The Durban Chamber Choir (DCC) will be performing on Sunday, May 28 at St Thomas's Church, Musgrave, and on Sunday, June 11 at the Lutheran Church, Hayfields, honouring the birthday of Claudio Monteverdi, and the works of another great transitional composer, Gabriel Fauré.
This year marks the 450th anniversary of Monteverdi's birth in Cremona, Italy, in 1567. The DCC will celebrate this landmark with a performance of one of the great composer's three Latin Mass settings (identified by its date of publication, 1650).
"Monteverdi lived through one of the most significant periods of change in the history of Western music-the transition from the Renaissance to the Baroque style," says Dr Christopher Cockburn, DCC's director.
"While his 1650 Mass primarily looks back to the Renaissance, it also includes many touches of the more modern style pioneered in such works as his Vespers and his famous opera, Orfeo. This blend of old and new makes this Mass a unique musical experience.
"The French composer, Gabriel Fauré, whose music makes up the rest of the programme, lived during another of the great transitional moments in Western music, as Romanticism gave way to Impressionism in the second half of the 19th century.
The works by Fauré will be accompanied by Jacques Heyns, and Cockburn conducts the whole programme.
Refreshments will be served after each performance. Both performances start at 3pm, and last about an hour.
Tickets are available at the door from 2pm, and admission to the venues is from 2.30pm. Tickets are R70 per person (student/scholar OAP concessions R40). School groups are admitted free, and group bookings may be made, by prior arrangement with the Choir (nicole@NGOLawSA.co.za).
The choir is looking to add a few tenors to their number, for an exciting programme to be presented later this year- those who are interested in joining are invited to attend a concert, and talk to the committee members, or email Nicole.


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Q: Join function in the query making the query run slower I have two table both contains data around 200,000. I have write a query like below to retrieve data using some joins.
This is the query I have tried
$dbYTD = DB::table('stdtsum as a')
            ->join(DB::raw("(select distinct s_id, c_cod, compid from stdcus) b"), function($join){
                $join->on('a.compid', '=', 'b.compid')->on('a.c_cod', '=', 'b.c_cod');
            })
        ->select('b.s_id', DB::raw('sum(turnover) as sumturn'))
        ->whereBetween('date', [$startYTD, $endYTD])
        ->groupBy('b.s_id')
        ->get()
        ->toArray();

This query is giving result correct but the process time it takes is very long, sometimes it even timesout.
Can anybody help me how can I optimize this query?

A: You'll need to index all your columns on which join condition is applied. In your case : "compid" , "c_cod" in both your tables.
Generally, "Primary Key constraint" columns are automatically indexed in databases, though you'll have to index the "Foreign Key constraint" columns manually.
Some Indexing Tips :

*

*Create an index on the field that has the biggest variety of values first. This will result in the "most bang for your buck", so to speak.

*Keep indexes small. It's better to have an index on just zip code or postal code, rather than postal code & country. The smaller the index, the better the response time.

*For high frequency functions (thousands of times per day) it can be wise to have a very large index, so the system does not even need the table for the read function.

*For small tables an index may be disadvantageous. The same can also be said for any function where the system would be better off by scanning the whole table.

Remember that an index slows down additions, modifications and deletes because the indexes need to be updated whenever the table is. So, the best practice is to add an index for values that are often used for a search, but that do not change much. As such, an index on a bank account number is better than one on the balance.

Tips reference : https://www.databasejournal.com/features/mysql/article.php/3840606/Maximizing-Query-Performance-through-Column-Indexing-in-MySQL.htm

redpajama_set_name: RedPajamaStackExchange


America. Fed ha aumentato i tassi dal 2.25% al 2.5%.
Se è vero che gli Stati Uniti avrebbero un disperato bisogno di mantenere bassi i tassi di interesse per facilitare il sistema economico, è altrettanto vero che le motivazioni della Fed all'aumento dei tassi sono più che motivate.
Sia la componente politica sia quella finanziaria e monetaria hanno le loro buone ragioni, e trovare un accordo è virtualmente impossibile.
Come al solito, qualsiasi scelta avrebbe sollevato un polverone di critiche e sembrerebbe essere impossibile trovare commenti equilibrati, non partigiani.
Gli aumenti dei tassi di interesse erano noti e programmati da tempo.
I mercati se ne diano una ragione.
– The Fed take the target range for its benchmark funds rate to 2.25 percent to 2.5 percent.
– Central bank officials now forecast two hikes next year, down from three rate raises previously projected.
– However, the Fed continues to include in its statement that further "gradual" rate hikes would be appropriate.
– GDP is now seen as rising 3 percent for the full year of 2018, down one-tenth of a percentage point from September, and 2.3 percent for 2019, a 0.2 percent point reduction.
The Federal Reserve on Wednesday raised its benchmark interest rate a quarter-point but lowered its projections for future hikes.
As markets had expected, the central bank took the target range for its benchmark funds rate to 2.25 percent to 2.5 percent. The move marked the fourth increase this year and the ninth since it began normalizing rates in December 2015. It came despite President Donald Trump's tweets against rate hikes.
Officials, though, now project two hikes next year, which is a reduction but still ahead of current market pricing of no additional moves next year.
The language in the post-meeting statement was not entirely dovish. The committee continued to include a statement that more rate hikes would be appropriate, though it did soften the tone a bit.
"The Committee judges that some further gradual increases in the target range for the federal funds rate will be consistent with sustained expansion of economic activity, strong labor market conditions, and inflation near the Committee's symmetric 2 percent objective over the medium term," the statement said.
Along with the hike, investors had been keyed on where the Federal Open Market Committee, which sets rates, expected to go in the future. Heading into this week's two-day meeting, the committee had been pointing to three more moves in 2019 and possibly another one in 2020.
That changed amid tightening financial conditions and worries that the Fed was moving too quickly. Stock indexes have moved into correction territory and are largely negative for the year.
That essentially reinforces recent public statements from Fed officials that they will be data dependent when making future rate decisions.
The FOMC also lowered its outlook for the long-run funds rate, from 3 percent in the September forecast to 2.8 percent this month. The 2019 estimate declined to 2.9 percent from 3.1 percent and both 2020 and 2021 dropped to 3.1 percent from 3.4 percent.
The funds rate is tied to most consumer debt, particularly credit cards and adjustable-rate loans.
There were no dissents in the vote to hike, but the "dot plot" of individual committee members' estimates show some division among members. Six still see three increases next year, down from nine in September, when officials last released their projections. Sixteen members in all submitted dots at this week's meeting.
Along with the tempered estimates for rates, the committee nudged lower its projections both for GDP and inflation.
GDP is now seen as rising 3 percent for the full year of 2018, down one-tenth of a percentage point from September, and 2.3 percent for 2019, a 0.2 percent point reduction. However, officials took up their long-run estimates, to 1.9 percent from 1.8 percent in September.
Overall, though, Fed officials expressed little worry about economic growth. GDP gains have averaged 3.3 percent per quarter this year, and the Atlanta Fed is forecasting a 2.9 percent increase in the fourth quarter.
Where the market is worried that the U.S. might be infected by a global slowdown, the FOMC statement showed little concern.
Officials continued to describe economic growth as "rising at a strong rate" and left descriptions of other parts of economic activity unchanged as well.
The summary of economic projections did note that headline inflation is expected to grow less quickly than the September estimate, slipping to 1.9 percent from 2.1 percent in 2018 and to 1.9 percent from 2 percent in 2019. The longer-run expectation remains 2 percent.
Articolo precedente Italia. Rinvio dell'adesione al Global Compact. Il caso del Belgio.


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Burqa, bombs and intolerance
Why has our so-called open-minded progressive class forgotten the values of tolerance that they themselves preach?
Naureen Aqueel January 04, 2011
A look through the timeline of bomb blasts and terrorist attacks indicate that a majority of attacks in Pakistan are carried out by young men - some wearing vests, others using cars laden with explosives. I believe this spells out a legitimate case to ban young men, vests and cars from public places. After all, in a country like ours which is always on high alert for terrorist attacks, we can't allow such security risks to roam about freely, can we?
If you find my logic ludicrous, you might want to take a look at the recent debate on banning the burqa or niqab due to security concerns. In an article in The Express Tribune titled "A mark of separation," Mr George Fulton makes an interesting case for banning the burqa or niqab in Pakistan based on one case of a female burqa-clad suicide bomber killing 47 people and injuring over 100 in an attack on the World Food Programme distribution point in Bajaur.
As unjust as focusing on one incident against the dozens of other attacks carried out by young men may be, what is more astonishing is the sweeping generalisations Mr Fulton makes to depict the veil as a symbol of suppression because in his view "there is little proof that women actively decide to adopt the veil" and he finds it difficult to find any woman of substance who wears the burqa or niqab.
I am amazed at these views. Why is it so difficult for our so-called progressive class which claims to espouse the values of freedom, liberty, tolerance and personal choice to understand that there are perfectly normal, educated and 'enlightened' people out there who choose to follow their religion? That there are women who have the freedom to fling off their covers and be fashionably under-dressed, yet they choose not to? That there are men and women who have the liberty to disassociate themselves from religion, yet they choose not to?
It has become fashionable nowadays for our 'intellectual elite' to pen an article or two reminiscing about the more secular days when religion was a private affair - when there were no beards, no hijabs, no burqas and no preachers on television. Excuse me for my flawed memory, but is it really true that there was no religiosity a decade ago or is it just that having let our religion be hijacked by militants today, we are now more bothered by its physical manifestations in the public sphere?
And for that matter, why has our so-called open-minded progressive class forgotten the values of tolerance that they themselves preached? Do I smell intolerant fanaticism here?
If there is one attitude that marks this age it is extremism – and that extremism is present not only in the religious class, but in the so-called 'liberals' as well. Until we all step into each other's shoes and understand the other person's perspective, there is no hope for ending the violence we see plaguing our society.
Making a case to ban what many consider a religious practice (and even if we assume it is a cultural practice, there are still no grounds to ban it) only breeds intolerance. Restricting women from wearing the veil is as much a violation of their rights as forcing them to wear one. We must understand that in the case of the burqa bombing in Bajaur and many others, the issue was not the item of clothing, but the lax in security. Where there are legitimate needs for identification, appropriate measures can be taken to satisfy the requirement by asking the woman to take off the veil in private.
The burqa or niqab has as much potential of being misused as say cars, cellphones and the Internet. What is needed is a proper strategy to prevent this misuse rather than banning these things altogether.
Naureen Aqueel
Bin Laden killing: Whither objective journalism?
Media sensationalism trumped by online insensitivity
Drone attacks: Unlawful killings, double standards
Nobody | 10 years ago | Reply @hassan: Correction, niqab and burka are NOT compulsory. Head covering (hijab) and modest garb (which can be anything that covers the important parts of your body and conceals your shape, doesnt specifically have to be an abaya) are compulsory. That being said, forcing any kind of dress or outlawing any kind of dress seems to be the same thing to me: trampling civil liberties. Let men and women decide what they do or do not want to wear....tragically that's too simplistic for today's world. @Naureen: well done!
hassan | 10 years ago | Reply As per the saying in Quran/Hadiths/Sunnah, the women should wear the burqa/hijab/niqab....this is compulsory, and not voluntary. I assume you support the above, since you have not protested strongly against the forcing of veil on women, as mentioned in the Holy Books. But You say: *"Restricting women from wearing the veil is as much a violation of their rights as forcing them to wear one."* What you mean is this: Forcing someone to wear a dress is NOT a violation of rights. But, forcing someone NOT to wear a dress IS a violation of rights. Can't you see some logical fallacy over there?
sameen | 10 years ago | Reply an excellent article........keep up the good work!..........if there is no such term as liberal extremism then there is no such term as fundamentalists or relgious extremists either ........if anyone is interested then read this as well...http://www.thenews.com.pk/TodaysPrintDetail.aspx?ID=229506&Cat=9&dt=3/17/2010.
Maha | 11 years ago | Reply If there is one attitude that marks this age it is extremism – and that extremism is present not only in the religious class, but in the so-called 'liberals' as well. Naureen! As a Pakistani woman, i don't observe pardah but i have no problem with those who observe it. You talked about liberal extremism. There is no such term as that. Liberals don't use guns to make their point heard neither they pass fatwas. If someone is against burqa then it's an opinion. Take it as that. Noone is forcing you not to wear it. Rest assured liberals will not go to extreme length otherwise it will negate the very idea of liberalism.. If liberals can stand people opinions about them for being immoral then people dislike for burqa can be taken in the same vein. What do you think?
Shariq Qureshi | 11 years ago | Reply Kudos to the writer for picking up and addressing an interesting subject in an interesting manner.
Haajar Eberle | 11 years ago | Reply @S.Haque: I'd like to know where the writer suggested that ALL the women who wear a burqa/niqab do so of their free will? Besides, which women are you speaking of? The women in Pakistan? India? Or the women all over the world. You must have traveled extensively and interviewed many niqabi women to speak on their behalf. What about the many women in the US who cover? There are no "creatures coming out of madrassas" roaming about here. And the hundreds of women in the Middle East and Europe who cover only because of personal religious beliefs and free will and not because of "ogling" men? Thank you Naureen for the well-written article! I give you the authority to speak on my behalf as a peaceful, free and educated niqabi woman :). I would like to keep my niqab on!!!
maria ishtiaq | 11 years ago | Reply if your articl makes taking hijab any easier and even if it does not, i say jazakillahu khair :) god bles u
Fazl-e-Azeem Bari | 11 years ago | Reply great work naureen
Someone | 11 years ago | Reply @faraz: The writer didnt label Fulton as a liberal extremist. She just spoke about an attitude of intolerance which is a form of extremism, nor did she compare an opinion with murder or say liberal extremists are as dangerous as militants. There is a difference between terrorists/militants and just an extremist/intolerant mindset. Stop jumping to conclusions people!
Hasan | 11 years ago | Reply Motorcycles banned in DIK: DPO DERA ISMAIL KHAN: Motorcycles would be banned in Dera Ismail Khan, District Police Officer said Sunday.He said even motorcycle rickshaws would be banned in the district until further orders.
Ayesha Ijaz Khan | 11 years ago | Reply @ Sara Abdul Khaliq---your definition of "extremist" is flawed. Most people, whether liberal or not, would like to impose their views on others. Those who favour the burqa believe that others should wear it too while those who don't favour it, mock it. That does not make one an extremist. What makes one an extremist is to what lengths they would go to impose their view. So the entire issue is the methodology, my dear. If someone talks, or even shouts about something that does not make them an extremist, but if they hold a gun to your head, and force you to do it or throw acid on your face if they see it uncovered then that does make them an extremist. @Asad Shairani---the reason I said that very few women wear the burqa out of choice is because I lived in Saudi Arabia for 12 years, where I was forced to wear either a chador or an abaya in public even though I had no desire to do so. The overwhelming majority of Saudi women are forced to wear the abayas and you can see this on a flight to London when come landing time, most women take them off. Many who keep them on do so out of fear of the brothers, husbands, etc. Only a small minority really do it out of choice. My Iranian roommate in college had exactly the same view. Speak to women in Afghanistan and certain parts of Pakistan where it is imposed upon women as well and you will find the same answer. I do not deny that there are women who do it out of choice but they are very few in comparison and mostly those who have not lived in conditions where they have been forced to do it but are suddenly fascinated by this new thing and hence like to experiment. I see some young girls in London who take on the hijab, then take it off---that is choice. But in Kabul or Miranshah or Riyadh or Tehran it is generally not choice. S Haque makes a really good point above. And, my reason for banning the burqa is not only that it is a security risk---that is only one aspect. Mostly my reason for banning the burqa is because it is a symbol of oppression and societies where it is commonplace are oppressive for women---it is used to hold them back in education, professionally, legally. When Iran imposed a burqa, it also got rid of women judges. In Saudi, women can't drive. In both countries, women need the permission of their husbands to travel abroad. In Turkey, on the other hand, when Ataturk got rid of the burqa, the literacy rate among women shot up astronomically. Ataturk may be responsible for human rights violations, but at least he had something to show for it. When the Taliban types engage in those violations, they have nothing to show either. So my view is that when France bans the burqa, where only about a handful of women are taking it on, it is doing it out of Islamophobia and hence it is the wrong reason. But when Syria does it, I support it---because there the reason is not Islamophobia---it is a concern for women's rights. I think we need to make these distinctions and speak to women who live in societies where these things are imposed in order to come to a realistic assessment.
faraz | 11 years ago | Reply @Sara Abdul Khaliq Everyone has the right to develop his own point of view regarding issues. I have no problem if a person holds extremist views and practices those views upon his own life. But the problem arises when a person imposes his views upon others. So, how did Fulton impose his views on others through an article? As an observor coming from a different cultural background, he simply made an observation. The entire world gives views about each others socio-cultural norms. But the author here is comparing him to religous extremists who impose thier views upon others through brute force. She is labelling a columnist as liberal extremists, and comparing it to a religious extremist who blows himself up in a market. You are comparing an opinion with murder? Isnt this tolerance at your part that you cant even to listen to a different point of view. And you are making assumptions from your ownself and then commenting on it; show some objectivity.
jahil_awam | 11 years ago | Reply Master piece Naureen!, and befitted reply for liberal fascists like Fulton and Fasi Zaka and few others...these bunch of ignorants need excuses to liberate Pakistani women with g-strings, tanga bikinis and what not! @Muhammed Ali Please liberate us with your views! go ahead! or else I have every reason to call you 'extremist liberal'
Majid Urrehman | 11 years ago | Reply JazakAllah Naureen!
Sara Abdul khaliq | 11 years ago | Reply enjoyed the article, i was getting quite nauseated by reading all the burqa/religion/beard (or anything related to religion) bashing articles on ET, it seems like ppl just never get tired of writing about how much they HATE the burqa/beard/ISLAM... The only problem i have with these so called 'liberal' minded people is that they claim to fight for women's rights (by being anti-burqa) but they can't seem to realized that the thing that oppresses the women (and men) of this country is POVERTY!!! but they really could care less about whether the poor of this country have food to eat or anything of that sort, all they care about is that they better not being wearing a beard, burqa or a be religious!!! These people actually can tolerate a burqa clad woman if she happens to fit into their 'oppressed' category, but they cant stand seeing someone who is educated and of their status and wearing a burqa!!!! what kind of hypocritical liberal mentality is this?!?!?! @Ayesha Ijaz Khan and @ Faraz: i think by 'extremist liberals' the author meant to refer to a group who is out to impost its point of view on other, an extremist group thinks that their opinion is the only valid one, whether other like it or not. the religious extremists are out to do the same (they think that their opinion is the only valid one and they want to impose it on others). the only difference between the two groups is in their methodology
Just Another Ayesha | 11 years ago | Reply You've got it covered, Naureen =) I honestly think that there are more pressing issues in the world than what a woman wears or doesn't wear. As far as clothes are concerned, only a liberated woman can choose to hide her beauty behind layers of cloth. Real beauty is not Photoshopped, airbrushed or plastered on billboards. It's underneath that skin. That's why many girls cover the irrelevant. And I wish that was why we'd spend more time saving Muslim women in persecution rather than questioning their choices of living.
Hira Mujahid | 11 years ago | Reply Well done naureen!
Hina | 11 years ago | Reply Great work Naureen! Putting a ban has never solved problems and it never will. As the writer said, appropriate measures and proper strategy is required to uproot such problems.
Ali | 11 years ago | Reply Burqa or jeans are not indiginous to Pakistan - our traditional dress is more than adequate for religious purposes so why do we need to adopt foreign (middle eastern or western) attire? The shalwar kameez is useful in the heat, and a properly placed dupatta is an adequate covering of the head as per religious and cultural requirements. It lends a gracefullness to the lady that no jeans/hijab/burqa ever can. It is the mark of a respectable Pakistani woman - if it was and still is good enough for my 5 time namazi mother, then why is it not good enough for the girls nowadays? Why is just being Pakistani not good enough for anyone anymore?
Tanzeela Afzal | 11 years ago | Reply Good article Naureen!
amoghavarsha.ii | 11 years ago | Reply This Blog is purely a persons opinion from the eyes through the prism of her own choice. there have been many article comommarted for burqa/niqab. All have been only for or against its wearing. THERE IS NO ARTICLE ON PRO'S AND CON'S OF WEARING OF BURQA. Hope somebody will enlighten.
Junaid Dar | 11 years ago | Reply Nice work sister! in lights of article from George I had already kept my facebook status to "I'm sensing a ban on 'Islamic Full Veil in Public' in our country very soon! And our so called beloved Media will take a big part in this Sacrilegious campaign" Now, feeling glad that can change it .... May ALLAH Bless You n Entire Ummah
yousaf | 11 years ago | Reply Its intersting to see how liberal facists are reacting to this article.... they simply cant tolerate half page response in the name of freedom of expression but they would definitlely fight for some one of their own who would proliferate humiliation against any thing related with religion. By the way how about Mr. Bush & Tony Blair, just name two liberal bigots, who killed millions just to 'protect their way of life'
SBK | 11 years ago | Reply It has become fashionable nowadays for our 'intellectual elite' to pen an article or two reminiscing about the more secular days when religion was a private affair – when there were no beards, no hijabs, no burqas and no preachers on television Simple and to the point......
IZ | 11 years ago | Reply And for that matter, why has our so-called open-minded progressive class forgotten the values of tolerance that they themselves preached? Do I smell intolerant fanaticism here? Yes, George Fulton, who, for some strange reason suddenly represents every single liberal person in Pakistan (their views are exactly the same because they belong to a group hive mind - with Mathira as the borg queen) is definitely doing EXACTLY the same thing as murdering men, women and children, blowing up schools, whipping women, burning shops and gunning people down in the streets. Writing an article suggesting a burqa ban is EXACTLY the same thing. We should immediately round up all these so-called progressives and kill them all before their intolerant fanaticism does any more harm.
Asad Shairani | 11 years ago | Reply @Ayesha Ijaz Khan: Even if we accept the statement that 'very few women wear the burqa out of choice' - which in my opinion is highly subjective - banning the burqa would not promote choice, which is the intention, but simply would change the face of enforcement. Stopping suicide bombings without proper intelligence is almost impossible, banning pillion riding, burqa, shawls etc. won't help.
S.Haque | 11 years ago | Reply Complete nonsense and twisting of ground realities at best!! Writer needs to know that many women wear that burka just to avoid the ugly stares of creatures coming out of madrassas, people roaming around in the markets and streets, ogling at the waiting areas etc. She has all the right to wear whatever she feels like, but thinking that all women wear burks, niqab, or hijab out of their own free will is as if the writer has been living in a paradise all these years and does not know the kind of society we live in.
Asad Shairani | 11 years ago | Reply @faraz: Liberal extremists are not dangerous like the traditional ones, and I don't think the writer hinted anything like that. The point I see is the intolerance towards the other belief. The right-wingers want everything "non-Islamic" banned, and it seems now the liberals are following suit on the other end. Since rational thinking is something liberals are proud of (taking the liberty since I consider myself as one), one would not expect them to "ban" stuff like this, when the logic behind it is extremely weak.
Asad Shairani | 11 years ago | Reply @Emmon Khan: This article is (probably) in response to George's article, and hence the repetition of the views was necessary.
Tyrone | 11 years ago | Reply Well written and the issue is that we are all afraid of the 'other' read To Kill a Mockingbird which shows clearly what fear does. Whether we're liberal or extreme we seem to ignore the dangers posed by men we see women as threats whether they wear burqas or skimpy clothes we're all guilty of misogyny. So Pakistanis are united in one thing the obsession over the female. This obssession takes many forms from people against burqas to people villifying prostitutes and killing women for adulter and immorality etc while men go scot free! When wil we see a campaign we can all subscribe to against the unfair treatment of women who are not naturally fair?
Rabiya | 11 years ago | Reply "Restricting women from wearing the veil is as much a violation of their rights as forcing them to wear one." Well said!! =)
yousafzai | 11 years ago | Reply excellent article!
Deen Sheikh | 11 years ago | Reply Why doea nO one write about the day to day intolerance shown by religious people towards non practising muslims. For some reason, they are unable to digest the fact that there are non practising muslims around them?
Ayesha Ijaz Khan | 11 years ago | Reply Actually Naureen, the attack in Bajaur is not the only time the burqa was used to cover up illicit activity. If you recall, the Lal Masjid maulvi also tried to escape with the help of the burqa. Moreover, there was an incident in Karachi when men carrying guns tried to create mayhem in Korangi but were fortunately stopped short of their plans when the police spotted so many burqas on a motorbike. BTW, very few women wear the burqa out of choice. The overwhelming majority are forced to wear it. So then should one be more concerned for the few who are deprived of wearing a burqa or the many who are forced to do so? Faraz makes a good point asking what do you mean by "liberal extremists"? Expressing an opinion does not make one an extremist unless the opinion incites people to violence---that then is extreme. @Reema----yes, people do need to get out of their shells and understand the peoples' perspectives and that includes the bulk of our population living in villages where women work in the field to earn a living---try doing that in a burqa!
Yusaf Khan | 11 years ago | Reply I think rather than banning the burqa we need to have a large womens police force that can do the full body checks to look for suicide bombers. I also think that the face is our identity. I don't understand why anyone would want to wear a burqa because the sole purpose of hiding the face is to hide someones identity. Even during Hajj a women is not allowed to hide her face so I cannot comprehend why some think it is an Islamic requirement to wear a burka. Still I dont believe in banning a burka.
Farah Qayum | 11 years ago | Reply In truth there isn't anything revolutionary in this article, but I'm more perturbed that people expect one article to provide an answer to a deep-rooted and complex problem. We need continuous debating about the niqaab in order to reach a sensible consensus on it. Articles like this keep the subject in the public domain, and for that reason I am glad you wrote it. I am in favour of a developed strategy for niqaab-wearing. And anyone with half a braincell can see that equates to those who choose to wear it, being legally able to wear it, whilst sustaining laws which mean it doesn't compromise safety and security. This isn't a question of religion, it's one of freedom of choice and expression. The world lacks common sense, quite tragic really.
Shabbir | 11 years ago | Reply This article is a pack of logical fallacies..!
Ammar R | 11 years ago | Reply well done naureen! murderous fanatics who kill thousands upon thousands in the name of God are the Exact moral equivalents of liberals who write their opinions about dress codes in little-read english newspapers. In fact i'm sure these liberals are more harmful. The only major cause of the relentless daily bombings is lax security after all. Certainly not socially pervasive, state-sponsored religious fanaticism. You are truly a Philosopher Queen! Don't let these 'liberal extremists' win! they might even try to stop the Righteous Fanatics from exploding their bombs every day. then who's going to do all the moral policing and save people from hellfire in this country?
Hassan Talal Maitla | 11 years ago | Reply Very well written article. Some liberals present themselves as knowledgeable, but as Stephen Hawking remarked "Its not ignorance which we should dread, its the illlusion of knowledge." Mr Fulton also depicted illusion of knowledge when he wrote that article.
Reema | 11 years ago | Reply VERY well written Naureen! Thank you for composing the perfect reply to Mr Fulton's piece. Our elites need to step out of their shells for a bit and understand other people's perspectives as well.
Muhammed Ali | 11 years ago | Reply Worst article ever !!!!
Burqa Clad | 11 years ago | Reply Excellent.. people use buzz word "Extremism" for winning any debate. Extremism is also sticking to ur notion and being intolerant let alone not knowing ur religious values to be in position to ACTually Debate on it. The burqa or niqab , cars, cellphones and the Internet are the tools that will change with time. People have enuf time to write provocative things about religious values. Had it have been expended on shedding light over root cause, security issues and other laxness issues, more awareness could have been spread. I second the writer for being so WISE and intellectual irrespective of whether she puts on a veil .Strange to read views on Mr George fulton's article. Suppressed and unwise thoughts were shared by people who were not even BURQA CLAD... amazing!!!
nida iqbal | 11 years ago | Reply @Emmon Khan: please do point out the half-baked theories(if any)!
nida iqbal | 11 years ago | Reply excellent work naureen!! simple and to the point!! and the point is crystal clear!!
Ali Hassan | 11 years ago | Reply "That there are women who have the freedom to fling off their covers and be fashionably under-dressed, yet they choose not too?" So the opposite of burqa is under-dressed.
Ameera Khan | 11 years ago | Reply Thank you SO MUCH! Totally behind you on this! Totally! :D two thumbs up I took the headcover purely out of choice and I was the first in my family to do so. Say what you will, I know I take it out of love for my religion. And what's this about covered women not being "women of substance"? Puh-lease! Give me a break! I'm graduating from med school in two months time, Insha'Allah. If you want to have this discussion, please visit my clinic Insha'Allah... you might find a GP or a resident or a consultant, Insha'Allah... and she'll be glad to have this "women of substance" discussion with you. Over and out! Hijabi and totally-not-pressured-into-it Muslimah, Ameera Khan
Farah Zahidi | 11 years ago | Reply Naureen you have brought to light a very important issue. Extremism is simply when we have no tolerance against views opposing ours and take extreme measures to lambast others with our perceptions and ideas. Stopping someone from the right to dress as they see fit is a vioaltion of a basic human right, may it be a Jew, a Hindu, a Muslim and may it be a man or a woman. Excellent write-up. Heartfelt & thought-provoking
Erfan | 11 years ago | Reply Naureen Zindabaaad !!!
Hania | 11 years ago | Reply Thank you Naureen Very rightly said, we cannot make generalizations about Islam just by looking at one two events in isolation another thing, will we ever accept a prescription from a mechanic to treat our tonsils? To get knowledge in a particular field we go to their experts Similarly, when it comes to Islamic laws, we should seek help from the experts in this field, and not accept the comments of every other layman
vikas ranjan | 11 years ago | Reply Far as I can make out the lady is a votary of complete freedom of choice in matters sartorial for all. If she is advocating the freedom of donning the all encompassing burqa and all revealing nothingness I think one has to support her. However, if the freedom of choice is defined only as the freedom of full cover and not the freedom of full exposure then she is not talking about freedom of choice as such. This then becomes the proverbial case of the devil quoting the scripture. And as the lady has so mavellously pointed out misuse of cars does not lead to banning of the car,misuse f the burqa likewise should not lead to banning of the burqa,but then this argument can be extended to anything , arms, weapons, polythene bags, hate literature, salacious matter even murder, afterall armies commit permitted murder so why ban it to the general populace, why not leave it as a matter of choice, albeit sans the misuse.
hira jeddy | 11 years ago | Reply well done naureen, iam glad that you wrote this. mr fulton must hear the other perspective too... i was very disappointed when i read what he wrote.
Emmon Khan | 11 years ago | Reply It is such articles which through half-baked theories promote and strengthen the cause of the extremists, who then get more people to go for the kill! Let them discuss issues of entertainment and gossip but do not let them destroy the minds of young people and the ultimately the future of this hapless country!
parvez | 11 years ago | Reply You have cleverly over simplified the case to make your point. But then in your last para you admit that the burqa has the potential of being misused and a strategy is required to prevent this, so there in two lines you have endorsed the case of the other side. The debate on this complicated issue involves religion,societal norms,culture and politics will continue for decades.
faraz | 11 years ago | Reply I agree that burqa ban would voilate the right to freedom of action. But what exactly do you mean by "liberal extremism"? You are equating suicide bombers and murderous gangs with a suggestion in a newspaper article! So Fulton is a liberal extremists because he wrote something in an article! Can you give any example of a "liberal extremist" who murdered innocent people? So people cant even write whats on their mind.
Emmon Khan | 11 years ago | Reply Useless, tired views expressed a million times. A break plz!
Sheila | 11 years ago | Reply THANK YOU Naureen! Finally a logical response to George Fulton's patronising previous article.

redpajama_set_name: RedPajamaCommonCrawl


text:
#import <CoreGraphics/CoreGraphics.h>
#import "ViewController.h"
#import "MGSequentialCommandGroup.h"
#import "PKMoveAnimation.h"
#import "PKEaseLinear.h"
#import "PKScaleAnimation.h"
#import "PKFadeAnimation.h"
#import "PKEaseBack.h"
#import "PKEaseQuadratic.h"

@interface ViewController ()

@property(nonatomic, strong) UIImageView *rectangle;
@end

@implementation ViewController

- (void)viewDidLoad {
    [super viewDidLoad];

    [self createRectangle];
    [self createAnimations];
}

- (void)createRectangle {
    self.rectangle = [[UIImageView alloc] initWithImage: [UIImage imageNamed: @"rectangle.png"]];
    [self centerRectangle];

    [self.view addSubview: self.rectangle];
}

- (void)createAnimations {
    MGSequentialCommandGroup *sequence = [[MGSequentialCommandGroup alloc] init];

    [sequence addCommand: [[PKMoveAnimation alloc] initWithView: self.rectangle
                                                       duration: 1.0f
                                                             by: CGPointMake(0.0f, 100.0f)
                                                        options: @{@"ease" : [[PKEaseQuadraticInOut alloc] init], @"delay" : [NSNumber numberWithFloat: 1.3f]}]];

    [sequence addCommand: [[PKScaleAnimation alloc] initWithView: self.rectangle
                                                        duration: 1.3
                                                            from: 1.0f to: 1.3f]];

    [sequence addCommand: [[PKFadeAnimation alloc] initWithView: self.rectangle
                                                        duration: 2.0
                                                            from: 1.0f to: 0.2f]];

    [sequence addCommand: [[PKFadeAnimation alloc] initWithView: self.rectangle
                                                       duration: 0.3
                                                           from: 0.2f to: 1.0f
                                                        options: @{@"delay" : [NSNumber numberWithFloat: 1.3f]}]];

    [sequence addCommand: [[PKMoveAnimation alloc] initWithView: self.rectangle
                                                       duration: 1.0f
                                                             by: CGPointMake(0.0f, -100.0f)
                                                        options: @{@"ease" : [[PKEaseBackOut alloc] init]}]];

    [sequence addCommand: [[PKScaleAnimation alloc] initWithView: self.rectangle
                                                        duration: 0.8
                                                            from: 1.3f to: 1.0f]];

    [sequence execute];
}

- (void)centerRectangle {
    CGSize winSize = self.view.frame.size;
    CGSize size = self.rectangle.frame.size;
    self.rectangle.frame = CGRectMake(winSize.width/2 - size.width/2, winSize.height/2 - size.height, size.width, size.height);
}

@end

redpajama_set_name: RedPajamaGithub


Have you at any time stopped to think about the variety of fabric? There are so numerous types of material. From content, to weave, to knit – from color to finishes.the selection is really amazing! Becoming able to function with all of this beauty and variety is 1 of the best factors to learn how to sew.
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Although mattress covers for mattress bugs produced of vinyl are a great deal less expensive, individuals nonetheless choose the types produced of fabric because of the ease and comfort they provide. Vinyl encasements have a tendency to tear up too effortlessly as well. This indicates that even if you are in a position to arrive up cheaper, you would have to purchase again in no time. The good thing about vinyl encasements although is its simplicity of cleaning. You can just use a moist fabric and you're prepared to go. Also, they arrive in different forms and sizes to snuggly match your mattress no make a difference what type it is.
Empty plastic Milk Jugs are perfect to use as a vase insert in a bowl. You can slip them into almost something big and spherical to produce a unique foundation to your centerpiece. It should fit most terra cots as nicely.
However, there is another kind of carpet that is utilized for flooring mats is produced from different supplies. These supplies are obtained straight from character. For example, the supplies that are used are woven material of hyacinth leaves, midrib of coconut, pendants, seaweed, and mats. Additionally, the very best place to place these kinds of mats is on the floor that is not humid. Splashing or spillage of drinking water could create a tough stains disappear. When there are spills, clean them instantly and sunbathe under the sunlight heat directly.


text:
\section{Introduction.}

For a system whose configuration space has dimension $n$, integrability in the sense of Liouville-Arnold requires the existence of a number $n$ of constants of motion which should be in involution (the Hamiltonian itself is one of them). Integrable systems are rare, yet many systems which are important from a physical standpoint turn out to be integrable, with the outstanding examples of the harmonic oscillator and the Kepler problem, which in fact are superintegrable (they have more than $n$ functionally independent constants of motion, albeit not all of them are in involution).

A type of integrability appears to be particularly relevant: the {\em quadratic integrability}.
A standard natural Hamiltonian has a kinetic part which is {\it quadratic} in the momenta. If all $n-1$ additional constants of motion have a similar kind of structure, we will speak of quadratic integrability (understanding quadratic as `at most quadratic', i.e., allowing possibly for constants of motion which are linear in the momenta).

For a 2D system in any constant curvature configuration space, quadratic integrability means the existence of a single constant of motion quadratic in the momenta; this case is simpler as there is no any extra condition ensuring the additional constants are in involution \cite{BaHeSaSa03,SaSa05}. For any space of constant curvature, integrable systems have a Hamiltonian which is  separable in confocal (or general elliptic) coordinates in that space \cite{Benen97, Cram03}. These coordinates include generic ones (as the elliptic coordinates in the sphere $S^2$ or the Euclidean plane $E^2$) as well as all its possible degenerations or limiting cases. This also holds in the 3D case \cite{HoMcLSm05, Ev90}. It is generally believed that a similar connection among quadratic integrability and the $n$D version of confocal coordinates does somehow hold, but to the best of our knowledge a proof is not available in the literature.

In this work we explore some aspects of quadratic integrability for a system in 3D configuration spaces $E^3$ or $S^3$. The results should be considered as a stage towards studying  how precisely integrable systems in Euclidean or in any constant curvature 3D space are related to separable systems in the generic version of ellipsoidal coordinates, associated to confocal quadrics \cite{Ka99,Ka02}. We discuss mainly the Euclidean case, but extension of these results to spaces with any constant curvature seems to be possible, and as a hint in that direction, we discuss the integrability with axial symmetry and the Kepler problem in a space with positive constant curvature $S^3$ (see also \cite{HeBa06}).

\section{3D systems with axial symmetry.}

In this first part, we consider a three dimensional system in
Euclidean space with axial symmetry around an axis given by the unit
vector ${\bf n}=(n_1,n_2,n_3)$. The classical analysis of this
system makes use of canonical coordinates $({\bf p},{\bf
x})=(p_i,x_i)$, $i=1,2,3$ and Poisson brackets
$\{A,B\}=\frac{\partial A}{\partial p_k}\frac{\partial B}{\partial
x_k}-\frac{\partial A}{\partial x_k}\frac{\partial B}{\partial
p_k}$. We want to characterize those systems with axial symmetry
which are completely integrable. For these systems dynamics is described by a Hamiltonian of the form
\begin{equation}\label{1}
    H= \frac{{\bf p}^2}{2m} + U({\bf x})\,,
\end{equation}
where $U(\bf x)$ is a time independent potential. Axial symmetry requires that the angular momentum $L$ around the symmetry axis
\begin{equation}\label{2}
L = {\bf n}\cdot\left({\bf x}\times {\bf p}\right)
\end{equation}
be a constant of motion, i.e., $\{H,L\}=0$. Complete quadratic integrability
requires  the existence of a third independent constant of motion $H_1$,
which is in involution with $H$ and $L$, i.e.,
$\{H,H_1\}=\{L,H_1\}=0$ and has a `Hamiltonian form' in general position dependent yet quadratic in the momenta,
plus a time independent `potential', with form
\begin{equation}\label{3}
 H_1=\frac{1}{2m}\,p_i\,g^{ik}({\bf x})\,p_k + \Phi({\bf x})\,.
\end{equation}
The tensor $g^{ik}({\bf x})$, which determines the `kinetic' part of
the constant of motion $H_1$ has to be a Killing tensor for the Euclidean metric. Its determination can be done by assuming that the `kinetic' terms in (\ref{1}) and (\ref{3}) commute among themselves
and also with the angular momentum $L$, with respect to the Poisson
bracket. If we choose the angular momentum direction as the $z$ axis, then the most general expression for $\,p_i\,g^{ik}({\bf x})\,p_k$ is
\cite{Wi67}:
\begin{equation}\label{1a}
    \,p_i\,g^{ik}({\bf x})\,p_k = L_1^2 + L_2^2 + \alpha p_3^2 +\beta (p_1L_2 - p_2L_1) + \gamma p_3L_3
\end{equation}
to which of course any linear combination of the quadratic Casimirs of the Euclidean algebra, $p_1^2 + p_2^2 + p_3^2$ and $p_1L_1 + p_2L_2 + p_3L_3$ can be added. Here we will discuss mainly the family with $\alpha<0, \beta=0, \gamma=0$, for which
\begin{equation}\label{1b}
   \,p_i\,g^{ik}({\bf x})\,p_k = L_1^2 + L_2^2 -a^2 p_3^2
\end{equation}
with $a$ a positive constant. If now the angular momentum direction is along a general unit vector ${\bf n}=(n_1,n_2,n_3)$, by  direct computation we get for the Killing tensor $g^{ik}({\bf x})$ the expression
\begin{equation}\label{4}
g^{ik}({\bf x})= \delta^{ik} ({\bf x}\cdot{ \bf n})^2 - {\bf x}\cdot
{\bf n} (x_i n_k + x_k n_i) + ({\bf x}^2- a^2)n_i n_k\,,
\end{equation}
For more details, see \cite{GaNePr07}. The commutation
of $L$ with $H$ and $H_1$ restricts the form of the functions $U$ and $\Phi$, collectively called `potential' terms to:

\begin{equation}\label{5}
U({\bf x}) = U({\bf x}^2,({\bf x}\cdot{ \bf n})^2),\qquad \Phi({\bf
x}) = \Phi({\bf x}^2,({\bf x}\cdot{ \bf n})^2)
\end{equation}
(note that the condition of axial symmetry around the $\bf n$ axis implies that $U$ and $\Phi$ cannot depend on the azimuthal angle $\varphi$) and the commutation of $H$ and $H_1$ leads to the equations:

\begin{equation}\label{6}
\partial_{i} \Phi({\bf x}) =
g^{ik}({\bf x})\partial_{k} U({\bf x})\,,\qquad i,k=1,2,3\,,
\end{equation}
where we always assume summation over repeated indices. In order to
obtain the solutions of (\ref{6}), we diagonalize the matrix
with components $g^{ik}({\bf x})$. Its eigenvalues $\lambda({\bf
x})$ and eigenvectors $A({\bf x})$ are obtained from the matrix
equation

\begin{equation}\label{7}
    \left(g^{ik}({\bf x})- \lambda({\bf x}) \delta^{ik}\right)
A_k({\bf x})=0\,.
\end{equation}

The eigenvalue equation (\ref{7}) has the following solutions:

\begin{equation}\label{8}
    \lambda_{\pm}({\bf x}) = \frac{{\bf
x}^2-a^2}{2}\pm\sqrt{\left(\frac{{\bf x}^2-a^2}{2}\right)^2+ a^2\,
{\bf x \bf n}^2},\qquad \lambda_0({\bf x})= {\bf x \bf n}^2\,,
\end{equation}
with corresponding eigenvectors given by

\begin{equation}\label{9}
A^\pm_i({\bf x})=\partial_i \lambda_\mp({\bf x})\,,\qquad A^0_i({\bf
x})= \frac{{\bf n}\times{\bf x}}{{\bf x}^2-{\bf x \bf n} ^2}=
\partial_{i}\varphi({\bf x})\,,
\end{equation}
where the index $i=1,2,3$ labels the component of the corresponding
vector field and $\varphi$ is the azimuthal angle around the ${\bf
n}$-axis. These vectors form a new coordinate basis, $\{
\mathbf{\nabla}\lambda_-,\mathbf{ \nabla}\lambda_+,\mathbf{
\nabla}\varphi\}$ referred to which the matrix $g^{ik}({\bf
x})$ is diagonal. Taking into account that, due to the geometric
symmetry, $U$ and $\Phi$ cannot depend on $\varphi$, (\ref{6})
becomes

\begin{equation}\label{10}
\partial_i\lambda_+\,\partial_+\Phi +
 \partial_i\lambda_-\,\partial_-\Phi=
g^{ik}\left(\partial_k\lambda_+\,\partial_+U +
\partial_k\lambda_-\,\partial_-U\right)\,,
\end{equation}
where $\partial_\pm$ stand for
$\frac{\partial}{\partial\lambda_\pm}$. In the basis $\{\mathbf{\nabla}\lambda_-,\mathbf{ \nabla}\lambda_+,\mathbf{
\nabla}\varphi\}$ that
diagonalizes the matrix $g^{ik}({\bf x})$, equation (\ref{10}) decouples in

\begin{equation}\label{11}
    \partial_+ \Phi = \lambda_- \partial_+U\,,\qquad \partial_- \Phi =
 \lambda_- \partial_+U
    \,.
\end{equation}
This means that

\begin{equation}\label{12}
\Phi - \lambda_+ U = -f(\lambda_+),\qquad   \Phi - \lambda_- U =
-g(\lambda_-)\,,
\end{equation}
where $f(\lambda_+)$ and $g(\lambda_-)$ are arbitrary functions on
their respective variables. These equations give an expression for
the `potentials' as follows:

\begin{equation}\label{13}
U=\frac{f(\lambda_+)-g(\lambda_-)}{\lambda_+-\lambda_-},\qquad
\Phi=\frac{\lambda_-}{\lambda_+-\lambda_-}\,f(\lambda_+) -
\frac{\lambda_+}{\lambda_+-\lambda_-}\,g(\lambda_-)\,.
\end{equation}
These are the most general expressions for the potentials $U({\bf
x})$ and $\Phi({\bf x})$ compatible with our choice for $H_1$ and with the requirement of $H, H_1, L$ being in involution.

A comment is here in order: the presence of a constant of motion which is first-order in the momenta, i.e. is a Noether constant, means that the system is invariant under rotations around the axis ${\bf n}$. Then this situation reduces to one that is essentially two-dimensional, with a single azimutal coordinate $\varphi$ added to the ordinary 2D elliptic coordinates in a fixed plane containing the axis ${\bf n}$. In the 3D Euclidean space these coordinates are the spheroidal (oblate) cordinates \cite{MiPaZa81}, which play the role of 3D separation coordinates for the system, as we will show explicitly in the next section.

\subsection{Integration.}

The integration of the classical system with axial symmetry with
respect to the $\bf n$ axis is based in the following idea: If
$e_i=h({\bf p},{\bf x})$, $i=1,2,3$ are three independent functions
with $\{e_i,e_j\}=0$, then, there exists a function $F({\bf x},{\bf
e})$ with ${\bf e}=(e_1,e_2,e_3)$ and

\begin{equation}\label{14}
p_k=\frac{\partial F({\bf x},{\bf e})}{\partial x_k}\,.
\end{equation}
The function $F({\bf x},{\bf e})$ is the characteristic function in
the Hamilton-Jacobi approach.

In our case, the chain rule and (\ref{14}) gives the following
expression for the momentum ${\bf p}=(p_1,p_2,p_3)$:

\begin{equation}\label{15}
\fl
{\bf p} = \mathbf{\nabla}\lambda_+\,
\partial_+F(\lambda_+,\lambda_-,\varphi)+ \mathbf{\nabla}\lambda_-\,
\partial_-F(\lambda_+,\lambda_-,\varphi)+ \frac{{\bf n}\times{\bf
x}}{{\bf x}^2-({{\bf x}\cdot{\bf n}})^2}\,
\partial_\varphi F(\lambda_+,\lambda_-,\varphi)\,.
\end{equation}
Then, the kinetic part of $H$ can be written in the following form:

\begin{equation}\label{16}
\fl
{\bf p}^2 = (\mathbf{\nabla}\lambda_+)^2 (\partial_+ F)^2+
(\mathbf{\nabla}\lambda_-)^2 (\partial_- F)^2+ \frac{\ell^2}{{\bf
x}^2-({{\bf x}\cdot{\bf n}})^2}= 2m\left(E-U\right)\,,
\end{equation}
where $\ell$ is the value of the integral of motion $L$
corresponding to the angular momentum around ${\bf n}$, i.e.,

\begin{equation}\label{17}
\left({\bf n}\times{\bf x}\right)\cdot {\bf p} = {\bf n}\cdot {\bf
L} = \ell=\partial_\varphi F(\lambda_+,\lambda_-,\varphi)
\end{equation}
and $E$ is a given value of the constant of motion $H$.

For the kinetic term of $H_1$, we obtain a similar expression:

\begin{equation}\label{18}
\fl
p_ig^{ik}p_k = \lambda_-(\nabla \lambda_+)^2 (\partial_+ F)^2+
\lambda_+(\nabla\lambda_-)^2 (\partial_- F)^2+ \frac{({{\bf
x}\cdot{\bf n}})^2\,\ell^2}{{\bf x}^2-({{\bf x}\cdot{\bf n}})^2}=2m(
E_1 - \Phi)\,,
\end{equation}
where $E_1$ is a constant value of $H_1$.

With some calculations \cite{GaNePr07}, we obtain the partial derivatives
of the function $F(\lambda_+,\lambda_-,\varphi)$ in terms of its
arguments:
\begin{eqnarray}
\fl
\displaystyle
 (\partial_{+}F)^2
=
\frac{1}{4\lambda_{+}(\lambda_{+}+a^2)}\left\{2m\left(\lambda_{+}(E-U)-(
E_1-\Phi)\right)
-\frac{ \lambda_{+}l^2}{\lambda_{+}+a^2}\right\}\,, \label{19}\\[2ex]
\fl
\displaystyle
 (\partial_{-}F)^2
=
\frac{1}{4\lambda_{-}(\lambda_{-}+a^2)}\left\{2m\left(\lambda_{-}(E-U)-(
E_1-\Phi)\right) -\frac{ \lambda_{-}l^2}{\lambda_{-} +a^2}\right\}
\,. \label{20}
\end{eqnarray}

Due to (\ref{12}), (\ref{19}) does not depend on $\lambda_-$ and
(\ref{20}) does not depend on $\lambda_+$. From this simple idea, we
conclude that the function $F(\lambda_+,\lambda_-,\varphi)$ is of
the form:

\begin{equation}\label{21}
F(\lambda_+,\lambda_-,\varphi)=A(\lambda_+)+B(\lambda_-)+\ell\varphi\,,
\end{equation}
where $A(\cdot)$ and $B(\cdot)$ are functions of one variable only.
Thus, we have separated variables in the Hamilton-Jacobi generating
function. This permits us to obtain the equations of motion in terms
of the variables $(\lambda_+,\lambda_-,\varphi)$. The final result
is \cite{GaNePr07}
\begin{eqnarray}
\displaystyle &\frac{\dot
\lambda_{+}}{\lambda_{+}\partial_{+}A(\lambda_{+})} +\frac{\dot
\lambda_{-}}{\lambda_{-}\partial_{-}B(\lambda_{-})}
=\frac{4}{m} \nonumber\\[2ex]
\displaystyle &\frac{\dot
\lambda_{+}}{\lambda_{+}(\lambda_{+}+a^2)\partial_{+}A(\lambda_{+})}
+\frac{\dot
\lambda_{-}}{\lambda_{-}\partial_{-}(\lambda_{-}+a^2)B(\lambda_{-})}=0
\nonumber\\[2ex]
\displaystyle &-\frac{la^2}{4}\left(\frac{\dot
\lambda_{+}}{\lambda_{+}(\lambda_{+}+a^2)^2\partial_{+}A(\lambda_{+})}
+\frac{\dot
\lambda_{-}}{\lambda_{-}\partial_{-}(\lambda_{-}+a^2)^2B(\lambda_{-})}\right)=\dot
\varphi\,, \label{22}
\end{eqnarray}
where the upper dot represents the derivative with respect time.
These equations are at least formally integrable.

\subsection{Quantum case.}

Canonical quantization of the functions $H,L,H_1$ give respective
operators that we represent with the same symbols. Then, complete
integrability means that

\begin{equation}\label{23}
    [H,L]=[H_1,L]=[H,H_1]=0\,,
\end{equation}
where $[A,B]=AB-BA$ denotes the commutator of the operators $A$ and
$B$. Equation (\ref{23}) implies the existence of simultaneous
eigenfunctions $\psi({\bf x})$ of these three operators:

\begin{equation}\label{24}
(H-E)\psi = (H_1-  E_1)\psi = (L - \ell)\psi=0\,,
\end{equation}
or equivalently,
\begin{eqnarray}
-\Delta \psi({\bf x}) = 2m(E- U({\bf x}))\psi({\bf x})
\label{25}\\[2ex]
 -\Delta_1 \psi({\bf x}) =
2m( E_1- \Phi({\bf x}))\psi({\bf x})\label{26} \\[2ex]
-i\partial_{\varphi} \psi({\bf x}) = \ell\, \psi({\bf
x})\,,\label{27}
\end{eqnarray}
where,

\begin{equation}\label{28}
\Delta = \partial_k\partial_k\,, \qquad  \Delta_1 = \partial_j
g^{jk}({\bf x}) \partial_k\,.
\end{equation}

We can express (\ref{25}) and  (\ref{26}) in terms of the variables
$\lambda_\pm$. Using (\ref{27}) we obtain respectively,
\begin{eqnarray}
\fl
-\left[4\lambda_{+}(\lambda_{+}{+}a^2)\psi_{++}
+2(a^2{+}3\lambda_{+})\psi_{+} - \frac{l^2 \lambda_{+} }{\lambda_{+}
{+}a^2}\psi \right
= 2m \left[(E{-}U)\lambda_{+} {-} (
E_1 {-}\Phi)\right]\psi\, ,\label{29}\\[2ex]
\fl
-\left[4\lambda_{-}(\lambda_{-}{+}a^2)\psi_{--}
+2(a^2{+}3\lambda_{-})\psi_{-} - \frac{l^2 \lambda_{-} }{\lambda_{-}
{+}a^2}\psi \right
 = 2m \left[(E{-}U)\lambda_{-} {-}
(E_1{-}\Phi)\right]\psi\,. \label{30}
\end{eqnarray}
Note that (\ref{29}) and (\ref{30}) depend only on $\lambda_+$ and
$\lambda_-$ respectively. Then, the wave function $\psi({\bf x})$
can be factorized as

\begin{equation}\label{31}
\psi({\bf x})=\psi^+(\lambda_+)\,\psi^-(\lambda_-)\,e^{i\ell\varphi}
\end{equation}
so that (\ref{28}), (\ref{29}) and (\ref{27}) can be written as
equations depending solely on the variables
$(\lambda_+,\lambda_-,\varphi)$ respectively. Once we have solved
equations (\ref{28}) and (\ref{29}), we have solved the problem of
finding solutions of (\ref{24}) or, equivalently,
(\ref{25}-\ref{27}). In order to finding solutions, we have to give
explicit expressions for the potentials $U$ and $\Phi$. These
expressions must be obtained from (\ref{13}) and the form of the
functions $f(\lambda_+)$ and $g(\lambda_-)$.

In order to illustrate the search for solutions of (\ref{29}) and
(\ref{30}), let us propose a simple although nontrivial choice of
$f(\lambda_+)$ and $g(\lambda_-)$ so that (\ref{28}) and (\ref{29})
are exactly solvable. Here, we propose

\begin{equation}\label{32}
f(\lambda_+) = \Phi-\lambda_+U = 0\,,\qquad g(\lambda_-) =
\Phi-\lambda_-U = - Q (\lambda_-+a^2)\,,
\end{equation}
where $Q$ is a constant. This choice give the following form for the
potentials:

\begin{equation}\label{33}
U({\bf x}) = -Q\, \frac{\lambda_-+a^2}{\lambda_+-\lambda_-}\,,\qquad
 \Phi({\bf  x}) = - Q\,
\frac{\lambda_+(\lambda_-+a^2)}{\lambda_+-\lambda_-}\,.
\end{equation}

The steps to solve (\ref{28}) and (\ref{29}) with these potentials
are the following: i) From (\ref{8}), we conclude that $\lambda_+$
is always positive meanwhile the values of $\lambda_-$ lie on the
interval $[-a^2,0]$. This suggest the following change of variables:
$\lambda_+ = a^2 \sinh^2\alpha$ and $\lambda_- = -a^2 \sin^2 \beta$.
The new coordinates $(\alpha,\beta,\varphi)$ are known as the {\it
oblate} spherical coordinates. This change of variables transforms
(\ref{28}) and (\ref{29}) into two new wave functions \cite{GaNePr07}.
ii) The new change of variables $t:=\sinh\alpha$ and $u:=\sin\beta$
and the introduction of the new parameters, ${\cal E}:=2ma^2E,
q:=2ma^2Q, {\cal E}_1:=\ell^2+2mE_1, G:={\cal E}_1+{\cal E}$ and
$q'=q-{\cal E}$, with $t=i\alpha$ transform equations (\ref{28}) and
(\ref{29}) respectively into
\begin{eqnarray}
\fl
 (1-\alpha^2) \frac{d^2\psi^+(\alpha)}{d \alpha^2} -
2\alpha\,\frac{d\psi^+(\alpha)}{d \alpha}
+\left\{G-\frac{\ell^2}{1-\alpha^2}-{\cal
E}(1-\alpha^2)\,\right\}\psi^+(\alpha)\,, \label{34}\\[2ex]
\fl
(1-u^2) \frac{d^2\psi^-(u)}{d u^2} - 2u\,\frac{d\psi^-(u)}{d u}
+\left\{ G- \frac{\ell^2}{1-u^2}  +q'(1-u^2)\, \right\}
\psi^-(u)=0\,.\label{35}
\end{eqnarray}

Equations (\ref{34}-\ref{35}) are spheroidal equations, a type
of second order differential equation which has been studied
\cite{MeScWoBKMaFu}. A brief discussion of the solutions for
(\ref{34}-\ref{35}) and therefore for (\ref{29}-\ref{30}) can be
found in \cite{GaNePr07}.

\section{General 3D integrable systems.}

In this section, we intend to generalize the previous discussion to
the case in which no (Noether) symmetry is present. As in the previous case,
we shall discuss the classical point of view first and then its
quantum counterpart. The key of the solution to this general case
will be the choice of a proper system of coordinates, as we shall
see. This justifies the next subsection.

\subsection{The ellipsoidal coordinates.}

The generic ellipsoidal coordinate system in the Euclidean three
dimensional space is determined as follows. Fix three positive
numbers $a$, $b$ and $c$ with $a>b>c$, and consider the one-parameter family of quadrics with equation
\begin{equation}\label{370}
 \frac{x^2}{a^2+\xi}+ \frac{y^2}{b^2+\xi}+
    \frac{z^2}{c^2+\xi}=1\,,
\end{equation}
Through any given point $(x,y,z)$ in Euclidean 3D space, there passes precisely three such quadrics, corresponding to the values $\lambda, \mu, \nu$ of the parameter $\xi$ which lie in the intervals
\begin{equation}\label{36}
-a^2< \nu < -b^2 < \mu < -c^2 < \lambda \, .
\end{equation}
One quadric is an ellipsoid (because $a^2+\lambda>b^2+\lambda>c^2+\lambda>0$)
\begin{equation}\label{37}
 \frac{x^2}{a^2+\lambda}+ \frac{y^2}{b^2+\lambda}+
    \frac{z^2}{c^2+\lambda}=1\, ,
\end{equation}
other is an one sheeted hyperboloid (because $a^2+\mu>b^2+\mu>0>c^2+\mu$)

\begin{equation}\label{38}
\frac{x^2}{a^2+\mu}+ \frac{y^2}{b^2+\mu}+
    \frac{z^2}{c^2+\mu}=1\,,
\end{equation}
and the last is a two sheeted hyperboloid (because $a^2+\nu>0>b^2+\nu>c^2+\nu$)

\begin{equation}\label{39}
\frac{x^2}{a^2+\nu}+ \frac{y^2}{b^2+\nu}+
    \frac{z^2}{c^2+\nu}=1\,.
\end{equation}

Solving (\ref{37}-\ref{39}) for $x^2$, $y^2$ and $z^2$, we obtain the parametrization of 3D Euclidean space in terms of ellipsoidal coordinates, $\lambda$, $\mu$ and $\nu$:
\begin{eqnarray}
x^2=\frac{(\lambda+a^2)(\mu+a^2)(\nu+a^2)}{(b^2-a^2)(c^2-a^2)}\,,
 \label{40}
\\[2ex]
 y^2=\frac{(\lambda+b^2)(\mu+b^2)(\nu+b^2)}{(c^2-b^2)(a^2-b^2)}\,,
 \label{41}
 \\[2ex]
z^2=\frac{(\lambda+c^2)(\mu+c^2)(\nu+c^2)}{(a^2-c^2)(b^2-c^2)}\,.\label{42}
\end{eqnarray}
Inversion of (\ref{40}-\ref{42}) gives ellipsoidal coordinates in
terms of Cartesian coordinates. Note that ellipsoidal coordinates are only one-to-one in the interior of each octant of the Euclidean space determined by the three coordinate planes $x=0, y=0, z=0$ through the origin.

\subsection{Classical 3D integrable systems.}

Let us study the situation from classical point of view first. Consider a three dimensional system with Hamiltonian given by
$H=\frac{{\bf p}^2}{2m}+U({\bf x})$ and look for two additional
independent integrals of motion of the form:

\begin{equation}\label{43}
\fl
H_1=\frac{1}{2m}p_ig_1^{ij}({\bf x})p_j+U_1({\bf x})\,, \qquad
H_2=\frac{1}{2m}p_ig_2^{ij}({\bf x})p_j+U_2({\bf x})\,,
\end{equation}
with $\{H,H_1\}=\{H,H_2\}=\{H_1,H_2\}=0$, where again the brackets
are the Poisson brackets. The tensors $g_1^{ij}({\bf x})$ and
$g_2^{ij}({\bf x})$ can be obtained using the hypothesis of
commutativity of the kinetic parts of $H$, $H_1$ and $H_2$ (due to
their structure, we could extend the name `Hamiltonian' to $H_i$,
$i=1,2$ too). The condition

\begin{equation}\label{44}
    \{p_i g_1^{ik}({\bf x})p_k,\ {\bf p}^2\}=0
\end{equation}
implies that $p_i g_1^{ik}({\bf x})p_k$ should be a quadratic
function of the momenta. A particular choice follows from a further restriction of $H_1, H_2$ being invariant under any reflection, so that both can be expressed as a
linear combination of the squares $L_i^2$ of the components $L_i$ of the angular momentum
and $p_i^2$ of the linear the momentum $p_i$. There is some arbitrariness in the choice of $H_1$, for which we set

\begin{equation}\label{45}
p_i g_1^{ik}({\bf x})p_k: ={\bf L}^2-p_1^2(b^2+c^2)-p_2^2(a^2+c^2)
-p_3^2 (a^2+b^2)\,,
\end{equation}
where $a$, $b$ and $c$ are the constants which specify the system of
ellipsoidal coordinates, see (\ref{36}).  In matrix form, the tensor
$g_1^{ik}({\bf x})$ defined by (\ref{45}) can be written as

\begin{equation}\label{46}
\fl
g_1^{ik}({\bf x})=\left(
\begin{array}{ccc}
  z^2+y^2-(a^2+c^2) & -xy & -xz \\
  -xy & x^2+z^2-(a^2+c^2) & -yz \\
  -xz & -yz & x^2+y^2-(a^2+b^2) \\
\end{array}
\right)\,.
\end{equation}

Analogously, the commutativity of the kinetic part of $H_2$ with
those of $H$ and $H_1$,

\begin{equation}\label{47}
\{p_i g_2^{ik} ({\bf x}) p_k,\ p_i g_1^{ik}({\bf x}) p_k\}=0
\end{equation}
fixes the linear
combination of the components of ${\bf L}^2$ and of ${\bf p}^2$ in $H_2$ as
\begin{equation}\label{48}
p_i g_2^{ik}({\bf x}) p_k=-
(L_1^2a^2+L_2^2b^2+L_3^2c^2)+p_1^2b^2c^2+p_2^2a^2c^2+p_3^2a^2b^2\,,
\end{equation}
with matrix form given by

\begin{equation}\label{49}
\fl
g_2^{ik}({\bf x})=\left(
\begin{array}{ccc}
  -c^2y^2-b^2z^2+b^2c^2 & c^2xy & b^2xz \\
  c^2xy & -c^2x^2-a^2z^2+a^2c^2 & a^2yz \\
  b^2xz & a^2yz & -b^2x^2-a^2y^2+a^2b^2 \\
\end{array}
\right)\,.
\end{equation}

Matrices (\ref{46}) and (\ref{49}) are symmetric and therefore
diagonalizable. In addition, these matrices commute and therefore
admit a diagonal form in the same basis. This basis is given by

\begin{equation}\label{50}
\fl
E_\lambda:= ( \partial_x\lambda, \partial_y\lambda,
\partial_z\lambda)\,,\;\; E_\mu:=  ( \partial_x\mu, \partial_y\mu,
\partial_z\mu)\,,\;\;E_\nu:= ( \partial_x\nu, \partial_y\nu,
\partial_z\nu)\,.
\end{equation}
These vectors are mutually orthogonal \cite{GaIoNePr07}. Note that
$\lambda$, $\mu$ and $\nu$ are functions of the Cartesian
coordinates, so that the partial derivatives in (\ref{50}) make
sense. The eigenvalues of (\ref{46}) and (\ref{49}) are easy to
obtain and are given in the following table:

\begin{equation}\label{51}
\begin{array}{cccc} \hspace{2cm}   & \qquad g^{ij}({\bf x}) &\qquad
 g^{ij}_1({\bf x}) &\qquad
 g_2^{ij}({\bf x}) \\[2ex]
  E_\lambda & 1 & \mu+\nu & \mu \nu \\
  E_\mu & 1 & \lambda+\nu & \lambda \nu \\
  E_\nu & 1 & \lambda+\mu & \lambda \mu
\end{array}
\end{equation}

Thus, we have determined the kinetic parts of $H_1, H_2$ in terms of ellipsoidal
coordinates. Our next goal is to obtain the most general form of the
`potential' terms $U({\bf x})$, $U_1({\bf x})$ and $U_2({\bf x})$.
From the commutation relations $\{H,H_1\}=\{H,H_2\}=\{H_1,H_2\}=0$,
we obtain the relations,

\begin{equation}\label{52}
\partial^i U_1 =  g_1^{ik}\partial_k U\,,\quad \partial^i U_2 =
 g_2^{ik}\partial_k
U\,, \quad  g_1^{ik}\partial_k U_2 = g_2^{ik}\partial_k  U_1\,.
\end{equation}

Then, after a calculation that makes use of the chain rule involving
ellipsoidal and Cartesian coordinates, the form (\ref{50}) of the
eigenvectors of matrices $g_1^{ik}({\bf x})$ and $g_2^{ik}({\bf x})$
and the orthogonality of these vectors, we can obtain the most
general form of the potentials $U$, $U_1$ and $U_2$ satisfying
$\{H,H_1\}=\{H,H_2\}=\{H_1,H_2\}=0$ \cite{GaIoNePr07}. This is:
\smallskip
\begin{eqnarray}
  U &=& \frac{l(\lambda)}{(\lambda-\nu)(\lambda-\mu)} +
   \frac{m(\mu)}{(\mu-\nu)(\mu-\lambda)} +
   \frac{n(\nu)}{(\nu-\mu)(\nu-\lambda)}\,,\label{53} \\ [2ex]
   U_1 &=& \frac{(\mu+\nu)\, l(\lambda)}{(\lambda-\nu)(\lambda-\mu)} +
   \frac{(\nu+\lambda)\, m(\mu)}{(\mu-\nu)(\mu-\lambda)} +
   \frac{(\lambda+\mu)\, n(\nu)}{(\nu-\mu)(\nu-\lambda)}\,,\label{54}
   \\[2ex]
 U_2 &=& \frac{(\mu\nu)\, l(\lambda)}{(\lambda-\nu)(\lambda-\mu)} +
   \frac{(\nu\lambda)\, m(\mu)}{(\mu-\nu)(\mu-\lambda)} +
   \frac{(\lambda\mu)\, n(\nu)}{(\nu-\mu)(\nu-\lambda)}\,,\label{55}
\end{eqnarray}
where $l(\lambda)$, $m(\mu)$ and $n(\nu)$ are arbitrary functions of
their arguments. This is the most general form of the potentials
compatible with complete integrability.

\subsection{The quantum case.}

We shall briefly comment the procedure here; details can be found in
\cite{GaIoNePr07}. First of all, we obtain the quantum operators $H$,
$H_1$ and $H_2$ by canonical quantization of their classical
counterparts (\ref{43}). Then, complete integrability means that
these three Hamiltonians commute with each other. Therefore, we can
find wave functions $\psi({\bf x})$ such that

\begin{equation}\label{56}
    (H-E)\psi=(H_1-E_1)\psi=(H_2-E_2)\psi=0\,.
\end{equation}
We can prove that the use of ellipsoidal coordinates $\lambda$,
$\mu$ and $\nu$, the form for the potentials given in (\ref{55}) and
the factorization $\psi({\bf x})=\psi(\lambda)\phi(\mu)\varphi(\nu)$
show that equations (\ref{56}) are equivalent to three wave
functions solely in the variables $\lambda$, $\mu$ and $\nu$
respectively, so that complete separation of variables is also
achieved in the quantum case \cite{GaIoNePr07}. The equation for $\lambda$
gives ($\psi_\lambda$ denotes the derivative of $\psi$ with respect to $\lambda$):
\begin{eqnarray}
\fl
4(\lambda+a^2)(\lambda+b^2)(\lambda+c^2)\psi_{\lambda\lambda}
+2[a^2b^2+a^2c^2+b^2c^2
+2\lambda(a^2+b^2+c^2)+3\lambda^2]\psi_\lambda\nonumber \\ \hskip10pt
  +2m (\lambda^2 E-\lambda  E_1+
  E_2-l(\lambda))\psi=0\,.\label{57}
\end{eqnarray}
The other two equations are just obtained by replacing $\lambda$ by
$\mu$ and $\nu$ respectively. Equation (\ref{57}) can be written in
a more compact form by means of an of a change of variable to the
auxiliary variable $t$. This is given by means of the condition:

\begin{equation}\label{58}
    \lambda'(t)=2\sqrt{(\lambda+a^2)(\lambda+b^2)(\lambda+c^2)}\,.
\end{equation}

In terms of $t$, equation (\ref{57}) has the follwing form

\begin{equation}\label{59}
 \frac{d^2\psi}{dt^2}+2m[\lambda^2(t)E-\lambda(t)
    E_1+ E_2-l(\lambda(t))]\psi=0\,,
\end{equation}
where $l(\lambda)$ has already appeared in (\ref{53}-\ref{55}). In
addition to (\ref{59}), there are two other equations, one for $\mu$
and the other for $\nu$. These three equations are similar and, in
particular, all depend on the function $\lambda(t)$ even in the free
particle case, $l(\lambda)=m(\mu)=n(\nu)=0$.

\section{Integrability in spaces of constant curvature.}

There are three types of homogeneous three dimensional manifolds with constant
curvature and Riemannian positive definite metric, which are the sphere $S^3$, the Euclidean plane $E^3$ and the hyperboloid $H^3$ (for general references on integrability in spaces of constant curvature, see \cite{Wozmis03}). In this section, we shall deal with the standard $S^3$ with curvature equal to $1$. As is well known, $S^3$ can be realized as the submanifold $X_1^2 +X_2^2 +X_3^2 +X_4^2 =1$ in a 4D Euclidean ambient space, with the induced metric. A convenient coordinate system for $S^3$ is provided through stereographic coordinates, the point on $S^3$ parametrized by ${\bf x}\in{\mathbb R}^3 \cup \infty$ being given by:
\begin{equation}\label{60}
    X = ({\bf X}, X_4) = \left( \frac{2\chi {\bf x}}{{\bf x^2}+\chi^2},\quad
\frac{{\bf x^2}-\chi^2}{{\bf x^2}+\chi^2}\right)\,.
\end{equation}
where $\chi$ is a parameter which should be different from zero. With this choice ${\bf x}={\bf 0}$ corresponds to the sphere's `South Pole' $(0,0,0,-1)$, and the projection is made from the North Pole $(0,0,0,1)$ to the plane $X_4=1-\chi$. No actual generality would be lost if we fix a particular value for $\chi$; the preferred choice $\chi=2$ corresponds to projecting over the plane tangent to the sphere's south pole and expressions are generally more clear for this choice, because in this case near the South Pole the coordinates ${\bf x}$ approach the ordinary Cartesian coordinates in the tangent $E^3$ with the same scaling as lengths on the sphere (a rescaling is required for other values of $\chi$) and neglecting terms which are higher order in ${\bf x}$ brings the Euclidean expression with the correct factors, as obvious for instance in (\ref{61})-(\ref{61b}).

The Lagrangian for the free evolution on $S^3$ can be taken in terms of the ambient space coordinates as:
\begin{equation}\label{61a}
L_0=\frac{m}{2} \left( {\dot
{\bf X}^2} + {\dot{X_4^2}}\right)\,,
\end{equation}
which can be expressed in stereographic coordinates as
\begin{equation}\label{61}
L_0=\frac{m}{2}4\chi^2 \frac{{\bf\dot
x^2}}{({\bf x^2}+\chi^2)^2}\,,
\end{equation}
Alternatively, we may start from the angular momentum tensor in ambient space, with components
$M_{\alpha\beta}:= X_\alpha \dot X_\beta-\dot X_\alpha X_\beta$ and
define the free lagrangian in $S^3$ as (proportional to) the square of this angular momentum tensor; this leads to the same free lagrangian. The canonical conjugate momenta associated to the stereographic coordinates $x_i$ is
\begin{equation}\label{61b}
p_i=m\frac{4\chi^2}{({\bf x^2}+\chi^2)^2}\,\dot x_i.
\end{equation}
and the Legendre transformation provides the following free Hamiltonian:

\begin{equation}\label{62}
H_0=\frac{1}{2m}\frac{1}{4\chi^2}{\bf p^2}({\bf x^2}+\chi^2)^2\,.
\end{equation}

The symmetry group for $S^3$ is $SO(4)$ and (\ref{62}) is invariant
with respect to the sphere isometries in $SO(4)$. Then, $H_0$ has 6
`kinematic' integrals of motion, which are generators of $SO(4)$; in terms of the stereographic parametrization these generators are:
\begin{equation}\label{63}
L_{i}=\epsilon_{ijk}x_j p_k\,,\qquad K_{i}=\frac{1}{2\chi}\Big(
2 x_i {\bf p}\cdot{\bf x} - p_i({\bf x^2}-\chi^2)
\Big)\,,
\end{equation}
with Poisson brackets

\begin{equation}\label{64}
\{L_i,L_j\}=-\epsilon_{ijk}L_k\,,\quad
\{K_i,K_j\}=-\epsilon_{ijk}L_k\,, \quad
\{L_i,K_j\}=-\epsilon_{ijk}K_k\,.
\end{equation}
(these close also an $so(4)$ algebra as expected).
There are two Casimirs given by ${\bf K}^2+{\bf L}^2$ and ${\bf
K}\cdot{\bf L}$. The free Hamiltonian is proportional to the first
Casimir:

\begin{equation}\label{65c}
H_0=\frac1{2m}({\bf K}^2+{\bf L}^2)\,.
\end{equation}
while the second Casimir vanishes.

Then, we are going to pose in $S^3$ the same question studied for the flat case
in section 1. Assume the existence of a symmetry axis, that,
for the moment, we can identify with the $z$-axis. Then, consider a
Hamiltonian with kinetic term given by $H_0$ plus a time independent
potential $U({\bf x})$,
\begin{equation}\label{66}
H= \frac1{2m}({\bf K}^2+{\bf L}^2) +U({\bf x})\,.
\end{equation}

Then, if this system has to be completely integrable, we need finding a new `Hamiltonian' $H_1$ with `kinetic' part
$T_1=(2m)^{-1}p_ig^{ij}({\bf x})p_j$ and `potential' $\Phi({\bf x})$,
$H_1=T+\Phi({\bf x})$, independent of $H$ and such that
$\{H,L_3\}=\{H_1,L_3\}=\{H,H_1\}=0$. As in Section 1, the kinetic term $T_1$ is obtained under the condition that it commutes with $H_0$ and $L_3$. There is an $S^3$ analog of the most general `kinetic' term (\ref{1a}), but we simply write down the version for $S^3$ of the Euclidean Killing tensor (\ref{1b}) leads to the constant of motion:

\begin{equation}\label{67}
H_1=\frac1{2m}\,(L_1^2+L_2^2- a^2 K_3^2)+\Phi({\bf
x})\,,
\end{equation}
where $a$ is a constant playing a role fully similar to those in (\ref{1b}). If the symmetry axis were arbitrary in
the direction given by the unitary axis ${\bf n}=(n_1,n_2,n_3)$, the
matrix $g^{ij}$ has the following form in terms of the stereographic parametrization:
\begin{eqnarray}
g^{ij}({\bf x})&=&\delta_{ij}({\bf x}\cdot{\bf n})^2-{\bf
x}\cdot{\bf n}(x_i n_j+x_j n_i) +
n_in_j{\bf x}^2\nonumber\\[1ex]
 &-&\frac{a^2}{4\chi^2}\Big({2 x_i}\,{\bf x}\cdot{\bf n}-n_i({\bf
x}^2-\chi^2)\Big)\Big({2 x_j}\,{\bf x}\cdot{\bf n}-n_j({\bf
x}^2-\chi^2)\Big)\,.\label{68}
\end{eqnarray}

The requirement $\{H,H_1\}=0$ leads to the version for $S^3$ of equation
(\ref{6}), which in this case reads

\begin{equation}\label{69}
\partial_i \Phi({\bf x})=G^{ij}({\bf x})\partial_j U({\bf
x})\,, \quad{\rm with}\quad G^{ij}({\bf x})=\frac{g^{ij}({\bf
x})}{({\bf x^2}+\chi^2)^2}\,.
\end{equation}

We shall denote the eigenvalues of $G^{ij}({\bf x})$  by
$\{\mu_+,\mu_-,\mu_3\}$. The two former are the roots of the
quadratic equation $z^2-Pz-Q=0$, where

\begin{equation}\label{70}
    P:= \frac{{\bf x}^2-\alpha^2 [({\bf x}^2-\chi^2)^2+4({\bf
x}\cdot{\bf n})^2]}{({\bf x}^2+\chi^2)^2}\,,\qquad
Q:=\frac{({\bf x}\cdot{\bf n})^2\alpha^2}{({\bf x}^2+1)^2}
\end{equation}
and $\mu_3=Q/\alpha^2$. As in the flat case described in section 1,
the eigenvalues $\mu_\pm$ together with  the azimuthal angle $\varphi$ define
a coordinate system, which is the analogous for the sphere $S^3$ of the spheroidal oblate coordinates \cite{Olevski}. As in the flat case, their corresponding
eigenvectors have components $\partial_i\mu_-$ and
$\partial_i\mu_+$, $i,j=1,2,3$ respectively, i.e.,

\begin{equation}\label{71}
G^{ij}\partial_j\mu_-=\mu_+\partial_i\mu_-\,,\qquad
G^{ij}\partial_j\mu_+=\mu_-\partial_i\mu_+\,.
\end{equation}

Finally, following a procedure similar to that studied in section 1,
we can show that equations (\ref{71}) provide the most general form
for the `potentials' $U({\bf x})$ and $\Phi({\bf x})$ on the variables
$\mu_\pm$. The final result is:
\begin{equation}\label{72}
U=\frac{f(\mu_+)-g(\mu_-)}{\mu_+-\mu_-}\,, \qquad
\Phi=\frac{\mu_-\,f(\mu_+)-\mu_+\,g(\mu_-)}{\mu_+-\mu_-}\,,
\end{equation}
where $f(\mu_+)$ and $g(\mu_-)$ are arbitrary functions. Note that
symmetry prevents that $U$ and $\Phi$ to depend on the azimuthal angle
$\varphi$, as constancy of $L_3$ means invariance under rotations with $z$-axis.

\subsection{Application to the Kepler problem.}

In this subsection, we shall discuss a particular case of very
special importance: the Kepler problem in the standard sphere $S^3$ with curvature equal to 1, which provides  an example of the type of situation just described (recall a factor which dimensionally is a lenght square has been taken as equal to 1, and hence is invisible in the expressions). For more details on the Kepler problem in spaces with constant curvature, see \cite{CRS05,CRS07c,DoZi91,GaSa07,Ko92,NiRoSa99,Pr07}.

In the sphere $S^3$ (and also in any dimension $n>3$), this system is an example of a maximally superintegrable system. Restricting attention to the 3D case, the Kepler Hamiltonian is
\begin{equation}\label{65}
H=\frac1{2m}({\bf K}^2+{\bf L}^2) - \frac{k}{\tan(r)}
\end{equation}
where the kinetic part is (\ref{66}) and the potential term depends only on the
intrinsic distance $r$ in $S^3$ to the potential center and $k>0$ for the atractive case. The potential, with center at sphere's south pole, can be expressed in terms of the ambient space coordinates as
\begin{equation}\label{65a}
- \frac{k}{\tan(r)} = k\frac{X_4}{\sqrt{X_1^2 + X_2^2 + X_3^2}}
\end{equation}
and in terms of stereographic coordinates the Kepler Hamiltonian is:
\begin{equation}\label{73}
H=\frac1{2m}({\bf K}^2+{\bf L}^2)+{k}\frac{1}{2\chi}\frac{{\bf
x}^2-\chi^2}{|{\bf x}|}\,.
\end{equation}
where the generators in the kinetic part are given in (\ref{63}).
This Hamiltonian possesses 3 integrals of motion linear in the momenta (the three components of the
angular momentum ${\bf L}$) and a further 3 integrals of motion which are quadratic in the momenta (the components of the Laplace-Runge-Lenz vector $\bf
A$):
\begin{equation}\label{74}
{\bf A}={\bf K}\times{\bf L}+m k \frac{\bf x}{|{\bf x}|},
\end{equation}

Of course there are 2 independent relations among the seven constants of motion $H, {\bf L}, {\bf A}$, reducing to five functionally independent constants, which entitles the Kepler problem in $S^3$ to be maximally superintegrable. These relations are:
\begin{equation}\label{75z}
{\bf A}^2 = m^2 k^2 + \Big( 2m H - {\bf L}^2 \Big) {\bf L}^2, \qquad {\bf A}\cdot{\bf L}=0
\end{equation}

The algebra of Poisson brackets for the components of ${\bf
L}$ and $\bf A$ has the following form:
\begin{equation}\label{75}
\{A_i,A_j\}=\epsilon_{ijk}L_k(2 m H -2{\bf L}^2)\,,\qquad
\{L_i,A_j\}=-\epsilon_{ijk}A_k\,.
\end{equation}

Should these commutation relations be computed in a sphere with curvature $\kappa$, the term $-2{\bf L}^2$ in the first Poisson bracket would appear as
$-2\kappa {\bf L}^2$, making expressions dimensionally correct. This term displays clearly the effects due to the curvature of the configuration space, when it is compared (\ref{75}) with the commutation relations for
the corresponding algebra in the flat case, which are
\begin{equation}\label{75y}
\{A_i,A_j\}=\epsilon_{ijk}L_k 2 m H\,,\qquad
\{L_i,A_j\}=-\epsilon_{ijk}A_k\,.
\end{equation}

From (\ref{75}) we see that the components of ${\bf L}$ and $\bf A$
do not have the commutation relations of a Lie algebra, because of the presence of higher order terms in the Poisson bracket; the quadratic term $L_k H$ is already present in the flat case, but on the curved sphere $S^3$ cubic terms $L_k {\bf L}^2$ appear as well. In the flat Euclidean case, this algebra has been studied under the name of Higgs algebra \cite{GrKu00}.

Then we consider this system as an example of the situation discussed in the previous section: the Kepler Hamiltonian admits two additional constant of motion which are also in involution: the components of the angular momentum anf of the Runge-Lenz vector along any fixed direction. If we take the $z$ axis, then we get $H, L_3, A_3$ as three constants of motion in involution.

Now we note that the Kepler problem has axial symmetry around any axis. In particular, the third constant $A_3$ in the Kepler problem in Euclidean  belong indeed to the family (\ref{4}) with the values $\alpha=0, \beta=1, \gamma=0$ for the parameters. While the coordinate system behind the family $\alpha\neq0, \beta=0, \gamma=0$ were oblate spheroidal coordinates in $E^3$ and their analogous in $S^3$, the family $\alpha=0, \beta\neq0, \gamma=0$ turns out to be separable in parabolic coordinates in $E^3$ and in its analogous for $S^3$.

To end, we mention the possibility, which exists in $S^3$, to rescale the Runge-Lenz vector by a factor which depends on the integrals of motion (so the property of being a constant is not disturbed) in such a way that the Poisson brackets of new components, together with those of angular momentum closes a {\it Lie algebra} $so(4)$. A rescaling ensuring this \cite{Pr07} is:
\begin{equation}\label{76}
{\bf R}={\bf A}\left[{\bf
L}^2 + m \left(\sqrt{ H^2 + k^2} - H \right)\right]^{-1/2}\,,
\end{equation}
(the rescaling factor is always positive, no matter neither the value nor the sign of $H$) and the components of $\bf L$ and $\bf R$ turn out to have the $so(4)$ commutation relations with respect to the Poisson bracket (compare
to (\ref{64})):

\begin{equation}\label{77}
\{L_i,L_j\}=-\epsilon_{ijk}L_k\,,\quad
\{R_i,R_j\}=-\epsilon_{ijk}L_k\,, \quad
\{L_i,R_j\}=-\epsilon_{ijk}R_k\,.
\end{equation}

For these commutation relations (\ref{77}) the first Casimir reads ${\bf
R}^2+{\bf L}^2$. Direct computations reveals that this Casimir is:
\begin{equation}\label{78a}
{\bf R}^2+{\bf L}^2 = m \Big( H + \sqrt{H^2 + k^2}\Big)
\end{equation}
Either by solving this equation for $H$ or by a direct tedious but straightfoward computation, this allows us to write Hamiltonian (\ref{73}) in
terms of this Casimir:

\begin{equation}\label{78}
2 m H ={\bf R}^2+{\bf L}^2-\frac{m^2k^2}{{\bf R}^2+{\bf L}^2}\,.
\end{equation}

This discussion has been purely classical. We simply mention that this last property can be also discussed in the quantum case, where now the new global $so(4)$ Lie algebra symmetry allows us to derive the energy spectrum first found by Schr\"odinger for the Kepler problem in $S^3$ \cite{Sch40,Sch41,St41,IS45}. Full details on this new derivation can be found in \cite{Pr07}; we must recall this approach is conceptually very different from the standard Pauli discussion for the Euclidean Kepler problem, where there is a different Lie algebra in each energy eigenspace, with its isomorphism class depending on the energy sign.

\section{Concluding remarks.}

We have discussed a special class of three dimensional completely integrable either in the flat Euclidean 3D space as well as in the standard sphere $S^3$; these systems are required to have a rotational symmetry axis, so they have axial symmetry. Further to the Hamiltonian $H$, there are two additional constants for such a system: $L$, the angular momentum with respect the symmetry axis, and $H_1=1/(2m)p_ig^{ij}_2({\bf x})p_j+\Psi({\bf x})$. The general form for the tensor $g^{ij}_2({\bf x})$ which guarantees  that the `kinetic' parts of $H$ and $H_2$ commute among themselves and with $L$ is given, and two special cases are mentioned; we discuss one of them, corresponding geometrically to separability in oblate spheroidal coordinates in full detail; the other corresponds to separation of variables in parabolic coordinates.  Integrability determines the `potentials' $V({\bf x}), \Psi({\bf x})$ to have a particular form in terms of undetermined functions of the coordinates, and the
equations of motion are written as three equations each one
involving one coordinate (separation of variables). In the quantum
case, one of the three separated wave equations is trivial and each
of the other two can be easily transformed into a spheroidal wave
equation, for which the solutions have been studied.

If no symmetry conditions are imposed a similar study can be
performed. Separation of variables can be achieved in this case, as one could expect,  in terms of the (general) ellipsoidal coordinates. In the quantum case, we obtain three
similar wave equations, one for each of the variables, each one in
terms of a different function. These functions appear in the form of
the potentials as a consequence of the integrability condition.

Finally, we have carried  the analysis of the systems with axial symmetry to the three dimensional sphere $S^3$ in the classical case. The results obtained are quite similar to the flat ones, in agreement to the idea that results in this area for the constant curvature spaces are essentially `the same' as in the flat case. This is illustrated with the particular case of the
three dimensional Kepler problem.
Extension to higher dimensions should be also possible \cite{DeKuOnVe02,KuTeV01}

\section*{Acknowledgements.}

Partial financial support is acknowledged to the Junta de Castilla y
Le\'on Project VA013C05, the Ministry of Education and Science of
Spain projects MTM2005-09183 and FIS2005-03988 and Grant
SAB2004-0169 and the Russian Science Foundation Grant 04-01-00352.
G.P. was also supported by the programme ENTER-2004/04EP-48,
E.U.-European Social Fund (75\%) and Greek Ministry of development -
GSRT (25\%) and  by RFFI grant 07-01-00234.

\def\otherrefs#1{\null}

\section*{References}

redpajama_set_name: RedPajamaArXiv


text:
Rickie Fowler (63) takes 2-shot lead at CJ Cup
Rickie Fowler shot a blistering 9-under 63 on Saturday to take a two-shot lead over Rory McIlroy, who carded the low round of the day with a 62 at the CJ Cup in Las Vegas.
Fowler sits at 21 under heading into Sunday's final round at the Summit Club. McIlroy made the biggest leap on moving day, shooting up 17 spots for sole possession of second at 19-under 197.
Robert Streb (65), Australia's Adam Scott (67) and Mexico's Abraham Ancer (63) are tied for third at 18 under, three shots back.
Keith Mitchell, who owned a five-stroke lead after 36 holes, dropped into a tie for sixth after a 73. He sits at 17 under, tied with England's Tyrell Hatton (67), four shots back.
Fowler followed opening 66s with a clean round, posting five birdies on the front nine and four more coming in.
"I've done everything I really wanted to do the first three days. It's nice to be executing the way we are," Fowler said. "It's just been a long time coming. I haven't been in this position a whole lot in the last couple years, so to feel the emotions again on the back nine being around and then in the lead, you know, kind of dealing with that again and kind of making sure that I was going through the right process and committing to what I was ultimately trying to do, it was great to see that I was still executing and hitting the shots that I wanted to."
But as good as Fowler's round was, McIlroy's was even better, thanks to an eagle on the par-5 18th hole. McIlroy of Northern Ireland recorded six birdies on the front nine, including five in a row, and finished with eight.
"I played an eight-hole stretch yesterday in even par, I made eight pars in row. Then today I played a stretch of golf, played nine holes in 6 under and all of a sudden you feel a little better about yourself," McIlroy said. "The big thing is I've played the last two days keeping it in the short stuff and I haven't made any bogeys. That's huge. ... I can't remember the last time I played a round of golf without a bogey and now I've played two rounds in a row, so that's nice to see."
--Field Level Media


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