text:
More than half of UK consumers and a third in the US in favor of plastic packaging tax
By Jenny Eagle
04-Sep-2018 - Last updated on 04-Sep-2018 at 21:08 GMT
Widespread support for plastic packaging tax. Photo: Ingredient Communications.
More than half of consumers in the UK and a third in the US are in favor of a tax on all plastic packaging on food products, according to Ingredient Communications.
Over a third of the food sold in the EU is packaged in plastic, and UK supermarkets create more than 800,000 tonnes of plastic packaging waste every year. (The Grocer, Plastic packaging increases food waste, April 2018), (The Guardian, Nearly 1m tonnes every year: supermarkets shamed for plastic packaging, 17 January 2018).
Challenge is to find alternatives
The survey for Ingredient Communications, carried out by Surveygoo, shows significant support for economic incentives to reduce the use of plastic food packaging.
A charge for plastic bags has reduced their use in the UK by over 80% since its introduction in 2015. (Gov.uk, Policy paper Carrier bags: why there's a charge, 11 January 2018).
Over half of 500 consumers in the UK (52%) said they were in favor of a tax on all plastic packaging of food products and in the US, although support was lower, 33% of American consumers were in favor of such a tax.
In the UK, 41% of consumers said they would be more likely to buy produce from supermarkets that sold fruit and vegetables without plastic packaging, with 3% saying this would make them less likely to shop there.
In the US, 38% of consumers said they would be more likely to shop in stores selling fruit and vegetables without plastic packaging, although the number saying they were less likely to do so was higher than in the UK (13%).
"There is high demand for food manufacturers to use more plastic-free packaging, and for supermarkets to introduce plastic-free features into their stores," said Richard Clarke, MD, Ingredient Communications.
"However, the benefits of plastic packaging for food and beverage products are often overlooked. It helps protect goods from damage, extends shelf life and creates a brand identity, which influences consumers' purchasing decisions.
"A key challenge for manufacturers and retailers is to find alternatives to plastic packaging that maintain these benefits for the supply chain and consumers."
In the UK, companies creating packaging waste are currently obliged to buy a 'packaging recovery note' (PRN), which effectively constitutes a tax.
As part of its 25-year environmental plan, the UK government is reported to be considering an increase in the cost of PRNs for businesses that package their goods in unrecyclable plastic.
The UK Treasury recently revealed it had received more than 162,000 responses from the public to a consultation on how taxes could help tackle plastic waste and promote recycling. The vast majority said they would back a tax on single-use plastic items, including cups, plastic cutlery, lids and straws. (The Telegraph, Tax on 'bad plastic' expected to be signed off by Government to boost recycling, 18 August 2018).
Support for action against plastic waste varies in the US, where laws are enforced state by state. California is currently the only state with a state-wide ban on plastic bags, and several other counties and cities have plastic bag bans or fees.
A law in the state of Michigan, however, prohibits local governments from banning, regulating or imposing fees on the use of plastic bags or other plastic containers.
Single-Use Plastics Directive
To address the problem of plastic pollution EUROPEN (The European Organisation for Packaging and the Environment), and 67 other European and national associations, representing the packaging value chain, have announced joint recommendations on the Commission's proposal for a Directive on the reduction of the impact of certain plastic products on the environment, i.e. the Single-Use Plastics Directive (SUP).
The 68 organizations acknowledge the problem of plastic pollution and are calling on the EU to make sure all intended and unintended consequences of the proposal are evaluated, and not to sacrifice legislative quality, effectiveness and workability for speed.
To help inform negotiations, the co-signatories make nine recommendations to improve and clarify the text in line with core policy principles such as policy coherence, Better Regulation and proportionality.
The recommendations want to ensure: a coherent EU Policy framework for packaging is maintained, the Internal Market is safeguarded, the root causes of marine litter are addressed holistically and meaningful innovation and investments are incentivised with sufficient development time from R&D to commercialization.
"Some aspects of the proposal challenge core policy principles of Better Regulation, which is of concern for all materials and sectors in the packaging value chain," said Hans van Bochove VP, public affairs and government relations, Coca-Cola European Partners and chairman, EUROPEN.
"With this cross-industry statement we want to clarify our common overarching objectives as value chain partners.
"It is crucial that policies contain clear definitions and are based on a complete evidence-based impact assessment to maintain confidence in an informed law-making process and avoid possible unintended consequence."
Virginia Janssens, MD, EUROPEN, added industry players are affected in different ways and at different levels, directly and indirectly by the SUP.
Pack2Go Europe
And last month, Pack2Go Europe, (Association of food and beverage service and convenience packaging manufacturers in Europe), officially complained to the European Ombudsman, Emily O'Reilly, about the way the Commission prepared the EU Plastic Strategy (released 16 January 2018) and then drafted the proposal for a directive on single use plastic items (released 28 May 2018).
In its submission to the EU Ombudsman, the Association raises questions about the appointment of Eunomia, a British environmental consulting firm, as the Commission's objective advisors in the consultation and policy development process around the EU Plastics Strategy.
It did this because the consultancy allegedly campaigned for a range of restrictive measures against so-called single use packaging and other single use items over several years, before being appointed as "neutral" advisors to the Commission.
Pack2Go Europe says the consultancy's objectivity was entirely compromised, that the Commission must have known this and therefore been aware of the associated conflict of interest.
The association claims at the same time Eunomia was working with the Commission on proposals to restrict single use plastics in 2017, the Commission also paid the consultancy via the intermediary of lobbying NGO Seas at Risk to research and prepare an advocacy report on single use plastics and the marine environment.
'The report was then used by Seas at Risk in the autumn of 2017 to lobby the Commission and other EU institutions at a crucial moment in the EU policy process,' it said.
'Many of the measures advocated by Eunomia in its campaigning and in the report written for Seas at Risk have since found prominence, directly or indirectly, in the proposed single use plastics Directive.'
Mike Turner, MD, Graphic Packaging International Foodservice Europe, and president, Pack2Go Europe, said it is the Commission employed as 'objective advisors' a consultancy that was evangelical about its own policy views.
"The Commission then paid the same people to lobby it at the very moment they were sitting together at the same table and drafting new EU policy together," he added.
The complaint to the EU Ombudsman then goes on to state the Commission ignored its own rules on better regulation by rushing through a partial and inadequate impact assessment on the measures to be applied to specific products.
Food hygiene
It claims the Commission carried out 'stakeholder consultation, a very partial impact assessment exercise and the legal drafting of the proposal' all at the same time.
'The impact assessment appears to have been tailored to fit the policy decisions that had already been taken, notably by omitting to address some vital elements that might have mitigated some of the proposed measures.'
As a result, Pack2Go says the draft law proposes to ban some products and restrict access to a range of food and beverage service packaging and related items (made both from plastics and combinations of plastics with other materials) that are vital to guarantee food hygiene, to safeguard public health and to provide consumer safety for food and drinks consumed out of home and on-the-go.
'The impact assessment includes no consideration of the impact of the proposal on these issues which are of vital concern to consumers, public health professionals and the food service industry.'
The risks to human health posed by cross contamination within the food chain are well-known and long-established. It is of considerable concern and a grave omission that the Commission failed to subject the proposal to a detailed impact assessment for the hygiene, health and safety risks to which it may unintentionally subject EU consumers.
The consultation is supposed to feed into the impact assessment which in turn is intended to inform the drafting of the law before the legal texts are penned. The Commission's own Scrutiny Board raised the fact that this had not happened properly but the short-circuiting of the better regulation rules seems to have been countenanced if not promoted by the commissioner for better regulation himself, Vice-President Timmermans.
"The impact assessment for the so-called plastic bags directive took two years to do properly and it was just one type of product," said Turner.
"The proposed single use plastics directive will covers hundreds of product types. The impact assessment on this is totally inadequate. It misses out vitally important analysis that will result in consumers being put at risk."
Pack2Go Europe also criticizes the impact assessment's analysis on the employment effects of the proposal, calling it "anecdotal and often based on conjecture with little or no clear factual evidence."
At no point did the Commission approach Pack2Go Europe – an association the Commission knows well – for information on the number and nature of manufacturing jobs in Europe.
"We were flabbergasted to hear Vice-Presidents Timmermans and Katainen declare in public on multiple occasions that there are no European companies and no European jobs at risk," says Turner,
"The Commission falsely suggested one of our members was manufacturing its products for Europe in India. Our members manufacture in Europe and employ Europeans."
Pack2Go Europe believes this complaint is a matter of public interest because of the risks to food hygiene, public health and consumer safety that are likely if this legislation is adopted as it is proposed and because of the direct and indirect impact on European companies and employment.
DairyReporter has contacted Eunomia for comment.
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Pop Art > Painting > Yankel Ginzburg
The Candy Store
Yankel Ginzburg
Kazakastani/American (1945)
Acrylic on Canvas, signed u.r.
Size: 38 in. x 46 in. (96.52 cm x 116.84 cm)
About The Artist: Yankel Ginzburg
Yankel Ginzburg was born Yuri Zhukov in 1945 in Alma-Ata, capital of the Kazakhstan Republic near the Chinese border. His parents were both Russian Army officers. His mother was the daughter of Marshal Georgi Zhukov, chief of the general staff of the Red Army and after Stalin's death, briefly, minister of defense. His father, a lawyer, fled from Poland and became an officer in the Russian Army. Ginzburg did not even know he was a Jew until he...
About The Medium: Acrylic
A fast-drying paint made of pigment suspended in acrylic polymer emulsion. Acrylic paints are water-soluble, but become water-resistant when dry.

redpajama_set_name: RedPajamaCommonCrawl


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\section{Madelung Fluid: Generalized Schr\"odinger Equation}

Let us consider a quantum system of a single free particle with mass $m$ whose dynamics is confined in two dimensional space, ${\bf q}=\{x,y\}$. The complex-valued wave function $\psi({\bf q};t)$, where $t$ is time, is governed by the linear Schr\"odinger equation
\begin{equation}
i\hbar\partial_t\psi({\bf q};t)=-\frac{\hbar^2}{2m}\partial_q^2\psi({\bf q};t),
\label{Schroedinger equation}
\end{equation}
where $\partial_q^2=\partial_{\bf q}\cdot\partial_{\bf q}$ is two dimensional Laplace operator.

Let us project the above dynamics onto real-space. To do this, let us write the wave function in polar form, $\psi({\bf q};t)=R({\bf q};t)\exp(iS({\bf q};t)/\hbar)$, where the quantum amplitude $R$ and the quantum phase $S$ are real-valued functions. Inserting this into Eq. (\ref{Schroedinger equation}) and separating into the real and imaginary parts, one gets
\begin{eqnarray}
\partial_tS+\frac{\partial_{\bf q}S\cdot\partial_{\bf q}S}{2m}+U=0,\hspace{10mm}\nonumber\\
\partial_t\rho+\partial_{\bf q}\cdot\Big(\frac{\partial_{\bf q}S}{m}\rho\Big)=0,\hspace{15mm}
\label{HJ-C equation}
\end{eqnarray}
where $U$ is the so-called quantum potential generated by the quantum probability density $\rho=\psi\psi^*=R^2$ as
\begin{equation}
U({\bf q};t)=-\frac{\hbar^2}{2m}\frac{\partial_q^2R}{R}.
\label{quantum potential}
\end{equation}

Next, let us postulate a velocity flow generated by the quantum phase as follows\begin{equation}
{\bf v}({\bf q};t)=\frac{1}{m}\partial_{\bf q}S({\bf q};t).
\label{velocity field-quantum phase}
\end{equation}
Using this, the coupled equations in (\ref{HJ-C equation}) can be rewritten into the following form
\begin{eqnarray}
m\frac{d{\bf v}}{dt}=-\partial_{\bf q}U,\hspace{5mm}\nonumber\\
\partial_t\rho+\partial_{\bf q}\cdot({\bf v}\rho)=0.
\label{Madelung fluid dynamics}
\end{eqnarray}
The upper equation can then be seen as an Euler equation for the velocity flow ${\bf v}({\bf q};t)$ dragged by a special kind of force field generated by the quantum amplitude, $-\partial_{\bf q}U$, later on to be referred to as quantum force. On the other hand, the lower equation can be regarded as the continuity equation for the quantum probability density, $\rho({\bf q};t)$. We have thus a non-linear fluid dynamics in real-space. This real-space hydrodynamical interpretation of Schr\"odinger equation is introduced by Madelung as soon as 1926 \cite{Madelung paper}. Moreover, the same mathematical formalism is also used to based what is later called as pilot-wave interpretation, initiated by de Broglie and expanded by Bohm \cite{de Broglie papers,Bohm paper,Bohm-Hiley's book,Holland's book}.

If we assume that the quantum phase $S$ is twice differentiable, then the velocity flow defined in Eq. (\ref{velocity field-quantum phase}) is irrotational
\begin{equation}
\partial_{\bf q}\times {\bf v}=\frac{1}{m}\Big(\frac{\partial^2S}{\partial_x\partial_y}-\frac{\partial^2 S}{\partial_y\partial_x}\Big){\bf \hat{z}}={\bf 0},
\label{irrotational flow}
\end{equation}
where ${\bf \hat{z}}$ is a unit vector orthogonal to the $xy-$plane. In this present work, we shall follow the suggestion made by Takabayashi \cite{Takabayashi paper} to consider the pair of equations in (\ref{Madelung fluid dynamics}) as a direct generalization of Schr\"odinger equation by allowing rotational flow. The special case of Madelung fluid with only irrotational flow will then reduce into the Schr\"odinger equation of (\ref{Schroedinger equation}). In other words, we shall consider Eqs. (\ref{Madelung fluid dynamics}) as describing the quantum system of a single free particle, allowing a class of wave functions whose quantum phase possesses singularity such that Eq. (\ref{irrotational flow}) is no more valid.

One of the important feature of the above Madelung fluid dynamics is that it is  self-referential. One can see that the quantum probability density $\rho({\bf q};t)$ will generate quantum potential $U({\bf q};t)$ through Eq. (\ref{quantum potential}), and in turn $U({\bf q};t)$ will tell $\rho({\bf q};t)$ the way it must evolve through Eqs. (\ref{Madelung fluid dynamics}), and so on and so forth. There is a dynamics circularity between the object to be ruled, $\rho$, and the rule, $U$. Self-referential property is a sympton of complex non-linear systems, and is argued to be the general origin for the emergence of many interesting phenomena observed in Nature \cite{self-referential dynamics and complexity}. Hence, it is of great interest to evaluate the fixed point of the self-referential Madelung fluid dynamics.

In this paper, we would like to address the following simple yet fundamental question: What is the most likely wave function of a single free particle with the average quantum mechanical energy $\langle E\rangle$ given by
\begin{equation}
\langle E\rangle=\int dq\hspace{1mm}\psi^*({\bf q})\Big(-\frac{\hbar^2}{2m}\partial_q^2\Big)\psi({\bf q}),
\label{average quantum mechanical energy}
\end{equation}
where $dq=dxdy$. We shall show that there exist a class of solutions which turns out to be the fixed point of the self-referential Madelung fluid dynamics described above.

\section{A Class of Self-Trapped Quantum Probability Densities}

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=10cm]{self-trappedQPD.eps}
\includegraphics*[width=10cm]{self-trappedQP.eps}
\end{center}
\caption{The profile of a self-trapped quantum probability density (upper) and the corresponding quantum potential it generates (lower).}
\label{2D QP-QPD}
\end{figure}

To investigate the question set up at the end of the previous section, let us discuss the following readily executable problem. First, in quantum mechanics $\rho({\bf{q}})$ provides the essential information about the position of the particle \cite{Copenhagen-ontological interpretation,Isham's book,Bohm-Hiley's book, Bell's unspeakable}. Let us quantify this information using differential entropy or Shannon information entropy for continuous random variable as \cite{Shannon entropy,Cover's book}
\begin{equation}
H[\rho]=-\int dq\hspace{1mm}\rho({\bf q})\ln\rho({\bf q}).
\label{Shannon entropy}
\end{equation}
It measures the degree of spatial localization of the quantum probability density, $\rho({\bf q})$. Now, let us search for a class of wave functions which maximizes the above Shannon information entropy, given a finite value of the average quantum potential
\begin{equation}
{\bar U}=\int dq\hspace{1mm}U({\bf q})\rho({\bf q}).
\label{average quantum potential}
\end{equation}

This is the so-called maximum entropy principle \cite{Jaynes-MEP}. It has been argued as the only way to infer from an incomplete information which does not lead to logical inconsistency \cite{Shore-Johnson-MEP}. Hence, it will give us the most likely $\rho({\bf q})$ whose average quantum potential is equal to $\bar{U}$. Maximizing Eq. (\ref{Shannon entropy}) with constraint of Eq. (\ref{average quantum potential}), one directly obtains the following relation \cite{Mackey-MEP}
\begin{equation}
\rho({\bf q};t)=\frac{1}{Z(\beta)}\exp\big(-\beta U({\bf q};t)\big),
\label{canonical quantum probability density}
\end{equation}
where $\beta$ is a real-valued constant (Lagrange constant), and $Z$ is a normalization factor given by
\begin{equation}
Z=\int dq \exp(-\beta U).
\label{normalization constant}
\end{equation}
Notice that Eq. (\ref{canonical quantum probability density}) together with the definition of quantum potential given in Eq. (\ref{quantum potential}) comprise a differential equation for $\rho({\bf q})$ or $U({\bf q})$, subjected to the condition that $\rho({\bf q})$ must be normalized. $\beta$ can then be calculated by inserting Eq. (\ref{canonical quantum probability density}) back into Eq. (\ref{average quantum potential}). Hence, it is a function of the average quantum potential, $\beta=\beta(\bar{U})$. Below, instead of using $\bar{U}$, we shall consider $\beta$ as parameter of the class of quantum probability density $\rho({\bf q};\beta)$ maximizing Shannon information entropy. Furthermore, for later convenient, we shall limit ourselves to positive definite $\beta$.

Let us show that there are infinitely many quantum probability densities satisfying Eq. (\ref{canonical quantum probability density}). To do this, applying Laplace operator to both sides of Eq. (\ref{canonical quantum probability density}), using the fact that $\rho=R^2$ and the definition of quantum potential given in Eq. (\ref{quantum potential}) one obtains the following non-linear partial differential equation
\begin{equation}
\partial_q^2U=\frac{\beta}{2}\partial_{\bf q}U\cdot\partial_{\bf q}U+\frac{4m}{\hbar^2\beta}U.\label{NPDE for quantum potential}
\end{equation}
Let us remark first that the above differential equation is invariant under the following rotation of coordinate system  \begin{eqnarray}
x'=x\cos\theta-y\sin\theta,\nonumber\\
y'=x\sin\theta+y\cos\theta,
\label{rotation}
\end{eqnarray}
where $\theta$ is the angle of rotation. Hence, if $U({\bf q})$ is a solution then so is $U({\bf q}')$.

We shall resort to numerical methods to solve Eq. (\ref{NPDE for quantum potential}). For simplicity, let us confine ourselves to a class of solutions in which the quantum probability density is separable
\begin{equation}
\rho({\bf q})=\rho_x(x)\rho_y(y),
\label{separable quantum probability density}
\end{equation}
where $\rho_i(i)$ depends only on $i=(x,y)$. In this case, the quantum amplitude is also separable $R({\bf q})=R_x(x)R_y(y)$ such that the quantum potential is decomposable as
\begin{equation}
U({\bf q})=U_x(x)+U_y(y),
\label{decomposible quantum potential}
\end{equation}
where each term on the right hand side is given by
\begin{equation}
U_i(i)=-\frac{\hbar^2}{2m}\frac{\partial_i^2R_i}{R_i},\hspace{2mm}i=x,y.
\label{partial quantum potential}
\end{equation}

Using Eq. (\ref{decomposible quantum potential}), Eq. (\ref{NPDE for quantum potential}) can be collected as
\begin{equation}
\partial_x^2U_x-\frac{\beta}{2}(\partial_xU_x)^2-\Lambda^2U_x=-\partial_y^2U_y+\frac{\beta}{2}(\partial_yU_y)^2+\Lambda^2 U_y.
\label{separation of variable 1}
\end{equation}
Hence, both sides must be equal to a constant, say $E$. Let us choose the case when $E=0$. One therefore obtains the following two de-coupled ordinary differential equations
\begin{eqnarray}
\partial_x^2U_x=\frac{\beta}{2}(\partial_xU_x)^2+\Lambda^2U_x,\nonumber\\
\partial_y^2U_y=\frac{\beta}{2}(\partial_yU_y)^2+\Lambda^2 U_y.
\label{un-coupled differential equation}
\end{eqnarray}
Fig. \ref{2D QP-QPD} shows the numerical solution of Eqs. (\ref{un-coupled differential equation}) with the boundary conditions $\partial_xU_x(0)=\partial_yU_y(0)=0$ and $U_x(0)=U_y(0)=1$, for $\beta=1$. For convenient, all numerical calculations in this paper is done by putting $m=\hbar=1$. The profile of quantum probability density (upper) is plotted together with the profile of quantum potential it itself generates (lower). We can see clearly from the numerical solution that globally the quantum probability density is being trapped by its own quantum potential.

The global self-trapping property can be justified for any positive value of $U_i(0)>0$ as follows. First, from the upper part of Eq. (\ref{un-coupled differential equation}), applying the boundary conditions at $x=0$, one gets $\partial_x^2U_x(0)=\Lambda U_x(0) > 0$, such that $U_x(x)$ is locally convex at $x=0$. Hence, since $\partial_xU_x(0)=0$, at spatial points nearby $x=0$ one has $U_x(x)>U_x(0)>0$. Moreover, since the first term on the right hand side is always non-negative, at this region one has $\partial_x^2U_x>0$. This geometrical reasoning can be extended such that for the whole spatial points one gets
\begin{equation}
\partial_x^2U_x>0.
\label{convex quantum potential}
\end{equation}
The same thing applies for $U_y$. On the other hand, from Eq. (\ref{decomposible quantum potential}) one has $\partial_x\partial_yU=0$. In general one therefore obtains $\partial_i\partial_jU\geq 0$. Hence, the case when $U_i(0)> 0$ will give an everywhere positive and convex quantum potential. ${\bf q}={\bf 0}$ turns out to be the global minimum of $U({\bf q})$. One can thus conclude that quantum probability density, $\rho({\bf q})$, is being trapped by its own self-generated quantum potential, $U({\bf q})$. Moreover, the rotational invariant of the differential equation (\ref{NPDE for quantum potential}) guarantees that one can generate a new solution by rotating the solution given in Fig. \ref{2D QP-QPD}. We have thus a class of infinitely many self-trapped quantum probability densities characterized by $\beta$.

Next, using the fact that $U_i(i)$ is positive everywhere and Eq. (\ref{partial quantum potential}), one gets
\begin{equation}
\partial_i^2R_i \le 0,\hspace{2mm}i=(x,y).
\label{concave quantum amplitude}
\end{equation}
Hence, $R_i(i)$ is everywhere concave. Since $R_i(i)$ is finite and possesses symmetrical property $R_i(i)=R_i(-i)$, then $R_i(i)$ must cross the $i-$axis at finite value of $i=\pm i_m$. In other words, $R_i(i)$ possesses only finite support on $i-$axis, namely $[-i_m,i_m]$. One can thus conclude that $R({\bf q})$ has only a finite support on $xy-$plane, that is all points ${\bf q}$ belonging to the rectangle $[-x_m,x_m]\otimes[-y_m,y_m]$. At the boundary of this rectangle, namely the lines $i=\pm i_m$, $i=(x,y)$, $U(x,y)$ is infinite, yet $R(x,y)$ is vanishing. We shall show later that for a subclass of solutions, $\bar{U}$ remains finite for non-vanishing $\beta$. Moreover, at $i=\pm i_m$, $\partial_iR_i$ is not continuous but $\partial_i\rho_i=2R_i\partial_iR_i$ is continuous and equal to zero.

We have thus developed a class of quantum probability densities satisfying Eq. (\ref{canonical quantum probability density}) which is self-trapped by its own self-generated quantum potential. Yet, we have not specified the quantum phase or the velocity vector field, to completely identify our Madelung fluid dynamics. Nevertheless, one can see that due to the global trapping property of the quantum potential, any initial velocity vector field will be dragged by the quantum force to localize even further. In the next section, we shall specify a specific velocity vector field which generates a force that exactly cancels the attractive quantum force of the trapping quantum potential to create a stationary Madelung flow.

\section{A Class of Stationary-Spinning Wave Functions}

Let us find a class of self-trapped wave functions satisfying the differential equation (\ref{NPDE for quantum potential}) which is rotationally symmetric. They are the eigenfunctions of the rotation operator. To do this, it is convenient to use polar coordinate $\{r,\theta\}$, where $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}(y/x)$. For a rotationally symmetric solution, one imposes $\partial_{\theta}U=0$. This condition leads to $\{\partial_x,\partial_y\}=\{\cos\theta\partial_r,\sin\theta\partial_r\}$. Inserting these into Eq. (\ref{NPDE for quantum potential}), one finally has to solve
\begin{equation}
\partial_r^2U+\frac{2}{r}\partial_rU-\frac{\beta}{2}(\partial_rU)^2-\frac{4m}{\hbar^2\beta}U=0.
\label{NPDE for quantum potential-rotationally symmetric}
\end{equation}
Fig. \ref{rotationally symmetric self-trapped QPD} shows the numerical solutions of Eq. (\ref{NPDE for quantum potential-rotationally symmetric}) for $\beta=1$. $\rho(r)$ (solid line) is plotted together with the corresponding $U(r)$ (dashed line). We can see clearly that globally the quantum probability density is indeed being trapped by the quantum potential it itself generates. Moreover, for a rotationally symmetric quantum potential, the attractive quantum force vector field, $-\partial_{\bf q}U$, is directing toward the origin, ${\bf q}={\bf 0}$.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=10cm]{rotationsymmQPD-QP.eps}
\end{center}
\caption{The radial profile of a self-trapped quantum probability density (solid line) trapped by its own quantum potential (dashed line). The quantum potential is shifted down such that its minimum is equal to zero and the quantum probability density is rescaled by a constant factor. See text for detail.}
\label{rotationally symmetric self-trapped QPD}
\end{figure}

Numerical solutions of the differential equation (\ref{NPDE for quantum potential-rotationally symmetric}) show that again the quantum probability density possesses a finite support, $\mathcal{M}$. Since the quantum probability is rotationally symmetric, then the support is a disk with finite radius. Let us denotes the radius as $r=r_m$. At the boundary of the support, $\partial\mathcal{M}$, which makes a circle $x^2+y^2=r_m^2$, the quantum potential is infinite, $U(r_m)=\infty$. See Fig. \ref{rotationally symmetric self-trapped QPD}. In Fig. \ref{radius of support}, we plot the radius of the support $r_m(\beta)$ and the second moment defined as $\bar{r^2}(\beta)=\int dq r^2\rho(r)$ against the value of $\beta$. Both are evaluated numerically. We can see that both quantities behave almost similarly as the function of $\beta$. First, both increase very quickly for small $\beta$ and then both seem to converge toward certain finite values for infinite value of $\beta$
\begin{equation}
\lim_{\beta\rightarrow\infty}r_m\equiv r_{\infty},\hspace{2mm}\lim_{\beta\rightarrow\infty}\bar{r^2}\equiv r^2_{\infty}.
\label{radius of support at infinite beta}
\end{equation}
In the next section, we shall discuss in more detail the behavior of $\rho(r;\beta)$ and $U(r;\beta)$ for $\beta\rightarrow\infty$.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=10cm]{spatiallocalization.eps}
\end{center}
\caption{$r_m$ (upper) and $\bar{r^2}$ (lower) against $\beta$.}
\label{radius of support}
\end{figure}

For the opposite limiting case, $\beta\rightarrow 0$, both quantities, $r_m$ and $\bar{r^2}$, are approaching zero. One can thus conclude that the rotationally symmetric self-trapped quantum probability density is converging toward a delta function,
\begin{equation}
\lim_{\beta\rightarrow 0}\rho({\bf q};\beta)\rightarrow\delta({\bf q}).
\label{delta function at vanishing beta}
\end{equation}
In Fig. \ref{small beta self-trapped QPD-QP}, we confirm the above conclusion numerically by plotting the radial profile of the quantum probability densities for decreasing small values of $\beta=10^{-4},10^{-5},5\times 10^{-6},10^{-6}$. One can also see that for small $\beta$ the quantum probability density is approximately taking a form of a rectangular function.

Next, let us write the coupled dynamical equations of (\ref{Madelung fluid dynamics}) in polar coordinate, $\{r,\theta\}$, to give us \cite{classical mechanics book}
\begin{eqnarray}
m\Big(\frac{dv_r}{dt}-r\omega^2\Big)=-\partial_rU,\hspace{2mm} \nonumber\\
m\Big(r\frac{d\omega}{dt}+2v_r\omega\Big)=-\partial_{\theta}U,\nonumber\\
\partial_t\rho+\partial_r(\rho v_r)=0.\hspace{12mm}
\label{Madelung's fluid in polar coordinate}
\end{eqnarray}
where $\omega=d\theta/dt$ is the angular velocity and $v_r=dr/dt$. In the last line we have used the fact that our dynamical system is rotationally symmetric. Let us now impose a stationary condition $v_r=0$. First, from the up most equation, the angular velocity is related to the quantum potential as
\begin{equation}
\omega=\sqrt{\partial_rU/(rm)}.
\label{spinning-stationary condition}
\end{equation}
Hence, the angular velocity depends only on the distance, $\omega=\omega(r)$. From the middle equation, since $\partial_{\theta}U=0$, one gets $d\omega/dt=0$, such that the angular velocity is constant. Finally, from the lower most equation, one gets $d\rho/dt=\partial_t\rho=0$. Hence, $\rho(r;t)$ is stationary and spinning around its center with a stationary angular velocity field, $\omega(r)$.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=9cm]{smallbetaQPD.eps}
\end{center}
\caption{The profile of $\rho(r)$ for various small values of $\beta$.}
\label{small beta self-trapped QPD-QP}
\end{figure}

Going back to Cartesian coordinate, the rotating velocity vector field can be obtained as
\begin{equation}
{\bf v}=\omega(r)\times {\bf q}=\{-\omega(r) y,\omega(r) x \}.
\label{rotating velocity vector field}
\end{equation}
This velocity vector field is divergence-less
\begin{equation}
\partial_{\bf q}\cdot{\bf v}=-\frac{xy}{r}\frac{\partial \omega}{\partial r}+\frac{yx}{r}\frac{\partial\omega}{\partial r}=0.
\label{divergence-less velocity vector field}
\end{equation}
This fact guarantees the existence of a twice differentiable scalar-valued function $\varphi({\bf q})$, a stream function \cite{vector calculus}, such that
\begin{equation}
{\bf v}=\{\partial_y\varphi,-\partial_x\varphi\}.
\label{Poincare theorem: the existence of stream function}
\end{equation}
This means that the tangent of a curve $\varphi({\bf q})=\varphi_0$, where $\varphi_0$ is arbitrary constant, is parallel to the velocity vector ${\bf v}$. The set of curves $\varphi({\bf q})=\varphi_0$ with various values of $\varphi_0$ then make concentric circles with different radius all enclosing ${\bf q}={\bf 0}$. See Fig. \ref{singular phase}.

If one further defines the quantum phase corresponding to this rotational flow as in Eq. (\ref{velocity field-quantum phase}), one will have the following differential equations
\begin{equation}
\partial_xS=m\partial_y\varphi,\hspace{3mm}\partial_yS=-m\partial_x\varphi.
\label{quantum phase rederived}
\end{equation}
From Eqs. (\ref{Poincare theorem: the existence of stream function}) and (\ref{quantum phase rederived}), one gets
\begin{equation}
\partial_{\bf q}S\cdot\partial_{\bf q}\varphi=m(v_x(-v_y)+v_yv_x)=0.
\label{stream function and stream function are orthogonal}
\end{equation}
Hence all curves $S({\bf q})=S_0$ are orthogonal to all curves $\varphi({\bf q})=\varphi_0$, where $S_0$ and $\varphi_0$ are arbitrary constants. Since the curves $\varphi({\bf q})=\varphi_0$ are a set of concentric circles with different radius depending on the value of $\varphi_0$, then the set of curves $S({\bf q})=S_0$ are line rays emanating from the origin ${\bf q}={\bf 0}$. See Fig. \ref{singular phase}. This means that at ${\bf q}={\bf 0}$, $S({\bf q})$ is not uniquely defined. Hence, at ${\bf q}={\bf 0}$, we have a phase defect or singular point such that Eq. (\ref{irrotational flow}) is not valid.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=9cm]{singularphase.eps}
\end{center}
\caption{The set of concentric circles are the curves of $\varphi({\bf q})=\varphi_0$, whereas the set of lines emanating from ${\bf q}={\bf 0}$ are the curves of $S({\bf q})=S_0$. $\varphi_0$ and $S_0$ are arbitrary constants.}
\label{singular phase}
\end{figure}

Finally, let us notice that Eq. (\ref{quantum phase rederived}) leads to an important fact that $S({\bf q})$ satisfies Laplace equation
\begin{equation}
\partial_{q}^2S=m\partial_x\partial_y\varphi-m\partial_y\partial_x\varphi=0.
\label{Laplace equation}
\end{equation}
Hence, the quantum phase $S({\bf q})$ of our stationary wave function is a two dimensional harmonic function. On the other hand, the stream function satisfies the following Poisson equation
\begin{equation}
\partial_q^2\varphi=-\partial_{\bf q}\times {\bf v}\neq {\bf 0}.
\label{Poisson equation}
\end{equation}

Keeping in mind all the above facts about the quantum phase, let us  calculate the average quantum mechanical energy. Writing the wave function in polar form, $\psi=R\exp(iS/\hbar)$, one obtains
\begin{eqnarray}
\langle E\rangle
=\int_{\mathcal{M}} dq\hspace{1mm}R\Big(-\frac{\hbar^2}{2m}\partial_q^2\Big)R
-i\frac{\hbar}{m}\int_{\mathcal{M}} dq\hspace{1mm}R\partial_{\bf q}R\cdot\partial_{\bf q}S\nonumber\\
+\frac{1}{2m}\int_{\mathcal{M}} dq\hspace{1mm}R^2\partial_{\bf q}S\cdot\partial_{\bf q}S
-i\frac{\hbar^2}{2m}\int_{\mathcal{M}} dq\hspace{1mm}\psi^*\partial_q^2S.\hspace{3mm}
\label{average quantum mechanical energy revisited}
\end{eqnarray}
Notice that we have made clear that all the spatial integration is taken over a finite support $\mathcal{M}$ inside which $R({\bf q})\neq 0$.
The first term on the right hand side is but equal to the average of quantum potential
\begin{equation}
\int_{\mathcal{M}} dq\hspace{1mm}R\Big(-\frac{\hbar^2}{2m}\partial_q^2\Big)R=\int_{\mathcal{M}} dq\hspace{1mm}\rho({\bf q})U({\bf q})=\bar{U}.
\end{equation}
Next, recalling the fact that $R$ depends only on $r$, and Eqs. (\ref{velocity field-quantum phase}), (\ref{rotating velocity vector field}), one has
\begin{equation}
\partial_{\bf q}R\cdot\partial_{\bf q}S=-m\frac{xy}{r}\omega\partial_rR+m\frac{yx}{r}\omega\partial_rR=0.
\end{equation}
Hence, the second term on the right hand side of Eq. (\ref{average quantum mechanical energy revisited}) is vanishing. Further, using again Eq. (\ref{velocity field-quantum phase}), the third term on the right hand side of Eq. (\ref{average quantum mechanical energy revisited}) can be rewritten as
\begin{eqnarray}
\frac{1}{2m}\int_{\mathcal{M}} dq\hspace{1mm}R^2\partial_{\bf q}S\cdot\partial_{\bf q}S=\int_{\mathcal{M}} dq\hspace{1mm}\rho\Big(\frac{1}{2}m{\bf v}\cdot {\bf v}\Big) \nonumber\\
\equiv \int_{\mathcal{M}} dq\hspace{1mm}\rho({\bf q})K({\bf q})= \bar{K}.\hspace{20mm}
\end{eqnarray}
Here, we have defined a new quantity $K=(1/2)m{\bf v}^2$ whose averaged over $\rho({\bf q})$ is denoted by $\bar{K}$. Namely, $\bar{K}$ can be interpreted as the average kinetic energy of the Madelung fluid. Notice that since ${\bf v}(r)$ depends on $U(r)$, so will $\bar{K}$. We shall show later that $\bar{K}$ depends only on the value of $\beta$, $\bar{K}=\bar{K}(\beta)$. Since $\beta$ depends on $\bar{U}$, one then conclude that $\bar{K}$ depends on $\bar{U}$. Finally, by Eq. (\ref{Laplace equation}), the last term on the right hand side of Eq. (\ref{average quantum mechanical energy revisited}) is vanishing. In total, we thus have the following decomposition of average quantum mechanical energy into the average quantum potential and average kinetic energy
\begin{equation}
\langle E\rangle=\bar{U}+\bar{K}.
\label{internal energy}
\end{equation}

All the above facts tell us that the self-trapped, rotationally symmetric, spinning yet stationary quantum probability density we developed in this section comprises the class of wave functions that maximizes the Shannon information entropy, given the average quantum mechanical energy of the free particle. \textit{It is thus the most likely wave function of a single free particle in two dimensional space with quantum mechanical energy $\langle E\rangle=\bar{U}+\bar{K}$}. Of course we have to remind ourselves that we have worked in the frame work of Madelung fluid which we assume as the generalization of quantum mechanics allowing rotational flow.

\section{A Class of Self-Trapped, Non-Spinning Yet Stationary Wave Functions}

Let us discuss the behavior of $\bar{U}$ and $\bar{K}$ of our stationary-spinning wave function as we vary $\beta$. From the stationary condition of Eq. (\ref{spinning-stationary condition}), it is clear that for finite value of $\beta$, the rotating velocity of the Madelung fluid is infinite at the boundary of the support, $\partial\mathcal{M}$. Yet, one should again keep in mind that along this circle of radius $r_m$, $r^2=x^2+y^2=r_m^2$, the quantum probability density is vanishing. For our stationary state, one can then show that the average kinetic energy is finite for non-vanishing value of $\beta$ as follows
\begin{eqnarray}
\bar{K}=\int_{\mathcal{M}} dxdy\hspace{1mm}(1/2)m{\bf v}^2\rho(r)\hspace{40mm}\nonumber\\
=m\pi\int_0^{r_m}dr\hspace{1mm}r^2\partial_rU\rho(r)
=-m\pi\beta^{-1}\int_0^{r_m}dr\hspace{1mm}r^2\partial_r\rho(r)\nonumber\\
=2m\pi\beta^{-1}\int_0^{r_m}dr\hspace{1mm}r\rho(r)=\frac{m}{\beta},\hspace{30mm}\label{kinetic energy versus beta}
\end{eqnarray}
where in the second equality we have used Eq. (\ref{spinning-stationary condition}), in the third equality we have used Eq. (\ref{canonical quantum probability density}) and in the fourth equality we used the partial integration. We have thus an important result which shows that the average kinetic energy $\bar{K}$ depends only on $\beta$, inversely. For non-vanishing $\beta$, assuming a system with finite average quantum mechanical energy $\langle E\rangle$, then the average quantum potential, $\bar{U}=\langle E\rangle-m/\beta$, is also finite. Moreover, given a finite value of average quantum mechanical energy, the definite positivity of $\bar{U}$ leads to the existence of a lower bound for the possible values of $\beta$ as $\beta>m/\langle E\rangle$, such that $r_m$ is also non-vanishing.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=9cm]{energyversusbeta.eps}
\end{center}
\caption{The average quantum potential, $\bar{U}$, and average kinetic energy $\bar{K}$ versus $\beta$. $U$ is obtained for the solution of the differential equation (\ref{NPDE for quantum potential}) with fixed boundary conditions, while $\bar{K}$ is obtained from Eq. (\ref{kinetic energy versus beta}).}
\label{energy versus beta}
\end{figure}

Fig. \ref{energy versus beta} shows the numerical plots of the average kinetic energy $\bar{K}$ and the average quantum potential $\bar{U}$ againstf $\beta$ for the stationary-spinning wave functions which satisfy the differential equation (\ref{NPDE for quantum potential}) with fixed boundary conditions and the stationary condition given by Eq. (\ref{kinetic energy versus beta}). One can see that, like $\bar{K}$, $\bar{U}$ decreases as we increase the value of $\beta$. Two limiting cases are of great interest. First is the case when $\beta$ is vanishing, $\beta\rightarrow 0$. As discussed in the previous section, in this case the quantum probability density is approaching a delta function with vanishing support. From Eq. (\ref{kinetic energy versus beta}) one has $\lim_{\beta\rightarrow 0}\bar{K}\rightarrow \infty$. Numerical simulation also shows that the average quantum potential is approaching infinity, $\lim_{\beta\rightarrow 0}\bar{U}\rightarrow\infty$. Hence, this limiting case is not a good model or irrelevant for a single free particle of finite average quantum mechanical energy.

Let us proceed to discuss the other extreme limiting case of infinite $\beta$, $\beta\rightarrow \infty$. Again, as discussed in the previous section, in this case, the radius of the support $r_m$ and the second moment $\bar{r^2}$ of the quantum probability density are converging toward some certain finite values. This shows that the stationary quantum probability density and its corresponding quantum potential are converging toward some functions
\begin{eqnarray}
\lim_{\beta\rightarrow\infty}\rho({\bf q};\beta)\equiv\rho_{\infty}({\bf q}),\nonumber\\
\lim_{\beta\rightarrow\infty}U({\bf q};\beta)\equiv U_{\infty}({\bf q}).
\label{non-spinning QPD-QP}
\end{eqnarray}
This also tells us that the average quantum potential, $\bar{U}$, is converging toward a finite value, $\lim_{\beta\rightarrow\infty}\bar{U}\equiv\bar{U}_{\infty}$, which is confirmed  by numerical simulation. See Fig. \ref{energy versus beta}. On the other hand, from Eq. (\ref{kinetic energy versus beta}), in this extreme case, the average kinetic energy is vanishing, $\lim_{\beta\rightarrow\infty}\bar{K}\rightarrow 0$. Hence in this limiting case, the average quantum energy is exactly equal to the average quantum potential and thus finite
\begin{equation}
\lim_{\beta\rightarrow\infty}\langle E\rangle=\lim_{\beta\rightarrow\infty}\bar{U}= \bar{U}_{\infty}.
\end{equation}
Moreover, since the average kinetic energy is vanishing, the quantum probability density is no more spinning. This situation is of great interest physically, since we have a lump of self-trapped quantum probability density which is non-spinning yet is still stationary.

Let us investigate this last situation in more detail. Fig. \ref{the quantum potential at the boundary} shows the profile of the quantum potential at increasing values of large $\beta$, $\beta=1, 5,10,100$. One can see that the profile of quantum potential is approaching the form of a cylindrical tube with infinite wall, as we increase $\beta$. Hence, for infinite value of $\beta$, the quantum potential is finite and constant inside the support $\mathcal{M}$, and infinite at $\partial{\mathcal{M}}$. Inside the support $\mathcal{M}$, the quantum force is thus vanishing. On the other hand, along the boundary $\partial\mathcal{M}$, the quantum force is infinite. Yet one should keep in mind that along this circle boundary $\partial\mathcal{M}$, the quantum probability density is vanishing. Hence, inside $\mathcal{M}$, both the quantum force and the centrifugal force are vanishing. This ensures that $\rho_{\infty}({\bf q})$ is a non-spinning-stationary solution of the two dimensional Madelung fluid.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=9cm]{largebetaQP.eps}
\end{center}
\caption{The radial profile of quantum potential for various large values of $\beta$.}
\label{the quantum potential at the boundary}
\end{figure}

Let us now proceed to discuss the profile of the wave function at the limit $\beta\rightarrow\infty$. As discussed in the previous paragraph, in this limit the quantum potential inside the support is constant. Let us denotes the constant value of $U$ inside $\mathcal{M}$ by $U_{c}$ (See Fig. \ref{the quantum potential at the boundary}). One therefore has
\begin{eqnarray}
\langle E\rangle=\bar{U}_{\infty}=\int dq\hspace{1mm}\rho_{\infty}({\bf q})U_{c}=U_{c}.
\end{eqnarray}
Next, recalling the definition of quantum potential given in Eq. (\ref{quantum potential}), inside $\mathcal{M}$, one has
\begin{equation}
-\frac{\hbar^2}{2m}\partial_q^2R_{\infty}({\bf q})=U_{c}R_{\infty}({\bf q})=\langle E\rangle R_{\infty}({\bf q}),
\label{infinite beta wave function 1}
\end{equation}
where we have denoted the quantum amplitude at infinite $\beta$ by $\lim_{\beta\rightarrow\infty}R({\bf q;\beta})\equiv R_{\infty}({\bf q})$. The above differential equation must be subjected to the boundary condition that along the boundary line of the support, $\partial\mathcal{M}$, the quantum amplitude is vanishing. Moreover, since in the limit $\beta\rightarrow\infty$ the phase $S$ is constant inside the support $\mathcal{M}$, multiplying both sides with $\exp(iS/\hbar)$, one gets
\begin{equation}
-\frac{\hbar^2}{2m}\partial_q^2\psi_{\infty}({\bf q})=\langle E\rangle \psi_{\infty}({\bf q}),
\label{infinite beta wave function 2}
\end{equation}
where we have denoted $\psi_{\infty}({\bf q})\equiv R_{\infty}({\bf q})\exp(iS/\hbar)$.

Eq. (\ref{infinite beta wave function 2}) with the boundary condition described above is nothing but the stationary (time-independent) Schr\"odinger equation for a particle trapped inside a cylindrical tube classical potential whose bottom is flat and boundary is infinitely high. In polar coordinate, recalling the wave function is rotationally symmetric, one obtains the following differential equation
\begin{equation}
\partial_r^2\psi_{\infty}(r)+\frac{2}{r}\partial_r\psi_{\infty}(r)=-\frac{2m\langle E\rangle}{\hbar^2}\psi_{\infty}(r).
\label{infinite beta wave function 3}
\end{equation}
The solution of which is given by
\begin{equation}
\psi_{\infty}(r)=AS_c(kr)\exp(iS_0/\hbar),
\label{infinite beta wave function}
\end{equation}
where $S_c(kr)=\sin(kr)/r$ is the Sinc function, $k=\sqrt{2m\langle E\rangle/\hbar^2}$, $A$ is a normalization constant and $S_0$ is an arbitrary phase constant. $k$ is thus a quantity of momentum dimensional. Fig. \ref{profile of infinite beta wave function} shows the quantum probability densities for various increasing large values of $\beta$ obtained by solving Eq. (\ref{NPDE for quantum potential}) with fixed boundary conditions, and its limiting case $\rho_{\infty}(r)$ given by Eq. (\ref{infinite beta wave function}). One can see that as $\beta$ is getting larger, $\rho(r;\beta)$ is converging toward $\rho_{\infty}(r)$. Already at $\beta=50$, $\rho(r;\beta)$ is difficult to be distinguished from $\rho_{\infty}(r)$. Next, from the boundary condition that the quantum probability density must be vanishing along the boundary line of the support, the Sinc function $S_c(kr)$ must reach its first zero point at $r=\lim_{\beta\rightarrow\infty}r_m(\beta)\equiv r_{\infty}$. One therefore has the relation between the average quantum mechanical energy of the single free particle and the radius of the wave function as $kr_{\infty}=\pi$.

\begin{figure}[tbp]
\begin{center}
\includegraphics*[width=9cm]{largebetaQPD.eps}
\end{center}
\caption{$\rho({\bf q};\beta)$ for various values of $\beta$ which solve Eq. (\ref{NPDE for quantum potential}) with fixed boundary conditions and its limiting case for infinite $\beta$, $\rho_{\infty}(r)$, given by Eq. (\ref{infinite beta wave function}).}.
\label{profile of infinite beta wave function}
\end{figure}

\section{Conclusion and Discussion}

To conclude, we have first assumed the Madelung fluid dynamics as a generalization of Schr\"odinger equation for a single free particle moving in two dimensional space, allowing a rotational flow. We then ask for a class of wave functions whose quantum probability density maximizes Shannon information entropy given the average value of quantum mechanical energy. We showed that there exist a class of solutions in which the quantum probability density is self-trapped, spatially localized with finite support, rotationally symmetric, spinning around its center and yet stationary. Hence it is a quantum vortex in $xy-$space. The stationarity  comes from the balance between the attractive quantum force field of the trapping quantum potential generated by the quantum amplitude and the fictitious repulsive centrifugal force field of the rotating velocity field generated by the quantum phase.

We then proceeded to show an asymptotic class of solutions in which the quantum probability density is no more spinning and yet still stationary. In this case, the quantum potential takes the form of a cylindrical tube with flat bottom and infinitely high wall. Moreover, we also showed that the wave function turns out to be the lowest eigenstate of the time-independent Schr\"odinger equation for a single particle trapped inside a cylindrical tube classical potential. In this sense, one can consider the class of spinning-stationary solutions of the Madelung fluid dynamics as the generalization of the notion of stationary states of the ordinary quantum mechanics. Yet, one should recall that in constrast to the ordinary quantum mechanics, we did not assume any classical potential to trap the particle.

Some immediate interesting problems are then ready to be mentioned for future works. It is first interesting to study the case when the initial quantum phase does not satisfy the stationarity condition given in Eq. (\ref{spinning-stationary condition}). In particular, it is instructive to investigate the stability of the stationary solutions with respect to a small perturbation either on the quantum amplitude or quantum phase. In this context, it is interesting to compare the class of localized, stationary and spinning solutions developed in this paper with other localized wave phenomena like soliton etc. The case of three spatial dimension is then of great interest. Moreover, one is tempted to recover the quantization of energy in ordinary quantum mechanics using the formalism developed in this work.

\begin{acknowledgments}

One of the author, A. B., would like to thank Masashi Tachikawa for useful and stimulating discussion.

\end{acknowledgments}

redpajama_set_name: RedPajamaArXiv


text:
Greg Yetman
Greg Yetman is associate director for the Geospatial Applications Division at CIESIN. He is a geographer specializing in the application of geographic information system (GIS) technologies in applied and research fields, including population geography, natural disasters, and environmental assessment. He has acted as a technical advisor on GIS to government and international organizations, including the United Nations Geographic Information Working Group (UNGIWG), the Global Spatial Data Infrastructure Association (GSDI), the National Center for Atmospheric Research (NCAR) climate data working group, and the Canadian province, Newfoundland and Labrador. He has led implementation of geospatial services for the NASA Socioeconomic Data and Applications Center (SEDAC) that have been used in initiatives such as the Global Earth Observation System of Systems (GEOSS) architecture implementation pilots.
Yetman has played an integral role in CIESIN activities including the development of the Gridded Population of the World (GPW) data set, the Natural Disaster Hotspots project, and the Global Rural-Urban Mapping Project (GRUMP). He has taught courses as an adjunct professor for Columbia University's Department of Earth and Environmental Engineering and Mailman School of Public Health. He previously worked as a senior consultant and business analyst for Esri Canada and as a consultant to the North American Commission for Environmental Cooperation.
Balk, D., U. Deichmann, G. Yetman, F. Pozzi, S. Hay, and A. Nelson. 2006. "Determining Global Population Distribution: Methods, Applications and Data." In Advances in Parisitology: Global Mapping of Infectious Diseases, ed. S. Hay, A. Graham, and D. Rogers. London: Elsevier.
de Sherbinin, A., M. Levy, S. Adamo, K. MacManus, G. Yetman, V. Mara, L. Razafindrazay et al. 2012. Migration and risk: Net migration in marginal ecosystems and hazardous areas. In Environmental Research Letters 7 (4): 1–14. http://dx.doi.org/10.1088/1748-9326/7/4/045602
Dilley, M., R.S. Chen, U. Deichmann, A.L. Lerner-Lam, M. Arnold with J. Agwe, P. Buys, O. Kjekstad, B. Lyon, and G. Yetman. 2005. Natural Disaster Hotspots: A Global Risk Analysis. Washington, D.C.: World Bank.
Meredith, T., G. Yetman, and G. Frias. 2002. "Mexican and Canadian case studies of community-based spatial information management for biodiversity conservation." In Community Participation and Geographic Information Systems, ed. W. Craig, T. Harris, and D. Weiner. New York: Taylor and Francis.
This page last modified: Oct 10, 2014


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Of course, the restaurant is not named after the chart-topping hit, "Gangnam Style," but rather the Gangnam District, an affluent part of Seoul.
Park and Cheong, who hail from South Korea, felt it was a good time to open a Korean restaurant in Boise — one that favors traditional fare over trendy fusion cuisine. They even brought in a chef from Seoul, Helena Jeong, Sukie's sister, to run the kitchen.
Korean food, with all its fire and ferment, is strange and wonderful when not dumbed down to appease American palates.
While Gangnam does offer a few holdout Japanese dishes from Wasabi's menu, it primarily focuses on Seoul-inspired cuisine, which is friendly to the gluten-free crowd.
Once seated, I was happy to receive an array of traditional banchan plates. You know, those gratis tiny dishes of kimchi and other fermented goodies meant to get things started.
Gangnam stays the traditional course with its banchan selection, and they keep them coming throughout the meal. Be prepared to taste a fiery napa cabbage kimchi, cucumber kimchi, pickled broccoli, squid in chili sauce, sesame glass noodles and a savory pancake flecked with carrot and scallion, to name a few.
With banchan like this, who needs appetizers? I didn't find many Korean starters on the menu anyway, except for the yang nyum tofu ($5), a shallow bowl of silky steamed tofu triangles draped in a spicy fish sauce with hints of sweetness.
Japanese gyoza, Korean mandu or the ubiquitous Chinese pot sticker, it's all the same thing, really. Gangnam calls its dumplings gyoza ($4), as did Wasabi. Regardless of the name, expect to get five steamed wonton wrappers pinched tightly around a lump of gingery ground pork and chopped scallion. These pedestrian dumplings are nothing to write home about, but the vinegary chili dipping sauce made a definite pronouncement from the Korean peninsula.
Koreans love beef — especially in the form of barbecue — yet I found the kalbi short ribs ($18/dinner) to be on the tough side. To the kitchen's credit, the flavor of the marinade (redolent of soy sauce, garlic and brown sugar) was spot on, giving the barbecued ribs a slightly fruity essence.
Bibimbap ($11/dinner) is not only fun to say, but it's fun to eat as well. A large stainless steel bowl contains steamed white rice arranged with a composition of seasoned minced beef, fermented bean sprouts, daikon radish, wood mushrooms, wilted spinach and walnuts, drizzled with a piquant chili paste-based sauce. A raw egg shimmers on top, just waiting to have its yolk popped.
Thick soups and hot pots bring families together around the communal dinner table. It's comfort food in South Korea.
One day, I went for the kimchi jigae ($9/lunch), a nose-clearing stew thick with napa cabbage, tofu, shredded pork, serrano pepper and scallion in a kimchi-kicked broth. It was served in a little cauldron next to a metal canister brimming with steamed white rice.
With cephalopods on our minds, we quickly changed the order to a squid stir-fry ($19/dinner), a wok-seared tangle of mostly tender tentacles, cabbage and scallion in a fiery chili sauce, next to a mound of rice noodles and a tin of steamed rice. It was a good substitute for the stew, albeit expensive for what we got.
Other seafood dishes include a Japanese-inspired chirashi ($13/lunch). This is more of a salad than a rice dish, as the menu states, but it's a refreshing pick on a hot day. Essentially, chirashi is a bed of warm rice topped with mixed greens, little cubes of sashimi (yellowtail and salmon), sliced cucumber, long strands of imitation crab and bright orange beads of flying fish roe. A drizzle of spicy sesame dressing accented the oceanic flavors, as did a gingery chili sauce that comes in a big squirt bottle.
Like all new restaurants, Gangnam is still working out some kinks on the menu. But it's the most traditional Korean food Boise has seen in a long time.


text:
Q: Get value of at least one coincidense between related tables postgres I have 3 tables:
CREATE TABLE public.student (id serial NOT NULL PRIMARY KEY,"student" varchar(30) NOT NULL) WITH (OIDS = FALSE);

  CREATE TABLE public."class" (id serial NOT NULL PRIMARY KEY,"class" varchar(30) NOT NULL) WITH (OIDS = FALSE);

 CREATE TABLE public.student_class (student_id integer, class_id integer, /* Keys */ CONSTRAINT student_class_index01
    PRIMARY KEY (student_id, class_id)) WITH (OIDS = FALSE);

With these values:
INSERT INTO public.student_class ("student_id","class_id") VALUES(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(2,1),(2,2),(2,3),(3,4),(3,5),(3,6),(3,7),(3,8);

I want to now the names of the students that have at least one coincident class with other one:

Student 1 has got at least one coincidence with 2 and 3. Student 2 has
  got at least one coincidence with 1. Student 3 has got at least one
  coincidence with 1.

I have tried this, but it is not doing it by student.
SELECT distinct a.id,a.student from public.student a
LEFT OUTER JOIN public.student_class b ON b.student_id=a.id
LEFT OUTER JOIN public.class d ON d.id=b.class_id
WHERE class_id IN(select class_id from student_class where student_id not in(1))
AND a.id not in(1);

I think taht is easier than I think. Thanks.

A: Try this:
select *
from   student
where  exists (select null
                 from student_class sc1
                where sc1.student_id = student.id
                  and exists (select null
                                from student_class sc2
                               where sc2.class_id   =  sc1.class_id
                                 and sc2.student_id != sc1.student_id))

A: What do you think about this :
with student_lst_classes as (
    select student_id,array_agg(class_id order by class_id) lst_classes
      from student_class  group by 1
 )
select st1.student,string_agg(st2.student,';' order by st2.student ASC) lst_student_coincident from
 (student_lst_classes l1 join student st1 on (st1.id=l1.student_id))
 ,(student_lst_classes l2 join student st2 on (st2.id=l2.student_id))
where l1.student_id!=l2.student_id and l1.lst_classes && l2.lst_classes
group by 1 order by 1

A: Besically a cartesian join of student X student, so there are no duplicates to take care of. (except for the symmetry s1 <--> s2)
SELECT s1.id AS a_id, s1.student AS a_student
        , s2.id AS b_id, s2.student AS b_student
FROM student s1
JOIN student s2 ON EXISTS (
        SELECT *
        FROM student_class x1
        JOIN student_class x2 ON x2.class_id = x1.class_id
        WHERE x1.student_id = s1.id
        AND x2.student_id = s2.id
        )
WHERE s1.id < s2.id -- tie-breaker
        ;

redpajama_set_name: RedPajamaStackExchange


Born Enrico Raphael Quiogue Nacino on March 21, 1987 in Baguio City, Benguet, Philippines, Rocco Nacino is one of the most talented Filipino actors of his generation. He rose to fame after joining the fifth season of StarStruck, a defunct reality show on GMA Network. He was declared the Second Prince of that season. Before joining Starstruck V, Rocco Nacino tried his luck in the Myx VJ Search but failed.
Rocco Nacino spent most of his younger years in Singapore where he studied Nursery up to Primary 6. When he returned to the Philippines, he went to Makati Hope Christian School for his secondary education and in Mapúa Institute of Technology for college. He took up Nursing in Mapua and eventually passed the Board Exam, making him a registered nurse.
While in college, Rocco joined the Mister and Miss Cardinals and won as Mister Cardinals 2005. This opened new opportunities in the world of TV commercials and print media. Among his projects were TV commercials and print media for Vaseline, Skyflakes, Goya, Jollibee, Chowking, Smart, Pepsi and STI. His sports are jiujitsu, muay thai, mixed martial arts, boxing, arnis and basketball. He also trains wing chun regularly. Aside from these sports, he also plays drums and a little of guitar.
Rocco Nacino's biggest break on TV came when he auditioned for and bagged the role for Sine Novela Presents: Gumapang Ka sa Lusak, opposite Jennylyn Mercado and Dennis Trillo. More projects followed including Dramarama sa Hapon's Koreana, opposite Kris Bernal. His excellent performance in Koreana gave him his first acting nomination from the Entertainment Press Society (Enpres) as Finalist for Outstanding Performance by an Actor for the Golden TV Awards in 2011. With the large viewership and high ratings garnered by Koreana, Nacino was given a more challenging project and an even better exposure via the first-ever dance serye Time of My Life, with Kris Bernal and Mark Herras.
He was then cast inthe weekly fantasy sitcom, Kaya ng Powers, with Rhian Ramos, Sheena Halili, Elmo Magalona, Joey Marquez and Rufa Mae Quinto and did a cameo in the finale week of Amaya, where he played the role of Adult Banuk. He also top-billed a 5-episode weekly series, Love Bug presents: Mistaken Identity, where he was paired with Lovi Poe and was shot in Palawan.
Rocco Nacino received his first acting awards via the indie film, Ang Sayaw ng Dalawang Kaliwang Paa, an entry at the 2011 Cinemalaya Film Festival. He co-starred with Paulo Avelino and Jean Garcia in the said indie film which showcased his acting chops and got him recognized by award giving bodies. The Philippine Movie Press Club hailed him as the "New Movie Actor of the Year" in the 28th PMPC Star Awards for Movies while the Entertainment Press Society (Enpres) lauded him with the "Best Breakthrough Performance by an Actor" in the 9th Golden Screen Awards.
He was also given the honor to portray Philippines' national hero Jose Rizal in the multi-awarded documentary narrated by Howie Severino, Pluma: Ang Dakilang Manunulat. The documentary was also nominated as Best Documentary at the New York Film Festival. He then starred in The Good Daughter, opposite Kylie Padilla. He also played the title role, Lam-ang – an epic movie that was shot in Ilocos, with Rochelle Pangilinan as his leading lady. He was also part of Party Pilipinas, Bubble Gang, Sunday All Stars, Bride, Akin Pa Rin Ang Bukas, and Genesis.
Rocco played the title role in Pedro Calungsod: Batang Martir, an official entry to the Metro Manila Film Festival 2013. The role landed on his lap when JM de Guzman, who was originally cast to play Pedro Calunsod, was pulled out of the cast due to drug problems.


In Episode 29 of the MBS podcast, Greg and Andy are updating a couple of topics they last examined a year ago! First, the do an update on "Class B" games, and add a couple to the list as well. Later, they go back to the movies to discuss 4 new naval-themed films.


The census figures for 2006 have been broken down to show which nationalities live in the larger towns (poulation over 1500). It makes interesting reading.
Overall - 11.24% of the population are not Irish (457000 out of a pupulation of 4.06 million).
The second in the list of towns with the most immigrants is Ballyhaunis in County Mayo - 36% foreign nationals there.
The town with the least foreign nationals is Clara in Kildare - with just 5.29% (156 people) - but 45 didn't state nationality - so the figures could be wrong.
The town with the lowest number of non irish is Portrane in County Dublin - with just 88 non irish (6 people didn't state what nationality they were).
The town with the highest percentage of Asians is Ballyhaunis Co mayo (just under 10%) - the next highest is Ballaghadereen in Mayo too - with 7.5%. The presence of asylum seeker hostels in both these towns might be a contributory factor.
Irish Economy still looking Good?


The K-lab space in the town of Reggio Emilia is the brainchild of a very special mum who has a child with disabilities. It is a place that seeks to transform what disabled children can express and do into a resource, an asset and possibly even a profession.
The "silent children" do not speak due to physical and behavioural issues. Often they are only able to express themselves and tell their stories in writing with the help of a special technique called "Augmentative and Alternative Communication", which after training, hard work and lots of practice allows them to find a way out of the silence. With the help of educators, facilitators and Neuro-Psychomotor specialists on the one hand, and designers, stylists, illustrators and graphic designers on the other, K-lab has created a network of professionals and high-profile creative people who can offer their services to companies and institutions. They are reviving the ethical values at the heart of the economy and of businesses.
There are wooden frames and objects that were designed and made in Indaco's woodworking workshop, where the children participate in practical and creative activities. The lamps (designed by Attica Design) and the plate collection distributed by Rossana Orlandi (designed by Stefania Vasquez), together with the paper envelopes by Essential and the t-shirts are all stamped with the words and thoughts of these special children.
"Clouds, I aspire to be the sun" or "Only if you light the way can we turn off the dark" are some of the phrases printed and framed in recycled wooden frames that we saw hung on a wall in one of the rooms at the Cascina Cuccagna during the Milano Design Week. The thoughts of "the silent children, who are without words but otherwise display normal intelligence".
Not "disabled" but "differently abled". Disability as a resource.


Patient history. The physician will ask the patient for a description of symptoms, how they started, and what triggers them.
Physical exam. The wrist and forearm will be examined for dislocation or discoloration. Physicians may also manipulate the wrist or ask patients to perform certain hand or wrist movements, if they are able.
An MRI or CT test can be helpful to examine complex fractures or look for additional injury to nearby soft tissues. MRIs and CTs may be done with injected contrast dye (called MRI-arthrograms and CT-arthrograms) to help show damage to ligaments or other soft tissues.
Once physicians take into account factors like the age and activity level of the patient and the nature of the fracture, they can determine if the fracture can be reset without surgery (closed reduction) or requires surgery (open reduction).


A dress code shall be applicable for all the students and following of the dress code on all the working days will be compulsory.
2. Final year students are required to wear black coat both in summer and winter.
All pre-final year students will have to buy shirt/pants/salvar/kurta/blazer cloth and tie from college only to maintain uniformity of shade and clothes quality.


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text:
ACCEPTED

#### According to
Index Fungorum

#### Published in
Opredelitel' gribov roda Mycosphaerella Johans 294 (1979)

#### Original name
Sphaerella myrtillina Pass.

### Remarks
null

redpajama_set_name: RedPajamaGithub


text:
Married Biography / Biography / Eva Marcille
Eva Marcille Bio
Last Modified January 11, 2018
Facts of Eva Marcille
5 feet 6 inches (1.69m)
$ 3 million
Afro-American
Waist Size:
Bra Size:
Hip Size:
I've learned that I can only live for myself. I cannot be focused on the world's idea of who they think I am or who I'm supposed to be. I can only be the best me. And if that means that even though I seem eligible and should be in a relationship, maybe I shouldn't be right now, because I am not emotionally available. It's knowing yourself and being more in tune with who you are vs. who the world wants you to be
So if I can give anyone advice in this business on love and balancing, it's that you truly have to take a second, step back, to figure out who you are and what you want. And it is okay if those people around you don't fit into that. Because what you don't want to do is end up living your life for someone else
I've been blessed, because the type I started in was competition based. It was about being witty and smart and innovative and different. I don't get backlash like some of the reality shows today because that's not who I am.
Relationship Statistics of Eva Marcille
What is Eva Marcille marital status ? (single, married, in relation or divorce):
Is Eva Marcille having any relationship affair ?:
Is Eva Marcille lesbian ?:
Regarding her personal life, In 2006, she dated Lance Gross, whom she met at Tyler Perry's House of Payne. The two became engaged in 2008 but broke-up in 2010.
She then had a relationship with rapper Flo Rida for two years. In 2013, she began dating singer-songwriter, Kevin McCall. This beautiful couple has a beautiful daughter named as Marley Rae McCall.
She was Marcille's first child and McCall's second. McCall and Marcille broke up in early 2015. She currently dating Michael T. Sterling. She has never been married.
1 Who is Eva Marcille?
2 Eva Marcille: Birth Facts, Family, and Childhood
3 Eva Marcille: Education History
4 Eva Marcille: Professional Life and Career
5 Eva Marcille: Salary and Net Worth
6 Eva Marcille: Rumors and Controversy
7 Eva Marcille: Body Measurements
8 Social Media Profile
Who is Eva Marcille?
Eva Marcille is an African American actress as well as a multitalented television host. She is also a beautiful and attractive fashion model who recently played the role of Tyra Hamilton on the Soap Opera The Young and the Restless. Her real name is Eva Marcille Pigford but changed her name for acting. She started her career by winning the third cycle of American's Next Top Model.
Eva Marcille: Birth Facts, Family, and Childhood
Marcille was born on October 30, 1984, in Los Angeles, California. She belongs to the ethnicity Afro-American and nationality American.
He was born in the Pigford family as a daughter of Evan Pigford (father) and Michelle Pigford (Mother). Her middle name, Marcille, is an amalgam of "Marjorie" and "Lucille," her grandmother's names.
Eva Marcille: Education History
She got her high school education from Washington Preparatory High school and graduated from Clark Atlanta University in Atlanta Georgia; however, she left the School Shortly after winning the third cycle of America's Next Top Model.
Eva Marcille: Professional Life and Career
Eva Marcille won the third Cycle of America's Next Top Model, beating fellow contestants Yaya DaCosta, Amanda Swafford, and Ann Markley and got a modeling contract with Ford Models as well as Cover Girl cosmetics contract, and a spread in Elle magazine.
This win launched her modeling career, soon gracing the covers of Women's Health and Fitness, Cosmopolitan, IONA, King, and Essence magazines. She was also featured in CoverGirl, Samsung, Red by Marc Ecko, Jewel magazine, In touch weekly magazine, King magazine, WeTheUrban Magazine, IONA, Unleashed magazine, Star Magazine, Elle, Elle Girl, Applebottoms, Lerner Catalog, Avon Campaign 4 and Rolling Out.
Her runway shows include Marc Bouwer Fall, ELLE girl presents Dare To Be You: Wal-Mart meets America's Next Top Models, Deborah Lindquist Spring, Naqada Spring, and 8th Annual Models of Perfection Show. In addition to her modeling career, she left her managers, Benny Medina and Tyra Banks.
Marcille has guest-starred on several series on UPN and The CW Networks. Her guest-star credits include two episodes of Kevin Hill, one episode of Smallville, one episode of Everybody Hates Chris and one episode of the Game As herself. Marcille has also appeared on Tyler Perry's House of Payne. She also won a Young Hollywood Award.
She was also nominated for two NAACP Image Award for Outstanding Actress. Marcille also performed as host for "Rip the Runway", "My Model Looks Better Than Your Model" and "Hair Battle Spectacular". Her movie credits now include "The Walk", "Crossover", and" I Think I Love My Wife".
She hosted the first season of the Oxygen reality series Hair Battle Spectacular in 2010. In 2015, Marcille stars in the main cast as the role of Tara in the TV one series Born Again Virgin.
Eva Marcille: Salary and Net Worth
Her net worth is $ 3 million but her salary is not revealed yet.
Eva Marcille: Rumors and Controversy
According to rumors, she was engaged with Lance Gross but they split up before her marriage.
Eva Marcille: Body Measurements
Eva Marcille has a good height of 5 feet 6.5 inches with body weight 56kg. He has light brown hair color and her eye color is hazel green. Her bra size is 32 B, her shoe size is 8(US) and her dress size is 2(US).
Social Media Profile
She is active on Facebook, Instagram, and Twitter. She has more than 34,633 followers on Facebook, she has more than 1.8 million followers 0n 1.8m followers on Instagram and has more than 259k followers on Twitter.
Tags : Actress Multitalented television host
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text:
bp and Oman form strategic partnership to progress significant renewable energy and hydrogen development
bp and Oman partner to progress world-class scale renewable energy and green hydrogen development in support of Oman's 2040 Vision
Integrated project expected to include onshore wind, solar and green hydrogen production
Successful execution could trigger significant investment in Oman, and support creation of new industries and skilled jobs
The agreement was signed by His Excellency Mohammed Al Rumhy, Minister of Energy and Minerals of Oman, and Stephen Willis, bp senior vice president in Middle East
Press release (English) pdf / 131.9 KB
Press release (Arabic) pdf / 246.7 KB
bp and the Ministry of Energy and Minerals in Oman today signed a Strategic Framework Agreement (SFA) and a Renewables Data Collection Agreement which will support the potential development of a multiple gigawatt, world-class renewable energy and green hydrogen development in Oman, by 2030.
As part of the agreement, bp will capture and evaluate solar and wind data from 8,000km2 of land – an area more than five times the size of Greater London. The evaluation will then support the government of Oman in approving the future developments of renewable energy hubs at suitable locations within this area to take advantage of these resources. The renewable energy resources could also supply renewable power for the development of green hydrogen, targeting both domestic and global export markets.
"Today's agreement represents what bp is able to offer as an integrated energy company. These projects will build on our gas business, and bring wind, solar and green hydrogen together in a distinctive and integrated way supporting Oman's low carbon energy goals."
Bernard Looney, chief executive officer
This partnership represents a significant evolution of bp's business in Oman and is aligned with bp's strategy, which includes rapidly growing our developed renewable generating capacity and to take early positions in hydrogen.
bp chief executive Bernard Looney said: "Today's agreement represents what bp is able to offer as an integrated energy company. These projects will build on our gas business, and bring wind, solar and green hydrogen together in a distinctive and integrated way supporting Oman's low carbon energy goals.
"And we're not just investing in energy. We are investing in Oman to create and develop infrastructure, support local supply chains and cultivate the skills and talent needed to usher in this next generation of energy leaders. We look forward to working closely with the Omani government to take this forward."
His Excellency Dr Mohammed Al Rumhy, Minister of Energy & Minerals of the Sultanate of Oman, said: "This is a proud moment for Oman and a significant step towards delivering our 2040 Vision. In partnership with bp, we will progress the development of new, world-class solar and wind resources – generating renewable power for the grid and powering the manufacture of green hydrogen to supply domestic demand and to export to global customers. Over the past 50 years, we've advanced our hydrocarbon production. Today's agreement signals the next step in our energy journey – unlocking the potential for Oman as a low-carbon energy hub."
The UK's Minister for Investment, Lord Gerry Grimstone, added: "Following the signing last week of the UK-Oman Sovereign Investment Partnership, this investment by bp into Oman's renewable energy sector is a shining example of our countries' joint ambition to facilitate strategic and commercial bilateral investment. The project demonstrates our shared vision for future prosperity through clean growth, further strengthening the partnership between the United Kingdom and the Sultanate of Oman."
Under the SFA, bp and Oman will also consider ways to collaborate in a number of areas, including a renewables strategy, regulation, the establishment of a renewable energy hub and the development and reskilling of the local workforce.
Oman has a strong track record in the oil and gas industry, which it has grown over recent years. Today's announcement, which is subject to final agreement of commercial terms, is an important step towards the country's 2040 Vision and an opportunity to become a leading low-carbon energy hub. And it would further support the Oman government's goals of diversifying the economy and bolstering investment.
bp is committed to growing its business and building on its 15-year history in Oman, where it operates Block 61, which produces a third of the country's gas demand. In 2020, bp's Oman business spent $610 million with Omani-registered companies – 90% of its total spend. And in 2021, bp joined Oman's national hydrogen alliance, Hy-Fly, to promote the hydrogen industry in Oman, and established a net zero taskforce to help develop a 'roadmap' for bp in Oman.
bp press office, London: +44 20 7496 4076, bppress@bp.com
Also on bp.com
bp in Oman


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text:
The Best New Music in Rock, Alternative, Hip-Hop, and More
Expertly Curated by the CoS Staff
She & Him cover Carole King-penned "Oh No, Not My Baby" — listen
by Chris Coplan
on November 04, 2014, 9:50am
With their upcoming covers album Classics, the duo of M. Ward and Zooey Deschanel (tka She & Him) are taking a trip back through music's rich past like an indie pop Mr. Peabody and Sherman. First, they landed in 1964, delivering a lush and ornate version of Dusty Springfield's "Stay Awhile". From there, they took the Wayback Machine to the mid-'40s to give the Frank Sinatra-sung "Time After Time" a shimmery update.
For their latest destination, they arrive at the heyday of the hit-making factory the Brill Building, tackling the Gerry Goffin and Carole King-penned classic "Oh No, Not My Baby". Already the track's been performed by pop greats like Cher, Linda Ronstadt, and Rod Stewart, among others, but Ward and Deschanel take a slightly more laidback, understated approach. Lithe and breezy, this rendition feels like a mix of easy listening rhythms and minimalist synthpop grooves, with Deschanel's satiny croon sounding extra beguiling amid all that sweet, lulling noise. Listen in below.
Classics is due out December 2nd via Columbia Records.
Classics Tracklist:
01. Stars Fell On Alabama
02. Oh No, Not My Baby
03. It's Not For Me To Say
04. Stay Awhile
05. This Girl's In Love With You
06. Time After Time
07. She
08. Teach Me Tonight
09. It's Always You
10. Unchained Melody (Feat. The Chapin Sisters)
11. I'll Never Be Free
12. Would You Like To Take A Walk?
13. We'll Meet Again
M. Ward
She and Him
Stevie Nicks delivers a solid gold performance on Fallon — watch
Typefighter share new song "Wall Song"— listen