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The Unofficially Official 2014 Sports Radio KJR Dance Preview
April 11, 2014 Alex	3 Comments
Once again, we've reached that point in the year when sports take a back seat to women, as they so often tend to do. In this instance, however, you have the opportunity to win a trip to any sporting event in the world by accurately selecting the hottest women on the planet.
Because the stakes are so high, we've decided to help you out on your way to the rugby World Cup in Australia, the Jai Alai championship in Malaysia, or the Beer Olympics in Europe. At the same time, you're more than welcome to play along with Seattle Sportsnet and a number of our closest friends by filling out a bracket at SportsRadioKJR.com, then navigating here, clicking "Join," and entering the password, "sonics."
Below you'll find projections for all 64 matchups in this year's tournament. We've done all the dirty work for you because we care. And also because we got to scour Google Images for pictures of these women. But mostly because we care.
Region 1: Your Fortune Awaits
Katy Perry vs. Rihanna
For the second year in a row, Perry and Rihanna are matched up against one another in the tournament's opening round. A season ago, it was Perry who triumphed over her Barbadian foe and there's no reason to believe the outcome will be any different in 2014. Take a bow, Rihanna, it's over. Winner: Perry.
Nina Agdal vs. Samantha Ponder
Youthful and exuberant, Agdal is the Kentucky of this year's Dance. The Sports Illustrated swimsuit model may not be without her flaws (or, more accurately, maybe she is without her flaws), but she's certainly hot at the right time. On top of that, if Samantha Ponder's husband is any indication, Agdal should have no problem disposing of her snakebitten opponent. Winner: Agdal.
Sara Jean Underwood vs. Jessica Alba
A former Queen of the Hardwood, Jessica Alba has fallen on hard times in Dances of recent past. While former Oregon State Beaver and frequent Playboy Playmate Sara Jean Underwood will put up a solid fight, Alba should return to her winning ways of days gone by and survive this first-round barnburner. Winner: Alba.
Kerry Washington vs. Alex Morgan
Forget all her other movie and TV appearances. If you've ever seen I Think I Love My Wife, you know how sizzling hot Kerry Washington truly is. That said, Washington is a Dance newcomer and still not necessarily a household name.
While Morgan also lacks in the all-important name-recognition department, this is her second Dance go-round and she's just so darn cute. Could be a close battle, but America's soccer sweetheart will prevail. Winner: Morgan.
Erin Andrews vs. Olivia Munn
Props to Olivia Munn. She used to be just another under-the-radar name in this competition, but then went and single-handedly made Magic Mike watchable (for reluctant boyfriends and husbands dragged along to witness their significant others' ogle over Channing Tatum, at least) by getting naked in the movie's opening scene. Bless her heart.
Unfortunately, Munn is going up against the perennially-attractive Erin Andrews, who will never not be hot. Andrews is a University of Florida alum, and we all know that the Gators do well in these types of bracket-bound scenarios. Winner: Andrews.
Peta Murgatroyd vs. Kate Upton
Last year's Queen of the Hardwood, Kate Upton is all but a lock to repeat as Dance champion. With tremendous size up front, Upton will be a tough matchup for any of her competitors throughout the tournament. This is your classic 1-16 pairing, with poor Peta Murgatroyd (who?) the hapless loser in Round 1. Winner: Upton.
Jennifer Lopez vs. Ashley Greene
With Jessica Alba likely laying claim to an opening-round victory, the 2014 Dance may just be the year of the veterans. Jennifer Lopez should help the cause for the old-timers if she can take down Ashley Greene, a repeat Dancer who has a knack for losing year after year. Winner: Lopez
Beyonce vs. Malin Akerman
Once upon a time, Malin Akerman was a movie star. We saw her in films like Harold and Kumar Go to White Castle (where she took her top off), Watchmen, The Proposal, and 27 Dresses before she seemingly disappeared from our collective consciousness. As it turns out, Akerman has recently left behind feature films for TV sitcoms, which means you probably never see her anymore because, let's face it, no one really watches sitcoms.
Sadly for Akerman, that career shift will probably cost her a shot at immortal Dance glory. Then again, losing to the wife of the man responsible for bringing Robinson Cano to Seattle is nothing to be ashamed about. Winner: Beyonce.
Region 2: Seize the Night
Amy Adams vs. Alessandra Ambrosio
Did you see Amy Adams in American Hustle? Was there any single point in that film during which she wore a bra? The transcendent blockbuster could have been nominated for an Academy Award solely on the grounds of Adams' cleavage show, which was by itself Oscar-worthy. In addition to her good looks, Adams is one of the best (and perhaps most underrated) actresses of our generation.
In spite of all that, the shallowness of voters will likely allow the less-talented but just-as-attractive Ambrosio to prevail because, you know, this is basically a petty, heartless battle of erections. Winner: Ambrosio.
Hannah Ferguson vs. Hayden Panettiere
The Missouri of the Dance, Hayden Panettiere may not possess the size of some of her competitors, but she is small, quick, and feisty. Poor Hannah Ferguson will be forced to run to keep up with the gritty little gym rat who made one of those Bring It On sequels almost worth watching. Alas, Panettiere will outpace Ferguson to the final buzzer. Winner: Panettiere.
Molly Mesnick vs. Jen Selter
Between Twitter and Instagram, Jen Selter has nearly 4,000,000 followers who cling to her solely for the plethora of half-naked selfies she bestows upon the internet. This proves something about society (horndogs), but I'm not sure what, exactly.
Molly Mesnick, on the other hand, is the wife of one-time Bachelor bachelor Jason Mesnick. Fun story, I've actually met Molly before. At the Richard Sherman Charity Softball Classic, as a matter of fact. We share a mutual friend, and both Molly and Jason seemed like genuinely nice people. If everyone had enjoyed the same brief encounter I had with the Mesnicks, maybe Molly could beat this Selter chick. Unfortunately, we all know that won't happen. Winner: Selter.
Scarlett Johansson vs. Jennifer Lawrence
Scarlett Johansson is what my girlfriend calls "a classic hottie." Sexy. At the other end of the spectrum Jennifer Lawrence is fun, perky, attractive, and really went to town on this competition last year. But that was last year.
Here in 2014, Lawrence has drawn one of the more intriguing matchups of the first round, taking on a Dance veteran who has wiped the floor clean with more than a handful of opponents in years past. This game might need overtime, but it should be Johansson celebrating when the clock finally hits all zeroes. Winner: Johansson.
Miranda Kerr vs. Lea Michele
Ever since Lea Michele's Glee run ended, she's fallen off the face of the earth. Apparently she's trying her hand at music these days (ask Jennifer Love Hewitt how that works out), but the reality is her stock is fading at exactly the wrong time.
Kerr, meanwhile, is peaking. After failing to impress in her first Dance appearance a year ago, Kerr has worked her way into a favorable pairing with the underwhelming Michele. This is the year the Australian model breaks through with a victory. Winner: Kerr.
Eva Mendes vs. Margot Robbie
For those who paid witness to the masterpiece that was Martin Scorsese's The Wolf of Wall Street, we all shared a deep appreciation for the woman providing the ass upon the bed of Leonardo DiCaprio's character. That ass belonged to none other than Margot Robbie, who makes her Dance debut riding a wave of momentum that can only be quelled by a savvy, veteran adversary like Eva Mendes.
Mendes is a pro's pro, having been here before, destroying competitor after competitor who has tried to outmaneuver her on the hardwood. Mendes is a force on defense and takes her time when the ball is in her hands. Like your classic Big 10 program, Mendes will wear Robbie down and outlast her to the final buzzer. Better luck next year, Margot. Winner: Mendes.
Blake Lively vs. Belle Knox
Belle Knox, for those who don't know, is a porn star who has risen to prominence of late because, lo and behold, she's a real-life freshman student at Duke University. Yes, believe it or not, Knox (that's her stage name, by the way) apparently assumed she'd get through college going unnoticed as a beautiful co-ed who also moonlit as the object of EVERY COLLEGE GUY'S FANTASIES — and you may find this mind-boggling, but it was indeed a male student who uncovered Knox's dual persona.
This barely-legal dirty, dirty girl pays her hefty tuition with cash earned from taking balls to the chin, which is sort of admirable when you consider the otherworldly cost associated with paying for college. We always hear about lowlifes who suck dick for drugs, but how often do we hear about upstanding citizens who suck dick for education? God bless America, am I right? Anyway, there's no way Knox wins this because a) she's going up against perennial Final Four contender Blake Lively, and b) she's a porn star, and porn stars never do well in these types of things. Winner: Lively.
Arianny Celeste vs. Emma Watson
Unless the Harry Potter Fan Club takes a sudden, keen interest in sports radio and its associated contests, Emma Watson should stand no chance of knocking off anyone, let alone Arianny Celeste. Winner: Celeste.
Region 3: Move Boldly
Emilia Clarke vs. Paulina Gretzky
Once upon a time, Wayne Gretzky, the greatest hockey player who ever lived, happened upon a striking American actress by the name of Janet Jones. The pair initially met while filming an episode of the hit TV show Dance Fever, but didn't really strike up an acquaintanceship until later rendezvousing at a Los Angeles Lakers game that was also attended by Gretzky's good friend, Alan Thicke. Perhaps it was the presence of Thicke, who played the charming head of the Seaver household on ABC's Growing Pains, that served as the necessary aphrodisiac to kindle a romance, but either way Gretzky and Jones took a liking to one another and later wed.
Shortly after their marriage was consummated, the couple gave birth to their first child, a girl they named Paulina. As the years passed, Paulina grew to become a lovely young woman, so lovely in fact that she gained fame and notoriety simply for being beautiful. Thus we find ourselves here, where the gorgeous daughter of a hockey great and his starlet bride will destroy some poor woman by the name of Emilia Clarke, who has nary a clue that Alan Effing Thicke of all people is at least partially responsible for what is about to transpire. Winner: Gretzky.
Liz Dueweke vs. Brooklyn Decker
Shout to Kevin Shockey, producer extraordinaire at Sports Radio KJR, who I'm quite sure lobbied successfully to have the inimitable Liz Dueweke included in the 2014 Dance field. For those unaware of Liz's existence, I'll briefly fill you in. A news reporter formerly of local ABC affiliate KOMO, Dueweke hails from Oklahoma, where she previously read the news for KOKH in Oklahoma City. In just a few short days, however, she'll migrate over to South Lake Union and begin reading the news for Seattle's Q13 Fox.
So where does Shockey fit in, you ask? Great question.
Dueweke became famous among Shockey's jackass friends, which may or may not include yours truly, when she began tweeting out KOMO's daily "Who Knew" trivia question. For those who play along on Twitter and also have the misfortune of following me, you're well aware that the answer to about 90-percent of the "Who Knew" questions could be representative of a form of sexual intercourse and/or a small cavity on one's body. Because Shockey and his friends are immature idiots, we grew to adore Liz thanks to her professionalism in the face of sheer stupidity. And other reasons, too, I'm sure.
Anyway, you don't care about any of that. You just want to know who will win. It'll be Brooklyn Decker. Sorry, Liz. Who knew? Winner: Decker.
Ronda Rousey vs. Britney Palmer
F**k my life. Who pits a UFC champion against a UFC octagon girl? That's all kinds of messed up. The selection committee must be stroking their massive, throbbing egos over this one. While Palmer is undeniably hot, Rousey is sexy, strong, and kind of scary. She also possesses a hefty dose of name recognition over her pacifist of an opponent. Go ahead and give this one to Rowdy Ronda in a coin flip. Winner: Rousey.
Miesha Tate vs. Kristen Bell
Back to that selection committee and their egos. Someone on the board wants to see a cage match between real-life UFC combatants Miesha Tate and the aforementioned Rousey in the Round of 32. The one thing possibly preventing that from occurring, though, is the always perky Kristen Bell. Men know Kristen Bell. Women know Kristen Bell. Frankly, everyone knows Kristen Bell. Tate, on the other hand? Despite hailing from Tacoma, she's a relative unknown. And as we all know, this entire competition is predicated on name recognition. Hence, we probably won't see that all-UFC throwdown after all. Winner: Bell.
Lily Aldridge vs. Shakira
I've heard rumors that Shakira is legally considered a dwarf. I don't know if those rumors are substantiated or not, but who really cares, they're funny. We may not know much about this Lily Aldridge character, but we know enough weird stuff about Shakira that we can't seriously pick her as our victor. Winner: Aldridge.
Mila Kunis vs. Jessica Gomes
Jessica Gomes is exotic and beautiful. Mila Kunis is also exotic and beautiful. Personally, I think Gomes is substantially more attractive than Kunis. But this isn't about what I believe, it's about what society believes. And society will always side with the Mila Kunises of the world. Winner: Kunis.
Naya Rivera vs. Holly Sonders
I didn't watch Glee because I'm a guy with self-respect, so I only first became aware of Naya Rivera when I found out she was the attractive sister of Oakland Raiders tight end Mychal Rivera. By the looks of it, Mr. and Mrs. Rivera have an unusual hard-on for the letter "Y," which means nothing to anybody, but I still wanted to point it out.
Holly Sonders, on the other hand, is a sideline reporter for the Golf Channel, which means you usually fall asleep on weekends to the sound of her voice. Hopefully having your naps narrated by the busty ex-golfer is enough to make you vote for her. I'm betting on it. Winner: Sonders.
Katherine Webb vs. Allie LaForce
Damn you, A.J. McCarron. You up and asked Katherine Webb to marry you last week and for some strange reason she said "Yes." She can probably do better. You, however, cannot. Allie LaForce is merely collateral damage standing in the way of Webb's impending tournament run. Winner: Webb.
Region 4: Spirited Golden Lager
Anna Kendrick vs. Bar Refaeli
That stupid, stupid Cups song is all the reason anyone should ever need to vote for Bar Refaeli. Winner: Refaeli.
Charissa Thompson vs. Sea Gal Jessica
Sea Gal Jessica
I'm going out on a limb here. Charissa Thompson rarely does well in this competition and Sea Gal Jessica just happens to dance for the Super Bowl Champion Seattle Seahawks. Seems like everything the Seahawks are associated with lately (not you, Ricardo Lockette) turns to gold. Don't let me down, 12s. Winner: Sea Gal Jessica.
Lucy Pinder vs. Olivia Wilde
Lucy Pinder has won this competition before and her boobies remain huge. Meanwhile, Olivia Wilde sleeps with quirky old Jason Sudeikis. Winner: Pinder.
Sofia Vergara vs. Karlie Kloss
If you know who Karlie Kloss is, you know more about life than I do. We all know Sofia Vergara, though. And she is really, really, steamy hot. Winner: Vergara.
Skylar Diggins vs. Chrissy Teigen
There's a subset of guys who, despite never watching a lick of women's basketball, really love them some Skylar Diggins. Don't get me wrong, she's certainly attractive. But more than that, she is the real-life incarnation of Love & Basketball's Monica Wright. And every guy who has ever watched that movie and played a little hoop has yearned for a Monica of his own. Of course, the trade off for landing a dime-dishing dime piece like Diggins is that you'll absolutely be obligated to attend her games. Her WNBA games. Enjoy the crap out of that, friend.
And then there's Chrissy Teigen, who I adore. Chrissy is from Snohomish and we're mutual followers of one another on Twitter. That's my claim to fame. Hers is much more expansive than that, what with being married to Grammy-winning artist John Legend, hosting her own TV show, and appearing on the cover of Sports Illustrated's 2014 Swimsuit Edition.
Like the sparse crowds that greet her at most WNBA venues, no one will show up to support poor Skylar Diggins. This is Teigen in a blowout. Winner: Teigen.
Megan Fox vs. Paula Patton
Once upon a time, I wrote a little blurb about Paulina Gretzky, her famous parents, and her famous parents' friend, Alan Thicke. Wouldn't you know it, Thicke had a child of his own once, a boy he named Robin. So what if Thicke wasn't great at naming children? He was still a darn good sitcom dad on Growing Pains.
As fate would have it, Robin Thicke would grow up to become famous in his own right, performing music for audiences that semi-reluctantly warmed to his songs over time. As Robin's star rose, he found himself in the company of increasingly attractive women. One woman in particular caught his eye and they eventually married. That woman's name? Paula Patton. Sadly, our story has no happy ending, as Patton and Thicke would separate (though not officially divorce … yet) in 2014.
Either way, none of this really matters since the lovely Ms. Patton stands almost no chance against the wife of Brian Austin Green and sizzling star of the forthcoming Teenage Mutant Ninja Turtles reboot, Megan Fox. But at least you got to read another fun story about Alan Thicke, right? Winner: Fox.
Janina Gavankar vs. Candice Swanepoel
In the battle of difficult names, Candice Swanepoel gets the nod over Janina Gavankar simply because I have no idea who Janina Gavankar is. Winner: Swanepoel.
Emma Stone vs. Stacy Keibler
I've never found Emma Stone to be all that attractive. Her voice is deep and raspy and her physical attributes leave something to be desired. I'd like to think everyone agrees with me, but would we really be in this position if you did? Of course not, so if you're like me, we'll just have to hope and pray that George Clooney's former love interest can send Stone to an early exit. Winner: Keibler.
Katy Perry vs. Nina Agdal
Nina Agdal is our Florida Gulf Coast. The slipper fits! Winner: Agdal.
Jessica Alba vs. Alex Morgan
The Jessica Alba renaissance claims another victim. Winner: Alba.
Erin Andrews vs. Kate Upton
Erin Andrews runs into the Kate Upton buzzsaw. Winner: Upton.
Jennifer Lopez vs. Beyonce
Let's have an on-stage booty-shaking competition to determine a winner, shall we? Winner: Beyonce.
Alessandra Ambrosio vs. Hayden Panettiere
I swear, this competition is single-handedly keeping Alessandra Ambrosio relevant. Winner: Ambrosio.
Jen Selter vs. Scarlett Johansson
One of these women takes pictures of herself while the other once dated Ryan Gosling. Winner: Johansson.
Miranda Kerr vs. Eva Mendes
Veteran guile prevails once again. Winner: Mendes.
Blake Lively vs. Arianny Celeste
All Lively. Winner: Lively.
Paulina Gretzky vs. Brooklyn Decker
Not even Alan Thicke can get her out of this mess. Winner: Decker.
Ronda Rousey vs. Kristen Bell
Name recognition, name recognition, name recognition. Winner: Bell.
Lily Aldridge vs. Mila Kunis
You're this competition's Mercer Bears, Aldridge. Be grateful you made it this far. Winner: Kunis.
Holly Sonders vs. Katherine Webb
Goodbye, Holly Sonders. Winner: Webb.
Bar Refaeli vs. Sea Gal Jessica
The power of the 12s should not be trifled with. That is, unless you're an international super model. Winner: Refaeli.
Lucy Pinder vs. Sofia Vergara
They'll go toe-to-toe, head-to-head, chest-to-chest, but in the end it will be Pinder left standing. Winner: Pinder.
Chrissy Teigen vs. Megan Fox
I will miss you, Chrissy. Winner: Fox.
Candice Swanepoel vs. Stacy Keibler
How did you get here again, Candice? Winner: Keibler.
Nina Agdal vs. Jessica Alba
Nina Agdal, thy name is Cinderella. Winner: Agdal.
Kate Upton vs. Beyonce
Kate Upton is on a mission. Winner: Upton
Alessandra Ambrosio vs. Scarlett Johansson
America loves Scarlett Johansson. Winner: Johansson.
Eva Mendes vs. Blake Lively
High-flying Blake Lively is too much for the grind-it-out fightin' Mendeses. Winner: Lively.
Brooklyn Decker vs. Kristen Bell
Bell's cakewalk into the Sweet 16 is all for naught. Winner: Decker.
Katherine Webb vs. Mila Kunis
This is where that whole "getting married to A.J. McCarron" thing finally catches up with her. Winner: Kunis.
Bar Refaeli vs. Lucy Pinder
Such a big frontcourt. Winner: Pinder.
Megan Fox vs. Stacy Keibler
This year is all about the renaissance. Winner: Fox.
Nina Agdal vs. Kate Upton
I hear these two are friends. They might be enemies after Upton ruthlessly disposes of her SI swimsuit buddy. Winner: Upton.
Scarlett Johansson vs. Blake Lively
Lively makes a triumphant return to the Final Four. Winner: Lively.
Brooklyn Decker vs. Mila Kunis
We'll look past her marriage to Andy Roddick for a moment. Winner: Decker.
Lucy Pinder vs. Megan Fox
No one can compete with Pinder's twin towers. Winner: Pinder.
Kate Upton vs. Blake Lively
This is Upton's (repeat) year. Winner: Upton.
Brooklyn Decker vs. Lucy Pinder
Pinder in a coin flip. Winner: Pinder.
Kate Upton vs. Lucy Pinder
And your 2014 Queen of the Hardwood is … Kate Upton!
950 KJRBigger DanceDanceMitch in the MorningQueen of the HardwoodSports Radio KJR
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3 thoughts on "The Unofficially Official 2014 Sports Radio KJR Dance Preview"
Maple Valley Mike says:
I was wondering if it would happen this year. Last year, it was barely noticed as a good percentage of the voting was online. When the contest started, it was "The Bigger Dance" but the NCAA claimed a copywrite violation. In addition, when the contest was started, Mitch spent his shows talking about girls more than sports. He was doing a Howard Stern type program. Now, the focus is back on sports as it should be.
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"Making the Connection: How Facebook is Changing the Supervisory Relationship" survey from Russell Herder found younger workers (age 18 to 34) are more likely to be Facebook friends with their bosses at 26%, compared to just 10% of those 35 or older. In addition, 38% of survey respondents said their bosses initiated the relationship and 29% felt "pressured" to accept the request.
And let's say you are on a job interview and the person behind the desk asks you for your Facebook login and password, should you object? Also getting media attention is the debate over whether such companies that ask potential hires for their social media login information are out of bounds. Sen. Richard Blumenthal (D-CT) and Sen. Charles Schumer (D-NY) have called for an immediate end to this practice, and have asked the U.S. Equal Employment Opportunity Commission and the Department of Justice to look into the matter.
While asking for social media logins may not be illegal, labor and employment attorney Lori Adelson of Arnstein and Lehr said small business owners should not open themselves up to risk by requesting such information from applicants. If a potential hire willingly gives their social media logins at your request, that is one thing, but it's not an area to really press a person on, she said.
Asking for the login information may be tempting for small businesses to do because they don't have the capacity to conduct in-depth background checks for new hires, Adelson said.
One social media site where requesting a potential employee's login is more acceptable in Adelson's opinion is LinkedIn.
She also does not recommend Googling potential hires, she said.
As for hitting that "friend request" button on Facebook, or "follow" on Twitter, Adelson has one word for business owners—don't.
Also, be sure you have a clearly drafted social media policy in place for your workers. This way it's clear to them what they can and cannot do on these sites in the office, and what they can and cannot post about work on their profiles.
Pfadenhauer echoed Adelson, and said if you wouldn't hang out with your workers at the local watering hole or have regular social engagements with them, then resist the urge to be friends on the Web.


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Ventrispina otagoensis är en insektsart som först beskrevs av Brittin 1938.

Ventrispina otagoensis ingår i släktet Ventrispina och familjen ullsköldlöss. Inga underarter finns listade i Catalogue of Life.

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redpajama_set_name: RedPajamaWikipedia


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Explainer United States
Chabad, the movement whose synagogue was targeted in the Poway shooting, explained
By	Ben Sales	 April 28, 2019 12:42 am
Some 4,700 Chabad-Lubavitch emissaries from 100 countries pose for the annual group photo in front of the organization's Brooklyn headquarters, Nov. 4, 2018. (Courtesy Chabad.org)
(JTA) — The shooting on Saturday that left one woman dead and three injured happened in a synagogue focused on welcoming and outreach.
The shooter, who has been apprehended by police, entered the synagogue, Chabad of Poway, at approximately 11:30 a.m. and began shooting. He is 19 years old and appeared to have had anti-Semitic intent. The shooting took place on the last day of Passover, and six months to the day after the synagogue shooting in Pittsburgh that left 11 dead.
Chabad of Poway is a branch of the Chabad-Lubavitch movement, a Hasidic group with emissaries across the globe who are focused on reaching out to unaffiliated Jews. While its rabbis are haredi Orthodox, its congregants are often secular or nonobservant Jews seeking to learn more about Judaism and its rituals.
Here's what you need to know about the movement at the center of Saturday's tragedy near San Diego.
Chabad focuses on encouraging Jews to connect to their tradition.
Chabad was founded in the 18th century in Lubavitch, a small town in Russia. It is a Hasidic movement, which means that its followers have traditionally focused on Jewish spirituality and mysticism. Chabad is a Hebrew acronym that stands for "intelligence, understanding and knowledge."
Its grand rabbi and his followers moved from Europe to the United States following the outbreak of World War II and the Holocaust. Since then, it has been based in the Brooklyn neighborhood of Crown Heights.
The movement was transformed in 1951 when its most recent grand rabbi, Menachem Mendel Schneerson, took the helm. Following the Holocaust, he expanded a program that sent rabbis as emissaries to Jewish communities in cities around the world. The rabbis' goal is to teach Jews about Jewish history, philosophy and practice, and to engage them in Jewish life.
Schneerson died in 1994, but the emissary program has only grown. There are now nearly 5,000 emissary families (generally a rabbi, his wife and children) in 100 countries, as well as all 50 states and 500 college campuses.
The rabbis are Orthodox, but the followers often are not.
Chabad rabbis fit the profile of haredi Jews: They wear black hats, long beards and dark suits, and stringently follow Orthodox Jewish law. But the emissaries' target audience is usually not other Orthodox Jews.
Many Chabad centers, including the synagogue in Poway, explicitly aim to reach Jews with little background in Judaism. Their modus operandi is to teach and model Orthodox Judaism in a nonjudgmental and welcoming way through Friday night dinners, classes and activities for families. Chabad of Poway's website calls the synagogue a "popular center for Jews of all backgrounds who want to learn more about their Jewish roots," and advertises "the Chabad-Lubavitch approach of joy, scholarship and unconditional acceptance and love for every Jew."
Chabad is also the movement behind the young men and boys who will set up shop on street corners in New York City and elsewhere, asking people if they are Jewish and offering men the opportunity to put on tefillin, or phylacteries.
There are also Chabads in places with few Jews, like in Kinshasa, Congo. The Chabad house in Mumbai, India, was targeted in a 2008 terror attack that killed the emissary couple there.
Some Chabad followers believe Schneerson, the late grand rabbi, is the Messiah.
Schneerson, who led Chabad for more than 40 years before his death in 1994, was a charismatic and influential leader who transformed the movement. His legacy has remained so powerful within Chabad that the movement has not named a new grand rabbi since his passing. Some Chabad Hasidim believe he is the Messiah and will return to the world.
The belief spawned a "messianist" faction within the movement. But many emissaries do not believe in Schneerson as Messiah and do not teach that belief at their Chabad centers.
Chabad rabbis have become connected to powerful officials globally, including Vladimir Putin and Jared Kushner and Ivanka Trump.
Chabad operates as a loose network, and emissaries for the most part tend to work independently — responsible for their own programming, fundraising and priorities. But some Chabad rabbis have become connected to powerful people.
Berel Lazar, for instance, one of two chief rabbis of Russia, is an ally of Russian President Vladimir Putin. Chabad ran a clandestine network of emissaries in the Soviet Union and has grown in Russia since the USSR's collapse three decades ago.
And in Washington, D.C., Jared Kushner and Ivanka Trump attend a Chabad synagogue. Kushner, President Donald Trump's son-in-law and senior adviser, has been involved with Chabad since he was a student at Harvard University.
Relations between Chabad and other Jewish religious denominations are sometimes strained over turf and philosophical issues, but the other movements have also looked to Chabad as an example in outreach programming, and using the internet and social media to engage unaffiliated Jews.

redpajama_set_name: RedPajamaCommonCrawl


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Big Midtown Projects Update-o-Rama: Still Slow Going!
By Joey Arak Sep 21, 2007, 10:05am EDT
The Farley Post Office to Moynihan Station transition may have been taken off life support when Eliot Spitzer was crowned, but since then the constant back-and-forth of the proposals have been buried in the linkage dump. No longer, at least for today. The most current vision included building two towers on the site of the current Madison Square Garden (which would be rebuilt on the Ninth Avenue side of Moynihan), but now the Post reports that the Empire State Development Corp., Vornado and Related have dropped the tower plans. Instead, the 4.5 million square feet of office space will be, um, "sprinkled" around the neighborhood. Yeah, we're not sure how that's going to work either.
It was a busy day for the Empire State Development Corp., which also issued a statement on another Spitzer albatross, the Javits Center expansion. The holdup on the ridiculously delayed Javits plan has to do with, obvs, money. A new porposal was due this summer, but now it's been pushed back to, well, that's not clear exactly. The Sun quotes the ESDC chairman as saying, "It is clear that Javits will require more money than has been expected." By the way, it was expected to cost $1.8 billion, so apparently the new Javits Center will be constructed entirely out of diamonds and cocaine.
· Tower Play at Farley Station [NYP]
· Javits Expansion Delays Blamed on Pataki [NYSun]
Next Up In Moynihan Station
See the transformation of a NYC landmark into the Moynihan Train Hall
Port Authority pledges $150M to Moynihan Station development
Port Authority Pulls Out of Its Planned Moynihan Station Space
Developers Behind Penn Station Overhaul Might Get the Boot
Transportation Watch
Long-Awaited Moynihan Station Is Actually Making Progress


text:
Q: Having the names of the fields, how can I initialize, with blank data, the undefined fields of an object? In Javascript, I have an array of strings, that represent all the possible names of an object's fields.
var fields = ["name", "age", "address"];

The server's response is an array of objects that may, or may not, contain all the fields from fields.
var response = [
  {"name" : "Tom"}, {"name" : "Jenny", "age" : 25}, ...
];

I need to populate all the missing fields with a an empty string or something so that there will be no more undefined fields (I can't do this on the server).
So far I have this
jQuery(fields).each(function(fieldKey, field){

  jQuery(response).each(function(resultKey, result){

    if (result[field] == undefined) result[field] = "";

  });

});

Is there a better and more efficient way ?

A: Use typeof to check a variable presence.
jQuery(fields).each(function(fieldKey, field){
  //fieldKey: index (number); field: array item (string)

  jQuery(response).each(function(resultKey, result){
    //resultKey: index (number); result: array item (object)

    if (typeof result[field] == 'undefined') result[field] = "";

  });

});

A: No need to use jQuery. The following code assumes the response object is present.
var fields = ["name", "age", "address"],
    i, j;

// for each response object, check the fields
for (i = 0; i < response.length; i++) {
    // for each field look if it is present in current response object
    for (j = 0; j < fields.length; j++) {
        // if not present, create with empty string
        if (typeof response[i][fields[j]] === 'undefined') {
            response[i][fields[j]] = '';
        }
    }
}

redpajama_set_name: RedPajamaStackExchange


text:
require 'byebug'
require 'simple_icloud_calendar'
require 'webmock/rspec'
require_relative 'helpers/application_helper'

WebMock.disable_net_connect!(allow_localhost: true)
include ApplicationHelper

redpajama_set_name: RedPajamaGithub


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text:
\section{Introduction}
The on-shell approach to computing scattering amplitudes had led to tremendous advances in our understanding of gauge theory and gravity \cite{Elvang:2013cua,TASI,Benincasa:2013}. Weinberg's seminal papers from the 60's \cite{Weinberg:1964ew,Weinberg:1965rz} showed that the combination of Poincar\'{e} invariance and unitarity uniquely picks out Maxwell's equations as the description of a massless spin-$1$ particle (photon) and Einstein's equations as the description of a massless spin-$2$ particle (graviton), at tree-level. In addition, charge conservation and the equivalence principle follow from consistency of scattering amplitudes involving photons and gravitons, respectively. \\

Since then, there have been many attempts to constrain or \textit{bootstrap} interactions using a purely on-shell approach, by invoking the principles of unitarity, locality and causality. One of these attempts is the technique known as BCFW deformations \cite{BCFW}, in which some of the external momenta are deformed by a complex parameter $z$. Consideration of the analytic structure of the amplitude as a function of $z$ often imposes non-trivial constraints on interactions \cite{Benincasa:2007xk}, as we will explain in detail in Section \ref{sec:SH}. Constraints can also be derived in another way \cite{NAH,McGadyRodina,Schuster:2008nh}, without deforming the amplitude, but rather by demanding consistent factorisation. Here one simply writes down the most general form of the tree-level four-particle amplitude, and demands that poles only arise due to intermediate (exchanged) particles going on-shell, with the corresponding residues given by a product of on-shell three-particle amplitudes. \\

Notable results of this bootstrap programme include the observations that long-range forces cannot be mediated by particles with spin $\geq 3$, Yang-Mills (YM) is the unique theory of multiple massless spin-$1$ particles, General Relativity (GR) is the unique low-energy theory of a massless spin-$2$ particle, and the existence of massless spin-$3/2$ particles requires gravity and supersymmetry leading uniquely to supergravity\footnote{One can also show that there is an upper limit on the number of spin-$3/2$ particles that a unitary theory can contain \cite{McGadyRodina}.} \cite{Benincasa:2007xk,McGadyRodina}.  This on-shell approach shows that the low-energy properties of Poincar\'{e} invariant theories are virtually inevitable.  \\

Given such spectacular success, attention has recently turned to applying similar methods to spatial correlation functions which are the fundamental observables in cosmology, see \cite{Arkani-Hamed:2018kmz,Baumann:2019oyu,Arkani-Hamed:2015bza,Sleight:2019mgd,Sleight:2019hfp,Baumann:2020dch,Green:2020ebl,Mata:2012bx,Ghosh:2014kba,Kundu:2014gxa,Arkani-Hamed:2017fdk,Arkani-Hamed:2018bjr,Benincasa:2018ssx,Benincasa:2019vqr,Hillman:2019wgh,Benincasa:2020aoj,COT} and references therein. These cosmological correlators live on the late-time boundary of an approximate de Sitter spacetime and encode details about the bulk spacetime through their dependence on spatial momenta. A key ingredient in this cosmological bootstrap is the fact that correlators contain flat-space scattering amplitudes in the residues of their total-energy poles \cite{Maldacena:2011nz,Raju:2012zr} (for a detailed proof see \cite{COT}). These amplitudes form part of the theoretical data required to bootstrap the corresponding correlators, and therefore the cosmological bootstrap programme requires us to have a solid understanding of flat-space scattering amplitudes. \\

Much effort has so far focused on correlators fixed by de Sitter symmetries (or conformal symmetries on the boundary). Intuitively, taking the cosmological bootstrap programme beyond exact de Sitter symmetries in order to construct inflationary correlators, requires taking the $S$-matrix bootstrap beyond exact Poincar\'{e} symmetries. Motivated by this, a recent paper \cite{PSS} derived the singular parts of tree-level four-particle amplitudes where the free propagators are assumed to be massless and Poincar\'{e} invariant, i.e. all dispersion relations are linear with each particle propagating at the same speed, but with interactions that are allowed to break Lorentz boosts. Within this set-up, analytically continued three-particle amplitudes are fixed by the helicities of the external particles up to an almost arbitrary function of their energies and these functions are then constrained by demanding consistent factorisation, but without making use of BCFW shifts. The most interesting results from this \textit{boostless bootstrap} are \cite{PSS}:
\begin{itemize}
\item When there is a graviton in the spectrum, all three-particle interactions must reduce to their Poincar\'{e} invariant form, even those that do not involve the graviton itself\footnote{Related results based on field-theoretic methods have been derived in \cite{Emergent1,Emergent2,Emergent3}. See also \cite{SoftGravitons} where Lorentz boosts arise in the presence of soft gravitons.}. Universal coupling of gravity to all other particles is then recovered.
\item Low-energy self-interactions of a photon must vanish. The leading allowed operator has six derivatives and is therefore mass dimension-$9$.
\item There exists at least one large class of boost-breaking theories involving scalars, spin-$1/2$ fermions and photons (\textit{boost-breaking massless QED}). The corresponding Lagrangians contain generalised, boost-breaking gauge redundancies.
\end{itemize}

In the Poincar\'{e} invariant cases, all the results derived by imposing consistent factorisation have also been reached via the BCFW formalism. Likewise, in this paper we show that the results of \cite{PSS} can be derived using BCFW shifts as a tool to automate consistent factorisation and therefore provide a neat consistency check. In Section \ref{sec:SH} we briefly introduce the spinor helicity formalism that we will use throughout, present general three-particle amplitudes for boost-breaking theories, and briefly review BCFW momentum shifts. In Section \ref{sec:Constructibility} we argue that BCFW shifts remain a useful tool for non-constructible theories by making a distinction between \textit{accessible} and \textit{inaccessible} singularities. In this case, while the four-particle amplitudes are not completely fixed by the three-particle ones, we can still use BCFW shifts to constrain the latter. We illustrate this in Section \ref{sec:4p} for boost-breaking theories of self-interacting spin-$S$ particles. We end the paper with some concluding remarks.

\section{Spinor helicity formalism and BCFW deformations} \label{sec:SH}

We work in four spacetime dimensions and use the spinor helicity formalism to present amplitudes in a compact form. In this formalism, a complex null four-momentum $p^{\mu}$ is represented as a product of two-component spinors as\footnote{The inverse of this equation is $p^{\mu} = \frac{1}{2}(\bar{\sigma}^{\mu})^{\dot{\alpha}\alpha} p_{\alpha \dot{\alpha}}$ and so the energy of a particle in terms of the spinors is $E = \frac{1}{2}(\bar{\sigma}^{0})^{\dot{\alpha}\alpha} \lambda_{\alpha} \tilde{\lambda}_{\dot{\alpha}}$.}
\begin{equation}
p_{\alpha \dot{\alpha}} = \sigma^{\mu}_{\alpha \dot{\alpha}}p_{\mu} = \lambda_{\alpha} \tilde{\lambda}_{\dot{\alpha}},
\end{equation}
where $\sigma^{\mu}_{\alpha \dot{\alpha}}$ are the Pauli matrices and the undotted and dotted indices transform in the spinor representations of the Lorentz group i.e. $(1/2,0)$ and $(0,1/2)$ respectively. Throughout we follow the conventions of \cite{Dreiner} and we assume that each particle satisfies $p^{\mu}p_{\mu} = E^2 - \bvec{p}^2= 0$ on-shell.
When boosts are broken, amplitudes are constructed from $SO(3)$ invariant quantities rather than $SO(1,3)$ invariant ones. In \cite{PSS}, it was shown that such three-particle amplitudes are functions of the following objects:
\begin{itemize}
\item ``angle'' brackets: $\langle i  j \rangle  =  \epsilon^{\alpha \beta} \lambda^{(i)}_{\alpha} \lambda^{(j)}_{\beta}$,
\item  ``square'' brackets: $[ i  j ]  =  \epsilon^{\dot{\alpha} \dot{\beta}} \tilde{\lambda}^{(i)}_{\dot{\alpha}} \tilde{\lambda}^{(j)}_{\dot{\beta}}$,
\item  energies: $E_i$.
\end{itemize}
Here latin indices label the external particles: $i,j = 1,2,3$. We remind the reader that the spinors are \textit{not} grassmanian and therefore the angle and square brackets are anti-symmetric. By demanding that the amplitudes scale in the appropriate way under helicity transformations, on-shell, non-perturbative three-particle amplitudes take the form \cite{PSS}
\begin{equation}
\mathcal{A}_3 = \begin{cases}
\langle 1 2 \rangle^{ d_3} \langle 2 3 \rangle^{d_1} \langle 3 1 \rangle^{d_2}F^H(E_1, E_2, E_3),  &h <  0,  \\
[ 1 2 ]^{ -d_3} [ 2 3 ]^{-d_1} [ 3 1 ]^{-d_2}F^{AH}(E_1, E_2, E_3) , & h > 0,
\end{cases}
\label{eq:3pFinalResult}
\end{equation}
where $d_i = 2 h_i - h$, $h_{i}$ is the helicity of the $i^{\text{th}}$ particle and $h$ is the sum of the helicities (if $h=0$, the amplitude can be a sum of the two expressions.) The presence of functions\footnote{The superscripts refer to holomorphic and anti-holomorphic kinematic configurations \cite{Benincasa:2007xk}.} $F^H, F^{AH}$ (which depend on the helicities) reflects the fact that Lorentz boosts are no longer an assumed symmetry. The Poincar\'{e} invariant amplitudes are recovered when these functions are constant. As energy is conserved, we will often consider the $F$'s as functions of two arguments only. As an example, the three-particle amplitude for three incoming gravitons, two with negative helicity and one with positive helicity, is
\begin{align}
\mathcal{A}_{3}(1^{-2},2^{-2},3^{+2}) = \left(\frac{\langle 12 \rangle^3}{\langle 23 \rangle \langle 31 \rangle}\right)^{2}F^{H}(E_{1},E_{2}).
\end{align}
In this case, Bose symmetry dictates that $F^{H}(E_{1},E_{2}) = F^{H}(E_{2},E_{1})$. \\

Now, to be consistent with unitarity and locality, a tree-level four-particle amplitude must factorise into a product of on-shell three-particle amplitudes on each of its poles. Poles correspond to exchanged particles going on-shell and consistent factorisation dictates that, for example,
\begin{equation}
\lim_{s \to 0} \left(s \mathcal{A}_4 \right) = \mathcal{A}_3(1,2, -I) \times \mathcal{A}_3(3,4, I),
\end{equation}
where $s  = (p_1 + p_2)^2$ is the propagator of the exchanged particle which is labelled by $I$. Analogous relations hold when $t = (p_1 + p_3)^2 \to 0$ and $u =  (p_1 + p_4)^2 \to 0$. For future reference, in terms of the spinors these Mandelstam variables are given by $s = \langle 12 \rangle [12] = \langle 34 \rangle [34]$, $t = \langle 13 \rangle [13] = \langle 24 \rangle [24]$ and $u = \langle 14 \rangle [14] = \langle 23 \rangle [23]$. Requiring the amplitude to factorise correctly on each pole is often highly non-trivial \cite{PSS,Benincasa:2007xk,NAH,McGadyRodina,Schuster:2008nh} since, for example, the $s$-channel residue can contain poles in $t$ and $u$ which then need to be interpreted as propagation of a particle in those channels. This is beautifully illustrated for multiple spin-$1$ particles where consistency in each channel requires the coupling constants to satisfy the Jacobi identity. \\

In \cite{Benincasa:2007xk}, Benincasa and Cachazo elegantly used BCFW shifts \cite{BCFW} to formally assess this consistency for a number of tree-level four-particle amplitudes. The simplest BCFW shift takes two particles $i$ and $j$ and deforms their energies and momenta according to
\begin{equation}
\lambda^{(i)}(z) = \lambda^{(i)} + z \lambda^{(j)}, \quad \tilde{\lambda}^{(j)}(z) = \tilde{\lambda}^{(j)} - z \tilde{\lambda}^{(i)},
\end{equation}
with all other spinors kept fixed. This choice preserves the on-shell conditions for both particles as well as energy-momentum conservation. The deformed amplitude $\mathcal{A}_{4}^{(i,j)}(z)$ is a rational function of the complex parameter $z$ and thus can be fully deduced from knowledge of its poles, residues and behaviour at infinity. It takes the form
\begin{equation}
\mathcal{A}_{4}^{(i,j)}(z) = \sum\limits_n \frac{res_{z = z_n} \mathcal{A}_{4}^{(i,j)}(z)  }{z-z_n} + B^{(i,j)}(z),
\end{equation}
where the boundary term $B^{(i,j)}(z)$ is regular in the entire complex plane. For four-particle amplitudes, only two poles can be reached by a given deformation (since $p_{i} + p_{j}$ is independent of $z$) and as we remarked above, the corresponding residues are evaluated from the three-particle amplitudes alone. For example, summing over the possible helicities of the exchanged particle, for $\mathcal{A}_{4}^{(1,2)}(z)$ the two poles that can be reached are $z_{t}$ and $z_{u}$ and we have
\begin{align} \label{Deformation12}
\mathcal{A}_{4}^{(1,2)}(z) = & \sum\limits_{h_I} \frac{A_3(\hat{1},3,-\hat{I}) A_3(\hat{2},4, \hat{I}) }{t(z)} \nonumber \\
+ &   \sum\limits_{h_I} \frac{A_3(\hat{1},4,-\hat{I}) A_3(\hat{2},3, \hat{I}) }{u(z)} \nonumber  \\
+ & B^{(1,2)}(z),
\end{align}
where a hat indicates that the particle has had one of its spinors deformed and evaluated at the appropriate pole. We have $t(z) = (p_{1}(z) + p_{3})^2 = \langle 13 \rangle [13] + z \langle 23 \rangle [13]$ and $u(z) = (p_{1}(z) + p_{4})^2 = \langle 14 \rangle [14] + z \langle 24 \rangle [14]$ and therefore the locations of the poles are
\begin{align}
z_{t} = -\frac{\langle 13 \rangle}{\langle 23 \rangle}, \qquad z_{u} =- \frac{\langle 14 \rangle}{\langle 24 \rangle}.
\end{align}
$\mathcal{A}_{4}^{(i,j)}(0)$ corresponds to the amplitude for unshifted momenta, and then constraints on three-particle couplings can be derived by demanding that distinct $\mathcal{A}_{4}^{(i,j)}(0)$ coincide at $z = 0$ \cite{Benincasa:2007xk} i.e.
\begin{equation}
\mathcal{A}_{4}^{(i,j)}(0) = \mathcal{A}_{4}^{(k,l)}(0) \quad \forall \quad i,j,k,l.
\label{eq:MijMkl}
\end{equation}
This is the \textit{four-particle test}.
\section{Constructibility criterion} \label{sec:Constructibility}
In its original formulation, the above described method is reserved for \textit{constructible} theories for which $B^{(i,j)}(z)$ vanishes. In this case the singular parts of undeformed amplitudes can be compared with one another and the full four-particle amplitude is determined by the three-particle ones. Since $B^{(i,j)}(z)$ is regular, it is sufficient to prove that the amplitude tends to zero as $z \to \infty$. This is usually a non-trivial matter, necessitating a reference to the Lagrangian and a detailed counting of powers of momenta. Fortunately, many theories describing nature are constructible including, most notably, YM \cite{BCF,BCFW} and GR \cite{Benincasa:2007qj} (scalar field theories are not constructible in the sense described above. This has lead to new, interesting momentum shifts and on-shell recursion relations being derived for scalar theories with non-linearly realised symmetries \cite{ScalarRecursion1,ScalarRecursion2}). \\

However, for the boost-breaking amplitudes of interest here, it is unlikely that $B^{(i,j)}(z)$ would vanish, since the unknown functions of energies will in general contribute positive powers of $z$ to the tree-level amplitude. Indeed, for both particles $i$ and $j$, the deformation of their energies is linear in $z$ and so the divergence at large $z$ gets worse as additional powers of energy are included:
\begin{align}
E_i(z) & = E_i(0) + \frac{z}{2}(\bar{\sigma}^{0})^{\dot{\alpha}\alpha} \lambda^{(j)}_{\alpha} \tilde{\lambda}^{(i)}_{\dot{\alpha}}, \\
E_j(z) & = E_j(0) - \frac{z}{2}(\bar{\sigma}^{0})^{\dot{\alpha}\alpha} \lambda^{(j)}_{\alpha} \tilde{\lambda}^{(i)}_{\dot{\alpha}}.
\end{align}

The BCFW method can still be useful for non-constructible theories, however. One possibility is to introduce a distinction between \textit{accessible} and \textit{inaccessible} singularities of $\mathcal{A}_{4}^{(i,j)}(z)$, along the lines of \cite{Feng:2014pia}. We say a singularity is \textit{accessible} via a deformation of momenta $i$ and $j$ if this singularity is approached as $z \to z_*$ for some $z_*$. Otherwise we say it is \textit{inaccessible}. The regular term $B^{(i,j)}(z)$, by definition, cannot have any singularities in the $z-$plane and therefore cannot contribute to any residues of the accessible singularities of $\mathcal{A}_{4}^{(i,j)}(z)$. But it may exhibit inaccessible singularities. As an illustration of this distinction, consider a single scalar theory which is famously non-constructible. In the absence of additional global charges, the three-particle amplitude is a non-zero constant, $\mathcal{A}_3 = g$, and so we have
\begin{eqnarray}
\mathcal{A}_{4}^{(1,2)}(0) &=& g^2 \left(\frac{1}{t} + \frac{1}{u}\right) + B^{(1,2)}(0), \\
\mathcal{A}_{4}^{(1,4)}(0) &=& g^2 \left(\frac{1}{s} + \frac{1}{t}\right) + B^{(1,4)}(0).
\end{eqnarray}
The consistency condition $\mathcal{A}_{4}^{(1,2)}(0)  = \mathcal{A}_{4}^{(1,4)}(0) $ can be satisfied by choosing $ B^{(1,2)}(z) = \frac{g^2}{s}$ and $B^{(1,4)}(z) = \frac{g^2}{u}$, since these two functions do not have any accessible singularities with regards to their own deformations. \\

In the following section we will constrain boost-breaking amplitudes using the fact that the regular term $B^{(i,j)}(z)$ does not have any accessible singularities. We will see that for spinning particles, we can derive the highly non-trivial constraints first found in \cite{PSS}.

\section{Constraining three-particle interactions} \label{sec:4p}
In this section we will constrain three-particle interactions for theories of a single spin-$S$ particle with integer $S$. We will derive the constraints first presented in \cite{PSS}. We also checked that the BCFW techniques allow us to recover other results in \cite{PSS}, namely those of (gravitational) Compton scattering and the full analysis for a scalar or a photon coupled to gravity. Those calculations contain only minor differences compared with what is presented below so we omit the details in favour of brevity. We remind the reader that we do not impose boost invariance, but only demand that the free theory is Poincar\'{e} invariant, with the on-shell condition $E^2 - \bvec{p}^2 = 0$ for each particle. \\

Consider the amplitude $\mathcal{A}_4(1^{+S}2^{-S}3^{+S}4^{-S})$, where superscripts denote the helicities of incoming particles of some integer spin $S$. We will impose matching conditions between deformations $(1,2)$ and $(1,4)$. First consider $(1,2)$. Using the expressions for\footnote{These amplitudes arise from the leading order couplings. Higher-dimension operators give rise to the $A_{3}(1^{+S},2^{+S},3^{+S})$, $A_{3}(1^{-S},2^{-S},3^{-S})$ amplitudes but we don't consider these here. We refer the reader to \cite{PSS} for a discussion on these amplitudes.}  $A_{3}(1^{+S},2^{+S},3^{-S})$ and $A_{3}(1^{-S},2^{-S},3^{+S})$ given in \eqref{eq:3pFinalResult}, and\footnote{Here we have assumed $[13]$ and $[14]$ are non-zero, and therefore $t=0$ and $u=0$ are approached as $\langle 13 \rangle = 0$ and $\langle 14 \rangle = 0$ respectively (or as $[24]=0$ and $[23]=0$ respectively, by momentum conservation).}
\begin{eqnarray}
p_{1}(z_t)+p_{3} & = & \frac{[13]}{[14]} \lambda^{(3)} \tilde{\lambda}^{(4)}, \\
p_{1}(z_u)+p_{4} & = & \frac{[14]}{[13]} \lambda^{(4)} \tilde{\lambda}^{(3)}, \label{IntermediateU}
\end{eqnarray}
to eliminate all copies of $\lambda^{(I)}$ and $\tilde{\lambda}^{(I)}$, which are the spinors associated with the exchanged particle\footnote{For example, in the $t$-channel we set $\lambda^{(I)} = \alpha \lambda^{(3)}$ and $\tilde{\lambda}^{(I)} = \beta \tilde{\lambda}^{4}$ with $\alpha \beta = \frac{[13]}{[14]}$. When computing the residue, $\alpha$ and $\beta$ only appear in the product $\alpha \beta$.}, we find
 \begin{eqnarray}
&& \mathcal{A}_{4}^{(1,2)}(0) =  B^{(1,2)}(0) + \nonumber \\
&&  \left( \frac{1}{t} F_{\hat{1},3}F_{\hat{2},4} + \frac{1}{u}  F_{\hat{1},-\hat{1}-4}F_{\hat{2},-\hat{2}-3}\right) \left(\frac{ [13]^2 \langle 24 \rangle^2}{s} \right)^S.
\end{eqnarray}
In the $u$-channel we have summed over the two possibilities for the helicity configuration of the exchanged particle but given \eqref{IntermediateU}, only one of these is non-zero.
To keep formulae compact, here we have introduced subscripts to the $F$'s to denote their arguments e.g. $F(E_{i},E_{j}) \equiv F_{i,j}$ and $F(E_{i},E_{j}+E_{k}) \equiv F_{i,j+k}$. Again, hats denote deformed objects evaluated at the appropriate points e.g. in the $1/t$ coefficient, $F_{\hat{1},3} \equiv F(\hat{E}_{1}(z_t),E_{3})$ where $\hat{E}_{1}(z_t)$ is the deformed energy of particle $1$ evaluated at $z = z_t$. Likewise, in the $1/u$ coefficient, hatted energies are evaluated at $z=z_{u}$. We have also removed the $H/AH$ superscripts since the functions are identical, due to parity, up to an inconsequential overall sign \cite{PSS}. \\

Now, we can also write
 \begin{eqnarray} \label{M12Final}
&& \mathcal{A}_{4}^{(1,2)}(0) =  \tilde{B}^{(1,2)}(0) + \nonumber \\
&&  \left( \frac{1}{t} F_{1,3}F_{2,4} + \frac{1}{u}  F_{1,-1-4}F_{2,-2-3}\right) \left(\frac{ [13]^2 \langle 24 \rangle^2}{s} \right)^S,
\end{eqnarray}
where here we dropped the hat above all the energies, which indicates that the expression is evaluated at their \textit{undeformed} values. This can be justified as follows. We assume that the $F$'s can be Taylor expanded around the undeformed energies. The deformed energies are
\begin{align}
&\hat{E}_{1}(z_{t}) = E_{1} - \frac{t}{2 \langle 23 \rangle [13]} (\bar{\sigma}^0)^{\dot{\alpha} \alpha}\lambda^{(2)}_{\alpha}\tilde{\lambda}^{(1)}_{ \dot{\alpha}}, \\
&\hat{E}_{2}(z_{t}) = E_{2} + \frac{t}{2 \langle 23 \rangle [13]} (\bar{\sigma}^0)^{\dot{\alpha} \alpha}\lambda^{(2)}_{\alpha}\tilde{\lambda}^{(1)}_{ \dot{\alpha}},
\end{align}
with similar expressions evaluated at $z = z_u$.  Here we see that potential new singularities generated by the deformed energies are all \textit{inaccessible}, as they correspond to the vanishing of $\langle 23 \rangle $ or $[13]$, but these do not depend on $z$. Moreover, only the leading term in the Taylor expansion will exhibit accessible singularities, since in all subleading terms $t$ and $u$ will be cancelled out. We can therefore simply absorb all subleading terms into $B^{(1,2)}$, thus introducing $\tilde{B}^{(1,2)}$ that still does not contain any terms singular in $t$ or $u$. Although it could become singular in some kinematic configurations, especially at $s=0$, that is not a problem, because this singularity is \textit{inaccessible} and we only demand that $\tilde{B}^{(1,2)}$, for those configurations for which it can be defined, does not have any singularities as a function of $z$. \\

We now play the same game for the $(1,4)$ deformation which amounts to interchanging particles $2$ and $4$. We have
\begin{eqnarray} \label{M14Final}
&& \mathcal{A}_{4}^{(1,4)}(0) = \tilde{B}^{(1,4)}(0) +  \nonumber \\
&& \left( \frac{1}{t} F_{1,3}F_{4,2} + \frac{1}{s}  F_{1,-1-2} F_{4,-4-3} \right) \left(\frac{ [13]^2 \langle 24 \rangle^2}{u} \right)^S.
\end{eqnarray}
We discussed the $S=0$ case earlier where we showed that equating $\mathcal{A}_{4}^{(1,2)}(0)$ and $\mathcal{A}_{4}^{(1,4)}(0)$ requires us to make certain choices for the boundary terms. Let us now consider $S >0$ with $S$ an integer. We see that $\mathcal{A}_{4}^{(1,2)}(0)$ in \eqref{M12Final} contains terms proportional to $1/(ts^S)$ and $1/(us^S)$ which are both singular in more than one Mandelstam variable and thus cannot be accounted for or modified by $\tilde{B}^{(1,2)}(0)$ nor $\tilde{B}^{(1,4)}(0)$. A similar observation applies to $\mathcal{A}_{4}^{(1,4)}(0)$ in \eqref{M14Final}.
Thus, by matching the amplitudes we find the necessary condition
\begin{equation}
\frac{a}{s^{S}t} + \frac{b}{s^{S}u} = \frac{c}{u^{S}t} + \frac{d}{u^{S}s},
\end{equation}
where
\begin{eqnarray}
a & = &  F_{1,3}F_{2,4}, \\
b & = &  F_{1,-1-4} F_{2,-2-3}, \\
c & = & F_{1,3}F_{4,2}, \\
d & = & F_{1,-1-2}F_{4,-4-3}.
\end{eqnarray}
Recalling that $s+t+u = 0$, this constraint, given that it must be valid for all kinematics, is equivalent to
\begin{align}
a u^{S} - b (s+u) u^{S-1} - c s^{S} + d (s+u) s^{S-1} =0.
\end{align}
For $S=1$ we therefore have $a = (b-d) = - c$, or equivalently,
\begin{align}
F_{1,3} F_{2,4} & - F_{1,-1-4}F_{2,-2-3} + F_{1,-1-2} F_{4,-4-3} = 0,
\end{align}
which is simply an alternative form of (4.25) from \cite{PSS}. Assuming that the $F$'s are polynomials, in \cite{PSS} it was shown that the only solution to this system is $F \equiv 0$ once we impose that the $S=1$ functions are alternating polynomials as dictated by Bose symmetry\footnote{The spinor helicity parts of the $S=1$ three-particle amplitudes are odd under the exchange of identical particles so if the amplitudes are to be even by Bose symmetry, the $F$'s must be alternating.}. We therefore see that the leading order three-particle interactions for three-photons must vanish, as is the case for Poincar\'{e} invariant theories. For $S=2$ we require $a=b =c=d$, or equivalently,
\begin{align}
F_{1,3}F_{2,4} = F_{1,-1-4} F_{2,-2-3} = F_{1,-1-2}F_{4,-4-3},
\end{align}
which gives rise to the constraints $(4.31)-(4.33)$ from \cite{PSS} once we use the fact that the $S=2$ functions are symmetric in their arguments by Bose symmetry\footnote{For $S=2$, the spinor parts of the three-particle amplitudes are even under the exchange of identical particles and so the $F$'s are symmetric polynomials.} (this also makes the $a=c$ constraint trival). In \cite{PSS} it was shown that the only solution to this system is $F=constant$ and so again the three-particle interactions are reduced to their Poincar\'{e} invariant form, but this time the amplitudes are non-zero and are those of GR. Finally, for $S > 2$, it is simple to see that $a=b=c=d=0$ is required and therefore there are no consistent three-particle interactions for these massless, higher-spin particles even when boosts are broken, as was also concluded in \cite{PSS}.

\section{Summary}\label{sec:Discussion}

Very recently, the singular parts of four-particle amplitudes were bootstrapped in \cite{PSS} by demanding that they factorise into a product of on-shell three-particle amplitudes on simple poles. In that work, consistent factorisation was implemented directly without making use of BCFW momentum shifts. In this short note, we have shown that the same results can be derived by using BCFW shifts to automate consistent factorisation. We presented full details for the illustrative cases of single spin-$S$ particle amplitudes but have also checked that the procedure produces the expected results for Compton scattering, and its gravitational analogue, as well as for scalars or photons coupled minimally to gravity. For single spin-$S$ particles, the boostless bootstrap teaches us that the leading three-particle couplings for a photon must vanish, the leading three-particle couplings for a graviton must be those of GR, while massless higher-spinning particles do not self-interact. For photon Compton scattering, boost-breaking interactions between the photon, scalars and spin-$1/2$ fermions are allowed and can be described by Lagrangians with generalised boost-breaking gauge redundancies. For gravitational Compton scattering, all couplings must reduce to their boost-invariant counter-parts with universal couplings of all particles to gravity. Finally, scalars and photons that are minimally coupled to gravity are forced to have Poincar\'{e} invariant self-interactions (constant or vanishing, respectively for the scalar and the photon). For full details see \cite{PSS}.  \\

Although the theories we have considered are not \textit{a priori} constructible, in the sense that the boundary terms do not necessarily vanish at large $z$, we have still been able to use BCFW shifts to constrain the three-particle couplings contributing to particle exchange. This does mean, of course, that the three-particle amplitudes themselves do not fully fix the four-particle ones. Indeed, all of the four-particle amplitudes we have constructed are defined up to the presence of ``contact" terms that are regular for all kinematic configurations. It would be very interesting to investigate the possibility of using generalised momentum shifts, possibly along the lines of \cite{Raju2}, to recursively derive exact higher-point amplitudes even if only for a subset of boost-breaking theories. It would also be very interesting to investigate the generalised on-shell recursion relations introduced in \cite{Benincasa:2011kn}, where boundary terms are fixed with additional knowledge of a subset of the zeros of the deformed amplitude, in our boost-breaking setting.

\section*{Acknowledgements}
We would like to thank Paolo Benincasa, Tanguy Grall, Sadra Jazayeri and Enrico Pajer for useful discussions and comments on a draft of this paper. D.S. has been supported in part by the research program VIDI with Project No. 680-47-535, which is (partly) financed by the Netherlands Organisation for Scientific Research (NWO). J.S. has been supported by a grant from STFC.

\bibliographystyle{apsrev4-1}

redpajama_set_name: RedPajamaArXiv


text:
Na-Allah had travelled to Maiduguri to attend a two-day workshop, and was asking for directions to a restaurant when he was attacked along with his three companions.
As they were talking, one of the mob suddenly claimed to have lost his penis and accused Na-Allah and his companions of being responsible � providing an excuse to attack the men, police said.
iafrica.com...
Sadly, he died.
Hey, I don't make the news, I just report it.
[edit on 1-10-2004 by Jamuhn]
posted on Oct, 1 2004 @ 09:40 PM
That is just downright bizarre! First that a guy would claim he lost his penis and then that the mob would actually believe it and act on it!!
Unbelievable!
I couldn't get that link to work, can you try to repost it?
[edit on 1-10-2004 by Jemison]
dr_strangecraft
. . . depending on the region of the world,
The neurosis is where a male imagines his penis is beginning to shrink, and contract up into his body. In traditional societies, witchcraft is usually to blame. I have heard of this kind of mob-hysteria in subsaharan Africa. Local reporters usually leave out the witchcraft angle, for fear of making their countrymen seem naive.
urbanlegends.about.com...
www.xent.com...
content.karger.com...
LordGoofus
posted on Oct, 2 2004 @ 01:21 AM
i keep hearing strange reports like that from africa etc where there is still a strong believe in witchcraft... out of curiosity, when someone over there claims some witch made his danglybits disapear, is there anyone AT ALL who think hmmm, maybe I should examine him and see what he's talking about? Or do they all just take the guys word for it and go hack someone to death?
doctorduh
I think that's the strangest thread title I've ever read. The story really didn't make sense. A man notices that his penis had disappeared? Then said man blames another man for his absent penis by accusing him of stealing it? Angry mob seeks justice by beating penis thief to death.
A little Shakespearian, if you ask me. Definintely Freudian.
zerotime
Africa is a messed up place. I really think it is a doomed country. You can find stories like this on almost a weekly basis. Diseases are ripping across the continent unchecked - 70% of the worlds AIDS population is in Africa. The average person in Africa has the education intelligence of a normal second grade school child. In South Africa alone there are 21,000 babies raped a year because the men believe that having sex with an infant will cure AIDS. When I hear these bizarre awful stories that come out of Africa I can't even comprehend what could be going through these sick minds at the time. It is pure insanity. It's like Hell cracked open and millions of demons spilled out.
feyd rautha
In South Africa alone there are 21,000 babies raped a year because the men believe that having sex with an infant will cure AIDS.
well, any sources for this statement???
I found it in 2 minutes w/ Google: " babies, cure, aids"
South African men rape babies as 'cure' for Aids
By Jane Flanagan in Johannesburg
(Filed: 11/11/2001)
THE alleged rape of a nine-month-old baby girl by six men in a remote part of rural South Africa last week has focused the nation on an 80 per cent rise in child sexual abuse over a year, much of it connected with the country's Aids pandemic.
More than 67,000 cases of rape and sexual assaults against children were reported last year, compared with 37,500 in 1998. Child welfare groups believe that the number of unreported incidents could be up to 10 times that number.
Some of the victims were as young as six-months-old, a number of whom died from their injuries, while others contracted HIV. The largest increase in attacks has been against children under seven.
Although rises in poverty, violent crime and unemployment are said to have contributed to the escalation in child abuse, the most significant and worrying factor is the widespread myth sweeping the country that having sex with children provides a cure for Aids.
Cati Vawda, the director of the Children's Rights Centre in Durban, said: "There is a belief across South Africa that a virgin will cure a man of HIV or Aids. We have no idea where this idea has come from, but it has been around for a few years and has certainly taken hold."
Kelly Hatfield, of People Opposed to Women Abused , added: "South Africa has reached a new low. This rumour certainly has become a common belief."
Police say at least one of the men who attacked the nine-month-old girl is HIV-positive. The baby has also been tested for the virus and given anti-retroviral drugs as a precaution.
The girl - whose name has not been released - remains in intensive care at the Kimberley Hospital. While doctors said it was likely that her physical wounds would heal, experts remained concerned about psychological trauma.
Even in a country that is accustomed to violent crime, the attack has shocked the nation. Pictures of the six accused appeared on front pages of newspapers across South Africa and radio phone-in shows were inundated with callers demanding to know what the police and Government were doing to stop child abuse.
More than 1,000 people, mainly women and children, gathered at the magistrate's court in the remote town of Upington in the north of the country to vent their anger as the trial opened against the men, aged between 24 and 66, accused of assaulting the infant. The crowd chanted slogans demanding more protection for children, and waved banners and posters.
Dozens of police and soldiers had to be deployed to stop the crowd from surging into the courtroom, where the case was postponed until later this month. The six accused men were remanded in custody.
South Africa has the highest number of HIV-positive citizens in the world. According to official figures, one in nine South Africans are infected with the virus, although many health workers believe the rate of infection to be closer to one in eight. Schools often harbour the worst sex offenders.
Next month, a deputy headmaster in the Eastern Cape goes on trial accused of raping schoolgirls over a 10-year period, resulting in 20 of them becoming pregnant.
Joan van Niekerk, the chairman of Childline South Africa, said: "How can we hope that a school will teach children to behave responsibly when the very people who are educating them are themselves abusers?"
The 185 child abuse cases that are reported to the police each day represent just a fraction of the true number committed, according to child welfare groups. Of these, just five per cent result in a successful conviction. For those mothers aware that their children have been abused, the decision to make a complaint is rarely easy.
Edith Kriel, a social worker who helps child victims in the Eastern Cape, said: "Child abusers are often relatives of their victims - even their fathers and providers. For women, making a complaint is a hard choice because they end up losing their husbands, the breadwinners and the roof over their heads."
IAF101
What the?????
What kind of sane being would do a thing like that ?
Downright hideous if you ask me!!
James the Lesser
First,
Ok, he stole my penis, so lets go kill him, and hope no one steps on my penis so I can sow it back on. Then the babies cure AIDS?
Damn, and people say Africa is becoming the new China when from the looks of it Africa won't have any people left alive on it soon, not that that is a bad thing, if it a disease, it came from there. never heard of the west Mississippi River virus, or the Ohioan Myangis Ebola virus, or AIDS from Europe, nope, it all came from Africa. At least the # Europe made was cureable, like Polio(polio shots) small pox, chicken pox, common cold, so forth.
JediMaster
I think we can all agree that Africa is not only scrwed bu incrediably messed up. Babies curing rapes via rape, and penis shrinking politicians
The sad thing was when I read the title I was thinking "Great what member of Congress is in trouble now".
Minime
Africa needs to seriously shape up, I would'nt want to know what other stuff have happened there in the past.
bjamison
How many of you people have actually been to Africa?
I have, Egypt
It is a bit much for a bunch of fat westerners to poop on them for being who they are. Especially when we have our OWN pecadillos that seem so barbaric to the rest of the planet.
Oh well. Provincialism begins at home.
[edit on 2-10-2004 by dr_strangecraft]
Chuck Stevenson
Yes ... Africa has a right to it's ignorant beliefs and backward ways ... after all they only got out of the bush 200 ... 150 .... 100 ... 50 ... last year .... yesterday!
Intelearthling
Originally posted by Chuck Stevenson
But blaming someone for making one's 'member' disappear is going a little bit too far!
And of course raping of babies to cure AIDS is right up there with it.
[edit on 3/10/04 by Intelearthling]
esdad71
(insert sarcasm)Damn those baby rapists! AIDS cases! THey might have WMD's in those huts in the middle of the desert.
We should invade now before the threat of baby rape comes to our shores.
I have read alot about Africa and none of it is good. There are poor native blacks, a few rich white people, diamonds/ore and LOTS of oil/pertroleum resources, as well as one of the biggest suppliers of rubber. One country that does nothing really except register ships for sailing around the world yet is run by unlruly factions and was founbded by slaves from the US. Read the history. A little country called LIberia. Might tell you why we cannot change that country, and why they still believe in witchcraft and dissapearing penises....
Paranoid thoughts like this is exactly what leads to mass slaughter of thousands of innocent civilians.
Yes, that makes sense. America invades Iraq and now realises that it bit off more than it can chew.. So, hell, lets just invade somebody else. The Africans could have WMDs. Lets blast them. If we don't find anything, we'll just say it was bad intelligence.
I have read alot about Africa and none of it is good. There are poor native blacks, a few rich white people, diamonds/ore and LOTS of oil/pertroleum resources, as well as one of the biggest suppliers of rubber. One country that does nothing really except register ships for sailing around the world yet is run by unlruly factions and was founbded by slaves from the US. Read the history
I don't know what book you read, but please throw it away. You have know idea what you are talking about. "Do nothing but register ships.....Founded by slaves from the US....." Do me a favour and REREAD the history.
Might tell you why we cannot change that country
Who the hell do you think you are that you have the audacity to have to change that country? Those resources you were alluding to is the only reason America would want to "invade" any country. Come in, kill the bad, bad people, and hey, while we here lets rape the land for its mineral wealth. We did'nt come here for it, but as we said, while we here, you understand.
From your post I now realise why the American Indians love the whites so much. Wake up and stop talking about something you have no concept about.
Originally posted by bjamison
Don't get me wrong, the country is beutiful (landscape wise), but the people there can be a bit.. ya know.
No I don't know, tell me please.
If you have'nt noticed, I'm one of "those" people.
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Update February 19, 2018: The boot kit that AMD have been offering to send out to users who are stuck in an infinite boot cycle is actually a Bristol Ridge APU.
Here are the very best AMD gaming motherboards on the market.
A user over on the Ars Technica forum received word, through the boot kit confirmation email, that the APU supplied would be AMD's Bristol Ridge A6 9500. This two-core Bulldozer-based part is seemingly abundant around AMD HQ, and will allow Raven Ridge desktop APU customers to install the latest version of their motherboard BIOS and install their Ryzen 2000-series chip. You are supposed to return it once the short update procedure is complete, however.
For users in the UK facing the issue, we spoke with OverclockersUK on AM4 motherboard support. They indicated that they would be "happy to support any of our customers who need an update", which offers another potential route for PC builders in need of an easy fix.
Original story February 16, 2018: AMD have updated their support article for users struggling to boot the latest Ryzen 2000-seriesRaven Ridge desktop APUs. If you are stuck with a never-ending boot issue, you can now request a boot kit straight from AMD to get setup.
AMD have been quite clear since the get-go that, while all their AM4 motherboards will be compatible with the new Raven Ridge APUs, such as the impressive Ryzen 5 2400G, and upcoming Ryzen 2 CPUs, you'll need a BIOS update to do so.
AMD and their board partners have started to indicate these updated boards with a sticker saying "AMD Ryzen Desktop 2000 Ready". Of course, if you missed this, or your hand was forced due to stock or availability, then you are left in a bit of a pickle. Your processor won't boot because your motherboard isn't updated, but your motherboard isn't updated because your processor won't boot. Just don't say you weren't warned.
Luckily, AMD are offering a get-out-of-jail-free card to users suffering in Dante's first boot cycle of hell, you can request a boot kit that will let you install the compatible BIOS onto your system. It's not yet clear exactly what the kit entails, but if it works, it works.
This boot kit is available through AMD's support page 'fee of charge', but we can only assume AMD meant free, right? You'll also need to provide a valid AMD processor OPN and serial number, but with all this to hand, AMD will ship the kit out to you that will allow you to boot your system and get everything updated. There is no word of regional restrictions on AMD's support page, but their RMA page seems to indicate the kit is available to most locales.
Otherwise, you could head down to your local PC store and ask if they have a compatible first-generation Ryzen chip and would be willing to help you out, or send it out to the manufacturer for a potentially costly fix or replacement – depending on the brand and your location.


German supplier Benteler Automotive opened a factory in the northeast China city of Shenyang to supply suspension parts to BMW Group's Chinese joint venture.
The $20 million (132 million yuan) plant will employ more than 140 people.
The 9,400-square-meter (101,181-square-foot) plant is Benteler's ninth in China. The other eight factories build parts for chassis, bodies, engines and exhaust systems.
BMW Brilliance Automotive, of Shenyang, is a 50-50 partnership between BMW and Brilliance China Automotive Holdings. It produces the BMW 3- and 5-series sedans and the X1 crossover.
This year, the joint venture opened an engine plant and foundry in Shenyang to build turbocharged three- and four-cylinder engines for locally produced BMW cars.


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Born in 1985 Ben lived in Peterborough most of his civilian life. He joined the Army at 16 and served in29 Commando Regiment as a Bombardier. Then he became a builder and carpenter.
In February 2017 on a trip to the Lake District, Ben could not afford a painting he loved in a local gallery, so decided on his return, to paint one for himself. this sparked a passion and a desire to paint. Being self-taught, and now venturing into art full time, he is greatly appreciative of his supportive family.


Whenever we see penguins, it will be either floating in the water or sliding with their bellies. Naturally a question will come in our mind, whether or not penguin is a bird or fish. We have to describe penguin as a sort of flightless bird that devote big component of its life in the sea. That is the purpose why advertisers use penguins as their models in advertisements associated to cooling equipments.
Some of the physique features that support the bird to adapt to the aquatic lifestyle in the sea are its wings, feet and tail. The wings of penguins are just like airplane wings and are named flippers. They use flippers to propel through water and at that time penguin got an appearance of a bird flying in the water. Meanwhile, penguin's feet are internet shaped and quite short, so they can move in the water very rapidly. The game they play in the ice is known as tobogganing. They use its feet and flippers for this, which will support them to move swiftly and glide smoothly over the ice making use of their belly.
They look just like a man wearing a dark coat, as their back portion is black, even though bellies are white. The top layer of a penguin's skin got thick layers of stiff feathers and this assists it to really feel warm, even they are gliding by means of ice covered mountains. Just like peacocks, penguins also go by means of molting course of action, exactly where they shed their feathers and grow new ones. They even got an oil secreting gland above their tail that aids it to make its outer thick coat more wind proof and water proof.
The wings of penguins are just like airplane wings and are named flippers.
Primarily based on their size, color, habits and habitats nearly seventeen penguin species are identified in the world. Amongst them, Emperor penguins are the largest and they will have a black cap, blue-grey neck, orange colored patch on the ear and yellow breasts. Meanwhile, the most common penguin in Antarctica is the king penguin and is regarded as the second largest one particular in that area. Some other penguin varieties are Galapagos penguin, Adelie penguin,Chinstrap penguin, Gentoo penguin, Rockhopper penguin and Macaroni penguin.
Although motion pictures based on penguins are quite rare, 'Happy Feet' is a incredibly good film to watch in a digital tv powered by Dish Tv. The story is primarily based on a tiny penguin named Mumble, who got an wonderful talent for tap dancing. All emperor penguins are supposed to have a heart song to express their true like in their country. But poor Mumble got a terrible voice and that grow to be a dilemma for it to come across its correct love. As it does not have a heart song it became an unacceptable bird in his neighborhood. The movie got a excellent clapping from little ones about the planet.
Most of these elfin creatures are located in the southern hemisphere. The Galapagos penguins can be discovered in the Galapagos Islands, although certain other species can be observed in the islands near New Zealand and Australia, the coast of South America and the icy waters close to Antarctica. Lately a container ship Rena was grounded on a reef on Mount Maunganui beach in New Zealand and the oil spilled from it affected some 360 birds living there, like practically 40 penguins. Wildlife rescuers did a good job there by cleaning and scrubbing these penguins and releasing it to its beach habitat immediately after clearing the oil coat.
Whenever we see penguins, it will be either floating in the water or sliding with their bellies. Naturally a query will come in our mind, no matter if penguin is a bird or fish. We have to describe penguin as a form of flightless bird that commit key portion of its life in the sea. That is the explanation why advertisers use penguins as their models in advertisements associated to cooling equipments.
Insects Lower Classifications – The best layer of a penguin's skin got thick layers of stiff feathers and this helps it to really feel warm, even they are gliding by means of ice covered mountains.
Some of the body functions that support the bird to adapt to the aquatic life-style in the sea are its wings, feet and tail. The wings of penguins are just like airplane wings and are called flippers. They use flippers to propel by means of water and at that time penguin got an appearance of a bird flying in the water. Meanwhile, penguin's feet are web shaped and pretty short, so they can move in the water incredibly swiftly. The game they play in the ice is known as tobogganing. They use its feet and flippers for this, which will help them to move swiftly and glide smoothly over the ice using their belly.
Anytime we see penguins, it will be either floating in the water or sliding with their bellies. Naturally a question will come in our thoughts, regardless of whether penguin is a bird or fish. We have to describe penguin as a sort of flightless bird that invest big aspect of its life in the sea. That is the cause why advertisers use penguins as their models in advertisements related to cooling equipments.
They look just like a man wearing a dark coat, as their back portion is black, while bellies are white. The prime layer of a penguin's skin got thick layers of stiff feathers and this assists it to really feel warm, even they are gliding by means of ice covered mountains. Just like peacocks, penguins also go via molting method, exactly where they shed their feathers and develop new ones. They even got an oil secreting gland above their tail that helps it to make its outer thick coat additional wind proof and water proof.
Though motion pictures based on penguins are extremely uncommon, 'Happy Feet' is a pretty good movie to watch in a digital tv powered by Dish Tv. The story is based on a small penguin named Mumble, who got an wonderful talent for tap dancing. All emperor penguins are supposed to have a heart song to express their correct appreciate in their nation. But poor Mumble got a terrible voice and that come to be a issue for it to obtain its true adore. As it does not have a heart song it became an unacceptable bird in his neighborhood. The film got a very good clapping from little ones around the globe.
Primarily based on their size, colour, habits and habitats nearly seventeen penguin species are identified in the globe. Amongst them, Emperor penguins are the biggest and they will have a black cap, blue-grey neck, orange colored patch on the ear and yellow breasts. Meanwhile, the most common penguin in Antarctica is the king penguin and is regarded as the second largest a single in that region. Some other penguin varieties are Galapagos penguin, Adelie penguin,Chinstrap penguin, Gentoo penguin, Rockhopper penguin and Macaroni penguin.
Most of these elfin creatures are found in the southern hemisphere. The Galapagos penguins can be found in the Galapagos Islands, though certain other species can be observed in the islands near New Zealand and Australia, the coast of South America and the icy waters close to Antarctica. Not too long ago a container ship Rena was grounded on a reef on Mount Maunganui beach in New Zealand and the oil spilled from it impacted some 360 birds living there, such as practically 40 penguins. Wildlife rescuers did a terrific job there by cleaning and scrubbing those penguins and releasing it to its beach habitat soon after clearing the oil coat.
Some of the body attributes that aid the bird to adapt to the aquatic way of life in the sea are its wings, feet and tail. The wings of penguins are just like airplane wings and are named flippers. They use flippers to propel through water and at that time penguin got an appearance of a bird flying in the water. Meanwhile, penguin's feet are web shaped and very short, so they can move in the water extremely speedily. The game they play in the ice is known as tobogganing. They use its feet and flippers for this, which will help them to move swiftly and glide smoothly over the ice applying their belly.
How Many Species Of Insects Are There As Of 2017 – If you do not reside in the city and you have observed the mess that the globe is becoming, you might have believed about raising your own chickens to reduce your meals bill a bit.
Some of the physique attributes that aid the bird to adapt to the aquatic lifestyle in the sea are its wings, feet and tail. The wings of penguins are just like airplane wings and are called flippers. They use flippers to propel through water and at that time penguin got an look of a bird flying in the water. Meanwhile, penguin's feet are internet shaped and incredibly quick, so they can move in the water really immediately. The game they play in the ice is known as tobogganing. They use its feet and flippers for this, which will aid them to move swiftly and glide smoothly more than the ice using their belly.
They appear just like a man wearing a dark coat, as their back portion is black, though bellies are white. The top rated layer of a penguin's skin got thick layers of stiff feathers and this aids it to really feel warm, even they are gliding by means of ice covered mountains. Just like peacocks, penguins also go through molting approach, where they shed their feathers and develop new ones. They even got an oil secreting gland above their tail that helps it to make its outer thick coat extra wind proof and water proof.
Even though films primarily based on penguins are incredibly uncommon, 'Happy Feet' is a quite good film to watch in a digital tv powered by Dish Television. The story is primarily based on a tiny penguin named Mumble, who got an incredible talent for tap dancing. All emperor penguins are supposed to have a heart song to express their accurate really like in their country. But poor Mumble got a terrible voice and that become a challenge for it to uncover its correct adore. As it doesn't have a heart song it became an unacceptable bird in his neighborhood. The film got a good clapping from kids around the globe.
Whenever we see penguins, it will be either floating in the water or sliding with their bellies. Naturally a query will come in our mind, regardless of whether penguin is a bird or fish. We have to describe penguin as a variety of flightless bird that devote major element of its life in the sea. That is the cause why advertisers use penguins as their models in ads associated to cooling equipments.
Recently a container ship Rena was grounded on a reef on Mount Maunganui beach in New Zealand and the oil spilled from it impacted some 360 birds living there, which includes nearly 40 penguins. Wildlife rescuers did a terrific job there by cleaning and scrubbing those penguins and releasing it to its beach habitat immediately after clearing the oil coat. Most of these elfin creatures are discovered in the southern hemisphere. The Galapagos penguins can be identified in the Galapagos Islands, though certain other species can be noticed in the islands close to New Zealand and Australia, the coast of South America and the icy waters near Antarctica.
Primarily based on their size, colour, habits and habitats almost seventeen penguin species are identified in the planet. Amongst them, Emperor penguins are the biggest and they will have a black cap, blue-grey neck, orange colored patch on the ear and yellow breasts. Meanwhile, the most typical penguin in Antarctica is the king penguin and is regarded as the second biggest 1 in that region. Some other penguin varieties are Galapagos penguin, Adelie penguin,Chinstrap penguin, Gentoo penguin, Rockhopper penguin and Macaroni penguin.
Gardner Snakes – If you do not live in the city and you have noticed the mess that the planet is becoming, you could possibly have thought about raising your personal chickens to reduce your food bill a bit.
Although films primarily based on penguins are pretty uncommon, 'Happy Feet' is a very good film to watch in a digital television powered by Dish Television. The story is primarily based on a tiny penguin named Mumble, who got an awesome talent for tap dancing. All emperor penguins are supposed to have a heart song to express their correct adore in their nation. But poor Mumble got a terrible voice and that turn out to be a problem for it to come across its true really like. As it does not have a heart song it became an unacceptable bird in his neighborhood. The film got a great clapping from youngsters about the world.
Some of the physique options that enable the bird to adapt to the aquatic life style in the sea are its wings, feet and tail. The wings of penguins are just like airplane wings and are named flippers. They use flippers to propel through water and at that time penguin got an appearance of a bird flying in the water. Meanwhile, penguin's feet are internet shaped and incredibly quick, so they can move in the water incredibly immediately. The game they play in the ice is called tobogganing. They use its feet and flippers for this, which will support them to move swiftly and glide smoothly more than the ice working with their belly.
They appear just like a man wearing a dark coat, as their back portion is black, when bellies are white. The top layer of a penguin's skin got thick layers of stiff feathers and this helps it to really feel warm, even they are gliding through ice covered mountains. Just like peacocks, penguins also go through molting approach, where they shed their feathers and grow new ones. They even got an oil secreting gland above their tail that aids it to make its outer thick coat much more wind proof and water proof.
Based on their size, color, habits and habitats almost seventeen penguin species are identified in the world. Among them, Emperor penguins are the biggest and they will have a black cap, blue-grey neck, orange colored patch on the ear and yellow breasts. Meanwhile, the most frequent penguin in Antarctica is the king penguin and is regarded as the second largest 1 in that area. Some other penguin varieties are Galapagos penguin, Adelie penguin,Chinstrap penguin, Gentoo penguin, Rockhopper penguin and Macaroni penguin.
Whenever we see penguins, it will be either floating in the water or sliding with their bellies. Naturally a query will come in our thoughts, whether or not penguin is a bird or fish. We have to describe penguin as a kind of flightless bird that commit big aspect of its life in the sea. That is the reason why advertisers use penguins as their models in advertisements associated to cooling equipments. Most of these elfin creatures are located in the southern hemisphere. The Galapagos penguins can be found in the Galapagos Islands, although specific other species can be seen in the islands near New Zealand and Australia, the coast of South America and the icy waters near Antarctica.
Though movies based on penguins are very uncommon, 'Happy Feet' is a incredibly good movie to watch in a digital tv powered by Dish Tv. The story is primarily based on a tiny penguin named Mumble, who got an wonderful talent for tap dancing. All emperor penguins are supposed to have a heart song to express their correct adore in their nation. But poor Mumble got a terrible voice and that turn into a dilemma for it to find its true really like. As it doesn't have a heart song it became an unacceptable bird in his neighborhood. The movie got a very good clapping from little ones about the planet.
Black Snakes In Florida – If you do not reside in the city and you have seen the mess that the globe is becoming, you may well have believed about raising your personal chickens to decrease your food bill a bit.


text:
import { Box, Flex } from "@artsy/palette"
import AutocompleteListMetaphysics from "client/components/autocomplete2/list_metaphysics"
import React from "react"

export const AdminAppearances = () => {
  return (
    <>
      <Flex flexDirection={["column", "row"]}>
        <Box width={["100%", "50%"]} pr={[0, 2]}>
          <AutocompleteListMetaphysics
            field="fair_programming_ids"
            label="Fair Programming"
            model="fairs"
            placeholder="Search by fair name..."
          />
        </Box>

        <Box width={["100%", "50%"]} pl={[0, 2]}>
          <AutocompleteListMetaphysics
            field="fair_artsy_ids"
            label="Artsy at the Fair"
            model="fairs"
            placeholder="Search by fair name..."
          />
        </Box>
      </Flex>

      <Flex flexDirection={["column", "row"]}>
        <Box width={["100%", "50%"]} pr={[0, 2]}>
          <AutocompleteListMetaphysics
            field="fair_about_ids"
            label="About the Fair"
            model="fairs"
            placeholder="Search by fair name..."
          />
        </Box>

        <Box width={["100%", "50%"]} pl={[0, 2]}>
          <AutocompleteListMetaphysics
            field="biography_for_artist_id"
            label="Extended Artist Biography"
            model="artists"
            placeholder="Search by artist name..."
            type="single"
          />
        </Box>
      </Flex>
    </>
  )
}