Too little and too late for hundreds of thousands of dead and wounded, not to mention the refugees. The cancer must be killing the dominate (in his case) scumbag-liar part of the brain.
Maybe he will come clean about Vietnam POWs left to rot.
Republican Sen. John McCain of Arizona admitted Friday that the Iraq War was a mistake.
This entry was posted on May 27, 2018 at 9:29 am	and is filed under Uncategorized. Tagged: Iraq, John McCain, POW Cover-Up, WMD. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.


text:
Steamboat female athlete of the year: Natalie Bohlmann thrives in team, individual settings
Ben Ingersoll
Steamboat Springs' Natalie Bohlmann, left, battles Glenwoood Springs' Carol Stepniewski in April.
Natalie Bohlmann enjoys her two sports for very different reasons.
The Sailors sophomore thrives on the individualism of Nordic skiing. She used self-motivation to push herself along cross-country courses to the best season of any Steamboat girl this year.
And on the soccer field — her favorite sport, Bohlmann said — team camaraderie resonates deeply with the two-year letter winner.
Regardless of the differences between Nordic skiing and soccer, Bohlmann knows her way around each, notching one of the top years a Steamboat Springs High School girl had in 2013-14, making her shine as the Sailors' female athlete of the year.
In the winter, Bohlmann skied her way to 10th overall at the state championship classic skiing race. She followed up that performance with a fourth-place finish in the state skate ski finals.
The sophomore capped off the year by leading the Sailors in goals scored on the soccer field, helping Steamboat finish 9-6.
But again, the young leader is quick to deflect attention in a team-sport atmosphere. She said next year isn't about scoring the most goals or being the top athlete, it's about getting the Sailors back to the playoffs in soccer after missing out on postseason play for the first time in a decade.
"I think it would be cool to build off of what we did this season and work even harder to accomplish more," she said. "I'd like to lead that."
Halfway through her high school career, she's already racking up the accolades, including a Class 4A Western Slope first-team all-league selection this year.
"I don't think about (accolades) while they're happening," she said. "But someday, I'll look back and see it's really cool what I've accomplished."

redpajama_set_name: RedPajamaCommonCrawl


text:
Cannabis use and earlier onset of psychosis: a systematic meta-analysis
Arch Gen Psychiatry. 2011 Jun;68(6):555-61. doi: 10.1001/archgenpsychiatry.2011.5. Epub 2011 Feb 7.
Matthew Large 1 , Swapnil Sharma, Michael T Compton, Tim Slade, Olav Nielssen
1 Mental Health Service, Prince of Wales Hospital, Sydney, Australia. mmbl@bigpond.com
DOI: 10.1001/archgenpsychiatry.2011.5
Context: A number of studies have found that the use of cannabis and other psychoactive substances is associated with an earlier onset of psychotic illness.
Objective: To establish the extent to which use of cannabis, alcohol, and other psychoactive substances affects the age at onset of psychosis by meta-analysis.
Data sources: Peer-reviewed publications in English reporting age at onset of psychotic illness in substance-using and non-substance-using groups were located using searches of CINAHL, EMBASE, MEDLINE, PsycINFO, and ISI Web of Science.
Study selection: Studies in English comparing the age at onset of psychosis in cohorts of patients who use substances with age at onset of psychosis in non-substance-using patients. The searches yielded 443 articles, from which 83 studies met the inclusion criteria.
Data extraction: Information on study design, study population, and effect size were extracted independently by 2 of us.
Data synthesis: Meta-analysis found that the age at onset of psychosis for cannabis users was 2.70 years younger (standardized mean difference = -0.414) than for nonusers; for those with broadly defined substance use, the age at onset of psychosis was 2.00 years younger (standardized mean difference = -0.315) than for nonusers. Alcohol use was not associated with a significantly earlier age at onset of psychosis. Differences in the proportion of cannabis users in the substance-using group made a significant contribution to the heterogeneity in the effect sizes between studies, confirming an association between cannabis use and earlier mean age at onset of psychotic illness.
Conclusions: The results of meta-analysis provide evidence for a relationship between cannabis use and earlier onset of psychotic illness, and they support the hypothesis that cannabis use plays a causal role in the development of psychosis in some patients. The results suggest the need for renewed warnings about the potentially harmful effects of cannabis.
Age of Onset*
Marijuana Abuse / complications*
Marijuana Abuse / epidemiology
Marijuana Abuse / psychology*
Psychotic Disorders / epidemiology
Psychotic Disorders / etiology*
Psychotic Disorders / psychology


text:
Oklahoma faces backlash over botched executions, history of lethal injection
Zaria Oates
Oklahoma was the first U.S. state to legalize lethal injection for executions as an alternative to electrocution, hanging and the firing squad, but lethal injections are now being referred to as just as cruel and unusual as the other methods.
Abe Bonowitz, director of Death Penalty Action, provided the history behind Oklahoma's 1977 switch to lethal injection as Julius Jones faces this execution style on Nov. 18 if he is not granted clemency.
"The idea behind lethal injection was to make it easier for the people witnessing," Bonowitz said. "It's not about the prisoner. It's about the witnesses and are they going to be traumatized?"
In the five years before Oklahoma's death penalty was suspended in 2015, there were nearly four thousand fewer violent crimes than in the six years executions were paused. But Bonowitz asserts that executions don't impact crime rates.
"The death penalty is not a detterent," Bonowitz said. "You have to be paying attention to the news to even know that executions are happening."
Bonowitz said even when people are aware of executions, the practice doesn't stop violent crimes.
"There was a woman named Karla Faye Tucker, this was a woman being executed," Bonowitz said. "You had to be living under a rock not to know that that execution was taking place. The very next day, a policeman was murdered just a couple hundred miles from where Karla Faye Tucker was executed."
Robert Dunham works for the non-partisan Death Penalty Information Center. The DPIC says Jones' scheduled execution should be postponed after last week's execution of John Grant. Witnesses said he vomited and convulsed before he died.
"When it comes to Oklahoma, one would hope that with a fourth botched execution they'll step back and say OK, 'we're going to put this back on hold again'," Dunham said.
But so far, there is no indication Oklahoma will suspend the execution procedure it pioneered 44 years ago.
Gaylord News is a Washington-based reporting project of the University of Oklahoma Gaylord College of Journalism and Mass Communication. For more stories from Gaylord News visit gaylordnews.net.
is a student at the University of Oklahoma Gaylord College of Journalism and Mass Communication and a Washington, D.C. reporter for Gaylord News.
https://twitter.com/ZARIAOATESTV
https://www.linkedin.com/in/zaria-oates-864241178
Oklahoma tribal leaders say synergy felt at Tribal Summit
House committee meets to consider historic Cherokee delegate
Thousands of Native Americans Attend Long Awaited Veteran's Memorial Dedication
The future of Native children's custody rests with divided Supreme Court
Tribes say Supreme Court adoption case could upend decades of federal Indian law
Stitt's reelection hopes uncertain in dead heat Oklahoma governor's race
Gen Z impact and female registration surge unsettling Tuesday's outcome
Candidates counting on young people to go to the polls
Tribes and organizations seek to maximize Native voter turnout
Tribes seek to lead way on opioid treatment
All Gaylord News Picks Reader Picks Sort: Newest


text:
Marvel At The Deep Discounts And Slick Deals On EVs That Too Few Folks Want
If you have been looking for a cheap green car, or if you love bargains on cars that hardly anyone wants, listen up.…
José Rodríguez Jr.
The 2021 Toyota Mirai Will Be Over $9,000 Cheaper Than 2020 Models
The second-generation Toyota Mirai is a car to consider if you can get it where you live and have a slightly twisted…
The 2021 Toyota Mirai Deserves More Than A Little Consideration
We've already mostly seen the second-generation Toyota Mirai, but now we know most of its features, too. It has a…
Toyota Pivots To Target Coveted Pope Market With The Mirai
It's safe to say that Toyota's hydrogen-powered cars haven't exactly been flying off dealer lots. This has a lot to…
Jason Torchinsky
Beyond CarsBoatlopnik
Hydrogen Is A Bad Car Fuel, But It's The Perfect Boat Fuel
There are issues inherent with hydrogen as a fuel for cars. It is incredibly expensive and energy intensive to…
Bradley Brownell
Here's Why James May Thinks There's Room In Enthusiasts' Hearts For Electric And Hydrogen Cars
Of the Three Amigos, I've always found James May to be the most progressive. His interests are wide-ranging and he…
Kristen Lee
The 2021 Toyota Mirai Is Much Less Ugly, Rear-Wheel Drive And Still Runs On Hydrogen
There's a new Toyota Mirai out for 2021, and visually it's a vast improvement over the previous Mirai, which was, to…
Behold, The Most Mysterious Used Car For Sale In America
Our world is generally pretty well understood. There are no more undiscovered continents, we're in pretty much…
Toyota Mirai Goes 312 Miles Before Having To Hunt For A Hydrogen Station
Toyota's hydrogen fuel cell-powered Mirai is going on sale later this fall, and for $58,325, you can drive the fuel…
How Do You Build The Hydrogen-Powered Toyota Mirai? By Hand.
In the same factory where they used to construct the spectacular $400,000 Lexus LFA, workers now assemble the…
Raphael Orlove
The Next Prius? What It's Like To Drive Toyota's $57,000 Hydrogen Car
Hydrogen fuel cells are a cool technology that were perfected to enable humans to go to the moon. Now that they're…
The Hydrogen Toyota Mirai Gets Three Years Of Free Fuel, Is Crazy Ugly
Toyota will launch their mid-size hydrogen sedan next fall, with a range of up to 300 miles, 153 horsepower and a…
Máté Petrány


text:
Q: A word for a rule that isn't strictly enforced since proper enforcement would result in the expulsion/paralysis of too many members? Is there a word for a rule that is not strictly enforced since proper enforcement of that rule would result in the expulsion/paralysis of too many members?
E.g., In "Game of Thrones" there is a part where a character says something to the effect of "If we beheaded all the watchmen who go to Molestown we'd only have ghosts manning the wall.".

A: One term is dead letter.

A law or treaty which has not been repealed but is ineffectual or defunct in practice.

'Environmental law, just like any other, is a dead letter if not enforced.'

Oxford

OED has the term first used with this meaning in 1663 following an earlier more literal meaning, "A writing, etc. taken in a bare literal sense without reference to its 'spirit', and hence useless or ineffective."
It's likely that letter here means much the same as letter in letter of the law, rather than "item of mail."

A: a loosely enforced rule is:  TDF
made less strict; relaxed:
Thus:

The rule is loosely enforced, as proper enforcement would result
  in the expulsion/paralysis of too many members?

redpajama_set_name: RedPajamaStackExchange


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text:
Q: Docker for Azure I have a docker image which is to be moved to Azure but I am confused with options as of now Azure has provided forllowing option
1: Docker VM extension(which require TLS certs)
2: Docker Container instance(still in preview)
3. Docker Azure Driver(looks like an easy to configure docker over VM)
Which option should I use and is there is any more easy to use option for docker

A: The Docker Azure driver and docker-machine options seem to be straight forward for your case; here's a step by step guide on how to get started with both: https://learn.microsoft.com/en-us/azure/virtual-machines/linux/docker-machine , and the Docker Azure driver documentation can be found here: https://docs.docker.com/machine/drivers/azure/  .
-Adam

A:
I have a docker image which is to be moved to Azure

If you just want to do some test with Azure docker, as Adam said, we can use Docker Azure driver or docker machine, create an Azure VM and install docker on it. You can follow Adam's links to do that. We can use VM's public IP address to access the docker, works like bridge.

Docker Container instance(still in preview)

You are right, for now, Azure container service(AKS) still in preview, AKS is a managed kubernetes in Azure.
We can use Azure container service(unmanaged,ACS) to create k8s cluster or swarm or DC/OS. In the cluster, we can expose the service to internet with Azure Load Balancer.
So, if you just want to test docker with Azure, maybe we should use Docker machine or Docker Azure Driver. (low cost)
If you want to create a docker for production, maybe we should use ACS cluster to avoid single point of failure.
Hope this helps:)


text:
(, ; plural ,  ) was an honorific used by the ancient Greeks to render the Roman imperial title of . The female form of the title was  (). It was revived as an honorific in the 11th-century Byzantine Empire, and came to form the basis of a new system of court titles.  From the Komnenian period onwards, the Byzantine hierarchy included the title sebastos and variants derived from it, like , , , and .

History
The term appears in the Hellenistic East as an honorific for the Roman emperors from the 1st century onwards, being a translation of the Latin . For example, the Temple of the Sebastoi in Ephesus is dedicated to the Flavian dynasty. This association also was carried over to the naming of cities in honor of the Roman emperors, such as Sebaste, Sebasteia and Sebastopolis.

The epithet was revived in the mid-11th century—in the feminine form —by Emperor Constantine IX Monomachos () for his mistress Maria Skleraina, to whom he accorded quasi-imperial honours. A number of individuals were qualified as  thereafter, such as Constantine Keroularios, or Isaac Komnenos and his brother, the future emperor Alexios I Komnenos ().

Under the Komnenian emperors
When the latter assumed the Byzantine throne in 1081, he set about to reorganize the old system of court dignities, with  as the basis for a new set of titles—,  and , ,  and —which primarily signalled the closeness of their holders' familial relationship to the emperor, either by blood or by marriage. This process profoundly transformed the very nature of Byzantine aristocracy, with the imposition of an entire class of imperial relatives and associates superimposed on the "traditional" administrative system and the higher officialdom that constituted the Senate. In the words of historian Paul Magdalino, this move further isolated the imperial family from the common people and made them "partners in, rather than executives of, imperial authority". In this context, the scholar L. Stiernon calculated that in the period from the late 11th to the end of the 12th century, 30% of all  belonged to the ruling Komnenos family, 20% to the closely allied Doukas clan, and another 40% to other families of the high aristocracy who intermarried with the Komnenoi, the remaining 10% encompassing both Byzantines as well as foreigners who either intermarried with the imperial family or received the title as an honorific distinction.

Initially the  formed the basis of this new familial aristocracy, with sons of a , a , or a  being  themselves; due to the proliferation of the title, however, under Manuel I Komnenos () a new class of dignitaries was created for the emperor's nephews and cousins, i.e., the sons of higher dignitaries; and the  were relegated to a grade below them, above the . The  were further divided in two groups: the simple  and the . The latter were members of various aristocratic families tied to the emperor via marriage to his female relatives ( means 'son-in-law in Greek). The  thus formed the upper layer of the  class, but should not be confused with the imperial , the actual sons-in-law of the emperor, who were even higher in the hierarchy, ranking above the cousins and nephews and just below the . The forms  ('venerable by all'), and  are also found in seals, inscriptions, and correspondence of the period, but they are merely rhetorical augmentations of the original title , and do not, as was believed by earlier scholars like Gustave Schlumberger, represent distinct and superior ranks. It is notable that among Byzantine , their precedence was not determined by the offices they might bear, but by the degree of their kinship to the emperor.

Later usage

The title was also conferred to foreign rulers, and spread to neighboring, Byzantine-influenced states, like Bulgaria, where a  was the head of an administrative district, and Serbia, where the title was employed for various officials.

In Byzantium itself, the title lost its pre-eminence in the late 12th century, and in the following centuries the  was a title reserved for commanders of ethnic units. By the time pseudo-Kodinos wrote his Book of Offices, shortly after the middle of the 14th century, the sebastos occupied one of the lowest rungs in the imperial hierarchy, coming 78th between the  and the . His court dress was a white  hat with embroideries, a long  of "commonly used silk", and a  hat covered in red velvet and topped by a small red tassel. He bore no staff of office. Earlier lists of offices, such as the appendix to the Hexabiblos, give slightly different ranks, placing him above the governor () of a fortress and of the , and after the .

References

Sources

Further reading

Byzantine imperial titles
Byzantine court titles
Augustus

redpajama_set_name: RedPajamaWikipedia


"John Whitmore delves into his postbag to find widespread dissatisfaction with the motor insurance and repair industry but great praise for mobile services that offer convenience and much lower cost.
Mobile repairers attract near-universal praise for their quality, convenience, service and price. Customers are frequently delighted and recommend them to friends. Their prices work out at approximately half the price of the small repairers. Some are franchises or similar, of which the best known are Trimline Systems (Autopaint 4U) and CPS (Cosmetic Paint Systems); there are a number of others, their names often beginning with the word Dent.
"By entering a number found under my bonnet into his on-board computer, a machine would dispense the exact match of paint to my car. Returning 90 minutes later, the marks were completely untraceable."
"With almost religious reverence, he massaged, cajoled, teased and persuaded not only the dent to disappear but rendered the minor scratch almot invisible. The cost was less than £50."
Autopaint 4U came to my home and managed to paint and repair my wife's Mercedes, my own Land Rover Discovery, my sons VW Golf and daughters Nissan Micra all done on the same day! Excellent, fast and reliable service at a very reasonable cost.
"Recent paint/repair to my 307 Peugeot front wing/bumper. Peter has twice now repaired and painted some unfortunate scrapes. On both occasions I have been able to drop off the car at his unit, then walk a short way to Ditchling garden centre to browse, have a cup of tea in the café, then return to pick up the car all within 2-3 hours. Splendid service, fully recommended."
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Just to say I had a stone chip on the tailgate of my Vauxhall Astra and what a brilliant job you have done Peter. I cannot tell its been repainted as it matches the car perfectly and done within three hours!!. Many thanks.
I am so pleased indeed with the job you did on my KA on Wednesday (25/6/08). I think you are a magician! I will keep your details in case for future use either by me or others.
Autopaint sorted out my annoying stone chips on my Porsche Carrera and also carried out paint repairs on my wife's BMW 23, all done whilst we were away on holiday. We found Peter and Martyn to be extremely reliable, trustworthy and work completed was of the highest standard.
Peter carried out quality paint repairs to my badly damaged rear bumper (split) and side moulding (scuffed), on my BMW touring which is now up for sale and looking great again.
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Autopaint have now carried out very successful repairs to my Mercedes SLK on two separate occasions at my home address. Highly recommended.
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text:
Like Glue
"Like Glue" is a song recorded by Jamaican dancehall artist Sean Paul, from his second album Dutty Rock (2002). The song was produced by Tony "CD" Kelly, and written by Kelly and Sean Paul. Lyrics from "Like Glue" were originally the intro to "Gimme the Light" until Sean Paul expanded it and made it into a full song. Lyrically, the song refers to how Sean Paul doesn't care what people say and that he has to stick to his girlfriends "like glue".
"Like Glue"
Single by Sean Paul
from the album Dutty Rock
"Remix"
(international single release)
CD single, CD maxi
7" single, 12-inch single
K-Licious Music (Jamaica)
VP/Atlantic Records
(U.S. and international)
Anthony Kelly, Sean Paul, Xavier "Flexx" Davidson, Craig "Craig T" Thompson, Alistaire "Alex" McCalla, Roshaun "Bay-C" Clarke
Tony "CD" Kelly
Sean Paul singles chronology
"Get Busy"
(2003) "Like Glue"
(2003) "Breathe"
Following the successful chart performances of "Gimme the Light" and "Get Busy", "Like Glue" was released worldwide on 1 July 2003 as the album's third single. It was distributed through K-Licious Music in Jamaica and through VP and Atlantic Records internationally. The single peaked at #13 on the U.S. Billboard Hot 100 and achieved success worldwide, becoming a top ten hit in Switzerland, UK, Canada and Ireland. The song was listed as the 325th best song of the 2000s by Pitchfork Media.
The single's accompanying music video was directed by Benny Boom. The video earned Paul a "Best International Artist Video" award at the 2003 MuchMusic Video Awards.
The song also has been featured on the 2005 video game, Midnight Club 3: DUB Edition.
Background and video
When Sean Paul appeared on the music channel Flava, he revealed that the song was originally going to be the intro of "Gimme the Light" until he expanded on the lyrics. The song was originally released in 2001 as part of the "Buy Out" rhythm, but became a hit on the strength of Sean Paul's popularity from "Gimme the Light". The song samples T.O.K.'s song "Money 2 Burn" from their album My Crew, My Dawgs. According to Allmusic, the song is influenced by the song "Louie Louie."
The song has two censored words in the edited version: "Trees" in the chorus, when he says "Need a lot of trees up in my head", and "chronic" in verse 2, when he says "After a chronic, we take a drag", both referring to cannabis.
The music video was directed by Benny Boom, and instead of Sean Paul singing verse 3, Paul's brother Jason shouts out dances, and the dancers in the video perform them accordingly. The fourth verse is the repeat of the first verse. Also, there's a verse in Jamaican patois. The song is also played at the end of the "Get Busy" video.
This song is also featured on the soundtrack of the video game Midnight Club 3: DUB Edition. Ivy Queen's single from 2004, "Papi Te Quiero" samples "Like Glue" but blends a reggaeton beat produced by Rafi Mercenario. The English version of "Papi Te Quiero", however, features the original beat of "Like Glue". Both can be found on the platinum edition of Queen's third studio album Diva (2003).
The song ranked #70 on About.com's list of the "Top 100 Pop Songs of 2003".[1] It was also listed as the 325th best song of the 2000s by Pitchfork Media. The music video won the 2003 MuchMusic Video Awards for Best International Artist Video.
"Like Glue" (album version) – 3:53
"Like Glue" (instrumental) – 4:01
CD maxi
"Like Glue (album version)" – 3:53
"Like Glue (instrumental)" – 4:01
"Like Glue (acapella)" – 4:05
"Like Glue" (video version) – 4:14
"Get Busy (Clap Your Hands Now Remix)" (street club long version) – 4:18
Charts and certifications
Weekly charts
Chart (2003/04)
Australia (ARIA)[2] 20
Austria (Ö3 Austria Top 40)[3] 17
Belgium (Ultratop 50 Flanders)[4] 21
Belgium (Ultratop 50 Wallonia)[5] 25
Canada (Nielsen SoundScan)[6] 6
Denmark (Tracklisten)[7] 20
Finland (Suomen virallinen lista)[8] 17
France (SNEP)[9] 31
Germany (Official German Charts)[10] 20
Hungary (Dance Top 40)[11] 11
Ireland (IRMA)[12] 4
Netherlands (Single Top 100)[14] 17
New Zealand (Recorded Music NZ)[15] 18
Switzerland (Schweizer Hitparade)[17] 6
UK Singles (Official Charts Company)[18] 3
US Billboard Hot 100[19] 13
US Hot R&B/Hip-Hop Songs (Billboard)[20] 9
US Rhythmic (Billboard)[21] 10
US Mainstream Top 40 (Billboard)[22] 29
Swiss Singles Chart[23] 38
U.S. Billboard Hot 100[24] 66
Canada (Music Canada)[25]
Gold 5,000^
*sales figures based on certification alone
^shipments figures based on certification alone
Lamb, Bill. "Top 100 Pop Songs of 2003". About.com. Retrieved 27 September 2011.
"Australian-charts.com – Sean Paul – Like Glue". ARIA Top 50 Singles.
"Austriancharts.at – Sean Paul – Like Glue" (in German). Ö3 Austria Top 40.
"Ultratop.be – Sean Paul – Like Glue" (in Dutch). Ultratop 50.
"Ultratop.be – Sean Paul – Like Glue" (in French). Ultratop 50.
"Like Glue", in various singles charts Acharts.us (Retrieved 17 May 2009)
"Danishcharts.com – Sean Paul – Like Glue". Tracklisten.
"Sean Paul: Like Glue" (in Finnish). Musiikkituottajat – IFPI Finland.
"Lescharts.com – Sean Paul – Like Glue" (in French). Les classement single.
"Musicline.de – Sean Paul Single-Chartverfolgung" (in German). Media Control Charts. PhonoNet GmbH.
"Archívum – Slágerlisták – MAHASZ" (in Hungarian). Dance Top 40 lista. Magyar Hanglemezkiadók Szövetsége.
"The Irish Charts – Search Results – Like Glue". Irish Singles Chart.
"Italiancharts.com – Sean Paul – Like Glue". Top Digital Download.
"Dutchcharts.nl – Sean Paul – Like Glue" (in Dutch). Single Top 100.
"Charts.nz – Sean Paul – Like Glue". Top 40 Singles.
"Swedishcharts.com – Sean Paul – Like Glue". Singles Top 100.
"Swisscharts.com – Sean Paul – Like Glue". Swiss Singles Chart.
"Sean Paul: Artist Chart History". Official Charts Company.
"Sean Paul Chart History (Hot 100)". Billboard.
"Sean Paul Chart History (Hot R&B/Hip-Hop Songs)". Billboard.
"Sean Paul Chart History (Rhythmic)". Billboard.
"Sean Paul Chart History (Pop Songs)". Billboard.
2003 Swiss Singles Chart Hitparade.ch Archived 27 December 2013 at the Wayback Machine (Retrieved 17 May 2009)
Pedro. "Longbored Surfer - 2003". LongboredSurfer.com. Retrieved 22 October 2014.
"Canadian single certifications – Sean Paul – Like Glue". Music Canada.
8. Tocy tocy refers to a Austin mini painted in Tango livery and the driver of the mini,James Shaw's nickname's were Jimbo and Jimmy, he decided to turn the washer jets round on the mini and squirt anyone and everyone, hence the verse "wet them up like a fool"
"Like Glue" music video on YouTube
Lyrics of this song at MetroLyrics
Stage One (2000)
Dutty Rock (2002)
The Trinity (2005)
Imperial Blaze (2009)
Tomahawk Technique (2012)
Full Frequency (2014)
The Odyssey Mixtape (2009)
Duttyology (2004)
"Deport Them"
"Gimme the Light"
"I'm Still in Love with You"
"We Be Burnin'"
"Ever Blazin'"
"Temperature"
"Never Gonna Be the Same"
"(When You Gonna) Give It Up to Me"
"Watch Dem Roll"
"So Fine"
"Press It Up"
"Hold My Hand"
"Got 2 Luv U"
"She Doesn't Mind"
"Dream Girl"
"How Deep Is Your Love"
"Touch the Sky"
"Other Side of Love"
"Turn It Up"
"Crick Neck"
"No Lie"
"Body"
"Mad Love"
"Feels Like Home"
"Hey DJ"
"Boasty"
"Fuego"
Featured singles
"Money Jane"
"Bossman"
"Make It Clap (Remix)"
"Come Get Some"
"Breathe"
"Baby Boy"
"Shoomp"
"Slow Wind (Remix)"
"Cry Baby Cry"
"Break It Off"
"Give It to You"
"Push It Baby"
"Come Over (Remix)"
"Feel It"
"Do You Remember"
"Shake Señora"
"Summer Paradise"
"Wine It Up"
"She Makes Me Go"
"What About Us"
"Come On to Me"
"Dangerous Love"
"Bailando"
"Make My Love Go"
"Cheap Thrills"
"Hair"
"Trumpets"
"Rockabye"
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text:
O Nono Doutor é uma encarnação do Doutor, o protagonista da série britânica de ficção científica Doctor Who. Ele é interpretado por Christopher Eccleston durante a primeira temporada de ressurgimento da série, em 2005. Dentro narrativa do programa, o Doutor é um alienígena humanoide viajante do tempo de uma raça conhecida como Senhores do Tempo. Quando ele está gravemente ferido, pode regenerar seu corpo, mas ao fazer isso, ganha uma nova aparência física e, com ela, uma nova personalidade.

A abordagem da equipe de produção para o personagem e a interpretação de Eccleston foram destacadas como sendo intencionalmente diferentes das suas antecessores, com o ator afirmando que seu personagem seria menos excêntrico. Para adaptar-se a um público do século XXI, o Doutor ganhou uma nova companheira, Rose Tyler, que foi projetada para ser tão independente e corajosa como ele mesmo. Ele também viaja brevemente com Adam Mitchell, um jovem gênio e interesseiro, que funciona como uma contraste para os companheiros anteriores, e Jack Harkness, um agente do tempo do século LI. O Doutor, Rose e Jack formam uma equipe unida, mas são separados no final da temporada na qual cada personagem tem que tomar decisões difíceis e enfrentar sacrifícios.

Em 2006, os leitores da Doctor Who Magazine votaram o Doutor de Eccleston como o terceiro Doutor mais popular de toda a série. Tanto a imprensa como os críticos de ficção científica em geral creditam o personagem e Eccleston como essenciais para o restabelecimento da série após o seu hiato entre 1996 e 2005. A interação do personagem com seus arqui-inimigos, os Daleks, foi particularmente elogiada. Eccleston ganhou vários prêmios por sua única temporada, incluindo o National Television Award de melhor ator em 2005.

Aparições

Televisão

O Nono Doutor aparece pela primeira vez no episódio "Rose", onde ele salva uma vendedora de 19 anos de idade chamada Rose Tyler de um ataque Auton na loja de departamentos onde ela trabalha. Depois de Rose ajudar o Doutor a derrotar a Consciência Nestene, ele a convida para viajar com ele na TARDIS.  Em sua primeira viagem, ele a leva para testemunhar a destruição do planeta Terra no ano de cinco bilhões. E é revelado que própria espécie do Doutor, os Senhores do Tempo, foram destruídos e que ele é o último de sua espécie. Na sequência eles visitam Cardiff, em 1869, onde eles encontram o autor Charles Dickens, do qual o Doutor afirma ser um grande fã. Quando confrontado com uma situação de quase morte, o Doutor diz a Rose que ele estava feliz por tê-la conhecido. Ao levar Rose para casa o Doutor acidentalmente retorna à Terra 12 meses depois de quando eles partiram. Por causa de suas ações, ele é tratado como um predador da Internet pela mãe de Rose, Jackie e o namorado dela, Mickey que se tornou suspeito de assassinar Rose.  Depois de Mickey ajuda o Doutor e Rose a derrotar os Slitheen com um disparo de míssil em sua base em 10 Downing Street, o Doutor oferece a Mickey um lugar na TARDIS com eles, mas ele se recusa.No episódio "Dalek" o Doutor encontra um Dalek, embora tivesse acreditado que a raça estaria extinta quando a "Guerra do Tempo" entre os Senhores do Tempo e os Daleks terminou com a aniquilação mútua de ambas as raças, um evento pelo qual o próprio Doutor era responsável. O Doutor tortura o Dalek sobrevivente e, no final do episódio tenta matá-lo a sangue frio. Mas ele se abstém quando Rose chama sua atenção sobre isto.

Adam Mitchell junta-se ao Doutor e Rose, como companheiro no final de "Dalek". No entanto, quando ele tenta contrabandear conhecimento futuro do Satélite Cinco no ano 200.000 de volta para seu próprio tempo em "The Long Game " o Doutor expulsa-o da TARDIS. O Doutor fica irritado com Rose, depois que ele a leva para o evento da morte do seu pai Pete Tyler e ela salva sua vida, causando um paradoxo. No entanto, quando Pete morre para restaurar a linha do tempo, o Doutor mostra compaixão e encoraja Rose a sentar-se ao seu lado enquanto ele morre.

Depois de conhecer o Capitão Jack Harkness, um e ex Agente do Tempo do século LI, em 1941, o Doutor percebe que Jack foi o responsável pela libertação de uma praga nanotecnológica mortal que varreria a raça humana transformando os seres humanos em zumbis com máscaras de gás. Após a resolução da situação Jack se prepara para se sacrificar, mas o Doutor o salva e o convida para a bordo da TARDIS.  Quando o Doutor encontra Blon, a única Slitheen sobrevivente à explosão em Downing Street, ele tem dúvidas sobre se deve ou não mandá-la ao seu planeta natal para ser executada. Durante este episódio o Doutor percebe que ele e Rose se mantiveram em contato com as palavras "Bad Wolf".  No episódio "Bad Wolf" o Doutor, Rose, e Jack encontram-se à mercê da Bad Wolf Corporation com sede no Satélite Cinco. No entanto, o verdadeiro inimigo é revelado como sendo os Daleks, o Dalek Imperador tendo também sobrevivido à Guerra do Tempo e tinha reconstruído a raça Dalek.

O Doutor envia Rose, de volta para o século XXI para protegê-la antes de tentar destruir o exército Dalek. Quando ele percebe que ao fazer isso iria destruir a maior parte do planeta Terra ele é incapaz de fazer isso, proclamando que ele preferia ser um covarde do que um assassino. E Rose tendo absorvido as energias do vortex temporal, é capaz de retornar para o Doutor e destruir os os Daleks. E para salvar Rose de ser morta por abrigar o vórtice temporal, o Doutor remove os efeitos nocivos ao beijá-la. No entanto, os danos a suas células fazem com que ele se regenere e o Décimo Doutor (David Tennant) toma o seu lugar.

As origens do Nono Doutor não são originalmente exploradas em 2005, mas foram exploradas no especial de 50 anos da série, "The Day of the Doctor", em 2013. Como consequência da Guerra do Tempo, o War Doctor (John Hurt) sucumbe à idade. A cena de regeneração é interrompida antes da imagem de Eccleston poder ser totalmente vista, como o ator se recusou a voltar para o episódio. Mas Eccleston continua a ser exibido no banco de videos e fotos ao lado de outros Doutores passados.

Referências

Ligações Externas

"O Nono Doutor", Doctor Who, Reino Unido: BBC.
Doctor Who 2005 Série Trailer

Personagens de ficção científica
Doutores de Doctor Who


text:
\section{Introduction}

Saddle points of the energy functional can play an important role in
non-perturbative dynamics, as they describe the minimal barrier to be taken to
go from one to the next minimum energy configuration. In a heat bath it
determines the critical temperature where barrier crossing is no longer
suppressed.  In a strongly interacting theory it determines the coupling
(or volume) where barrier crossing - quantum mechanically always allowed -
is no longer exponentially suppressed. The saddle point we have in mind here
is the sphaleron~\cite{kli}, which has precisely one unstable mode. This comes
about since one determines the minimal barrier height from a mini-max
procedure~\cite{tau}. When one has a path in field space from one minimum
to the other, one first finds the maximal energy along the path. Then
one minimizes this maximum over the space of all such paths. The unstable
mode will correspond to the decrease along the path that goes through
this mini-max configuration. Given such a saddle point, it is often difficult
to prove in all rigour that it is the one with lowest energy, as there
may be local minima when varying over the space of paths.

Quite often a sphaleron lies on a particular instanton path (i.e. the
action associated to the path is also minimal), but this is by no means
guaranteed. When the instanton has non-trivial parameters (moduli),
mini-max with respect to those parameters provides a candidate sphaleron.
For $S^3$ one can analytically verify that this candidate is a saddle
point of the energy functional with one unstable mode~\cite{baa1}.
For $T^3$ this was studied numerically~\cite{gar1,gar2}. It was not entirely
conclusive from that study if the sphaleron was indeed associated to
the top of an instanton path. With the new algorithm we have been able
to address this issue more directly and conclude that it is the case
for both periodic and twisted~\cite{gar3} boundary conditions.

We will only consider the case of pure SU(2) gauge theories. In an infinite
volume the mini-max procedure is easily seen to lead to zero sphaleron
energy, as instantons can have an arbitrarily large scale parameter.
In a finite volume, the scale parameter can not be larger than the
volume, which stabilizes the mini-max procedure at a non-zero energy.
In the presence of a Higgs field, like in the standard model~\cite{kli},
the scale is set by the Higgs mass.

Before discussing the numerical results we will first describe the
algorithm for finding saddle points and discuss the convergence.
This study is specific for SU(2), but the principle of taking the square
of the equations of motion as the action to be minimized will work equally
well for SU(N) gauge theories, with or without scalar fields. In section 4 we
will discuss results for the class of analytically known solutions with
constant field strength, to test the various aspects of the method.
In section 5 the results for the sphalerons are presented in detail.
We end the paper with some concluding remarks. Readers only interested
in the results should directly go to section 5, perhaps after section 2.

\section{The algorithm}

Consider the square of the equations of motion summed over its
variables as a functional. Its absolute minima are by construction
solutions of the equations of motion. Stability is now obvious, but
also follows from the fact that at such a solution the Hessian of the
new functional is (twice) the square of the Hessian of the energy (or action)
functional. In the $n$-dimensional continuum we have the following expressions
for the two functionals
\begin{equation}
S=-{\scriptstyle{{1\over 2}}}\int d_nx~{\rm Tr}(F_{\mu\nu}^2(x))\quad,\label{eq:Scon}
\end{equation}
\begin{equation}
\hat S=-2\int d_nx~{\rm Tr}(({\cal D}_\mu F_{\mu\nu}(x))^2)\quad\label{eq:hScon},
\end{equation}
with $F_{\mu\nu}(x)=\partial_\mu A_\nu(x)-\partial_\nu A_\mu(x) +[A_\mu(x),
A_\nu(x)]$ the Yang-Mills field strength in terms of the anti-hermitian
vector potential $A_\mu(x)$ and ${\cal D}_\mu=\partial_\mu+{\rm ad} A_\mu(x)$ the
covariant derivative.

This was considered in the past for constructing monopole solutions
in three dimensions~\cite{dun}, but the process of finding the minimum
of $\hat S$ was not gauge invariant. Like for the Wilson action~\cite{wil}
\begin{equation}
S=N^{4-n}\sum_{x,\mu,\nu} {\rm Tr}\left(1-\plaq\right)=N^{4-n}\sum_{x,\mu,\nu}
{\rm Tr}\left(1-U_\mu(x)U_\nu(x+\hat{\mu})U_\mu^{\dagger}(x+\hat{\nu})
U_\nu^{\dagger}(x)\right)\quad,\label{eq:Swil}
\end{equation}
a gauge invariant lattice action for the square of the equations of motion
is not too difficult to write down~\cite{sij}
\begin{equation}
\hat S=N^{6-n}\sum_{x,\mu}
{\rm Re~}{\rm Tr}(\tilde{U}_\mu(x)\tilde{U}_\mu^{\dagger}(x)-(U_\mu(x)
\tilde{U}_\mu^{\dagger}(x))^2)\quad,\label{eq:Sarj}
\end{equation}
where $U_\mu(x)$ is a group element on the link that runs from $x$ to
$x+\hat{\mu}$ ($\hat\mu$ is the unit vector in the $\mu$ direction),
$N$ is the number of lattice points in one of the directions
and $\tilde{U}_\mu(x)$ is the sum over $2(n-1)$ staples
\begin{equation}
\tilde{U}_\mu(x)=\sum_{\nu\neq\mu}\left(\stapup+\stapdw\right)=
\sum_{\nu\neq\mu}(U_\nu(x)U_\mu(x+\hat{\nu})U_\nu^{\dagger}(x+\hat{\mu})
+U_\nu^{\dagger}(x-\hat{\nu})U_\mu(x-\hat{\nu})U_\nu(x+\hat{\mu}-\hat{\nu}))
,\label{eq:Util} \end{equation}
To derive the lattice equations of motion, we observe that $S$ depends on
$U_\mu(x)$ through the expression:
\begin{equation}
S(U_\mu(x))=2N^{4-n}{\rm Re~}{\rm Tr}(1-U_\mu(x)\tilde{U}_\mu^{\dagger}(x))
\quad.\label{eq:Ssin}
\end{equation}
As $\tilde{U}_\mu(x)$ is independent of $U_\mu(x)$,
it follows that
\begin{equation}
U_\mu(x)\tilde{U}_\mu^{\dagger}(x)-\tilde{U}_\mu(x)
U_\mu^{\dagger}(x)=0\quad.\label{eq:leqm}
\end{equation}
Its square, summed over the lattice, yields $\hat S$. For SU(2) one can
also use the fact that the equations of motion are solved by
\begin{equation}
U_\mu(x)=\pm\tilde{U}_\mu(x)/\|\tilde{U}_\mu(x)\|\quad,\quad
\|\tilde{U}_\mu(x)\|^2\equiv {\scriptstyle{{1\over 2}}}{\rm Tr}(\tilde{U}_\mu(x)\tilde{U}_\mu^{\dagger}
(x))\quad,\label{eq:Ssol} \end{equation}
where only the positive sign will allow for solutions that have a smooth
continuum limit. We can now introduce an error functional~\cite{gar2}
\begin{equation}
\tilde{S}=N^{6-n}\sum_{x,\mu}{\rm Tr}(1-U_\mu(x)\tilde{U}_\mu^{\dagger}
(x)/\|\tilde{U}_\mu(x)\|)\quad.\label{eq:Sus}
\end{equation}
In the continuum limit one finds $\tilde{S}={\scriptstyle{{1\over 16}}}(n-1)^{-2}\hat S$.
This follows in the usual way~\cite{gar1}
from expanding the links in $a=1/N$, defining the relation between
the vector potential in the continuum and the link variables on the
lattice by
\begin{equation}
U_\mu(x)={\rm Pexp}(\int_0^a A_\mu(x+s\hat{\mu})ds)\quad.\label{eq:link}
\end{equation}
One finds after some algebra
\begin{equation}
U_\mu(x)\tilde{U}_\mu^{\dagger}(x)/\|\tilde{U}_\mu(x)\|=\exp\left(-{1\over
{2(n-1)N^3}}\sum_\nu{\cal D}_\nu F_{\mu\nu}(x)+{\cal O}(N^{-5})\right)\quad.\label{eq:exp}
\end{equation}

We can instead of the standard Wilson action use any improved action, containing
$n\times m$ Wilson loops with appropriate couplings, such as the
over-improved actions we have used in ref.~\cite{gar1}
\begin{equation}
S(\varepsilon)=\frac{4-\varepsilon}{3}\sum_{x,\mu,\nu}{\rm Tr}\left(1-\plaq\right)
+\frac{\varepsilon-1}{48}
\sum_{x,\mu,\nu}{\rm Tr} \left(1-\twoplaq\right)\quad.\label{eq:Seps}
\end{equation}
All that needs to be modified in the above is the definition of
$\tilde U_\mu(x)$, which will now also contain the staples associated
with the larger Wilson loops appearing in eq.~(\ref{eq:Seps})
\begin{equation}
\tilde{U}_\mu(x)=\sum_{\nu\neq\mu}{4-\varepsilon\over 3}\left(\stapup+\stapdw\right)+
{\varepsilon-1\over 48}\left(\twoone+\twotwo+\twothree+\twofour\right)\quad.
\label{eq:stap2}
\end{equation}
Like for $S(\varepsilon)$, the $\varepsilon$ dependence drops out in the continuum limit
for $\hat S(\varepsilon)$. However, for $\tilde S$ one finds in this limit
$\tilde S(\varepsilon)=36((n-1)(31-7\varepsilon))^{-2}\hat S$.

The ordinary cooling algorithm~\cite{ber} for both of these actions is
based on the observation that the replacement
\begin{equation}
U_\mu^\prime(x)=\tilde{U}_\mu(x)/\|\tilde{U}_\mu(x)\|\quad,\label{eq:slup}
\end{equation}
minimizes $S(U_\mu(x))$. Note that for $S$ (resp. $S(\varepsilon)$) cooling we should
require the lattice to be at least two (resp. three) sites in each direction.
To completely specify the algorithm one also should prescribe in what order
every link is being updated. One sweep is the process of updating each link
precisely once in that particular order. This defines a mapping
($U\equiv\{U_\mu(x)\}$)
\begin{equation}
U^\prime=T(U)\quad,\quad T=T_{\cal N}\circ T_{{\cal N}-1}\circ\cdots\circ T_2\circ T_1
\quad,\label{eq:swee}
\end{equation}
where $T_i$ is the single-link update of eq.~(\ref{eq:slup}) and the
label $i$ stands for the order in which the ${\cal N}=n N^n$ different links
are being updated. It is actually not difficult to compute exactly by
which amount the action is decreasing. For the single-link update we
find
\begin{equation}
S(U^\prime_\mu(x))-S(U_\mu(x))=-2N^{4-n}\|\tilde{U}_\mu(x)\|~\|U^\prime_\mu(x)-
U_\mu(x)\|^2\quad,\label{eq:delS}
\end{equation}
where we made use of eq.~(\ref{eq:slup}). To find the change
of the action after one sweep, one has to simply add the contributions of
each single-link update keeping in mind, however, that the value of
$\|\tilde{U}_\mu(x)\|$ depends on which links had been updated before.
In the continuum limit $\|\tilde{U}_\mu(x)\|\rightarrow 2(n-1)$ and
to a good accuracy the action changes from one sweep to the next
by the amount
\begin{equation}
S(T(U))-S(U)\sim-4(n-1)N^{4-n}\|T(U)-U\|^2
\equiv-4(n-1)N^{4-n}\sum_{x,\mu}\|U^\prime_\mu(x)-U_\mu(x)\|^2\quad.
\label{eq:dSsw}\end{equation}
For saddle points, which are fixed points of this algorithm, one can
of course lower the action further and the algorithm is necessarily
unstable.

For the functional $\hat S$ any solution corresponds to the absolute
minimum $\hat S=0$, and cooling with this functional should hence not
lead to any instability for the saddle points
of the original action, eqs.~(\ref{eq:Swil},\ref{eq:Seps}).
Unfortunately, it can
be proven that one cannot analytically minimize $\hat S$ as a function
of one of its links. All that is required, however, is that the functional
is lowered if and only if we are not at a fixed point. We will see, as
our intuition tells us, that this is a sufficient condition for
convergence of the algorithm (necessity should be obvious). Let us nevertheless
write down the necessary ingredients to minimize the single-link
dependence of $\hat S$, which was already considered by Van der
Sijs~\cite{sij}. We restrict ourselves in this paper to the
version of $\hat S$ derived from the Wilson action, as the analysis
is prohibitively more complicated for improved actions (for $\tilde S$ the
situation is in either case intractable, because of the appearance of
$\|\tilde U_\mu(x)\|$ in the denominator of eq.(\ref{eq:Sus})).
The analogue of eq.~(\ref{eq:Ssin}) becomes
\begin{equation}
\hat S(U_\mu(x))=N^{6-n}{\rm Re~}{\rm Tr}\left(U^\dagger_\mu(x)W_\mu(x)-3
\sum_{\alpha=1}^{2(n-1)} (U^\dagger_\mu(x)V^\alpha_\mu(x))^2-
(U^\dagger_\mu(x)\tilde{U}_\mu(x))^2\right)~,\label{eq:hSsin}
\end{equation}
where the index $\alpha$ runs over the $2(n-1)$ different staples of
eq.~(\ref{eq:Util}), such that $\tilde{U}_\mu(x)=\sum_\alpha V_\mu^\alpha(x)$.
The expression for $W_\mu(x)$ is a complicated combination of planar and
non-planar Wilson loops, which are best exhibited in terms
of unit vectors $\hat a,~\hat b\in\{\pm\hat 1,\cdots,\pm\hat n\}$
\begin{eqnarray}
W_\mu(x)&=&2\sum_{\stackrel{a\neq-b}{a,b\neq\pm\mu}}\wplaqone-\wplaqonem+
2\sum_{\stackrel{a\neq-\mu}{b\neq\pm\mu,\pm a}}
\wplaqtwo-\wplaqtwom+\wplaqthree-\wplaqthreem\quad,\nonumber
\end{eqnarray}
or in explicit form, using the convention that $U_a(x)\equiv
U_{-a}^\dagger(x-\hat a)$:
\begin{eqnarray}
& &\hskip-16mm W_\mu(x)={{2\sum}_{}}_{a,b\neq\pm\mu;a\neq-b}\{U_a(x)
U_b(x+\hat a)U_\mu(x+\hat a+\hat b)U_b^\dagger(x+\hat a+\hat\mu)
U_a^\dagger(x+\hat\mu)\nonumber\\ & &\hskip-4mm-
U_a(x)U_\mu(x+\hat a)U_b(x+\hat a+\hat\mu)
U_\mu^\dagger(x+\hat a+\hat b)U_b^\dagger(x+\hat a)
U_\mu(x+\hat a)U_a^\dagger(x+\hat\mu)\}\nonumber\\
&&\hskip-4mm+{{2\sum}_{}}_{b\neq\pm\mu,\pm a;a\neq-\mu}\{
U_b(x)U_\mu(x+\hat b)U_a(x+\hat b+\hat\mu)
U_b^\dagger(x+\hat a+\hat\mu) U_a^\dagger(x+\hat\mu)
\nonumber\\ & &\hskip-4mm-U_b(x)U_\mu(x+\hat b)U_b^\dagger(x+\hat\mu)
U_a(x+\hat\mu)U_b(x+\hat a+\hat\mu)U_a^\dagger(x+\hat b+\hat\mu)
U_b^\dagger(x+\hat\mu)\nonumber\\
&&\hskip-4mm+U_a^\dagger(x-\hat a)U_b(x-\hat a)U_a(x+\hat b-\hat a)
U_\mu(x+\hat b) U_b^\dagger(x+\hat\mu)\nonumber\\ & &\hskip-4mm-
U_b(x)U_a^\dagger(x+\hat b-\hat a)U_b^\dagger(x-\hat a)U_a(x-\hat a)
U_b(x)U_\mu(x+\hat b)U_b^\dagger(x+\hat\mu)\}\quad.
\label{eq:mess}
\end{eqnarray}
The sum in this expression is restricted to those combinations of $\hat a$
and $\hat b$ that do not lead to any backtracking in the Wilson loop
defined by the products of the links.
If one tries to minimize $\hat S(U_\mu(x))$ with respect to $U_\mu(x)$
under the constraint $\|U_\mu(x)\|=1$, one finds
\begin{equation}
\lambda U_\mu(x)=M(U_\mu(x))-W_\mu(x)\quad.
\end{equation}
Note that the lattice should now be at least three sites in each direction.
The operator $M$, whose $\mu$ and $x$ dependence was suppressed, is
defined for SU(2) through the equation
\begin{equation}
M(U_\mu(x))\equiv 6\sum_\alpha{\rm Tr}\left(U_\mu(x)V_\mu^\alpha(x)^\dagger\right)
V_\mu^\alpha(x)+2{\rm Tr}\left(U_\mu(x)\tilde{U}_\mu^\dagger(x)\right)
\tilde{U}_\mu(x)\quad,\label{eq:Mdef}
\end{equation}
An attempt to solve for $\lambda$, using the condition that $U_\mu(x)$ is
unitary, leads to a complicated eighth order polynomial, and solving it
numerically for each update would be too costly. Instead, we define
our update by the simple equation
\begin{equation}
U^\prime_\mu(x)={M(U_\mu(x))-W_\mu(x)\over{\|M(U_\mu(x))-W_\mu(x)\|}}\quad,
\label{eq:hSit}
\end{equation}
The new feature is that the r.h.s. still depends on $U_\mu(x)$ and that
we do not obtain a minimum of eq.~(\ref{eq:hSsin}). Nevertheless, this is
harmless since some algebra will show that our update does lower $\hat S$
and as we argued above, and shall show in some more detail in the next
section, this is sufficient. One finds
\begin{equation}
\delta\hat S(U_\mu(x))=-N^{6-n}\left(\|M(U_\mu(x))-W_\mu(x)\|~
\|\delta U_\mu(x)\|^2+<\delta U_\mu(x),M(\delta U_\mu(x))>\right),
\label{eq:dShs}
\end{equation}
where, of course, $\delta U_\mu(x)\equiv U_\mu^\prime(x)-U_\mu(x)$,
$\delta\hat S( U_\mu(x))\equiv \hat S(U_\mu^\prime(x))-\hat S(U_\mu(x))$
and
\begin{equation}
<\delta U_\mu(x),M(\delta U_\mu(x))>=3\sum_\alpha|{\rm Tr}(\delta U_\mu^\dagger(x)
V_\mu^\alpha(x))|^2+|Tr(\delta U_\mu^\dagger(x)\tilde{U}_\mu(x))|^2\quad.
\label{eq:rest}
\end{equation}
In the continuum limit one has $\|M(U_\mu(x))-W_\mu(x)\|
\rightarrow 8(n-1)(2n+1)$.

We note that lattice artefact solutions for $\hat S$
will not be eliminated by eq.~(\ref{eq:hSit}). For example,
putting one link to minus the identity (while keeping all others equal to the
identity) is a fixed point of eq.~(\ref{eq:hSit}). It is, however, not a fixed
point for eq.~(\ref{eq:slup}). In practise we first use
$S$ cooling to bring the energy (or action) down to values not too much
above where we expect a saddle point to occur, after which the $\hat S$
cooling will in general bring it to a smooth solution. One can also
combine the $S$ and $\hat S$ cooling, by replacing $U_\mu(x)$ on the r.h.s.
of eq.~(\ref{eq:hSit}) with $\tilde{U}_\mu(x)/\|\tilde{U}_\mu(x)\|$.
We have not proven that this always lowers $\hat S$ (although it did
in all cases we considered).

\section{Convergence}

We will study the convergence of the algorithm by considering the
linear approximation for the mapping $T$ in eq.~(\ref{eq:swee}).
For $S$ (or $S(\varepsilon)$) cooling, $T_i$ will correspond to eq.~(\ref{eq:slup}),
while for $\hat S$ cooling it will correspond to eq.~(\ref{eq:hSit}).
To warm up we first consider the single-link convergence of the latter,
while keeping all links (except $U_\mu(x)$) fixed. For notational convenience
we drop the $\mu$ and $x$ dependence. Let us assume that $U_o$ is a
fixed point, and $X$ is a small element of the SU(2) Lie-algebra. We now
consider $U=e^XU_o$ and compute $X^\prime$ to linear order in $X$
from $U^\prime\equiv e^{X^\prime} U_o$. After some algebra one finds
\begin{equation}
X^\prime=\|M(U_o)-W\|^{-1}\left(6\sum_\alpha{\rm Tr}(XY_\alpha^\dagger)Y_\alpha
+2{\rm Tr}(X\tilde{Y}^\dagger)\tilde{Y}\right)\quad,\label{eq:scon}
\end{equation}
where $\tilde Y\equiv\sum_\alpha Y_\alpha$ and $Y_\alpha={\rm Im~}(U_o^\dagger
V_\alpha)\equiv{\scriptstyle{{1\over 2i}}}(U_o^\dagger V_\alpha-V_\alpha^\dagger U_o)$.
In the continuum limit this is easily seen to imply that $\|X^\prime\|/\|X\|=
{\cal O}(N^{-4})$ and we have verified that the  single-link iteration is indeed
extremely efficient. One iteration typically brings $U$ already
very close to the minimum of eq.~(\ref{eq:hSsin}) (although we would like to
emphasize again that this is not really important, as the other links
that contribute to $\hat S$ will be changed during the sweep).

With respect to the SU(2) Lie-algebra basis $i\tau_k/2$
($\tau_k$ the Pauli matrices)
eq.~(\ref{eq:scon}) is symmetric, but this is in general no longer true for
the linear approximation of $T$, which we will denote by $dT$. The latter is
determined exactly as above by evaluating $U^\prime=T(e^XU_o)$ and
extracting $X^\prime_\mu(x)$ to linear order in $X$ from
$\exp(X^\prime_\mu(x))=U^\prime_\mu(x)U^o_\mu(x)^\dagger$, where
$U_o$ is the fixed point. Numerically this is easily achieved by
first obtaining the fixed point to a high accuracy, assuming the
last value of $U$ can be equated to $U_o$, and then looking at
the response of one complete sweep to $U=e^XU_o$, if necessary doing
extrapolation in $X\rightarrow 0$ to extract the part linear in $X$.
One might be tempted to prove that $T$ is a so-called contraction,
for which it is required that $\|T(V)-T(U)\|\leq\kappa\|V-U\|$, with
$\kappa<1$. The contraction theorem asserts that in such a case
there is a unique
fixed point of the mapping $T$. At the linear level, near the fixed
point, this would imply that $dT^tdT$ (which is symmetric and positive
definite) has no eigenvalues bigger or equal to one. We know, however,
that the fixed point is neither globally nor locally unique.
Indeed, for all cases we studied there are plenty of eigenvalues bigger
than one. If $dT$ would have been symmetric, there would have also
been eigenvalues of $dT$ that are bigger (in absolute value) than one,
which would clearly be in contradiction with the fact that the action
is lowered with each update.

As $dT$ is not a symmetric matrix, we can not diagonalize it. Still, it
can over the complex numbers be brought to the so-called Jordan-normal
form by a non-singular (complex) transformation $A$
\begin{eqnarray}
dT=A^{-1}J~A\equiv
A^{-1}\pmatrix{J_1& &\ominus\cr &\ddots& \cr\ominus & &J_\ell\cr}A\quad,
\end{eqnarray}
where $J_i$ is the smallest Jordan block, of the form
\begin{eqnarray}
J_i\equiv\pmatrix{\lambda_i&1& &\ominus\cr &\ddots&\ddots& \cr
 & & \ddots&1\cr\ominus& & &\lambda_i\cr}\quad,
\end{eqnarray}
with $\lambda_i$ the eigenvalues. When complex, they occur in conjugate
blocks. Different blocks in our conventions may have the same $\lambda_i$,
and blocks of dimension 1 correspond to eigenvalues with equal left and
right eigenvectors. As $dT^k=A^{-1}J^kA$, any block with $|\lambda_i|<1$
will approach zero. Blocks with $|\lambda_i|>1$ would grow without bound
and can never be consistent with the fact $\hat S$ is a decreasing
functional as a function of the sweeps. What remains are
the eigenvalues $|\lambda_i|=1$. They can, and will, occur for the
following reason. Suppose $X$ is a zero mode for the Hessian of $\hat S$
at the fixed point $U_o$, then $\hat S(e^XU_o)-\hat S(U_o)={\cal O}(X^4)$.
On the other hand using eq.~(\ref{eq:dShs}),
$\hat S(e^{X^\prime}U_o)-\hat S(e^XU_o)={\cal O}(\|X^\prime-X\|^2)$,
which can only be consistent if $X^\prime=X$. Consequently
$X$ is an eigenvector of $dT$ with eigenvalue 1. The converse is also true,
i.e. an eigenvector with eigenvalue 1 is a zero-mode of the Hessian.
Consequently, there are at least as many unit eigenvalues as
there are gauge degrees of freedom. To be precise, there are $3N^n$ minus
the dimension of ${\rm Stab}(U_0)$ unit eigenvalues.
(${\rm Stab}(U_0)$ is the stabilizer of $U_o$, which is
the subgroup of the gauge group that leaves $U_o$ unchanged.  It will
only for very special configurations be non-trivial). It is slightly more
subtle to argue that $|\lambda_i|=1$ is only compatible with eq.~(\ref{eq:hSit})
if $\lambda_i=1$ (one uses the fact that there can be no linear relation
among the eigenvectors, because $A$ is non-singular, which implies that
after each sweep  the amount by which $\hat S$ is lowered is bounded
away from zero if $\lambda_i\neq 1$). It is more easy to rule out that
$\lambda_i=1$ can occur in blocks $J_i$ of dimension greater than 1.
In other words, left and right eigenvectors for the eigenvalue equal to
1 always will coincide.

Clearly, the fixed point is only unique up to the gauge
degrees of freedom and the moduli of the fixed point (by moduli we
mean the gauge invariant parameter space along which $\hat S=0$ and
consequently along which $S$ is degenerate), and there is a one to
one relation with the $\lambda_i=1$. These do not influence the
convergence of the algorithm. But it is important to realize that
one has no control over which fixed point will be obtained.
(The gauge in which the fixed point
is obtained will entirely be determined by the starting configuration, as
$T$ commutes with applying a gauge transformation, i.e. $T({}^gU)={}^gT(U)$.)
It can happen that there are zero-modes of the Hessian that are
not associated to a moduli parameter of the fixed point, i.e. $\hat S$
is no longer degenerate to fourth (or higher) order. In that case one needs
to go beyond the linear approximation of $T$ and convergence is worse than
exponential. In the generic case, however, the rate in which $\hat S$
decreases is determined by the gap $\hat\mu$ in $|\lambda_i-1|$, or after
$k$ sweeps $\hat S$ decreases the next sweep by an amount
$\delta\hat S\sim\exp(-2k\hat\mu)$. The same argument holds of course for
$S$ cooling to a (local) minimum, replacing the gap $\hat\mu$
by its appropriate value ($\mu$) obtained from the mapping $T$ for $S$ cooling.

We can find a relation between the values of these gaps and the minimal
non-trivial
curvature of $S$ at the fixed point if we assume that the gap is due
to a real eigenvalue and its eigenvector coincides with the fluctuation along
the direction of minimal curvature. The associated zero-point frequency will
be denoted by $\omega$ (scaled so as to have the appropriate continuum limit).
At a fixed point $U_o$ one has for
$X_\mu(x)\equiv i\delta A_\mu^a(x)\tau_a/(2N)$
\begin{equation}
S(e^XU_o)=S(U_o)+{\scriptstyle{{1\over 2}}} N^{-n}\sum_{x\mu a;y\nu b}\delta A^a_\mu(x)
H^{ab}_{\mu\nu}(x,y)\delta A^b_\nu(y)+{\cal O}(\delta A^3)\quad,
\label{eq:Hess}
\end{equation}
which provides the definition for the Hessian $H^{ab}_{\mu\nu}(x,y)$
as a $3nN^n$ square matrix, whose eigenvalues will scale to the appropriate
continuum limit for $N\rightarrow\infty$. Hence, $\omega^2$ equals the lowest
non-zero eigenvalue ($\lambda_m$) of the Hessian. It can be easily shown that
for any saddle point of $S$, such that $\hat S(U_o)=0$, the functional
$\hat S$ has the expansion
\begin{equation}
\hat S(e^XU_o)={\scriptstyle{{1\over 2}}} N^{-n}\sum_{x\mu a;y\nu b}\delta A^a_\mu(x)
\hat H^{ab}_{\mu\nu}(x,y)\delta A^b_\nu(y)+{\cal O}(\delta A^3)~,\quad
\hat H = 2 H^2~,\label{eq:hHess}
\end{equation}
where by $H^2$ we mean of course the matrix multiplication of the Hessian of S
with itself. In the same shorthand notation (i.e. dropping the indices),
we assume that under the $S$ cooling $H\cdot X=\lambda_m X$ {\bf and}
$X^\prime\equiv dT\cdot X=(1-\mu) X$, which allows us to calculate the change
in the action both from eq.~(\ref{eq:Hess}), giving $\delta S=-4
\omega^2\mu(1-{\scriptstyle{{1\over 2}}}\mu)N^{2-n}\|X\|^2$ and from eq.~(\ref{eq:dSsw}),
giving $\delta S=-4(n-1)\mu^2N^{4-n}\|X\|^2$. Equating these results gives
\begin{equation}
\mu\sim{\omega^2\over (n-1)N^2}\equiv{\lambda_m\over (n-1)N^2}\quad.
\label{eq:Sgap}
\end{equation}
Similarly, using eqs.~(\ref{eq:dShs},\ref{eq:hHess}) one finds for
cooling with the functional $\hat S$
\begin{equation}
\hat\mu\sim{\omega^4\over (n-1)(2n+1)N^4}\equiv{\lambda_m^2\over (n-1)(2n+1)N^4}
\quad.\label{eq:Shgap}
\end{equation}
This is of course in accordance with ones intuition that the algorithm
can slow down considerably if the functional has directions in
which it is very shallow. Note that our results indicate that slowing
down of the algorithm goes proportional to $N^{2+n}$ for $S$ cooling and
proportional to $N^{4+n}$ for $\hat S$ cooling (in both cases a factor
$N^n$  comes from the number of links to be updated in a single sweep).

Finally, most of the moduli for a continuum solution
are removed by lattice artefacts, although we will discuss in the
next section that in some cases a few of them can survive on a lattice.
An important example of moduli that are removed by lattice artefacts are
those that are related to the translational invariance of the continuum
theory on a torus. As this breaking is due to the fact that integrals in
the continuum are replaced by sums on the lattice, one can easily argue,
as in ref.~\cite{gar1},
that the breaking is very mild and goes as $\exp(-\rho N)$ where $\rho$
is the typical scale of the continuum solution in units of the length $L$ of
the torus. For small lattices these translational modes can
cause some trouble, but those lattices are usually anyhow too small
to extract reliable information about the continuum limit. In ref.~\cite{gar1}
we made judicious use of the dependence of the lattice action on the
moduli parameters, by using the over-improved action $S(\varepsilon=-1)$ (see
eq.~(\ref{eq:Seps})) so as to ensure that the instanton with the maximal
scale parameter allowed by the finite volume is obtained as a
local minimum for the action. With $S$ cooling we could then find rather
accurate lattice solutions, whose scaling to a continuum solution was
studied in detail. One of the advantages of the $\hat S$ cooling is that
one can verify if the fixed point has been reached, as $\hat S$ should
approach zero. This can, of course, also be used as a check under $S$
cooling. For example, we performed for an instanton on an $8^3\times 24$
lattice cooling for about $6600$ sweeps, which allowed for a very precise
exponential fit to the decay of the action. The effective value for $\mu$
deduced from this fit is $9.8\times10^{-5}$ (see fig.~3 of ref.~\cite{gar1}).
We find for the best configuration of this four dimensional instanton (see
table I of ref.~\cite{gar1}) $\hat S(\varepsilon=-1)=4.8\times10^{-4}$ and
for $\tilde S(\varepsilon=-1)=1.3\times10^{-6}$. The results obtained with $S(\varepsilon=0)$
cooling on this configuration (see the same table), lead to
$\hat S(\varepsilon=0)=2.0\times10^{-6}$, $\tilde S(\varepsilon=0)=8.2\times10^{-9}$,
and effectively $\mu\sim10^{-6}$ (which can be understood since for $S(\varepsilon=0)$
the ${\cal O}(a^2)$ corrections to the action are canceled, removing to this order
the dependence of the action on the scale parameter. Using the heuristic
estimates of $\mu$ discussed above we find $\mu\sim\omega^2/(3N^2)$. Since
$\omega(\varepsilon\neq 0)={\cal O}(N^{-1})$, whereas $\omega(\varepsilon=0)={\cal O}(N^{-2})$, one
obtains with $N=8$ the right orders of magnitude for $\mu$).
In the remainder we will discuss the new results for $n=3$,
i.e. results for the Yang-Mills energy functional.

\section{Results: Constant solutions}

We will first describe the case of constant curvature solutions, which are
known analytically both on the lattice and in the continuum. We take from
ref.~\cite{baa2} the following results. If we introduce the so-called twist
tensor $n_{ij}$, which is antisymmetric and takes integer values
(equivalently specified by the magnetic flux $m_i={\scriptstyle{{1\over 2}}}\varepsilon_{ijk}n_{jk}$), the
following configuration is a solution on a symmetric three torus of length
$L=1$, or a symmetric lattice with $N^3$ sites
\begin{equation}
A^o_j(x)=i(-\pi n_{jk}x_k+C_j){\tau_3\over 2}\quad,\quad U^o_i(x)=
\exp(A^o_i(x/N)/N)
\end{equation}
These fields are periodic up to a gauge transformation, to be
specific~\cite{tho1}
\begin{equation}
A_j(x+\hat k)=\Omega_k(x)(A_j(x)+\partial_j)\Omega_k^{-1}(x)\quad,\quad
U_j(x+N\hat k)=\Omega_k(x/N)U_j(x)\Omega_k^{-1}((x+\hat j)/N)\quad,
\label{eq:twist}
\end{equation}
where
\begin{equation}
\Omega_k(x)=\exp({\scriptstyle{{1\over 2}}} i\pi x_j n_{jk}\tau_3)\quad.
\end{equation}
If some of the components of the twist matrix are odd, quantities that
transform in the fundamental representation of SU(2) are anti-periodic,
which can be most easily seen from the Polyakov line
\begin{equation}
P_j(x)={\scriptstyle{{1\over 2}}}{\rm Tr}\{{\rm P}\exp(\int_0^1~ds~A^o_j(x+s\hat j))\Omega_j(x)\}=
\cos({\scriptstyle{{1\over 2}}} C_j-\pi n_{jk}x_k)\quad.\label{eq:Poly}
\end{equation}
The identical result follows on the lattice, after replacing $x$ by $x/N$.
For the continuum Yang-Mills and lattice Wilson actions (energies) we find
\begin{equation}
S=\pi^2\sum_{ij}n_{ij}^2\quad,\quad S=4N^4\sum_{ij}\sin^2
\left({\pi n_{ij}\over 2 N^2}\right)\quad,\label{eq:Sanal}
\end{equation}
which are constant and independent of $\vec C$. The latter form the gauge
invariant moduli of these solutions, both on a lattice and in the continuum.
If the action is non-zero, i.e. the twist is non-trivial, two of
the moduli can also be identified with translational invariance, whereas
translation in the other independent direction (which forms a zero eigenvector
of the twist matrix) leaves the configuration invariant (on the lattice
there is of course no continuous translation invariance). The stabilizer
of this configuration is the subgroup of constant abelian gauge transformations,
which is one dimensional. On the lattice this means that the Hessian has
$3N^3+2$ exact zero-modes and that $dT$ has the same number of unit eigenvalues
(which we verified numerically). In the continuum we can even compute the
Hessian exactly~\cite{baa2}. The eigen modes are products of a
theta function (compare the study in the context of the ``Copenhagen
vacuum''~\cite{amb}) and a plane wave (purely plane waves for abelian
fluctuations). One easily shows that the gauge field is (anti) periodic
in the $\vec m$ direction, i.e.
$\vec A(\vec x+\vec m/e)=\tau_3^\Delta\vec A(\vec x)\tau_3^{-\Delta}$,
with $e\equiv|{\rm g.c.d. }(m_i)|$ and $\Delta(\vec m)=0$ or 1, its value
being determined from (repeated use of) eq.~(\ref{eq:twist}).
The spectrum is now given by
\begin{equation}
\lambda_\pm(n,\ell)=2\pi\|\vec m\|(2n+1\pm2)+e^2
((2\ell+\Delta)\pi+\vec C\cdot\vec m/e)^2/\|\vec m\|^2
~,\quad \lambda_o(\vec k)=(2\pi\vec k)^2~,
\end{equation}
The multiplicities are $2|e|$ for $\lambda_\pm$ and 2 for $\lambda_o$.
The parameters $(n,~\ell~,\vec k)$ run over $(\relax{\rm I\kern-.18em N},{\rm Z \!\! Z},{\rm Z \!\! Z}^3)$.
On the lattice $\lambda_o(\vec k)$ will be replaced
by $4N^2\sum_i\sin^2(\pi k_i/N)$, but the effect of the lattice
on $\lambda_\pm$ is not expected to admit an analytic expression.
In fig.~1a-c we plot the first few eigenvalues of the Hessian for
resp. $\vec m=(0,0,1)$, $\vec m=(0,0,2)$ and $\vec m=(1,1,1)$, computed
numerically for $N=4$ and 6 in comparison with the exact continuum result
as a function of $\vec C$, of which only the component along $\vec m$ is
relevant. Only for $\vec m=(1,1,1)$ is $\Delta$ non-trivial.

A new feature that we believe not to have been observed so far, is
that there are constant ($\neq0$) curvature solutions on a torus that are
marginally stable. To be precise, for $\|\vec m\|=1$ and
$|\vec C\cdot\vec m-\pi|\leq\pi-\sqrt{2\pi}$ (using the $2\pi$ periodicity
of the spectrum in $\vec C\cdot\vec m$ we can restrict to
$\vec C\cdot\vec m\in[0,2\pi]$) there are no negative eigenvalues for the
Hessian. We call this marginal stability, as one can first bring
$\vec C\cdot\vec m$ outside the above range without changing the energy, and
then let the configuration decay along one of the two unstable modes.
For $\|\vec m\|\neq1$ one can easily see that this effect doesn't occur.

Nevertheless, in all cases considered in fig.~1 the Hessian can have accidental
zero eigenvalues for particular values of $\vec C$, such that potentially the
algorithm can slow down dramatically. In table I we compare the gap
$\hat\mu$ (the distance of the largest non-unit eigenvalue of $dT$ to 1) with
the predicted value $\hat\mu=\lambda_m^2/(14 N^4)$, determined from the
smallest non-zero (absolute) eigenvalue $\lambda_m$ of the Hessian for $S$
(see eq.(\ref{eq:Shgap}). The agreement is (perhaps surprisingly) very good.
For $\vec m=(1,1,1)$ and $N=4$ there is considerable slowing down. What is
called slowing down is of course dependent on the available computer power. We
consider something slowed down considerably if (several) 10.000 or more sweeps
are required to half the value of $\hat S$. In the case of
slowing down, we could generally achieve good results by performing a small
number of $S$ coolings, when the $\hat S$ cooling is being ``stuck'',
sometimes after a (very) small random update on the links.
That $S$ cooling can sometimes help can be argued for the case that
$\lambda_m>0$ as follows. We have seen that to
some degree the $\lambda_m$ eigenvector is also an eigenvector
of $dT$, corresponding to the eigenvalue $\lambda=1-\hat\mu$. This should
be the dominant mode along which the configuration differs from the true
fixed point. All other modes, in particular the unstable ones, should have
decreased significantly under the $\hat S$ cooling. If we don't cool too long
with $S$, the unstable mode does not get a chance to grow, whereas the
direction of the stable mode keeps on contracting (faster by a factor
$14N^2/\lambda_m$). With $\hat S$ cooling one can re-contract the unstable
mode, which might have been grown during the $S$ cooling. We expect that
the extra computational effort will not exceed by much what one would have
to spend in trying to accelerate the algorithm, e.g. using
fast-fourier~\cite{dav,dun} or multi-grid~\cite{hul} methods.

\vskip5mm
\hskip-8mm\hbox to \hsize{\hfil\vbox{\offinterlineskip
\halign{&\vrule#&\ $#{\vrule height 15pt depth 4pt width 0pt}$\hfil\ \cr
\noalign{\hrule}
&N&&\quad~\vec m&&~\vec C\cdot\vec m&&\quad\quad\hat S
&&\quad\quad~S&&\quad\quad S_{\rm th}&&\quad\lambda_m&&\quad\quad\hat\mu&&
\quad\quad\hat\mu_{\rm th}&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
&4&&(1,1,1)&&3.472&&2.90\cdot10^{-4}&&59.02774641&&59.02761860&&
\phm0.051&&6.49\cdot10^{-7}&&7.31\cdot10^{-7}&\cr
&6&&\omit&&2.517&&1.60\cdot10^{-2}&&59.17251733&&59.18005528&&-0.574&&
1.74\cdot10^{-5}&&1.82\cdot10^{-5}&\cr
&8&&\omit&&3.175&&2.65\cdot10^{-6}&&59.20573728&&59.20573660&&
\phm1.906&&\omit&&6.34\cdot10^{-5}&\cr
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\noalign{\hrule}
&6&&(0,0,2)&&4.229&&5.24\cdot10^{-7}&&78.75660845&&78.75660838&&
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height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}}}\hfil}
\vskip3mm
{\narrower\narrower{\noindent
Table I: Results for the constant curvature solution on a $N^3$ lattice with
magnetic flux $\vec m$. $S_{\rm th}$ is the value obtained from the exact
lattice result given in eq.~(\ref{eq:Sanal}), whereas $\hat\mu_{\rm th}=
\lambda_m^2/(14N^4)$, the predicted value for the distance of the largest
non-unit eigenvalue of $dT$ to unity. The smallest (in absolute value)
non-zero eigenvalue for the Hessian of $S$ is indicated by $\lambda_m$.
This, as well as $\hat\mu$, was obtained from the exact lattice solution
with the tabulated values of $\vec C\cdot\vec m$ extracted from the data.
}\par}
\vskip5mm

Not only $\hat S$ itself is a good indication for how far one is from
the fixed point. Also $dT$ can, and usually does, have an eigenvalue bigger
than 1 if one is not close enough to the fixed point (the gauge modes
always correspond to unit eigenvalues), whereas only {\bf at}
a fixed point will the gauge modes be exact zero-modes for the Hessian.
All this provides us with more than enough criteria to be sure that we have an
accurate lattice solution of the equations of motion. In table I we also
compare the energy of the constant curvature solutions with the exact
lattice result of eq.~(\ref{eq:Sanal}). The agreement is excellent,
except perhaps for $N=6$ and $m=(1,1,1)$, but in that case we did not
try to cool the configuration down to a very small value of $\hat S$.
Nevertheless, the error in the energy is smaller than one part in 6000. Also
quantities like the Polyakov loops agree very well with the exact lattice
expressions, eq.~(\ref{eq:Poly}).

\section{Results: The sphaleron}

We now discuss the results for the new sphaleron solutions. Some of the
results are collected in table II. Also here we see that convergence of the
$\hat S$ cooling algorithm can be well predicted in terms of $\lambda_m$.
In figs.~2a-c we show scatter plots for the (complex) eigenvalues of $dT$ for
the $\vec m=\vec 0$ sphaleron, using a checkerboard updating for fig.~2a
(to allow for vectorization of the algorithm) and a sequential updating for
fig.~2b, both at $N=4$. We see that the checkerboard updating makes the bulk
of the modes contract much faster (in addition to the gain of computational
speed due to the vectorization). This remains true on larger lattices.
In fig.~2c we show for comparison the checkerboard result for $N=8$.
The eigenvalues that are responsible for the slowing down of the algorithm
near the fixed point, in particular the eigenvalue closest to 1, will only
very weakly depend on the order in which one updates the links.
Table II shows that the sphaleron with periodic boundary conditions
($\vec m=\vec 0$) where $\lambda_m\sim1.3$, has considerable slower cooling than
the sphaleron with twisted boundary conditions, for which $\lambda_m\sim-10.8$.
Nevertheless, we managed to obtain rather accurate results.
In table II we also list $S$, $\hat S$ and the blocked
Wilson action $S_{2\times2}$, for which the lattice spacing is effectively
twice as large. Like the instantons of maximal size studied in
ref.~\cite{gar1,gar2}, the sphaleron is a very smooth solution.
We do not require huge lattices to approach the continuum limit.
Together with easily obtainable expansions in the lattice spacing $a=1/N$,
we have a large enough window (available values for $N$) to get accurate
results. We can extract, as for the instantons (see ref.~\cite{gar1} for
details), the continuum action by fitting to the formula
\begin{equation}
S_{n\times n}=S_0\left(1-\alpha n^2/N^2-(\beta n^4+\gamma n^2+\delta)/N^4
\right)+{\cal O}(n^6/N^6)\quad.
\end{equation}
We find for the twisted continuum sphaleron ($\vec m=(1,1,1)$) an energy
${\cal E}=34.148(2)$ and for the periodic one ($\vec m=\vec 0$) we find
${\cal E}=72.605(2)$, with conservative error estimates.
\vskip5mm
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&\pho6&&\omit&&\hphantom{-}\pho1.415&&1.08\cdot10^{-4}&&1.10\cdot10^{-4}&&
70.70174&&64.36296&&2.11\cdot10^{-5}&\cr
&\pho8&&\omit&&\hphantom{-}\pho1.301&&2.41\cdot10^{-5}&&2.95\cdot10^{-5}&&
71.55060&&68.16790&&6.93\cdot10^{-9}&\cr
&12&&\omit&&\omit&&\omit&&5.83\cdot10^{-6}&&72.14095&&70.70631&&
4.13\cdot10^{-5}&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}}}\hfil}
\vskip3mm
{\narrower\narrower{\noindent
Table II: Results for the sphalerons on a $N^3$ lattice with twisted and
periodic boundary conditions. The smallest (in absolute value) non-zero
eigenvalue for the Hessian of $S$ is indicated by $\lambda_m$. The
predicted value for the distance of the largest non-unit eigenvalue of
$dT$ to unity is given by $\hat\mu_{\rm th} =\lambda_m^2/(14N^4)$. We also
give the value of the Wilson action, its blocked value $S_{2\times 2}$
and $\hat S$.
}\par}
\vskip5mm

The evidence that these configurations are sphalerons follows from a study
of the eigenvalues for the Hessian of $S$, whose low-lying values are listed
in table III for $N=4,~6$ and $8$.
We have verified that there are exactly $3N^3+3$ zero-modes for both
sphalerons, corresponding to the gauge modes and the three translational
moduli (for $N>6$ the latter are indistinguishable from the gauge zero-modes,
for $N=4$ their eigenvalues are not bigger than 0.01). Most importantly we
have verified that there is precisely one unstable mode. Furthermore, we have
not been able to find any other saddle point with a lower (non-zero) energy
than the energies of the present configurations. For all practical purposes
we thus consider the newly found solutions to be sphalerons.

We now wish to show that like on $S^3$~\cite{baa1}, these sphalerons are
at the top of a tunnelling path. For the case of twisted boundary conditions
this is the easiest to demonstrate, since the instanton with a continuum
action of $4\pi^2$, which was extensively studied on the lattice
in ref.~\cite{gar3}, has only the trivial translational and some
discrete moduli. This means we only have to demonstrate that the
top of the instanton path is identical to the sphaleron. There are two reasons
why the top of an instanton path might not correspond exactly to a sphaleron.
First, the lattice has a finite resolution and second, the extent in the time
direction is finite. In particular the finite resolution causes problems
because one needs to extrapolate along the tunnelling path the value
of the energy and of $\hat S$. In particular the latter has a sharp dip
near the top. A very large value of $N$ would be required to obtain
a sufficient resolution. The effect of the finite extent in time
is expected to be small since the instanton path approaches the vacua
exponentially in time~\cite{gar3}.  In fig.~3 we plot $\tilde S
\sim\hat S/64$ for each time-slice along the tunnelling path,
cmp. ref.~\cite{gar2}.

\vskip5mm
\hbox{\hskip1.5cm$\vec m=(1,1,1)$ sphaleron\hskip2.5cm$\vec m=\vec 0$
sphaleron\hskip3cm index 2}
\hbox to\hsize{\hfil\vbox{\offinterlineskip\halign
{&\vrule#&\ $#{\vrule height 15pt depth 4pt width 0pt}$\hfil\ \cr \noalign{\hrule}
&\quad N=4&&\quad N=6&&\quad N=8&&\quad&&\quad N=4&&\quad N=6&&\quad N=8&&
\quad&&\quad N=8&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
&-11.1280&&-10.9003&&-10.8174&&\omit&&-13.1960&&-13.3935&&-13.4432&&\omit&&
-13.1170&\cr
&\hphantom{-}\pho0.0074&&\hphantom{-}\pho0.0001&&\hphantom{-}\pho0.0000 &&\omit&&\hphantom{-}\pho0.0073&&
\hphantom{-}\pho0.0006&&\hphantom{-}\pho0.0000 &&\omit&&-\pho0.7466&\cr
&\hphantom{-}\pho0.0091&&\hphantom{-}\pho0.0001&&\hphantom{-}\pho0.0000 &&\omit&&\hphantom{-}\pho0.0124&&
\hphantom{-}\pho0.0010&&\hphantom{-}\pho0.0000 &&\omit&&\hphantom{-}\pho0.0000&\cr
&\hphantom{-}\pho0.0101&&\hphantom{-}\pho0.0001&&\hphantom{-}\pho0.0000 &&\omit&&\hphantom{-}\pho0.0139&&
\hphantom{-}\pho0.0011&&\hphantom{-}\pho0.0000 &&\omit&&\hphantom{-}\pho0.0000&\cr
&\phm13.1667&&\phm13.3371&&\phm13.3934 &&\omit&&\hphantom{-}\pho1.7041&&
\hphantom{-}\pho1.4151&&\hphantom{-}\pho1.3009 &&\omit&&\hphantom{-}\pho0.0000&\cr
&\phm13.1667&&\phm13.3371&&\phm13.3934 &&\omit&&\hphantom{-}\pho1.7070&&
\hphantom{-}\pho1.4303&&\hphantom{-}\pho1.3012 &&\omit&&\hphantom{-}\pho3.2618&\cr
&\phm13.1667&&\phm13.3372&&\phm13.3934 &&\omit&&\hphantom{-}\pho7.9677&&
\hphantom{-}\pho8.0984&&\hphantom{-}\pho8.1260 &&\omit&&\hphantom{-}\pho5.8733&\cr
&\phm16.0475&&\phm16.8663&&\phm17.1597 &&\omit&&\hphantom{-}\pho7.9682&&
\hphantom{-}\pho8.1008&&\hphantom{-}\pho8.1261 &&\omit&&\hphantom{-}\pho5.8807&\cr
&\phm16.0478&&\phm16.8663&&\phm17.1597 &&\omit&&\hphantom{-}\pho7.9684&&
\hphantom{-}\pho8.1279&&\hphantom{-}\pho8.1266 &&\omit&&\hphantom{-}\pho6.1738&\cr
&\phm16.0484&&\phm16.8663&&\phm17.1597 &&\omit&&\hphantom{-}\pho8.9984&&
\phm10.2977&&\phm10.6604 &&\omit&&\hphantom{-}\pho8.5986&\cr
&\phm21.4686&&\phm22.9444&&\phm23.4481 &&\omit&&\hphantom{-}\pho9.0189&&
\phm10.3306&&\phm10.6611 &&\omit&&\hphantom{-}\pho8.6069&\cr
&\phm21.4747&&\phm22.9444&&\phm23.4481 &&\omit&&\hphantom{-}\pho9.9919&&
\phm10.8287&&\phm11.0701 &&\omit&&\phm15.0281&\cr
&\phm21.8119&&\phm23.6594&&\phm24.2198 &&\omit&&\phm10.0256&&
\phm10.8312&&\phm11.0701 &&\omit&&\phm16.7321&\cr
&\phm21.8260&&\phm23.6594&&\phm24.2198 &&\omit&&\phm10.0354&&
\phm10.8594&&\phm11.0707 &&\omit&&\phm19.7489&\cr
&\phm25.0998&&\phm26.1760&&\phm26.4962 &&\omit&&\phm24.6675&&
\phm26.4047&&\phm26.9618 &&\omit&&\phm20.6039&\cr
&\phm25.1050&&\phm26.1760&&\phm26.4962 &&\omit&&\phm24.6703&&
\phm26.4205&&\phm26.9621 &&\omit&&\phm21.3605&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}}}\hfil}
\vskip3mm
{\narrower\narrower{\noindent
Table III: The lowest lying eigenvalues for the Hessian of $S$ for the
twisted and periodic sphaleron on a $N^3$ lattice for $N=4,~6$ and 8.
Also is given the spectrum for the periodic solution with two unstable
modes (index 2), close in energy to the sphaleron, discussed in the text
(see also table IV).
}\par}
\vskip3mm

We take the configuration at the top of the instanton path, where
$\tilde S$ is minimal, and
cool it with $\hat S$. This way we easily obtain the sphaleron solution.
As $\hat S$ decreases monotonically under the $\hat S$ cooling, and
as $\hat S$ is already small at the top of the instanton path,
we expect that the instanton path goes through a sphaleron. Further
evidence is provided by calculating the quantity
\begin{equation}
D_N(t)=4\|U_t-U_s\|^2/N\sim\int d_3x (\delta A^a_i(\vec x))^2\quad,
\label{eq:dist}
\end{equation}
where the latter expression is obtained in the continuum limit.
$U_t$ is the configuration at the time $t$ along the instanton path and $U_s$
is the sphaleron configuration, obtained after cooling from the configuration
$U_t$ (for this, $t$ should be chosen not too far away from the top of the
barrier, as otherwise cooling will bring the configuration to the vacuum).
As the cooling algorithm commutes with gauge transformations, this gives
a gauge invariant measure for the distance of the two configurations. Indeed,
computing $\|U_t-U_{t^\prime}\|$ gives huge values, as the gauge between
different time slices will differ randomly. For the configurations of fig.~3
we find $D_8(0)=0.0246$, $D_8(1/8)=1.162$, $D_8(-1/8)=1.5713$ and
$D_{12}(0)=0.0340$. We have indicated also the values at either side of the
top of the barrier for $N=8$ to show that, not being exactly at the top
can have a sizable effect on $D_N(0)$ (and on $\tilde S(t=0)$). We believe
the numerical evidence is sufficiently convincing to conclude that the
instanton path does go through a sphaleron. In fig.~4 we give for $N=12$ the
two dimensional cross sections of ${\cal E}_B(x,y,z)$ at the twelve lattice
values of $x$. Because of the cubic symmetry of the sphaleron, the other
cross-sections have identical profiles (modulo translations).

For the case of periodic boundary conditions the instantons have
additional moduli.
We did provide evidence in ref.~\cite{gar1} that for $T\rightarrow\infty$
the moduli are described by the scale, translational and vacuum parameters.
The latter being specified by the Polyakov loops $P_i(t=\infty)=
-P_i(t=-\infty)$ (we have not yet proven or disproven if this relation
at $t=\pm\infty$ can be relaxed).
In particular we probed the dependence of the tunnelling
path on $P_3\equiv P_3(t=-\infty)$, keeping the other two equal to
unity. In practise we can of course not reach $|t|=\infty$, and this
is a problem for $P_3=\pm1$. In that case one can show that the approach to
the vacuum for large $t$ is no longer exponential due to the quartic
behaviour of the potential in the zero-momentum sector. We demonstrate
the sensitivity of the instanton path on the extent in the time direction
for $N=8$ and $N_t=24,~48$ in fig.~5. We see that for large $N_t$ the tails
in the electric energy~\cite{gar1,gar2} are lowered considerably, while the
magnetic energy at the top of the barrier (not visible at the scale of fig.~5)
is lowered from $S(\varepsilon=-1)=74.041$ at $N_t=24$ to 73.465 at $N_t=48$,
actually below the value of 73.923 obtained for $P_3=0.5$ in
ref.~\cite{gar2} (for which we did not observe such a sensitivity in
$N_t$). It means that there is near degeneracy in energy at the top of the
barrier as a function of $P_3$. The variation of the energy as a function
of $P_3$ can be so small, because the energy functional near the sphaleron
has two rather shallow directions (associated to the two eigen modes with
$\lambda_m\sim1.3$). To investigate if our sphaleron indeed corresponds to the
top of a tunnelling path, presumably the one with $P_3=1$, we took the
configuration at the top of the energy barrier for the values of $P_3$ that
were considered in ref.~\cite{gar2} and applied $\hat S$ cooling for 8000
sweeps on each, such that the energy and $\hat S$ stabilized, and further
cooling would make the configuration only move (very slowly) along the slowest
mode in the cooling algorithm. We discovered a very useful set of parameters
to distinguish the various configurations thus obtained. It turns out that
the energy profile is almost identical for each configuration, that the
profiles for the Polyakov lines only vary weakly, but that the vector
\begin{equation}
\vec b\equiv\left({\rm Tr}(B_1^2),{\rm Tr}(B_2^2),{\rm Tr}(B_3^2)\right)/
\sqrt{{\scriptstyle{{1\over 3}}}\sum_i{\rm Tr}(B_i^2)^2}\quad,\label{eq:orpa}
\end{equation}
averaged over the lattice ($<\vec b>$), clearly differentiates the various
configurations. The sphaleron solution distinguishes itself by a symmetric
value $\vec b=(1,1,1)$ at each point independently. In all other cases the
variance of the parameter $\vec b$ is non-zero. To a good degree
$<b_1>=<b_2>$, which is of course a direct consequence of the symmetry of
the lattice and of the fact that $P_1=P_2$. We also compute $D^\prime_8$,
which is defined as in eq.(\ref{eq:dist}) but with $U_t$ the configuration
obtained after the $\hat S$ cooling. For this it is important to note
that the instantons in table IV were all generated from the same random start
at $P_3=1$ (or $C_3=0$; see ref.~\cite{gar2}). This means that each is in
the same gauge and $D^\prime_8$ is a good gauge invariant measure for the
distance of the various cooled configurations to the sphaleron.
In table IV we collect the results thus obtained, and
see that after 8000 cooling sweeps $\hat S$ is dramatically lowered
also for $P_3=0$. If we compute the Hessian for this configuration we find
a Morse index of two, i.e. there are two unstable modes
(see table III for the low-lying eigenvalues).
Note that $S$ for this configuration is only slightly higher than
for the sphaleron, and that $\lambda_m=-0.74655$ will cause an even slower
cooling than for the sphaleron. We have performed 8000 extra $\hat S$ cooling
sweeps before computing the Hessian, but the quality of its zero modes were
sufficiently good to trust the cooled configuration for $P_3=0$ to be close
enough to a solution. We notice that except near $P_3=1$, the
value of $<\vec b>$ at the top of the barrier (denoted by $<\vec b_t>$ in
table IV) is changed after 8000 $\hat S$ cooling sweeps. We conclude that the
index 2 solution is not at the top of any of the instanton paths we
considered here.

\vskip5mm
\hbox to \hsize{\hfil\vbox{\offinterlineskip
\halign{&\vrule#&\ $#{\vrule height 15pt depth 4pt width 0pt}$\hfil\ \cr
\noalign{\hrule}
&\hphantom{1} C_3&&\quad\quad<\vec b_t>&&\quad\quad<\vec b>&&\quad~ S&&
\quad S_{2\times 2}&&\quad\quad\hat S&&D^\prime_8&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
&\pho0&&(1.01,1.01,0.99)&&(1.01,1.00,0.99)&&71.5507&&68.1680&&
2.08\cdot10^{-4}&&0.000&\cr
&\hphantom{1}\pi/5&&(1.02,1.02,0.96)&&(1.03,1.02,0.95)&&71.5547&&68.1732&&
9.77\cdot10^{-3}&&0.056&\cr
&2\pi/5&&(1.05,1.05,0.89)&&(1.07,1.06,0.85)&&71.5913&&68.2205&&
7.53\cdot10^{-2}&&0.100&\cr
&2\pi/3&&(1.10,1.10,0.77)&&(1.14,1.13,0.65)&&71.7569&&68.4322&&
1.58\cdot10^{-1}&&0.560&\cr
&4\pi/5&&(1.11,1.11,0.73)&&(1.15,1.15,0.60)&&71.8301&&68.5250&&
1.18\cdot10^{-1}&&0.894&\cr
&\hphantom{1}\pi&&(1.12,1.12,0.69)&&(1.17,1.16,0.53)&&71.9318&&68.6537&&
2.60\cdot10^{-4}&&2.186&\cr
height 4pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}}}\hfil}
\vskip3mm
{\narrower\narrower{\noindent
Table IV: Results for $N=8$ associated to the top of the instanton path
(obtained with $S(\varepsilon=-1)$ cooling, see ref.~\cite{gar2}) that interpolates
from the vacuum specified by $P_1=P_2=1$ and $P_3=\cos(C_3/2)$ at
$t=-\infty$, to the vacuum specified by $P_1=P_2=-1$ and $P_3=-\cos(C_3/2)$
at $t=\infty$. The vector $\vec b$ is the parameter defined in
eq.~(\ref{eq:orpa}). From the third column on are tabulated the results
after 8000 $\hat S$ cooling sweeps. $D^\prime_8$ is defined as in
eq.~(\ref{eq:dist}), with $U_t$ the cooled configuration. Both from the first
and last case one finds solutions of the equations of motion, resp. the
sphaleron and the index 2 saddle point. See table III for their Hessians.
}\par}
\vskip5mm

Finally, we calculated $D_N(t)$ from the configuration in table IV with
$C_3=0$ ($P_3=1$) after further $S(\varepsilon=-1)$ cooling on the instanton, as
represented in fig.~5 (at $N_t=24$). We find $D_8(0)=0.0496$, $D_8(-1/8)=2.212$,
$D_8(1/8)=3.369$ and $D_{12}(0)=0.0368$ (not represented in fig.~5).
Taking into account the time-tail effects discussed before and
the fact that the instanton path was generated with $S(\varepsilon=-1)$ cooling,
whereas $\hat S$ cooling was performed at $\varepsilon=1$, the quality of these
results are sufficient to conclude that the sphaleron lies on the instanton
path that has $P_i(t=-\infty)=-P_i(t=\infty) =\pm1$. In fig.~6 we give
the results for the energy distribution on a $12^3$ lattice.

\section{Conclusions}

The main motivation for this paper was to find the sphaleron solution for
pure SU(2) gauge theories in a finite volume and to show that it is at
the top of the barrier that separates two classical minima,
connected by an instanton solution with minimal action. We studied both
the case of twisted and periodic boundary conditions. For the latter the
situation was rather subtle because of the many additional moduli in the
instanton parameter space. Furthermore, at the sphaleron there are two very
flat directions for the energy functional, and we found a saddle point with
two unstable modes not more than $\delta{\cal E}=0.5$ above the sphaleron.
The sphaleron solution
for the periodic case has some very special properties. Like for the twisted
case~\cite{gar3} only low fourier components seem to dominate, but
especially the property that locally $Tr(B_i)^2$ is independent of $i$
is intriguing. There is good hope that in the future this problem might be
tractable analytically. In any case it presents an interesting challenge.

To achieve all these results we built on the idea that one can find
unstable solutions to the equations of motion by minimizing the functional
obtained by squaring the gradient of the field equations~\cite{dun,sij}.
What we have added is a specific algorithm for SU(2) to minimize this
functional in a deterministic way. This allowed us to analyse in
detail the convergence of the algorithm, also providing results for the
standard cooling algorithm~\cite{ber}. It is likely that the algorithm
can be extended to include a scalar sector so that we can also study
the sphaleron in the standard model~\cite{kli}. In any case one can
always try to minimize the functional by doing random updates.
One can easily think of many more applications worthwhile pursuing with
these methods.

\section{Acknowledgements}

We have benefited from discussions with Jim Hetrick, Jan Smit, Arjan van der
Sijs and Jeroen Vink. The eigenvalues of the linearized algorithm and Hessian
were determined using the EISPACK~\cite{eisp} routines. We thank Willem Vermin
for providing the source codes. Figs.~4 and 6 were produced using
Mathematica~\cite{math}, its programs with the raw data for the energies are
attached to the electronic hep-lat bulletin board version of this paper.
This work was supported in part by grants from ``Stichting voor Fundamenteel
Onderzoek der Materie (FOM)'' and ``Stichting Nationale Computer
Faciliteiten (NCF)'' for use of the CRAY Y-MP and C98/4256 at SARA.
M.G.P. was supported by a Human Capital and Mobility EC fellowship.
\eject

redpajama_set_name: RedPajamaArXiv