I wasn't at all worried about this week's topic to 'Speak Compassionately'. I was actually so unworried that I didn't even think I needed to come up with an 'assignment' to do. I guess because I've always thought of myself as a pretty compassionate person, I just assumed speaking compassionately for a week would not be too much of a problem.
Well, as it turned out, it was a little harder than I expected! In fact, and this is pretty embarrassing to admit, I totally flunked the first night of 'Speak Compassionately' week! I'm not kidding…within about 15 minutes of sitting down to watch the news, I found myself saying some pretty uncompassionate things about how a couple of the reporters looked. I didn't like one of their outfits. And another's teeth looked a little funny to me. I couldn't believe I was being so mean!
I guess I had somehow forgotten the old adage that 'if you don't have anything nice to say, don't say anything at all'. And even though it wasn't like I was saying these things about people I knew, or actually hurting anyone's feelings, the fact that I felt compelled to make such comments at all was pretty troublesome to me.
So I decided that this would be the perfect week to put those wise old words into practice. I would watch what came out of my mouth very closely, and if I caught myself about to say something mean or unkind, I would either not say anything, or try to find something nice to say instead.
Fortunately, I realized pretty quickly that I don't really say unkind things very often, at least about anyone other than television news people! I have no idea why I am so critical of them…maybe I'm just trying to be funny, or somehow make conversation. But I did find, the next time I was watching the news, that by holding my tongue whenever I felt the urge to say something mean, I felt much better. Almost lighter. It was as though the whole energy in the room shifted because I decided to keep those negative words to myself.
And as it turns out, words really do have the power to change the energy in a room, or in anything for that matter.
In the 1990's, Dr. Masuru Emoto performed a series of experiments observing the physical effect of words on the crystalline structure of water. He hired photographers to take pictures of water after being exposed to different types of words, and then froze them so that they would form crystalline structures.
Amazingly, when exposed to prayer, or positive words like 'love' and 'appreciation', the crystals formed beautiful patterns that were pure in color, and in perfect symmetry. On the other hand, when exposed to negative words like 'anger' and 'hate', the crystals were rough, dull, and in a state of disarray.
And because humans are made up of almost 60% water, it just goes to reason that the words we use to communicate with each other could have as big an impact on the people we're talking to (not to mention ourselves!), as they did on the water in Dr. Emoto's experiment.
So whether I'm talking to myself or someone else, it seems to me that it's really important that I try to pay attention to the words I'm choosing to use. Because when I'm more conscious of what I'm saying, the greater chance I'll have of choosing words that are positive, rather than negative. And that can only be good, for me and anyone else I'm talking to, or about. Even television news reporters.
Because if there's one thing we could all use a little more of in this world, it's kindness.
This entry was posted in Compassion, Personal growth, Raising Consciousness and tagged Compassion, consciousness, Dr. Masuro Emoto, spiritual growth. Bookmark the permalink.
I am reminded of a Sufi saying that says we should place three gates in front of our mouth that our words must pass through before we speak. The first gate is asking ourself if what we are about to say is true, the second gate is asking if what we are about to say is necessary and the third gate is asking ourselves if it is kind.
Have a great time in Florida with Sal and your mom.
A loving hello from a childhood family friend. When I read Paul's comment above with "Sal and your Mom" I was immediately swept into a sweet spot of memory and decided I must reach through this cyberspace place to "touch" you with a fond hello.
I've read quite a bit of your writing and have found myself nodding right along with it. Happily, I also really enjoy your writing and the way the words are strung together. That part of the writing process always fascinates me since we all have the same access to words and yet, one person's combination can make all the difference in how we absorb another person's ideas and thoughts.
Oh my gosh Karen…how great to hear from you! And I am so glad to know you are following along! I am guessing Sal sent you the link? I am having fun with my project and so happy when I hear that my experiences are resonating with others. Thank you so much for your kind words! I will definitely pass along your love and hugs to Mom and Sal. We had a great time in Florida celebrating Mom's bday last week and Libby surprised us all and showed up too. I miss your Dad every time I am in Darien and will always look fondly down St. Nicholas road when I drive by it. Lots of love to you!


Did you like the campsite from Thursday? Today's photos come from that same trip.
You might not be able to tell it at first glance. But you are looking at the trail leading away from the site. 😀 Head down the break in the trees and you will quickly be swallowed up by the forest, losing site of the site after about 15 feet. Isn't it wonderful? We love it and can't wait to go back!


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Highest Individual Scores
Simmons out of World T20 due to back injury
Posted By admin on March 5, 2016
Since his last T20I appearance, Lendl Simmons has appeared in T20 leagues around the world © Pakistan Cricket Board
West Indies batsman Lendl Simmons will miss the World Twenty20 after suffering a back injury. Although the West Indies Cricket Board has not made the news public yet, it is understood that no replacement has been named as the selectors, lead by former captain Clive Lloyd, cannot agree on a unanimous choice.
Dwayne Smith, who won the Man of the Match in the final of the Pakistan Super League after helping Islamabad United take the title last month, remains a favourite.
Simmons is the fourth player from the West Indies squad to be ruled out of the World T20 in India. Sunil Narine, Kieron Pollard and Darren Bravo had earlier withdrawn from the team.
Narine, who had been banned from bowling in internationals in November last year due to a suspect action stated cited "insufficient progress in the rehabilitative work" on his action as the reason for his withdrawal. Pollard had not rehabilitated sufficiently from a knee injury he suffered last November, while Bravo pulled out by stating he wanted to focus on Test cricket.
Simmons was a part of the West Indies squad for the 2009, 2012 and 2014 editions of the World T20, although he did not play a game in 2012 when the side won the tournament. He has played 34 T20Is since his debut in June 2007, scoring 761 runs at an average of 25.36 with four fifties, and his last game in the format was against South Africa in January 2015.
Since his last international appearance, Simmons has played in T20 leagues around the world, including the Indian Premier League, the Caribbean Premier League, the Bangladesh Premier League, the Big Bash League and the inaugural Pakistan Super League. In that period, he has scored 1406 runs in 46 matches at an average of 32.69 with 13 fifties.
Source: ESPN Crickinfo
← Bangladesh take on India in Asia Cup Twenty20 final
Bangladesh hopeful of Shakib's fitness for final →
Adam Milne takes it one step at a time but dreams of New Zealand comeback
Action tweak the key to Andrew Tye's new found speed
Nathan Coulter-Nile expected his release from Mumbai Indians: 'They paid overs for me'
Baroda suspend Deepak Hooda for rest of season for 'indiscipline'
Rajasthan Royals trade Robin Uthappa to CSK in all-cash deal
Pawan Kumar on Sydney Sixers vs Brisbane Heat, BBL 2020-21, Fantasy Pick, team predictions
visualize on Lewis Gregory joins Somerset team-mate Tom Banton at Brisbane Heat
Hectorpem on Finch, Head star in Australians' tense win
Avon on Lewis Gregory joins Somerset team-mate Tom Banton at Brisbane Heat
Elizabet Gosling on Kumble's future uncertain after Kohli's feedback
© T20 IPL 2020 | Indian Premier League 2020

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Sabrina has always had a passion for animals. Born and raised in London, she regularly helped out and gained pet handling experience on her uncle's farm in Ireland. Having worked as a PCA with Ben Shorten for three years, Sabrina has developed her nursing skill set and has been a crucial part of the Kensal Green practice.
On Ben's retirement, Sabrina joined the Village Vet team and continues to provide a friendly, experienced service to our clients and their beloved pets. In 2017, she commences her vet nursing training to become fully qualified.
Her beloved pet Python 'Dinamite' gives her a special interest in reptiles. Outside of work, Sabrina loves travelling and learning new languages; she speaks Portuguese and basic Spanish.


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Senate diverges over renewal of internet spying law
By Dustin Volz
WASHINGTON (Reuters) - A Senate panel on Tuesday approved legislation to renew the National Security Agency's internet surveillance program, while other lawmakers pushed a competing measure seeking to end the ability to search for data on Americans without a warrant.
The competing plans were likely to complicate congressional renewal of that law, known as Section 702 of the Foreign Intelligence Surveillance Act, due to expire at the end of the year. The Trump administration supports a permanent renewal of the program without any changes.
The Senate Intelligence Committee voted 12-3 to advance legislation renewing Section 702 until Dec. 31, 2025. The panel voted privately, meaning it did not immediately share the bill text, a common practice for the committee.
Senators Richard Burr and Mark Warner, the panel's Republican and Democratic leaders, said in a statement the bill would protect national security while improving privacy protections for Americans and adding transparency requirements about who can be targeted.
The three no votes came from Democrats, including Senator Ron Wyden, an author of the alternative measure.
U.S. intelligence officials value Section 702, calling it a vital tool for fighting national and cyber security threats and helping protect American allies. It allows U.S. intelligence agencies to eavesdrop on and store vast amounts of digital communications from foreign suspects living outside the United States.
The panel unanimously adopted an amendment from Warner, requiring the Federal Bureau of Investigation to send any queries it makes for U.S. data to the Foreign Intelligence Surveillance Court, according to a source familiar with the vote. The court would have two days to review the query for legality, the source said.
Privacy advocates have blasted the bill, saying it does not contain enough safeguards. Some complained that a version of the bill that had leaked might even expand the U.S. government's surveillance powers.
Fourteen other senators introduced legislation that would require the NSA to obtain a warrant for queries of data on Americans under an internet surveillance program.
The effort, led by Wyden and Republican Rand Paul, would reform other aspects of the warrantless program.
The surveillance program, classified details of which were exposed in 2013 by former NSA contractor Edward Snowden, also incidentally scoops up communications of Americans, including if they communicate with a foreign target living overseas.
Those communications can then be subject to searches without a warrant, including by the FBI. The USA Rights Act authored by Wyden and Paul would end that practice.
The measure was introduced with support from more than 40 civil society groups, including the American Civil Liberties Union and FreedomWorks. A companion bill was introduced in the House of Representatives.
It would renew Section 702 for four years with additional transparency and oversight provisions, such as making it easier for individuals to raise legal challenges against the law and expand the oversight jurisdiction of the Privacy and Civil Liberties Oversight Board, a government watchdog.
Earlier this month, a bipartisan group in the House introduced separate legislation to add privacy protections to Section 702, including partially restricting the FBI's ability to access U.S. data when investigating a crime. Privacy groups criticized that plan as too narrow.
Reporting by Dustin Volz; editing by Leslie Adler and David Gregorio


Sad news to report that Dean Summers was rushed into hospital and was critically ill for several days; his condition has now improved. We wish him well and our thoughts are with him, his family and close friends.
At Birmingham National the judge was Meg Purnell-Carpenter. 1st JD Cullen's Veneze Erik. 1st GD Swift's Veneze Dino. 4th GD Ball's Channbelle Chelsea Boy. 1st LD&RCC Andrews & Rees Nikonis Arthur. 1st OD&DCC&BOB&Gp1&BIS Cullen's Elmo vom Huhnegrab. 1st LB Hensley's Fahra vom Amulree. 1st OB&BCC Cullen's Veneze Chaos.
At the back to back show Allertonshire and North Eastern, alternating judges Ludger Goke and Bernhard Norda. 2nd PD both days Belfield & Mackenzie's Kassieger Cello. 3rd JD&2nd JD Cullen's Veneze Erik. 1st YD both days Kessel & Brimacombe's Conbhairean Ursus at Kesenburg. 3rd LD Bowen & Julian's Lararth Sonny. 4th OD Cullen's Elmo vom Huhnegrab. 3rd MPB&5th MPB Veneze Furby. 4th MPB Cullen's Veneze Frankie. 5th PGB Kessel & Brimacombe's Conbhairean Nathalie at Kesenburg. 1st LB&RCC & 2nd LB Sheffield's Willowdale Kiefa. 4th LB both days Cullen's Veneze Chaos.
The South Western GSD Club is holding a Breed Specific Judging Seminar. The morning session is 'understanding the Breed Standard and the KC Hands on Assessment.' The afternoon session in 2 groups; group 1 Judging the German Shepherd Dog; group 2 (B list judges only restricted to 6 candidates) Judging Assessment and Exam. The KC HOA, judging assessment and exam all on one day and about 25 miles away. What a contrast to taking the Breed Council route. I travelled to Essex twice, once for the exam and another day for an assessment at Essex Champ. Show. I travelled to Wales twice to take the KC HOA and another day for an assessment at Wales Champ. Show. I think a prospective judge's knowledge would be more thoroughly tested the BC way, by spending a day taking the exam, with multiple choice questions in the morning, and questions requiring fuller written answers in the afternoon for the theory part. And for the practical part being assessed at Champ. Shows by well-respected and experienced breed judges. I suspect the fast track candidates who get it all done in a day will be judging Champ. Shows long before we that have chosen to take the long and winding road.
At Southern Counties the judge was Brigid Fahy-Newlands. 1st PD Belfield & Mackenzie's Kassieger Cello. 2nd PGD Swift's Veneze Dino. 3rd LD Hensley, Hiscox & Vessey's Zumander Bacchus. 2nd PGB Julian's Gazjackhof Carlie at Lindmark. 2nd LB Sheffield's Willowdale Kiefa.


European made from European leathers and finished on a low wedge, K-LEVINA is a great summer sandal for all occasions.
Halo from Colchester is right, I too have been buying these woderful sandals for years! Comfortable and smart.
These were definitely not of the same quality. The Leather felt much harder on the foot. They were definitely not exactly the same shape either - very fine detail differences, but apparent none the less. I returned them.
Having contacted customer services I was advised that sandals from the original supply source will be in stock during May.
I look forward the opportunity to purchase sandals as good as my first pair.
Good in every respect- comfort, fit, colour. Have used this style for years and always good, except for one period when a different finish was used on the lining if the sole and this caused the black finish to stain the soles of the feet. This appears to gave been rectified over the last year. Will be ordering the same shoe in blue as soon as it is available; it would be good if Jones could let me know as soon as these are available.
Very prompt and efficient service. I have bought this sandal for a number of years from a shop and this is the first time on line. Extremely satisfied. Would like other available colours as before.


Yellow Matter Entertainment is proud to announce that it is now shipping Steven Slate Drums 1.5 and Steven Slate Drums 1.5 Elite Mixer Edition.
NEW DEMOS!! These are perhaps my favorite demos thus far, with more coming each day.


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The hexagon is one of the basic shapes in crochet and just like the classic granny square, once mastered, it will allow you to create wonderful items, from blankets and home decor to bags and afghans.
There are many variations and patterns, but today I will show you how to crochet a basic solid hexagon.
I like this pattern because it's easy and quick and I can crochet it while watching a movie without having to count stitches and rows.
I used it to make the honeycomb baby blanket and I am sure i will use it again as it's such a versatile pattern.
Ch 3 (counts as a dc), dc into the ring,ch 2, (2 dc, ch2) 5 times into the ring, sl st.
Do not turn, pull the yarn tail to close the magic ring and weave in the end.
You should have 6 clusters divided by chains of 2.
Ch 3 (counts as a dc), dc into the next stitch.
You have now reached the chain of two from previous round in which you will create the corner: dc into the chain, chain 2, dc again into the chain.
Work 1 dc into next 2 stitches then create the corner ( dc, ch 2, dc) into the chain of 2 from previous round.
Continue until the end of the row, closing with a slip stitch. Do not turn.
You should have 6 clusters of 4 dc each separated by chains of 2.
Ch 3 (count as a dc), 2 dc in the next 2 stitches, once you reach the chain of 2 from previous round, create the corner (dc, ch 2, dc).
Dc in the next 4 stitches, make a corner into the previous corner (dc, ch 2, dc), 4 dc in the next 4 stitches, another corner, continue until the end of the row. You should end with a dc into the last stitch and close with a sl st.
You should have 6 clusters of 6 dc each separated by chain of 2.
Chain 3 (counts as a dc), 3 dc in the next stitches, form the corner into the previous corner (dc, ch 2, dc). continue with dc in each stitch followed by the corner stitches until the end. Sl st and close weaving in the end.
You should end up with 6 clusters of 8 dc separated by chain of 2.


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[Photo News] How was holiday way home like for Koreans in the past?
Seollal is one of Korea's two main holidays along with Chuseok. Both kick off a mass movement of people going to visit their hometowns.
The race to return home involves people competing for bus, train and air tickets online, as well as at ticket booths.
Just a few years ago, however, people lined up at train stations and bus terminals, with thousands queueing up for tickets. Some even camped outside, while thousands more braved the expressways in their cars.
With the network of expressways much more limited back then compared to now, traveling to the provinces involved grueling hours, sometimes taking 20 hours from Seoul to Busan and 16 hours to Gwangju. These routes now take less than six hours on average.
The following old photos show how Koreans managed their trip to home for traditional holidays.
Ticket boxes are crowded with citizens seeking Seollal train tickets in Seoul Station on Jan. 8 in 2014. (Herald DB)
The highway is clogged with cars heading out of Seoul at the Seoul Tollgate on Seollal eve in 2013. (Herald DB)
Cars are clogged in a bottleneck at the Pyeongsa Service Area in Gyeongbu Expressway during the Chuseok holidays in 1994. (Bae Young-chan)
Seollal homegoers in Yeouido are seen waiting in line for bus to other provinces in 1981. (Herald DB)
Homegoers gather at Seoul Station on Seollal holiday in 1977. (National Archives of Korea)
Citizens wait in lines at Seoul Station to buy a Seollal train ticket to their hometowns in the 1970s. (Korail)
Homegoers are seen in a packed Chuseok holiday train in 1968. Some who could not get a seat are standing or lying in overhead compartments. (Seoul Museum of History)
Lost articles are seen at a stairway to platform in Seoul Station after 31 people died from a crowd crush on Jan. 26, two days before Seollal, in 1960. (news report)


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Austin FC Signs Forward Alfonso Ocampo-Chávez
Austin FC announced today the Club reached an agreement to sign forward Alfonso Ocampo-Chávez. Ocampo-Chávez joins the Verde & Black on a one-year guaranteed deal through the end of the 2023 season, with Club options for the 2024 and 2025 seasons.
"Alfonso is an exciting young attacking player who proved his ability at the youth international level," said Austin FC Sporting Director Claudio Reyna. "We wish him a warm welcome to Austin, and hope to help him continue to develop and showcase his talent."
Ocampo-Chávez, a 20-year-old California native, spent the 2022 season with Seattle Sounders FC's MLS NEXT Pro affiliate Tacoma Defiance. He made 23 appearances for Tacoma in 2022, scoring nine (9) goals and tallying four (4) assists. He spent the latter part of the 2021 season on loan from Seattle with FC Pinzgau Saalfelden of the Austrian Regionalliga Salzburg, where he made 13 appearances and scored one (1) goal.
At the youth international level, Ocampo-Chávez represented the United States at the U-14, U-15, and U-17 levels. He competed for the U.S. at the FIFA U-17 World Cup in 2019, and was a key part of the team's qualifying campaign as one (1) of two (2) players to appear in all seven (7) of the team's matches at the 2019 Concacaf U-17 Championship.
Austin originally claimed Ocampo-Chávez off the MLS End-of-Year Waivers list on November 17, meaning Ocampo-Chávez retains his MLS Homegrown designation. Ocampo-Chávez will also occupy a spot on the Club's Supplemental roster.
Transaction: Austin FC signs forward Alfonso Ocampo-Chávez to a one-year guaranteed contract through the end of the 2023 season, with Club options for the 2024 and 2025 seasons.
Player Details:
Name: Alfonso Ocampo-Chávez
Pronunciation: ahl-FOHN-soh oh-CAHM-poh SHA-ves
Born: Merced, California
Last club: Seattle Sounders FC
How acquired: Claimed off the MLS End-of-Year Waivers list on November 17, signed on November 29.
Roster Designations: Homegrown, Supplemental


text:
By Laxman Kafle
Kathmandu, Dec. 19: The Nepal Rastra Bank (NRB) has asked Nepali citizens not to panic over a possible shortage of Indian currency (IC) as the central bank has sufficient reserve to meet the IC demand for the next six months in the country.
The reassuring remark of the NRB has come amidst reports that India has unofficially stated it is not in a position to supply Indian banknotes to Nepal.
According to Narayan Poudel, spokesperson at the NRB, it has around 4.5 billion Indian rupees in stock - sufficient to provide to the needy people under the new provision of the NRB.
After India banned the transaction of IC 500 and 1,000 notes, the NRB reduced the limit on Indian banknotes to be provided from its counters to individuals seeking exchange services while going to India.
"Currently, we are providing IC to people in need under three categories, considering the people's necessity and the currency stock. We provide IC 2,000 to people seeking money exchange in general. Besides, the NRB provides IC 10,000 to those who submit the required documents citing the need for Indian banknotes," he said.
Earlier, the NRB used to provide IC 5,000 to individuals seeking exchange, he said and added that the NRB slashed the exchange limit due to the decline in the stock following the ban on high denomination Indian currency notes.
As per the new provision, the NRB provides up to IC 25,000 to those who need Indian banknotes for treatment on submission of medical papers.
To control the misuse of Indian currency notes, the NRB is offering exchange facilities to the people on the basis of citizenship cards and the required documents, Poudel said.
"During talks with NRB officials, officials of the Reserve Bank of India (RBI) have informally told us that they would supply banknotes in due course of time. It means that the RBI is not in a position to supply Indian banknotes immediately," Poudel said while talking with The Rising Nepal on Sunday.
He said that the NRB had doubts that the RBI would supply the banknotes to the NRB by the end of January as there was high pressure to exchange the demonetised notes in India itself.
He, however, said that an agreement had been reached between the NRB and RBI to supply 6 billion Indian banknotes to the NRB annually.
The NRB has received only IC 1.2 billion lately, he said.
He said that there would be no problem if the RBI did not provide Indian banknotes for a few months owing to IC scarcity in India because the NRB could manage from the present reserve.
Spokesperson Poudel urged the people to exchange Indian currency notes if it was urgent, and not hoard them at home.
Talking about the progress in the exchange facilities for the 500 and 1,000 Indian banknotes demonetised recently, he said that the NRB was holding discussion with representatives of the RBI to sort out the Indian currencies crisis at the earliest.
"We have not received any decision from the RBI over the issue till date. But we expect the RBI to take a positive decision in this regard," he said.


Insanity Workout Review - What Is The Insanity Workout and Should You Try It?
What Is Insanity All About?
The program costs $144, and includes two phases over a roughly two-month period. In the first phase, you get five workouts, including sequences focused on lower-body plyometrics (explosive movements), upper-body strength, cardio and abs, plus a recovery workout.
Insanity results: What can you expect?
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text:
Prize for ice cream with veggies
Do you remember an article we published on ice cream with veggies? Not too enthusiastic about it? Well, maybe this product will suit you better. There's a new ice with veggies available in the US and the product just won a price.
Peekaboo Organics' ice cream featuring hidden vegetables was named the grand prize winner today of the Real California Milk Snackcelerator dairy snack innovation competition created by the California Milk Advisory Board (CMAB) and VentureFuel.
The competition will provide Peekaboo with $200,000 in resources and funding to bring a snack-sized version of their innovative products to market and expand the overall market for their products. This is the second event of its kind held by the California dairy industry designed to inspire ideas integrating the values of fluid milk and dairy ingredients into snacks to meet growing consumer demand and providing resources to help bring them to market.
As presented by Founder and CEO, Jessica Levison, Peekaboo's mission is to solve the parental dilemma of getting children to consume more vegetables by putting them into a format they can't get enough of – ice cream. As the owner of a popular ice cream scoop shop in Miami, she turned her frozen dairy knowhow into flavorful super-premium ice creams hiding a nutritious, vegetable secret. The products are currently available in pints at select retailers but Levison is looking forward to introducing the single serve snack line and expanding distribution with Real California Milk at the core.
"Like many of our fellow startups, we confronted what seemed like insurmountable challenges this year. This win validates our innovation, our mission and our team's hard work and resilience," said Levison. "We learned so much from this Snackcelerator experience and cannot wait to launch Peekaboo minis into the market – made with Real California Milk, of course! Much gratitude to everyone involved."
Peekaboo is one of four innovators selected from a field of 16 sweet and savory semi-finalist competitors to present to a field of judges from the CPG, retail, venture capital and food science world representing such companies as Clorox, Mondelez, Whole Foods, CircleUp and more in a live pitch (and sample) event tapping into future technologies and emerging startups like OS Studios, Perksy and more to transform what was an in-person sampling experience into a state-of-the-art virtual event. Alongside three other finalists, Peekaboo was judged on uniqueness or innovation, taste, the ability to scale and the founder team. Yummy Industries came in second place with their Cheese-Bits premium snacking cheese packs and will receive $100,000 in support for growth and distribution.
"Being selected as a semi-finalist really was significant for us because we were associated with a tremendous caliber of founders and snack concepts," said Yummy Co-Founder Stefen Choy. "Throughout this process, the feedback and mentorship sessions have really help move our business forward significantly. We have changed our strategic plan and even added a broker based on this competition. The bonus of continued support as a runner up is icing on the cake."
The Real California Milk Snackcelerator competition was designed to inspire innovation and investment in dairy based snack products, packaging and capacity within California by connecting manufacturers, producers, investors, ideas and entrepreneurs for high quality, sustainable products. The competition received 76 entries, which inspired expansion from the original eight finalists to 16 representing both sweet and savory snacking startups in the Sweet & Savory 16 virtual semi-finals the week of November 9th.
"This was a tough decision, all four of these finalists presented products that fit the needs of our current consumer and each with their own unique take on snacking. Peekaboo just had a slight edge because of its unique concept marrying two things that normally don't go together like ice cream and vegetables and then executing the product perfectly in terms of creative flavor combinations," said John Talbot, CEO of the CMAB. "Not only are we excited to grow the dairy snack portfolio offered with Real California Milk by helping to expand distribution for Peekaboo ice creams, we look forward to continuing our relationship with all of our finalists and semi-finalists to see what they are able to achieve."
"Innovation is the ultimate growth accelerator for any business or category and this competition continues to deliver a flywheel of new ideas, new products and new ways to think about dairy from a consumer perspective," said Fred Schonenberg, CEO and Founder of VentureFuel. "When my family begged me for more samples between the semi-finals and finals, I knew we had chosen four winners all, of which can make an impact on-shelf. Peekaboo's cotton-candy with beets flavor dominated the pre-judge chatter as one of the most delicious and unique desserts the judges had ever tasted, meanwhile Cheese-Bits was the product everyone was snacking on behind the scenes. They are making premium cheese easy to snack on and introducing new styles and flavors that will stand out."
All sixteen semi-finalists received $10,000 worth of support each, to develop an edible prototype, while receiving a suite of resources including graphic design, lab or kitchen time and elite mentorship from global marketing, packaging, and distribution experts from companies including Diageo, Nestle, Mondelez, My/Mo Mochi Ice Cream, Bristol Farms, Rosa Brothers, Food Network/Hearst Magazines, Pod Foods, Latitude Ventures, The Marketing Greenhouse, The Creative Pack, Whole Brain Consulting and Power Brands. Semi-finalist rounds were judged by a prestigious group of snack experts from Conagra, SnackRiot, Hershey's, Bristol Farms, General Mills, Unilever, Mondelez, Rosa Brothers, RangeMe, Acreage Holdings, Better Food Ventures and Trail Mix Ventures. The total value of competition prizing is over $800,000.
California, known for innovation, has a reputation for quality dairy products. As the number one producer of fluid milk in the nation, California also leads the nation in sustainable dairy farming practices. More than 1,200 family dairy farms produce the California milk found in fluid milk, cheese, butter, yogurt, ice cream and other dairy products identified by the Real California Milk seal.
Peekaboo Organics
www.eatpeekaboo.com


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text:
NBC/Universal Breaks the Seal, Set to Shatter the Theatrical Window
By Dustin Rowles | Film | March 16, 2020 |
We all assumed that this moment would arrive soon enough, and now it has.
While the release of a number of films have been delayed over COVID-19 concerns, it appears that NBC/Universal is going to take a different approach, at least with many of its titles. It's going to make them available VOD, beginning on Friday. Henceforth, Universal will release movies at home at the same time as their theatrical releases (or supposed theatrical releases) throughout the duration of the Coronavirus pandemic.
Those titles include The Hunt and The Invisible Man, as well as Focus Features' Emma, which will all be made available on Friday. Trolls World Tour will also be made available on April 10th, the day it was scheduled for theatrical release.
The catch? VOD prices will be $19.99. This is great for a movie like Trolls World Tour, which would have cost parents $25-$40 for tickets for the family anyway, depending on the size, and we get to watch at home, where we don't have to pay for expensive concessions (expensive concessions that provide jobs to a lot of people).
Notwithstanding the expense — will I pay $20 to watch The Hunt by myself? I don't know! Probably not! But I would have for The Invisible Man — I think this is a great idea, if only because it allows us to maintain as much a sense of normalcy as possible. If NBC/Universal can continue to release films online, and continue to make money doing so, then they will continue to make movies, which keeps actors and crew members employed, and they will continue to advertise, which keeps media organizations employed.
The question is whether things will return to normal after the pandemic has passed? Is this the end of the theatrical industry, beyond the tentpole movies from big studios that can eventually buy their own theater chains? Will audiences begin to expect to be able to watch movies at home? It's too early to say. In the meantime, if you live with friends, check out The Invisible Man or Emma! And if you have a family, well, if I'm still stuck inside, we'll be watching Trolls World Tour from the couch on April 10th!
← A Post Script on Evan Olson and that Outstanding 'Reply/All' Episode, 'The Case of the Missing Hit' | This Year's Met Gala Indefinitely Postponed Due to Coronavirus →
Header Image Source: Universal


text:
\section{INTRODUCTION}
In a recent  paper  \cite{1} an alternative  global description of the  Schwarzschild black hole  has been proposed. The motivation behind such an endeavor  was, on the one hand    the fact that the space--time within the horizon, in the classical picture,  is necessarily non--static or, in other words, that  any transformation that maintains the static form of the Schwarzschild metric (in the whole space--time) is unable to remove the coordinate singularity appearing on the horizon in the line element  \cite{rosen}.
Indeed, as  is well known,  no static observers can be defined inside the horizon (see \cite{Rin,Caroll} for a discussion on this point). This conclusion becomes intelligible if we recall  that the Schwarzschild horizon is also a Killing horizon, implying that the time--like Killing vector existing outside the horizon, becomes space--like inside it.

On the other hand, based on the physically reasonable point of view that  any equilibrium final state of a physical process should  be static, it would be desirable  to have  a static solution over the whole space--time.

Based on the arguments above, the following model was proposed in Ref. \cite{1}.

Outside the horizon ($R >2M$) one has the usual Schwarzschild  line element corresponding to the spherically symmetric vacuum solution to the Einstein  equations, which in polar coordinate reads (with signature $+2$)
\begin{eqnarray}
ds^2&=&-\left(1-\frac{2M}{R}\right)dt^2+\frac{dR^2}{\left(1-\frac{2M}{R}\right)}+R^2d\Omega^2, \nonumber \\ d\Omega^2&=&d\theta^2+\sin^2 \theta d\phi^2.
\label{w2}
\end{eqnarray}

This metric is static and spherically symmetric, meaning that it admits four Killing vectors:
\begin{eqnarray}
\mathbf{\xi}_{(\mathbf{0})} = \partial _{\mathbf{t}}, \quad {\bf \xi_{(2)}}=-\cos \phi \partial_{\theta}+\cot\theta \sin\phi \partial_{\phi},\nonumber \\
{\bf \xi_{(1)}}=\partial_{\phi}, \quad {\bf \xi_{(3)}}=\sin \phi \partial_{\theta}+\cot\theta \cos\phi \partial_{\phi}.
\label{2cmh}
\end{eqnarray}

The solution proposed for $R <2 M$  (with signature $(+ - - -)$) is
\begin{eqnarray}
ds^2&=&\left(\frac{2M}{R}-1\right)dt^2-\frac{dR^2}{\left(\frac{2M}{R}-1\right)}-R^2d\Omega^2, \nonumber \\ d\Omega^2&=&d\theta^2+\sinh^2 \theta d\phi^2.
\label{w3}
\end{eqnarray}

This is a static solution, meaning that it admits the time--like Killing vector  $\mathbf{\xi }_{(\mathbf{0})}$, however  unlike (\ref{w2})  it is not spherically symmetric, but hyperbolically symmetric, meaning that it admits  the three Killing vectors
\begin{eqnarray}
{\bf K_{(2)}}=-\cos \phi \partial_{\theta}+\coth\theta \sin\phi \partial_{\phi},\nonumber \\
{\bf K_{(1)}}=\partial_{\phi}, \quad {\bf K_{(3)}}=\sin \phi \partial_{\theta}+\coth\theta \cos\phi \partial_{\phi}.
\label{2cmhy}
\end{eqnarray}

Thus  the situation may be summarized as follows: if one wishes to keep sphericity within the horizon, one should abandon staticity, if one wishes to keep staticity within the horizon, one should abandon sphericity.

The classical picture of the black hole entails sphericity within the horizon, instead in \cite{1} we have proceeded differently and have assumed staticity within the horizon.

The three Killing vectors above define the hyperbolical symmetry.
Space--times endowed with hyperbolical symmetry have  previously  been the subject of research  in different contexts (see \cite{Ha}--\cite{pak3} and references therein).

In \cite{2nc} a general study of  geodesics in the spacetime described by (\ref{w3}) was presented, leading to some interesting conclusions about the behavior of  a test   particle in this new picture of the Schwarzschild  black hole, namely:
\begin{itemize}
\item the gravitational force inside the region $R<2M$ is repulsive.
\item test particles cannot reach the center.
\item test particles can cross the horizon outward, but only along the $\theta=0$ axis.
\end{itemize}

Motivated by these intriguing results we embarked into a general study of fluid distributions endowed with hyperbolical symmetry \cite{st1,hd,hd2}.

It is the purpose of this paper to present the main results concerning dynamic and dissipative fluids endowed with hyperbolical symmetry. This   includes the general equations  governing the behavior of such fluids as well as  a selection of some exact analytical solutions.  We shall   discuss about  prospective applications of these results to the study of some astrophysical and cosmological problems. We shall not consider here the static case analyzed in \cite{st1}.

Finally, it is worth stressing the fact that although the main  motivation to undertake the study of fluids endowed with hyperbolical symmetry was the intriguing properties of  the black hole model briefly described above, the  results we are going to exhibit in the sections below are completely  independent on such a model.

\section{Basic equations and variables}
We consider  hyperbolically symmetric  distributions  of evolving
fluids, which may be  bounded, or not, from outside by a surface  $\Sigma^{e}$. On the other hand, as we shall see below, fluids endowed with hyperbolical symmetry cannot  fill the central region, therefore they should be  bounded from inside by a  surface   $\Sigma^{i}$.  The fluid is assumed to be locally anisotropic (principal stresses unequal) and undergoing dissipation in the form of heat flow (diffusion approximation).

Choosing comoving coordinates, the
interior metric, admitting the three Killing vectors (\ref{2cmhy}), may  be written as
\begin{equation}
ds^2=-A^2dt^2+B^2dr^2+R^2(d\theta^2+\sinh^2\theta d\phi^2),
\label{1}
\end{equation}
where  $A$, $B$ and $R$ are assumed
positive, and  due to the hyperbolical symmetry  are functions of $t$ and $r$. We number the coordinates $x^0=t$, $x^1=r$, $x^2=\theta$
and $x^3=\phi$.

The energy momentum tensor $T_{\alpha\beta}$ of the fluid distribution
may be written in its canonical form as
\begin{equation}
T_{\alpha \beta} = {\mu} V_\alpha V_\beta + P h_{\alpha \beta} + \Pi_{\alpha \beta} +
q \left(V_\alpha \chi_\beta + \chi_\alpha V_\beta\right) \label{Tab}
\end{equation}
with
$$ P=\frac{P_{r}+2P_{\bot}}{3}, \qquad h_{\alpha \beta}=g_{\alpha \beta}+V_\alpha V_\beta,$$

$$\Pi_{\alpha \beta}=\Pi\left(\chi_\alpha \chi_\beta - \frac{1}{3} h_{\alpha \beta}\right), \qquad
\Pi=P_{r}-P_{\bot}.$$

$$V^{\alpha}V_{\alpha}=-1, \;\; V^{\alpha}q_{\alpha}=0, \;\; \chi^{\alpha}\chi_{\alpha}=1,\;\;
\chi^{\alpha}V_{\alpha}=0$$.

$$
V^{\alpha}=A^{-1}\delta_0^{\alpha}, \;\;
q^{\alpha}=qB^{-1}\delta^{\alpha}_1, \;\;
\chi^{\alpha}=B^{-1}\delta^{\alpha}_1,
$$

where $\mu$,  $P_r$,
$P_{\perp}$,  $q^{\alpha}$, $V^{\alpha}$ have the usual meaning,
and $\chi^{\alpha}$ is unit four--vector along the radial direction.  It is worth noticing that  bulk or shear viscosity could be introduced by redefining
 the radial and tangential pressures.  In addition, dissipation in the free streaming approximation can be absorbed in $\mu, P_r$ and $q$.

 Since the Lie derivative  $\mathcal{L}$ and the partial derivative commute, then
\begin{equation}
\mathcal{L}_K (R_{\alpha \beta}-\frac{1}{2}g_{\alpha \beta}{\cal R})=8\pi \mathcal{L}_KT_{\alpha \beta}=0,
\label{ccm1}
\end{equation}
implying that all physical variables only depend on $t$ and $r$.

Finally, it is worth mentioning that because of the hyperbolical symmetry there are only two possible unequal principal stresses  ($P_r, P_\bot$).
\subsection{Einstein equations and conservation laws}

The Einstein equations for (\ref{1}) and (\ref{Tab}) are
\begin{eqnarray}
  8\pi \mu&=& -\frac{1}{R^2}-\frac{1}{B^2}\left
  [-\frac{2B^\prime}{B}\frac{R^\prime}{R}+\left(\frac{R^\prime}{R}\right)^2+\frac{2R^{\prime\prime}}{R} \right ]\nonumber\\
   &+& \frac{1}{A^2}\left ( \frac{2\dot B}{B}\frac{\dot R}{R}+\frac{\dot R^2}{R^2} \right ),\label{EE00}
\end{eqnarray}

\begin{equation}\label{EE01}
 4\pi q=-\frac{1}{AB}\left ( \frac{R^\prime }{R}\frac{\dot{B}}{B}+\frac{A^\prime}{A}\frac{\dot
 R}{R}-\frac{\dot{R}^\prime}{R}\right),
\end{equation}

\begin{eqnarray}
8\pi P_r &=& \frac{1}{R^2}+\frac{1}{B^2}\left [\frac{2A^\prime}{A}\frac{R^\prime}{R}+\left(\frac{R^\prime}{R}\right)^2 \right
]\nonumber \\
   &+& \frac{1}{A^2}\left ( \frac{2\dot{A}}{A}\frac{\dot{R}}{R}-\frac{\dot R^2}{R^2}-\frac{2\ddot{R}}{R}\right
   ),\label{EE11}
\end{eqnarray}

\begin{eqnarray}
 8\pi P_\bot&=& \frac{1}{B^2}\left (
 -\frac{A^\prime}{A}\frac{B^\prime}{B}+\frac{A^\prime}{A}\frac{R^\prime}{R}-\frac{B^\prime}{B}\frac{R^\prime}{R}
 +\frac{A^{\prime\prime}}{A}+\frac{R^{\prime\prime}}{R} \right ) \nonumber\\
   &+& \frac{1}{A^2}\left (
   \frac{\dot{A}}{A}\frac{\dot{B}}{B}+\frac{\dot{A}}{A}\frac{\dot{R}}{R}-\frac{\dot{B}}{B}\frac{\dot{R}}{R}
   -\frac{\ddot B}{B}-\frac{\ddot R}{R}  \right ),\label{EE22}
\end{eqnarray}
where dots and primes denote derivative with respect to $t$ and $r$ respectively.
It is worth noticing the difference between these equations and the corresponding to the spherically symmetric case (see for example Eqs.(7)--(10) in \cite{epjc}).

The conservation laws $T^{\mu \nu}_{;\mu}=0$, as in the spherically symmetric case,  have only two independent components, which are displayed in the Appendix.
\subsection{Kinematical variables}

The four--acceleration $a_{\alpha}$ and the expansion $\Theta$ of the fluid are
given by
\begin{equation}
a_{\alpha}=V_{\alpha ;\beta}V^{\beta}, \;\;
\Theta={V^{\alpha}}_{;\alpha}. \label{4b}
\end{equation}
From  which we obtain  for the  four--acceleration  and its scalar $a$,
\begin{equation}
a_1=\frac{A^{\prime}}{A}, \;\; a=\frac{A^{\prime}}{AB}\Rightarrow a^\alpha=a\chi^\alpha, \label{5c}
\end{equation}
and for the expansion
\begin{equation}
\Theta=\frac{1}{A}\left(\frac{\dot{B}}{B}+2\frac{\dot{R}}{R}\right).
\label{5c1}
\end{equation}

The   shear tensor $\sigma_{\alpha\beta}$ is defined by (the vorticity vanishes identically)
\begin{equation}
\sigma_{\alpha\beta}=V_{(\alpha
;\beta)}+a_{(\alpha}V_{\beta)}-\frac{1}{3}\Theta h_{\alpha\beta},
\label{4a}
\end{equation}
its non zero components are
\begin{equation}
\sigma_{11}=\frac{2}{3}B^2\sigma, \;\;
\sigma_{22}=\frac{\sigma_{33}}{\sinh^2\theta}=-\frac{1}{3}R^2\sigma,
 \label{5a}
\end{equation}
and its scalar
\begin{equation}
\frac{3}{2}\sigma^{\alpha\beta}\sigma_{\alpha\beta}=\sigma^2,
\label{5b}
\end{equation}
reads
\begin{equation}
\sigma=\frac{1}{A}\left(\frac{\dot{B}}{B}-\frac{\dot{R}}{R}\right).\label{5b1}
\end{equation}

All the expressions above are the same, in terms of the metric functions, as in the spherically symmetric case.
\subsection{The Weyl tensor}
The magnetic part of the Weyl tensor for  our metric (\ref{1}) vanishes, whereas  its electric part may be written as

\begin{equation}\label{weyl1}
  E_{\alpha\beta}=\mathcal{E}\left(\chi_\alpha \chi_\beta-\frac{1}{3}h_{\alpha\beta}\right),
\end{equation}
with
\begin{eqnarray}
\mathcal{E}&=&\frac{1}{2B^2}\left[
-\frac{A^\prime}{A}\frac{R^\prime}{R}-\frac{A^\prime}{A}\frac{B^\prime}{B}+\frac{R^\prime}{R}\frac{B^\prime}{B}+
\left(\frac{R^\prime}{R}\right)^2+\frac{A^{\prime \prime}}{A}-\frac{R^{\prime \prime}}{R}\right]\nonumber \\
&+& \frac{1}{2A^2}\left[ \frac{\dot A}{A}\frac{\dot B}{B}-\frac{\dot A}{A}\frac{\dot R}{R}+\frac{\dot
R}{R}\frac{\dot B}{B}-
\left(\frac{\dot R}{R}\right)^2-\frac{\ddot B}{B}+\frac{\ddot R}{R}\right]+\frac{1}{2R^2}.\label{escE}
\end{eqnarray}

\subsection{The mass function}
Following \cite{Misner,Cahill}  we may define the mass function  as
\begin{equation}
m(r,t)=-\frac{R}{2}R^{3}_{232}=\frac{R}{2}\left[\left(\frac{R^{\prime}}{B}\right)^2-\left(\frac{\dot
R}{A}\right)^2+1\right],
 \label{fmasa}
\end{equation}
where the Riemann tensor component $R^{3}_{232}$ is now calculated for (\ref{1}).

Introducing the proper time derivative $D_T$, and the proper radial derivative $D_R$  by
\begin{equation}
D_T=\frac{1}{A}\frac{\partial}{\partial t}, \label{16}
\end{equation}
and
 \begin{equation}
D_R=\frac{1}{R^{\prime}}\frac{\partial}{\partial r},\label{23a}
\end{equation}
we may  define the ``areal'' velocity $U$ as
\begin{equation}
U=D_TR, \label{19}
\end{equation}
which must be smaller than $1$ (in relativistic units).

Then, since $U <1$, it follows at once from (\ref{fmasa}) that $m$ is a positive defined quantity.
Also, (\ref{fmasa}) can be rewritten as
\begin{equation}
E \equiv \frac{R^{\prime}}{B}=\left(\frac{2m}{R}+U^2-1\right)^{1/2}.
\label{20x}
\end{equation}

Using (\ref{fmasa})  with (\ref{16}), (\ref{23a}), and the field equations, we obtain

\begin{eqnarray}
D_Tm=4\pi\left(P_rU+ qE\right)R^2,
\label{22Dt}
\end{eqnarray}
and
\begin{eqnarray}
D_Rm=-4\pi\left(\mu+q\frac{U}{E}\right)R^2,
\label{27Dr}
\end{eqnarray}
producing
\begin{equation}
m=-4\pi\int^{r}_{0}\left( \mu +q\frac{U}{E}\right)R^2R^\prime dr, \label{27intcopy}
\end{equation}
satisfying the regular condition  $m(t,0)=0$.

Integrating (\ref{27intcopy}) we find
\begin{equation}
\frac{3m}{R^3} = -4\pi {\mu} +\frac{4\pi}{R^3} \int^r_0{R^3\left(D_R{ \mu}-3 q \frac{U}{RE}\right) R^\prime dr}.
\label{3m/R3}
\end{equation}

From (\ref{27intcopy})  two important general properties of hyperbolically symmetric fluids, follow.

On the one hand, $\mu$ is necessarily negative. This follows from the (physically meaningful) condition that  the fluid is not very far from thermal equilibrium, implying $q<<\vert\mu\vert$, and   furthermore $R^\prime>0$ to avoid shell crossing, $m>0$ and $E$ is a regular function within the fluid.

 On the other hand, we see from  (\ref{27intcopy}) that whenever the energy density  is regular, then $m\sim r^3$ as $r$ tends to zero. However, in this same limit $U\sim 0$, and $R\sim r$ implying because of (\ref{20x}) that the central region  cannot be filled with our fluid distribution.  In other words, there is a central region of finite dimensions, which cannot be filled with a fluid endowed with hyperbolical symmetry. At this point many different scenarios may be envisaged. We shall assume here that the center is surrounded by a vacuum cavity described by a Minkowski space--time. It is worth emphasizing that such a choice does not affect the general properties of the fluids endowed with hyperbolic symmetry.

 The two above mentioned features of the fluid appear also in the static case \cite{st1}.

Before concluding this section it is worth  discussing with some detail  on equation (\ref{pprimaU2}), and compare it with the corresponding equation for the spherically symmetric case (see eq.(C6) in \cite{epjc}).

Equation (\ref{pprimaU2}) is particularly appealing because it  has  the ``Newtonian'' form
$Force = Mass \ density \times \ Acceleration$.  Let us now analyze its  different terms. The first term on the right represents the gravitational interaction,  it is the product of the passive gravitational mass density (p.g.m.d) ($\mu+P_r$), which due to the fact that the energy density is negative, would be negative,  and the active gravitational mass (a.g.m) ($4\pi P_r R^3-m$)  which would also be negative for most equations of state. Thus  the gravitational term has the same sign as in the spherically symmetric case. However its effect  is the inverse of this latter case. Indeed, since the p.g.m.d is negative so is the inertial mass density, implying that  the gravitational term tends to increase $D_TU$, i.e. it acts as a repulsive force, instead of  an attractive one as in (C6) of \cite{epjc}. Also,  we see that a negative pressure gradient which produces a force pointing outward, would tend to push any fluid element inwardly.

\section{THE TRANSPORT EQUATION}
Since we are considering dissipative systems we have to adopt a transport equation. In order to ensure causality we shall resort to   the  transport equation obtained form
the  M\"{u}ller--Israel--Stewart  theory \cite{Is,Is2,Is3}.

Then, the  corresponding  transport equation for the heat flux reads

\begin{equation}
\tau
h^{\alpha\beta}V^{\gamma}q_{\beta;\gamma}+q^{\alpha}=-\kappa h^{\alpha\beta}
(T_{,\beta}+Ta_{\beta}) -\frac 12\kappa T^2\left( \frac{\tau
V^\beta }{\kappa T^2}\right) _{;\beta }q^\alpha ,  \label{21t}
\end{equation}
where $\kappa $  denotes the thermal conductivity, and  $T$ and
$\tau$ denote temperature and relaxation time respectively.

There is only one non-vanishing independent component of  Equation (\ref{21t}),  which may be written as
\begin{equation}\label{ecTra}
  \tau D_Tq=-q-\frac{\kappa}{AB}(AT)^\prime-\frac{1}{2}\tau \Theta q-\frac{1}{2}\kappa T^2D_T\left(\frac{\tau}{\kappa T^2}\right)q.
\end{equation}
In the case $\tau=0$ we recover the Eckart--Landau equation \cite{17T}.

Under some circumstances it is possible to adopt  the so called ``truncated'' version where the last term in (\ref{21t}) is neglected \cite{PAN},
\begin{equation}
\tau
h^{\alpha\beta}V^{\gamma}q_{\beta;\gamma}+q^{\alpha}=-\kappa h^{\alpha\beta}
(T_{,\beta}+Ta_{\beta}) \label{V1},
\end{equation}
and whose    only non--vanishing independent component becomes
\begin{equation}
\tau \dot q+qA=-\frac{\kappa}{B}(TA)^{\prime}. \label{V2}
\end{equation}

Two important thermodynamical properties of  hyperbolically symmetric fluids can be inferred from (\ref{ecTra}).

Let us first consider the condition of thermal equilibrium.

As it was pointed out by Tolman many years ago \cite{Tol}, the condition of thermal equilibrium in the
presence of a gravitational field must change with respect to its form in the absence
of gravity since  thermal energy tends to displace to regions of lower gravitational
potential  (independently on any temperature gradient).  Thus, a temperature gradient is necessary in thermal equilibrium in order
to prevent the flow of heat from regions of higher to lower gravitational potential.

Indeed, as it follows at once from (\ref{ecTra}), thermal equilibrium implies
\begin{equation}
\left(TA\right)^\prime=0 \Rightarrow T^\prime=-\frac{T}{A}A^\prime=-TaB.
\label{t1}
\end{equation}
However in our case $a<0$, (the four--acceleration  is now directed radially inwardly), implying the existence of a repulsive   gravitational force, leading to a positive temperature gradient in order to assure thermal equilibrium. This situation is at variance with the spherically symmetric case, where a negative temperature gradient is required to assure thermal equilibrium.

A second interesting property appears from the combination of (\ref{pprimaU2}) with (\ref{ecTra}), producing (\ref{pprimaU3}).  It brings out the effect of dissipative processes  on the p.g.m.d., and by virtue of the equivalence principle, on the effective inertial mass density as well. This kind of effect was pointed out for the first time for the spherically symmetric case in \cite{1tef} (see also \cite{ther}  for a discussion on this effect). In our case the term $\frac{\kappa T}{\tau}$ increases the absolute value of the effective p.g.m.d (which is negative), thereby increasing the absolute value of the effective inertial mass density (the term in the bracket on the left of (\ref{pprimaU3})), as a result of which  any hydrodynamic force directed outward tends to push  the fluid element  inward but  weaker  than in the non--dissipative case, due to the term $\frac{\kappa T}{\tau}$. Thus in the hyperbolically symmetric  case the thermal effect on the inertial mass density enhances the tendency to expansion,  as in the spherically symmetric case, but comes about  from different way.

In order to obtain specific solutions to the Einstein equations we shall need to impose additional restrictions. In this work we shall assume two different types of restrictions.  On the one hand, we shall assume that the fluid evolves in the quasi--homologous regime and satisfies the vanishing complexity factor condition. The next  section is devoted to explain these conditions in some detail. On the other hand we shall seek for spacetimes which could be considered as hyperbolically symmetric versions of Lemaitre--Tolman--Bondi spacetimes, for doing so we shall assume that the fluid is geodesic, shearing, non--conformally flat and the energy density is inhomogeneous.

\section{COMPLEXITY FACTOR AND QUASI--HOMOLOGOUS EVOLUTION}
The complexity factor is a scalar function intended to measure the degree of complexity of a self-gravitating system  (in some cases more than one scalar function may be required).
For a static, hyperbolically symmetric fluid distribution it was assumed in \cite{st1} (following the arguments developed in \cite{c1}) that the simplest system corresponds  to a
 homogeneous (in the energy density), locally  isotropic fluid distribution (principal stresses equal). Thus, a zero value of  the
 complexity factor is assigned  to  such a distribution. Furthermore, it was shown that   a single scalar function (hereafter referred to as $Y_{TF}$) describes the modifications introduced by the energy density inhomogeneity and pressure anisotropy   to the Tolman mass, with respect to its value  for the vanishing  complexity factor.

This  scalar belongs to   a set of variables named  structure scalars, defined in \cite{20} and which appear in the orthogonal splitting of the Riemann tensor \cite{18,19,20}.  For our purpose here we shall need only one of the five structure scalars characterizing   our fluid distribution, namely  $Y_{TF}$ .

The scalar $Y_{TF}$  defines the trace--free part of the  electric Riemann tensor  $Y_{\alpha\beta}$  given by

\begin{eqnarray}
Y_{\alpha\beta} = R_{\alpha \gamma \beta \delta}V^\gamma
V^\delta.\label{Telectric} \end{eqnarray}

In our case the expression for $Y_{TF}$ reads (see \cite{20} for details)

\begin{eqnarray}
Y_{TF}={\cal E}-4\pi \Pi.\label{TEY}
\end{eqnarray}

Using (\ref{EE11}), (\ref{EE22}) and  (\ref{escE}), we may express $Y_{TF}$ in terms of the metric functions and their derivatives as

\begin{eqnarray}
  Y_{TF}&=&\frac{1}{B^2}\left (\frac{A^{\prime\prime}}{A}-\frac{A^\prime}{A}\frac{R^\prime}{R}-\frac{A^\prime}{A}\frac{B^\prime}{B}\right)\nonumber\\
 &+& \frac{1}{A^2}\left (\frac{\dot A}{A}\frac{\dot B}{B}- \frac{\dot A}{A}\frac{\dot R}{R} -\frac{\ddot{B}}{B}+\frac{\ddot{R}}{R}\right ).\label{ecYTF}
\end{eqnarray}

In the dynamic case, we still need to provide a criterion for the definition of complexity  of the pattern of evolution.

 We  shall consider the  quasi-homologous evolution defined in \cite{epjc} as the simplest mode of evolution.

The quasi--homologous condition  reads
\begin{equation}
U=R\frac{U_{\Sigma^e}}{R_{\Sigma^e}},
 \label{qsh4bis}
 \end{equation}
implying

 \begin{equation}
\frac{4\pi q }{E}+\frac{\sigma}{R}=0.
 \label{qsh4}
 \end{equation}

The above condition will be used to obtain specific models, the rationale behind such a choice is based on the fact that it represents  the relativistic version of the well-known homologous condition widely used in classical  astrophysics, furthermore as shown in \cite{epjc},  it qualifies as   one of the simplest  patterns of evolution.

\section{JUNCTION CONDITIONS}
In the case that the fluid is bounded then junction conditions on the boundary have to be imposed \cite{Darmois}, otherwise   we have to cope with  the presence of thin shells \cite{17}.

Thus, outside $\Sigma^e$  we assume that we have the hyperbolic version of the Vaidya
spacetime, described by
\begin{equation}
ds^2=-\left[\frac{2M(v)}{\bf r}-1\right]dv^2-2 d{\bf r} dv+{\bf r}^2(d\theta^2
+\sinh^2\theta
d\phi^2) \label{1int},
\end{equation}
where $M(v)$  denotes the total mass,
and  $v$ is the retarded time.

The continuity of the first and the second fundamental forms imply, on
$\Sigma^e$,
\begin{equation}
q\stackrel{\Sigma^e}{=}P_r.\label{j3}
\end{equation}
 and
 \begin{equation}
m(t,r)\stackrel{\Sigma^e}{=}M(v), \label{junction1}
\end{equation}
where $\stackrel{\Sigma^e}{=}$ means that both sides of the equation
are evaluated on $\Sigma^e$.

In the cases where the central region is surrounded by an empty vacuole bounded by a surface $\Sigma^i$, junction conditions at the inner boundary of the fluid distribution read
\begin{equation}
P_r\stackrel{\Sigma^i}{=}0,\label{j3b}
\end{equation}
and
 \begin{equation}
m(t,r)\stackrel{\Sigma^i}{=}0. \label{junction1b}
\end{equation}
\section{SOME EXACT SOLUTIONS}

In the following subsections we shall exhibit several families of solutions to the Einstein equations for hyperbolically symmetric fluids. More specifically, we shall consider  two families of solutions. One of them  will be obtained  by assuming quasi--homologous evolution and the vanishing of the complexity factor. The other family of solutions corresponds to hyperbolically symmetric versions of Lemaitre--Tolman--Bondi (LTB) space--times.  In each case dissipative and non--dissipative models were obtained. The purpose of the presentation of these models is not only the  potential application of some of them to the study of specific astrophysical scenarios, but also to  illustrate the richness of fluid distributions endowed with hyperbolical symmetry.

\subsection{ Models with vanishing complexity factor and evolving in the quasi--homologous regime}

In this case we shall exhibit only non--dissipative models (for dissipative models see \cite{hd}). Then, excluding dissipative processes, and assuming the quasi--homologous condition (\ref{qsh4}) we may write

\begin{equation}\label{qcero}
  q=0\,\,\Rightarrow\,\, \sigma=0\,\,\Rightarrow\,\, \frac{\dot B}{B}=\frac{\dot R}{R}\,\,\Rightarrow\,\, R=rB,
\end{equation}
and

\begin{equation}\label{chomo}
  U=\frac{\dot R}{A}=\frac{r\dot B}{A}={\tilde a}(t)rB.
\end{equation}
Imposing next the condition  $Y_{TF}=0$ we have

\begin{equation}\label{YTFcero1}
  \frac{A^{\prime\prime}}{A}-\frac{A^\prime}{A}\frac{B^\prime}{B}-\frac{A^\prime}{A}\frac{R^\prime}{R}=0.
\end{equation}

In order to exhibit specific  solutions, we shall further assume some additional restrictions.

\subsubsection{ $\mathcal{E}=0$, $\Pi=0$ }
We shall assume here that the fluid is conformally flat ($\mathcal{E}=0$) and the pressure is isotropic ($\Pi=0$), which combined with $Y_{TF}=0$ produces $\mu^\prime=0$ (i.e. the energy density is homogeneous).

In this case the metric functions are (see \cite{hd} for details)
\begin{eqnarray}\label{nds1}
  R &=& \frac{\tilde R(t)}{\cos[c_1(t)+\ln r]}, \\
  B &=&  \frac{\tilde R(t)}{r \cos[c_1(t)+\ln r]},  \\
  A &=&\gamma(t) \tilde R^2(t) \tan [c_1(t)+\ln r]+b(t), \label{nds7}
\end{eqnarray}
where $\tilde R(t), c_1(t), \gamma(t), b(t)$ are arbitrary functions of their argument.

We shall further  specify the  solution by choosing the above functions as follows

\begin{equation}\label{condFRW}
  \dot c_1=\frac{\dot{\tilde R}}{\tilde R}, \qquad b(t)=\gamma(t)\tilde R^2,
\end{equation}
producing

\begin{eqnarray}
  \frac{\dot R}{R} &=& \frac{\dot{\tilde R}}{\tilde R}(1+\tan u), \\
  A &=& \gamma(t)\tilde R^2(1+\tan u),\quad \Rightarrow \quad A= \frac{\tilde a\dot R}{R}\label{condA},
\end{eqnarray}
with  $\tilde a=\frac{\gamma(t)\tilde R^3}{\dot{\tilde R}}$ and  $u=c_1(t)+\ln r$.
From  the above expressions we found for the physical variables and the mass function,

\begin{eqnarray}
  8\pi \mu &=& -\frac{3}{\tilde R^2}+\frac{3}{\tilde a^2}, \\
  8\pi P_r =8\pi P_\bot&=&-\frac{3}{\tilde a^2}+\frac{3\tan u+1}{\tilde R^2(\tan u+1)}\nonumber
  \\ &+&\frac{2\tilde R \dot{\tilde a}}{\tilde a^3 \dot{\tilde R}(\tan u+1)},\label{nds2}
  \\
  m &=& \frac{\tilde{R}}{2 \cos ^3 u}\left(1-\frac{\tilde {R}^2}{\tilde{a}^2}\right).
\end{eqnarray}

If we choose  $\tilde R(t), c_1(t), \gamma(t)$ such that they tend to a constant as $t\rightarrow \infty$, then the above solution tend to the incompressible isotropic solution found in  \cite{mimc2}, which is a particular case of the hyperbolically symmetric Bowers--Liang solution found in \cite{st1}.

This model  might be considered as a hyperbolically symmetric version of the Friedman--Robertson--Walker space--time (FRW), since both  share some  similar properties e.g. $\mathcal{E}=\Pi=\mu^\prime=\sigma=0$. However our solution  is not geodesic as in the spherically symmetric case. We shall next find another  version of the hyperbolically symmetric  FRW space--time, but satisfying the geodesic condition $A^\prime=0$.
\subsubsection{$A=1$, $\mathcal{E}=0$}

If we further impose the geodesic condition on the fluid ($A=1$),  then  the quasi--homologous condition implies
\begin{equation}
\frac{R_I}{R_{II}} = constant,
\label{g1}
\end{equation}
where $R_I$ and $R_{II}$ denote the areal radii of two shells $(I, I I)$ described by $r = r_I = constant$, and $r = r_{II} = constant$, respectively. In the notation of \cite{hd},  conditions (\ref{chomo}) and (\ref{g1}) define the homologous evolution.

From (\ref{g1}) it follows at once that  $R$ is a separable function.

The conditions $A=1$ and  $q=0$  imply

\begin{equation}\label{ecA1q0}
\frac{\dot{B}}{B}=\frac{\dot{R}^\prime}{R^\prime},
\end{equation}
where (\ref{EE01}) has been used.
From the shear--free  condition  and the separability of $R$ it can be easily shown that $B$ becomes a function of $t$ alone $B=B(t)$, i.e.
\begin{equation}\label{YTF0A1}
 R=rB(t).
\end{equation}
Then (\ref{ecA1q0}) is automatically satisfied,  as well as  $Y_{TF}=0$ as it follows from  (\ref{YTFcero1}).

The physical variables and the mass function  for this model read
\begin{eqnarray}
8\pi\mu&=&-\frac{2}{r^2 B^2}+\frac{3 \dot{B}^2}{B^2},
\label{1b}\\
8\pi P_r&=&\frac{2}{r^2 B^2}-\frac{\dot{B}^2}{B^2}-\frac{2\ddot{B}}{B},
\label{2b}\\
8\pi P_{\bot}&=&-\frac{2\dot{B}^2}{B^2}-\frac{2\ddot{B}}{B},
\label{3b}\\
m&=&\frac{rB}{2}(2-r^2 \dot {B}^2).
\label{4b}
\end{eqnarray}
Thus the fluid  is conformally flat, shear--free, geodesic, evolves homologously and satisfies the  vanishing complexity factor condition. So, it also qualifies as a hyperbolically symmetric version of FRW space--time.  However, unlike the spherically symmetric case, it is anisotropic in the pressure and the energy--density is inhomogeneous.

It is worth analyzing with some detail the differences between this case and the situation in  the spherically  symmetric case (FRW). In the latter case we have seen \cite{c2} that for a non--dissipative fluid satisfying the homologous condition, the complexity factor vanishes and there is a single solution characterized by $\Pi=\mu^\prime=a={\cal E}=0$ (FRW).

However in the present case, imposing homologous condition on a geodesic non--dissipative fluid we get a conformally flat, shear--free fluid  with  $\Pi, \mu^\prime\neq 0$.
If we want to describe an isotropic, homogeneous, shear--free non--dissipative  fluid, then we have to relax the geodesic condition.

Finally,  let us build a toy model with the above solution, by choosing a particular form of  $B$ such that asymptotically it leads to a static regime.

Thus, let us assume.
\begin{equation}
B=\beta\left(1+e^{-\alpha t}\right),
\label{t1}
\end{equation}
where $\alpha, \beta$ are two positive constants.

Then it  is a simple matter to check that as $t\rightarrow \infty$ we get
\begin{eqnarray}
  8\pi \mu &=& -\frac{2}{r^2 \beta^2},
  \\
  8\pi P_r &=& \frac{2}{r^2 \beta^2},\\
  8\pi P_\bot &=&0,
\end{eqnarray}
and for the mass function we get asymptotically $m=r\beta$.

As we see, our toy model converges to the static solution corresponding to the stiff equation of state $(P_r=\vert \mu \vert)$ found in \cite{st1} (Eqs.(138-139) in that reference).

\subsection{Hyperbolically symmetric versions of Lemaitre--Tolman--Bondi spacetimes}
We shall now consider a family of solutions sharing some basic properties with LTB space--times, namely.

\begin{itemize}

\item $\mu^\prime, \sigma, {\cal E} \neq 0$, $A=1$.
\item We shall NOT require homologous condition.

\end{itemize}
Let us first consider the simplest case (non--dissipative dust) and afterward we shall consider extensions to the dissipative anisotropic case.
\subsubsection{The non--dissipative dust case}

In this case it follows from the field equations
\begin{equation}
B(t,r)=\frac{R^\prime}{\left[k(r)-1\right]^{1/2}},\label{BTBh}
\end{equation}
\noindent where $k$ is an arbitrary function of $r$.

Next, from the definition of the mass function we may write
\begin{equation}\dot R^2=-\frac{2m}{R}+k(r),
\label{intltb1h}
\end{equation}

 implying  $k(r)>\frac{2m}{R}$. Thus unlike the spherically symmetric LTB space--time we now have only one case $k(r)>0$.

The solution to (\ref{intltb1h}) may be written as:

\begin{equation}
 R=\frac{m}{k}(\cosh \eta+1), \qquad  \frac{m}{k^{3/2}}(\sinh \eta+\eta)=t-t_{0}(r),
\label{int5h}
\end{equation}
and the line element reads
\begin{equation}
ds^2=-dt^2+\frac{(R^{\prime})^2}{k(r)-1}dr^2+R^2(d\theta^2+\sinh^2\theta
d\phi^2).\label{mTB}
\end{equation}

In order to prescribe an explicit model we have to provide the three functions $k(r)$, $m(r)$ and $t_{0}(r)$. However, since (\ref{mTB}) is invariant under  transformations of the form $r=r(\tilde r)$, we only need two functions of $r$.

Assuming  as example $m_0=\frac{m}{k}=constant$, and $t_{0}(r)=constant$ the expressions for $\Theta$ and  $\sigma$ read
\begin{eqnarray}
  \Theta &=& \frac{\sqrt{k}}{m_0}\left ( \frac{\sinh \eta}{\sinh \eta +\eta}+\frac{\cosh \eta+2\sinh\eta+1}{(\cosh \eta+1)^2} \right ), \\
  \sigma &=& \frac{\sqrt{k}}{m_0}\left ( \frac{\sinh \eta}{\sinh \eta +\eta}+\frac{\cosh \eta-\sinh\eta+1}{(\cosh \eta+1)^2} \right ),
\end{eqnarray}
from where it is clear that the expansion is always positive.

The only non--trivial conservation law in this case reads
\begin{equation}
\dot \mu+\mu \Theta=0,
\label{cl1}
\end{equation}
or

\begin{eqnarray}
\dot \mu+\mu\left(\frac{\dot B}{B}+2\frac{\dot R}{R}\right)=0,\label{an'}
\end{eqnarray}
producing
\begin{equation}
 \mu=\frac{h(r)}{BR^2},\label{nltb}
\end{equation}
or, using (\ref{BTBh})
\begin{equation}
 \mu=\frac{3 h(r)\left[k(r)-1\right]^{1/2}}{(R^{3})^\prime},\label{n1ltb}
\end{equation}
where $h(r)$ is a function of integration, which due to the fact that the energy density is negative, must be necessarily negative. For this family of solutions it can  be shown the the complexity factor cannot vanish.

\subsubsection{Anisotropic dissipative models}

We shall now consider the possibility that the system radiates, and the pressure is non--vanishing and may be anisotropic.  Then following the scheme developed in  \cite{LTBd}, we write
\begin{equation}
B(t,r)=\frac{R^\prime}{\sqrt{K(t,r)-1}},\label{ecB}
\end{equation}
where
\begin{equation}\label{K1}
  K(t,r)-1=\left [ \int 4\pi {q}Rdt  +C(r) \right ]^2,
\end{equation}
and $C(r)=\sqrt{k(r)-1}$.

Thus, the general form of the line element may be written as
\begin{equation}
  ds^2=-dt^2+\frac{(R^\prime)^2 dr^2}{\left [ \int 4\pi {q}Rdt  +C(r) \right ]^2}+R^2(d\theta^2 +\sinh ^2\theta d\phi^2).
\end{equation}

A first class of models of this family will be obtained from conditions on the complexity factor, while a second one  will be obtained by a specific restriction on the function $B$, particularly suitable for describing situations  where a cavity surrounding the central region appears \cite{16n}.

Let us first consider models obtained upon conditions on the complexity factor.

In our case the complexity factor $Y_{TF}$ may be written as
\begin{eqnarray}
  Y_{TF} &=& \frac{\ddot{R}}{R}-\frac{\ddot{R}^\prime}{R^\prime} +\frac{\ddot{K}}{2(K-1)}\nonumber \\
  &+&\frac{\dot{K}}{K-1}\left (\frac{\dot{R}^\prime}{R^\prime}-\frac{3}{4}\frac{\dot K}{K-1} \right)
\end{eqnarray}
\noindent In order to obtain specific models we shall assume that the complexity factor  has the same form as in the non-dissipative case, implying
\begin{equation}
\frac{\ddot{K}}{2(K-1)}
  +\frac{\dot{K}}{K-1}\left (\frac{\dot{R}^\prime}{R^\prime}-\frac{3}{4}\frac{\dot K}{K-1} \right)=0.\label{ecK}
\end{equation}

\noindent The integration of  (\ref{ecK}) produces

\begin{equation}
\frac{R^\prime \sqrt{\dot{K}}}{(K-1)^{\frac{3}{4}}}=C_1(r)\label{pik},
\end{equation}
where $C_1$ is an integration function.

From the above, using the field equations we may write for $q$
\begin{equation}\label{ecK4}
  2\pi {q}=\frac{1}{R\left ( R^\prime C_1(r) \int \frac{dt}{(R^\prime)^2}\right)^2}.
\end{equation}

Now, it is worth noticing that this class of solutions does not admit pure dust condition (pressure cannot vanish).

Indeed, from the dust condition and (\ref{ecK}) we obtain

\begin{equation}\label{K1122}
 \frac{\dot{R}}{R} \frac{C_1^2\sqrt{K-1}}{2(R^\prime)^2}=0,
\end{equation}

\noindent which cannot be satisfied, implying that there are no radiating dust solutions in this case. Therefore we have to relax the dust condition, and we shall consider models of radiating anisotropic fluids. A simple  model of this kind, may be obtained by assuming
{$P_\bot=0$,  $P_r\neq 0$ }.

 Then, from the condition $P_\bot=0$, we obtain

\begin{equation}\label{BPbot}
  \ddot{B}=0,\quad \Rightarrow B=b_1(r) t+b_2(r),
\end{equation}
  where $b_1$ and $b_2$ are two arbitrary functions.

Using  (\ref{ecB}) we find for  $R$
\begin{equation}\label{RPbot}
  R^\prime-B\dot{R}=0, \quad \Rightarrow\quad R=\Phi\left[a_1(r) t+a_2(r)\right],
\end{equation}
where $\Phi$ is an arbitary function of its argument
\noindent and
\begin{equation}
a_1(r)=e^{\int b_1(r)dr},\qquad a_2(r)=\int b_2(r)e^{\int b_1(r)dr}dr.\label{a1a2}
\end{equation}

For specifying further the model let us choose  $b_1(r)$ and $b_2(r)$ as

\begin{equation}\label{bb1}
  b_1(r)=\frac{\beta_1}{r+\beta_2},\qquad b_2(r)=(r+\beta_2)^\alpha,
\end{equation}

\begin{eqnarray}
  B &=& \frac{\beta_1 t}{r+\beta_2}+(r+\beta_2)^{\alpha},\\
  R &=& (a_1t+a_2)^n,
\end{eqnarray}
where $\beta_1$, $\beta_2$, $\alpha$ and $n$ are arbitrary constants, chosen positive to ensure the positivity of $B$ and $R$.

Then  the physical variables and the complexity factor for this model read

\begin{eqnarray}
  8\pi \mu &=&-\frac{1}{(a_1t+a_2)^{2n}} -\frac{2(n-1)na_1^2}{(a_1t+a_2)^2},\label{d2}\\
  4\pi q &=& \frac{(n-1)na_1^2}{(a_1t+a_2)^2},\label{qn1} \\
  8\pi P_r &=&\frac{1}{(a_1t+a_2)^{2n}} -\frac{2(n-1)na_1^2}{(a_1t+a_2)^2},\label{p2}\\
m&=&\frac{(a_1t+a_2)^n}{2},
\end{eqnarray}

\begin{eqnarray}
  \Theta &=& \frac{b_1}{b_1t+b_2}+\frac{2n a_1}{a_1t+a_2} ,\\
 \sigma&=& \frac{b_1}{b_1t+b_2}-\frac{n a_1}{a_1t+a_2},
\end{eqnarray}

\begin{equation}
Y_{TF}=\frac{(n-1)na_1^2}{(a_1t+a_2)^2}.
\label{cf1}
\end{equation}
It is worth noticing that $Y_{TF}$ has exactly the same expression as $q$, as given by (\ref{qn1}). Therefore any solution of this family satisfying the vanishing complexity factor is necessarily non--dissipative.
On the other hand  $Y_{TF}$  is zero  if  $n=1$. Thus the solution of this family with vanishing complexity factor is characterized by $n=1$, which using (\ref{d2}) and (\ref{p2}) produces
\begin{equation}
P_r=-\mu.
\label{st}
\end{equation}
Finally, let us mention that the the expansion scalar is always positive.

We shall next obtain some models by imposing  a specific condition on $B$ ($B=1$). The reason to assume such a condition comes from the fact that, as shown in \cite{16n}, it is particularly suitable for describing situations where a vacuum cavity surrounds the central region. In this case the field equations read

\begin{eqnarray}\label{EEB1}
 8\pi  \mu &=& -\frac{1}{R^2}-\frac{2 R^{\prime\prime}}{R}-\left(\frac{R^\prime}{R}\right)^2+\frac{\dot{R}^2}{R^2}, \\
  4\pi q &=& \frac{\dot{R}^\prime}{R}, \label{101}\\
  8\pi  P_r &=& \frac{1}{R^2} +\left(\frac{R^\prime}{R}\right)^2-\left [\left(\frac{\dot R}{R}\right)^2+\frac{2\ddot{R}}{R}\right],\label{102}\\
 \end{eqnarray}
\begin{eqnarray}
  8\pi P_\bot &=&  \frac{R^{\prime\prime}}{R}-\frac{\ddot{R}}{R}.\label{103}
\end{eqnarray}

Let us first consider  the non--dissipative case ($q=0$). Then it follows at once from (\ref{101}) that

\begin{equation}
R=R_1(t)+R_2(r),
\label{se1}
\end{equation}
where $R_1$ and $R_2$ are arbitrary functions of their  arguments.

In order to exhibit an exact solution let us further assume $P_\bot=0$. Using this condition in (\ref{103}) produces
\begin{equation}
R_1(t)=a_1 t^2+b_1t +c_1,\qquad R_2(r)=a_1 r^2+b_2 r+c_2,
\label{se3}
\end{equation}
where $a_1, b_1, c_1, b_2, c_2$ are positive arbitrary constants.

The physical and kinematical variables for this model are
\begin{equation}
8\pi \mu=\frac{1}{\alpha^2}\left(-1-4a_1 \alpha-\beta^2+\gamma^2\right),
\label{se4}
\end{equation}

\begin{equation}
8\pi P_r=\frac{1}{\alpha^2}\left(1-4a_1 \alpha+\beta^2-\gamma^2\right),
\label{se5}
\end{equation}

\begin{equation}
P_r+\mu=-\frac{a_1}{\pi \alpha},
\label{semp}
\end{equation}

\begin{equation}
\Theta=\frac{2 \gamma}{\alpha}\label{se6},
\end{equation}

\begin{equation}
\sigma=-\frac{\gamma}{\alpha}\label{se7},
\end{equation}

\begin{equation}
m=\frac{ \alpha}{2}\left(\beta^2-\gamma^2+1\right),\label{se8}
\end{equation}

where
\begin{eqnarray}
\alpha&\equiv& a_1(t^2+r^2)+b_1t+b_2r+c_1+c_2;\qquad \beta\equiv 2a_1r+b_2, \nonumber \\ &&\gamma\equiv 2a_1t+b_1.
\label{se9}
\end{eqnarray}

For this model the expression for $Y_{TF}$ reads
\begin{equation}
Y_{TF}=\frac{2a_1}{\alpha}.
\label{cf3}
\end{equation}

Therefore the vanishing complexity factor condition implies $a_1=0$, producing because  of (\ref{se4})  and (\ref{se5})
\begin{equation}
P_r=-\mu.
\label{st3}
\end{equation}
Thus the solution of this family with the vanishing complexity factor  condition is also characterized by the stiff equation of state. Besides, the expansion scalar is always positive.

Finally, let us now consider the dissipative case ($q\neq0$). If we impose the condition $P_\bot=0$, then we get the equation $\ddot R=R^{\prime \prime}$, whose general solution is of the form
\begin{equation}
R(t,r)=c_1 \Psi(t+r) +c_2\Phi(t-r),
\label{se10}
\end{equation}
where $c_1, c_2$ are arbitrary constants, and $\Psi, \Phi$ arbitrary functions.

As an example let us choose

\begin{equation}
R=a_1(t-r)^n,
\label{se18}
\end{equation}
where $a_1, n$ are positive constants, and the solution applies in the region $t>r$.

The ensuing physical and kinematical variables are in this case:
\begin{equation}
8\pi \mu=-\frac{1}{a_1^2\left(t-r\right)^{2n}}-\frac{2n\left(n-1\right)}{\left(t-r\right)^2},
\label{se19}
\end{equation}
\begin{equation}
4\pi q=-\frac{n\left(n-1\right)}{\left(t-r\right)^2},
\label{se20}
\end{equation}

\begin{equation}
8\pi P_r=\frac{1}{a_1^2\left(t-r\right)^{2n}}-\frac{2n\left(n-1\right)}{\left(t-r\right)^2},
\label{se21}
\end{equation}
\begin{equation}
\Theta=\frac{2 n}{t-r}, \quad \sigma=-\frac{ n}{t-r},
\label{se22}
\end{equation}

\begin{equation}
m=\frac{a_1\left(t-r\right)^n}2,
\label{se24}
\end{equation}

\begin{equation}\label{T2}
  T(t,r)=\frac{n(n-1)}{4\pi \kappa (t-r)}-\frac{n(n-1)\tau}{4\pi \kappa (t-r)^2}+T_0(t),
\end{equation}

where the temperature has been calculated used the truncated transport equation (\ref{V2}).

The corresponding expression of the complexity factor for this case reads
\begin{equation}
Y_{TF}=\frac{n(n-1)}{(t-r)^2}.
\label{cf5}
\end{equation}

Thus, the vanishing complexity factor  conditions requires $n=1$, implying because of (\ref{se20}) that the fluid is non--dissipative, and because of (\ref{se19}) and (\ref{se21}) that the fluid satisfies the stiff equation of state $P_r=-\mu$.

As in the previous model, the expansion scalar is always positive.

\section{CONCLUSIONS}
The  general approach hereby described to analyze  the dynamics of hyperbolically symmetric fluids, including dissipative processes, brings out some  remarkable features of hyperbolically symmetric fluids, namely:
\begin{enumerate}
\item The energy density is necessarily negative.
\item The Tolman condition for thermodynamic equilibrium implies in this case the presence of a positive temperature gradient, unlike the negative temperature gradient required in the spherically symmetric case.
\item The  thermal modification of  the inertial mass density  reported for the spherically symmetric case in \cite{1tef}, produces an effect that is similar   to the one  obtained in the spherically symmetric case (to  enhance the tendency to expansion) but  comes about through  the increasing of the absolute value of the effective inertial mass density.
\item The fluid cannot fill the central region.
\end{enumerate}

The violation of the weak energy condition ($\mu<0$) should not scare us. Indeed,  while it is true that at classical level we do not expect negative energy density in a realistic fluid, the situation is quite different in presence of  quantum effects. In fact there is abundant theoretical evidence illustrating the appearance of negative energy density  in some astrophysical and cosmological  scenarios  (see \cite{we1,we2,we3,cqg,pav} and references therein). Thus the type of fluids considered in this manuscript  might be useful for studying systems under extreme conditions where quantum effects are expected to play a relevant role.

This negative energy density implies the appearance of a repulsive  gravitational force which has two important thermodynamic consequences mentioned in points 2 and 3 above.

Finally, the fact that  the fluid distribution cannot  fill the central region is consistent with the result obtained in \cite{2nc}, indicating that  test particles are not allowed to reach the center for the line element (\ref{w3}). Here we  have assumed that the central region is surrounded by  an empty  vacuole, however it could be assumed as well that  the central region is filled with a fluid endowed with a different type of symmetry.

Once  the set of equations  describing the dynamics of hyperbolically symmetric fluids have been obtained, we have presented a selection of  exact solutions. These were found under two types of conditions. A first class of solutions emerge from the condition of the vanishing complexity factor defined in \cite{c1} ($Y_{TF}=0$) and the quasi--homologous evolution  (\ref{qsh4}) as defined in \cite{20}. A second class of solutions was found  by looking for models characterized by non--vanishing shear, inhomogeneous energy--density and vanishing four--acceleration (geodesics). This class of solutions are entitled to be considered as  hyperbolically symmetric versions of LTB space--times.

For the first class of models  two exact solutions were found in the  the non--dissipative case. One of them, (\ref{nds1})--(\ref{nds2}), describes a fluid distribution satisfying conditions ${\cal E}=\sigma=0=\Pi=\mu^\prime=0$, which is a reminiscence of the usual FRW space--time. However, unlike the latter it is not geodesic. The second one appears from the geodesic condition,  then the quasi--homologous evolution  becomes homologous, and the solution is described by (\ref{YTF0A1})--(\ref{4b}). This is a geodesic fluid, satisfying also the conditions ${\cal E}=\sigma=0$, and therefore is also a good candidate to be regarded as a  hyperbolical version of the FRW space--time, however unlike the latter, it is anisotropic in the pressure and inhomogeneous in the energy--density.

Dissipative models  belonging to this first class  of solutions may be found in \cite{hd}.

Among the second class of solutions we have presented first, solutions corresponding  to non--dissipative dust configurations. Comparing with the spherically symmetric case we observe that only one family of solutions   ($k(r)>0$) exists, instead of the three  families existing in this latter case ($k(r)\lesseqqgtr 0$).

These solutions cannot satisfy the vanishing complexity factor, neither can they evolve in the quasi--homologous regime. Also, the scalar expansion is positive as expected for pure dust submitted to a repulsive gravity.

Next we have analyzed the case of dissipative anisotropic fluids. For doing this we generalized the expression (\ref{BTBh})  by assuming (\ref{ecB}). Different specific models were found from two different conditions. One family of solutions was obtained from a condition imposed on the complexity factor (\ref{ecK}). In this case   the pressure must be anisotropic. A specific solution  was found assuming $P_\bot=0$. The subclass of this type of solution satisfying the vanishing complexity factor is necessarily non--dissipative, and satisfies the stiff equation of state $P_r=-\mu$.

The other family  of solutions was found under the condition $B=1$. For the non--dissipative case a family of solutions was found under the additional condition $P_\bot=0$. In this case too, the vanishing complexity factor condition implies the stiff equation of state $P_r=-\mu$. In the dissipative case,  assuming (\ref{se18}) we may identify two types of contributions in the expression of the temperature. On the one hand   contributions in the stationary dissipative regime (non containing $\tau$)  and on the other hand, contributions from the transient regime (terms proportional to $\tau$).
If we assume further the vanishing complexity factor condition then the fluid becomes non--dissipative and satisfies the  stiff equation of state $P_r=-\mu$.

The expansion scalar  for all the above models are always positive.

Finally we would like to conclude with some general remarks:
\begin{itemize}

\item All the models exhibited above, although not directly related to   specific scenarios, suggest their potential applications in cosmology and astrophysics. We have in mind the construction of more sophisticated models of the Universe  beyond the  standard FRW  (see for example \cite{cm1,cm2}). More specifically, the repulsive character of gravitation exhibited in hyperbolically symmetric fluids, suggests its possible use in the modeling of very early stages of the Universe.

\item The particular behavior of test particles  within the horizon, assuming the line element (\ref{w3}), suggests a possible model of relativistic jets (see \cite{2nc} for a discussion on this issue).

\item It is worth mentioning that, unlike the spherically symmetric case, there is not a Birkhoff  theorem for hyperbolical symmetry, and therefore neither is the metric (\ref{w3}) unique, nor is the line element (\ref{1}) the most general to describe hyperbolically symmetric fluids. It would be very interesting to delve deeper into this issue, considering other hyperbolically symmetric space--times.
\end{itemize}

\section*{Acknowledgments}

This work was partially supported by Ministerio de Ciencia, Innovacion y Universidades. Grant number:
PGC2018096038BI00. I also wish to thank  Universitat de les  Illes Balears  for financial support and hospitality.

redpajama_set_name: RedPajamaArXiv


text:
Q: Servlet Request parameters with utf-8 i am trying to get UTF-8 string with using servlet, my code block is showing below;
 request.setCharacterEncoding("UTF-8");
 response.setContentType("text/html;charset=UTF-8");

but I could not get the utf-8 string , interesting chars is apperaing...
Is there any way to get UTF-8 string ?

A: The problem could be that the string received by the servlet is not really in UTF-8. Make sure that the JSP calling the servlet is also in UTF-8. Try adding this at the top of your JSP:
<%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%>

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text:
February 14, 2020 One Guys Opinion
Producers: Jason Blum, Marc Toberoff and Jeff Wadlow Director: Jeff Wadlow Screenplay: Jeff Wadlow, Chris Roach and Jillian Jacobs Cast: Michael Peña, Maggie Q, Lucy Hale, Austin Stowell, Jimmy O. Yang, Portia Doubleday, Ryan Hansen, Parisa Fitz-Henley and Michael Rooker Distributor: Sony Entertainment/Columbia Pictures
Grade: F
The original ABC TV series of "Fantasy Island," which ran from 1978 to 1984, might have been a pretty horrible program, but it wasn't a genuine horror. That's what producer Jason Blum and writer-director Jeff Wadlow, who previously collaborated on the awful "Truth or Dare," have made of it—a horror movie that's genuinely horrible, its stupidity being exceeded only by its dullness.
In a move that recalls ABC's attempted 1998 reboot of the original, which bombed of course, Wadlow and his co-writers have ratcheted up the supernatural elements of the premise while emphasizing their potentially fatal consequences. The fantasies here turn out to be more nightmarish than satisfying as they collide and replace ostensible hopefulness into what emerges as nothing more than a contrived revenge plot, complete with zombie-like villains.
The picture begins with Julia (Parisa Fitz-Henley and her boss, the mysterious Mr. Roarke (colorless Michael Peña), welcoming their five new guests, who have each won trips to the purported paradise. Brothers JD (Ryan Hansen, obnoxious in Dax Shepard mode) and Brax (goofy Jimmy O. Yang), who want the "time of their lives;" Patrick (stolid Austin Stowell), who wants to play soldier to honor his dead soldier dad; Melanie (Lucy Hale), a hottie who, in turns out, fantasizes about humiliating Sloane (Portia Doubleday), the mean girl from her past; and Gwen (subdued Maggie Q), a reserved woman still grieving her failure to accept the marriage proposal offered by Rocklin (Robbie Jones) years before.
The fantasies—which, Roarke warns them, will have to follow the "natural course" determined by the island to their ends—begin, but they quickly grow awry and begin to intersect in weird ways. Other figures intrude on the action, among them a grizzled guy in the forest (Michael Rooker) who shows up when characters get into trouble and a character called Devil Face (Kim Coates), who leads a squad of masked gunmen. Even Patrick's dead father shows up.
As things grow more and more complicated, the movie becomes decidedly chaotic, and it takes more attention than the material deserves to keep things straight, especially since the script is constantly tossing in contrived plot curveballs, which in the last act culminate in a series of revelations and resolutions so absurd that they leave the movie a complete mess. Along the way, there are a few gross moments (like a "Hostel" reminiscence early on), but generally the picture moseys along surprisingly pokily, overstaying its welcome by clocking in at nearly two full hours.
You have to give a certain degree of credit, though, to the behind-the-camera craftsmen (save for Wadlow, of course, whose direction is pedestrian). Marc Fisichella's production design has some elegance, and cinematographer Toby Oliver provides glossy widescreen images; and one feels sorry for editor Scott Albertson, who tries desperately to give shape and coherence to the constantly shifting storylines, even if he doesn't always succeed. Bear McCreary's score, though, bangs away mercilessly.
As usual nowadays, the movie concludes with the suggestion of forthcoming sequels (along with an especially lame "reveal"). By the close of the movie though, you'll definitely be inclined to agree with the survivor who acidly remarks that she just wants to get off this damned "Island."
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