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2010 Google Zeitgeist South Africa
Rafiq Phillips 10 December 2010 Hey Google
Despite being the year of the 2010 FIFA World Cup™, the soccer spectacle was outranked in Google's fastest rising searches by teen sensation, Justin Bieber, as well as South Africa's latest rap-rave hotshots, Die Antwoord.
The 'Bieber' also tops the list of fastest rising people searches this year, with Afrikaans 'zef' rapper, Jack Parow, coming in at number three. Music is clearly more important than politics as Parrow is placed ahead of ANC Youth League leader, Julius Malema, who claims the number six spot. Reality TV queen, Kim Kardashian and controversial rap artist, Eminem, are placed seventh and ninth respectively.
Malema also features on the list of fastest rising politician searches at pole position, followed by perennial favourites, Nelson Mandela and President Zuma. Although he's been keeping a fairly low profile, former South African President, Thabo Mbeki, comes in at number four followed by Western Cape Premier and DA (Democratic Alliance) leader, Helen Zille. Patricia de Lille, whose party (the Independent Democrats, ID) merged with the DA this year, is placed seventh on the list. She is outranked by Fikile Mbalula, the recently appointed Minister of Sports and Recreation, who fills the number six spot.
The devastating earthquake in Haiti dominated world headlines earlier this year, as well as Internet news searches. 'Haiti' tops the list of fastest rising news searches for 2010 ,followed by another appearance by Julius Malema, 'earthquake', 'Bafana' and 'Nelson Mandela'. Sport still resonates strongly with South Africans, as reflected by 'soccer', 'World Cup' and 'Kaizer Chiefs' featuring on the list.
Our boys, Bafana Bafana, do however claim the number one spot on the list of fastest rising World Cup-related searches, with the Brazilian soccer team coming in second, followed by 'flags of countries'. Shakira's Waka Waka comes in at number five and Paul the late psychic World Cup octopus, fills the number six position – wonder if he saw that coming? The vuvuzela was certainly the rising star of the football tournament ,with the most searches for the instrument originating from Kwa-Zulu Natal, followed by Gauteng and the Western Cape. Searches focused on pictures of the vuvuzela, mp3 downloads of vuvzela noises and vuvuzela manufacturers.
Web lyric searches also reveal what South Africans have been singing along to during the past twelve months. Lyrics to 'Airplanes' the smash hit by B.o.B, featuring Hayley Williams from Paramore and Eminem, seem to be the most popular with South African music lovers, alongside 'Bedrock' by Lil' Wayne and 'Deuces' sung by Chris Brown.
With more and more South Africans turning to their phones to browse the Internet it is not surprising that Facebook and MXit top the list of mobile searches, along with song lyrics and games. Interestingly, it seems that mobile users have been looking for 'love' in 2010 with the search term appearing in fifth place.
In addition to telling us what people are curious about in South Africa, Google Zeitgeist tools also offer insight into which topics were of interest across the globe. Top search terms from over fifty countries appear on our Google Zeitgeist webpage,http://www.google.com/intl/en/press/zeitgeist2010/— from political figures in the United States, to news searches in the United Kingdom.
This year's global fastest rising searches show interest across social networking, technology and pop culture.
Global fastest rising queries, 2010
myxer
gamezer
To see more information about Google Zeitgeist across the globe, take a look at the site: http://www.google.com/intl/en/press/zeitgeist2010/
Fastest rising searches, South Africa, 2010
locnville
powerball results
news24.com south africa
Popular searches, South Africa, 2010
Fastest rising searches, people, South Africa, 2010
jack parow
julius malema
Fastest rising politician searches, South Africa, 2010
1. Julius Malema
2. Nelson Mandela
3. Jacob Zuma
4. Thabo Mbeki
5. Helen Zille
6. Fikile Mbalula
7. Patricia de Lille
8. Siyabonga Cwele
9. Nathi Mthethwa
10. Siphiwe Nyanda
Fastest rising news searches, South Africa, 2010
malema julius
Top mobile searches, South Africa, 2010
Mxit
Fastest rising song lyrics searches, South Africa, 2010
Airplanes lyrics
Bedrock lyrics
Deuces lyrics
Justin Bieber lyrics
Lyrics waka waka
not afraid lyrics
omg lyrics
pyramid lyrics
rude boy lyrics
stereo love lyrics
Fastest rising World Cup searches, South Africa, 2010
Bafana bafana squad
brazil soccer team
flags of countries
shakira waka waka
world cup octopus
world cup concert
fan parks
world cup flags
world cup songs
Zeitgeist Explained
Google reveals the internet "Zeitgeist" (German for "the spirit of the times") through an exploration of the billions of search queries we receive each year. In addition to the Year-end Zeitgeist, which highlights the top trends of 2010, we also have several tools that give insight into global, regional, past and present search trends. Google Zeitgeist tools can never be used to identify individual users because we rely on anonymized, aggregated counts of how often certain search queries occur over time. These tools are available year-round for you to play with, explore, and learn from: www.google.com/zeitgeist
👐SA Innovation Summit The 2014 world design capital and arguably the world's most beautiful city — Cape Town — will this year be honoured to…
👐How Would Nelson Mandela Measure Success? Day 2 of the #30daychallenge where I choose to change everything in 30 days. How will reaching the goal of changing everything in 30…
👐The Most Searched for Car Brands in South Africa South Africans bought 22 452 light vehicles during May 2017, a figure that reflects an ongoing trend of slumping new vehicle sales,…
📝 Have you thought about... ? Hey Google ? 2010 Google Zeitgeist South Africa


"Creativity is your best makeup skill, don't be afraid to experiment", words of the most influential makeup artist Pat McGrath are words every makeup artist must live by.
Known for her creativity, army of assistants and experimental bold looks, she is someone who has been creating the most iconic over the past decades. She is someone who has given the makeup world most ingenious makeup tricks like adding a white curve to the cupid's bow to make your lips look fuller, applying lip liner before a lipstick and blotting lips with a tissue and dusting translucent powder for long lasting wear and more.
Inspired by the looks she has created over the years I collaborated with some experienced talents in London like Aarti Kumar, Arthy Really, Desiree By Natasha, Sejal Anujah Shah and Aarsim Official to create some experimental looks that are going to be trending this season.
I wanted to get bold this New Year with my look and get out of my comfort zone. You know the whole New Year New Me kind of a feeling. So I decided so get bold with my makeup with some products from Pat Mcgrath Labs. I was really happy with the bold and beautiful look created by Aarti Kumar. Aarti is a vibrant and talented makeup artist based in London who loves to play around and create unique looks for her clients. She specialises in Bridal makeup and enjoys experimenting with new products and techniques. The first look she created focused on the glitter red lips. I was looking for a playful red tint to match my embellished skirt and crop top. For the second outfit, I wanted a fierce and feisty look, hence, the golden eye shadow and the funky lines on the eyes. The makeup made me look like a lioness and the sleek high ponytail made the whole look more intense.
The makeup did justice to the elegant handcrafted outfits by Sejal Anujah Shah. This clothing brand brings the best of both the worlds with its contemporary designs. The sleeveless white crop top adds structure to the high waisted skirt with intricate zardozi work. It is an outfit which can be worn for a sangeet or a fancy work event, party, or any other celebration. The skirt and the top can easily be matched with other outfits in your closet to create everyday styles.
Cold shoulders are really in this season. The second outfit is a twist from the regular off-shoulder or cold shoulder trend. The asymmetrical one-shoulder top is a great addition to the party wardrobe. I paired the top with skinny black pants and black shoes. I let my top and the makeup be the hero of the entire look. You can even layer up the top with a funky jacket for evenings.
No outfit is complete without the right kind of accessories. My accessories were styled by DesireebyNatasha. The brand is known for its exquisite collection of jewellery and watches made from semi-precious stones and pearls. They are elegant, modern and make any outfit pop instantly. They are also a great gifting option for someone who loves to accessorise.
I love statement bags which are fun, like the one I have incorporated in my look. The bags by Aarsim is just perfect for me since I have decided that this year I am going to focus on being happy and content with my life.
But a picture is only as good as its model….Hahaha, just kidding.. a picture is only as good as the photographer, and I love the picture play that's been created with my look.
Thanks to my lovely photographer Arthy Really. From travel to editorial photo shoots, Arthy has the talent to capture the essence of the subject strikingly in her images.
I had a great time collaborating with some young and vibrant talent in the city to create my experimental bold looks. Let me know in the comments below if you enjoyed this fun outfit and challenge yourself to go bold and beautiful this year!


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Wilfried "Willi" Schneider (* 12. März 1963 in Mediaș, Rumänien) ist ein ehemaliger deutscher Skeletonfahrer und aktueller Skeletontrainer. Gegen Ende des 20. Jahrhunderts war Schneider einer der erfolgreichsten Skeletonsportler weltweit.

Willi Schneider debütierte im Januar 1992 bei einem Rennen in Königssee als 16. im Weltcup. In seinem dritten Rennen in Cortina d'Ampezzo kam er als Neunter erstmals unter die Top-10. In der Folgezeit konnte Schneider seine Leistungen immer weiter konsolidieren und verbessern. Im Dezember 1997 gewann er in La Plagne und Winterberg seine ersten Weltcuprennen. Insgesamt konnte Schneider vier Weltcuprennen gewinnen. Die Saison 1993/94 schloss er als Vierter des Gesamtweltcups ab, 1994/95 als Fünfter, 1995/96 erneut als Vierter und 1996/97 als Dritter. In der Saison 1997/98 gewann er den Gesamtweltcup. 1998/99 wurde er nochmals Vierter und 1999/2000 Sechster.

1992 nahm Schneider erstmals auch an Weltmeisterschaften teil. In Calgary erreichte er den 14. Platz. Ein Jahr später erreichte er in La Plagne schon den fünften Platz. 1995 in Lillehammer und 1997 in Lake Placid erreichte er diesen Platz erneut. 1998 gewann Schneider seinen ersten Weltmeistertitel in St. Moritz. Ein Jahr später in Altenberg gewann er Bronze. 2000 in Igls kam nochmals ein guter fünfter Rang hinzu. Karrierehöhepunkt war für Schneider die Teilnahme an den Olympischen Winterspielen 2002 in Salt Lake City, wo Skeleton zum ersten Mal seit 1948 wieder olympisch war. Schneider belegte hier den neunten Platz.

1992 konnte Schneider seinen ersten Titel bei Deutschen Meisterschaften gewinnen. 1994 bis 1997, 2000 und 2003 so konnte er gesamt neunmal den Titel gewinnen. Hinzu kommen 1998 und 1999 Vizemeistertitel hinter Andy Böhme und ein Dritter Platz 2001.

Nach seiner aktiven Karriere wurde Willi Schneider Trainer der kanadischen Nationalmannschaft. Hier war er als Trainer beispielsweise für drei Medaillen bei den Olympischen Spielen in Turin verantwortlich. Bei den Herren für Gold für Duff Gibson, Silber für Jeff Pain und bei den Damen für Bronze für Mellisa Hollingsworth. Bis auf Duff Gibson, der seine Karriere nach den Olympischen Spielen beendete, betreut Schneider die kanadischen Spitzenathleten auch nach den Spielen weiter. Bei den Olympischen Winterspielen 2010 in Vancouver gewann der von ihm betreute Jon Montgomery weiteres Gold. Nach Vancouver konnte Willi Schneider als Technikcoach für das US-Skeleton Team arbeiten und holte mit Katie Uhlander einen weiteren Weltmeistertitel. In der Saison 2012/13 wechselte Schneider als Cheftrainer nach Russland und gewann gleich im ersten Jahr mit Alexander Tretjakow für Russland den ersten Weltmeistertitel und bei den Frauen mit Elena Nikitina den Junioren-Weltmeistertitel und gleich darauf mit ihr auch den Europameistertitel. Im Olympiajahr 2013/14 gewannen Schneiders Athleten mit Alexander Tretjakow und Jelena Nikitina Gold und Bronze bei ihren Heimspielen. Aufgrund der Erfolge wurde Schneider 2014 der Orden der Freundschaft als staatliche Auszeichnung der Russische Föderation verliehen.

Weblinks

   (englisch)

Skeletonfahrer (Deutschland)
Olympiateilnehmer (Deutschland)
Teilnehmer der Olympischen Winterspiele 2002
Weltmeister (Skeleton)
Deutscher Meister (Skeleton)
Skeletontrainer
Träger des Ordens der Freundschaft
Deutscher
Geboren 1963
Mann

redpajama_set_name: RedPajamaWikipedia


This is the first rifled musket of the British army, very similar to the pattern 1842 except it is a little slimmer in size. The barrel caliber was reduced to .70 with 4 grooves and 39" in length. The barrel was also retained with keys. There were approximately 35,000 made. The long range rear sight was sighted for 900 yards. This rifle made a grand showing in the Crimean War where it preformed exceedingly well. Lock parts will be available by mid to late summer.


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Hill of Hell
By Laura van den Berg, Illustrations by Ryan Floyd Johnson
I had traveled up the Hudson Line at my friend's invitation to deliver a lecture to his literature students at the college where he taught. There had been three people in attendance and one had fallen asleep halfway through. My friend had treated me to lunch before the talk and to a drink afterward, so that by the time we hit the train back into the city, where we both lived, we had sailed through the small talk and were ready for the blood and guts.
After we opened the second bottle of wine, which he'd been keeping in his satchel, I told him about the worst thing that had happened to me in the last three years, as this was the period of time that had elapsed since we last saw each other. We sat at a table in the café car, the panoramic windows looking out on the vast sweep of the Hudson. At first, I was surprised that we could drink openly on the train, but my friend assured me that we could eat and drink whatever we wanted because the café car was closed on this route—and besides, he had been taking this train three days a week for a decade and he knew every conductor on it and could get away with anything.
"It was around this time last year when everything came apart," I said, turning my plastic cup on the table.
Last September, I was pregnant. My husband had been the one wracked with longing for a child and I had allowed myself to be carried along by the tide of his enthusiasm, but once it was underway I felt like I had been conned into a heist for which, as the plans came into focus, I was woefully unprepared. You're talking about robbing the Louvre and I'm just a common criminal! In those early weeks, I willed my body to show up with the getaway car and then four months later, after I had forgotten all about getaway cars, I was standing in Ikea, of all the undignified places, waving a spatula and lecturing my husband about how our dairy products were teeming with opiates, when my shorts filled with blood and I fainted. While I was unconscious, I had a dream that men in white coats were elbow-deep in me and then I awoke in a hospital bed to find a doctor elbow-deep in me, working on my body with the grave air of an executioner. The baby had ten fingers and ten toes, the only thing that many a stranger had told me I should care about. Eyelids as thin as organza.
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On Death and Decluttering
By Michelle Orange
Two women share a hospital room, separated by a green-blue curtain, at the end of a brief, beige hallway. Their prospects foreclosed by illness, the women have agreed to enter this room, if not to share it, and to find what peace is possible. Separated always by the green-blue curtain, the women receive their respective family members, making few requests, their needs muted in a way that humbles and vexes their visitors. They are aware of each other's presence, vaguely yet certainly, in the way of animals on opposite sides of an open field, at night.
The Will of the Water
By Lacy M. Johnson
The bayous fill, and the water runs into the streets; the streets fill, and the water fills the highways and the underpasses. The water swallows cars and trucks and entire families of people. It swallows fathers and mothers and babies. The water turns the highway into an ocean; the white peaks of waves crest and crash against the sides of buildings. People wade out of their houses, through the water, toward one another and dry land. They climb to the second floor, and then the third; they scramble to their roofs and wave white T-shirts or towels toward the rescue they believe will come. Cages like open coffins descend from helicopters, and people climb into them, one at a time or as an inseparable group. A mother clings to her children as they ascend from the water toward safety. She never lets them go.
My husband and I watch the rescues on the news. There aren't enough helicopters for everyone who needs saving, aren't enough high-water vehicles, or boats, or flashlights, or meals, or warm beds. We watch the water rising in our own neighborhood, filling the streets up to our ankles, our knees, up to our waists. We are trapped here, on the little island of our address. We occupy ourselves and the children in the ways we can: we eat, we drink, we play board games and curl together in the bed. My husband and I take turns going outside to check the water, watch it rise. When we wake on the fourth day of rain, it is still rising.
By Rachel Eliza Griffiths
In the evenings, we watched Jeopardy.
Wore surgical masks once she got sick.
Before that my mother sent me to the store
for cigarettes all the time. Pack of Salem Lights.
I'd come into the room & try to write
a different ending on those anonymous walls.
There was less time all the time
until time changed. You know what I mean.
There is no title. There is no title. The body is content. The body is window.
The body is container, curtain, chair, grid. Do you see? Bones & shoulders, a spine
My Father's Toe
By Sean Mills, Illustrations by Michelle Thompson
My father recently lost a toe. The second one on his right foot, lopped off in an outpatient procedure, quick and painless. Such a funny thing to lose, everybody thought—my mother, sisters, brother, the grandkids all finding much levity in the situation. They call him "Nine-toed Joe" now, and for his birthday his granddaughters gave him customized white tube socks, the ghoulish gap of his little amputation rendered with a red Sharpie. My father found the gift hilarious, and wore the socks proudly with his new sandals right through to Halloween. I laughed, too, pretending not to find it disturbing and macabre. His toes had become grotesque with old age, as toes do when you approach eighty, after decades of punishing footwear: Army boots, oxfords, wingtips, Chuck Taylor Converse All Stars on the basketball court, running shoes in which my father pounded the pavement, training for marathons he never ran. Now he's barely able to get any shoes onto his feet in order to make it to church.
By Deborah Eisenberg, Illustrations by Nicole Rifkin
Thundering down, a cataract from a high plateau, raising billows of dust, manes, tails, whinnies rippling like banners, a glamorous species, captive yes, but not entirely subdued, they—oh, no, a fellow in that ridiculous getup pops up from behind a rock and pulls out a—bink! That's enough, goodbye stupid old show, time for a cup of tea. Pulls out—bang, bang, bang. Yes, sensible Cordis decides, not a drink, time for a nice cup of tea.
The dog, a parting so-called gift from unfortunate Mrs. Munderson, peers at the blank screen, baffled, then paws at Cordis. Moppet is not glamorous, except in the most trivial sense; Moppet is cute. What does Moppet want? A treat? A tickle? A furlough?
The Pardner
By Koye Oyedeji, Illustrations by Anna Schuleit Haber
It has been a year and five days since Mayowa lost her daughter—lost, because she cannot say the other word: suicide.
By Elizabeth Weld, Illustrations by Lauren Nassef
Two years after her mother's death, Jane's boyfriend asked her to marry him, and nine months later, they moved across the country to start their new life. Jane was twenty-nine, ready to step away from Phoenix after a hard few years. Ryan had taken a job at a recording studio in Tennessee, and he pointed out that the public schools there were as bad as the ones in Arizona, so she could easily fail teaching fourth graders in either place. Her father was a kind, if distant, ichthyologist, and he seemed to think the move was maybe not ideal, but maybe not a bad idea. Jane was excited to start over. She'd been adopted when she was six, and she thought of six as the beginning of her real childhood. As they drove out of town, she decided twenty-nine was the beginning of her real adult life.

redpajama_set_name: RedPajamaCommonCrawl


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Best VPN apps for Android : BROWSE SAFE!
Damian Yan Chu September 24, 2017 Android
Keep your browsing safe and anonymous with a quality VPN service. Using a virtual private network (VPN) when you are online is becoming more important these days. While the Internet is a public space, a VPN acts as a cloak of invisibility and makes your online activity virtually anonymous, making it difficult for someone to track you when you are online, whether you are concerned about your service provider Internet,, malicious hackers. We have previously examined the best VPN services on all platforms, but here we will specifically look at VPNs for your Android devices. These are applications that are free to download but usually require you to pay for a monthly or yearly subscription. We've got the best, so take a look!
IPVanish VPN
1. IPVanish VPN
There are a number of great reasons to go with IPvanish for your VPN services - and one reason they might give you pause. First, the good: IPVanish features solid encryption, unlimited bandwidth and allows you to connect up to five devices simultaneously. The Android app offers a clean and dark theme and is easy to use for beginners, with features that advanced users will appreciate. Connection and switching between servers is quick and easy, as is the overall user experience. The only drawback is that everything is based in the United States, which means that there is a possibility that IPVanish will be pressured to release user information to an outside agency with little or no prior notice to users.
Download: IPVanish VPN (Free, $10/month, $26.99/3 months, $77.99/year subscription)
2. Private Internet Access
For those who are new to using a VPN, one of the most important features will be an easy-to-use interface. You want something as easy as moving a switch. Private Internet access offers that, with an extremely simple Android application. Best of all, an annual subscription is around $ 40, which is one of the lowest prices you'll find for a premium VPN. Private Internet access uses more than 3,000 servers in 24 different countries and has received numerous nominations as one of the best commercial VPNs you can use. They do not register any activity online and the application is compatible not only with their Android devices but also with any computer running Windows, MacOS or Linux
Download: Private Internet Access VPN (Free, $39.99/year or $6.95/month subscription)
ExpressVPN is another highly valued Premium VPN service that has been praised for its customer service and includes all the features you would expect from a premium VPN, such as unlimited bandwidth, unlimited speed and strong encryption. You can connect up to three devices simultaneously through ExpressVPNs thousands of servers in over 94 different countries.
The Android app is easy to use for beginners, with settings and features that users will appreciate. Subscriptions are a little more expensive but ExpressVPN offers a free 7 day trial and a 100% money back guarantee for the first 30 days of service so you can try it out for yourself.
Download: ExpressVPN (Free, $99.95/year, or $12.95/month subscription)
NordVPN is a great choice if you are looking for an encrypted connection for up to six devices. Operated outside Panama, NordVPN has no legal obligation to register the activity of its users, so it can confidently connect to one of more than 1000 servers in 57 different countries. Also, it comes highly recommended from a variety of retail outlets including CNET and PCMag, with an Android app that is absolutely simple to use. Connect to a server at the push of a button and keep your phone protected while browsing public Wi-Fi.
Download: NordVPN (Free, $69/year or $11.95/month subscription)
5. TunnelBear VPN
Do not confuse the cartoony design of the TunnelBear VPN Android application, it's an easy-to-use application that offers serious VPN protection. You can create an account for free, which gives you 500 MB of secure data per month, but we recommend that you review a paid subscription that allows you to connect up to five different devices at a time to servers in 20 different countries. The Android application is really fun to use, and you have to give TunnelBear credit to really double-down on your behalf by actually running with the concept. TunnelBear does not currently support torrenting, so if that's an important feature it's best to look elsewhere. If you are just looking for a casual VPN for your phone, TunnelBear is a great free option with affordable prices to upgrade to premium service.
Download: TunnelBear VPN (Free, $59.88/year or $9.99/month subscription)
Please Write a Comment as suggestion or anything ......
About Damian Yan Chu
Damian Yan Chu is a fictional character creaeted by me.
By Damian Yan Chu - September 24, 2017
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The Living Room, located in Manhattan's Lower East Side, has long been a favorite venue for aficionados of folk, country and pop/rock singer/songwriters.
The Living Room, located in Manhattan's Lower East Side, has long been a favorite venue for aficionados of folk, country and pop/rock singer/songwriters. Co-owner Steve Rosenthal recently installed studio-quality ATC reference monitors as the sole speaker system.
"I worked with [ATC founder] Billy Woodman and [Sony's SACD director] David Kawakami in setting up the new sound system for the new Living Room," reported Rosenthal, an engineer/producer and owner of New York's Magic Shop Recording Studio, who is partners in the club with Jennifer Gilson. "I do a lot of surround sound mixing for Sony. I've done the Rolling Stones and a bunch of stuff. I'm a big ATC fan, so I wanted to see if we could have ATCs in the performance space. We put in the big ATC SCM150s in the main room, and then I have a separate lounge and DJ area, and there I have a pair of ATC SCM20s. I also have a pair of SCM20s that I use as pinpoint monitors for the stage for recording.
"I thought it was a pretty revolutionary thing to put a large pair of studio monitors in a performance space. After working on ATCs in the studio for a while, I thought it would be an interesting switch to use them in a live setting. After all, it's still the same question of performance, sound quality and accuracy. So far, they've been getting an incredibly great reception. We're probably the only live room that actually has no EQ on the house speakers."
The new Living Room location is slightly larger than the previous venue, seating 100 with standing room for another 35 to 40. Rosenthal worked with engineer Joe Warda to tune the room. "He did all the acoustic work," added Rosenthal, "and did a really great job."
The equipment list at the new location is rounded out with a Midas console and a variety of processing, including Anthony DeMaria compressors. "I don't know a lot of clubs that have tube compressors on the vocals," commented Rosenthal. Microphones include models from Neumann, Sennheiser, Shure and Crown.
"We're going to be doing a lot of surround sound recording in The Living Room using a Decca Tree mic setup. I intend to record everything on DSD [Sony and Philips' proprietary 1-bit Direct Stream Digital format]." A compilation album, The Living Room: Live in NYC, Vol.1, produced and mixed by Rosenthal and featuring performances by Norah Jones and Jesse Harris plus Malcolm Holcombe, Rachel Loshak and Chris Moore, was released last year on Stanton Street Records.
For more on the loudspeakers, visit distributor Transamerica Audio Group at www.transaudiogroup.com. For more on the club, visit www.livingroomny.com.


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6 Prominent Technology Trends for 2023
#1 Artificial Intelligence
AI is constantly evolving and finds newer applications in various industries every day, with ChatGPT being one of the latest AI disruptors
#2 Quantum Computing
Based on the principles of quantum theory, quantum computing can easily query, analyze and take action in a much faster rate than normal computers, gaining prominence in 2023
#3 DevOps
DevOps combines the development and operations departments within an organization to help enhance and automate the process of software development
#4 Predictive analytics
Predictive analytics helps businesses predict future customer behaviors and trends with some of its most common applications being in risk management, marketing and operations
#5 5G
5G technology has the potential to change the way we interface and interact online by integrating AR and VR technology and better cloud-based gaming experiences
#6 Internet of Behaviors
One of the fastest-growing industry trends with application in nearly every industry, IoB collects and analyzes data from users connected to the internet to better understand human behavior
Revolutionize Candidate Engagement & Experience with Hyreo.


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Lazlo's Clicker – True Blood, Mad Men Finale, Beyond Jumping The Shark
Posted by Rob Lazlo
So what's with the latest season of True Blood? To listen to the chatter on the internet, you'd think the show had gone completely off the rails, but I don't see some massive drop in quality here. Are people just tired of the story?
Hit the title/continue reading to read more. . .
Season 6 begins right where Season 5 left off. Bill (Stephen Moyer) has consumed the blood of the first vampire Lilith, and has been imbued with superhuman, er supervampire type powers. He's also gone a bit cracked, lapsing into trances and openly stating he doesn't know what he is anymore. This has caused everyone but his progeny Jessica (Deborah Ann Woll) to back way off of him. The SuperBill storyline has been OK so far – the trancelike scenes with Lilith are way too long, but it has given Woll a chance to really shine.
Much more interesting is the development of the Tara (Rutina Wesley) and Pam (Kristin Bauer von Straten) relationship, and these two have become the most interesting characters on the show (save for Nelsan Ellis' ongoing brilliant portrayal of Lafayette, who steals every scene he's in). Sadly, the same cannot be said for the lead character of the show, Sookie (Anna Paquin), whose latest romance with a fellow half fairy has not resonated at all with an audience that's used to Sookie cozying up with vampires and werewolves.
Season 6 also introduces two new characters. Rutger Hauer is having some obvious fun playing Sookie and Jason's (Ryan Kwanten) bizarre fairy godfather Niall, while Arliss Howard has been set up as an ambiguous Season 6 villain, the sleazy Governor Burrell.
Rutger Hauer rockin' the crazy hair
It's true that True Blood Season 6 isn't a whole lot like vintage True Blood Season 2 or 3, but the show hasn't lost its sense of humor or sense of fun, it's simply channeling it in a different way. Moreover, many of the characters have grown immensely from their more stock origins in the early years. I admit I am not enamored of the two storylines headlining the show – Bill's and Sookie's – but the season is early, and all of the rest of the plotlines are humming along nicely. True Blood is still well worth your viewing time.
THERE'S LITTLE DOUBT that the just concluded Season 6 of Mad Men took the show in as dark a direction as we've seen, but in true Mad Men style, the season ended on a note that was simultaneously crushing and momentously uplifting in the same breath.
Season 6 both wrapped (and set up for next season) it's three big story arcs. First there was Pete Campbell (Vincent Kartheiser). Pete was been on a downward spiral all year, and the final episode saw him learning that his mother had been murdered (in a great twist, Mad Men managed to turn this into a brilliant bit of black comedy and Pete and his brother commiserate). Pete is now truly a man without a country. He's been boxed out at work, his home life is non-existent. But as his estranged wife Trudy (Alison Brie) put it, he's now finally "free."
The ongoing flirtation between Ted (Kevin Rahm) and Peggy (Elisabeth Moss) was finally consummated, and we've seen Peggy now come full circle. Ted, of course, could not bring himself to leave his wife, so he's leaving New York altogether to get away from Peggy, whom he can't resist. The final scene of Peggy had her sauntering into Don's office and chilling in his chair Mad Men opening style. With Ted bound for California and Don on the outs (more on that later), Peggy appears to be poised to make the full climb of the corporate ladder. All it cost her was her soul.
As iconic as it gets: Peggy as the new Don
Finally, there's Don Draper (Jon Hamm). He truly hit rock bottom this week (and this season). His marriage to Megan (Jessica Pare) is on life support. He wound up in the drunk tank after slugging a minister. The partners had an intervention-style meeting and told him he was "on leave." On his way out the door, he met his replacement. And, of course, he's still dealing with his daughter Sally (Kiernan Shipka) having seen him for what he is, in flagrante delicto with Sylvia Rosen (Linda Cardellini). Where was there left for Don to go but to the truth. Mad Men's awesome final scene had Don taking his kids to the dilapidated former whore house where he grew up as Dick Whitman. As Judy Collins' Both Sides Now filled the speakers, Don and Sally exchanged a look that drew goosebumps.
The look that framed a season: Hamm & Shipka's wordless masterpiece
There's just one season of Mad Men left. I suspect Season 7 will skip ahead in time a bit. . .to the '70s or just before it maybe? About the only thing show creator Matthew Weiner has promised is that the show won't end with Don falling out a window (as is depicted in the opening credits). Where this show end is anyone's guess, but it's not speculation to already know that once all is said and done, Mad Men's place as one of the greatest television shows ever is secure.
BY NOW, EVERYONE knows the phrase "Jumping the Shark" and maybe most have also heard of "Growing The Beard", but I think we need a lot more phrases to describe what happens on our favorite TV shows. Here are some suggestions (along with the original two which started it all):
Refers to: The moment when a TV show gets bad or ridiculous
Named For: The episode of Happy Days when Fonzi goes water skiing and (you guessed it) jumps over a shark
Example: Roseanne Jumped The Shark in that last season where they won the lottery
The scene that started it all: Fonzi (Henry Winkler) on water skis
Growing the Beard
Refers to: The moment when a TV show really comes into its own
Named For: Jonathan Frakes' Commander William Riker showing up on Season 2 of Star Trek The Next Generation with a full beard
Example: Parenthood Grew The Beard this season with the storyline about Kristina's (Monica Potter) cancer battle
Putting on the Hat
Refers to: A character embracing his or her dark nature
Named For: Walter White's (Bryan Cranston) signature black porkpie hat, which he wears to embody his personna as drug mastermind Heisenberg on Breaking Bad.
Example: On Dexter, Deborah Morgan (Jennifer Carpenter) Put On The Hat when she shot LaGuerta.
Yeah, you bad now
Moving to Los Angeles:
Refers to: A major upheaval in setting, format or cast
Named For: When Laverne and Shirley moved from Milwaukee to Los Angeles in Season 6
Example: Is Grey's Anatomy Moving To Los Angeles with that plane crash episode?
Introducing Randy:
Refers To: Replacing a major character with a lame-oid and/or trying to keep a show going after all the best characters leave
Named For: Randy (Josh Meyers) on That '70s Show. Really, Foreman & Kelso leave and you give us this guy?
Example: Body of Proof tried bringing in a new cop/boyfriend character, but Introducing Randy is not going to save that show.
Munching:
Refers To: A character showing up on another show or shows
Named For: Richard Belzer's ubiqitious Detective John Munch, who has appeared on nine different television shows
Example: CSI and CSI New York were Munching this week
Dancing With The Midget:
Refers To: When a show just goes over the top with its imagery, mythology or symbolism
Named For: Twin Peaks, and one of its signature images in the muddled and confusing Season 2.
Examples: Bringing Starbuck back on BSG was Dancing With the Midget. Lost Danced with the Midget for the last 9 episodes of its run.
Shit just got weird
What was your favorite gross-out TV moment of the week? Linden & Holder find a severed finger on The Killing, a cow get cut neatly in half on Under The Dome or Bill telekinetically doing a mouth to mouth blood feeding on True Blood?
Reality Check: America's Got Talent revamped its judging panel in the absence of Sharon Osbourne, adding another spot, and balancing out Howard Stern and Howie Mandel with the serious Heidi Klum and the enthusiastic Mel B. Bottom line, of course, is that on this show, more than any other reality competition, the judging panel isn't that vital. The show works because it's largely immune to the personality of its panel and relies instead on the variety and occasional absurdity of the talent on display. America's Got Talent remains the heir apparent to the old Ed Sullivan Show, and is the only place you can see animal acts, daredevils, magicians, stand up comics, and unicyclists spinning plates on sticks along with the singing and dancing.
Has there been a more contrived reality show than ABC's Whodunit. I thought perhaps the show would assemble a group of ordinary people and ask them to watch a dramatization of a mystery and then ask them (and the audience) to solve it. Instead, it's an elaborate haunted house game with numerous Survivor-like asides. Been there, done that.
Educational TV. Things we learned from watching TV this week: 1) The Portugese Language sounds like a dog being raped by a bagpipe (The Colbert Report); 2) Scalia is the Supreme Court Justice whose name most sounds like a GI Joe villain (The Daily Show); 3) Creepy Ass Crackers are racist but delicious (Real Time With Bill Maher).
So You Think You Can Dance is just a dance show right? They come out, they dance. Like singing, only with your feet, right? Wrong. The show found a completely new and original way to open its first live show of the season, presenting a brilliantly choreographed routine to Herb Alpert's version of Puttin On The Ritz that took place entirely backstage (and in the dressing room, control room, cafeteria, etc.) and even featured cameos from Exec Producer Nigel Lythgoe, judge Mary Murphy and choreographers Jason Gilkison and Tabitha and Napoleon D'umo. I daresay it was the very best and most original large group routine the show has ever done.
Dexter returns to Showtime on June 30, 2013.
The series premiere of Ray Donovan airs on Showtime on June 30, 2013.
The season finale of The Goodwin Games airs on Fox on July 1, 2013.
The series premiere of Siberia airs on NBC on July 1, 2013.
The season finale of Inside Amy Schumer airs on Comedy Central on July 2, 2013.
BBC America will run a Star Trek The Next Generation marathon on July 4, 2013.
SyFy will run a Twilight Zone marathon on July 4, 2013.
USA will run an NCIS marathon on July 4, 2013.
TV's a big place and I haven't been to all of it yet. Got a favorite show you'd like me to comment on? Post a comment below, contact me on twitter @RobLazlo. or shoot me an email: RobNJ564@yahoo.com. I welcome your input!
Posted by Rob Lazlo | Jun 30, 2013 | GuysGirl
Rob Lazlo
Rob Lazlo lives in Atlantic City, NJ and is a lifelong Philadelphia sports fan, following the Phillies, Eagles, Flyers and Sixers. A former professional musician, he toured with the USO and visited Greenland, Japan, Korea, the Philippines, Bahrain and many other locales. Rob is a big science fiction fan and loves movies, television, cooking, music and, of course, sports of all kinds. He works as a business consultant and is happily divorced with three children. For other hobbies, Rob plays sports titles video games and dabbles in online games. He also enjoys reading and following politics.
Landing & Taking Off From a Nuclear Aircraft Carrier for Navy's Distinguished Visitor Embark Program
How I Livestream My Radio Show
Why a Blowout Should Be Part of Your Game Day Routine
NFL anthem protests, another Louisville basketball scandal plus more on this week's show


text:
\section{\label{sec:level1}INTRODUCTION\protect\\ \lowercase{}}

It is well known that the spin splitting (SS) of energies in various quantum systems is intimately related to their symmetry. According to the Kramers theorem, the energy of an electron in a periodic system satisfies the equality: $E_{{\bf k}\uparrow} = E_{-{\bf k}\downarrow}$, where $\bf k$ is the wave vector and the arrows $\uparrow$ and $\downarrow$ signify spin-up and spin-down projections, respectively. If, in addition, a system is characterized by the
inversion symmetry, the two spin states are degenerate for any $\bf{k}$ value: $E_{{\bf k}\uparrow} = E_{{\bf k}\downarrow}$. In general, however, for a given ${\bf k}$ direction and value SS may occur even without an external magnetic field. It was shown by Dresselhaus [1] that in bulk semiconductors of the zinc blende structure, which have no inversion symmetry, the conduction bands of III-V compounds are characterized by an anisotropic spin splitting
proportional to $k^3$ for small $k$ values. This splitting was extensively investigated in bulk semiconductors. With the advancement of semiconductor quantum structures a new type of inversion asymmetry became possible, namely the structure inversion asymmetry (SIA). The SIA mechanism also leads to SS of electron energies. The SS related to SIA, often called "the Rashba splitting", is of interest due to possible applications, as it can be influenced by an external electric
bias. This interest is a part of the wider movement aimed to use properties of electron spin for technical use.

The subject of SS related to SIA has a controversial history, as reviewed by the present authors [2]. In a widely quoted paper, Bychkov and Rashba [3] wrote
down the following Hamiltonian for SS in heterostructures, in which the growth direction is parallel to a high symmetry axis
\begin{equation}
{\hat H} = \alpha ({\bm {\sigma}} \times {\bf k})\cdot {\bm{\nu}}\;\;.
\end{equation}
Here $\bm{\sigma}$ are the Pauli matrices, $\bf{k}$ is the 2D electron wave vector transverse to the growth direction, $\bm{\nu}$ is the unit vector in the growth direction, and $\alpha$ is a coefficient. In their paper, Bychkov and Rashba did not mention the inversion asymmetry, but it is recognized by now that the
coefficient $\alpha$ has a nonzero value only if the system is characterized by a structure inversion asymmetry along the growth direction. The Hamiltonian given by Eq. (1) resembles that of the spin-orbit interaction (SOI), but one should not confuse it with the SOI in which an electric field $\bf {\cal E}$ appears instead of $\bm{\nu}$. As follows from the description given in [2], a consensus on the theoretical treatment of the problem of SS due to SIA has been reached, see also [4-6]. Namely, it has been demonstrated that the SS of conduction energies in asymmetric quantum wells (QWs) is mostly related to asymmetric offsets of the conduction and valence bands, both due to different energy gaps and spin-orbit energies. In particular, the SS of a conduction band is not proportional to the electric field in this band because the average electric field in a bound state is zero if one neglects a difference of the effective masses in the well and barriers [7].

It was recognized a few years ago that the SOI can be used to manipulate electron spins in semiconductor heterostructures. In particular, it was proposed to fabricate spin filters by either driving electrons through inhomogeneous heterostructures with different strengths of the SOI [8, 9] or by reflecting 2D electrons from a lateral potential barrier in an asymmetric quantum well [10 - 15]. Recently, the present authors described spin-flip and spin-conserving electron reflections from a potential barrier in a vacuum using the Dirac equation for relativistic electrons [16].

In this paper we are concerned with the spin-flip reflection processes in heterostructures. It was exprimentally demonstrated by Chen et al [10, 12, 14] that, in the presence of the SOI, an opaque reflection from a potential barrier separates electrons undergoing spin-flip reflections from those experiencing spin-conserving reflections.  In their system, the electron spin splitting was caused by SIA of asymmetric QWs and the interaction with the barrier.
In our paper we describe the spin-flip and spin-conserving reflections introducing three new elements. First, we consider the simplest system in which \emph{the only} source of electron spin-splitting is its interaction with the barrier. This can be realized either by using a symmetric QW or by performing experiments without 2D heterostructures. Second, we actually calculate the spin splitting of unbound electron energies due to the interaction with the barrier, which was not attempted in previous works. Third, we point out quite unusual possibilities of non-attenuated electron propagation in the barrier whose height exceeds the energies of incoming electrons and the total reflection of electrons whose energies exceed barrier's height.

Our paper is organized as follows. In Section II we consider the spin splitting of electron energies due to the interaction with the barrier. Section III contains analysis of spin-conserving and spin-flip opaque reflections: their kinematics and amplitudes. Section IV gives our main results. In Section V we discuss our approach as well as those of other authors. The paper is concluded by a Summary.
\section{\label{sec:level1} Spin splitting due to potential barrier\protect\\ \lowercase{}}

We first calculate the spin splitting of electron energies caused by the spin-orbit interaction between the incoming electron and the potential barrier. The interaction is of the general form given by Eq. (1) but the problem in our case is not trivial since the energy spectra of unperturbed and perturbed energies are continuous. We consider a finite potential barrier at $z = 0$. The starting point for the theory is the multiband {\kp}
formulation including the potential $V(\textbf{r})$ on the diagonal
\begin{equation}
\sum_l
\left[\left(-E+V+\varepsilon_{l0}+\frac{p^2}{2m_0}\right)\delta_{l'l} +\frac{1}{m_0}\textbf{p}_{l'l} \cdot \textbf{\textbf{p}} \right] f_l(\textbf{r}) = 0\;\;\;,
\end{equation}
where l'=1, 2, ..., $E$ is the electron energy, $\varepsilon_{l0}$ are the
band edge energies, $m_0$ is the free electron mass, $f_l(\textbf{r})$ are the envelope functions, and $\textbf{p}_{l'l}$ are the
interband matrix elements of momentum taken between the Luttinger-Kohn periodic amplitudes $u_l(\textbf{r})$. The potential
$V(\textbf{r})$ and the envelope functions $f_l(\textbf{r})$ are assumed to be slowly varying within the unit cell (which is not
always a realistic assumption). If the potential varies only in one dimension: $V(\textbf{r}) = V(z)$, one may separate the variables
by looking for solutions in the form $f_l(\textbf{r}) = exp(ik_x x+ik_y y) \chi_l(z)$. This gives $p_x\rightarrow \hbar k_x$,
$p_y\rightarrow \hbar k_y$, and the only nontrival variable is $z$. Within the three-level model (3LM) one deals with 8 bands arising from $\Gamma_6^c$,
$\Gamma_8^v$ (double degenerate) and $\Gamma_7^v$ levels. Thus the sum over $l$ in Eq. (2) runs from 1 to 8 and $l'$
=1...8, so that one deals with 8 coupled differential
equations for the functions $\chi_l(z)$. The parameters in set (2) are the band edge energies and the momentum matrix elements. Within 3LM there is one nonvanishing matrix element $P_0$ between $\Gamma^c_6$ and $\Gamma^v_8, \Gamma^v_7$ symmetry functions and one spin-orbit energy $\Delta$ splitting the $\Gamma^v_8$ and $\Gamma^v_7$ functions.
Using minor approximations one reduces set (2) by
substitution to two equations for the effective spin-up and spin-down conduction states, see Ref. 16. The resulting eigenvalue problem for the growth direction $z$ parallel to [001] crystal axis is
\ \\
\begin{equation}
 \left( \begin{array}{cc} \hat{A}- E&\hat{K}_{SIA} \\
\hat{K}^\dagger_{SIA}&\hat{A}- E \end{array} \right)
 \left( \begin{array}{c} \chi_1(z) \\ \chi_2(z) \end{array} \right) =0,
\end{equation}
\ \\
where
\ \\
\begin{equation}
\hat{A}=-\frac{\hbar^2}{2}\frac{\partial}{\partial z}
\frac{1}{m^*}\frac{\partial}{\partial
z}+\frac{\hbar^2k^2_\perp}{2m^*}+V(z)\;\;,
\end{equation}
\ \\
in which $k^2_{\perp} = k^2_x + k^2_y$ and $m^*(E, z)$ is the effective mass for the conduction electrons
\ \\
\begin{equation}
\frac{m_0}{m^*(E, z)}=1+C-\frac{E_P}{3}(\frac{2}{\tilde
\varepsilon_i}+ \frac{1}{\tilde f_i})\;\;,
\end{equation}
\ \\
where we explicitly indicate by the subscripts that $\tilde
\varepsilon_i$ and $\tilde f_i$ are different in various parts of the system.
Further
\ \\
\begin{equation}
\hat{K}_{SIA}=\frac{-ik_-}{\sqrt 2}\frac{\partial \eta(z)}{\partial
z}\;\;,
\end{equation}
where
\begin{equation}
\eta(z)=\frac{2P_0^2}{3}(\frac{1}{\tilde
\varepsilon_i}-\frac{1}{\tilde f_i})\;\;,
\end{equation}
\ \\
in which $k_- = (k_x - ik_y)/{\sqrt 2}$, while $\tilde \varepsilon_i(z) = \varepsilon_i(z)+V(z)-E$ and
$\tilde f_i(z) = \varepsilon_i(z)+\Delta_i(z)+V(z)-E$. Here $\varepsilon_i(z)$ are the energy gaps, $\Delta_i(z)$ are the
spin-orbit energies, $E_P = P^2_0 2m_0/\hbar^2$ and $C$
represents far-band contributions to the effective mass. The functions $\tilde
\varepsilon_i(z)$ and $\tilde f_i(z)$ depend on $z$ not only via
$V(z)$, but also due to the jumps of $\varepsilon_i$ and
$\Delta_i$ at the interfaces. We consider the case of steep barrier: $V(z) = 0$, $\varepsilon_i(z) = \varepsilon_A$, $\Delta_i(z) = \Delta_A$ and $\eta(z) = \eta_A$ for $z\le 0$, while $V(z) = V_b$, $\varepsilon_i(z) = \varepsilon_B$, $\Delta_i(z) = \Delta_B$ and $\eta(z) = \eta_B$ for $z > 0$, see Fig. 1. We denote $m^*(E, z) = m^*_A$ on the left of the barrier and $m^*(E, z) = m^*_B$ on the right. We consider an electron coming to the barrier from an oblique direction. Without loss of generality one can choose the coordinate system in such a way that $k_y = 0$, while $k_x \ne 0$ and $k_z \ne 0$.

It is seen that for $K_{SIA} = 0$ the two spin states have the same energy. We will solve set (1) using the well known approximate procedure for the perturbation of two degenerate states by calculating the matrix element of the nondiagonal term $K^{\uparrow\downarrow}_{SIA}$ between the solutions of diagonal terms. Explicitly
\ \\
\begin{equation}
E_{1, 2}=\hat{A}^{\uparrow\uparrow} \pm|\hat{K}_{SIA}^{\uparrow\downarrow}|\;\;.
\end{equation}
\ \\
The matrix element is taken between two orthogonal spin states $\chi_{\uparrow}(z) = (1\;,\;0)^T \chi(z)$ and $\chi_{\downarrow}(z) = (0\;,\;1)^T \chi(z)$, where $\chi(z)$ is solution of the Schrodinger equation $\hat{A}\chi(z) = E\chi(z)$, see Eq. (3). The unperturbed and perturbed electron spectrum is continuous and the wave functions are plane waves. This means that the wave functions along the $z$ direction must be normalized to the delta function $\delta(k_z - k'_z)$. In consequence, the matrix elements $A^{\uparrow\uparrow}$ and $K^{\uparrow\downarrow}_{SIA}$ have the dimension of [energy $\times$ length]. In order to obtain the correct dimension of [energy] one has to integrate Eq. (8) over $k'_z$. There is, for each spin, an incoming and a reflected plane wave for $z \le 0$ and a decaying wave for $z > 0$. The incoming wave vector is $k_z > 0$, the reflected one is $-k_z$ and the decaying wave is described by the imaginary wave vector $q_z$. The value of $k_x$ does not change since there is no force acting along the $x$ direction. Explicitly, the unperturbed spin-up and spin-down wave functions have the standard form
\ \\
\begin{equation}
\chi_{\uparrow} = \frac{\sqrt{N}}{\sqrt{2\pi L_x}} e^{ik_x x }\left(\begin{array}{c}1\\0\end{array}\right) \chi(z)
\;\;,
\end{equation}
\\
\begin{equation}
\chi_{\downarrow} = \frac{\sqrt{N}}{\sqrt{2\pi L_x}} e^{ik_x x }\left(\begin{array}{c}0\\1\end{array}\right) \chi(z)
\;\;,
\end{equation}
\ \\
where
\ \\
\begin{equation}
\chi(z) = (e^{ik_z z} + R e^{-ik_z z})_{| z\le 0} + (T e^{iq_z z})_{| z > 0}
\;\;.
\end{equation}
\ \\
Along the $x$ direction the functions are normalized to the length $L_x$, while along the $z$ direction we aim at the normalization to the delta
function $\delta(k_z - k'_z)$ and keep the normalization coefficient $\sqrt N$, where $N$ is to be determined.

We first consider the boundary conditions
at the interface. There is
\ \\
\begin{equation}
\chi|_{+_{0}}=\chi|_{-_{0}}\;\;\;,
\end{equation}
\ \\
and
\ \\
\begin{equation}
\left(\frac{1}{m^*_{B}}\frac{\partial \chi}{\partial
z}\right)_{+_{0}} = \left(\frac{1}{m^*_A}\frac{\partial
\chi}{\partial z}\right)_{-_{0}}\;\;\;.
\end{equation}
\ \\
This gives for both spin functions
\ \\
\begin{equation}
T = 1+R\;\;\;,
\end{equation}
\ \\
\begin{equation}
\frac{ik_z(1-R)}{m^*_A} = \frac{iTq_z}{m^*_B}\;\;\;,
\end{equation}
\ \\
and finally
\ \\
\begin{equation}
R=\frac{(k_z-q_z M)}{(k_z+q_z M)}\;\;\;.
\end{equation}
\ \\
where $ M = m^*_A/m^*_B$.
\ \\

We assume that initially the electrons are characterized by the energy $E$ and the wave vector $k_x$. Since in the region
 A there is $V(z) = 0$, the energy of incoming electrons is $E = \hbar^2(k^2_x + k^2_z)/2m^*_A$, so we have
\ \\
\begin{equation}
k_z^2=\left(E\frac{2m^*_A}{\hbar^2}-k^2_x\right)\;\;\;,
\end{equation}
\ \\
while in the barrier region B there is
\ \\
\begin{equation}
q_z^2=\left[(E-V_b)\frac{2m^*_B}{\hbar^2}- k_x^2\right]\;\;\;.
\end{equation}

Now we calculate the normalization of $\chi(z)$.
\ \\
$$
\frac{N}{2\pi}\int_0^{\infty}dk'_z\int_{-\infty}^0dz (e^{ik_z z}+Re^{-ik_z z})(e^{-ik'_z z} + R'^*e^{ik'_z z}) +
$$
\ \\
\begin{equation}
+ \frac{N}{2\pi}\int_0^{\infty}dq'_z\int_0^{\infty}dz Te^{iq_z} T'^*e^{-iq'_z z} = 1\;\;.
\end{equation}
\ \\
The final formula for $N$ depends on the relation between the electron energy $E$ and $V_b + \hbar^2k_x^2/2m^*_B$.

a) The first case is: $E < V_b + \hbar^2k_x^2/2m^*_B$, so that $q_z = i|q_z|$ according to Eq. (18). The function in the barrier $Te^{-|q_z|z}$ decays fast and we can neglect the term proportional to $T$. Consequently, $\chi(z)$ is nonzero for $-\infty < z \le 0$. The normalization condition is
\ \\
$$
\frac{N}{2}\int_0^{\infty} dk'_z\left( (1 + R R^{'*})\delta(k_z - k'_z)+\right.
$$
\begin{equation}
\left. - (R + R^{'*})\delta(k_z + k'_z)\right) = 1
\end{equation}
\ \\
The second term gives no contribution to the integral since both $k_z$, $k'_z > 0$. The first term imposes $k'_z = k_z$, then from Eq. (16) there is $1 + R R'^* = 2$, so that finally $N = 1$.
It should be noted that the integration over $k'_z$ is related to the correct dimension of energy in Eq. (3), as mentioned above.

b) The second case is, $E > V_b + \hbar^2k_x^2/2m^*_B$, so that $q_z, R, T$ are real. The function
$\chi(z)$ is nonzero for $-\infty < z < \infty$. The normalization condition is
\ \\
$$
\frac{N}{2}\left[\int_0^{\infty} dk'_z (1 + R R')\delta(k_z - k'_z) + \right.
$$
\begin{equation}
\left. +\int_0^{\infty} dq'_z T T'\delta(q_z - q'_z)\right] = 1\;\;.
\end{equation}
\ \\
The first term gives $1+R^2$ and the second one gives $T^2 = (1+R)^2$. Finally there is $N = (k_z+Mq_z)^2/(3k_z^2+M^2q_z^2)$.
\begin{figure}
\includegraphics[scale=0.45,angle=0, bb = 700 20 202 540]{fig1.ps}
\caption{\label{fig:epsart}{One-dimensional potential barrier used for the calculation of spin-orbit energy (schematically). Energy gaps and spin-orbit energies in regions A (InSb) and B (InAlSb) are indicated.}} \label{fig1th}
\end{figure}
\ \\

To calculate the spin splitting determined by the average value of $|\hat{K}^{\uparrow \downarrow}_{SIA}|$ = $\Delta$, we take into account the
valence offsets contained in the $\tilde \varepsilon_i(z)$ and $\tilde f_i(z)$ functions. We will assume here that the potential barrier is a steep but linear function of $z$, extending from  $-a/2$ to $a/2$, see Fig. 1. After some manipulation the expression for $\Delta$ is brought to the form
$$
\Delta = \int^{\infty}_0 dk'_z|\int^{a/2}_{-a/2}dz \chi^*_{\downarrow}(z,k'_z)\hat{K}_{SIA}\chi_{\uparrow}(z,k_z)|  =
$$
\ \\
$$
= \frac{Nk_x}{2}\frac{(\eta_B-\eta_A)}{a}
\int^{\infty}_0dk'_z [1+R^*(k'_z)][1+R(k_z)]\cdot
$$
$$
\cdot\frac{2}{2\pi}\frac{\sin[(k_z-k'_z)a/2]}
{(k_z-k'_z)} =
$$
\ \\
$$
=\frac{Nk_x}{2}\frac{(\eta_B-\eta_A)}{a}\int_0^{\infty}[1+R^*(K'_z)][1+R(K_z)]\cdot
$$
\begin{equation}
\cdot\frac{\sin[(K_z-K'_z)D]}{\pi(K_z-K'_z)}dK'_z
\end{equation}
\ \\
By changing the variable $k_z$ into $K_z = k_z a/2D$, $k'_z$ into $K'_z = k'_z a/2D$, $|q_z|M$ into $|Q_z| = |q_z| M a/2D$ and $|q'_z|$ into $|Q'_z| = |q'_z| M a/2D$, where $D$ is a large number and $R(K_z) = ((K_z-Q_z)(K_z+Q_z)$, the last factor in the integrand of Eq. (22) can be replaced to a very good approximation by $\delta(K_z - K'_z)$. Then the integration over $dK'_z$ in Eq. (22) can be  carried out and the final result for the case $E < V_b + \hbar^2k_x^2/2m^*_B$ is
\begin{equation}
\Delta_{E <} = \frac{k_x(\eta_B-\eta_A)}{2a}
\frac{4k_z^2}{(k_z^2 + |q_z|^2 M^2)}\;\;,
\end{equation}
\ \\
while for the case $E > V_b + \hbar^2k_x^2/2m^*_B$ there is
\begin{equation}
\Delta_{E >} = \frac{k_x(\eta_B-\eta_A)}{2a}
\frac{4k_z^2}{(3k_z^2 + q_z^2 M^2)}\;\;.
\end{equation}
\ \\
The spin splitting is $2\Delta$, see Eq. (8) and Fig. 2.

It can be
seen that the spin splitting of the conduction band due to the SIA mechanism is proportional to the spin-orbit energies in the
valence bands contained in $\eta_B - \eta_A$ term. Also, $\Delta$ depends on $k_z$ and $q_z$, which characterize the incoming electrons. In the published literature one deals with the Bychkov-Rashba spin splitting in the quantum wells, where $\Delta$ is related to the electron wave function in the well. The unusual feature of our situation is that we do not deal with the well, so the wave function is replaced by the incoming and returning waves. According to the two-state perturbation procedure, as formulated in Eqs. (3) and (8), the resulting wave functions for the effective spin-up and spin-down have the form $(1, 1)^T$ and $(1, -1)^T$, respectively.
\section{\label{sec:level1} Spin-conserving and spin-flip reflection\protect\\ \lowercase{}}

The off-diagonal terms $\hat{K}_{SIA}$ mix the spins in the wave functions due to the SOI in the complete solutions of Eq. (3). The resulting spin-mixed functions correspond to the \emph{effective} spin-up and spin-down states, which we label by subscripts $1$ and $2$, respectively.
\begin{widetext}
\begin{equation}
\Psi_1=\frac{e^{ik_x x}}{\sqrt{4\pi L_x}}\left\{\left[e^{ik'_z z}\left(\begin{array}{c}1\\1\end{array}\right)+Re^{-ik'_z z}\left(\begin{array}{c}1\\1\end{array}\right)+R'e^{-ik''_z z}\left(\begin{array}{c}1\\-1\end{array}\right)\right]_{|z\le 0}+
\left[Te^{iq'_z z}\left(\begin{array}{c}1\\1\end{array}\right)+T'e^{iq''_z z}\left(\begin{array}{c}1\\-1\end{array}\right)\right]_{|z>0}\right\}
\;,
\end{equation}
\ \\
\begin{equation}
\Psi_2=\frac{e^{ik_x x}}{\sqrt{4\pi L_x}}\left\{\left[e^{ik''_z z}\left(\begin{array}{c}1\\-1\end{array}\right)+
Pe^{-ik''_z z}\left(\begin{array}{c}1\\-1\end{array}\right)+P' e^{-ik'_z z}\left(\begin{array}{c}1\\1\end{array}\right)\right]_{|z\le0}+
\left[Fe^{iq''_z z}\left(\begin{array}{c}1\\-1\end{array}\right)+F'e^{iq'_z z}\left(\begin{array}{c}1\\1\end{array}\right)\right]_{|z>0}\right\}
\;,
\end{equation}
\end{widetext}
\ \\
Function $\Psi_1$ describes a spin-up electron with the incoming wave vector $k'_z > 0$ which, if reflected without the change of spin, is returning with the wave vector $-k'_z$ and has the penetrating component characterized by $q'_z$. If the electron is reflected with the change of spin direction, the returning wave is characterized by $-k''_z$ and the penetrating one by $q''_z$. Things are analogous for the electron with the initial effective spin down.
Suppose the incoming electron is in the state $\Psi_1$, exactly speaking in the state $(1, 1)^T$, so that its energy is
\ \\
\begin{equation}
E_1 = \frac{\hbar^2 {k'}^2}{2m^*_A} + \Delta
\;\;,
\end{equation}
\ \\
where ${k'}^2 = k_x^2 + {k'}_z^2$. After an elastic reflection from the barrier there are two possibilities. If the reflected electron is still in the $(1, 1)^T$ state (spin-conserving process) its energy is again given by Eq. (27), the resulting $k_x$ remains the same while $k'_z$ changes sign. This means that, in a spin-conserving reflection, the outgoing and incoming electron directions form the same angles with the normal to the barrier. If, on the other hand, the reflected electron is in the $(1, -1)^T$ state (spin-flip process), its energy is
\ \\
\begin{equation}
E_2 = \frac{\hbar^2 {k''}^2}{2m^*_A} - \Delta
\;\;.
\end{equation}
\ \\
where ${k''}^2 = k_x^2 + {k''}_z^2$. In an elastic reflection the total electron energy must remain the same, so we have from Eqs. (27) and (28)
\begin{equation}
\frac{\hbar^2}{2m^*_A}({k''}^2 - {k'}^2) = 2\Delta\;\;.
\end{equation}
\begin{figure}
\includegraphics[scale=0.45,angle=0, bb = 700 20 202 540]{fig2.ps}
\caption{\label{fig:epsart}{Electron energy versus wave vector $k_z$ for spin-up and spin-down states (schematically). The spin-orbit energy is $\Delta$. For a spin-conserving elastic reflection of spin-up electron the outgoing $k'_z$ is equal to $-k'_z$, for a spin-flip reflection the outgoing wave vector is $-k''_z$ such, that $|k''_z| > |k'_z|$. Situation for an initial spin-down electron with the initial wave vector $k''_z$ is also shown.}} \label{fig2th}
\end{figure}
Since $k_x$ does not change, there is
\begin{equation}
{k'}_z^2 = (E - \Delta)\frac{2m^*_A}{\hbar^2} - k_x^2\;\;,
\end{equation}
\ \\
and
\begin{equation}
{k''}_z^2 = (E + \Delta)\frac{2m^*_A}{\hbar^2} - k_x^2\;\;.
\end{equation}
\ \\
This means that, because of the spin splitting related to the SOI, in a spin-flip reflection the outgoing and incoming directions do \emph{not} form the same angles with the normal to the barrier. The above reasoning is illustrated in Fig. 2. Clearly, there are limits to the initial and final directions: they should not be parallel to the barrier. The incoming angle with the normal to the barrier is
$\alpha'$ = ctg$^{-1}(k'_z/k_x)$ and the outgoing angle after spin-flip process is: $\alpha''$ = ctg$^{-1}(k''_z/k_x)$ and the difference of angles between spin-conserving and spin-flip reflected electron beams is $\alpha' - \alpha''$.

Now we turn to the amplitudes of functions (25) and (26). Let us again consider an incoming spin-up electron. There exist, in addition to the waves characterized by $k'_z$ and $k''_z$ considered above, waves penetrating the barrier. For the spin-up penetrating wave the wave vector $q'_z$ is
\begin{equation}
q'^2_z=(E - \Delta - V_b)\frac{2m_B}{\hbar^2}-k^2_x\;\;,
\end{equation}
\ \\
while for the spin-down penetrating wave the wave vector $q''_z$ is
\begin{equation}
q''^2_z=(E + \Delta - V_b)\frac{2m_B}{\hbar^2}-k^2_x\;\;\;.
\end{equation}
\ \\
The spin-conserving and spin-flip waves penetrating the barrier form the angles $\beta'$ = ctg$^{-1}(|q'_z|/k_x)$ and $\beta''$ = ctg$^{-1}(|q''_z|/k_x)$ with the normal to the step potential, respectively, see Fig. 3.
\begin{figure}
\includegraphics[scale=0.45,angle=0, bb = 600 20 202 540]{fig3.ps}
\caption{\label{fig:epsart}{Geometry of spin-polarized electrons reflected from a
semiconductor potential barrier (schematically). In a spin-conserving reflection of a spin-up electron the
outgoing angle is equal to the incoming angle $\alpha'$, in a spin-flip
reflection the outgoing angle $\alpha''$ is smaller: $\alpha'' < \alpha'$. The waves penetrating the barrier at angles $\beta'$ and $\beta''$ are also shown. The effect is due to the spin-orbit interaction manifested in
oblique motion, see Fig. 1.
}} \label{fig3th}
\end{figure}
In the following we consider explicitly an incoming spin-up electron characterized by $\Psi_1$.
The amplitudes of incoming, reflected and penetrating parts of the function can be determined from the boundary conditions for the wave functions and their derivatives at $z = 0$. For the upper-spin components in Eq. (27) the continuity condition is
\ \\
\begin{equation}
 {\Psi_1^{up}}_{| z = 0^-} = {\Psi_1^{up}}_{| z = 0^+}\;\;,
\end{equation}
\ \\
while for the lower spin components there is
\begin{equation}
{\Psi_1^{low}}_{| z = 0^-} = {\Psi_1^{low}}_{| z = 0^+}\;\;\;.
\end{equation}
\ \\
The remaining boundary conditions come from the integration of Eqs. (3) across the interface at $z = 0$. We have
\ \\
$$
\frac{\hbar^2}{2m^*_A}\frac{\partial {\Psi_1^{up}}}{\partial z}_{| z = 0^-} = \frac{\hbar^2}{2m^*_B}\frac{\partial {\Psi_1^{up}}}{\partial z}_{| z = 0^+} +
$$
\ \\
\begin{equation}
i\frac{k_x(\eta_B-\eta_A)}{2}{\Psi_1^{low}}_{| z = 0}\;\;\;,
\end{equation}
\ \\
$$
\frac{\hbar^2}{2m^*_A}\frac{\partial {\Psi_1^{low}}}{\partial z}_{| z = 0^-} = \frac{\hbar^2}{2m^*_B}\frac{\partial {\Psi_1^{low}}}{\partial z}_{| z = 0^+} - $$
\ \\
\begin{equation}
i\frac{k_x(\eta_B-\eta_A)}{2}{\Psi_1^{up}}_{| z = 0}\;\;\;,
\end{equation}
\ \\
which gives
\begin{equation}
 \frac{\partial {\Psi_1^{up}}}{\partial z}_{| z = 0^-} = \frac{m^*_A}{m^*_B}\frac{\partial {\Psi_1^{up}}}{\partial z}_{| z = 0^+} + iS{\Psi_1^{low}}_{| z = 0}\;\;\;,
\end{equation}
\ \\
\begin{equation}
 \frac{\partial {\Psi_1^{low}}}{\partial z}_{| z = 0^-} = \frac{m^*_A}{m^*_B}\frac{\partial {\Psi_1^{low}}}{\partial z}_{| z = 0^+} - iS
{\Psi_1^{up}}_{| z = 0}\;\;\;,
\end{equation}
\ \\
where $S = k_x(\eta_B-\eta_A)m^*_A/\hbar^2$. The terms with $\partial \Psi_1^{up/low}/\partial z$ come from the integration of the diagonal terms in Eq. (3), while the terms with $S$ result from the integration of the nondiagonal terms related to the SOI. We set, as before, $k_y = 0$.
From Eqs. (34) and (35) one gets have: $1 + R + R' = T + T'$ and $1 - R - R'= T - T'$, which gives $R' = T'$ and $R = T - 1$.
From Eqs. (38) and (39) we have
\ \\
$$
ik'_z(1-R)-ik''_zR'=
$$
\begin{equation}
iMq'_zT+iMq''_zT'+iS(T-T')\;\;\;,
\end{equation}
\ \\
$$
 ik'_z(1-R)+ik''_zR'=
 $$
 \begin{equation}
 iMq'_zT-iMq''_zT'-iS(T+T')
\;\;\;,
\end{equation}
\ \\
where $M = m^*_A/m^*_B$. After some manipulations we finally obtain from Eqs. (40) and (41)
\ \\
\begin{equation}
R = \frac{[(k'_z-Mq'_z)(k''_z + Mq''_z)-S^2]}{[(k'_z+Mq'_z)(k''_z+Mq''_z)+S^2]}
\end{equation}
\ \\
\begin{equation}
R' = \frac{-2S k'_z}{[(k'_z+Mq'_z)(k''_z+Mq''_z)+S^2]}
\end{equation}
\ \\
\begin{equation}
T  = \frac{2k'_z(k''_z+Mq''_z)}{[(k'_z+Mq'_z)(k''_z+Mq''_z)+S^2]}
\;\;\;,
\end{equation}
\ \\
\begin{equation}
T' = R'\;\;\;.
\end{equation}
\ \\
It is seen that the amplitudes of spin-flip reflected and penetrating waves, proportional to $R'$ and $T'$ terms in Eq. (25), do not vanish because of the $S$ terms in Eqs. (43) and (45). The latter are due to the spin-orbit energies in the valence bands and, technically, come about from the spin-dependent boundary conditions for $\Psi_1$ function in $z = 0$.

A similar analysis can be carried out for the incoming spin-down electron with the use of Eq. (26). For such electrons one has $P = R$, $P' = -R'$, $F = T$ and $F' = -T'$. At the same time one should replace in the formulas $k'_z$ by $k''_z$, $q'_z$ by $q''_z$, $k''_z$ by $k'_z$, and $q''_z$ by $q'_z$. If we use Fig. 3 then, qualitatively speaking, for the initial spin-up electron, the spin-flip process results in a reflection angle smaller than the incoming one, while for the initial spin-down electron the spin-flip process results in a larger reflection angle than the incoming one.

Since, as follows from Eqs. (23 - 24), the spin splitting 2$\Delta$ is characterized by $k_x$, $k_z$ and $q_z$ and, in turn, the kinematics in determined by $\Delta$, one should analyze how the final results are related to the initial conditions. Suppose one begins with the spin-up electron having the energy $E$ and coming to the barrier at the angle $\alpha'$. Then one gets $k_x = k'_z (tg{\alpha'})$. One can now use Eq. (17) to obtain $k_z^2 = \left(E\frac{2m^*_A}{\hbar^2}-k'^2_z tg^2{\alpha'}\right)$ and Eq. (18) to get $q_z^2 = \left[(E-V_b)\frac{2m^*_B}{\hbar^2}- k'^2_z tg^2{\alpha'}\right]$. Putting the above expressions for $k_x$, $k_z$, $q_z$ to Eq. (23) for $\Delta_{E < V_b}$ or to Eq. (24) in order to calculate $\Delta_{E > V_b}$, and using Eq. (30) we obtain an equation for one unknown $k'_z$. Having determined $k'_z$ we calculate $k_x$, $k_z$ $q_z$, then $\Delta$ and finally $k''_z$, $q'_z$ and $q''_z$. This way one can calculate the angle $\alpha''$ and then the wave amplitudes given by the above formulas.
\section{\label{sec:level1} Results\protect\\ \lowercase{}}

When performing experiments one can imagine two different situations. In the first, a source produces spin-polarized electrons which come to the barrier from an opaque direction and are reflected in the spin-conserving or spin-flip processes. This case is illustrated in Fig. 3. In the second situation, realized in the experiments of Chen et al [10, 12, 14], electrons come to the barrier from a defined direction but they are not spin polarized. In this case one can consider the spin to be a combination of spin-up and spin-down components. For the spin-flip processes, the reflected spin-down component has the direction further from normal (see Fig. 3), the spin-up component has the direction closer to the normal, and the spin-conserving processes give the reflection having the same angle as the incoming beam. Thus the reflected beams on both sides contain spin polarized electrons while the middle beam contains unpolarized electrons.
\begin{table}[h]
\caption{Material parameters used for InSb/In$_{0.91}$ Al$_{0.09}$Sb barrier. The lattice constant $a$ is taken as an average value between InSb and In$_{0.91}$Al$_{0.09}$Sb.}
\begin{tabular}{c|c|c|c|c}

\hline &&InSb&In$_{0.91}$Al$_{0.09}$Sb&\\
\hline
\hline
&$E_g$(eV)& -0.2466 & -0.426&\\
&$\Delta_0$(eV)& -0.8419 & -0.777 &\\
&$m^*_0/m_0$& 0.0143 & 0.0234 &\\
&C& -1.496 & -1.3685 &\\
&$E_p$(eV)& 23.4 & 23.4 &\\
&$a$($10^8$cm)& 6.4&6.4&\\
&$V_b$(eV)& - & 0.111 &\\\hline
\end{tabular}
\end{table}
\ \\
\begin{table}[h]
\caption{Heterojunction parameters used for InSb/In$_{0.91}$ Al$_{0.09}$Sb barrier. The effective masses $m^*_A/m_0$ and $m^*_B/m_0$ are calculated for given electron energies $E$.}
\begin{tabular}{c|c|c|c|c}
\hline &&InSb&In$_{0.91}$Al$_{0.09}$Sb&\\
\hline
\hline
&E(eV)& 0.08 & 0.08 &\\
&$m^*_A/m_0$&0.01854 & - &\\
&$m^*_B/m_0$& - & 0.02184 &\\ \hline
&E(eV)& 0.13 & 0.13 &\\
&$m^*_A/m_0$& 0.02113 & - &\\
&$m^*_B/m_0$& - & 0.02435 &\\\hline
\end{tabular}
\end{table}

Now we estimate numerical values of the reflected directions and amplitudes by considering specific barrier: InSb/In$_{0.91}$Al$_{0.09}$Sb. In Table I we quote the band parameters of the involved narrow-gap semiconductor materials plus the offset $V_b$ of the heterojunction. All the other parameters can be inferred from the given values. In Table II we quote the effective masses calculated for the electron energies $E$ in regions A and B. In Tables III and IV we quote  characteristics of reflected and penetrating waves, as calculated for different initial sets of parameters. The initial sets are the electron energy $E$ for $E < V_b + \hbar^2k_x^2/2m^*_B$ (Table III) and $E > V_b + \hbar^2k_x^2/2m^*_B$ (Table IV)) and the wave vector $k_x$ (two values) determining the incoming angle $\alpha'$ (for spin-up electrons) and $\alpha''$ (for spin-down electrons). In entries for $k_x = 0.35 \cdot 10^6$ cm$^{-1}$ we deal with the "standard" case in which $k'_z$ and $k''_z$ have real values, while $q'_z$ and $q''_z$ have imaginary values for $E < V_b + \hbar^2k_x^2/2m^*_B$ and real values for $E > V_b + \hbar^2k_x^2/2m^*_B$. An interesting and unexpected situation is described in the entries for $k_x = 1\cdot10^6$ cm$^{-1}$. It follows from Tables I and III that, for the incoming energy $E$ = 80 meV of the spin-down electron and $V_b$ = 111 meV, there is $E < V_b + \hbar^2k_x^2/2m^*_B$, i.e. the kinetic electron energy is smaller than the barrier height. According to the intuitive expectations, it should be a "normal" case in which the wave penetrating the barrier quickly decays with the distance $z$. However, the calculated spin-orbit energy for this case is $\Delta$ = 58 meV, so that $E + \Delta$ = 138 meV is higher than $V_b$. As a consequence, it follows from Eq. (33) that the resulting $q''_z$ = 0.736 $\cdot 10^6$ cm$^{-1}$ is \emph{real}. This means that the penetrating spin-down wave $exp(iq''_z z)$\emph{ propagates in the barrier without attenuation}!
\begin{table}[h]
\caption{Case $E < V_b + \hbar^2k_x^2/2m^*_B$. Spin-orbit energies and reflection characteristics for spin-up electron ($\alpha'$) and spin-down electron ($\alpha''$) calculated for $E$ = 0.08 eV and $V_b$ = 0.111 eV, and given values of $k_x$.}
\begin{tabular}{c|c|c|c|c}
\hline
&$k_x$($10^{-6}$/cm)& 0.35&1.0&\\
&$\Delta$(eV)& 0.025 &0.058&\\
\hline
\hline
&$\alpha'$& 12.37$^o$ &74.63$^o$ &\\
&$q'_z$($10^{-6}$/cm)&i1.828 &i2.47& \\
&$R$&0.028-i0.999&=0.97-i0.24& \\
&$R'$&-0.008+i0.014&i0.006& \\
&$RC_1$& 1&0.9994& \\
&$TC_1$& 0&0.0006& \\
\hline
&$\alpha''$& 8.90$^o$ &22.71$^o$ &\\
&$q''_z(10^{-6}$/cm)&i0.676&0.736&\\
&$P$& 0.89-i0.45 &0.585& \\
&$P'$& 0.013-i0.02 &0.006-i0.05& \\
&$RC_2$&1&0.343& \\
&$TC_2$&0&0.657& \\
\hline
\end{tabular}
\end{table}
\ \\
\begin{table}[h]
\caption{Case $E > V_b + \hbar^2k_x^2/2m^*_B$. The same as in Table III but for $E$ = 0.13 eV and $V_b$ = 0.111 eV.}
\begin{tabular}{c|c|c|c|c}
\hline
&$k_x(10^{-6}$/cm)& 0.35&1.0&\\
&$\Delta$(eV)& 0.0085 & 0.025 &\\
\hline
\hline
&$\alpha'$& 7.75$^o$&24.48$^o$&\\
&$q'_z(10^{-6}$/cm)&0.74&i1.176& \\
&$R$&-0.600&0.645-i0.764& \\
&$R'$& -0.0098&-0.03+i0.01& \\
&$RC_1$&0.361&0.9997& \\
&$TC_1$&0.639&0.0003& \\
\hline
&$\alpha''$&7.25$^o$&19.94$^o$ &\\
&$q''_z(10^{-6}$/cm)&1.279&1.346&\\
&$P$& 0.425&0.405& \\
&$P'$&0.0104&0.035-i0.016& \\
&$RC_2$&0.18&0.164& \\
&$TC_2$&0.82&0.836& \\
\hline
\end{tabular}
\end{table}

An "inverse" situation is created by the incoming spin-up electron with the energy $E$ = 130 meV, see Table IV. For the barrier $V_b$ = 111 meV there is $E > V_b + \hbar^2k_x^2/2m^*_B$, so that the incoming electron moves in principle above the barrier. However, for this case there is $\Delta$ = 25 meV. So that $E-\Delta < V_b + \hbar^2k_x^2/2m^*_B$ and, according to Eq. (32), the resulting $q'_z$ is imaginary. This means that the continuing spin-up component \emph{quickly decays above the barrier} and there is an almost total reflection of the spin-up wave.

It is instructive to calculate not only the amplitudes, but also reflection and transmission coefficients of various waves. A transmission coefficient TC is defined as TC=$|J_{tr}/J_{in}|$ and the reflection coefficient as RC=$|J_{re}/J_{in}|$, where $J_{in}$ is the incoming current.
According to Eqs. (25 - 26), we have for the incoming spin-up electron
\begin{equation}
TC_1 = (|T|^2q'_z + |T'|^2q''_z)(\frac{ m^*_A}{k'_z m^*_B})\;\;,
\end{equation}
\ \\
\begin{equation}
RC_1 = |R|^2+|R'|^2\frac{k''_z}{k'_z}\;\;,
\end{equation}
\ \\
while for the incoming spin-down electron
\begin{equation}
TC_2 = (|F|^2q''_z + |F'|^2q'_z)(\frac{ m^*_A}{k''_z m^*_B})\;\;,
\end{equation}
\ \\
\begin{equation}
RC_2 = |P|^2+|P'|^2\frac{k'_z}{k''_z}\;\;,
\end{equation}
where the second terms in the above equalities correspond to spin-flip amplitudes.
For imaginary $q'_z$ or $q''_z$ we have the decaying waves in the barrier, i. e. $J_{tr}$ = 0 and TC = 0.
In the last entries of Tables III and IV we give the values of RC and TC, as calculated from the above formulas. It can be seen that RC + TC = 1, as it should be. One can also see that in the interesting two cases, i. e. the non-attenuated transmitted wave for $E < V_b + \hbar^2k_x^2/2m^*_B$ (Table III) has large transmission coefficient TC$_2$ = 0.657, while for electron with $E > V_b + \hbar^2k_x^2/2m^*_B$ (Table IV) the transmitted wave is attenuated in the barrier and has the transmission coefficient TC$_1$ = 0.0003. Small values of TC$_1$ = 0.0006 in Table III and TC$_1$ = 0.0003 in Table IV merit special comments. Concerning Table III, there is a small component of the spin-flip wave which is transmitted without attenuation (TC$_1$ = 0.0006), while the spin-conserved wave is completely reflected (RC$_1$ = 0.9994), so that the sum TC$_1$ + RC$_1$ = 1, as it should be. Similar reasoning applies to TC$_1$ = 0.0003 and RC$_1$ = 0.9997 in Table IV.

It is clear that the considered system can serve as a source or a filter of spin-polarized electron beams since it spatially separates electrons with specifically oriented effective spins. One can also say that this arrangement realizes the Stern-Gerlach experiment for mobile electrons, which is not possible for free electrons in a vacuum (see discussion in Ref. 15) and very difficult for electrons in semiconductors (Wrobel et al. [16]).
\section{\label{sec:level1} Discussion\protect\\ \lowercase{}}

We first examine the approximations used in our approach. Our description of the potential barrier is somewhat ambiguous since we sometimes use a vertical barrier and, when calculating the spin-orbit energy for unbound electrons, a barrier of finite slope. As we mentioned in Section II, the latter assumption was necessary in order to avoid an infinite electric field. However, this ambiguity is not troublesome since one needs a \emph{steep} barrier to obtain sizable spin splittings. On the other hand, a narrow barrier (on the order of interatomic distance) is in a certain contradiction with the validity of the {\kp} theory. The latter requires that the potential changes be slow compared to the lattice period. Fortunately, it is known that the {\kp} theory, usually gives better results than it is supposed to. Also, when calculating SOE we assume in Eq. (22) that the reflection amplitudes $R$ do not depend on $z$. All in all, it is possible that our calculations of SOE for unbound electrons are not very precise. This, however, should not influence qualitatively our results. To the best of our knowledge, a calculation of the Bychkov-Rashba spin splitting for unbound electrons has not been attempted before.

As it was mentioned in the Introduction, the spin-flip ballistic reflection of electrons from a potential barrier was experimentally realized by Chen and coworkers in InSb/InAlSb and InAs/AlGaSb heterostructures [10, 12, 14]. In their experiments the authors used a "triangular geometry": two sides of a triangle with slits were employed to inject and detect electron beams, the third side was utilized as a barrier. An external magnetic field transverse to the 2D plane of the heterostructure was used to direct the reflected beams to the detecting slit. A theoretical description of these experiments was presented mostly in Ref. 11. The authors assumed that the spin splitting of electron energies due to the SOI had two origins: the structural asymmetry of the quantum well and the barrier. However, it was then incorrectly concluded that the Bychkov-Rashba spin splitting of electron energies due to the asymmetric well was proportional to the electric field in the conduction band and that the spin splittings caused by the two origins were governed by the same material parameter. A calculation of spin splitting due to the SO interaction of unbound electrons with the barrier was not attempted. Teodorescu and Winkler [17] in their analysis of spin-dependent reflections from an impenetrable barrier used as an origin of the Bychkov-Rashba spin splitting only the inversion asymmetry of the well. Also in the work on the spin separation in 2D cyclotron motion (Ref. 18) and transverse electron focusing in 2D systems with the SOI (Ref. 19), the Bychkov-Rashba splitting for unbound electrons was taken into account only phenomenologically.

It should be emphasized that, as follows from our Eq. (23) and Ref. 2, the Bychkov-Rashba splitting for unbound and bound electrons bear some similarities but are also marked by important differences. In both cases it is necessary to have the SOI and structure inversion asymmetry. In both cases it is necessary to have a transverse motion and the resulting spin splitting is proportional to the transverse momentum. On the other hand, for the unbound electrons the spin splitting depends on the longitudinal momentum $\hbar k_z$, while for the bound electrons this momentum is "frozen" in the electron wave function of the quantum well and it does not explicitly appear. Also, as follows from Eqs. (23) and (24), for the unbound electrons even the functional forms of $\Delta$ are different for $E < V_b + \hbar^2k_x^2/2m^*_B$ and $E > V_b + \hbar^2k_x^2/2m^*_B$ situations.
\section{\label{sec:level1} Summary\protect\\ \lowercase{}}

We described spin-conserving and spin-flip opaque reflections of 2D electrons from a barrier in InSb/InAlSb heterostructure. In such a system the electric field of the barrier is the only source of spin splitting in the presence of spin-orbit interaction. Formulas for the  Bychkov-Rashba spin splitting are calculated for unbound electrons. Angles and amplitudes are calculated for reflected and penetrating electron beams. Two unexpected possibilities are predicted due to the spin-orbit interaction: a) non-attenuated electron propagation in the barrier when the latter is higher than the kinetic energy of incoming electrons, b) total reflection of electrons whose energies exceed barrier's height. It is shown that the considered system can serve as a source or filter of spin-polarized electron beams.
\ \\

{\bf Acknowledgments}
\ \\
We are grateful to Dr T. M. Rusin for elucidating discussions.
\ \\

redpajama_set_name: RedPajamaArXiv


Upcoming nerf to Amaya Lake Lodge?
Agree, but probably more ridiculous will be thier explanation.
@ZOS_RichLambert WAKE UP THE DAMN HOUSING ART TEAM, or whoever does this crap and tell them to stop making changes. Or at least, tell Gina so she gets them in the update notes, which really wont make those who buy crowns to buy these houses any happier..
Or, is Zenimax telling players, yeah.. "We want to *** of the whales who buy our most expensive houses for cash."?
Has anyone here ever owned (or lived in) a real-life house? Things change. Things break. Things disappear. In real life we can't even report it to a Higher Power and hope that God Magically Fixes It for us.
If the planter disappears, I will put a pot in there and call it day.
You're very optimistic, and that's good for you, but things in real-life houses don't break or disappear the way they do in ESO despite what you say. Even when this is just a game, people are still spending quite a wealth in ingame gold (that comes from long hours of gameplay and efforts)and real-life money (crowns) to buy those products, and liking them as they were "advertised", just to find out a month later that the house they bought and probably decorated to the fullest no longer looks as it was intended, overnight.
If it were a single case in a single home, then it would be "fine" (only if you don't own or don't enjoy the house as much as others do), but it's been happening "consistently" for quite some time. It seems every single update brings a broken house with it, along with a growing list of dissatisfied customers that aren't happy with the bugs anymore.
The problem I would guess is developers aren't aware when they break something (which is a shame), because if they did a fix would surely be included in next week's patch. As they keep adding more content, and using pre-existing content as templates, and then changing textures to give the impression of "new content", things will keep breaking on and on until they start designing original objects for whatever new content they want to bring to the game.
Intended feature or a bug?
Will this change go to live server or no?
I actually prefer the now uncluttered beach at Hundings, so a few pebbles and bits of seaweed vanishing doesn't worry me at all.
But I do feel for those owners of Earthtear Cavern who lost whole areas which they had spent time and effort developing.
SH*T like this is why I won't ever touch this games housing system. It's overpriced and lackluster at best but seeing this kind of crap being pulled now too makes me glad I didn't invest at all in housing.
As an owner of Earthtear Cavern you have my sympathy. I personally like the look without the box, but I don't own this house so I don't have much of a say. Since it appears that the planter is in the coding they should offer it as a free piece, that way you can put it back or take it out as you see fit. Changes to housing shouldn't be made unless the customers are made aware. Keep at em, it worked for the cavern. Best of luck to you.
Since it appears that the planter is in the coding they should offer it as a free piece, that way you can put it back or take it out as you see fit.
Interesting proposition. I would +1 that. It would be a good time to add planters with soil as new furniture as well (instead of the empty vases and whatnot even devs have been using since forever.
You all complaining about some missing planter while nobody notices you literally do not have Red Mountain in the background.
Nobody has been missing that (perhaps), except you, but good point. Time to add it to the list of bugs then. (It was added to Tel Galen, though).
Agree. Option to purchase for free 1 planter from crown store (for all Amaya Lake Lodge owners) could solve the problem.
Hey everyone, wanted to touch on the issues mentioned in this thread.
First, the planter. This was actually never supposed to go live in the house where it was placed, which is why console players don't even have the planter to begin with; it was only seen on PC. This was a very nice planter, though, so we're figuring out a way to provide the planter as a furnishing for those that previously had it. Details to come as we're still working everything out, but expect to have this back one way or another.
As for Red Mountain, this was never viewable from the home and wasn't a change that was recently made. We'll see if we're able to add this to the distance, but there are a lot of variables that go into what can be viewed from homes.
Is this the new marketing plan. Put something in the houses that wasn't "supposed" to be there then remove it. Only to sell it back to the person who already bought it. Same with Hundings Palatial Hall.
We wouldn't sell it to those that previously had it - we would give it to them.
Did you even read the post you quoted?
I am ok with this. And with buying MORE planters! I was disappointed to find that I couldn't have indoor planters like so many NPC buildings, now I can!
I've seen some people made out of other things that look absolutely amazing. Sink a barrel or a crate in the ground and it gives a pretty cool effect.


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<string name="low_memory_error">"Den tidligerere handling kunne ikke fuldføres på grund af manglende hukommelse"</string>
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<string name="copy_to_clipboard_failure_message">"Der kunne ikke kopieres til udklipsholder"</string>
<string name="accessibility_date_picker_month">"Måned"</string>
<string name="accessibility_date_picker_year">"År"</string>
<string name="date_picker_dialog_set">"Angiv"</string>
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redpajama_set_name: RedPajamaGithub


text:
PARENT QUESTIONS GRADING POLICY
SPOTSYLVANIA PROPOSAL WOULD LEVEL PLAYING FIELD
WHEN GRADES ARE FIGURED FOR COLLEGE
Susan Sarver moved to Spotsylvania County knowing its students have to work hard for an A.
That didn't bother her until she compared her son Wilson's grade-point average with those of students from other school districts. He had a lower grade-point average for the same grades.
"I was told: 'Mrs. Sarver, don't worry about that. The colleges and universities recompute the GPAs,'" she recalled recently.
But as her son neared graduation, Sarver found out that many colleges didn't have the manpower to recalculate every applicant's GPA to a uniform standard. When a Maryland college her son applied to recalculated his GPA, it rose from a 3.2 to a
It meant an additional $8,000 in scholarship money.
"That's huge," Sarver said. "It does equate to missed opportunities, and it equates to dollars."
Sarver is one of several parents who approached school officials with a proposal to level the field for Spotsylvania students. Tomorrow, school administrators will present Sarver's proposal to the School Board.
A HALF-POINT DIFFERENCE
As getting into college and earning scholarships has become increasingly competitive, more parents are questioning their school systems' grading policies.
Fairfax County parents formed the research/advocacy group Fairgrade to push for a change in grading policies. School officials in Loudoun County also are looking into the issue.
Under Spotsylvania's current policy, first adopted in 1987 and revised six times since, students need a 94 or higher to earn an A.
An A is valued at 4.0 on a four-point scale. An extra point is given for each advanced-placement and dual-enrollment course, so they're calculated on a five-point scale.
No extra credit is given for honors classes, and a 98 counts the same toward the GPA as does a 94.
In Sarver's proposal, the value of an A would increase in 0.25-point increments as you go up the scale.
A 96, for example, would count as 4.25 points toward the GPA, and a 98 would count as 4.5 points. Under the current system, 100 is a 4.0. Under Sarver's proposal, it would be worth 4.5. That half-point, she said, could make the difference in college admissions.
"The difference in how your GPA shows is the difference between the 'yes' or 'maybe' pile, and the 'no' pile," Sarver said.
GRADING DISPARITY
Spotsylvania school officials say changing the grading policy won't eliminate the disparity.
"No matter what you change to, there's always going to be somebody different. That's exactly the problem," said Assistant Superintendent of Instruction Edlow Barker.
Still, Barker said administrators want to make Spotsylvania students' grades comparable to those in other divisions. Because there's no state law regulating grading policy, even neighboring counties may use different methods.
Stafford County has a grading scale similar to Spotsylvania's, but uses a plus or minus system that awards a half-point for pluses. In effect, an A-plus in an advanced-placement class in Stafford results in a higher GPA than the same grade in the same course in Spotsylvania.
Sarver's proposal also includes a plus or minus system.
If her proposal is approved, it's mathematically impossible for students' current GPAs to decrease, she said. They will either increase or stay the same. Parents will be notified once a change is approved, Barker said.
Sarver's son Wilson chose Old Dominion University. Her younger son, Steven, a ninth-grader, may benefit from the new policy when it's his turn to apply.
But that's not why she spoke out, Sarver said.
"When it came down to a difference of $8,000 per year because of a mathematical recalculation, I'm thinking to myself, 'How many students out there aren't going on to four-year schools because their GPA's not high enough?'" she said.
>>> If Susan Sarver's proposal is approved, would Spotsylvania students appear smarter than they are? No, says Sarver.
Many school divisions use a 10-point grading scale (90–100 for an A, 80–89 for a B, etc.). Spotsylvania uses a seven-point grading scale. Because of this, Sarver says, it will always be harder for Spotsylvania students to earn high GPAs than for students on a 10-point scale.
Spotsylvania students need at least a 94 to earn an A and be credited with a 4.0. Those on a 10-point scale need only a 90 to be credited with a 4.0.


Now some longtime McCainiacs are bitter that they didn't just stick with what made the candidate in the first place. "They set up a national campaign, a la Bush 2000, and that won't work," said Steve Duprey, a New Hampshire backer since that first campaign and former state party chair there. "McCain works best with the 2000 approach."
"After 2000, there were such hard feelings about him from the Bush campaign," said a McCain adviser. "We were able to patch that up here in Washington. But out in the hinterlands, they didn't get over it."


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