwise, the global Hollywood machine will destroy our film industry." The main concern seems to be that Hollywood will destroy Russia's deep film culture. "I was indignant to see nearly two-thirds of all screenings be given to one movie," he added. "What is it, other than a monopoly, a total monopoly of Hollywood studios who dictate their conditions to the cinema chains?" Indeed, Russia has extensive film history, which sparked after the Lumière brothers brought the first films into the Russian Empire back in 1896.
Although tensions between the US and Russia may be somewhat strained, this move by Netflix could create a bridge between countries, at least in the digital content market. According to Eduard Iloyan, general producer of Yellow, Black and White, "The 'Netflix Original' banner means large-scale promo activities, thanks to which, we believe, the Russian content will be watched by millions of viewers worldwide." It has been reported that it took $1 million to close the deal on this series, making it the most expensive sale of Russian television content to an international market. To this, Iloyan noted that "While previously, international distribution was just a status thing for Russian producers, now it's been turning into a real business."
The sci-fi series will be set in the near future and focus on android robots. It has already aired in Russia on their largest free service online network Channel One and on the online video service, Start. No release date for Netflix has been announced yet.
Better Than People
Rachel Beede


It's been a while since I've written what's happening with our family and our dreams on my blog. We've been staying in a garden cottage for the past few months and are moving again in 2 weeks. To where we don't yet know. Its been an interesting, stretching time and quite challenging for us as we've squeezed into a small space that hasn't been the most comfortable or convenient for our little family of 5 but a lot of good has come out of it. We are learning to rely completely on God, to live lighter and keep our eyes on Him in all situations. As we prepare to pack we've collected a small pile of boxes which were too tempting for us not to make something with so this is what we got up to yesterday. Eth made a garage and I made Gianna a doll's house. Hopefully it should keep them entertained while I try to get some packing done.
Posted in Dreams, family, Home School Fun, PhotographyBookmark the permalink.
AWESOME – love your creativity!


[Z-Interview] 'LEGGO' WANNA.B "Hero? Women can be heroes too"
'The Helmet' Kim Bo-jeong, "Action scenes? I'm still growing up"
'Red' Jeong Bo-seok "I think I now know a little bit about Rothko's worries"
[Stars' Favorite Restaurants] What is BTS's all-time favorite restaurant?


text:
Turkey's AI roadmap looks to boost economy and add thousands of jobs
"Robots will be your colleagues not your replacement"
Turkey plans AI institute to tap new technology
India's actions in Kashmir and the muted response from Arab Gulf states
Braving bullets to resist Turkey's July 15 failed coup
Arab tribes to resume protests against US ally the SDF post-Ramadan
A roadmap for an artificial intelligence sector is part of Turkey's digital transformation that seeks to make the country a global competitor in technology.
Turkey aims to push the artificial intelligence sector to contribute 5 percent towards the country's GDP by 2025, it has revealed in its first strategic artificial intelligence (Al) plan.
The strategy was launched in a public event held by the Presidency's Digital Transformation Office and the Industry and Technology Ministry on Tuesday.
The strategy foresees employing at least 50,000 people in the sector as one among 24 other objectives in its plan by 2025.
The ceremony was attended by artificially intelligent characters and holograms where Turkey's plans were presented. The five-year roadmap was launched at Turkey's Informatics Valley in Gebze, Kocaeli, and was developed with the collaboration of private, public, and academic groups.
In August, the National Artificial Intelligence Strategic Plan report was published with a preface from Turkish President Recep Tayyip Erdogan, stating, "The field of artificial intelligence is not a matter of choice, but one of the biggest bearers of our development goals."
The major tenets of the artificial intelligence (Al) strategy include realising actions of decision-making, exploring meaning, and learning in dynamic conditions unique to intelligent creatures by computer or computer-controlled machines, according to the official report.
"We have made the national artificial intelligence (Al) strategy to ensure that the digital transformation is most beneficial to society," said Musyafa Varank, Minister of Industry and Technology, indicating that Turkey is one of the few countries to develop such a plan..
"The artificial intelligence will be an indispensable part of economic and social fields in the very near future," Varank also said while explains the plan's socio-economic aims.
The plan outlines six priorities primarily centered around supporting Turkey's Al initiatives and strengthening international cooperation in this field by training artificial intelligence specialists and increasing employment , encouraging research, entrepreneurship, and innovation, reaching to quality data and technical infrastructure, and expediting structural and workforce transformation.
The government would like to commercialise applications developed and prioritise in public procurement and boost the number of MSc graduates specialized in AI technology by at least 10,000.
"We will enter the top 20 countries in the international artificial intelligence indices," Ali Taha Koc, Head of the Digital Transformation Office of the Turkish Presidency said during the ceremony, highlighting the importance of becoming an accurate Al data-sharing country.
The strategy forecasts a two-fold increase in market size and global AI expenditures within five years. It also strives to enhance Turkey's economic and technological independence by reinforcing indigenous infrastructure and a structural transformation.
Efforts are still ongoing to establish institutions and 'unicorns' similar to The Scientific and Technological Research Council of Turkey (TUBITAK) Artificial Intelligence Institute . When compared to the 2,000 US and Chinese startups, the number in Turkey is low at approximately 200 companies.


It's that time of year again – parties, resolutions, and, of course, predictions for how the year ahead will unfold.
Recently, I read a prediction list for digital marketing in 2017. It was very informative. Except for one thing… The prediction list was geared to big brands and Fortune 500 companies with a team of digital experts and big budgets.
Which is super awesome if you're one of those companies or brands. For the rest of the 99%, who deal with bean-counting bosses and decision makers still stuck in the '90s, here are my FIVE recommendations on where to take your digital marketing in 2017.
1. Integrate Digital with your Marketing Campaign.
Say goodbye to the days when a digital campaign was considered a tactic, a test, or came from a different budget.
All of the research is out there. You don't really need for me to quote stats, do you? Transform your organization's entire thinking. Digital marketing should serve as a centralized, integrated component to your entire marketing message and plan. In 2017, digital initiatives will need to help drive the organization's strategic direction.
2. Leverage Social Media Listening and Digital Reporting to Listen, Respond, and Personalize.
Big Data, right? Every blog is about data, data, data. It's all helpful, and if you have a lot of data, then use it.
But first, get started with the simple. Listen to your customers. It's one of the oldest and most battle-tested business strategies. Your current 2016 digital campaign should net you plenty of data on what your customers are thinking, buying, and wanting.
So in 2017….listen, respond, and personalize. For social efforts, use social media listening tools. For digital, use the plethora of tools that measure clicks, interests, conversions, etc.
Once you learn what your customers want, evolve your messaging and product offerings, then personalize and dial in on what your customers are asking for. Watch as these personalization efforts lead to greater loyalty, increased retention, and more revenue for the organization.
3. Smash the Digital & Social Silos.
The digital transformation is not coming. It's already here. Similar to recommendation #1 above, stop "siloing" your digital efforts and your social efforts.
Your digital footprints should walk and work together. Your customers are going to find you, shop you, and make decisions based on what they learn from your website and your social content. So why are you doing different things in these two mediums and having different people manage them?
Social content allows for expanded brand storytelling and secondary and tertiary messaging, but bring these two worlds together — it's all digital now.
The next big trend in digital and social is augmented reality. But we're a few years from this technology being affordable and from the bean-counting bosses even understanding it.
So Go Video. Now, you could take this recommendation as "so last year," and you'd be right. But there are companies and organizations who still aren't on this bandwagon. Videos have the most engagement, click-through, and conversion when looking at digital + social. The social platforms are expanding their tools to use video including longer videos and more exposure to your own social audience. Combine these trends with a superior ad recall, and the choice is simple. So make 2017 the year of video.
5. Kill the CPM Model.
So measure it. We're not at 100% trackability, but we're pretty close in the digital world. Kill the CPM model and start focusing on how much you have to spend to acquire a new customer or sale.
And there you go.. Another 2017 list. This time for the rest of us.


text:
NA-16 Basic Trainer
The last prototype airplane from North American Aviation was the NA-16 trainer. It was initially built at North American's General Aviation Manufacturing Corp., located at the Curtiss-Caproni plant at Dundalk, Md. Later, it was produced in California. The low-wing monoplane had open cockpits in tandem and a fixed, unfaired undercarriage. Made mostly of metal, but with some fabric on the rear fuselage, it was submitted to the U.S. Army Air Corps for evaluation within a month of its first flight.
The design was selected for production as a basic trainer, although the Air Corps requested that the cockpits be enclosed, fairing be installed on the undercarriage and the engine changed to a 600-horsepower Pratt & Whitney R1340 engine. With these modifications, and a new designation of NA-18, the prototype was eventually sold to Argentina.
The production versions launched North American as a manufacturer of training aircraft, starting with 267 BT-9s and 340 BC-1 "basic combat" trainers. In all, more than 17,000 derivatives of the NA-16 were built including 13,685 of the famous T-6/SNJ series built at North American's Los Angeles, Calif., and Dallas, Texas, plants during the 1940s.
First flight April 1, 1935
Span 42 feet
Weight 3,078 pounds
Power plant 400-horsepower Wright Whirlwind engine
Range 700 miles
Crew Two


If you are looking for the most suitable means to ensure you can practice your accuracy with a gun, the Do All Outdoors Accu Blue Diamond Paper Target is the target for your needs. These Targets from the industry experts at Do All Outdoors are produced using the highest quality substances that will ensure that you get years of effective use.
Do All Outdoors has been making some of the best targets for shooters for quite a long time, and the Do All Outdoors Accu Blue Diamond Paper Target is the product of their determination and dedication to giving shooters the methods to practice their craft with the targets they'll want to allow them to thrive in almost any setting.


text:
Members-only cannabis club opens for business
A businesses believed to be the first of its kind in Massachusetts is open for business.For $50 a month, a new social club called The Summit Lounge allows people to smoke marijuana, consume edibles and hang out with their friends.
WORCESTER, Mass. —
A businesses believed to be the first of its kind in Massachusetts is open for business.
For $50 a month, a new social club called The Summit Lounge allows people to smoke marijuana, consume edibles and hang out with their friends.


GetMoreOffers.com is the #1 source to list a house on MLS for a flat fee in Peoria.
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There is no love in this world greater than that of a mother for her child. Celebrating this precious and unconditional love of a mother, special assemblies marking 'Mother's Day', were held for the whole week across all schools of Presidium. Children from all grades participated in various activities organised for them. As a token of love and gratitude for their mothers, they made special hand crafted gifts and bouquets for them.The children also performed on heart-touching songs and dance sequences. It was a memorable moment indeed to see children valuing the greatest bond of love on earth-a mother's love!


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Each week, we pick one winner at random to receive free ad space. This week our winner is Allyson from Domestic Superhero! Thanks so much Allyson and everyone else who linked up – we're honored that you spend your time here.
By linking up, you are granting us permission to use picture(s) and text link on the blog and social media.
Want a quick reminder about The Humble Brag?
As Always, thanks for reading…and linking up!
Thank you so much for the feature, we really appreciate it!
Thanks so much for the feature! I'm honored.
Thanks for the party Jocie!
Hey! Thanks so much for featuring my Chicken Tortilla Soup and I am psyched to be your weekly winner!!! Do you need me to send you anything? Thanks!!!
All great ideas! I love the Christmas sign and the Thanksgiving name cards! Thanks for hosting.
Thanks so much for hosting! I hope you've had a good week.
Hi Jocie! thanks for the party. Sweet features, that berry wreath is so pretty!
Wow so many fun Thanksgiving and gift ideas Jocie! I think you've outdone yourself in finding the best of the best. I especially love the little kitchen with the LED heating elements and the crocheted turkey. I can't wait until our Granddaughter is a little older so I can make her a kitchen set from an old ET center. Thanks so much for all this inspiration each week. You Rock!


Using the latest update (64 bit) running under Windows 7 Professional.
I want to create a paragraph border for two single line paragraphs. Here is what I do and the result.
2. Select line style. I want single line, 0.5 point.
3. Select bottom of paragraph 1.
6. Repeat the above for paragraph 2.
8. Re-open document. The line styles have changed.
I cannot get the line styles to stick to the chosen one. I see double line, 3 point.
Does the style stay unchanged when you save as tmdx? If so, this could be a docx compatibility issue.
No. It exhibits the same behavior. The single 0.5 point lines become 3 point double lines.
I cannot reproduce this specific issue, but something is definitively fishy with the border function.
1. Create a new document with two single line paragraphs.
2. Set the cursor somewhere in the first paragraph, go to Format > Paragraph..., tab "Borders", and set a single line border (0.5 pt black) at the bottom. Line is created and correct.
3. Set the cursor somewhere in the second paragraph, go to Format > Paragraph..., tab "Borders", and set a single line border (0.5 pt black) at the bottom. Line is created and line at the bottom of paragraph 1 vanishes.
I checked and re-checked that these are really separate paragraphs. Looks like a bug.
I think I found a clue to my specific problem, although I agree that paragraph borders are prone to problems.
In my case the document was originally created in MS Word and I was applying borders to it using TextMaker. I now see why this problem cannot be replicated in TextMaker alone. When I create such a document with TextMaker and edit only using TextMaker, then it works as it should. But since compatibility is emphasized, I think this should work.
Could you please send us a sample file (the file that was originally created in MS Word) to forum(at)softmaker.com? Thank you!
The file was sent. Please let us know your findings.


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text:
Fiji to move more than 40 villages inland as seas rise
The country is also looking at ways to help – including potentially taking in – migrants from other threatened Pacific island nations
* Fiji has moved 3 villages, to move 43 more
* Working on ways to accept future Pacific migrants
By Alister Doyle Environment Correspondent
BONN, Germany, Nov 17 (Reuters) – Fiji plans to move more than 40 villages to higher ground to escape coastal floods and is also working on ways to help future migrants from other Pacific island nations as sea levels rise, Fiji's attorney general said.
Fijian fishermen stand in their boats as they work in front of an island's mountain range near Suva April 29, 2004. REUTERS/David Gray
Fiji, presiding at negotiations among 200 nations in Germany on ways to strengthen action to curb climate change that is melting ice from the Alps to the Andes, is working to resolve issues such as land rights and citizenship.
"We have had to relocate people to higher ground — we have moved three villages to higher ground and there are another 43 that are earmarked," Aiyaz Sayed-Khaiyum told Reuters during the two-week talks that end on Friday.
The first was Vunidogoloa, a village on Fiji's second-largest island, Vanua Levu, where 26 homes were moved in 2012, he said. Apart from homes, authorities are also moving graves as settlements are moved inland.
"Of course there are people who are uncomfortable about leaving their traditional, ancestral ares," he said. He said it was unclear how many people would be covered by the programme.
Fiji has also offered to help other Pacific islanders on low-lying island coral atoll nations. Fiji, with 900,000 people on hundreds of islands, is larger and has mountains rising to 1,300 metres.
"We were the first to openly say we will give them residency or refuge in Fiji should rising sea levels make it inevitable," he said.
Worldwide, sea levels have risen 26 centimetres (10 inches) since the late 19th century, driven up by melting ice and a natural expansion of water in the oceans as they warm, U.N. data show. Seas could rise by up to a metre by 2100.
The rise aggravates the impact of storm surges such as Cyclone Winston in 2016 that killed 44 people in Fiji and caused $1.4 billion in damage, a third of Fiji's Gross Domestic Product. It also washes more salt water onto cropland.
The Pacific state of Kiribati bought 6,000 acres (2,400 hectares) of forest land on Fiji in 2014. It might help grow crops in future, or provide a refuge if seas rise.
Sayed-Khaiyum, who is also the Minister of Economy, said Fiji has allocated $50,000 to look into what he called the "peripheral legal issues" about climate change, such as how to grant residency or citizenship.
Another is whether a nation's exclusive economic zone, which stretches 200 nautical miles offshore, shrinks if a remote, outlying island is washed off the map.
So far, he said, there had not been negative reactions among Fiji's population to the idea of receiving migrants.
(Additional reporting by Laurie Goering, Editing by Angus MacSwan)
Our Standards: The Thomson Reuters Trust Principles.
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US » Minnesota » St. Marys Point
Minnesota State Demographics
St. Marys Point, Minnesota Population History 1990 - 2021
St. Marys Point Population By Year
What is the current population of St. Marys Point?
Based on the latest 2022 data from the US census, the current population of St. Marys Point is 349. St. Marys Point, Minnesota is the 14,615th largest city in the US.
What was the peak population of St. Marys Point?
The peak population of St. Marys Point was in 2010, when its population was 368. In 2010, St. Marys Point was the 14,640th largest city in the US; now its fallen to the 14,615th largest city in the US. St. Marys Point is currently 5.2% smaller than it was in 2010.
How quickly is St. Marys Point growing?
St. Marys Point has shrunk 0.0% since the year 2000. St. Marys Point, Minnesota is growing somewhat quickly; it is growing faster than 71% of similarly sized cities since 2000.
Similarly Sized Cities In Minnesota
14507 Lancaster 360 -0.5%
14511 New Munich 360 5.8%
14533 Jeffers 357 -9.6%
14558 Newfolden 355 -1.3%
14573 Bricelyn 353 -6.6%
14591 Russell 351 -5.3%
14615 St. Marys Point 349 0.0%
14631 Vergas 348 12.2%
14639 Garfield 347 57.0%
14650 Darwin 346 25.3%
14704 Long Beach 341 28.1%
14722 Lowry 339 25.5%
14734 Lakeland Shores 338 -4.7%
14609 Coyle, Oklahoma 349 3.5%
14610 Coyote Flats, Texas 349
14611 Edom, Texas 349 25.0%
14612 Herndon, Pennsylvania 349 -8.6%
14613 Lake Bridgeport, Texas 349 -7.1%
14614 Oktaha, Oklahoma 349 6.4%
14615 St. Marys Point, Minnesota 349 0.0%
14616 Templeton, Iowa 349 4.8%
14617 Tennyson, Wisconsin 349 -5.4%
14618 Wales, Utah 349 55.8%
14619 Wattsburg, Pennsylvania 349 -7.4%
14620 Allen, Nebraska 348 -15.1%
14621 Declo, Idaho 348 3.8%
What is the voting age population of St. Marys Point, Minnesota?
The total voting age population of St. Marys Point, Minnesota, meaning US citizens 18 or older, is 260. The voting age population is 53.8% male and 46.2% female.
What percentage of St. Marys Point, Minnesota residents are senior citizens?
According to the latest census statistics, 24.9% of the residents of St. Marys Point are 65 or older.
What are the racial demographics of St. Marys Point, Minnesota?
The racial demographics of St. Marys Point are 97.2% White, 2.2% Asian and 0.6% Two or more races.
What percentage of St. Marys Point, Minnesota residents are below the poverty line?
In St. Marys Point, 3.7% of residents have an income below the poverty line, and the child poverty rate is 9.8%. On a per-household basis, 1.1% of families are below the poverty line in St. Marys Point.
What percentage of St. Marys Point, Minnesota residents are in the labor force?
Among those aged 16 and older, 59.7% of St. Marys Point residents are in the labor force.
What are the education levels among St. Marys Point, Minnesota residents?
Among the adult population 25 years old and over, 98.3% of St. Marys Point residents have at least a high school degree or equivalent, 36.3% have a bachelor's degree and 12.4% have a graduate or professional degree.
What percentage of St. Marys Point, Minnesota residents speak a non-English language at home?
Among St. Marys Point residents aged 5 and older, 4.9% of them speak a non-English language at home. Broken down by language: 2.3% of residents speak Spanish at home, 1.9% speak an Indo-European language, and 0.6% speak an Asian language.
What is the unemployment rate in St. Marys Point, Minnesota?
The unemployment rate in St. Marys Point is 1.3%, which is calculated among residents aged 16 or older who are in the labor force.
What percentage of St. Marys Point, Minnesota residents work for the government?
In St. Marys Point, 5.1% of the residents in the non-military labor force are employed by the local, state and federal government.
What is the median income in St. Marys Point, Minnesota?
The median household income in St. Marys Point is $75,781.
What percentage of housing units are owner-occupied in St. Marys Point, Minnesota?
In St. Marys Point, 81.7% of housing units are occupied by their owners.
What percentage of housing units are rented in St. Marys Point, Minnesota?
Renters occupy 18.3% of housing units in St. Marys Point.
What percentage of St. Marys Point, Minnesota housing units were built before 1940?
Of all the housing units in St. Marys Point, 26.9% of them were build before 1940.
What percentage of St. Marys Point, Minnesota housing units were built after 2000?
In St. Marys Point, 19.4% of the total housing units were built after the year 2000.
What is the median monthly rent in St. Marys Point, Minnesota?
The median gross monthly rent payment for renters in St. Marys Point is $1,281.
What percentage of households in St. Marys Point, Minnesota have broadband internet?
In St. Marys Point, 92.4% of households have an active broadband internet connection.


Chews up any terrain at 360!! Much bigger and more sensation and more acrobatic flips with its 2.4Ghz control system on two sides. A new driving experience!
Batteries: 8 x AA Batteries Included (Car) 1 x 9V Included (Remote).
Measures approximately: 14.00 x 10.50 x 8.00 inches..