text:
'Pokémon' for the Wii is a lackluster effort
Pokemon attacks looks cool in the Wii game — it's just too bad the Pokemon themselves never really tussle.Nintendo
July 27, 2007, 7:59 PM UTC / Source: msnbc.com contributor
By By Levi Buchanan
Nintendo must think Pokémon fans will buy anything — that's the only explanation for the lackluster effort put forth in "Pokémon Battle Revolution," the very first Pokémon game for the Nintendo Wii.
Coming on the heels of "Pokémon Diamond" and "Pearl" for the Nintendo DS doesn't help "Battle Revolution's" weak cause. Those superlative adventure games were so feature-rich that this Wii game is almost pointless by comparison. Sure, "Pokémon Battle Revolution" allows you to battle online with friends — this is the first online multiplayer game for Wii — but so did the DS duo.
So what does $50 buy you? A few hours of frustratingly dull Pokémon battles that just happen to play out in some pretty attractive 3D graphics. Big deal.
To be fair, "Pokémon Battle Revolution" does look good at first blush. The 3D Pokémon models are colorful and nicely detailed. Some of the Pokémon attacks also employ nice effects that simply cannot be seen in the Nintendo DS games.
But "Pokémon Battle Revolution" doesn't aspire to be anything more than a 3D battle simulator — these are the exact same fights you've played a zillion times since the original Game Boy titles. After nine years, these Pokémon still never truly mix it up in battle. Even with the horsepower of the Wii goading them on, the Pokémon models rarely even touch each other, much less have special battle animations that might reveal what a true Pokémon dust-up looks like.
Once you're no longer wowed by the flashy visuals, the value of "Pokémon Battle Revolution" drops faster than a Chimchar squaring off against an Empoleon. (Ask your kids.) Hopefully, you already have "Pokémon Diamond" or "Pearl" on the Nintendo DS and can import your hard-earned monsters, because "Pokémon Battle Revolution" practically punishes you if you don't. You are force fed a beginner's set of six Pokémon and must play through a series of tournaments to unlock additional passes worth six more Pokémon apiece. Want to play against a friend? Then you better bite down and earn that second pass, because the game doesn't even offer a courtesy pass for another player.
What's so particularly galling about "Pokémon Battle Revolution" is that's not even remotely as full-featured as the original "Pokémon Stadium" games on the old Nintendo 64 — and those were released in 2000 and 2001, respectively. Sure, there was no online multiplayer back then, but at least the "Stadium" games let you play with all of the available Pokémon without forcing you to sync up the portable games. From the get-go, you could play with "rental" versions of all Pokémon available (up to 251 in "Pokemon Stadium 2"). And there was no horrible single-player game that twisted your arm behind your back and made you earn Pokémon you may or may not like.
The "Stadium" games also included a series of mini-games for just goofing off and the ability to play the old Game Boy games on your television. "Pokémon Battle Revolution" manages neither. The only really "goodie" in Pokémon battle Revolution is the ability to earn credits and dress up your in-game Pokémon trainer. If you do have a Nintendo DS, you can use those credits to buy mystery gifts and forward them to your trainer in "Pokémon Diamond" or "Pearl."
"Pokémon Battle Revolution" does indeed go online, allowing you to battle against other Pokémon trainers across the nation. There are random battles, but you cannot communicate during the game. Should you want to seek out a specific player, it is quite easy to set up a match once you have exchanged Friend Codes specific to "Pokémon Battle Revolution." Nintendo's online service does not encourage random online encounters; you must contact your friends in the real world and register each other on your Wiis. This might sound cumbersome, but any parent that's heard an earful of the wretched filth spewed over headsets on Xbox Live will be grateful for Nintendo's efforts here.
However, you could go online and battle with other Pokémon trainers in "Diamond" and "Pearl" on the DS just as easily, and while there are no 3D battles, you certainly get a lot more game for a lot less money. (The DS games retail for approximately $35 each.)
The DS titles also offer an enormous single-player adventure and a robust cataloging tool for sorting your Pokémon, both of which shame the experience of "Pokémon Battle Revolution." Even after playing against a friend via "Battle Revolution," there are no statistics or online leaderboards to proclaim your dominance to the world.
If you are any sort of Pokémon fan, you've already invested in "Diamond" or "Pearl" for the Nintendo DS and have likely earned yourself some rare, powerful monsters. The idea of seeing those Pokémon in full 3D might be tempting, but resist and send a message to Nintendo. Just because you are a fan, you will not accept an expensive watered-down experience that lacks essential features — features Nintendo saw fit to include in seven-year-old games.
By Levi Buchanan


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Das Erzbistum Bordeaux (-Bazas) (), bis 1937 Erzbistum Bordeaux, ist ein im Südwesten Frankreichs gelegenes Erzbistum der römisch-katholischen Kirche. Bischofssitz der Erzbischöfe von Bordeaux ist die Kathedrale Saint-André in Bordeaux. Sein heutiges Gebiet entspricht dem Département Gironde.

Geschichte

Das Bistum wurde im 314 errichtet und der erste Bischof war Oriental. Die Heiligen Delphin (380–404), Amandus (404–410 und 420–432) und Severin von Köln (410) wirkten als Bischöfe in der Stadt. Im 5. Jahrhundert wurde Bordeaux durch die Westgoten, kurz darauf durch die Franken eingenommen. 732 verwüstete Abd ar-Rahman während seines Feldzugs die Stadt (siehe Schlacht von Tours und Poitiers). Im 9. Jahrhundert fielen die Normannen ein und plünderten die Stadt erneut. Erst danach begann sich Bordeaux zu erholen. Gérard d'Angoulême (de Blaye) war von 1131 bis 1135 erster Erzbischof des Bistums.

Der alten romanischen Bischofskirche folgte die gotische Kathedrale Saint-André, die im 11. bis 14. Jahrhundert erbaut wurde. Sie ist einschiffig und auffallend breit mit einem reich mit Statuen geschmückten, von zwei 50 m hohen Türmen flankierten Portal und dem Glockenturm Peyberland.

1305 wurde Erzbischof Bertrand de Goth als Clemens V. zum Papst gewählt. Im März des Jahres 1309 bestimmte er Avignon zum neuen Sitz der Päpste (siehe Avignonesisches Papsttum). Weitere erwähnenswerte Erzbischöfe waren François Hugotion de Aguzzoni (1389–1412) und Jean du Bellay (1544–1553). In seiner über 30-jährigen Amtszeit wirkte Henry de Béthune ganz im Sinne der tridentinischen Kirchenreform.

Das Erzbistum war auch von den Wirren der Französischen Revolution betroffen, die Kathedrale wurde in eine Scheune umgewandelt. Erzbischof Jérôme Marie Champion de Cicé (seit 1781) wurde durch die Konstitutionellen Bischöfe Pierre Pacareau und Dominique Lacombe zum Amtsverzicht gedrängt, verweigerte diesen aber und emigrierte. Mit der Unterzeichnung des Konkordats von 1801 durch Napoléon und Kardinal Consalvi als Vertreter von Papst Pius VII. gelangte Charles-François d'Aviau Du Bois de Sanzay (1802–1826) in das Amt des Erzbischofs. Ihm folgten Jean-Louis Anne Madelain Lefebvre de Cheverus (1826–1836), François-Auguste-Ferdinand Donnet (1836–1882) und Aimé-Victor-François Guilbert (1883–1889). Erzbischof Victor Lécot (1890–1908) war für sein soziales Engagement, z. B. mit der Schaffung von Arbeiterküchen, und sein Eintreten für die Arbeiterklasse bekannt. Sein Nachfolger Pierre-Paulin Andrieu (1909–1935) ging als strikter Vertreter des Antimodernismus in die französische Geschichte ein. Unter Maurice Feltin (1935–1949) erfolgte am 20. November 1937 die Umbenennung in Erzdiözese in Bordeaux (-Bazas); das Bistum Bazas war im Jahr 1802 aufgehoben worden. Feltin, der danach Erzbischof von Paris wurde, folgten Kardinal Paul Richaud (1950–1968), Marius Maziers (1968–1989), Kardinal Pierre Eyt (1989–2001) und Jean-Pierre Ricard (2001–2019), der 2006 zum Kardinal ernannt wurde. Ricards Nachfolger wurde nach einer kurzen Sedisvakanz Jean-Paul James.

Suffragansitze

 Vor 1801
 Bistum Agen
 Bistum Angoulême
 Bistum La Rochelle
 Bistum Luçon
 Bistum Périgueux
 Bistum Poitiers
 Bistum Saintes
 Bistum Sarlat

 Von 1822 bis 2002
 Bistum Agen
 Bistum Angoulême
 Bistum La Rochelle-Saintes
 Bistum Luçon
 Bistum Périgueux
 Bistum Poitiers

 Seit 2002
 Bistum Agen
 Bistum Aire und Dax
 Bistum Bayonne
 Bistum Périgueux

Siehe auch
 Liste der Erzbischöfe von Bordeaux

Weblinks

 Homepage des Erzbistums Bordeaux (französisch)
  (englisch)

Bordeaux
Organisation (Bordeaux)
Christentum in Nouvelle-Aquitaine

Bordeaux
Bordeaux
Gegründet 314

redpajama_set_name: RedPajamaWikipedia


text:
Q: Can't install Internet Explorer 9 in Windows 7 x86 I need Internet Explorer 9 for developing. So I go here and I can download it only for Windows Vista and Windows Server 2008. However when I try to install I get the error (translated):
This version of IE install program doesn't support this version of Windows.

How can I fix that?

A: After some research on the web I found the offline installer download link of Internet Explorer 9 for Windows 7. And the localizations download link.
Also be sure to check and install the Internet Explorer 9 prerequisites.

redpajama_set_name: RedPajamaStackExchange


text:
Q: Sitecore Remote Publishing Targets - One of Two Fails Silently I have configured two Remote Publishing Targets in the connection string config and web.config file and in the content tree per the Sitecore docs.
Target 1 is on the same physical machine and uses the same database server as the authoring site. Target 2 is on a different physical machine with a different database server from the authoring site. I used the Data Source configuration utility to insure that the authoring server could connect to the database server for Target 2 and validated the server name, database name, user name, and password are all correct. Connection works just fine.
I modified the security on the remote publishing targets in the content tree to allow my publishing lead to select the target for publishing. Both targets have the same security settings applied.
Publishing works to target #1 but not to target #2. The "Click here for more information" says that there were 2 updates made to both Target #1 and Target #2 so it looks on the face to be working but it really isn't.
How do I get more information about what exactly is failing so that I can get this resolved?
Many thanks!

A: Have you looked at the Sitecore logs? They will likely contain publish info. Check them on both the auth side and the slave side.


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\section{Introduction}
Counter-streaming instabilities in pair plasmas play a key role as the trigger of collisionless shock formation in such media \cite{SilvaApJ,BretPoP2013,BretPoP2014}. Because many instabilities such as two-stream, Weibel or oblique, compete in the linear regime, it is important to know which one grows the most in terms of the system parameters. For the unmagnetized case, counter-streaming pair plasmas are equivalent to counter-streaming electron beams, for which the instability hierarchy map has been established in terms of the beams temperatures, Lorentz factor and density ratio \cite{BretPRL2008,BretPoPReview}.

Progresses are slower for the magnetized case, due to the complexity of the analytical calculations involved when implementing a relativistic kinetic theory. The search for the most unstable mode requires sweeping the full \textbf{k}-spectrum, rendering inoperative the simplifications which can be done when the wave-vector is parallel or normal to the flow. To our knowledge, this kind of relativistic calculation has only been performed twice in the literature \cite{timofeev2009,Timofeev2013}.

In a recent paper, the hierarchy map of unstable modes has been derived for two cold colliding symmetric pair plasmas over a flow-aligned magnetic field \cite{bretPoP2016}. For such a system, the linear spectrum only depends on 2 parameters which are the strength of the field and the initial Lorentz factor of the shells. The goal of this article is to extend this previous work to an arbitrary obliquity between the flow and the field.

The system considered is pictured on Fig. \ref{fig:setup}. The initial flow is along the $z$ axis, and the field $\mathbf{B}_0$ lies in the $(z,x)$ plane, with $(\widehat{\mathbf{B}_0,\mathbf{e}_z})=\theta$. The wave-vector $\mathbf{k}$ of the perturbations applied to the system lies in the same $(z,x)$ plane. We therefore implement a 2D model for a direct comparison with 2D PIC simulations of such systems. The most general case would require considering an $k_y \neq 0$ component for $\mathbf{k}$. Nevertheless, previous 3D studies of the Weibel instability ($k_z=0$) found that the maximum growth-rates are to be found precisely for $k_y=0$ \cite{bretPoP2014a,Stockem2016}. The present choice of the wave-vector orientation is therefore likely to render the largest growth-rates over the full 3D \textbf{k}-space.

  \begin{figure}
  \begin{center}
   \includegraphics[width=.5\textwidth]{fig1.eps}
  \end{center}
  \caption{System considered. Two  counter-streaming pair beams over an oblique magnetic field $\mathbf{B}_0$. The beams are initially cold and symmetric, with electronic density $n_0$ and positronic density $n_0$.}\label{fig:setup}
 \end{figure}

The same 4-fluids model than in Ref. \cite{bretPoP2016} is implemented, where 2 fluids stand for the electrons and positrons of one beam, and 2 more fluids for the electrons and positrons of the other beam. For the present article to be self-contained, the main lines of the calculation are reproduced here. We write the 4 matter conservation equations,
\begin{equation}\label{eq:matter}
  \frac{\partial n_i}{\partial t} + \nabla \cdot (n_i \mathbf{v}_i) =0,
\end{equation}
for $i=1\ldots 4$, and the 4 momentum conservation equations,
\begin{equation}\label{eq:momentum}
  \frac{\partial \mathbf{p}_i}{\partial t} + (\mathbf{v}_i\cdot\nabla) \mathbf{p}_i = q_i \left( \mathbf{E} + \frac{\mathbf{v}_i \times (\mathbf{B} + \mathbf{B}_0)}{c} \right),
\end{equation}
again with $i=1\ldots 4$. These equations are then linearized considering small departures from the initial conditions of the form $\exp(i \mathbf{k}\cdot \mathbf{r} - i \omega t)$. The first order density perturbations $n_{1i}$ are derived from Eq. (\ref{eq:matter}). Inserting them in Eq. (\ref{eq:momentum}) and using $\mathbf{B}_1 = (c/\omega)\mathbf{k}\times \mathbf{E}_1$ allows to express the first order velocity perturbations $\mathbf{v}_{1i}$ as a function of the first order field $\mathbf{E}_1$. This allows to write the first order current as,
\begin{equation}\label{eq:J1}
  \mathbf{J}_1 = \sum_{i=1}^4 q_i n_0 \mathbf{v}_{1,i} + \sum_{i=1}^4 q_i n_{1,i} \mathbf{v}_{0,i} \equiv \mathbf{J}_1 (\mathbf{E}_1).
\end{equation}
This expression is then inserted into a combination of Maxwell-Faraday's and Maxwell-Amp\`{e}re's equations, yielding
\begin{equation}\label{eq:Max}
 \mathbf{k} \times (\mathbf{k} \times \mathbf{E}_1) + \frac{\omega^2}{c^2}\left(\mathbf{E}_1 + \frac{4 \imath \pi}{\omega} \mathbf{J}_1 \right) \equiv \mathcal{T} (\mathbf{E}_1) = 0.
\end{equation}

The dielectric tensor $\mathcal{T}$ has been computed analytically with the \emph{Mathematica} tensor associated with this article and described in Ref. \cite{BretCPC}. The dispersion equation is a 16 degree polynomial which has been transferred to \emph{Matlab} for numerical analysis using a \emph{Mathematica} Notebook described in Ref. \cite{BretMM2010}. It is expressed in terms of the dimensionless parameters,
\begin{equation}\label{eq:var}
x = \frac{\omega}{\omega_p},~~\mathbf{Z} = \frac{\mathbf{k} v_0}{\omega_p},~~\beta = \frac{v_0}{c},~~\gamma_0 = \frac{1}{\sqrt{1-\beta^2}},~~\sigma = \frac{B_0^2/4\pi}{\gamma_0 (2n_0) m c^2},
\end{equation}
with
\begin{equation}
\omega_p = \sqrt{\frac{4 \pi n_0 q^2}{m}}.
\end{equation}
Note that the magnetic field is measured through $\sigma$ in Eq. (\ref{eq:var}), which is the parameter typically used in collisionless shocks physics \cite{Marco2016}. The factor ``2'' at the denominator aims at accounting for the total kinetic energy of the shells. Some authors sometimes use the parameter $\epsilon_B$ instead of $\sigma$, with $\epsilon_B=\sigma/2$ \cite{SilvaApJ,Sironi2011ApJ}.

An important point for the present study is that the state which is perturbed is not an equilibrium for $\theta\neq 0$. As  emphasized in previous studies \cite{BretPoPOblique,BretPoP2013a,bretPoP2014a}, the forthcoming calculations are valid as long as the response of the system to growing harmonic perturbations, is faster than its response to the tilted magnetic field. Since the latter response develops on time scales comparable to the cyclotron frequency of the charges, while the former response follows the growth-rate $\Delta$, we need,
\begin{equation}
\Delta > \frac{q B_0 \sin\theta}{\gamma_0 m c},
\end{equation}
where the $\sin\theta$ factor accounts for the fact that only the normal component of the field is involved in this respect. Dividing both sides by $\omega_p$ gives,
\begin{equation}\label{eq:validity}
\frac{\Delta}{\omega_p} \equiv \delta > \sin\theta \sqrt{\frac{2\sigma}{\gamma_0}}.
\end{equation}
The growth-rates we are about to compute are therefore relevant only if they satisfy the condition above. The threshold so defined will be numerically  computed in the sequel, and an analytical counterpart will be given when possible.

For a given set of parameters $(\gamma_0,\sigma,\theta)$, we compute the growth-rate $\delta(Z_z,Z_x)$. We then determine the ``hierarchy map'' under the form,
\begin{eqnarray}
\delta_{max}(\gamma_0,\sigma,\theta) &=& \max \{ \delta(Z_z,Z_x), (Z_z,Z_x) \in \mathbb{R}^2 \},\nonumber\\
\mathbf{Z}_{max}(\gamma_0,\sigma,\theta)      &=& (Z_z,Z_x) / \delta(Z_z,Z_x) = \delta_{max}.
\end{eqnarray}
A hierarchy map gives therefore the most unstable $\mathbf{Z}$ and its growth-rate for any set of parameters $(\gamma_0,\sigma,\theta)$.

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig2.eps}
  \end{center}
  \caption{Hierarchy map for $\theta=0$. This figure is the counterpart of Figs. 1 \& 7 of Ref \cite{bretPoP2016}, with the mapping (\ref{eq:map}). The upper-plot shows the largest growth-rate  for any given couple $(\sigma,\gamma_0)$. The lower-left plot shows the $Z_z$ component of the most unstable $\mathbf{Z}$, and the lower-right plot shows its $Z_x$ component. The 4 modes governing the system are Weibel, two-stream (TS), oblique and upper-hybrid-like \cite{Godfrey1975} (UHL). }\label{fig:theta0}
 \end{figure}

\section{Hierarchy map for $\theta=0$}
The case $\theta=0$ has been explored in Ref. \cite{bretPoP2016}. However, the field strength was parameterized by the $\Omega_B$ parameter, different from $\sigma$ given in Eqs. (\ref{eq:var}). The correspondence between them reads,
\begin{equation}\label{eq:map}
\Omega_B = \frac{1}{\omega_p}\frac{q B_0}{m c} ~~ \Rightarrow   ~~  \Omega_B = \sqrt{2\gamma_0\sigma}.
\end{equation}
We therefore display on Fig. \ref{fig:theta0} the hierarchy map for $\theta=0$ in terms of $\gamma_0$ and $\sigma$. The equations for the frontiers are here translated from Fig. 7 of Ref. \cite{bretPoP2016} to the new mapping.

In the cold limit, some modes, like the Weibel modes for example, have their growth-rate saturating at large $Z_x$ \cite{califano3,BretPoPReview}. When such a mode governs the spectrum, the corresponding $Z_{x,max}$ is set to 10. We recover a hierarchy map governed by 4 different kind of modes, with a triple point at $(\sigma,\gamma_0)=(16.48,2.27)$.

We now turn to the hierarchy map for $\theta=\pi/2$ before exploring intermediate obliquities.

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig3.eps}
  \end{center}
  \caption{Hierarchy map for $\theta=\pi/2$. The white line has been numerically determined, and shows where equality is reached for the validity condition (\ref{eq:validity}). Calculations are valid to the left of this line which intersects the axis $\gamma_0=1$ for $\sigma=1/4$ (see Eq. \ref{eq:validpisur2}).}\label{fig:thetapisur2}
 \end{figure}

\section{Hierarchy map for $\theta=\pi/2$}
The result of the numerical computation of the hierarchy map for $\theta=\pi/2$ is displayed on Fig. \ref{fig:thetapisur2}. The validity condition (\ref{eq:validity}) arising from the field obliquity is fulfilled to the left of the white line, which intersects the axis $\gamma_0=1$ for $\sigma=1/4$ (see Eq. \ref{eq:validpisur2} below). For the sake of the figure, it has been numerically determined although we derive below its exact analytical expression.

The computation unravels an extremely simple hierarchy map: it simply does not depend on $\sigma$. Previous works already found that the Weibel growth-rate becomes independent of $\sigma$ for $\theta=\pi/2$ because the instability has the charges moving sideways, that is, parallel to the field \cite{Stockem2016}. Other $\sigma$-dependent modes grow (not shown), but they do not outgrow the Weibel mode.

The hierarchy map shows that while the fastest growing mode always has $Z_x=10$, that is $Z_x=\infty$, the $Z_z$ component is not always 0, as expected for Weibel. We now discuss this point analytically by solving the problem in the limit $Z_x=\infty$. The dispersion function $P$ is here a 4th degree polynomial in $Z_x$. The dispersion function $P_{Z_x\infty}$ for $Z_x=\infty$ is therefore the coefficient of $Z_x^4$ in $P$. It reads,
\begin{eqnarray}\label{eq:disperpi/2}
  P_{Z_x\infty} &=& QR^2, \\
  Q             &=& -16 \beta^2 - 4(x^2+Z_z^2)/\gamma_0+ (x^2-Z_z^2)^2\gamma_0^2 \nonumber \\
  R             &=& 4 \sigma ^2+\gamma_0^6 \left(x^2-Z_z^2\right)^2-4\gamma_0^3 \sigma  (x^2+Z_z^2) \nonumber \\
\end{eqnarray}
Some simple algebra shows that the equation $R=0$ does not yield any unstable mode. We are thus left with $Q=0$, which, as expected from Fig. \ref{fig:thetapisur2}, is independent of $\sigma$. Solving $Q=0$ gives,
\begin{equation}\label{eq:tauxpisur2}
\delta^2 = \frac{2+\gamma_0^3 Z_z^2-2 \sqrt{4 \beta ^2 \gamma_0^4+2 \gamma_0^3 Z_z^2+1}}{\gamma_0^3},
\end{equation}
with $\delta^2 < 0$ for $Z_z < 2/\sqrt{\gamma_0}$. The growth-rate $\delta$ reaches its maximum for,
\begin{eqnarray}\label{eq:Zzmaxpisur2}
Z_{z,max} &=& \frac{\sqrt{3/2-2 \gamma_0^4+2 \gamma_0^2}}{\gamma_0^{3/2}},~~\mathrm{for}~~\gamma_0 < \sqrt{3/2}, \nonumber\\
Z_{z,max} &=& 0, ~~\mathrm{for}~~ \gamma_0 > \sqrt{3/2}
\end{eqnarray}
because the expression under the square-root turns negative for $\gamma_0 > \sqrt{3/2}$. The maximum growth-rate then reads,
\begin{eqnarray}\label{eq:Tauxmaxpisur2}
\delta_{max} &=& (1+\beta^2)\sqrt{\frac{\gamma_0}{2}},~~\mathrm{for}~~ \gamma_0 < \sqrt{3/2}, \nonumber\\
\delta_{max} &=& 2\frac{\beta}{\sqrt{\gamma_0}}, ~~\mathrm{for}~~ \gamma_0 > \sqrt{3/2}
\end{eqnarray}

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig4.eps}
  \end{center}
  \caption{Growth-rate (\ref{eq:tauxpisur2}). For $\theta=\pi/2$, the maximum growth-rate is given by a dispersion equation which does not depend on $\sigma$. At any rate, the dominant mode has $Z_{x,max}=\infty$. For $\gamma_0 > \sqrt{3/2}$, $Z_{z,max}=0$ and the Weibel instability governs the linear phase.}\label{fig:Wthetapisur2}
 \end{figure}

These different regimes are illustrated on Fig. \ref{fig:Wthetapisur2}, explaining the hierarchy map (\ref{fig:thetapisur2}). The validity condition (\ref{eq:validity}) reads here,
\begin{eqnarray}\label{eq:validpisur2}
  \gamma_0 &>& \sqrt{\frac{2}{2-\sigma }}, ~~\gamma_0 > \sqrt{3/2}  ~~ (\sigma > 2/3) \nonumber\\
  \gamma_0 &>& \frac{\sqrt{\sigma +\sqrt{\sigma  (\sigma +2)}+1}}{\sqrt{2}}, ~~\gamma_0 < \sqrt{3/2}~~ (\sigma < 2/3),
\end{eqnarray}
where the validity threshold $\sigma<1/4$ is analytically recovered for $\gamma_0=1$, together with its vertical asymptote at $\sigma=2$.

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig5.eps}
     \end{center}
  \caption{Hierarchy maps for intermediate obliquities. Only the part of the maps satisfying the validly condition (\ref{eq:validity}) is displayed. It has been numerically determined. An alternative presentation of these maps is available in the Supplementary Material file ``Hierarchies.pdf''.}\label{fig:inter}
 \end{figure}

\section{Hierarchy maps for intermediate obliquities}
While the hierarchy map for $\theta=0$ is quite involved, the one for $\theta=\pi/2$ is remarkably simple. We now examine how one map evolves into the other when the field obliquity  progressively increases from $\theta=0$ to $\pi/2$.

Figure \ref{fig:inter} pictures the numerical calculation of the hierarchy maps for intermediate values of $\theta$, together with the domain where condition (\ref{eq:validity}) is fulfilled for each (also numerically determined). An alternative presentation of these maps is available in the Supplementary Material file ``Hierarchies.pdf''. A few points are worth commenting:

\begin{itemize}
  \item In the relativistic regime, the fastest mode always has $Z_x=\infty$. Yet, its $Z_z$ component may be 0 (Weibel), or finite. We will therefore study the border between these two regimes.
  \item Around $\theta=0.45$ $(25^\circ)$, we find a region of the phase space centered around $(\sigma,\gamma_0) \sim (1,1.7)$ with both $Z_x \neq 0$ and $Z_z \neq 0$. We will comment on this oblique modes regime.
  \item The non-relativistic regime can be analyzed analytically because it is  equivalent to the field-free limit. We will therefore comment it in details.
\end{itemize}

\subsection{Relativistic regime}
The hierarchy maps indicate that in the relativistic regime, the fastest mode always has $Z_x=\infty$. We thus derive the dispersion equation in this limit. Its exact expression is reported in Appendix \ref{ap:GdZx}.

Expanding it in powers of $\gamma_0$ and keeping the three largest powers, namely 10, 9 and 8, gives the following approximate dispersion equation in the $Z_x=\infty$ and large $\gamma_0$ limit,
\begin{equation}\label{eq:DisperGdZxGdGam}
  \gamma_0^8 (x^2-Z_z^2)^2 \left(4 \sigma ^2 \cos ^4\theta+\gamma_0^2 (x^2-Z_z^2)^2-4 \gamma_0 \sigma  \cos^2\theta (x^2+Z_z^2)-16\right) = 0,
\end{equation}
which solution yielding unstable modes gives,
\begin{equation}
\delta^2 \sim \frac{\sigma  \cos (2 \theta )+\sigma +\gamma_0 Z_z^2 -2 \sqrt{2 \gamma_0 \sigma  \cos^2\theta Z_z^2+4}}{\gamma_0},
\end{equation}
which is not an exact expression due to our large $\gamma_0$ expansion.

The profile $\delta(Z_z)$ is still similar to the one pictured on Fig. \ref{fig:Wthetapisur2}. Hence, the largest $\delta$ will have $Z_z =0$ or not, depending on the sign of $\partial^2\delta/\partial Z_z^2$ in $Z_z=0$. A little algebra shows this second derivative vanishes in $Z_z=0$ for,
\begin{equation} \label{eq:WregimeOblique}
\sigma \sim 1/\cos^2\theta.
\end{equation}
We thus find that as expected from Fig. \ref{fig:inter}, the level of magnetization required to exit the Weibel regime increases with the field obliquity.

Another feature of Fig.  \ref{fig:inter} is that for large $\gamma_0$'s, condition (\ref{eq:validity}) translates to a condition of the type $\sigma<\sigma_T(\theta)$, where $\sigma_T$ is a magnetization threshold beyond which the present calculations are no longer relevant.

We can therefore expect that beyond a certain obliquity $\theta_W$, the critical sigma given by Eq. (\ref{eq:WregimeOblique}) becomes larger than $\sigma_T$, so that the Weibel instability governs the relativistic regime whenever the present theory is relevant.

An analytical determination of $\theta_W$ has not been possible. Yet, examining Fig. \ref{fig:inter} shows $\theta_W \sim 0.78$ (44$^\circ$). While for $\theta \lesssim 44^\circ$, various modes are likely to govern the linear phase, the Weibel instability always govern the relativistic unstable spectrum for $\theta \gtrsim 44^\circ$.

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig6.eps}
  \end{center}
  \caption{Calculations of $\delta (Z_z,Z_x)$ for $\sigma=1.2$, $\gamma_0=1.7$ and 3 values of $\theta$ bracketing $\theta=0.48$ (25$^\circ$). The white spot localizes the most unstable $\mathbf{Z}$.}\label{fig:obli}
 \end{figure}

\subsection{Oblique regime near $\sigma=1.2$, $\gamma_0=1.7$ and $\theta=0.48$ (25$^\circ$)}
Figure \ref{fig:inter} shows that there is a regime fully governed by oblique modes with $Z_z\neq 0$ and $Z_x\neq 0$, around $\theta=0.48$.

Figure $\ref{fig:obli}$ shows 3 calculations of $\delta (\mathbf{Z})$ for $\sigma=1.2$, $\gamma_0=1.7$ and 3 values of $\theta$ bracketing $\theta=0.48$. In each case, the white spot localizes the most unstable $\mathbf{Z}$. Although it is definitely oblique for $\theta=0.48$, the growth-rate gradient toward large $Z_x$'s is extremely weak. The $Z_x$ component of the fastest growing mode switches therefore quickly from a finite value to $Z_x=\infty$. Such is the reason why the transition away from this oblique regime looks discontinuous on the hierarchy maps. It is indeed continuous, but extremely steep.

\subsection{Non-relativistic regime}
A series of hierarchy maps for various intermediate orientations has been computed emphasizing the non-relativistic regime $\beta \ll 1$. The results are displayed in the Supplementary Material file ``Hierarchies\_NR.pdf''.

A clear picture emerges: in this regime, the $Z_z$ component of the fastest growing mode $\mathbf{Z}_{max}$, together with its growth-rate, settles to some constant values with $Z_{z,max} \sim 1.2$ and $\delta_{max} \sim 0.7$. As for the $Z_x$ component of $\mathbf{Z}_{max}$ its behaviour is not so monotonic.

These patterns can be understood noting that the Lorenz force $\mathbf{F}_L$, through which the field $\mathbf{B}_0$ acts on the system, decreases with $\beta$ because $F_L \propto \beta B_0$. Since the maximum $B_0$ we can explore is bounded by the validity condition (\ref{eq:validity}), the non-relativistic limit is also the field-free, $B_0=0$, limit.

What is then the non-relativistic behavior of the system for $B_0=0$? First, it becomes independent of $\theta$ since there is no longer any $\mathbf{B}_0$, and second, the dominant instability is the two-stream instability \cite{BretPRL2008,BretPoPReview}.

Setting therefore $B_0=0$, we derive the dispersion equation for the two-stream instability,
\begin{equation}\label{eq:ts}
  1-\frac{2}{(x+Z_z)^2}-\frac{2}{(x-Z_z)^2} = 0.
\end{equation}
Solving for $x$ gives the exact growth-rate,
\begin{equation}\label{eq:TauxTS}
\delta_{TS}^2 = 2 + Z_z^2-2 \sqrt{2 Z_z^2+1},
\end{equation}
reaching its maximum,
\begin{equation}\label{eq:TauxmaxTS}
\delta_{TS,max} = \frac{1}{\sqrt{2}} \sim 0.7,
\end{equation}
for,
\begin{equation}\label{eq:ZzTS}
Z_{z,max} = \sqrt{\frac{3}{2}} \sim 1.22.
\end{equation}
We thus find that these $Z_{z,max}$ and $\delta_{TS,max}$ fit precisely what is observed in the hierarchy maps.

  \begin{figure}
  \begin{center}
   \includegraphics[width=\textwidth]{fig7.eps}
  \end{center}
  \caption{Three maps $\delta(\mathbf{Z})$ for $\beta=10^{-2}$, and various combinations of $(\sigma,\theta)$, the first one having $\sigma=0$. All maps tend to map (a) in the $\beta \rightarrow 0$ limit. The white point shows the location of the fastest growing mode.}\label{fig:NR}
 \end{figure}

Let us finally turn to the behavior of $Z_{x,max}$. Figure \ref{fig:NR} displays three maps $\delta(\mathbf{Z})$ for $\beta=10^{-2}$, and various combinations of $(\sigma,\theta)$, the first one having $\sigma=0$. This latter map has been known for long \cite{Watson,Bludman,fainberg}. The two-stream instability is found for $Z_x=0$, and there is nearly no $Z_x$ dependance of the growth-rate.

Since for $\sigma=0$ and $\beta \ll 1$, neither the maximum growth-rate, nor its location along the $Z_z$ axis, depend on $\beta$ [see Eqs. (\ref{eq:TauxmaxTS},\ref{eq:ZzTS})], this map is the limit of the growth-rate when $\beta \rightarrow 0$. As a consequence, any map $\delta(\mathbf{Z},\sigma,\theta,\beta)$ has to tend to this one when $\beta \rightarrow 0$. The convergence is simple for $\delta_{max}$ and $Z_{z,max}$. However, the behavior of $Z_{x,max}$ is more involved. The reason for this is that $Z_{x,max}$ must definitely tend to $\infty$, and indicated by Fig. \ref{fig:NR}(a), but the gradient $\partial\delta_{max}/\partial Z_x$ in nearly zero. Therefore, while we still have $\lim_{\beta \rightarrow 0}Z_{x,max} = \infty$, the path to the limit is much more varied that it is for the $Z_{z,max}$ component.

Although temperature effects are beyond the scope of this work, they are likely to ``regularize'' the convergence of $Z_{x,max}$. The reason for this is that temperature always tend to stabilize large $\mathbf{k}$'s because pressure opposes the formation of small density bumps \cite{Silva2002,BretPoPMagne}. For the field-free case, it has been found that this effect localizes $Z_{x,max}$ to a finite value \cite{BretPRL2005}. The same has been found for the magnetized case  \cite{timofeev2009,Timofeev2013}. Hence, for realistic settings, or even in the case of small temperatures, the convergence of $Z_{x,max}$ is likely to resemble the one of $Z_{z,max}$.

\section{Conclusion}
We have considered a counter-streaming pair plasmas system, over an arbitrarily tilted magnetic field. The two colliding plasma shells are initially both cold and of the same density. The unstable linear spectrum is determined by only 3 parameters: the initial Lorentz factor of the shells $\gamma_0$, the strength of the magnetic field as measured by the $\sigma$ parameter, and its obliquity $\theta$.

The wave-vector of the perturbations pertains to the plane where the fastest growing modes are expected. For field obliquities varying from $0$ to $\pi/2$, we numerically determined the hierarchy map of linear instabilities, namely, the fastest growing mode $\mathbf{Z}_{max}(\sigma,\gamma_0)$ and its growth-rate $\delta_{max}(\sigma,\gamma_0)$.

The case $\theta=0$ has already been treated in Ref. \cite{bretPoP2016} and shows 4 different kinds of modes are likely to govern the linear phase. For finite obliquities, the hierarchy maps tends to simplify for 2 reasons:
\begin{enumerate}
  \item For these calculations to be relevant, the growth-rate must be larger than the cyclotron frequency of the charges in the field component normal to the initial flow. This sets condition (\ref{eq:validity}), which restricts the relevant $(\sigma,\gamma_0)$ domain.
  \item While the Weibel instability can be perfectly cancelled for $\theta=0$ \cite{Stockem2006ApJ}, it cannot for $\theta \neq 0$ \cite{BretPoP2011,BretPoP2013a,bretPoP2014a}. When $\theta=\pi/2$ is reached, it is simply unaffected because it has the particles moving sideways, that is, parallel to the field. As a result, this instability gains robustness and tend to govern a larger part of the map.
\end{enumerate}

In the relativistic regime, upper-hybrid-like (UHL) and Weibel modes mostly share the map until $\theta \sim 0.78$ (44$^\circ$). Beyond this critical obliquity, the UHL domain is ruled out by the validity condition (\ref{eq:validity}), and the Weibel instability always governs.

In the non-relativistic limit, the map progressively becomes $\theta$-independent simply because the dynamics becomes field-independent. In such conditions, the interaction is governed by the two-stream instability, even if the normal component of the most unstable $\mathbf{Z}$ may tend to 0 is a complicated way.

Our results show that the Weibel instability still dominates for $\sigma = 0.1$, while  Ref. \cite{Sironi2009Apj} suggests it does not. This seeming contradiction can be resolved by a comparison of the assumptions taken by us and in Ref. \cite{Sironi2009Apj}. We assume that we are in the reference frame in which the magnetic field is at rest and where the total momentum of the cold beams vanish. We analyze the instability spectrum in the full $\mathbf{k}$-space. The initial conditions used in Ref. \cite{Sironi2009Apj} introduce an oblique magnetic field into a moving pair cloud. The magnetic field has a component along the plasma's propagation direction and one that points out of the two-dimensional simulation plane. A filamentation of the pair cloud along the direction of the magnetic field is excluded by the simulation geometry. However, this is the direction along which the instability would develop, because the magnetic field does not affect the particle mobility along its direction. Reference \cite{Sironi2009Apj} examines the shock formation also in 3 spatial dimensions, but only for the quasi-parallel magnetic field directions that let the filamentation instability grow also in the 2D simulations. The rapid heating of the inflowing upstream plasma furthermore implies that the interacting beams cease to be cold in the simulation after a short time. We do not take into account thermal effects in our analysis.

However, the main reason for the difference between the results obtained by us and those of Ref. \cite{Sironi2009Apj} could be that in the simulation, the magnetic field  is moving with the pair cloud. This is achieved by the introduction of a convective electric field. The pair cloud that is reflected by the wall has the same density, speed modulus and temperature than the inflowing plasma. This symmetry implies that the magnetic field in the beam overlap layer, which is occupied by the inflowing and by the reflected particles, has to be at rest in the reference frame of the wall; the convective electric field in this interval must vanish. The convective electric field points along the simulation direction, which is orthogonal to the beam direction, and its amplitude varies along the beam direction. This electric field is thus rotational, which lets a magnetic field grow at the interface between the inflowing pair cloud and the overlap layer. The energy density of the initial magnetic field is not small compared to the particle's kinetic energy and its amplification will extract a substantial amount of energy from the upstream plasma. We neglect this important nonlinear effect in our analytic work.

In addition, the instability which is first triggered when the two shells overlap may not be the one eventually forming, and/or sustaining the shock. Such in the case for example when two electron/ion plasmas collide \cite{Bret2015ApJL}. In that case, electrons turn Weibel unstable first, but the field they grow is found unable to form a shock. It takes the ions to turn Weibel unstable, and then their Weibel filaments to merge, in order to start the shock formation. For the present case, further works will be needed to discriminate the instability first triggered, from the mechanism responsible for the shock formation and/or sustainment.

\section{Acknowledgments}
This work was supported by grants ENE2013-45661-C2-1-P and ENE2016-75703-R from the Ministerio de Educaci\'{o}n y Ciencia, Spain and grant PEII-2014-008-P from the Junta de Comunidades de Castilla-La Mancha. The authors thank Lorenzo Sironi and Asaf Pe'er for fruitful discussions.

redpajama_set_name: RedPajamaArXiv


text:
20th century births, Battle of Hogwarts participants, Bones family,
British or Irish individuals
Dumbledore's Army
Gilderoy Lockhart's fans
Half-bloods
Hufflepuffs
Hufflepuffs Sorted in 1991
Splinchees
Susan Bones
between 1 September 1979 and 31 August 1980[1]
Blood status
Half-blood[2]
British or Irish[3]
Female[4]
Relationship information
Mr Bones (grandfather) †
Mrs Bones (grandmother) †
Amelia Bones (aunt) †[5]
Edgar Bones (uncle) †
Mrs Bones (aunt) †
Bones children (cousins) †
Magical characteristics
Unknown length, wood and core[7]
Hufflepuff[4][5]
Bones family[5]
Hogwarts School of Witchcraft and Wizardry
Hufflepuff[4]
Dumbledore's Army[5]
"There were relatives of their victims among the Hogwarts students, who now found themselves the unwilling objects of a gruesome sort of reflected fame as they walked the corridors: Susan Bones, whose uncle, aunt, and cousins had all died at the hands of one of the ten, said miserably during Herbology that she now had a good idea what it felt like to be Harry."
—Susan as the subject of much unwanted attention[src]
Susan Bones (b. 1979/1980) was a half-blood witch[2] and a Hufflepuff student at Hogwarts School of Witchcraft and Wizardry from 1991-1998.[5] She was a member of the prominent Bones family, a family with strong connections to the Ministry of Magic.
Susan's entire family, including her grandparents, uncle, aunt, and cousins were all murdered personally by Lord Voldemort during the First Wizarding War,[8] and her other aunt Amelia Bones was murdered by Voldemort in July of 1996.[9] In her fifth year, Susan became a member of Dumbledore's Army, an organisation taught and led by Harry Potter.[5] Susan fought along side her fellow D.A. members in the Battle of Hogwarts.
It can be assumed that Susan survived the Second Wizarding War, as her name was not listed amongst the casualties of the fallen deceased. What happened to Susan afterwards is unknown.
1.2 Hogwarts years (1991-1998)
1.2.1 First year
1.2.2 Second year
1.2.3 Third year
1.2.4 Fourth year
1.2.5 Fifth year
1.2.6 Sixth year
1.2.7 Seventh year and Battle of Hogwarts
3 Magical abilities and skills
4 Possessions
5 Relationships
5.1 Family
5.2 Dumbledore's Army
Little is known about Susan's early life, other than the fact that she was born to at least one wizard parent, and presumably named after her aunt Amelia's middle name. Susan came from a largely magical family background, and also a rather powerful and respected family.
Due to her family's position, they were sought after during the First Wizarding War by Death Eaters and a large majority were killed. Susan was born in or around 1980, a year before the war ended.
Hogwarts years (1991-1998)
Minerva McGonagall: "Bones, Susan."
Sorting Hat: " HUFFLEPUFF!"
— Susan Bones being sorted[src]
Susan in the boat with Seamus Finnigan and Hermione Granger
Susan Bones started her education at Hogwarts School of Witchcraft and Wizardry in 1991 the same year as the famous Harry Potter. Susan shared a boat with Seamus Finnigan and Hermione Granger.[10] She was the third student to be called upon by Minerva McGonagall for the Sorting ceremony and was Sorted into Hufflepuff House[4] the Head of Hufflepuff House at the time was Pomona Sprout, the Herbology Professor. She was sorted into this house along with Hannah Abbott, Justin Finch-Fletchley, Wayne Hopkins, Megan Jones, and Ernie Macmillan all of whom she presumably became friends with. It is also assumed that Hufflepuff house is the house that many of her relatives were Sorted into as well.
Susan being Sorted into Hufflepuff House
During her first year, Susan would learn all about the basics of magic and spell performance, like her other classmates. She was also present at the End-of-Term Feast during which Gryffindor won the House Cup breaking Slytherin's seven year winning streak.
"Hannah, he's a Parselmouth. Everyone knows that's the mark of a dark wizard. Have you ever heard of a decent one who could talk to snakes? They called Slytherin himself Serpent-tongue."
—Ernie Macmillan shares his opinion of Harry with the Hufflepuffs[src]
Susan with fellow students during the first and last meeting of the Duelling Club
During her second year, Susan seemed to develop a crush on then-Defence Against the Dark Arts Professor Gilderoy Lockhart, which caused her to be off-task daydreaming during his lessons. She participated in the Duelling Club, as a precaution due to the Basilisk attacks that occurred after the Chamber of Secrets had been opened. During the meeting, Susan's friend Justin Finch-Fletchley was almost attacked by a serpent, which was acting under the orders of Harry Potter via the use of his Parseltongue ability. Many assumbed that Harry was egging the snake on, but in reality he was the only thing that kept the snake from attacking Justin.[11]
Susan in her second year Herbology class with Hermione Granger
Susan seemed to get along well with Hermione Granger, a Gryffindor, and the two shared an interest in the new Defence Against the Dark Arts teacher Gilderoy Lockhart and the two would be usually seated together in their Defence Against the Dark Arts classes, both listening intently on Lockhart's every word. She and Hermione seemed to become quite good friends this year and Herbology both worked next to one another when potting Mandrakes, and it must have come as a shock when later on in the year Hermione was petrified by the Basilisk that resided in the Chamber of Secrets.[11]
Later in the year, Lockhart was shown to be a fraud. Presumably, after knowing all the truth about Lockhart in the end of the school year, she lost her interest in him as he no longer was a great wizard who had done heroic deeds and performed extraordinary magic. When Hagrid returned to Hogwarts, Susan applauded him along with a majority of the school as they celebrated the close of a terrifying year.[11]
During her third year, she would have to select what subjects she wished to take. The school would also be in believed danger from Azkaban escapee, Sirius Black, who was suspected to be looking for Harry Potter.
During Susan's fourth year Hogwarts held the Triwizard Tournament and Hufflepuff's Cedric Diggory was one of the Triwizard champions. It is likely that Susan Bones supported Cedric Diggory in favour of Harry Potter, as he was in her house and both her best friends Hannah Abbott and Ernie Macmillan were strong supporters of Diggory.
The memorial feast of Cedric Diggory that Susan attended
Cedric Diggory was technically the joint-winner with Harry Potter after he and Harry both reached the Triwizard Cup, thus making them the champions. Sadly Diggory lost his life to Lord Voldemort after being taken to Little Hangleton and the Second Wizarding War began. The students were horrified to learn of Cedric's death, mainly students of Hufflepuff, who were closest to him, but the relationship of Susan with Cedric is unknown. It is more than likely that Susan attended the Memorial feast to Cedric Diggory at the end of the school year.
Susan: "My aunt says you can produce a fully corporeal Patronus, is that true?"
Harry Potter: "Er — you don't know Madam Bones, do you?"
Susan: "She's my auntie. I'm Susan Bones."
— Susan Bones asking Harry about his Patronus[src]
In 1995, Hogwarts was to be under the watchful eye of the Minister and his personally appointed Defence Against the Dark Arts teacher Dolores Umbridge. Umbridge worked with Susan's aunt in the Ministry, and both served on the court for Harry Potter's trial. When Umbridge came to Hogwarts, she ensured that no practical defensive spells would be taught in her lessons, despite the claims that Lord Voldemort had returned. Thus she did not teach anything worth learning.
Susan in the Room of Requirement after a D.A. meeting
With the students looking for a teacher, and Harry with the best knowledge of defensive and offensive spells, Hermione Granger pledged the idea of forming an alliance to learn spells and rebel against Umbridge. This would be called Dumbledore's Army, which Susan would become a member of. During the first meeting, Susan was curious to know whether Harry could produce a corporeal Patronus, a fact which she had been told by her aunt; when Harry asked if she knew Madam Bones, Susan confirmed the familial relationship.[5]
Harry later gave lessons on Patronuses, but it is unknown whether Susan produced one or what form it took. Later on in the year, the Death Eaters who murdered her uncle Edgar and his family escaped from Azkaban, the wizarding prison, and many students questioned Susan for information about this. After the Death Eaters' escape, Susan told Harry that she understood how he felt, knowing that the person who had murdered his family was at large and constantly being asked questions about it. Harry finally had someone who could almost relate to his state of mind. Susan was pestered throughout the year by students asking about her family and its murdered members, which she found distressing.[8]
Susan being bullied by Crabbe and Goyle for being a member of the Bones family
On top of dealing with the murders of her family members, this year was also her O.W.L. year at Hogwarts, which left her in even greater stress as she attempted to keep it all together and do well in her exams. Also, on at least one occasion Vincent Crabbe and Gregory Goyle took it upon themselves to bully Susan, making fun of her aunt and complaining that she was interfering with the Ministry of Magic's agenda. They went so far as to take her wand, and Harry, Ron, and Hermione had to come to her aid. The trio duelled the two Slytherins, defeating them and forcing them to leave Susan alone. Susan got her wand back and learned about the D.A. meeting in the Room of Requirement.[7]
At the end of the year, Susan was among the D.A. members who came to Harry's aid when he was attacked on the Hogwarts Express by Malfoy, Crabbe and Goyle, her own erstwhile bullies. Susan and her friends all unleashed a volley of curses which reduced the three to gigantic slugs.[12]
Sixth year
Girl 1: "Did you hear about her aunt? Poor Susan!"
Girl 2: "Yes, it's awful, I didn't want to leave my family behind this summer."
— Two students conversing about Susan Bones[src]
Susan Bones Splinching herself in an Apparition lesson
During the summer holidays before her sixth year at Hogwarts, one of Susan's last living relatives and one of the greatest witches of her time, Amelia Bones, was personally murdered by Lord Voldemort in her own home, though to her credit she did not go down without a fight.[9] This caused Susan some distress and worry, as most of her family had been killed during the First Wizarding War and now that the Second Wizarding War had been declared official it was only a matter of time before Death Eaters sought out after the remaining members of her family. When Susan returned to Hogwarts, she was learning how to Apparate and on one occasion Splinched herself during one of the lessons, losing one of her legs. Her leg was successfully reattached, but the incident appeared to leave Susan somewhat shaken.[13]
Later on in the school year, Hogwarts fell under attack by Death Eaters. The D.A. members were called to arms via their coins, and despite the murder of her aunt, Susan did not show up for battle, and possibly try and avenge her, it is possible that since the D.A. meetings were no longer held that Susan no longer carried her coin around with her and did not notice the call to arms message.
Seventh year and Battle of Hogwarts
"We have already placed protection around the castle, but it is unlikely to hold for very long unless we reinforce it. I must ask you therefore to —"
—McGonagall preps the students before battle[src]
Susan returned to Hogwarts for her seventh year, at which point the school was under the indirect control of Lord Voldemort. If she did not, she would have been a fugitive, since attendance was compulsory for children of magical families that year. Susan would have been in danger anyway, what with her being a member of the Bones family. A wizarding family high on Voldemort's list to eliminate, as most of them were stanch supports of Dumbledore and the Order of the Phoenix.[14]
Susan and the other defenders of Hogwarts witness Harry Potter's final defeat of Voldemort
She rejoined Dumbledore's Army, which had been reformed in a protest against Snape and the Carrows and fought in the Battle of Hogwarts alongside members of the D.A. and her professors. She was most likely in the Great Hall during the second half of the battle and witnessed Harry Potter's final defeat of Lord Voldemort and therefore the end to the Second Wizarding War, as she was not listed among the deceased.[14]
Susan was known for wearing her hair in a long plait down her back.[5] In 1996, Susan splinched herself whilst practising Apparition, and in doing so she lost one of her legs. However Susan was able to have her leg successfully reattached by the Heads of Houses but the ordeal left her confidence somewhat shaken.[13]
Magical abilities and skills
Defence Against the Dark Arts: Susan was a member of Dumbledore's Army in her fifth year,[5] and learned how to cast many defensive and offensive spells under Harry Potter's exemplary teaching.[12][7]
Duelling: She also fought in, and most likely survived, the Battle of Hogwarts, given how her name was not those among the fallen deceased, showing she had a level of skill in martial magic.[14] Susan and her friends also attacked Draco Malfoy, Gregory Goyle and Vincent Crabbe with a variety of dark charms that made them resemble slugs, showing she had skill in martial magic.[12]
Wand: Like all wizards and witches it can be assumed that Susan's wand was among her most valued possessions and that she purchased it from Ollivanders at the age of eleven. Her wand was of unknown length, wood and core material.[7]
"Susan Bones' grandparents were killed by Voldemort."
—J. K. Rowling in a interview[src]
Amelia Bones, her paternal aunt
Not much is known about the relationship of Susan with her family. Susan was a member of the Bones family, a family of many great witches and wizards. The Bones family supported the Order of the Phoenix during the First Wizarding War. That was the reason why the most of them were murdered by Lord Voldemort or his Death Eaters. Her grandparents, uncle Edgar, his wife, and their children. Amelia Bones, who participated in Harry Potter's hearing was Susan's aunt. They seemed to have a close relationship. However, Amelia was killed by Voldemort in 1996.
Gregory Goyle : "My father reckons your aunt's a troublemaker, someone will teach her a lesson!"
Susan: "What do you want? Harry! Hermione, Ron hey! Please, help!"
Vincent Crabbe: "Aw look, they've come to save their ickle friend."
— The Trio rush to Susan's defence[src]
Dumbledore's Army, her allies
Dumbledore's Army was an organisation founded by Harry Potter, Ron Weasley, and Hermione Granger, and led by Harry. Susan became a member during her fifth year, in order to be taught Defence Against the Dark Arts by Harry, as Dolores Umbridge, the new D.A.D.A. professor, didn't teach them how to properly defend themselves, thinking they would oppose the Ministry of Magic.[5]
Susan became a member, along with her fellow Hufflepuffs Hannah Abbott, Ernie Macmillan, Justin Finch-Fletchley and Zacharias Smith.[5] Susan shared a dormitory with Hannah Abbott (a fellow Hufflepuff) and was most probably invited to join Dumbledore's Army by her. She may have been also a close friend of Ernie and Justin. It is unknown if she was a friend of Zacharias, too. Susan and her fellow housemates (except Zacharias) fought for Hogwarts at the Battle of Hogwarts.
She was very impressed by learning that Harry could cast a Patronus, and she also told him how she felt about her relatives' deaths during the First Wizarding War.[5]
She was also seemed to be a friend of Seamus Finnigan, whom she shared a boat with on her first year and was often seen talking to him in lessons and in Dumbledore's Army meetings.[10]
Other members of the army she possibly knew are Ginny Weasley, Parvati Patil, Lavender Brown, Dean Thomas, Padma Patil, Luna Lovegood, Neville Longbottom, Alicia Spinnet, Angelina Johnson, Katie Bell, Fred and George Weasley, Colin and Dennis Creevey, Terry Boot, Anthony Goldstein, Michael Corner, Cho Chang and Marietta Edgecombe.[5] The majority of them fought at the Battle of Hogwarts and their fate is ambiguous.
"Susan" is the Anglicised form of Susanna, meaning Lily or Rose.[15] The lily is often a symbol of purity. It is noted that Susan is most likely named after her Aunt Amelia Bones, whose middle name is Susan.
"Bones" is the word for the skeletal form of a being, many members of Susan's family are murdered or have been killed during the books and this may be in reference to this.
Susan Bones as depicted on Pottermore
Susan Bones as a LEGO mini-figure
In the first two films, she was played by Eleanor Columbus, the daughter of director Chris Columbus, making her one of the few American actresses to appear in the films as one of the students. However, she had no dialogue. She was portrayed by Emma Jayne-Corboz in the film adaptations of Harry Potter and the Deathly Hallows.
She is one of the Original Forty students listed by J. K. Rowling along with Harry Potter, Hermione Granger and Ronald Weasley.
In the video game adaptation of Harry Potter and the Order of the Phoenix, Susan was voiced by Alex Tregear.
Susan wears her hair in a long plait down her back, however in the films and video games she is shown to leave her hair down.
The middle name of Amelia Bones is Susan, which could imply that Susan was named after her aunt, or that the name Susan runs in the Bones family.
In the earliest draft of the fifth film, it is Susan who says "That is most remarkable, Mr Potter".
In the film adaptation of Harry Potter and the Order of the Phoenix, Susan is omitted and one of Susan's lines is spoken instead by Luna Lovegood.
In the books, the sorting is done in alphabetical order, but in the films, it is not, and in 1991, Susan is the third sorted. She shows relief when put into Hufflepuff, suggesting that her family may all be in Hufflepuff. Susan is the only non-major character in the series whose Sorting is shown onscreen (the others being Harry, Ron, Hermione, Draco, and Lily, though the latter was in another year and shown in a memory.)
The Harry Potter Wiki has 79 images related to Susan Bones.
Susan Bones as she appears in LEGO Harry Potter: Years 1-4
Harry Potter and the Philosopher's Stone (First appearance)
Harry Potter and the Philosopher's Stone (film)
Harry Potter and the Chamber of Secrets (film)
Harry Potter and the Order of the Phoenix (film) (Appears on Parchment)/(Possible appearance)[16]
Harry Potter and the Order of the Phoenix (video game)
Harry Potter and the Half-Blood Prince (video game) (Mentioned only)[17]
Quidditch Through the Ages (Mentioned only)
Harry Potter: Quidditch World Cup
Harry Potter: The Creature Vault
↑ As she was sorted in 1991, at which point she would be eleven years old. Thus, she had to have been born between September 1979 and August 1980.
↑ 2.0 2.1 Harry Potter and Me (see this image)
↑ "World Exclusive Interview with J K Rowling," South West News Service, 8 July 2000 - "Hogwarts just serves Britain and Ireland."
↑ 4.0 4.1 4.2 4.3 4.4 Harry Potter and the Philosopher's Stone, Chapter 7 (The Sorting Hat)
↑ 5.00 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 5.10 5.11 5.12 Harry Potter and the Order of the Phoenix, Chapter 16 (In The Hog's Head)
↑ LEGO Harry Potter: Years 1-4
↑ 7.0 7.1 7.2 7.3 Harry Potter and the Order of the Phoenix (video game)
↑ 8.0 8.1 Harry Potter and the Order of the Phoenix, Chapter 25 (The Beetle at Bay)
↑ 9.0 9.1 Harry Potter and the Half-Blood Prince, Chapter 1 (The Other Minister)
↑ 10.0 10.1 Harry Potter and the Philosopher's Stone (film)
↑ 11.0 11.1 11.2 Harry Potter and the Chamber of Secrets (film)
↑ 12.0 12.1 12.2 Harry Potter and the Order of the Phoenix, Chapter 38 (The Second War Begins)
↑ 13.0 13.1 Harry Potter and the Half-Blood Prince, Chapter 18 (Birthday Surprises)
↑ 14.0 14.1 14.2 Harry Potter and the Deathly Hallows: Part 2
↑ Ancestry.com etymologies
↑ Unidentified Hufflepuff girl (I) might be her.
↑ During game play some students when you walk past will say "Did you hear about her Aunt? Poor Susan", referring to the death of Susan's Aunt, Amelia Bones.
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Dreamland Summary
August 14, 2020 September 19, 2022 Luke Rowley Culture, Health, History, Politics, Society
1-Sentence-Summary: Dreamland blows open the story of the United States' opioid crisis, from the frustrating greed and oversight that created it, how drug dealers accelerated it's spread, and what we're doing now to stop it.
Everyone knows that the opiate crisis in the United States is serious. Even the top leaders of the country, including President Donald Trump, are aware of it. And yet it's still here, threatening people's lives and society as a whole.
To call it an epidemic seems an understatement almost. But that's just the title that the US Drug Enforcement Administration, back in 2015, used to describe overdose deaths, particularly from heroin and prescription drugs.
Interestingly, this problem seems to only be limited to America. Although that's not a huge surprise when you consider the country's outlook on science, crime, the media, poverty, addiction, and especially health care.
Where did it all go wrong and what is being done to curb the devastating effects of this epidemic? You'll find these answers and more in Sam Quinones' Dreamland: The True Tale of America's Opiate Epidemic.
Here are just 3 of the many mind-blowing lessons I've learned from this book:
OxyContin is a small pill that's had a significant impact on the opiate crisis.
The statistics surrounding this epidemic are grim.
Many affected by opiate addiction have been suffering in silence for decades, but some are finally speaking out and making a difference.
Let's dive right in and see what Dreamland has to teach us!
Lesson 1: The massive opiate crisis is a result of deception and greed around one small pill called OxyContin.
Although the widespread effects of this dilemma are obvious now, it began slowly back in the 1980s.
It was 1984 when a drug company named Purdue came out with MS Contin, a morphine-based, time-release pain treatment pill. While it was originally for people on their death beds or those just out of surgery, it's success soon got Purdue carried away.
In 1996, the company released a similar pill by the now infamous name of OxyContin. Its active ingredient was the opium derivative called oxycodone, which is similar to heroin.
The company again touted the pill's time-release technology. Purdue got the FDA to approve the drug on the belief that this mechanism would limit addiction by restricting the highs and lows that opiates normally brought.
This became the hallmark of Purdue's marketing tactics for OxyContin, bringing wild success.
Lurking underneath, however, was a lot of bad science. For one, the perceived addiction rates were so low because it was only used for testing in controlled hospital conditions instead of by the doctors outside hospitals that the company marketed the drug to.
What's worse, one doctor who was simply compiling a hospital database found some information that would be misconstrued for years. Looking at the roughly 12,000 patients, he noticed only four developed painkiller addictions.
After publishing the finding in a short paragraph in the New England Journal of Medicine, the information spread like cancer. Some even mistakenly called it a "landmark study."
It wasn't long before the crisis was in full swing, only for drug dealers to make it worse.
Lesson 2: Grim statistics are just one marker of the awful effect this epidemic is having on individuals and families.
By the time 2000 rolled around the opiate crisis was in full swing. Of the roughly 100 million Americans suffering from chronic pain, almost all of them were on opiates of some kind.
Of the world's entire supply of oxycodone, the US alone was consuming 86%. And the use of hydrocodone, another opiate, was up to 99% and the most prescribed individual drug in the country.
Still today there are about 136 million prescriptions for these kinds of drugs every year.
Along with a rise in prescriptions has also come a rise in recreational use as well. From 2002 to 2011 alone, 25 million Americans consumed prescription pills in a way other than their intended use.
And the age at which people were abusing these drugs was dropping, too. According to the 2010 National Survey on Drug Use and Health, 2.4 million people 12 years old and up had consumed prescription pills nonmedically for the first time in the previous year.
Deaths from overdose have shot up as well. In 1999 a normal average was just 10 a day, but by 2012 the number was up to 48, almost five times higher.
What's worse, just 10 years after OxyContin was first prescribed, 6.1 million United States citizens had abused it. That's 2.4% of the entire population, or one in 40 people!
Lesson 3: Although many have suffered from this crisis, some are beginning to take a stand against it and are making changes happen.
Sadly, these issues have been happening for decades but the only real change has come in the last few years. With over 16,000 fatal overdoses yearly, it's a wonder that there weren't more than only a few groups raise awareness of this issue.
There is hope, though, and we're beginning to make progress.
Jo Anna Krohn's story is just one horrific example of the effects of this epidemic on families. Her son was playing with a gun while high on OxyCodone and shot himself.
As he was dying in a hospital bed, Krohn committed to using her voice to speak up about this issue. Her hope was that she might be able to spare another family from the pain she was going through.
She later began SOLACE to help parents grieving the deaths of their children who died of opiates. With her efforts, it quickly grew to have chapters in 16 counties.
Brad Belcher is another who has worked to raise awareness. After having been sober for five years, he decided to plaster signs all over his town that read "HEROIN IS MARION'S ECONOMY".
Although the authorities got ahold of most of the signs the night he put them up, enough remained that a picture of one made its way onto the internet. The image went viral and the town finally admitted it had to act.
While we're far from done with this epidemic, awareness is increasing and hopefully one day it will be over.
Dreamland Review
Wow, Dreamland unexpectedly had me on the edge of my seat while reading it! This story is crazy and disappointing and enraging all at the same time. I'm glad we're starting to get control of this epidemic and feel for all the individuals and families who have had to suffer for the laziness and greediness of a few pill manufacturers.
Who would I recommend the Dreamland summary to?
The 64-year-old European that doesn't understand the United States' opiate issues, the 34-year-old who is just beginning their career in the medical field, and anyone that wants to become more aware of the devastating effects of OxyContin and other pain pills.
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