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The Sports Scouting Report Podcast With Lee Brecheen Episode 60: New Orleans Saints Hall of Fame Linebacker Rickey Jackson
The Sports Scouting Report Podcast…
Baton RougeBelow HomepageCenLa NatchDailyHouma-LafourcheLake Charles-LafayetteLatestNew OrleansNewsNewsflashNorth LaNorth Shore-TangiRecruit SpotlightRecruitingRegionRight of HomepageSoutheast LATop Home PageTop of HomepageTop Story
For Monday's episode of the Sports Scouting Report, we have a very special guest as Lee catches up with New Orleans Saints Hall of Fame linebacker Rickey Jackson. Rickey talks about his radio show with Ro Brown, why he chose to play for the Pittsburgh Panthers in college, his thoughts on playing on one of the greatest NFL linebacking corps of all time in the Dome Patrol, winning a Super Bowl with the San Francisco 49ers, the best quarterback he has ever played against, what he is doing now a days, getting into the NFL Hall of Fame, his thoughts on what is going on now in the NFL, and much more!
Click link below to check out this week's podcast! Subscribe to wherever you listen to your podcasts below in order to keep up with our episodes each week! Please keep in mind that it may take a couple of hours for the podcast to be uploaded to iHeart Radio, Google Play, and iTunes/Apple Podcast.
Podcast Host Link: https://www.lafootballmagazine.com/podcasts/
Also, we would love your feedback! Tell us what you think or if you have a question that you want Lee to answer on our next Podcast Episode, please email Lee at Lbrecheen@aol.com or at https://www.lafootballmagazine.com/contact-us/ !
Hey football fans. Like our recruiting interviews, stories, and other content on our website? Click on our website daily to view articles on recruits, coaches, and so much more at our website: https://www.lafootballmagazine.com/.
Also, we have exciting news that for the fourth straight year, we will have our North and South LA issues and both are now out! Both issues will continue to cover all the Louisiana colleges as well as all the high schools in state, but there will be 16 pages that will focus on North Louisiana as well as South Louisiana. Along with our print editions of both magazine issues, our digital magazine editions will also be available to pre-order. Just like with our print versions, our digital editions will be available to you when both magazines are available.
For more information, please check the following link to where you can either order our print or digital version of the magazine of your choice. Also, below are the two covers as well as where you can order both magazines on our website.
Buy Print/Digital Magazine Here: https://www.lafootballmagazine.com/magazine/
Jace is from Addis, LA but resides in Port Allen, LA. Jace attends Southeastern Louisiana University and is majoring in Communications. This is Jace's fourth year as Associate Editor at Louisiana Football Magazine. He is also a graduate from Catholic High School in Baton Rouge where he was the editor of the school newspaper the "Bearly Published." While at CHS, Jace also announced their football, basketball, and baseball games for the Bruin Broadcast Network which is the Catholic High's streaming website for their events. On weekends, Jace is the PA announcer for his middle school football games. At home, Jace enjoys time with his family and friends. Besides being a huge football fan, Jace is also an avid NHRA drag racing fan.
Categories: Baton Rouge, Below Homepage, CenLa Natch, Daily, Houma-Lafourche, Lake Charles-Lafayette, Latest, New Orleans, News, Newsflash, North La, North Shore-Tangi, Recruit Spotlight, Recruiting, Region, Right of Homepage, Southeast LA, Top Home Page, Top of Homepage, Top StoryBy Staff September 21, 2020 Leave a comment
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This Heritage BC commissioned report was written as our submission to the CleanBC Roadmap to 2030 – Buildings Pathway. Heritage BC's submission – Leverage Existing and Historic Buildings for Zero Net Carbon makes the argument that existing buildings are a vast resource that we can leverage, through proper stewardship. The report demonstrates how using existing buildings reduces major amounts of carbon emissions, increases overall sustainability and resilience, improves community place-making and continuity, and more.
» Leverage Existing and Historic Buildings for Zero Net Carbon Report, 2021.
Prepared By
Mark Thompson Brandt, Emily Guy, Catherine McBain, Melissa Lengies, of TRACE Architecture (formerly MTBA Associates) on behalf of Heritage BC.
Suggested Use & Distribution
This report provides information on how existing buildings can be leveraged to assist BC in meeting its CleanBC goals. It is relevant for heritage advocates, climate advocates, planners, architects and policy makers at all levels of government. It can and should be used to develop cases for adaptive re-use and other conservation plans for building re-use at the community, municipal, provincial, and federal levels.
» Webinar: Bringing the Power of Heritage to Climate Action (with Mark Thompson Brandt, 2021)
Read Without Downloading
Making a case to leverage existing and historic buildings for zero net carbon

redpajama_set_name: RedPajamaCommonCrawl


iWellness or iHealth – There is a difference. Can Apple change healthcare forever? Yes! Will they?
The rumors are pretty strong that Apple will introduce Healthkit technology in it next IOS 8 operating system for iPhones and iPad's..and maybe iWatch this October. It will enable software, apps and hardware creations that will monitor your health. …and hopefully affect your wellness.
The people at Apple have always listened to what people say and then do it their way. This backhanded compliment is from a guy that once lived in the world that Apple calls home…Silicon Valley.
I must admit that I feel guilty that I was paid during the 80's working in Silicon Valley. It was way too much fun to have been a real job. Frankly, I am not sure anyone in "The Valley" felt they were really working in those days. It was a giant sandbox to play in.
As I rose to become a CEO in the valley the memories of those garage shop times were never lost. The industry had changed by the mid 90s and the spirit of entrepreneurial adventure was still the primary element, like oxygen, in the air we all breathed.
Apple was and is an icon that many believe sets the stage for innovation. Hated or loved, there was no middle ground when it came to Apple. Equally, the culture at Apple was bipolar compared to the mainstream computer companies trying to dabble in the PC industry. Apple had created a culture under the tutelage of Steve Jobs that was unique among competitors like IBM, Texas Instruments and Digital Equipment Corporation. I clearly remember it was a west coast vibe where the key phrase was "can," versus the competitors irritating "can't" point of view.
Yes there was Steve Jobs, but he stood on the shoulders of Hartmut Esslinger, Jay Chiat, Andy Hertzfeld and the rest of the Macintosh team. They were building the "computer for the rest of us." They forever change the world. It was the combined brilliance of Hartmut Esslinger's "Snow White" industrial design theme, the Mac teams software genius and the marketing revelations of Jay Chiat that clearly made the Mac. Jay's quote, "I'm uncomfortable when I'm comfortable." says it all. Anyone that knows, knows that it took years for the Mac to be successful. It was the Apple way and that continues today.
Apple changed the world then and Tim Cook's Apple is about to change it again in an even more profound way. I just want to make sure he gets it really right.
Now Mr. Cook and ensemble have before them something more bombasitic and more relevant than anything that came before if they take the big risk of bucking the system. It is currently referred to as Healthkit and it is a suite of software, in the new IOS 8 that will facilitate new apps and hardware. Healthkit has the potential to save millions of lives by focusing mere mortals on their health and wellness. Yes, the next generation of IOS will have the ability to monitor your heart rate, blood pressure, glucose levels as well as apps that will prescreen for skin cancer. Apps will even be able to send your vital information to the Mayo Clinic and other healthcare providers. You name it. For as many imaginations that can concoct uses, Healthkit will have apps and devices that will prescreen for just about anything. Mark my words, this will forever change the world and forever change medicine. It might not be right away, but it is certain.
So why the open letter to Tim Cook from an ex-Silicon Valley CEO turned natural products formulator. It is because if Tim really wants to make a difference, he must change the way most people think and focus on wellness and not just a facilitator for our already dysfunctional medical system. Just remotely offering what is already provided in an in-office visit is geekily interesting, but who cares if Healthkit's results only end up creating medical tests, more tests and a prescription for ten different drugs. Turning your iPhone into a remote sensor for national healthcare is not my idea of a step ahead. Yes, yes it is cool to be able to tell your heart rate, blood pressure, distance walked or run and your blood sugar levels all on a wrist watch like device. …but, what if it also offered a wellness curriculum with monitoring and a performance system that shows your hormone levels, aerobic state and dietary requirements So that you can change your lifestyle to optimize your health.
There is a fundamental difference between screening for health and apps to facilitate wellness. Think of it this way. Today, most of the medical system and doctors operate the way a car mechanic did in the 50s-60s. In those days, you bought a car, put gas and oil in it and when it broke you took it to the mechanic. It was the days when brake pads had little metal tabs on them so that when they wore down, you heard the awful squeal of metal against metal and you then took it to the mechanic for new brakes. Our medical system and our "helpless patient" mentality are fundamentally the same. We cruise through life and when something breaks, we go to the doctor to find out what is wrong and get something, usually a prescription pill, to fix it. Almost never do we or our doctors search for the cause. The irony today is that we take our cars in for periodic maintenance to prevent bad stuff from happening. So to me, the old way where we wait for something to break is akin to the current health care system and working to prevent problems is wellness. Tim, it is not a nuance it is a state change and your decisions will indeed change the healthcare system, but will it change the way we observe our wellness, well-being and extend our lives.
I became a reluctant formulator of natural skin care and natural medicines because of, my wife, Wendy's Melanoma diagnosis. As an engineer/scientist, I wondered what was the cause of her skin cancer? We still have no idea. What caused her distress after the disease were the chemicals in her prescription drug regimen. What we know for sure is that the shock of the Melanoma is that it began a wellness lifestyle for us both. What can we change to prevent disease before it happens? That is wellness!
Wellness is different from healthcare in many ways. Simply, it is challenging the conventional thinking and focusing on our diet, lifestyle, exercise and supplemental factors that extend the quality of life and reduce our "real-age" by changing habits early and not waiting for something to break. Equally, as we age, what Apple can actually do is to provide us tools that we didn't even know we needed that will extend our youthfulness. Yes, knowing your glucose levels may help diagnose diabetes, but a tool that regulates how you avoid high glucose levels by monitoring your food selection is really cool. Can you imagine a sensor on your iPhone that would track and diagnose your hormone levels offering foods, supplements and activities that would enhance the production of these hormones. Imagine the same device when you hit your 50's or 60's and it recommends a Bioidentical Hormone Replacement Therapy (BHRT) with suggested dosage. The list is only as long or as limitless as your imagination can handle. Now maps these ideas on any number of wellness factors that can be monitored. Same is true for disorder prevention and lifestyle.
Yes, Healthkit will bring great diagnostic tools. Just imagine the world with a division of Apple called iHealth & Wellness powered by Healthkit. Let us not lose the wellness part.


Once again the webmaster community has been misinformed about what Google will be updating this side of the new year after ripples of discussion have begun to started to fluctuate surrounding a recent Google Pagerank update to the toolbar data that shows through various tools and plugins.
Matt Cutts, head of Google Webspam team said that there were no plans for Google to update this information following reports that there was a broken aspect of the data stream that pushes this information out to the tools that display the metric.
Although the data seems to have been updated, questions are being asked surrounding the reliability of the data after many complained that they have failed to see the impact to their site that they believed they would receive (increased Pagerank score).
Speculation that the data could be as old as insight taken from September 2013 is being passed around and it seems that there are quite a few losers in the vast scale of things, especially the dropped domainers that had previously purchased domains with a view to make use of former toolbar Pagerank indications.
With another curveball thrown into the pre-Christmas reviews for website owners the question has to be whether those that have been devalued will look to act on the drop or whether they too will see that Pagerank is not a ranking factor that dictates the success of their website.
How did your websites measure up?
I was looking for pr update from past few months and now i get pr for my website. I am so happy with this update even hummingbird update loose my keywords rank I am trying to get out.
In fact data of the last page rank "update" are not up to date 🙂 That's not speculations, that's really true and very simple to verify. The last toolbar page rank update (6th dec 2013) correspond to 31th August Data. So the new datas provided by Google are motre than 4 month old. I don't know if I m able to post that link here, but I tried to explai everything my SEO Blog http://www.lightonseo.com/actualite/870-1207-fake-google-dance/ (This a french SEO Blog but the post is short and could be translated online).


Joey Tenute was born April 2, 1983 in Hamilton, Ontario. The eighth round, 261st overall selection of the New Jersey Devils in the 2003 NHL Entry Draft, Tenute is a graduate of the OHLs Barrie Colts and Sarnia Sting.
A diminutive, yet offensively gifted forward, Tenute played four seasons in the OHL registering 271 points (95-176-271) before making his professional debut with the ECHLs South Carlolina Stingrays in 2004-05.
Following an impressive rookie season in the ECHL, Tenute signed as a free-agent with the Washington Capitals organization in the summer of 2005. Upon his arrival with the Caps organization, Tenute spent the majority of his first season with the club's AHL affiliate in Hershey, however, impressed the Caps brass enough to earn his first taste of the NHL in the latter stages of the 2005-06 season.
A key player in Hershey's run to the 2006 Calder Cup title, Tenute tallied five points (2-3-5) in the club's playoff drive. The following year as a free agent, Tenute signed with the Phoenix Coyotes.
ECHL All-Rookie Team (2005) ECHL Second All-Star Team (2005) ECHL - Rookie of the Year (2005) Signed as a free agent by Washington, November 21, 2005. Signed as a free agent by Phoenix, July 9, 2007. Signed as a free agent by Malmo (Sweden), May 16, 2013.


City League is a beginners league that emphasizes having fun while teaching the basics of the game of lacrosse. It is a developmental program that is made up of 10U players.
Practices will start in mid-late February. Teams will practice up to two times per week.
If more than one team is needed, the players are placed on a City League team determined by the coaches. No preferential treatment is given. Parent volunteers will be asked to coach. Training and support will be provided by the Folsom Lacrosse Association. If you are interested in being a coach, please fill out our Coaching Application by clicking HERE.
Games will be held on Fridays (late afternoons/early evenings) starting in early to mid-March. There are no games during Spring Break (FCUSD schools). The season will end before Memorial Day Weekend in May. All games will be played locally, either in Granite Bay, El Dorado Hills or Folsom.
Registration typically opens in September. Cost is $250 + US LACROSSE annual membership ($30) is required of all players in FLA.
All players are required to purchase a club uniform through a link to a 3rd party vendor provided by the club.
Uniform Fitting and an Equipment night held in October to buy used or new equipment. FLA tries to host a free custom mouth guard event (cannot fit players with braces unfortunately) each year in December, if available.
Boys require stick, helmet, shoulder pads, elbow pads, gloves, mouth guard and cup.
Girls require stick, mouth guard and goggles.


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« Franco Center presents "Just Us" Family Christmas Show
LA 9-1-1 earns second reaccreditation »
Mason, Crafts present Joint Resolution to Israeli Knesset
December 10, 2015 | Author editor
Sen. Garret Mason and Rep. Dale Crafts present a Joint Resolution of the 126th Maine Legislature affirming its support of Israel to Revital Swid of the Israeli National Legislature.
Maine Senate Majority Leader Garrett Mason (R-Androscoggin) and Representative Dale Crafts (R-Lisbon) met with a member of the Israeli National Legislature, known as the Knesset, recently to present a Joint Resolution from the 126th Maine Legislature.
"It was an honor to meet with Revital Swid of the Israeli Knesset to present this Joint Resolution supporting Israel," said Senator Mason. "To do so in Israel was especially momentous. The Joint Resolution describes Maine's connections to and support of Israel, a message I was proud to convey."
"Bringing a goodwill message from Mainers to Israel was an incredible experience," added Representative Crafts. "It was also remarkable to visit Jaffa, a colony in Israel founded by Mainers in 1866, and to see Mainers' direct impact on the Holy Land."
The Joint Resolution was passed by the 126th Maine Legislature in 2013. It received unanimous support in the Maine Senate and passed the House with a bipartisan margin of 124-11.


Colima in western Mexico is one of the most beautiful and most active volcanoes in the world. Apart from highly interesting volcanic features to admire, it has been in near constant eruptive activity for decades.
There are often lava flows, strombolian and vulcanian explosions to observe. We regularly offer this Volcano Special Tour, both at announced group tour dates, or - on request - at custom dates for small groups or individuals. These tours are perfect occasions to learn and improve on volcano photography!
Insights into geologic and volcanological processes, explanations of outcrops and volcanic land forms, but also tips for volcano photography are provided by the accompanying volcanologist (Tom Pfeiffer).
The tour starts normally in Colima (CLQ airport); typically, you fly to Mexico City and from there on to Colima with Aeromexico. Transfers from and to Guadalajara airport can be arranged in some cases (request).
Price (p.p.): 1750 € ?


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ARC Review: Sacrificial Magic (Downside Ghosts #4) by Stacia Kane
When Chess Putnam is ordered by an infamous crime boss—who also happens to be her drug dealer—to use her powers as a witch to solve a grisly murder involving dark magic, she knows she must rise to the challenge. Adding to the intensity: Chess's boyfriend, Terrible, doesn't trust her, and Lex, the son of a rival crime lord, is trying to reignite the sparks between him and Chess.
Plus there's the little matter of Chess's real job as a ghost hunter for the Church of Real Truth, investigating reports of a haunting at a school in the heart of Downside. Someone seems to be taking a crash course in summoning the dead—and if Chess doesn't watch her back, she may soon be joining their ranks.
As Chess is drawn into a shadowy world of twisted secrets and dark violence, it soon becomes clear that she's not going to emerge from its depths without making the ultimate sacrifice. ~ goodreads
Note: There are no spoilers for Sacrificial Magic but there are a few for the previous books in the series. IMO they must be read in order. You can read my review for the first three books in the series here.
Life…work... love... these things have never been easy for Chess Putnam. In Sacrificial Magic, the fourth book in the Downside Ghosts series, all three collide with shocking and often heartbreaking results.
It's hard to read about a character that struggles and f*cks up as much as Chess. This poor girl has had the worst of life thrown at her and she numbs the pain with drugs and self-destructive behavior. Kane has the herculean task of pushing us out of our comfort zone while still keeping us rooting for Chess. She accomplishes that task.
I was definitely pushed out my comfort zone in this book. Chess is working on yet another case for the Church with help from Lex, her ex- lover, while she battles her insecurities in her relationship with her current lover, Terrible. Just when Chess and Terrible take one giant leap forward they do something that makes them fall two steps back. I was really frustrated with Chess for a lot of this book but man the things she confronts are huge and ultimately she's growing as a character and I was proud of her.
There are several big ( and I mean BIG) developments on different fronts that all come to a head at the end of Sacrificial Magic. I would say that this book is a game changer. Chess will no longer be able to hang back and work for all the various sides - Lex, Bump, and the Church. She's going to have some tough choices to make. This couldn't come at a worse time for her since she's finally learning to let people show her some affection and realizes that she can have people in her life that don't want anything in return from her.
Out of all the ghosts and personal issues Chess has to face the thing that creeps me out the most continues to be the Church. I chafe at anything that requires that kind of blind loyalty. Chess still follows the Churches teachings without question. I'm really hoping that changes at some point.
Despite the physical & emotional turmoil Chess goes through in this book, I felt that Sacrificial Magic had a lot of lighter moments. Chess actually smiles a few times. I particularly love her new friendship with Beaulah, a woman she meets during the case she's working on. It's an interesting dynamic in Chess's mostly testosterone filled world. I also enjoyed her conversations with Elder Griffin. Of course the big draw of this series for most people is the relationship between Chess & Terrible (& Lex?) but this is about the growth of Chess and adding these different types of relationships in her life helps her confront her issues from all sides.
As wonderful as the fantasy idea of falling in love with a guy and having that solve all our problems sounds, it's just that - a fantasy. Love does help and that relationship can give you the strength & support you need but it's never a magic fix and Stacia explores that even though it's painful to read at times. We want the fantasy. As they say it takes a village. Chess has a strange, wacky band of villagers as her support group but at least she's not on her own anymore.
Rating 3.5 out of 4 This is the most important book of the series IMO. It almost hurt to read at times but ultimately you will come out the other side smiling. I did have a hard time with the "downside speak" at first. Maybe it's been awhile since I had read the last book or maybe because Lex, Bump & Terrible were in this book more so the dialect was more prevalent but it took me some time to get the feel of it again.
Stacia Kane: website | Twitter | Facebook
Source: Netgalley / Del Rey for my honest review
Buy the book! Sacrificial Magic - Release date - March 27, 2012
Jen at Red Hot Books March 22, 2012 at 11:14 AM
My favorite book in the series so far. It's amazing how Kane made me feel what Chess was feeling --especially that scene right in the middle of the book.
Julie@My5monkeys March 22, 2012 at 1:18 PM
Loved your review and I do agree with you that this was a big book for Chess. Yes I was smiling and love her band of villiagers.
Blodeuedd March 22, 2012 at 4:55 PM
I just did not fall for book 1 :/ or book 2..or 3
The series isn't for everyone that's for sure. I'm surprised you kept going!
© For What It's Worth: ARC Review: Sacrificial Magic (Downside Ghosts #4) by Stacia Kane


It has 601 horsepower and a 200-mile range, but only 155 examples are slated for production.
Our attention is admittedly focused on the 2019 New York Auto Show, but Auto Shanghai is also shaking the automotive pillars this week and Aston Martin is doing some shaking of its own. The automaker is ready to step into the electric world with the Rapide E – the first fully electric production car from the posh British manufacturer.
Yes, the shape is familiar to Aston fans. The four-door Rapide is faithfully recreated here, but gone is the aurally delightful 6.0-liter V12. Instead, the Rapide E turns just the back wheels with a pair of rear-mounted electric motors. In the place normally occupied by the V12, you'll find a bespoke 800V battery pack wrapped in Kevlar and carbon fiber, consisting of more than 5,600 lithium-ion cells. It's enough to produce a combined 601 horsepower (448 kilowatts) and 701 pound-feet (950 Newton-meters) of torque – numbers which eclipse all iterations of the V12-powered Rapide by a fair margin.
Pin your foot to the floor of a fully charged Rapide E, and Aston Martin says you'll hit 60 mph in less than four seconds. For the record, that's nearly a full second quicker than the Rapide S, and even eclipses the crazy AMR edition. Top speed is limited to 155 mph, so as far as accelerative performance goes, Aston's first electric production vehicle doesn't disappoint.
As for range, that varies on how often you exercise that substantial forward thrust but Aston Martin estimates a range of over 200 miles under WLTP standards. It's capable of fast charging using a 400V 50kW charger, or the car's high-voltage battery system allows for even faster 800V 100kW charging, which can add 310 miles of range in just an hour. A more modest on-board AC charger can completely refill the battery pack in three hours.
There's far more to the Rapide E than just an electric powertrain, however. Inside, the traditional gauge cluster behind the wheel is replaced with a 10-inch digital display. Outside, the underbody is completely redesigned to provide smoother airflow front-to-back, including a new rear diffuser that focuses on aero efficiency. It rides on new aluminum wheels designed to increase efficiency as well as brake cooling. Speaking of cooling, since there's no longer a beefy V12 up front that requires airflow, engineers have streamlined the Rapide E's nose to help the four-door better slip through the air.
Yes, the Rapide E is Aston Martin's first step into the electric realm, but admittedly it's not a big one. Only 155 models are slated for production, but it's seen as a pivotal step towards the automaker's new Lagonda brand, which will be a full-on zero-emission operation. As it stands now, the Rapide E is available for order, but the company restricts pricing information to those who apply for an opportunity to own one.
16 April 2019, Shanghai, China: Aston Martin Lagonda has revealed the final production-ready iteration of its first all-electric production car, with the spectacular Rapide E making its world debut at China's Auto Shanghai motor show.
Unveiled by Aston Martin Executive Vice President & Chief Creative Officer, Marek Reichman, Rapide E is a pivotal model for the luxury marque. Following the success of previous special-edition models, including Aston Martin Vulcan, the Vanquish Zagato family and DB4 GT Continuation, Rapide E continues to deliver a superior level of performance, craftmanship and exclusivity, while also propelling Aston Martin into the all-electric realm.
The first car to be built at Aston Martin's state-of-the-art St Athan production facility – the brand's Home of Electrification – Rapide E represents a pioneering first step towards achieving the company's wider electrification strategy and the successful fruition of Lagonda – the world's first zero-emission luxury brand. A special edition with a production run strictly limited to 155 units, Rapide E has been developed in collaboration with Williams Advanced Engineering (WAE).
Shown for the first time here in Shanghai today, Aston Martin's design and engineering teams have worked in partnership to extract the optimal aerodynamic performance available from the Rapide's elegant form. Traditional metal vanes give way to a striking honeycomb front grille. With lower cooling requirements demanded of the EV powertrain, Aston Martin's aerodynamic engineers have been given the freedom to optimize the aperture of the frontal area, minimising airflow through the car's body, improving the car's aerodynamic efficiency and increasing range as a result.
A re-designed underfloor streamlines airflow from the front splitter right the way through to Rapide E's new larger rear diffuser – a feature that is now wholly dedicated to aero efficiency due to the removal of the exhaust system required before. The model's forged aluminium aerodynamic wheels – shod with bespoke low rolling-resistance Pirelli P-Zero tyres – have also been re-designed to give greater efficiency, without compromising brake cooling capability. The sum of these changes gives Rapide E's aerodynamic package an 8% improvement over the previous internal combustion model.
Inside and out, Rapide E is equipped with the materials and technology befitting of the marque's first EV model. Gone are the analogue displays of the past. A 10" digital display now sits in their place, delivering all key information to the driver while on the move, including the battery's state of charge, current motor power levels, regenerative performance and a real-time energy consumption meter. Swathes of carbon fibre have been deployed throughout, assisting in delivering the strict weight targets set by Aston Martin's engineering team from the outset.
Extending the experience beyond the car, a dedicated app allows key information from the car to be remotely monitored. This powerful dashboard displays remaining range, battery status, and time to charge, as well as allowing navigation destinations to be sent from the app to the car for easy route planning. The car's parking spot can also be seen via the app, with walking directions provided to help re-locate it in busy locations. Finally, the health and status of Rapide E can also be tracked from the app, providing reminders such as when a key service is due.
Throughout the development process, the combined AML/WAE engineering team first focussed their efforts on finding packaging solutions for the battery system and motors within the Rapide's architecture before working to achieve effective cooling of the batteries and motors.
Rapide E is powered by an 800V electrical architecture battery – encased in carbon fibre and Kevlar casing – with a 65kWh installed capacity using over 5600 lithium-ion 18650 format cylindrical cells. This bespoke battery pack lies where the original 6.0-litre V12, gearbox and fuel tank were located, with the 800V system allowing for more efficient charging and greatly improved thermal characteristics over existing electrical architectures. This battery system powers two rear mounted electric motors producing a combined target output of just over 610PS and a colossal 950Nm of torque.
Rapide E's range is over 200 miles under the Worldwide Harmonised Light Vehicle Test Procedure (WLTP), while the model is capable of charging at a rate of 1851,2 miles of range per hour using a typical 400V 50kW charger. However, the car's 800V high-voltage battery system enables faster charging of 3101,2 miles of range per hour, using an 800V outlet delivering 100kW or higher. For destination charging, the car also features an industry-leading high-power AC on-board charger capable of recharging the battery in as little as 3 hours.
Top speed for Rapide E is 155mph, with a sub-4.0 second 0-60mph time and a 50-70mph time of just 1.5 seconds. These figures are not restricted to a narrow window of battery charge or climatic conditions, instead, thanks to the cutting-edge 800V architecture, the Rapide E will deliver its performance in a consistent and repeatable way as would be expected from a traditional Aston Martin product. This includes the ability to drive a full lap of the Nürburgring with no performance derating of the battery or the motors.
Rapide E also enhances the feel, character and delivery of the V12-engined recently acclaimed Rapide AMR, thanks to careful attention paid to the tuning of both the electric powertrain and the chassis across three driving modes – GT, Sport and Sport + – which are selected for both the powertrain and handling. The rear-wheel drive Rapide E's twin electric motors drive through a Limited-Slip Differential, which combined with a revised spring and damper rates, ensures the pure handling characteristics of the petrol-powered Rapide AMR are retained.
Rapide E is available to order now, with prices available on application.


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The very important reason this woman's bikini pic is going viral
Nikita Richardson
The fact that we're already halfway through the summer season may come as a relief to some women. Because with the end of summer comes the end of so-called "bikini season," which tends to separate the "body confident" from the "body insecure" in a truly startling way. But, if 21-year-old Michelle Elman has anything to say about it (and she does), bikini season shouldn't be a time for high anxiety and low self-esteem, it should be a time for celebration. Especially when you've gone through as much as she has.
This week, Elman took to Instagram to prove that there's no one body type for bikinis, by sharing a picture of herself in swimwear. Since birth, she has struggled with hydrocephalus, a condition that causes fluid to build up in the brain, forcing Elman to endure, "15 surgeries, a brain tumor, a punctured intestine, an obstructed bowel, [and] a cyst in brain." As a result, Elman's abdomen has been covered with scars for most of her life, scars that she desperately tried to hide even as a child.
"At age seven, I tried on my first bikini and after receiving a range of reactions from disgust to pity, it soon became easier to hide away and be doomed to a life of tankinis and one-pieces," she wrote in a piece for Huffington Post UK. "I had become ashamed of my body and soon other people's disgust became my own and this was allowed to be the case because of one simple reason – I had no one to talk to about it."
Now, she's clearly taking a different path. Here's the picture she posted.
PEOPLE WITH SCARS CAN'T WEAR BIKINIS This is what I have believed for the last 21 years of my life and when asked in January why I never wear bikinis, this horrible sentence came out of my mouth. I was shocked – at myself! I had had 15 surgeries, a brain tumour, a punctured intestine, an obstructed bowel, a cyst in brain and have lived for the last 21 years with a condition called Hydrocephalus yet somehow my scars have always been the hardest part. They make already awkward moments in your adolescence even more uncomfortable – like taking your top off for the first time in front of your boyfriend, and made me feel even more isolated in a world where I felt no one could understand. At age 7, I tried on my first bikini and after receiving a range of reactions from disgust to pity, it soon became easier to hide away and be doomed to a life of tankinis and one-pieces. Why did I believe this? Because over the years, I have learnt that my scars make people uncomfortable. I had become ashamed of my body and soon other people's disgust became my own and this was allowed to be the case because of one simple reason – I had no one to talk to about it. Well in January, I started to talk about it – all of it, and I want other people to join in on the conversation. Every human has scars, whether there are emotional and physical – they are part of our story and we should be proud of them. About a month ago, I finally faced up to wearing a bikini and although, I love my body thoroughly and have for many years – this was one of the most difficult things I have ever done, yet once it was on, it was one of the most liberating feelings to know that I wasn't letting two pieces of material stop me from being comfortable in my own body. My belief is that no one should have to feel ashamed of their body, whether you have stretch marks or a C-section scar so… THIS summer, let's stand up and be proud of our scars and what they represent – a story! Tag a friend below and lets make this the summer of scars! #scarrednotscared
A photo posted by Michelle Elman (@mindsetforlifeltd) on Jul 13, 2015 at 5:32am PDT
These days, Elman isn't just ready to talk about body positivity, she's ready to shout about it from the rooftops. She's started her own company, Mindset For Life, where she works as a "body confidence coach" working to serve a growing community of those who are #scarrednotscared. She's also launched a video series offering advice on everything from telling boys about your scars to — of course — how to wear a bikini with confidence.
"Your confidence is actually costing you a lot more than you think," Elman says in one video, touching on asking for raises or buying clothes that you ultimately don't feel comfortable wearing. "Every investment you make in yourself is the best investment you'll make. You're your best asset."
You tell 'em, girl.
Why this teen girl's bikini pic is inspiring the world
This woman's bikini pic went viral for the best reason, ever
[Image via Instagram]


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text:
\section{Introduction and Summary}
Holography is very useful for the analysis of properties of strongly coupled
quantum field theories. Recently these ideas were extended to non-relativistic
theories since today it is well known that   non-relativistic holography  is very useful tool for the study of strongly correlated systems in condensed matter, for recent review see \cite{Hartnoll:2016apf}. Non-relativistic symmetries also have fundamental meaning in the recent proposal of renormalizable quantum theory of gravity known today as Ho\v{r}ava-Lifshitz gravity \cite{Horava:2009uw}, for recent review and extensive list of references, see \cite{Wang:2017brl}. There is also an interesting connection between Ho\v{r}ava-Lifshitz gravity
and Newton-Cartan gravity \cite{Hartong:2016yrf,Hartong:2015zia}. Newton-Cartan gravity is covariant and geometric reformulation of Newton gravity that is now very intensively studied, see for example
\cite{Bergshoeff:2017dqq,Bergshoeff:2017btm,Bergshoeff:2016lwr,
    Hartong:2015xda,Bergshoeff:2015uaa,Andringa:2010it,Bergshoeff:2015ija,Bergshoeff:2014uea}
\footnote{See also
	\cite{Banerjee:2014pya,Banerjee:2014nja,Banerjee:2015rca,Banerjee:2016laq,Banerjee:2016bbm}.}

Concept of non-relativistic physics also  emerged
in string theory  when strings and branes were analyzed at special backgrounds. At these special points of string moduli spaces  non-relativistic symmetries emerge in natural way
 \cite{Gomis:2000bd,Danielsson:2000gi} . These actions were obtained by non-relativistic "stringy" limit where time direction and one spatial direction along the string are large. The stringy limit of superstring in $AdS_5\times S^5$ was also formulated in
\cite{Gomis:2005pg} and it was argued here that it provides another soluble sector of AdS/CFT correspondence, for related work, see \cite{Sakaguchi:2006pg,Sakaguchi:2007zsa}.  Non-relativistic limit was further extended to the case of higher dimensional objects in string theory, as for example p-branes
\cite{Gomis:2004pw,Kluson:2006xi,Brugues:2004an,Gomis:2005bj}\footnote{For recent works, see
        \cite{Batlle:2016iel,Gomis:2016zur,Batlle:2017cfa,Kluson:2017djw,Kluson:2017ufb,Kluson:2017vwp}.}.

It is important to stress that these constructions of non-relativistic
objects are based on manifest separation of directions along which  the non-relativistic limit is taken and directions that are transverse to them. The first step for the more covariant formulation which corresponds to the particle in Newton-Cartan background was performed in \cite{Bergshoeff:2014gja} and further elaborated in \cite{Bergshoeff:2015uaa}. The structure of this action is very interesting and certainly deserves further study. In particular, it would be very useful to find Hamiltonian for this particle.
Our goal in the first part of this article
is to find the Hamiltonian formulation of the particle in Newton-Cartan and in Newton-Cartan-Hooke background. It turns out that this is non-trivial task
even in the bosonic case due to the complicated structure of the action. On the other hand when we determine Hamiltonian constraint we find that the canonical structure of this theory is trivial due to the fact that there is only one scalar first class constraint. A more interesting situation occurs when
we consider supersymmetric generalization of the non-relativistic particle.
As the first case we study Galilean  superparticle whose action was proposed in   \cite{Bergshoeff:2014gja}. We find its Hamiltonian form and identify
primary constraints. We show that fermionic constraints are the second class constraints that can be solved for the momenta conjugate to fermionic variables when we also obtain non-trivial Dirac brackets between fermionic variables.
Next we consider more interesting case corresponding to the $\kappa-$symmetric
non-relativistic particle action \cite{Bergshoeff:2014gja}. We again determine all primary constraints. Then the requirement of the preservation of the fermionic primary constraints determines corresponding Lagrange multipliers. In fact we find a linear combination of the fermionic constraints that is the first class constraint and that can be interpreted as the generator of the
$\kappa-$symmetry. Since it is the first class constraint it can be fixed
by imposing one of the fermionic variables to be equal to zero and we return to the previous case.

Finally we perform Hamiltonian analysis of superparticle in Newton-Cartan  background. This action was found in \cite{Bergshoeff:2015uaa} up to terms quadratic in fermions. We again determine Hamiltonian constraint which however
has much more complicated form  due to the presence of fermions . We also determine two sets of primary fermionic constraints. As the next step we study the requirement of the preservation of these constraints during the time evolution of the system. It turns out that this is rather non-trivial and complicated task in the full generality and hence we restrict ourselves to the simpler case of the background with the flat spatial sections. In this case we
show that the Hamiltonian constraint is the first class constraint with
vanishing Poisson brackets with fermionic constraints. We also identify linear combination of the fermionic constraints which is the first class constraint and that can be interpreted as the generator of $\kappa-$symmetry.

The extension of this paper is obvious. It would be very interesting to analyze constraint structure of superparticle in Newton-Cartan background in the full generality. Explicitly, we should analyze the time evolutions of all constraints and determine conditions on the background fields with analogy with the case of relativistic superparticle as was studied in \cite{Shapiro:1986xp}. We hope to return to this problem in future.

The organization of this paper is as follows. In the next section
(\ref{second}) we perform Hamiltonian analysis of particle in
Newton-Cartan background. Then in section (\ref{third}) we
generalize this analysis to the case of particle in
Newton-Cartan-Hooke background. In section (\ref{fourth}) we perform
Hamiltonian analysis of Galilean superparticle. Finally in section
(\ref{fifth}) we analyze Non-Relativistic Superparticle in
Cartan-Newton Background.
\section{Hamiltonian of Newton-Cartan Particle}\label{second}
By Newton-Cartan Particle we mean a particle that moves in
Newton-Cartan background with an action invariant under general
coordinate transformations. In order to describe Newton-Cartan
background we need a temporal vielbein $\tau_\mu$ and spatial
vielbein $e_\mu^{ \ a} \  , \mu=0,\dots,d-1 \ , a=1,\dots,d-1$. We
also need a central charge gauge field $m_\mu$. Then the action for
particle in Newton-Cartan background   has the form
\cite{Bergshoeff:2015uaa}
\begin{equation}\label{actnonBerg}
S=\frac{m}{2}\int d\lambda
\left[\frac{\dot{x}^\mu e_\mu^{ \ a}\dot{x}^\nu e_\nu^{ \ b}\delta_{ab}}
{\dot{x}^\rho \tau_\rho}-2m_\mu\dot{x}^\mu\right] \ ,
\end{equation}
where $\lambda$ is a parameter that labels the world-line of the particle and $\dot{x}^\mu=\frac{dx^\mu}{d\lambda}$.

 From (\ref{actnonBerg})  we derive following
conjugate momenta
\begin{equation}\label{momenta}
p_\mu= m\frac{e_\mu^{ \ a}\dot{x}^\nu e_\nu^{ \ b}\delta_{ab}}{
    \dot{x}^\rho \tau_\rho}-mm_\mu
-\frac{m}{2}\frac{\dot{x}^\rho e_\rho^{ \ a}\dot{x}^\nu
    e_{\nu}^{ \ b}\delta_{ab}}{(\dot{x}^\rho \tau_\rho)^2}\tau_\mu \ .
\end{equation}
Using (\ref{momenta}) it is easy to see that the bare Hamiltonian is equal to
zero
\begin{equation}
H_B=p_\mu \dot{x}^\mu-L=0
\end{equation}
as we could expect since the action (\ref{actnonBerg}) is manifestly invariant under world-line reparameterization
\begin{equation}
\lambda'=\lambda'(\lambda) \ , \quad  x'^\mu (\lambda')=x^\mu(\lambda) \ .
\end{equation}
The fact that the theory is invariant under world-line diffeomorphism suggests that there should exist corresponding Hamiltonian constraint. In order to find it we use following
 relations
\begin{equation}
\tau^\mu\tau_\mu=1 \ , \quad \tau^\mu e_\mu^{ \ a}=0
\end{equation}
in (\ref{momenta}) and
we obtain
\begin{eqnarray}
\tau^\mu (p_\mu+mm_\mu)&=&-\frac{m}{2}
\frac{\dot{x}^\mu e_\mu^{ \ a}\dot{x}^\nu
    e_{\nu}^{ \ b}\delta_{ab}}{(\dot{x}^\rho \tau_\rho)^2} \ , \nonumber \\
e^\mu_{ \ a}(p_\mu+mm_\mu)&=&
m\frac{\delta_{ab}e_\nu^{ \ b}\dot{x}^\nu}{\tau_\rho \dot{x}^\rho} \ .
\nonumber \\
\end{eqnarray}
If we combine these results together we obtain following primary constraint
\begin{eqnarray}\label{Hamconorg}
\mathcal{H}_\lambda=
e^\mu_{ \ a}(p_\mu+mm_\mu)e^\nu_{ \ b}(p_\nu+mm_\nu)\delta^{ab}+2m \tau^\mu(p_\mu+mm_\mu) \approx 0
\nonumber \\
\end{eqnarray}
which is desired result. Then extended Hamiltonian has the form
\begin{equation}
H_T=\lambda^\lambda \mathcal{H}_\lambda \ ,
\end{equation}
where $\lambda^\lambda$ is a Lagrange multiplier corresponding to the first class
constraint $\mathcal{H}_\lambda$. The equations of motion for $x^\mu,p_\mu$ have the form
\begin{eqnarray}
\dot{x}^\mu&=&\pb{x^\mu,H_T}=
2\lambda^\lambda \left( h^{\mu\nu}(p_\nu+mm_\nu)+2m\tau^\mu \right)\ ,  \nonumber \\
\dot{p}_\mu&=&\pb{p_\mu,H_T}=
\lambda^\tau\left(\partial_\mu h^{\rho\sigma}(p_\rho+mm_\rho)(p_\sigma+mm_\sigma)
-2m\partial_\mu m_\rho h^{\rho\sigma}(p_\sigma+mm_\sigma)\right.-\nonumber \\
&-&\left.
2m\partial_\mu \tau^\rho (p_\rho+mm_\rho)-2m^2\tau^\rho \partial_\mu
m_\rho\right)  \ , \nonumber \\
\end{eqnarray}
where
\begin{equation}
h^{\mu\nu}=e^\mu_{ \ a}e^\nu_{ \ b}\delta^{ab} \ .
\end{equation}
As the next step  we would like to present an alternative way how to derive  an action for Newton-Cartan particle. We start with an action for  relativistic particle
in general background
\begin{equation}\label{actrel}
S=-M\int d\lambda( \sqrt{-\eta_{AB}E_\mu^{ \ A}E_\nu^{ \ B}\dot{x}^\mu
    \dot{x^\nu}}-M_\mu\dot{x}^\mu) \ ,
\end{equation}
where $E_\mu^{ \ A}$ is vielbein where $A=0,\dots,d-1$ and where exists inverse vielbein $E^\mu_{ \ A}$ defined as
\begin{equation}
E_\mu^{ \ A}E^\nu_{ \ B}=\delta_B^A \ ,  \quad
E_\mu^{ \ A}E^\nu_{ \ A}=\delta_\mu^\nu \ .
\end{equation}
Now we find corresponding canonical form of the action (\ref{actrel}). To do this we derive conjugate momenta from (\ref{actrel})
\begin{equation}
p_\mu=M\frac{\eta_{AB}E_\mu^{ \ A}E_\nu^{ \ B}\dot{x}^\nu}
{\sqrt{-\eta_{AB}E_\mu^{ \ A}E_\nu^{ \ B}\dot{x}^\mu \dot{x}^\nu}}+MM_\mu
\end{equation}
that implies following primary constraint
\begin{eqnarray}\label{constrel}
\mathcal{H}_\lambda\equiv   (p_\mu-MM_\mu)E^\mu_{ \ C}\eta^{CA}E^\nu_{ \ A}
(p_\nu-MM_\nu)+M^2\approx  0  \ .
\nonumber \\
\end{eqnarray}
Now we are ready to find Hamiltonian constraint for Newton-Cartan particle
using following form of the vielbein $E_\mu^{ \ A}$ that was introduced
in \cite{Bergshoeff:2015uaa}
\begin{equation}\label{bergans}
E_\mu^{ \ 0}=\omega \tau_\mu+\frac{1}{2\omega}m_\mu \ , \quad  E_\mu^{ \ a}=
e_\mu^{ \ a} \ ,
\end{equation}
where $a=1,\dots,d-1$ and where $\omega$ is a free parameter and we  take  $\omega\rightarrow \infty$
in the end.
An inverse vielbein to (\ref{bergans}) has the form
\begin{eqnarray}\label{inbergans}
E^\mu_{ \ a}=e^\mu_{ \ a} \  , \quad  E^\mu_{ \ 0}=\frac{1}{\omega}\tau^\mu \ ,
\nonumber \\
\end{eqnarray}
where $e^\mu_{ \ a},\tau^\mu$ are defined as
\cite{Bergshoeff:2015uaa}
\begin{equation}\label{etau}
e^\mu_{ \ a}e_\mu^{ \ b}=\delta^a_b \ , \quad  \tau^\mu\tau_\mu=1 \ , \quad
\tau^\mu e_\mu^{ \ a}=\tau_\mu e^\mu_{ \ a}=0  \ , \quad
e^\rho_{ \ a}e_\mu^{ \ a}=\delta^\rho_\mu-\tau_\mu \tau^\rho \ .
\end{equation}
Note that (\ref{inbergans}) contain term linear in $\frac{1}{\omega}$. In principle it contains infinite number of terms of corrections $\frac{1}{\omega^n}$ that will give vanishing contributions in the limit $\omega\rightarrow \infty$.
Finally the gauge field has the form   \cite{Bergshoeff:2015uaa}
\begin{equation}\label{gaugeberg}
M_\mu=\omega \tau_\mu-\frac{1}{2\omega}m_\mu \ .
\end{equation}
Inserting (\ref{inbergans}) together with (\ref{gaugeberg}) into
(\ref{constrel}) and rescaling $M$ as $M=\omega m$ we obtain
\begin{eqnarray}
\mathcal{H}_\lambda
&=&-\frac{1}{\omega^2}p_\mu \tau^\mu p_\nu \tau^\nu+p_\mu e^\mu_{ \ a}\delta^{ab}e^\nu_{ \ b}p_\nu+2mp_\mu \tau^\mu +m p_\mu h^{\mu\nu}m_\nu-\frac{1}{\omega^2}
mp_\mu\tau^\mu \tau^\nu m_\nu-\nonumber \\
&-&m^2\omega^2+m^2m_\mu \tau^\mu-\frac{m^2}{4\omega^2}(m_\mu \tau^\mu)^2
+\frac{m^2}{4}m_\mu e^\mu_{ \ a}\delta^{ab}e^\nu_{ \ b}m_b+m^2\omega^2\approx 0
\nonumber \\
\end{eqnarray}
that in the limit $\omega\rightarrow \infty $ simplifies considerably
\begin{equation}\label{Hamconsalt}
\mathcal{H}_\lambda=2m(p_\mu+\frac{m}{2}m_\mu)\tau^\mu+(p_\mu+\frac{m}{2}m_\mu)
h^{\mu\nu}(p_\nu+\frac{m}{2}m_\nu)\approx 0 \ .
\end{equation}
After trivial redefinition $\frac{m_\mu}{2}\rightarrow
m_\mu$ the Hamiltonian constrain (\ref{Hamconsalt}) coincides with the
constraint (\ref{Hamconorg}). Using (\ref{Hamconsalt}) we find the canonical
form of the Newton-Cartan particle action
\begin{equation}\label{Sactcan}
S=\int d\lambda(p_\mu \dot{x}^\mu-v^\lambda \mathcal{H}_\lambda) \ .
\end{equation}
It is instructive to find corresponding Lagrangian. Using (\ref{Sactcan})
we obtain
\begin{equation}\label{dotxcan}
\dot{x}^\mu=\pb{x^\mu,H_T}=2v^\lambda (m\tau^\mu+h^{\mu\nu}(p_\nu+mm_\nu))
\end{equation}
so that
\begin{equation}\label{Ldotx}
L=p_\mu\dot{x}^\mu-H_T=v^\lambda(p_\mu+mm_\mu)h^{\mu\nu}(p_\nu+mm_\nu)-
mm_\mu \dot{x}^\mu \ .
\end{equation}
Of course, this is not final form of the Lagrangian since it has to be function
of $x^\mu$ and $\dot{x}^\mu$ instead of $p^\mu$. In order to find this form we
 multiply (\ref{dotxcan}) with
 $\tau_\mu$ and use  (\ref{etau}) so that we can express $v^\lambda$ as
\begin{equation}\label{vlambda}
v^\lambda=\frac{1}{2m}\tau_\mu \dot{x}^\mu  \ .
\end{equation}
Further, if we multiply  (\ref{dotxcan}) with $h_{\mu\nu}$ we obtain
\begin{equation}
\frac{1}{2}h_{\mu\nu}\dot{x}^\nu=v^\lambda (p_\mu+mm_\mu)-v^\lambda \tau_\mu
\tau^\rho (p_\rho+mm_\rho)
\end{equation}
Inserting this result into (\ref{Ldotx}) and using again (\ref{etau})
together with (\ref{vlambda})
we finally obtain
\begin{equation}
L=\frac{m}{2}\frac{\dot{x}^\mu h_{\mu\nu}\dot{x^\nu}}{\dot{x}^\rho \tau_\rho}-mm_\mu\dot{x}^\mu
\end{equation}
that coincides with the Lagrangian defined in (\ref{actnonBerg}).
 \section{The Newton-Cartan-Hooke Particle}\label{third}
 In this section we perform Hamiltonian analysis of the Newton-Cartan-Hooke
 particle which is cosmological extension of the Newton-Cartan particle
 whose action was derived  in
\cite{Bergshoeff:2014gja}
 \begin{equation}\label{NCHaction}
 S=\int d\lambda \frac{m}{2}
 \left[\frac{\dot{x}^\mu e_\mu^{ \ a}\dot{x}^\nu
    e_{\nu}^{ \ b}\delta_{ab}}
 {\dot{x}^\rho \tau_\rho}-2m_\mu \dot{x}^\mu
 -\dot{x}^\rho \tau_\rho \frac{x^\mu e_\mu^{ \ a}
 x^\nu e_\nu^{ \ b}\delta_{ab}}{R^2}\right] \ ,
 \end{equation}
 where the $AdS$ radius $R^2$ is related to the cosmological constant
 $\Lambda, \Lambda<0$ as $R^2=-\frac{1}{\Lambda}$.
From
 (\ref{NCHaction}) we find conjugate momenta
 \begin{equation}
 p_\mu=m\frac{e_\mu^{ \ a}\dot{x}^\nu e_\nu^{ \ b}\delta_{ab}}
 {\dot{x}^\rho \tau_\rho}-\frac{m}{2}
 \frac{\dot{x}^\rho e_\rho^{ \ a}\dot{x}^\nu
    e_{\nu}^{ \ b}\delta_{ab}}
 {(\dot{x}^\rho \tau_\rho)^2}\tau_\mu-mm_\mu-\frac{m}{2R^2}\tau_\mu
 x^\rho e_\rho^{ \ a}x^\nu e_\nu^{ \ b}\delta_{ab} \ .
 \end{equation}
 Following analysis performed in previous section we determine various
 projectors of the combination $p_\mu+mm_\mu$
 \begin{eqnarray}
\tau^\mu (p_\mu+mm_\mu)&=&-\frac{m}{2}
 \frac{\dot{x}^\rho e_\rho^{ \ a}\dot{x}^\nu e_\nu^{ \ b}\delta_{ab}}{
 (\dot{x}^\rho \tau_\rho)^2}-\frac{m}{2R^2}x^\rho e_\rho^{ \ a}x^\nu e_\nu^{ \ b}
\delta_{ab} \ , \nonumber \\
e^\mu_{ \ a}(p_\mu+mm_\mu)&=&m\frac{\dot{x}^\nu e_\nu^{ \ b}\delta_{ba}}{\dot{x}^\rho
\tau_\rho} \ .
\nonumber \\
\end{eqnarray}
If we combine these two relations we obtain following primary
Hamiltonian constraint for the Newton-Cartan-Hooke particle
 \begin{equation}
 \mathcal{H}_\lambda=2m\tau^\mu (p_\mu+mm_\mu)+(p_\mu+mm_\mu)h^{\mu\nu}
 (p_\nu+mm_\nu)+\frac{m^2}{R^2}h_{\mu\nu}x^\mu x^\nu \approx 0 \ .
 \end{equation}
  \section{Supersymmetric Generalization }\label{fourth}
 In this section we proceed to the Hamiltonian analysis of
 non-relativistic superparticles whose actions were derived in
 \cite{Bergshoeff:2014gja} and in \cite{Bergshoeff:2015uaa}.
 We begin with the simplest case which is Galilean Superparticle.
 We restrict ourselves to the case of three dimensions as in
 \cite{Bergshoeff:2014gja}.
 \subsection{The Galilean Superparticle}
 The action for Galilean Superparticle has the form
  \footnote{Note that we use Majorana representation where
    all gamma matrices are real $\gamma^\mu=(i\sigma_2,\sigma_1,\sigma_3)$, or explicitly
    $\gamma^0=\left(\begin{array}{cc}
    0 & 1 \\
    -1 & 0 \\ \end{array}\right) \ ,
    \gamma^1=\left(\begin{array}{cc}
    0 & 1 \\
    1 & 0 \\ \end{array}\right) \ , \gamma^2=
    \left(\begin{array}{cc}
    1 & 0 \\
    0 & -1 \\ \end{array}\right)$ that obey the standard relation
    $\gamma^a\gamma^b+\gamma^b\gamma^a=2\eta^{ab}\mathbf{I} \ ,
    \eta^{ab}=\mathrm{diag}(-1,1,1), \mathbf{I}=\left(\begin{array}{cc}
    1 & 0 \\
    0 & 1 \\ \end{array}\right)$.
    Then we presume that
    $\theta_-$ is real Majorana spinor so that
    \begin{equation}
    \bar{\theta}_-=\theta^T_-\gamma^0 \ .
    \end{equation}}
  \begin{equation}\label{Sgall}
 S=\int d\lambda \frac{m}{2}
 \left[\frac{\dot{x}^i\dot{x}_i}{\dot{t}}-\bar{\theta}_-
 \gamma^0 \dot{\theta}_-\right] \ .
 \end{equation}
 From (\ref{Sgall}) we determine corresponding conjugate momenta
\begin{eqnarray}
p_i&=&\frac{\partial L}{\partial \dot{x}^i}=m\frac{\dot{x}_i}{\dot{t}} \ ,  i=1,2 \ ,  \quad
p_t=\frac{\partial L}{\partial \dot{t}}=
-\frac{m}{2}\frac{\dot{x}^i\dot{x}_i}{\dot{t}^2} \ , \nonumber \\
 p_\alpha^-&=&\frac{\partial^L L}{\partial \dot{\theta}_-^{\alpha}}=
\frac{m}{2}(\bar{\theta}_-\gamma^0)_\alpha \ , \alpha=1,2  \ ,
\nonumber \\
\end{eqnarray}
 where $\partial^L$ means left-derivative. As usually we find
 that the bare Hamiltonian is zero
 \begin{equation}
 H_B=\dot{x}^ip_i+\dot{t}p_t+\dot{\theta}^\alpha_-p_\alpha^--L=0
 \end{equation}
 while the theory possesses three primary constraints
 \begin{eqnarray}
 \mathcal{G}_\alpha^-&\equiv& p_\alpha^--\frac{m}{2}(\bar{\theta}_-\gamma^0)_\alpha=
 p_\alpha^-+\frac{m}{2}\theta^\beta \delta_{\beta\alpha} \approx 0 \ , \nonumber \\
 \mathcal{H}_\lambda&\equiv& p_ip^i+2mp_t\approx 0 \ .
 \nonumber \\
 \end{eqnarray}
 It is important to stress that $p_\alpha^-$ and $\theta^\beta_-$ are
 Grassmann odd  variable with the graded Poisson  brackets
 \begin{equation}
 \pb{\theta^\alpha_-,p_\beta^-}=\pb{p_\beta^-,\theta^\alpha_-}=
 \delta_\alpha^\beta \ .
 \end{equation}
 Then it easy to find following
  non-zero Poisson bracket
 \begin{equation}
 \pb{\mathcal{G}_\alpha^-,\mathcal{G}_\beta^-}= m\delta_{\alpha\beta} \ .
 \end{equation}
 In other words  $\mathcal{G}_\alpha^-$ are second class constraints. As a result we can eliminate all conjugate momenta $p_\alpha^-$ with the help of $\mathcal{G}_\alpha^-=0$. Then of course we have to replace Poisson brackets between $\theta^\alpha_-,\theta^\beta_-$ with corresponding  Dirac brackets
 \begin{eqnarray}
 \pb{\theta^\alpha_-,\theta^\beta_-}_D=\pb{\theta^\alpha_-,
 \theta^\beta_-}-\pb{\theta^\alpha_-,\mathcal{G}_\gamma^-}
 \frac{1}{m}\delta^{\gamma\delta}\pb{\mathcal{G}_\delta^-,\theta^\beta_-}=
 -\frac{1}{m}\delta^{\alpha\beta} \ .
 \end{eqnarray}
  \subsection{$\kappa-$symmetric Galilean Superparticle}
 It is well known that the relativistic superparticle is invariant under additional fermionic symmetry called
 $\kappa-$symmetry \cite{deAzcarraga:1982dhu,Siegel:1983hh}.
 In case of strings and branes an existence of $\kappa-$
 symmetry is necessary for the correct counting of degrees of freedom. In the non-relativistic case it is known from
 the work \cite{Gomis:2004pw} that the $\kappa-$symmetry is just a St\"{u}ckelberg symmetry that acts as a shift on one of the fermionic coordinates, see also
\cite{Bergshoeff:2014gja}. Then this symmetry can be easily fixed by setting this fermionic coordinate to be equal to zero. It is instructive to see how it works in the canonical description so that we now perform Hamiltonian analysis of $\kappa-$symmetric  version of the flat Galilean superparticle action that depends on an additional fermionic coordinate $\theta_+$. The action has
the form \cite{Bergshoeff:2014gja}
 \begin{equation}\label{actkappagal}
 S=\int d\lambda\frac{m}{2}
 \left[\frac{\pi^i\pi_i}{\pi^0}-\bar{\theta}_-\gamma^0
 \dot{\theta}_-\right] \ ,
 \end{equation}
 where the line elements $\pi^0,\pi^i$ are defined as
 \begin{equation}
 \pi^0=\dot{t}+\frac{1}{4}\bar{\theta}_+\gamma^0
 \dot{\theta}_+ \ , \quad
 \pi^i=\dot{x}^i+\frac{1}{4}\bar{\theta}_-\gamma^i
 \dot{\theta}_++\frac{1}{4}\bar{\theta}_+\gamma^i
 \dot{\theta}_-
  \ .
  \end{equation}
 Now we proceed to the canonical formulation of this theory. From
 (\ref{actkappagal}) we obtain
 \begin{eqnarray}
p_i&=&\frac{\partial L}{\partial \dot{x}^i}=m\frac{\pi_i}{\pi^0} \ ,
 \quad
p_t=\frac{\partial L}{\partial \dot{t}}=-\frac{m}{2}\frac{\pi_i\pi^i}{(\pi^0)^2} \ ,
\nonumber \\
p_\alpha^-&=&\frac{\partial^L L}{\partial \dot{\theta}^\alpha_{-}}=
-\frac{m}{4}(\bar{\theta}_+\gamma^i)_\alpha\frac{\pi_i}{\pi^0}+
(\bar{\theta}_-\gamma^0)_\alpha \ , \nonumber \\
p_\alpha^+&=&\frac{\partial^L L}{\partial \dot{\theta}^\alpha_{+}}=
\frac{m}{8}(\bar{\theta}_+\gamma^0)_\alpha\frac{\pi^i\pi_i}{(\pi^0)^2}
-\frac{m}{4}\frac{\pi^i}{\pi^{0}}(\bar{\theta}_-\gamma^i)_\alpha  \ .
\nonumber \\
\end{eqnarray}
Last two equations imply two sets of primary constraints
\begin{eqnarray}
\mathcal{G}_\alpha^-=
p_\alpha^-+\frac{1}{4}(\bar{\theta}_+\gamma^i)_\alpha p_i-\frac{m}{2}
(\bar{\theta}_-\gamma^0)_\alpha \approx 0 \ , \nonumber \\
\mathcal{G}_\alpha^+=p_\alpha^++\frac{1}{4}(\bar{\theta}_+\gamma^0)_\alpha p_t+\frac{1}{4}
(\bar{\theta}_-\gamma^i)_\alpha p_i \approx 0 \
\nonumber \\
\end{eqnarray}
 with following non-zero Poisson brackets
\begin{eqnarray}
\pb{\mathcal{G}_\alpha^-,\mathcal{G}_\beta^-}=m\delta_{\alpha\beta} \ , \quad
\pb{\mathcal{G}_\alpha^-,\mathcal{G}_\beta^+}=\frac{1}{2}(\gamma^0\gamma^i)_{\alpha\beta}p_i \ ,
\quad
\pb{\mathcal{G}_\alpha^+,\mathcal{G}_\beta^+}=-\frac{1}{2}\delta_{\alpha\beta}p_t \ .
\nonumber \\
\end{eqnarray}
 Finally the theory possesses the Hamiltonian constraint in the form
 \begin{equation}
 \mathcal{H}_\lambda=2mp_t+p_ip^i\approx 0
 \end{equation}
 that is clearly first class constraint
 \begin{equation}
 \pb{\mathcal{H}_\lambda,\mathcal{G}_\alpha^+}=\pb{\mathcal{H}_\lambda,\mathcal{G}_\alpha^-}=0 \ .
 \end{equation}
Again the bare Hamiltonian is equal to zero so that the extended Hamiltonian
with the primary constraints included has the form
 \begin{equation}
 H_T=\lambda^\lambda \mathcal{H}_\lambda+\lambda^\alpha_+\mathcal{G}_\alpha^++
 \lambda^\alpha_-\mathcal{G}_\alpha^-  \ ,
 \end{equation}
 where $\lambda^\lambda,\lambda^\alpha_+,\lambda^\alpha_-$ are corresponding
 Lagrange multipliers.

 Now we have to check  consistency of all constraints. $\mathcal{H}_\lambda\approx 0$
 is clearly preserved and it is the generator of world-line reparameterization. In case of the fermionic constraints  we obtain
 \begin{eqnarray}\label{eqpresG}
 \frac{d}{d\lambda} \mathcal{G}_\alpha^-&=&\pb{\mathcal{G}_\alpha^-,H_T}=
 \frac{1}{2} \lambda^\beta_+ (\gamma^0\gamma^i)_{\alpha\beta}p_i+
\lambda^\beta_- \delta_{\alpha\beta}m=0 \ , \nonumber \\
\frac{d}{d\lambda} \mathcal{G}_\alpha^+&=&\pb{\mathcal{G}_\alpha^+,H_T}=
-\lambda^\beta_+ \frac{1}{2}\delta_{\alpha\beta} p_t+\lambda^\beta_-
\frac{1}{2}(\gamma^0\gamma^ i)_{\alpha\beta}p_i=0 \ .
 \nonumber \\
 \end{eqnarray}
 From the last equation we obtain
 \begin{equation}
 \lambda^\alpha_+=\frac{1}{p_t}\delta^{\alpha\beta}(\gamma^0\gamma^ i)_{\beta\gamma}
 \lambda^\gamma_- p_i \ .
 \end{equation}
 Inserting this result
 to the first equation (\ref{eqpresG})  we find that it is obeyed since it is proportional to
 \begin{equation}
 (p_ip^i+2mp_t)=\mathcal{H}_\lambda\approx 0
 \end{equation}
 so that $\lambda^\beta_-$ is not specified and it is a free parameter. Then
 it is natural to introduce following  linear combination of the fermionic constraints
\begin{equation}
\Sigma_\alpha=
\mathcal{G}_\gamma^+ \delta^{\gamma\beta}(\gamma^0\gamma^i)_{\beta\alpha}\frac{1}{p_t}+\mathcal{G}^-_\alpha \ .
\end{equation}
Then it is easy to see that
 \begin{eqnarray}
 \pb{\Sigma_\alpha,\mathcal{G}_\beta^-}=
 \frac{1}{2p_t}
 (p_ip^i+2mp_t)\delta_{\alpha\beta}\approx 0 \ ,  \quad
  \pb{\Sigma_\alpha,\mathcal{G}_\beta^+}=0
 \nonumber \\
 \end{eqnarray}
 which implies that $\pb{\Sigma_\alpha,\Sigma_\beta}\approx 0$ and hence
 $\Sigma_\alpha\approx 0$ is the first class constraint while $\mathcal{G}_\alpha^-\approx 0$ is the second class constraint. Let us then  calculate the Poisson brackets between all canonical variables and $\Sigma_\alpha$
 \begin{eqnarray}\label{pbSigma}
 \pb{\Sigma_\alpha,\theta^\beta_{+}}&=&(\gamma^0\gamma^i)_\alpha^\beta\frac{p_i}{p_t} \ ,  \quad
 \pb{\Sigma_\alpha,\theta^\beta_{-}}=\delta_\alpha^\beta \ ,
 \nonumber \\
 \pb{\Sigma_\alpha,x^i}&=&-\frac{1}{4p_t}(\theta_-\gamma^i\gamma^j)_\alpha p_j-\frac{1}{4}(\bar{\theta}_+ \gamma^i)_\alpha \ , \quad
 \pb{\Sigma_\alpha,t}=\frac{1}{4}(\bar{\theta}_+\gamma^j)_\alpha
 \frac{p_j}{p_t} \ . \nonumber \\
 \end{eqnarray}
 With analogy with the Lagrangian description we would like to define
 $\kappa-$transformation
 that acts on the $\theta_+$  as a shift. For that reason we  define $\kappa-$transformation  in the following way
 \begin{equation}
 \delta_\kappa X=\frac{p_j}{2m}(\kappa \gamma^j\gamma^0)^\alpha\pb{ \Sigma_\alpha,X}\equiv \kappa^\alpha\pb{\Theta_\alpha,X} \ .
 \end{equation}
 Then using (\ref{pbSigma}) we easily find
 \begin{eqnarray}\label{kappatr}
 \delta_\kappa \theta^\alpha_+&=&\kappa^\alpha \ , \quad
 \delta_\kappa\theta^\alpha_-=\frac{p_j}{2m}(\kappa
 \gamma^j\gamma^0)^\alpha \ , \nonumber \\
 \delta_\kappa x^i&=&\frac{1}{4}
 \bar{\kappa} \gamma^i\theta_--\frac{p_j}{8m}
 \kappa \gamma^j\gamma^i\theta_+  \ , \quad
 \delta_\kappa t=-\frac{1}{4}\kappa \theta_+ \  \nonumber \\
  \end{eqnarray}
  that are correct $\kappa-$symmetry transformations. Fixing the
  $\kappa-$symmetry can be done by imposing the gauge fixing condition
  \begin{equation}
  \Omega^\alpha=\theta^\alpha_+=0
  \end{equation}
  so that
  \begin{equation}
  \pb{\Theta_\alpha,\Omega^\beta}=\delta_\alpha^\beta
  \end{equation}
  and hence we see that they are second class constraints that can be explicitly solved for $\theta_+^\alpha$ and $p_\alpha^+$. Further,
  the constraints $\mathcal{G}_\alpha^-\approx 0$ are two second class constraints
  that can be solved for $p_\alpha^-$ exactly in the same way as in
  previous section.
  \section{Non-Relativistic Superpraticle in Newton-Cartan Background}\label{fifth}
  In this section we perform Hamiltonian analysis of the non-relativistic
  superparticle in Cartan-Newton Background. The action for this particle
  was derived in    \cite{Bergshoeff:2015uaa}
 and it has the form
\begin{eqnarray}
S=\frac{m}{2}\int d\lambda\left[\frac{\hat{\pi}^a\hat{\pi}^b\delta_{ab}}{\hat{\pi}^0}
-2\dot{x}^\mu (m_\mu-\bar{\theta}_-\gamma^0\psi_{\mu-})
-\bar{\theta}_-\gamma^0\hat{D}\theta_--\frac{1}{2}\dot{x}^\mu
\omega_\mu^{ \ a}\bar{\theta}_+\gamma_a \theta_-\right] \ ,
\nonumber \\
\end{eqnarray}
where we have following supersymmetric line elements
\begin{eqnarray}
\hat{\pi}^0&=&\dot{x}^\mu (\tau_\mu-\frac{1}{2}\bar{\theta}_+\gamma^0
\psi_{\mu+})+\frac{1}{4}\bar{\theta}_+\gamma^0\hat{D}\theta_+ \ ,
\nonumber \\
\hat{\pi}^a&=&\dot{x}^\mu(e_\mu^{ \ a}-\frac{1}{2}\bar{\theta}_+
\gamma^a \psi_{\mu -}-\frac{1}{2}\bar{\theta}_-\gamma^a
\psi_{\mu +})+\frac{1}{4}\bar{\theta}_+\gamma^a
\hat{D}\theta_-+\frac{1}{4}\bar{\theta}_-\gamma^a
\hat{D}\theta_++\nonumber \\
&+&\frac{1}{8}\bar{\theta}_+\gamma^a \gamma_{b0}
\theta_+ \dot{x}^\mu \omega_\mu^{ \ b} \ ,
\nonumber \\
\end{eqnarray}
 where $\hat{D}$ is covariant derivative with respect to spatial
 rotation
 \begin{equation}
 \hat{D}\theta_-=\dot{\theta}_--\frac{1}{4}\dot{x}^\mu
 \omega_{\mu}^{\ ab}\gamma_{ab}\theta_- \ .
 \end{equation}
Finally it is understood that we are interested in terms up to second order in fermions in $\hat{\pi}^a\hat{\pi}^b$. It is important to stress that spin connections
$\omega_\mu^{ \ a}, \omega_{\mu}^{ \ ab}$  depend on $e,\tau,m,\psi$. The explicit form of this dependence can  be found in \cite{Bergshoeff:2015uaa}.

 Now we can proceed to the Hamiltonian formalism. First of all we rewrite
 the line elements $\hat{\pi}^a$ and $\hat{\pi}^0$ as
 \begin{eqnarray}
 \hat{\pi}^a&=&\dot{x}^\mu \mathbf{M}_\mu^{ \ a}+\frac{1}{4}\bar{\theta}_+\gamma^a \dot{\theta}_-+
 \frac{1}{4}\bar{\theta}_-\gamma^a\dot{\theta}_+ \ , \nonumber \\
 \hat{\pi}^0&=&\dot{x}^\mu \mathbf{V}_\mu+\frac{1}{4}\bar{\theta}_+\gamma^0\dot{\theta}_+ \ ,
 \nonumber \\
 \end{eqnarray}
 where
 \begin{eqnarray}
 \mathbf{M}_\mu^{ \ a}&=&e_\mu^{ \ a}
 -\frac{1}{2}\bar{\theta}_+
 \gamma^a \psi_{\mu -}-\frac{1}{2}\bar{\theta}_-\gamma^a
 \psi_{\mu +}-\frac{1}{16}\bar{\theta}_+\gamma^a \omega_\mu^{ \ cd}\gamma_{cd}\theta_-
 -\frac{1}{16}\bar{\theta}_-\gamma^a \omega_\mu^{\ cd}\gamma_{cd}\theta_++
 \frac{1}{8}\bar{\theta}_+\gamma^a\gamma_{b0}\theta_+\omega_\mu^{ \ b} \ ,
 \nonumber \\
 \mathbf{V}_\mu&=&\tau_\mu-\frac{1}{2}\bar{\theta}_+\gamma^0\psi_{\mu+}-\frac{1}{16}
 \bar{\theta}_+\gamma^0\omega_\mu^{ \ ab}\gamma_{ab}\theta_+ \ . \nonumber \\
 \end{eqnarray}
 To proceed further we now  presume an existence of the inverse matrix $\mathbf{M}^\mu_{ \ b}$ to $\mathbf{M}_\mu^{ \ a}$ that obeys the relation
 \begin{equation}
 \mathbf{M}^\mu_{ \ b}\mathbf{M}_\mu^{ \ a}=\delta_b^a
 \ .
 \end{equation}
 In fact, since we are interested in terms up to second order in fermions
 we derive this matrix as follows. Let us denote original matrix and its inverse as
 \begin{equation}
 \mathbf{M}_\mu^{ \ a}=e_\mu^{ \ a}+\mathbf{V}_\mu^{ \ a}  \ ,  \quad
 \mathbf{M}^\mu_{ \ b}=e^\mu_{ \ b}+\mathbf{W}^\mu_{ \ b}  \ .
 \end{equation}
 Then the condition $\mathbf{M}^\mu_{ \ b}\mathbf{M}_\mu^{ \ a}=\delta_b^a$ implies
 \begin{eqnarray}
 \mathbf{M}^\mu_{ \ b}\mathbf{M}_\mu^{ \ a}=
 e^\mu_{ \ b}e_\mu^{ \ a}+e^\mu_{ \ b}\mathbf{V}_\mu^{ \ a}+
 \mathbf{W}^\mu_{ \ b}e_\mu^{ \ a}+O(\theta^4)=\delta_b^a
 \end{eqnarray}
 so that we obtain the condition
 \begin{equation}
 \mathbf{W}^\mu_{ \ b}e_\mu^{ \ a}=-e^\mu_{ \ b}\mathbf{V}_\mu^{ \ a} \ .
 \end{equation}
 In order to determine $\mathbf{W}^\mu_{ \ a}$ we multiply the last equation with  $e^\nu_{ \ a}$ and we obtain
\begin{eqnarray}
\mathbf{W}^\mu_{ \ b}(\delta_\mu^\nu-\tau_\mu \tau^\nu)=-e^\mu_{ \ b}\mathbf{V}_\mu^{ \ a}e^\nu_{ \ a}
\nonumber \\
\end{eqnarray}
 that has solution
 \begin{equation}
 \mathbf{W}^\mu_{ \ a}=-e^\nu_{ \ a}\mathbf{V}_\nu^{ \ b}e^\mu_{ \ b}
 \end{equation}
 since $e^\mu_{ \ b}\tau_\mu=0$.
  Using this notation we determine corresponding conjugate momenta
 \begin{eqnarray}\label{pmupalpha}
 p_\mu&=&m\frac{\mathbf{M}_\mu^{ \ a}\delta_{ab}\hat{\pi}^b}{\hat{\pi}^0}
 -\frac{m}{2}\frac{\hat{\pi}^a \hat{\pi}^b \delta_{ab}}{(\hat{\pi}^0)^2}\mathbf{V}_\mu-\nonumber \\
 &-&mm_\mu+\frac{m}{8}\bar{\theta}_-\gamma^0
 \omega_\mu^{ \ ab}\gamma_{ab}\theta_--\frac{m}{4}\omega_\mu^{ \ a}
 \bar{\theta}_+\gamma_a \theta_- \ ,
 \nonumber \\
  p_\alpha^-&=&\frac{\partial^L L}{\partial \dot{\theta}^\alpha_-}=
 -\frac{m}{4}(\bar{\theta}_+\gamma^b)_\alpha \delta_{ba}\frac{\hat{\pi}^b}{
    \hat{\pi}^0}-\frac{m}{2}\theta^\beta_{-}\delta_{\beta\alpha} \ ,
 \nonumber \\
 p_\alpha^+&=&\frac{\partial^L L}{\partial \dot{\theta}^\alpha_+}=
 -\frac{m}{4}(\bar{\theta}_-\gamma^a )_\alpha
 \frac{\delta_{ab}\hat{\pi}^b}{\hat{\pi}^0}+\frac{m}{8}\frac{\hat{\pi}^a
    \hat{\pi}^b\delta_{ab}}{(\hat{\pi}^0)^2}(\bar{\theta}_+\gamma^0)_\alpha \ .  \nonumber \\
 \end{eqnarray}
 To proceed further we have to introduce vector $\mathbf{T}^\mu$ that obeys the condition
 \begin{equation}\label{bTbM}
 \mathbf{T}^\mu \mathbf{M}_\mu^{ \ a}=0
 \ .
 \end{equation}
With analogy with the pure bosonic case we presume that it has the form
 \begin{equation}
 \mathbf{T}^\mu=\tau^\mu+\mathbf{W}^\mu \ ,
 \end{equation}
 where $\mathbf{W}^\mu$ is quadratic in fermions. Then the condition
 (\ref{bTbM}) implies
 \begin{equation}
 \tau^\mu \mathbf{V}_\mu^{ \ a}=-\mathbf{W}^\mu e_\mu^{ \ a}
 \end{equation}
 that can be again solved as
 \begin{equation}
 \mathbf{W}^\mu=-\tau^\nu \mathbf{V}_\nu^{ \ a}e^\mu_{ \ a} \ .
 \end{equation}
 Finally note that generally
 \begin{equation}
 \mathbf{T}^\mu \mathbf{V}_\mu\neq 1 \ .
 \end{equation}
 With the help of the knowledge of $\mathbf{T}^\mu \ , \mathbf{M}^\mu_{ \ a}$ we
 can proceed to the derivation of corresponding Hamiltonian constraint.
 Using $\mathbf{T}^\mu$ and $\mathbf{M}_\mu^{ \ a}$ we find
 \begin{eqnarray}\label{bMPi}
& & \frac{m}{2}\frac{\hat{\pi}^a \hat{\pi}^b\delta_{ab}}{(\hat{\pi}^0)^2}=
 -\frac{1}{\mathbf{T}^\mu \mathbf{V}_\mu} \mathbf{T}^\mu \Pi_\mu
 \ , \nonumber \\
& & \mathbf{M}^\mu_{ \ a}\Pi_\mu
=
 m\frac{\delta_{ab}\hat{\pi}^b}{\hat{\pi}^0}
 -\frac{m}{2}\frac{\hat{\pi}^a\hat{\pi}^b\delta_{ab}}{(\hat{\pi}^0)^2}
 \mathbf{M}^\mu_{ \ a}\mathbf{V}_\mu \ ,  \nonumber \\
 \end{eqnarray}
 where we defined $\Pi_\mu$ as
\begin{equation}
\Pi_\mu=
p_\mu+mm_\mu+\frac{m}{4}\omega_\mu^{ \ a}
\bar{\theta}_+\gamma_a \theta_--\frac{m}{8}\bar{\theta}_-
\gamma^0\omega_\mu^{ \ ab}\gamma_{ab}\theta_- \ .
\end{equation}
Inserting the first  equation in (\ref{bMPi}) into the second one
we obtain
 \begin{eqnarray}
\left( \mathbf{M}^\mu_{ \ a}-\frac{\mathbf{T}^\mu}{\mathbf{T}^\nu \mathbf{V}_\nu}\mathbf{M}^\rho_{ \ a} \mathbf{V}_\rho\right)\Pi_\mu
= m\frac{\delta_{ab}\hat{\pi}^b}{\hat{\pi}^0}
 \nonumber \\
 \end{eqnarray}
 so that  we obtain following Hamiltonian constraint
 \begin{eqnarray}
 \mathcal{H}_\lambda=
  ( \mathbf{M}^\mu_{ \ a}\mathbf{T}^\sigma \mathbf{V}_\sigma-\mathbf{T}^\mu\mathbf{M}^\rho_{ \ a} \mathbf{V}_\rho)\Pi_\mu
 \delta^{ab}( \mathbf{M}^\nu_{ \ a}\mathbf{T}^\sigma \mathbf{V}_\sigma-\mathbf{T}^\nu\mathbf{M}^\sigma_{ \ a} \mathbf{V}_\sigma)
 \Pi_\nu+2\mathbf{T}^\rho\mathbf{V}_\rho\mathbf{T}^\mu \Pi_\mu \approx 0 \ . \nonumber \\
  \end{eqnarray}
  Further, using (\ref{bMPi}) in the second and third equation in
 (\ref{pmupalpha})   we also find  fermionic primary constraints
 \begin{eqnarray}
 \mathcal{G}^-_\alpha&=&p^-_\alpha+\frac{m}{2}\theta^\beta_-\delta_{\beta\alpha}
  +\frac{1}{4}(\bar{\theta}_+\gamma^a)_\alpha
 (\mathbf{M}^\mu_{ \ a}-\frac{\mathbf{T}^\mu}{\mathbf{T}^\nu \mathbf{V}_\nu}\mathbf{M}^\rho_{ \ a}\mathbf{V}_\rho)\Pi_\mu\approx 0  \ ,
 \nonumber \\
 \mathcal{G}^+_\alpha&=&p^+_\alpha +\frac{1}{4}(\bar{\theta_-}\gamma^a)_\alpha
 ( \mathbf{M}^\mu_{ \ a}-\frac{\mathbf{T}^\mu}{\mathbf{T}^\nu \mathbf{V}_\nu}\mathbf{M}^\rho_{ \ a} \mathbf{V}_\rho)
\Pi_\mu
+\frac{1}{4}(\bar{\theta}_+\gamma^0)_\alpha \mathbf{T}^\mu \Pi_\mu  \approx 0 \ .  \nonumber \\
 \end{eqnarray}
 As the next step we should analyze the requirement of the preservation
 of all constraints during the time evolution of the system. It is a difficult
 task in the full generality due to the complicated form of these constraints.
 For that reason we restrict ourselves to the special case when some of the background fields vanish
\begin{equation}
 \psi_{\mu-}=\psi_{\mu+}=\omega_\mu^{ \ cd}=\omega_\mu^{ \ a}=0 \
 \end{equation}
 when we have
 \begin{equation}
 \mathbf{M}_\mu^{ \ a}=e_\mu^{ \ a} \ , \mathbf{V}_\mu=\tau_\mu \  .
 \end{equation}
 Then  all constraints simplify considerably
 \begin{eqnarray}
 \mathcal{H}_\lambda&=&(p_\mu+mm_\mu)h^{\mu\nu}(p_\nu+mm_\nu)+2m \tau^\mu(p_\mu+mm_\mu)\approx 0
 \ , \nonumber \\
 \mathcal{G}_\alpha^-&=&p_\alpha^-+\frac{m}{2}\theta^\beta_-\delta_{\beta\alpha}+
 \frac{1}{4}(\bar{\theta}_+
 \gamma^a)_\alpha e^\mu_{ \ a}(p_\mu+mm_\mu)\approx 0 \ ,
 \nonumber \\
 \mathcal{G}_\alpha^+&=&p_\alpha^++\frac{1}{4}(\bar{\theta}_-\gamma^a)_\alpha e^\mu_{ \ a}(p_\mu+mm_\mu)
 +\frac{1}{4}(\bar{\theta}_+\gamma^0)_\alpha \tau^\mu(p_\mu+mm_\mu) \approx 0 \ .
 \nonumber \\
 \end{eqnarray}
 Now we have to calculate Poisson brackets between all constraints. We start with the following one
 \begin{eqnarray}
 \pb{\mathcal{G}_\alpha^-,\mathcal{H}_\lambda}&=&

 %
 \frac{1}{4}(\bar{\theta}_+\gamma^a)^\alpha
 ( (\partial_\rho e^\mu_{ \ a}+\partial_\rho e^\sigma_{ \ a}) h^{\rho\sigma}
 -e^\rho_{ \ a}\partial_\rho h^{\mu\sigma})(p_\mu+mm_\mu)(p_\sigma+mm_\sigma)+
 \nonumber \\
 &+&  \frac{m}{2}(\bar{\theta}_+\gamma^a)^\alpha (\partial_\rho e^\mu_{ \ a}\tau^\rho-e^\rho_{ \ a}
 \partial_\rho \tau^\mu)(p_\mu+mm_\mu) \ . \nonumber \\
 \end{eqnarray}
 It is important to stress that the fact that we restrict to the case $\omega_\mu^{\ ab}=
 \omega_\mu^{ \ a}=0$ implies, since they depend on spatial derivatives of $e_\mu^{ \ a},\tau^\mu$,  that we should impose the condition that $e_\mu^{ \ a},m_\mu,\tau^\mu$ do not depend on $x^\mu$ as well. Then the  equation given above  implies that    $\pb{\mathcal{G}_\alpha^-,\mathcal{H}_\lambda}=0$. In the same way we find that
 \begin{equation}
 \pb{\mathcal{G}_\alpha^+,\mathcal{H}_\lambda}=0 \
 \end{equation}
 and also
 \begin{eqnarray}
 \pb{\mathcal{G}_\alpha^-,\mathcal{G}_\beta^-}&=&m\delta_{\alpha\beta} \ ,
 \quad
 \pb{\mathcal{G}^+_\alpha,\mathcal{G}^+_\beta}=-\frac{1}{2}\delta^{\alpha\beta}\tau^\mu
 (p_\mu+mm_\mu) \ ,
 \nonumber \\
 \pb{\mathcal{G}^-_\alpha,\mathcal{G}^+_\beta}&=&\frac{1}{2}(\gamma^0\gamma^a)_{\alpha\beta}
 e^\mu_{ \ a}(p_\mu+mm_\mu) \ . \nonumber \\
 \end{eqnarray}
Finally we have to determine the time evolution of these constraints when we take into account that the total Hamiltonian is the sum of all primary constraints
 \begin{equation}
 H_T=\lambda^\lambda \mathcal{H}_\lambda+\lambda^\alpha_- \mathcal{G}_\alpha^-+\lambda^\alpha_+
 \mathcal{G}_\alpha^+ \ .
 \end{equation}
 $\mathcal{H}_\lambda$ is preserved automatically while the time evolution of $\mathcal{G}^\alpha_-,\mathcal{G}^\alpha_+$ is governed by following equations
 \begin{eqnarray}
 \dot{\mathcal{G}}^-_\alpha&=&\pb{\mathcal{G}_\alpha^-,H_T}=
\lambda^\beta_+ \frac{1}{2}(\gamma^0\gamma^a)_{\beta\alpha}e^\mu_{ \ a}
 (p_\mu+mm_\mu)+m\lambda^\beta_- \delta_{\beta\alpha}=0  \ ,
 \nonumber \\
 \dot{\mathcal{G}}_\alpha^-&=&\pb{\mathcal{G}_\alpha^-,H_T}
 =\frac{1}{2}\lambda^\beta_-(\gamma^0\gamma^a)_{\alpha\beta}e^\mu_{ \ a}
 (p_\mu+mm_\mu)-\frac{1}{2}\lambda^\beta_+ \delta_{\alpha\beta}
 \tau^\mu (p_\mu+mm_\mu)=0 \ .  \nonumber \\
 \end{eqnarray}
 From the last equation we express $\lambda^\beta_+$ as
 \begin{equation}
 \lambda^\beta_+=\delta^{\beta\gamma}(\gamma^0\gamma^a)_{\gamma\delta}
 \lambda^\delta_- e^\mu_{ \ a}
 (p_\mu+mm_\mu)\frac{1}{\tau^\mu(p_\mu+mm_\mu)}
 \end{equation}
 and inserting back to the first equation we obtain
 \begin{equation}
 \dot{\mathcal{G}}_\alpha^-=
((p_\mu+mm_\mu)h^{\mu\nu}(p_\nu+mm_\nu)+2m\tau^\mu (p_\mu+mm_\mu))\lambda^\beta_-\delta_{\beta\alpha}=
\mathcal{H}_\lambda \lambda^\beta_- \delta_{\beta\alpha}\approx 0
\end{equation}
 and we see that this equation is obeyed for all $\lambda^\beta_-$ on the constraint surface $\mathcal{H}_\lambda\approx 0$.  This fact again
 implies an existence of the first class constraint in the form
 \begin{equation}
\Sigma_\alpha=\mathcal{G}_\alpha^-
\tau^\mu (p_\mu+mm_\mu)
+\mathcal{G}_\beta^+\delta^{\beta\gamma}(\gamma^0\gamma^a)_{\gamma
\alpha}e^\mu_{ \ a}(p_\mu+mm_\mu)
 \end{equation}
 that has following Poisson brackets with canonical variables
 \begin{eqnarray}\label{pbSigmagen}
 \pb{\Sigma_\alpha,x^\mu}
 &=&-\frac{1}{4}(\bar{\theta}_+\gamma^a)_\alpha
 e^\mu_{ \ a}\tau^\nu(p_\nu+mm_\nu)+\nonumber \\
 &+&\frac{1}{4}\tau^\mu(\bar{\theta}_+\gamma^b)_\alpha
 e^\nu_{ \ b}(p_\nu+mm_\nu)-\frac{1}{4}(\theta_-\gamma^a\gamma^b)_\alpha e^\mu_{ \ a}
 e^\nu_{ \ b}(p_\nu+mm_\nu) \ , \nonumber \\
 \pb{\Sigma_\alpha,\theta_+^\beta}&=&(\gamma^0\gamma^a)_{\alpha\omega}\delta^{\omega\beta}
 e^\mu_{ \ a}(p_\mu+mm_\mu) \ , \nonumber \\
 \pb{\Sigma_\alpha,\theta^\beta_{-}}&=&\delta_\alpha^{\beta}\tau^\mu (p_\mu+mm_\mu) \ .
 \nonumber \\
 \end{eqnarray}
 Now we define $\kappa$ variation as
 \begin{equation}
 \delta_\kappa X=
 -\frac{1}{2m\tau^\mu(p_\mu+mm_\mu)}(\bar{\kappa}\gamma^a)^\alpha e^\mu_{ \ a}(p_\mu+mm_\mu)
  \pb{\Sigma_\alpha,X} \ .
 \end{equation}
Then with the help of   (\ref{pbSigmagen}) we obtain following
transformation rules
 \begin{eqnarray}\label{kappagen}
 \delta_\kappa \theta^\alpha_+&=&
 \kappa^\alpha \ ,  \quad
 \delta_\kappa \theta_-^\alpha=-(\bar{\kappa}\gamma^a)^\alpha e^\mu_{ \ a}(p_\mu+mm_\mu) \ ,
 \nonumber \\
 \delta_\kappa x^\mu&=&
\frac{1}{8m}(\kappa \gamma^a\gamma^b\theta_+)e^\nu_{ \ a}(p_\nu+mm_\nu)e^\mu_{ \ b}-\frac{1}{4}\tau^\mu \kappa \theta_+
-\frac{1}{4}(\bar{\kappa}\gamma^a\theta_-)e^\mu_{ \ a}
\nonumber \\
 \end{eqnarray}
 that are covariant form of   $\kappa-$transformations
 (\ref{kappatr}). To see this explicitly note that in the
 flat non-relativistic space-time we can choose $\tau^0=1 \ , \tau^i=0,
 e_\mu^{ a}=\delta_i^a$ and the $\kappa-$transformations
 given in (\ref{kappagen}) have explicit form
   \begin{eqnarray}
 \delta_\kappa \theta_+^\alpha&=&
 \kappa^\alpha \ ,  \quad
 \delta_\kappa \theta^\alpha_-=-(\bar{\kappa}\gamma^i)^\alpha (p_i+mm_i) \ ,
 \nonumber \\
 \delta_\kappa x^i&=&
 \frac{1}{8m}(\kappa \gamma^j\gamma^i)(p_j+mm_j)
 -\frac{1}{4}(\bar{\kappa}\gamma^i\theta_-) \ ,  \quad
 \delta_\kappa t=-\frac{1}{4}\kappa \theta_+
 \nonumber \\
 \end{eqnarray}
 that coincide with the transformations (\ref{kappatr}) which is a  nice
 consistency check.

 In this section we performed Hamiltonian analysis of the non-relativistic
 superparticle in Newton-Cartan background. We derived general form of the Hamiltonian and fermionic constraints. Then we studied their properties
 for special configurations of background fields. It would be extremely interesting to analyze this theory for general background. We expect that the
 requirement of the existence of three first class constraints, where one is Hamiltonian constraint while remaining two constraints correspond to generators of $\kappa-$symmetry, will impose some restriction on the background fields.
 We hope to return to this analysis in near future.

\acknowledgments{This  work  was
    supported by the Grant Agency of the Czech Republic under the grant
    P201/12/G028. }

redpajama_set_name: RedPajamaArXiv


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text:
American Crime Story: Impeachment, Awards, Breaking News, Emmy, Emmys, Obi-Wan Kenobi, Predictions, Ted Lasso, Television, The Morning Show, The White Lotus, TV News
Emmy Predictions 2022: What Could Be Grabbing The Gold A Year From Now?
With the 2021 Emmys in the rear view and the Fall television season is in full swing, what better time is there to look forward and make a few bold predictions about what we might see being honored when the Emmys roll around again come September of 2022?
If this sounds like fool's folly, think again. The Television Academy actually starts official screenings and "campaign" season right after the first of the year, and already there are a number of logical contenders rolling out. In fact even now at the beginning of October, I have already found the show I will vote for in the all-important Outstanding Limited Series category.
That is because I was — and am — among the many who found HBO's summer series, The White Lotus simply irresistible.
For me, having been to the Four Seasons in Maui where the fictional White Lotus was filmed, it was like taking a vacation every week. (I didn't binge the show but savored the thought instead that a new episode of the six-part limited series would arrive every Sunday in time for my fix.)
It certainly felt surreal at times, but there was just something so brilliantly original about what writer/director Mike White created, and the acting so brilliant, that I just can't imagine what will come along that is more memorable. I am also heartened that HBO, after enjoying increased ratings every week and based no doubt on positive word of mouth, has ordered a new White Lotus set in another location. Ah, something to look forward to.
Right now I say — and am somewhat confident about it — that The White Lotus will grab multiple acting nominations, first and foremost for veteran Jennifer Coolidge whose time has finally come.
That was obviously evident when she was introduced as a presenter on the Emmys this year and drew the biggest ovation of the night. That tells you everything.
Murray Bartlett also will be nominated as the beleaguered hotel manager, and I would find places as well for Connie Britton, Steve Zahn, Alexandra Daddario, Sydney Sweeney, Jake Lacy, Natasha Rothwell and Brittany O'Grady. Give an Emmy nomination to everyone, and that includes you too, Fred Heichinger.
Although the latest limited series edition of American Crime Story, Impeachment, drew mixed reviews and lower ratings for FX than previous editions centered on O.J. Simpson and Gianni Versace, I found its telling of the Bill Clinton/Monica Lewinsky scandal to be eminently watchable. That's due to a superior cast who also deserve multiple nominations come Emmy time.
It's led of course by Sarah Paulson, with her stunning portrayal of Linda Tripp. Beanie Feldstein also confounded my initial expectations and really captured the beating heart and eventual terror of Monica Lewinsky (maybe it helped that Lewinsky was there as producer to help guide her off camera). Clive Owen's Bill Clinton was not the usual impression of the 42nd President — and for that alone he should be considered. Annaleigh Ashford's Paula Jones and Edie Falco's Hillary Clinton are likely to be in the mix as well.
Jessica Chastain and Oscar Isaac are likely to figure heavily with their HBO limited series Scenes From A Marriage that also benefits from having the pedigree of being based on an Ingmar Bergman classic that not only was a much-acclaimed movie, but actually started out as a limited six-hour series in Sweden.
And we already know Hulu's Nine Perfect Strangers made waves with its summer arrival, so the likes of Nicole Kidman, Melissa McCarthy and Bobby Cannavale among other perfect strangers could compete.
That the drama series category will continue to see a shakeup is a pretty safe prediction. There will be no new season of this year's big winner The Crown until after Emmy eligibility ends on May 31, 2022. That opens up the field once again, especially for shows returning after taking last season off.
Succession, which won the Emmy in the Drama Series category in its second season, is likely to find itself a front runner for season three.
Better Call Saul is also coming back after being off last season and should figure strongly in nominations. At the very least Bob Odenkirk should have a real shot at finally winning in Lead Actor in a Drama Series, especially after his medical problems caused a shutdown on filming and great concern in the industry for the star. He's back at work, but the incident can only serve to remind us that time can be fleeting and he is long overdue for this recognition.
Stranger Things has been a player for Netflix in past seasons and its return to the contest could be well timed, as will the reappearance of another show from the streamer, Ozark , which also returns to Emmy contention. Both those series are past Emmy winners, but have not grabbed the brass ring. The Crown was in the same boat for its first three seasons, but proved that when your time comes, your time comes.
Also look out for Apple TV+'s The Morning Show — which won an Emmy in its first season for supporting actor Billy Crudup — to considerably up the ante with its season two. It could be a major contender as the plotlines heat up. There is already talk that Apple is considering moving Crudup to Lead Actor next time around since his role has been expanded.
And there will be more Star Wars action for Disney+ with Ewan McGregor (an Emmy winner this year for Halston) taking on the title role in Obi- Wan Kenobi, which could give their own Mandalorian a run for its money.
And could the sixth and final season for This Is Us actually put a broadcast network back in the Drama Series Emmy game? Finales for beloved series can be tempting for voters.
In the race for Comedy Series, we have had four different winners in four years, but there are extenuating reasons for that. For instance both Fleabag and Schitt's Creek came out of nowhere in their final season to sweep the Emmys, but obviously could not repeat. The Marvelous Mrs. Maisel which won big in its first season will be returning to the race this year after sitting last season out, but it faces super stiff competition from the second seasons of both Ted Lasso and Hacks. The former took seven Emmys in its first year on Apple TV+ including winning Best Comedy Series. Its second season played as voting was going on, but will it be forgotten when time comes to judge the 2022 crop? There was also some criticism that it didn't live up to expectations caused by the beloved first season.
HBO Max's Hacks looks like it is just getting started, and wininng both writing and directing Emmys against Lasso (not to mention Jean Smart's lead actress Emmy) means it is obviously well liked in the industry. That could have it poised to take the top prize for its second season.
Expect HBO Max's The Flight Attendant and Netflix's final season of Dead To Me to also be contenders. Hulu's nominated Pen15 seems to be building as well, so if it is back in time it could also compete.
This is all just a snapshot at this point. Look to this space for further predictions as the season rolls along.
Phoebe Bridgers, Blake Mills, Jackson Browne, and More Contribute to New Daisy Jones & the Six Album
Same Same But Different: Editor-Loved Wellness Essentials
Slowthai Announces New Album Ugly, Shares New Song "Selfish": Listen
Barbarian's Director Just Got A Massive Payday To Make His Second Horror Feature


Discussion in 'Designers and Collections' started by Inaya, Mar 6, 2009.
Fantastic! Best show so far this year.
this is wonderful, i want so many of the pieces. those sunglasses are intense.
^eta: agree with inaya totally. it uses many miyake tropes but in a fresh way.
I like it but I don't love it....some fantastic pieces tho.
oh god, this one just justifiably blew me out of the water. def, the fav so far. thank you issey.
talk about color, shape and texture. i cannot wait to see this up close and personal.
Last dress in #6... the first Photoshop effect I was taught.
The pleats never get old.
Hooray Hooray Hooray! I love it all!!!!
It's like elegant raver chic.
Seems a little unfocused but everything is incredibly lovely.


News Center –TDO- A reception was held in Ankara HiltonSA Hotel in consequence of 71st independence day of Sri Lanka. The event was hosted by Sri Lanka's Ambassador to Turkey Pakeer Mohideen Amzaand his spouse Saleema Amza.
Turkish Minister of Commerce Ruhsar Pekcan, diplomatic mission chiefs performing their duty in Ankara and many other guests attended to the reception. Ambassador Amza and some other guests lit a traditional oil lamp. Also a round trip ticket to Sri Lanka was given as a present to one of the guests after a draw.
Minister of Commerce Ruhsar Pekcan and Sri Lanka's Ambassador to Turkey cut the celebration cake.
The reception ended after Sri Lankan tea and dinner served to guests.


text:
Q: Why the image is not displayed? I have searched the topic in the Google, but it wasn't very clear about what I should do.
my question is:

*

*What's wrong with my code? How to revise it.

*The answer in many websites always use URL, how can I know the URL of my image?
Here is the code:
import java.awt.BorderLayout;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class TwoPanel {
    public static void main(String[] args) {
        JFrame frame = new JFrame ("Key test");
        MyDrawPanel1 dp1 = new MyDrawPanel1();
        //MyDrawPanel2 dp2 = new MyDrawPanel2();
        //MyDrawPanel3 dp3 = new MyDrawPanel3();
        //MyDrawPanel4 dp4 = new MyDrawPanel4();
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.setVisible (true);
        JPanel p = new JPanel ();
        p.setLayout(new BorderLayout());
        p.add (dp1);
        //p.add (dp2);
        //p.add(dp3);
        //p.add(dp4);
        frame.getContentPane().add(p);
        frame.pack();
    }
}

then the drawpanel:
public class MyDrawPanel1 extends JPanel {
    /**
    *
    */
    private static final long serialVersionUID = 1L;
    Image image = new ImageIcon("Koala.jpg").getImage();/*something new: if i use the

    path to the disk name(D:/Java/workspace/firstJava/src/com/founder/panel/Koala.jpg'),

    it can work.*/

    public void paintComponent (Graphics g) {
        super.paintComponent(g);
        //Graphics2D g2 = (Graphics2D) g;
        g.drawImage(image, 3, 4, null);
    }
    public Dimension getPreferredSize() {
        if (image != null) {
            return new Dimension(image.getWidth(null), image.getHeight(null));
        }
        return super.getPreferredSize(); // default
    }
}

And this panel can work, although i cann't understand why other doesn't work
package com.founder.panel;

import java.awt.Graphics;

import javax.swing.JPanel;
import java.awt.*;
import java.awt.image.*;
import java.io.IOException;
import java.net.MalformedURLException;
import java.net.URL;

import javax.imageio.ImageIO;

public class MyDrawPanel4 extends JPanel{

/**
 *
 */
private static final long serialVersionUID = 1L;

private static final String URL_PATH = "file:///D:/Java/workspace/firstJava/src/com/founder/panel/" +
        "Koala.jpg";

BufferedImage image = null;

public MyDrawPanel4() {
    // TODO Auto-generated constructor stub

      setBackground(Color.white);
      try {
         image = ImageIO.read(new URL(URL_PATH));
      } catch (MalformedURLException e) {
         e.printStackTrace();
      } catch (IOException e) {
         e.printStackTrace();
      }
   }

   @Override
   public Dimension getPreferredSize() {
      if (image != null) {
         return new Dimension(image.getWidth(), image.getHeight());
      }
      return super.getPreferredSize(); // default
   }

   @Override
   protected void paintComponent(Graphics g) {
      super.paintComponent(g);
      if (image != null) {
         g.drawImage(image, 0, 0, null);
      }
   }
  }

A: The issue is ImageIcon(String) is looking for a file named Koala.jpg within the current directory that the program is been executed in.
Your image is what is known as an embedded resource, you can not access it like it is a file, instead you need use Class#getResource, for example...
Image image = new ImageIcon(MyDrawPanel1.class.getResource("Koala.jpg")).getImage();

Preferrably, ImageIO would be a better choice...
class MyDrawPanel1 extends JPanel {
    private BufferedImage image;

    pubic MyDrawPanel1() {
        try {
            // If the image resides in the same package as the class
            image = ImageIO.read(getClass().getResource("Koala.jpg"));
            // If the image resides somewhere else...
            image = ImageIO.read(getClass().getResource("/path/to/image/Koala.jpg"));
        } catch (IOException exp) {
            exp.printStackTrace();
        }

ImageIO has the capacity to read more formats than ImageIcon and will also throw an IOException if the file can not be read for some reason (like it doesn't exist)

A: My guess is that the image is not loaded, use the path of the file as an argument to ImageIO.read(new File("Koala.jpg")). Using ImageIO:
import java.awt.*;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;
import javax.swing.*;

public class TwoPanel {
    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            public void run() {
                showMainWindow();
            }
        });
    }

    public static void showMainWindow() {
        JFrame frame = new JFrame ("Key test");
        MyDrawPanel1 dp1 = new MyDrawPanel1();
        //MyDrawPanel2 dp2 = new MyDrawPanel2();
        //MyDrawPanel3 dp3 = new MyDrawPanel3();
        //MyDrawPanel4 dp4 = new MyDrawPanel4();
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        JPanel p = new JPanel ();
        p.setLayout(new BorderLayout());
        p.add (dp1);
        //p.add (dp2);
        //p.add(dp3);
        //p.add(dp4);
        frame.getContentPane().add(p);
        frame.setLocationRelativeTo(null); // center frame
        frame.setMinimumSize(new Dimension(300,400)); // make the frame at least 300x400
        frame.pack();
        frame.setVisible(true);
    }
}

The MyDrawPanel1:
 class MyDrawPanel1 extends JPanel {

    private static final long serialVersionUID = 1L;
    //Image image = new ImageIcon("Koala.jpg").getImage();
     BufferedImage image;

     MyDrawPanel1() {
         super();
         try {
             this.image = ImageIO.read(getClass().getResource("com/founder/pane/Koala.jpg"))
         } catch (IOException e) {
             System.err.println("Koala.jpg does not exist!");
             e.printStackTrace();
         }
     }
     @Override
     public void paintComponent (Graphics g) {
        super.paintComponent(g);
        g.drawImage(image, 3, 4, null);
    }
}

redpajama_set_name: RedPajamaStackExchange


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