The Sig Sauer SP 2022 is a modern Polymer handgun that has been adopted by law enforcement agencies, militaries and civilians around the world. This is an article about the author's impressions of the firearm and the basic information about it.
The Sig Sauer SP2022 is a very good bargain, sometimes priced better than S&W, Glock, HK and Springfield market competitors, but it isn't Sig's finest gun, despite being an approachable, well adopted handgun suitable for everyday carry and home defense, and having been adopted by militaries and law enforcement agencies scattered throughout the globe. It comes in 9mm & .40S&W chamberings.
NOTE: THIS ARTICLE WAS ORIGINALLY WRITTEN MORE THAN TWO YEARS AGO, AS A RESULT, SOME INFORMATION MAY NOT BE ACCURATE AT THE TIME YOU ARE READING IT. ARMSCART.COM MAINTAINS A MORE ACCURATE PAGE ON THIS ITEM (IF IT IS A PRODUCT) YOU CAN VIEW THAT PAGE HERE ARMSCART SIG SAUER SP2022.
Sig doesn't occupy a position in the gun world that causes people to find excuses to buy other brands. Sig has people trying to find excuses to buy Sigs. Well, they gave people another good reason by making the Sig Sauer SP2022 the descendent of the original Sig Pro model 2340 released in the late 1990's. It's a gun that has already enjoyed much success on the international stage in law enforcement and military and should be considered as a better alternative to many of the higher priced polymer framed pistols. It's not typical that Sig is the low price leader in any market segment; but then again, they have never really had to be. They have enjoyed success worldwide because of their heavy R&D, innovative designs and (usually) flawless execution on model runs. The talisman of the Sig line in the United States was the original alloy frame handgun which felt as though it was manufactured in a precision watch factory, with its ultra smooth actions and the incredibly high tolerances as well as the ridiculous reliability, which seems unmatched in most circles. Sig has a pretty good history with law enforcement as well; they are on the shortlist for just about every agency on the planet each time there is a contract to be fulfilled for a duty weapon. It is this relationship that probably provided a catalyst for the SP2022 to be made. There are rumors, that I have not verified with Sig factory that state the SP2022 was named such as it was a gun made in the year 2000 and slated to be part of the pistol lineup for 20 years, hence the service life ending in (SP)2022. The SP stands for SigPro. There are also rumors that the SP2022 were discontinued; that is not true. As you will see in the remainder of this specific article, the business of the SP2022 is far too lucrative for Sig to discontinue such a gun at this point. There were suggestions from the Sig factory that the debut of the Sig P250 would make the SP 2022 obsolete in some markets, but the only thing that suffered was the sales of the P250. In one of Sig's biggest markets and one of the biggest in the world for guns: California, the Sig SP2022 is still alive and well; thriving as part of the handgun roster.
Since Sig was known for its alloy framed service weapons and the venerable Sig P210 All steel target model, as well as other special pieces scattered around the archives at factory, it was finely tuned to receive the indications that it was losing the law enforcement war to newcomer Glock, and stalwart competitor Heckler and Koch in the 1980's and 1990's. Both of whom were utilizing polymer frames at the time with excellent profits and wonderful success in large domestic U.S. law enforcement contracts. Sig couldn't compete on price, and so they decided to join the polymer fray.
Originally Sig came out with the SP 2340 in 1998. A year later they began seeing interest from Law enforcement and military procurement teams for their now increasingly important SP2022. It was a gun which enjoyed an incredible reputation, without ever having to fire a shot; had a parent company full of LE and MIL experience, and had all the right features at the right price point.
The SP2022 has been the subject of some pretty interesting and high profile agency purchases, including the National police force of France which has order totaling more than a quarter of a million SP2022 handguns. The US DEA also chose the SP2022, buying up more than 60K units.
Colombia's Policia Nacional purchased over 55,000 units on the recommendation of the US DEA that was training them for continued success in the region in the war on drugs.
Special forces groups and police agencies in Switzerland (by the way they have a long and storied with Sig pistols including the P210), Portugal and Malaysia also can be counted amongst the users and "early" adopters of the SP2022 as a duty gun.
How much will the Sig SP 2022 cost to own?
Many others have begun to research and take the SP2022 on trial in the US and abroad. As the economy worsens globally (with the exception of certain European, Middle Eastern and Asian countries) HK has seen a price increase, and is therefore on the lower end of the shortlists in some cases; Glock experienced some difficulties in the first several production runs with their new 4th Generation pistols, and now that the Sig is so widely adopted, it makes the (approximately) $500 SP2022 a more attractive option, even though it has always matched the Glock and HK on reliability and performance.
It's not that an agency endorsement guarantees performance for a civilian user of a specific firearm, but the good performance of such an agency gun does impact its marketing and help people determine if it has the capabilities to meet the demand they will expect it to handle. The real benefit to the SP2022 is that you get EVERYTHING for the price it costs to get a much lesser gun usually.
The ergonomics and the control of the SP2022 are spot on, and it is customizable without altering the firearm too much. You also get the Sig name; brand image; research and development team, state of the art materials and the backing of one of the biggest names in firearms when you buy an SP2022.
The SP2022 You get a gun made in the image of greatness (the other sig pistols from various stages of their history), with an attached rail for accessory mounting; a great trigger and the lightweight of a polymer pistol. There is nothing in the SP2022 or any Sig for that matter that leaves you feeling remorseful about a purchase.
Out of the box, you get a quality firearm every time with few exceptions. In the event of an exception, Sig takes care of you quite well.
Sig has finally gotten to the point where they can offer the polymer weapons at a steep discount to their competitors, which proves something telling: Guns can be had for less money which can perform above their pay grade.
The combination of the massive purchase orders from LE and MIL, with the tooling having already largely been paid for, as well as the now ancient (though the gun doesn't present this) R&D costs, give a great indication as to the reasons Sig can provide the SP2022 for such a low price.
Usually, we are convinced that if something is too good to be true, it probably is; the SP2022 isn't too good to be true. The SP2022 is a great value.
The SP2022 isn't perfect, but it's VERY nice. The tolerances aren't QUITE what Sigs other guns are, it might look tactical, but it's nowhere near as pretty as some of the other Sig guns; and it is a bit top heavy compared to other pistols in the line. The last comment is a disease of polymer framed guns, and not indicative of a shortcoming of the SP2022.
The SP2022 is the lowest priced full power Sig; a unique and wonderful alternative to the Glock and HK pistols, as well as the Springfield and S & W polymer guns (amongst others). It's a gun equally at home in a woman's hand as a man's, and one which is so good you'll find yourself going to it time and again to use for carry and target shooting.
You can find it in three calibers if you look hard enough: .357 Sig; 9 mm and .40 S & W. The SP2022 is a perfect first handgun; it's inexpensive, infinitely usable and highly regarded. The SP2022 is not going anywhere and it provides a solution to a problem that cannot be matched at the same price point by many competitors.
If you need a no mess handgun, take a look at the venerable Sig SP2022.
Armscart.com offers comprehensive buying information, reviews and expert commentary about products in the firearms and shooting industry.


In this article we will examine the steps needed to build an Event App on The Velocevent platform. The platform will do all of the heavy lifting for you, so by following these simple steps you will be able to build and publish an app for your event on the same day. If your event consists of less than 100 guests, you can use our Launch It package! It's free and no credit card is required.
– Event branding: It's important that your users can easily recognize your event in the app, you want all of the icons to display your logo, and all of the headers to display your banners.
– Venue: This way attendees will be able to find your event using Google Maps.
– Sponsor information: Show your sponsors how much you value all of their support by making sure all of your guests have easy access to their information in any device.
– Speaker/Exhibitor information: The app for your event should have all of its speaker's information, plus the location on an exhibitor map, so users can find any session they may like to attend. Velocevent will build this map automatically.
– Attendee list: The most important part of your event are the people that are going to enjoy it. We keep our contact information on Infusionsoft but we noticed that migrating all of those contacts to the app's database for a large event could be annoying. So, we decided to implement a feature that allows you to upload and update all of the relevant information from a Google Sheets file.
– Social Media information: It's important that your app allows easy access to your social media channels. Link your event to your Facebook page, a Twitter hashtag, or to your Instagram account.
– Color Palette: A crucial part of the branding is the color palette of your event, choose carefully.
Start with the basic set up: a name description and the event email. This email will be the subject to all of the notifications Velocevent will send to your guests.
If you're using any of our premium packages, just insert your key here to upgrade your event app.
It's crucial that your attendees can find your venue in time for your event. Velocevent will automatically sync the location of your venue with Google Maps.
Whatever industry you're in, chances are you're promoting it on social media. If you want to take full advantage of your social media strategy you need to make it easy for your guests to participate. Velocevent integrates social media feeds in your app dashboard making it very simple for your participants to share your event.
Here you can customize your app to use the corporate colors. You can make it easier for your attendees to relate your app to your event by customizing it to match your branding.
Your event app should have an easy to navigate interface. Using icons instead of plain text will make your app more user-friendly.
You can upload custom icons to match your event tone and audience, or choose from our standard icons.
One of the most important aspects of any event app is making it easier for attendees to interact with your content. In this section, you can choose specific settings to allow your guests to create custom agendas.
Allowing your viewers to rate the different exhibits and lectures will be very helpful to other attendees.
This is where you'll upload the logo and main banner for your event.
Then you have the option of uploading the exhibitor map and agenda locations. Velocevent will automatically set up a dynamic exhibitor map so your attendees can find any of your exhibits.
Velocevent will automatically build a speaker photo collage, you just need to choose your favorite layout.
Finally, we have some specific settings about the general details of your app and organization.
That's it, by this point your event app is ready to go live!
Don't overthink it, making changes to a published app is as simple as updating Google Sheets. If you still have any doubts, please request a free demonstration. Just fill out the form and we will contact you as soon as possible.


text:
Q: Powershell. Выборка файлов по самой новой дате Необходимо из одной папки (с кучей подпапок) выбрать самые новые файлы и скопировать их в другую папку. С копированием я разобрался, а вот как выбрать именно те файлы что нужны - не пойму.
Поиск подсказал: Get-ChildItem | Where-Object{$_.CreationTime.Day -like "23"}. Но это именно по конкретному дню выборка.
Не подскажете куда поглядеть, чтобы не брать все даты файлов, сортировать, а потом выбирать? Если по другому никак то как это сделать быстрее? Просто файлов крайне много, поэтому такой скрипт будет обрабатываться очень долго.
Использовать что-либо, кроме Powershell и cmd - нельзя.

A: Так устроит?
setlocal EnableDelayedExpansion

set idx=0
for /f %%i in ('dir /A:-D /B /O:-D') do (
  copy %%i 1\
  set /a idx= !idx! + 1
  if !idx!==5 exit
)

Выполняем команду dir с сортировкой файлов по убыванию даты и после первых 5 файлов выходим

A: PowerShell

*

*Самый новый файл(!) в текущей папке.
dir -File | sort LastWriteTime -Descending | select -First 1

*Файлы за последние 2 дня:
dir -File | where LastWriteTime -gt (Get-Date).AddDay(-7)

Для Get-ChildItem (она же - dir) можно указать путь, даже сетевой: D:\Папка с архивами, \\Server\Mail\Incoming\

*Комбинируем примеры 1 и 2, чтобы выбрать самые новые, но только за последнюю неделю:
dir -File | where LastWriteTime -gt (Get-Date).AddDay(-2) | sort LastWriteTime -Descending | select -First 1

*Самый новый файл в каждой подпапке 1-го уровня (непосредственно дочерняя в текущей)
dir -Directory -Recurse -Depth 1 | %{dir $_ -File | sort LastWriteTime -Descending | Select -First 1}

*Для копирования добавьте конвеер команды копирования:
<# что-то из примеров выше #> | cp -Destination "D:\Папка со свежими файлами\$(Get-Date -Format 'yyyy-MM-dd')\"

Создаст в нужной папке подпапку с сегодняшней датой и переложит все свежие файлы туда.

*Для правильного создания путей лучше всего использовать Join-Path:
❯ Join-Path 'D:\Где всё лежит' (Get-Date -Format 'yy-MM-dd HH:mm')
D:\Где всё лежит\20-10-26 15:51

*По поводу где посмотреть:
alias dir,ls,sort,select,where,cp | Get-Help -Online

redpajama_set_name: RedPajamaStackExchange


To initiate this public consultation, and on the basis of the SET Plan Integrated Roadmap, the European Commission developed an Issues Paper that addressed the R&I challenges driving ambition in carbon capture storage and use deployment. The first draft of this Issues Paper was sent to SET Plan countries and relevant research and industrial stakeholders requesting them to take position on the proposed targets/priorities, in accordance with the guidelines set out in the document entitled ''the SET Plan actions – Implementation process and expected outcomes'' (see box below).
Relevant stakeholders and SET Plan countries were invited to provide their comments for this consultation round in the form of "input papers" by 25th April 2016. Following those exchanges the European Commission organised a dedicated SET Plan Steering Group meeting on 24th May 2016 whereby, invited stakeholders and SET Plan countries discussed the Issues Paper and reached an agreement on common targets that took the form of a Declaration of Intent (DoI). Following this meeting, the DoI for carbon capture storage and use deployment was endorsed by written procedure and presented to the Steering Group meeting on 14th September 2016.
In November 2016, interested Member states/Associated countries, stakeholders and the European Commission held the kick off meeting of the SET Plan temporary working group (TWG) on driving ambition in carbon capture and storage deployment. This TWG, chaired by national representatives from Norway and the Netherlands and co-chaired by Zero Emissions Platform (ZEP), met regularly times during 2016 and 2017 to draft an Implementation plan that contains concrete R&I activities, and proposes relevant funding opportunities, which are considered essential for achieving the agreed set of targets. The Implementaion plan was then ensorsed by the SET Plan Steering Group members at the meeting held on 27 September 2017.


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Limoncello, Lemon Blossom and Lookouts: Amalfi Coast, Italy
A heady cocktail of limoncello, pastel-hued villas and sun-drenched Mediterranean vistas, the Amalfi Coast is the pinnacle of the Italian dream
Sara D'Souza and Carlos Souza
Positano bites deep. It is a dream place that isn't quite real when you are there and becomes beckoningly real after you have gone.
The The Amalfi Coast is the pinnacle of the Italian dream, a line of thirteen towns tucked among lemon blossom and bougainvillea. Known for its cascading landscape, these settlements are only accessible via the Strada Statale 163, more commonly known as the Amalfi Drive. Forming a near-magical cliffside strip between the mountains and the Mediterranean Sea below, it's a narrow, winding route that, while being unsettling to traverse, offers unparalleled views.
For decades, aesthetes from around the world - Jacqueline Kennedy, John Steinbeck, Gore Vidal - have been lured by the Amalfi coast's quaint pebbled beaches, scenic hikes, perfect climate and legendary establishments. Such attractions aren't limited to storied hotels and restaurants; the Amalfi Cathedral, Ravello's Villa Rufolo, the gardens of Villa Cimbrone all conjure the culture and spirit of bygone centuries.
A heady cocktail of limoncello, pastel-hued villas and sun-drenched Mediterranean vistas, the Amalfi Coast is in a class of its own.
Amalfi Coast, Italy
Amalfi Coast is published by Assouline
Sea, Sun and Spaghetti: Amalfi Coast, Italy
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The New Yorker's SecureDrop (formerly called Strongbox) is a method for you to share newsworthy tips, information, and files whose importance or sensitivity demands a greater degree of anonymity and security than is afforded by conventional e-mail.
To help protect your anonymity, SecureDrop is only accessible using the Tor network (https://torproject.org). When using the SecureDrop, The New Yorker will not record your I.P. address or information about your browser, computer, or operating system, and will not embed third-party content or deliver persistent cookies to your browser. No method of communication, however, is completely secure.
You can read the full privacy statement here.
TO GET TO SECUREDROP AND BEGIN USING IT TO CONTACT WRITERS AND EDITORS AT THE NEW YORKER, FOLLOW THE THREE STEPS BELOW.
Download and install software to access the Tor network: https://www.torproject.org. This should only take a few minutes.
Once you load the Tor browser, copy and paste the URL http://icpozbs6r6yrwt67.onion into the Tor address bar. When the page loads, you will find further instructions on how to submit files and messages to The New Yorker.
You will be assigned a randomly generated and unique code name. If a writer or editor at The New Yorker wants to contact you about the information you have submitted, he or she will leave a message for you in SecureDrop. These messages are the only way we will be able to reach you, and can only be accessed using your code name.
Please note: general fiction, poetry, art, and PR submissions sent via SecureDrop will not be assessed. To contact The New Yorker for general inquiries, please read our Contact Us page. Please do not include links to Google Documents or Dropbox files—files should be submitted through SecureDrop directly.
For more about The New Yorker's SecureDrop project, watch this video, and read the introductory posts written by Amy Davidson, Kevin Poulsen, and Joshua Rothman.
The New Yorker's SecureDrop is designed to let you communicate with our writers and editors with greater anonymity and security than is afforded by conventional e-mail.
When you visit or use our public SecureDrop server at http://strngbxhwyuu37a3.onion, The New Yorker and our parent company, Condé Nast, will not record your I.P. address or information about your browser, computer, or operating system, nor will we embed third-party content or deliver cookies to your browser.
The The New Yorker'sSecureDrop servers are under the physical control of The New Yorker and Condé Nast.
SecureDrop is designed to be accessed only through a "hidden service" on the Tor anonymity network, which is set up to conceal both your online and physical location from us and to offer full end-to-end encryption for your communications with us. This provides a higher level of security and anonymity in your communication with us than afforded by standard e-mail or unencrypted Web forms. SecureDrop does not provide perfect security. Among other risks, if you share your unique code name, or if your computer is compromised, any activities, including communications through SecureDrop, should be considered compromised as well. The system is provided on an "as is" basis, with no warranties or representations, and any use of it is at the user's own risk.
On October 24, 2017, the Freedom of the Press Foundation announced that it had discovered a vulnerability in the SecureDrop code, which it is taking steps to address. The New Yorker has no reason to believe it was affected by this vulnerability, and will update this page as needed.


Even after Tara Webb signed her letter of intent Wednesday morning, it still hasn't quite hit her.
Mountainburg coach Dan Garringer sent off a DVD of Webb playing to Arkansas Tech and within days, the Golden Suns offered a scholarship.
Webb signed a Division II letter of intent to play basketball for the Golden Suns as part of the early signing period.
"I was still very nervous about signing," said Webb after the brief ceremony at Mountainburg. "Now, I don't have to worry. It takes the weight off my shoulders."
Webb is the middle sister of Destiny and Kristen Webb. Destiny was a former standout for the Lady Dragons and is attending college. Kristen is a freshman at Mountainburg.
The trio had more than their share of trials, but with the support of the community, the school administrators, coaches, and friends, the Webb sisters not just survived, but they've thrived.
"It would have been easy for Tara or her sisters to fall back on their problems and setbacks. They're all great kids," said Garringer. "Tara is obviously a good athlete. She's very quiet. I think the first year I was here, she didn't say one word to me, unless I asked her something.
Webb, who scored a 27 on her ACT, is one of six seniors for the Lady Dragons. That group have been together since seventh grade and has compiled a 109-9 record since that time, and 57-4 as sophomores and juniors.
Garringer, who was a coach at Cedarville, briefly retired, but returned to coaching at Decatur before coming back to Crawford County.
"Working with Tara, her sisters and these kids have kind of rejuvenated me and my wife, Kathy. This is supposed to our golden years," said Garringer. "Tara's signing, and Maddie Backus' signing last season, and watching the other girls succeed. It's rewarding for us to work with such great people."
Backus signed with UA-Fort Smith, another Division II school, last season and has started for the Lady Lions this season.
"Both of these signings have been great for Mountainburg," Garringer said. "It's very meaningful and brings a lot of pride for the school, teachers, coaches and the fans."
Arkansas Tech finished 17-3 in the Great American Conference last season, but fell in the first round of the NCAA Tournament. The Golden Suns are picked to finish third this season.


text:
Q: Changing the text color of a react-bootstrap Navbar using webpack I'm new to web development, and I'm trying to customize the colors of a react-bootstrap Navbar.
This is the contents of my NavbarComponent.cs file:
var React = require('react');
var ReactDOM = require('react-dom');

import Navbar from 'react-bootstrap/lib/Navbar';
import Nav from 'react-bootstrap/lib/Nav';
import NavItem from 'react-bootstrap/lib/NavItem';
import Button from 'react-bootstrap/lib/Button';
import styles from './Navbar.css'

export default class NavigationBar extends React.Component {

  render() {
    return (
      <Navbar brand="react-bootstrap" className={styles.navbar}>

          <NavItem className={styles.navitem} eventKey={1} href="#">Thing 1</NavItem>
          <NavItem className={styles.navitem} eventKey={2} href="#">Thing 2</NavItem>

      </Navbar>
    );
  }
}

and this is my css override file, Navbar.css:
:local(.navbar) {
  background:#2E5F91;
  color:white !important;
}

:local(.navitem) {
  color:white !important;
}

The background color of the Navbar changes just fine, but I can't get the text color to change, even using the !important tag (which is dangerous, I know).
Looking at the elements in the resulting page:

  Thing 1

it looks like my class isn't getting deep enough. How can I fix this?
Thank you!

A: Welcome to web development, hope you will like as much as we do !
I am not familiar with the css pseudo-class :local, I am curious why not use usual css class ?
.navItem {
  color: white;
}

I think the component NavItem does not have a className property
(source : https://react-bootstrap.github.io/components.html#navs-props-navitem)
So maybe that's why it doesn't work, not sure though... I would have done it like this :
<NavItem eventKey={2} href="#">
  Thing 2
</NavItem>

Here's a jsfiddle: http://jsfiddle.net/u4g5x93w/1/

A: you can solve this problem by putting
<link rel="stylesheet" href="../Navbar.css">

to the index.html in your project. And also give specific path for Navbar.css in your index.html. then edit what you like.

A: <Navbar collapseOnSelect expand="lg" bg="light" variant="light" >

A: .navbar-nav .nav-item a {
  color: white !important;
}


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

In a previous paper \cite{Rahmfeld1}, we argued that the spectrum of
elementary BPS ($N_R=1/2$) states of compactified heterotic strings could be
identified with extremal electrically charged black holes. Further evidence
for this interpretation was supplied in
\cite{Sen2,Peet,Khurimyers,Callan,Shiraishi}. In particular, the $N_L=1$ states
and the $N_L>1$ states (with vanishing left-moving internal momentum) admit a
single scalar/Maxwell interpretation with parameters $a=\sqrt{3}$ or $a=1$
respectively.  In other words, by choosing appropriate combinations of dilaton
and moduli fields to be the scalar field $\phi$ and appropriate combinations
of the field strengths to be the Maxwell field $F$, the field equations can be
consistently truncated to a form given by the Lagrangian  \begin{equation}} \newcommand{\ee}{\end{equation} {\cal L}=
\frac{1}{32\pi}\sqrt{-g}\left [R-\frac{1}{2}(\partial \phi)^2-\frac{1}{4}e^{-a
\phi}F^2 +...\right] \la{actiona}  \ee
for these two values of $a$, these combinations being just those
corresponding to the quantum numbers of the string states\footnote{A {\it
consistent} truncation is defined to be one for which all solutions of
the truncated theory are solutions of the original theory. The dots in
(\ref{actiona}) refer to terms involving a pseudoscalar combination of axion and
moduli fields which are in general required for consistency but which do not
contribute to non-rotating black hole solutions.  We are grateful to R. Myers
for pointing out their omission in the original version of this paper.}. In
the case of zero angular momentum, the ADM mass $M{}_{black}$ of the extremal
black hole solutions of (\ref{actiona}) is given by  \begin{equation}} \newcommand{\ee}{\end{equation} M_{black}^2=Q^2/4(1+a^2)
\la{2} \ee  where $Q=\int \tilde F/8\pi=\int e^{-a\phi}{}{*F}/8\pi$ is the
electric charge, where  $*$ denotes the Hodge dual and where, for simplicity, we
have set the asymptotic value of $\phi$ to zero. The $a=1$ case yields the
supersymmetric dilaton black hole \cite{Gibbons}. The $a=\sqrt{3}$ case
corresponds to the Kaluza-Klein black hole and the ``winding" black hole
\cite{Khurinew} which are related to each other by $T$-duality. The Kaluza-Klein
solution has been known for some time \cite{Gibbons} but only recently
recognized \cite{Khurinew} as a heterotic string solution. We also argued that
the corresponding solitonic magnetically charged and dyonic spectrum, predicted
by S-duality \cite{Schwarz1}, is also described by extremal black holes. Indeed,
we were first motivated to make the black hole conjecture for the elementary
states by first noting that the ``winding'' magnetic monopoles are extremal
black holes \cite{Khurinew} and then noting that string/string duality
interchanges the roles of $S$-duality and $T$-duality and therefore that the
solitonic monopoles play the same role for the dual string as the elementary
winding states play for the fundamental string \footnote{Of course this involves
extending the classical notion of a black hole from the weak coupling to the
strong coupling regime.  We will therefore take the liberty of  describing a
state by the words {\it black hole} if there exists at least one string picture
in which its mass exceeds the Planck mass for weak coupling.}.  By allowing $F$
to describe combinations of field strengths and their duals, other dyons,
preserving fewer supersymmetries and therefore not predicted by S-duality alone,
can also be assigned $a$ values. One finds $a=1/\sqrt{3}$ and $a=0$. The $a=0$
case yields the Reissner-Nordstrom solution\footnote{The Reissner-Nordstrom
black hole is not a solution of dimensionally reduced pure gravity but has long
been known to be a solution of $M$-theory \cite{Duffpope,Pope}.} which,
notwithstanding contrary claims in the literature, does solve the low-energy
string equations \cite{Khurinew,Rahmfeld1}. The $a=1/\sqrt{3}$ black hole
\cite{Horowitz2} was identified as a dyonic solution in \cite{Rahmfeld3}. These
four values of $a$ yield solutions which are special cases of the most general
solution subsequently found in \cite{Cveticyoum}, and shown to be exact to all
orders in $\alpha'$ in \cite{Cvetictseytlin}. In the $N=2$ theory, black holes
with $a=\sqrt 3,1,1/\sqrt 3,0$ all preserve $1$ supersymmetry and therefore
belong to fundamental supermultiplets with maximum spins $1/2$ \cite{Rahmfeld3}.
In the $N=4$ theory, black holes with $a=\sqrt{3},1,1/\sqrt{3},0$ preserve
$2,2,1,1$ supersymmetries, respectively, and therefore belong to fundamental
supermultiplets with maximum spins $1,1,3/2,3/2$
\cite{Kalloshpeet,Rahmfeld1,Rahmfeld3}.  This same black hole interpretation
could be extended to the spectrum of BPS states of toroidally compactified Type
$II$ strings \cite{Hull} where the supersymmetric black holes admitting a single
scalar/Maxwell interpretation correspond once again to the same four values of
$a=\sqrt{3},1,1/\sqrt{3},0$. They preserve $4,2,1,1$ of the $N=8$
supersymmetries, respectively \cite{Popelu1,Khuriortin}, and therefore belong to
fundamental supermultiplets with maximum spins $2,3,7/2,7/2$.

On the basis of their mass and charge assignment, it was further
suggested \cite{Rahmfeld1,Rahmfeld3} that we interpret these four values
of $a$ as $1-, 2-, 3-$ and $4$-particle bound states with zero binding
energy. This is reviewed in section (\ref{bound}).  For example, the
Reissner-Nordstrom ($a=0$) black hole equals four Kaluza-Klein
($a=\sqrt{3}$) black holes! This zero-binding-energy bound-state
conjecture can, in fact, be verified in the classical black hole picture
by finding explicit $4$-centered black hole solutions which coincide
with the $a=\sqrt{3},1,1/\sqrt{3},0$ solutions as we bring $1,2,3,4$
centers together and take the remaining $3,2,1,0$ centers out to infinity
\cite{Rahmfeld4}. Such a construction is possible because of the
appearance of four independent harmonic functions
\cite{Cvetictseytlin}. Moreover, this provides a novel realization of the {\it
no-force} condition in that the charge carried by each black hole corresponds
to a different $U(1)$. Thus the gravitational attraction cannot be cancelled
by an electromagnetic repulsion but rather by a subtle repulsion due to scalar
exchange. This phenomenon was also observed in \cite{Kalloshcancel}. In
section (\ref{supermultiplets}) we shall provide further evidence for the
bound state interpretation in the quantum string state picture by showing that
not only the masses, electric charges and magnetic charges but also the spins,
and supermultiplet structures are consistent with this interpretation of the
$a=0,1/\sqrt 3,1$ string states being merely bound states of the fundamental
$a=\sqrt{3}$ states. This is entirely consistent with the claim of
\cite{Holzhey}, using completely different reasoning, that only $a>1$ dilaton
black holes can be interpreted as {\it elementary} particles. Can this
interpretation also apply to the recently discussed {\it massless} black holes
\cite{Strominger2,Cveticyoum2,Behrndt,Kalloshcancel,Chancvetic}? Classically, the
answer is yes in that there exist 2-centered solutions which coincide
with the massless black hole as we bring the two centers together. In
this case, however, it is necessary to assume that one of the constituents
has a negative mass and it therefore seems unlikely that this bound state
interpretation can survive quantum-mechanically.

For the purely electric elementary string states, where it makes sense to
assign an oscillator number, the massive BPS states in the
heterotic theory are given by the $(N_R=1/2,N_L\geq1)$ states
\cite{Rahmfeld1} and the massless  BPS states belong to the $(N_R=1/2,N_L=0)$
sector \cite{Behrndt}. Curiously, however, it is possible to extend the black
hole bound state interpretation to non-BPS states, for example the
non-supersymmetric $a=1$ dilaton black hole of \cite{Garfinkle} corresponds to
$(N_R=3/2,N_L=1)$.

There is now a consensus that all of string theory and its duality
properties follow from an underlying eleven-dimensional theory
\cite{Howe,Townsend,Witten,Duffliuminasian,Schwarz2,Horava}, now known as
{\it $M$-theory}. In section (\ref{multi}) we turn to the black branes of
$M$-theory \cite{Dufflupope2,Klebanov1,Mukherji2,Muto}, where (\ref{actiona})
is now in an arbitrary spacetime dimension $D\leq11$ and the $F$ is now a
$(p+2)$-form.  The parameter $a$ can be conveniently re-expressed as
\cite{Popestainless}  \begin{equation}} \newcommand{\ee}{\end{equation} a^2 = \Delta - \frac{2 (p+1)(D-p-3) }{D-2} , \ee
since $\Delta$ is a parameter that is preserved under dimensional reduction
\cite{Popestainless}.  Originally, attention was focussed on the $\Delta=4$
solutions \cite{Strominger1,Dufflu,Khuristring} but various new supersymmetric
solitons with $\Delta\neq 4$ have recently been studied
\cite{Popestainless,Popelu1,Popelu2}.  These authors proposed to classify
$p$-branes into ``rusty'' or ``stainless'' according as they can or cannot be
``oxidized'' to isotropic brane solutions of a higher-dimensional
supergravity. (Oxidation is the opposite of double dimensional reduction
\cite{Howe,fifteen}.) Examples of new stainless solutions included a $\Delta=2$
$5$-brane in $D=9$ and  $\Delta=2,4/3$ strings in $D=5$ \footnote{The
$\Delta=4$ $6$-brane in $D=9$ can, in fact, be oxidized to a Type $IIB$
$7$-brane in $D=10$.}. The authors of \cite{Popestainless} then raised the
question of whether these new $\Delta \neq 4$ branes deserve to be treated as
fundamental in their own right. In this section we shall generalize the
treatment of extremal black holes \cite{Rahmfeld4} and confirm on the level of
classical solutions that these new $\Delta=4/n$ $p$-branes can also be
regarded as bound states with zero binding energy of $n$ fundamental
$\Delta=4$ branes.   We find new $1\leq n \leq m$-centered $p$-brane solutions
which reproduce the $\Delta=4/n$  solutions of \cite{Popelu2} as we allow $n$
of the centers to coincide and take the remaining $(m-n)$ out to infinity. In
particular, the $\Delta=2$ fivebrane is a bound state of two $\Delta=4$
fivebranes and the  $\Delta=2$ and $\Delta=4/3$ strings are respectively bound
states of two and three fundamental $\Delta=4$ strings.

In section (\ref{entropy}), we also discuss the temperature and entropy
of the extreme $a=\sqrt{3}$, $1$, $1/\sqrt{3}$, $0$ black holes. Here a good
deal more needs to be understood since only for the $a=0$ solution is the
dilaton vanishing and so the classical entropy prediction for the other
three cases (namely zero) is unreliable.  Similar remarks apply to the
macroscopic entropy and temperature of the other $p$-branes.

\section{The bound state conjecture}
\la{bound}

Let us begin by recalling the bound state conjecture in the context of
the four-dimensional heterotic string obtained by toroidal
compactification. At a generic point in the moduli space of vacuum
configurations the unbroken gauge symmetry is $U(1)^{28}$ and the low
energy effective field theory is described by $N=4$ supergravity
coupled to $22$ abelian vector multiplets. Using the canonical metric,
the bosonic part of the Lagrangian is given by \cite{Sen}
\begin{equation}
{\cal{L}}={1\over32\pi}\sqrt{-g}\left[R-\frac{1}{2}(\partial\eta)^2
-\frac{1}{12}e^{-2\eta}H^2+{1\over8}{\rm Tr} } \newcommand{\NP}{Nucl. Phys. (\partial  {\cal M}
L\partial {\cal M} L)-{1\over4}e^{-\eta}F{}^T
( L {\cal M} L)F\right],
\la{full}
 \end{equation}
where $L$ is the metric of $O(6,22)$. ${\cal M}={\cal M}^T\in O(6,22)/O(6)\times
O(22)$ parametrizes the scalars in the sigma model, the 28
$F_{\mu\nu}$'s are the $U(1)$ fields strengths, $\eta$ is the four
dimensional dilaton and $H$ the $3$-form field strength with the usual
Chern-Simons terms. The string theory has a perturbative $O(6,22;Z)$ $T$-duality
that transforms Kaluza-Klein states into winding states, and a non-perturbative
$SL(2,Z)$ $S$-duality that transforms electric states into magnetic states.
This is reflected in the $O(6,22;R)$ invariance of the Lagrangian (\ref{full})
and the $SL(2,R)$ invariance of its equations of motion.

Let us work at a special point in the moduli space and set the
asymptotic value of ${\cal M}$ to $I$, and the asymptotic dilaton field
to zero. We shall return to the general case later.
Let us define $\vec{Q}_{R,L}$ and $\vec{P}_{R,L}$ by
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
\vec{Q}_{R,L}&=&\frac{1}{2}(I\pm L)\vec{Q} \nonumber} \def\bd{\begin{document} \\
\vec{P}_{R,L}&=&\frac{1}{2}(I\pm L)\vec{P}\ ,
\eea
where $\vec{Q}$ and $\vec{P}$ are the $28$-dimensional electric and
magnetic charge vectors. Denoting by $N_L$ and $N_R$ the number of
left and right oscillators respectively, the mass of an elementary
(purely electric) string state is given by
\begin{equation}} \newcommand{\ee}{\end{equation}
M_{string}^2=\frac{1}{8}(\vec{Q}_R^2+2 N_R-1)=\frac{1}{8}\
  (\vec{Q}_L^2+2 N_L-2)
\la{stringmass}
\ee
See, for example, \cite{Sen}.  In this $N=4$ theory, states (whether
elementary or solitonic) fall into $3$ categories according as they are
annihilated by $q=2,1,0$ supersymmetries, in which case their masses are
given by \cite{Rahmfeld3}
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
M_{central}&=&Z_1=Z_2~~~~~~~q=2\nonumber} \def\bd{\begin{document}\\
&=&Z_1>Z_2~~~~~~~q=1\nonumber} \def\bd{\begin{document}\\
&>&Z_1 \geq Z_2~~~~~~~q=0
\eea
where $Z_1$ and $Z_2$ are the moduli of the
central charges in the supersymmetry
algebra given by \cite{Cveticyoum,Rahmfeld3}
\begin{equation}
Z_{1,2}^2=\frac{1}{8}\left[\vec{Q}_R^2+\vec{P}_R^2\pm
2\left(\vec{Q}_R^2\vec{P}_R^2-(\vec{Q}_R\vec{P}_R)^2\right)^{\frac{1}{2}}\right]\
. \la{central}
\end{equation}
It follows by comparing $M_{string}$ of (\ref{stringmass}) and
$M_{central}$ of (\ref{central}) that the elementary string states, being
purely electric, are either BPS states with $N_R=1/2$ or else
non-supersymmetric states with $N_R >1/2$.

The $N_R=1/2$ states correspond to that subset of the full spectrum that
is annihilated by half of the supersymmetry generators ($q=2$), belongs
to short representations of the $N=4$ supersymmetry algebra and
saturates the Bogomol'nyi bound $M_{central}=Z_1=Z_2$. The basic
superspin $L=0$ multiplet is the $16$ dimensional ($J_{max}=1$)
multiplet $(1,4,5)$. This is the only multiplet appearing for $N_L=0$.
However, higher values of superspin $L$ may appear for higher $N_L$.
Since the left moving oscillators have spins $0$ if they lie in the
$22$ compact dimensions or $1$ if they lie in the spacetime dimensions,
the superspin obeys the bound $L\leq N_L$. In particular for $N_L=1$, we
have in addition to the above $L=0$ multiplet the $48$ dimensional
($J_{max}=2$) multiplet $(1,4,6,4,1)$. For $N_L>1$ we have $J_{max}
=L+1$.

The $N_R >1/2$ states, belonging to the long
representations, are annihilated by no supersymmetries ($q=0$) and
satisfy $M_{central}>Z_1 \geq Z_2$. The $L=0$ multiplet is $256$
dimensional with ($J_{max}=2$).  No elementary states belong to
the intermediate  representation,
which are annihilated by one supersymmetry ($q=1$) and satisfy
$M_{central}=Z_1>Z_2$. These $L=0$ multiplets are $64$ dimensional
with ($J_{max}= 3/2$).  The situation changes when we allow for
solitonic states of the string theory which carry magnetic charge. Then
we can have all three categories of supermultiplet
\cite{Rahmfeld1,Rahmfeld3}, and indeed we predicted the existence of
new dyonic ($J_{max}=3/2$) multiplets in the string spectrum, over and
above the ($J_{max}=1$) dyonic states related by $S$-duality to the
elementary states and predicted by Schwarz and Sen \cite{Schwarz1}.

In \cite{Rahmfeld1} we considered these elementary electrically charged
massive $N_R=1/2$ states, and showed that the spin zero, superspin
zero, states correspond to extreme limits of non-rotating black hole
solutions which preserve $1/2$ of the spacetime supersymmetries. By
supersymmetry, the black hole interpretation then applies to all
members of the $N=4$ supermultiplet \cite{Gibbonsperry,Aichelburg},
which has $J_{max}=1$. For a subset of states the low-energy string
action can be truncated to (\ref{actiona}). The scalar-Maxwell
parameter is given by $a=\sqrt 3$ for $N_L=1$ and $a=1$ for $N_L>1$
(and vanishing left-moving internal momenta). The other superspin
zero states with $N_L>1$ are extremal non-rotating black holes too, but
are not described by a single scalar truncation of the type
(\ref{actiona}). We also made a similar identification for the dyonic
states \cite{Rahmfeld3}. To see this, let us recall that general black
hole solutions are determined by the $56$ components of the electric and
magnetic charge vectors $\vec{Q}$ and $\vec{P}$. We can simplify the
problem by applying an $O(6)\times O(22)$ T-duality transformation
which eliminates all but four electric and four magnetic components.
This corresponds to a truncation involving just four field strengths:
$F_1, \ F_2, \ F_3$ and $F_4$, and just three complex scalars:
$S=a+ie^{-\eta}$, $T=b+ie^{-\sigma}$, $U=c+ie^{-\rho}$.  \begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
{\cal L}&={1\over 32\pi}\sqrt{-g}\biggl \{
&\hspace{-0.4truecm}R-\frac{1}{2}\left[(\partial\eta)^2
+(\partial\s)^2+(\partial\rho)^2\right]-\nonumber} \def\bd{\begin{document}\\ & &
-\frac{1}{2}\left[e^{2\s}(\partial b)^2 +
e^{2\r}(\partial c)^2\right]-\frac{1}{12}e^{-2\eta}H^2- \nonumber} \def\bd{\begin{document} \\ &&
\hspace*{-0.5truecm} -\frac{e^{-\eta+\s+\r}}{4} \biggl[
|T|^2 |U|^2 F_1^2 +|T|^2F_2^2+ F_3^2+ |U|^2 F_4^2
+2bF_2F_3+2cF_3F_4 +\nonumber} \def\bd{\begin{document}\\ &&\hspace{20mm}-2c
|T|^2F_1F_2 -2b|U|^2 F_1F_4 -cb (F_1F_3-F_2F_4)\biggr]\biggr\}, \la{action1}
\eea
It is noteworthy that
this procedure keeps only the degrees of freedom which arise from the
toroidal compactification from six to four dimensions: $N=2$
supergravity coupled to three vector multiplets.  $F_1,F_2$ are the
Kaluza-Klein fields, $F_3,F_4$ are the winding fields: $S$ is the
axion/dilaton, $T$ the Kahler-form and $U$ the complex structure. The axion
field is obtained by dualizing $H$, which satisfies the standard Bianchi
identity. The Lagrangian (\ref{action1}) was thoroughly analyzed in
\cite{Rahmfeld3} where, in particular the triality of the three duality
groups, $SL(2;Z)_S$, $SL(2;Z)_T$, $SL(2;Z)_U$, was emphasized: (\ref{action1})
is obviously invariant under $T\leftrightarrow U, \ F_2\leftrightarrow F_4$
exchange, but the equations of motion are also invariant under
$S\leftrightarrow T,\ S\leftrightarrow U $ exchange (accompanied by an
appropriate electric/magnetic transformation of the field strengths), which
lead to the interpretation of a triality of the $S$, $T$ and $U$ strings.
To simplify further, we shall here consider solutions with vanishing
pseudoscalars so that the reduced Lagrangian is
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
{\cal L}&={1\over 32\pi}\sqrt{-g}\biggl \{
&\hspace{-0.4truecm}R-\frac{1}{2}\left[(\partial\eta)^2
+(\partial\s)^2+(\partial\rho)^2\right]-\nonumber} \def\bd{\begin{document}\\ & & \hspace*{-0.5truecm}
-\frac{e^{-\eta}}{4}\left[e^{-\s-\r}F_1^2+e^{-\s+\r}F_2^2
+e^{\s+\r}F_3^2 +e^{\s-\r}F_4^2\right] \biggr\}, \la{action}
\eea
where
one has to keep in mind the constraints imposed by the requirement of
vanishing pseudoscalars.

The extremal black holes we have in mind to illustrate the point,
namely those that allow for a description by an action of type
(\ref{actiona}) with $a=\sqrt{3},1,1/\sqrt{3},0$, can be obtained
from (\ref{action}) by making various truncations:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
a=\sqrt{3}: & F_{1}\neq0,  &F_2=F_3=F_4=0,~~Q^2=1 \nonumber} \def\bd{\begin{document} \\
a=1: & F_{1}=F_3\neq0, & F_2=F_4=0,~~~~~~~~~Q^2=2 \nonumber} \def\bd{\begin{document} \\
a=1/\sqrt{3}: & F_{1}=F_3={\tilde F}_2\neq0, & F_4=0,~~~~~~~~~~~~~~~~Q^2=3
\nonumber} \def\bd{\begin{document} \\ a=0: & F_{1}=F_3={\tilde F}_3 ={\tilde F}_4
\neq0, &~~~~~~~~~~~~~~~~~~~~~~~~~~Q^2=4 \eea
Table 1 summarizes the
charge and mass quantum numbers for
those and a few more
black holes and string states in the heterotic string theory. (If so desired,
one may then use  $S-T-U$ triality to describe them in the dual Type
$II$ string pictures \cite{Rahmfeld3}.) Based on this Table it can be easily
verified \cite{Rahmfeld1,Rahmfeld3} that the mass and charge quantum numbers
of the $a=0,1/\sqrt{3},1,\sqrt{3}$ black holes admit the interpretation  of
$4,3,2,1$ particle bound state with zero binding energy. For the purposes of
illustration we have chosen the special cases where all non-zero charges are
equal to unity but it is easily generalized to the case of different charges
$Q_1$, $P_2$, $Q_3$ and $P_4$ \cite{Cveticyoum} where the interpretation is
that of a $(Q_1+P_2+Q_3+P_4)$-particle bound state with zero binding energy.

Note, by the way, that even the extremal, but {\it non-supersymmetric},
$a=1$ black hole with electric charges $(1,0,-1,0)$ fits into the string
spectrum with the assignments $N_L=1,N_R=3/2$. (Since the masses of
non-supersymmetric states are not protected from quantum corrections by
any non-renormalization theorems we do usually not expect
(\ref{stringmass}) to give the correct answer for arbitrary black
holes.)  This solution is related by T-duality \cite{Rahmfeld1} to the
non-supersymmetric black hole of \cite{Garfinkle} which has just one
non-vanishing charge corresponding to one of the $16$ $U(1)$s in the
Yang-Mills sector. It is frequently claimed that extremal black holes and
supersymmetric black holes are synonymous, but while
$M_{string}^2={\vec{Q}}_L^2/8$ and  $M_{string}^2={\vec{Q}}_R^2/8$ are
both extremal, only $M_{string}^2={\vec{Q}}_R^2/8$ is supersymmetric,
owing to the left/right asymmetry of the heterotic string.  (Note,
however, that this solution preserves one quarter of the supersymmetries
when embedded into maximal $N=8$ supergravity coming from the Type $II$
string). {\footnotesize \begin{table} \begin{center}
\begin{tabular}{|c|c|c||c|c|c|c||c|c|} \hline
    \multicolumn{3}{|c||} {\em Quantum Numbers} & \multicolumn{4}{c||}
   {\em String  States} & \multicolumn{2}{c|} {\em Black Holes }\\
  \hline \hline
     {$(Q_1,Q_2,Q_3,Q_4)$} & {$
 (P_1,P_2,P_3,P_4)$} &q &$N_L$ &$N_R$ &$M_{string}^2$
     & $M_{central}^2$ & a &  $M_{black}^2$   \\ \hline
\hline
   & & & & & & & & \\
   (1,0,0,0)  &
   (0,0,0,0) & 2 &1 & $\frac{1}{2}$&
    $\frac{1}{16}$ & $\frac{1}{16}$ &  $\sqrt{3}$ & $\frac{1}{16}$\\
   (1,0,1,0)  & (0,0,0,0) &2 &2 & $\frac{1}{2}$&
    $\frac{1}{4}$ & $\frac{1}{4}$ & 1 & $\frac{1}{4}$ \\
   (2,1,1,1)  & (0,0,0,0) &2  &4 & $\frac{1}{2}$&
    $\frac{13}{16}$ &  $\frac{13}{16}$ &x& $\frac{13}{16}$ \\
 (1,0,-1,0)  & (0,0,0,0) & 2& 0 & $\frac{1}{2}$&
     $0$ & 0 & x &$0$ \\
 (1,0,-1,0)  & (0,0,0,0) & 0& 1 & $\frac{3}{2}$&
     $\frac{1}{4}$ & 0 & 1 &$\frac{1}{4}$ \\
   & & & & & & & & \\
     (1,0,0,0)  & (0,1,0,0) & 1&x & x&
     x & $\frac{1}{4}$ & 1& $\frac{1}{4}$ \\
   (1,0,1,0)& (0,1,0,0)& 1& x& x& x
    & $\frac{9}{16}$ &  $\frac{1}{\sqrt{3}}$ &
    $\frac{9}{16}$ \\
    (1,0,1,0)  & (0,1,0,1) &1& x & x & x & $1$ & 0 & 1 \\
 (1,0,-1,0)  & (0,1,0,-1) & 0& x & x &
     x &  0 & 0 &
     1 \\\hline
\end{tabular}
\end{center}
\normalsize
\begin{center}Table 1: Masses and charges of black holes and
string states. \end{center}
\end{table}
\normalsize}

As discussed in \cite{Rahmfeld1}, although the superpartners of the
non-rotating black holes are themselves black holes, they are {\it not}
rotating black holes in the sense of Kerr. On the contrary, as
explained in \cite{Aichelburg}, it is their fermionic hair that carries
the angular momentum in contrast to conventional rotating black holes
where the angular momentum is bosonic. Rather this bosonic angular
momentum is supplied by the left moving oscillators, which leads us to
identify the Kerr-type angular momentum with the superspin $L$.
However, as also discussed in \cite{Rahmfeld1}, these $N_R=1/2$ string
states cannot then be rotating {\it black holes} since these
mass=charge solutions have event horizons only for vanishing Kerr angular
momentum.

\section{Black
hole supermultiplets}  \la{supermultiplets}

\begin{table}
\begin{center}
\begin{tabular}{|c||ccccc|}\hline
 & &$a=0$&$a=1/\sqrt{3}$&$a=1$&$a=\sqrt{3}$\\ \hline\hline
$N=2$&&1&1&1&1\\
$N=4$&&1&1&2&2\\
$N=8$&&1&1&2&4\\ \hline
\end{tabular}
\end{center}
\normalsize
\begin{center}Table 2:
Number of preserved supersymmetries, $q$, for black holes with
parameters $a=0,1\sqrt{3},1,\sqrt{3}$ in $N=2,4,8$ theories.
\end{center}
\end{table}
\normalsize
Since the basic quantities like mass and charge are not going to change if we
move to the $N=8$ or $N=2$ theories, we may as well extend the conjecture to
these theories also.  The number of preserved supersymmetries is given in
Table 2.  In this section we check that the supermultiplet structure of the black
holes is consistent with this bound state interpretation. The relevant group
theory may be found in \cite{Ferrarasavoy}. Let us recall that an $N$-extended
supersymmetry algebra admits $N/2$ central charges $Z_1,...,Z_{N/2}$. States
fall into $1+N/2$ categories according as they are annihilated by $ N/2 \geq q
\geq 0$ supersymmetry generators. $q$ also counts the number of $Z$s that obey
the bound $M_{central}=Z_{max}$. We are primarily interested in multiplets with
states with spin $J=0$ and superspin $L=0$, which we identify with the
non-rotating black hole solutions. The rest of the $L=0$ supermultiplet may then
be filled out using the fermionic zero modes \cite{Aichelburg}. In the
fundamental multiplets the spin will run from $J=0$ up to $J= (N-q)/2$ whereas
for multiplets with non-zero superspin $L$, the spin will run from $J=0$ to
$J=L+(N-q)/2$ for $L<(N-q)/2$ and  from $J=L-(N-q)/2$ to $J=L+(N-q)/2$ for $L
\geq (N-q)/2$. The multiplets are constructed in the spirit of
\cite{Ferrarasavoy} by combining massless supermultiplets and then employing the
Higgs mechanism to obtain the massive multiplet. For each number $q$ of
preserved supersymmetries we give the results up to $J_{max}=4$ (implying
superspins $L=0,\frac{1}{2},1,\frac{3}{2}$ and $L=2$ for the case of four
preserved supersymmetries and so on). The results are shown in Tables 3, 4 and 5
for the $N=8, \ N=4$ and $N=2$ theories, respectively.

{\footnotesize
\begin{table}
\begin{center}
\begin{tabular}{|r |r| r r r r r |}
\hline $q$&$J$&$L=0$&$L=\frac{
1}{2}$& $L=1$&$L=\frac{3}{2}$&
$L=2$\\
\hline\hline
4 & 4                 &     &   &      &    &1\\
 & $\frac{7}{2}$  &    &    &      &1   &8\\
 &3                    &     &   & 1   &8   &28\\
 &  $\frac{5}{2}$ &    & 1 &8    & 28&56\\
 & 2                   & 1  & 8 & 28 & 56&70\\
 &  $\frac{3}{2}$ & 8  & 28& 56& 70&56\\
 & 1                   & 27 & 56& 70& 56&28\\
 & $\frac{1}{2}$  &48  & 69& 56& 28&8\\
 &0                    & 42 & 48& 27&8   &1\\ \hline
3 & 4                      &    &      &    &1&\\
 & $\frac{7}{2}$      &     &      &1   &10&\\
 &3                         &    & 1   &10   &45&\\
 &  $\frac{5}{2}$     & 1   & 10    & 45&120&\\
 & 2                       & 10 &  45 & 120&210&\\
 &  $\frac{3}{2}$     & 44 & 120& 210&252&\\
 & 1                       & 110& 209& 252&210&\\
 & $\frac{1}{2}$  &     165& 242& 209&120&\\
 &0                        & 132& 165&110   &44&\\ \hline
2 & 4                     &&   &1&&\\
 & $\frac{7}{2}$      &  &   1   &12&&\\
 &3                       & 1   &12  &66&&\\
 &  $\frac{5}{2}$   & 12    & 66&220&&\\
 & 2                     &  65 & 220&495&&\\
 &  $\frac{3}{2}$ &     208& 494&792&&\\
 & 1                      & 429& 780&923&&\\
 & $\frac{1}{2}$    & 527& 858&780&&\\
 &0                     & 429 & 572   &429&&\\ \hline
1 & 4                      &    &1&&&\\
 & $\frac{7}{2}$  &   1   &14&&&\\
 &3                        &14  &91&&&\\
 &  $\frac{5}{2}$   & 90&664&&&\\
 & 2                      & 350&1000&&&\\
 &  $\frac{3}{2}$    & 910&1988&&&\\
 & 1                        & 1638&2912&&&\\
 & $\frac{1}{2}$       & 2002 & 3068&&&\\
 &0                   &1430 &2002 &&&\\ \hline
 0 & 4                 &    1& & & &\\
 & $\frac{7}{2}$  &    16& & & &\\
 &3                    &   119 & & & &\\
 &  $\frac{5}{2}$    &544& & & &\\
 & 2                   &    1700& & & &\\
 &  $\frac{3}{2}$ &    3808& & & &\\
 & 1                   &6188 & & & &\\
 & $\frac{1}{2}$  &   7072 & & & &\\
 &0                 &   4862 & & & &\\ \hline
\end{tabular}
\end{center}
\normalsize
\begin{center}Table 3: Massive supersymmetry
representations of $N=8$\end{center}
\end{table}
\normalsize}
\begin{table}
\begin{center}
\begin{tabular}{|r |r| r r r|}
\hline $q$&$J$&$L=0$&$L=\frac{1}{2}$& $L=1$\\
\hline\hline
2 & 2                 &     &   & 1     \\
 & $\frac{3}{2}$  &    & 1   & 4    \\
 &     1              & 1    & 4  &  6  \\
 &  $\frac{1}{2}$ & 4   & 6 &  4 \\
 &0                    & 5 & 4&   1 \\ \hline
1 & 2                 &     &1   &    \\
 & $\frac{3}{2}$  & 1   &6    & \\
 &     1              &  6   & 15  & \\
 &  $\frac{1}{2}$ &  14  & 20 &    \\
 &0                    &14 & 14&\\ \hline
0 & 2                 &  1   &   &      \\
 & $\frac{3}{2}$  &  8  &    &      \\
 &     1              & 27    &   &    \\
 &  $\frac{1}{2}$ & 48   &  &    \\
 &0                    &  42& & \\ \hline
\end{tabular}
\end{center}
\normalsize
\begin{center}Table 4: Massive supersymmetry representations
 for $N=4$\end{center}
\end{table}

\begin{table}

\begin{center}
\begin{tabular}{|r |r| r r|}
\hline $q$&$J$&$L=0$&$L=\frac{1}{2}$\\
\hline\hline
1 &     1              &     & 1   \\
  &  $\frac{1}{2}$ & 1   & 2 \\
 &0                    & 2 & 1\\ \hline
0 &      1              &  1   &    \\
 &  $\frac{1}{2}$ &  4  &      \\
 &0                    &5 & \\ \hline
\end{tabular}
\end{center}
\normalsize
\begin{center}Table 5: Massive supersymmetry representations
 for $N=2$\end{center}
\end{table}
\normalsize

Let us begin by considering black hole solutions of the $N=8$
supergravity theory with $Z_1 \geq Z_2\geq Z_3 \geq Z_4$. Here we have
the five categories:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
M_{central}&=&Z_1=Z_2=Z_3=Z_4~~~~~~~~~~~~~~~~~~~~~~~~~~q=4 \nonumber} \def\bd{\begin{document} \\
&=&Z_1=Z_2=Z_3>Z_4~~~~~~~~~~~~~~~~~~~~~~~~~~q=3  \nonumber} \def\bd{\begin{document} \\
&=&Z_1=Z_2>Z_3\geq Z_4~~~~~~~~~~~~~~~~~~~~~~~~~~q=2    \\
&=&Z_1=Z_2\geq Z_3\geq Z_4~~~~~~~~~~~~~~~~~~~~~~~~~~q=1 \nonumber} \def\bd{\begin{document} \\
&>&Z_1\geq Z_2 \geq Z_3 \geq Z_4~~~~~~~~~~~~~~~~~~~~~~~~~~q=0  \nonumber} \def\bd{\begin{document}
\eea
In particular, the $a=\sqrt{3}$ black holes preserve $q=4$
supersymmetries and must belong to short supermultiplets which we
assume to be the maximum spin $J=2$, superspin $L=0$ multiplet
$(1,8,27,48,42)$ appearing in Table 3.  The bound state
interpretation requires that the $a=1$ black holes preserving $q=2$
supersymmetries appear in the product of two $a=\sqrt{3}$
representations.  Ignoring the internal quantum numbers, this product
will decompose into multiplets with $4 \geq J_{max} \geq 2$ as follows

\begin{equation}} \newcommand{\ee}{\end{equation}
\begin{array}{ll}
1\times\left[ 4\right]\oplus 8\times\left[ \frac{7}{2}\right]
\oplus27\times\left[3\right]\oplus48\times\left[\frac{5}{2}\right]
       \oplus42\times\left[2\right]&~~~q=4\\
1\times \left[4\right] \oplus 6\times \left[\frac{7}{2}\right]
                \oplus14\times\left[3\right]\oplus14\times
\left[\frac{5}{2}\right]&~~~q=3 \\
1\times \left[4\right]\oplus 4\times \left[\frac{7}{2}\right]
                              \oplus 5\times \left[3\right]&~~~q=2 \\
1\times \left[4\right]\oplus 2\times \left[\frac{7}{2}\right]&~~~q=1\\
1\times \left[4\right]&~~~q=0
\end{array}
\ee

Which of the above possibilities is actually realized, however, will
depend on the charge assignments of the two constituents. Suppose each
is singly charged under just one of the Kaluza-Klein $U(1)$s, then the
bound state will again belong to a $q=4$ multiplet if the $U(1)$s are
the same. On the other hand, if one
carries a Kaluza-Klein charge and the other a winding
charge in the same dimension,
then we get $q=2$. Since these are precisely the
quantum numbers of the $a=1$ black hole, which has twice the mass of the
$a=\sqrt{3}$ black hole, this is entirely consistent with our hypothesis.
Note, however, that although one might also expect to obtain $q=3$ multiplets,
a closer look at the internal quantum numbers shows that these do not in fact
arise, since the $M_{central}=Z_1=Z_2=Z_3>Z_4$ configuration can never
be achieved with $2$-particle states.

The same arguments go through for an embedding of the black holes into
$N=4$ or $N=2$ supergravity with an appropriate number of matter
multiplets. For the $N=4$ case the compositions for the product of two
short multiplets are given by \footnote{ Since the same group theory applies,
this suggests that higher superspin multiplets also appear in the spectrum
of global $N=4$ Yang-Mills theories, where traditionally attention is
focussed only on maximum spin $1$.}
 \begin{equation}} \newcommand{\ee}{\end{equation} \begin{array}{ll} 1\times
\left[2\right]\oplus 4\times \left[\frac{3}{2}\right]
       \oplus5\times\left[1\right]& ~~~q=2\\
1\times \left[2\right] \oplus 2\times \left[\frac{3}{2}\right]
            &~~~q=1 \\
1\times\left[ 2\right]&~~~q=0
\end{array}
\ee
For example, the appearance of the $q=0$ multiplet is consistent with
the interpretation of the non-supersymmetric $a=1$ black hole as a bound state
of a supersymmetric $a=\sqrt{3}$ positively charged Kaluza=Klein black hole and
a supersymmetric $a=\sqrt{3}$ negatively charged winding black hole.  Similarly,
for $N=2$ we find
\begin{equation}} \newcommand{\ee}{\end{equation} \begin{array}{ll} 1\times\left[ 1\right]\oplus
2\times\left[ \frac{1}{2}\right]
       & ~~~q=1\\
1\times\left[ 1\right]&~~~q=0
\end{array}
\ee
Therefore, we have shown that the supermultiplet structures are
consistent with our hypothesis in the case of $2$-particle bound states.
It is straightforward to show by taking further tensor products that
things go through in a similar way for the $3$- and $4$-particle states.

It should be stressed that we are not claiming that all bound states can
be identified with oscillator states. In the $N=4$ table, for example, $L=1/2,
J_{max}=3/2$ states appear in the tensor product, whereas there are no $L=1/2$
supermultiplets in the oscillator spectrum of the $N=4$ heterotic string because
the left-moving sector is purely bosonic.  Similarly, the $L=1,J_{max}=2$
states appearing in the tensor product will be black holes whose angular
momentum is fermionic in origin \cite{Aichelburg}, whereas the $L=1$ oscillator
states have a bosonic Kerr-type angular momentum and cannot therefore be
black holes.

So far we have put ourselves at the special point in moduli space $S=T=U=i$. At
generic points, the mass formula becomes \cite{Ceresole,Rahmfeld3}
\begin{equation}} \newcommand{\ee}{\end{equation}
M_{central}=\frac{1}{{\rm Im}~S~ {\rm Im}~T~{\rm Im}~U}
{\rm
max}|(\alpha_1+U\alpha_2+T\alpha_4-TU\alpha_3)\pm(S\beta_1-SU\beta_2-ST\beta_4
-STU\beta_3)| \ee
where $\alpha$ and $\beta$ are the quantized electric and magnetic charge
vectors. Thus, although the bound state interpretation continues to apply, the
zero binding energy phenomenon does not. The triangle inequality ensures that
at generic points the bound states have non-zero binding energy (except of
course when the charges are not relatively prime).

\section{Multi-black brane solutions of $M$-theory}
\la{multi}

\vspace{1truecm}

\begin{center}
\begin{tabular}{|l||l|l|l|l|}
\hline & 4-form & 3-form & 2-form & 1-form \\
\hline\hline
D=11&4&&&\\ \hline
D=10 &$4$& $4$ &$4$ & \\ \hline
D=9 &$4$&$ 4$ &$4,2$ &$4$ \\ \hline
D=8 &$4$&$ 4$ &$4,2$ &$4,2$ \\ \hline
D=7 & &$4 $ &$4,2$ &$ 4,2$\\ \hline
D=6 && $4,2$ & $4,2$&$4,2,\frac{4}{3},1$ \\ \hline
D=5 && &$4,2,\frac{4}{3}$ &$4,2,\frac{4}{3},1$ \\ \hline
D=4 &&  &$4,2,\frac{4}{3},1$ &$4,2,\frac{4}{3},1,\frac{4}{5},
\frac{2}{3},\frac{4}{7}$  \\
\hline
\end{tabular}
\la{brane}
\end{center}
\begin{center}Table 6: $\Delta$ values for
supersymmetric $p$-branes\end{center}

\vspace{1truecm}
Let us now turn to the black branes of $M$-theory
\cite{Dufflupope2,Klebanov1}.
Starting from eleven dimension, toroidal compactification gives rise to a
variety of $(d +1)$-form field strengths and hence fundamental
$(d -1)$-brane and solitonic $(\tilde d-1)$-brane solutions in the lower
dimension $D=d+\tilde d+2$.  The compactified eleven-dimensional
supergravity theory admits a consistent truncation to the following set
of fields: the metric tensor $g_{MN}$, a set of $n$ scalars
$\vec{\phi}=(\phi_1,...\phi_n)$, and $n$ field strengths $F_{\alpha}$ of
rank $(\tilde d +1)$.  The $D$-dimensional action describing the
$p$-branes under consideration  is then given by \cite{Popelu2}
\begin{equation}} \newcommand{\ee}{\end{equation}
\lag=\sqrt{-g}\left[R - \frac{1}{2} (\partial \vec{\phi})^2 -\frac{1}
{2\times (d+1)!}
  \sum_{\a=1}^{n} e^{\vec {a}_\a \vec{\phi}} F_{d+1}^{\a \ 2}\right]\ ,
\label{actionp}
\ee
where $n$ is the number of participating field strengths. If all active
charges are equal, this can be
further truncated to the Lagrangian (\ref{actiona}) involving a single scalar
and single field strength where $a$, $\phi$ and $F$ are given by

\[ a^{-2}=\sum_{\alpha\beta}(M^{-1})_{\alpha\beta}\]

\[\phi=a\sum_{\alpha\beta}(M^{-1})_{\alpha\beta}\vec {a}_\a \vec{\phi}\]

\begin{equation}} \newcommand{\ee}{\end{equation} (F_{\alpha})^2=a^2\sum_{\alpha\beta}(M^{-1})_{\alpha\beta}F^2 \ ,
\ee
where the matrix $ M_{\a \b}$ is given by
\begin{equation}} \newcommand{\ee}{\end{equation}  M_{\a \b}=\vec{a}_{\a} \vec{a}_{\b}.
\la{matrix}
\ee
The parameter $a$ can conveniently be expressed as
\begin{equation}} \newcommand{\ee}{\end{equation}
a^2=\Delta-\frac{2d\tilde d}{d+\tilde d}
\ee
As discussed in \cite{Popelu2},
supersymmetric $p$-branes solutions can arise only when the value of $\Delta$ is
given by $\Delta=4/n$.  This occurs when   \begin{equation}} \newcommand{\ee}{\end{equation}
 M_{\a \b}=4\delta_{\a \b} -2\frac{d\tilde{d}}{d+\dt}.
\ee

Originally, attention was focussed on the
$\Delta=4$ solutions \cite{Strominger1,Dufflu,Khuristring} but various new
supersymmetric solitons with $\Delta\neq 4$ have recently been studied
\cite{Popestainless,Popelu1,Popelu2}. In this section we shall
generalize the treatment of extremal black holes \cite{Rahmfeld4} and
confirm on the level of classical solutions that these new $p$-branes
can also be regarded as bound states with zero binding energy of
fundamental $\Delta=4$ branes.  We find new $1\leq n \leq m$-centered
$p$-brane solutions which reproduce the $\Delta=4/n$  solutions of
\cite{Popelu2} as we allow $n$ of the centers to coincide and take the
remaining $(m-n)$ out to infinity. Table 6 summarizes in which
dimensions the various solitons in maximal supergravities arise. Below
eight (six)  dimensions the four(three)-form field strengths are
dualized.

For the purposes of exhibiting the multi-centered solutions, we shall work
with (\ref{actionp}).  The equations of motion  are
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
  \frac{1}{\sqrt{-g}}\partial_{M}\left(\sqrt{-g}  e^{\vec{a}_{\a} \vec\phi}
F_{d+1}^{\a \  MN}\right)&=&0
                       \la{eom1}\\
 \frac{1}{\sqrt{-g}}\partial_{M}\left(\sqrt{-g} \partial^{M} \phi_i\right)
&=&
  \sum_{n}\frac{a_{i\a}}{2\times (d+1)!}F_{d+1}^{\a \ 2}
\la{eom2}\\
  R_{MN}&=&\frac{1}{2}\partial_{M}\vec{\phi}\partial_{N}\vec{\phi} + \nonumber} \def\bd{\begin{document} \\
& & \qquad\quad
+\frac{1}{2 d !}\sum_{n}e^{\vec{a}_{\a} \vec{\phi}}\biggl(F^{\a \
2}_{MN}-\frac{d}{n(d+\tilde{d})}
 F^{\a \ 2}
  g_{MN}\biggr) \la{eom3}
\eea
The one-center solutions of these equations have been intensively studied.
Here we find the {\it multi-center} solution which reads for the solitonic
case:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
f_{\a}&=&(1+\frac{\l_\a}{d|\vec{y}-\vec{y}_{0,\a}|^{d}}) \la{factor}\\
ds^2&=&\prod_{\a=1}^{n}f_{\a}^{-\frac{d}{d+\tilde{d}}}dx^{\mu}dx_{\mu} +
\prod_{\a=1}^{n}f_{\a}^{\frac{\tilde{d}}{d+\tilde{d}}}dy^m dy_m \\
e^{-\vec{a}_{\a} \vec{\phi}}&=& f_{\a}^2 \prod_{\b=1}^{n}f_\b^{-\frac{d\tilde{d}}{d+\tilde{d}}} \\
F^\a_{m_1...m_{d+1}}&=&\l_\a\e_{m_1 ...m_{d+1}p} \frac{y^p}{|\vec{y}-
\vec{y}_{0,\a}|^{d+2}}
\eea
where $\mu$ refers to the $\tilde{d}$ world-volume coordinates of the solitonic
$(\tilde{d}-1)$-brane and $m$ to the
$D-\tilde{d}=d+2$ transverse coordinates. In the special case $d=0$ we
choose
\begin{equation}} \newcommand{\ee}{\end{equation}
f_{\a}=(1+\l_\a\ln{\frac{|\vec{y}-
\vec{y}_{0,\a}|}
{r_{0,\a}}})
\ee
instead of (\ref{factor}). The solutions  for elementary
multi-$p$-branes are easily obtained by generalizing the
single-centered solutions of \cite{Popelu2} along the lines above.
Since (\ref{eom1}) and (\ref{eom2}) are
essentially linear in the contributions of each field strength they
are obviously satisfied for the multi-center solutions. The only
slightly non-trivial check comes from the Einstein equations, since
here the scalar fields and $R_{mn}$ involve non-linearities in the
individual soliton contributions. If we plug the ansatz into the
field equations we find the following condition for vanishing
non-linearities:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray} \frac{d\tilde{d}^2}{2(d+\tilde{d})}+\tilde{d}
M^{-1}_d\left[(n-2)(\frac{d\tilde{d}}{d+\dt})^2-2 (2-\frac{d\tilde{d}}{d+\dt}) \frac{d\tilde{d}}{d+\dt}\right] + \ \ \ \ \ \
\ \ & & \nonumber} \def\bd{\begin{document} \\
\hspace*{-0.5truecm}+\tilde{d} M^{-1}_n\biggl[(2-\frac{d\tilde{d}}{d+\dt})^2
+(\frac{d\tilde{d}}{d+\dt})^2-2(n-2) (2-\frac{d\tilde{d}}{d+\dt}) \frac{d\tilde{d}}{d+\dt}+ \ \ & & \nonumber} \def\bd{\begin{document} \\ +2(n-2)
(\frac{d\tilde{d}}{d+\dt})^2+(n-2)(n-3)(\frac{d\tilde{d}}{d+\dt})^2\biggr]&=&0,
\eea
where $M^{-1}_{d(n)}$ denote the (off)-diagonal elements of the matrix
$M_{\a \b}$ defined in (\ref{matrix}). Indeed, for all the new
supersymmetric $p$-branes the above condition holds, allowing us to
generalize the single-center solutions to multi-center ones. However,
there is one subtlety, for certain values of $d,n,D$, $M$ is singular
and cannot be inverted.  The problem arises for ($D=4, n=4, d=1$) and
($D=5, n=3, d=1$). The first case was considered in \cite{Rahmfeld4}
and shown to be a valid solution. The second one can also be shown to
work by an independent calculation.  If so desired, one may now consider the
special case of coincident centers and equal charges to obtain the
$\Delta=4/n$ solutions, thus confirming that these admit the interpretation
of bound states with zero binding energy of $n$ fundamental $\Delta=4$
branes.

Most of our results are not based on the fact that we consider maximal
supergravities. So we can ask ourselves what kind of bound states
survive in the heterotic theory. It appears that all $1$-form and $2$-form
types do. In four dimensions we also have the four classic black hole
types and also strings with up to seven participating field
strengths. In the heterotic theory the number seven finds a very
natural explanation in the presence of one $S$-field 3 $T$-fields and
3 $U$-fields of section (2) \cite{Rahmfeld3,Rahmfeld2}. The multi-string
solution  found in \cite{Rahmfeld2} with non-vanishing $S$ and $T$
fields is given by
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
ds^2&=&-dt^2+dz^2+S_2T^{(1)}_2T^{(2)}_2T^{(3)}_2(dx^2+dy^2) \nonumber} \def\bd{\begin{document} \\
S&=&S_1+iS_2=\sum_{i=1}^{n}s_i\ln{\frac{x+iy-w_i}{r_{i}}}\\
T^{(a)}&=&T^{(a)}_1+iT^{(a)}_2=
\sum_{i=1}^{n}t^{(a)}_i\ln{\frac{x+iy-w^{(a)}_i}{r^{(a)}_{i}}}
\eea
where $w_i=x_i+iy_i$ and $r_{i}$ denote the positions and sizes of
the sources. The solution
preserves $1/2$, $1/4$ and $1/8$ of the supersymmetries
for one, two and three $T$-fields but without an $S$ charge. In the
presence of the $S$ field the supersymmetries get halved with the
exception of the configuration with 3 $T$ and one $S$ field which
breaks either breaks either seven eighth or all supersymmetries, depending
on the chirality choice.  It is straightforward to generalize these
results to include the $U$ fields.

The presence of only one 3-form in all dimensions (above five)
forbids the $D=6,d=2,n=2$ solution. Nevertheless, we have a very
interesting solution in $D=6,d=2$, which also can be viewed as a bound
state: the dyonic string of \cite{Rahmfeld2,Dufflupope1}:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
\Phi&=&\Phi_E+\Phi_M, \nonumber} \def\bd{\begin{document} \\
ds^2&=&e^{\Phi_E}(-dt^2+dz^2)+e^{\Phi_M}dx^idx_i \label{dyonstring}\\
e^{-\Phi_E}&=& 1+\frac{Q}{y^2} \qquad
e^{-\Phi_M}= 1+\frac{P}{y^2} \nonumber} \def\bd{\begin{document} \\
H_3&=& 2Q\e_3+2Pe^{\Phi}*\e_3 \nonumber} \def\bd{\begin{document}
\eea
with $y^2=x^ix_i, \ \ i=1,2,3,4$ and (in general)
\begin{equation}} \newcommand{\ee}{\end{equation}
e^{\Phi_E}\partial^2 e^{-\Phi_E}=e^{\Phi_M}\partial^2 e^{-\Phi_M}=0
\ee
Since the  electric as well as the
magnetic part are determined by two independent harmonic functions,
we can easily generalize (\ref{dyonstring}) to a multi-string
solution. The same holds for the ten dimensional dyonic string
also found in \cite{Rahmfeld2}:
\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}
ds^2&=&e^{\Phi_{E_1}+\Phi_{E_2}}
(-dt^2+dz^2)+e^{\Phi_{M_1}}\delta^{ij}dy_idy_j
+e^{\Phi_{M_2}}\delta^{ab}dy_ady_b \nonumber} \def\bd{\begin{document} \\
e^{\Phi_{E_\a}} \partial^2_1 e^{-\Phi_{E_\a}}&=&0 \qquad
e^{\Phi_{M_\a}} \partial^2_2 e^{-\Phi_{M_\a}}=0 \la{tendyon}
\eea
with $i,j=2,3,4,5; \ \ a,b=6,7,8,9$ and $\a=1,2$. $\partial^2_{1,2}$
denote the d'Alemberts in the $(2,3,4,5)$ and $(6,7,8,9)$ subspaces
respectively. The antisymmetric
tensor field is given as
\begin{equation}} \newcommand{\ee}{\end{equation}
B_{01}=e^{\Phi_{E_1+E_2}}, \ \ H_{ijk}=\e_{ijkl}\partial^l\Phi_{M_1},
\ \ H_{abc}=\e_{abcd}\partial^d\Phi_{M_2}.
\ee
(\ref{tendyon}) is written in ten dimensional string coordinates.

Another interesting aspects of the multi-$p$-brane solutions is that
the charge parameters of the individual constituents are
independent. In particular, if we have only two $p$-branes, we can
choose their charges to be of equal magnitude but different sign, which
leads to massless solutions
\cite{Cveticyoum2,Behrndt,Kalloshcancel,Chancvetic,Popelu1,Ortin,Dufflupope1}.
The interpretation is that two supersymmetric $p$-branes, one with positive and
one with negative mass, can combine to form a supersymmetric massless
$p$-brane. Certainly, an isolated brane with negative mass does not make sense
quantum-mechanically but there may be some quantum confinement mechanism
that allows it to exist only as a bound state.

\section{Entropy and temperature}
\la{entropy}

In this section, we ask whether the entropy and temperature of these
black $p$-branes \cite{Dufflupope2,Klebanov1,Mukherji2} is consistent with the
bound state interpretation.  The mass per unit volume and the charges of the
multi black $p$-branes may be written in terms of parameters $k$ and
$\mu_{\alpha}$ as \cite{Dufflupope2}
\[
M_{black}=k(\tilde d \sum_{\a=1}^{n}\sinh^2\mu_{\alpha} +\tilde d +1)
\]
\begin{equation}} \newcommand{\ee}{\end{equation}
\lambda_{\alpha}=\frac{1}{2}\tilde d k \sinh 2\mu_{\alpha}
\ee
The Hawking temperature $T$ and entropy $S$ of these multi black
$p$-branes in the case where all centers are coincident are given by
\cite{Dufflupope2}
\[
T=\frac{\tilde d}{4\pi y_+} \prod_{\a=1}^{n}(\cosh\mu_{\alpha})^{-1}
\]
\begin{equation}} \newcommand{\ee}{\end{equation}
S=\frac{1}{4}{y_+}^{\tilde d +1} \omega_{\tilde d +1}
\prod_{\a=1}^{n}(\cosh\mu_{\alpha})
\ee
where the event horizon is located at $y={y_+}=k^{1/\tilde d}$ and where
$\omega_{\tilde d +1}$ is the volume of the unit $(\tilde d +1)$-sphere.

The form of the total entropy as a product of the individual entropies is
puzzling from the  bound state interpretation. It remains a puzzle when we take
the extremal limit. To illustrate this, let us consider the special case where
each of the $n$ field strengths is equal. Then we have
\cite{Dufflupope2}
\[
T=\frac{\tilde d}{4\pi y_+}(\cosh\mu)^{-n}
\]
\begin{equation}} \newcommand{\ee}{\end{equation}
S=\frac{1}{4}{y_+}^{\tilde d +1} \omega_{\tilde d +1}(\cosh\mu)^n
\ee
The extremal limit corresponds to $k\rightarrow 0$, $\mu\rightarrow
\infty$, holding $\lambda=\tilde d \sqrt{n/2}ke^{\mu}$ constant. Thus
the entropy vanishes unless the constant $a$ is zero and $d=1$. This
happens only for $(D=4,n=4,d=1)$, which is just the Reissner Nordstrom
black hole, and for the five-dimensional black hole $(D=5,n=3,d=1)$.
Remarkably, these were precisely the two cases where the matrix $M$
of (\ref{matrix}) was singular.  At first sight this result seems strange: In
$D=4$, for example, it seems very unnatural to combine three black holes and
still the entropy is zero; but adding a fourth one suddenly forces it to be
finite. To see in more detail how this comes about, let us
invoke \cite{Ferrarakallosh}, where a new recipe for the
calculation of the horizon area was given: the scalar fields on the horizon
(in our case $\eta$, $\sigma$ and $\rho$) are determined by the requirement
that the central charge becomes extremal. The entropy is then given by the
value of the central charge evaluated with the scalar fields at the
horizon.  For the standard black holes with charges $Q_1,Q_3,P_2$ and $P_4$ the
scalar fields on the horizon are fixed to \cite{Ferrarakallosh} \begin{equation}} \newcommand{\ee}{\end{equation}
e^{-2\eta}\rightarrow|\frac{P_2 P_4}{Q_1Q_3}| \ , \ \ \ \
 e^{-2\s}\rightarrow|\frac{P_2Q_3}{Q_1 P_4}| \ ,\ \ \ \
e^{-2\r}\rightarrow|\frac{Q_3 P_4}{Q_1 P_2}|.
\ee
For each ``incomplete'' black hole, i.e. a state with not all four
charges non-zero, at least one of the scalars blow up, either to plus
or minus infinity. This ties in very nicely with our bound state
hypothesis. For an $a=1/{\sqrt{3}}$ with (for example)
$Q_1=Q_3=P_2=1$ all three scalars diverge:
\begin{equation}} \newcommand{\ee}{\end{equation}
\eta\rightarrow\infty \ , \ \ \ \ \s\rightarrow-\infty \ , \ \ \ \
\r\rightarrow \infty.  \ee The elementary magnetic black hole with
$P_4=1$ on the other hand has diverging scalars with the opposite
sign: \begin{equation}} \newcommand{\ee}{\end{equation} \eta\rightarrow-\infty \ , \ \ \ \ \s\rightarrow\infty \ , \
\ \ \ \r\rightarrow-\infty!
\ee
Since we know from \cite{Rahmfeld4,Cvetictseytlin} that there is a
multi-black hole solution and the scalars are additive we can safely
conclude that the $a=1/\sqrt{3}$ and $a=\sqrt{3}$ state
conspire to force the scalars and therefore the
entropy to be finite.

All this suggests that it is only the $a=0$ non-dilatonic $p$-brane
entropies that can be trusted since when the dilaton is non-trivial
there will be strong coupling effects that we do not yet know how to
handle.  However, it was shown in \cite{Mukherji2} that if quantum corrections
smooth  out singularities, even black holes with $a \neq 0$ may have
non-vanishing entropy.

\section{Conclusions}

We have reexamined the suggestion in \cite{Rahmfeld1} that, in the
extremal limit, the non-rotating black hole solutions of string theory may
be identified with both elementary and solitonic BPS string states, and
have confirmed that the interpretation of certain multiply charged black
holes as bound states at threshold of singly charged black
holes is consistent with the masses, charges, spins and supermultiplet
structure of the string states. We also confirm that the bosonic Kerr-type
angular momentum, arising from the left-moving sector of the heterotic
string, corresponds to the superspin $L$ of the oscillator states and hence that
only $L=0$ BPS oscillator states can be black holes. One is tempted to conclude
that the $L>0$ oscillator states must therefore be described by naked
singularities, but there also exist solutions that are asymptotically identical
to such solutions but near the core have a much milder singularity and whose
angular momentum is naturally Regge-bounded \cite{Harveyblack}.  Moreover, this
bound state interpretation generalizes to super $p$-branes in $D$ dimensions. In
doing so, of course we have to put ourselves at special points in moduli space.
In section (\ref{bound}), for example, we set the asymptotic value of ${\cal M}$
to $I$ and the asymptotic value of the dilaton to zero. For generic points
in moduli space, the bound state interpretation would continue to apply but we
would no longer have the zero binding energy phenomenon (except of course if
the charges are not relatively prime). We might add that these results are also
consistent with the recent recognition that some $p$-branes carrying
Ramond-Ramond charges admit an interpretation as Dirichlet-branes, or
$D$-branes, and are therefore amenable to the calculational power of conformal
field theory \cite{Polchinski}. Bound states of $p$-branes have been discussed
from the somewhat different perspective of Dirichlet branes in
\cite{Schwarz3,Witten2,Li,Sen3,Vafa,Tseytlin}.  Apart from their intrinsic
importance, therefore, these black holes and black $p$-branes have recently come
to the fore as away of providing a microscopic explanation of the Hawking
entropy and temperature formulae which have long been something of an enigma.
See \cite{Horowitz} for a recent review. For reasons discussed in section
(\ref{entropy}), however, only the $a=0$ branes are amenable to these
calculations, given our current technology.

An interesting special case is provided by the $a=\sqrt{3},1,1/\sqrt{3},0$
black holes which admit the interpretation as $1,2,3,4$-particle bound
states at threshold \cite{Rahmfeld1,Rahmfeld3}. One feature which appeared
mysterious to us at the time was: Why the unit charge solutions singled
out {\it four} values of $a$ and hence why only $1,2,3,4$- and
not $5,6,...$-particle bound states? An explanation of this has recently been
given  \cite{Townsendpapa,Klebanov2,Behrndt2,Gauntlett,Larsen} in terms of
intersecting membranes \cite{Duffstelle} and fivebranes \cite{Guven} in
$D=11$. This opens up a new direction for the study of black hole and black
$p$-brane bound states.

Finally, another crucial consistency check on the black hole, bound state,
string state picture is supplied by comparing gyromagnetic and
gyroelectric ratios.  This turns out to be quite subtle \cite{Russo} and will
treated in a separate publication \cite{Duffliurahmfeld}.

\section{Acknowledgements}

We have enjoyed useful conversations with Jim Liu and Massimo Porrati
on string state gyromagnetic ratios and with Sudipta Mukherji, Hong Lu and
Chris Pope on $p$-brane entropy.

\newpage

\bibliographystyle{preprint}

redpajama_set_name: RedPajamaArXiv


How much power, in watts, does the brain use?
When IBM's Watson won on Jeopardy a few years ago, it did so using a room full of servers with a cooling system and a fat power feed, competing against a couple of humans powered by the equivalent of a tuna fish sandwich every several hours. How much power does the human brain actually use?
Notes: If other assumptions are necessary, just state what you're answering for (e.g. healthy adults). If you know of differences between groups (e.g. people who are awake vs. asleep), please mention both.
References to scientific papers, especially reviews, are helpful.
This paper, which might be the one discussed here, discusses power usage of individual neurons but not the whole brain.
Watson mentioned this in 2011: "The human brain only requires 20 watts of power to operate ... Watson? About 20,000 watts." An interview with John Kelly (an IBM senior vice president and director of IBM Research) also puts the brain at "about 20 watts" (but puts the machine at 85 kW; the inconsistency doesn't matter for this question).
Popular Science also says 20 watts in an article explaining how simulating it in 2009 technology would take tens of megawatts.
Glenn Elert's high school students cite a collection of resources concluding 20W, as 20% of a 100W full-body power consumption. That's the source cited on Wikipedia.
The 20W figure is cited like common knowledge from "several different sources" here, but those sources aren't named.
Not the answer you're looking for? Browse other questions tagged neuroscience or ask your own question.
What is the brain power devoted to vision and haptics?
How much information does the somatosensory system produce?
Does the "use it or loose it" principle apply to the brain?
How does the brain learn abstract maths?
How does neurotransmitter distribution and usage vary thought the brain?
How long does it take for electrical impulses from neurons to travel in the human brain?
How much pleasure can brain endure?


It isn't hard to create your own sophisticated invitations. On Heidi Swapp's blog I share ways you can make foiled wedding invitations if you do not know how to use Photoshop. In this tutorial, I will share how easy it is to create them if you know basic Photoshop skills.
1. Create a new 5x7 document with a white background at 300dpi.
3. Next, you will want to open the same document you just printed and add the names and date.
4. In your layers, hide the items you don't want foiled (in this case everything besides the names and date). Click on the "eye" as shown below to hide layers. Flatten and save.
You will take this image to your local copy shop and have them print it on the color copy you just did with a laser printer.
5. While you are at it, have them cut your paper too!
6. Now you are ready to foil your wedding invitation! There will be no waste in foil as you can now just place foil over the names and date. This part was the only part printed with laser and will only be the part that will react with the foil inside the Minc! Watch how to do that here!
Learn more budget friendly ways to add foil to your wedding right here! It is amazing how custom and expensive this invite looks in person and the price was pennies to make!
Create Well: Get a foiled invitation for less using the Heidi Swapp Minc Foil Applicator!


text:
The Wirefliez
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Strangers in life....partners in music.
Return to The Wirefliez's profile
One-Sheet
Hometown Joplin, MO
Genres Alt. Rock, Indie Rock
Band Members Anthony Mignogno; Jacqueline Durbin; Mark Nisbet; Neil ... » more
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The Wirefliez are an unlikely group of passionate creatives sharing complimentary talent and a unique friendship. Strangely, each of the 4 members, from 4 different regions of the US and UK, have never met in person. Yet they stand united in their...
The Wirefliez are an unlikely group of passionate creatives sharing complimentary talent and a unique friendship. Strangely, each of the 4 members, from 4 different regions of the US and UK, have never met in person. Yet they stand united in their musical works of art, each telling a unique story, driven by diverse rock and folk laden influences.
The band met on a collaboration website after working on multiple other projects as individuals. Establishing a virtual group who worked exclusively together, as a "band", was the brainchild of drummer, Anthony Mignogno. He started searching for members and came across vocalist/songwriter Jacqueline Durbin. After a fair amount of consideration (or stalking as he would say) he approached her with the idea. They immediately clicked and started recording. A short time later they completed the package by acquiring lead guitarist, Neil Schmidt and bass guitarist, Mark Nisbet. They both possessed the talent to propel the group, but most importantly great attitudes and shared enthusiasm for the project.
In 2012 they put out their first EP, entitled The First Five, receiving impressive ratings and positive comments from professionals, as well as online exposure, airplay, and featured artist spots. Seeing the impact their music could make, they vowed to complete their next album. During this time they endured many personal trials and set-backs, in the midst of all the unique challenges that go along with a long-distance creative process. Nevertheless, they stayed firmly committed to the cause, and have continued to move forward. They now announce their second release, first full-length album, Uncommon Ground. The tunes are available for download on The Wirefliez website and will soon be available on CD.
Up to this point they've done everything themselves and are proud to have come this far. Performing live together has never been logistically possible, but they dream of a day it will be. The group is open to whatever the future holds as they promote their new album, look for new opportunities and create new music.
Special thanks and recognition goes to:
Jon Yanes – First guitarist for the group, who was instrumental in the development of their first few songs. These were eventually reworked and cut on The First Five album.
Pat Felitti - A true professional who graciously embellished "If We Hitch a Ride" with his piano mastery, from the new album, Uncommon Ground.
Mike Durbin – Husband to Jacqueline, who offers ongoing tech support and computer expertise to the project.
4 Strangers Productions – Fondly embraced as the name under which they operate as producers and publishers of their own work.
Recent comments from people in the industry….."Love the sound. Reminds me of Evanescence & Nightwish. You guys will blow the music industry away once you get together. Your great now hate to see the power you have when working togeth
"Very nice song, great vocals and guitars, song is well structured, has good progression, can imagine the new record will be great if the tracks are like this o
The Wirefliez –
For list of accomplishments Since July 2012….. PLEASE VISIT OUR WEBSITE!!!
New Woman
The War's Not Over
Invisible Connection
Stronger Than the Monster
Broken Soldier
Green Light Go
Another Year Gone
Too Much Too Little
Folk Winner, January 2017
Top 40 in Folk, May 2019
Top 40 in Folk, April 2019
Top 40 in Folk, March 2019
Top 40 in Folk, February 2019
Top 40 in Folk, January 2019
Top 40 in Folk, August 2018
Top 40 in Folk, July 2018
Top 40 in Folk, June 2018
Top 40 in Folk, November 2016
Top 40 in Folk, October 2015
Top 40 in Folk, September 2015
Top 40 in Folk, December 2014
Top 40 in Rock, August 2013
Top 40 in Rock, July 2013
Top 40 in Indie Pop, March 2013
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text:
Der Coastal Spirit Football Club ist ein neuseeländischer Fußballklub aus Christchurch in der Region Canterbury.

Geschichte
Der Klub wurde im Jahr 2007 gegründet. Er entstand als Ergebnis der Fusion von New Brighton und Rangers AFC. Ursprünglich hatte der Klub Einrichtungen im Bexley Park, welche aber bei dem Erdbeben im Jahr 2011 zerstört wurden. Somit bezog man danach den in der Nähe gelegenen Linfield Park.

Die Männer-Mannschaft wurde zur Saison 2021 aus der Mainland Premier League in die neue Southern League hochgezogen und spielt seitdem durchgehend im System der National League. In beiden Spielzeiten verpasste man jeweils mit einem 6. Platz aber immer die Championship-Runde.

Die Frauen-Mannschaft spielt derzeit in der Mainland Women's Premier League und gewann den Women's Knockout Cup im Jahr 2013.

Weblinks

 Website des Klubs

Einzelnachweise

Neuseeländischer Fußballverein
Gegründet 2007
Sportverein (Christchurch)

redpajama_set_name: RedPajamaWikipedia


To rephrase an old saying: 'It takes a village to raise an Analyst.' Data Analysts and Scientists are working in teams delivering insight and analysis on an ongoing basis. So how do you get the team to support experimentation and insight delivery without ending up in an IT Engineer vs Analyst vs Data Governance war? We present 5 shocking steps to get these teams of people working together with practical, doable steps that can help you achieve data agility.
Paweł Machnik , Software Developer at Azimo - Send Money for Less!
5. SO WHAT IS THE PROBLEM? In one word ….