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

Lorentz symmetry is one of the most tested symmetries of nature. Nevertheless, it is nowadays believed not to be an exact symmetry. In fact, it has been argued in~\cite{Kos1} that tiny spontaneous breaking of Lorentz symmetry might arise in the context of string theory. Thus, if we add Lorentz-violating terms to the standard model, this new theory, namely the standard-model extension (SME)~\cite{Kos2,Kos3}, can be viewed as the low-energy limit of a physically relevant fundamental theory in which spontaneous Lorentz violation occurs. Many experiments and observations have been conducted to detect physical effects due to the Lorentz violation and some constraints on the Lorentz-violating terms were obtained, as we can see in~\cite{Kos4}.

The most studied part of the SME presents renormalizable operators of mass dimension $d=3$ and $d=4$, consequently, the coefficients contracted with these operators have mass dimension $d=1$ or they are dimensionless, respectively. This part of the SME is known as the minimal SME. However, in the SME there are operators with mass dimension $d \geq 5$, which are nonrenormalizable, contracted with coefficients of mass dimension $d\leq -1$. Although these operators are likely to be specially relevant in searches involving very high energies, providing new corrections, there are few studies on them.

Recently, some studies on higher-derivative Lorentz-violating terms, described by higher mass dimension operator ($d\geq 5$), have been made, such as~\cite{Gam,Alf,Sud,Mye,Bol,Jac,Bol1,Rey,Kos5,Rey2,Mar,Lei,Kos6,Bri}. In particular, the work~\cite{Kos5} on operators of arbitrary mass dimension in the QED extension have shown that the general form of the higher-derivative terms in the photon sector, is given by
\begin{eqnarray}\label{hdd}
S_{(d)}=\int{d^4x}\, \mathcal{K}^{\alpha_1 \alpha_2 \cdots \alpha_d}_{(d)} A_{\alpha_1}\partial_{\alpha_3}\cdots \partial_{\alpha_d} A_{\alpha_2},
\end{eqnarray}
where $d$ is the dimension of the tensor operator and $\mathcal{K}^{\alpha_1 \alpha_2 \cdots \alpha_d}_{(d)}$ has mass dimension $4-d$.

On the one hand, for CPT-even terms, the first four indices of the coefficient $\mathcal{K}^{\alpha_1 \alpha_2 \cdots \alpha_d}_{(d)}$ have the symmetries of the Riemann tensor, and there is total symmetry in the remaining $d-4$ indices. On the other hand, for CPT-odd terms, the coefficient $\mathcal{K}^{\alpha_1 \alpha_2 \cdots \alpha_d}_{(d)}$ is antisymmetric in the three first indices and symmetric in the last $d-3$.

In this work, we study the fermion sector of the quantum electrodynamics (QED) extended by a CPT-odd term, governed by the coefficient $b_\mu$, namely $b_\mu\bar{\psi}\gamma_5\gamma^\mu\psi$. From the fermion sector of this minimal QED extension, at zero temperature, we know that we can generate through radiative corrections the higher-derivative Chern-Simons term,
\begin{eqnarray}\label{hd5}
S_{(5)}=\int{d^4x}\, \mathcal{K}^{\mu \nu \rho \alpha \beta}_{(5)} A_{\mu}\partial_{\rho} \partial_{\alpha}\partial_{\beta} A_{\nu},
\end{eqnarray}
with $\mathcal{K}^{\mu \nu \rho \alpha \beta}_{(5)}\propto \epsilon^{\mu \nu \rho \sigma} b_\sigma g^{\alpha \beta}$, as can be seen by expanding the results in~\cite{Kos3,JacKos} or directly in~\cite{Bon,Mar2}. Our goal here is to calculate the generation of the above term (\ref{hd5}) using the method of derivative expansion \cite{Ait,Fra,Vai,Bab}, as a preparation for the finite temperature analysis, which has never been studied. As a complement, we also address the question of large gauge invariance of this higher-derivative term as well as the conventional Chern-Simons term, both at finite temperature. For issues related to the Chern-Simons term, ${\cal L}_{(3)} =\mathcal{K}^{\mu \nu \rho}_{(3)} A_{\mu}\partial_{\rho} A_{\nu}$, see \cite{Col,Chu,Per,Cha,Ada,And,Bel,Mar3}, and references therein.

It is interesting to mention that the term (\ref{hd5}) arises when the Myers-Pospelov bosonic term \cite{Mye} is radiatively induced from the fermion sector \cite{Mar2}. Therefore, effects of finite temperature on the higher-derivative Chern-Simons term are somehow related to the effects on the Myers-Pospelov term. Thus, studies of this nature deserve to be considered in order to better characterize the higher-derivative Lorentz-violating models.

This paper has the following structure. In Sec.~\ref{zero}, we analyze the fermion sector of the Lorentz-violating QED with the coefficient $b_\mu$, performing radiative corrections and using derivative expansion to induce the higher-derivative Chern-Simons term at zero temperature. In the end of the section, we look for constraints on the coefficient of the induced term imposed by experimental and observational data. In Sec.~\ref{finite}, we keep looking at the same theory of the previous section, but now we are interested on the behavior of the term (\ref{hd5}) at finite temperature. The issue of large gauge invariance is discussed in Sec.~\ref{large}. Finally, we present a summary in Sec.~\ref{summary}.

\section{Radiative correction at zero temperature}
\label{zero}

As our aim is to study the generation of higher-derivative CPT-odd terms (\ref{hd5}), we consider the fermion sector of the Lorentz-violating QED, given by
\begin{eqnarray}\label{lg}
\mathcal{L}=\bar{\psi}(i\slashed{\partial}-m-\slashed{A}-\gamma_5\slashed{b})\psi,
\end{eqnarray}
where the three first terms are the usual ones, and the last and unusual term, which is governed by the coefficient $b^\mu$, is that responsible for the Lorentz and CPT violation.

In order to perform this task, we prefer to use the method of derivative expansion (for details see~\cite{Baz}), which is more suitable to calculate the sum-integrals of the finite-temperature approach.

Then, from (\ref{lg}), the effective action $S_{\mathrm{eff}}$ is defined as
\begin{eqnarray}
Z = \int D \bar{\psi} D \psi e^{i\int d^4 x \bar{\psi}(i\slashed{\partial}-m-\slashed{A}-\gamma_5\slashed{b})\psi} = e^{i S_{\mathrm{eff}}},
\end{eqnarray}
in which
\begin{equation}\label{seff}
S_{\mathrm{eff}}=-i\mathrm{Tr}\ln(\slashed{p}-m- \slashed{A}-\gamma_5 \slashed{b}).
\end{equation}
Now, we can easily rewrite the effective action as $S_{\mathrm{eff}}=S^{(0)}_{\mathrm{eff}}+S^{(1)}_{\mathrm{eff}}$, where $S_\mathrm{eff}^{(0)}=-i \mathrm{Tr} \ln{(\slashed{p}-m- \gamma_5 \slashed{b})}$ and
\begin{eqnarray}\label{seff1}
S^{(1)}_{\mathrm{eff}}=i \mathrm{Tr} \sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{\slashed{p}-m-\gamma_5\slashed{b}} \slashed{A}\right]^n .
\end{eqnarray}
As the term $S^{(0)}_{\mathrm{eff}}$ does not depend on the gauge field, we must focus only on the term $S^{(1)}_{\mathrm{eff}}$. To generate the higher-derivative Chern-Simons term it is necessary to have two contributions of the gauge field. Thus, we take $n=2$ in the expression (\ref{seff1}) and use the following expansion for the fermionic propagator,
\begin{eqnarray} \frac{1}{\slashed{p}-m-\gamma_5\slashed{b}}&=&\frac{1}{\slashed{p}-m}+\frac{1}{\slashed{p}-m}\gamma_5\slashed{b}\frac{1}{\slashed{p}-m}\nonumber\\
&&+\frac{1}{\slashed{p}-m}\gamma_5\slashed{b}\frac{1}{\slashed{p}-m}\gamma_5\slashed{b}\frac{1}{\slashed{p}-m}+\cdots,
\end{eqnarray}
to single out contributions of first order in $b_\mu$. It is worth emphasizing that the higher-derivative Chern-Simons term (and the Myers-Pospelov term) is also induced when we consider contributions of third order in $b_\mu$, as it has been shown in~\cite{Mar2}. Following with the derivative expansion approach \cite{Ait,Fra,Vai,Bab,Baz}, we find
\begin{eqnarray}\label{Sb}
S_{b}^{(1,2)}&=&\frac{i}{2} \mathrm{Tr} [S(p)\gamma^\mu S(p-k)\gamma_5 \slashed{b}S(p-k)\gamma^\nu+S(p)\gamma_5 \slashed{b}S(p)\gamma^\mu S(p-k)\gamma^\nu ]A_\mu A_\nu,
\end{eqnarray}
where $S(p)=(\slashed{p}-m)^{-1}$. Finally, expanding the propagator up to third order in $k_\mu$,
\begin{eqnarray}
S(p-k)&=&S(p)+S(p)\slashed{k}S(p)+S(p)\slashed{k}S(p)\slashed{k}S(p)+S(p)\slashed{k}S(p)\slashed{k}S(p)\slashed{k}S(p)+\cdots,
\end{eqnarray}
we obtain
\begin{eqnarray}\label{S5}
S_{(5)}&=&\frac{i}{2}\int{\frac{d^4k}{(2 \pi)^4}\Pi^{\mu\nu}_{(5)}\, A_\mu(k) A_\nu(-k)},
\end{eqnarray}
with
\begin{eqnarray}\label{pi1}
\Pi^{\mu\nu}_{(5)} &=&\int\frac{d^4 p}{(2\pi)^4} \mathrm{tr}[S(p)\gamma^\mu S(p)\gamma_5 \slashed{b}S(p)\slashed{k} S(p)\slashed{k}S(p)\slashed{k} S(p)\gamma^\nu \nonumber\\
&&+ S(p)\gamma^\mu S(p)\slashed{k} S(p)\gamma_5 \slashed{b}S(p)\slashed{k} S(p)\slashed{k} S(p)\gamma^\nu \nonumber \\
&&+ S(p)\gamma^\mu S(p)\slashed{k} S(p)\slashed{k} S(p)\gamma_5 \slashed{b}S(p)\slashed{k} S(p)\gamma^\nu \nonumber\\
&&+ S(p)\gamma^\mu S(p)\slashed{k} S(p)\slashed{k} S(p)\slashed{k} S(p)\gamma_5 \slashed{b}S(p)\gamma^\nu  \nonumber \\
&&+ S(p)\gamma_5 \slashed{b}S(p)\gamma^\mu S(p)\slashed{k} S(p)\slashed{k} S(p)\slashed{k} S(p)\gamma^\nu],
\end{eqnarray}
where we have calculated the trace over the coordinate and momentum spaces and considered cubic terms of $k_\mu$.

Before evaluating the integrals in~(\ref{pi1}), we first calculate the trace over the Dirac matrices, so that we have
\begin{eqnarray}\label{pi2}
\Pi^{\mu\nu}_{(5)}&=& 4i \epsilon^{\mu \nu \sigma \rho} b_\sigma k_\rho k^2\int{ \frac{d^4 p}{(2\pi)^4}}\frac{1}{(p^2-m^2)^3} - 16i  \epsilon^{\mu \nu \sigma \rho} b_\sigma k_\rho k^\alpha k^\beta \int{ \frac{d^4 p}{(2\pi)^4}}\frac{p_\alpha p_\beta}{(p^2-m^2)^4}\nonumber \\
&&+ 24i m^2 \epsilon^{\mu \nu \sigma \rho}  b_\sigma k_\rho k^2 \int{ \frac{d^4 p}{(2\pi)^4}}\frac{1}{(p^2-m^2)^4} + 24i \epsilon^{\mu \nu \rho \alpha} k_\rho b^\beta k^2 \int{ \frac{d^4 p}{(2\pi)^4}}\frac{p_\alpha p_\beta}{(p^2-m^2)^4}\nonumber \\
&&+ 32i \epsilon^{\mu \nu \rho \alpha }k_\rho (b\cdot k)k^\beta \int{ \frac{d^4 p}{(2\pi)^4}}\frac{p_\alpha p_\beta}{(p^2-m^2)^4} - 128i \epsilon^{\mu \nu \sigma \rho} b_\sigma k_\rho k^\alpha k^\beta \int{ \frac{d^4 p}{(2\pi)^4}}\frac{p_\alpha p_\beta}{(p^2-m^2)^5}\nonumber\\
&&- 128i \epsilon^{\mu \nu \rho \alpha} k_\rho b^\beta k^\delta k^\gamma \int{ \frac{d^4 p}{(2\pi)^4}}\frac{p_\alpha p_\beta p_\delta p_\gamma}{(p^2-m^2)^5}.
\end{eqnarray}
Now, by analyzing the above integrals we observe that, after a simple power counting, we do not need to use regularization schemes because all integrals are convergent. Therefore, we can solve (\ref{pi2}) directly by using well-known solutions in which the higher-derivative Chern-Simons term is given by
\begin{equation}\label{pi0}
\Pi^{\mu\nu}_{(5)}=-\frac{\epsilon^{\mu \nu \sigma \rho} b_\sigma k_\rho k^2}{12 m^2 \pi^2},
\end{equation}
or through the following Lagrangian as
\begin{equation}\label{lg0}
\mathcal{L}_{(5)}=\frac{e^2}{24 m^2 \pi^2} \epsilon^{\mu\nu\rho\sigma} b_\sigma A_\mu \partial_\rho \Box A_\nu,
\end{equation}
with $\mathcal{K}^{\mu \nu \rho \alpha \beta}_{(5)}= \frac{e^2}{24 m^2 \pi^2}\epsilon^{\mu\nu\sigma\rho} b_\sigma g^{\alpha \beta}$, where we have re-instaled the electron charge.

The higher-derivative Chern-Simons term (\ref{lg0}) is also induced through radiative corrections from another CPT-odd term of the SME, namely governed by the coefficient $g^{\mu\nu\rho}$, when $g^{\mu\nu\rho}$ is totally antisymmetric \cite{Mar}.

Numerical estimates for the coefficient $b_\mu$ of the higher-derivative Chern-Simons term can be obtained from bounds on the coefficient $\mathcal{K}^{\mu\nu\rho\alpha\beta}_{(5)}$ of the photon sector (see~\cite{Kos4}, table XIX). From observational data related to the cosmic microwave background (CMB) polarization, we can estimate $b\sim 10^{-24}$. Moreover, from systems related to the astrophysical birefringence, we estimate that the coefficient is $\sim 10^{-36}$. In these estimates we have considered $m$ as being the electron mass, $m\simeq0.5\times10^{-3}$GeV, and $e\simeq10^{-1}$ for the electron charge. In this way, we observe that we have found tiny values for the coefficient $b_\mu$ compatible with maximal sensitivities for the electron sector, table II of Ref.~\cite{Kos4}.

\section{Radiative correction at finite temperature}
\label{finite}

In this section, we are interested in the finite temperature behavior of the higher-derivative term (\ref{pi0}). For this, we take the Eq.~(\ref{pi2}) and change it from Minkowski space to Euclidean space. We would like to emphasize here that the method of derivarive expansion used above, in general, is rather delicate at finite temperature, due to the fact that the limits $k_0\to0$ and $\vec k\to0$ do not commute, arising from the non-analyticity of thermal amplitudes \cite{Das}. In the expression~(\ref{pi2}) we have considered the limits $k_0\to0$ and $\vec k\to0$. Then, by performing the following procedure: $p_0 \rightarrow i p_0$ ($g^{\mu\nu}\rightarrow -\delta^{\mu\nu}$), $d^4p\rightarrow i d^4p_E$, and $p^2\rightarrow -p_0^2-\vec p^2 = -p^2_E$, we obtain
\begin{eqnarray}\label{pi3}
\Pi^{\mu\nu}_{(5)}&=& - 4 \epsilon^{\mu \nu \sigma \rho} b^\sigma_E k^\rho_E k^2_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{1}{(p^2_E + m^2)^3} + 16 \epsilon^{\mu \nu \sigma \rho} b^\sigma_E k^\rho_E k^\alpha_E k^\beta_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^\alpha_E p^\beta_E}{(p^2_E + m^2)^4}\nonumber \\
 &&+ 24 m^2 \epsilon^{\mu \nu \sigma \rho}  b^\sigma_E k^\rho_E k^2_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{1}{(p^2_E + m^2)^4} - 24 \epsilon^{\mu \nu \rho \alpha} k^\rho_E b^\beta_E k^2_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^\alpha_E p^\beta_E}{(p^2_E + m^2)^4}\nonumber \\
 &&- 32 \epsilon^{\mu \nu \rho \alpha }k^\rho_E (b_E\cdot k_E)k^\beta_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^\alpha_E p^\beta_E}{(p^2_E + m^2)^4} - 128 \epsilon^{\mu \nu \sigma \rho} b^\sigma_E k^\rho_E k^\alpha_E k^\beta_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^\alpha_E p^\beta_E}{(p^2_E + m^2)^5} \nonumber \\
 &&+ 128 \epsilon^{\mu \nu \rho \alpha} k^\rho_E b^\beta_E k^\delta_E k^\gamma_E \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^\alpha_E p^\beta_E p^\delta_E p^\gamma_E}{(p^2_E + m^2)^5}.
\end{eqnarray}

In order to effectively tackle the issue of the finite temperature behavior of the above expression, we separate the space and time components of the four-momentum $p^\sigma_E=(p_0,\vec{p})$ as $p^\sigma_E\rightarrow \hat{p}^\sigma +p_0\delta^{\sigma 0}$, so that $ \hat{p}^\sigma=(0,\vec{p})$. Also, due to the symmetry of the integral under spacial rotations, it is possible to use the substitutions
\begin{eqnarray}
\hat{p}^\alpha \hat{p}^\beta \rightarrow \frac{\hat{p}^2}{3}(\delta^{\alpha\beta}-\delta^{\alpha 0} \delta^{\beta 0})
\end{eqnarray}
and
\begin{eqnarray}
\hat{p}^\alpha \hat{p}^\beta \hat{p}^\delta \hat{p}^\gamma &\rightarrow& \frac{\hat{p}^4}{15}[(\delta^{\alpha\beta}-\delta^{\alpha 0}\delta^{\beta 0})(\delta^{\delta\gamma}-\delta^{\delta 0}\delta^{\gamma 0})+(\delta^{\alpha\delta}-\delta^{\alpha 0}\delta^{\delta 0})(\delta^{\beta\gamma}-\delta^{\beta 0}\delta^{\gamma 0})\nonumber\\
&&+(\delta^{\alpha\gamma}-\delta^{\alpha 0}\delta^{\gamma 0})(\delta^{\beta \delta}-\delta^{\beta 0}\delta^{\delta 0})].
\end{eqnarray}
After applying these changes into Eq.~(\ref{pi3}), we find seven different tensorial structures. The first two structures, $\epsilon^{\mu \nu \sigma \rho} k^{\sigma}_E k^{\rho}_E b^0_E k^0_E$ and $\epsilon^{\mu \nu \sigma \rho} k^{\sigma}_E k^{\rho}_E (b_E \cdot k_E)$, obviously vanish by antisymmetry of $\epsilon^{\mu \nu \sigma \rho}$. In addition, the third contribution, $\epsilon^{\mu \nu 0 \sigma} k^{0}_E k^{\sigma}_E (b_E \cdot k_E)$, goes to zero after the integration in the space components of $p_E^\alpha$. Therefore, we are left with four different structures to analyze, namely $\epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E k^2_E$, $\epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E k^2_E$, $\epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E (k^0_E)^2$ and $\epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E (k^0_E)^2$. Thus, we have $\Pi^{\mu\nu}_{(5)}=\Pi^{\mu\nu}_{(a)}+\Pi^{\mu\nu}_{(b)}+\Pi^{\mu\nu}_{(c)}+\Pi^{\mu\nu}_{(c)}$, with the following expressions for these four tensorial structures:
\begin{eqnarray}
\Pi^{\mu\nu}_{(a)}&=& -4 \epsilon^{\mu \nu \sigma \rho}  b^{\sigma}_E k^{\rho}_E k^2_E \int\frac{dp_0}{2 \pi}\left[ \frac{32}{15}I_1(p_0,m)+\frac{32 m^2}{3} I_2(p_0,m)-\frac{10}{3} I_4(p_0,m)\right. \nonumber\\
&&- 6 m^2 I_5(p_0,m)+ I_6(p_0,m)\bigg], \\
\Pi^{\mu\nu}_{(b)}&=& 8 \epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E k^2_E \int\frac{dp_0}{2 \pi} \left[ \frac{16}{15}I_1(p_0,m)-\frac{16}{3}p_0^2 I_2(p_0,m) - I_4(p_0,m)\right. \nonumber \\
&&\left.+ 3p_0^2 I_5(p_0,m)\right], \\
\Pi^{\mu\nu}_{(c)}&=& -16  \epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E  (k^0_E)^2\int\frac{dp_0}{2 \pi}\left[ -\frac{8}{15}I_1(p_0,m)+\frac{8}{3}(p_0^2-m^2)I_2(p_0,m) \right. \nonumber\\
&&\left.+8 p_0^2 m^2 I_3(p_0,m) + \frac{1}{3} I_4(p_0,m)-p_0^2I_5(p_0,m)\right], \\
\Pi^{\mu\nu}_{(d)}&=& -128\epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E (k^0_E)^2 \int\frac{dp_0}{2 \pi}\left[ \frac{1}{5}I_1(p_0,m)-2 p_0^2 I_2(p_0,m)+p_0^4 I_3(p_0,m)\right],
\end{eqnarray}
where the integrals $I_{1,2,3,4,5,6}(p_0,m)$ are given by
\begin{eqnarray}
I_{1,2,3}(p_0,m)&=& \int\frac{d^3p_E}{(2 \pi)^3}\frac{\alpha_{1,2,3}}{(\vec{p}^2+p_0^2+m^2)^5}, \\
I_{4,5}(p_0,m)&=& \int\frac{d^3p_E}{(2 \pi)^3}\frac{\alpha_{4,5}}{(\vec{p}^2+p_0^2+m^2)^4}, \\
I_6(p_0,m)&=& \int\frac{d^3p_E}{(2 \pi)^3}\frac{1}{(\vec{p}^2+p_0^2+m^2)^3},
\end{eqnarray}
with $\alpha_1=\vec{p}^4,\alpha_2=\alpha_4=\vec{p}^2$ and $\alpha_3=\alpha_5=1$.

Now, we calculate the integrals over the space components $\vec p$. This can be done without using regularization schemes, because all the integrals are convergent. Then, we find
\begin{eqnarray}
\Pi^{\mu\nu}_{(a)}&=& \frac{m^2}{ 8\pi}\epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E  k^2_E \left[\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}}-\frac{2}{3m^2}\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{3}{2}}\right], \nonumber\\
&&+\frac{1}{12\pi}\epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E  k^2_E \int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{3}{2}} \\
\Pi^{\mu\nu}_{(b)} &=& -\frac{m^2}{8 \pi}\epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E k^2_E \left[\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}}-\frac{2}{3m^2}\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{3}{2}}\right],\\
\Pi^{\mu\nu}_{(c)}&=& \frac{ m^2}{ \pi}\epsilon^{\mu \nu \sigma \rho} b^{\sigma}_E k^{\rho}_E (k^0_E)^2\left[\frac{5 m^2}{4} \int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{7}{2}}-\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}}\right], \\
\Pi^{\mu\nu}_{(d)}&=& -\frac{m^2}{\pi}\epsilon^{\mu \nu 0 \rho} b^{0}_E k^{\rho}_E (k^0_E)^2 \left[\frac{5 m^2}{4} \int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{7}{2}}-\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}}\right].
\end{eqnarray}
Note that $\Pi^{\mu\nu}_{(b)}$ and $\Pi^{\mu\nu}_{(d)}$ cancel the corresponding time components $b_E^0$ of $\Pi^{\mu\nu}_{(a)}$ and $\Pi^{\mu\nu}_{(c)}$, respectively. Therefore, by summing all expressions, we obtain
\begin{eqnarray}\label{piBC}
\Pi^{\mu\nu}_{(5)} &=& -\frac{1}{m^2}\epsilon^{\mu \nu \sigma \rho } b^{\sigma}_E k^{\rho}_E  k^2_E B(m) - \frac{1}{m^2}\epsilon^{\mu \nu i \rho} b^{i}_E k^{\rho}_E k^2_E C_1(m)\nonumber\\
&&- \frac{1}{m^2}\epsilon^{\mu \nu i \rho} b^{i}_E  k^{\rho}_E (k^0_E)^2 C_2(m),
\end{eqnarray}
where
\begin{subequations}
\begin{eqnarray}
B(m) &=& -\frac{m^2}{12 \pi} \int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{3}{2}},\\
C_1(m) &=& \frac{m^2}{12\pi}\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{3}{2}} - \frac{m^4}{8 \pi}\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}}, \\
C_2(m) &=& \frac{m^4}{\pi}\int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{5}{2}} - \frac{5 m^6}{4 \pi} \int\frac{dp_0}{2 \pi}(p_0^2+m^2)^{-\frac{7}{2}},
\end{eqnarray}
\end{subequations}
with $i=1,2,3$.

The finite temperature behavior of the above terms can be obtained by using the Matsubara formalism. For this, we write $p_0=(n +1/2) \frac{2\pi}{\beta}$, which are the Matsubara frequencies, and change the integral over $p_0$ into a sum, $\frac{1}{2\pi}\int{dp_0}\rightarrow \frac{1}{\beta}\sum_n$, where $T=\beta^{-1}$ is the temperature of the system. Assuming that $\xi=\frac{\beta m}{2 \pi}$, we can write $B(m)\rightarrow B(\xi)$, $C_1(m)\rightarrow C_1(\xi)$, and $C_2(m)\rightarrow C_2(\xi)$, so that
\begin{subequations}\label{BCC1}
\begin{eqnarray}
\label{B}B(\xi) &=& -\frac{\xi^2}{24 \pi^2}\sum_n{[(n+1/2)^2+\xi^2]^{-3/2}},\\
\label{C1}C_1(\xi) &=& \frac{\xi^2}{24\pi^2}\sum_n{[(n+1/2)^2+\xi^2]^{-3/2}} - \frac{\xi^4}{16\pi^2} \sum_n{[(n+1/2)^2+\xi^2]^{-5/2}} \\
\label{C2}C_2(\xi) &=& \frac{\xi^4}{2 \pi^2} \sum_n{[(n+1/2)^2+\xi^2]^{-5/2}} - \frac{5 \xi^6}{8\pi^2}\sum_n{[(n+1/2)^2+\xi^2]^{-7/2}}.
\end{eqnarray}
\end{subequations}

As the above summations are all convergent, they can be easily numerically evaluated. Nevertheless, the asymptotic limits, zero and high temperatures, are not easy to obtain. One way to do this is to convert the sums into integrals, by using the expression~\cite{Ford}
\begin{eqnarray}\label{sum}
\sum\limits_{n}[(n+b)^2+a^2]^{-\lambda}&=&\frac{\sqrt{\pi}\Gamma(\lambda - 1/2)}{\Gamma(\lambda)(a^2)^{\lambda-1/2}}+ 4\sin(\pi \lambda) f_\lambda (a,b),
\end{eqnarray}
where
\begin{eqnarray}\label{intsum}
f_\lambda(a,b)= \int\limits_{|a|}^{\infty} \frac{dz}{(z^2-a^2)^\lambda} Re \left( \frac{1}{e^{2\pi(z+ib)}-1} \right),
\end{eqnarray}
which is valid for $Re\,\lambda < 1$, asides from the poles at $\lambda = 1/2, -1/2, -3/2, \cdots.$ However, by analyzing the equations (\ref{B}), (\ref{C1}), and (\ref{C2}), we have the values $\lambda=3/2$, $\lambda=5/2$ and $\lambda=7/2$, i.e. they are all out of range of validity. Luckily, we can use a recurrence relation, given by
\begin{eqnarray}\label{rr}
f_{\lambda -1}(a,b)=\!-\frac{1}{2a^2}\frac{2\lambda-5}{\lambda-2}f_{\lambda-2}(a,b)-\frac{1}{4a^2}\frac{1}{(\lambda-3)(\lambda-2)}\frac{\partial^2}{\partial b^2}f_{\lambda-3}(a,b),
\end{eqnarray}
in order to decrease these values of $\lambda$. In this way, for $\lambda=3/2$ we must use the recurrence relation once, for $\lambda=5/2$  twice, and for $\lambda=7/2$ thrice. We must also write $\lambda=3/2\to D/2$, $\lambda=5/2\to D/2+1$, and $\lambda=7/2\to D/2+2$, to avoid the poles $\lambda=1/2,-1/2$ in the first term of the solution (\ref{sum}), so that the expressions (\ref{BCC1}) become
\begin{subequations}
\begin{eqnarray}
B(\xi) &=& -\frac{m^{3-D}}{24\pi^2}\xi^{D-1}\sum_n{[(n+1/2)^2+\xi^2]^{-\frac D2}},\\
C_1(\xi) &=& \frac{m^{3-D}}{24\pi^2}\xi^{D-1}\sum_n{[(n+1/2)^2+\xi^2]^{-\frac D2}} \nonumber \\
&&- \frac{m^{3-D}}{16\pi^2}\xi^{D+1}\sum_n{[(n+1/2)^2+\xi^2]^{-\frac D2-1}}, \\
C_2(\xi) &=&  \frac{m^{3-D}}{2 \pi^2}\xi^{D+1}\sum_n{[(n+1/2)^2+\xi^2]^{-\frac D2-1}} \nonumber\\
&&- \frac{5m^{3-D}}{8\pi^2}\xi^{D+3}\sum_n{[(n+1/2)^2+\xi^2]^{-\frac D2-2}}.
\end{eqnarray}
\end{subequations}
Then, by using the recurrence relation (\ref{rr}) and considering the sum (\ref{sum}), after we take the limit $D\to3$, we obtain
\begin{eqnarray}\label{pift}
\Pi^{\mu\nu}_{(5)} &=&  \frac{1}{m^2}\epsilon^{\mu \nu \sigma \rho } b_{\sigma} k_{\rho} k^2 B(\xi) + \frac{1}{m^2} \epsilon^{\mu \nu i \rho } b_{i} k_{\rho}  k^2 C_1(\xi) \nonumber\\
&&+\frac{1}{m^2}\epsilon^{\mu \nu i \rho} b_{i} k_{\rho} k_0^2 C_2(\xi),
\end{eqnarray}
with
\begin{subequations}\label{BCC2}
\begin{eqnarray}
\label{G}B(\xi) &=& \frac{1}{6}\int\limits_{|\xi|}^{\infty}dz \sqrt{z^2-\xi^2}\tanh(\pi z)\mathrm{sech}^2(\pi z)-\frac{1}{12 \pi^2},\\
\label{H1}C_1(\xi) &=&\frac{\xi^2}{12}\int\limits_{|\xi|}^{\infty}dz\frac{ \tanh(\pi z)\mathrm{sech}^2(\pi z)}{ \sqrt{z^2-\xi^2}}, \\
\label{H2}C_2(\xi) &=& \frac{\pi^2 \xi^2}{6}\int\limits_{|\xi|}^{\infty}dz\sqrt{z^2-\xi^2}\mathrm{sech}^5(\pi z)[\sinh(3\pi z)-11\sinh(\pi z)],
\end{eqnarray}
\end{subequations}
where we have returned to the Minkowski space.

By analyzing the above equations we observe that when $T\to0$ ($\xi\to\infty$) all the integrals vanish, so that $\Pi^{\mu\nu}_{(5)}$ goes to $-\frac{1}{12 \pi^2 m^2}\epsilon^{\mu \nu \sigma \rho } b_{\sigma} k_{\rho} k^2$. This is in fact the result of the temperature zero, obtained previously in~(\ref{pi0}). On the other hand, when $T\to\infty$ ($\xi\to0$) the expression (\ref{G}) goes to zero, as well as the expressions (\ref{H1}) and (\ref{H2}). This happens mainly because all the summations in (\ref{BCC1}) are strongly suppressed by the temperature, thus suggesting that in general the operators of mass dimension $d\ge5$ vanish in the limit of high temperature.

The plot of the functions $B(\xi)$, $C_1(\xi)$, and $C_2(\xi)$, which can be numerically calculated from either expressions (\ref{BCC1}) or (\ref{BCC2}), are presented in Figs.~\ref{fig1},~\ref{fig2}, and~\ref{fig3}. It is interesting to note here that these functions are the same as those obtained in the context of the coefficient $g^{\mu\nu\rho}$ \cite{Lei}. This is because under a certain fermion field redefinition \cite{Kos2,McD} the totally antisymmetric component of $g^{\mu\nu\rho}$ is completely absorbed into coefficient $b^\mu$.

\begin{figure}[h]
\includegraphics[scale=.8]{B}
\caption{Plot of the function $B(\xi)$} \label{fig1}
\end{figure}
\begin{figure}[h]
\includegraphics[scale=.7]{C1}
\caption{Plot of the function $C_1(\xi)$} \label{fig2}
\end{figure}
\begin{figure}[h]
\includegraphics[scale=.7]{C2}
\caption{Plot of the function $C_2(\xi)$} \label{fig3}
\end{figure}

\section{Large gauge invariance}
\label{large}

Let us now briefly discuss the question of large gauge invariance, at finite temperature, in the context of Lorentz- and CPT-violating QED. This issue has been extensively studied in three-dimensional Chern-Simons theories \cite{Dun,Des1,Des2,Fos1,Fos2,Fos3,Bra1,Bra2}. In our case, we want to examine the invariance of the higher-derivative Chern-Simons term (\ref{hd5}) under large gauge transformation, as well as the four-dimensional Chern-Simons term, namely,
\begin{eqnarray}\label{hd3}
S_{(3)}=\int{d^4x}\, \mathcal{K}^{\mu \nu \rho}_{(3)} A_{\mu}\partial_{\rho} A_{\nu},
\end{eqnarray}
with $\mathcal{K}^{\mu \nu \rho}_{(3)}\propto \epsilon^{\mu \nu \rho \sigma} b_\sigma$, where both are dynamically generated by the effective action (\ref{seff}). It would be interesting to consider the other terms with operators of mass dimension $d=7,9,\cdots$, however, this is a challenging task. In fact, in the literature we find only models with Lorentz-violating operators of mass dimension $d=5$, for CPT-odd terms. In particular, we can argue that, as ${\cal K}_{(d)}\sim M_{Pl}^{4-d}$, these operators with $d\geq 7$ are highly suppressed by the mass Planck $M_{Pl}$ and, therefore, are even more negligible.

To perform this investigation, we continue to use the method of derivative expansion, however, in a particular gauge field background, where we have a vanishing electric field and a time-independent magnetic field, as follows
\begin{equation}\label{bg}
A_0=A_0(t), \hspace{0.5cm} A_i=A_i(x,y,z).
\end{equation}
It is worth mentioning that this specific choice of gauge background is consistent with the static limit, as shown in~\cite{Bra1}, although it is not a static background. The point here is that in this static limit we can avoid the non-analyticity of thermal amplitudes, which can appear when the effective action is expanded in powers of derivatives. Furthermore, in this background (\ref{bg}) we believe that the effective action can be calculated exactly.

Finally, by considering the appropriate gauge transformation \cite{Fos1,Fos2},
\begin{equation}
A_\mu \to A_\mu + \partial_\mu \Omega, \hspace{0.5cm} \Omega(t)=\left(-\int_0^t+\frac t\beta\int_0^\beta\right)dt' A_0(t'),
\end{equation}
we obtain
\begin{equation}
A_0 \to \frac{a_0}\beta = \frac1\beta \int_0^\beta dt\, A_0(t), \hspace{0.5cm} A_i \to A_i(x,y,z),
\end{equation}
so that under large gauge transformation,
\begin{equation}\label{lgt}
a_0 \to a_0 + 2\pi n,
\end{equation}
where $n$ is an integer. Then, by taking into account these constraints, we can rewrite the effective action (\ref{seff}) in the Euclidean space as
\begin{equation}\label{seffe}
S_{\mathrm{eff}}=-\sum_n\mathrm{Tr}\ln(\vec{\slashed{p}}+\tilde\omega_n\gamma^0+m-\vec{\slashed{A}}-\gamma_5 \slashed{b}_E),
\end{equation}
in which we have introduced the definitions: $\vec{\slashed{p}}=p^i\gamma^i$, $\slashed{b}_E=b^0\gamma^0+b^i\gamma^i$,  and
\begin{equation}
\tilde\omega_n=\omega_n-\frac{a_0}\beta=(n +1/2) \frac{2\pi}{\beta}-\frac{a_0}\beta.
\end{equation}
Now, following the procedure performed in Ref.~\cite{Fos3}, let us first derivate the effective action (\ref{seffe}) with respect to $a_0$, and later consider only the linear term in $A^i$, so that we can write
\begin{equation}
\frac{\partial S_{\mathrm{eff}}^{(1)}}{\partial a_0} = \sum_n \mathrm{Tr}\, \frac{1}{\vec{\slashed{p}}+\tilde\omega_n\gamma^0+m-\gamma_5 \slashed{b}_E}\vec{\slashed{A}}  \frac{1}{\vec{\slashed{p}}+\tilde\omega_n\gamma^0+m-\gamma_5 \slashed{b}_E} \gamma^0.
\end{equation}
As we are interested only in contributions of first order in $b_\mu$, we single out the expression
\begin{equation}\label{Sb1}
\frac{\partial S_b^{(1)}}{\partial a_0} = \sum_n \mathrm{Tr}\, [S_E(\vec p)\gamma^i S_E(\vec p-i\vec\partial)\gamma_5 \slashed{b}_E S_E(\vec p-i\vec\partial)\gamma^0+S_E(\vec p)\gamma_5 \slashed{b}_E S_E(\vec p)\gamma^i S_E(\vec p-i\vec\partial)\gamma^0 ]A^i,
\end{equation}
where $S_E(\vec p)=(\vec{\slashed{p}}+\tilde\omega_n\gamma^0+m)^{-1}$. Note that the above equation is similar to Eq.~(\ref{Sb}), and therefore the calculations of Eq.~(\ref{Sb1}) are similar to those that have been performed previously.

In order to generate the Lorentz-violating Chern-Simons term (\ref{hd3}), we expand the propagator $S_E(\vec p-i\vec\partial)$ up to first order in $\partial^k$, so that we obtain
\begin{equation}
\frac{\partial S_{(3)}}{\partial a_0} = i \int d^3x\, \Pi_{(3)}^{i0} A^i
\end{equation}
with
\begin{eqnarray}\label{pi1}
\Pi_{(3)}^{i0} &=& \sum_n \int\frac{d^3 p_E}{(2\pi)^3} \mathrm{tr}[S_E(\vec p)\gamma_5 \slashed{b}_E S_E(\vec p)\gamma^i S_E(\vec p)\vec{\slashed{\partial}} S_E(\vec p)\gamma^0 \nonumber\\
&&+ S_E(\vec p)\gamma^i S_E(\vec p)\vec{\slashed{\partial}} S_E(\vec p)\gamma_5 \slashed{b}_E S_E(\vec p)\gamma^0 \nonumber \\
&&+ S_E(\vec p)\gamma^i S_E(\vec p)\gamma_5 \slashed{b}_E S_E(\vec p)\vec{\slashed{\partial}} S_E(\vec p)\gamma^0].
\end{eqnarray}
Hereafter, we calculate the trace over the Dirac matrices and thus find the expression
\begin{eqnarray}
\Pi_{(3)}^{i0} &=& -12i \epsilon^{i0jk} b^j \partial^k \sum_n\int\frac{d^3 p_E}{(2 \pi)^3}\frac{1}{(\vec p^2 + \tilde\omega_n^2 + m^2)^2} \nonumber\\
&&-16i \epsilon^{0jkl} b^j \partial^k \sum_n\int\frac{d^3 p_E}{(2 \pi)^3}\frac{p^ip^l}{(\vec p^2 + \tilde\omega_n^2 + m^2)^3} \nonumber\\
&&+16i \epsilon^{ijk0} b^j \partial^k \sum_n\int\frac{d^3 p_E}{(2 \pi)^3}\frac{\tilde\omega_n^2}{(\vec p^2 + \tilde\omega_n^2 + m^2)^3} \nonumber\\
&&+16i \epsilon^{i0jl} b^j \partial^k \sum_n\int\frac{d^3 p_E}{(2 \pi)^3}\frac{p^k p^l}{(\vec p^2 + \tilde\omega_n^2 + m^2)^3},
\end{eqnarray}
where we have moved $\gamma_5$ to the very end of every expression, before calculating the trace. The next step is to calculate the space integral, by promoting the three- to D-dimension and using the substitution $p^ip^l\to \vec p^2\delta^{il}/D$, which leads to the following result:
\begin{eqnarray}\label{Pi3}
\Pi_{(3)}^{i0} &=&  -i\epsilon^{i0jk} b^j \partial^k 2^{1-D}m\,\pi^{-D/2}(\mu^2)^{\frac32-\frac D2} \Gamma\left(2-\frac D2\right)\nonumber\\
&&\times \sum_n\left[\frac{(D-3)}{(\tilde\omega_n^2+m^2)^{2-\frac D2}}-\frac{(D-4)m^2}{(\tilde\omega_n^2+m^2)^{3-\frac D2}}\right].
\end{eqnarray}
To perform the above summations, we cannot readily take the limit $D\to3$ in the expression because the first sum exhibits singularity. Therefore, we must use the Eq.~(\ref{sum}) in order to isolate the corresponding divergent term, which becomes finite due to the factor $(D-3)$ in the numerator. For the second sum, when $D\to3$, we have $\lambda \to 3/2$, i.e. we also want to use here the recurrence relation (\ref{rr}). Following these procedures, we arrive at
\begin{equation}
\Pi_{(3)}^{i0} = -i\epsilon^{i0jk} b^j \partial^k\,F'(\xi,a_0),
\end{equation}
where
\begin{equation}
F'(\xi,a_0) = \int_{|\xi|}^\infty dz \sqrt{z^2-\xi^2}\,\sinh(2\pi z)\frac{2\cos(a_0)\cosh(2\pi z)-\cos(2a_0)+3}{[\cos(a_0)+\cosh(2\pi z)]^3},
\end{equation}
or, in other words,
\begin{equation}\label{F}
F'(\xi,a_0) = \frac{\partial\;}{\partial a_0}F(\xi,a_0) = \frac{\partial\;}{\partial a_0} \int_{|\xi|}^\infty dz \sqrt{z^2-\xi^2}\,\frac{2\sin(a_0)\sinh(2\pi z)}{[\cos(a_0)+\cosh(2\pi z)]^2}.
\end{equation}
Therefore, after returning to the Minkowski space, we obtain the Lorentz-violating Chern-Simons action,
\begin{equation}\label{S3lgt}
S_{(3)} = \int d^3x\, \epsilon^{i0jk} b_j \partial_k A_i\, F(\xi,a_0),
\end{equation}
with
\begin{equation}
F(\xi,a_0) = \int_{|\xi|}^\infty dz \sqrt{z^2-\xi^2}\,\frac{2\sin(a_0)\sinh(2\pi z)}{[\cos(a_0)+\cosh(2\pi z)]^2},
\end{equation}
which, as expected, is invariant under large gauge transformation (\ref{lgt}). Note that a perturbative expansion in terms of $a_0$ yields the usual perturbative result,
\begin{equation}
F(\xi,a_0) = a_0 \int_{|\xi|}^\infty dz \sqrt{z^2-\xi^2}\,\mathrm{sech}^2(\pi z)\tanh(\pi z) + {\cal O}(a_0^2),
\end{equation}
that it was obtained previously in~\cite{Cer}, and in~\cite{Lei} from the derivative term $\bar\psi {i\over2}g^{\kappa\lambda\mu}\sigma_{\kappa\lambda}(\partial_\mu+iA_\mu)\psi$, governed by the coefficient $g^{\kappa\lambda\mu}$.

Let us finally consider the generation of the higher-derivative Chern-Simons term (\ref{hd5}), by expanding the propagator $S_E(\vec p-i\vec\partial)$, in Eq.~(\ref{Sb1}), up to three order in $\partial^k$. As we have mentioned, the corresponding action is similar to Eq.~(\ref{S5}), however, given by the expression
\begin{equation}
\frac{\partial S_{(5)}}{\partial a_0} = i \int d^3x\, \Pi_{(5)}^{i0} A^i,
\end{equation}
where
\begin{eqnarray}
\Pi^{i0}_{(5)}&=& 4i \epsilon^{i0jk} b^j \partial^k \vec\nabla^2 \sum_n \int{\frac{d^3 p_E}{(2 \pi)^3}}\frac{1}{(\vec p^2 + \tilde\omega_n^2 + m^2)^3} \nonumber\\
&&- 16i \epsilon^{i0jk} b^j \partial^k \partial^l \partial^m \sum_n \int{\frac{d^3 p_E}{(2 \pi)^3}}\frac{p^l p^m}{(\vec p^2 + \tilde\omega_n^2 + m^2)^4}\nonumber \\
 &&- 24i m^2 \epsilon^{i0jk}  b^j \partial^k \vec\nabla^2 \sum_n \int{\frac{d^3 p_E}{(2 \pi)^3}}\frac{1}{(\vec p^2 + \tilde\omega_n^2 + m^2)^4} \nonumber\\
 &&- 24i \epsilon^{i0jk} b^l \partial^k \vec\nabla^2 \sum_n \int{\frac{d^3 p_E}{(2 \pi)^3}}\frac{p^k p^l}{(\vec p^2 + \tilde\omega_n^2 + m^2)^4}\nonumber \\
&&+ 128i \epsilon^{i0jk} b^j \partial^k \partial^l \partial^m \sum_n \int{\frac{d^3 p_E}{(2 \pi)^3}}\frac{p^l p^m}{(\vec p^2 + \tilde\omega_n^2 + m^2)^5} \nonumber \\
 &&- 128i \epsilon^{i0kl} b^j \partial^k  \partial^m \partial^n \sum_n \int{\frac{d^4 p_E}{(2 \pi)^4}}\frac{p^j p^l p^m p^n}{(\vec p^2 + \tilde\omega_n^2 + m^2)^5}.
\end{eqnarray}
In the above expression we must also use the substitution $p^j p^l p^m p^n \to \vec p^4(\delta^{jl}\delta^{mn}+\delta^{jm}\delta^{ln}+\delta^{jn}\delta^{ml})/[D(D+2)]$, so that after we evaluate the space integrals and the sums, we obtain
\begin{equation}
\Pi^{i0}_{(5)} = -\frac{i}{m^2}\epsilon^{i0jk} b^j \partial^k \vec\nabla^2 G'(\xi,a_0),
\end{equation}
with
\begin{equation}
G'(\xi,a_0) = \frac1{12}\int_{|\xi|}^\infty dz \frac{2z^2-\xi^2}{\sqrt{z^2-\xi^2}}\,\sinh(2\pi z)\frac{2\cos(a_0)\cosh(2\pi z)-\cos(2a_0)+3}{[\cos(a_0)+\cosh(2\pi z)]^3} -\frac{1}{12 \pi^2},
\end{equation}
which can be written as
\begin{equation}
G'(\xi,a_0) = \frac{\partial\;}{\partial a_0}G(\xi,a_0) = \frac{\partial\;}{\partial a_0} \left(\frac1{12}\int_{|\xi|}^\infty dz \frac{2z^2-\xi^2}{\sqrt{z^2-\xi^2}}\,\frac{2\sin(a_0)\sinh(2\pi z)}{[\cos(a_0)+\cosh(2\pi z)]^2}-\frac{a_0}{12 \pi^2}\right).
\end{equation}
Then, the higher-derivative Chern-Simons action takes the form
\begin{equation}\label{S5lgt}
S_{(5)} = \int d^3x\, \epsilon^{i0jk} b_j \partial_k \vec\nabla^2A_i\, G(\xi,a_0),
\end{equation}
where
\begin{equation}
G(\xi,a_0) = \frac1{12}\int_{|\xi|}^\infty dz \frac{2z^2-\xi^2}{\sqrt{z^2-\xi^2}}\,\frac{2\sin(a_0)\sinh(2\pi z)}{[\cos(a_0)+\cosh(2\pi z)]^2}-\frac{a_0}{12 \pi^2},
\end{equation}
which is also invariant under large gauge transformation (\ref{lgt}). Observe that by expanding the above expression, up to first order in $a_0$, Eqs.~(\ref{G}) and (\ref{H1}) are readily recovered. This complete our analysis.

\section{Summary}
\label{summary}

In this work, we study the radiative generation of the higher-derivative Lorentz-violating Chern-Simons term, at zero temperature and at finite temperature. At zero temperature, we use the method of derivative expansion, as a preparation for the finite temperature analysis. We also present numerical estimations for the coefficient $b^\mu$, which are compatible with maximal sensitivities for the electron sector. At finite temperature, by using the Matsubara formalism, we observe that the higher-derivative Chern-Simons term vanishes when the temperature goes to infinity. This behavior has already been observed in the context of the coefficient $g^{\mu\nu\rho}$ \cite{Lei}, suggesting, therefore, that in general the operators of mass dimension $d\ge5$ vanish in the limit of $T\to\infty$. In the analysis of large gauge invariance, we compute the exact effective actions, Eqs.~(\ref{S3lgt}) and (\ref{S5lgt}), at finite temperature and in a specific gauge field background. We observe that, in fact, the conventional Chern-Simons (\ref{S3lgt}) and the higher-derivative Chern-Simons (\ref{S5lgt}) actions are invariant under large gauge transformation.

\vspace{.5cm}
{\bf Acknowledgements.} This work was supported by Conselho Nacional de Desenvolvimento Cient\'{\i}fico e Tecnol\'{o}gico (CNPq).
\vspace{.5cm}

redpajama_set_name: RedPajamaArXiv


text:
Appeals Civil Rights Criminal
Teen Sniper Pushes Supremes to Chuck Life Sentences
October 16, 2019 October 16, 2019 TIM RYAN
sentencing, Shooting, teens, U.S. Supreme Court
Lee Boyd Malvo enters a courtroom in Virginia Beach, Va., in 2003. (Photo by Davis Turner/AP)
WASHINGTON (CN) – Operating out of a blue 1990 Chevy Caprice with a hole drilled in the trunk to accommodate a rifle's barrel, the infamous Beltway Snipers terrorized the nation's capital for three weeks in the fall of 2002 with a string of shootings that left 10 people dead and three others injured.
Virginia executed the older of the two snipers, John Allen Muhammad, in 2009, while Lee Boyd Malvo is serving four life sentences without the possibility of parole.
On Wednesday, with his spree nearly 20 years in the past, the Jamaican-born Malvo urged the Supreme Court for a chance at freedom.
Malvo had been 15, abandoned by his mother and not in contact with his father for years, when he met Muhammad, then 39. The shootings erupted two years later, targeting public places like gas stations and parking lots, and causing school lockdowns and widespread fear.
Years after Malvo and Muhammad's convictions, however, the Supreme Court decided Miller v. Alabama, which held mandatory life sentences without the possibility of parole for juvenile offenders violate the Eighth Amendment, and Montgomery v. Louisiana, which gave Miller retroactive effect to cases on collateral review.
Virginia seeks a reversal now after the Fourth Circuit upheld a decision granting Malvo's request for a new sentence in light of Miller and Montgomery.
Arguing before the justices this morning, Virginia Solicitor General Toby Heytens said Malvo was reading into the precedent a principle the court never endorsed, mainly that the prohibition on life without parole sentences applies to those handed down in discretionary as well as mandatory sentencing schemes.
After all, the defendants in both of those cases received mandatory sentences, while Malvo's was not mandatory under Virginia law. That gave the sentencing judge the chance to consider Malvo's youth when crafting the appropriate sentence and put the state on the right side of the court's precedents.
"The court said that the sentencer was deprived of 'the opportunity to consider youth and its mitigating factors' and instead that the states at issue in that case had required that all children receive life without parole sentences," Heytens said. "As a matter of Virginia state law, that was not true here."
Danielle Spinelli, an attorney for Malvo at Wilmer Hale, argued the two precedents require more work than Heytens was letting on. She said they require trial court judges to consider the particular circumstances of the young person they are sentencing and only allows them to hand down sentences of life without parole to those who are "permanently incorrigible."
She also noted the only other sentence Malvo could have received was the death penalty and that there is no evidence the judge ever considered giving Malvo anything less than life without parole.
"The notion that somehow Miller was satisfied by the theoretical opportunity to consider youth, when it wasn't actually considered, simply can't be squared with the language of Miller itself or the language and reasoning of Montgomery," Spinelli said.
Justice Brett Kavanaugh was the most active questioner among the justices on Wednesday, appearing as a possible swing vote in the case.
Kavanaugh wondered if simply giving judges discretion not to impose mandatory life sentences without the possibility of parole is enough to satisfy the principles found in Miller and Montgomery.
"Why is something more procedurally not required?" Kavanaugh asked. "We know a finding of fact is explicitly ruled out by Montgomery and that's very important. But why isn't something more than just a discretionary sentencing regime necessary?"
He also pressed Spinelli on what exactly her proposal would require of judges, as most do not explicitly state all the factors they consider when making their sentencing decisions. He wondered if it would be enough to presume that the issue of a defendant's youth was considered so long as the defense counsel raised it during sentencing.
"There are lots of state cases and federal cases that say, so long as the issue has been raised, we assume the judge 'considered it,'" Kavanaugh said. "Now if that's true, and you can disagree with that, but if that's true, doesn't a discretionary regime where the argument can be raised necessarily satisfy Miller and Montgomery's requirement of consideration."
Malvo confessed to being the triggerman on 10 shootings when police tracked down the men and their Caprice, but later walked that back when testifying against Muhammad in his trial in Maryland.
A U.S. citizen more than twice Malvo's age, Muhammad had met Malvo after he lost custody of his three children but opted instead to flee with them to Antigua. He brought Malvo with him upon returning to the United States in 2001 and started "training" the boy for the shooting rampage after his children were returned to his ex-wife.
Muhammad strictly controlled Malvo's diet, taught him how to shoot a gun, trained him in military tactics and educated him about his beliefs — a mix of the teachings of the Nation of Islam and his own "violent" ideology, according to court briefs.
In 2002, Muhammad and Malvo trekked to Washington, D.C., on a "mission" to get Muhammad's kids back from his ex-wife. Muhammad's plan was to commit a series of random shootings with the goal of getting $10 million from the government, which he would use to establish a "utopian society" for black children.
Evidence later tied Malvo and Muhammad to a string of earlier shootings and crimes in other states. In addition to his life sentences in Virginia, Malvo separately received six life sentences in Maryland.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Peachy Places Around Perry, Georgia
Peach In Perry
Peachy places around Perry: Restaurants, events, museums, and shopping near FMCA's 99th International Convention and RV Expo.
By Peggy Jordan, Associate Editor
Location, location, location. It's a big factor when figuring a property's value, and it helps explain why Perry, Georgia, is super valuable. Situated along Interstate 75 in the center of the state, the city is home to the Georgia National Fairgrounds & Agricenter, where the state fair is held each year.
At the Go Fish Education Center in Perry, Georgia, kids and adults learn about fish and angling.
It's also where FMCA has held 10 international conventions since 1996. And where the association's 99th International Convention and RV Expo, dubbed "Peachy In Perry," will take place March 13 through 16, 2019.
FMCA has set a standard for putting on fantastic conventions. RVers can explore new RV exhibits and RV-related items for sale; attend seminars and craft-making sessions; take in live entertainment; and, best of all, make new friends who share the RV lifestyle. (Turn to page 62 in this issue to read about the entertainment lineup that has been scheduled for the convention.)
When it's time to check out the town and beyond, the following information should acquaint you with what awaits.
Perry's Places
Perry (population 16,000) has a cute downtown, with shops and boutiques lining its streets. A popular new store is Mossy Creek Natural (915 Carroll St.), which has locally made bath and body products. The wares include deliciously scented soap in bar and in liquid form; skincare and makeup items; bohemian-style clothing; and jewelry.
Two events will take place in downtown Perry Saturday, March 16, the final day of the FMCA convention. First is the 9:00 a.m. opening of the Perry Farmers' Market, full of locally made items, fresh eggs, and produce. Another is the first-ever Mustache and BBQ Bash, which will include delicious local foods.
Restaurants in Perry offer Southern-style comfort foods, barbecue and burgers, pizza and sandwiches, and Italian and Mexican fare. Many food franchises also are represented in town.
New in Perry is Stripling's General Store, with its Brunswick stew
Foods new to the Perry landscape are at Stripling's General Store, whose motto is "You never sausage a place." Stripling's started as a sausage shop in the town of Cordele in 1964. The Perry location, part store and part restaurant, is the family's fourth.
Stripling's is famous for Brunswick stew; flavorful beef jerky, sold in more than 300 retail locations around the United States; aged steaks; bacon; ribs; and, of course, sausage. At the store you also can choose from down-home sauces, seasonings, and meat rubs, plus mason-jar-style containers filled with pickled side veggies; jams; and jellies. Even fruit cobbler is sold in a jar, ready to be spooned out as you wish.
The restaurant inside Stripling's has a variety of hot and fresh items to choose from daily, so fill up a plate. You'll likely need nourishment while you try to decide what to buy. Stripling's is at 301 Agricultural Village Blvd.
Local lore and items from long ago, such as toys, books, and even medical equipment, are carefully stored at the Perry Area Historical Society Museum. The collection is open Tuesday through Thursday, and by appointment.
Perry has three golf courses, and several more can be found in the nearby Warner Robins area. Perry public parks welcome family activities; the biggest one, Rozar Park, offers an 18-hole disc golf course, picnic pavilions, and fishing ponds.
Fishing is emphasized even more at the Go Fish Education Center, located near the south entrance of the Georgia National Fairgrounds & Agricenter. It's close enough to FMCA activities to walk to; admission is $7 for adults, $6 for seniors (65 or older), and $5 for children 3 to 12. You can examine aquariums full of fish and other aquatic critters, and try out a shooting gallery and a fishing simulator. Fishing poles and bait are available at the museum (free to use) so you can try your luck at the catch-and-release pond outside.
Canines and their people love to visit dog parks, and one such place is next to the Perry Area Convention & Visitors Bureau building. It's right across the street from the Georgia National Fairgrounds & Agricenter. Even if you don't travel with a furry friend, head to this visitors center for info, because the above is just a taste of what you can do and see in town. FMCA also will have a fenced-in dog park on the fairgrounds for attendees' pups.
Beyond Perry
A B-29B "Superfortress" is in the collection at the Warner Robins Museum of Aviation.
Warner Robins: Admission and parking are free at the Museum of Aviation in Warner Robins. The impressive collection has many planes from World War II and Vietnam, plus an assortment of related items. The museum is also home to the Georgia Aviation Hall of Fame. Plan to spend several hours here; it's open daily from 9:00 a.m. to 5:00 p.m.
Also free in Warner Robins are the intriguing nods to history along Depot Row. They include a 1944 train depot now called the E.L. Greenway Welcome Center, which features a 1971 Southern Railroad caboose with an exhibit inside. The beautifully restored Elberta Depot Heritage Center, opened in March 2018, has historical displays inside, too. Depot Row is located at 99 Armed Forces Blvd. Warner Robins is approximately 14 miles (23 minutes) northeast of Perry.
Macon, right up Interstate 75, invites you to "Eat, Drink, And Be Cherry!" at its annual International Cherry Blossom Festival. The fest runs March 22 through 31, 2019. Parades, fireworks, live music, and vendor and craft booths all are part of the fun. An amazing 350,000 Yoshino cherry trees line the streets in these parts!
While you're in Macon, consider touring the Allman Brothers Band Museum, which honors the popular Southern rock group. Macon is 30 miles (32 minutes) north of Perry.
Fort Valley is home to Massee Lane Gardens, where you can walk the beautiful trails to see any remaining spring-blooming camellias. And inside, a lovely porcelain figurine collection awaits. It's open Tuesday through Sunday; admission is $5 for adults and $4 for seniors 55 and over.
Lane Southern Orchards' restaurant is open year-round and specializes in sweet treats.
Peaches and pecans grow very well around here, too, as evidenced at Lane Southern Orchards in Fort Valley. Get a piece of peach cobbler there at Peachtree Café & Bakery, which is open daily year-round. Explore the Lane Farm & Market Shop for tempting pies, candies, breads, jellies, syrups, and much more — even cider made from peaches, grapes, apples, and strawberries. Fort Valley is 12 miles (19 minutes) northwest of Perry.
Before you arrive in central Georgia, be sure to do some research online or by phone to find out what else can be explored. You don't want to miss out. As you expand your search even farther from Perry, the number of must-see stops increases exponentially. It's going to be a peachy time, that's for sure!
Perry Area Convention & Visitors Bureau
Email: info@perrycvb.com
www.perryga.com
Georgia Tourism
Email: travel@exploregeorgia.org
www.exploregeorgia.org
Georgia Roads and Travel
(highway construction, traffic info, weather alerts, and more)
www.511ga.org
The following may not be a complete list, so check your campground directory or the RV Marketplace, found at FMCA.com and in the January issue of Family RVing, for additional listings.
Boland's Perry Overnight Park
800 Perimeter Road
Perry, GA 31069
Crossroads Travel Park
1513 Sam Nunn Blvd.
www.crossroadstravelpark.com
Fair Harbor RV Park
515 Marshallville Road
www.fairharborrvpark.com
Perry Ponderosa Park
13841 State Route 96 E.
Fort Valley, GA 31030
Twin Oaks RV Park, C6864
305 State Route 26 E.
Elko, GA 31025
www.twinoaksrvpark.com
Volunteer And Have Fun
FMCA conventions would not be possible without the assistance of hundreds of volunteers. You may sign up to volunteer as an individual or with an FMCA chapter; either way, it's a great way to meet new friends. Volunteers are needed in many areas, including seminars, security, parking, trams, Information Center, and more. If you're interested in exploring volunteer possibilities in Perry, visit the Events/Information Central page at FMCA.com and click the "Volunteers" icon, or contact the FMCA Events Department (convention@fmca.com; 800-543-3622, 513-474-3622).
News & Notes: December 2018
Tech Talk: Tech Tip
Tech Talk: RV Doctor
Music, Magic, And More In Minot
Minot Magic — 100 Times The Fun —...
Gillette RV Expo
Magical Minot
Chapter Spotlight: Dogwood
The Weekender By Midwest Automotive Designs
Gillette Convention Youth Program
"Peachy" Perry Entertainment


text:
// RobotBuilder Version: 1.0
//
// This file was generated by RobotBuilder. It contains sections of
// code that are automatically generated and assigned by robotbuilder.
// These sections will be updated in the future when you export to
// Java from RobotBuilder. Do not put any code or make any change in
// the blocks indicating autogenerated code or it will be lost on an
// update. Deleting the comments indicating the section will prevent
// it from being updated in the future.
package org.usfirst.frc3566.OfficialCode;
import edu.wpi.first.wpilibj.IterativeRobot;
import edu.wpi.first.wpilibj.command.Command;
import edu.wpi.first.wpilibj.command.Scheduler;
import edu.wpi.first.wpilibj.livewindow.LiveWindow;
import edu.wpi.first.wpilibj.networktables.NetworkTable;
import edu.wpi.first.wpilibj.smartdashboard.SendableChooser;
import edu.wpi.first.wpilibj.smartdashboard.SmartDashboard;
import org.usfirst.frc3566.OfficialCode.commands.*;
import org.usfirst.frc3566.OfficialCode.subsystems.*;
/**
 * The VM is configured to automatically run this class, and to call the
 * functions corresponding to each mode, as described in the IterativeRobot
 * documentation. If you change the name of this class or the package after
 * creating this project, you must also update the manifest file in the resource
 * directory.
 */
public class Robot extends IterativeRobot {
    Command autonomousCommand;
    public static OI oi;
    public static Vision vision;
    // BEGIN AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=DECLARATIONS
    public static DriveTrain driveTrain;;
    public static ElToro elToro;
    public static Catapult catapult;
    public static Dogbox dogbox;
    public SendableChooser autonomousMode;
    // END AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=DECLARATIONS
    /**
     * This function is run when the robot is first started up and should be
     * used for any initialization code.
     */
    public void robotInit() {
        RobotMap.init();
        vision = new Vision();
        // BEGIN AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=CONSTRUCTORS
        driveTrain = new DriveTrain();
        elToro = new ElToro();
        catapult = new Catapult();
        dogbox = new Dogbox();
    // END AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=CONSTRUCTORS
        // This MUST be here. If the OI creates Commands (which it very likely
        // will), constructing it during the construction of CommandBase (from
        // which commands extend), subsystems are not guaranteed to be
        // yet. Thus, their requires() statements may grab null pointers. Bad
        // news. Don't move it.
        oi = new OI();
        // instantiate the command used for the autonomous period
        // BEGIN AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=AUTONOMOUS
        autonomousCommand = new Autonomous();
    // END AUTOGENERATED CODE, SOURCE=ROBOTBUILDER ID=AUTONOMOUS

        autonomousMode = new SendableChooser();
        autonomousMode.addDefault("1 Ball", new Autonomous(1));
        autonomousMode.addObject("2 Ball", new Autonomous(2));
        autonomousMode.addObject("Movement Only", new Autonomous(0));
        autonomousMode.addObject("Disabled", null);

        SmartDashboard.putData("Autonomous Mode", autonomousMode);
      }
    public void autonomousInit() {
        // schedule the autonomous command (example)
        autonomousCommand = (Command) autonomousMode.getSelected();
        if (autonomousCommand != null) {
            autonomousCommand.start();
        }
    }
    /**
     * This function is called periodically during autonomous
     */
    public void autonomousPeriodic() {
        Scheduler.getInstance().run();
    }
    public void teleopInit() {
        // This makes sure that the autonomous stops running when
        // teleop starts running. If you want the autonomous to
        // continue until interrupted by another command, remove
        // this line or comment it out.
        if (autonomousCommand != null) {
            autonomousCommand.cancel();
        }
    }
    /**
     * This function is called periodically during operator control
     */
    public void teleopPeriodic() {
        Scheduler.getInstance().run();
    }
    /**
     * This function called periodically during test mode
     */
    public void testPeriodic() {
        LiveWindow.run();
    }
}

redpajama_set_name: RedPajamaGithub


text:
Q: Checking if words with an accent was displayed In automated tests, I need to know how I can check if words with an accent was displayed. I performed automated tests as below:
file.feature
# encoding: utf-8

Feature: Cálculo de triângulo

Scenario: Calculando um triângulo equilátero
    Given que estou na tela do TrianguloApp
    When eu preencher o campo Lado 1 com "3"
        And eu preencher o campo Lado 2 com "3"
        And eu preencher o campo Lado 3 com "3"
        And eu clicar em Calcular
    Then a mensagem "O triângulo é Equilátero" será exibida

file_steps.rb
Given(/^que estou na tela do TrianguloApp$/) do
    element_exists("* text:'TrianguloApp'")
end

When(/^eu preencher o campo Lado 1 com "(.*?)"$/) do |lado1|
      enter_text "* id:'txtLado1'", "#{lado1}"
end

When(/^eu preencher o campo Lado 2 com "(.*?)"$/) do |lado2|
      enter_text "* id:'txtLado2'", "#{lado2}"
end

When(/^eu preencher o campo Lado 3 com "(.*?)"$/) do |lado3|
      enter_text "* id:'txtLado3'", "#{lado3}"
end

When(/^eu clicar em Calcular$/) do
      touch("* id:'btnCalcular'")
end

Then(/^a mensagem "(.*?)" será exibida$/) do |mensagem|
     element_exists("* O triângulo é Equilátero")
end

The result is as in the image below.

The error occurred because of trying to verify that the phrase O triângulo é Equilátero was displayed on the screen.

A: Well your statement is missing the text:
element_exists("* text:'O triângulo é Equilátero'")

That might be all you need to get this working, however if you problem is the accents then you could find the element first though a different identifier then check the text.
the_text = query("* id:some_id", "text")

expect(the_text).to eql 'O triângulo é Equilátero'

The expect is from rspec https://github.com/rspec/rspec-expectations

A: I was able to validate the text O triângulo é Equilátero
I changed the command element_exists to macro %Q|I should see "#{text}"|
See how it got:
Entao(/^a mensagem "(.*?)" será exibida$/) do |mensagem|
    macro %Q|I should see "#{mensagem}"|
end
Source: http://testmunk.readthedocs.io/en/latest/steps.html

redpajama_set_name: RedPajamaStackExchange


text:
Barnabas Suebu (born 29 April 1945) also known as Bas Suebu, is an Indonesian politician. He was the governor of the Indonesian province of Papua for two periods: 1988–1993 and 2006–2011. He wants to protect the province's forests, and has made plans to declare a moratorium on log exports and recommended that no new logging concessions be granted to timber companies. Suebu appeared on Time magazine's list of Heroes of the Environment October 2007.

Early life, education, and career
Barnabas Suebu was born on 29 April 1945 in Sentani, Papua, to Bonifasius Suebu and Salomi Monim. Suebu obtained a Bachelor of Laws at Cenderawasih University in 1988. He was also a graduate from National Resilience Institute (Lemhanas) in 1984.

He was the chairman of Irian Jaya KNPI (Indonesian Youth National Committee) in 1974 and 1975, speaker of Irian Jaya Legislative Assembly (1987–1988), adviser to the Minister for Research and Technology, a member of Congress (1997–2002) and Indonesian ambassador to Mexico, Honduras and Panama (1999–2002).

Personal life
Suebu is married and has six children.

References

1945 births
Governors of Papua (province)
Indonesian Christians
Living people
Papuan people
Cenderawasih University alumni

redpajama_set_name: RedPajamaWikipedia


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text:
Home » Weekly Torah Portion » Advanced » Shem MiShmuel
Efriam's Offering
Naso (Numbers 4:21-7:89)
May 16, 2010 | by Rabbi Zvi Belovski
The end of this, the longest sidrah in the Torah, describes at great length the offerings brought by the princes of the tribes of Yisrael at the dedication of the altar. Each of the twelve princes brought an identical set of presents, each of which is listed in the verses. The Midrash has much to say about these gifts. Let us look at one rather intriguing statement about the gift of Elishama, the prince of the tribe of Efraim:
On the seventh day, the prince of the children of Efraim - Elishama ben Amihud. (Bemidbar 7:48)
Efraim is the strength of my head... (Tehillim 60:9) - this refers to the prince of Efraim, who brought his offering at the dedication of the altar on Shabbos, as the verse says, On the seventh day, the prince of the children of Efraim. We know that it was Shabbos, for we have earlier demonstrated that the princes started to bring their offerings on Sunday... (Bemidbar Rabbah 14:1)
The verse quoted at the start of this midrash, "Efraim is the strength of my head," refers to the incident at the end of Yaakov's life when he blessed his grandsons, Menasheh and Efraim. He chose to bless Efraim before Menasheh, even though Menasheh was older. Thus "Efraim is the strength of my head" - Efraim was given precedence over Menasheh. This midrash implies that it was only because Yaakov favored Efraim that Elishama could offer his gift on Shabbos. For ordinarily the offering could not override the laws of Shabbos, but somehow the power which Yaakov had invested in the tribe of Efraim enabled their offering to be an exception. This fascinating midrash is complicated by the following:
This one pushed aside Shabbos and tumah [ritual impurity], although the offering of an individual pushes aside neither. (Sifri, Naso 51)
This indicates that had any of the princes come to offer their gift on Shabbos, their donation would have superseded the Shabbos regulations. (The same would also have applied to the laws of tumah.) What have the blessings of Yaakov to do with the ability of a prince to bring a gift on Shabbos centuries later, and how did this affect the other tribes?
EFRAIM AND MENASHEH
To begin our study, we must examine the natures of the two sons of Yosef, Menasheh and Efraim. As always, the name of a person reveals much about his essence, and in these two cases, we are given this information explicitly:
Yosef named the firstborn Menasheh, "for God has made me forget all of my trouble and all of my father's house." He named the second Efraim, "for God has made me fruitful in the land of my oppression."(Bereishis 41:51-52)
The name Menasheh finds its root meaning in forgetting, or distancing oneself from the past. This represents a particular sort of Divine service, in which one divests oneself of all manner of bad traits, hoping to achieve perfection in their stead. Efraim, on the other hand, finds its root meaning in fruitfulness. This is a different style of progression toward spiritual goals, in which one focuses on developing good traits and performing mitzvos. These two distinct types of service are concisely described by the verse:
Depart from evil and do good... (Tehillim 34:15)
In our context, Menasheh is "depart from evil," whereas Efraim is "do good."
If we recall the episode mentioned earlier, when Yaakov blessed Efraim and Menasheh, we remember that in actual fact Yosef, the father of the two brothers, wanted them to be blessed in order of age - Menasheh before Efraim. But Yaakov refused, instead blessing the younger before the older. Understanding this apparent dispute between Yosef and Yaakov will be crucial to the whole of our study.
Yosef wanted Menasheh to precede Efraim; that is, he wanted events to follow the order described by the verse: first "depart from evil" and only then "do good." This, as we would expect, reflected the character of Yosef, who had spent his whole life struggling against bad to achieve greatness. Yaakov, however, chose to bless Efraim before Menasheh. Within the context we have defined, this represents a lifestyle in which one first concentrates on performing good deeds. Then, due to the influx of holiness generated by one's new mode of life, any evil traits will automatically dissipate. In Yaakov's view, this approach to life was preferable to his son's mode of waiting until the bad has been destroyed before worrying about good deeds. In Yaakov's opinion (which we may assume is the norm) this is the general rule in Jewish life: we must begin our observance of the Torah by seeking mitzvos and learning, assigning a secondary role to eliminating evil. This will follow later, for as the holiness of a Torah lifestyle enters our beings, any bad will be consumed or expelled.
The problem with Yosef's approach to life is evident from examining its application to Shabbos. The Shabbos is very holy, and, as such, perhaps we should engage in great spiritual preparations to be ready to accept its holiness. The trouble with this is, who could ever say that he is ready? Surely, still more preparations could be performed. The inevitable result of this is that we would never consider ourselves ready, and hence, we would never have Shabbos! Instead, we each do our best during the week, and Shabbos just comes; somehow, we are ready to receive its majesty. This means that an ordinary person who follows Yosef's approach will spend his whole life attempting to eliminate bad traits from his personality, but will never reach the point when he feels that he is finished with that stage and ready to move on. As such, he will always remain stuck at the "depart from evil" stage, unable to "do good" at all. Instead, we follow Yaakov's view - getting on with a life of mitzvos and Torah study, confident that we will ultimately achieve success both as doers and departers.
THE SHABBOS OFFERING
Returning to our original subject, we can now appreciate why Efraim's offering pushed aside the Shabbos prohibitions. Yaakov's special selection of Efraim over Menasheh meant that the emphasis in Jewish life was forever placed on "doing good." When the prince of Efraim brought his offering, this was an act of great generosity and a shining example of the preference for starting the Divine worship of one's tribe with a positive act. This is indicated by the fact that Elishama brought his offering even though it was Shabbos. The Divine wisdom underscored the rationale behind this departure from normative halachah by arranging it that Efraim and no other tribe was scheduled to bring their offering on Shabbos.
Of course, this emphasis on doing good does not apply only to the members of Efraim - it is, since the blessings administered by Yaakov, a universal rule, applicable to every member of klal Yisrael. This is what is meant by the Sifri, which claims that any one of the tribes could have offered on Shabbos had it been necessary. Once Yaakov had determined the suitable path for all of his descendants, any one of them could have and would have brought their offerings on Shabbos.
Excerpted from Shem MiShmuel by the Sochatchover Rebbe, rendered into English by Rabbi Zvi Belovski, published by Targum Press. Click here to order.
Rabbi Zvi Belovski
Rabbi Zvi Belovski is the Rav of the Golders Green Synagogue in London. He received Semicha from Gateshead Yeshivah and attended University College, Oxford. A popular speaker and counsellor, Rabbi Belovski is active in Jewish outreach; he teaches at the Jewish Learning Exchange (Ohr Somayach) and the London School of Jewish Studies, and serves as Rabbinic Mentor for British campus rabbis across the country. He is the author of Shem MiShmuel: Selections on the weekly parashah and festivals. His special areas of interest include the study of Midrash, contemporary Jewish Law, Polish Chassidic thought, and he is particularly interested in the writings of the Rebbes of Sochaczew and Ishbitz. Rabbi Belovski lives in Golders Green with his wife and family.
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(3) Anonymous, June 16, 2016 3:11 PM
Efriam?
There are at least two places here where Efriam is spelled that way. They are both titles. It is spelled like that in the email as well.
All the other places here, it is spelled Efraim.
Which is correct?
(2) Avraham, May 30, 2015 12:51 AM
Good will displace evil
This Biblical principle is also reflected in nature: by cutting, trimming or "pruning" the bush (evil) gets bushier! It is the arrival of the new leaf on a tree that pushes out the old, thus, "good" displaces "evil." Only by killing the root will the evil die. That is to say that the origins of our bad behaviors must be dealt with else we will just be merely culling the symptoms & not the causes.
(1) Steve (Rashi ben Anshul), May 15, 2013 8:00 PM
Which could come first?
This is a dilemma. Is it necessary to depart from evil before doing good? Can one do good before departing from evil? It is a matter, perhaps, of striving for a higher level of the spiritual. Do both at the same time. Failing to do one without intent to fail is not terrible. It is the striving to do both at the same time that is the highest.
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I had my first real bad night. Devin was working a night shift and I was with Kassie all night. She was fussy and moaning and groaning. I felt really bad. So we took her to the Doctor and she said that she is collic. I am really sad. I still love her no matter what. I will just need a little help. She weighs 5lbs. 7 oz. She is still gaining so no worries there.
Ohhh, you have been lucky if you just had your first bad night. Don't worry it will get easier. On a side note: I didn't sleep all the way through the night until Vance was 9 months old. He often wakes up still but just once. Some kids are better than others, make sure she is full when she goes down.(not that it means she'll sleep through the night) I didn't have a collicky baby so I'm no help there. Can you use the drops to help her?
Bummer! Babies always have lil' surprises for you. Just remember it is OK to let her cry, so you can keep your sanity! And take your family & friends offers to help. Keep smiling, she is so sweet!!!
*Don't lay her down right after she eats. She may have some reflux causing her pain and keeping her upright will help keep things down.
*Exercise ball + bouncing + baby swaddling = possible solution.
*Sling or wrap-carrier really tightly and go for a walk.
*Soft music and low lights. Some people think "colic" is partially due to overstimulation.
Maybe that helps, maybe not - but I feel for you :( Sully was my toughest baby yet and there were some nights that I just had to hand him off to someone else and I would sit on my bed and cry. After a good cry I would always feel better and then hearing my baby's cries would re-motivate me to go try another round of soothing.


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Argyll Scott Singapore
Inspiring Business Women in APAC
TypePermanent Temporary Contract
SectorAccounting & Finance Advanced Analytics & Data Science Banking Business Support Business Transformation Corporate Governance Digital & e-Commerce ERP Financial Services Human Resources Insurance Legal Life Sciences Marketing Manufacturing & Engineering Procurement Retail Projects Retail Sales Sales & Marketing Strategy Supply Chain Technology
SalariesUp to 1M THB 1M - 1.5M THB 1.5M - 2M THB 2M - 2.5M THB 2.5M - 3M THB 3M - 4M THB 4M - 5M THB 5M+ THB Up to £25000 £25,001 - £40,000 £40,001 - £60,000 £60,001 - £80,000 £80,001 - £100,000 £100,000 - £150,000 £150,000 + 360,000 - 480,000 HKD 480,000 - 600,000 HKD 600,000 - 720,000 HKD 720,000 - 840,000 HKD 840,000 - 1M HKD 1M - 1.2M HKD 1.2M - 1.5M HKD 1.5M - 2M HKD 2M + HKD Up to MYR 75,000 75,000 - 100,000 MYR 100,000 - 140,000 MYR 140,000 - 180,000 MYR 180,000 - 220,000 MYR 220,000 - 260,000 MYR 260,000 - 300,000 MYR 300,000 - 360,000 MYR 360,000 - 420,000 MYR 420,000 - 500,000 500,000 + MYR
Inspiring Business Women In APAC: Dr. Bijna K.Dasani
25 Jun 01:00 by Peter Ison
Dr. Bijna K. Dasani is the Head of Architecture & Innovation at Lloyds Banking Group. With 19+ years of professional experience, she has worked across EMEA, the Americas and Asia, and has a robust track record of delivering successes for global Banking and Financial Services across the verticals of Strategy, Innovation and Technology. Bijna serves the Boards of FinTech Connect, CIONet, IPQC CDO Exchange, and the Inclusive Companies Network. She is a FinTech Mentor and has co-hosted the Women in Business TV series and films videos on Mentorship, Leadership & Technology.
Can you tell us about your career progression into your current role?
After work-experience with the Royal Bank of Scotland Group (during secondary education), my first tangible role in Financial Services was with Credit Suisse; I was instantly inspired and intellectually stimulated by my colleagues and the working environment alike. I later moved to Goldman Sachs in New York, transitioning from Operations into Change Management and later Strategy roles. As my career has since taken me to different roles and locations, Strategy has become my core skill set and Innovation my secondary; which encompasses verticals of data, core innovation, fintech, and applied sciences – all of which have been the pillars underpinning career over the past decade.
Looking back at your career to date, can you pinpoint when you first noticed an emphasis on diversity and inclusion around you?
This is very circumstantial at times; there have been moment of gaps and moments of emphasis; personally, it's always been a priority throughout my career. As a graduate entry into Financial Services, I was an equal in terms of gender parity and an evident minority in terms of culture and religion. I've observed labelling and globally, the statistical evidence on the lack of parity for women in Financial Services, as well as other sectors; women are typically 10%-20% of the C Suite in most sectors across geographies. A few years ago, I chose to write a paper exploring the reasons behind the lack of parity, particularly within the Financial Services C Suite. While historical, environmental and economic factors play a role, my research showed distinct attrition at the middle management level, when women's priorities often change at home and at work simultaneously; which has a fundamental impact on women reaching the boardroom. For the past decade, I've been sharing views about the importance of shared responsibility at home and at work, particularly through those increased responsibility phases. While things are improving with regards to awareness; the fact remains that we are still 200+ years away from gender parity when it comes to salary.
You have worked across EMEA, APAC and the Americas – what are the cultural differences, and have they affected your management style?
There are cultural differences not just between these regions, but also within them. One of my roles covered both North America and Latin America – both provided completely different experiences during the same tenure. In fact, no two states in America are the same. It's critical to understand the value-system of colleagues in any region, and thereafter understand how it drives their decision-making, thought-process and communication - and then it's really important to adapt your own style as necessary. My own leadership style is therefore adaptive; it supports this strategy and has served me well.
What is your approach to innovation?
Start small but think big. I see innovation as a series of small, incremental steps that lead to impactful change. My approach is to focus on a problem, develop a solution (often through technology) and create a business model around that solution.
How can we encourage people to be more innovative?
Encourage an innovative spirit and provide colleagues with safe environments to test-and-learn and experiment; if they can fail-fast, learn-fast and incrementally conquer challenges over time - these learnings underpin the makings of a solid foundation.
How has your field evolved through technology and innovation?
I often quote "Banking is necessary, banks are not." Banks are not simply competing with other banks today, they're competing with technology, apps and start-ups that often don't even play directly in the Financial Services spectrum – take WeChat for example! As non-financial services players are absorbing slices of the Financial Services pie, the banking industry is responding with rapid change. For example, we are seeing lots of transitions to cashless society's in the developed World and in comparison, 25% of the World's population in the developing World remains in the 'unbanked' or 'underbanked' segment. Innovation and technology are simply disrupting how we have traditionally done business.
Do you have a role model or mentor in your career?
Many; there are a lot of people who have influenced my leadership style over the years. I'm a people-watcher and I like to observe and understand why people make certain decisions or act in a certain way - so I'm always influenced by people who I observe doing things well and who have a sound value-system; these could be members of my personal or professional networks. Integrity is at the heart of everything I do. I have had some excellent managers over the years, who are now (informal) mentors from whom I can seek guidance as needed and at times they are my best sponsors.
How do you balance long hours with your personal life successfully?
Discipline. When I was 16, I started working 9am to 5pm whilst simultaneously studying at night school between 6pm-9pm. I learnt the value of time-management, multi-tasking and living a disciplined life early on! It was challenging to begin with, but it was also one of the best experiences as the routine really prepared me for the Investment Banking industry which I later joined. It is very important to have a healthy work-life balance, where one keeps you engaged with the other. Emotional, physical and mental wellbeing are all extremely important personally and for the teams I manage; they are always factors' I talk openly about and encourage colleagues to focus on. Building the foundation of my career was absolutely key in the early years, and my time was weighted towards that, I have become more aware of the balance over time and encourage it in the workplace – the industry has also evolved to more proactivity in this domain over time.
In your experience, what are the benefits of diverse teams and diverse organisations?
The key measurable is impact. The more diverse the thinking and representation around the room, the more diverse and creative the outcome is likely to be. If you hire people who look and sound like you, your results are limited (as the studies show). If you hire people who do things differently, there's learning for everyone. I've been hired because I do think differently, and to positively disrupt the status quo! There is power in inclusion. We're in the Fourth Industrial Revolution, a time of disruption and new technology; the emerging economies (E7) are catching up with, or in some cases surpassing the developed economies (G7) because they are agile, less regulated and can often adapt much quicker and harness their diverse thinking.
What is your advice to leaders who want to create a more diverse and inclusive culture?
Think broadly and adopt a lens beyond the day-to-day to understand the world from different perspectives like political, economic, technological, and sociological alike. Times have never been so interesting! Leaders today are in a position to shape the future of Banking and Financial Services and beyond, thereforeit's important to consider (beyond the status quo) future consumers and workforce trends and start to be inclusive of their needs, thinking and culture in order to survive and thrive.
Peter Ison
Associate Director | Argyll Scott Singapore
Peter began his recruitment career in London in 2007, focusing on the Technology sector, he worked in both financial and commercial fields.
Peter joined Argyll Scott Singapore in 2016 and is currently an Associate Director overseeing the Technology sector. His experience and industry knowledge has enabled him to better understand the different recruitment needs and in providing tailored consultative services to both his clients and candidates.
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