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Turkey after the Genocide and the Trials that Never Occurred. Red Cross. B ryce Toynbee Report. Baron Von Wangenheim the German Ambassador. Treaty of Versailles and the League of Nations created. Henry Morgenthau the US Ambassador. WWI Ends November 1918. The Evidence and Witnesses.
Henry Morgenthau US Ambassador in Ottoman Empire.
3. British and the Bryce Toynbee Report. Viscount James Bryce British Parliament and Arnold Toynbee a historian collect reports.
To the right you see excerpts from my cables sent to Washington during WWI.
Why the US did not enter?
Ambassadors are guests and should have respect for their host nation.
What a sovereign state does in its own borders is their business.
Turkish government had informed the State Department at Washington that the Red Cross would not be permitted to send surgeons and nurses to the aid of the Armenian people.
Organization donated $ 1.800.000 for relief works in the Near East. Later this amount rose up to $ 6.000.000.
American Red Cross personnel kidnapped Armenian children were restored to their parents and women in Muslim harems were freed.
How should the Young Turks be punished after WWI?
What was done to address the issue of mass killings after WWI?
Compare the end of the Holocaust to the end of the Armenian Genocide using one example that focuses on holding perpetrators accountable for their actions. | {
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Northrop Grumman Innovation Systems designs, builds and delivers Commercial aerospace structures, to customers around the world. Our main product at the ACCE facility is advanced aerospace composite structures.
Northrop Grumman is currently seeking a Sr. Principal Mission Assurance Engineer to provide support to our FCC COE Quality Team.
Candidate will work in an industrial environment and support solving complex engineering problems. Will effectively interact with design engineering, Operations team and provide support to Materials and Process experts, Process engineering, and Tool Design as required.
Routinely coordinates technical issues with various program teams and customers.
Strong knowledge of MRB (Material Review Board) standards.
Strong understanding of (Lean) principles with experience leading and implementing significant improvements. | {
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} |
Hennepin County Attorney > News > Minneapolis man charged with killing person living in his house
Minneapolis man charged with killing person living in his house
A Minneapolis man was charged with second-degree murder in the shotgun shooting death of a young man in front of four young children, Hennepin County Attorney Mike Freeman announced Tuesday.
Eddie Niles Hubbard, 52, was charged (PDF) in the Friday shooting of Jeremiah Curtis, 23, inside Hubbard's North Minneapolis home. He also was charged with four counts of second-degree assault for pointing the gun at Curtis' girlfriend and three of the children.
Hubbard made his first appearance today and bail was set at $500,000.
According to the criminal complaint, Curtis and his girlfriend had lived off and on at Hubbard's home with her four children, two of whom were his. On Friday, Hubbard sent a text message to her, asking them to move out.
While they were packing their belongings, Hubbard shouted up the stairs for Curtis to accompany him to the drugstore. Curtis said they were too busy packing. Hubbard then came to the upstairs bedroom and shot Curtis in the stomach before fleeing, the complaint states. | {
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Home Gun News Armscor's New RIA 5.0 American-Made 9mm Pistol
Armscor's New RIA 5.0 American-Made 9mm Pistol
From Armscor/Rock Island Armory . . .
Meet the latest American-made firearm from RIA-USA. The RIA 5.0 has a patented RVS recoil system to maximize barrel mass and linear movement for super soft recoil.
It also features a smooth trigger for greater accuracy and a two-piece modular grip frame for flexibility and control.
The RIA 5.0 is one of the most flawlessly engineered sporting pistols you'll find. Backed by extensive testing, its fit and performance will make it hard to shoot anything else. Plus, it's made right here in the USA.
The RIA 5.0 features a patented RVS recoil system engineered to maximize barrel mass and linear movement for super soft recoil. It has a buttery-smooth trigger for greater accuracy in rapid-fire situations, and a two-piece modular grip frame for flexibility and control. The level of research and precision engineering that went into the making of this sporting pistol is unmatched.
CALIBER – 9MM
CAPACITY – 17+1
BARREL – 4.91″
OVERALL LENGTH – 8.11 INCH / 205.99 MM
OVERALL WIDTH – 1.33 INCH / 33.78 MM
OVERALL HEIGHT – 5.17 INCH / 131.32 MM
WEIGHT – 2.47 LBS / 1.12 KG
FRONT SIGHT – DAWSON PRECISION, FIBER OPTIC ON DOVETAIL CUT
REAR SIGHT – LPA ADJUSTABLE ON DOVETAIL CUT (ALSO AVAILABLE WITH C MORE RTS II RED DOT OPTIC)
GRIPS – NYLON
GRIP FINISH – E100 CERAKOTE
ACTION – SEMI-AUTOMATIC
TRIGGER PULL – 4.5 LBS
PURPOSE – PERSONAL PROTECTION
MSRP = $998.88
Armscor RIA 5.0
RIA 5.0
Rock Island Armory RIA 5.0
Previous articleFranklin Armory Sues ATF Over 'Multiple Meritless ATF Actions and Inactions'
Next articleNew York Supreme Court Rules State's Red Flag Law Unconstitutional In As-Applied Case
Bravo Concealment Announces the New LINKed Holster With Mag Pouch
SHOT Show: Lionheart Industries Vulcan 9 Compact 9mm Pistol
Brownells Makes It Easy to Turn Your Tax Refund Into Shooty Fun
Peter Gunn January 5, 2023 At 09:41
M. Murcek January 5, 2023 At 09:42
Staff writer basically copied a company press release.
generate clicks = generate money
Dan Zimmerman January 5, 2023 At 10:41
Yes. And we marked it as a press release at the top of the post. We run press releases for products we think will be of interest to our readers. You're welcome.
Debbie W. January 5, 2023 At 11:20
I wonder if those who permeated yesterday's new release steel frame Canik post with beavis and butthead opinions will do the same for this new firearm? Will they lap dog this firearm or will they concede it has just as much RoboCop as the Canik, etc? Actually Chrome produces shadows enhancing cuts, the Black Canik which was not pictured looks very right, of course such little things like that don't matter to origin fixated beavis and buttheads.
The same beavis and buttheads hypocrites who will use their Made in China devices to praise this firearm because it is Made in America and denigrate equal or better firearms made elsewhere…pathetic.
As for this firearm it appears to be well built as was said for the Canik.
I'm not going to apologize for loving this country. As far as the rest of your screed, Pathetic …is right.
You should be happy to know that further down people are bashing this just as much as the Canik. Happy now? But then you did read my post further down but decided to post above. Someone before gave you crap about how you take your posts to the top and I thought meh, now I understand what they meant.
strike my hammer January 5, 2023 At 14:55
isn't the canik striker and this is hammer fired?
about the same as apples and oranges
tsbhoa.p.jr January 5, 2023 At 16:48
Art out West January 5, 2023 At 21:58
Is "made in America" a huge thing to brag about in terms of firearms? Most guns are made here.
I prefer to buy American made firearms to financially support my countrymen.
On the other hand, even many "less expensive" and basic guns are nonetheless made in America. Hi-Poimts are made here, as are Kel-Tecs and Heritage Rough Riders. My Marlin 60 and 795 were made here. My Taurus G3C was made here. My Palmetto State guns were made here. My Ruger LCP and Security Nine were made here. I think even my Maverick 88 was made here. None of those guns fall into the "fancy" category. They are all fairly basic.
Cooter E Lee January 6, 2023 At 18:40
I don't understand your love affair for Turkish firearms but most of us here are patriots and love and support this country. I don't think someone USA traitor to own a firearm from Turkey, ammunition from Russia or even a shotgun from China – I do. It's just my preference to buy American and support my friends, family, and neighbors and also to keep the R&D money here. Americans fall short at many things, but creativity isn't one of them.
Lost Down South January 5, 2023 At 11:39
It would be nice to add Country of Origin to all press releases, reviews, etc.
It make a difference to me…and probably many others.
red wolf January 5, 2023 At 11:54
Arrogant and as fat as all holy hell. Nice.
At least it won't cut you when you're trying to pick it up like that Canik…😁
Wally1 January 5, 2023 At 09:46
What is meant be "extensive testing"?. In this day and age, most products are tested by the public after the initial sale. This is now the norm for automobiles, It's cheaper for the manufacture to fix a few defects and cross their fingers that the product makes it through the warrnty period before failure. Then it's not their problem. I would think the same would be for other products to include firearms.
how does this magical revolutionary recoil system work?
rosignol January 5, 2023 At 11:28
Maybe a fancy spring, probably a heavier gun, so 'classic newtonian physics'.
That thing looks downright chonky, especially for a 9mm.
Sam I Am January 5, 2023 At 15:19
I tried google search; no joy. Many instances of the press release, but no explanation of the components, or operation. Maybe after ShotShow?
Ken January 13, 2023 At 02:35
Search for Patent No. 10018433 at the US Patent Office website.
https://ppubs.uspto.gov/pubwebapp/static/pages/ppubsbasic.html
Made in the USA!!!! 🇺🇸
Looks like a nice gun, how about a review?
Leigh January 5, 2023 At 10:57
What will street price be?
sound awake January 5, 2023 At 11:24
i need another 9mm pistol
like i need a hole in the head
and i especially dont need
a 9mm pistol
that costs more than an ar15
that i can build
thats capable of groups less than 1 moa
at 300 yards with factory match ammo
No one of Consequence January 5, 2023 At 11:57
Hmmm… Will a 22 TCM conversion kit be available?
Umm . . . January 5, 2023 At 19:37
I love TCM, but they don't seem to be supporting it anymore.
Namey Name January 5, 2023 At 12:23
Wow, another 9mm striker fired pistol. There were so few choices before this. Yawn.
tell me you didn't read the article without telling me you didn't read the article.
HINT – it's not striker fired.
DAL January 5, 2023 At 13:00
Looking at the pictures again, I just realized that this is a slide-inside-frame design (a la CZ), which isn't something I think I've seen before in a striker pistol.
Huh. Neat.
It's general shape reminds me of a Steyr.
well; it's not striker fired so . . . . .
Rokurota January 5, 2023 At 13:37
Looks like it has a cut-out in the back of the slide for a hammer. Is that possible with the low bore axis? Possibly a cut-out for the striker channel?
Anymouse January 5, 2023 At 18:54
It's a cutdown internal hammer that they call a micro hammer assembly. The external profile of the barrel is roughly trapezoidal and serves as a guide for the slide. There's some kind of delayed blowback device that wasn't detailed. The barrel doesn't tip, but I don't know if it's fixed or travels linearly with the slide. Definitely not a Gen3 Glock with the name filed off. Hopefully, it'll be more available than the FK BRNOs that also look interesting.
Hey that's nifty. So we could possibly get second strike AND a short trigger pull? Verrrrry intrigued…
J S Bryan January 5, 2023 At 14:07
Well, if you can wade through the somewhat slobbery article on Amerindian Handgunner, this does look to be something a bit different. It's not a conventional Steyr/Glock striker fired action, it's not exactly a hammer gun either, it doesn't tilt to unlock like a Browning style action but it's not just a blow-back style action either – designer Fred Craig (of .22TCM fame so I'd be shocked if Armscor doesn't have a version for that round in the works) appears to have ginned up something that, if it's not new, it's at least intriguing. https://americanhandgunner.com/handguns/the-rock-island-armory-5-0-radically-revolutionary/
Bloody [email protected]! ..lAMERICAN Handgunner…crummy touch keyboards and autocorrect…
Ok. The answer to what RVS is in the article; thanx!
The pistol review was intriguing. Maybe if it is made in .22 one day, I might push to afford it.
I would like this exact same pistol only not the Personal Defense model, do they have one in a "Let's have funm shootzing gunms" model?
I'm getting tired of killin. Besides my freezer is still full of last years mostly peaceful protesters.
Oh scratch the whole deal, I see it's in 9mm, to much lung blowing out and bloodshot meat with that caliber if I ever do have to go to filling up the freezer again.
Hopefully they'll have a .44Automag perfected that doesn't jam by that time.
Fingers X's 👍
Whitey January 8, 2023 At 13:12
So…@ 2 1/2lbs. isn't that = to 10 Quarter-Pounders w/CHEESE?
C.B Cochran Sr. January 11, 2023 At 09:24
You stupid ass self proclaimed "experts" are laughable. I've trained law enforcement for almost 50 years in firearms and survival skills and one thing is very clear. Those who talk the most usually know the least. Their knowledge base comes from articles written by authors such as Moosefart Habbibee , etc. I'm impressed.
Sgt. Davis January 12, 2023 At 14:47
Pass. What would I use it for? Looks clunky and if wanted s*** hanging off the side of slide I'd be cruisin' down the block in my '65.
Seriously, though. I own no Hi-Point but laugh at them all you want, simple blowback. I'll stick with my Smiths and the ONE RIA I actually do have, the M206 snub.
Fal Phil January 13, 2023 At 14:21
There was nothing in this article that I didn't already read on the RIA website.
Could you please re-write it with a detailed analysis of the RVS recoil system? What does RVS stand for anyway? | {
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Frank Giustra | What Does China's Rise Mean for the World?
What Does China's Rise Mean for the World?
It seems our world is as unstable as it has ever been. There are brewing conflicts everywhere, in many cases interconnected and with such a cacophony of players that even seasoned diplomats are confused. To make things worse, we have a non-functioning UN Security Council and a lack of will by the international community to resolve conflicts.
Many of these conflicts have the potential to flare up into truly global problems. The war in Syria, for example, has resulted in confrontations between Russia, Western powers, and competing governments and armed factions across the Middle East. The Syrian conflict has also contributed to the rise of the Islamic State, which in recent weeks claimed responsibility for consecutive terrorist attacks in Egypt, Lebanon and France.
Over the longer term, however, I am more concerned with the type of tectonic shift in world power that only happens every century or so. The contest between the U.S. and China is worth examining in this regard.
Unfortunately, the passing of the baton from one global power to the next rarely happens without a major conflict.
The rise and fall of empires follow familiar patterns. Unfortunately, the passing of the baton from one global power to the next rarely happens without a major conflict. It is not easy on the pride of any reigning nation to cede power. The only recent exception that I can think of was the transition from Britain to the U.S. as the dominant power following World War II.
The rise of the U.S. and China as industrial powers share similar features, and are separated by only 100 years. During its industrial build up, the U.S. mostly tried to mind its own business and stay out of global squabbles, much like China today. The U.S. focus was mostly economic superiority, again not dissimilar to China. It was not until the introduction of the Monroe Doctrine in 1823 that the U.S. began to flex its military muscle to warn off European powers from further colonising within the Americas. Any interference would be seen as an act of war. This was the first of a series of U.S. presidential doctrines that incrementally expanded circumstances in which the nation would engage against real or perceived external threats — culminating with the Bush Doctrine proclaiming that the U.S. has the right to attack preemptively if it feels its security is threatened. I wonder if China is about to proclaim its own Monroe Doctrine as its economic might grows.
Over the past 35 years, China has transformed itself from a communist society to a state-managed capitalist economy. During this period, it has maintained a fairly low-key foreign policy. Some analysts predict that China will surpass the U.S. as the leading global economic power within the next ten years — and by some measures, China is already ahead. More recently, Beijing has quietly built up its military strength beyond levels that are comfortable for the West. It is also working hard to catch up to the U.S. in its technological capabilities. Some may argue that the U.S. will always maintain its military and technological superiority. Perhaps, but I would never underestimate the Chinese.
History shows that when countries are busy trading with each other, there is little appetite for conflict.
The potential decline in cross-border trade, especially between China and the West, is worrisome. Over the past several decades, China achieved explosive growth by focusing on its export economy, with the U.S. as its largest trading partner. Going forward, the U.S. market may become less important. There is only so much that China can sell to an over-leveraged economy such as the U.S. The problem is compounded by the fact that China is the single largest holder of U.S. debt. China knows the risks, and is now attempting to move from an export-led economy to one driven by domestic consumption. History shows that when countries are busy trading with each other, there is little appetite for conflict. People on both sides may prosper by selling or be happy in their hedonistic consumption. It is usually when trade stops that wars start.
Recently, competing territorial and maritime claims in the South China Sea between China and its regional neighbours have prompted a U.S. response. In August, at a regional meeting in Kuala Lumpur, U.S. Secretary of State John Kerry sharply accused Beijing of violating international maritime law. "Let me be clear: The United States will not accept restrictions on freedom of navigation and overflight, or other lawful uses of the sea", Kerry said. In response, Chinese Foreign Minister Wang Yi issued a statement that, "China opposes any non-constructive words and acts which widen division, exaggerate antagonism or create tensions".
In October, a U.S. navy warship challenged the territorial limits around a Chinese-claimed reef in the Spratly archipelago as part of its Freedom of Navigation program. In November, U.S. President Barack Obama made a point of visiting a Philippine warship soon after arriving in Manila for the start of the Asia-Pacific Summit. He underscored the U.S. commitment to defend the Philippines, one of several South East Asian nations embroiled in the South China Sea dispute.
America's willingness to play global cop by way of warnings and displays of military might is something that China looks upon with a lot of trepidation. How far will the U.S. go in drawing a red line, given its alliances in the region? How far will the U.S. push its military might in the face of its economic decline versus China? And at what point does China feel sufficiently equipped to proclaim its own version of a Monroe Doctrine and protect its area of influence?
If China stays on track to emerge as the leading global economic power, these questions will take on new urgency. The trends may evolve gradually, and recent signs of an economic slowdown in China may mean that things remain quiet for years. However, if my concern proves correct that one day China will decide to pursue a more muscular foreign policy, it will take the involvement of the entire global community and a lot of luck to prevent a catastrophic outcome. | {
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Schedule an appointment using our convenient online form below or contact our Cary auto body repair shop by phone. We are located at 81 MacKenan Drive. | {
"redpajama_set_name": "RedPajamaC4"
} |
Jarßum is een dorp vlak bij de stad Emden in de Duitse deelstaat Nedersaksen. Het dorp ligt direct achter de dijk langs de Eems. Bestuurlijk maakt het deel uit van Emden.
Het dorp heeft een kerk uit 1797. Deze verving een oudere kerk uit de veertiende eeuw.
Plaats in Nedersaksen
Emden | {
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Business movers are known to ensure a smooth relocation of one's business. In this century, a large part of the population have opted to indulge in the business industry. Hence due to many people indulging in the business industry there have been many business enterprises that are upcoming. As most of the industries business also has some of the difficulties that face this specific industry. One of the main challenges that threaten the business industry is the lack of market. When there is no market for your business the downfall of the business is the next thing that ought to happen. Therefore the market is one of the crucial element to look into before setting up a business enterprise. As the market of the product or service shift from one place to another the business also relocate to close that market gap. The business will relocate to a place where the market is concentrated. For the activity of business relocation to be done efficiently, one needs to hire the most effective local business movers to do the business relocation. Any business owner in Maryland, Northern Virginia and DC in need of relocating his or her business ought to follow the following tips to get the most effective local business movers.
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The most effective business movers in DC, Maryland, and Northern Virginia ought to have positive reviews in the market. To get the reviews of that particular local business mover's company, one needs to question some clients earlier served those local business movers. Positive reviews will show that the company services are of the required quality. On the other hand local business movers in Maryland, DC and Northern Virginia with a negative reputation means the services offered by that particular company are not of the required standards. Therefore it is of great need to source for local business movers in Maryland, DC and Northern Virginia who has a good reputation in the market.
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Q: Base 2 exponential of native int Some algorithms (allocate a binary tree...) need to compute a base 2 exponential. How to compute it for this native type?
newtype {:nativeType "uint"} u32 =
x: nat | 0 <= x < 2147483648
This is an obvious try:
function pow2(n: u32): (r: u32)
requires n < 10
{
if n == 0 then 1 else 2 * pow2(n - 1)
}
It fails because Dafny doubts that the product stays below u32's max value. How to prove that it's value is below 2**10?
A: In this case, it is more convenient to first define the unbounded version of the function, and then prove a lemma showing that when n < 10 (or n < 32, even) it is in bounds.
function pow2(n: nat): int
{
if n == 0 then 1 else 2 * pow2(n - 1)
}
lemma pow2Bounds(n: nat)
requires n < 32
ensures 0 <= pow2(n) < 0x100000000
{ /* omitted here; two proofs given below */ }
function pow2u32(n: u32): u32
requires n < 32
{
pow2Bounds(n as nat);
pow2(n as nat) as u32
}
Intuitively, we might expect the lemma to go through automatically, because there are only a small number of cases to consider: n = 0, n = 1, ... n = 31. But Dafny will not perform such case analysis automatically. Instead, we have a couple of options.
First proof
First, we can prove a more general property, which, by the magic of inductive reasoning, is easier to prove, despite being stronger than what we need.
lemma pow2Monotone(a: nat, b: nat)
requires a < b
ensures pow2(a) < pow2(b)
{} // Dafny is able to prove this automatically by induction.
The lemma then follows.
lemma pow2Bounds(n: nat)
requires n < 32
ensures 0 <= pow2(n) < 0x100000000
{
pow2Monotone(n, 32);
}
Second proof
Another way to prove it is to tell Dafny it should unroll pow2 up to 32 times, using a :fuel attribute. These 32 unrollings are essentially the same as asking Dafny to do case analysis on each possible value. Dafny can then complete the proof without additional help.
lemma {:fuel pow2,31,32} pow2Bounds(n: nat)
requires n < 32
ensures 0 <= pow2(n) < 0x100000000
{}
The :fuel attribute is (lightly) documented in the Dafny Reference Manual in Section 24.
A: A bit of a cheat, but with so narrow a domain, this works very well.
const pow2: seq<u32> :=
[0x1, 0x2, 0x4, 0x8, 0x10, 0x20];
lemma pow2_exponential(n: u32)
ensures n == 0 ==> pow2[n] == 1
ensures 0 < n < 6 ==> pow2[n] == 2 * pow2[n - 1]
{}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} |
According to Simone Bartley, CEO at Saatchi & Saatchi, "The Make Your Move Campaign succeeded where so many before had failed. It emboldened V6 drivers to challenge their loyalty and make the move away from Holden and Ford.
Bold words indeed. What do you think? Is this campaign really changing the game? And can Toyota really shake the stranglehold Holden have on this market? | {
"redpajama_set_name": "RedPajamaC4"
} |
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Doctype or document type declaration in html defines the type of web document we are using. Doctype includes the version of HTML and its DTD. Website will not be fully functional without doctype in Internet Explorer lower versions.
Doctype is always declared before HTML opening tag. As per W3C, it is compulsory to define doctype of every webpage. Doctype is not a tag, its just document type representation of a webpage.
As per W3C, it is compulsory to defined Doctype of every webpage.
Doctype is web standard from HTML3 onwards. HTML3, HTML4, XHTML and HTML5 have doctype declaration. HTML5 Based Doctype is smaller than HTML4 and XHTML based doctypes. No need to specify the DTD and html version.
HTML5 Doctype is also known as DTD Less Doctype.
In HTML4, there are two doctype declarations, Strict Doctype and transitional Doctype. In strict doctype, presentational elements, like b, i, small, u, s are not allowed. But in transitional doctype, presentational can be used. Strict and Transitional are data type declarations. | {
"redpajama_set_name": "RedPajamaC4"
} |
Get this cute set of golf ball digital papers you can use for digital scrapbooking, journaling, planner decorating, card making, and more.
You'll find that these printable Father's Day theme scrapbook papers are large and of high quality at 300 DPI and 12×12 inches.
There are two papers in this set: Both have a repeating pattern of a large golf ball on a golf tee. One paper has a white background and the second has a bright green colored background. | {
"redpajama_set_name": "RedPajamaC4"
} |
The Hogget approached us wishing to update their current brand to include redecorating, a new menu, food photography, signage and of course a fresh new logo.
Businesses operating in any service industry ultimately must be judged on exactly that – SERVICE. We are happy to report the SERVICE we have received from The Typeface Group has been second to none in every way.
My pub business, The Hogget in Hook, Hampshire has gone from strength to strength since the work (online and offline) was carried out on our web site but also our signage plus a host of other areas.
Professional at every stage – well done TFG!!!
A half day photoshoot to include Headshots, interior shots and a large number of food images, to be used as part of the website design.
Creation of a relevant, timeless brand using original artwork, in line with the requirements of The Hogget and the image they wanted to portray.
Full keyword analysis for The Hogget website to ensure existing keywords were still relevant.
A full website build including updated copy in line with Google updates and their wider target audience.
Design and then print four new menus in line with the branding, with bespoke sizing and material choice.
Social Media training to ensure maximum gain from social media. | {
"redpajama_set_name": "RedPajamaC4"
} |
Positive Thursdays
BLACK AUGUST
LIWI
MARK STODDART
RANDELL ADJEI
SOLIDARITEES
LIVE IT. WEAR. IT. WITH PURPOSE!
COLLECTIONSNEW
Krystal Ball Going where her curiosity lies
'I believe in finding the beauty in things."
Dearly Beloved,
We want to thank you for joining us on this journey as we move from the past, through the present, and are now embarking on our future.
The future is bright, and during this time, communities, young people, and families have clung to the arts to bring a sense of togetherness during this time of increased isolation. We have found ways to use art, whether it is films, television shows, music, or poetry, to combat the anxiety and isolation that we are faced with in our day-to-day lives.
What I have learned is that creativity is the highest form of intelligence; the type of intelligence that cannot be replicated. Every year, a young creative springs out of the abyss, and demonstrates the uniqueness of their being. Our woman of the future Krystal Ball is no exception. She single handedly has ushered in a creative renaissance, that even 15 – 20 years from now, when industries are highly automated, her work will be of high value.
I was first introduced to Krystal's work through the Peel Art Gallery Museum + Archives Contemporary Caribbean Art exhibits early last year. We were at the height of the pandemic, and it was refreshing to see that the museum was doing something to keep the creative art scene going.
I was blown away with how her work captured the imagination. Yes, the colours were vivid; yes she utilized texture, but you could tell that her aim was to elicit a personal reaction from each person who viewed her work. Her art was created to express an idea, the emotions during the time of creation. It made me want to dig deeper, to explain how she had made the piece.
I finally had a chance to speak with Krystal, and when I did, it all began to make sense.
"My personality has grown into being comfortable with my artistic ability."
"My parents are not the conventional Jamaicans," Krystal began, "I was more of a parent then my mom was. I was very on top of things.
I started art at a very young age. We didn't really have art programs at school, but I somehow found a way to exercise my creative abilities.
I think I was 10 years old when I won an international art competition. I only did it because they offered $300.00 for local level, and then you had to compete at an international level and the prize money was $500.00. At 10 years old, $800.00 is a lot of money.
The money wasn't the best part. I also won a trip to Disneyworld. It was a mind-blowing experience. I was obsessed with Disneyland. It was not part of the package, so I was really surprised when I found out that I was going. They went to my mom, and asked her if there was something that I liked. She was like, Disneyland.
It was an amazing experience. It was nice to see that someone had built a world, and brought it to fruition. He had made it tangible. It was the first time that I travelled. It was the start of a love affair for me. Since then, travelling has become a part of my creative experience.
Krystal told me that she was very much a hustler. She had a lot of drive. She enjoyed doing her best, and her best turned into winning award after award
"I won a few art competitions. I ended up turning art competitions into a full time gig, and I used it to save up for college. It is a good thing that I did so well, because I could not see my parents being able to afford my education.
During high school, I took a break from the competitive art world. Along with painting and creating, I had been speaking at forums. Half the time I wasn't in class. I was spending so much time with adults that I had some issues speaking with kids in my cohort, but if you put me in a room of adults, I was great.
I got my freedom after high school. I had time to think. What was I doing? What did I want the world to see? It was an internal battle. It took some time to come into myself. Being good at a lot of things is very confusing. My parents weren't pushing me like that, but other forces were saying, you can do art on the side."
Those questions made her realize that she was not comfortable with herself. She shared with me that she didn't do anything that she wanted to do for herself until she was 21.
"I felt like I was living in a cage. A mental cage is a whole lot of work to get out of. It is different from a physical cage. I used art to break out of certain tendencies."
She began to travel a lot, and she was meeting different people. It was a time she told me, that made her introspective.
Before this transition, I didn't know who I was. I moved out at 17, and moved to Philadelphia. When I left Jamaica, I realized I didn't know anything about my country. Tourists began to tell me things about Jamaica, things that I vowed to discover when I returned, so when I came back I rediscovered Jamaica.
Life really began to change for Krystal during all of the Black Lives Matters Protest.
"It became very heavy. Social media can be heavy sometimes. It was a tough time for me.
I had an amazing opportunity that brought me to Manitoulin Island. It was during my Manitoulin island experience that my artistry came back to me. It was amazing to be somewhere, away from the lockdowns, and all the nonsense that surrounded Covid-19. While we were there, my partner came up with a brilliant idea. Since it was lockdown, let's go back to Jamaica. That is where I got most of my work done. Every time I go back, I view Jamaica with a different lens.
I put up a good fight trying to do other things, but art has always been with me. It has helped me get through a lot of bad situations. It has helped me with my mental health. It has helped me a lot. I also realize that I don't have to be one-dimensional. There are so many things that I can be doing. I now go where my curiosity lies."
This is only a snapshot of what lies in the mind of this creative genius. As an artist, she remains steadfast in working, finding some understanding, and envisioning a better world. Krystal has found a way to take in all the pain and struggles that exist in this world, and release beautiful works of art that heal and teach.
Simone Jennifer Smith
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Ontario's First Ever Poet Laureate
Using music as a tool for truth, justice, and soul The Queen of Soul Aretha Franklin
"Do you know who I am?"
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Paying Homage to an Icon - Cicely Tyson
Remembering "Hammerin Hank"
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Ontario's First Ever Poet Laureate April 28, 2021
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Natural Remedies for Alcohol Addiction Treatment
par admin7836 | Juil 22, 2020 | Sober living | 0 commentaires
Might help treat alcoholism
Herb to Cut Drinking: Study Details
Chinese Herb Kudzu May Help Drinkers Cut Down
Side effects of kudzu root
Kudzu extract treatment does not increase the intoxicating effects of acute alcohol in human volunteers.
Learn more about medications for alcohol cravings, and other ways to round out your support system https://ecosoberhouse.com/ in recovery. Cannabidiol, or CBD, is a compound that naturally occurs in the cannabis plant.
We have subsequently shown that puerarin is the major active isoflavone because 7 days treatment with this compound alone (1,200 mg/day) produced a similar reduction of binge drinking as the extract (Penetar et al., 2012). The present study provides further evidence that extracts of the kudzu root are effective in reducing alcohol consumption but unlike any other medication it does so after a single dose was taken shortly before a binge drinking opportunity. And, contrary to disulfiram treatment, the drinking that did occur after kudzu administration did not result in any noxious side effects, increases in subjective ratings of nausea, uncomfortable, or feeling terrible. The reduction in drinking was evident rather quickly as it was apparent for the second through sixth beers and no kudzu-treated participant drank five or six beers, which suggests that binge drinking was curtailed. In spite of the compelling preclinical and clinical evidence of its efficacy, the precise mechanism of action of kudzu in reducing alcohol consumption is not currently known. Prior studies of its antidipsotropic effect have focused on taste-aversion, alterations in alcohol metabolism or effects on neurotransmitters.
After completion of the program each subject re-took the AUD questionnaire and also filled out an exit interview. The Declinol Compound for the study was manufactured according to cGMP guidelines and was processed at an FDA registered, OTC, Rx facility. There are certain herbs and nutritional supplements that are best avoided by those who drink heavily.
Increases in 5-HIAL have been shown to be correlated with decreased alcohol consumption in hamsters (Keung et al., 1995).
Decreased drinking due to ALDH-2 inhibition is attributed to the aversive properties of acetaldehyde accumulated during alcohol consumption.
Certainly we caution any real interpretation which must await further study.
Importantly, individuals looking to discontinue alcohol use should considerseeking helpfrom health care providers and counseling services, who have access to better, prescription medications foralcoholism treatment.
These drugs are often prescribed to alcoholics to keep them abstinent.
Some of the most effective ways to reduce the cravings for alcohol include support groups and regular exercise.
The placebo-treated group opened 33 beers during baseline conditions and 38 following treatment whereas the kudzu-treated group opened 32 beers during baseline conditions and only 21 following treatment. Typically, alcohol withdrawal symptoms happen for heavier drinkers. Alcohol withdrawal can begin within hours of ending a drinking session. Importantly, individuals looking to discontinue alcohol use should considerseeking helpfrom health care providers and counseling services, who have access to better, prescription medications foralcoholism treatment.
The remains of a medieval skeleton has shown the first physical evidence that a fern plant could have been used for medicinal purposes in cases such as alopecia, dandruff and kidney … Rysuly MR, Azhibekova RJ Treatment of the Rennaissance-Iodine containing drug on patients with Hepatitis C. Amaty, Astana, Kazakhstan. Written informed consent was obtained from the patient for the publication of this pilot study and the accompanying data.
Is kudzu a flower?
Kudzu usually does not flower until its third year, with flowers and seeds forming only on vertical climbing vines. Kudzu produces clusters of 20 – 30 hairy brown seed pods, 1.6 – 2 inch (4 – 5 cm) long pods.
The leaves of this plant were smoked by Russian soldiers during World War II when there was a shortage of tobacco. kudzu root for alcohol cravings Is considered safe, with few side effects other than the potential for an allergic reaction to the plant.
Tangerine Peel has some bitter qualities as well, and is an excellent complement to the actions of Gentian. Tangerine Peel delivers several novel flavonoids that all offer numerous health advantages including enhancing metabolism, promoting detoxification, and protecting cells from free radical damage. Most importantly extracts of Tangerine Peel have been shown to enhance learning and memory. Kawahata et al. recently reported on the enhancement properties of Tangerine Peel extracts to facilitate potently- mediated transcription linked to the upstream cAMP/PKA/ERK/CREB pathway in hippocampal neurons. This may have important anti-alcohol relapse benefits based on dopaminergic genetics and its relationship to executive function and good decision making necessary for appropriate relapse prevention . Potential health perks of taking kudzu root include decreasing alcohol intake, easing menopause symptoms, and regulating blood sugar levels.
How much kudzu is needed for alcoholism?
The following doses have been studied in scientific research: BY MOUTH: For alcohol use disorder: 1.5-3 grams of kudzu root extract has been taken in 3 divided doses per day for 1-4 weeks. A single dose of 2 gram of kudzu extract has also been taken before a drinking episode.
« We suspect it may work because it increases blood flow, » he says. « It may deliver alcohol to the brain's reward center faster. So you get an effect sooner; therefore, you don't drink as much. » Before the second, they took either 1,200 milligrams of puerarin or identical-looking placebo pills for a week. They came to the lab and could drink as much as they wanted, up to six beers. Daily drinking can have serious consequences for a person's health, both in the short- and long-term. Many of the effects of drinking every day can be reversed through early intervention.
Several research groups have noted that adults also may change from nonproducer to producer status after heightened soy consumption, although not all studies concur. Franke et al. found that both post- and premenopausal women may begin to produce equol with increased isoflavone exposure. Changes in the chemistry of 6″-malonyl-7′-β-glucosyldaidzein, a major soybean component, during food processing. Dry heating of 6″-malonyl-7′-β-glucosyldaidzein leads to decarboxylation to 6″-acetyl-7′-β-glucosyldaidzein .
Dandelion may also be able to help with alcohol withdrawal symptoms. As with many other OTC herbal supplements, there's little convincing evidence that the aforementioned supplements are clinically effective treatments of alcohol dependency. There have been some scientific studies conducted, but a majority involve animal models and show modest findings that rarely translate into conclusive human trials.
Therapeutic lessons from traditional Oriental medicine to contemporary Occidental pharmacology.
For this reason, it's difficult to make recommendations for various uses. It's best to speak with your healthcare provider to determine whether kudzu root could interact with any medications you're taking.
Bouchery EE, Harwood HJ, Sacks JJ, Simon CJ, Brewer RD. Economic costs of excessive alcohol consumption in the U.S., 2006. Medically Reviewed By Benjamin Caleb Williams, RNA licensed behavioral health or medical professional on The Recovery Village Editorial Team has analyzed and confirmed every statistic, study and medical claim on this page. Take our short alcohol quiz to learn where you fall on the drinking spectrum and if you might benefit from quitting or cutting back on alcohol. Excessive drinking has numerous impacts on your body and mind, ranging from mild to severe. Learn which signs to look out for, and how to care for your well-being.
A copy of the written consent is available for review by the Editor- in-Chief of this journal. Due to the lack of data on the pre-intervention, it cannot be said statistically whether the results were affected by the intervention. A further clinical study with a larger sample size is required to confirm the results. "You want to make sure that your medication is not as bad or worse than the drug you're trying to treat," Lukas said. On blood pressure, fibrinolysis and oxidative stress in patients with stage 1 hypertension.
It's been used for health purposes in Traditional Chinese Medicine for a really long time.
Though KU inhibited the alcohol intake and alcohol tolerance development, it reduced ghrelin levels in alcohol-preferring rats, which could play a role as an indicator of the currently used drugs.
There is some evidence that milk thistle aids in the regeneration of liver cells.
Specifically, male and female "heavy" alcohol drinkers were treated with either placebo or a kudzu extract for 7 days and then given an opportunity to drink their preferred brand of beer while in a naturalistic environment.
While this drug is not yet FDA-approved to treatalcohol addictionand cravings, increasing amounts of evidence show that topiramate can be used to treat alcohol cravings if used as your doctor prescribes. Other supplements such as L-glutamine and milk thistle are thought to decrease cravings and aid in detoxifying the liver, respectively. There is some evidence that milk thistle aids in the regeneration of liver cells. However, these supplements have not been thoroughly substantiated with scientific research to confirm the efficacy of their actions and benefits. Nonetheless, they are safe to take and may help to alleviate some symptoms of withdrawal and cravings. Of course, these are only nine helpful options—there are many vitamins and supplements for alcoholics you can take to aid your system in recovery.
These services can also assist in the process of alcohol detox and help with the development of coping skills to prevent setbacks. Additionally, depending on the severity ofalcohol dependence, withdrawal from alcohol can have life-threatening side effects and should be completed under the supervision of medical staff in an alcohol rehab treatment facility. Ashwagondha is an important herb in traditional Ayurvedic medicine. | {
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We looked at shopping habits and personal incomes to determine who needs to tone down the shopping this Christmas.
Note: Desktop users: Use your mouse and hover over a state to see the stats in the map below. Mobile users, there's a smaller version a little lower on the page.
Can you afford to shop this holiday season? There's only 10 days left to get everything you need.
According to the U.S. Federal Reserve, Americans are in over their heads in credit card debt now more than ever before. The average U.S. household carries $7,281 in credit card debt alone.
So, we were curious – which states can shop guilt free this holiday season, and who can't afford it? We're about to find out, using science and data.
Read on below to see how we crunched the numbers and how your state ranked. If your state isn't on this list, then you might want to back away from the register a little bit this holiday season.
Our argument is states where residents earn higher salaries (adjusted for cost of living), where people don't already shop a lot, and where folks are in less credit card debt can afford to stack the piles high under the tree this year.
Nebraska residents can afford to shop the most during this Christmas. So get out there and stand in long lines. You've certainly earned it.
Why? First off, Nebraskans only have about $4,298 in credit card debt each. That might sound like a lot, but it's actually the 3rd lowest total in the nation. And, it's obvious that folks here aren't shopping nearly as much as people in other states. They only spend about 25 minutes a day buying things.
Lastly, households average almost $58,000 a year in combined incomes, after cost of living adjustments.
So, if anyone can afford to go all out this holiday season, it is the people in Nebraska.
You know who has been keeping their debt in check? Iowans, that's who. There isn't any other group in the nation with lower credit card debt than here. People average an astoundingly low $3,884 each in debt.
The next closest state is North Dakota, where people average a tad over $4,000 each in debt. The problem with North Dakota is they shop too often – for about 36 minutes each day, on average.
In Iowa, people spend only about 24 minutes a day, on average, purchasing things.
So, go ahead and buy the entire family a new set of overalls this year. You've certainly earned it.
We're going to see a lot of states from the midwest on this ranking. When looking at the data, it's the midwest that has the highest incomes and lowest debt.
In Minnesota, families bring in a combined $57,000 a year, and only shop for about 25 minutes a day as it is.
South Dakota is an interesting story. They only spend about 20 minutes a day shopping here, which is just about the least you'll find in the country.
And, as you might expect, their credit card debt is really low as a result. South Dakotans only have $4,238 each in debt. The problem? Their incomes are far lower than the other cities on this list.
So while they've been saving up, they don't have a lot of wiggle room here.
The only state that spends less time buying things is Mississippi, which spends money for about 19 minutes a day, on average.
Kansas is in the opposite position. They, too, don't shop nearly as often as most of the country. However, while they are high earners, they're also quite in over their heads in debt.
We're giving Kansas the nod to go ahead and make lots of purchases. However, the amount of debt here means Kansans should pay with cash; families here average $57,610, so they can certainly afford to put down the plastic.
Kentucky is in the same boat as South Dakota. Individuals here have managed their shopping habits (24 minutes a day), and their debt ($4,436). But they are also low earners.
However, the cost of living in Kentucky is far below the national average. So, if anyone in the south can afford to stack up the gifts under the tree this year, it's folks in Kentucky, who have been pretty frugal all year long.
Folks in Indiana have a bit of the shopping bug, but it looks like they're managing their spending wisely. While they take about three hours each week to shop, their debt is far below what most other states have accumulated.
And being relatively high earners means if they are using plastic for purchases, they can likely afford to keep up with the payments.
So we say, Santa Claus should definitely spend more than a fair amount of time in Indiana living rooms this year.
Good news, Arkansas! You're in the same boat as Kentucky. As families, you don't bring in huge salaries, but you're also not in over your heads at this point in the year.
We say you can shop guilt free down there in the Natural State this Christmas.
Families in Wisconsin only average about $8,600 each in credit card debt. That sounds high, but for a husband and wife, that's actually one of the lowest you'll find in America. A majority of the population in Wisconsin earns pretty decent salaries, too.
We noticed Wisconsin already has a big of the shopping bug as it is (27 minutes a day, on average.) So, they're used to managing their money effectively.
Pennsylvania is the only east coast state to make this list. While this state doesn't rank high in any one particular category, across the board, it's safe to assume that families here can afford to make those extra holiday purchases this year.
It that good news? We wonder how many PA wives are going to share this with their husbands today.
Here's a link to the data for all 50 states. | {
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Birding the OC: Eurasion Widgeon at Tewinkle Park and swallows at San Jaoquin wildlife refuge.
A Eurasian Widgeon had been reported at Tewinkle Park in Costa Mesa and that would be a life bird for us. Mike had to fly out of John Wayne so he stopped by before his flight. The bird was easy to spot in the "cement pond" and easy to photograph.
There's not much to do at Tewinkle other than see the Eurasian Widgeon, so there was a little time to drop by San Jaoquin Wildlife Refuge where the tree swallows were out in force.
When we lived on the east coast we never imagined that Robins or Waxwings feasted at Palm Trees.
We went to Bolsa Chica in Huntington Beach. The most interesting find was a light-footed clapper rail bathing in front of the footbridge.
We dropped by Yucaipa Regional Park on the way to a nursery. We got some good views of red-breasted sapsuckers. The female hooded merganser eating a fish was interesting too.
We stopped by Andulka Park and the Say's Phoebe was posing.
We went to Irvine Lake to find the bald eagles that had been reported. No luck there, but the painted redstart was nice. We would have gone just for that if we hadn't seen some in Arizona recently.
The western bluebird was nice too. | {
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Lula drops out of Brazil presidential race
Published : Wednesday, 12 September, 2018 at 9:56 AM Count : 189
Edited by Golam Yusuf
A supporter of Brazilian former president Luiz Inácio Lula da Silva holds a mask depicting him, as they gather outside the federal police building in Curitiba on Tuesday. Photograph: Nelson Almeida/AFP/Getty Images
Jailed former Brazilian president Luiz Inacio "Lula" da Silva has withdrawn his candidacy for the country's top seat ahead of a general election next month.
The switch was approved at a meeting of the Workers Party in the southern city of Curitiba — where Lula has been held since April for corruption — as the clock ticked down on a court-ordered deadline for him to name a stand-in.
"The decision has been made," a party official told AFP.
Hundreds of Lula supporters were gathered near the jail where he is being held. Haddad was expected to read a letter there from his mentor anointing him as his political heir, AFP reports.
The decision comes less than two weeks after Brazil's Superior Electoral Tribunal ruled that the popular but polarizing former president cannot run while serving his 12-year prison sentence.
Though jailed, the 72-year-old Lula was the frontrunner in polls, and his removal from the race has scrambled the field, catapulting right-wing populist Jair Bolsonaro to the fore.
Bolsonaro, a polarizing figure who has been criticized for outbursts deemed racist, misogynist and homophobic, was stabbed while on the campaign trail last week.
He is not expected to appear at any rallies before the October 7 polls, but remains on the ballot.
Haddad, a former mayor of Sao Paulo who also served as Lula's education minister, faces a race against time with the first round of voting less than a month away.
His ability to hold on to Lula's base will be key if he and his expected running mate, youthful communist Manuela d'Avila, are to reach the second round, set for October 28.
– Clean slate rules –
Haddad's political career put him at the center of the Workers Party, but without ever emerging from Lula's shadow — and the 55-year-old has displayed little of the star power of his mentor.
A poll released Monday by Datafolha shows Haddad with nine percent support, up five points from a month ago.
That places him in a mix of candidates aspiring to go to a second round of voting against Bolsonaro, who currently is out front at 26 percent.
Lula's supporters have been camped out outside the federal police headquarters in Curitiba since he was incarcerated.
The city is the epicenter of a sprawling corruption investigation that has brought to justice dozens of politicians and business leaders, including Lula, who was president from 2003 to 2011.
He was convicted in July 2017 of taking a bribe from a Brazilian construction company in the form of a luxury seaside apartment in return for contracts with state oil giant Petrobras.
Numerous appeals of the conviction and sentence have failed, and his lawyers also have been unable to get around rules that have kept Lula off the ballot.
He faces trial in five other cases, but insists he is the innocent victim of politically motivated prosecutions to keep him out of office.
A former metalworker, Lula rose as a union leader during Brazil's military dictatorship, co-founding the Workers Party in 1980.
His presidency was credited with lifting millions out of poverty through generous social programs, transforming his party into a political powerhouse.
It has won the last four presidential elections, the last two by Dilma Rousseff, Lula's handpicked successor who was ousted from power by Congress in 2016, accused of manipulating federal budgets. Current President Michel Temer, Rousseff's vice president, replaced her.
Lula drops out Brazil presidential
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Kristin Olness
Kristin Olness played the original Helga in the award winning 1998 Broadway revival of Cabaret and is a recipient of the Fred Astaire award. She recently reprised her role of Helga in the 2014 Revival of Cabaret (featuring Alan Cumming, Michelle Williams, Emma Stone, and Sienna Miller). In addition to her Broadway successes, she has enjoyed an extensive career as a singer, dancer and aerialist across the country and around the world. She can be seen in the PBS special " My Favorite Broadway" and other TV shows. She was a featured aerialist in the Westchester Broadway revival of Barnum, and has toured the U.S. and abroad as a principal aerialist with Cirque Le Masque, Cirque on Ice, Circus Nexus, and ImaginAerial. She has choreographed many NYC productions in venues such as La Mama, Dixon Place, and Theater for The New City, as well as productions in Maine, Philadelphia, and China. She co-founded and has performed with Suspended Cirque (an innovative NYC based theatrical aerial company, suspendedcirque.com), and is also a co-founder and aerial teacher at Aerial Arts NYC.
Her other interests include playing clarinet, saxophone, singing, dancing, and yoga.
Cabaret 2014 Broadway Revival "Also on board is at least one of the original 1998 Kit Kat chorus girls, Kristin Olness, who plays Helga…….. Mr. Cumming's M.C. noted that he and Ms. Olness's Helga went back a long way."
— Ben Brantley – New York Times
The Baby Janes [prod. by Kristin Olness] The flirty troupe makes your stockings hang a little straighter with this aerial burlesque, which soars well above the mantle height during a sexy quadruple-trapeze act."
— Time Out New York
Above the Belt performance: 'Above the Belt' is a stunning, terrifying, magical and sensual experience. Stand out performers include Kristin Olness…
— UnCoolkids.com
I also enjoyed the work of Kris Olness, whose striking beauty and balletic extensions made her work in the fabric cocoon lovely to watch.
— Spectacle Magazine
Kit Kat Girl Revue If you fancy kicking your evening of nightclub follies off with a Broadway-style extravaganza, the deluxe, velvet-curtain-heavy Show features just that on Fridays, with the va-va-va-voom dancers and singers.
(WHM) ©2021 Kristin Olness | Site by KPFdigital | Website Administrator Login | {
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The Republic of Bulgaria does not have a functioning embassy in Nairobi. Bulgarian citizens residing short-term or long-term in the country may address the Embassy of the Republic of Bulgaria in Addis Ababa, Ethiopia for consular matters.
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In extraordinary situations, such damaged, lost or stolen passport the Bulgarian citizens can apply for a temporary passport for an Emergency Travel Document at the embassy of any EU member state in Nairobi. | {
"redpajama_set_name": "RedPajamaC4"
} |
\section{Introduction}
After the realization of Bose-Einstein condensations (BECs) in trapped atomic vapors of $^{87}$Rb, $^7$Li, and $^{23}$Na \cite{WiemanCornell,Ketterle}, a new period of intense experimental and theoretical research has been initiated. The equilibrium properties of these novel systems have been quite well understood, but there are still several open questions concerning their nonequilibrium behavior. One of the most important questions concerns the behavior of the condensate after cooling a nondegenerate trapped Bose gas to a temperature below the BEC critical temperature. While the experimental research has, up to now, concentrated mainly on the initial formation of BECs, their theoretical behaviour at finite temperatures is a frontier of many-body physics. The theoretical description of BECs has to take into account the coupled nonequilibrium dynamics of both the condensed and noncondensed components of the gas under investigation, and has to involve the collisional processes of atoms between the two components. Such a quantum kinetic theory was inititated by Kirkpatrick and Dorfman \cite{KD1,KD2}, based on the rich body of research carried out in the period 1940-67 by Bogoliubov, Lee and Yang, Beliaev, Pitaevskii, Hugenholtz and Pines, Hohenberg and Martin, Gavoret and Nozi`eres, Kane and Kadanoff and many others. The terminology ``Quantum Kinetic Theory'' has been later introduced in a series of papers by Gardinier, Zoller and collaborators \cite{QK0,QK1,QK2,QK3}. After that, there has been an explosion of research on quantum kinetic theory (see \cite{josserand2001nonlinear,BaoCai, BPM, bijlsma2000condensate,KD1,KD2,Spohn:2010:KOT,GriffinNikuniZaremba:2009:BCG,QK0,QK1,QK2,QK3,ArkerydNouri:2012:BCI,ArkerydNouri:2015:BCI,ArkerydNouri:AMP:2013,ReichlGust:2013:RRA,pomeau1999theorie,HuJin:2011:OKF,FilbetHuJin:2012:ANS,DyachenkoNewellPushkarev:OTW:1992,ZakharovNazarenko:DOT:2005,zaremba1998two}, and references therein). We refer to the review paper \cite{anglin2002bose} and the books \cite{inguscio1999bose,ColdAtoms1}, for more discussions and a complete list of references on this rapidly expanding topic.
The current paper is devoted to the study of the hydrodynamic approximations of such a quantum kinetic system. The system contains two equations: a quantum Boltzmann equation describing the non-condensate atoms (with two types of collisions, one between excited atoms and one between condensate atoms and excited atoms), and a nonlinear Schr\"odinger (or Gross-Pitaevski) equation for the condensate. The hydrodynamic limits of the system is an interesting mathematical question, first studied in \cite{Allemand:Thesis:2009}, where an Euler limit has been derived. This derivation relies on the assumption that, in the considered trapped Bose gas, the noncondensate and condensate share the same
local equilibrium. It is known (cf. \cite{KD1,KD2}) that the condition of
complete local equilibrium between the condensate and the thermal cloud
requires the energy of a condensate atom in the
local rest frame of the thermal cloud to be equal to the local thermal cloud chemical
potential. When the condition is satisfied, there is no exchange of particles
between the condensate and the thermal cloud (cf. \cite{GriffinNikuniZaremba:2009:BCG}). As a consequence, in the derived fluid system, the mass of each component - condensate and non-condensate - does not exchange. Note that the two-fluid low-frequency dynamics of superfluid $^4$He was first developed
by Tisza and Landau \cite{lifshitz1987fluid}. Their description accounts for the characteristic features associated with
superfluidity in terms of the relative motion of superfluid
and normal fluid degrees of freedom, and was shown to be a consequence of a Bose broken symmetry (cf. \cite{bogoliubov1970lectures}). In the Landau two-fluid theory, the two components superfluid
and normal fluid exchange mass (cf. \cite{Allemand:Thesis:2009,lifshitz1987fluid,bogoliubov1970lectures}). In this paper, we revisit the derivation of the Euler hydrodynamic limit of the system by a different point of view: following \cite{KD1,KD2,GriffinNikuniZaremba:2009:BCG}, we assume that even if the thermal cloud atoms are in equilibrium among
themselves, the noncondensate and condensate
parts may not be in local equilibrium with each other. Moreover, the derivation of the Navier-Stokes approximation of the system is also provided via the
classical Chapman-Enskog expansion (cf. \cite{TruesdellMuncaster:FOM:1980}). In
such circumstance, the Euler limit includes the mass exchange between the
condensate and the non-condensate. Our Euler and Navier-Stokes approximations
agree with the Landau two-fluid theory (cf. \cite{lifshitz1987fluid,bogoliubov1970lectures}).
As an attempt to build a rigourous theory for quantum kinetic equations, some mathematical results have been obtained in \cite{AlonsoGambaBinh,CraciunBinh,Binh9,GambaSmithBinh,GermainIonescuTran,ToanBinh,nguyen2017quantum,SofferBinh1,ReichlTran,SofferBinh2}) . Note that quantum kinetic equations have very similar formulations with the so-called wave turbulence kinetic equations. We refer to \cite{buckmaster2016analysis,FaouGermainHani:TWN:2016,germain2015high,germain2015continuous,LukkarinenSpohn:WNS:2011,Nazarenko:2011:WT,Spohn:WNW:2010,zakharov2012kolmogorov,Zakharov:1998:NWA} for more recent advances on the rigorous theory of weak turbulence.
The plan of the paper is as follows. In Section \ref{Sec2} we introduce
the quantum kinetic system and the scalings that will lead to the
hydrodynamic approximation.
In Section \ref{Sec:Operators}, we list the most important features of the two collision operator $C_{12}$ and $C_{22}$.
The two-fluid Euler and Navier-Stokes limits are then derived in the two Sections \ref{Sec:Euler} and \ref{Sec:NavierStokes} respectively.
\section{The quantum kinetic system and scalings}\label{Sec2}
\subsection{The quantum kinetic system}
Let us consider a trap Bose gas, whose temperature $T$ is smaller than the Bose-Einstein transition temperature $T_{BEC}$ and strictly greater than $0$ K or $-273.15^oC$. Denote $f(t,r,p)$ to be the density function of the normal fluid at time $t$, position $r$ and momentum $p$ and $\Phi(t,r)$ be the wave function of the the condensated (or superfluid) phase. Employing the short-handed notation
$f_i=f(t,r,p_i)$, $i=1,2,3,4$,
we first recall the quantum kinetic - Schr\"odinger system describing the dynamics of a BEC and its thermal cloud. The Schr\"odinger (or the Gross-Pitaevski) equation for the condensates reads (cf. \cite{bijlsma2000condensate}):
\begin{equation}
\begin{aligned}\label{GP}
i \hbar {\partial_t \Phi(t,r)} =&\ \Big(-\frac{\hbar^2 \Delta_{{r}}}{2m}+g[n_c(t,r) + 2n_n(t,r)] -i\Lambda_{12}[f](t,r) +V(r)\Big)\Phi(t,r), \ \ (t,r)\in\mathbb{R}_+\times\mathbb{R}^3,\\
\Lambda_{12}[f](t,r) = &\ \frac{\hbar}{2n_c}\Gamma_{12}[f](t,r),\\
\Gamma_{12}[f](t,r)= &\ \int_{\mathbb{R}^{3}}C_{12}[f](t,r,p)\frac{d p}{(2\pi \hbar)^3},\\
n_n(t,r) \ = & \ \int_{\mathbb{R}^3}f(t,r,p)d p,
\\
~~~\Phi(0,r)=& \ \Phi_0(r), \forall r\in\mathbb{R}^3,
\end{aligned}
\end{equation}
where $n_c(t,r)=|\Phi|^2(t,r)$ is the condensate density, $\hbar$ is the Planck constant, $g$ is the interaction coupling constant proportional to the $s$-wave scattering length $a$, $V(r)$ is the confinement potential, and the operator $C_{12}$ can be found in
the quantum Boltzmann equation for the non-condensate atoms (cf. \cite{bijlsma2000condensate}), written below:
\begin{eqnarray}\label{QBFull}
{\partial_t f}(t,r,p)&+&\frac{p}{m}\cdot\nabla_{{r}} f(t,r,p) \ - \ \nabla_r U(t,r)\cdot \nabla_p f(t,r,p)\\\nonumber
&=&Q[f](t,r,p):=C_{12}[f](t,r,p)+C_{22}[f](t,r,p), (t,r,p)\in\mathbb{R}_+\times\mathbb{R}^3\times\mathbb{R}^3,\\\nonumber
C_{12}[f](t,r,p_1)&:=&\lambda_1n_c(t,r) \iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(mv_c+ {p}_1-{p}_2-{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{{p}_1}-\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3})\\\label{C12}
& &\times[(1+f_1)f_2f_3-f_1(1+f_2)(1+f_3)]d p_2d p_3\\\nonumber
&&-2\lambda_1n_c(t,r)\iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(mv_c+{p}_2-{p}_1-{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_1}-\mathcal{E}_{{p}_3})\\\nonumber
& &\times[(1+f_2)f_1f_3-f_2(1+f_1)(1+f_3)]d p_2d p_3,\\\label{C22}
C_{22}[f](t,r,p_1)&:=&\lambda_2\iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta({p}_1+{p}_2-{p}_3-{p}_4)\\\nonumber
& &\times\delta(\mathcal{E}_{{p}_1}+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3}-\mathcal{E}_{{p}_4})\times\\\nonumber
&&\times [(1+f_1)(1+f_2)f_3f_4-f_1f_2(1+f_3)(1+f_4)]d p_2d p_3d p_4,
\\\nonumber
f(0,r,p)&=&f_0(r,p), (r,p)\in\mathbb{R}^3\times\mathbb{R}^3,
\end{eqnarray}
where $\lambda_1=\frac{2g^2}{(2\pi)^2\hbar^4},$ $\lambda_2=\frac{2g^2}{(2\pi)^5\hbar^7}$, $m$ is the mass of the particles, $\mathcal{E}_{{p}}$ is the Hartree-Fock energy (cf. \cite{bijlsma2000condensate})
\begin{eqnarray} \label{def-E}
\mathcal{E}_p\ =\ \mathcal{E}(p)\ = \ \frac{|p|^2}{2m} + U(t,r).
\end{eqnarray}
Notice that $C_{22}$ is the Boltzmann-Norheim (Uehling-Ulenbeck) quantum Boltzmann collision operator.
If one writes
\begin{equation}\label{def-Phi}
\begin{aligned}
\Phi \ = & \ |\Phi(t,r)|e^{i\phi(t,r)},
\end{aligned}
\end{equation}
the condensate velocity can be defined as
\begin{equation}\label{def-vc}
v_c(t,r) = \frac{\hbar}{m}\nabla \phi(t,r),
\end{equation}
and the condensate chemical potential is then
\begin{equation}\label{def-muc}
\mu_c
=\ \frac{1}{\sqrt{n_c}}\left(-\frac{\hbar^2\Delta_r}{2m}+ V +g[2n_n+n_c]\right)\sqrt{n_c}.
\end{equation}
When $V=0$, the following system for the super-fluid of the condensate can be obtained
\begin{equation}\label{BECSuperFluid}
\begin{aligned}
{\partial_t n_c}\ + \ \nabla_r\cdot(n_cv_c)\
=&\ - \ \Gamma_{12}[f]\\
\partial_t v_c +\frac{\nabla_r v_c^2}{2} = & \ -\nabla_r\mu_c.
\end{aligned}
\end{equation}
The potential $U$ and the condensate energy $\mathcal{E}_c$ are written as follows
\begin{equation}\label{def-U}
U(t,r)\ = \ V(r) + 2g[n_c(t,r)+n_n(t,r)],
\end{equation}
and
\begin{equation}\label{def-Ec}
\mathcal{E}_c(t,r) \ = \ \mu_c(t,r) \ +\ \frac{mv_c^2(t,r)}{2}.
\end{equation}
For the sake of simplicity, we suppose that $V\equiv 0$ and define the differential quantity
\begin{equation}\label{Diff}
\bar{d}p=\frac{dp}{(2\pi \hbar)^3}.
\end{equation}
Notice that \eqref{C12} describes collisions between the condensate and the non-condensate atoms (condensate growth term) and \eqref{C22} describes collisions between non-condensate atoms.
\begin{remark}
At temperature $T$, bosons of mass $m$ can be regarded as quantum-mechanical
wavepackets which have an extent on the order of a thermal de Broglie wavelength $\lambda_{dB} =
\left(\frac{2 \pi \hbar^2}{m k_B T}\right)^\frac12$, where $k_B$ is the Boltzmann constant. The de Broglie wavelength $\lambda_{dB}$ describes the position uncertainty associated with the thermal momentum
distribution. When the gas temperature is high $T>T_{BEC}$, $\lambda_{dB}$ is very small and the weakly interacting gas can be treated
as a system of ``billiard balls'' (cf. \cite{durfee1998experimental,ketterle1999making}). The dynamics of the gas is described by the Boltzmann-Norheim (Uehling-Ulenbeck) equation, whose operator sometimes reads (cf. \cite{UehlingUhlenbeck:TPI:1933})
\begin{equation}\label{QBHightT}
\begin{aligned}
\mathcal{C}_{22}[f](t,r,p_1)\ =& \ \iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta({p}_1+{p}_2-{p}_3-{p}_4)\delta(\mathcal{E}_{{p}_1}+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3}-\mathcal{E}_{{p}_4})\times\\
\ & \ \times [(1+\vartheta f_1)(1+\vartheta f_2)f_3f_4-f_1f_2(1+\vartheta f_3)(1+ \vartheta f_4)]d p_2d p_3d p_4,
\end{aligned}
\end{equation}
where $\vartheta$ is proportional to $\hbar^3$. In the semiclassical limit, as $\vartheta$ tends to $0$, the quantum Boltzmann collision operator becomes the classical one. This means at high temperature, the behavior of the ``billiard balls'' Bose gas is, in some sense, still very similar to classical gases.
At the BEC transition temperature, $\lambda_{dB}$ becomes comparable to the distance between
atoms. As a result, the atomic wavepackets ``overlap'' and the indistinguishability of atoms becomes important. At this
temperature, bosons undergo a quantum-mechanical phase transition and the Bose-Einstein condensate is formed (cf. \cite{durfee1998experimental,ketterle1999making}). When the temperature of the gas is finite $T_{BEC}>T>0$K, the trapped Bose gas is composed of two distinct components: the high-density condensate, being localized at the center
of the trapping potential, and the low-density
cloud of thermally excited atoms, spreading over a much wider region. The dynamics of the thermal
cloud atoms is described by the kinetic equation \eqref{QBFull}. At this low temperature, the de Broglie wavelength of the excited atoms is very large, in comparison with the high temperature boson de Broglie wavelength. As a consequence, the thermal cloud kinetic equation cannot be treated
as a system of ``billiard balls'' anymore. This explains the difference between the forms of the two collision operators $C_{22}$ and $\mathcal{C}_{22}$.
Note that, different from classical Boltzmann collision operators, where the collision kernels are functions depending on the types of particles considered, the derived collision kernel for the quantum Boltzmann collision operator for bosons is 1 (cf. \cite{EscobedoVelazquez:2015:FTB}) when $T>T_{BEC}$.
\end{remark}
\subsection{Scalings}
Different from the thesis \cite{Allemand:Thesis:2009}, in which the two collision operators $C_{12}$ and $C_{22}$ are assumed to have the same equilibrium distribution function, we follow \cite{GriffinNikuniZaremba:2009:BCG} to consider the most general regime, where excited atoms in the condensate need not to be in local equilibrium with the condensate atoms. As a consequence, $C_{12}$ and $C_{22}$ in general do not share the same equilibrium distribution. A comparison between our results and the result of \cite{Allemand:Thesis:2009} will be discussed in details in Section \ref{Comparison}. Relying on these physical assumptions, we propose a new approach to obtain new Euler and Navier-Stokes approximations of the system.
It is known that the dynamics of the trapped Bose gases depends on its temperature $T$. Let us restrict our attention to the case where $T$ is smaller but very close to the Bose-Einstein critical temperate $T_{BEC}$. At this temperature regime, the collisions between excited atoms are rapid to establish a local equilibrium within the non-condensate component. As a consequence, the collision operator $C_{22}$ can be assumed to be stronger than the collision operator $C_{12}$. This regime is often called {\it the state of partial local equilibrium} which arises near $T_{BEC}$ when the density of the condensate is small.
Following \cite{GriffinNikuniZaremba:2009:BCG}, we define the {\it static equilibrium} of the system
\begin{equation}\label{StaticEquilibrium}
\mathcal{F}_0(p) \ = \ \frac{1}{e^{\beta_0[(p-mv_{n0})^2/(2m) + U_0 - \mu_0]}-1},
\end{equation}
where $\beta_0$ is the static temperature parameter, $v_{n0}$ is the static fluid velocity, $\mu_0$ is the static chemical potential, $U_0$ is the static mean field. We also set the static density to be
\begin{equation}\label{FluidDensity}
{n_n}_0 \ = \ \int_{\mathbb{R}^3}\mathcal{F}_0(p) \bar{d} p.
\end{equation}
Note that when $T$ is sufficiently close to $T_{BEC}$, the bosons are in the {\it particle-like regime}, i.e. they behave like particles. Let us also mention that when temperature $T$ is very close to $0$, the bosons will be in the phonon-like regime (cf. \cite{ReichlBook}). Since we are interested in the behavior of the particles when $T$ is close to $T_{BEC}$, let us define the collision frequency with respect to $C_{12}$
\begin{equation}\label{CollisionFrequencyC12}
\begin{aligned}
\nu_{12}(p_1) \ = \ & \frac{\lambda_1 n_c}{m^2}
\iint_{\mathbb{R}^3\times\mathbb{R}^3}\delta(mv_c + p_1 -p_2-p_3)\delta(\mathcal{E}_c+\mathcal{E}_{p_1}-\mathcal{E}_{p_2}-\mathcal{E}_{p_3})\times\\
&\times[\mathcal{F}_0(p_2)+\mathcal{F}_0(p_3)+1]dp_2dp_3\ + \ \\
& \ +2\frac{\lambda_1 n_c}{m^2}
\iint_{\mathbb{R}^3\times\mathbb{R}^3}\delta(mv_c + p_2 -p_1-p_3)\delta(\mathcal{E}_c+\mathcal{E}_{p_2}-\mathcal{E}_{p_1}-\mathcal{E}_{p_3})\mathcal{F}_0(p_3)dp_2dp_3,
\end{aligned}
\end{equation}
as well as the associated mean collision frequency:
\begin{equation}\label{MeanCollisionFrequencyC12}
\bar{\nu}_{12}\ = \ \frac{1}{{n_{n_0}}m}\int_{\mathbb{R}^3}\nu_{12}(p)\mathcal{F}_0(p)\bar{d}p.
\end{equation}
The inverse of $\nu_{12}(p) $ and $\bar{\nu}_{12}$ are defined to be, respectively, the free time $\tau_{12}(p)$ and the mean field time $\bar\tau_{12}$:
\begin{equation}\label{MeanFieldTimeC12}
\tau_{12}(p) \ = \ \frac{1}{\nu_{12}(p)}, \ \ \ \ \ \bar\tau_{12} \ = \ \frac{1}{\bar{\nu}_{12}}.
\end{equation}
We now determine the average speed of the particles
\begin{equation}\label{AvergeSpeed}
\bar{c} \ = \ \frac{1}{{n_n}_0m}\int_{\mathbb{R}^3}\sqrt{p^2}\mathcal{F}_0(p)\bar{d}p,
\end{equation}
and the mean free path
\begin{equation}\label{MeanFreePathC12}
l_{12} \ = \ \bar{c}\bar{\tau}_{12}.
\end{equation}
Similarly, the collision frequency and the mean collision frequency associated to $C_{22}$ can be defined
\begin{equation}\label{CollisionFrequencyC22}
\begin{aligned}
\nu_{22}(p_1) \ = \ & \frac{\lambda_2}{{n_n}_0m}
\iint_{\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3}\delta(p_1 + p_2-p_3-p_4)\delta(\mathcal{E}_{p_1}+\mathcal{E}_{p_2}-\mathcal{E}_{p_3}-\mathcal{E}_{p_4})\times\\
& \times \mathcal{F}_0(p_2)(1+\mathcal{F}_0(p_3))(1+\mathcal{F}_0(p_4))dp_2dp_3dp_4,
\end{aligned}
\end{equation}
and
\begin{equation}\label{MeanCollisionFrequencyC12}
\bar{\nu}_{22}\ = \ \frac{1}{{n_n}_0m}\int_{\mathbb{R}^3}\nu_{22}(p)\mathcal{F}_0(p)\bar{d}p.
\end{equation}
We also define the free time $\tau_{22}(p)$, the mean field time $\bar\tau_{22}$ and the mean free path $l_{22}$
\begin{equation}\label{MeanFieldTimeC12}
\tau_{22}(p) \ = \ \frac{1}{\nu_{22}(p)}, \ \ \ \ \ \bar\tau_{22} \ = \ \frac{1}{\bar{\nu}_{22}}, \ \ \ \ \ \ l_{22} \ = \ \bar{c}\bar{\tau}_{22}.
\end{equation}
Let $L$ and $\theta$ be the reference length and time, respectively. Following \cite{Sone:KTN:2002,BouchutGolse:2000:KEA}, we introduce the rescaled variables
\begin{equation}\label{RescaledVariables}
\tilde{r}=\frac{r}{L}, \ \ \ \tilde{t}=\frac{t}{\theta}, \ \ \ \tilde{p}=\frac{p}{P}, P=m\bar{c}, \ \ \ \tilde{v}_c=\frac{v_c}{\bar{c}}.
\end{equation}
Note that under this scaling,
\begin{equation}\label{nnnewscaling}
n_n(t,r) \ = \ \int_{\mathbb{R}^3}f(t,r,p)d p \ = \ P^3\int_{\mathbb{R}^3}f(t,r,\tilde{p})d \tilde{p}.
\end{equation}
We also rescale $U$ as $\tilde{U}=U/U_0$, where $U_0$ is the reference potential field. Define
\begin{eqnarray}\nonumber
\tilde{C}_{12}[f](t,r,\tilde{p}_1)&:=&\tilde{\lambda}_1n_c(t,r) \iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(\tilde{v}_c+ \tilde{p}_1-\tilde{p}_2-\tilde{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{\tilde{p}_1}-\mathcal{E}_{\tilde{p}_2}-\mathcal{E}_{\tilde{p}_3})\\\label{RescaledC12}
& &\times[(1+f_1)f_2f_3-f_1(1+f_2)(1+f_3)]d \tilde{p}_2d \tilde{p}_3\\\nonumber
&&-2\tilde{\lambda}_1n_c(t,r)\iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(\tilde{v}_c+\tilde{p}_2-\tilde{p}_1-\tilde{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{\tilde{p}_2}-\mathcal{E}_{\tilde{p}_1}-\mathcal{E}_{\tilde{p}_3})\\\nonumber
& &\times[(1+f_2)f_1f_3-f_2(1+f_1)(1+f_3)]d \tilde{p}_2d \tilde{p}_3,\\\label{RescaledC22}
\tilde{C}_{22}[f](t,r,\tilde{p}_1)&:=&\tilde{\lambda}_2\iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(\tilde{p}_1+\tilde{p}_2-\tilde{p}_3-\tilde{p}_4)\\\nonumber
& &\times\delta(\mathcal{E}_{\tilde{p}_1}+\mathcal{E}_{\tilde{p}_2}-\mathcal{E}_{\tilde{p}_3}-\mathcal{E}_{\tilde{p}_4})\times\\\nonumber
&&\times [(1+f_1)(1+f_2)f_3f_4-f_1f_2(1+f_3)(1+f_4)]d \tilde{p}_2d \tilde{p}_3d \tilde{p}_4,
\end{eqnarray} where
\begin{equation}\label{LambdaC12}
\tilde{\lambda}_{1}=P^2\lambda_{1}/\bar{c},
\end{equation}
and
\begin{equation}\label{LambdaC22}
\tilde{\lambda}_{2}=P^5\lambda_{2}/\bar{c}.\end{equation}
As a consequence, we can define the rescaled mean free paths and the rescaled mean field times to be
$$\tilde{l}_{22}=\frac{l_{22}}{P^5}, \ \ \tilde{\tau}_{22}=\frac{\bar{\tau}_{22}}{P^5},$$
and
$$\tilde{l}_{12}=\frac{l_{12}}{P^2}, \ \ \tilde{\tau}_{12}=\frac{\bar{\tau}_{12}}{P^2}.$$
We also set
\begin{equation}\label{Chat}
\hat{C}_{12}[f]: =\tilde{l}_{12} \tilde{C}_{12}[f], \ \ \ \hat{C}_{22}[f]: =\tilde{l}_{22}\tilde{C}_{22}[f].
\end{equation}
The following rescaled version of \eqref{QBFull} then follows:
\begin{equation}\label{RescaledQB1}
\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{\theta\bar{c}} \partial_{\tilde{t}}f \ + \ \frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{L}\frac{P}{m\bar{c}}\tilde{p}\cdot \nabla_{\tilde{r}}f \ - \ \frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{L}\frac{U_0}{P\bar{c}}\nabla_{\tilde{r}}\tilde{U}\cdot \nabla_{\tilde{p}} f \ = \ \sqrt{\frac{\tilde{l}_{22}}{\tilde{l}_{12}}}\hat{C}_{12}[f] \ + \ \sqrt{\frac{\tilde{l}_{12}}{\tilde{l}_{22}}}\hat{C}_{22}[f].
\end{equation}
Notice that $\frac{\tilde\tau_{22}}{\tilde\tau_{12}}=\frac{\tilde{l}_{22}}{\tilde{l}_{12}}$ is a dimensionless parameter and is proportional to $\frac{\tilde\lambda_1}{\tilde\lambda_2}$.
In this paper, we will consider two hydrodynamic approximations: Euler and Navier-Stokes.
\begin{itemize}
\item The Euler approximation is quite general and valid under a general physical situation. The collisions between excited atoms are fast to establish a local equilibrium within the non-condensate component, and the quantity $\tilde\tau_{22}$ is smaller than $\tilde\tau_{12}$ but the ratio between $ \tilde\tau_{22}$ and $\tilde\tau_{12}$ is not necessarily very small.
\item The Navier-Stokes approximation is valid under the physical assumption that the collisions between excited atoms are extremely rapid to establish a local equilibrium within the non-condensate component and $\tilde\tau_{22}<<\tilde\tau_{12}$.
\end{itemize}
We suppose $\frac{\tilde\tau_{22}}{\tilde\tau_{12}}=\epsilon^{2}$. {\it The Euler approximation is valid in any physical assumption and we do not need to impose the assumption that $\epsilon$ is small, then $\epsilon$ is just a parameter. In the Navier-Stokes approximation, we need to impose the assumption that $\epsilon$ is small and then we will use it as the small parameter in the usual Chapman-Enskog expansion process. }
The constants $\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{\theta\bar{c}}$, $\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{L}$ can be set to be $1$ by rescaling again the space and time variables $\tilde{t}\to \frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{\theta\bar{c}}\tilde{t}$, $\tilde{r}\to \frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{L}\tilde{r}$, and note that $\frac{P}{m\bar{c}}=1$, we obtain the following equation
\begin{equation}\label{RescaledQB2}
\partial_{\tilde{t}}f \ + \ \tilde{p}\cdot \nabla_{\tilde{r}}f \ - \ \frac{U_0}{m\bar{c}^2}\nabla_{\tilde{r}}\tilde{U}\cdot \nabla_{\tilde{p}} f \ = \ \epsilon\hat{C}_{12}[f] \ + \ \frac{1}{\epsilon}\hat{C}_{22}[f].
\end{equation}
Notice that $g$ is also the principle small parameter used in the derivation of the system \eqref{GP}-\eqref{QBFull}. Indeed, the derivation starts with the usual Heisenberg equation of motion for the quantum field
operator. The equation for the condensate wavefunction follows by averaging the Heisenberg equation with respect to a broken-symmetry nonequilibrium ensemble. Taking the difference between the Heisenberg equation and the equation for the condensate wavefunction and keeping only the terms of low orders with respect to $g$, we obtain the equation of the noncondensate field operator, which, by a Wigner transform, leads to the quantum Boltzmann equation. In this process, one computes the collision integrals $C_{12}$, $C_{22}$ to second order $O(g^2)$ in $g$ and keep interaction effects in the excitation energies and chemical potential only to first order $O(g)$. For a more detailed explanation of this procedure, we refer to, for instance, Sections 3.1, 3.2 and 5.3 of the book \cite{GriffinNikuniZaremba:2009:BCG}. {\it Since $U_0$ has to be chosen proportional to $g$, the dimensionless parameter $\frac{U_0}{m\bar{c}^2}$ might be considered to be small and set it to be $\tilde{g}=\epsilon^{\delta_0}$, $0<\delta_0<1$ in Section \ref{Sec:NavierStokes}, where the Chapman-Enskog expansion is used.}
The equation then follows, as a result of the previous scaling
\begin{equation}\label{RescaledQBFinal}
\partial_{\tilde{t}}f \ + \ \tilde{p}\cdot \nabla_{\tilde{r}}f \ - \ \tilde{g}\nabla_{\tilde{r}}\tilde{U}\cdot \nabla_{\tilde{p}} f \ = \ \ \epsilon\hat{C}_{12}[f] \ + \ \frac{1}{\epsilon}\hat{C}_{22}[f].
\end{equation}
Under this scaling, the Gross-Pitaevski equation also becomes
\begin{equation}
\begin{aligned}\label{RescaledQB4}
i \frac{\hbar}{\theta} {\partial_{\tilde{t}} {\Phi}(t,r)} =&\ \Big(-\frac{\hbar^{2} \Delta_{\tilde{r}}}{2m L^2}+g[n_c(t,r) + 2{n_n}(t,r)] -\frac{i}{\tilde\tau_{12}}\tilde{\Lambda}_{12}[f](t,r)\Big){\Phi}(t,r),
\end{aligned}
\end{equation}
where $$\tilde{\Lambda}_{12}[f]=\frac{\hbar}{2n_c}\int_{\mathbb{R}^3}\hat{C}_{12}[f]\bar{d}p.$$
By the same argument as above, we also obtain
\begin{equation}
\begin{aligned}\label{RescaledQB4b}
i \frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{\theta\bar{c}} {\partial_{\tilde{t}} {\Phi}(t,r)} =&\ \Big(-\frac{\hbar}{mL}\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}\Delta_{\tilde{r}}}{2 L\bar{c}}+\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}U_0}{\hbar\bar{c}}\tilde{U}_*(t,r)-\frac{i \sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{\tilde{l}_{12}}{\tilde{\Lambda}_{12}[f](t,r)}\Big){\Phi}(t,r),
\end{aligned}
\end{equation}
where $\tilde{U}_*=U_*/U_0$ and $U_*(t,r)=g[n_c(t,r) + 2{n_n}(t,r)]$, since $U_*(t,r)$ has the dimension of $U(t,r)$.
Notice that $\frac{\hbar}{m\bar{c}}$ has the dimensions of a length (Compton wavelength) and $\frac{\hbar}{mL\bar{c}}$ is dimensionless; hence the quantity $\frac{\hbar}{mL}\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{2 L\bar{c}}$ is dimensionless. Moreover, $\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}U_0}{\hbar\bar{c}}$ is the product of the three dimensionless parameters $\frac{\sqrt{\tilde{l}_{12}\tilde{l}_{22}}}{L}$, $\frac{mL\bar{c}}{\hbar}$ and $\frac{U_0}{m\bar{c}^2}=\tilde{g}$. Setting all of the dimensionless parameter to be $1$ by the same rescaling argument used for \eqref{RescaledQB2} and dropping the tilde and hat signs
\begin{equation}
\begin{aligned}\label{RescaledGPFinal}
i {\partial_{{t}} {\Phi}} =&\ \Big(-\frac{\Delta_{{r}}}{2}+{g}U_* -i\epsilon{\Lambda}_{12}[f]\Big){\Phi},
\end{aligned}
\end{equation}
where $g$ stands for the dimensionless parameter $\tilde{g}=\epsilon^{\delta_0}$. When $V=0$, the following system for the super-fluid of the condensate can be deduced
\begin{equation}\label{RescaledSystem2b}
\begin{aligned}
{\partial_t n_c}\ + \ \nabla_r\cdot(n_cv_c)\
=&\ - \epsilon \Gamma_{12}[f]\\
\partial_t v_c +\frac{\nabla_r v_c^2}{2} = & \ -\nabla_r\mu_c.
\end{aligned}
\end{equation}
we then obtain
the system
\begin{equation}
\begin{aligned}\label{RescaledSystem}
\partial_{{t}}f \ + \ {p}\cdot \nabla_{{r}}f \ - & \ g\nabla_{{r}}{U}\cdot \nabla_{{p}} f \ = \ \ \epsilon {C}_{12}[f] \ + \ \frac{1}{\epsilon}{C}_{22}[f], (0<\delta_0<1), \\
i {\partial_t {\Phi}} =&\ \Big(-\frac{\Delta_{{r}}}{2}+g[n_c + 2{n_n}] -i\epsilon\Lambda_{12}[g]\Big){\Phi}.
\end{aligned}
\end{equation}
We recall below the formulas for $C_{12}$, $C_{22}$ and $\Lambda_{12}$
\begin{eqnarray}\nonumber
{C}_{12}[f](t,r,{p}_1)&=&n_c(t,r) \iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(v_c+ {p}_1-{p}_2-{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{{p}_1}-\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3})\\\label{C12Dimless}
& &\times[(1+f_1)f_2f_3-f_1(1+f_2)(1+f_3)]d {p}_2d {p}_3\\\nonumber
&&-2n_c(t,r)\iint_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta(v_c+{p}_2-{p}_1-{p}_3)\delta(\mathcal{E}_c+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_1}-\mathcal{E}_{{p}_3})\\\nonumber
& &\times[(1+f_2)f_1f_3-f_2(1+f_1)(1+f_3)]d {p}_2d {p}_3,\\\label{C22Dimless}
{C}_{22}[f](t,r,{p}_1)&=&\iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta({p}_1+{p}_2-{p}_3-{p}_4)\delta(\mathcal{E}_{{p}_1}+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3}-\mathcal{E}_{{p}_4})\times\\\nonumber
&&\times [(1+f_1)(1+f_2)f_3f_4-f_1f_2(1+f_3)(1+f_4)]d {p}_2d {p}_3d {p}_4,\\
\Lambda_{12}[f](t,r) & = &\ \frac{1}{n_c(t,r)}\int_{\mathbb{R}^{3}}C_{12}[f](t,r,p){d p}.
\end{eqnarray}
We also define the differential operators
\begin{equation}\label{Df}
\mathcal{D}f \ = \ {\partial_t f}\ +\ {p}\cdot\nabla_{{r}} f \ - \ g\nabla_r U\cdot \nabla_p f\ - \epsilon C_{12}[f],
\end{equation}
\begin{equation}\label{Df2}
\mathbb{D}f \ = \ {\partial_t f}\ +\ {p}\cdot\nabla_{{r}} f \ - \ g\nabla_r U\cdot \nabla_p f,
\end{equation}
\begin{equation}\label{Df3}
\Pi f \ = \ {\partial_t f}\ +\ {p}\cdot\nabla_{{r}} f,
\end{equation}
and then get
\begin{equation}
\begin{aligned}\label{RescaledSystem2a}
\mathbb{D}f \ = & \ \ \epsilon {C}_{12}[f] \ + \ \frac{1}{\epsilon}{C}_{22}[f], (0<\delta_0<1).
\end{aligned}
\end{equation}
The new constant $\epsilon$ is the small parameter that we will use in the usual Chapman-Enskog expansion process in Section \ref{Sec:NavierStokes}.
\section{Properties of the collision operators}\label{Sec:Operators}
In this section, we study the main properties of the two collision operators $C_{12}$ and $C_{22}$.
\subsection{Collision invariants and equilibrium of $C_{22}$}
Let us start with $C_{22}$, which can be represented as:
\begin{equation}
\label{C22Presentation}
C_{22}[f] \ = \ B_1[f,f] \ + \ B_2[f,f,f],
\end{equation}
in which
\begin{equation}
\label{Q1}
\begin{aligned}
B_1[f,g] \ = & \ \frac{1}{2}\iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta({p}_1+{p}_2-{p}_3-{p}_4)\delta(\mathcal{E}_{{p}_1}+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3}-\mathcal{E}_{{p}_4})\times\\
&\times [f_3g_4+f_4g_3-f_1g_2-f_2g_1]d p_2d p_3d p_4,
\end{aligned}
\end{equation}
and
\begin{equation}
\label{Q2}
\begin{aligned}
B_2[f,g,h] \ = & \ \frac{1}{6}\iiint_{\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}}\delta({p}_1+{p}_2-{p}_3-{p}_4)\delta(\mathcal{E}_{{p}_1}+\mathcal{E}_{{p}_2}-\mathcal{E}_{{p}_3}-\mathcal{E}_{{p}_4})\times\\
&\times [f_3g_4h_1+f_4g_3h_1+f_3g_4h_2+f_4g_3h_2\\
&\ \ +f_1g_4h_3+f_1g_3h_4+f_2g_4h_3+f_3g_3h_4\\
&\ \ +f_4g_1h_3+f_3g_1h_4+f_4g_2h_3+f_3g_2h_4\\
&\ \ -f_1g_2h_3-f_2g_1h_3-f_1g_2h_4-f_2g_1h_4\\
&\ \ -f_3g_1h_2-f_3g_2h_1-f_4g_1h_2-f_4g_2h_1\\
&\ \ -f_1g_3h_2-f_2g_3h_1-f_1g_4h_2-f_2g_4h_1]d p_2d p_3d p_4,
\end{aligned}
\end{equation}
where we have used the same notations $f_1$, $f_2$, $f_3$, $f_4$, $g_1$, $g_2$ , $g_3$, $g_4$, $h_1$, $h_2$ , $h_3$, $h_4$ with the ones used in \eqref{QBFull}.
The operator $C_{22}$ shares some important features with the classical Boltzmann collision operator. Among these features, the following can be proved by switching the variables $(p_1,p_2)\leftrightarrow (p_2,p_1)$, $(p_1,p_2)\leftrightarrow (p_3,p_4)$, in the integrals of $B_1$ as in the classical case (cf. \cite{Villani:2002:RMT}):
\begin{equation}\label{Q1Conservation}
\int_{\mathbb{R}^3}\Psi_i(p)B_1[f,g](p)d p \ = \ 0, \ \ \ i=0,1,2,3,4,
\end{equation}
where
\begin{equation}\label{NullSpace}
\Psi_0(p) \ = \ 1, \ \ \Psi_i(p) \ = \ p^i, \ \ (i=1,2,3), \ \ \Psi_4(p) \ = \ |p|^2,
\end{equation}
are the collision invariants and $p^i$ is the $i$-th component of the vector $p=(p^1,p^2,p^3)$.
Moreover, we also have
\begin{equation}\label{Q1Conservation}
\int_{\mathbb{R}^3}\Psi_i(p)B_2[f,g,h](p)d p \ = \ 0, \ \ \ i=0,1,2,3,4.
\end{equation}
Similar as the classical Boltzmann collision operator, $C_{22}$ also has a local equilibrium of the form
\begin{equation}\label{Sec:EulerLimit:E27}
\mathcal{F}(t,r,p) \ = \ \frac{1}{e^{\beta[(p-v_n)^2/2 + U - \mu]}-1},
\end{equation}
where $\beta(t,r)$ is the temperature parameter, $v_n(t,r)$ is the local fluid velocity, $\mu(t,r)$ is the local chemical potential (which is different from the condensate chemical potential $\mu_c(t,r)$ defined in \eqref{def-muc}), $U(t,r)$ is the mean field. Then
$$C_{22}[\mathcal{F}] \ = \ 0.$$
Let us now define the following Gaussian
\begin{equation}\label{Gaussian}
\mathcal{M}(t,r,p) \ = \ \gamma(t,r)e^{-\frac{|p-u(t,r)|^2}{2\tau(t,r)}},
\end{equation}
where
\begin{equation}
\gamma(t,r) \ = \ e^{\beta(U(t,r)-\mu(t,r))},\ \ \
u(t,r) \ = \ v_n(t,r),\ \ \
\tau(t,r)\ = \ \frac{1}{\beta(t,r)}.
\end{equation}
The local equilibrium $\mathcal{F}$ can be expressed in terms of $\mathcal{M}$ as
\begin{equation}\label{Maxwellian}
\mathcal{F}(t,r,p) \ = \ \frac{\mathcal{M}(t,r,p)}{1-\mathcal{M}(t,r,p)}.
\end{equation}
Note that $u$ is a vector $u=(u_1,u_2,u_3)$.
\subsection{Linearized operator of $C_{22}$}
Let $L^2(\mathbb{R}^3)$ be the space of real, measurable functions, whose second power is integrable on $\mathbb{R}^3$, with the norm $\|\cdot\|_{L^2}$ and inner product $(,)_{L^2}$. We consider the linearized operator of $C_{22}$ around a fixed equilibrium $\mathcal{F}(t,r,p)$, which, by a classical process can be defined as
\begin{equation}\label{C22Linear1}
\mathcal{L} \ : = \ 2B_1(\mathcal{F},\cdot) \ + \ 3B_2(\mathcal{F},\mathcal{F},\cdot),
\end{equation}
or equivalently
\begin{equation}\label{C22Linear2}
\begin{aligned}
\mathcal{L}(\mathcal{F}f)(t,r,p_1) \ = & \ \int_{\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3}\delta(p_1+p_2-p_3-p_4)\delta(\mathcal{E}_{p_1}+\mathcal{E}_{p_2}-\mathcal{E}_{p_3}-\mathcal{E}_{p_4})\\
\ & \ \times \frac{\mathcal{M}_1\mathcal{M}_2}{(1-\mathcal{M}_1)(1-\mathcal{M}_2)(1-\mathcal{M}_3)(1-\mathcal{M}_4)}\times\\
\ &\ \times \Big[(1-\mathcal{M}_3)f(p_3)+(1-\mathcal{M}_4)f(p_4) - (1-\mathcal{M}_2)f(p_2)\\
\ & \ - (1-\mathcal{M}_1)f(p_1)\Big]d p_2d p_3d p_4,
\end{aligned}
\end{equation}
for some function $f(p)$ and fixed values $(t,r)\in\mathbb{R}_+\times\mathbb{R}^3$ and we employ the shorthand notations $\mathcal{M}_i=\mathcal{M}(t,r,p_i)$, $i=1,2,3,4$. Now, let us consider the inner product between the above linearized operator and some test function $\varphi$. The classical argument (cf. \cite{Villani:2002:RMT}) for the classical linearized Boltzmann collision operator can be applied and gives:
\begin{equation*}
\begin{aligned}
\left(\frac{\mathcal{M}}{\mathcal{F}}\varphi,\mathcal{L}(\mathcal{F}f)\right)_{L^2} \ =& \ -\frac{1}{4}\int_{\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3}\delta(p_1+p_2-p_3-p_4)\delta(\mathcal{E}_{p_1}+\mathcal{E}_{p_2}-\mathcal{E}_{p_3}-\mathcal{E}_{p_4}) \\
\ & \ \times \frac{\mathcal{M}_1\mathcal{M}_2}{(1-\mathcal{M}_1)(1-\mathcal{M}_2)(1-\mathcal{M}_3)(1-\mathcal{M}_4)}\times\\
\ &\ \times \Big[(1-\mathcal{M}_3)f(p_3)+(1-\mathcal{M}_4)f(p_4) - (1-\mathcal{M}_2)f(p_2)\\
\ & \ - (1-\mathcal{M}_1)f(p_1)\Big]\Big[(1-\mathcal{M}_3)\varphi(p_3)+(1-\mathcal{M}_4)\varphi(p_4)\\
\ &\ - (1-\mathcal{M}_2)\varphi(p_2) - (1-\mathcal{M}_1)\varphi(p_1)\Big]dp_1d p_2d p_3d p_4,\end{aligned}
\end{equation*}
which implies
\begin{equation}\label{C22Linearizedpositivity}
\left(\frac{\mathcal{M}}{\mathcal{F}}f,\mathcal{L}(\mathcal{F}f)\right)_{L^2} \ \le \ 0,
\end{equation}
and
\begin{equation*}
\left(\frac{\mathcal{M}}{\mathcal{F}}\varphi,\mathcal{L}(\mathcal{F}f)\right)_{L^2} \ = \ \left(\frac{\mathcal{M}}{\mathcal{F}}f,\mathcal{L}(\mathcal{F}\varphi)\right)_{L^2},
\end{equation*}
for all function $\varphi$ and $f$ such that the integrals are well-defined. The equality in \eqref{C22Linearizedpositivity} holds true if and only if $\frac{\mathcal{M}}{\mathcal{F}}f$ is identical to one of the five functions defined in \eqref{NullSpace}.
From the above observation, we are now able to define the kernel of the linearized collision operator $\mathcal{L}$ of $C_{22}$:
$$\mathcal{N} \ := \ \mathrm{ker}\mathcal{L} \ = \ \mathrm{span}\left\{\frac{\mathcal{F}^2}{\mathcal{M}}\Psi_i: \ i=0,\cdots,4\right\},$$
and its orthogonal space:
$$\mathcal{R} \ := \ \mathcal{N}^{\bot} \ = \ \left\{G\in L^2(\mathbb{R}^3)\ : \ \left(G,\frac{\mathcal{F}^2}{\mathcal{M}}\Psi_i\right)_{L^2}=0, \ i=0,\cdots,4\right\}.$$
On $L^2(\mathbb{R}^3)$, we also define the orthogonal projection operators $\mathbb{P}$ and $\mathbb{P}^{\bot}=1-\mathbb{P}$ on to $\mathcal{N}$ and $\mathcal{R}$. By normalizing $\{\Psi_i\}_{i=0,\cdots,4}$, we obtain the following orthonormal basis of the space $\mathcal{N}$
\begin{equation}\label{OrthonormalBasis}
\left\{\frac{\psi_i}{\sqrt{\omega_i}}\frac{\mathcal{F}^2}{\mathcal{M}}: \ \ \ i=0,\cdots,4\right\},
\end{equation}
with
$$\psi_0=1; \ \ \psi_{i}=p^i-u_i, \ \ i=1,2,3; \ \ \psi_4=|p-u|^2-6\tau\frac{\Omega_1(\gamma)}{\Omega_0(\gamma)},$$
\begin{equation*}
\begin{aligned}
\omega_0 \ = & \ \int_{\mathbb{R}^3}\frac{\mathcal{F}^2}{\mathcal{M}}d p \ = \ 2^{3/2}\pi\tau^{3/2}\gamma\Omega_0(\gamma);\\
\omega_i \ = & \ \int_{\mathbb{R}^3}\frac{\mathcal{F}^2}{\mathcal{M}}|\psi_i(p)|^2d p \ = \ 2^{5/2}\pi\tau^{5/2}\gamma\Omega_1(\gamma),\ \ i=1,2,3; \\
\omega_4 \ = & \ \int_{\mathbb{R}^3}\frac{\mathcal{F}^2}{\mathcal{M}}|\psi_4(p)|^2d p \ = \ 2^{7/2}\pi\tau^{7/2}\gamma\Sigma(\Omega_0(\gamma),\Omega_1(\gamma),\Omega_2(\gamma));
\end{aligned}
\end{equation*}
where
$$\Sigma(x,y,z) \ = \ \frac{5xz-9y^2}{x},$$
and
\begin{equation}\label{sigmafunction}
\Omega_k(\gamma) \ = \ \int_{0}^\infty \frac{y^{k-1/2}}{e^y+\gamma} dy, \ \ \ \ k>-1/2.
\end{equation}
\subsection{Hydrodynamics quantities}
In order to study the hydrodynamics limit of the system, let us define the following moments of the function $f(t,r,p)$:
\begin{equation}\label{FluidDensity}
{n_n}[f](t,r) \ = \ \int_{\mathbb{R}^3}f(t,r,p){d} p,
\end{equation}
\begin{equation}\label{FluildVelocity}
u[f](t,r)\ = v_n(t,r)[f](t,r)\ = \ \frac{1}{{n_n}[f](t,r)}\int_{\mathbb{R}^3} {p}f(t,r,p){d} p,
\end{equation}
\begin{equation}\label{FluidEnergy}
{\mathbb{E}_n}[f](t,r) \ = \ \frac{1}{2}\int_{\mathbb{R}^3}f(t,r,p)|p-v_n[f](t,r)|^2 {d} p,
\end{equation}
\begin{equation}\label{FluidEnergye}
{\tilde{\mathbb{E}}_n}[f](t,r)\ = \ \frac{2{\mathbb{E}_n}[f](t,r)}{3}, \ \ \ e_n[f](t,r) \ = \ \frac{{\tilde{\mathbb{E}}_n}[f](t,r)}{ {n_n}[f](t,r)}.
\end{equation}
Replacing $f$ by $\mathcal{F}$, we obtain
\begin{equation}\label{FluidDensityM}
\begin{aligned}
{n_n}[\mathcal{F}] \ = & \ 2^{5/2}\pi\tau^{3/2}\Omega_1(\gamma),\\
{\mathbb{E}_n}[\mathcal{F}] \ = & \ {2^{5/2}}\pi \tau^{5/2}\gamma\Omega_2(\gamma),
\end{aligned}
\end{equation}
where $\Omega_1,$ $\Omega_2$ are defined in \eqref{sigmafunction}.
For the sake of simplicity, we denote ${n_n}[\mathcal{F}], v_n[\mathcal{F}],$ $ u[\mathcal{F}], $ ${\mathbb{E}_n}[\mathcal{F}],$ $\tilde{\mathbb{E}}_n[\mathcal{F}]$, $ e_n[\mathcal{F}]$ by ${n_n}, v_n, u, {\mathbb{E}_n}, \tilde{\mathbb{E}}_n$, and $e_n$.
We indeed can compute $\gamma$ and $\tau$ as
\begin{equation}
\label{gamma}
\gamma \ = \ \left(\frac{\mathrm{Id}\Omega_2}{\Omega_1^{5/3}}\right)^{-1}\left(\frac{2^{5/3}\pi^{2/3}{\mathbb{E}_n}}{{n_n}^{5/3}}\right),
\end{equation}
and
\begin{equation}
\label{tau}
\tau \ = \ \left(\frac{{n_n}}{2^{5/2}\pi\Omega_1\left(\left(\frac{\mathrm{Id}\Omega_2}{\Omega_1^{5/3}}\right)^{-1}\left(\frac{2^{5/3}\pi^{2/3}{\mathbb{E}_n}}{{n_n}^{5/3}}\right)\right)}\right)^{2/3}.
\end{equation}
\subsection{Computing $\Gamma_{12}[\mathcal{F}]$}
Now, let us consider the collision operator $C_{12}$. This operator also has the collision invariant property:
\begin{equation}\label{C12Conservation}
\int_{\mathbb{R}^3}(\Psi_i(p) - v_{ci}) C_{12}[f]d p \ = \ \int_{\mathbb{R}^3}\left(\Psi_4(p)+ {2}U- {2}\mu_c-{v_c^2}\right)C_{12}[f]d p\ = \ 0, \ \ \ i=1,2,3.
\end{equation}
An important property of $C_{12}$ is that $\mathcal{F}$ is not an equilibrium of $C_{12}$. We have:
\begin{equation}\label{Gamma12a}
\begin{aligned}
&\Gamma_{12}[\mathcal{F}]\ := \ \int_{\mathbb{R}^3}C_{12}[\mathcal{F}]{d}p \ \ \\
& = \ -n_c[1-e^{-\beta(\mu-\mu_c -(v_n-v_c)^2/2)}]\iiint_{\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3}\delta(v_c+p_1-p_2-p_3)\times\\
& \ \times \delta(\mathcal{E}_c+\mathcal{E}_{p_1}-\mathcal{E}_{p_2}-\mathcal{E}_{p_3})(1+\mathcal{F}(t,r,p_1))\mathcal{F}(t,r,p_2)\mathcal{F}(t,r,p_3)dp_1dp_2dp_3.
\end{aligned}
\end{equation}
Expanding $\mathcal{F}$ into Taylor series of $\mathcal{M}$, we can simplify the above integral as
\begin{equation}\label{Gamma12b}
\begin{aligned}
&\ \Gamma_{12}[\mathcal{F}] \ \ \\
& = \ -n_c[1-e^{-\beta(\mu-\mu_c -(u-v_c)^2/2)}]\sum_{k_2,k_3\in\mathbb{N}\cup\{0\},k_1\in\mathbb{N}}\gamma^3 e^{-\frac{|v_c-u|^2(k_1+k_2+k_3)}{2\tau}}\times \\
& \times e^{\frac{(-2\mathcal{E}_c+2U+v_c^2)k_1}{2\tau}}\int_{x\cdot y = \frac{v_c^2}{2}+U-\mathcal{E}_c}e^{-(k_1+k_2)[|x|^2+x\cdot(v_c-u)]-(k_1+k_3)[|y|^2+y\cdot(v_c-u)]/(2\tau)}dxdy,
\end{aligned}
\end{equation}
with the notice that from \eqref{gamma} and \eqref{tau}, $\gamma$ and $\tau$ are functions of ${n_n}$ and ${\mathbb{E}_n}$.
\section{The two-fluid Euler quantum hydrodynamic limit}\label{Sec:Euler}
In this section, we will derive a two-fluid Euler quantum hydrodynamic limit from \eqref{BECSuperFluid} - \eqref{RescaledSystem2a}. In this case, $\epsilon$ is a constant, so we will set it to be $\epsilon=1$. Choose $\tilde\epsilon$ to be any small parameter.
In order to obtain the Euler hydrodynamics limit, let us start with the following Hilbert expansion using $\tilde\epsilon$ as the small parameter (cf. \cite{Caflisch:TFD:1980}):
\begin{equation}\label{Hilbert}
f \ = \ \sum_{i=0}^n \tilde\epsilon^i f^{(i)} \ + \ \tilde\epsilon^l \varsigma,
\end{equation}
in which $n$ and $l$ are positive integers. As a consequence, we can replace $f$ by its Hilbert expansion into
$$\mathcal{D}f \ = \ C_{22}[f],$$
to get a linear system of equations and a weakly nonlinear equation for the remainder $\varsigma$, which reads as:
\begin{eqnarray}\label{HilbertSystem1}
B_1(f^{(0)},f^{(0)}) \ + \ B_2(f^{(0)},f^{(0)},f^{(0)}) & = & 0,\\\label{HilbertSystem2}
2B_1(f^{(0)},f^{(1)}) \ + \ 3B_2(f^{(0)},f^{(0)},f^{(1)}) & = & \mathcal{D}f^{(0)},\\\nonumber
2B_1(f^{(0)},f^{(i)}) \ + \ 3B_2(f^{(0)},f^{(0)},f^{(i)}) & = & \mathcal{D}f^{(i-1)}\ - \ \sum_{j=1}^{i-1}B_1(f^{(i)},f^{(i-j)})\\\label{HilbertSystem3}
& & - \ \sum_{j,k=1, 0<j+k<i}^{i-1}B_2(f^{(i)},f^{(k)},f^{(i-j-k)}),
\end{eqnarray}
for $i=2,3,\cdots,n$.
The equation for the remainder $r$ is as follows:
\begin{equation}\label{RemainderEquation}
\begin{aligned}
\mathcal{D}\varsigma \ =& \ \frac{1}{\tilde\epsilon}\mathcal{L}\varsigma \ + \ 2\sum_{i=1}^n\tilde\epsilon^{i-1}B_1(f^{(i)},\varsigma)\ + \ \tilde\epsilon^{l-1}B_1(\varsigma,\varsigma)\ + \ 3\sum_{i=1}^nB_2(\mathcal{F},f^{(i)},\varsigma)\\
& \ +3\sum_{i,j=1}^{n}\tilde\epsilon^{i+j-1}B_2(f^{(i)},f^{(j)},\varsigma) \ + \ 3\tilde\epsilon^{(l-1)}B_2(\mathcal{F},\varsigma,\varsigma) \ + \ 3\tilde\epsilon^{l-1}\sum_{i=1}^n \tilde\epsilon^{i}B_2(f^{(i)},\varsigma,\varsigma)\\
& \ + \tilde\epsilon^{2l-1}B_2(\varsigma,\varsigma,\varsigma) \ + \ \tilde\epsilon^{n-1}\mathfrak{Q},
\end{aligned}
\end{equation}
where $\mathfrak{Q}$ is an operator of $\mathcal{F}, f^{(1)},\cdots, f^{(n)}$.
Let us now consider each equation in the above system. From the first equation \eqref{HilbertSystem1}, we deduce that $f^{(0)} $ has to be a Bose-Einstein distribution:
\begin{equation}\label{Eulerf0}
f^{(0)} \ = \ {\mathcal{F}}.
\end{equation}
The equations \eqref{HilbertSystem2} and \eqref{HilbertSystem3} lead to linear integral equations for $f^{(1)},\cdots, f^{(i)}$. Thanks to Fredhom's theory, these linear integral equations are solvable if the right hand sides are orthogonal to $\mathcal{N}$ in $L^2(\mathbb{R}^3)$. As a consequence, $f^{(1)}$ can be solved from \eqref{HilbertSystem2}, if the following condition is satisfied
\begin{equation}\label{EulerSolvable}
\mathbb{P}\mathcal{D}{\mathcal{F}} \ = \ 0.
\end{equation}
We recall that $\mathbb{P}$ and $\mathbb{P}^{\bot}=1-\mathbb{P}$ are the orthogonal projection operators onto $\mathcal{N}$ and $\mathcal{R}$ in $L^2(\mathbb{R}^3)$.
\subsection{The Euler quantum hydrodynamic limit of the thermal cloud kinetic equation}\label{Sec:EulerLimit}
Integrating Equation \eqref{QBFull} in $p$, we obtain
\begin{equation}\label{Sec:EulerLimit:E1}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}f(t,r,p)dp \ + \ \nabla_r\cdot\int_{\mathbb{R}^3} {p}f(t,r,p)dp - \int_{\mathbb{R}^3}\nabla_r U (t,r,p)\cdot \nabla_p f(t,r,p)dp\\
=&\ \int_{\mathbb{R}^3}C_{12}[f](t,r,p)dp \ + \ \int_{\mathbb{R}^3}C_{22}[f](t,r,p)dp.
\end{aligned}
\end{equation}
Using the fact that
$$\int_{\mathbb{R}^3}\nabla_r U(t,r) \cdot\nabla_p f(t,r,p)dp \ = \ \int_{\mathbb{R}^3}C_{22}[f](t,r,p)dp \ = \ 0,$$
we get
\begin{equation}\label{Sec:EulerLimit:E2}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}fdp \ + \ \nabla_r\cdot\int_{\mathbb{R}^3}{p} fdp \
=&\ \Gamma_{12}[f].
\end{aligned}
\end{equation}
Equation \eqref{Sec:EulerLimit:E2} can be rewritten as
\begin{equation}\label{Sec:EulerLimit:MomentEq1}
\partial_t n_n \ + \ \nabla_r\cdot(n_n v_n)\
=\ \Gamma_{12}[f].
\end{equation}
For an arbitrary momentum vector $p=(p_1,p_2,p_3)$, we choose $p_j$, $j\in\{1,2,3\}$ as a test function for \eqref{QBFull} and obtain
\begin{equation}\label{Sec:EulerLimit:E4}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}f(t,r,p)p_j dp \ + \ \nabla_r\cdot\int_{\mathbb{R}^3} {p}f(t,r,p)p_j dp - \int_{\mathbb{R}^3}\nabla_r U(t,r) \cdot \nabla_p f(t,r,p) p_jdp\\
=&\ \int_{\mathbb{R}^3}p_j C_{12}[f](t,r,p)dp \ + \ \int_{\mathbb{R}^3}p_j C_{22}[f](t,r,p)dp.
\end{aligned}
\end{equation}
Due to the conservation of momentum for $C_{12}$ and $C_{22}$,
$$\int_{\mathbb{R}^3}(p_j - v_{cj}) C_{12}[f]dp \ = \ \int_{\mathbb{R}^3}p_jC_{22}[f]dp \ = \ 0,$$
we get
\begin{equation}\label{Sec:EulerLimit:E5}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}f(t,r,p)p_j dp \ + \ \nabla_r\cdot\int_{\mathbb{R}^3} {p}f(t,r,p)p_j dp - \int_{\mathbb{R}^3}\nabla_r U(t,r)\cdot \nabla_p f(t,r,p) p_jdp\\
=&\ \int_{\mathbb{R}^3}v_{cj}(t,r) C_{12}[f](t,r,p)dp\\
=&\ v_{cj}(t,r) \Gamma_{12}[f](t,r).
\end{aligned}
\end{equation}
Let us look at the first term on the left hand side of \eqref{Sec:EulerLimit:E5}
\begin{equation}\label{Sec:EulerLimit:E6}
\begin{aligned}
\partial_t \int_{\mathbb{R}^3}fp_j dp \ = &\ \partial_t(n_nv_{nj})\ = \ \partial_t n_n v_{nj}\ + \ n_n\partial_t v_{nj},
\end{aligned}
\end{equation}
in which $v_{nj}$ is the component of $v_n = (v_{n1}, v_{n2}, v_{n3})$.
\\ By using \eqref{Sec:EulerLimit:MomentEq1}, we can deduce from \eqref{Sec:EulerLimit:E6} that
\begin{equation}\label{Sec:EulerLimit:E6a}
\begin{aligned}
\partial_t \int_{\mathbb{R}^3}fp_j dp \ = &\ \Gamma_{12}[f]v_{nj}\ - \ v_{nj}\nabla_r \cdot (n_n v_n) \ + \ n_n\partial_t v_{nj},
\end{aligned}
\end{equation}
\\ Now, let us look at the second term on the left hand side of \eqref{Sec:EulerLimit:E5},
\begin{equation}\label{Sec:EulerLimit:E7}
\begin{aligned}
\nabla_r\cdot\int_{\mathbb{R}^3}{p} fp_j dp\
= & \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3}{p_ip_j}fdp\\
= & \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3}[{(p_i-v_{ni})(p_j-v_{nj}) + p_iv_{nj} + p_jv_{ni} -v_{ni}v_{nj}}]fdp\\
= & \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3}{(p_i-v_{ni})(p_j-v_{nj})}fdp \\
&\ + \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3}({p_iv_{nj} + p_jv_{ni}})fdp \ - \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3} v_{ni}v_{nj}fdp.
\end{aligned}
\end{equation}
By observing that
\begin{equation}\label{Sec:EulerLimit:E8}
\begin{aligned}
\int_{\mathbb{R}^3}({p_iv_{nj} + p_jv_{ni}})f(t,r,p)dp \ = & \ 2v_{nj}(t,r)v_{ni}(t,r)n_n(t,r) \\
\int_{\mathbb{R}^3} v_{ni}v_{nj}f(t,r,p)dp \ = & \ v_{nj}(t,r)v_{ni}(t,r)n_n(t,r),
\end{aligned}
\end{equation}
we infer from Identity \eqref{Sec:EulerLimit:E7}
\begin{equation}\label{Sec:EulerLimit:E9}
\begin{aligned}
\nabla_r\cdot\int_{\mathbb{R}^3}{p} fp_j dp
= & \ \sum_{i=1}^3 \partial_{r_i}\int_{\mathbb{R}^3}{(p_i-v_{ni})(p_j-v_{nj})}fdp \ + \ \sum_{i=1}^3 \partial_{r_i}[v_{nj}v_{ni}n_n].
\end{aligned}
\end{equation}
The last term on the left hand side of \eqref{Sec:EulerLimit:E5} can be rewritten in the following form, by integration by parts and the definition of $n_n$
\begin{equation}\label{Sec:EulerLimit:E10}
\begin{aligned}
-\int_{\mathbb{R}^3}\nabla_r U\cdot \nabla_p fp_j dp \ = & \ \int_{\mathbb{R}^3}\partial_{r_j} U fdp
\ = \ n_n\partial_{r_j} U.
\end{aligned}
\end{equation}
Putting the three terms \eqref{Sec:EulerLimit:E6a}, \eqref{Sec:EulerLimit:E9} and \eqref{Sec:EulerLimit:E10} together, we find
\begin{equation}\label{Sec:EulerLimit:MomentEq2}
\begin{aligned}
n_n\left(\partial_t + v_n\cdot \nabla \right) v_{nj}\ = & \ -\sum_{i=1}^3 \partial_{r_j} \mathcal{P}[f]_{ij} -n_n\partial_{r_j}U \ -(v_{nj}-v_{cj})\Gamma_{12}[f],
\end{aligned}
\end{equation}
where
\begin{equation}\label{Sec:EulerLimit:E11}
\mathcal{P}[f]_{ij} \ = \ \int_{\mathbb{R}^3} \left({p_i} -v_{ni}(t,r)\right)\left({p_j} -v_{nj}(t,r)\right)f(t,r,p)dp.
\end{equation}
Choosing $|p|^2$, as a test function for \eqref{QBFull} yields
\begin{equation}\label{Sec:EulerLimit:E12a}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}f(t,r,p)|p|^2 dp \ + \ \nabla_r\cdot \int_{\mathbb{R}^3} {|p|^2p}f(t,r,p) dp - \int_{\mathbb{R}^3}\nabla_r U(t,r) \cdot\nabla_p f(t,r,p) |p|^2dp\\
=&\ \int_{\mathbb{R}^3}|p|^2 C_{12}[f](t,r,p)dp \ + \ \int_{\mathbb{R}^3}|p|^2 C_{22}[f](t,r,p)dp.
\end{aligned}
\end{equation}
Let us recall the conservation of energy for $C_{12}$ and $C_{22}$
$$\int_{\mathbb{R}^3}|p|^2 C_{22}[f]dp=0,$$
and
\begin{equation*}\label{Sec:EulerLimit:E12a}
\begin{aligned}
0\ =&\ 2\int_{\mathbb{R}^3}\left(\mathcal{E}_p-\mathcal{E}_c\right)C_{12}[f]dp\ = \int_{\mathbb{R}^3}\left({|p|^2}+ {2}U- {2}\mu_c-{v_c^2}\right)C_{12}[f]dp,
\end{aligned}
\end{equation*}
which leads to
\begin{equation}\label{Sec:EulerLimit:E12}
\begin{aligned}
&\ \partial_t \int_{\mathbb{R}^3}f(t,r,p)|p|^2 dp \ + \ \nabla_r\cdot\int_{\mathbb{R}^3} {|p|^2p}f(t,r,p) dp - \int_{\mathbb{R}^3}\nabla_r U(t,r)\cdot \nabla_p f(t,r,p) |p|^2dp\\
=&\ \left(- {2}U + {2}\mu_c + {v_c^2}\right) \Gamma_{12}[f](t,r).
\end{aligned}
\end{equation}
Similar as above, we consider each term on the right and left hand sides of \eqref{Sec:EulerLimit:E12}. Let us start with the first term on the left hand side
\begin{equation}\label{Sec:EulerLimit:E13}
\begin{aligned}
\partial_t \int_{\mathbb{R}^3}f|p|^2 dp \
=&\ \partial_t \left(\int_{\mathbb{R}^3}f|p-v_n|^2 dp \right) \ + \ \partial_t \left(\int_{\mathbb{R}^3}f2p\cdot v_n dp \right)\\
&\ - \partial_t \left(\int_{\mathbb{R}^3}f|v_n|^2 dp \right),
\end{aligned}
\end{equation}
where we have used the identity
\begin{equation}\label{Sec:EulerLimit:E13a}
|p-v_n|^2 \ + \ 2p\cdot v_n \ - \ |v_n|^2 \ = \ |p|^2.
\end{equation}
Since
$$\left(\int_{\mathbb{R}^3}fp\cdot v_n dp \right)\ = \ |v_n|^2n_n \ = \ \left(\int_{\mathbb{R}^3}f|v_n|^2 dp \right),$$
we obtain from \eqref{Sec:EulerLimit:E13} that
\begin{equation}\label{Sec:EulerLimit:E14}
\begin{aligned}
\partial_t \int_{\mathbb{R}^3}f|p|^2 dp \
=&\ \partial_t \left(\int_{\mathbb{R}^3}f|p-v_n|^2 dp \right) \ + \ \partial_t \left(|v_n|^2n_n\right)\\
=&\ 2\partial_t \mathbb{E} \ + \ \partial_t \left(|v_n|^2n_n\right).
\end{aligned}
\end{equation}
Expanding the second term on the right hand side of \eqref{Sec:EulerLimit:E14} gives us
\begin{equation}\label{Sec:EulerLimit:E16}
\begin{aligned}
\partial_t \int_{\mathbb{R}^3}f|p|^2 dp \
=&\ 2\partial_t \mathbb{E} \ + \ 2 n_n v_n\cdot \partial_t v_n \ + \ |v_n|^2 \partial_t n_n,
\end{aligned}
\end{equation}
which, by \eqref{Sec:EulerLimit:MomentEq1} and \eqref{Sec:EulerLimit:MomentEq2}, can be rewritten as
\begin{equation}\label{Sec:EulerLimit:E17}
\begin{aligned}
& \partial_t \int_{\mathbb{R}^3}f|p|^2 dp \\
=&\ 2\partial_t \mathbb{E} \ + \ \sum_{j=1}^3 2 v_{nj}\left[ -\sum_{i=1}^3 \partial_{r_j} \mathcal{P}[f]_{ij} -n_n\partial_{r_j}U \ -(v_{nj}-v_{cj})\Gamma_{12}[f]- n_nv_n\cdot \nabla_r v_{nj}\right] \\
&\ + \ |v_n|^2 [\Gamma_{12}[f] \ - \ \nabla_r\cdot(n_nv_n)].
\end{aligned}
\end{equation}
Now, for the second term on the left hand side of \eqref{Sec:EulerLimit:E12}, it is straight forward that
\begin{equation}\label{Sec:EulerLimit:E18}
\begin{aligned}
\nabla_r\cdot\int_{\mathbb{R}^3} {|p|^2p}f dp \
=&\ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}f dp\right)\ + \ \nabla_r \cdot\left(\int_{\mathbb{R}^3}|v_n|^2v_nfdp\right)\\
&\ -3\nabla_r\cdot \left(\int_{\mathbb{R}^3} |v_n|^2p fdp\right)\ + \ 3\nabla_r\cdot \left(\int_{\mathbb{R}^3}|p|^2v_nf dp\right),
\end{aligned}
\end{equation}
which, as a view of the identity
$$\int_{\mathbb{R}^3}|v_n|^2v_nfdp \ = \ \int_{\mathbb{R}^3} |v_n|^2p fdp \ = \ |v_n|^2v_nn_n,$$
can be expressed as
\begin{equation}\label{Sec:EulerLimit:E19}
\begin{aligned}
&\ \nabla_r\cdot\int_{\mathbb{R}^3} {|p|^2p}fdp \\
=&\ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}fdp\right)\ - \ 2\nabla_r\cdot \left(|v_n|^2v_nn_n\right)\\
&\ + \ 3\nabla_r\cdot \left(\int_{\mathbb{R}^3}|p|^2v_nf dp\right).
\end{aligned}
\end{equation}
Using \eqref{Sec:EulerLimit:E13a}, we can rewrite \eqref{Sec:EulerLimit:E19} as
\begin{equation}\label{Sec:EulerLimit:E20}
\begin{aligned}
\nabla_r\cdot\int_{\mathbb{R}^3} {|p|^2p}f dp \
=&\ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}f dp\right)\ - \ 2\nabla_r\cdot \left(|v_n|^2v_nn_n\right)\\
&+3\nabla_r \cdot\left(\int_{\mathbb{R}^3}|p-v_n|^2v_nf dp + 2|v_n|^2\int_{\mathbb{R}^3}pf dp -|v_n|^2v_n\int_{\mathbb{R}^3}f dp\right),
\end{aligned}
\end{equation}
which can be reduced to
\begin{equation}\label{Sec:EulerLimit:E21}
\begin{aligned}
\nabla_r\cdot\int_{\mathbb{R}^3} {|p|^2p}f dp \
=&\ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}f dp\right)\ - \ 2\nabla_r \cdot\left(|v_n|^2v_nn_n\right)\\
&+3\nabla_r\cdot \left(\int_{\mathbb{R}^3}|p-v_n|^2v_nf dp + |v_n|^2v_nn_n\right)\\
=&\ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}f dp\right)\ + \ \nabla_r\cdot \left(|v_n|^2v_nn_n\right)\\
&+3\nabla_r\cdot \left(v_n\int_{\mathbb{R}^3}|p-v_n|^2fdp\right),
\end{aligned}
\end{equation}
The last term on the left hand side of \eqref{Sec:EulerLimit:E12} can be rewritten in a straightforward manner as follows:
\begin{equation}\label{Sec:EulerLimit:E22}
\begin{aligned}
\int_{\mathbb{R}^3}|p|^2\nabla_r U \cdot\nabla_p f dp\
=&\ - 2\nabla_r U \cdot \int_{\mathbb{R}^3} p f dp.
\end{aligned}
\end{equation}
Notice that the right hand side of \eqref{Sec:EulerLimit:E22} can be expressed in terms of $n_n$ and $v_n$ as
\begin{equation}\label{Sec:EulerLimit:E23}
\begin{aligned}
- 2\nabla_r U \cdot \int_{\mathbb{R}^3} p f dp\
=&\ - 2\nabla_r U \cdot\left(n_nv_n\right).
\end{aligned}
\end{equation}
As a consequence, we find
\begin{equation}\label{Sec:EulerLimit:E24}
\begin{aligned}
\int_{\mathbb{R}^3}|p|^2\nabla_r U \cdot \nabla_p fdp\
=&\ - 2\nabla_r U \cdot\left(n_nv_n\right).
\end{aligned}
\end{equation}
Combining \eqref{Sec:EulerLimit:E12}, \eqref{Sec:EulerLimit:E17}, \eqref{Sec:EulerLimit:E21} and \eqref{Sec:EulerLimit:E24}, yields
\begin{equation}\label{Sec:EulerLimit:E25}
\begin{aligned}
&\ 2\partial_t \mathbb{E} \ + \ \sum_{j=1}^3 2 v_{nj}\left[ -\sum_{i=1}^3 \partial_{r_j} \mathcal{P}[f]_{ij} -n_n\partial_{r_j}U \ -(v_{nj}-v_{cj})\Gamma_{12}[f]- n_nv_n\cdot \nabla_r v_{nj}\right] \\
&\ + \ |v_n|^2 [\Gamma_{12}[f] \ - \ \nabla_r\cdot(n_nv_n)]\\
&\ + \ \nabla_r\cdot\left(\int_{\mathbb{R}^3} {|p-v_n|^2(p-v_n)}f dp\right)\ + \ \nabla_r \cdot\left(|v_n|^2v_nn_n\right)\\
&+3\nabla_r \cdot\left(v_n\int_{\mathbb{R}^3}|p-v_n|^2f dp\right) - 2\nabla_r U \cdot\left(n_nv_n\right)\\
=&\ \left(- {2}U + {2}\mu_c + {v_c^2}\right) \Gamma_{12}[f],
\end{aligned}
\end{equation}
which leads to
\begin{equation}\label{Sec:EulerLimit:MomentEq3}
\begin{aligned}
&\ \partial_t \mathbb{E} \ + \ \nabla_r\cdot (\mathbb{E}v_n)\\
=&\ -\nabla_r \cdot \mathcal{R}[f] -\sum_{i,j=1}^3\frac{1}{2}\left(v_{ni}\partial_{r_j}+v_{nj}\partial_{r_i}\right)P_{ij}+\left(\frac{(v_n-v_c)^2}{2}+\mu_c-U\right) \Gamma_{12}[f],
\end{aligned}
\end{equation}
where
\begin{equation}\label{Sec:EulerLimit:E26}
\mathcal{R}[f](t,r) \ = \int_{\mathbb{R}^3}\frac{|p-v_n|^2(p-v_n)}{2}f(t,r,p)dp.
\end{equation}
The three equation \eqref{Sec:EulerLimit:MomentEq1}, \eqref{Sec:EulerLimit:MomentEq2} and \eqref{Sec:EulerLimit:MomentEq3} lead to the following system of moment equations for the kinetic equation of the thermal cloud:
\begin{equation}\label{Sec:EulerLimit:MomentEqFinal}
\begin{aligned}
\partial_t n_n \ + \ \nabla_r\cdot(n_nv_n)\
= &\ \Gamma_{12}[f],\\
n_n\left(\partial_t + v_n\cdot \nabla \right) v_{nj}\ = & \ -\sum_{i=1}^3 \partial_{r_j} \mathcal{P}[f]_{ij} -n_n\partial_{r_j}U \ -(v_{nj}-v_{cj})\Gamma_{12}[f],\\
\ \partial_t \mathbb{E} \ + \ \nabla_r\cdot (\mathbb{E}v_n)
=&\ -\nabla_r \cdot \mathcal{R}[f] -\sum_{i,j=1}^3\frac{1}{2}\left(v_{ni}\partial_{r_j}+v_{nj}\partial_{r_i}\right)P_{ij}\\
& \ +\left(\frac{(v_n-v_c)^2}{2}+\mu_c-U\right) \Gamma_{12}[f].
\end{aligned}
\end{equation}
Replacing $\cal F$ into \eqref{Sec:EulerLimit:MomentEqFinal}, we obtain $R[\mathcal{F}]=0$,
\begin{equation}\label{Sec:EulerLimit:E28}
P_{ij}(t,r) \ = \ \delta_{ij} \tilde{\mathbb{E}}(t,r) \ \equiv \ \delta_{ij} \int_{\mathbb{R}^3}\frac{|p|^2}{3}f_\infty(t,r,p) dp.
\end{equation}
Moreover, we note that
\begin{equation}\label{Sec:EulerLimit:E29}
\mathbb{E}(t,r) \ = \ \frac{3}{2}\tilde{\mathbb{E}}(t,r) .
\end{equation}
As a consequence, we can close the system \eqref{Sec:EulerLimit:MomentEqFinal} to obtain
\begin{equation}\label{Sec:EulerLimit:EulerEqbis}
\begin{aligned}
{\partial_t n_c}\ + \ \nabla_r\cdot(n_cv_c)\
=&\ - \ \Gamma_{12}[\mathcal{F}],\\
{n_n}\left(\partial_t + v_n\cdot \nabla \right) v_{n}\ = & \ - \nabla_r \tilde{\mathbb{E}}_n-{n_n}\nabla_r U \ -(v_{n}-v_{c})\Gamma_{12}[\mathcal{F}],\\
\ \partial_t \tilde{\mathbb{E}}_n \ + \ \nabla_r\cdot (\tilde{\mathbb{E}}_nv_n)
=&\ -\frac{2}{3}\tilde{\mathbb{E}}_n\nabla_r \cdot v_n \ +\frac{2}{3}\left(\frac{(v_n-v_c)^2}{2}+\mu_c-U\right) \Gamma_{12}[\mathcal{F}]
.
\end{aligned}
\end{equation}
\subsection{Comparison with a previous result}\label{Comparison}
Putting the two systems \eqref{BECSuperFluid} and \eqref{Sec:EulerLimit:EulerEqbis} together, one finds the following two-fluid Euler quantum hydrodynamic
\begin{equation}\label{EulerTwoFluidG1}
\begin{aligned}
{\partial_t n_c}\ + \ \nabla_r\cdot(n_cv_c)\
=&\ - \ \Gamma_{12}[\mathcal{F}],
\\
\partial_t v_c +\frac{\nabla_r v_c^2}{2} = & \ -\nabla_r\mu_c,\\
\partial_t {n_n} \ + \ \nabla_r\cdot({n_n}v_n)
= &\ \Gamma_{12}[\mathcal{F}],\\
{n_n}\left(\partial_t + v_n\cdot \nabla \right) v_{nj}\ = & \ - \nabla_r \tilde{\mathbb{E}}_n-{n_n}\nabla_r U \ - (v_{nj}-v_{cj})\Gamma_{12}[\mathcal{F}],\\
\ \partial_t \tilde{\mathbb{E}}_n \ + \ \nabla_r\cdot (\tilde{\mathbb{E}}_nv_n)
=&\ -\frac{2}{3}\tilde{\mathbb{E}}_n\nabla_r \cdot v_n +\frac{2}{3}\left(\frac{(v_n-v_c)^2}{2}+\mu_c-U\right) \Gamma_{12}[\mathcal{F}].
\end{aligned}
\end{equation}
In \eqref{EulerTwoFluidG1}, the condensate and non-condensate parts are coupled through both $\mu_c$ and $\Gamma_{12}$. Notice that $\Gamma_{12}$ is already computed in \eqref{Gamma12a} and \eqref{Gamma12b}.
In the thesis \cite{Allemand:Thesis:2009}, the author has derived the following hydrodynamic limit:
\begin{equation}\label{EulerTwoFluideG1}
\begin{aligned}
{\partial_t n_c}\ + \ \nabla_r\cdot(n_cv_c)& \
=\ 0,\\
\partial_t {n_n} \ + \ \nabla_r\cdot({n_n}v_n)&\
= \ 0,\\
\partial_t(n_nv_n) \ + \ \nabla_r\cdot(n_nv_n \otimes v_n + \mathbb{E}_n I_3)& \ = \ - gn_n\nabla_r\cdot(2n_n+n_c),\\
\partial_t(n_cv_c) \ + \ \nabla_r\cdot(n_cv_c \otimes v_c)& \ = \ - \frac{g}{2}n_c\nabla_r\cdot(2n_n+n_c),\\
\partial_t(\frac12 n_n|v_n|^2+\frac12 n_c|v_c|^2+\frac32\mathbb{E}_n +\frac{g}{4}(2n_n+n_c)^2)) \ & \\
+ \nabla_r\cdot\Big(\frac12 n_n|v_n|^2v_n+\frac12 n_c|v_c|^2v_c+\frac52\mathbb{E}_n v_n \ & \\
+ \frac{g}{2}(2n_n+n_c)(2n_nv_n+n_cv_c) \Big) & \ = 0,
\end{aligned}
\end{equation}
where $I_3$ is the identity $3\times3$ matrix. It is also mentioned \cite{Allemand:Thesis:2009} that this is a two-phases Euler system, the second fluid (the superfluid) being pressureless and they do not exchange mass, contrary
to what occurs the Landau two-fluid theory \cite{bogoliubov1970lectures,lifshitz1987fluid}.
On the contrary, our limit \eqref{EulerTwoFluidG1} agree with the Landau two-fluid theory \cite{bogoliubov1970lectures,lifshitz1987fluid}. The main reason is that, following \cite{GriffinNikuniZaremba:2009:BCG}, in general excited atoms in the condensate need not to be in local equilibrium with the condensate atoms. As a consequence, $C_{12}$ and $C_{22}$, in most of the cases, do not share the same equilibrium distribution. Our equilibrium distribution $\mathcal{F}$ is the natural equilibrium used in most physical contexts \cite{GriffinNikuniZaremba:2009:BCG} and $C_{22}[\mathcal{F}]=0$ but $C_{12}[\mathcal{F}]\ne 0$. Therefore, the two fluids are coupled.
In \cite{Allemand:Thesis:2009}, the author considers a very special choice of $\mathcal{F}$
$$\mathcal{F}(t,r,p) \ = \ \frac{1}{e^{\beta[(p-v_n)^2/2 - |v_c-v_n|^2/2-U/2]}-1},
$$
where the temperature parameter $\beta$ is a constant, instead of being a function of $(t,r)$. Moreover, the effect of the chemical potential $\mu(t,r)$ is also ignored. This special choice of the distribution $\mathcal{F}$ implies $C_{22}[\mathcal{F}]=C_{12}[\mathcal{F}]=0$. The two fluids are then decoupled, that is in contradiction with the Landau two-fluid theory \cite{bogoliubov1970lectures,lifshitz1987fluid}, as the author pointed out.
\section{The two-fluid Navier-Stokes quantum hydrodynamic approximations}\label{Sec:NavierStokes}
This section is devoted to the derivation of the Navier-Stokes approximation of the system \eqref{RescaledSystem2b} - \eqref{RescaledSystem2a} through the Chapman-Enskog expansion, under the assumption $g=\epsilon^{\delta_0}$. Similar as in Section \ref{Sec:EulerLimit}, we also have the expansion:
\begin{equation}\label{Sec:NavierStokes:E3}
f \ = \ \sum_{i=0}^n \epsilon^i f^{(i)} \ + \ \epsilon^l \varsigma,
\end{equation}
in which $n$ and $l$ are positive integers.
Arguing similarly as above, we deduce that $f^{(0)} $ has to be a Bose-Einstein distribution:
\begin{equation}\label{Eulerf0q}
f^{(0)} \ = \ {\mathcal{F}}.
\end{equation}
Decompose $f^{(i)}$ into two parts
\begin{equation}\label{Sec:NavierStokes:E8}
f^{(i)} \ = \ h^{(i)} \ + \ k^{(i)},
\end{equation}
where
$$h^{(i)} \ \in \mathcal{R}, \ \ \ k^{(i)} \ \in \mathcal{N}.
$$
From \eqref{HilbertSystem2}, one has
\begin{equation}\label{Sec:NavierStokes:E9}
h^{(1)}\ = \ \mathcal{L}^{-1}\mathcal{D}\mathcal{F}.
\end{equation}
Adopting the same techniques used in \cite{ArlottiLachowicz:EAN:1997,KawashimaMatsumuraNishida:OTF:1979}, we decompose $h^{(1)}$ into the sum of $h'$ and $h''$:
$$h^{(1)} \ = \ h' \ + \ h'',$$
where $h'$ and $h''$ satisfy the following system of equations:
\begin{eqnarray}\label{Sec:NavierStokes:E10}
\mathcal{L}h' & = & \mathbb{P}^{\bot}\mathcal{D}\mathcal{F},\\\label{Sec:NavierStokes:E11}
\mathbb{P}\mathcal{D}\mathcal{F} & = & -\epsilon \mathbb{P}\mathcal{D}h',\\\label{Sec:NavierStokes:E12}
\mathcal{L}h'' & = & \epsilon\mathbb{P}^{\bot}\mathcal{D}k^{(1)},\\\label{Sec:NavierStokes:E13}
\mathbb{P}\mathcal{D}k^{(1)} & = & -\mathbb{P}\mathcal{D}h'',
\end{eqnarray}
and
\begin{equation}\label{Sec:NavierStokes:E14}
\mathcal{L}h^{(i)} \ = \ \epsilon\mathbb{P}^{\bot}\mathcal{D}k^{(i)} \ + \ \mathbb{P}^{\bot}\mathcal{D}h^{(i-1)}\ - \ \sum_{j=1}^{i-1}{Q}_1(f^{(j)},f^{(i-j)}) \ - \ \sum_{j,k=0,0<j+k<i}^{i-1}{Q}_2(f^{(j)},f^{(k)},f^{(i-j-k)}),
\end{equation}
\begin{equation}\label{Sec:NavierStokes:E15}
\mathbb{P}\mathcal{D}k^{(i)}\ = \ - \mathbb{P}\mathcal{D}h^{(i)}.
\end{equation}
By the Fredholm theory, the system \eqref{Sec:NavierStokes:E10}-\eqref{Sec:NavierStokes:E13} can be solved in $L^2(\mathbb{R}^3)$, if
\begin{equation}\label{Sec:NavierStokes:E17}
\begin{aligned}
h' \ = & \ \mathcal{L}^{-1}(\mathbb{P}^\bot \mathcal{D}\mathcal{F}),\\
h'' \ = & \ \mathcal{L}^{-1}(\epsilon\mathbb{P}^\bot \mathcal{D}k^{(1)})
\end{aligned}
\end{equation}
and
\begin{equation}\label{Sec:NavierStokes:E18}
\begin{aligned}
h^{(i)}\ = & \ \mathcal{L}^{-1}\Big(\epsilon\mathbb{P}^{\bot}\mathcal{D}k^{(i)}\ + \ \mathbb{P}^\bot \mathcal{D}h^{(i-1)}\ - \ \sum_{j=1}^{i-1}{B}_1(f^{(j)},f^{(i-j)})\\
& \ - \ \sum_{j,k=0;0<j+k<i}^{i-1}{B}_2(f^{(j)},f^{(k)},f^{(i-j-k)})\Big),
\end{aligned}
\end{equation}
for $i=2,3,\cdots$
The equation \eqref{Sec:NavierStokes:E17} yields
\begin{equation}\label{Sec:NavierStokes:E19}
\mathbb{P}\mathcal{D}\mathcal{F} \ = \ -\epsilon \mathbb{P}\mathcal{D}h'\ = \ -\epsilon \mathbb{P}\mathcal{D}\mathcal{L}^{-1}\mathbb{P}^\bot\mathcal{D}\mathcal{F}.
\end{equation}
Equation \eqref{Sec:NavierStokes:E19} leads to the Navier-Stokes approximation, which will be computed in Section \ref{Sec:NavierStokesLimit}.
\subsection{Inversion of the linearized operator of $C_{22}$}
Define
\begin{equation}\label{Symmetries1}
\begin{aligned}
\mathcal{A}(p)=p \otimes p -\frac{1}{3}|p|^2Id, \ \ \ \ \ \mathcal{B}(p)=\frac{1}{2}p(|p|^2-5),
\end{aligned}
\end{equation}
clearly,
\begin{equation}\label{Orthogonal}
\mathcal{A}_{jk}\perp \mathrm{ker}\mathcal{L}, \ \ \ \ \mathcal{B}_{l}\perp \mathrm{ker}\mathcal{L}, \ \ \ \ \mathcal{B}_{l}\perp \mathcal{A}_{jk}, \ \ \ \ j,k,l =1,2,3.
\end{equation}
By the same algebraic argument as the one used for the classical Boltzmann collision operator (cf. page 64-65 \cite{BouchutGolse:2000:KEA}), one can deduce that there exists scalar-valued functions $\alpha_0(|p|)$, $\alpha_1(|p|)$ such that
\begin{equation}\label{Symmetries2}
\begin{aligned}
\mathcal{L}^{-1}\left(\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{A}(p)\right)= \alpha_0(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{A}(p), \ \ \ \ \ \mathcal{L}^{-1}\left(\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{B}(p)\right)= \alpha_1(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{B}(p).
\end{aligned}
\end{equation}
A direct consequence of \eqref{Symmetries2} is the existence of scalar-valued functions $\beta_0(|p|)$ and $\beta_1(|p|)$ such that
\begin{equation}\label{Symmetries3}
\begin{aligned}
\mathcal{L}^{-1}\left(\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}p^ip^j\right)\ = &\ \beta_0(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{A}_{ij}(p), \\ \ \mathcal{L}^{-1}\left(\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\left(|p|^2-\frac{10\tau\Omega_2(\gamma)}{\Omega_1(\gamma)}\right)p^i\right)\ = &\ \beta_1(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{B}_i(p),
\end{aligned}
\end{equation}
where $p^i$, $\mathcal{B}_i(p)$ are the $i$-th component of the vectors $p$ and $\mathcal{B}(p)$ respectively. In addition, $\mathcal{A}_{ij}(p)$ is the $(i,j)$-th element of the matrix $\mathcal{A}(p)$. Note that these symmetry invariances are very similar to the ones obtained in the context of the classical Boltzmann collision operator (cf. Equation $(2.100)$, page 64-65 \cite{BouchutGolse:2000:KEA}); we then denote
\begin{equation}\label{LinearizedInverse}
\begin{aligned}
\mathfrak{C}_{ij}(p):= \beta_0(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{A}_{ij}(p), \ \ \ \ \ \mathfrak{C}_{i}(p):= \beta_1(|p|)\frac{\mathcal{F}^2(p)}{\mathcal{M}(p)}\mathcal{B}_i(p).
\end{aligned}
\end{equation}
\subsection{Navier-Stokes quantum hydrodynamic approximation of the thermal cloud}\label{Sec:NavierStokesLimit}
In this subsection, we will derive the Navier-Stokes system resulting from \eqref{Sec:NavierStokes:E19}. First, observe that
\begin{equation}\label{Sec:NavierStokes:E20}
\begin{aligned}
\mathbb{P}^\bot\Pi\mathcal{F} \ = & \ \frac{\mathcal{F}^2}{\mathcal{M}}\sum_{i,j=1}^3 \left\{(p^i-{v_n}_i)(p^j-{v_n}_j) \ - \ \frac{1}{3}|p-{v_n}|^2\delta_{i,j}\right\}\frac{1}{\tau}\frac{\partial {v_n}_j}{\partial x_i}\\
&\ + \ \frac{\mathcal{F}^2}{\mathcal{M}}\left\{|p-{v_n}|^2 \ - \ \frac{10\tau \Omega_2(\gamma)}{3\Omega_1(\gamma)}\right\}\sum_{i=1}^3(p^i-{v_n}_i)\frac{1}{2\tau^2}\frac{\partial \tau}{\partial r_i}.
\end{aligned}
\end{equation}
Classical techniques for the classical Boltzmann collision operator can be applied (cf. \cite{TruesdellMuncaster:FOM:1980} - pp. 456-457 and \cite{KawashimaMatsumuraNishida:OTF:1979,ArlottiLachowicz:EAN:1997}), to get
\begin{equation}\label{Sec:NavierStokes:E21}
\begin{aligned}
-\mathbb{P}\Pi\mathcal{L}^{-1}\mathbb{P}^\bot\Pi\mathcal{F} \ = & \ \frac{\mathcal{F}^2}{\mathcal{M}}\left(\sum_{k=1}^3\frac{\psi_k}{\omega_k}\left(\sum_{i=1}^3\frac{\partial}{\partial r_i}\left(\varpi(\gamma,\tau)\left(\frac{\partial {v_n}_k}{\partial r_i}+\frac{\partial {v_n}_i}{\partial r_k}\right)\right)\right)\right.\\
&\ -\frac{2}{3}\frac{\partial}{\partial r_k}\left.\left(\varpi(\gamma,\tau)\sum_{i=1}^3\frac{\partial {v_n}_i}{\partial r_i}\right)\right) \ + \ 2\frac{\psi_4}{\omega_4}\left(\sum_{i=1}^3\frac{\partial}{\partial r_i}\left(\varrho(\gamma,\tau)\frac{\partial \tau}{\partial r_i}\right)\right.\\
&\ -\frac{2}{3}\varrho(\gamma,\tau)\left(\sum_{i=1}^3\frac{\partial {v_n}_i}{\partial r_i}\right)^2 \ + \ \varpi(\gamma,\tau)\sum_{i,k=1}^3\frac{\partial {v_n}_k}{\partial r_i}\left.\left.\left(\frac{\partial {v_n}_k}{\partial r_i}+\frac{\partial {v_n}_i}{\partial r_k}\right)\right)\right),
\end{aligned}
\end{equation}
where
\begin{equation}\label{Sec:NavierStokes:E22}
\varpi(\gamma,\tau) \ = \ -\frac{1}{\tau} \int_{\mathbb{R}^3}\xi_1\xi_2\mathfrak{C}_{12}(\xi)d\xi,
\end{equation}
\begin{equation}\label{Sec:NavierStokes:E23}
\varrho(\gamma,\tau) \ = \ -\frac{1}{4\tau^2} \int_{\mathbb{R}^3}|\xi|^2\xi_1\mathfrak{C}_{1}(\xi)d\xi,
\end{equation}
with $\xi_1$, $\xi_2$ are the components of the vectors $\xi=(\xi_1,\xi_2,\xi_3)$ and $\mathfrak{C}_1$, $\mathfrak{C}_{12}$ are defined in \eqref{LinearizedInverse}.
Notice that
$$\mathcal{D}\mathcal{F}=\Pi \mathcal{F}+O(\epsilon^{\delta_0}).$$
The first order approximation in terms of $\epsilon$ of the quantity $\epsilon \mathbb{P}\mathcal{D}\mathcal{L}^{-1}\mathbb{P}^\bot\mathcal{D}\mathcal{F}$ is then $\epsilon \mathbb{P}\Pi\mathcal{L}^{-1}\mathbb{P}^\bot\Pi\mathcal{F}$. The Navier-Stokes system \eqref{Sec:NavierStokes:E19} becomes
\begin{equation}\label{Sec:NavierStokes:E24}
\mathbb{P}\mathcal{D}\mathcal{F} \ = \ -\epsilon \mathbb{P}\Pi\mathcal{L}^{-1}\mathbb{P}^\bot\Pi\mathcal{F},
\end{equation}
which, thanks to the identity \eqref{Sec:NavierStokes:E21}, leads to
\begin{equation}\label{Sec:EulerLimit:NavierStokesEq}
\begin{aligned}
\partial_t {n_n} \ + \ \nabla_r\cdot({n_n}v_n)\
= &\ \epsilon{\Gamma}_{12}[\mathcal{F} ],\\
{n_n}\left(\partial_t + v_n\cdot \nabla \right) v_{nj} \ + \partial_{r_j} ({n_n}e_n)\ = & \ -{n_n}\nabla_r \epsilon^{\delta_0}{U} \ -(v_{nj}-v_{cj})\epsilon{\Gamma}_{12}[\mathcal{F} ]\\
& + \ \epsilon\left[\sum_{i=1}^3\frac{\partial}{\partial r_i}\left(\bar{\varpi}({n_n},e_n)\left(\frac{\partial {v_n}_j}{r_i} + \frac{\partial {v_n}_i}{r_j}\right)\right)\right.\\
& - \ \left.\frac{2}{3}\frac{\partial}{\partial r_j}\left(\bar{\varpi}({n_n},e_n)\sum_{i=1}^3\frac{\partial {v_n}_i}{r_i}\right)\right],\\
\ \partial_t e_n \ + \ \nabla_r\cdot (e_nv_n) \ + \ \frac{2}{3}e_n\nabla_r \cdot v_n
=&\ \frac{1}{{n_n}}\Big[ \frac{2}{3}\left(\frac{(v_n-v_c)^2}{2}+\mu_c-\epsilon^{\delta_0}{U}+e_n\right) \epsilon{\Gamma}_{12}[\mathcal{F} ]\Big]\\
& \ + \frac{\epsilon}{\mathcal{G}({n_n},e_n)}\left[\sum_{i=1}^3\frac{\partial}{\partial r_i}\left(\varrho_1({n_n},e_n)\frac{\partial e_n}{\partial r_i}+\varrho_2({n_n},e_n)\frac{\partial {n_n}}{\partial r_i}\right)\right.\\
&\ +\bar{\varpi}({n_n},e_n)\sum_{i,k=1}^3\frac{\partial {v_n}_k}{\partial x_i}\left(\frac{\partial {v_n}_k}{\partial x_i}\right.+ \left.\frac{\partial {v_n}_i}{\partial x_k}\right)\\
&\ \left. -\frac{2}{3}\varpi({n_n},e_n)\left(\sum_{i=1}^3\frac{\partial {v_n}_i}{\partial x_i}\right)^2\right],
\end{aligned}
\end{equation}
where
\begin{equation}\label{Sec:EulerLimit:NavierStokesEqConstants}
\begin{aligned}
\bar{\varpi}({n_n},e_n) \ = & \ \varpi(\gamma,\tau),\\
\mathcal{G}({n_n},e_n) \ = & \ 2^{5/2}\pi \tau^{3/2}\gamma \Omega_1(\gamma),\\
\varrho_1({n_n},e_n) \ = & \ \varrho(\gamma,\tau)\frac{\partial \tau}{\partial e_n},\\
\varrho_2({n_n},e_n) \ = & \ \varrho(\gamma,\tau)\frac{\partial \tau}{\partial {n_n}}.
\end{aligned}
\end{equation}
Combining \eqref{BECSuperFluid} and \eqref{Sec:EulerLimit:NavierStokesEq}, we get the ``closed system''.
Moreover,
The Navier-Stokes system of the excitations is very different from the Navier-Stokes system obtained from the classical Boltzmann equation (cf. \cite{TruesdellMuncaster:FOM:1980}) in several points:
\begin{itemize}
\item First, in the classical Navier-Stokes system, the viscosity coefficient $\bar{\varpi}$ and the heat conduction coefficient $\varrho_1$ depend only on $e_n$. In the above quantum Boltzmann system, they depend on both $e_n$ and $n_n$.
\item Second, different from the classical Navier-Stokes system, the second derivatives of $n_n$ also appear in the system.
\item Third, the Navier-Stokes system of the excitations is coupled with the system of the BEC super fluid via the quantity $\epsilon{\Gamma}_{12}[\mathcal{F} ]$, computed in \eqref{Gamma12b}.
\end{itemize}
The Navier-Stokes system for the excitations therefore has a completely different nature in comparison with the classical Navier-Stokes equation. And, hence, one could expect more complicated behaviors, that would be a subject of our future studies.
~~ \\{\bf Acknowledgements.} This research was supported by NSF grants DMS-1522184 and DMS-1107291: RNMS KI-Net, by NSFC grant No. 91330203, and by the Office
of the Vice Chancellor for Research and Graduate Education at the University of Wisconsin–Madison with funding from the Wisconsin Alumni Research
Foundation.
M.-B. Tran was supported partially by ERC Advanced Grant DYCON and NSF grant DMS-1814149. Tran would like to thank Prof. Linda Reichl and Prof. Jay Robert Dorfman for the discussions on the topic. The authors would also to express gratitude to the referee for very fruitful comments and suggestions that help to improve the quality of the paper.
\bibliographystyle{plain}
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All-rounder Moeen Ali - who followed up his 38 in the first innings with 40 in the second - was frank about England's deficiencies after they matched Australia for much of the first three days at the Gabba . "The first three days, we were in the game but today we let ourselves down with the bat especially", Moeen said after play. The way our boys bowled against them in the last home Ashes , we've used that as a bit of a blueprint.
World Rugby Breakthrough Player Of The Year Has Been Announced
New Zealand once again also took the World Team of the Year award - although this time it was their women's team, who defeated England 41-32 in an enthralling World Cup final at the Kingspan Stadium in Belfast. "I am just one player amongst a great team". "They do all the work and I am out there on the sideline just waiting for the ball". England boss Eddie Jones was also perhaps a surprise victor in Monaco as he beat All Black mentor Steve Hansen to the Coach of the Year award, ...
Wolverines QB Wilton Speight will transfer for final season
Enrolling in January of 2014 as a kid from Virginia I didn't know what to expect. "Four years later I leave a Michigan Man". Speight was injured in the Wolverines' win over Purdue in September and did not play the rest of the season. He thanked the Michigan fans who stuck with the team "through thick and thin" and thanked former Michigan coach Brady Hoke, who gave "a kid with no offers a chance" for recruiting him to Michigan.
Tottenham looking at Bologna's Adam Masina amid Danny Rose exit reports
Manchester United have been given encouragement in their pursuit of Danny Rose after the Spurs man was omitted from the squad again yesterday. It is understood that United want to sign Rose, 27, in the January transfer window and Mauricio Pochettino has already made plans for his exit by setting his sights on the Italy Under-21 worldwide flyer, who is playing regularly in Serie A with Bologna .
Sabalenka surges to Mumbai Open title
Once she had a 5-2 lead, the 19-year old's excellent forehand bailed her out of a couple of precarious moments in the eighth game. Jakupovic, on the other hand, started the next must-win game in the set with a double-fault and was down set-point following a splendid forehand on-the-run victor by Sabalenka who then broke the Slovenian's serve when Jakupovic buried her forehand into the net.
Rangers reach deal with veteran righty Fister
The rest of the details of this signing are still unknown at this time. The 33-year-old finished with a 4.88 ERA in 18 games, 15 of them starts. He spent last season with the Boston Red Sox after being claimed off waivers from the Los Angeles Angels, and was last seen giving up three runs in 1.1 innings against the Houston Astros in Game 3 of the ALDS.
Brawl in Oakland: Crabtree, Talib ejected from game after fight
That prompted multiple Broncos to chase Crabtree away. Raiders guard Gabe Jackson was also ejected from the game after being flagged for shoving an official in the aftermath of the play. "You snatch a chain in front of the police and run off?" I could've hurt him, but I'm playing football. You're acting. You're snatching chains up on the field.
Results: Hugh Bowman and Cheval Grand win Japan Cup
Having been third in the race 12 months ago to Japanese Horse of the Year Kitasan Black , Cheval Grand emerged from the chasing pack under Bowman to run on strongly and overhaul the former, who was to also lose second place to Japanese Derby victor Rey De Oro close to the finish line.
Panthers TE Greg Olsen leaves game with a foot injury
The game also marked the first one back from injury for the Carolina tight end Greg Olsen . Olsen caught Carolina's first pass of the day for a 10-yard pick-up but was shut out on his other three targets and never seemed to be on the same page with Cam Newton .
Bills improve to 6-5 with 16-10 win over KC
Perhaps the loss to Buffalo is still too fresh in my mind, but this Kansas City Chiefs team looks completely broken. Overall this season, the 33-year-old has thrown for 2,674 yards, 18 touchdowns and three picks while completing 69.4 percent of his passing attempts and registering a passer rating of 107.6.
Julio Jones has big game as Falcons beat Bucs
Freeman will miss his second consecutive game with a concussion. Freeman is in the concussion protocol. His touchdown total this season is three after having just one in the first 11 weeks. Barber had five carries for 7 yards. The Falcons will receive the ball to start the second half. Coleman had 97 yards on 19 carries, both season-highs.
Panthers' Greg Olsen (foot) will not return to Sunday's game
There was no immediate word on whether he reinjured the same foot. The 32-year-old veteran is the only tight end in National Football League history to post three consecutive 1,000-yard receiving seasons. Olsen had a solid rookie campaign, recording 39 catches for 391 yards and two scores, but it wasn't until his third year that he really broke out, snaring 60 balls for 612 yards while reaching the end zone eight times.
AP Top 25 poll: Alabama, Miami plummet after upset losses
The latest AP poll has Clemson ranked No. 1. The Canes will next play Clemson in the ACC Championship game December 2 in Charlotte. Clemson and Oklahoma take over the top two spots, respectively, after blowout wins in their final regular-season matchups.
Gators working to finalize deal with Dan Mullen as new head coach
Mullen, who most recently led the Mississippi State program, would UF's 25th football coach. At Mississippi State , Mullen had a record of 69-46. When the school hired away Mississippi State defensive coordinator Geoff Collins in 2015, Mullen called it a "lateral move". Mullen brings a lot to the table for the Gators, including being a now-proven program builder and quarterback-developer, two things Florida has sorely missed since Meyer's departure.
Tennessee to hire Greg Schiano
The head coach position at Tennessee has been open since athletic director John Currie fired Butch Jones on November 12. His Ohio State defense has since piqued the interest of Tennessee, which has a history of making such brilliant coaching hires as Lane Kiffin and Derek Dooley.
Manchester City defeat Huddersfield, set record
Manchester City have defeated Huddersfield 2-1 in the last hour in an English Premier League match. David Wagner is happy with Huddersfield Town's performance in their 2-1 loss to Manchester City but believes that the visitors' winning goal was unlucky for the Terriers.
Ian Wright comments on David Unsworth
The caretaker manager has been in the role since Ronald Koeman was sacked over a month ago, and Pardew believes the 44-year-old has ruined his chances at the job due to his results . "Because this group of players is under-performing and whether it's Ronald in charge, whether it's myself in charge it doesn't seem to be working so big decisions have to be made".
Pouille defeats Darcis in Davis Cup decider
It is France's tenth Davis Cup, but their first since 2001 and Pouille explained it was a momentous moment for him and his team-mates. Updates with France winning Davis Cup title. Goffin, the best player of the final, had earlier pushed the tie into a fifth match with an impressive demolition of Tsonga. Goffin, who outclassed Lucas Pouille in Friday's opening rubber, saved a set point while serving at 5-6, as Tsonga was left to rue his inability to capitalise on any of his six break points.
Darren McFadden: Dallas Cowboys waive veteran RB
His role on that committee is no longer required. It was a mutual agreement between the two parties, and with McFadden owning a reasonable contract, he could get claimed by another team. The team has chose to release kicker Mike Nugent after fellow kicker Dan Bailey returned to the team and proved his health in practice, according to David Moore of the Dallas Morning News .
Centre should take final call on bilateral series with Pakistan: Dhoni
A video circulating on social networking website Facebook shows the crowd raising slogans like "Bhoom Bhoom Afridi", in praise of the former Pakistan cricket superstar, Shahid Afridi. India have not played any bilateral series with Pakistan since the limited-overs series featuring two T20Is and three ODIs in India in 2012-13.
Arizona State to fire Todd Graham
Arizona State Sun Devils football coach Todd Graham expects to be fired after Saturday's Territorial Cup game vs. the Arizona Wildcats, according to a report Saturday. If Arizona State fires Graham, Wolken listed Texas A&M head coach Kevin Sumlin, Michigan passing game coordinator Pep Hamilton, Detroit Lions defensive coordinator Teryl Austin and Vanderbilt head coach Derek Mason as potential replacements.
Former Blazers player and broadcaster Steve 'Snapper' Jones dies at 75
He then played at the University of OR in college and played eight professional seasons in the American Basketball Association. Following the end of his playing career, Jones transitioned courtside, becoming a successful sports broadcaster.
Swansea boss Clement hails Mesa, Bony for Bournemouth stalemate
Striker Jordan Ayew was adjudged to have fouled Cherries defender Nathan Ake , but Clement strongly disagreed. The decision also angered the players, and had to be dragged away from Stuart Atwell as the referee blew the whistle seconds later to end the half.
National Bank of Canada (NTIOF) Analysts See $1.08 EPS
Laurentian Bank of Canada (TSE:LB) has 0.00% since November 26, 2016 and is. It has underperformed by 16.70% the S&P500. Over the last three months, insiders bought 16,900 shares of company stock worth $988,018 and sold 17,100 shares worth $980,254.
Liberty ready to reveal new F1 logo
New F1 supremo Chase Carey confirmed that Liberty Media is now ready to unveil the new logo. The new logo is supposed to represent "two cars racing across the finish line" according to managing director of marketing Sean Bratches. "It's served Formula One extremely well for the past 23 years but in terms of where we're taking the business and our vision for the business, it's the negative space in the "1" doesn't come through candidly in digital.
Jaguars downgrade Parnell, Omameh; Ramsey questionable
Bouye have quickly become a shutdown cornerback duo. Ramsey reportedly "jammed" his hand, so it doesn't sound like the type of thing that will sideline him for long. Three other players are also listed as questionable, including starting offensive linemen Jermey Parnell (knee) and Patrick Omameh (quadriceps). Wide receiver Allen Hurns , who has not practiced since he suffered an ankle injury against the Los Angeles Chargers on November 12, was listed as out.
England star investigated over Bancroft headbutt
The alleged incident happened in Perth earlier this month, with claims emerging Down Under overnight. Bancroft was not a member of the Australia squad at the time. England were in Perth for a two-day match against a Western Australia XI, their first outing of the tour before heading to Adelaide and Townsville for four-day matches against a Cricket Australia XI.
Ben Simmons out Saturday vs. Magic, report says
The Orlando Magic lost to the Philadelphia 76ers 130-111 to suffer an eighth consecutive defeat in National Basketball Association on Saturday. McConnell was also on-fire, scoring 15 points, 13 assists and seven rebounds. The elbow injury isn't considered serious, with Simmons set to be re-evaluated on Monday. Orlando, which was playing on the second day of a back-to-back, never led after the first quarter.
Charred Body Found Atop SEPTA Train
His charred body was discovered on the train's roof after it pulled into the station at 7:49 a.m., police said. Philadelphia police are trying to determine how the man died, and where. Police believe the person may have been electrocuted by high voltage equipment which connects the trains to power lines and keeps them running. SEPTA did not provide details about the incident, initially describing it as a medical emergency in a tweet.
Ovechkin's hat trick powers Capitals past Maple Leafs
Russian Alexander Ovechkin, who plays for the Washington ice hockey club, managed to score the fifteenth goal in this season, equalized with rivals from Tampa already in the 19th minute. Though the Capitals scored again in the second to take a 3-0 lead, and carried that advantage into the third period, Toronto mounted a comeback, getting goals by defensemen Jake Gardiner Nikita Zaitsev to cut its deficit to one.
Steve Kerr regrets playing Kevin Durant in 'marquee' game to appease National Basketball Association
The Warriors also learned in their 110-95 victory they can also manage just fine when their other star players have off nights. "You have to block it out the best you can, have a sense of amnesia nearly, and just rely on the repetitions and work you put into it".
Winnipeg Jets vs. San Jose Sharks NHL Odds, Prediction
The Jets are now 6-2 against the Pacific division, and 10-3 against the Western conference. The Sharks were outshot 16-5 in the second period, but added to their lead, thanks to outstanding play on the penalty kill. The Sharks worked it around on top of the circles before Kevin Labanc took a pass from Joe Thornton and backhanded a pass to Heed, who one-timed a slap shot past Steve Mason for his third goal of the season.
No 21 Stanford pulls away from No 8 Notre Dame
After linebacker Curtis Robinson intercepted Notre Dame's Brandon Wimbush on the game's next snap after the go-ahead score, Costello found his fourth different target for a touchdown, hitting Brandon Schultz with a 19-yard pass for a 31-20 lead with 12:21 left in the game.
Carey Price Returns To Montreal Canadiens Lineup
The last time he played was in a 6-3 loss against the Wild in Minnesota. With a 4-4-2 record, Montreal actually earned more points when Price was out (10) than it did when he was in for the first 13 games of the season (nine). Galchenyuk helped set up Jeff Petry's power play goal in the first period before putting his 11-game goal drought to bed in the second frame. 6 - Paul Byron scored his sixth goal of the season to put the Canadiens up 3-0 in the third.
Shurmur, Vanderbilt carve up Tennessee
Vanderbilt (5-7, 1-7 SEC) has beaten Tennessee (4-8, 0-8) four of the last six years after going 1-28 against the Volunteers from 1983-2011. The Vols have endured a tumultuous season filled with injuries, transfers and unprecedented defeat that resulted in the firing of Butch Jones.
Real Madrid coach Zidane: Valencia vs Barcelona will be decisive
They have not yet been beaten in the league and are on a nine-game winning streak in all competitions. "I hope the best team wins, I have a lot of love for my old clubs". He took a long, staggered run-up only to balloon the ball over the bar as Italy were eliminated. The anxiety spilled over into his brief and unhappy loan spell at West Ham United , where he failed to score in 11 games , which he said resulted in him "going insane".
Roberto Firmino, Sadio Mane on bench for start of Liverpool-Chelsea
It was a unusual line-up that left fans scratching their heads, but Klopp almost got away with it thanks to the ever-dependable Mohamed Salah , who popped up in the second half to the opening goal of the game. Klopp said after the game: "I was shouting at [Mane] because he was in the centre, but no player is happy about that". They believe that Mane was simply asking Klopp why he start - or come on sooner - as Liverpool threw away another lead.
Oklahoma captains take Baker Mayfield's jersey with them to coin toss
West Virginia's first touch consumed almost half the first quarter, moving 64 yards in 13 plays before settling for a 28-yard Evan Staley field goal. Mayfield ran over to the sideline, jumping up and down, waving his arms up and exhorting the crowd. The Jayhawks (1-10, 0-8) will have a hard time avoiding a fourth 0-for-conference play in that span.
Browns defense wants to tighten pass coverage in rematch with Bengals
And the way they did it. They're having a really hard time simply staying on the field. If he wants to stand any chance of fending off competition next season he has to put in some good performances and most importantly take care of the football.
Rashaad Penny reaches 2000-yard mark on big TD run
The Aztecs become the first program in FBS history to have back-to-back seasons of 2,000-yard rushers in Pumphrey and Penny. San Diego State Quarterback Christian Champman threw 2 TD's and Penny's backup, Juwan Washington had a 60 yard TD run in the 4th quarter to round out the scoring for the Aztecs.
Win over Everton will give us belief, says Pellegrino
Now sat 14th in the table, the South Coast club have been a team in freefall since the end of September, picking up a 1-0 win over West Brom last month, that is Southampton's only win in their last seven attempts. Leighton Baines (L) of Everton celebrates scoring his side's first goal to make it 1-1 with his team mates during the Premier League match between Crystal Palace and Everton at Selhurst Park on November 18, 2017 in London, England.
Trump supporters confuse LeVar Burton with LaVar Ball online
The feud with Trump featured Trump mentioning LaVar's name a few times to his 43.3 million followers - tweets that have been valuable free advertising for LaVar's Big Baller Brand. LaVar is just a poor man's version of Don King, but without the hair. According to the Congressional Research Service , Trump would be allowed to keep them.
Champions League: Ibrahimovic makes history as Basel shock Manchester United
Speaking recently, Eric Cantona stated that, although he is a huge admirer of Mourinho's, he feels that the Portuguese coach should be more adventurous with the players he is overseeing. "We managed to give him 15/20 minutes in both matches, he needs more of that, so hopefully tomorrow we play him again". "We played against Newcastle with (Paul) Pogba, (Marcus) Rashford, (Anthony) Martial, (Juan) Mata, ( Romelu) Lukaku , the full-backs are not full-backs, ( Antonio) Valencia and (Ashley) ...
Former Phillies pitcher Miguel Alfredo González dies in vehicle accident
He was released from the team in April of 2016 after spending the bulk of his career in the minor leagues. Gonzalez died in a motor vehicle accident in Havana, Cuba. Gonzalez only appeared in six games with Philadelphia due to arm trouble, allowing four runs in 5 1/3 innings in 2014. According to his official age in major league baseball, Gonzalez turned 31 in September.
Arsene Wenger reveals who will replace him at Arsenal
The Brazil global made 64 appearances after joining the Gunners in January 2015, but injuries and stiff competition for places meant he was unable to cement a regular starting place in Wenger's strongest XI. 'I think I played in four or five games en route to the FA Cup semi-final against Manchester City and I played brilliantly in that game. "But it's hard when you only play every three weeks and now I want to play as much as I can".
Emery backs Mbappe to shine on Monaco return
Cal will be in the hot-seat for kick off but we'll be building up to the action throughout the day with team news, interesting tidbits and occasional attempts at humour. Claude Puel is his nearest rival, the former midfielder - who two Ligue 1 titles as a player and another as a manager with Monaco - playing in 22 matches.
Bayern Munich drop points in loss to Borussia Monchengladbach
A Thorgan Hazard penalty and Matthias Ginter strike in the first half saw Bayern beaten at Borussia-Park despite Arturo Vidal's 74th-minute goal. Bayern managed to pull a goal back when Chile global Arturo Vidal fired in a long-range shot on 74 minutes which Yann Sommer got a glove to, but couldn't prevent from rolling into the net.
How to Watch Penn State vs. Maryland
I think the week before we had a special teams takeaway. Senior teammates Simone Lee , Ali Frantti , and Haleigh Washington helped Penn State knock off the Badgers for the second time this season - 18-25, 25-16, 25-20, 25-20. But after two heartbreaking losses to Ohio State and Michigan State, they look like they are going to fall short on both of those hopes. Ty Johnson leads the way on the ground with 805 yards and five TDs, but he has failed to reach the 100-yard mark in the ...
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One of the most common international political clichés is the bilateral relations between China and the United States. However, what is not very clear is the exact nature of the relations, which until recently, observers had settled for an inelegant but accurate description: the countries are neither foes nor friends. This seemed quite a reasonable designation: for starters, they are clearly not allies.
The United States and China have never had much in common. There are countless China and America differences. One of the most significant is that the two countries do not share political values or overriding security interests.
Their fundamental perception of the world order clashes awfully. While China is working towards a multi-polar post-American world, the United States is tirelessly working towards preserving the liberal order despite its waning power. Additionally, other issues such as the East Asia tensions over Taiwan and the frequent disputes between Japan and China increase the friction between these two countries.
These two countries, despite their numerous differences, cannot really be termed as adversaries. They really don't regard each other as security or ideological threats. Besides, their economies are intertwined so deeply that they will do anything to avoid any conflict that would interrupt China and America trade ties.
China's remarkable economic expansion and the global financial meltdown that pushed many economies to their knees give the impression that the West is slowly going down as the others rise. The previously inconceivably large economic gap between the two countries has significantly narrowed down.
The relations between these two countries are bound to change should a substantial shift in power balance occur. This, although not surprising, will come with additional strains to the already struggling relationship.
China is rapidly changing its stand on critical issues as it asserts its power in the East. For starters, it has adopted a more assertive policy on foreign issues and employed tough measures in dealing with maritime and territorial disputes with neighboring countries.
China has also adopted a program of rapid modernization of military actions and cyber attacks, a move that has unsettled the US and their Eastern Asia allies. China, on the other hand, interprets this response to their tough policies as a subtle attempt to contain their power.
What Are the Deeper Issues?
The US-China relations are tough, but tougher times are in the offing since the two countries can't understand each other, especially on the critical issues that are vital to stable relations between them. Here is a simplified dissection of the strained relationship.
China's political system hasn't changed over the years, which has resulted in a surge in nationalism. This is one of the main reasons why there is some friction. Further differences in its military action modernization, regional security, and human rights policies have also raised eyebrows with both China and American allies. China's beliefs on human rights imply that they are exempt from complying with universal norms as they believe in codes of ritual behaviors as opposed to individual rights.
Due to its modernization of military policy, China today is safe from inland invasion, with none of the neighbors even trying to match its power. The country also basks in military and technological superiority, with the army boasting of better capabilities, which include mobile nuclear missiles, quiet submarine attacks, and super advanced jet fighters. This is also the case with the US, which also enjoys military superiority with no neighbors to compete with.
Beijing and Washington have almost nothing in common when it comes to policies. Some of the main contentious policies include climate change, security, and economic and global responsibilities. The fundamental differences in the policies have fuelled the distrust between the two countries. These disputes are likely to remain unsolved for some time. Since it is almost hopeless to assume that the differences will be resolved, it is far more conceivable to have the differences managed instead.
The International Monetary Fund has estimated that the Chinese economy is likely to overtake that of America by 2025, all this due to the registered remarkable growth. This rise of the economic power in developing countries inverts the existing world order and suggests that China should take the opportunity to modify the international mechanisms on international laws and regimes that they deem unreasonable. However, all this talk of a shifting power balance and the inevitable rise of China is yet to convince America.
The difference between China and America policy development goes far beyond their borders. It is each country's belief that the other lacks legitimacy in its policies. The US asserts that China gives territorial; no strings attached resources to developing countries. China, on the other hand, gives the assertion back at them, arguing that in their rise to power, America had the same mercantilist trade policies. Both countries claim that the other underwrites the global order out of sheer self-interest and not altruism.
There are liberalism and realism assumptions that America's policies on China are based upon.
The US assumes that until China agrees to be a stakeholder, and even after it does, America will maintain its military might and alliances which they will use to deter China should it threaten to weaken the existing world order.
Initially, the liberal assumption seemed likely to take the day, especially since its military might was almost negligent compared to that of the US. However, given China's current acquiescence, the realism assumption seems more likely. The existing incompatibilities in their international systems will keep these two countries in strained relations. As China's power continues to grow, and America fails in their attempt to control them, the balance and their relations will remain strained. | {
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Durable rubber construction. U-Gauge hose suitable for any manometer.
2 metre length of rubber hose for use with all U-gauge or digital manometers.
Used for pressure testing applications and can be utilised for a variety of tasks. | {
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} |
Next Left
How Bernie Sanders Is Making the Next Left Possible
The roots of the Next Left podcast go back almost exactly a year—energized by the wave of progressive activism sweeping the country, we wanted to take a deep dive into the new politics of this moment. The idea was to talk politics with the people who were upending primaries and general elections across the country... by challenging incumbents, taking on party establishments and, above all, bringing fresh ideas to the campaign trail and to governance.
Over the past six months, we've gone to Capitol Hill, where we met with members of "the squad," like Ilhan Omar and Rashida Tlaib, and with people who are changing debates about foreign policy, like Congressional Progressive Caucus stalwarts Pramila Jayapal, Mark Pocan and Ro Khanna.
We went to the basement of the state Capitol in Wisconsin, where newly-elected state Treasurer Sarah Godlewski explained how she's putting economics on the side of the people. We spoke with judges and district attorneys and city council members and mayors. We turned up the volume on messages from Texas and Mississippi and Puerto Rico and North Dakota.
We talked mostly to political newcomers who had won elections against the odds, like Anna Eskamani in Florida, but also to activist officials who are building movements, such as Helen Gym in Philadelphia.
We talked to new leaders who had won landslide victories, like Jackson Mayor Chokwe Antar Lumumba, and to new leaders who suffered narrow defeats but are not going anywhere, such as Tiffany Caban in Queens. We followed candidates who were up for election in 2019 and won epic victories, like Lee Carter in Virginia, Kshama Sawant in Seattle and Chesa Boudin in San Francisco.
In every case, our conversations were about the personal and the political. Candidates talked about their ancestors and their children, about their communities, about the music they listen to-Ilhan Omar's a country fan-and about their role models and heroes.
We decided to finish the season by interviewing a pair of political veterans who were frequently mentioned by the young candidates and officials we interviewed. California Congresswoman Barbara Lee joined us last week. This week, for the final episode of the podcast, Vermont Senator Bernie Sanders is our guest for a compelling conversation about his own early campaigns, about the importance of Jesse Jackson's "Rainbow Coalition," about how he makes endorsements, about the way in which media treats insurgent candidates-and about the inspiration he has taken from the Next Left.
When Bernie Sanders Endorsed Jesse Jackson for President The Nation Steve Cobble
Bernie Sanders Is Back The Nation John Nichols
Bernie Sanders: The 'Nation' Interview The Nation Katrina vanden Heuvel and John Nichols
Subscribe to The Nation to support all of our podcasts: thenation.com/podcastsubscribe
New from The Nation: More Than Enough
Ilhan Omar: "There's a Reason That I Got Elected to Be in Congress and it has Nothing to do with the Fact that I'm a Refugee"
by Next Left
Franklin Bynum Is a Texas Judge Who Wants to Abolish Prisons
Pramila Jayapal Is Not Backing Down | {
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Adobe Photoshop Tip for True Color. Adobe Photoshop has come a long way since its exclusive introduction for Apple Macintosh computers by Adobe Systems in 1990. Its origin actually dates back to 1987 when Thomas Knoll, "a PhD student at the University of Michigan, began writing the program so he could display grayscale images on his monochrome display." Since then, Adobe Photoshop has undergone major version upgrades and it's utilized on many millions of computers with color monitor displays for business and consumer use worldwide.
Unfortunately, though many users fail to get the full-color benefit that Adobe Photoshop offers, even those practicing some form of color management, because of their cluttered computer wallpaper backgrounds when editing photos and in this article, I'll prove to you how perceptions affect our eyes in when working in Adobe Photoshop, or any photo imaging program. This is a common problem even with some professional photographers and many blame it on poor color management when a simple keystroke can solve their color interpretation problems with photos.
Unlike the monochrome days of computer displays where desktop backgrounds were virtually non-existent, today a computer desktop background is usually a favorite scene chosen from the default selections of our computers, or a scene captured by ourselves that usually features our loved ones or a favorite vacation photo. Most of these computer desktop backgrounds are filled with brilliant colors projected through color monitors capable of producing millions of colors, hues, shades and superb saturations — all affecting how we perceive a photo on our computer, especially during the postproduction workflow.
For the most part, there are two types of photographers when it comes to postproduction, the diehard color management professional that fills their entire screen with the photo they're editing that leave room only for the working window pallets and tool selector menu. The other photographers are those that don't utilize their entire desktop workspace and allow plenty of their background wallpaper to surround their postproduction photo, thus throwing off to some degree their perception of color. Don't despair, here's the simple workaround to prevent that, just use the "F" key function (not the keyboard function keys, just the character "F" key) when working in Adobe Photoshop.
But first, go to your Adobe Photoshop menu, Photoshop >> Preferences >> Interface, and make sure that under "Standard Screen Mode" and "Full Screen Mode," you have selected "medium grey." This medium grey is more representative to the common 18% gray standard used in photographic light meters and gray is neutral in general.
Merely pushing the F key on your computer keyboard will instantly change your desktop background to a neutral medium gray, if you followed the preferences changes, then when you strike the F key again and it now goes to a neutral black desktop. Don't panic though, simply strike the F key again while in Adobe Photoshop and you'll find your background will reset to the original wallpaper. Obviously if you close Adobe Photoshop it will also return your desktop back to its original background again.
The reason behind using the F key function in Adobe Photoshop is simple, the human eye and brain "perceive" color based on surrounding colors in your work area — even from the walls in your computer work area, not just the computer desktop. But for this article we'll focus on your computer monitor as it's the immediate area where your eyes are focused during postproduction.
Basically, if you have a dark color around your main photo, your photo will appear lighter and if you have a light color around the photo you're editing, it will appear darker. This is one reason why professional photographers match the tone in their finished print to the matte if their photos are framed for display. Colors are decoded differently in our brains when surrounded by other dominant colors.
Is the green blocks on the left darker than the ones on the right? They are both equal.
As an example, look at the image above. We see two vertical rows of green surrounded between yellow and magenta and though none of the green rows are connected, our eyes will perceive the green rows on the left darker than the green rows on the right — in this environment our eyes together with our brain perceive that the green on the right is a lighter shade than the green on the left.
Though this appears correct, all the green rows in the entire image are identical shades of green — download the image and view the color values with the densitometer tool in Adobe Photoshop and you'll see they are identical. The same goes for the image below, all rows and columns of green are identical, but our eyes see one thing and our brain perceives another — our visual sensory is tricked.
All the green areas in this image are the same color, but do you see that or a difference in green?
During my postproduction of photos in Adobe Photoshop, my preference is using the F key function in the medium gray mode. The ideal neutral gray is known in the industry as Munsell gray, usually ranked as N5 through N7, however I don't recommend changing the color of your computer room walls unless you're working in a professional prepress environment as our monitors today are much larger than the monochrome monitors of the past.
All areas of green are equal in color values.
The F key function makes photo postproduction more accurate and when combined with the "tab" key, even more accurate and less distracting. Basically, strike the tab key while working in Adobe Photoshop so the tool bars and pallet windows temporarily disappear. Simply striking the tab key again, they will reappear, just like circulating through the F key, so you can always return back to your original workspace environment.
While John Knoll eventually convinced his brother Thomas to turn his original "Display" program for monochrome monitors into the what would later become "Image Pro," then evolve into Adobe Photoshop, one of the original concepts then is still the same now, correctly displaying images on a computer monitor. Though as photographers, without utilizing the F key function during postproduction, chances are, we're not really seeing the true colors we captured so I highly recommend you give it a try, especially with images where correct color is critical. | {
"redpajama_set_name": "RedPajamaC4"
} |
Guests are choosing rooms based on the images shown on your website or booking platform in conjunction with other factors such as location and price. You're choosing the best shots and hopefully updating them occasionally. What about the actual rooms?
It's common to leave a guest questionnaire to elicit useful management feedback. These usually address the efficiency and courtesy of reception, the quality of breakfast and the general cleanliness of the guest accommodation. While these issues are absolutely critical to the hotel's success and repeat custom it is partially the 'room shot' which has attracted new guests. By amending your questionnaire to prompt comments on the décor this information will be at hand when you need to establish refurbishment criteria in the future. | {
"redpajama_set_name": "RedPajamaC4"
} |
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