Mount Fromme sits just east of Grouse Mountain and is home to a large number of mountain biking and hiking trails. At an elevation of 1,185 m, the view from the summit of Mount Fromme is rewarding, but is by no means the best around. Mount Fromme offers views of Lynn Ridge, Coliseum Mountain, Mount Burwell, and Mount Seymour.
Close to the summit of Mount Fromme is another peak – Senate Peak – which also offers partially obstructed, yet beautiful views. Senate Peak overlooks Grouse Mountain and the Burrard Inlet.
The hike to Mount Fromme is worth the effort. The views are good, the trails up are enjoyable, and the trail sees less traffic than other nearby hikes.
There are several possible starting points for the hike to Mount Fromme. This guide starts at the St. Mary's Trail at the end of St. Mary's street in North Vancouver. Another regularly used access point is at the end of St. Georges Street. It is also possible to start at the top of Grouse Mountain and access Mount Fromme via Old Mountain Highway, or via the Thunderbird Ridge and Erik the Red trails.
There are countless mountain biking trails on Fromme. For this reason there is no single way to get up to the summit of Fromme from the bottom. It also makes for many junctions, and opportunities to get off track. If you are not familiar with the trails on the bottom of Mount Fromme, it can be confusing trying to stay on the route. At many points, a single trail will briefly split into two, before merging again. It is a good idea to study the route for the hike to Mount Fromme and have a map handy. A GPS device may be helpful if you are not familiar with the trails.
For this guide, we will be taking three main trails to get to the Mount Fromme summit: Dreamweaver > Executioner > Per Gynt. Along the way, two key landmarks are crossing over the Baden-Powell Trail, and meeting Old Mountain Highway twice.
The good news is the mountain biking community does an amazing job of maintaining these trails and they are very well marked. There is a good online map of the Mount Fromme mountain biking trails here.
As many of trails are shared-use, you'll want to be ready for mountain bikers zooming down the mountain. If you hear mountain bikers coming, step off the trail and give them plenty of room (you will not win a head-to-head collision with a well-armored rider). Often, the lead rider will tell you how many other mountain bikers you can expect behind him or her in their group (ie, they'll yell 'two more!' or 'three more!'). Stay aside until the group passes and carry on.
Park on St. Mary's Street and you will see the trailhead on the side of the road. Starting at the St. Mary's trailhead, within a few minutes you will pass under the power line and cross over Power Line Trail. You will enter the dense forest of Mount Fromme and quickly find a marked junction. Stay along St. Mary's Trail. Carry along the trail a few minutes further and you will come to another junction for Boundary Trail and Dreamweaver (it will be marked 'climbing route'). Go left here, and generally speaking, you'll want to follow any of the signs marked 'climbing route' along the hike.
You'll soon come to another junction at a wide double-track trail. Stay in the direction marked for Dreamweaver, which is directly straight ahead on the other side of the wide trail.
Moving along Dreamweaver trail, you will come to a bridge with wooden 'Dreamweaver' sign hanging from a tree. Stay along this trail until you reach a junction for Executioner trail, around 1.9 km into the hike. Here you will switch-back onto Executioner trail. The trail here becomes steep and technical. The trail climbs, switchbacks once, and within 1 km you will reach Old Mountain Highway – a large dirt road – and you will go left onto the road.
At the switchback in the corner of the road, you will see a 4×4 wood post with a ribbon tied around it and the words 'Per Gynt' written in felt marker. At this marker go right, heading uphill. It is flagged here as being at 710 m elevation. Per Gynt Trail will take you the rest of the way to the summit of Mount Fromme. The trail is marked with orange markers and orange flagging. Per Gynt is a fairly difficult trail, but it is easy to follow without going astray from this point.
Hike another 600 m until you once again intersect Old Mountain Highway at 3.5 km. Walk almost directly straight across Old Mountain Highway and you will see Per Gynt trail off to the side of the road, marked with ribbon. Once you're here, it's basically a straight shot to the summit. The trail is less technical and much flatter – however you have a solid 2 km of hiking.
Follow the trail through the forest, and as you begin to approach the summit it will begin to climb upwards again. You will reach one final junction, and go right here. The summit is just a few minutes away.
At the summit, you can see The Needles and Lynn Ridge, Coliseum Mountain, and Mount Seymour and Mount Elsay off to the east. To the north, through an opening in the trees, you can also see Crown Mountain.
To the north, you will also find another trail that drops downward. This will take you to nearby Senate Peak, which is worth the 15 minute (each way) detour. From Senate Peak you can see Grouse Mountain and get a great view of Crown Mountain. It is also possible to do a loop between Senate Peak and Mount Fromme down to a tarn, or just return the way you came from.
Once you're back at Mount Fromme and all rested, retrace your steps back down to your car.
From the Upper Levels Highway (Highway #1), exit at Lonsdale heading north. Drive up Lonsdale for a few minutes and take a left on Braemar Street. Take a left of St. Georges, a right on Balmoral, and a left on St. Mary's. The trailhead is up the street on your right-hand side.
Mount Fromme is a nice hike with solid views at the summit. It's not the best hike on the North Shore, but it's worth exploring at least once, and has the benefit of small crowds.


text:
Ливенский уезд — название двух уездов (исторических административных единиц в составе наместничеств и губерний) Российской империи и первых лет РСФСР.

 Ливенский уезд — в составе Воронежского наместничества (1779—1796) и Слободско-Украинской губернии (1796—1802).
 Ливенский уезд — в составе Орловского наместничества (1778—1796) и Орловской губернии (1796—1928).

redpajama_set_name: RedPajamaWikipedia


Students must qualify for WEX12, by matching a selected menu of courses in six Focus area – Business, Fine Arts, Fitness/Recreation, Health & Human Services; Liberal Arts & Humanities; and Science. Refer to the course selection book to see if you qualify. Then contact your counsellor to discuss.
Students will earn four Grade 12 credits for completing 100 hours of work experience in the community. These hours are done during the school day and are scheduled into a student's timetable so as not interfere with other studies. All work sites are open to students regardless of their Focus Area. Most students will most likely attend at least three placements depending on their daily hours worked and how many placements are needed to total 100 hours.
This course allows students an opportunity to test their future occupational choices: the nature of that work, the people that usually work in that field, and other aspects. As you think about post secondary studies, some institutions (university, college, technical and trade schools) are asking students to have previous volunteer experience in related areas, WEX 12 will satisfy that requirement. You can download WEX 12 Course Info.
Special Education students access work experience through their school programs. Students are usually placed for 2 days a week, 2 hours daily over a four to five week period. Several placements per year can be organized.


Updated Monday February 11, 2019 by Fred Steigerwalt.
Babe Ruth and leagues across the country have instituted new background check policies based on new federal laws passed in wake of the Dr. Larry Nassar, Michigan State, US Gymnastics' team abuse cases.
Also, the State of PA (meaning government) has placed a prevision where in addition to the Federal process we also must complete Background checks, Sexual abuse check and FBI finger print specific to the State of Pennsylvania.
All past Background checks submitted to Hilltop are now obsolete due to the new Federal Laws beyond any organizations current policies.
The law requires all our managers, coaches and volunteers to complete the new process and the cost will be $25.50. Sorry, for the additional cost but again it's the new federal law. The cost increase from the previous process is $5.50.
Hilltop has always taken our background check process seriously and we will continue to do so by adopting the new Federal and State laws immediately.
Thank you for your cooperation and the new process is good for two years.
First - Complete these three forms and place on your PC in .pdf format. This is the state of PA requirement. Backgournd check and Child Abuse are free, the third one is the FBI check. Now if you have lived in the PA for ten years you can complete the "Disclouse statement" in lieu of the FBI fingerprint I have highlighted the "disclose link" for you to click on in step three below. Your witness for the "discloser statement" can be any family member that can vouch for you. Once that statement is complete save it in .PDF format with your background and Sexual abuse clearance.
Pennsylvania ONLY – these three forms must be completed and saved on your PC as they will have to be uploaded as a PDF's when you create your Sports Engine background check account.
3. Federal Bureau of Investigations (FBI) Criminal Background Check or Disclosure Statement Application for Volunteers. Who have lived in PA for more than 10 years.
3) Than you will be asked to add your Background check, Sexual abuse check and Discloser statement as .pdf attachements.
4) You will have three pages of acknowledgement that a national background check will be completed.
5) At the conclusion of the background screen, you will be prompted to complete payment ($25.50) *there are no codes to enter at checkout as Hilltop does not have a PIN.


text:
Одриска краљевина ( [-{Basίleion Odrysōn}-]) је био савез трачких племена, који први пут спомињу антички писци крајем 6. века п. н. е. Према утицајном бугарском тракологу Димитру Попову, оно је настало као "логичан резултат материјализације државотворних процеса, општег и закономерног целокупног развоја и праксе у трачким земљама".

Настанак "државне" организације у југоисточној Тракији

Када је реч о томе шта је претходило Одриској федерацији, ваља бити на опрезу, јер је то ствар о којој је, како наводи академик Бенјамин Исак (, -{Benjamin Isaac}-), професор античке историје на Универзитету у Тел Авиву, подоста шпекулисано у савременој литератури.

Према настаријим очуваним подацима везаним за југоисточну Тракију, тамошњи житељи били су за ондашњи свет најмногољуднији после Индијаца. Међутим, неслога их је учинила разједињеним а самим тим им је онемогућила стварање неког јачег савеза који би превазилазио оквире месних династија. На основу археолошких налаза са тог подручја, констатовано је да су заиста вођени ратови међу завађеним племенима пропраћени у исто време заплењивањем робља, трговином и пустошењем.

Персијском инвазијом на Тракију (крајем 6. столећа п. н. е.) и успостављањем сатрапије Скудре на њеном тлу, подстакнут је даљи развој државотворног уређења код доморадачког живља. Међу њима најдоминантнији су били Одришани, који су убрзо након Даријевог пораза задобијеног у рату са Скитима кренули да запоседају области напуштене од Персијанаца. Под вођством краља Тереса I, они су запоседајући широко пространство североисточно од планине Балкан дошли у додир са Скитима. Међутим, уместо отвореног сукоба дошло је до преговора који је уродио плодом. Прихваћено је да се за границу узме река Дунав, а да би се тај споразум учврстио склопљен је и династички брак између једне од Тересових кћерки (није јој забележено име) и скитског владара Аријапита.

Према Тукидиду, Одризи су имали своје "владаре" још пре појаве Тереса, који је наметнуо власт многим трачким племенима. Ипак, његовим ступањем на политичку сцену процес стварања одришке државе, који је трајао више деценија ако не и векова, учвршћен је и тиме су коначно створени услови за даље ширење у правцу Дунава односно Црног мора.

Исти аутор нам на једном месту даје исцрпан податак на основу којег се могу одредити приближне границе ове државе за време њеног потоњег успона:
Наравно да до њеног формирања, као и успона који је следио пар деценија потом, није дошло тек тако, преко ноћи и без уплитања каквог страног елемента. Истина, низ позитивних фактора је суделовало у томе, од којих вреди издвојити погодан стратешки положај и слободан приступ мореузима (Босфорском и Хелеспонтском) и Пропонтидском приморју. Исто тако, одлучујућу улогу у настанку каквог-таквог облика "државотворног" уређења код Одришана имале су не само грчке колоније и уопште процес колонизације северног Егеја, већ и Стари Исток, на којем је све већи утицај имало Персијско царство под династијом Ахеминида. Међутим, несумњиво је да је источњачко-персијски утицај однео превагу у односу на хеленски. То се најбоље огледа у чињеници да су државна уређења народа са Истока, за разлику од полисног уобичајеног за Старе Грке, по свој прилици више импоновала специфичностима друштвеног уређења заступљеним код Трачана, као и његовој организацији и структури.

 Економија

Тракија је у античко доба обиловала сребром, премда постоји мало доказа о његовој експлоатацији из времена пре персијских освајања. Основна грана привреде свакако је било катунско сточарство, а искоришћавани су и ресурси из дивљине. Успостављање односа средиште-периферија () са егејско-анадолијским светом довело је до експортне трговине. Као посебну субкатегорију она је обухватала и трговину са младим женским робљем, о којој је Херодот оставио један занимљиви податак. Уз то, такође за трговинске сврхе, узгајана је и једна врста профитабилне конопље (-{cannabis sativa}-), која је коришћена за израду одеће и била исто тако цењена као и египатски лан.

Што се пак тиче интерних односа (), особит је престиж размене (). Тукидид наводи да је било потпуно немогуће успоставити било какав договор генерално са свим Трачанима који би уродио плодом, уколико им том приликом нису били уручени дарови (δώρα, -{dora}-) у злату или сребру, извезеним и глатким тканинама и покућству. Тукидодова тврдња наилази на своју потврду код Ксенофонта, који описује једну гозбу на коју га је био позвао Сеут II приликом проласка кроз његову земљу. Уочи обеда гостима је дато на знање да је код краља обичај да му се уручују дарови. Том приликом краљ је дариван белим коњем, младим робом, сребрном чинијом и ћилимом вредним десет мина, а сваки од дариваоца му је обавезно по уручивању дара наздрављао. Овакав начин доласка до материјалних средстава претходио је успостављању монетарне економије и ковању домаћег новца по узору на грчки, што је нарочито било практиковано од стране одришке династије која је ионако убирала висок порез како од подређених племена тако и од грчких приморских апоикија.-{Mitchell.134-35. стр. 134-35}-

Ископавањем којим је почетком 90-их руководио пољски археолог и траколог, Мечислав Домарадски (), откривен је недалеко од данашњег села Ветрен (код Пловдива), на горњем току реке Марице, локалитет Пистирос. Он датира из средине 4. столећа п. н. е. и представља трговиште (εμπόριον, -{emporion}-'') на којем су се окупљали Грци са Тасоса, Маронеје и Аполоније (није искључено да је реч о оној са обали Понта). Важан је због тога што је тамо, између осталог, пронађен двојезични натпис дужине 46 редака настао на двору Котиса I. На основу њега, грчки трговци који су обитовали у Пистиросу имали су да плаћају порез на сва добра која су увозили односно извозили из Тракије.

Види још
 Одризи

Референце

Литература

Примарна литература
Од Хомера

Од Херодота

Од Тукидида

Од Ксенофонта

Секундарна литература

Спољашње везе

Тракија
Историја Балкана
Бивше државе Европе
Бивше монархије
Антика на тлу данашње Бугарске
Бивше државе на Балканском полуострву
Римска вазална краљевства


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Home / Awards / Our 2021 Gadget of the Year is…
Our 2021 Gadget of the Year is…
It's time to reveal our overall Gadget of the Year...
Earlier we revealed the winners of our Stuff Gadget Awards 2021 as well as the winner of our Readers' Gadget of the Year as voted by you.
Now it's time to reveal our choice for Gadget of the Year as chosen by us here at Team Stuff.
There was a decent amount of competition this year and our runners up – which you can read about below – were very close to winning. And if there was a fourth spot up for grabs it was probably taken by the 2021 Apple MacBook Pro.
There were other very good contenders as well. But this year's Gadget of the Year is…
GADGET OF THE YEAR – Pixel 6 Pro
As you may have noticed, we love the iPhone 13 Pro. We've also fawned over the Galaxy S21 Ultra. We're big fans of the Oppo Find X3 Pro. But nobody has done it better than Google over the last 12 months.
The Pixel 6 Pro represents everything a flagship phone should be. It offers brilliant specs, a fun design and some excellent photography smarts – including the Magic Eraser to remove unwanted people from your images, Stalin-style. And it's all powered by a new Google-designed Tensor chip that should mean the company can roll out, even more, features over the next couple of years (though Android 12 is a doozy already).
And this is all before you get to the price, which is superb for a top-of-the-line handset (the lower-priced Pixel 6 is even more impressive on that score). It must be said that Google's sales stats for Pixel phones aren't great, but if there's a handset that can take the fight to Apple and Samsung, this is the one.
Read our Pixel 6 Pro review
SECOND PLACE – Apple iPad Mini (2021)
This is an amazing device. Don't mention the high price or lack of an official keyboard; just think about the incredible power inside such a tiny package. It's the new darling of the iPad range.
Read our Apple iPad Mini review
THIRD PLACE – Sony WF-1000XM4
Last year it was the over-ears that hit an aural high. This year, Sony tightened its grip on the true wireless in-ear crown with these stunners. How do they do it? Rivals would love to know.
Read out Sony WF-1000XM4 review
Check out our other 2021 Stuff Gadget Awards Winners and the Readers' Gadget of the Year
How to watch Samsung Galaxy Unpacked 2023: see the S23 reveal live
At last year's Galaxy Unpacked we got the S22 and Tab S8 series – what will this year's have in store?

redpajama_set_name: RedPajamaCommonCrawl


David found a perfectly intact, dead horseshoe crab.
Morning Glories on Walker Street– no filter, just Mother Nature.
Love the color: flower boxes at Anejo.
All alone in the parking lot.
Posted in Cape, create, Explorer of the World, simplify, Walkdoc. Bookmark the permalink.
Keri Smith's How to Be An Explorer of the World has left its mark on me. I've been taking photos of "stuff" that I might have previously missed while taking my daily walk. I'm learning to take the time to see and document what catches my eye, and then think about why these things capture my attention.
The tourists are gone and life is back to normal: less traffic, no crowds, more quiet. I love it!
Walker Street has eye candy that has weathered storms and time.
For the birds, and me, to appreciate.
Morning Glories: my absolute favorite.
And the simple, but colorful leaves of coleus, never disappoint.
Posted in Cape, create, exercise, Explorer of the World, simplify and tagged eye candy, iphone, keri smith. Bookmark the permalink.


text:
The Cost waren eine Emocore-Band aus Oakland, Kalifornien / USA, die von 1999 bis 2004 existierte.

Geschichte
Alle Mitglieder der Band waren schon vor der Gründung der Band enge Freunde, so dass die Band für sie auch immer ein Freundschaftsprojekt war.

Die Band tourte oft ausgiebig an der Westküste sowie auch überall in den USA.

2002 kam mit Chimera das erste und einzige Album der Band beim Label Lockout! Records heraus. Insgesamt veröffentlichte die Band noch eine EP sowie eine Split-EP und war auf einigen Samplern vertreten.

In einem Bericht mit Interview über die Band 2002 wird über das Album gesagt:

Im Juli 2003 spielte die Band ihr letztes Konzert und löste sich danach auf.

Stil
Die Band spielt einen teilweise recht chaotischen Emo-Stil. Charakteristisch sind die langsamen, melancholischen Parts in ihren Stücken – die durch kurz auftauchende atmosphärisch wirkenden Gitarrenwände unterstützt werden – sowie schnellere, härtere Ausbrüche mit emotionalen Schreien und geschrienen oder halb-geschriebenen Textteilen. Weiterhin finden sich auch eher sehr schnelle Hardcore-Punk typische Stücke mit emotionalem – teilweise geschrienem – Gesang.

Als Einflüsse werden neben vielen Emo- und Hardcore-Punk-Bands die Post-Hardcore-Legende Fugazi oder die No Wave / Noiserock-Band Sonic Youth genannt.

Diskografie

EPs/Splits
 Split mit d.b.s., 7" (1999, Sellout Records)
 El Sueno De La Razon Produce Monstrous (1999, New Disorder Records)

Alben
 Chimera, CD (2002, Lookout! Records)

Samplerbeiträge
 A Little Something for Everyone (New Disorder Records, 2000)
 Another Juan Rides the Bus (Stickrod Tapes, 2000)
 New Disorder Soda (New Disorder Records, 2001)
 Lookout! Freakout Episode 3 (Lookout! Records, 2002)

Weblinks
 Die Seite der Band bei MySpace
 Die Seite der Band beim Label New Disorder Records
 Die Band beim Label Lockout! Records* Interview mit der Band aus dem Jahr 2002

Quellen

US-amerikanische Band
Emoband


The Queensguard captain has a long respawn timer. Form a group to be sure you get the tap (the limit is 5 people).
this has got to be the most frustrating world quest to complete!
For some otherworldly reason these captains seem to have group tag instead of the more popular alternative of faction tag, making these world quests incredibly dull to complete, combined with a fairly long respawn timer (a few minutes).
The captain's respawn timer has been adjusted for quicker respawns.
Fly to Hafr Fjall in Stormheim. Point towards the quest on the map, jump off the ledge next to the flight master, and use a kite. You can land on the top of the tower and kill Gilneas Brigade Captain like this.
Short Video of the World Quest Starting at Hafr Fjall.
This is considered a PvP world quest and will flag you for PvP even if you don't kill other players. First off fly to Hafr Fjall if you have it and run down or glide on over to the tower.. You need to kill the captain that's on top of the tower and get 100% complete by killing the troops/brigade. You can also use the explosive barrel on the ground and it will help stun/kill the troops much easier.
If you don't have Hafr Fjall or Dreadwake's Landing flight points, you can fly to Cullen's Post, go by the ledge that's very close, then use a glider like Rocfeather Skyhorn Kite to glide down to Dreadwake's Landing, then run along the coastline until you get to the area.


Liberty Wellbeing is very proud to be working in partnership with Yoga Reloaded.
Starting in January 2020, Yoga Reloaded will be launching their full 200 hour Yoga Teacher Training Programme for those curious about beginning to teach yoga, with an updated understanding of movement.
From the roots of yoga, you will be introduced to a modern approach to movement where you will look at the science, not the lineage, and break free from the outdated dogma that is not designed for the modern human.
Be part of the revolution of yoga.


P4B Law has extensive experience in preparing cases and conducting the advocacy in Employment Tribunals across the UK. We regularly appear in employment tribunals for employers.
We offer a free 30 minute consultation with regards to any employment law tribunal claim defence. At the end of this consultation we will be able to advise on the appropriate fee structure and projected costs.
Being advised by an experienced employment lawyer who understands the language and procedures of an employment tribunal can greatly improve your chances of success of defending a claim at an employment tribunal.
We are also able to advise on offers of settlement and whether an offer is worth making or accepting.
For an initial assessment of your claim contact us at info@p4b-law.co.uk outlining in as much detail as possible the context of your complaint.


Difference between revisions of "Achievo"
|Full description=Achievo is a Web-based project management and tracking tool for small- to medium-sized companies, aiming to become a fully featured ERP-like system in the future. Features include a time registration module with several statistical tools. Achievo is multi-language, fully customizable, and uses a modular system to allow future extensions. The package is currently available in 13 languages.
It's main website went down in mid-2015. As of 9/2016 it is only available via an archive.org copy of [https://web.archive.org/web/20150319081753/http://www.achievo.org/ achievo.org] and its [https://github.com/atkphpframework/achievo GitHub page], and has [https://github.com/atkphpframework/achievo/issues/21 publicly announced security vulnerabilities] that have been declared [https://github.com/atkphpframework/achievo/issues/17#issuecomment-64930877 won't fix due to being semi-abandoned from lack of time and maintainers].
Achievo is a Web-based project management and tracking tool for small- to medium-sized companies, aiming to become a fully featured ERP-like system in the future. Features include a time registration module with several statistical tools. Achievo is multi-language, fully customizable, and uses a modular system to allow future extensions. The package is currently available in 13 languages.
It's main website went down in mid-2015. As of 9/2016 it is only available via an archive.org copy of achievo.org and its GitHub page, and has publicly announced security vulnerabilities that have been declared won't fix due to being semi-abandoned from lack of time and maintainers.
This entry (in part or in whole) was last reviewed on 30 September 2016.


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The first couple to honeymoon in space will be George and Loretta Whitesides. They are both 100 Virgin Galactic 'Founders' which means they have paid in full to be among the first to fly on Richard Branson's Virgin Galactic Spaceline. A website to mark the occasion has been set up at SpaceLove.org – it will document the preparation and lead-up to their honeymoon flight.
It seems like a great idea and no doubt will be a once in a lifetime trip but I just wonder how romantic space will be as a honeymoon destination?


The Journey Bible Project guides you on a path through the Bible in order to take next steps with Jesus.
MY FAVORITE PART OF THE RETREAT!
I am grateful with everyone who helped make the two weeks I spent in Tulsa, memorable! I will cherish them always.
Join The Journey Bible Project and receive regular blogs that guide you to connect with God in scripture, next steps, and prayer.
Searching for Tulsa Churches or Churches of Christ?
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Sizes Estimate or measure anything's dimensions Body Length (Back)Measure from the back high point of shoulder seam (or from the natural shoulder fold) straight down to the bottom of garment, keeping the tape measure parallel to center back. Body Length (Front)Place the tape measure at the highest point of the shoulder, where the Merchandise joins the neck seam and then the tape measure right down the front of your body to the desired length. BustPut the measuring tape below the under arms and around the fullest part of the bust.
I recently bought this tank in black red color and love them both. They are thick, flattering, and perfect for yoga. As I have a small chest, the front of the tank is lower than my bra, but it doesnt bother me. Very soft, comfortable and breathable. I love these tanks as they are eco-friendly and not harmful to me and for the environment. Would highly recommend.
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Love these tanks! soft and light, But like the other reviewer, I ordered the royal blue and received a royal blue. I am happy about this tank, Thanks.


text:
2018 December 7 Fine European Art - DallasAuction #5384
William Adolphe Bouguereau (French, 1825-1905). La leçon de flûte, 1868. Oil on canvas. 46 x 35 inches (116.8 x 88.9 cm)...
William Adolphe Bouguereau (French, 1825-1905)
La leçon de flûte, 1868
46 x 35 inches (116.8 x 88.9 cm)
Signed and dated lower right: W. Bouguereau 1868
PROPERTY FROM A PROMINENT HOUSTON ESTATE
The artist;
Goupil & Cie, Paris, acquired from the above, December 15, 1868;
Hollender of Brussels, acquired from the above, February 19, 1869;
Hammer Galleries, New York;
Acquired by the present owner from the above, 1973.
Bouguereau's accounts as La leçon de flûte;
Goupil's stockbooks and sales records (no. 3792) as La leçon de musique;
C. Vendryès, William Bouguereau, Paris, 1885, p. 38 as La leçon de flûte;
C. Franquet (Comte de Franqueville), Le premier siècle de l'Institut de France, Paris, 1895, p. 370 as La leçon de flûte;
M. Vachon, William Bouguereau, Paris, 1900, p. 149 as La leçon de flûte;
D. Bartoli and F. Ross, William Bouguereau, Catalogue Raisonné of his Painted Work, Vol. II, New York, 2010, p. 112, no. 1868/15 as La leçon de flûte, pl. 89 illustrated.
In the 1860s, Bouguereau beguiled his audience with Italian pastoral subjects, many featuring the feminine charms of modern day Madonnas. Differing from these many works is The flute lesson where the artist has chosen instead to depict a father introducing his young child to the delights of music. As the patient teacher gently embraces the young one and steadies the instrument, the mastery of Bouguereau's brush breathes such life into the figures that the viewer can almost hear the first tentative notes from the little student's efforts. Bouguereau's accomplished handling can also be seen in the description of the various textures: the hardness of stone, the softness of the child's chemise. Equally compelling is the artist's sureness in the harmony of line and color that produces a wonderful balance between these country folk and the rugged mountain setting.
In its discussion of the musical instruments depicted in this painting, the catalogue raisonné of the artist notes that Bouguereau has portrayed not a flute but a piffero, a double-reed oboe-type instrument played in the mountains of the Apennines and elsewhere in Italy. Also pictured to the right of the young father is a zampogna, a double-chantered bagpipe, a traditional instrument played in The Marches and southward into Sicily. These instruments were often played together and were the livelihood of traveling musicians who were frequently depicted entertaining city folk in popular contemporary pictures. In the present picture, however, the viewer is treated to a tender moment between the musicians at home, playing the traditional music of Italy just for themselves.
Condition Report*: Glue-lined canvas. Under UV light there appears to be two periods of restoration to the man's face. The older period: small touches of inpaint to his left eye and cheek, the more recent period are small touches to his jawline and to the hair above his right eye. There are a few tiny touches of inpaint in and around the child's eyes and right cheek and a few further tiny touches to child's chest, shift, and legs. There is a 1 1/4 x 3/4in area of inpaint below child's left foot. There are tiny, discrete touches of inpaint (to the fine craquelure) scattered in sky and background, minor inpaint to frame abrasion in corners of canvas. Fine craquelure scattered throughout. Framed Dimensions 54 X 43.5 Inches
It would indeed have been much simpler to work with a local auction house, and we had local options. But we never once regretted going with Heritage.
Virgene S.,


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Directions: I-71 South To Glencoe Exit, Turn Right On 127, Left On Spencer.


Pokemon GO Download came in that we are eagerly waiting for. But it does not appear in the Apple store now in our country. when he has entered in the Android store, but there is also an attempt to download the credibility "is not supported for your country" as we get a warning.
Current was officially activated in a number of countries to be active in all countries remained very low.
Important: Download the APK file and install who can play smoothly on your android device, but we do not recommend it at the moment. Risky business is a little more sense to wait for the download store itself.


text:
Passy () is a commune in the Yonne department in Bourgogne-Franche-Comté in north-central France.

Geography
The altitude of the village of Passy is of approximately 95 meters. The surface of Passy is 5.74 square kilometers. The latitude and longitude of Passy are 48.109 degrees North and 3.304 degrees East. Towns in the vicinity of Passy are :

Population
The population of Passy was 304 inhabitants in 2007, a very small increase compared to 1999 ( 301 ). There were 223 inhabitants in 1990, 197 inhabitants in 1982, 219 inhabitants in 1975 and 169 inhabitants in 1968.

See also
Communes of the Yonne department

References

Communes of Yonne


Snow Leopards inhabit cold, rocky, mountaintops across the high elevations of Central Asia. They span from Afghanistan to Kazakstan and Russia in the north to India and China in the east. Although they once populated Mongolia, they have disappeared from certain parts of that area.
Humans are the leading cause in the disappearance of this shy creature, thanks to hunting, habitat loss and retaliatory killings. Climate change also poses the greatest long-term threat to the animal because it could reduce up to 30 percent of the snow leopard's habitat in the Himalayas alone.
In an effort to conserve the snow leopards, many organizations, such as the World Wildlife Fund and the Snow Leopard Trust work with local communities to reduce the number of retaliatory killings.
World Wildlife Fund: Learn more about the Snow Leopard, conservation efforts, and what you can do to help ensure the species' survival.
Snow Leopard Trust: Discover how you can contribute directly to the conservation of Snow Leopards.
Defenders of Wildlife: Get fast facts about this species and conservation efforts currently underway.


It would be really easy if you could look in a book and learn all about how to be the best parent. You can't, though. It would be great if children fit into some textbook mold so you can know what to expect. But they don't. The reality is parenting is about the unexpected and walking around like you have no clue what's going on most of the time, but are just really happy that everyone survived the day. Even if it did require a Deathwish french press coffee in the morning and only God knows how many other cups of instant caramel latte you consumed throughout the day.
I had my first child at 18. Imagine all of that insane wisdom that came with it for my friends in their future. When they mentioned a birth plan, I laughed. There is no planning it. There's a "dream"; not a plan. I guarantee for the most part, nothing will go as planned. My birth plan at 18? Not to die in childbirth. I like to set the bar low. My birth plan at 29? Can't say that it was much different. Just simply having a healthy baby and not dying in a pool of my own blood was sufficient enough. When I hear people start planning out their home birth, I do roll my eyes. I do judge. That's great and all, but what about the unexpected? People need to learn to not be so rigid if they are going to have kids. Because the unexpected is your life now.
I have been very open about the differences between my oldest and my youngest. My youngest, who seems to be in an endless loop of observations and interventions. When he went to preschool for his IEP, I was happy that he was going to get the help that he needed. Or at least that first year, that they would be able to see that he needed speech therapy and set it up. They did. He got it for his second year of preschool. Then he satisfied his IEP and he went into Kindergarten with no plan. Aside from some anxiety issues, he didn't do terribly in Kindergarten. For a kid who never spoke before 4, he was not only on par with his class in speech, he was exceeding some. He was a little "active" and had "attention" problems, but we all attributed them to his sensory disorder. It was fine.
Then at his parent teacher conference back in November, it was suggested that they do an occupational therapy evaluation on him because of some issues such as his handwriting and need for noise cancelling headphones in class. His teacher has been great. Eventually I finally got a letter a couple of months later. I was anxious to see what the results were. If they mailed it, that means they didn't need to setup a meeting for an IEP, right? It turns out, it was a letter to inform me that there was more testing needed. It was stressful. When you have been dealing with evaluations for practically his whole life for various things, it does get into your head a little. It isn't about me thinking he's less than anything because of these interventions. It isn't about what other kids will think of him. It's all about wanting to do the best for your child. If he needs the help, I'm going to make sure that he gets what he needs. I want him to succeed. I want him to thrive. I just hate the waiting game, because that means you just sit around worrying until the results come in.
No one expects watching their young child go into surgery when they are planning out the nursery. No one expects struggling through evaluations when they feel that first kick. No one expects that they are going to agonize over every decision that they make because who knows what the repercussions are for their future. There is no greater responsibility in the world than being a parent. My best advice that I have ever given to any other parent is: "At the end of the day, as long as you did everything that you could to make sure everyone made it out alive, you did exactly what you were supposed to." You can't control what happens, but you can give them the best odds possible.
This entry was posted in Writing and tagged Parenting, Writing.


We have found 3 chiropractors in Marina and chiropractic clinics in Marina California.
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"There's a piece of me missing, because of the stress," she says, standing in t he middle of her dimly-lit kitchen, the air thick and sticky in the overpowering August heat.
The kitchen sink is about half full of dirty, brownish water; the drain hasn't worked for the past year, and despite putting in a request to have it repaired five months ago, it's still blocked.
The bathroom light fixture has been broken for months, while the outer glass of a double-paned window is missing off the first-floor room Mohamed's disabled 18-year-old daughter sleeps in, in a hospital bed. In the winter, cold air seeps in, making the teenager ill.
They are now one of more than 100 families facing eviction in Heron Gate, a neighbourhood in the south end of Ottawa, in what a local tenants' rights group says is the "largest forced displacement in Canada" in recent history.
Several families have found new homes, but despite seeing other places and scouring the listings every day, Mohamed hasn't found anything affordable that can accommodate her large family and its needs.
"I came to Canada thinking it was an escape, only to realise it's like torture," she says in Somali through a translator.
"People are playing with our lives," adds her husband, Jamale Ibrahim.
"Winter and the co ld is coming," Mohamed says, "and we have nowhere to go."
Heron Gate is one of Ottawa's most diverse communities, home to a large number of residents of Somali and Arab backgrounds. It comprises a mix of townhouses, medium-sized buildings and tall apartment towers.
The company, which operates in 27 cities across Canada, said the townhouses are no longer viable and it gave tenants 120 days - until September 30 - to find alternative housing.
That's the required notice period under the law in the province of Ontario, as Timbercreek moves ahead with a mass development project for the neighbourhood.
Under the provincial Residential Tenancies Act, a landlord can issue eviction notices for the purpose of demolishing a pr operty or conducting extensive renovations, among other reasons.
The company is offering displaced tenants three months' rent, as well as an additional $1,530 ($2,000 Canadian) compensation, up from an original offer of $1,150 ($1,500 Canadian), to move. It is also negotiating reduced rates with moving companies, and employing a relocation team to help tenants find other properties, the company said.
"Timbercreek is going beyond the requirements of the law in providing relocation assistance," its statement read.
"Ti mbercreek's vision has always been for Heron Gate to be a diverse and sustainable community. A project of this magnitude, however, requires that tenants relocate during the revitalisation process. We will offer all impacted residents the right to return to Heron Gate when the redevelopment is completed."
However, several families who remain in Heron Gate say they're struggling to find alternative housing that meets their needs and doesn't exceed their budgets.
I came to Canada thinking it was an escape, only to realise it's like torture."
A recent article in the Ottawa Citizen bore the headline, "Rising rents and cutthroat competition". The newspaper reported that "Ottawa's residential rental market has recently become red-hot, with demand and prices ballooning and vacancies increasingly scarce".
Asked to comment on the complaints, the company said it "continues to maintain Heron Gate units and quickly respond to repair requests".
But a resident, Amina, who didn't give Al Jazeera her last name out of fear of reprisals, keeps her food in closed rubbish bags because her kitchen is overrun by cockroaches.
"What we're living right now is not a life," the mother of four says.
At that time, dozens of families were displaced when about 80 townhouses were razed. Today, a crane and a construction site stand in their place, soon to be transformed into "three, six-storey multi-residential buildings&quo t;.
"They didn't tell us they intended to demolish [this home, too]," Ali tells Al Jazeera, just outside his front door, the construction site where his previous home stood towering over his right shoulder in the distance.
If they had, Ali says he wouldn't have stayed in Heron Gate.
"It's Canada, it's a highly respected nation for human rights. This is worrying ... This is something we never expected to happen in Canada," he says.
For many, especially new immigrants who don 9;t speak English and lack a local support network, the community is a lifeline.
"We support each other. This is our culture, and this is how our community works," Ali says.
It's Canada, it's a highly respected nation for human rights. This is worrying ... This is something we never expected to happen in Canada.
"A lot of them are dependent upon the community via translation work, for moral support, for babysitting while they go out to find jobs or go to English class," says Egal, who grew up in the neighbourhood.
< p>"Taking them away from this neighbourhood and moving them to the east end or moving them to the west end - where they don't have those community ties for the day-to-day necessities of life - is extremely debilitating," she tells Al Jazeera.
He hasn't found another place y et, and Mohamed says he can't imagine living anywhere else.
"My community is here," he tells Al Jazeera. "It's very stressful. It's not easy to move."
If the residents that remain in Heron Gate refuse to move by September 30, Timbercreek can go to Ontario's Landlord and Tenant Board to apply for a formal eviction order. It must then apply to the city for permits to demolish the units.
The company says it will "design a community that fits the needs of the people in the community", but it doesn't have a plan in place yet for the type of units that will replace the townhouses.
Because landlord and tenant issues fall under provincial laws in Ontario, the City of Ottawa says it can't intervene in the eviction process. The city only gets involved once a landlord applies for a demolition permit or files a development proposal.
A spokesperson for Mayor Jim Watson said his focus has been "to encourage Timbercreek to increase compensation to affected households, and to provide families with more time to look for a new home".
"Mayor Watson sought, and received reassurance from Timbercreek representatives, that the company is focused first and foremost on helping families relocate and not on the eviction process," Livia Belcea told Al Jazeera in an email.
About 10,500 families are currently on a subsidised housing waiting list, which gives them access to "rent-geared-to-income assistance and associated housing b enefits", said Shelley VanBuskirk, director of the city's housing department.
Additionally, low vacancy rates, coupled with high demand in the private rental sector and higher rental costs, "are placing additional demands on the wait list for social housing and for affordable housing options", VanBuskirk said.
Back in Heron Gate, residents have asked Timbercreek for more time â€" as much as a year's leeway - to find other places to go.
The residents who remain are in a "desperate situation", says Abdullahi Ali. They called on the city to do more to support the people facing eviction.
Daniel Tucker-Simmons, a lawyer representing some of the residents, said a request for accommodation was sent to Timbercreek's lawyer and to Mayor Watson.
These mass evictions are having a differential, a discriminatory, impact on my client s because of their race, because of their ethnicity, because of their religion.
They're asking that if parts of the neighbourhood need to be demolished, that the current tenants receive more relocation assistance, and be guaranteed a right to return to units with similar rents once the redevelopment is finished. It also asks that units be preserved if they can be.
The request is currently being considered, Tucker-Simmons said. If he doesn't receive a response, he said the complaint could be sent to the Ontario Human Rights Tribunal, which adjudicates on alleged human rights code violations.
"These mass evictions are having a differential, a discriminatory, impact on my clients because of their race, because of their ethnicity, because of their religion," he said.
While they wait to see what happens next, a sense of anxiety runs through every h ome and every family, says Ali's wife, Saido.
"We don't say hi [to each other]. We say, 'Did you find a house?' … Our minds [are] always [on this]."


text:
\section{Introduction}

Fluctuation-dissipation theorems (FDT) \cite{fdt} establish a beautiful and useful connection between response functions and noise.
They have a long history beginning with the Einstein relations and Nyquist formula and culminating in Kubo's linear response theory.
The standard FDT applies in thermal equilibrium only and much attention has been focused on its violations in nonequilibrium conditions. It became gradually clear that the FDT forms a special case of more general fluctuation theorems valid for various classes of nonequilibrium systems \cite{fdt}. Well-known examples are the Jarzynski equality \cite{J} and the Agarwal formula \cite{agarwal}.

The foundations of the linear response theory and the FDT are the Gibbs distribution and causality. According to the causality principle, there is a fundamental asymmetry between the past and the future since the future depends on the past but the past is not influenced by future events. This imposes crucial restrictions on response to any perturbations. In this paper we address chiral systems \cite{footnote1} which possess a similar asymmetry between left and right so that what happens on the right affects what later happens on the left but not {\it vice versa}. Obviously, this may only be possible if excitations can propagate just in one direction. Such chiral transport can occur in topological states of matter, a primary example being a low-temperature 2D electron gas in the conditions of the quantum Hall effect \cite{wen} (QHE). The gas is gapped in the bulk and its low-energy physics is determined by 1D chiral edge excitations. In the simplest QHE states all edge modes have the same chirality and hence the current flows in one direction only, e.g., clockwise. We show that in such chiral systems a Nyquist-type formula (\ref{1}) holds for the low-frequency current noise and {\it nonlinear} conductance even far from equilibrium. This far-from-equilibrium FDT is different from a more general Agarwal formula \cite{agarwal} for non-chiral systems which connects quantities that do not generally have an obvious physical meaning and cannot be easily extracted from experiment.

Our results apply beyond QHE. As usual in statistical mechanics, the simplest example of a chiral system comes from the physics of ideal gases. Consider a large reservoir filled with an ideal gas. A narrow tube with smooth walls and an open end is connected to the reservoir. Molecules can leave the reservoir through the tube. The projections of their velocities on the tube axis cannot change. Hence, they can only move from the reservoir to the open end of the tube and the system is chiral. Imagine now that molecules can escape through the walls of the tube with the probability depending on their position and velocity. A relation similar to Eq. (\ref{1}) can then be derived for the particle flux through the walls and the fluctuations of the fluxes through the walls and the open end of the tube. We discuss that relation in the appendix. The gas example is one-dimensional. Chiral systems are also possible in 2D. Indeed, topological states of matter with gapless chiral excitations on a 2D surface of a 3D system are possible (e. g., Ref. \onlinecite{FL} and related systems). In such systems, charge can propagate in both directions along one of the coordinate axes but only in one direction along the second axis. Moreover, chiral models can emerge beyond conventional condensed matter physics. For example, statistical mechanics has been used to describe traffic \cite{traffic}. A chiral model describes traffic on a network of one-way streets with no parking as long as no traffic jams form.

Chiral edge states in QHE are of particular interest. It was proposed that non-Abelian anyons exist in QHE at some filling factors \cite{pf}, such as 5/2 and 7/2. If the prediction is true this will have major implications for fundamental physics and quantum information technology \cite{pf}. However, the nature of the 5/2 state remains an open question. Competing theories predict both Abelian and non-Abelian statistics [see Ref. \onlinecite{wb} for a review of proposed states].  Some of the proposed states have chiral edges and others do not. In particular, all published proposals for Abelian states are chiral \cite{footnote3}. Thus, testing chirality of QHE edges is important in this context \cite{feld-li,heiblum} and the theorem (\ref{1}) will be useful for that purpose. On the other hand, it is generally believed that the edges of the Laughlin states at $\nu=1/(2p+1)$ are chiral. This expectation is supported by the chiral Luttinger liquid model \cite{wen} (CLL). However, CLL faces major challenges from experiment (for a review, see Ref. \onlinecite{hr}). For example, it cannot explain observed quasiparticle transmission through an opaque barrier without bunching into electrons \cite{comforti,kf}. Thus, it is important to test major assumptions of CLL. One of them is chirality. Our theorem can be used for testing that assumption. Eq. (\ref{1}) has already been verified \cite{privite} in the limiting cases of $T\gg V$ and $V\gg T$.

\begin{figure}[h]
\includegraphics{fig1_fdt.eps}
\caption{Three-terminal setup. A quantum Hall bar is connected to source S at the voltage $V$. Charge tunnels into terminal C. The arrows represent the directions of the chiral edge modes.}
\end{figure}

A nonequilibrium FDT can be formulated for chiral systems in various geometries. Below we focus on the simplest geometry illustrated in Fig. 1. We consider a quantum Hall bar connected to the source (S) and drain (D) terminals. The impedance between the bar and the outside world is small. Long-range Coulomb forces are screened by a gate (this also ensures the absence of bulk currents \cite{bulk-curr}). Excitations propagate from the right to the left on the lower edge and in the opposite direction on the upper edge. The system size is much greater than the magnetic length; we assume that the chiral edges are far apart and do not influence each other. A third terminal C is connected to the lower edge through a tunneling contact. The details of the contact are unimportant and our results apply no matter how high or low the tunneling current $I_{\rm T}$
into terminal C is. A voltage bias $V$ is applied between the source and C.
The temperature $T$ of the source and drain reservoirs may be different from the temperature of reservoir C. Our results do not depend on the latter temperature or the nature of conductor C. $V$ and $T$ should be much lower than the QHE gap (otherwise, QHE is absent and the system is not chiral).
We consider the current noise in terminal C,
$S_{\rm C}=\int dt \langle \Delta I_{\rm T}(t)\Delta I_{\rm T}(0)+\Delta I_{\rm T}(0)\Delta I_{\rm T}(t)\rangle$, and in the
 drain,
$S_{\rm D}=\int dt \langle \Delta I_{\rm D}(t)\Delta I_{\rm D}(0)+\Delta I_{\rm D}(0)\Delta I_{\rm D}(t)\rangle$, where  $I_{\rm D}$ is the electric  current in the drain and $\Delta I=I-\langle I\rangle$, and derive the relation
\begin{equation}
\label{1}
S_{\rm D}=S_{\rm C}-4T\frac{\partial I_{\rm T}}{\partial V}+4GT,
\end{equation}
where $G$ is the Hall conductance of the quantum Hall bar without a tunneling contact. This is the main result of the paper.

A similar formula in a different geometry was obtained in Refs. \onlinecite{a,b,c} for the exactly solvable CLL model. As seen below, Eq. (\ref{1}) holds independently of the integrability of a model and does not rely on CLL. Only chirality matters. This point is disguised in the model \cite{a,b,c} since the same solvable Hamiltonian describes a chiral system with tunneling between quantum Hall edges and a nonchiral quantum wire with an impurity.

We derive Eq. (\ref{1}) below. In the next section we give a simple heuristic derivation. Section III contains a full quantum proof of the fluctuation-dissipation theorem. We summarize our results and discuss their experimental implications in Section IV. Appendix contains a derivation of Eq. (\ref{1}) in an ideal gas model.

\section{Heuristic derivation}

Equation (\ref{1}) does not contain the Planck constant and so we first give its   heuristic classical derivation. The current $I_{\rm D}=I_{\rm L}-I_{\rm U}$ is composed of the current $I_{\rm L}$, entering the drain along the lower edge, and the current
$I_{\rm U}$ on the upper edge. These currents are uncorrelated and hence the noise in the drain is the sum of the noises of $I_{\rm U}$ and $I_{\rm L}$:
\begin{equation}
\label{2}
S_{\rm D}=S_{\rm U}+S_{\rm L}.
\end{equation}
The noise on the upper edge is the same as in the symmetric situation without tunneling into C. In the latter case the noise $S_{\rm D}$ is given by the Nyquist formula. Thus, $S_{\rm U}$ is one half of the equilibrium Nyquist noise,
$S_{\rm U}=2GT$.
In order to evaluate $S_{\rm L}$ we note that in a steady state there is no charge accumulation on the lower edge and hence the low-frequency component of the current, absorbed by the drain, $I_{\rm L}=I_{\rm S}-I_{\rm T}$, where $I_{\rm S}$
is the low-frequency part of the current, emitted from the source. Thus,
\begin{equation}
\label{3}
S_{\rm L}=S_{\rm C}+S_{\rm S}-2S_{\rm ST},
\end{equation}
where the cross-noise $S_{\rm ST}=\int dt \langle \Delta I_{\rm T}(t)\Delta I_{\rm S}(0)+\Delta I_{\rm S}(0)\Delta I_{\rm T}(t)\rangle$ and the noise of the emitted current equals one half of the Nyquist noise because of chirality,
\begin{equation}
\label{4}
S_{\rm S}=S_{\rm U}=2GT.
\end{equation}
We are left with the calculation of the cross-noise. The tunneling current depends on the average emitted current $GV$ and its fluctuations $I_{ \omega}$.
We assume that the central part of the lower edge has a relaxation time $\tau$. It is convenient to separate the fluctuations of $I_{\rm S}$ into fast, $I^{>}$, and slow, $I^{<}$, parts.
$I^{<}$ contains only frequencies below $1/\tau$. An instantaneous value of the tunneling current $I_{\rm T}(t)$
depends on the emitted current within the time interval $\tau$.  $I^{<}$ does not exhibit time-dependence within such time interval. Hence, it enters the expression for the tunneling current in the combination $GV+I^{<}$ only,
$I_{\rm T}=\langle I(GV+I^{<},I^{>})\rangle$, where the brackets denote the average with respect to the fluctuations of $I_{\rm S}$. According to the Nyquist formula for the emitted current, its  harmonics with different frequencies have zero correlation functions: $\langle I_\omega I_{-\omega'}\rangle\sim\delta(\omega-\omega')$.
 For the sake of the heuristic argument we will assume a Gaussian distribution of $I_S$ and hence independence of its high and low frequency fluctuations (no such assumptions are needed in a general proof).
After averaging with respect to the fast fluctuations of $I_{\rm S}$ we can write $I_{\rm T}=\langle J(GV+I^{<})\rangle$, where $J$ is obtained by averaging over $I^{>}$. $I^{<}$ corresponds to a narrow frequency window and can be neglected in comparison with $GV$, i.e., $I_{\rm T}=
J(GV)$. For the calculation of the cross-noise we expand $J(GV+I^{<})$ to the first order in $I^{<}$ and obtain
\begin{equation}
\label{5}
S_{\rm ST}=\langle I_{T,\omega}I_{-\omega}+I_{T,-\omega}I_{\omega}\rangle=\frac{\partial J(GV)}{G\partial V}\langle I_\omega I_{-\omega}+I_{-\omega}I_{\omega}\rangle=2T\frac{\partial I_{\rm T}}{\partial V},
\end{equation}
where we used the Nyquist formula for the fluctuations $I_{\omega}$ of $I_{\rm S}$. A combination of Eqs. (\ref{2}-\ref{5}) yields the desired result (\ref{1}) for nonequilibrium steady states.
Note that only low-frequency particle current conserves
along the edge in our screened conductor and hence it is plausible to expect that $I_{\rm T}$ depends only on harmonics $I_\omega$  with $\hbar\omega<\hbar\omega_{\rm cutoff}\ll T $.  In such case the assumption
of the Gaussian distribution and independence for $I_{<,>}$ is not needed since the same results can be obtained from the lowest order expansion in $I_{<,>}$.

The above argument easily generalizes to other geometries with many terminals and/or tunneling between QHE edges.

One can estimate noises $S_{\rm D}$ and $S_{\rm C}$. We assume that $V\sim T\approx 100mK$. These are realistic parameters for noise experiments. The injected current $I_S\sim e^2V/h$ since $G\sim e^2/h$. We assume that the transmission probability to conductor C is of the order of $1/2$. Thus, $I_T\sim I_S$. Then the second and third terms on the right hand side of Eq. (1) are of the order of $e^2[kT/h] \sim e^2[eV/h] $. The noise $S_C$ can be estimated from the fact that its physical meaning corresponds to the ratio of the fluctuation $\langle\Delta Q^2\rangle$ of the transmitted charge to
the time $\Delta t$ over which the charge was transmitted to C. The probability of a single tunneling event during the time interval $\Delta t \sim h/eV$ is of the order of $1/2$. Thus $S_C\sim e^2/\Delta t\sim e^2 [eV]/h$ and has the same order of magnitude as the second and third terms on the right hand side of Eq. (1). Finally, $S_D$ has the same order of magnitude. This corresponds to the noises of the order of $10^{-28}A^2/Hz$. This is a very small number but even lower noises are measured in the state of the art experiments in the field.

Counter-propagating edge modes break Eq. (\ref{1}). Imagine that an ``upstream" neutral modes propagates in the direction, opposite to that of the charged mode. Each tunneling event into C creates a neutral excitation
that brings energy $\sim V$ back to the source and heats it. This increases the noise, generated by the source, and raises the effective temperature above $T$ in Eq. (\ref{1}).  We would like to emphasize that the bulk of the source remains at the temperature $T$ since the heat capacity of the source is large. However, the bulk plays relatively little role in noise generation in our system due to a relatively low resistance of a massive bulk conductor.

\section{Proof of the nonequilibrium FDT}

We now turn to a full quantum derivation of Eq. (\ref{1}). To simplify notations we will omit the Boltzmann constant and $\hbar$ from the equations below.
As is clear from the above classical argument, the drain potential has no effect on the noises. It will be convenient to assume that the source and drain potentials are equal. It will also be convenient to assume that at the initial moment of time $t=-\infty$ there was no tunneling or other interactions between conductor C and the QHE subsystem containing the quantum Hall bar and the source and drain reservoirs. Thus, the system is initially described by the Hamiltonian $H_0=H_{\rm C}+H_{\rm H}$, where $H_{\rm C}$ and $H_{\rm H}$ denote the Hamiltonians of the two subsystems: conductor C and the QHE subsystem respectively. At later times the Hamiltonian includes an interaction term, $H=H_0+H_{\rm I}(t).$
One of the effects of the interaction is charge tunneling between the QHE subsystem and conductor C.
We assume that the interaction $H_{\rm I}$ becomes time-independent well before the moment of time $t=0$ and the system is in its nonequilibrium steady state at $t=0$. The steady state depends on the voltage bias $V$, temperature $T$
and the temperature $T_{\rm C}$ of conductor C. All these energy scales must be lower than the QHE gap. Otherwise, the system is unlikely to allow a chiral description. We do not assume any special relation between the energy scales.
If $T=T_C$ and $V=0$ then the system is in equilibrium. If $T-T_C\ll T$ and $V\ll T$ then the system is close to equilibrium. Otherwise, the system is far from equilibrium. Our main result (\ref{1}) applies in all those cases.

\subsection{Chiral systems}

A general definition of a chiral system is the following: Consider a system     whose Hamiltonian has a time-dependent contribution
$H_t=\int_{-\infty}^ydxh(x,t)$, where the integration extends to the left of point $y$. In a chiral system, local observables to the right of  point $y$
do not depend on the form of  $h(x,t)$ for any initial conditions.

In what follows, we will not need the most general definition. Instead, we will focus on one particular observable, current $I_{\rm S}$, emitted from the source to the lower edge. We are only interested in the low-frequency regime.
In that limit, the precise choice of the point, where the current $I_{\rm S}$ is measured, is unimportant due to the charge conservation. The same low-frequency current flows in all points of the lower edge between the source and conductor C. Let us select a point A in the gapped region in the bulk of the QHE liquid and a point B below the QHE bar (Fig. 2). The same current $I_{\rm S}$ must flow through any line, connecting A and B. This remains true, even if the line does not cross the lower edge and instead goes through the source (and the boundary between the source and the QHE liquid). Thus, $I_{\rm S}$ can be defined both in terms of the edge and source physics.

\begin{figure}[h]
\includegraphics{fdt_fig2.eps}
\caption{The same low-frequency current $I_{\rm S}$ flows through both dashed lines.}
\end{figure}

The chirality assumption means that the average emitted current $I_{\rm S}(t)$ does not depend on the presence of conductor C for any initial conditions. In other words, the expression ${\rm Tr}\rho(t=-\infty)I_{\rm S}(t)$, where $\rho$ is the initial density matrix,
is the same when the Heisenberg operators

\begin{equation}
\label{defISt}
I_{\rm S}(t)=[{\rm T}\exp(-i\int_{-\infty}^t V(t)dt)]^{-1}I_{\rm S}(-\infty)[{\rm T}\exp(-i\int_{-\infty}^t V(t)dt)],
\end{equation}
where $I_{\rm S}(-\infty)$ is a Schr\"odinger operator,
 are defined in terms of the Hamiltonians $V=H_0$ and
$V(t)=H_0+H_{\rm I}(t)=H$. The latter can be true for a general $\rho(-\infty)$ only if the Heisenberg operators
$I_{\rm S}(t)$ are the same in the presence and absence of conductor C. Note that this property is satisfied in  the chiral Luttinger liquid model. To the right of conductor C, the CLL  action
assumes the form $L=m\hbar/(4\pi)[\partial_t\phi\partial_x\phi-v(\partial_x\phi)^2]$,
where $v$ is the velocity of the edge excitations, the charge density $q=e\partial_x\phi/(2\pi)$ and the current operator $I_{\rm S}=vq$. From the equation of motion, $\partial_t\partial_x\phi-v\partial_x^2\phi=0$, we find that
the electric current on the right of conductor C, $I_{\rm S}=I_{\rm S}(t+x/v)$, depends only on the initial conditions on the right
and is not affected by the form of $H_{\rm I}(t)$.
The same property is satisfied in any other chiral conformal theory and in many other situations.
For example, the chirality property of the operator $I_{\rm S}$ survives, if any changes are introduced into the above CLL action to the left of the point, where $I_{\rm S}$ is measured.
The chirality assumption also holds for QHE edges with several modes of the same chirality but breaks down, generally, if counter-propagating modes are present, as e.g., in the anti-Pfaffian state \cite{apf1,apf2} proposed at $\nu=5/2$.

Edge reconstruction \cite{Wen-Chamon} may result in ``net chiral" edges that are not chiral. For example, a pair of counter-propagating integer QHE modes can emerge on a $\nu=1/3$ edge. In general, this breaks chirality.
However, in practice, disorder is likely to localize such mode pairs and restore chirality.

\subsection{Initial density matrix and Heisenberg current operator}

At the time $t=-\infty$ there is no interaction between conductor C and the QHE subsystem that includes the source, drain and 2D electron gas.
Hence, the initial density matrix $\rho(-\infty)=\rho_{\rm H}\rho_{\rm C}$ factorizes into a product of the initial density matrix $\rho_{\rm C}$ of conductor $C$ and the initial density matrix $\rho_{\rm H}$ of the QHE subsystem.
Each of them corresponds to an independent Gibbs distribution determined by an appropriate reservoir. At later times the subsystems interact, the steady state depends on both reservoirs, and the factorization property no longer holds
in the Schr\"odinger representation.
Thus, it will be convenient for us to perform calculations in terms of the initial density matrix because of its simpler structure \cite{foot3}. This means that we will use the Heisenberg formalism so that all time dependence is placed into
the operators of observables. The chirality property will allow us to extract considerable information about the matrix elements of the Heisenberg operator $I_{\rm S}(t)$ and prove Eq. (\ref{1}).
We would like to emphasize that the Hamiltonian has a time-dependent piece $H_{\rm I}(t)$ and this piece enters the definition of all Heisenberg operators.
The presence of that piece is crucial for the difference between the density matrices in the Heisenberg and Schr\"odinger representations.
We will omit the time argument
in $\rho_{\rm H}$ and $\rho_{\rm C}$. It will be always understood that these are initial density matrices at $t=-\infty$. In all calculations below, $\rho$ is also taken at $t=-\infty$.
Certainly, in the Heisenberg representation, the density matrix does not depend on time.

Our approach resembles the Keldysh formalism, where all correlation functions are also expressed in terms of the initial density matrix. In the Keldysh technique, if the interaction is adiabatically turned on the initial density matrix describes free particles and hence factorizes into a product of single-particle density matrices. A difference from our approach consists in the application of the interaction representation in the Keldysh perturbation theory. The average of any properly time-ordered product of creation and annihilation operators is known exactly in the interaction representation. This allows development of a diagrammatic technique. We use the Heisenberg representation instead and rely on special properties of the matrix elements of the operator $I_{\rm S}$
in the basis, in which the initial density matrix is diagonal.

Any density matrix is Hermitian and can be diagonalized. Hence,

\begin{equation}
\label{er1}
\rho_{\rm H,C}=\sum \rho_{{\rm H,C}n}|n_{\rm H,C}\rangle\langle n_{\rm H,C}|
\end{equation}
 and

\begin{equation}
\label{er2}
\rho(-\infty)=\sum \rho_{n}|n\rangle\langle n |,
\end{equation}
 where

\begin{equation}
\label{er3}
\rho_n=\rho_{{\rm H}n'}\rho_{{\rm C}n''}
\end{equation}
and

\begin{equation}
\label{er4}
|n\rangle=|n_{\rm H}'\rangle|n_{\rm C}''\rangle,
\end{equation}
where the states $|n_{\rm H}'\rangle$ and $|n_{\rm C}''\rangle$ are selected from the Hilbert spaces of the QHE system and conductor C respectively.
$\rho_{\rm H}$ is a Gibbs distribution, $\rho_{{\rm H}n}\sim \exp(-E_n/T-|e|VN_n/T)$, where $N_n$ is the number of electrons in the QHE subsystem, $T$ the temperature of the source and drain reservoirs and $E_n$ are the eigenenergies of the eigenstates $|n_{\rm H}\rangle$ of the quantum Hall subsystem  {\it before} the tunneling contact was turned on, {\it i.e.}, $|n_{\rm H}\rangle$ are eigenstates of the Hamiltonian $H_{\rm H}$ with particle numbers $N_n$. We do not make assumptions about $\rho_{\rm C}$.
Our proof applies as long as the initial density matrix $\rho(-\infty)$ factorizes and the initial density matrix of the QHE subsystem $\rho_{\rm H}$ is given by the Gibbs distribution. In practice, the initial density matrix
$\rho_{\rm C}$ is also likely to be a Gibbs distribution. To avoid a possibility of confusion, we emphasize that all states in the bases $|n_{\rm H}\rangle$ and $|n_{\rm C}\rangle$ are time-independent. Thus, they are no longer eigenstates of the time-dependent Hamiltonian after
the interaction $H_{\rm I}$ has been turned on.

If the interaction $H_{\rm I}(t)$ is never turned on then $I_{\rm S}(t)$ acts in the Hilbert space of the QHE subsystem and hence its nonzero matrix elements are always diagonal in the basis of $|n_{\rm C}\rangle$.
The chirality property means that the same restriction applies to nonzero matrix elements of $I_{\rm S}(t)$ even {\it after} the interaction $H_{\rm I}(t)$ has been turned on since $I_{\rm S}(t)$ must be the same in the presence and absence of the interaction $H_{\rm I}(t)$.  The emitted current operator commutes with the number $N$ of the particles in the quantum Hall subsystem since it describes particle transfer between the source and the edge. Thus, in the absence of the interaction $H_{\rm I}(t)$, it has nonzero matrix elements only between states $|n_{\rm H}\rangle$  with the same $N_n$. Again, the chirality property means that the same restriction on nonzero matrix elements applies even {\it after} the interaction has been turned on.
  Before the interaction between the quantum Hall bar and subsystem C has been turned on, it is easy to write the time-dependence for matrix elements of any operator acting in the Hilbert space of the QHE subsystem: $\langle n|O_{\rm H}(t_1)|m\rangle=\exp(i[E_n-E_m](t_1-t_2))\langle n|O_{\rm H}(t_2)|m\rangle$.
The same relation would apply at all times, if the interaction $H_{\rm I}(t)$ were never turned on.
The chirality property means that the emitted current operator $I_{\rm S}(t)$ exhibits exactly the same time-dependence at any times, if the interaction $H_{\rm I}(t)$ is turned on and if it is not. This applies both before and
{\it after} the tunneling between two subsystems has been turned on. Setting $t_2=0$ in the above relation, we obtain

\begin{equation}
\label{ISt}
\langle n|I_{\rm S}(t)|m\rangle=\exp(i[E_n-E_m]t)\langle n|I_{\rm S}(0)|m\rangle.
\end{equation}

\subsection{Voltage bias}

A standard way to include voltage bias in mesoscopic systems is based on the Landauer-B\"uttiker formalism: One assumes that the tunneling term is initially absent and then turned on and that the lower edge is initially at equilibrium with the reservoir with the chemical potential $V$. We will use a mixed Kubo-Landauer formalism to determine the response of $I_{\rm T}$ to a small change $\delta V$ of the voltage bias. It will allow us to reduce the problem of nonlinear response to $V$ to the linear response to $\delta V$.
In the mixed Kubo-Landauer language, an additional electromotive force $\delta V$ is generated by an infinitesimal time-dependent vector potential $\delta {\bf A}$.

\begin{figure}[h]
\includegraphics{fdt_fig3.eps}
\caption{(color online). Illustration of the bias voltage. $\delta{\bf A}$ and $\delta {\bf E}$ are applied in the region between two solid vertical lines. In the example in the figure the region with $\delta{\bf E}$ crosses both the source (shaded) and the gapped QHE region (white). $\delta\phi$ is constant on the vertical dashed line. $\delta\phi=0$ in point Q and $\delta\phi=\delta V$ in point P.}
\end{figure}

We assume that different contacts are connected with infinite reservoirs at different electrochemical potentials. Their difference determines the voltage bias $V$: the source electrochemical potential is $V$ and the potential of
conductor C is 0.
 A small change of the bias $\delta V$ can be introduced with an electric field described by a time-dependent vector potential, $\delta {\bf E}=-1/c d\delta {\bf A}/dt$. The electric field is applied in a finite part of the source terminal (Fig. 3) and cannot affect chemical potentials of the infinite reservoirs. The chemical potentials determine electric potentials of the reservoirs because of charge neutrality. The distribution of charges certainly changes in the middle of the conductor in the presence of $\delta {\bf A}$, so the electrostatic potential $\phi$ also changes. However, the potential difference between the reservoirs does not. The magnetic field must be time-independent inside the sample as required by the restrictions on e.m.f. sources in the circuit theory. In other words, $\delta{\bf B}= {\rm curl} \delta{\bf A}=0$ inside the conductor.
Hence, the integral of $\delta {\bf E}$ does not depend on the choice of a path inside the conductor at fixed positions of its ends. If a path PQ begins in the infinite source reservoir and ends on the opposite side from the region with the field $\delta{\bf E}$ (Fig. 3) then $\int_P^Q d{\bf r}\delta{\bf E}=\delta V$. Hence, $\delta{\bf A}=ct\times{\rm grad}\delta\phi$, where $\delta\phi=\delta V$ in the source reservoir and $\delta\phi=0$ far on the left in the quantum Hall region (Fig. 3). As a consequence, the vector potential $\delta{\bf A}$ can be gauged out inside the conductor at the expense of changing the electrostatic potential $\phi\rightarrow\phi+\delta\phi$.
This means a change of $\delta V$ in the electrochemical potential of the source reservoir and no change in the potential of conductor C.
Thus, one can see that the Kubo formalism is equivalent to the Landauer-B\"uttiker approach in the presence of infinite electrically neutral reservoirs.

$\delta{\bf A}$ generates a correction to the Hamiltonian: $\delta H=-\int d^3 r \delta{\bf Aj}/c$, where ${\bf j}$ is the current density. Consider an arbitrary surface of constant $\delta\phi$ in the region with nonzero $\delta {\bf E}$ (Fig. 3). Similar to the discussion of Fig. 2, in the low-frequency limit, the total current through any such surface is the same and equals the total current through the source

\begin{equation}
\label{totalcurrent}
I=I_{\rm U}-I_{\rm S},
\end{equation}
the signs in front of $I_{\rm U,S}$ reflecting our conventions about current directions, Fig. 1. This allows rewriting

\begin{equation}
\label{deltaH}
\delta H=-I\delta \tilde A/c,
\end{equation}
where
$d\delta \tilde A/dt=c\delta V$.
The same approach can be used to describe small changes of the drain potential but they are irrelevant for our purposes.

In what follows it will be convenient to consider the case of $\delta \tilde A$ oscillating with a low frequency $\omega$, $\delta \tilde A=c\delta V{\sin\omega t}/{\omega}$.

\subsection{Main argument}

We now give a full quantum derivation of Eq. (\ref{1}). The arguments leading to Eqs. (\ref{2}-\ref{4}) do not change compared to Section II and we concentrate on Eq. (\ref{5}). The cross-noise can be expressed as
\begin{align}
\label{6}
& 2S_{\rm ST}=\int dt \langle I_{\rm T}(0)I_{\rm S}(t)+h.c.\rangle(\exp(i\omega t)+\exp(-i\omega t)) = \nonumber\\
& \int dt \sum_{mn} (\exp(i\omega t)+\exp(-i\omega t))[\langle m|I_{\rm T}(0)|n\rangle\langle n|I_{\rm S}(t)|m\rangle\rho_m(-\infty)+\langle n|I_{\rm S}(t)|m\rangle\langle m|I_{\rm T}(0)|n\rangle\rho_n(-\infty)],
\end{align}
where a low frequency $\omega<1/\tau$ ($\tau$ is the relaxation time, Section II), $\rho(-\infty)$ is the initial density matrix and $I_{\rm T,S}$ are Heisenberg operators [see Eq. (\ref{defISt})].
As usual, introducing a small nonzero frequency allowed us to write the expression in terms of $I_{\rm S,T}$ and not $\Delta I_{\rm S,T}=I_{\rm S,T}-\langle I_{\rm S,T}\rangle$.
Eq. (\ref{6}) gives the noise at $t=0$, when the system is in a steady state.
As discussed in Section III.B it is convenient to use the Heisenberg representation in which the density matrix is not the steady state density  matrix $\rho(t=0)$ but the  initial $\rho(t=-\infty)$ since $I_{\rm S}(t)$  exhibits remarkable properties in such representation. Inserting the time dependence (\ref{ISt}) of the matrix elements $\langle n|I_{\rm S}(t)|m\rangle$, one finds
\begin{equation}
\label{6a}
2S_{\rm ST}=2\pi\sum_{mn}[\rho_m(-\infty)+\rho_n(-\infty)]\langle m|I_{\rm T}(0)|n\rangle\langle n|I_{\rm S}(0)|m\rangle[\delta(E_n-E_m+\omega)+\delta(E_m-E_n+\omega)].
\end{equation}

Next, we need to compute $R_{\rm T}=\partial I_{\rm T}/\partial V$. As discussed in Section III.C, this is a linear response problem with respect to $\delta V$.
Similar to Ref. \onlinecite{safi}, $R_{\rm T}$ is given by the same Kubo formula as in equilibrium. Indeed,

\begin{equation}
\label{current1}
\langle I_{\rm T}(t=0)\rangle=
{\rm Tr}[\rho(-\infty)S_{\rm A}(-\infty,0)I_{\rm T}^s S_{\rm A}(0,-\infty)],
\end{equation}
 where $I_{\rm T}^s$ is a Schr\"odinger operator, $S_{\rm A}(t_2,t_1)$ the evolution operator, $S_{\rm A}(0,-\infty)={\rm T}\exp(-i\int^0_{-\infty}H_{A}(t)dt)$,
the Hamiltonian $H_A(t)=H-I\delta\tilde A/c$ and $H=H_{\rm C}+H_{\rm H}+H_{\rm I}(t)$. The expansion to the first order in $\delta\tilde A$ yields

\begin{equation}
\label{current2}
R_{\rm T}\times\delta V=
i\int_{-\infty}^0 dt{\rm Tr}[\rho(-\infty)\{S(-\infty,t)\delta H^s S(t,0)I_{\rm T}^sS(0,-\infty)-S(-\infty,0)I_{\rm T}^sS(0,t)\delta H^sS(t,-\infty)\}],
\end{equation}
where $\delta H^s$ is the Schr\"odinger operator (\ref{deltaH}) and $S(b,a)={\rm T}\exp(-i\int_{a}^b dt [H_{\rm C}+H_{\rm H}+H_{\rm I}(t)])$.
Substituting (\ref{deltaH}) in the above equation we see that the response of $I_{\rm T}$ to $\delta V$ expresses as the sum of the responses of $I_{\rm T}$ to the perturbations $I_{\rm S}\delta \tilde A/c$ and
and $-I_{\rm U}\delta\tilde A/c$. The latter response is zero since the edges are far apart and perturbations on the upper edge have no effect on the lower edge.
With this in mind, we rewrite the
nonlinear response to $V$ in the form
\begin{equation}
\label{7}
R_{\rm T}=\partial I_{\rm T}/\partial V=i\lim_{\omega\rightarrow 0}\int^0_{-\infty} dt \sum_{mn} \frac{\exp(i\omega t)- \exp(-i\omega t)}{2i\omega}[\langle n|I_{\rm S}(t)|m\rangle\langle m|I_{\rm T}(0)|n\rangle\rho_n(-\infty)-\langle m|I_{\rm T}(0)|n\rangle\langle n|I_{\rm S}(t)|m\rangle\rho_m(-\infty)].
\end{equation}
In the above equation we absorbed evolution operators into the Heisenberg current operators.

It is convenient to combine the above response with the response $R_{\rm S}$ of $I_{\rm S}$ to the perturbation $\delta V I_{\rm T}\sin (\omega t)/\omega$ in the Hamiltonian. Certainly, that response is zero because of chirality.
Indeed, we consider a perturbation, acting on the left of the point, where $I_{\rm S}$ is measured.
 We get an expression of the same structure
as above with the indices $\rm S$ and $\rm T$ exchanged. In a steady state we expect that $\langle I_{\rm S}(0) I_{\rm T}(t)\rangle=\langle I_{\rm S}(-t) I_{\rm T}(0)\rangle$. This allows rewriting $R_{\rm S}$ in the form
\begin{equation}
\label{8}
R_{\rm S}=i\lim_{\omega\rightarrow 0}\int_0^{+\infty} dt \sum_{mn}\frac{\exp(i\omega t)- \exp(-i\omega t)}{2i\omega}[\langle n|I_{\rm S}(t)|m\rangle\langle m|I_{\rm T}(0)|n\rangle\rho_n(-\infty)-\langle m|I_{\rm T}(0)|n\rangle\langle n|I_{\rm S}(t)|m\rangle\rho_m(-\infty)].
\end{equation}
We next compute $R_{\rm T}=R_{\rm T}+R_{\rm S}$:
\begin{equation}
\label{9}
R_{\rm T}=2\pi\lim_{\omega\rightarrow 0}\sum_{mn}\delta(E_n-E_m+\omega)\frac{\rho_n(-\infty)-\rho_m(-\infty)}{2\omega}[\langle n|I_{\rm T}(0)|m\rangle\langle m|I_{\rm S}(0)|n\rangle+\langle m|I_{\rm T}(0)|n\rangle\langle n|I_{\rm S}(0)|m\rangle],
\end{equation}
where we used the time-dependence (\ref{ISt}). The above equation contains the initial density matrix at time $t=-\infty$ and the Heisenberg current operators (\ref{defISt}) at time $t=0$.
Finally we apply the results of Section III.B for $I_{\rm S}(t)$ and $\rho(-\infty)$. We notice that nonzero matrix elements $\langle m|I_{\rm S}(0)|n\rangle$ correspond to $N_n=N_m$ and $|n_{\rm C}\rangle=|m_{\rm C}\rangle$.
Hence, in the limit of low frequencies in Eq. (\ref{9}), $[\rho_n(-\infty)-\rho_m(-\infty)]/\omega=-\rho_{{\rm C}n}d\rho_{{\rm H}n}/dE_n=\rho_n(-\infty)/T$, where we used the factorization property (\ref{er3})
which is only valid for the initial density matrix. Comparison of Eqs. (\ref{6a}) and (\ref{9}) at small $\omega$ establishes Eq. (\ref{5}).

The above calculation relies on the structure of the initial density matrix $\rho(-\infty)$. This does not mean that the steady state depends on minor details of the initial state. Only the temperatures and chemical potentials of the large reservoirs are important. Those temperatures and potentials remain the same in the initial and steady state. If, on the other hand, one of the reservoirs is not large then the steady state does not depend on the initial density matrix
of that reservoir.
This can be easily seen from Eq. (\ref{1}) in the limit of a small reservoir attached to conductor C. Indeed, in that case, $I_{\rm T}=S_{\rm C}=0$ in a steady state since conductor C cannot accumulate charge. Thus, Eq. (\ref{1})
reduces to $S_{\rm D}=4GT$. This is a usual Nyquist formula, valid for a system in thermal equilibrium at the temperature $T$ and a uniform chemical potential. Obviously, the steady state is indeed an equilibrium state with the temperature $T$, if conductor C is attached to a finite reservoir.  In this example, the final state does not depend on the initial density matrix $\rho_{\rm C}$.

A similar argument does not work for a finite source reservoir. Indeed, the derivation of Eq. (\ref{1}) relies on the assumption that $I_{\rm U}$ is uncorrelated with $I_{\rm S}$. If the source reservoir is not large then the assumption is violated in the steady state and $I_{\rm U}=I_{\rm S}$ instead.

\section{Discussion}

The focus of the preceding section is on QHE, but similar non-equilibrium FDT apply in many other systems. The simplest example of a chiral system, based on an ideal gas, is considered in the appendix. Our results can also be generalized beyond 1D, for example, for the surface transport in a 3D stack of QHE systems.

The geometry of Fig. 1 allows only electron tunneling to conductor C. FDT's, similar to (\ref{1}), can also be derived in other geometries, where fractionally charged anyons tunnel: One can consider tunneling between two edges of the same QHE liquid.

Eq. (\ref{1}) does not contain the temperature $T_{C}$ of conductor C. This, certainly, does not mean that the properties of the system do not depend on it. The current $I_{\rm T}$ and the noises $S_{\rm D}$ and $S_{\rm C}$
are all affected by the temperature of C. The general relation (\ref{1}), however, remains the same. We would like to emphasize that our derivation does not contain any assumptions about the character of the dependence of $I_{\rm T}$
and $S_{\rm C}$ on the temperature and voltage. An interesting situation is possible, if conductor C is chiral and $T_C\ne T$. One can then derive two equations of the structure
(\ref{1}) with two different temperatures in them.

\begin{figure}[h]
\includegraphics{fdt_fig4.eps}
\caption{A possible experimental setup. Charge carriers, emitted from the source, can either tunnel through the constriction Q and continue towards the drain or are absorbed by the Ohmic contact C.}
\end{figure}

Our main result, Eq. (\ref{1}), applies in chiral systems and can be used for an experimental test of chirality. A convenient measurement setup is illustrated in Fig. 4. Several mechanisms break chirality and can lead to the violation of Eq. (\ref{1}). One mechanism involves long range forces in the 2D electron gas. Our discussion assumed that a gate screens long-range Coulomb interaction. This allowed us to assume that the tunneling Hamiltonian
$H_{\rm I}$ does not depend on the voltage bias and the bias manifests itself in the 2D electron gas only through the chemical potential of the lower edge. Without screening, $H_{\rm I}$ may depend explicitly on the voltage and this must be taken into account at the calculation of $\partial I_{\rm T}/\partial V$. Strong interaction of edge modes with non-chiral bulk modes may also break chirality.

The most interesting mechanism of chirality breaking involves ``upstream"
modes \cite{feld-li,heiblum}, Fig. 5. In the simplest example, two charged modes carry charge in the opposite directions. Let us imagine that the two chiral channels do not interact and all charge tunneling into C occurs due to particles, populating the upstream channel, directed from D to S. Then the noise in the drain is the same as in the absence of C, in contradiction with Eq. (\ref{1}).

\begin{figure}[h]
\includegraphics{fdt_fig5.eps}
\caption{A non-chiral system. The solid line along the lower edge illustrates the ``downstream mode", propagating from the source to the drain. The dashed line shows a counter-propagating ``upstream" mode.}
\end{figure}

This discussion neglects a possible heating effect. To illustrate it, let us assume that the upstream mode is neutral. It cannot carry charge but carries energy. In general, the tunneling operator into C includes a product of operators creating charged and neutral excitations on the edge. A neutral excitation of the energy $\sim V$ travels to the source and heats it. This affects noise, generated by the source, and leads to the violation of Eq. (\ref{1}). The details of the interaction of a neutral quasiparticle and the source are poorly understood theoretically. The experiment suggests that the heating effect will be strong \cite{heating}. Thus, large deviations from Eq. (\ref{1})
can be expected in the presence of ``upstream" neutral modes.

Our only assumption about $V$, $T$ and $T_{C}$ was that they are much lower than the QHE gap. Otherwise, a chiral description is unlikely to apply. If the system is chiral we make no assumptions about the relation between $V$ and $T$.
Nevertheless, our main focus was on the regime with $V\sim T$. Eq. (\ref{1}) greatly simplifies and becomes less interesting in the opposite limits $V\gg T$ and $T\gg V$. In the former case, let us set $T$ to zero. Then
Eq. (\ref{1}) reduces to $S_{\rm D}=S_{\rm C}$. This relation reflects noiseless character of the emitted current. In the opposite limit, let us assume that $V=0$ and $T=T_{\rm C}$. Then the equilibrium FDT applies.
$\partial I_{\rm T}/\partial V$ is now linear response. Hence, $S_{\rm C}=4T\partial I_{\rm T}/\partial V$. Finally, Eq. (\ref{1}) reduces to $S_{\rm D}=4GT$. This simple relation reflects the fact that the lower edge is in thermal equilibrium on both sides of the contact with C.

In conclusion, we established a non-equilibrium FDT (\ref{1}) for chiral systems,  both close ($V\ll T$) and far ($V>T$) from equilibrium. The result does not apply to non-chiral conductors and can be helpful in the search of counter-propagating modes on
quantum Hall edges \cite{feld-li,heiblum}.

\acknowledgments

We acknowledge helpful discussions with
M. P. A. Fisher, M. Heiblum, S. Kehrein and A. W. W. Ludwig.  This work
was supported by NSF under Grant No. DMR-0544116 and
BSF under Grant No. 2006371.

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