text:
\section{Introduction}

	Thermoelectricity is both of fundamental interest and technologically promising, because it allows to convert waste heat to useful electric power without moving parts. It was observed for the first time almost two centuries ago by T.~J.~Seebeck in bismuth~\cite{Seebeck1825}. The room temperature Seebeck coefficient is smaller in bismuth than in germanium or silicon. But along the trigonal axis, it is as large as $S \approx 100\ \mu\mathrm{V\,K^{-1}}$, which combined with an electrical resistivity of $\rho\approx 135\ \mu\Omega\,\mathrm{cm}$ and a thermal conductivity of $\kappa \approx 6\ \mathrm{W\,K^{-1}\,m^{-1}}$~\cite{Gallo1963} leads to a thermoelectric figure of merit $ZT=\frac{S^2 T}{\kappa \rho}\approx 0.37$, the largest in the periodic table. Bi-Sb alloys have the largest known thermoelectric figure of merit of any solid at cryogenic temperatures and applying a small magnetic field allows to significantly increase $ZT$ further~\cite{Yim1972}. Despite many investigations over a long period of time~\cite{Seebeck1825,Yim1972,Chandrasekhar1959,Gallo1963,Wolfe1963,Michenaud1970,Uher1974,Lenoir1996,Collaudin2014,Hansen1978,Heremans1979,Mikhail1980,Teramoto2008,Popescu2012,Popescu2012a}, the Seebeck effect of bismuth, especially in presence of a finite magnetic field, is far from being understood up to the present day.

	Bismuth has extraordinary electronic properties, which give rise to the large, negative and anisotropic Seebeck coefficient. It is a semimetal, i.e.\ the electron density~$n$ equals the hole density~$p$. At low temperatures, they amount only to $n=p=3.0\cdot10^{-17}\ \mathrm{cm^{-3}}$~\cite{Bhargava1967}, being equivalent to one carrier of each sign per $10^5$ atoms as well as a very small Fermi energy. The very large magnetoresistance reflects the extremely high mobility of the charge carriers, which are ballistic at low temperatures~\cite{Prakash2016,Kang2021}. The Fermi surface consists of one hole pocket with parabolic dispersion and three cigar-shaped electron pockets containing Dirac fermions with an extremely anisotropic band structure, the lowest effective mass being equivalent to approximately $10^{-3}$ bare electron masses~\cite{Zhu2011}. The valley degeneracy of the three electron pockets can be lifted by a magnetic field~\cite{Zhu2011a}. High fields even dry up one or two Fermi seas~\cite{Zhu2017}.

	During the last two centuries, the Seebeck effect of bismuth has been studied intensively~\cite{Chandrasekhar1959,Gallo1963,Wolfe1963,Michenaud1970,Yim1972,Uher1974,Collaudin2014}. However, surprisingly, no systematic experimental investigation of the Seebeck effect in magnetic field can be found in literature and the highest magnetic field reported amounts only to 5.5~T~\cite{Uher1974}.

	Here, we report on a systematic study of the magneto-Seebeck effect of bismuth from room temperature down to 20~K under a magnetic field of up to 13.8~T in the binary-bisectrix plane. We find that the Seebeck coefficient displays a non-trivial evolution with temperature, magnetic field and the orientation of the magnetic field. To explain the experimental results, we developed a model based on semiclassical transport theory. In doing so, we approximated the well established band structure~\cite{Liu1995} by the Lax model~\cite{Lax1960} to account for the non-parabolicity of the electron bands. The scattering time was treated as in Ref.~\cite{Mikhail1980}, implying an energy dependence compatible with electron-phonon scattering. Phonon drag was not included, since it is relevant only at temperatures below 10~K, which is out of the scope of this work~\cite{Mikhail1980,Uher1978,Issi1979}. Because of the very low Fermi energy, Landau quantization is important in bismuth already at comparably low magnetic fields~\cite{Zhu2011}. Therefore, the semiclassical model was extended as to include the effects of Landau quantization. The goal of our theoretical work was to identify the physical mechanisms playing an important role with regard to the magneto-Seebeck effect of bismuth. This is why we aimed at a model, which is as simple as possible and contains as few unknown parameters as possible, instead of perfect agreement with experimental data. Nevertheless, the model reproduces well the observed behavior in a large part of the ($T,B,\Theta$)-space.

	We identify two mechanisms which contribute unexpectedly strongly to the magneto-Seebeck effect of bismuth. Firstly, the transverse Nernst response gives rise to a longitudinal Seebeck voltage via the Hall effect. Secondly, Landau quantization significantly affects the thermopower of bismuth up to temperatures as high as 120~K. These effects could also be relevant to other materials with low carrier concentration.

	Our experimental and theoretical results are to be compared with previous theoretical studies of the magneto-Seebeck effect in bismuth~\cite{Hansen1978,Heremans1979,Mikhail1980,Teramoto2008,Popescu2012,Popescu2012a}. Our experimental results disagree with the predictions of Ref.~\cite{Popescu2012}. In contrast, for low magnetic fields, our work confirms the formalism of Mikhail et al.~\cite{Mikhail1980}, which is based on a theoretical framework developed by Heremans and Hansen~\cite{Heremans1979}.

	\section{Theory}\label{sec:theory}
	\subsection{General}\label{sec:theory:general}

	An electric current $\mathbf{j}$ can be generated not only by an electric field $\mathbf{E}$, but also by a thermal gradient $\boldsymbol\nabla T$. This is expressed by
	\begin{equation}\label{eq:j}
		\mathbf{j} = \hat{\sigma} \mathbf{E} -\hat{\alpha}\boldsymbol\nabla T,
	\end{equation}
	where $\hat{\sigma}$ and $\hat{\alpha}$ are the electrical and the thermoelectric conductivities, respectively~\cite{Behnia2015}. For zero current and diagonal conductivity tensors, equation~\eqref{eq:j} leads to
	\begin{equation}
		S_{zz} = \frac{E_z}{\partial_z T} = \frac{\alpha_{zz}}{\sigma_{zz}},
	\end{equation}
	where the first equation is the definition of the Seebeck coefficient $S_{zz}$. In general, however, the tensorial nature of $\hat{\sigma}$ and $\hat{\alpha}$ has to be taken into account:
	\begin{equation}\label{eq:S_tensorial}
		\hat{S} = \hat{\sigma}^{-1} \hat{\alpha} = \hat{\rho} \hat{\alpha}
	\end{equation}
	$\hat{\rho}$ denotes the electrical resistivity tensor. For a magnetic field $\mathbf{B}$ parallel to the $x$-axis, one obtains
	\begin{equation}\label{eq:S_contributions}
		S_{zz} = \rho_{zz} \alpha_{zz} + \rho_{zy} \alpha_{yz}.
	\end{equation}
	Note that the off-diagonal component of the thermoelectric conductivity $\alpha_{yz}$ is commonly associated with the Nernst effect. In the following, we will refer to the first summand of equation~\eqref{eq:S_contributions} as diagonal or longitudinal, whereas the product of the Hall resistivity $\rho_{zy}$ and $\alpha_{yz}$ will be called off-diagonal or transversal contribution.

	Time reversal symmetry implies $\hat{\sigma}(\mathbf{B}) = \hat{\sigma}^\mathsf{T} (-\mathbf{B})$ and $\hat{\alpha}(\mathbf{B}) = \hat{\alpha}^\mathsf{T} (-\mathbf{B})$~\cite{Behnia2015}. Therefore, the diagonal entries of $\hat{\sigma}$, $\hat{\alpha}$ and $\hat{\rho}$ are symmetric functions of $\mathbf{B}$. However, time reversal symmetry is not violated by
	\begin{equation}
		\sigma_{ij} (\mathbf{B}) \neq -\sigma_{ij} (-\mathbf{B})\qquad (i\neq j).
	\end{equation}
	The same is true for $\alpha_{ij}$ and $\rho_{ij}$~\cite{Akgoz1975}. Hence, in equation~\eqref{eq:S_contributions}, the term $\rho_{zz} \alpha_{zz}$ is symmetric in $\mathbf{B}$, whereas there are no restrictions on the symmetry of the term $\rho_{zy} \alpha_{yz}$. Therefore, depending on the crystal symmetry, it can happen that $S_{zz} (\mathbf{B}) \neq S_{zz} (-\mathbf{B})$. This behavior is dubbed Umkehr effect~\cite{Akgoz1975}.

	\subsection{The case of bismuth}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_band_structure.pdf}
		\caption{\textbf{Fermi surface and band structure of bismuth:} a) The Fermi surface of bismuth consists of one hole pocket at the $T$-point of the Brillouin zone and three electron pockets at the $L$-points which are tilted by about 6\degree\ out of the binary-bisectrix plane. The Fermi surface is very small as there are only one free electron and one hole per $10^5$ atoms. The Fermi surface is enlarged for better visibility. Adapted from Ref.~\cite{Zhu2018}. b) Dispersion relation of bismuth at 0 K according to the Lax model~\cite{Lax1960}. The valence band at the $T$-point is parabolic, whereas the electrons at the $L$-points have a Dirac-like dispersion due to the energy gap of only 13.6 meV (at zero temperature). Note the very small Fermi energy $\epsilon_F$.}
		\label{fig:Fermi_surface}
	\end{figure*}

	In order to calculate the Seebeck coefficient $S_{zz}$, the conductivity tensors $\hat{\sigma}$ and $\hat{\alpha}$ have to be determined. In the case of bismuth, several subtleties of this material have to be considered. In the following, we will focus on the description of these subtleties, whereas a more detailed derivation of the model can be found in the supplement~\cite{Supplement}.

	The starting point is the band structure (see Fig.~\ref{fig:Fermi_surface}). The Fermi surface of bismuth consists of one hole pocket at the $T$-point, which is symmetric with respect to the trigonal axis ($z$-axis), and three equivalent electron pockets at the $L$-points of the Brillouin zone. The electron pockets are perpendicular to the binary axis ($x$-axis) and tilted by about 6\degree\ with regard to the bisectrix axis ($y$-axis)~\cite{Zhu2018}. The Fermi surface shows threefold symmetry with the trigonal axis as symmetry axis. This means that there are three binary and three bisectrix axes in the binary-bisectrix (i.e.\ trigonal) plane.

	The hole band at the $T$-point has an ordinary parabolic dispersion. In contrast, the dispersion of the electron bands is mostly linear, because at the $L$-points, there is only a very small energy gap $\epsilon_g$ between the conduction and valence bands (see Fig.~\ref{fig:Fermi_surface}b). An appropriate way to describe these Dirac electrons is the Lax model~\cite{Lax1960}:
	\begin{equation}\label{eq:Lax}
		\epsilon (\mathbf{k}) = \pm \frac{1}{2}\left(\epsilon_g^2+2\epsilon_g\hbar^2 \mathbf{k}^\mathsf{T} \hat{m}_{be}^{-1} \mathbf{k}\right)^{1/2}-\frac{1}{2}\epsilon_g
	\end{equation}
	In this context, it is useful to define the quantity $\gamma$ and its derivative with respect to energy~\cite{Heremans1979}:
	\begin{equation}\label{eq:Lax_gamma}
		\gamma (\epsilon) = \epsilon \left(1+\frac{\epsilon}{\epsilon_g}\right) = \frac{\hbar^2}{2} \mathbf{k}^\mathsf{T} \hat{m}_{be}^{-1} \mathbf{k}
	\end{equation}
	\begin{equation}\label{eq:Lax_dgamma}
		\gamma' (\epsilon) = \frac{\partial \gamma}{\partial \epsilon} = 1+2\frac{\epsilon}{\epsilon_g}
	\end{equation}
	In the limit $\epsilon_g \to \infty$, equation~\eqref{eq:Lax} reduces to a quadratic dispersion and can therefore be used to describe the hole band at the $T$-point. In this case, $\gamma$ equals the energy~$\epsilon$ and $\gamma' = 1$.

	The inverse mass tensor at the band edge reads
	\begin{equation}\label{eq:inverse_mass_L}
		\hat{m}_{be,L}^{-1}= m_0^{-1}
		\begin{pmatrix}
			806 & 0 & 0\\
			0 & 7.95 & 37.6\\
			0 & 37.6 & 349\\
		\end{pmatrix}
	\end{equation}
	for one of the electron pockets and has to be rotated by 120\degree\ and 240\degree, respectively, around the trigonal axis ($z$-axis) for the other two pockets~\cite{Zhu2011}. $m_0$ denotes the bare electron mass. Note the presence of off-diagonal components in the mass tensor, which is a consequence of the tilt of each electron pocket off the trigonal plane of the crystal~\cite{Brown1968}. This tilt is at the origin of the Umkehr effect in bismuth~\cite{Akgoz1975}.

	According to Ref.~\cite{Zhu2011}, the inverse mass of the holes at the $T$-point amounts to
	\begin{equation}\label{eq:inverse_mass_T}
		\hat{m}_{be,T}^{-1}= m_0^{-1}
		\begin{pmatrix}
			14.75 & 0 & 0\\
			0 & 14.75 & 0\\
			0 & 0 & 1.387\\
		\end{pmatrix}.
	\end{equation}

	The density of states $D(\epsilon)$ follows from equation~\eqref{eq:Lax}:
	\begin{equation}\label{eq:DOS}
		D(\epsilon) = \sqrt{\frac{2}{\det \hat{m}_{be}^{-1}}} \frac{1}{\pi^2\hbar^3} \sqrt{\gamma (\epsilon)} \gamma'(\epsilon)
	\end{equation}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_theory_v3.pdf}
		\caption{\textbf{Carrier density and band structure vs.\ temperature as well as energy- and temperature-dependence of the scattering time:} a) Carrier density $n=p$ as a function of temperature. b) Temperature-dependence of the chemical potential $\mu$ and the bottom and top of the electron and hole bands, respectively. c) Scattering time $\tau_{zz}$ vs.\ energy at $T=60$~K. The divergence at the band edge does not strongly influence the observables, because the density of states is close to zero at these energies. d) Scattering time $\tau_{zz}$ at the chemical potential vs.\ temperature as determined from zero-field resistivity measurements and used for all calculations. For comparison, the dotted lines depict the values found by Hartman~\cite{Hartman1969}.}
		\label{fig:theory}
	\end{figure*}

	Unlike other semimetals, the carrier density of bismuth is not constant, but changes by more than a factor of eight between zero and room temperature~(see Fig.~\ref{fig:theory}a)~\cite{Michenaud1972,Issi1979}. This is due to two reasons: Firstly, the thermal broadening of the Fermi-Dirac distribution has a large impact because of the small Fermi energy. Secondly, the band structure is strongly temperature-dependent (see Fig.~\ref{fig:theory}b). For example, the energy gap at the $L$-point almost triples upon heating from 4~K to room temperature~\cite{Vecchi1974,Otake1980}.

	In general, the scattering time $\tau$ constitutes the most uncertain ingredient in the description of transport phenomena, because, apart from some proportionalities, it can be directly determined neither theoretically nor experimentally. Here, we assume that it can be described as a product of an energy-dependent scalar $b(\epsilon)$, a temperature-dependent scalar $c(T)$ and a second-order tensor $\hat{a}_p$ capturing the anisotropy, where the index $p$ refers to the $T$- and $L$-points (i.e.\ holes and electrons), respectively:
	\begin{equation}\label{eq:tau_factorization}
		\hat{\tau}_p(\epsilon,T) = \hat{a}_p b(\epsilon) c(T)
	\end{equation}
	Concerning the energy-dependence of the relaxation time, we follow Refs.~\cite{Heremans1979,Mikhail1980,Zawadzki1971,Ravich1971}. Assuming acoustic electron-phonon (and hole-phonon) scattering, one finds Fermi's golden rule
	\begin{equation}\label{eq:scattering_time_general}
		\frac{1}{\tau} \propto D(\epsilon)W^2(\epsilon),
	\end{equation}
	where
	\begin{equation}
		W^2(\epsilon) \propto {\gamma'}^{-2}(\epsilon)
	\end{equation}
	holds for the squared scattering matrix element $W^2$. Inserting these proportionalities into equation~\eqref{eq:tau_factorization} leads to
	\begin{equation}
		\hat{\tau}_p(\epsilon,T) = \hat{a}_p \frac{\gamma'(\epsilon)}{\sqrt{\gamma(\epsilon)}} c(T).
	\end{equation}
	This energy-dependence of $\tau$ is equivalent to an energy-independent mean free path $l$ as predicted for scattering on acoustic phonons\footnote{$l = v\tau = \hbar^{-1} \frac{\partial \epsilon}{\partial k} \tau \propto \frac{\partial \epsilon}{\partial \gamma} \frac{\partial \gamma}{\partial k} \frac{\gamma'(\epsilon)}{\sqrt{\gamma(\epsilon)}} \propto \frac{1}{\gamma'(\epsilon)}\sqrt{\gamma(\epsilon)}\frac{\gamma'(\epsilon)}{\sqrt{\gamma(\epsilon)}} \propto \epsilon^0$}~\cite{Ioffe1960}. For $\epsilon_g \to \infty$, i.e.\ parabolic bands, the energy-dependence reduces as expected to $\tau \propto \epsilon^{-1/2}$~\cite{Heremans2012}. Fig.~\ref{fig:theory}c depicts the scattering time as a function of energy. Once $b(\epsilon)$ is determined, $c(T)$ can be calculated from zero-field resistivity measurements\footnote{Note that $c$ has the dimension M$^{1/2}$L and could be rewritten e.g.\ as $c(T) = \tau_0 (k_B T)^{1/2} \tilde{c}(T)$, where $\tilde{c}(T)$ is a dimensionless function of $T$ and $\tau_0$ a constant with the dimension of time.}. The resulting temperature-dependence of the scattering time is shown in Fig.~\ref{fig:theory}d.

	The scattering time $\hat{\tau}_p$ in Eq.~\eqref{eq:tau_factorization} is a tensor. In order to keep time reversal symmetry, it has to be chosen such that $\hat{m}^{-1}_{be,p}\hat{\tau}_p$ is a symmetric tensor~\cite{Mikhail1980}. The five independent variables in $\hat{a}_{T}$ and $\hat{a}_{L}$ were the only arbitrary parameters used to adjust the model to all experimental results (Seebeck effect, resistivity and Hall data in the whole accessible $(T,B,\Theta)$-space). The best set of parameters found is
	\begin{equation}
		\hat{a}_{L}=
		\begin{pmatrix}
			0.538 & 0 & 0\\
			0 & 0.610 & -2.64\\
			0 & -0.0180 & 1\\
		\end{pmatrix},
	\end{equation}
	\begingroup\abovedisplayshortskip=-1em
	\begin{equation}
		\hat{a}_{T}=
		\begin{pmatrix}
			2.21 & 0 & 0\\
			0 & 2.21 & 0\\
			0 & 0 & 8.04\\
		\end{pmatrix}.
	\end{equation}\endgroup
	Due to computation time limitations, these values were not determined by an automatic fitting procedure, but by means of a manual heuristic approach with few iterations. Thus, it is very likely that a better agreement between theoretical and experimental curves could be achieved by refining the values of $\hat{a}_{L}$ and $\hat{a}_{T}$. Note that the off-diagonal entries $\hat{a}_{L,23}$ and $\hat{a}_{L,32}$ have to be non-zero in order to fulfill the requirement $\hat{m}^{-1}_{be,L}\hat{\tau}_L = (\hat{m}^{-1}_{be,L}\hat{\tau}_L)^\mathsf{T}$.

	In the framework described above, the conductivity tensors of bismuth are given by
	\begin{widetext}
		\begin{equation}\label{eq:sigma}
			\hat{\sigma} = \sum_{\mathrm{pockets}} -\sqrt{\frac{2}{\det \hat{m}_{be}^{-1}}} \frac{2e}{3\pi^2\hbar^3} \int \gamma^{3/2} (\epsilon)
			\left(\left(\frac{e}{\gamma'(\epsilon)}\hat{m}_{be}^{-1}\hat{\tau}(\epsilon,T)\right)^{-1}-\hat{B}\right)^{-1} \frac{\partial f^0}{\partial\epsilon}\ \mathrm{d}\epsilon,
		\end{equation}
		\begin{equation}\label{eq:alpha}
			\hat{\alpha} = \sum_{\mathrm{pockets}} -\sqrt{\frac{2}{\det \hat{m}_{be}^{-1}}} \frac{2}{3\pi^2\hbar^3} \int \frac{\epsilon-\mu}{T} \gamma^{3/2} (\epsilon)
			\left(\left(\frac{e}{\gamma'(\epsilon)}\hat{m}_{be}^{-1}\hat{\tau}(\epsilon,T)\right)^{-1}-\hat{B}\right)^{-1} \frac{\partial f^0}{\partial\epsilon}\ \mathrm{d}\epsilon,
		\end{equation}
	\end{widetext}
	where the sum is taken over the hole pocket at the $T$-point and the three electron pockets at the $L$-points. It was checked that the holes at the $L$-points only contribute negligibly to $\hat{\sigma}$ and $\hat{\alpha}$. $f^0$ and $\mu$ denote the Fermi-Dirac distribution and the chemical potential, respectively. From equations~\eqref{eq:sigma} and~\eqref{eq:alpha}, the zero-field and low-field Seebeck coefficient $S_{zz}$ is determined via equation~\eqref{eq:S_tensorial}. This formalism is equivalent to the one used by Mikhail et al.~\cite{Mikhail1980}. We extended this work by including Landau quantization into the model. In order to do so, the dispersion relation~\eqref{eq:Lax} has to be replaced by
	\begin{widetext}
		\begin{equation}\label{eq:dispersion_Landau}
			\epsilon (j,k_\parallel) = \pm \frac{1}{2}\left(\epsilon_g^2+4\epsilon_g \left(j \hbar \omega_c+\frac{\hbar^2 k_\parallel^2}{2m_{be,\parallel}}\right)\right)^{1/2}-\frac{1}{2}\epsilon_g + g' \mu_B s B.
		\end{equation}
	\end{widetext}
	This equation contains the quantum number $j=n+s+1/2$, where $n \in \mathbb{N}$ and $s=\pm 1/2$ is the spin quantum number~\cite{Wolff1964}, and the cyclotron frequency $\omega_c = eB/m_c$. The term $g' \mu_B s B$ accounts for the effect of the outside bands on spin splitting~\cite{Maltz1970}. The cyclotron mass $m_c$ and the longitudinal effective mass $m_{be,\parallel}$ are calculated from the effective mass tensors according to Ref.~\cite{Zhu2011}, where the values of $g'$ were also taken from. Note that the non-parabolicity of the energy band leads to unequal spacing of the Landau levels.

	From equation~\eqref{eq:dispersion_Landau}, the density of states follows as
	\begin{widetext}
		\begin{equation}\label{eq:DOS_Landau}
			D(\epsilon) = \sum_{s=\pm 1/2} \sum_{n=0}^{\infty} \frac{\abs{eB} (2m_{be,\parallel})^{1/2}}{4\pi^2\hbar^2}\frac{\gamma'(\epsilon^\ast)}{(\gamma(\epsilon^\ast) - j\hbar\omega_c)^{1/2}}.
		\end{equation}
	\end{widetext}
	Here, we use $\epsilon^\ast = \epsilon-g' \mu_B s B$ for better readability. As will be shown in section~\ref{sec:results_Landau}, there is an accumulation of electrons in the lowest Landau level leading to a significant change of the carrier density of both electrons and holes in order to keep charge compensation~\cite{Zhu2011,Zhu2017}.

	When considering Landau quantization, the conductivity tensors read
	\begin{widetext}
		\begin{equation}\label{eq:sigma_Lax_Landau}
			\hat{\sigma} = \sum_{\mathrm{pockets}} \frac{(2m_{be,\parallel})^{1/2} \abs{eB} e}{2\pi^2\hbar^2} \sum_{s=\pm 1/2} \sum_{n=0}^{\infty} \int (\gamma(\epsilon^\ast)-j\hbar\omega_c)^{1/2} \left(\left(\frac{e}{{\gamma'}(\epsilon^\ast)}\hat{m}_{be}^{-1}\hat{\tau}(\epsilon,T)\right)^{-1}-\hat{B}\right)^{-1} \frac{\partial f^0}{\partial\epsilon}\ \mathrm{d}\epsilon,
		\end{equation}
		\begingroup\abovedisplayskip=-1em
		\begin{equation}\label{eq:alpha_Lax_Landau}
			\hat{\alpha} = \sum_{\mathrm{pockets}} \frac{(2m_{be,\parallel})^{1/2} \abs{eB}}{2\pi^2\hbar^2} \sum_{s=\pm 1/2} \sum_{n=0}^{\infty} \int \frac{\epsilon-\mu}{T} (\gamma(\epsilon^\ast)-j\hbar\omega_c)^{1/2} \left(\left(\frac{e}{{\gamma'}(\epsilon^\ast)}\hat{m}_{be}^{-1}\hat{\tau}(\epsilon,T)\right)^{-1}-\hat{B}\right)^{-1} \frac{\partial f^0}{\partial\epsilon}\ \mathrm{d}\epsilon.
		\end{equation}\endgroup
	\end{widetext}
	Refer to the supplement for a more detailed derivation and a discussion of the scattering time in presence of Landau quantization~\cite{Supplement}.

	\section{Experimental details}

	\begin{figure}
		\centering
		\includegraphics[width=1\linewidth]{Fig_Setup_v2.pdf}
		\caption{\textbf{Experimental setup:} A thermal gradient was applied along the trigonal axis and measured with a thermocouple. The Seebeck effect leads to a voltage $V_S$ parallel to the thermal gradient. The Seebeck coefficient is given by $S=-V_S/\Delta T$. The magnetic field was oriented parallel to the binary-bisectrix plane, i.e.\ perpendicular to the thermal gradient.}
		\label{fig:setup}
	\end{figure}

	The Seebeck coefficient $S_{zz}$ was measured with a homemade sample holder in a Quantum Design PPMS. As shown in Fig.~\ref{fig:setup}, a thermal gradient was applied along the trigonal axis using a RuO$_2$ heater and a cold finger made out of copper. The resulting temperature difference $\Delta T$ was determined with a type E thermocouple. The voltage contacts, which were made out of silver paste, were connected to the sample holder with manganin wires. Therefore, the resulting Seebeck coefficient $S=-V_S/\Delta T$ was corrected for the contribution of manganin according to Ref.~\cite{Rathnayaka1985}. The bismuth sample was obtained commercially through MaTecK GmbH. For angle dependent measurements, the rotator option of the PPMS was used. The sample was rotated such that the magnetic field was always lying in the binary-bisectrix plane (i.e.\ perpendicular to the thermal gradient) and the angle $\Theta$ is defined such that $\Theta = 0\degree$ for $B\parallel \mathrm{binary}$.

	\section{Results}\label{sec:results}
	\subsection{Zero-field Seebeck effect}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_SvsT_v3.pdf}
		\caption{\textbf{Zero-field Seebeck coefficient vs.\ temperature:} a) Seebeck coefficient parallel to the trigonal axis ($S_{zz}$) and in the binary-bisectrix plane as a function of temperature at zero magnetic field. At low temperatures, the amplitude of the Seebeck effect is increasing linearly with temperature. Starting from $\approx 80$ K, it saturates to a very large value of $S_{zz}\approx -100\ \mu\mathrm{V\,K}^{-1}$. The theoretical curve is in very good agreement with the experimental data. Additional data is taken from Gallo et al.~\cite{Gallo1963}, Yim et al.~\cite{Yim1972} and Collaudin~\cite{Collaudin2014}. b) Contributions of holes and electrons to the zero-field thermoelectricity. It is dominated by the electrons due to their higher mobility.}
		\label{fig:SvsT}
	\end{figure*}

	The Seebeck coefficient $S_{zz}$ of bismuth is depicted in Fig.~\ref{fig:SvsT}a as a function of temperature $T$. It is negative, at $T < 80\ \mathrm{K}$ almost proportional to the temperature and shows a plateau-like behavior upon further heating. In this temperature range, $S_{zz}$ amounts to approximately $-100\ \mu\mathrm{V\,K^{-1}}$, which is a very large absolute value for a conducting material.

	How can these striking features of the zero-field Seebeck effect be understood? Firstly, the contributions of the different pockets have to be separated. The three electron pockets can be treated together, because regarding $S_{zz}$, they are equivalent when no magnetic field is applied. This is done theoretically in Fig.~\ref{fig:SvsT}b. As can be seen there, $S_{zz}$ is dominated by the electrons, which results in the negative sign of the Seebeck coefficient. The reason behind this is the mobility, which is much higher for the electrons than for the holes. Accordingly, the smaller difference of the mobilities in the binary-bisectrix plane leads to a less pronounced domination of the electrons and therefore to a lower absolute value of $S_{xx}$. Secondly, a qualitative understanding of the temperature dependence and the order of magnitude of the Seebeck coefficient can be reached by considering the Mott formula
	\begin{equation}\label{eq:Mott}
		S_{zz}^p = \frac{\pi^2k_B^2}{3e}T\left.\frac{\partial \ln(\sigma_{zz}^p)}{\partial \epsilon}\right|_{\epsilon=\epsilon_F} \propto \frac{T}{T_F^p}
	\end{equation}
	even though the condition $k_B T\ll \epsilon_F$ is not fulfilled here~\cite{Behnia2015}. The low Fermi temperatures $T_F^p$ of bismuth lead to its large Seebeck effect. Moreover, from the change of the band structure with temperature shown in Fig.~\ref{fig:theory}b follows that $T/T_F^p$ is almost constant between 100~K and 300~K for both electrons and holes. This explains why $S(T)$ is almost flat in this temperature range.

	\subsection{Magneto-Seebeck effect}
	\subsubsection{General behavior, Umkehr effect}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_umkehr_effect_v2.pdf}
		\caption{\textbf{Umkehr effect:} a) Seebeck coefficient vs.\ absolute magnetic field for $B\parallel\mathrm{binary}$ at $T=60$ K. A huge Umkehr effect can be observed, i.e.\ the Seebeck coefficient is not the same for positive and negative fields. b) As expected from symmetry considerations, the Umkehr effect is absent for $B\parallel\mathrm{bisectrix}$.}
		\label{fig:umkehr}
	\end{figure*}

	Fig.~\ref{fig:umkehr}a shows the Seebeck coefficient $S_{zz}$ at $T=60$~K as a function of the magnetic field applied parallel to a binary axis. The magnetic field strongly influences the Seebeck effect, e.g.\ a field of $B=-3$~T leads to a fivefold increase of $S_{zz}$. Furthermore, there is a large difference between positive and negative magnetic fields. At $B=+10$~T, the Seebeck coefficient amounts to $+191\ \mu\mathrm{V\,K^{-1}}$, whereas at $B=-10$~T, its value is $-277\ \mu\mathrm{V\,K^{-1}}$.

	At this point one could argue that the difference between positive and negative fields is due to a misalignment of the voltage contacts, which would then lead to a contamination of the signal by the extremely large Nernst effect~\cite{Korenblit1969,Mangez1976,Behnia2007}. However, Fig.~\ref{fig:umkehr}b indicates that this is not the case. When the magnetic field is applied along a bisectrix axis, the difference between $S_{zz} (B)$ and $S_{zz} (-B)$ is much smaller than for $B\parallel\mathrm{binary}$.

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_SvsB_all.pdf}
		\caption{\textbf{Symmetrized Seebeck coefficient vs.\ magnetic field:} Symmetrized Seebeck coefficient $S_{sym}$ as a function of the magnetic field for $20~\mathrm{K} \leq T \leq 180~\mathrm{K}$ (upper panels) and $200~\mathrm{K} \leq T \leq 300~\mathrm{K}$ (lower panels). Measurements for $B\parallel\mathrm{binary}$ and $B\parallel\mathrm{bisectrix}$ are shown on the left and on the right, respectively.}
		\label{fig:SvsB_all}
	\end{figure*}

	As explained in section~\ref{sec:theory:general}, the Seebeck effect can show an Umkehr effect if the crystal symmetry is sufficiently low. This is possible for $B\parallel\mathrm{binary}$, whereas for $B\parallel\mathrm{bisectrix}$, the Umkehr effect is forbidden due to the crystal symmetry of bismuth~\cite{Wolfe1963,Akgoz1975}. The fact that the measured difference between $S_{zz} (B)$ and $S_{zz} (-B)$ is much smaller for $B\parallel\mathrm{bisectrix}$ than for $B\parallel\mathrm{binary}$ therefore strongly suggests that this unexpected behavior can be explained by the Umkehr effect. Moreover, the good theoretical description of the observed Umkehr effect also indicates that this effect is real. In this case, the difference between $S_{zz} (B)$ and $S_{zz} (-B)$ is given by the part of the off-diagonal contribution to equation~\eqref{eq:S_contributions}, which is odd in $B$. Hence, it becomes obvious from Fig.~\ref{fig:umkehr}a that the transversal contribution to the magneto-Seebeck effect is of great importance, at least for $B\parallel\mathrm{binary}$.

	For the small difference between the two experimental curves in Fig.~\ref{fig:umkehr}b, there are three conceivable reasons: a slight misalignment of the contacts, a small misalignment of the magnetic field and the bisectrix axis or an intrinsic Umkehr effect due to lifting of the twofold symmetry by magnetostriction~\cite{Michenaud2004}. In the following, we will only show data on and discuss the symmetrized Seebeck coefficient
	\begin{equation}
		S_{sym}(B) = \frac{S_{zz}(B)+S_{zz}(-B)}{2}
	\end{equation}
	in order to facilitate the comparison of the two field directions and to minimize the effect of a potential small misalignment of the voltage contacts.

	\subsubsection{Symmetrized Seebeck coefficient}

	Fig.~\ref{fig:SvsB_all} shows the full data set on $S_{sym}(B)$ for $B<13.8$~T oriented along the binary and bisectrix directions at $20\ \mathrm{K}\leq T \leq 300\ \mathrm{K}$. At temperatures above 200~K (panels c and~d), the symmetrized Seebeck coefficient gets monotonically more negative with increasing magnetic field. Moreover, the absolute value increases when lowering the temperature. Below 200~K (panels a and~b), on the other hand, lower temperatures lead to higher values of $S_{sym}$ and $S_{sym}$ is no longer a monotonic function of the magnetic field. As we will see below, this non-monotonic behavior is a consequence of Landau quantization. In general, the field dependence of the symmetrized Seebeck coefficient is more pronounced when the field is aligned with a bisectrix axis as when it is parallel to a binary axis. Note that this statement only holds true for $S_{sym}$, but not for the actual Seebeck coefficient $S_{zz}$ including the Umkehr effect (cf. Fig.~\ref{fig:umkehr}).

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_SvsB_Nernst_v2.pdf}
		\caption{\textbf{Seebeck effect at low magnetic fields:} Symmetrized Seebeck coefficient as a function of the magnetic field for $B<0.5$ T. Experimental data is depicted in black. Coloured lines show the theoretical prediction (red) as well as its two components: The ordinary longitudinal component (blue) and the transversal contribution (green). The latter is important for both field orientations and even dominating for $B\parallel\mathrm{bisectrix}$.}
		\label{fig:SvsB_Nernst}
	\end{figure*}

	\subsubsection{Longitudinal and transversal contribution}

	The symmetrized Seebeck coefficient at low magnetic fields is depicted in Fig.~\ref{fig:SvsB_Nernst} for both field orientations and two temperatures (60~K and 120~K). It exhibits a non-trivial behavior: $S_{sym}$ is flat in a very narrow field window around 0~T, then the absolute value increases and at slightly higher fields, it starts to flatten again. In the case of $B\parallel\mathrm{binary}$ at 60~K, there is even a plateau in between (around $B=60$~mT). Increasing the temperature leads to a less pronounced response to the magnetic field. All of these features are captured by the calculations. To reach a better understanding of the underlying physics, the diagonal and off-diagonal components of the theoretical result are also plotted in Fig.~\ref{fig:SvsB_Nernst} (cf.\ Eq.~\eqref{eq:S_contributions}). When the magnetic field is applied along a bisectrix axis, the transversal contribution is clearly dominating as it sets the sign of the slope and at higher fields also the sign of the Seebeck coefficient itself. For $B\parallel\mathrm{binary}$, the absolute value of the transversal contribution to $S_{sym}$ is lower than the one of the longitudinal contribution, but it sets its slope at very low fields and is needed to explain the plateau close to $B=60$~mT. Taking into account the relevance of the off-diagonal component for the Umkehr effect, it becomes clear that the off-diagonal component, which is commonly ascribed to the Nernst and Hall effects, is indispensable to explain the magneto-Seebeck effect for both field orientations.

	\subsubsection{Angular dependence}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_SvsTheta_v2.pdf}
		\caption{\textbf{Angle-dependent magneto-Seebeck effect:} a) Symmetrized Seebeck coefficient $S_{sym}$ along the trigonal axis vs.\ orientation of the magnetic field ($B=1\ \mathrm{T}$). Note that at high temperatures, $S_{sym}$ is maximal for $B\parallel \mathrm{binary}$ ($\Theta = 0\degree$) and minimal for $B\parallel \mathrm{bisectrix}$ ($\Theta = 30\degree$). This disagrees with the prediction of Ref.~\cite{Popescu2012}. b) Corresponding theoretical curves.}
		\label{fig:SvsTheta}
	\end{figure*}

	Having understood the low field magneto-Seebeck effect in the cases when the magnetic field is oriented parallel to a main crystallographic axis, it is straightforward to compute the angular dependence of the Seebeck coefficient by choosing the tensor $\hat{B}$ in equations~\eqref{eq:sigma} and~\eqref{eq:alpha} accordingly. The result, which is depicted in Fig.~\ref{fig:SvsTheta}b, reproduces well the experimental data displayed in Fig.~\ref{fig:SvsTheta}a. The symmetrized Seebeck coefficient shows sixfold symmetry as expected from the symmetry of the Fermi surface. For $B=1$~T, at low temperatures, it is minimal for $B\parallel\mathrm{binary}$. Upon heating, the maxima and minima get inverted at 80~K (theory:\ 130~K) and the angular dependence is most pronounced roughly around 200~K. Further heating reduces the difference between peaks and dips, but $S_{sym} (\Theta)$ still shows maxima for $B\parallel\mathrm{binary}$.

	\subsubsection{Landau quantization}\label{sec:results_Landau}

	\begin{figure*}
		\centering
		\includegraphics[width=1\linewidth]{Fig_SvsB_Landau_v2.pdf}
		\caption{\textbf{Seebeck effect at high magnetic fields:} Symmetrized Seebeck coefficient as a function of the magnetic field for $B<13.8$ T. Experimental data is depicted in black ($B\parallel\mathrm{binary}$) and red ($B\parallel\mathrm{bisectrix}$). Blue and green lines show the theoretical prediction. Dashed lines correspond to the purely semiclassical theory without Landau quantization, whereas solid lines include it. The magneto-Seebeck effect of bismuth is clearly affected by Landau quantization at both 60~K and 120~K.}
		\label{fig:SvsB_Landau}
	\end{figure*}

	So far, we focused on the Seebeck effect in low magnetic fields up to 1~T. But what happens if the field strength is raised further? As shown in Fig.~\ref{fig:SvsB_Landau}, $S_{sym}$ approaches zero at high magnetic fields and even becomes slightly positive above 7.2~T for $B\parallel\mathrm{bisectrix}$ at 60~K. The minimum of the Seebeck coefficient is located at lower fields for $B\parallel\mathrm{bisectrix}$ than for $B\parallel\mathrm{binary}$ and shifts for both orientations to higher fields upon heating. Moreover, $S_{sym}$ is larger for $B\parallel\mathrm{bisectrix}$ than for $B\parallel\mathrm{binary}$ at high magnetic fields, whereas it is the other way round at low fields.

	The results of the purely semiclassical model are indicated by dashed lines in Fig.~\ref{fig:SvsB_Landau}. They are obviously not appropriate to describe the experimental data both at 60~K and 120~K as they quickly saturate at strongly negative values.

	\begin{figure*}
		\centering
		\includegraphics[width=0.86\linewidth]{Fig_mu_n_vsB_v2.pdf}
		\caption{\textbf{Field-dependence of chemical potential and carrier density:} Chemical potential (top) and carrier density (bottom) as a function of the magnetic field for $B\parallel \mathrm{binary}$ (left) and $B\parallel \mathrm{bisectrix}$ (right) at several temperatures. Landau quantization changes the density of states. Therefore the chemical potential has to shift in order to keep the electron density $n=n_{1}+n_{2}+n_{3}$ equal to the hole density $p$. Note that only one third of the hole density is plotted in the lower panels.}
		\label{fig:mu_nvsB}
	\end{figure*}

	In contrast, if Landau quantization is taken into account, the theory qualitatively captures the features described above (see solid lines in Fig.~\ref{fig:SvsB_Landau}). The quantitative differences between model and theory are smaller at 60~K than at 120~K, but in both cases Landau quantization is essential to reproduce the experimentally observed non-monotonic behavior. This shows that Landau quantization significantly affects the Seebeck effect at both temperatures. At first glance, it is very surprising that a transport property is strongly influenced by Landau quantization at temperatures as high as 120~K, because generally it only plays a role at much lower temperatures. Let us see in the following why this is the case for the Seebeck effect in bismuth.

	Because of the extremely low carrier density and the strong anisotropy of the electron pockets in bismuth, the quantum limit of electrons is unusually small. This means that a magnetic field of only 1.3~T and 1.6~T along the bisectrix and binary direction, respectively, is sufficient to confine all electrons of at least one pocket to the lowest Landau level (see Eq.~\eqref{eq:dispersion_Landau} and Ref.~\cite{Zhu2011}). Further increasing the magnetic field above the quantum limit leads to an accumulation of electrons in the lowest Landau level. In order to keep charge compensation, the chemical potential decreases (see upper panels of Fig.~\ref{fig:mu_nvsB}). This change of about 15~meV (from 0~T to 14~T) affects the carrier density of all pockets and not only of the one which has reached the quantum limit (see lower panels of Fig.~\ref{fig:mu_nvsB}). The Mott formula~\eqref{eq:Mott} gives a qualitative account of the impact on the hole pocket's contribution to thermoelectricity: As $T_F$ increases at constant $T$, its Seebeck coefficient decreases.

	For the electrons in the quantum limit, another mechanism is crucial. In general, the Seebeck effect measures the difference between the density of states above and below the chemical potential~\cite{Behnia2015}. Roughly speaking, the thermoelectric counductivity~$\hat{\alpha}$ (and hence the Seebeck coefficient) is given by the integral over the kernel $-\frac{(\epsilon - \mu)}{k_B T}\frac{\partial f^0}{\partial \epsilon}$ times the density of states (see Fig.~\ref{fig:DOS} and eqs.~\eqref{eq:alpha} and~\eqref{eq:alpha_Lax_Landau}). Now, the density of states is fundamentally changed by Landau quantization: Electrons are moved from above the chemical potential to the lowest Landau level (far below the chemical potential). Therefore, in the vicinity of the chemical potential~$\mu$, the difference between $D(\epsilon)$ for $\epsilon > \mu$ and for $\epsilon < \mu$ essentially disappears and thus the contribution to the Seebeck effect of the respective pocket almost vanishes. This still holds true at $T=120$~K even though the term $\frac{\partial f^0}{\partial \epsilon}$ is thermally broadened, because the sign change of the term $\epsilon - \mu$ at the chemical potential $\mu$ and thus the shape of the curve depicting the kernel in Fig.~\ref{fig:DOS} is independent of temperature. Since the Seebeck effect is dominated by the electrons, the measured Seebeck coefficient also approaches zero.

	\begin{figure}
		\centering
		\includegraphics[width=1\linewidth]{Fig_DOS_kernel.pdf}
		\caption{\textbf{Vanishing of the Seebeck effect in the quantum limit:} Density of states (black) in a) zero magnetic field and b) a field of 6~T along the bisectrix axis. The red curves depict the kernel $-\frac{(\epsilon - \mu)}{k_B T}\frac{\partial f^0}{\partial \epsilon}$ of the integral used to determine the thermoelectric conductivity~$\hat{\alpha}$ (cf.\ eqs.~\eqref{eq:alpha} and~\eqref{eq:alpha_Lax_Landau}) for $T=60$~K. Put simply, the Seebeck effect measures the integral of the product of this kernel and the density of states, i.e.\ the difference between the density of states above and below the chemical potential $\mu$ indicated by the broken line~\cite{Behnia2015}. In the quantum limit, the Seebeck effect vanishes upon increasing the magnetic field, because all Landau levels except the lowest Landau level (LLL) move to higher energies and the integral therefore approaches zero.}
		\label{fig:DOS}
	\end{figure}

	\section{Discussion}

	The experimental data on the zero-field Seebeck effect is in very good agreement with the values reported by Gallo et al.~\cite{Gallo1963} and Collaudin~\cite{Collaudin2014}. A small systematic error could be introduced by a misalignment of voltage and temperature contacts or by a slightly inhomogeneous heat flow.

	In a large part of the $(T,B,\Theta)$-space, the theory developed in section~\ref{sec:theory} is in good overall agreement with the experimental results presented in section~\ref{sec:results}. This is also true for the angular dependence of magnetoresistance and the Hall effect as shown in the supplement~\cite{Supplement}. For the zero-field Seebeck effect, the theoretically obtained values perfectly agree with the measurements of $S_{zz}$ and also match well the temperature dependence of $S_{xx}$ observed by Yim et al.~\cite{Yim1972} and Gallo et al.~\cite{Gallo1963}. In magnetic fields, the model works well above 50~K except for high magnetic fields at temperatures above 200~K (not shown). The agreement between experiment and theory is not as perfect as at zero field, but this would be rather mysterious given the choice we made in developing the model: We aimed at an understanding of the physical mechanisms via a model which is as simple as possible instead of perfectly reproducing experimental data by introducing a lot of adjustable parameters.

	One potential reason for the mismatch between theory and experiment at high magnetic fields above 200~K is the fact that it is not clear to what extent the Lax model (cf.\ Eq.~\eqref{eq:Lax}) is appropriate to approximate the band structure at high temperatures. Tight-binding calculations~\cite{Liu1995,Fuseya2015} suggest that the electron band bends within an energy window which could be relevant to the magneto-Seebeck effect at room temperature. We checked that taking Landau quantization out of the model does not solve this problem.

	At temperatures below 50~K, the main problem seems to be to get correct values for the Hall conductivities $\sigma_{ij}\ (i\neq j)$ which contribute to the Seebeck effect via the transversal component (cf.\ Eq.~\eqref{eq:S_contributions}). In general, it is quite difficult to predict the Hall effect of bismuth, because due to compensation the total Hall conductivity is the tiny difference of two very large values (for holes and electrons). For example, according to our model, at 60~K and 14~T, the total Hall conductivity amounts to less than $5\cdot 10^{-5}$ of the value for one carrier type. Hence, the predicted Hall conductivity is very susceptible to any changes to the model. Nevertheless, we achieve a good match between theory and the measured Hall effect above 50~K~\cite{Supplement}. At lower temperatures, one problem of our theory could be that we consider only electron-phonon scattering, but not electron-electron scattering. Furthermore, there is a recent report on a difference between bulk and surface conductance at low temperatures in bismuth~\cite{Kang2021}, which is out of the scope of the theory developed here. Lastly, phonon drag is important to the thermoelectricity of bismuth at very low temperatures~\cite{Mikhail1980,Uher1978,Issi1979}, but not taken into account here.

	As mentioned above, the Hall conductivity is very susceptible to changes to the model. This is particularly true for deviations from compensation. It was checked that tiny differences between the hole and the electron density affect the outcome of the calculations enormously. Therefore our results strongly suggest that bismuth is a perfectly compensated material. This implies that, contrary to what is sometimes assumed~\cite{Feng2021}, absence of compensation is not a prerequisite for a large Seebeck effect.

	We note that Popescu and Woods~\cite{Popescu2012} calculated the angle-dependent magneto-Seebeck effect of bismuth for $100~\mathrm{K} < T < 300~\mathrm{K}$ and $B<2$~T. They predicted the Seebeck coefficient to be minimal for $B\parallel\mathrm{binary}$ and to increase with increasing magnetic field. Both these features are in contradiction with what was observed here. In addition, they failed to predict the Umkehr effect. From our point of view, the main reason why the model of Popescu and Woods conflicts with the experimental observations is the fact that they did not take into account the transversal contribution to the magneto-Seebeck effect.

	This transversal contribution was already implicitly included in the theory of Mikhail et al.~\cite{Mikhail1980}. The present work confirms their results for low magnetic fields. However, we explicitly point out the importance of the transversal contribution to the magneto-Seebeck effect: An applied thermal gradient gives rise to a transversal electric current. But due to the boundary conditions, this current cannot flow and instead, an electric field develops both in the transversal and the longitudinal direction. The one mentioned second, which is caused by the Hall resistivity, significantly impacts the measured Seebeck voltage and hence the value of the Seebeck coefficient. Since the transversal contribution increases the absolute value of the Seebeck coefficient, it is responsible for the strong increase of the thermoelectric figure of merit $ZT$ at low magnetic fields. This explanation probably also holds true for the doubling of $ZT$ under magnetic fields of a few hundred~mT in Bi-Sb alloys, which have the largest known thermoelectric figure of merit of any solid at cryogenic temperatures~\cite{Yim1972}.

	Furthermore, this work presents for the first time correct theoretical results on the angular dependence and the high-field behavior of the magneto-Seebeck effect of bismuth. The latter is achieved by including Landau quantization into the semiclassical transport theory, mainly by changing the dispersion relation and using the appropriate density of states. A priori, it was not clear if this procedure is allowed, but a posteriori, our results strongly suggest that it is a valid approach.

	Lastly, let us highlight the conceptual importance of the Umkehr effect. We observed a huge Umkehr effect for magnetic fields along the binary axis and were able to explain it theoretically. Although the Umkehr effect has been known in principle for decades~\cite{Akgoz1975,Wolfe1963,Michenaud1970}, it seems to us as this knowledge has got lost in parts of the community. Feng and Skinner recently wrote that because of Onsager reciprocity ``the value of the Seebeck coefficient is independent of the sign of the magnetic field''~\cite{Feng2021}. As we saw above, this is not the case here and in perfect agreement with Onsager reciprocity. This also means that extracting the value of the magneto-Seebeck effect by symmetrizing with respect to field inversion~\cite{Stockert2017,Gourgout2021} is only justified when there is no uncertainty about the alignment of the symmetry axes of the electron fluid and the underlying lattice.

	\section{Summary}

	We reported for the first time on a systematic study on the magneto-Seebeck effect of bismuth. In order to understand our experimental results, we developed a model based on semiclassical transport theory to which we added Landau quantization. In a large part of the $(T,B,\Theta)$-space, the calculations are in good agreement with experimental data on the zero-field Seebeck effect, the magneto-Seebeck effect, magnetoresistance and the Hall effect.

	We found that the large difference between the mobilities of electrons and holes as well as the temperature dependence of the band structure are essential to explain the zero-field Seebeck effect of bismuth. In magnetic fields, the tranverse contribution, which is composed of conductivity tensor entries that are commonly ascribed to the Nernst and Hall effects, plays an important role in setting the amplitude of the longitudinal Seebeck effect. It also gives rise to a large Umkehr effect, i.e.\ an odd-in-$B$ component of the magneto-Seebeck effect. At high magnetic fields, the Seebeck effect of bismuth is strongly affected by Landau quantization up to temperatures as high as 120~K.

	\section{Acknowledgements}

	F.S. thanks Liangcai Xu for his support on experimental issues related to thermocouples. This work was supported by the Agence Nationale de la Recherche (ANR-18-CE92-0020-01; ANR-19-CE30-0014-04), by Jeunes Equipes de l'Institut de Physique du Coll\`ege de France and by a grant attributed by the \^Ile-de-France regional council.

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text:
Finland, Sweden and Norway will eliminate the wolf population | Conservation
Finland joins Sweden and Norway in culling wolves this winter to control their population, as conservation groups call on the European Union to take action against the killing.
Hunters in Sweden have already shot most of their annual target of 27 wolves, while Finland will authorize the killing of 20 wolves in its first "population management cull" in seven years.
Norway will kill around 60% of its wolves this winter (51 animals) to keep a maximum of just three breeding pairs in the country, and its population, including animals living between Sweden and Norway, is limited to four or six breeding pairs. .
Conservationists accuse the Nordic nations of creating the most hostile environment for wolves in Western Europe and flouting EU laws protecting the species, which has made a comeback in recent years but remains endangered in many countries.
"It's a horrible situation," said Siri Martinsen, executive director of Noah, an animal rights group that is challenging Norway's wolf hunt in its courts. "Wolf management in Norway is out of control and they are only shooting wolves because some people don't like them. It is outrageous to keep a species at a critically endangered level."
In Norway, 5% of the country is designated as a wolf protection zone, where wolf protection is a priority. Despite this, 25 wolves will die within the protection zone this winter, unless Noah's legal action, together with WWF Norway and Our Predators Association, is successful.
Wolves outside the protection zone cannot breed and are killed if a regional committee decides they "may pose a threat" to semi-domesticated cattle or reindeer.
Although Norway is not a member of the EU, wildlife groups say its killing of wolves violates the Bern Convention on the Conservation of European Wildlife and Natural Habitats.
Christian Anton Smedshaug, state secretary to the Norwegian minister for climate and environment, said: "Keeping the Norwegian wolf population at this level is a political commitment reached by the majority in parliament in 2016 to keep both wolves and livestock production in Norway and save different social visions in Norway.
"The main concern for the management of large carnivores in Norway is to maintain livestock grazing, with as little loss as possible. In addition, agriculture also contributes to common goods such as cultural landscapes and biological diversity.
"Wolves feed on game animals and consequently the presence of wolves can affect local hunting. Wolves can also pose a threat to dogs used for hunting small and large game. However, reducing effects on hunting is not a primary goal behind population targeting or a primary focus in managing large carnivores."
In Sweden, wildlife groups say the population estimate of 395 for 2020-21 could have fallen below 300 by the end of that winter.
"Sweden has promised the EU that we should not go below 300, that is the minimum," said Magnus Orrebrant, president of the NGO. Swedish Predator Association. "We have informed the EU that 300 is too low. We have a habitat that could house more than 1,000 wolves."
"The common denominator in Norway, Sweden and Finland is strong hunting organizations that politicians worry about," Orrebrant added. "There are no farms near some of the packs that are hunting this winter. The wolves haven't created any problems, but it's an important place to hunt elk and hunters want a large elk population."
Wolves are also opposed by hunters because they kill highly prized hunting dogs, widely used in Nordic nations to track game and deer.
Finland's wolf population of 300 is the highest in a century, according to Sami Niemi, an official with the Ministry of Agriculture and Forestry in charge of wolf management.
modeled by Natural Resources Institute of Finland says that a genetically healthy wolf population should be over 500.
"The long-term goal is to reach genetic viability of the wolf population," Niemi said. "When we set the target for the management hunt, we kept in mind that we were not aiming for a population reduction. The goal of hunting management is to increase tolerance to the wolf population, especially among people who share their environment with wolves."
On the argument that killing wolves assures anxious rural communities that wolf populations are under control and thus reduces illegal killing, Sami Säynevirta of nature association, a Finnish wildlife charity, said: "This argument has been made for many years, but we still have the problem of poaching. The authorities really should act to prevent it.
"There has to be a change in attitude towards wildlife. It is important to talk about the benefits of the wolf: they play a key role in a healthy ecosystem, but the news about wolves focuses on the negative side.
Professor Fiona Matthews, founding chair of Mammal Conservation Europe, said: "It seems extraordinary that countries are brazenly doing things that are illegal under the EU habitats directive. One would think that these countries could live with their predators, particularly given their low population density. It appears to be driven by hunting interests and the argument that wolves are a danger to hunting dogs."
Wildlife groups in Finland and Sweden have appealed to the European Commission and the European Court of Justice to make wolf killings illegal, but both national governments maintain that derogations from the habitats directive allow legal kills.
In Norway, Martinsen called on other European countries "to step in and complain to the Berne Convention so that we can stop this situation where Norway is leading the way in tolerating a policy of extinction and making these conventions not worth the role in which are written".
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text:
William III of England
William III and II
Portrait by Sir Godfrey Kneller, 1680s
King of England, Scotland, and Ireland
(more ...)
1689[a] – 8 March 1702
James II & VII
Co-monarch
Mary II
Stadtholder of Holland, Zeeland, Utrecht, Guelders, and Overijssel
4 July 1672 – 8 March 1702
William II
Second Stadtholderless period
Prince of Orange
4 November 1650[b] –
John William Friso (titular)
[NS: 14 November 1650][b]
Binnenhof, The Hague, Dutch Republic
8 March 1702 (aged 51)
[NS: 19 March 1702]
Kensington Palace, Middlesex, Kingdom of England
Westminster Abbey, London
Mary II of England
(m. 1677; died 1694)
Orange-Nassau
William II, Prince of Orange
Mary, Princess Royal
William III and II (4 November 1650 – 8 March 1702) was King of England and Ireland from 13 February 1689 (as William III) and he was king of Scotland from 11 April 1689 (as William II). He remained king until his death on 8 March 1702.
William was born in the Netherlands as Prince William Henry of Orange. His mother was Mary Stuart. Mary was the sister of the King of England, James II, so King James was William's uncle. William married King James's daughter Mary (his own first cousin) on 4 November 1677.
The Protestant majority in England did not like King James, and he was overthrown in the "Glorious Revolution" of 1688. William landed in England (at Brixham) with a Dutch army. When William landed, James's support melted away. James was allowed to leave for France, and William became the last person to successfully invade England by force. The events are known as the Glorious Revolution.
The English Parliament offered the English crown to William and Mary jointly (and when one died the other would be be monarch), making them William III and Mary II. This later came to be known as the reign of William and Mary. Laws were passed which protected Parliament from unreasonable acts of the Sovereign, guaranteed religious toleration to Protestant nonconformists, but restricted the religious liberty of Roman Catholics and non-Christian faiths.
William summoned a Convention of the Estates of Scotland in 1689 and sent them a conciliatory (~friendly) letter, unlike James, who had tried to give them orders. On 11 April, the day of the English coronation, the Convention finally declared that James was no longer King of Scotland. William and Mary were offered the Scottish crown, and they accepted on 11 May. The Jacobites tried for over 50 years to have James and his heirs reinstalled.
William's enemy was the King of France, Louis XIV, who protected and supported James in his efforts to return. In Ireland, the majority Roman Catholics were reinforced by French forces led by Jacobites. William personally led his army to victory at the Battle of the Boyne in 1690. James fled back to France.
There was next a Nine Years' War between a coalition led by William, and France. The Anglo-Dutch Alliance did well on sea and defeated a French fleet at La Hogue in 1692. On the other hand, the alliance lost Namur in 1692, and the Battle of Landen on land, in 1693.
Mary died on 12 December 1694 of smallpox, leaving William to rule alone. He was eventually succeeded by his sister-in-law, Queen Anne. One more fact is of great importance. In 1701 an Act of Parliament was passed which settled the succession of the thrones of England and Ireland on Protestants only. Scotland was not part of this arrangement until the parliamentary union of the two kingdoms in 1707.
↑ William was declared King by the Parliament of England on 13 February 1689 and by the Parliament of Scotland on 11 April 1689.
↑ 2.0 2.1 During William's lifetime, two calendars were in use in Europe: the Old Style Julian calendar in Britain and parts of Northern and Eastern Europe, and the New Style Gregorian calendar elsewhere, including William's birthplace in the Netherlands. At the time of William's birth, Gregorian dates were ten days ahead of Julian dates: thus William was born on 14 November 1650 by Gregorian reckoning, but on 4 November 1650 by Julian reckoning. At William's death, Gregorian dates were eleven days ahead of Julian dates. He died on 19 March 1702 by the Gregorian calendar, and on 8 March 1702 by the standard Julian calendar. (However, the English New Year fell on 25 March, so by English reckoning of the time, William died on 8 March 1701.) Unless otherwise noted, dates in this article follow the Julian calendar with New Year falling on 1 January.
William III (Britain) Citizendium
Wikiquote has a collection of quotations related to: William III of England
English monarchs
Athelstan • Edmund I • Edred • Edwy • Edgar I • Edward the Martyr • Ethelred • Sweyn I • Edmund II • Canute • Harold I • Harthacanute • Edward the Confessor • Harold II • Edgar II • William I • William II • Henry I • Stephen • Matilda • Henry II • Richard I • John • Henry III • Edward I • Edward II • Edward III • Richard II • Henry IV • Henry V • Henry VI • Edward IV • Edward V • Richard III • Henry VII • Henry VIII • Edward VI • Jane • Mary I • Elizabeth I • James I • Charles I • Interregnum • Charles II • James II • Mary II • William III • Anne
Scottish monarchs
Kenneth I • Donald I • Constantine I • Áed • Giric • Donald II • Constantine II • Malcolm I • Indulf • Dub • Cuilén • Amlaíb • Kenneth II • Constantine III • Kenneth III • Malcolm II • Duncan I • Macbeth • Lulach • Malcolm III • Donald III • Duncan II • Edgar • Alexander I • David I • Malcolm IV • William I the Lion • Alexander II • Alexander III • Margaret, Maid of Norway • First Interregnum • John Balliol • Second Interregnum • Robert I the Bruce • David II • Robert II • Robert III • James I • James II • James III • James IV • James V • Mary, Queen of Scots • James VI • Charles I • Charles II • James VII • Mary II • William III • Anne
Retrieved from "https://simple.wikipedia.org/w/index.php?title=William_III_of_England&oldid=7610508"
Kings and Queens of England
Dutch royalty
House of Orange-Nassau
Kings and Queens of Scotland
This page was last changed on 12 June 2021, at 04:21.


The King Research Institute selected ten of the largest cities in the United States of America and has ranked them based on the King Research Institute Urban Well-Being Index. Cities are changing rapidly in light of technological innovation, globalization and demographic shifts. Residents of cities desire enhanced security, job opportunities and a healthy environment. In the future the King Research Institute will post articles on the dynamics of contemporary urban society in America.


text:
Q: Permission 'documentai.processors.processOnline' denied on resource (or it may not exist) I am trying to send a POST request to the Cloud Document AI API using Postman. I have tried sending a POST request with the API key included, along with providing an OAuth access token as the OAuth 2.0 Authorization (generated using gcloud auth application-default print-access-token). However, this error is returned:
{
"error": {
    "code": 403,
    "message": "Permission 'documentai.processors.processOnline' denied on resource '//documentai.googleapis.com/projects/<project id>/locations/us/processors/<processor id>' (or it may not exist).",
    "status": "PERMISSION_DENIED",
    "details": [
        {
            "@type": "type.googleapis.com/google.rpc.ErrorInfo",
            "reason": "IAM_PERMISSION_DENIED",
            "domain": "documentai.googleapis.com",
            "metadata": {
                "resource": "projects/<project id>/locations/us/processors/<processor id>",
                "permission": "documentai.processors.processOnline"
            }
        }
    ]
}

I think this this a problem with the service account permission. If so, is there any way I can resolve this if I don't have the access to change roles?

A: Just to give an update to this question. The problem was related to the service account permission. There is no way to resolve this without setting up your service account with the correct permission. Once you have correctly set up your service account, by using the service account key of said account, you should be able to resolve this problem.
TLDR; Follow the documentation (https://cloud.google.com/document-ai/docs/setup#auth) properly. If you don't have access to the Google Account (like I did), try to get access to it. If not, I don't think there is another way around it.

A: I struggled with this question as well. To resolve the issue, go to IAM roles and change the role of your service account to Document AI service user. The default is Document AI service Agent

redpajama_set_name: RedPajamaStackExchange


Even if culture has been kidnapped as faire valoir by a boring and bored bourgoisie, it was each time the sparkle of a new thinking, a new feeling a revolt against the mainstream mind.
I hear the wind of revolt in Mahler, Beethoven, Schubert. I read it in the antic Greek philosophy, in medieval poetry. I see it in the impressionist, in the first hand spit on a wall, in each expression of human creativity who was a risk at his time and not a flattery of the wealthy.
Use culture shamelessly for your inner strength, listen to his hidden neglected power. Don't stop at the surface concept using it for class division.
Lets us re-appropriate this wild deep spirit in it back! It is our all hard gain heritage who might help us to survive the driven short thought determinist.
They don't fear anything more than creative sparkles of intelligence trespassing there own mind limitations. And nothing protects one more than the wisdom to know to be in a long lineage of courageous human.
Culture is a superbly dangerous revolutionary tool!
How many "uncultivated" workers are needed to finance the art bubble by limiting their earn to less more than nothing to finance the arbitrary nouveaux riche taste of those assuming that it needs something upon the sofa and that art is just another speculation resource more.
Art talk…is at the end, only the wrapping paper of the real deal.
If you ask children at pre-school who is an artist, all children say "me!", if years later you ask older children, you still get some voices, as adults the most pretend to not know how , and to leave that to "professionals". WHAT WENT WRONG?
My mother(painter and art professor) had show me since early age to see instead of only to "look", by stepping a bit away from the picture and closing slightly the eyes.
She used to pretend, everybody can paint, once they forget that they have been told that they cant. Creativity is natural to humans, the division from the own potential, comes later as societal dividing "pet training".
To go through "art training" is one option, but far more important is to be aware of the conditioning who happen, who destroyed the capacity to perceive the world and to communicate it expressively back .
This non perception assuming that art is for "special people", a restrict boheme and some "experts" , during the rest of humanity has more "serious" things to do, shows how dangerous creativity&perception is to a dumped downing engine who fears to be seen for what it is.inhuman!
Don t restrain yourself by forcing yourself into "being good" in whatever domain, try to let go and enjoy the playful child in yourself so long forgotten.The inner joy is the way to YOUR "great art"!
Une bonne occasion d entrer par la grande porte pour rencontrer notre histoire.
Etait proche d un sentiment "loin du monde, dans le vrais monde "expirime aussi par Mahler.
Que de bien, nous est offert a travers temps par le courage de soi de certains.


Sautée corn and fennel until tender in 1 tbsp of butter, cool, and set aside. Combine crab, 2 eggs, vegetables, mayonnaise, tarragon, salt, and pepper until everything is well incorporated. Form into 4 equal size portions and create a puck shape. Chill pucks in the refrigerator for 10 minutes to set. Dredge in flour, egg, and then into panko breadcrumbs. Sautée in 1 tbsp of butter on medium-heat on both sides until golden brown, approximately 3 minutes per side. Serve with a green salad for an entrée or make mini cakes for a great hors d'oeuvre.
The crisp fruit flavours in a Pinot Gris or Pinot Grigio are natural pairings with seafood.


Jennifer Poindexter-Runge and Jay Poindexter, Ree Heights, South Dakota, bought 11 bulls.
The whole crew at Eagle Pass Ranch presented an outstanding set of bulls and heifers for their annual production sale. This was a fantastic sale, with a huge crowd, and many bidders buying several bulls. Eagle Pass Ranch furnished the whole crowd with a delicious steak dinner before the sale.
Lot 94 for $85,000, EGL Firesteel 103F (G A R Sure Fire x MR NLC Upgrade U8676), sold to Lazy C Diamond Simmental Ranch – Kintyre, North Dakota, C Diamond Ranch – Dawson, North Dakota, Hook Farms – Tracy, Minnesota, Clear Springs Cattle Company – Starbuck, Minnesota, Schooley Cattle – Bloomfield, Iowa, and J/C Simmentals – Clare, Michigan (Black SimAngus).
Lot 100 for $19,000, EGL Firesteel 116F (G A R Sure Fire x MR NLC Upgrade U8676), sold to Gibbs Farms – Ranburne, Alabama (Black SimAngus).
Lot 101 for $12,500, EGL Firesteel 111F (G A R Sure Fire x MR NLC Upgrade U8676), sold to All Beef and ABS Global – DeForest, Wisconsin (Black SimAngus).
Lot 102 for $12,500, EGL Firesteel 115F (G A R Sure Fire x MR NLC Upgrade U867), sold to Hamilton Farms – Hitchcock, South Dakota (Black SimAngus).
Lot 98 for $12,000, EGL Firesteel 110F (G A R Sure Fire x MR NLC Upgrade U8676), sold to Werning Cattle Company – Emery, South Dakota (Black SimAngus).
Lot 1 for $11,000, EGL Forum F412 (EGL Game Changer D136 x Wingman 639T), sold to Robert and Jeremy Haselhorst – Mansfield, South Dakota (Black Balancer).


text:
Q: How to send FCM Token ID to PHP server? The FCM Token ID has been generated and I want it to send to the PHP server and then store it in a variable. What should be the approach?
    @Override
    public void onNewToken(String token) {
        Log.d(TAG, "Refreshed token: " + token);

        // If you want to send messages to this application instance or
        // manage this apps subscriptions on the server side, send the
        // Instance ID token to your app server.
        sendRegistrationToServer(token);
    }

    private void sendRegistrationToServer(String token) {
    }

PHP Code
<?php

$token = $_POST["tokenid"];
echo ($token);

?>

A: You can store you FCM-Id in Preference and then pass this FCM-Id to backend pass it as a parametr using API calling. here below i'm get FCM-Id and pas to PHP using API.
MyFirebaseInstanceIDService.java
public class MyFirebaseInstanceIDService extends FirebaseInstanceIdService {

    private static final String TAG = "MyFirebaseIIDService";
    Context context;

    /**
     * Called if InstanceID token is updated. This may occur if the security of
     * the previous token had been compromised. Note that this is called when the InstanceID token
     * is initially generated so this is where you would retrieve the token.
     */
    // [START refresh_token]
    @Override
    public void onTokenRefresh() {
        // Get updated InstanceID token.
        String refreshedToken = FirebaseInstanceId.getInstance().getToken();
        Log.d(TAG, "Refreshed token: " + refreshedToken);
        context = getApplicationContext();

        AppPreference.setStringPref(context, AppPreference.PREF_SIGNUP_FCM_ID, AppPreference.PREF_KEY.PREF_KEY_FCM_ID,
                refreshedToken);
        // If you want to send messages to this application instance or
        // manage this apps subscriptions on the server side, send the
        // Instance ID token to your app server.
        sendRegistrationToServer(refreshedToken);
    }

    // [END refresh_token]

    /**
     * Persist token to third-party servers.
     *

     * Modify this method to associate the user's FCM InstanceID token with any server-side account
     * maintained by your application.
     *
     * @param token The new token.
     */
    private void sendRegistrationToServer(String token) {
        // TODO: Implement this method to send token to your app server.

        Map<String, String> params = new HashMap<String, String>();
        String device_id = Common.getDeviceId(this);
        params.put(FCM_TOKEN, token);
        params.put(DEVICEID, device_id);
        params.put(DEVICE_TYPE, device_type);
        JsonObjectRequest request = new JsonObjectRequest(FCM_TOKEN_URL, new JSONObject(params),
                new Response.Listener<JSONObject>() {

                    @Override
                    public void onResponse(JSONObject response) {
                        try {
                            parseJsonPersonalDetail(response);
                        } catch (Exception e) {
                            e.printStackTrace();
                        }
                    }
                },
                new Response.ErrorListener() {

                    @Override
                    public void onErrorResponse(VolleyError error) {
                        if (error.networkResponse != null) {
                            int statusCode = error.networkResponse.statusCode;
                            NetworkResponse response = error.networkResponse;

                            Log.d("testerror", "" + statusCode + " " + response.data);
                        }
                    }
                }) {
            @Override
            public Map<String, String> getHeaders() {
                Map<String, String> headers = new HashMap<String, String>();
                headers.put("User-agent", "Mozilla/5.0 (TV; rv:44.0) Gecko/44.0 Firefox/44.0");
                return headers;
            }
        };

        Common.setVolleyConnectionTimeout(request);
        ApplicationClass.getInstance().getRequestQueue().add(request);

    }

    /**
     * Description - Get back response for calling  callUserDetailSave API
     *
     * @param jsonObject - Pass API response
     */
    private void parseJsonPersonalDetail(JSONObject jsonObject) {
        try {
            Log.i("get response", "get response" + jsonObject);
            if (jsonObject.toString().contains(Constant.JSON_KEY.MSG)) {
                String message = jsonObject.getString(Constant.JSON_KEY.MSG);
                String status = jsonObject.getString(Constant.JSON_KEY.CODE);
            }
        } catch (Exception e) {
            e.getStackTrace();
        }
    }
}

Here first i'm get FCM id then call API method  sendRegistrationToServer and pass token in API as a paramter so back-end developer get this token from API parameter.
Here i'm pass three parameters

*

*params.put(FCM_TOKEN, token);

*params.put(DEVICEID, device_id);

*params.put(DEVICE_TYPE, device_type);

device_id and device_type pass because it's my requirment.

Add dependency in app level gradle file for calling Volley API call :
implementation 'com.android.volley:volley:1.1.0'

Checkout i'm created Demo for you: Demo
Volley Libraries Example :
Tutorial 1
Tutorial 2
Tutorial 3


This mandala greeting card is a Milkweed Rising exclusive and is printed from an original work of digital art created by Linda S. Gribko. Circle Dance is a tribute to the half of me that sprang from Celtic ancestors, who began streaming over from the Fair Isles in the early 1600's and continued crossing the pond well into the late 1800's. Four hundred years of American roots does nothing to diminish the pull on my heart of a Celtic knot or a circle dance. I pay homage to both with this mandala created from a photo of a little portion of one of my Pleasant Hill butterfly gardens bright with orange sneezeweed and lavender butterfly bush blooms. And, yes, those definitely are garden sprites holding hand in dance as they contemplate the web of live.
This blank card makes a wonderful Birthday or Everyday greeting. It's 5″ wide and 7″ tall and is printed on premium 100 lb. white matte paper, suitable for framing. A white envelope is included. The original color photo and story of the piece are printed on the back of the card.
Please note the watermark shown in the photo does not appear on the printed card.


As the summer turns and the shadows begin to lengthen, the soundtrack to the approaching autumn has already arrived. We Are Millionaires is the second offering from former Adorable mainman Pete Fij and Terry Bickers, the highly-rated guitarist of The House Of Love. The sleeve of the record says everything, a trail of rusting cars abandoned in a forest: a symbol of decay amid a backdrop of beauty. For this represents life for the duo, who once were on the brink of stardom before they crashed under the intensity of its particularly unsettling light. Nothing is without blemish, nothing endures. Their experiences of the music business have left Pete and Terry harbouring an obsession with failure, with might-have-beens, where melancholia is a constant companion, almost comforting in its intimacy. What soundtrack, then, could be better to lead you into the season of decay and the failing of the light? Of course there is re-birth and new life in the spring, but autumn always follows, and all things fade away. This is how it sounds.
The album follows on from 2014's Broken Heart Surgery, a gloriously maudlin collection of downbeat ballads that marked Pete's return to the music scene for the first time since Polak split in 2002. A move to the Sussex coast had seen the singer-guitarist record a solo album in 2004, but the tapes of this were buried in a kitchen drawer for many years until the recruitment of Terry, who was residing in Brighton, gave him the enthusiasm to resurrect the songs, record them as a duo, and finally release them into the community.
Fij and Bickers. Photo by Guy Christie.
The new album opens with 'Let's Get Lost Together', a song that dwells on the relationship between the two musicians. It's affectionate and witty, and in scope mirrors closely the sound of Broken Heart Surgery thus making it an excellent bridge between the two records.
The whole tone of the album then expands with 'If The World Is All We Have', which explores new depths, emerging rather like a downbeat Bond theme from the Connery era. But the track has more sinister roots.
As the mind boggles, 'Love's Going To Get You' shows Pete's voice dominate; surely this is one of his best ever vocals? There's a real retro pop sound to this, with its resonating keyboard sound and one of his most obviously tongue-in-cheek lyrics including a surprise "OMG" from the literary sophisticate.
'We Are Millionaires' takes a wry look at the duo's career, gently feedbacking for half a minute before the song is finally shaken awake, though it seldom breaks out of its resigned torpor, the sound of a life support machine working at reduced power. Terry's guitars are gloriously anaesthetised, his vocal harmonies barely conscious. As he narrates, Pete wonders whether even success would break the grip of melancholy that holds the duo firmly in its grasp.
embarrassment. The lyrics stand out for their positivity, a rarity in the duo's music, though the song concludes with Pete singing, "It's been a long, long winter" for nearly a minute. It's a hint at emerging into the light, but it's more of a dream than a statement of fact.
'Marie Celeste' again clings to the idea of positivity, the song lapping at the shore like a summer sea. It's a pure love song, but again it is the thought that attracts the singer more than a real situation.
Terry Bickers. Photo by Guy Christie.
Perhaps this is the crux of the matter, Pete and Terry have an inability to escape the darkness that constantly threatens to overwhelm them. Album closer 'Sometime Soon' is almost a plea for redemption, casting a longing look at a future where genuine happiness dwells. You cannot escape the feeling that this is what Pete is really searching for.
It's a sombre conclusion to a record that can tug at your emotions whilst still making you smile through its nicely balanced wit and helpless charm. There are flickers of light here, but you have to wonder whether they are the first glimpses of the dawn or sparks from the accelerant that will burn your house to ashes. There's no doubt that Pete was looking for more positivity in this album, but only he can make positivity sound like regret, and only Terry can frame his words with the saddest guitars you have ever heard. Don't come here for salvation, it doesn't live here, but if you like your music to be intelligent and beautifully dressed Fij & Bickers always deliver.


Our 2oz. salted almond bar features our super-smooth dark chocolate generously studded with raw organic almonds, and just the right amount of Himalayan sea salt. The result is bursts of sea salt and almonds – the ultimate sweet and salty crunch in perfect balance.
Ingredients: Organic Cacao Liquor, Organic Cacao Butter, Organic Cacao Powder, Organic Honey, Organic Almonds, Himalayan Pink Salt.
Contains almonds. Made in a facility that handles tree nuts. Products that contain raw honey should not be fed to children less than 1 year of age.


I was really excited to get to watch Tomorrowland. This is a movie that both my daughter and I wanted to see and so glad we got to watch it together. It definitely did not disappoint. Tomorrowland is a movie to inspire everyone that anything is possible and that if you set your mind to it, you can achieve it. Not only is it inspiring, but it's also filled with tons of adventure that will keep you on your feet from beginning to end. I absolutely loved this movie. It was great to see both my kids really into the movie and I loved how much it inspired them and got their creativity juices flowing. Both George Clooney and Brit Robertson hit their characters spot on and worked so well with one another.
Definitely a movie to watch with the entire family. Tomorrowland is now out on both Blu-ray and DVD.
Disclosure: All opinions are my own. I received Tomorrowland on Blu-ray at no cost for the purpose of this review. No other compensation was received.


text:
Thyreus axillaris is een vliesvleugelig insect uit de familie bijen en hommels (Apidae). De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1903 door Vachal.

axillaris

redpajama_set_name: RedPajamaWikipedia


text:
Special Needs Services
Destination Weddings/Vow Renewals
Honeymoon/ Anniversary >
Romance Gift Certificates
Destination Videos
Inspirations for Travel
Travel With Us!
AMA Waterways
Tulip Time River Cruise March 20-28, 2020
Alaska Aug. 23, 2020
Spain Sept. 13-27, 2020
Peru & Machu Picchu, December 7-13, 2020
Top 10 Travel Deals
Tulip Time
7 NIGHT RIVER CRUISE with AMAWATERWAYS
CRUISE FROM : AMSTERDAM TO AMSTERDAM from Austin WITH OUR GROUP!
Celebrate spring in the Netherlands and Belgium when the Keukenhof Gardens and Floralia come alive. Experience Amsterdam's colorful canals, the whimsical Kinderdijk windmills and the art of Rubens and Michelangelo. Discover the timeless traditions of Delft porcelain and revel in the medieval architectural treasures of Bruges, Middelburg and Ghent. Throughout your journey indulge in Belgian chocolates, warm waffles and delicious cheeses—all while savoring the storybook atmosphere of the region.
COUNTRIES: , BELGIUM, NETHERLANDS
RIVERS: DUTCH AND BELGIAN WATERWAYS
March 20-ARRIVE IN AMSTERDAM, THE NETHERLANDS from Austin– EMBARKATION
Welcome to Amsterdam, where you'll board your luxurious river cruise ship. (D)
March 21-AMSTERDAM – HOORN
Enjoy a scenic cruise through the IJsselmeer to Hoorn, where you have a choice of three excursions. Visit a family-run tulip farm with over one million tulips, each in different phases of growth and in many species and colors. For those seeking to travel back in time, you can visit the Dutch East India Museum, providing insight into the Dutch Golden Age between 1500 and 1800. And for more active adventurers, go for a guided bike ride through the Dutch countryside. (B,L,D)
March 22- MIDDELBURG
After enjoying a scenic cruise through the Delta, you can explore medieval Middelburg, which was the second largest trading city of the Dutch East India Company. Or, if you feel like exploring the city on two wheels, a guided bike ride is the perfect choice. Alternatively, you may venture off to the Delta Works storm surge barrier, an impressive series of dams and Dutch hydraulic engineering. More than half of the Netherlands lies below sea level and the Delta Works were built to prevent a recurrence of the 1953 flood disaster. (B,L,D)
March 23-GHENT, BELGIUM
There are several choices of excursions, depending on your interests. For those who wish to discover two medieval cities, Ghent and Bruges, you'll wish to go on a highlights tour. Admire Ghent's canal-side architecture and some of its famous sights, including the Castle of the Counts and Ghent Cathedral, and then visit Bruges, one of the best preserved medieval cities in Europe. Alternatively, you may go on a full-day excursion to Bruges, which includes a visit to the Church of Our Lady, home to the beautiful Carrara marble Madonna and Child sculpture by Michelangelo. Those more inclined to learn about the art of Belgian chocolatemaking can join a master chocolatier as he shares his insights. And for those who wish to enjoy more free time, join a walking tour of Ghent, followed by leisure time. (B,L,D)
March 24-ANTWERP
Explore this trendy city on a walking tour where you'll see Antwerp's Steen Castle, Grote Market and Brabo Fountain, along with the UNESCO-designated Cathedral of Our Lady. For a more active exploration of the area, join a guided bike tour. Or you can visit the Rubens House, the former home of the famous 17th-century Flemish Baroque painter. Later in the afternoon, you have a choice of two excursions. Foodies will want to go on a culinary tour to indulge in chocolate, waffles and beer. Flower lovers and gardeners can roam through the Floralia, an annual spring flower show featuring approximately one million bulbs that are handplanted each year by the castle staff. (B,L,D)
March 25-ROTTERDAM, THE NETHERLANDS – KINDERDIJK - SCHOONHOVEN
During the morning, you can go on a walking tour of Rotterdam, nicknamed the "Gateway to Europe," and taste local specialties. Or visit Delfshaven, where, on August 1, 1620, the Pilgrim fathers departed on their epic journey aboard the Speedwell. Stopping in Southampton, many then boarded the Mayflower and concluded their journey at Plymouth Rock in Massachusetts. Alternatively, go on an excursion to Delft, known as the home of the painter Johannes Vermeer, Delft Blue earthenware and the Royal House. While here, you'll visit the Royal Delft Porcelain Factory and Museum. If you prefer a more active adventure, then go on a bike ride through Rotterdam and past some of its famous landmarks and architecture. In the afternoon, venture to the UNESCO World Heritage Site of Kinderdijk, a lovely village renowned for having the country's greatest concentration of windmills, all dating from the 18th century. You may also pedal past the windmills from the seat of a bike. In the evening, head to Schoonhoven, the "Silver City," known for its silversmiths and clock makers, and enjoy free time to stroll among its shops and boutiques, or along the charming canal. (B,L,D)
March 26-AMSTERDAM
One of the most impressive public gardens in the world, Keukenhof presents a spectacular display of millions of flowering tulips spread out over 70 acres. Spend time ambling through this legendary garden, open only nine weeks a year and sometimes referred to as the "Garden of Europe."Later in the day, admire Amsterdam's historic harbor and canals, where you'll see beautiful homes lining the water and the fabled Skinny Bridge. (B,L,D)
March 27-DISEMBARKATION – DEPART AMSTERDAM to Austin
Disembark the ship and prepare for your flight home. (B)
7 nights luxury accommodation in an outside stateroom (most with twin balconies)
Sun deck pool with swim-up bar and walking track
Sip & Sail Cocktail Hour with complimentary wine, beer, spirits and soft drinks
Tapas, snacks and refreshments served daily in the Main Lounge
Category E stateroom with fixed windows is in the cost of the cruise. These rooms look above the water.
Category C French Balcony.
Inclusive Cruise with room, meals, drinks, excursions per port and other amenities
Cost: Deposit of $400 pp to claim cabin on or before April 30, 2019
Get first dibs at your cabin the sooner you book!
Prices are based on double occupancy
Cost for the cruise starts at $1949 per person. (Savings of $750 pp)
Single guest paying 200% supplement count as two guest.
You can upgrade your cabin.
You can add 2 nights in Amsterdam pre-cruise for $660 per person. (Includes hotel and transfers)
International Air and transfers. Transfers can be purchased for guests with air schedules that coincide with the cruise-land departure and arrival times. Guest that purchase Amawaterways airfare plus land package will receive transfers provided. There are no deviations from the published cruise-land schedule. Transfers for deviated air schedules may be available for an additional charge. Air can be added in Mid May once flights are available.
Airline baggage fees and seat assignments
Baggage handling at airport
Gratuities for service personnel (i.e. guides, drivers, crew members) $141 can be pre-paid
Port charges are additional $189.00 per person
Entry and or departure fees (if applicable)
Travel insurance (strongly recommended)
NOTE: Quantity is limited
Any questions please contact us via phone at 512.968.9914 or email at tina@experiencetravelin.com
Managed by Voyager Websites


Ortiz was driving a green 2004 BMW 330 on northbound Interstate 5 about 1 mile south of Blaine when her car left the freeway and drove in the median for several hundred feet, the state patrol said in a report on the crash.
Her car struck a guardrail on the overpass and sailed into the air, crashing onto West Road about 20 feet below I-5. The car continued north on West Road for 100 feet, landing on its top engulfed in flames.
Ortiz was not wearing a seat belt, the report said.


Dr. Marie Nkodo is a caring and polite gynecologist serving the women of the communities of Temple Hills and Greenbelt, MD. At present, she works at Dr. Navita Modi's practice and along with her team members strives to deliver quality care to all visiting individuals. Dr. Nkodo attended medical school at Howard University and then completed additional training in obstetrics and gynecology at the same institute. She also holds certification from the American Board of Obstetrics and Gynecology. Trained to address a broad range of medical conditions, she sees patients for conditions like pregnancy cramps, uterine fibroids, anorgasmia, urinary tract infection, and miscarriage. Dr. Nkodo enjoys being an OB-GYN as she is passionate about guiding women during the most important times of their lives, which include adolescent stages, pregnancy, and menopause. She believes that it is essential for all women to understand how the reproductive system functions and actively invests in providing them with the necessary information. Her goal during patient visits is to educate women about their bodily functions and get them to be actively involved in the upkeep of their well-being.
The Rockville office is pretty run down looking. The exam room was quite gross looking. The staff and over all office atmosphere seems disorganized.
This was BY FAR the WORST experience I have ever had going to a doctor's office. My appointment was at the Temple Hills location in March 2019. The clinic appears to be pretty run down. I was 40 minutes early andthe check in process was confusing and nearly non-existent. After I wrote my name down on the check-in form as I was told, no one called my name at any point (though no one was being attended to). I had to return to the front desk and declare that I was a new patient and was there any paperwork I needed to complete. I was told there wasn't but then given a questionnaire, the I was told not to work too hard on. (The front desk person noted they are not typically running the front desk, so perhaps that is why the interaction was disoriented). I was taken in by the medical asst, 5 minutes after my appointment time, despite arriving before other patients, and despite having an earlier appointment time than those taken in earlier. After seeing the medical assistant (by the way, there was only one electronic pressure cuff to go around to check patient's blood pressure), I was told I could get undressed, wear the paper coverage, and the doctor would be in to see me shortly. Now I have waited in an exam room for 20-30 minutes before, waiting for a doctor, so I waited. I remained naked with paper draped over me (because they didn't have paper dress), and waited for OVER an hour. You may ask why I didn't check for help. I tried. After 40 minutes of waiting, 50 minutes, I kept stepping out of my room (with only paper to cover me) to inquire what was going on, when the doctor would see me, but no one was available in the hall. I said, "HELLOOOO", and got no respond. Finally after over an hour after I had been seen by the medical asst, I heard someone in the hall. I opened my door and asked where the doctor was. It was the medical asst. She appeared surprised to see me-- like they'd forgotten I was there (so much so, when I returned to the front desk, the front desk lady was surprised I was still in the office and asked me if they had forgotten about me. And everyone who arrived after me were gone). Of course, in the haste, none of my paperwork was reviewed, none of what I had told the MA was reviewed, and I had to go through the process of retelling my concerns for the 3rd time while being present (not including the Zocdoc forms). And all of what I had actually come for, didn't get discussed. (I was frustrated and exhausted and realized I would not be returning). Nothing there is electronic and apparently their paper system is ineffective since I was forgotten about and had to repeat myself several times (perhaps this is why I was told not to put too much energy into the questionnaire). The doctor did apologize to me but it was a quick sorry, and then getting down to inquiring about my complaints (which I had already told the MA). When I was leaving, I overheard the doctor complaining and expressing concern about the lack of charting in the office, which adds to the inefficiency and questionable practices of that particular office. The bathroom reeked of urine (which most medical offices manage to cloak with some aromatics). Everyone who worked there had wonderful bedside manner. They were very nice and kind. I wish that had been enough. The poor office organization and inefficiency, the limited medical supplies and treatment of them (my doc came back holding an instrument, ungloved), limited physicians, the dilapidated conditions of the center-- something about the entire experience felt disturbing and traumatizing. I don't know if the doctor I saw was Dr. Nkodo because she said her name so fast, but while she was knowledgeable and generally did a good job, this medical site should be closed. I would consider working with her if she worked somewhere of quality (assuming I knew her name) Perhaps that is why the front desk woman had asked me why I had scheduled to come there rather than the Greenbelt location. This is my WORST experience with a medical office in my life and worst experience through Zocdoc. NEVER AGAIN.
The worst! Do not waste yourtime. Your wellbeing is not important to these people at all.
It was great !!! Always leave smiling the staff are nice there ??
She was nice and helpful with questions.
I waited a full 2.5 hours before I was seen. The doctor seemed rushed. My blood pressure or weight were not taken. Doctor did not address any of my basic concerns, not one! She spoke to me for less than 5minutes. This was a total waste of time. I will be filing a complaint with my insurance company.
Please save your time and book your appointment somewhere else! Couple things bothered me. first there is no blood pressure meter at the office, they had to get it from the secretary's car because they only use one for 3 differentoffices. Second, they donnot call you about your tests. I took a pap smear test and they said "the result will be in office in 2 weeks and we will call you.". Third, bed side manner was like my mom was testing me instead of a dr. She didn't know where all the equipments were. Very unprofessional and unorganized.
Great doctor ! She answered all my questions that I had concerning my baby , made me feel really comfortable . It was little to no wait to be seen . I really will feel confident with her delivering my baby .
The one of the worst. I really don't want to go back. Never heard back from anyone for another appointment after I had my ultrasound. I caught a cold at work and when I called to get information on what Ishould take I never heard back from anyone. I was sick for 3 weeks. Lastly I needed an emergency dental appointment and when I called to get a fax for what can and cannot be given to me the lady who answered the phone was very unpleasant to speak to. I was on hold and later on hung up on. When I called back they said they will send it and my dentist never received anything! My appointment was at 9:30! And they still haven't faxed anything. Very unprofessional.


BOSKRUIN: 3 Bedroom Family home in the heart of Boskruin. 2 baths, fitted bar with fireplace, sunken lounge, dining room leading onto a patio. Well maintained garden and sparkling pool, Fitted kitchen with gas stove. Maids quarters/teenage pad. double garage. Pan handle close to Boskruin Shopping centre, Lifestyle, Norhtgate and Bellairs shopping centres, Very close to Curro Aurora, easy excess to N1 Highways..
We will e-mail you when another property of this kind becomes available in Boskruin.


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