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Lucky Town brewery closing early March
BY DUNCAN DENT
Thursday, January 24, 2019 12:00 PM
Lucky Town Brewing Company was founded in Gluckstadt by a group of beer enthusiasts. They later opened a brewery in Jackson, which is set to close early-March.
Lucky Town Brewing Company announced last week that after six years of operation, the company that was founded in Madison County would be closing its doors in early March.
Founder and co-owner Lucas Simmons said that they were hampered by a series of recurring issues.
"We had some ongoing issues and did not see it as something we could sustain long-term," Simmons, who helped start the business in Gluckstadt, said.
Simmons and four of his friends debt financed the business six years ago. Many founders maintained day jobs and worked long hours after work to make the brewery a reality. An announcement on the brewery's Facebook page last week said that they had been "undercapitalized" since day one.
"It is unfortunate but I keep telling people that we don't regret anything. We were part of building something we could be proud of, helped push game-changing legislation for brewers in the state and had a lot of fun doing it," Simmons said. "We hope we have convinced a few more people to drink beer brewed locally in Mississippi. We are friends with a lot of the other breweries in the state and they are all locally owned so that money stays in Mississippi."
Lucky Town founders and employees, as well as brewers and enthusiasts, statewide were instrumental in pushing a Cory Wilson co-sponsored bill that passed in the summer of 2017 that allowed brewers in state to sell their product for onsite consumption, allowing Lucky Town to open their tap room.
Simmons and company started in Gluckstadt, where they took their name.
The goodbye post featured an outpouring of support from both within the state and outside.
Jan Richardson of Ridgeland wrote, "Oh Lucky town! We will miss you!"
Carrigan Mitchell of Madison left a message saying how the brewery had opened her up to her new favorite variety of beer, the sour.
Simmons said to unload their remaining inventory they have extended their tap room hours from 4 p.m. to 8 p.m.Tuesday to Friday and 1 p.m. to 8 p.m. on Saturdays. On March 9, they will host their last Sippin' Saturday, a regular event the brewery with food trucks and live music.
New beer law allows brewers to sell drinks to on-site patrons


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Our Mother of Sorrows OFS is a Fraternity of the Secular Franciscan Order meeting in Grand Rapids, MI. We gather as community generally every last Wednesday of the month at St. Thomas the Apostle Parish.
Is there going to be a Transitus service in Grand Rapids in October?
We are not able to have hold a Transitus this year.


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Lead Editorials
EDITORIAL: Tuition will continue to rise if the General Assembly doesn't act
By Editorial Board | 01/15/2020
Failing to address these issues adequately is a startling omission from a political party that claims to be a proponent of greater college affordability. READ MORE
EDITORIAL: Bring transparency to the recycling process at U.Va.
By Editorial Board | 11/20/2019 10:00pm
This crusade to combat climate change can't be successful if it's all for show.
EDITORIAL: Virginia's newly elected policymakers must keep students in mind
Given that students represent an increasingly important constituency, elected officials must focus on the issues that disproportionately affect them. READ MORE
EDITORIAL: Ryan's minimum wage announcement puts U.Va. on the right track
The University's implementation of a living wage is the most important aspect of his ongoing goal to create better relations with the Charlottesville area. READ MORE
EDITORIAL: Adopting the New College Curriculum was a mistake
If the New College Curriculum is as great as its advocates contend, then there should be no reason to mandate enrollment in it. READ MORE
EDITORIAL: Democratize the BOV
Considering the vast array of issues the Board votes on, it is unclear why the student and faculty members' opinions are not considered beyond a mere advisory role. READ MORE
EDITORIAL: U.Va. must continue to reform the Hospital's debt collection policies
We urge the administration to work diligently to end all debt collection lawsuits. READ MORE
EDITORIAL: It's time for Student Council to step up to the plate
The Council must do something to begin fulfilling its stated purpose. READ MORE
EDITORIAL: Honor should reinstate the Alternative Sanction Working Group
The Alternative Sanction Working Group continued into the spring until after student elections when the newly elected Honor Committee decided not to renew it. READ MORE
EDITORIAL: Second-years shouldn't be required to live on Grounds
While this plan could lead to a much needed increase in on-Grounds housing, the difficulties associated with its implementation outweigh the perceived benefits. READ MORE
EDITORIAL: End exclusionary zoning in Charlottesville
It is important to acknowledge how unaffordability and unavailability have steepened housing inequality in the City for years. READ MORE
EDITORIAL: End restrictions on public drinking
There is no need to fear droves of drunkards descending upon Charlottesville and disrupting peaceful enjoyment of common spaces. READ MORE
EDITORIAL: Student Council should not hold special elections
While the advocacy in favor of special elections is undertaken in good faith, it would be a mistake for a number of reasons to allow them. READ MORE
EDITORIAL: Beware the health impacts of vaping
The Centers for Disease Control and Prevention has reported that e-cigarettes have erased relatively recent decreases in youth tobacco use. READ MORE
EDITORIAL: Support free printing on Grounds
Student Council should be lauded for their efforts to make printing more accessible — however, printing should be free in all libraries at the University. READ MORE
EDITORIAL: Stop politicizing U.Va. men's basketball
In creating this petition, Kohlmoos is attempting to politicize a spectacular moment in Virginia men's basketball history. READ MORE
EDITORIAL: Increase oversight of Opportunity Zones
While this plan may be an important tool to economic revitalization, this can only be so if it is implemented in a way that curbs potential abuses. READ MORE
EDITORIAL: Abolish the Echols Scholars program
Even if the Echols Scholars are academically superior to their peers, it is unclear why they are given the opportunity to circumvent courses that will challenge them. READ MORE
EDITORIAL: Trump must revise his Higher Education Act proposal
While Trump claims these changes are an attempt to repair the financing higher education, many of these proposed changes will actually harm those who depend on financial assistance. READ MORE
EDITORIAL: Reform the class sign-up process
The particulars of the University's course registration process make it a substantial burden for students that we believe must be addressed. READ MORE

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RamPage Features
Meet the CSU Gymnastics Club
Rachel Rasmussen
The CSU Club Gymnastics Club is one of the newest clubs at CSU. It was founded last semester by first-year Chloe Brekhus and Jess Moser, one of the coaches of the team. Last September, Moser's boss decided they wanted a gymnastics team at the club level so that people could continue to do gymnastics after high school without worrying about getting scholarships for colleges with D1 gymnastics teams.
Currently, there are 13 women on the team and two coaches, Moser and Tariq Syed. Moser has been a coach for 10 years now and Syed has been coaching for 20 years. Each athlete comes to the team with varying years in the sport.
The practices, which are from 2-3:30 p.m. every Tuesday and Thursday at Timberline Gymnastics, have less structure than some teams and gyms. But they still have a general plan for the athletes to follow for each practice.
The team is part of the National Association of Intercollegiate Gymnastics Clubs and they compete against CU, DU, Utah State, Arizona State and 5280 Alumni. Not every athlete will compete in all four events (Balance Beam, Uneven Bars, Floor and Vault) for women's gymnastics, so during practices, they will usually just practice on the events that they compete.
The goal of the team is to "provide an opportunity for current and former gymnasts to keep doing what they love in a welcoming and fun environment while representing CSU."
For more information on the team, you can email them at csuclubgymnastics@gmail.com.
A Brief History of Earth Day
Companies using Recycled Products
Safety Tips for Hiking
What's it like to be a climbing guide? Ryan Dresser shares insight
Things to do to Stay Occupied During Quarantine
How to Train for a Triathlon
Helpful Tips for Your Next Run
Recreational Games to Play When the Weather Warms Up
Different Types of Slacklining to Try


Conscious Parenting: Mindful Living Course for Parents - Nataša Nuit Pantović - Google Books. Nataša Pantović Nuit is a poet, a writer, and a spiritual researcher that lives and works in Malta.
Conscious Parenting: Mindful Living Course for Parents (Alchemy of Love Mindfulness Training): Ms Nataša Nuit Pantovic, Ms Ivana Milosavljevic: 9789995754044: Amazon.com: Books. Conscious Parenting: Mindful Living Course for Parents by Ms Nataa Nuit Pantovic, Paperback. Conscious Parenting: Mindful Living Course for Parents (Alchemy of Love Mindfulness Training, #5) by Natasa Pantovic Nuit. Conscious Parenting - Nataša Pantović Nuit. Parenting is the most amazing spiritual experience one can go through.
It exercises our Will Power, and strengthens our capability to Love. It is free, it is constant, and it challenges our existing beliefs, and breaks our Ego Structures. And it is always extremely rewarding! We have designed... show more Parenting is the most amazing spiritual experience one can go through. It is in our nature to learn and grow. Conscious Parenting by Natasa Pantovic Nuit. Conscious Parenting Course Mindful Living Course for Parents Self-Development, Conscious Parenting, Alchemy of Love, Mindfulness, Online Life Coaching, Personality Tests Alchemy of Love Mindfulness Training Book #5 by Nataša Pantović Nuit Conscious Parenting is the Alchemy of Love mindfulness Training book designed for parents.
We use Transformation Tools and Spiritual Exercises that will help you get in touch with your Soul, with Love, and with Patience when dealing with kids. Concious Parenting Course Description Within 12 Modules Examine 12 areas of Your Life and Implement various Spiritual Transformation Tools to help you Live your Highest Potential as a Parent. Conscious Parenting: Mindful Living Course for Parents: Volume 5 (Alchemy of Love Mindfulness Training): Amazon.co.uk: Ms Nataša Nuit Pantovic, Ms Ivana Milosavljevic: 9789995754044: Books. Conscious Parenting: Mindful Living Course for Parents (Alchemy of Love Mindfulness Training Book 5) eBook: Nataša Nuit Pantović: Amazon.co.uk: Kindle Store. Conscious Parenting: Mindful Living Course for Parents. Author: Nataša Pantović Nuit Product Details Series: Alchemy of Love Mindfulness Training (Book 5) Paperback: 226 pages Publisher: Artof4elements; 2 edition (March 23, 2015) Language: English Product Dimensions: 8 x 0.5 x 10 inches Available through Amazon.
Conscious Parenting Course 16 intro and content smaller. Conscious Parenting: Mindful Living Course for Parents - Nataša Nuit Pantović, Ivana Milosavljevic - Google Books. Ivana is Mgr. of Education with the Masters in Special Needs Education.
Conscious Parenting Self Development Mindfulness Training. Conscious Parenting and Mothering as Spiritual Quality. How to Live Your Highest Potential. Do rewards, punishment work? In an experiment, children's brains reacted strongly to positive feedback and scarcely responded at all to negative feedback.
Our basic strategy for raising children or teaching students is a reward and punishment strategy. We are what we believe we are - C.S. Lewis From an early age we introduce Santa Claus, who carefully observes what children do and appears from the heavens to deliver gifts for the 'good ones' or to punish the 'bad ones'. While manipulating children with incentives seems to work in the short run, it is a strategy that ultimately fails. Praise is frequently a judgement and a kind of bribe as the child must 'earn' points by doing the 'right' thing. Conscious Parenting: Mothering as Spiritual Quality. Why Conscious Parents or How to Be a More Loving Parent? Conscious Parenting Because the world belongs to our children and we are their soil, their water, their air... If we are full of love, understanding, and wisdom, our children will grow into free and inspired adults.
The parenting IS the most difficult job in the world The art of parenting is extremly complicated. There are many variables that influence our kids growth and development and parents wish to understand them all. Education of the future. In a typical classroom in Finland, students work in small groups.
The teacher nurtures independence and active learning, allowing them to develop skills to understand and solve problems. "You may give them your love but not your thoughts, for they have their own thoughts. You may house their bodies but not their souls, for their souls dwell in the house of tomorrow, which you cannot visit, not even in your dreams. You may strive to be like them, but seek not to make them like you. For life goes not backward nor tarries with yesterday." There is a strong emphasis on relaxed schools that nurture creativity, questioning and in-depth subject analysis.
Conscious Parenting: Mindful Living for Parents. Conscious Parenting Best Quotes Video. Positive Parenting: Mindful Parenting in Action (Tips) Positive Parenting.
Arts make students smart. When children are physically active and creative, they tend to focus better and work more enthusiastically with the rest of the curriculum.
Arts, sport, music and drama are often viewed as fun extra-curricular activities for children but are given less importance compared with core subjects such as English, science, or mathematics. The arts should be taken seriously as a source of inspiration, as a way of life. Conscious Parenting and Kids Happiness: - Shattering The Matrix. Nuit is the author of seven mindfulness books: A Guide to Mindful Eating, Mindful Eating with Delicious Raw Vegan Recipes (Artof4Elements, 2014), Chanting Mantras with Best Chords, Mindful Being towards Mindful Living Course (Artof4Elements, 2015), Conscious Parenting (Artof4Elements, 2015), Art of 4 Elements: Discover Alchemy through Poetry (Artof4Elements, 2012), A-Ma Alchemy of Love (Artof4Elements, 2016).
Q. You say that the parenting is the most difficult job in the world. Conscious Parenting Best Quotes Video. Conscious Parenting [Article] 7 Main Children Development Tips. Famous Education Quotes from Einstein and Steiner. Quotes on Education Famous Quotes on Education Conscious Parenting, Spiritual Quotes, Online Life Coaching Quotes on Education Famous Quotes About Education Alber Einstein Famous Quotes About Education "Education is what remains after one has forgotten everything he learned in school.
" famous quotes on education Einstein "Never regard study as a duty, but as the enviable opportunity to learn to know the liberating influence of beauty in the realm of the spirit for your own personal joy and to the profit of the community to which your later work belongs. " educationsl quotes on study as a duty Einstein "The point is to develop the childlike inclination for play and the childlike desire for recognition and to guide the child over to important fields for society.
Conscious Parenting Booklike Review: Conscious Parenting teaches mindfulness. The Alchemy of Love Mindfulness Training is about our alchemy transformation, the alchemy of love, the alchemy of soul. Conscious Parenting is the Alchemy of Love Mindfulness Training designed for parents. I had a very inspiring time exploring 100s of exercises designed to help parents come back to creativity and patience when dealing with kids. Nordic countries' teaching methods reaping rewards. Unesco statistics of tertiary education enrollment by country show that while 21.5 per cent of Maltese students successfully enrol into university each year, 70 per cent of the students in Nordic countries continuing their education after secondary school. Nordic countries use techniques and tools that encourage collaboration. Shifting education goals. Education should teach children how to ask appropriate questions, how to analyse a problem, stimulate a desire to learn, and flexibility to consider different points of view.
One of the biggest problems of education today is that the 'factory model' of teaching: the top-down approach and the rewards-and-punishments approach, limit students' ability to contribute with their imagination and creativity. In the Finnish educational model, active learning is taken seriously. Schoolchildren do not sit at their desks memorising - Natasa Pantovic The system needs a shift in focus: from one that teaches children a curriculum, to the one that inspires lifelong learning.
Both Waldorf and Montessori learning methods establish a collaborative environment without tests, with the child's learning and creativity at the centre of the focus. Create freedom in the learning environment. Allow your child to cook, wash dishes, and take care of plants and animals. These activities are extremely interesting for children because they can act as grown-ups. A child has a deep longing to discover that the world is based on truth.
Conscious Parenting Interview With Nuit. Conscious Parenting: Mindful Living Course for Parents - Nataša Nuit Pantović - Google Books.


Share your expertise and contribute to your profession by presenting at ASHRM 2019 in Baltimore. ASHRM is currently accepting presentation proposals for the conference education program.
Guidelines and evaluation criteria are available on the proposal submission website. ASHRM is seeking engaging and novel sessions that highlight models and strategies that risk managers and patient safety professionals can apply within their organizations.
This is your opportunity to give back to your profession and help ASHRM provide the best possible learning experience for health care risk managers everywhere.


Racing is pain, and that's why you do it, to challenge yourself and the limits of your physical and mental barriers. You don't experience that in an armchair watching television.
Today's stage was the last "mountain" stage, and arguably the hardest stage of the series. It was the one I was scared of the most. The "Alpine" run. Various experienced ENDURrunners had warned us rookies of the dangers of the course, that it was steep, that it shredded quads at whim, and that it humbles even the most elite of runners. The weeks leading up to the ENDURrun, I'd flipped through the book and looked at the course diagram, and thought... well... it doesn't LOOK bad, even through the elevation map basically shows four 60m vertical climbs per loop. It wasn't until I got to SEE the course with my own eyes that I realized what I was in for, and it was worse than I'd imagined.
Stage 5 is set in and around the Chicopee Ski Club. It consists of five 5km loops through the ski hill. The course winds its way through some switchback trails then takes a service road all the way to the summit. Some more switchbacks to the halfway down point, then straight up the face of the main ski run to the top. We then descend switchback style across the back of the hill, go back up to the top, then down the face. And do it all over again, four more times.
The weather gods didn't make things easier for us. Forecast was highs of 28 degrees, and sunny, with 80% humidity. Lovely. When we got to the hill, the sun was shining, and there wasn't a cloud to be seen for miles around. With the exception of the two small trail sections, the course is totally expsoed, no shade from the glaring ball of fire in the sky.
As I got myself ready before the start, I tried to calm myself down. Just take it easy, I told myself. Walk the hills if you have to; don't kill yourself, you still have a 10km and a marathon to run. Did it work? No. I also slept real bad the night before, worrying about this stage. I think someone said earlier that Stage 3 (Bechtel Park) was a kick in the gut, and Stage 4 was a sucker punch to the face, then this stage was a flying kick to the head by a mastodon.
The other thing I was worrying about was my groin. I'd pulled it ever so slightly going down one of the big hills in Stage 4. It didn't bug me then, but this morning, I could feel it. In my warm-up jog around the parking lot, I could feel it. And with this run containing so many up hills, I knew it would be sore by the end.
We all assembled at the start, Lloyd gave us the course low-down, took attendance, and with a "Mark, Set, GO!" we were off and running. The first flat switchback was good; nice and shady, and I was keeping pretty good pace, albeit conservative. It was fun running through the woods on these narrow tight switchbacks. After 1km, we came out to the first ascent. The road was large gravel and dirt, and a steady climber. I made it about half way up, before I started walking. Again, I was trying to be conservative, knowing that there was much more to be done.
At the top, we hit the first water station. The sun was brutal; it felt like the Sahara Desert out there. I downed some Gatorade and poured water on my head and soldiered on. The switchbacks again were not too bad. The descent back to the bottom was some long back and forth single track trails across the face of one of the hills. This was very tough, as the trail was sideways, and I found that I hadn't tied my shoes as tight as I should have, so my foot kept slipping in the shoe. That, and almost falling after hitting a root in the ground, made this very difficult. At the bottom, another water station, then the straight up climb up the front face. I managed to run the first quarter, then gave up. The heat was oppressive, and the bugs started to attack me wholeheartedly. I think there was a cloud waiting for every runner going up that hill. My groin was starting to hurt at this time, and I felt real tired. I pushed my legs with my hands to get up that hill.
From there, we descended down the back fairly quickly, then some more switchbacks to the bottom. Another road climb to the top, and descent, and the loop was done. That was all one loop! Four more to go.
With each loop, I got slower and slower, and the heat got hotter and hotter. My groin was starting to really hurt bad. By the 3rd loop, I was seriously wondering if I would finish this stage. I bumped into Duff, one of the ENDURrunners staying in the neighbouring suite. He hurt his knee badly, and had to DNF the stage. Another runner stepped into a pothole and twisted her ankle badly. She finished the stage, but only after downing 9 Advils. Other runners tripped and fell over roots and rocks, but no other injuries.
Stage 4 was when I started getting lapped by the leaders. The event leader, Bob, who ran by me as if on wings, said a couple of words of encouragement, and I could only envy his long gracefull stride, and compare it to my own stumpy bumpy gait down the hill. The going was in full pain mode at this point, and I was starting to limp. I think I was doing way more walking than running at this point, just trying to finish. As I came through the finish for loop 4 and on to loop 5, the elites were done, but still cheered for me. I grabbed some water and Gatorade, got a sponge for my head, and trudged on for my last lap.
I have never EVER wanted to quit as bad as this last stage. Every step forward (upward or downward) was excruciating. I couldn't see anyone I'd been running with; they were all way ahead of me. Again, I almost tripped and fell, mainly because I was so tired I couldn't lift my feet anymore. Every single slight uphill I walked. There was no hope for me to finish in 2:45, never mind the 2:30 that I thought I could do. As in the previous stages, I thought back to all the well wishes I'd been receiving from friends, running buddies and family, and dug deep. Making that last descent down the front face to the finish, I looked at my watch, and saw it said 2:56. I swore a blue streak, told myself I was going to come under 3 hours, and kicked it into high gear. I think I flew down the face at 4:15min/km pace (groin be damned) and burned it to finish in 2:59:30!
This stage really showed me how inadequate my training for this has been. You really have to do a lot of hills to even have a hope of running this stage well. To give you an idea, the winner of the stage finished in 2:06, almost a full hour before me. That's a decent time on FLAT GROUND, never mind up and down a ski hill 10 times. I was happy to finish relatively intact. The groin will need some attention (vitamin I, ice, and rest), but will be okay for the next two stages. The heat was also a very big factor in how I felt in this race, but I think the constant drinking of Gatorade and water helped a lot.
Tomorrow's stage is a flat net downhill 10km time trial. Everyone's saying it will be an easy stage, but I think there's something up Lloyd's sleeve with this one. Maybe it's that final tuneup for the marathon on Sunday. The plan will be to test the groin, and go easy but at a decent pace.


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Gophers take 4th in match play
Only one Minnesota golfer finished the event with a winning record.
John Hageman
The Minnesota menâÄôs golf team began its spring season with Big Ten Match Play this weekend in Coral Gables, Fla., finishing in fourth place. MinnesotaâÄôs top golfer, senior Ben Pisani, led the Gophers past Indiana 3.5 to 2.5 in the first round Friday, with freshman Erik van Rooyen and sophomore Derek Chang also taking victories. The Gophers fell to Purdue on Saturday 4.5 to 1.5, with junior Donald Constable managing a tie in his match. Constable transferred from the University of Texas after the fall season, returning to his home state where he won the state championship in 2005 and 2006 as a golfer for Minnetonka High School. The Gophers went on to face Iowa in the third-place consolation match, where they fell 3.5 to 2.5. Northwestern took down Purdue 6.0 to 0.0 to take first place in the tournament. Van Rooyen was the only Gophers competitor to come away with a winning record, finishing 2-1-0. Pisani and Chang both finished 1-1-1, while senior Thomas Campbell and Constable were the only Gophers to not win a match, both finishing 0-2-1. Also competing over the weekend was freshman Robert Bell, who finished 1-2-0. The Gophers will be back in action for the Alumni Challenge on Feb. 20 in Scottsdale, Ari.


Sign-off clip of Westinghouse's Channel 3 in Philly back when it was an NBC network affiliate (it is now owned and operated by CBS) that kicks off with a station ID and promo for The Newlywed Game. Following this is a Channel 3 Editorial, delivered on the street by Malcolm Poindexter. The editorial is on the subject of homeless youths in Philadelphia and promotes an upcoming KYW-TV news special on the subject. The station sign-off comes immediately at the end of the editorial, which includes a copyright and legal use statement. The SSB film is one produced by the United States Navy, the very same one used by another Westinghouse O&O station at the time, WJZ-TV 13 in Baltimore (which is also now O&O'ed by CBS).
KYW-TV "CBS 3" is the third oldest commercial TV station in the United States, signing on initially as WPTZ on 1 September 1941. It was originally owned by the Philco Corporation, a Philly-based radio and tube manufacturer. Westinghouse acquired Channel 3 from Philco in 1953 for a then-record price of $8.5 million. Two years later, the NBC network gets Channel 3 from Westinghouse in one of the most controversial deals in U.S. media history (read the Wikipedia entry on KYW-TV for the details), and changed the calls to WRCV-TV. Westinghouse regains control of the station in 1965 and rechristens it as KYW-TV, after its Philadelphia AM radio station's calls. Channel 3's long affilition with the NBC network ended in September 1995 when CBS obtained an interest in the station from Westinghouse, buying complete control in 1996.
Page created Sunday, 29 November 2015. This page replaces an earlier one featuring an embedded YouTube video of a 1986 KYW-TV sign-off that's no longer accessible.


I can see what he's saying – an yeah the world's crazy but you know what – there's a fine line between saying the Devil's in the world and saying all of mankind is stupid! Not all science or 'New age beliefs' (whatever that is) are at odds with Christianity you know! I don't know, I do wonder if we are living in an age of too many polarities – you're either this or that – no! That's not right. Maybe I'm just feeling cynical this evening. I've had a great day at bible by the beach – but I keep being treated very dismissively by some "Christians" and quite frankly it's starting to really …me off – I thought patience was a virtue! There's nothing worse than people cutting you off when you are trying to talk to them! Like you are beneath them or something. Well, I'm starting to see this a LOT and usually by well known Christians too! So what does that say? I feel God is speaking to me through the bible and my everyday interactions with people and the world – but today I feel certain Christians need to get down off their high horse and look at their attitude because they get people's backs up. We are all living in this world together aren't we? Jesus went where the people are – you're not going to win people over being snobby, know it all preachy, 2 faced… OK Rant over – sorry about that – having a vent!
But what happened is – she took my venting in the wrong light and started saying this is why she doesn't want to be labelled as a Christian and how it's just another belief and with each line of her response I could feel her moving further away – distancing herself – running away. And then I felt as if God was slapping me round the face with the fruit of the spirit – saying see! This is why you have to be what you want to see in others!
And I was like – Oh my goodness – what? I AM a Christian, don't get me wrong! I love Jesus and I believe he died on the cross for me (for MY sins) it's more than just a belief and I HAVE had the Holy Spirit working in my life – and you know I've had those miracles because I told you about them. Don't get me wrong I'm not saying I'm not a Christian, and things in this world, people in this world who are distracted from God's purpose that upsets me and infuriates me but what I'm angry with is Christians who think they are above every one else. There are always people who think they are better than the average person – that's got nothing to do with being a Christian or someone of another faith or no faith – that's just human nature! LOL I think I'm more upset with them because they ARE Christians and because as I'm getting closer to Jesus in my own walk through reading the bible daily etc. I feel close to him and then some Christians come along and pollute that. By not expressing the fruit of the spirit in their own interactions. .
So when I'm ignored, put down, not welcomed, dismissed by "Christians" I don't say – oh no Christianity must be wrong, God isn't here in this walk and the world is better!! NO! I say – these people who call themselves Christians are not demonstrating the Fruit of the Spirit and therefore they are not being a good example of what Christ wants to see in their Spirit – all of the above! When I come away feeling like that – I'm angry – but at the same time I am not angry at God – I'm angry at those people for not staying true to Jesus' teachings of the Fruit of the Spirit and I have to pray that God give me MORE of those = LOVE, JOY, PEACE, PATIENCE, KINDNESS, GOODNESS, FAITHFULNESS, GENTLENESS, SELF-CONTROL so that I can stand in truth myself.
Now I feel bad that I have not demonstrated it by getting angry, venting and putting doubt in my friend's mind that Jesus IS the way, the, truth, the life. God REALLY wanted to teach me a lesson today.


text:
\section{Introduction}
\label{sec:intro}

In general relativity, the exterior solution of vacuum, asymptotically flat, stationary and axisymmetric black hole is given by the Kerr spacetime, which is characterized by its mass and angular momentum.
This is a consequence of the unique theorem~\cite{Israel:1967wq,Robinson:1975bv,PhysRevLett.26.331}.
If we consider modified gravity theories,
black holes need not be described by the Kerr spacetime (see, e.g.,~\cite{Barack:2018yly} and references therein).
Most of the stationary and axisymmetric black hole solutions found so far satisfy the circularity condition.\footnote{
Roughly speaking, the circularity condition means that
$t-\phi$ part and $r-\theta$ part of the metric are orthogonal,
where $t$ and $\phi$ correspond to Killing coordinates of time translation and axisymmetry, respectively,
and $r$ and $\theta$ are the other two coordinates. See Appendix \ref{app:circularity}, for details.
}
In general relativity, it is shown that the circularity condition holds for the asymptotically flat spacetimes in vacuum system (see, e.g.,~\cite{heusler_1996,wald1984general}), the Einstein real scalar system~\cite{Herdeiro:2019oqp} and Einstein--Maxwell system~\cite{Carter:1969zz}.
Yet, in modified gravity theories,
it is not obvious whether spacetimes satisfy the circularity condition or not.
If we can observe the effect of non-circularity, it directly suggests the violation of general relativity
because the circularity condition should hold for general relativity.

While the no-hair theorem holds for Einstein real scalar systems~\cite{Herdeiro:2019oqp, Bekenstein:1995un}
and a class of modified gravity theories with stationary scalar fields~\cite{Hui:2012qt}, it is not obvious for other systems.
For some Einstein complex scalar systems,
hairy stationary circular black hole solutions have been found, for example,~\cite{Herdeiro:2014goa,Herdeiro:2014ima,Herdeiro:2015gia} (see also~\cite{Herdeiro:2015waa} and references therein), where scalar fields have the harmonic time dependence.\footnote{
In~\cite{Smolic:2015txa},
the possible time dependence of scalar fields whose
stress energy tensors are stationary is studied.
}
In case of modified gravity theories, the linear time dependence of the scalar field has been introduced to obtain hairy static black hole solutions~\cite{Babichev:2013cya}.
Moreover, in~\cite{VanAelst:2019kku}, it is indicated that time dependence of scalar fields is crucial to obtain non-circular black hole solutions.
Recently, non-circular stationary black hole solutions have been obtained by using the disformal transformation in DHOST theories~\cite{Anson:2020trg,BenAchour:2020fgy}, while the associated scalar fields have linear time dependence.

The aims of this paper are to explore the possibility of the existence of non-circular stationary black hole solutions
in string-inspired gravity, and to develop the analysis methods for the construction of such solutions.
To this end, we focus on a particular class of the quadratic gravity theories: dynamical Chern--Simons (dCS) gravity~\cite{Smith:2007jm,Alexander:2009tp} and shift symmetric Einstein scalar Gauss--Bonnet (ESGB) gravity~\cite{Sotiriou:2013qea,Sotiriou:2014pfa}.
These two gravity theories modify general relativity by introducing dynamical scalar fields coupled with curvature invariants.
In these quadratic gravity theories,
stationary axisymmetric circular black hole solutions that differ from the Kerr spacetime have been found in slow rotation approximation~\cite{Yunes:2009hc,Yagi:2012ya,Pani:2011gy,Ayzenberg:2014aka} as well as non-perturbative numerical approaches~\cite{Delsate:2018ome,Delgado:2020rev}.
We should note that these previous works assume stationarity of the scalar fields.
In this paper, we consider perturbative solutions around the Schwarzschild spacetime and find slowly rotating black hole solutions up to quadratic order in the spin and linear order in coupling constants without assuming the circularity condition and stationarity of the scalar fields.
We show that the stationarity of the effective stress energy tensors determines the time dependence of the scalar fields. From this condition,
the linear time dependence of the scalar fields is derived.
This result is natural because spacetimes can be stationary even if scalar fields have the linear time dependence due to the shift symmetry of the theories.
We note that our analysis explicitly show the other time dependence of the scalar fields is impossible.
However,
these time dependent scalar fields yield singular behavior of the metric functions at the black hole horizons in dCS gravity,\footnote{Note that the corresponding metric functions are non-circular but they are singular at the horizon. }
and they are incompatible with the stationarity of the spacetime in ESGB gravity.
Thus, these quadratic gravity theories admit only known solutions in~\cite{Yunes:2009hc,Yagi:2012ya,Pani:2011gy,Ayzenberg:2014aka} as regular stationary solutions. Our results suggest the non-existence of rotating non-circular black holes in these quadratic gravity theories.

This paper is organized as follows.
In Sec.~\ref{sec:ThyandApp},
at the beginning, we review dCS gravity and ESGB gravity briefly.
After that, we explain the approximation scheme.
In Secs.~\ref{sec:dCSgravity} and~\ref{sec:ESGBgravity},
we study slowly rotating black holes in dCS gravity and ESGB gravity, respectively.
Section~\ref{sec:conclusion} is devoted to summary and discussion.
In Appendix~\ref{app:circularity},
we review the circularity condition.
In Appendix~\ref{appB},
we review the linear stationary metric perturbation in the Regge--Wheeler gauge.
In Appendix~\ref{appC},
we discuss the homogeneous solutions for the Einstein equations.
In what follows, we use the geometric units: $c=G=1$ and parentheses in index lists for symmetrization:  $A_{(\mu \nu)}=(A_{\mu \nu}+A_{\nu\mu})/2$.

\section{Theories and approximation schemes}
\label{sec:ThyandApp}

\subsection{dynamical Chern--Simons gravity and shift symmetric Einstein scalar Gauss--Bonnet gravity}

In this paper, we focus on quadratic gravity theories, which introduce dynamical scalar fields coupling to curvature squared terms.
In particular,
we consider dynamical Chern--Simons (dCS) gravity and shift symmetric Einstein scalar Gauss--Bonnet (ESGB) gravity.
The actions are given by
\begin{align}
    S_{\mathrm{dCS}} &= \int \mathrm{d} x^4 \sqrt{-g} \left [
    \kappa R + \frac{\alpha}{4} \vartheta R_{\nu \mu \rho \sigma} \widetilde R^{\mu \nu \rho \sigma} - \frac{\beta}{2} \{
    \nabla _{\mu} \vartheta \nabla ^{\mu} \vartheta + 2 V_{\mathrm{dCS}}(\vartheta)
    \}
     \right ],
     \\
     S_{\mathrm{ESGB}} &= \int \mathrm{d} x^4 \sqrt{-g}
     \left [
    \kappa R + \alpha  \varphi \{ R^2 - 4R_{\mu \nu}R^{\mu \nu} + R_{\mu \nu \rho \lambda} R^{\mu \nu \rho \lambda}\} - \frac{\beta}{2} \{
    \nabla _{\mu} \varphi \nabla ^{\mu} \varphi + 2 V_{\mathrm{ESGB}}(\varphi)
    \}
     \right ],
\end{align}
respectively.
Here, $\kappa = 1/(16 \pi)$, $g$ denotes the determinant of $g_{\mu \nu}$, and $\widetilde R_{\mu \nu \rho \sigma}$ is the dual of the Riemann tensor defined by
\begin{align}
    \widetilde R^{\mu \nu \rho \sigma} = \frac{1}{2} \epsilon ^{\rho \sigma \alpha \beta } R^{\mu \nu} \, _{ \alpha \beta},
\end{align}
where $\epsilon ^{\mu \nu \rho \sigma}$ is the Levi--Civita tensor.
$\vartheta$ and $\varphi$ are scalar fields, and $V_{\mathrm{dCS}}(\vartheta)$ and $V_{\mathrm{ESGB}}(\varphi)$ are potentials of the each gravity theory, while $\alpha$ and $\beta$ are coupling constants.
In this paper, we set $V_{\mathrm{dCS}}(\vartheta) = V_{\mathrm{ESGB}}(\varphi) = 0$ which implies that those theories possess shift symmetry, i.e., theories are invariant under the transformation $\vartheta \to \vartheta + \mathrm{const}.$ or $\varphi \to \varphi + \mathrm{const}.$
We take $\vartheta$, $\varphi$, and $\beta$ are dimensionless and $\alpha$ have dimensions of $(\mathrm{length})^2$.
Hereafter, we set $\beta=1$.

\subsection{approximation schemes}

We consider the slow rotation and the weak coupling approximation.
The expansion parameters are the slow rotation parameter $\varepsilon (\ll 1) $ and the weak coupling parameter $\zeta (\ll 1)$ which are respectively defined by
\begin{align}
    \varepsilon \equiv \frac{a}{M}, \,\,\,
    \zeta \equiv \frac{\alpha^2}{\kappa M^4},
\end{align}
where $M$ is the mass of the system and $a$ is the spin parameter of the system defined as the angular momentum divided by $M$.
In this paper,
we study the slowly rotating solutions up to quadratic order in the spin and linear order in coupling constant.
The circular black hole solutions in dCS gravity theory have been discussed at linear order in the spin \cite{Yunes:2009hc} and quadratic order in the spin \cite{Yagi:2012ya}.
The non-perturbative spinning solution have been also obtained numerically in \cite{Delsate:2018ome}.

We employ the approximation schemes used in~\cite{Yunes:2009hc,Yagi:2012ya}.
We expand the metric in terms of the weak coupling parameter as
\begin{align}
    g_{\mu \nu} = g_{\mu \nu} ^{(0)} + \zeta g_{\mu \nu} ^{(1)} + \mathcal{O} (\zeta ^2),
    \label{eq:expandzeta}
\end{align}
where $g_{\mu \nu} ^{(0)}$ is the exact Kerr metric
\footnote{We note that introducing small correction terms to the Einstein-Hilbert action,
the theory may admit solutions which are very different from the case of general relativity (see an example in higher dimensional Einstein Gauss-Bonnet gravity \cite{Boulware:1985wk, Deser:2005jf}).
In this paper, we focus on solutions which become those of general relativity in $\zeta \to 0$ limit.}
, which is expressed in the Boyer-Lindquist coordinates $\{ t, r, \theta, \phi \}$:
\begin{align}
    \mathrm{d}s^2 = - \frac{\Sigma \Delta}{A} \mathrm{d}t^2
    + \frac{\Sigma}{\Delta} \mathrm{d}r^2
    + \Sigma \mathrm{d} \theta ^2
    + \frac{A}{\Sigma} \sin ^2 \theta \left [ \mathrm{d} \phi - \frac{a(r^2 + a^2 - \Delta)}{A} \mathrm{d}t \right]^2,
    \label{eq:kerrmetric}
\end{align}
with
\begin{align}
    \Sigma &\equiv r^2 + a^2 \cos^2 \theta,
    \\
    \Delta &\equiv r^2 + a^2 - 2Mr,
    \\
    A &\equiv (r^2 + a^2)^2 - \Delta a^2 \sin ^2 \theta.
\end{align}
Next, we re-expand the metric~\eqref{eq:expandzeta} in terms of the slow rotation parameter as
\begin{align}
   g_{\mu \nu} ^{(0)} &= g_{\mu \nu} ^{(0,0)} + \varepsilon g_{\mu \nu} ^{(1,0)} + \varepsilon ^2 g_{\mu \nu}^{(2,0)} + \mathcal{O} (\varepsilon ^3),
   \\
   \zeta g_{\mu \nu}^{(1)} &= \zeta g_{\mu \nu} ^{(0,1)} + \varepsilon \zeta g_{\mu \nu}^{(1,1)} + \varepsilon ^2 \zeta g_{\mu \nu} ^{(2,1)} + \mathcal{O}(\varepsilon ^3 \zeta).
\end{align}
The metric function $g_{\mu \nu} ^{(0,0)}, g_{\mu \nu} ^{(1,0)},$ and $g_{\mu \nu}^{(2,0)}$ are obtained from the expansion of the Kerr metric~\eqref{eq:kerrmetric} in terms of the slow rotation parameter up to quadratic order.
Note that $g_{\mu \nu} ^ {(m,n)} \propto \varepsilon ^m \zeta ^n$. We can express the metric function $g_{\mu \nu}^{(m,n)}$ in the Regge--Wheeler gauge\footnote{While we are interested in non-linear perturbations, at each order, the highest order corrections can be treated as linear perturbations with source terms from the lower order corrections.
Thus, we can use the form of metric functions in Appendix~\ref{appB} at each order.
We note that the definition of the functions $H_{0\ell}^{(m,n)}$ and $H_{2\ell}^{(m,n)}$ in Eq.~\eqref{eq:perturbation} are slightly different from those in Appendix~\ref{appB}.
}
\begin{align}
     g_{\mu \nu} ^{(m,n)} =
     \sum_{\ell = 0} ^{\infty}
     \left(
    \begin{array}{cccc}
      f(r) H_{0\ell}^{(m,n)}(r) & H_{1\ell}^{(m,n)}(r) & 0 & h_{0\ell}^{(m,n)}(r)\sin \theta \partial _{\theta} \\
        & H_{2\ell}^{(m,n)}(r)/f(r) & 0 & h_{1\ell}^{(m,n)}(r)\sin \theta \partial _{\theta} \\
       \mathrm{symm. } &   & r^2 K_{\ell}^{(m,n)}(r) & 0 \\
       & & & r^2 \sin ^2\theta K_{\ell}^{(m,n)}(r)
    \end{array}
  \right) Y_{\ell 0 }(\theta),
  \label{eq:perturbation}
\end{align}
where $Y_{\ell 0}$ is the spherical harmonics with vanishing azimuthal number.
Note that we consider only $Y_{\ell 0}$ perturbation because we assume the axisymmetry of the system.
In a similar way, we expand $\vartheta$ and $\varphi$ as follows
\begin{align}
    \vartheta &= \alpha \left[
    \vartheta^{(0,\frac{1}{2})} +
    \varepsilon   \vartheta^{(1,\frac{1}{2})} +
    \varepsilon^2   \vartheta^{(2,\frac{1}{2})}
    \right]
    + \mathcal{O} (\varepsilon ^3 \alpha ),
    \\
    \varphi &= \alpha \left[
    \varphi^{(0,\frac{1}{2})} +
    \varepsilon \varphi^{(1,\frac{1}{2})} +
    \varepsilon^{2} \varphi^{(2,\frac{1}{2})}
    \right] + \mathcal{O}(\varepsilon^3 \alpha).
\end{align}
Because the source terms of the equation of motion for scalar fields are the order $\mathcal{O}(\alpha)$ (see Eqs.~\eqref{eq:eomtheta} and~\eqref{eq:eomphi}), the leading terms of the scalar fields also are proportional to $\alpha$ in the above equations. Also, because ${\cal O}(\alpha) = {\cal O}(\zeta^{1/2})$,
we labeled the order of $\zeta$ on the scalar fields as $1/2$.
Here, we list up the scalar fields that can affect to the metric solution up to $\mathcal{O}(\varepsilon^2 \zeta)$.

\section{slowly rotating black holes in dCS gravity}
\label{sec:dCSgravity}

For dCS gravity, the field equations are given by
\begin{align}
    G_{\mu \nu}
    & =
    \frac{1}{2\kappa} T^{\vartheta}_{\mu \nu}
    - \frac{\alpha}{\kappa} \mathcal{C}_{\mu \nu}
    \equiv T^{\mathrm{dCS}}_{\mu \nu},
    \label{eq:fieldeq}
    \\
    \mathcal{C}^{\mu \nu}
    &\equiv
    (\nabla _{\sigma} \vartheta) \epsilon ^{\sigma \delta \alpha ( \mu} R^{\nu )}\,_{\delta} + (\nabla _{\sigma} \nabla _{\delta} \vartheta) \widetilde R ^{\delta (\mu \nu) \sigma},
    \label{eq:ctensor}
    \\
    T^{\vartheta}_{\mu \nu}
    &\equiv
    (\nabla_{\mu} \vartheta)(\nabla_{\nu} \vartheta) - \frac{1}{2} g_{\mu \nu}  (\nabla_{\delta} \vartheta)
    (\nabla ^{\delta} \vartheta),
    \\
    \Box \vartheta
    &=
    - \frac{\alpha}{4} R_{\nu \mu \rho \sigma} \widetilde R^{\mu \nu \rho \sigma}.
    \label{eq:eomtheta}
\end{align}
In Eq.~\eqref{eq:eomtheta},
we can expand the source terms by only spin parameter as follows
\begin{align}
    \alpha R_{\nu \mu \rho \sigma} \widetilde R^{\mu \nu \rho \sigma}
    =
    \alpha
    \left(
    \frac{288 M^3 \varepsilon \cos{\theta}}{r^7}
    \left( 1 - \frac{28M^2}{3 r^2} \varepsilon^2 \cos^2 \theta  \right)
    +\mathcal{O}(\varepsilon ^5)
    \right)
    + \mathcal{O}(\alpha^3),
\end{align}
where we used that the metric is expanded by $\zeta$ as in Eq.~\eqref{eq:expandzeta}.
Notice that the source term contains only odd power of $\varepsilon$.

We assume the background spherically symmetric spacetime is the Schwarzschild spacetime.
Under this assumption,
the only second term of the $\mathcal{C}$-tensor~\eqref{eq:ctensor} is relevant for the metric solution to $\mathcal{O}(\zeta)$:
\begin{align}
  \frac{\alpha}{\kappa}\mathcal{C}_{\mu \nu}
  =
  \frac{\alpha}{\kappa}(\nabla ^{\sigma} \nabla ^{\delta}\, \vartheta) \widetilde R _{\delta (\mu \nu) \sigma} + \mathcal{O}(\zeta ^2).
\end{align}
We expand the scalar filed with the spherical harmonics as $\vartheta^{(m,\frac{1}{2})} = \sum_{\ell}\Theta^{(m,\frac{1}{2})}_{\ell}(t,r) Y_{\ell 0}(\theta)$.
In the following subsections,
we see that the time dependence of $\vartheta$ is determined from the stationarity of the effective stress energy tensor $T_{\mu \nu}^{\mathrm{dCS}}$:
\begin{align}
    \partial_t T_{\mu \nu}^{\mathrm{dCS}}=0.
    \label{eq:stationarycondcs}
\end{align}
In addition,
we impose the following condition
\begin{align}
    T_{ti}^{\mathrm{dCS}} |_{\ell=0}=0,
    \label{eq:Tdcsti}
\end{align}
where $i=r,\theta$, and $\phi$, because these components of the Einstein tensor vanish for stationary spacetimes for $\ell=0$.

\subsection{$\mathcal{O}(\zeta)$ corrections}

Supposing that in the limit $\varepsilon \to 0$,
the spacetime possesses the spherical symmetry,
we need to consider only $\ell=0$ mode.\footnote{In dCS gravity, the static uniqueness of the metric with stationary scalar fields is proven in~\cite{Shiromizu:2013pna}.
}
Thus, we can express $\vartheta ^{(0,\frac{1}{2})} (t,r) = \Theta_{0}^{(0,\frac{1}{2})}(t,r) Y_{00}$.
Since on any spherical symmetric spacetime,
the $\mathcal{C}$-tensor with $\ell=0$ mode is identically zero,
the effective stress energy tensor contains only $T_{\mu \nu}^{\vartheta}$.
The $tr$-component of the stationary condition~\eqref{eq:stationarycondcs} becomes
\begin{align}
    \partial_t\Theta_{0}^{(0,\frac{1}{2})}
    \,\partial_t \partial_r \Theta_{0}^{(0,\frac{1}{2})} + \partial_r \Theta_{0}^{(0,\frac{1}{2})} \partial^2 _t \Theta_{0}^{(0,\frac{1}{2})} = 0.
\end{align}
The general solution is given by
\begin{align}
    \Theta_{0}^{(0,\frac{1}{2})}(t,r)
    =
    \mathcal{F}_{0}^{(0,\frac{1}{2})}(r) + t\, \mathcal{G}_{0}^{(0,\frac{1}{2})}(r) + \mathcal{H}_{0}^{(0,\frac{1}{2})}(t),
\end{align}
where $\mathcal{F}_{0}^{(0,\frac{1}{2})}$ and $\mathcal{G}_{0}^{(0,\frac{1}{2})}$ are arbitrary functions of $r$, while $\mathcal{H}_{0}^{(0,\frac{1}{2})}$ is an arbitrary function of $t$.
After substituting this into Eq.~\eqref{eq:stationarycondcs},
we solve $tt$-component and $tr$-component of Eq.~\eqref{eq:stationarycondcs} simultaneously with respect to $\mathrm{d} \mathcal{H}_{0}^{(0,\frac{1}{2})} / \mathrm{d} t$ and $\mathrm{d}^2 \mathcal{H}_{0}^{(0,\frac{1}{2})} / \mathrm{d} t^2$, then we have
\begin{align}
    \frac{\mathrm{d}^2 \mathcal{H}_{0}^{(0,\frac{1}{2})}(t)}{\mathrm{d} t^2}
    = \left ( 1 - \frac{2M}{r} \right) \frac{\mathrm{d} \mathcal{G}_{0}^{(0,\frac{1}{2})}(r)}{\mathrm{d} r}.
\end{align}
Since the left hand side is a function of only $t$ and the right hand side is a function of only $r$,
both sides have to be equal to a constant $\mathtt{d}_{0}^{(0,\frac{1}{2})}$.
We obtain the expressions for $\mathcal{G}_{0}^{(0,\frac{1}{2})}$ and $\mathcal{H}_{0}^{(0,\frac{1}{2})}$ as follows
\begin{align}
    \mathcal{G}_{0}^{(0,\frac{1}{2})}(r)
    &=
    \mathtt{a}_{0}^{(0,\frac{1}{2})}
    + \mathtt{d}_{0}^{(0,\frac{1}{2})} [r + \log (r-2M)],
    \\
    \mathcal{H}_{0}^{(0,\frac{1}{2})}(t)
    &=
    \frac{1}{2} \mathtt{d}_{0}^{(0,\frac{1}{2})} t^2 + \mathtt{b}_{0}^{(0,\frac{1}{2})} t + \mathtt{c}_{0}^{(0,\frac{1}{2})},
\end{align}
where $\mathtt{a}_{0}^{(0,\frac{1}{2})}$, $\mathtt{b}_{0}^{(0,\frac{1}{2})}$, and $\mathtt{c}_{0}^{(0,\frac{1}{2})}$ are integration constants.
Substituting $\mathcal{G}_{0}^{(0,\frac{1}{2})}$ and $\mathcal{H}_{0}^{(0,\frac{1}{2})}$ into Eq.~\eqref{eq:stationarycondcs},
we obtain $\mathtt{d}_{0}^{(0,\frac{1}{2})}=0$.
Consequently, we obtain the form of $\Theta_{0}^{(0,\frac{1}{2})}$
\footnote{
Solving the equation of motion~\eqref{eq:eomtheta} with~\eqref{eq:Theta01_20}, we obtain the configuration of the scalar field which is regular at $r=2M$: $\vartheta ^{(0,\frac{1}{2})} \sim q_{0} v$ where $v$ is an advanced null coordinate expressed as $v \sim t + 2M \log(r-2M)$.
However, we need to set $q_0=0$ to satisfy the condition~\eqref{eq:Tdcsti}. }
:
\begin{align}
    \Theta_{0}^{(0,\frac{1}{2})}(t,r) =
    \mathcal{F}_{0}^{(0,\frac{1}{2})}(r) +
    (\mathtt{a}_{0}^{(0,\frac{1}{2})} + \mathtt{b}_{0}^{(0,\frac{1}{2})}) t + \mathtt{c}_{0}^{(0,\frac{1}{2})}.
    \label{eq:Theta01_20}
\end{align}
Substituting this into Eq.~\eqref{eq:Tdcsti},
from the $tr$-component, we have
\begin{align}
    \frac{\mathrm{d}\mathcal{F}_{0}^{(0,\frac{1}{2})}(r)}{\mathrm{d}r}
    \left( \mathtt{a}_{0}^{(0,\frac{1}{2})} + \mathtt{b}_{0}^{(0,\frac{1}{2})} \right)
    =0.
\end{align}
There are two branches to satisfy this equation: $\mathcal{F}_{0}^{(0,\frac{1}{2})}(r) = \mathrm{const.}$ or $ \mathtt{a}_{0}^{(0,\frac{1}{2})} =- \mathtt{b}_{0}^{(0,\frac{1}{2})}$.
We choose the former branch and then the scalar field depends $t$ only
\begin{align}
    \vartheta^{(0,\frac{1}{2})} (t)
    =
    q_0 t + \mathrm{const.},
    \label{eq:vartheta1_20}
\end{align}
where $q_0(= [\mathtt{a}_{0}^{(0,\frac{1}{2})} + \mathtt{b}_{0}^{(0,\frac{1}{2})}]Y_{00})$.
The expression for the scalar field~\eqref{eq:vartheta1_20} is a solution of the equation of motion~\eqref{eq:eomtheta}.
We note that the scalar field $\vartheta ^{(0,\frac{1}{2})}$ becomes stationary when we choose the later branch.
Solving the equation of motion~\eqref{eq:eomtheta} with this stationary scalar field configuration and imposing the regularity condition at $r=2M$,
the scalar field becomes trivial: $\vartheta^{(0,\frac{1}{2})} = \mathrm{const.}$

Let us move on to the discuss on the metric.
For $\ell=0$ mode, the non-vanishing components of the metric perturbation are $H_{00}^{(0,1)}$ and $H_{20}^{(0,1)}$.
We solve the $tt$- and $rr$-components of the field equations~\eqref{eq:fieldeq} to find $H_{00}^{(0,1)}$ and $H_{20}^{(0,1)}$:
\begin{align}
    H_{00}^{(0,1)}(r)
    &=
    -\frac{M^4 \sqrt{\pi}}{3} q_{0}^{2} r^2 \left[ 1 + \frac{15M}{2r} - \frac{84M^3}{r^3} \log(r-2M) - \frac{6M^2}{r^2} (7 - 6 \log(r-2M)) \right] - \frac{c_1^{(0,1)}}{r f} + c_2^{(0,1)},
    \\
    H_{20}^{(0,1)}(r)
    &=
    \frac{M^4 \sqrt{\pi}}{6} q_{0}^2 r^2
    \left[
    1 + \frac{3M}{r} + \frac{12M^2}{r^2} + \frac{24M^3}{r^3} \log(r-2M)
    \right] - \frac{c_1^{(0,1)}}{r f},
\end{align}
where $c_1^{(0,1)}$ and $c_2^{(0,1)}$ are integration constants.
The logarithmic dependence in $H_{00}^{(0,1)}$ and $H_{20}^{(0,1)}$ leads to the divergence of the  Kretschmann invariant $R_{\mu \nu \rho \lambda}R^{\mu \nu \rho \lambda}$ at $r=2M$.
In order to remove this divergence,
we need to set $q_{0}=0$,
which implies $\vartheta^{(0,\frac{1}{2})} = \mathrm{const}$.
We set $c_1^{(0,1)}=c_2^{(0,1)}=0$ so that the metric to be asymptotically flat at the spatial infinity, and the mass of the system does not change from $M$.

\subsection{$\mathcal{O}(\varepsilon \zeta)$ corrections}

Because the stress energy tensor of the scalar field $T_{\mu \nu}^{\vartheta}$ is always of order of $\mathcal{O}(\varepsilon ^2 \zeta)$, $T_{\mu \nu}^{\mathrm{dCS}}$ contains only $\mathcal{C}$-tensor~\eqref{eq:ctensor}.
At the order $\mathcal{O}(\varepsilon \zeta)$,
$\mathcal{C}$-tensor is identically zero for $\ell=0$ mode.
This means that we cannot obtain the form of the scalar field with $\ell=0$ from the stationary condition~\eqref{eq:stationarycondcs}.
As shown later, the time dependence of $\vartheta^{(1,\frac{1}{2})}$ for $\ell=0$ is determined by the stationary condition of the order $\mathcal{O}(\varepsilon ^2 \zeta)$.

For $\ell \ge 1$, the $\theta \theta$-component of the stationary condition~\eqref{eq:stationarycondcs} becomes $\partial_t ^2 \Theta^{(1,\frac{1}{2})}_{\ell}(t,r)=0$.
The general solution is given by
\begin{align}
   \Theta^{(1,\frac{1}{2})}_{\ell}(t,r) = \mathcal{F}^{(1,\frac{1}{2})}_{\ell}(r) + t\, \mathcal{G}^{(1,\frac{1}{2})}_{\ell}(r).
\end{align}
Substituting this into Eq.~\eqref{eq:stationarycondcs}, we obtain
\begin{align}
  r\, \frac{\mathrm{d}\mathcal{G}^{(1,\frac{1}{2})}_{\ell}(r)}{\mathrm{d}r} -  \mathcal{G}^{(1,\frac{1}{2})}_{\ell}(r) = 0.
\end{align}
The general solution is
\begin{align}
 \mathcal{G}^{(1,\frac{1}{2})}_{\ell}(r) = \mathtt{a}^{(1,\frac{1}{2})}_{\ell} \,r,
\end{align}
where $\mathtt{a^{(1,\frac{1}{2})}_{\ell}}$ is an integration constant.
We consider the cases of $\ell=1$ and $\ell \ge 2$, separately.
For $\ell = 1$, the equation of motion for the scalar field~\eqref{eq:eomtheta} can be written as
\begin{align}
    \mathtt{a}^{(1,\frac{1}{2})}_{1}\, t = \frac{24 M^3 \sqrt{3 \pi}}{r^5} - \mathcal{F}^{(1,\frac{1}{2})}_{1}(r) + (r-M)\frac{\mathrm{d}\mathcal{F}^{(1,\frac{1}{2})}_{1}(r)}{\mathrm{d}r} + \frac{2(r-2M)}{2} \frac{\mathrm{d}^2 \mathcal{F}^{(1,\frac{1}{2})}_{1}(r)}{\mathrm{d}r^2},
    \label{eq:c1eqoml1}
\end{align}
which means $\mathtt{a}^{(1,\frac{1}{2})}_{1}=0$ so that the above equation holds for all $t$ and $r$.
By solving Eq.~\eqref{eq:c1eqoml1},
we obtain the form of $\mathcal{F}^{(1,\frac{1}{2})}_{1}$:
\begin{align}
    \mathcal{F}^{(1,\frac{1}{2})}_{1}(r) =
    \frac{5\sqrt{3 \pi}}{12r^2}
    \left(
    1 + \frac{2M}{r} + \frac{18M^2}{5r^2}
    \right).
\end{align}
We chose the integration constants as the scalar field is regular everywhere.
For $\ell \ge 2$,
we have $\mathtt{a}^{(1,\frac{1}{2})}_{\ell}=0$ from the similar analysis of the equation of motion for the scalar field~\eqref{eq:eomtheta}.
Then, Eq.~\eqref{eq:eomtheta} reduces to
\begin{align}
   \ell (\ell + 1) \mathcal{F}^{(1,\frac{1}{2})}_{\ell}(x) - (2x+1) \frac{\mathrm{d} \mathcal{F}^{(1,\frac{1}{2})}_{\ell}(x)}{\mathrm{d} x} - x(x+1) \frac{\mathrm{d}^2 \mathcal{F}^{(1,\frac{1}{2})}_{\ell}(x)}{\mathrm{d}x^2} = 0,
   \label{eq:Fequation}
\end{align}
where we introduce a new variable $x \equiv r/2M -1$.
The general solution is given by
\begin{align}
    \mathcal{F}^{(1,\frac{1}{2})}_{\ell}(x) = \mathtt{b}^{(1,\frac{1}{2})}_{\ell} P_{\ell}(1+2x) + \mathtt{c}^{(1,\frac{1}{2})}_{\ell}\, Q_{\ell}(1+2x),
    \label{eq;Fsolution}
\end{align}
where $\mathtt{b}^{(1,\frac{1}{2})}_{\ell}$ and $\mathtt{c}^{(1,\frac{1}{2})}_{\ell}$ are integration constants, and $P_{\ell}$ and $Q_{\ell}$ are Legendre polynomials of first kind and second kind, respectively.
Requiring that the scalar field is regular everywhere,
we should set $\mathtt{b}^{(1,\frac{1}{2})}_{\ell}=\mathtt{c}^{(1,\frac{1}{2})}_{\ell}=0$.\footnote{For large $x$, Legendre polynomial of first kind diverges as $P_\ell(1+2x) \sim x^\ell$.
For small $x$, Legendre polynomial of second kind diverge because it  behaves as $Q_{\ell}(1+2x) \sim \log\left( (x+1)/x \right)$.}
Therefore, the configuration of the scalar field $\vartheta^{(1,\frac{1}{2})}$ for $\ell \ge 1$ is given by
\begin{align}
    \vartheta^{(1,\frac{1}{2})} (r,\theta) = \frac{5}{8} \alpha \varepsilon \frac{\cos \theta}{r^2} \left( 1 + \frac{2M}{r} + \frac{18M^2}{5r^2} \right).
\end{align}
We note again that the time dependence of $\vartheta^{(1,\frac{1}{2})}$ for $\ell=0$ is determined by the stationary condition of the order $\mathcal{O}(\varepsilon ^2 \zeta)$.

We derive the metric solution of the order $\mathcal{O}(\varepsilon \zeta)$.
The the effective stress energy tensor has only non-trivial component $T_{t \phi}^{\mathrm{dCS}}$ for $\ell=1$.
The $t\phi$-component of the field equation~\eqref{eq:fieldeq} for $\ell=1$ becomes
\begin{align}
    r^2 \frac{\mathrm{d}^2 h_{01}^{(1,1)}}{\mathrm{d}r^2}
    - 2 \frac{\mathrm{d}h_{01}^{(1,1)}}{\mathrm{d}r} = - \frac{15\sqrt{3\pi} M^5}{2r^4}
    \left(
    1 + \frac{8M}{3r} + \frac{6M^2}{r^2}
    \right),
\end{align}
and the solution is given by
\begin{align}
    h_{01}^{(1,1)}(r) = -\frac{5 \sqrt{3\pi} M^5}{4r^4}
    \left(
    1 + \frac{12M}{7r} + \frac{27M^2}{10r^2}
    \right)
    +
    \frac{c_1^{(1,1)}}{r}
    +
    c_2^{(1,1)}r^2,
    \label{eq:oddh0111dcs}
\end{align}
where $c_1^{(1,1)}$ and $c_2^{(1,1)}$ are integration constants.
We set $c_1^{(1,1)}=c_2^{(1,1)}=0$ so that the metric is asymptotically flat at the spatial infinity and the amplitude of the angular momentum does not change from $Ma$.
For $\ell=0$ and $\ell \ge 2$, the field equations~\eqref{eq:fieldeq} are homogeneous equations, the solutions are discussed in Appendix~\ref{appC}.
The metric perturbations for $\ell=0$ and  $\ell \ge 2$ vanish by imposing that the spacetime is asymptotically flat and regular at the horizon.
The result~\eqref{eq:oddh0111dcs} with $c_1^{(1,1)}$ = $c_2^{(1,1)} = 0$ coincides with the metric solution obtained in~\cite{Yunes:2009hc}.
Note that the spacetime is circular up to this order.

\subsection{$\mathcal{O}(\varepsilon^2 \zeta)$ corrections}

For $\ell=0$ mode,
the $t\theta$-component of the stationary condition~\eqref{eq:stationarycondcs} is
\begin{align}
    2M \partial^2_t\Theta_{0}^{(1,\frac{1}{2})}(t,r) +
    r(r-2M) \partial^2_t \partial_r \Theta_{0}^{(1,\frac{1}{2})}(t,r) = 0.
\end{align}
The general solution is given by
\begin{align}
    \Theta_{0}^{(1,\frac{1}{2})}(t,r)
    =
    \mathcal{F}_{0}^{(1,\frac{1}{2})}(r)
    +
    t\, \mathcal{G}_{0}^{(1,\frac{1}{2})}(r)
    +
    \frac{r}{r-2M} \mathcal{H}_{0}^{(1,\frac{1}{2})}(t).
\end{align}
Substituting this into Eq.~\eqref{eq:stationarycondcs},
we solve the $r \theta$-component with respect to $\mathrm{d} \mathcal{G}_{0}^{(1,\frac{1}{2})} /\mathrm{d}r$:
\begin{align}
    \frac{\mathrm{d}\mathcal{G}_{0}^{(1,\frac{1}{2})}(r)}{\mathrm{d}r}
    =
    \frac{1}{(r-2M)^2}
    \left (
    2M \frac{\mathrm{d}\mathcal{H}_{0}^{(1,\frac{1}{2})}(t)}{\mathrm{d} t} - r^3 \frac{\mathrm{d}^3 \mathcal{H}_{0}^{(1,\frac{1}{2})}(t)}{\mathrm{d}t^3}
    \right).
    \label{eq:derivativec3}
\end{align}
Since the left hand side is a function of $r$,
the right hand side also has to be a function of $r$. From that condition, we have
\begin{align}
    \mathcal{H}_{0}^{(1,\frac{1}{2})}(t)
    =
    \mathtt{a}_{0}^{(1,\frac{1}{2})}
    +
    \mathtt{b}_{0}^{(1,\frac{1}{2})} t,
\end{align}
where $\mathtt{a}_{0}^{(1,\frac{1}{2})}$ and $\mathtt{b}_{0}^{(1,\frac{1}{2})}$ are constants.
Then we can integrate Eq.~\eqref{eq:derivativec3} to obtain the expression of $\mathcal{G}_{0}^{(1,\frac{1}{2})}$:
\begin{align}
    \mathcal{G}_{0}^{(1,\frac{1}{2})}(r) = - \frac{2M \mathtt{b}_{0}^{(1,\frac{1}{2})}}{r-2M} + \mathtt{c}_{0}^{(1,\frac{1}{2})},
\end{align}
where $\mathtt{c}_{0}^{(1,\frac{1}{2})}$ is an integration constant.
Thus, we have
\begin{align}
    \Theta_{0}^{(1,\frac{1}{2})}(t,r)
    =
    \mathcal{F}_{0}^{(1,\frac{1}{2})}(r)
    +
    \frac{\mathtt{a}_{0}^{(1,\frac{1}{2})} r}{r-2M}
    +
    \left( \mathtt{b}_{0}^{(1,\frac{1}{2})} + \mathtt{c}_{0}^{(1,\frac{1}{2})} \right) t.
\end{align}
{}From the regularity of the scalar field at $r=2M$, we need to set $\mathtt{a}_{0}^{(1,\frac{1}{2})}=0$.
From the $tr$-component of Eq.~\eqref{eq:Tdcsti},
we have
\begin{align}
    \frac{\mathrm{d}\mathcal{F}_{0}^{(1,\frac{1}{2})}(r)}{\mathrm{d}r}
    \left( \mathtt{b}_{0}^{(1,\frac{1}{2})} + \mathtt{c}_{0}^{(1,\frac{1}{2})} \right)
    =0.
\end{align}
We choose the branch in which $\mathcal{F}_{0}^{(1,\frac{1}{2})} = \mathrm{const. }$ for the same reason as for the case of $\vartheta ^{(0,\frac{1}{2})}$, i.e,
the other branch in which $\mathtt{b}_{0}^{(1,\frac{1}{2})} = - \mathtt{c}_{0}^{(1,\frac{1}{2})}$ makes the $\ell=0$ mode of the scalar field $\vartheta ^{(1,\frac{1}{2})}$ trivial: $\vartheta ^{(1,\frac{1}{2})}|_{\ell=0}=\mathrm{const.}$
Therefore, the scalar field $\vartheta^{(1,\frac{1}{2})}$ is given by
\begin{align}
    \vartheta^{(1,\frac{1}{2})} (t,r,\theta)
    =
    \alpha \varepsilon
    \left[
    q\, t
    +
    \frac{5 \cos \theta}{8r^2} \left(  1 + \frac{2M}{r} + \frac{18M^2}{5r^2} \right)
    +
    \mathrm{const}.
    \right],
    \label{eq:thetaqt}
\end{align}
where $q(=[\mathtt{b}_{0}^{(1,\frac{1}{2})} + \mathtt{c}_{0}^{(1,\frac{1}{2})}]Y_{00})$ is a constant.

We discuss the metric function $g_{\mu \nu}^{(2,1)}$.
First, we focus on the non-circular component $H_{1 \ell}^{(2,1)}$. From the $tr$-component of the field equation~\eqref{eq:fieldeq} for $\ell =1$, we obtain $H_{11}^{(2,1)}$ as follows
\begin{align}
    H_{11}^{(2,1)}(r) = \frac{5 M^4 \sqrt{3 \pi} q}{12 f r}
    \left( 1 + \frac{M}{r} + \frac{6M^2}{5r^2} - \frac{24 M^3}{5 r^3} \right).
\end{align}
Since $H_{11}^{(2,1)}$ is proportional to $q$,
we can see that the non-circularity comes from the linear time dependence of the scalar field $\vartheta^{(1,\frac{1}{2})}$.
For $\ell \ge 2$, the equations for $H_{1\ell}^{(2,1)}$ become homogeneous, and
these can be solved as $H_{1\ell}^{(2,1)} = 0$. We note that $H_{1\ell}^{(2,1)} = 0$ for $\ell = 0$ from the gauge condition.

Next, we discuss the other metric components.
For $\ell =0$, we solve $tt$-component of the field equation~\eqref{eq:fieldeq}, which is a first order ordinary differential equation for $H_{20}^{(2,1)}$ and use $H_{20}^{(2,1)}$ to obtain $H_{00}^{(2,1)}$.
The components $H_{00}^{(2,1)}$ and $H_{20}^{(2,1)}$ are given by
\begin{align}
    H_{00}^{(2,1)}(r) &=
    -\frac{5 \sqrt{\pi}}{192} \varepsilon ^{2} \zeta \frac{M^{5}}{r^{5} f}
    \left(
    1
    + \frac{100M}{r}
    + \frac{194M^{2}}{r^{2}}
    + \frac{2220 M^{3}}{7 r^{3}}
    - \frac{1512 M^{4}}{5 r^{4}}\right)
    - \frac{c_1^{(2,1)}}{rf}
    + c_2^{(2,1)}
    \notag\\
    &- \frac{M^4 q^2 \sqrt{\pi}}{3 f} r^2 \varepsilon ^2 \zeta \left[
    1
    + \frac{15 M}{2 r}
    - \left( 19 - 36 \log (r-2M) \right) \frac{M^2}{r^2}
    - \left( 46 + 84 \log (r-2M) \right) \frac{M^3}{r^3}
    \right],
    \\
    H_{20}^{(2,1)}(r) &=
    -\frac{25 \sqrt{\pi}}{192}  \varepsilon^{2} \zeta \frac{M^{4}}{r^{4} f}
    \left(1
    + \frac{3 M}{r}
    + \frac{322 M^{2}}{5 r^{2}}
    + \frac{198 M^{3}}{5 r^{3}}
    + \frac{6276 M^{4}}{175 r^{4}}
    - \frac{17496 M^{5}}{25 r^{5}}
    \right)
    - \frac{c_1^{(2,1)}}{rf}
    \notag\\
    & - \frac{M^4 q^2 \sqrt{\pi}}{6 f}r^2 \varepsilon ^2 \zeta
    \left[ 1
    + \frac{3 M}{r}
    + \frac{12 M^2}{r^2}
    + \frac{24 M^3}{r^3} \log (r-2M)
    \right],
\end{align}
where $c_1^{(2,1)}$ and $c_2^{(2,1)}$ are integration constants.
For $\ell =2$, there is no contribution from the time dependence of the scalar field because it appears only in $\ell=0$ mode (see Eq.~\eqref{eq:thetaqt}).
Thus, $H_{02}^{(2,1)}$, $H_{22}^{(2,1)}$, and $K_{2}^{(2,1)}$ are given by the same forms obtained in~\cite{Yagi:2012ya}.\footnote{
The functions $H_{02}^{(2,1)}$, $H_{22}^{(2,1)}$, and $K_{2}^{(2,1)}$
correspond to the functions $H_{020}, H_{220}$ and $K_{20}$ in Eqs.(A18)-(A20) in~\cite{Yagi:2012ya}, respectively.}

Because the components $H_{00}^{(2,1)}$ and $H_{20}^{(2,1)}$ include the logarithmic divergence as $r \to 2M$,
the non-circular spacetime obtained in this paper is singular at $r=2M$.
Indeed, the Kretschmann invariant $R_{\mu \nu \rho \lambda} R^{\mu \nu \rho \lambda}$ diverges at $r=2M$.
In dCS gravity, therefore, there does not exist the non-circular regular black hole solution at least around the Schwarzschild spacetime up to the order $\mathcal{O}(\varepsilon^2 \zeta)$.

\section{slowly rotating black holes in ESGB gravity}
\label{sec:ESGBgravity}

For ESGB gravity, the field equations are given by
\begin{align}
    G_{\mu \nu}
    &=  \frac{1}{2\kappa} T^{\varphi}_{\mu \nu}
    - \frac{\alpha}{\kappa} \mathcal{D}_{\mu \nu}
    \equiv T^{\mathrm{ESGB}}_{\mu \nu},
    \label{fieldeqesgb}
    \\
    \mathcal{D}_{\mu \nu}
    &\equiv -2 (\nabla_{\mu} \nabla_{\nu}\varphi) R
    + 2(\nabla ^{\sigma} \nabla _{\sigma} \varphi) (g_{\mu \nu}R - 2R_{\mu \nu})
    \notag\\
    &\,\,\,\,\,\,\,
    + 8 R_{\sigma (\mu} \nabla^\sigma \nabla_{\nu )} \varphi
    - 4 g_{\mu \nu} (\nabla_{\sigma} \nabla_{\delta} \varphi) R^{\sigma \delta}
    + 4 (\nabla^{\sigma} \nabla^{\delta} \varphi) R_{\mu \sigma \nu \delta} ,
    \label{eq:Dtensor}
    \\
    T^{\varphi}_{\mu \nu}
    &\equiv  (\nabla_{\mu} \varphi)(\nabla_{\nu} \varphi) - \frac{1}{2} g_{\mu \nu}  (\nabla_{\delta} \varphi) (\nabla ^{\delta} \varphi),
    \\
    \Box \varphi
    &= - \alpha(R^2 -4 R_{\mu \nu}R^{\mu \nu} + R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma}).
    \label{eq:eomphi}
\end{align}
We consider the perturbed solution around the Schwarzschild spacetime.
We can expand the source term of Eq.~\eqref{eq:eomphi} in terms of only the spin parameter as follows
\begin{align}
     \alpha(R^2 -4 R_{\mu \nu}R^{\mu \nu} + R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma})
    =
    \alpha
    \left(
    \frac{48 M^2}{r^6}
    \left( 1 - \frac{21  M^2 }{r^2} \varepsilon ^2 \cos^2 \theta \right)
    +\mathcal{O}(\varepsilon ^4)
    \right)
    + \mathcal{O}(\alpha^3),
\end{align}
where we used that the metric is expanded by $\zeta$ as in Eq.~\eqref{eq:expandzeta}.
Note that the source term contains the even power of $\varepsilon$.

When we focus on the perturbed solution around the Schwarzschild spacetime,
only the last term of $\mathcal{D}$-tensor~\eqref{eq:Dtensor} is relevant for the linear order of $\zeta$:
\begin{align}
    \frac{\alpha}{\kappa} \mathcal{D}_{\mu \nu}
    =
    \frac{4 \alpha}{\kappa} (\nabla^{\sigma} \nabla^{\delta} \varphi) R_{\mu \sigma \nu \delta}
    +
    \mathcal{O}(\zeta^2).
\end{align}
We expand the scalar field with the spherical harmonics as $\varphi^{(m,\frac{1}{2})} = \sum_{\ell} \Phi^{(m,\frac{1}{2})}_{\ell}(t,r) Y_{\ell0}(\theta) $.
As we discussed in the previous section,
we impose the stationary condition to the effective stress energy tensor $T_{\mu \nu}^{\mathrm{ESGB}}$:
\begin{align}
    \partial _t T_{\mu \nu}^{\mathrm{ESGB}} = 0.
    \label{eq:stationaryconesgb}
\end{align}
We also should impose
\begin{align}
    T_{ti}^{\mathrm{ESGB}} |_{\ell=0}=0,
    \label{eq:Tti}
\end{align}
where $i=r,\theta$, and $\phi$, because these components of the Einstein tensor vanish for stationary spacetimes for $\ell=0$.
While the stationary condition~\eqref{eq:stationaryconesgb} admits linear time dependence of the scalar fields, such behavior is prohibited from the condition~\eqref{eq:Tti} as shown in the following discussion.

\subsection{$\mathcal{O}(\zeta)$ corrections}

We assume that the spacetime possesses the spherical symmetry in the limit $\varepsilon \to 0$. This means that we can express $\varphi^{(0,\frac{1}{2})}(t,r) = \Phi^{(0,\frac{1}{2})}_{0}(t,r)Y_{00}$.
Combining the $tr$-component of Eq.~\eqref{eq:Tti} and its time derivative, we obtain\footnote{
First we solve $tr$-component of Eq.~\eqref{eq:Tti} with respect to $\partial_r \Phi^{(0,1/2)}$,
and substitute it to the time derivative of the $tr$-component of Eq.~\eqref{eq:Tti}. Then, we can solve the differential equation and the general solution is Eq.~\eqref{eq:Phi01_20tr}.
}
\begin{align}
    \Phi^{(0,\frac{1}{2})}_{0}(t,r)
    =
    \mathcal{I}^{(0,\frac{1}{2})}_{0}(r)
    + t \mathcal{J}^{(0,\frac{1}{2})}_{0}(r)
    + t^2 \mathcal{K}^{(0,\frac{1}{2})}_{0}(r)
    + \mathcal{L}^{(0,\frac{1}{2})}_{0}(t),
    \label{eq:Phi01_20tr}
\end{align}
where $\mathcal{I}^{(0,\frac{1}{2})}_{0}$, $\mathcal{J}^{(0,\frac{1}{2})}_{0}$, and $\mathcal{K}^{(0,\frac{1}{2})}_{0}$ are arbitrary functions of $r$, while $\mathcal{L}^{(0,\frac{1}{2})}_{0}$ is an arbitrary function of $t$.
Substituting this expression into Eq.~\eqref{eq:Tti} again,
we can solve the $tr$-component with respect to $\mathrm{d} \mathcal{L}^{(0,\frac{1}{2})}_{0}/\mathrm{d}t$ as
\begin{align}
    \frac{\mathrm{d}\mathcal{L}^{(0,\frac{1}{2})}_{0}(t)}{\mathrm{d}t}
    =
    -\frac{32M\sqrt{\pi}}{r^3 {\cal A}(t,r)}
    \left(
    \frac{\mathrm{d}\mathcal{J}^{(0,\frac{1}{2})}_{0}(r)}{\mathrm{d}r} + 2t \frac{\mathrm{d}\mathcal{K}^{(0,\frac{1}{2})}_{0}(r)}{\mathrm{d}r}
    \right)
    -
    \left(
    \mathcal{J}^{(0,\frac{1}{2})}_{0}(r) + 2t \mathcal{K}^{(0,\frac{1}{2})}_{0}(r)
    \right),
    \label{eq:LL01_20t}
\end{align}
where
\begin{align}
    {\cal A}(t,r) \equiv \frac{\mathrm{d}\mathcal{I}^{(0,\frac{1}{2})}_{0}(r)}{\mathrm{d}r} + t \frac{\mathrm{d} \mathcal{J}^{(0,\frac{1}{2})}_{0}(r)}{\mathrm{d}r} + t^2 \frac{\mathrm{d}\mathcal{K}^{(0,\frac{1}{2})}_{0}(r)}{\mathrm{d}r}-\frac{32M^2\sqrt{\pi}}{r^4(r-2M)}.
\end{align}
{}From the condition for the right hand side of Eq.~\eqref{eq:LL01_20t} to be a function of only $t$,
we obtain
\begin{align}
    \mathcal{K}^{(0,\frac{1}{2})}_{0}(r)
    &= \mathtt{f}^{(0,\frac{1}{2})}_{0},
    \\
    \mathcal{J}^{(0,\frac{1}{2})}_{0}(r)
    &= \mathtt{g}^{(0,\frac{1}{2})}_{0},
\end{align}
where $\mathtt{f}^{(0,\frac{1}{2})}_{0}$ and $\mathtt{g}^{(0,\frac{1}{2})}_{0}$ are constants.
Then, we integrate Eq.~\eqref{eq:LL01_20t} to obtain $\mathcal{L}^{(0,\frac{1}{2})}_{0}$:
\begin{align}
    \mathcal{L}^{(0,\frac{1}{2})}_{0}(t)
    =
    - t^2\, \mathtt{f}^{(0,\frac{1}{2})}_{0}
    - t\, \mathtt{g}^{(0,\frac{1}{2})}_{0}
    + \mathtt{h}^{(0,\frac{1}{2})}_{0},
\end{align}
where $\mathtt{h}^{(0,\frac{1}{2})}_{0}$ is an integration constant.
When we substitute this into Eq.~\eqref{eq:Phi01_20tr}, we obtain
\begin{align}
    \Phi^{(0,\frac{1}{2})}_{0}(r)
    =
    \mathcal{I}^{(0,\frac{1}{2})}_{0}(r) + \mathrm{const}.
    \label{eq:phi01_2r}
\end{align}
Notice that the time dependence of the scalar field is not allowed.

Substituting Eq.~\eqref{eq:phi01_2r} into the equation of motion for scalar field~\eqref{eq:eomphi} and impose the regularity condition at the horizon,
we obtain the configuration of the scalar field $\varphi^{(0,\frac{1}{2})}$
\footnote{
We imposed the regularity condition at the horizon $r = 2 M$ to obtain this expression.
Here, we implicitly assumed the existence of the black hole, i.e., $M \neq 0$,
and we need to discuss $M = 0$ case separately.
In fact, $\varphi^{(0,\frac{1}{2})}(r) = 0$ for $M = 0$ case.
Thus, we cannot take a smooth $M \to 0$ limit to this solution.
}~\cite{Yunes:2011we}:
\begin{align}
    \varphi^{(0,\frac{1}{2})}(r)
    =
    \alpha \frac{2}{Mr}
    \left(
    1 + \frac{M}{r} + \frac{4M^2}{3r^2}
    \right)
    + \mathrm{const}.
\end{align}

The metric of the order $\mathcal{O}(\zeta)$ can be obtained by solving the field equation~\eqref{fieldeqesgb}
\begin{align}
    g_{tt}^{(0,1)} (r)
    &=
    - \frac{\zeta}{3} \frac{M^3}{r^3}
    \left(
    1 + \frac{26 M}{r} + \frac{66M^2}{5r^2} + \frac{96M^3}{5r^3} - \frac{80M^4}{r^4}
    \right),
    \\
    g_{rr}^{(0,1)} (r)
    &=
    - \frac{\zeta}{f^2} \frac{M^2}{r^2}
    \left(
    1 + \frac{M}{r} + \frac{52M^2}{3r^2} + \frac{2M^3}{r^3} + \frac{16M^4}{5r^4} - \frac{368M^5}{3r^5}
    \right).
\end{align}
These expressions are same as those in~\cite{Yunes:2011we}.

\subsection{$\mathcal{O}(\varepsilon \zeta)$ corrections}

For $\ell=0$ mode,
from the $tr$-component of Eq.~\eqref{eq:Tti},
we obtain
\begin{align}
    \Phi^{(1,\frac{1}{2})}_{0}(t,r)
    =
    \mathcal{I}^{(1,\frac{1}{2})}_{0}(r)
    +
    e^{\frac{r(r+4M)}{16M^2}}  \sqrt{r-2M}
    \mathcal{L}^{(1,\frac{1}{2})}_{0}(t),
\end{align}
where $\mathcal{I}^{(1,\frac{1}{2})}_{0}$ and $\mathcal{L}^{(1,\frac{1}{2})}_{0}$ are arbitrary functions of $r$ and $t$, respectively.
From the $rr$-component and the $\theta \theta$-component of Eq.~\eqref{eq:Tti}, we obtain
\begin{align}
    \mathcal{L}^{(1,\frac{1}{2})}_{0} (t)
    =
    \mathtt{f}^{(1,\frac{1}{2})}_{0},
\end{align}
where $\mathtt{f}^{(1,\frac{1}{2})}_{0}$ is a constant.
In order to remove the divergence of the scalar field at spatial infinity, we set $\mathtt{f}^{(1,\frac{1}{2})}_{0} = 0$, which means
\begin{align}
    \Phi^{(1,\frac{1}{2})}_{0}(r) = \mathcal{I}^{(1,\frac{1}{2})}_{0}(r)
    \label{eq:Phi11_20r}.
\end{align}
Substituting Eq.~\eqref{eq:Phi11_20r} into the equation of motion for the scalar field~\eqref{eq:eomphi}, and imposing the regularity at $r=2M$, we have
\begin{align}
    \Phi^{(1,\frac{1}{2})}_{0}(r)
    =
    \mathrm{const}.
\end{align}

For $\ell \ge 1$,
the $t\theta$-component of the stationary condition~\eqref{eq:stationaryconesgb} is $\partial_t ^2 \Phi^{(1,\frac{1}{2})}_{\ell} = 0$.
The general solution is given by
\begin{align}
    \Phi^{(1,\frac{1}{2})}_{\ell}(t,r)
    =
    \mathcal{I}^{(1,\frac{1}{2})}_{\ell}(r)
    +
    t\, \mathcal{J}^{(1,\frac{1}{2})}_{\ell}(r),
\end{align}
where $\mathcal{I}^{(1,\frac{1}{2})}_{\ell}$ and $\mathcal{J}^{(1,\frac{1}{2})}_{\ell}$ are arbitrary functions of $r$.
When we substitute this into Eq.~\eqref{eq:stationaryconesgb},
we obtain the expression for $\mathcal{J}^{(1,\frac{1}{2})}_{\ell}$:
\begin{align}
    \mathcal{J}^{(1,\frac{1}{2})}_{\ell}(r)
    =
    \mathtt{f}^{(1,\frac{1}{2})}_{\ell}
    \left(
    r - 3M
    \right)^{\ell(\ell + 1)/2},
\end{align}
where $\mathtt{f}^{(1,\frac{1}{2})}_{\ell}$ is an integration constant.
Substituting this into Eq.~\eqref{eq:stationaryconesgb},
we obtain $\mathtt{f}^{(1,\frac{1}{2})}_{\ell}=0$.
Since the equation of motion of the scalar field~\eqref{eq:eomphi} becomes the same form as~\eqref{eq:Fequation},
the expression for $\Phi^{(1,\frac{1}{2})}_{\ell}$ is given by
\begin{align}
    \Phi^{(1,\frac{1}{2})}_{\ell}(x)
    =
    \mathtt{g}^{(1,\frac{1}{2})}_{\ell} P_{\ell}(1+2x)
    +
    \mathtt{h}^{(1,\frac{1}{2})}_{\ell}
    Q_{\ell}(1+2x),
\end{align}
where $\mathtt{g}^{(1,\frac{1}{2})}_{\ell}$ and $\mathtt{h}^{(1,\frac{1}{2})}_{\ell}$ are integration constants and $x=r/2M-1$. From the regularity of the scalar field,
we need to set $\mathtt{g}^{(1,\frac{1}{2})}_{\ell}=\mathtt{h}^{(1,\frac{1}{2})}_{\ell}=0$.
Therefore,
the scalar field $\varphi^{(1,\frac{1}{2})}$ has only $\ell=0$ mode and the configuration becomes
\begin{align}
    \varphi^{(1,\frac{1}{2})}(r)
    =
    \mathrm{const}.
\end{align}
As in the case of the order $\mathcal{O}(\zeta)$, the time dependence of the scalar field is not allowed at the order $\mathcal{O}(\varepsilon \zeta)$.

The metric of the order $\mathcal{O}(\varepsilon \zeta)$
can be obtained by solving the field equations~\eqref{fieldeqesgb}:
\begin{align}
    g_{t\phi}^{(1,1)}(r)
    =
    \frac{3}{5}\varepsilon \zeta \frac{M^4 \sin^2 \theta}{r^3}
    \left(
    1 + \frac{140M}{9r} + \frac{10M^2}{r^2} + \frac{16M^3}{r^3} - \frac{400M^4}{9r^4}
    \right).
\end{align}
This result is same from in~\cite{Pani:2011gy}.
Thus, the circularity condition holds at this order.

\subsection{$\mathcal{O}(\varepsilon^2 \zeta)$ corrections}

We can show that the scalar field is time independent:  $\varphi^{(2,\frac{1}{2})} = \varphi^{(2,\frac{1}{2})}(r)$ by the same way as the case of the order $\mathcal{O}(\varepsilon \zeta)$.
The configuration of the scalar field $\varphi^{(2,\frac{1}{2})}$ is given by~\cite{Pani:2011gy,Ayzenberg:2014aka}
\begin{align}
    \varphi^{(2,\frac{1}{2})}(r)
    =
    -  \frac{\varepsilon^2 \alpha}{2Mr}
    \left[
    1 + \frac{M}{r} + \frac{4M^2}{5r^2} + \frac{2M^3}{5r^3} +
    \frac{28M^2 \cos^2 \theta}{5r^2}
    \left(
    1 + \frac{3M}{r} + \frac{48M^2}{7r^2}
    \right)
    \right].
\end{align}
We choose the homogeneous integration constants so that the scalar field to be regular everywhere.
The metric of the order $\mathcal{O}(\varepsilon^2 \zeta)$ is given by the same form obtained in~\cite{Ayzenberg:2014aka}.\footnote{
The functions $H_{0\ell}^{(2,1)}$, $H_{2\ell}^{(2,1)}$, and $K_{\ell}^{(2,1)}$
correspond to the functions $H_{0\ell 0}, H_{2\ell 0}$ and $K_{\ell 0}$ in Eqs.(A15)-(A21) in~\cite{Ayzenberg:2014aka}, respectively
(see also the erratum in~\cite{Ayzenberg:2014aka} because there are some typos).}
Thus, the spacetime is circular at this order.

\section{summary and discussion}
\label{sec:conclusion}

We have studied slowly rotating black hole solutions up to linear order in coupling constant and quadratic order in the spin in dynamical Chern--Simons gravity and shift symmetric Einstein scalar Gauss--Bonnet gravity.
In our analysis, we did not assume the circularity condition for the spacetime and the time independence of the scalar fields.
We have shown that in both gravity theories, up to this order, such regular non-circular black hole solutions do not exist at least around the Schwarzschild spacetime.
We note that the linear time dependence of the scalar fields leads non-circular spacetimes in dynamical Chern--Simons gravity, while the linear time dependence makes the spacetime singular.
We also have shown that in dynamical Chern--Simons gravity, the time dependence of the scalar field is limited to the linear time dependence from the stationarity of the effective stress energy tensors.
On the other hand, we have shown in shift symmetric Einstein scalar Gauss--Bonnet gravity, any time dependence of the scalar field is prohibited.
This is the consequence of the conditions~\eqref{eq:stationaryconesgb} and~\eqref{eq:Tti} which mean that the effective stress energy tensor is stationary and the $ti$-component of the effective stress energy tensor should vanish for $\ell = 0$ mode.
Our results suggest the non-existence of rotating non-circular black holes in these gravity theories.

In our analysis,
we first determine the time dependence of the scalar fields from the stationarity of the metric functions and the effective stress energy tensors, and next we solve the field equations.
In both steps, we adapt the slow rotation approximation, and solve the equations at each order of the spin parameter.
Since our analysis method is widely usable,
it is interesting to use it to find non-circular black hole solutions in other string-inspired gravity theories,
but we leave this to future work.

\section*{ACKNOWLEDGMENTS}
We would like to thank K. Aoki, V. Cardoso, T. Harada, T. Ikeda, W. Ishizaki,  T. Kobayashi, Y. Koga, H. Motohashi, A. Naruko, K. Ogasawara, K. Takahashi, T. Tanaka, and K. Yamada for useful comments and discussions.
K.N. also thanks the Yukawa Institute for Theoretical Physics at Kyoto University, where this work was developed, for their hospitality.
This work was supported by the Rikkyo University Special Fund for Research (K.N.) and a MEXT Grant-in-Aid for Scientific Research on Innovative Areas 20H04746 (M.K.).

redpajama_set_name: RedPajamaArXiv


Driving along Highway 22 after a day with the eagles, I was traveling parallel to the Livingstone Range of the Canadian Rockies. The foothills lead up to it in a couple of rounded hills and then the mountains jut up sharply from there. It was an impressive scene with the sharp contrast between hills and mountains as well as light and shadow of the late afternoon.
The winds that came with the weather change last weekend were heavy when I left my home in Bragg Creek for the Banff National Park in the morning. When I got into the mountains, the Bow Valley was pretty calm but higher up on the slopes, the snow was blowing around in opaque sheets while the clouds raced by above. Watching from the Vermillion Lakes shoreline, I was mesmerized by the view of Mount Rundle. The sun catching the wispy snow drawn out over the slopes before fraying into the shadow as it flew over the cliffs was beautiful to watch.
I drove up to Apgar, a small village in Glacier National Park, this morning. I arrived at the southern edge of Lake McDonald in the dark and headed past the sleeping townsite for the rocky beach. The full moon provided a bit of light out over the water and I could see the mist was already rising up into the cold air. I started getting excited as the eastern edge of the sky brightened and silhouetted the mountain peaks above the north and east sides of the lake. The glow in the sky deepened and the colours came in beautifully.
As the intense colour began to fade, I was able to balance this great stem of autumn leaves with the lovely Grinnell argillite rocks under the water (the first image). A very beautiful morning in Montana's Glacier National Park.
My family spent the weekend at Lake Louise and I got out to greet the sunrise on the top of the rock pile at Moraine Lake on Sunday morning. As the eastern sky began to brighten clouds were swirling along the Valley of the Ten Peaks and I was hopeful for a nice backstop to develop above the mountains and catch the colourful light. When dawn was breaking the clouds had mostly cleared out around the lake but to the east a different set had anchored on the horizon and I worried that by the time the sun climbed that little bit higher and the light could paint the Ten Peaks, the colour may have faded to normal daylight. I waited, along with a few other photographers strung along the top of the trail, and we were granted a very short window where a beam of red light shot through a whole in the eastern cloudbank and painted the rocky slopes. The beam lasted well short of a minute but the valley transcended its normal beauty by a long margin while it lasted. Above is one of the images from this moment. I took the photograph below after the red light faded as the clouds returned and glided above the valley. The last sunrise I caught there was in July and by the time the light fought through the clouds stacked in front of the sun it had no colour left so I have no complaints with this weekend's weather.
Moraine Lake is one of the Canadian Rockies most iconic landscapes. I have been there many times and it continues to share new magic with each visit. I was up on top of the rock pile with a couple of good friends for a quiet evening and we returned a few hours later for a cloudy sunrise. Both times presented views of the Valley of the Ten Peaks and the lake that I had not seen previously. I enjoyed them all immensely.
The evening watched as the clouds ran towards the horizon leaving open sky above the peaks that loom above the lake and curl west down the valley. The soft light near sunset looked beautiful where it touched the peaks and provided a very subtle contrast to the deepening blues and greens that ushered in the night.
When I was crossing the stream where the lake most visibly drains out, the bright colors in the landscape's palette had been wrung out so I was drawn to the speck of orange upstream. I liked how this small information shelter's log frame stood defiantly against the gloom. At this point, some great clouds had stretched out above the water and they provided an abstract mirror of the river's folds as revealed in this 13 second exposure.
When we returned around 5am, the clouds had staked out all four corners of the sky. We watched breaks in the sky expectantly for more than an hour, taking us through sunrise without any light painting the peaks or the clouds curling around them. We were joined by a hopeful couple from Japan and two Chinese ladies on top of the moraine. Quiet chattering among the separate groups along with the occasional shutter click marking the time shuffling by. It was nice, not the dramatic alpen glow or early light that I have seen before but another interesting side of this valley.
Around 6:30 a large break in the clouds developed in the east and 15 minutes later the first shafts of sunlight hit the mountains. The light was still pretty warm and the drama I had been looking for unfolded for the next 45 minutes before the sun had risen too high for my landscape photography tastes. I enjoyed watching the color in the lake swirl and change as the house lights of the day came up. With stray clouds still wrapping peaks occasionally and the sunlight marching down the forest side of the lake, there was a lot to watch and to photograph.
Packing up, I retraced my steps down the path back towards the lodge. Crossing the river once more, I was drawn in again. This time the wet rocks were sparkling in the sunshine and I found the light on Yamnee (Mount Bowlen), Tonsa and Sapta (Mount Perren) particularly attractive. Breakfast was calling my friends (and me too – if I had been listening) and it was a good final image to complete this time with the lake, the valley and these wonderful peaks.
The light this evening was lovely and the lines of the hillsides of the Foothills towards Kananaskis Country looked like a water-colour painting.


text:
6 Russian Words You'll Struggle To Pronounce (If You're Not Russian)
Challenge number one: make it past 'hello.'
By Steph Koyfman
Of all the things that feel most daunting to beginner learners, Russian pronunciation looms the largest. What are those strange, inscrutable letters? Why do only some of them sound how you'd expect them to? How do those consonants exist in that combination and sound so effortless rolling off the tongues of native speakers?
First, there's the matter of familiarizing yourself with the Cyrillic alphabet, which is made up of a combination of Greek, Latin and Old Slavic symbols. Fortunately, 11 out of its 33 letters already match the English alphabet, so you're already a third of the way there! But learning which ones do and which ones only look like they do is another matter.
Getting a handle on the individual sounds is tricky enough, but then you've got to understand how they fit together.
In Russian pronunciation, there are hard consonants and soft consonants. For instance, the letter л (L) is pronounced like a hard л when it falls at the end of a word or it's followed by у, а, о, э or ы — as is the case with other consonants. When followed by ю, я, ё, е or и, or the soft sign ь, it's soft. A hard л is one that requires your tongue to make full contact with your teeth; a soft л is one that doesn't require any contact at all.
And don't worry — vowels change too depending on the context. They can be stressed (elongated) and they can be unstressed (shortened).
Are you stressed yet? Don't be! You just have to tackle it piece by piece. Below, each word has its Cyrillic spelling, as well as the Latin alphabet transliteration.
Teachable Moments In Russian Pronunciation
1. достопримечательность (sightseeing)
dastaprimechátel'nast'
Here's our first example of unstressed vowels. The letter "о" shows up three times here, and each of them are unstressed, or "reduced." Effectively, they're pronounced almost like an "a." Another aspect of Russian pronunciation that frequently trips up language learners? The soft sign, or the ь. What this does is indicate that the consonants immediately preceding them — the л and т, in this word — must be pronounced softly.
достопримечательность
2. Здравствуйте (hello)
zdrástvuyte
The formal "hello" is, perhaps ironically, not an easy word to get your tongue around, and it just so happens to be one of the first words that Russian learners have to confront. There are actually so many consonants in this word that even native Russian speakers skip the first в (which makes a "v" sound).
3. усовершенствование (improvement, perfection)
usavershénstvovaniye
Once again, we're looking at a complicated assortment of letters and sounds. And once again, we've got some unstressed "о"s that are pronounced like "a"s. The trickiest part? Pronouncing the "нств" ("nstv"), which features four consonants in a row. It helps if you break it down into two syllables: "shenst" and "vo." But when you pronounce it all as a word, they tend to create a tiny verbal obstacle course for your tongue.
усовершенствование
4. девятнадцать (nineteen)
divitnátsat'
Russian numbers ranging from 11 to 19 all feature the -надцать ending. The consonant combination дц (dts) is not an easy one — even Russians tend to blur out the "d" sound so that it effectively sounds like a "ts." In this case, we've also got a couple unstressed vowels. The e and the я are pronounced almost like an "i."
девятнадцать
5. сиреневенький (little lilac)
sirénivin'kiy
This is the diminutive form of сиреневый, which refers to a lilac shade of purple. Truthfully, you're unlikely to hear people use the diminutive version of this word in Russia unless they're trying to figure out how drunk you are. When even tipsy Russians struggle to pronounce this word, you know it's a challenge.
Let's break it down, shall we? Because all the consonants are followed by soft vowels like и and е, they must also be pronounced softly. Furthermore, the unstressed "и"s are pronounced with a reduced sound, and the Russian р (r) is rolled.
сиреневенький
6. потому что (because)
patamú shta
This is an important word to master. Not only is it an important word that can link two clauses together, but it can also stand alone. "Why did you do that?" "Because!"
You should be familiar with this concept by now, but once again, we have a series of unstressed "о"s. The curveball: the ч (ch) is pronounced like ш (sh) because of the following voiceless consonant т (t).
потому что
Need more Russian lessons?
Steph Koyfman
Steph is a writer, lindy hopper, and astrologer. She's also a language enthusiast who grew up bilingual and had an early love affair with books. She has mostly proved herself as a New Yorker, and she can introduce herself in Swedish thanks to Babbel. She also speaks Russian and Spanish, but she's a little rusty on those fronts.
Cursing Abroad: Russian Insults To Round Out Your Studies
If you like your Russian with a bit of spice, keep reading.
The 7 Best Films For Learning Russian
Here are seven amazing Russian movies for beginner and intermediate language learners.
By Richard Wess
How To Learn The Cyrillic Alphabet In Only 2 Days
The Cyrillic alphabet can seem intimidating, but don't be fooled. It only has 33 letters — just 7 more than the Latin alphabet! Here's how you can learn the Cyrillic alphabet in only 2 days.
By Arnaud Bernier


PURPOSE: Japanese hop (Humulus japonicus) is a major cause of weed pollinosis in East Asia. However, supplies of commercial allergen extract from this plant have not met clinical demand. The pollen of common hop (Humulus lupulus), a closely related species, may provide an alternative source if there is strong IgE cross-reactivity between these two species. We aimed to compare the IgE cross-reactivity and allergenicity of common hop and Japanese hop pollen. MATERIALS AND METHODS: Cross-reactivity was measured by inhibition ELISA. One- and two-dimensional (2D) gel analyses combined with IgE immunoblotting and mass spectrometry [liquid chromatography coupled to electrospray ionization tandem mass spectrometry (LC-ESI-MS/MS)] were performed to detect IgE-reactive pollen components. RESULTS: Up to 16.7% of IgE reactivity to Japanese hop was inhibited by common hop. A 12-kDa protein component of Japanese hop pollen that showed the most potent IgE reaction was absent from common hop. Six IgE-reactive components from Japanese hop were detected by 2D gel electrophoresis and LC-ESI-MS/MS, but showed low Mascot scores, preventing positive identification. CONCLUSION: No significant IgE cross-reaction was observed for Japanese and common hop pollen allergens. Development of allergy diagnostic and immunotherapeutic reagents based on Japanese hop pollen are urgently needed.


text:
Q: Do @NotNull annotations on getters have signifcant impacts on performance? Random example: In the source code of one of my programs I have this method:
    public @NotNull Currency[] getCurrencies() {
        return this.currencies.values().toArray(new Currency[0]);
    }

Which, after being compiled, gets turned into the following by my IDE (Intellij):
    @NotNull
    public Currency[] getCurrencies() {
        Currency[] var10000 = (Currency[])this.currencies.values().toArray(new Currency[0]);
        if (var10000 == null) {
            $$$reportNull$$$0(4);
        }

        return var10000;
    }

this.currencies refers to private final ConcurrentHashMap<String, Currency> currencies = new ConcurrentHashMap(); - which means that only via means of reflection or cosmic rays causing bit flips, the method getCurrencies() could ever return null.

*

*Should I - in such cases where there is no chance for a null being returned - forgo the usage of the @NotNull annotation?

*Does the annotation even have significant impact on performance, or do the benefits of having the annotation outweigh the performance impact?

Thank you
/edit:
I am using the annotation package org.jetbrains.annotations

A: @NotNull is not an annotation for your compiler. The compiler might use this annotation and do some additional checks. But @NotNull is an annotation for the humans reading your code that the compiler checks. Of Java's many problems, the biggest by far is the prevalence of unchecked null values floating around everywhere. In an otherwise statically-typed language, any variable anywhere in your program may simply not exist and violate every guarantee the type checker could otherwise provide.
Kotlin has proper nullable types. Recent versions of Scala have (opt-in) nullable types. Even new versions of C# have nullable types. In Java, @NotNull should be considered part of your function's contract, almost part of its type. You're not saying "hey, compiler, check if this is null". You're saying "I guarantee this function will never return null" and the compiler is checking it, in the same way that declaring a return type of String is making a guarantee that you're returning a String. Removing the @NotNull annotation because you're sure it's right is tantamount to changing all of your return types to Object because you know in your head what types the functions return. If you think
String getName(String prefix) { ... }

looks better than
Object getName(Object prefix) { ... }

then you should think
@NotNull String getName(@NotNull String prefix) { ... }

looks even better. It's a stronger guarantee.

A: Having a null check on your getter is a bit strange. At the time you call the getter your object is already in a state it shouldn't be in. You probably want to have the @NotNull on the parameter for your setter or constructor instead.
What you can do is write a JavaDoc for your getCurrencies.
/**
/* @return An array of currencies. Never null.
*/
public Currency[] getCurrencies() {}

But it's very uncommon to document that a method doesn't return null. Not returning null should be the gold standard and if a method can return null it should be documented, not the other way around. Also I wouldn't expect a getter that returns a set of data (arrays, lists etc.) to return null. Instead return an empty object.

redpajama_set_name: RedPajamaStackExchange