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Baron Alvanley, of Alvanley in the County Palatine of Chester, was a title in the Peerage of the United Kingdom. It was created on 22 May 1801 for Sir Richard Arden, the Chief Justice of the Common Pleas and former Master of the Rolls.  The title became extinct on the death of his second son, the third Baron (who had succeeded his elder brother), in 1857.

Barons Alvanley (1801)
Richard Pepper Arden, 1st Baron Alvanley (1744–1804)
William Arden, 2nd Baron Alvanley (1789–1849)
Richard Pepper Arden, 3rd Baron Alvanley (1792–1857)

Arms

References

Extinct baronies in the Peerage of the United Kingdom
Noble titles created in 1801
Noble titles created for UK MPs

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text:
I want my future sorted soon
Sam Marsden
Moises Llorens
Sergio Busquets hopes to have made a decision on his future by February and admits he is attracted by the possibility of playing in Miami.
Busquets, 34, is out of contract with Barcelona at the end of the season and is one of several big European-based players that is drawing interest from Major League Soccer side Inter Miami.
Sources have told ESPN that Barcelona are also open to extending Busquets' contract with the club, but they will leave the midfielder to decide what he feels is best for his career.
"I know it's the last year [of my deal with Barca] but I have always said that I have nothing signed with anyone," Busquets told Cadena Ser from Spain's World Cup camp in Qatar.
"I would like to have some certainty by February. I have always said I would like to play in the United States, above all in Miami, but from there I have not reached any agreement with any club — in the US or in any other league.
"Barcelona will be the first to know. When I have made a decision, I will tell them."
Busquets — who joined Barcelona in 2005 and has been in the first team since 2008 — is not the only player on Inter Miami's radar.
Former Barcelona teammate Lionel Messi has also been strongly linked with the MLS side. A report last week suggested a deal was already in place for him to move to Florida, although sources told ESPN nothing had been agreed and that Messi, like Busquets, would not make any decision on his future until after the World Cup.
Messi, 35, could also decide to extend his deal at Paris Saint-Germain, which expires in June, while there is also tentative interest from Barca in bringing him back to Camp Nou.
"If he wants to return at any moment, who would not like to coach him?" Barca coach Xavi Hernandez told reporters in Qatar on Tuesday. "He is still the No. 1."
However, as Messi prepares for Argentina's make-or-break group stage game against Poland on Wednesday, LaLiga president Javier Tebas poured cold water on a possible Barca return, saying he would be out of their price range.
"I don't know what [Messi] wants," Tebas said. "First of all, if he wants to leave PSG; secondly, if PSG will let him leave; and, thirdly, it then depends on what he earns, right?
"If all these three things fall into place and it's an affordable amount for Barcelona, inside [LaLiga's] financial regulations, he will sign. But it depends on what he earns. If it's what he earned when he left [in 2021]then no. With the amount Messi earns [at the moment]only PSG can afford him."
Barcelona's financial situation did improve last summer after they sold club assets worth over €700 million, but the signings of Robert Lewandowski, Jules Kounde and Raphinha, among others, means there is currently no money left for the January transfer window.
"Right now, we cannot sign in January," sporting director Jordi Cruyff said on Tuesday. "The fair play situation has changed for a cumulation of reasons.
"It is not a question of if a player goes it is all solved, either. There are different solutions. But we are clear on what we want to do if we can sign in January."
Cruyff also revealed Ajax and Ghana's Mohammed Kudus, one of the stars of the World Cup so far, has been tracked by the Catalan club, although he stopped short of saying there is any genuine interest at this stage. "I saw him play a year ago and have followed his progress at Ajax," he added. "But we're not at the point of saying 'Barca want this player,' not at all.
"He is a player that has caught the attention with his performances and goals at the World Cup. He's caught my attention especially because there is a debate in the Netherlands about his position, whether he's an advanced midfielder or a forward."
Published November 30, 2022 By mary0christian909
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Date: 06/27/2013 Category: General
Are You Undervaluing the Power of User-Generated Video? User-generated video can be a powerful marketing tool. Here are three ways to take advantage of this user-centric form of video to further your brand.
Date: 06/23/2013 Category: Video Production
The Power of Video Newsletters for Your Small Business Video newsletters should be part of your small businesses video marketing plan. Here's why.
For Web Video Success, Set Clear Video Campaign Goals What are the goals of your web video campaign? If you don't know, then you're doomed to fail.
YouTube Breaks Laws of Time and Space With 60 minutes in every hour, it seems unnatural for YouTube users to be loading 100 hours of video every minute.
Date: 06/04/2013 Category: Web Video
Top Brands on YouTube Get 884,000 Video Views Per Month A recent report complied by Outrigger Media shows that the top 500 brands on YouTube have 315,000 videos that are viewed 442 million times monthly.
Five Important Steps for Shooting an Effective Video Interview Vimeo's Video School recently released a list of top tips for shooting an interview. If you're thinking of shooting one, this is a must read.
Researchers: Digital Video Advertising Will Double by 2016 While spending on both television ads and online ads is expected to increase, online video advertising spending is expected to triple in the next three years.
Why Ignoring Mobile Video Viewers Is a Huge Mistake If your videos are not optimized for tablets and smartphones, you are missing out on a large segment of the online population.
Date: 05/25/2013 Category: Video Editing
New LinkedIn Profile Features Allow for Video LinkedIn has added a new feature that allows users to upload their photo or video projects to their profiles.
Google Video Results Beat Out Images, Shopping, Maps, and News A new study shows that video content is still leading the pack in Google's Universal Search results.
Optimizing Your YouTube Videos Is Necessary for Their Success Looking for the best ways to optimize your YouTube videos? Look no further. Our YouTube video team has answers here.
The Top 5 Reasons Online Video Marketing Is Crucial to Your Success There are a million reasons that video is critical to the success of your Virginia-based business. Here are the five most compelling.
Google Trends + YouTube Search Results = Fun, Useful Information YouTube and Google Trends have teamed up to show you just what YouTubers are looking for and when they are searching.
The Power of Video as Internal Communication in Your Small Business Video has the power to not only convert web visitors to customers, but to engage and inform your employees, as well.
Why We All Need to Know About YouTube's SMOSH Channel Find out some interesting information about YouTube's SMOSH channel and YouTube statistics in general in this news article.
"Harlem Shake" Gets Miners in Trouble A group of Australian miners are fired for performing the "Harlem Shake" underground and posting a YouTube video.
YouTube Testing New Pre-Video Online Video Advertising Idea YouTube is testing new online video ad choices for viewers of long content videos. How may it affect you and what exactly are your choices? Find out here.
Microsoft Opens Interactive Video Studio Microsoft's new L.A. studio is devoted to churning out footage for the Xbox Kinect platform.


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Home » Community » Pleasanton
Live the good life in Pleasanton
Pleasanton is a place that truly lives up to its name. Year-round balmy weather, picturesque neighborhoods, proximity to the great outdoors, and a wealth of community-focused events make it the perfect city to raise a family in. Is it any wonder why it's been named as one of Livability.com's Top 10 Places to Live in 2018? Located in Alameda County, this city is home to around 81,000 residents. Just 45 miles away from San Francisco, Pleasanton has easy access to the latter's high-tech job market while being far enough away to offer a serene, suburban vibe. If you're keen on purchasing Pleasanton, CA homes for sale, here's what you need to know about this premier city:
A vibrant city with a colorful history
Named after Major General Alfred Pleasonton (later spelled "Pleasanton") of the Union Army, this city has a deep and fascinating history. Inhabited as early as 5,000 years ago by Native Americans, the area was first sighted by the Spaniards in 1772. However, it was only in 1826 when Jose Amador arrived that the area would formally become a Spanish settlement. During that time, the settlement was known as Alisal and became a key stopover among miners during the California Gold Rush. The arrival of transcontinental railroads boosted development and in 1894 the area was incorporated as Pleasanton. Today, Pleasanton is one of the most affluent cities in California and one of the state's most sought-after real estate markets.
Overview of the real estate market
Given the high quality of life in Pleasanton, it's unsurprising that homes here command a premium. According to Realtor.com, the city's median listing price stands at $1.29 million—considerably higher than the national median of $329,000. However, you certainly get what you pay for as you get to live in a safe, wholesome city that has everything your family could need. There's a range of Pleasanton, CA real estate options to choose from, including condos, townhomes, single-family abodes, and luxury properties. Likewise, properties come in an array of architectural designs, including Mediterranean, Cape Cod, Craftsman, and contemporary styles. The typical family home in Pleasanton has three bedrooms and two bathrooms, with a floor space of around 1,500 square feet.
A peek into the Pleasanton lifestyle
Known as a bedroom community, Pleasanton offers the quintessential suburban atmosphere: peaceful, tight-knit, and family-friendly. One of the perks of living in Pleasanton, CA is the city's easy access to outdoor destinations. Enjoy hiking and camping in the Sierra Mountain Range, take a dip in the warm waters of Santa Cruz's beaches, or unwind in stunning Lake Tahoe, all of which are just a few hours' drive away. Of course, there's plenty of leisure to be had in the city, too. It's home to the Pleasanton Golf Center, which boasts three par-four holes. Fitness buffs will surely appreciate the various walking and biking paths that meander in and around the city. There are also several neighborhood parks that provide amenities such as tennis courts, dog parks, picnic areas, and more. And while Pleasanton has a small-town feel, it has an active social scene thanks to the Alameda County Fairgrounds and a plethora of top-notch bars and restaurants. And thanks to the city's very own shopping mall, Stoneridge Shopping Center, everything you need is right where you are.
Make the move today
Want to call Pleasanton, CA home? Allow DeAnna Armario to be your trusted guide to the city's real estate market. With her help, you can find a dream home in the perfect neighborhood that suits all of your needs and wants. Give her a call at 925.260.2220 or send an email to deanna(at)armariohomes(dotted)com to get started.
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Genavieve Linkowski Returns To American Idol After Losing Her Sister - Inspirational Videos
Genavieve Linkowski returns to American Idol for a touching performance after losing her sister in 2019.
Genavieve is no stranger to American Idol. In fact, she auditioned two years ago and made it all the way to Hollywood. Along the way, her family wholeheartedly supported her as she pursued her dream. And one standout is Genavieve's sister Corrin who was even at her very first audition.
Sadly, Corrin lost her life after a bad car accident at the age of 18. The death took a toll on Genavieve. "I felt really lost. I had no motivation to sing whatsoever. My passion was gone for a while and my voice was gone for a while. I just didn't know how to do life without her because I always did life with her. But I know that Corrin would still want me to push forward with this."
As she fought through tears before the audition, her mom comforted her by saying, "I know Corrin is just watching over you. She's singing with angels." Genavieve's story instantly resonated with the judges and her unique rendition of 'How Am I Suppose To Live Without You' was a big hit. Katy Perry said, "I just felt that song through and through. I mean, you were singing that song from your pain that you've been through."
Lionel Ritchie also raved about Genavieve's strength and vocal performance. "It took a lot of guts to come back and stand up, but I really must tell you, I love that performance." Ultimately, with Corrin's spirit by her side, Genavieve earned her ticket to Hollywood. Who else is rooting for this young girl?
Boy's Routine With Brother with Autism Inspires Entire School
godtube-inspiration 2455 views 2020-03-02T13:13:02 source: American Idol Genavieve Linkowski returns to American Idol for a touching performance after losing her sister in 2019.
'Unchained Medley' by 6-Year-Old Singer Evan Riley
Cheerleader Gets Homecoming Surprise From Hero Big Brother
Liquor Store Worker Gives Homeless Man Her Shoes, Gets Surprising Reward Back
Loretta Lynn's Granddaughter And Willie Nelson's Son Sing 'Lay Me Down'
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You are watching Genavieve Linkowski Returns To American Idol After Losing Her Sister on Godtube.com the largest video sharing platform offering online Christian videos with faith-based, family friendly content. Popular video sections on Godtube include; Christian bands and singers in Christian music videos, Christian comedians and comedy skits, spoofs and parodies in funny videos, cute videos featuring kids and animals, sports videos, Christian news videos and inspirational videos. Be inspired in your walk with Jesus Christ and grow in your knowledge of the Bible with videos highlighting inspirational messages and verses. Start uploading your Christian videos today!


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69% of Marketers Primarily Use Email for Lead Nurturing
Krystle Vermes
Lead nurturing is a critical part of marketers' strategies, and new research indicates that email is continuing to help them in their efforts to attract customers and prospects.
Ascend2 recently published "The State of Lead Nurturing" report, and statistics showed that out of all of the digital channels used during the lead nurturing process, email (69%) leads the way. This is followed by social media (67%), website personalization (38%) and paid advertising/retargeting (34%).
Although email is helping marketers in this area, many still think that there is room for improvement, according to the report. The majority (47%) at least "somewhat agree" that allocating more time/resources to lead nurturing efforts would improve their conversion rates. Thirty-seven percent "strongly agree" with this sentiment.
Driving Revenue via Email Marketing
When it comes to utilizing email, previous research suggests that it is also still a large driver of sales revenue.
Cheetah Digital conducted its "Digital Consumer Trends Index 2021" report, and data showed that email is still overwhelmingly the top driver of sales. It outperformed SMS by 92% and banner ads by 39%. The majority of respondents (45%) claimed that over the past months, they have primarily made purchases through email.
Social media, however, does not lag far behind. Nearly 43% stated that they had made a purchase via a post on social media over the past year as well.
KoMarketing B2B Marketing Industry News
KoMarketing industry news provides the latest information, news, and trends in online marketing for B2B marketers. We emphasize all facets of B2B demand generation including search, social media, content marketing, email, and marketing technology among other key categories and topics.
B2B Marketing News
Mobile Device Adoption
Building The Next B2B Digital Marketing Success Story: KoMarketing Acquired by Walker Sands
6 Key Takeaways: Migrating to GA4 for B2B Marketers
GA4 and Google Ads: What B2B Marketers Need to Know
GA4 Conversion Tracking: What A B2B Marketer Needs to Know
A B2B Marketers' Guide to GA4 Reporting & Measurement
How to Drive Engagement on Social Media for B2B Marketers
How to Use B2B Google Ads for Lead Generation: A 6-Step Guide
B2B Content Marketing Metrics: Tracking What Matters
Twitter Advertising for B2B Marketers: Twitter Ads Campaign Types
How to Run Paid Ads on LinkedIn: A 7-Step Guide
"The KoMarketing team took our PPC and SEM marketing efforts to the next level. They are a team of thoughtful, creative, and smart people, and they get downright scientific when it comes to measurement and attribution. A true pleasure to work with."
— Will Bernholz, Vice President, Marketing, Dropsource


Pyry is a hobbyist photographer who has been practising the use of one "Nikon fm2" for some 17 years now. His works can be found under pseudonyms such as "toistaitoinen" where a lot of his works can be found in this gallery on deviantart. Also, Pyry is one of the members behind a Finnish self-made photography magazine called Harmaa-Alue (grey zone).


Caffeine is one of the most loved stimulants in America! But now that you are pregnant, you may need to lighten up on the daily intake of your favorite drinks and treats.
Caffeine is a stimulant and a diuretic. Because caffeine is a stimulant, it increases your blood pressure and heart rate, both of which are not recommended during pregnancy. Caffeine also increases the frequency of urination. This causes reduction in your body fluid levels and can lead to dehydration.
Caffeine crosses the placenta to your baby. Although you may be able to handle the amounts of caffeine you feed your body, your baby cannot. Your baby's metabolism is still maturing and cannot fully metabolize the caffeine. Any amount of caffeine can also cause changes in your baby's sleep pattern or normal movement pattern in the later stages of pregnancy. Remember, caffeine is a stimulant and can keep both you and your baby awake.
Caffeine is found in more than just coffee. Caffeine is not only found in coffee but also in tea, soda, chocolate, and even some over-the-counter medications that relieve headaches. Be aware of what you consume.
Caffeine causes birth defects: Numerous studies on animals have shown that caffeine can cause birth defects, preterm delivery, reduced fertility, and increase the risk of low-birth weight offspring and other reproductive problems. There have not been any conclusive studies done on humans though. It is still better to play it safe when it comes to inconclusive studies.
Caffeine causes infertility: Some studies have shown a link between high levels of caffeine consumption and delayed conception.
Caffeine causes miscarriages: A few studies have shown that there may be an increase in miscarriages among women who consume more than 300 mg (three 5 oz cups of coffee) a day. Other outcomes include preterm labor and low-birth weight babies. Again, it is safer to avoid caffeine as much as possible.
A pregnant woman should not consume ANY caffeine: Experts and studies have stated that "moderate" levels of caffeine have not been found to have a negative effect on pregnancy. The definition of "moderate" varies anywhere from 150 mg – 300 mg a day.
How much caffeine is in your favorite drinks & snacks?
The less caffeine you consume, the better. Some experts say more than 150 mg of caffeine a day is too much, while others say more than 300 mg a day is too much. Avoiding caffeine as much as possible is your safest course of action. If you must get your fix, it is best to discuss this with your health care provider to make the healthiest choice for you and your baby.


The video begins with an informal exchange between Dr. Trappler and a student off-camera. The doctor then provides a summary of historical events which illustrate the concept of G-d retracting and then revealing his Inner Light. The "Body" of the Torah was originally expressed at the Revelation at Sinai. In contrast, the "Inner Soul" of the Light of the Torah was manifest on the day of the passing of the original Master of the Esoteric- Rabbi Shimeon Ben Yochai (The Rashbi), when he named the disciples gathered around his bed (as witnesses) to reveal, in an impromptu deliberation, what is now generically referred to as "The Zohar".


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Home » Specialities » Knee » Knee Ligament Injuries
Knee Ligament Injuries
The knee is made up of two sets of strong ligaments that help to stabilize the joint. The collateral ligaments include the medial (or inside) and lateral (or outside) collateral ligaments, which connect the thigh bone (femur) to the shin bone (tibia) and stabilize the knee. The cruciate ligaments (anterior or "front", and posterior or 'back") cross in the inside of the knee and prevent excessive front-to-back and twisting motions. Additional stabilizing structures also play a role in normal knee function.
Knee ligaments are usually injured by a sudden traumatic event such as a blow to the knee, twisting, falls, or motor vehicle accidents. The severity of the injury depends on the nature and force of the trauma, and may range from simple sprains to the ligament, with stretching of the tissue, to a complete tear of the ligament. A ligament may be injured in isolation, or as part of a combined injury termed a "multi-ligament knee injury." Multi-ligament injuries are generally severe injuries that result from a dislocation of near-dislocation (termed "subluxation") of the knee and may involve damage to other structures, such as ligaments or nerves around the knee.
Symptoms of knee ligament injuries involve pain at the time of the injury, with further stress on the knee. The knee may become swollen and bruising may occur. With severe injuries the knee will feel "unstable' and the patient will be unable to bear weight.
Diagnosis of knee ligament injuries involves review of the patient's history of the injury and symptoms, as well as a physical examination of the stabilizing structures around the knee. X-rays are reviewed to look for associated fractures and to look for signs of knee instability. A MRI scan provides detail on the extent of the injury and involved areas. Additional studies may be required outline complex fractures (CT scan) or to look for injury to blood vessels around the knee (arteriogram).
Treatment of knee ligament injuries depends on many factors including the nature and extent of the injury, areas of the knee involved, and the age and activity-level of the patient. Mild sprains may be treated with rest and bracing of the knee, crutches and limited weight bearing movement, icing, and pain medicines, followed by physical therapy to restore normal strength and motion of the knee after the injury has healed. More severe injuries that involve complete tears – particularly multiple ligament injuries – often require surgery to repair or reconstruct ligaments and restore stability and function of the knee. These may be extensive and complex surgeries and may even require a "staged" approach with more than one procedure. Dr. Boes will review the details of your particular injury with you and discuss the treatment processes.
For more information, visit http://orthoinfo.aaos.org/topic.cfm?topic=A00551.


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Q: Can the numerical method for a set of ordinary differential equations be non-conservative? I have a set of highly coupled ODE that I am solving. I'm just not understanding whether the quantities will be inherently conserved because it's an ODE. Everything I'm seeing on Google is for PDE's.

A: Numerical methods for ODEs are more often than not found to be non-conservative when solving systems having conserved quantities. How non-conservative they are depends on things like the step-size, the order of the method and how long "time" you integrate over.
A typical example of this is found when numerically solving for plantary orbits (e.g. the Earth around the Sun). Below are some plots showing the numerical solution of such a system for different ODE solvers taken from this paper:

All three methods deviate from the true analytical solution (in blue) after enough orbits. If we look at the total energy, which should be conserved, we see that the last integrator conserves it very well while the first two don't:

Integrators that have this property is often called symplectic integrators. Such integrators can always be constructed for systems that derive from a Hamiltonian (as is often the case for physical systems).
If you are solving a system that have conserved quantities then monitoring the conservation of these is often a good way of checking the "goodness" of the numerical solution.

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\section{Introduction}\label{intro}

Traversable Lorentzian wormholes \cite{motho} are solutions of the equations of gravitation representing geometries which have a throat that connects two regions of the same universe or two separate universes \cite{motho, visser}. For static wormholes, the throat is a minimal area surface satisfying a flare-out condition \cite{ hovis1}. In the context of the general relativity theory of gravitation, wormholes must be threaded by exotic matter that violates the null energy condition \cite{motho, visser, hovis1, hovis2}. The amount of exotic matter needed around the throat can be made arbitrarily small by an  appropriate choice of  the geometry of the wormhole \cite{viskardad}, although it may be at the expense of
large stresses at the throat \cite{dil,lst}.\\

Thin--shell wormholes are mathematically constructed by cutting and pasting two manifolds \cite{visser, mvis} to form a new one, which is geodesically complete and it has a shell placed  in the joining surface. The exotic matter required for their existence can be located at the shell, with normal matter outside this surface. Stability studies of thin--shell wormholes under perturbations preserving the original symmetries have been carried out in several works.  A linearized  stability analysis  of a thin--shell wormhole made by joining two Schwarzschild geometries was done in Ref. \cite{poisson}. The same method was applied to  wormholes constructed using branes with negative tensions \cite{barcelo}, and to transparent spherically symmetric thin--shells and wormholes \cite{ishak}. Later, it was extended to Reissner--Nordstr\"{o}m thin--shell wormholes \cite{eirom}, and to wormholes with a cosmological constant \cite{lobo}. Dynamical thin--shell wormholes were considered in Ref. \cite{lobo2}. The stability and energy conditions for five dimensional thin--shell wormholes with spherical symmetry in Einstein--Maxwell theory with a Gauss--Bonnet term were studied in Ref. \cite{marc}, while thin--shell wormholes in dilaton gravity were analyzed in Refs. \cite{dil, eir}. Thin--shell wormholes associated with cosmic strings have been treated in Refs. \cite{eisi}. Other recent related articles can be found in Refs. \cite{other}. \\

Observational data suggest an accelerated expansion of the Universe \cite{acc}, which within general relativity implies that the strong energy condition should be violated. Several models for the matter leading to this scenario have been proposed \cite{matt}. One of them is the Chaplygin gas \cite{chap}, a perfect fluid with an equation of state of the form $p\rho =-A$, where $A$ is a positive constant. The Chaplygin gas has the property that the squared sound velocity is always positive, even in the case of exotic matter. Although it was introduced for phenomenological reasons, not related with cosmology \cite{aero}, this equation of state can be obtained from string theory \cite{brane}. Models of exotic matter of interest in cosmology have already  been considered in wormhole construction. Wormholes supported by phantom energy, with equation of state $p=\omega \rho $, where $\omega<-1$, have been analyzed by several authors \cite{phantom}. A generalized Chaplygin gas, with an equation of state $p\rho^\alpha=-A$ which has two parameters $A>0$ and $0<\alpha\leq 1$, has been used in Ref. \cite{lobo73} as the exotic matter supporting a wormhole of the Morris--Thorne type \cite{motho}; there, the wormhole metric was matched to an exterior vacuum metric to keep the exotic matter within a finite region of space. Instead, in a thin--shell wormhole the exotic matter can be  restricted from the beginning to the shell located at the joining surface, as was previously done for the original Chaplygin gas ($\alpha =1$) in Ref. \cite{chaply}. Recently, other authors \cite{chapnew} have also considered wormholes with a Chaplygin gas. The purpose of the present work is the study of spherically symmetric thin--shell wormholes with matter in the form of a generalized Chaplygin gas. In Sec. 2, the equations that give the possible radii of wormholes  and determine their stability for a general class of geometries are obtained. In Secs. 3 and 4, the formalism is applied to Reissner--Nordstr\"om wormholes and to wormholes with a cosmological constant, respectively. Finally, in Sec. 5, the conclusions of this work are summarized. Units such that $c=G=1$ are adopted.\\

\section{General equations}\label{tswh}

We start from the spherically symmetric metric
\begin{equation}
ds^2=-f(r)dt^2+f(r)^{-1}dr^2+h(r) (d\theta ^2+\sin^2\theta d\varphi^2),
\label{e1}
\end{equation}
where $r>0$ is the radial coordinate, $0\le \theta \le \pi$ and $0\le \varphi<2\pi $ are the angular coordinates, $h(r)$ is always positive and $f(r)$ is a positive function from a given radius. We take a radius $a$, greater than the event horizon radius $r_{h}$ if the geometry (\ref{e1}) has any, to avoid the presence of horizons and singularities, we cut two identical copies of the region with $r\geq a$:
\begin{equation}
\mathcal{M}^{\pm }=\{X^{\alpha }=(t,r,\theta,\varphi)/r\geq a\},  \label{e2}
\end{equation}
and we paste them at the hypersurface
\begin{equation}
\Sigma \equiv \Sigma ^{\pm }=\{X/F(r)=r-a=0\},  \label{e3}
\end{equation}
to create a new manifold $\mathcal{M}=\mathcal{M}^{+}\cup \mathcal{M}^{-}$. If $h'(a)>0$ (condition of flare-out), this construction creates a geodesically complete manifold representing a wormhole with two regions connected by a throat of radius $a$, where the surface of minimal area is located. On this manifold it is possible to define a new radial coordinate $l=\pm \int_{a}^{r}\sqrt{1/f(r)}dr$ that represents the proper radial distance to the throat, which is situated at $l=0$; the plus and minus signs corresponding, respectively, to $\mathcal{M}^{+}$ and $\mathcal{M}^{-}$. On the wormhole throat $\Sigma $, which is a synchronous timelike hypersurface, we  define coordinates $\xi ^{i}=(\tau ,\theta,\varphi )$, with $\tau $ the proper time on the shell. We use the  Darmois-Israel formalism \cite{daris} and we let the throat radius be a function of time $a(\tau)$. The second fundamental forms (or extrinsic curvature) associated with the two sides of the shell are given by:
\begin{equation}
K_{ij}^{\pm }=-n_{\gamma }^{\pm }\left. \left( \frac{\partial ^{2}X^{\gamma
} } {\partial \xi ^{i}\partial \xi ^{j}}+\Gamma _{\alpha \beta }^{\gamma }
\frac{ \partial X^{\alpha }}{\partial \xi ^{i}}\frac{\partial X^{\beta }}{
\partial \xi ^{j}}\right) \right| _{\Sigma },
\label{e4}
\end{equation}
where $n_{\gamma }^{\pm }$ are the unit normals ($n^{\gamma }n_{\gamma }=1$) to $\Sigma $ in $\mathcal{M}$:
\begin{equation}
n_{\gamma }^{\pm }=\pm \left| g^{\alpha \beta }\frac{\partial F}{\partial
X^{\alpha }}\frac{\partial F}{\partial X^{\beta }}\right| ^{-1/2}
\frac{\partial F}{\partial X^{\gamma }}.
\label{e5}
\end{equation}
Adopting the orthonormal basis $\{ e_{\hat{\tau}}=e_{\tau }, e_{\hat{\theta}}=a^{-1}e_{\theta }, e_{\hat{\varphi}}=(a\sin \theta )^{-1} e_{\varphi }\} $, we obtain
\begin{equation}
K_{\hat{\theta}\hat{\theta}}^{\pm }=K_{\hat{\varphi}\hat{\varphi}}^{\pm
}=\pm \frac{h'(a)}{2h(a)}\sqrt{f(a)+\dot{a}^2},
\label{e6}
\end{equation}
and
\begin{equation}
K_{\hat{\tau}\hat{\tau}}^{\pm }=\mp \frac{f'(a)+2\ddot{a}}{2\sqrt{f(a)+\dot{a}^2}},
\label{e7}
\end{equation}
where a prime and the dot represent, respectively, the derivatives with respect to $r$ and $\tau$. With the definitions $[K_{_{\hat{\imath}\hat{\jmath}}}]\equiv K_{_{\hat{\imath}\hat{\jmath}}}^{+}-K_{_{\hat{\imath}\hat{\jmath}}}^{-}$, $K=tr[K_{\hat{\imath}\hat{\jmath }}]=[K_{\; \hat{\imath}}^{\hat{\imath}}]$ and introducing  the surface stress-energy tensor $S_{_{\hat{\imath}\hat{\jmath} }}={\rm diag}(\sigma ,p_{\hat{\theta}},p_{\hat{\varphi}})$, where $\sigma$ is the surface energy density and $p_{\hat{\theta}}$, $p_{\hat{\varphi}}$ are the transverse pressures, we have from the Einstein equations on the shell (or Lanczos equations):
\begin{equation}
-[K_{\hat{\imath}\hat{\jmath}}]+Kg_{\hat{\imath}\hat{\jmath}}=8\pi
S_{\hat{\imath}\hat{\jmath}},
\label{e8}
\end{equation}
that
\begin{equation}
\sigma=-\frac{\sqrt{f(a)+\dot{a}^2}}{4\pi }\frac{h'(a)}{h(a)},
\label{e9}
\end{equation}
and
\begin{equation}
p=p_{\hat{\theta}}=p_{\hat{\varphi}}=\frac{\sqrt{f(a)+\dot{a}^2}}{8\pi}\left[ \frac{2\ddot{a}+f'(a)}{f(a)+\dot{a}^2}+\frac{h'(a)}{h(a)}\right] .
\label{e10}
\end{equation}
The negative sign in Eq. (\ref{e9}) plus the flare-out condition $h'(a)>0$ implies that $\sigma <0$, indicating that the matter at the throat is exotic. We adopt a generalized Chaplygin gas as the exotic matter in the shell $\Sigma $. For this gas, the pressure has opposite sign to the energy density, resulting in a positive pressure. Then, the equation of state for the exotic matter at the throat can be written in the form
\begin{equation}
p=\frac{A}{|\sigma |^{\alpha}},
\label{e11}
\end{equation}
where $A>0$ and $0<\alpha\le 1$ are constants. When $\alpha=1$ the Chaplygin gas equation of state $p=-A/\sigma$ is recovered. Replacing Eqs. (\ref{e9}) and (\ref{e10}) in Eq. (\ref{e11}), we obtain the differential equation that should be satisfied by the throat radius of thin--shell wormholes threaded by exotic matter with the equation of state of a generalized Chaplygin gas:
\begin{equation}
\{ [2\ddot{a}+f'(a)]h(a)+[f(a)+\dot{a}^2]h'(a) \}[h'(a)]^{\alpha}-2A[4\pi h(a)]^{\alpha +1}[f(a)+\dot{a}^2]^{(1-\alpha)/2}=0.
\label{e12}
\end{equation}
In particular, for static wormholes, the surface energy density and pressure are given  by
\begin{equation}
\sigma_{0}=-\frac{\sqrt{f(a_{0})}}{4\pi}\frac{h'(a_{0})}{h(a_{0})},
\label{e13}
\end{equation}
and
\begin{equation}
p_{0}=\frac{\sqrt{f(a_{0})}}{8\pi}\left[ \frac{f'(a_{0})}{f(a_{0})}+\frac{h'(a_{0})}{h(a_{0})}\right] .
\label{e14}
\end{equation}
From Eq. (\ref{e12}), the static solutions, if they exist, have a throat radius $a_{0}$ that should fulfill the equation
\begin{equation}
[f'(a_{0})h(a_{0})+f(a_{0})h'(a_{0})][h'(a_{0})]^{\alpha}-2A[4\pi h(a_{0})]^{\alpha +1}[f(a_{0})]^{(1-\alpha)/2}=0,
\label{e15}
\end{equation}
with the condition $a_{0}>r_{h}$ if the original metric has an event horizon. The existence of static solutions depends on the explicit form of the function $f$. The method developed in Ref. \cite{chaply} to study the stability of the static solutions under perturbations preserving the symmetry in the case of the Chaplygin gas is rather cumbersome to be extended to the generalized Chaplygin gas. Here we follow the standard ``potential" approach (both methods were shown equivalent for the metric given by Eq. (\ref{e1}) in the case of a linearized equation of state \cite{eir}). From Eqs. (\ref{e9}) and (\ref{e10}), it is easy to check the energy conservation equation:
\begin{equation}
\frac{d}{d\tau }\left( \sigma \mathcal{A}\right) +p\frac{d\mathcal{A}}{d\tau }=
\left\lbrace \left[ h'(a)\right]^2 -2h(a)h''(a)\right\rbrace \frac{\dot{a}\sqrt{f(a)+ \dot{a}^2}}{2h(a)},
\label{p1}
\end{equation}
where $\mathcal{A}=4\pi h(a)$ is the area of the wormhole throat. In Eq. (\ref{p1}), the first term in the left hand side represents the internal energy change of the throat and the second the work done by the internal forces of the throat, while the right hand side represents a flux. If $[h'(a)]^2 -2h(a)h''(a)=0$, the flux term is zero and Eq. (\ref{p1}) takes the form of a simple conservation equation. This occurs when $h(a)=C(a+D)^2$  ($C>0$ and $D$ constants) or  $h(a)=C$  (this case is unphysical, since there is no throat) \cite{eir}. It is straightforward to see that Eq. (\ref{p1}) can be written in the form
\begin{equation}
h(a)\dot{\sigma}+h'(a)\dot{a}(\sigma +p)=-\left\lbrace \left[ h'(a)\right]^2 -2h(a)h''(a)\right\rbrace
\frac{\dot{a}\sigma }{2 h'(a)},
\label{p2}
\end{equation}
which using that $\sigma '=\dot{\sigma }/\dot{a}$ gives
\begin{equation}
h(a)\sigma '+h'(a)(\sigma +p)+\left\lbrace \left[ h'(a)\right]^2 -2h(a)h''(a)\right\rbrace
\frac{\sigma }{2 h'(a)}=0.
\label{p3}
\end{equation}
The pressure $p$ is a function of $\sigma $ given by the equation of state, thus Eq. (\ref{p3}) is a first order differential that can be recast in the form $\sigma '(a)=\mathcal{F}(a, \sigma (a))$, for which a unique solution with a given initial condition always exists, provided that $\mathcal{F}$ has continuous partial derivatives. Then, Eq. (\ref{p3}) can be formally integrated to obtain $\sigma (a)$, so replacing $\sigma (a)$ in Eq. (\ref{e9}) and regrouping terms, the dynamics of the wormhole throat is completely determined by a single equation:
\begin{equation}
\dot{a}^{2}=-V(a),
\label{p4}
\end{equation}
with
\begin{equation}
V(a)=f(a)-16\pi ^{2}\left[ \frac{h(a)}{h'(a)}\sigma (a)\right] ^{2}.
\label{p5}
\end{equation}
A Taylor expansion to second order of the potential $V(a)$ around the static solution yields:
\begin{equation}
V(a)=V(a_{0})+V'(a_{0})(a-a_{0})+\frac{V''(a_{0})}{2}
(a-a_{0})^{2}+O(a-a_{0})^{3}.
\label{p6}
\end{equation}
From Eq. (\ref{p5}) the first derivative of $V(a)$ is
\begin{equation}
V'(a)=f'(a)-32\pi ^{2}\sigma (a) \frac{h(a)}{h'(a)} \left\lbrace  \left[ 1-\frac{h(a)h''(a)}{[h'(a)]^2}\right] \sigma (a) +\frac{h(a)}{h'(a)}\sigma '(a)\right\rbrace ,
\label{p7}
\end{equation}
which using Eq. (\ref{p3}) takes the form
\begin{equation}
V'(a)=f'(a)+16\pi ^{2}\sigma (a)\frac{h(a)}{h'(a)} \left[ \sigma (a)+ 2 p(a) \right] .
\label{p8}
\end{equation}
The second derivative of the potential is
\begin{eqnarray}
V''(a)&=&f''(a)+16\pi ^{2}\left\lbrace \left[ \frac{h(a)}{h'(a)}\sigma '(a)+\left( 1-\frac{h(a)h''(a)}{[h'(a)]^2} \right)  \sigma (a)\right] \left[ \sigma (a)+2p(a)\right] \right. \nonumber
\\
& & \left. +\frac{h(a)}{h'(a)}\sigma (a) \left[ \sigma '(a)+2p'(a)\right] \right\rbrace  .
\label{p9}
\end{eqnarray}
From Eq. (\ref{e11}) we have that $p'(a)=A\alpha|\sigma (a)|^{-\alpha -1}\sigma '(a)=\alpha p(a)\sigma '(a)/|\sigma (a)|$, then $\sigma '(a)+2p'(a)=\sigma '(a)[1+2\alpha p(a)/|\sigma (a)|]$, and using Eq. (\ref{p3}) again, we obtain
\begin{equation}
V''(a)=f''(a)-8\pi ^{2}\left\lbrace \left[ \sigma (a)+2p(a)\right] ^{2} +2\sigma (a)\left[ \left(  \frac{3}{2}-\frac{h(a)h''(a)}{[h'(a)]^2}\right) \sigma (a)+p(a)\right] \left[ 1+2\frac{\alpha p(a)}{|\sigma (a)|}\right] \right\rbrace .
\label{p10}
\end{equation}
Using Eqs. (\ref{e13}) and (\ref{e14}), it is not difficult to see that $V(a_{0})=V'(a_{0})=0$, so the potential is
\begin{equation}
V(a)=\frac{1}{2}V''(a_{0})(a-a_{0})^{2}+O[(a-a_{0})^{3}],
\label{p11}
\end{equation}
with
\begin{eqnarray}
V''(a_{0})&=& f''(a_{0})+\frac{(\alpha -1)[f'(a_{0})]^2}{2f(a_{0})}+\left[ \frac{(1-\alpha )h'(a_{0})}{2h(a_{0})}+\frac{\alpha h''(a_{0})}{h'(a_{0})}\right] f'(a_{0}) \nonumber \\
& & +(\alpha +1)\left[ \frac{h''(a_{0})}{h(a_{0})} -\left( \frac{h'(a_{0})}{h(a_{0})} \right) ^{2}\right] f(a_{0}).
\label{p12}
\end{eqnarray}
The wormhole is stable under radial perturbations if and only if $V''(a_{0})>0$.

\section{Reissner-Nordstr\"om wormholes}\label{rn}

The Reissner--Norsdtr\"om geometry, which represents a spherically symmetric charged object, have metric functions
\begin{equation}
f(r)=1-\frac{2M}{r}+\frac{Q^2}{r^2},\hspace{1cm} h(r)=r^2,
\label{rn1}
\end{equation}
where $M$ is the mass and $Q$ is the charge. For $0<|Q|<M$ this geometry has two horizons with radii
\begin{equation}
r_{\pm}=M\pm \sqrt{M^{2}-Q^{2}},
\label{rn2}
\end{equation}
where the minus sign corresponds to the inner one and the plus sign to the outer event horizon. When $|Q|=M$ the two horizons merge into one, and if $|Q|>M$ no horizons are present and there is a naked singularity in $r_{s}=0$. If $Q=0$ the Schwarzschild geometry is obtained, which has a horizon with radius $r_{h}=2M$. The thin--shell wormholes constructed from the Reissner--Norsdtr\"om metric consist of a charged surface of exotic matter (the throat) with an electric field in vacuum outside it. When $|Q|\le M$ the throat radius $a_{0}$ should be taken greater than $r_{h}=r_{+}$ so that the manifold  $\mathcal{M}$ has no horizons. If $|Q|>M$ the condition $a_{0}>0$ removes the naked singularity. By replacing Eq. (\ref{rn1}) in Eqs. (\ref{e13}) and (\ref{e14}), the energy density and pressure at the throat are obtained:
\begin{equation}
\sigma _{0}=-\frac{\sqrt{a_{0}^{2}-2Ma_{0}+{Q^2}}}{2\pi a_{0}^{2}},
\label{rn3}
\end{equation}
and
\begin{equation}
p_{0}=\frac{a_{0}-M}{4\pi a_{0}\sqrt{a_{0}^{2}-2Ma_{0}+{Q^2}}}.
\label{rn4}
\end{equation}
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{frn1.eps}
\vspace{-0.5cm}
\end{center}
\caption{Reissner--Nordstr\"{o}m wormholes supported by a generalized Chaplygin gas with $\alpha =0.2$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $Q$, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius of the original manifold.  }
\label{frn1}
\end{figure}
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{frn2.eps}
\vspace{-0.5cm}
\end{center}
\caption{Reissner--Nordstr\"{o}m wormholes supported by a generalized Chaplygin gas with $\alpha =0.6$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $Q$, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius of the original manifold. }
\label{frn2}
\end{figure}
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{frn3.eps}
\vspace{-0.5cm}
\end{center}
\caption{Reissner--Nordstr\"{o}m wormholes supported by a Chaplygin gas ($\alpha =1)$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $Q$, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius of the original manifold. }
\label{frn3}
\end{figure}
The throat radius should satisfy the equation
\begin{equation}
a_{0}-M-2A(2\pi )^{\alpha +1}a_{0}^{2\alpha +1}(a_{0}^{2}-2Ma_{0}+Q^{2})^{(1-\alpha)/2}=0.
\label{rn5}
\end{equation}
If $\alpha =1$ the equation is cubic and it can be solved analytically (see Ref. \cite{chaply}) to obtain $a_{0}$, while for other values of $\alpha $ it should be solved numerically. For the stability analysis of the solutions, we obtain from Eqs. (\ref{p12}) and (\ref{rn1}) the second derivative of the potential, which is
\begin{equation}
V''(a_{0})=\frac{2 \left\{ -(1 + \alpha ) a_{0}^{3} + M (3 + 4 \alpha ) a_{0}^{2} - \left[ 2 Q^2 \alpha + 3 M^2 (1 + \alpha )\right] a_{0} + M Q^2 (1 + 2 \alpha ) \right\} }
{a_{0}^{3}(a_{0}^{2} - 2 M a_{0} + Q^{2})}.
\label{rn6}
\end{equation}
The solutions of Eq. (\ref{rn5}) correspond to stable wormholes if replaced in Eq. (33) they satisfy $V''(a_{0})>0$. The calculations were done with standard software and the results for representative values of the parameters are shown in Figs. \ref{frn1}, \ref{frn2} and \ref{frn3}. When $0<\alpha <1$ we can summarize the results of Figs. \ref{frn1} and \ref{frn2} as follows:
\begin{enumerate}
\item When $0\le |Q|<1$ and $|Q|$ is not very close to $1$, for any value of $AM^{\alpha +1}$ there is always one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and tends to the horizon radius $r_{h}/M$ of the original manifold for large values of $AM^{\alpha +1}$.
\item If $|Q|<1$ and $|Q|$ is very close to $1$, for small values of $AM^{\alpha +1}$ there is one unstable solution. For intermediate values of $AM^{\alpha +1}$ there are three solutions: the larger and the smaller ones are unstable and the middle one is stable. For large values of $AM^{\alpha +1}$ there is one unstable solution, which is very close to the horizon of the original manifold.
\item When $|Q|>1$, for small values of $AM^{\alpha +1}$ there are two solutions: the larger one is unstable and the smaller one is stable, while for large values of $AM^{\alpha +1}$ there are no solutions.
\end{enumerate}
The case $\alpha =1$ corresponds to the Chaplygin gas studied in detail in a previous work \cite{chaply}. For comparison, the results are shown in Fig. \ref{frn3}. We see that
\begin{enumerate}
\item When $0\le |Q|<1$ and $|Q|$ is not very close to $1$, for small values of $AM^{\alpha +1}$ there is always one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and it cuts to the horizon radius $r_{h}/M$ of the original manifold for a finite value of $AM^{\alpha +1}$, so there are no solutions for large values of $AM^{\alpha +1}$.
\item If $|Q|<1$ and $|Q|$ is very close to $1$, for small values of $AM^{\alpha +1}$ there is one unstable solution. For intermediate values of $AM^{\alpha +1}$ there are two solutions: the larger one is unstable and the smaller one is stable. For large values of $AM^{\alpha +1}$ there are no solutions.
\item When $|Q|>1$, for small values of $AM^{\alpha +1}$ there are two solutions: the larger one is unstable and the smaller one is stable, while for large values of $AM^{\alpha +1}$ there are no solutions.
\end{enumerate}
We see that the main difference between the cases $0<\alpha <1$ and $\alpha =1$ is that in the former case there is an extra unstable solution for some values of the parameters.

\section{Wormholes with a cosmological constant}\label{schw-ds}

The vacuum solution of the Einstein equations with a cosmological constant has metric functions
\begin{equation}
f(r)=1-\frac{2M}{r}-\frac{\Lambda}{3}r^2, \hspace{1cm} h(r)=r^2,
\label{cc1}
\end{equation}
where $M$ is the mass and $\Lambda $ is the cosmological constant. If $\Lambda M^{2}>1/9$ the function $f(r)$ is always negative, so we take $\Lambda M^{2}\le 1/9$. When $0<\Lambda M^{2}\le 1/9$, i.e. the Schwarzschild--de Sitter case, the geometry has two horizons, which are placed at
\begin{equation}
r_{h}^{dS}=\frac{-1+i\sqrt{3}-(1+i\sqrt{3})\left(-3\sqrt{\Lambda}M+i\sqrt{1-9\Lambda M^{2}}
\right)^{2/3}}{2\sqrt{\Lambda}\left(-3\sqrt{\Lambda}M+i\sqrt{1-9\Lambda M^{2}}\right)^{1/3}},
\label{ds2}
\end{equation}
\begin{equation}
r_{c}^{dS}=\frac{1+\left(-3\sqrt{\Lambda}M+i\sqrt{1-9\Lambda M^{2}}\right)^{2/3}}
{\sqrt{\Lambda}\left(-3\sqrt{\Lambda}M+i\sqrt{1-9\Lambda M^{2}}\right)^{1/3}}.
\label{ds3}
\end{equation}
The event horizon radius $r_{h}^{dS}$ is a continuous and increasing function of $\Lambda $, with  $r_{h}^{dS}\rightarrow 2M$ when $\Lambda \rightarrow 0^{+}$ and $r_{h}^{dS}=3M$ when $\Lambda M^{2}=1/9$. The cosmological horizon radius $r_{c}^{dS}$ is a continuous and decreasing function of $\Lambda $, with $r_{c}^{dS}\rightarrow +\infty$ when $\Lambda \rightarrow 0^{+}$ and $r_{c}^{dS}=3M$ when $\Lambda M^{2}=1/9$. If $\Lambda =0$, the Schwarzschild geometry with horizon radius $r_{h}^{S}=2M$ is obtained. When $\Lambda <0$, i.e. the Schwarzschild--anti de Sitter case, the event horizon is placed at
\begin{equation}
r_{h}^{AdS}=\frac{1-\left(-3\sqrt{|\Lambda|}M+\sqrt{1+9|\Lambda|M^{2}}\right)^{2/3}}
{\sqrt{|\Lambda|}\left(-3\sqrt{|\Lambda|}M+\sqrt{1+9|\Lambda|M^{2}}\right)^{1/3}}.
\label{ads1}
\end{equation}
The horizon radius $r_{h}^{AdS}$ is a continuous and increasing function of $\Lambda$, with values in the interval $0<r_{h}<2M$, with $r_{h}^{AdS}=0$ when $\Lambda\rightarrow -\infty$ and $r_{h}^{AdS}\rightarrow 2M$ when $\Lambda \rightarrow 0^{-}$. Thin--shell wormholes constructed from Eq. (\ref{cc1}) have a throat of exotic matter and vacuum with a cosmological constant outside it. If $0<\Lambda M^{2}<1/9$ the wormhole throat radius should be taken in the range $r_{h}^{dS}<a_{0}<r_{c}^{dS}$, and if $\Lambda M^{2}=1/9$ the construction of the wormhole is not possible, because $r_{h}^{dS}=r_{c}^{dS}=3M$. If $\Lambda <0$ the wormhole throat radius $a_{0}$ should be greater than $r_{h}^{AdS}$. Using Eqs. (\ref{e13}) and (\ref{e14}), we obtain that the energy density and the pressure at the throat are given by
\begin{equation}
\sigma _{0}=-\frac{\sqrt{-\Lambda a_{0}^{3}+3a_{0}-6M}}{2\pi a_{0} \sqrt{3 a_{0}}},
\label{cc2}
\end{equation}
and
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{fcc1.eps}
\vspace{-0.5cm}
\end{center}
\caption{Wormholes with a cosmological constant, supported by a generalized Chaplygin gas with $\alpha =0.2$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $\Lambda $, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius or (if $\Lambda >0$) larger than the cosmological horizon radius of the original manifold.  }
\label{fcc1}
\end{figure}
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{fcc2.eps}
\vspace{-0.5cm}
\end{center}
\caption{Wormholes with a cosmological constant, supported by a generalized Chaplygin gas with $\alpha =0.9$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $\Lambda $, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius or (if $\Lambda >0$) larger than the cosmological horizon radius of the original manifold. }
\label{fcc2}
\end{figure}
\begin{figure}[t!]
\begin{center}
\vspace{0cm}
\includegraphics[width=12cm]{fcc3.eps}
\vspace{-0.5cm}
\end{center}
\caption{Wormholes with a cosmological constant, supported by a Chaplygin gas ($\alpha =1)$: the solid curves represent the static solutions with throat radius $a_{0}$ which are stable under radial perturbations for given parameters $A$, $M$ and $\Lambda $, and the dotted curves those unstable under radial perturbations. The gray zones are unphysical, corresponding to a throat radius smaller than the horizon radius or (if $\Lambda >0$) larger than the cosmological horizon radius of the original manifold. }
\label{fcc3}
\end{figure}
\begin{equation}
p_{0}=\frac{-2 \Lambda a_{0}^{3}+3a_{0}-3M}{4 \pi a_{0} \sqrt{3a_{0}(-\Lambda a_{0}^{3}+3a_{0}-6M)}}.
\label{cc3}
\end{equation}
From Eq. (\ref{e15}), the throat radius $a_{0}$ should satisfy the equation
\begin{equation}
-\frac{2\Lambda}{3}a_{0}^{3}+a_{0}-M-2A(2\pi )^{\alpha+1}a_{0}^{3(\alpha +1)/2}\left( -\frac{\Lambda}{3}a_{0}^{3}+a_{0}-2M\right)^{(1-\alpha)/2}=0.
\label{cc4}
\end{equation}
When $\alpha =1$ this equation is cubic in $a_{0}$ and it can be solved analytically \cite{chaply}, while for other values of $\alpha $ it should be solved numerically. By replacing Eq. (\ref{cc1}) in Eq. (\ref{p12}), we obtain
\begin{equation}
V''(a_{0})=\frac{2 \left[ -2 \alpha \Lambda a_{0}^{4} + 3 M (-1 + 2 \alpha) \Lambda a_{0}^{3} + 3 (1 + \alpha) a_{0}^2 -  3 M (3 + 4 \alpha )a_{0} + 9 M^2 (1 + \alpha) \right] }{a_{0}^3 (\Lambda a_{0}^3 -3 a_{0} + 6 M )}.
\label{cc5}
\end{equation}
The solutions of Eq. (\ref{cc4}) correspond to stable wormholes if $V''(a_{0})>0$.  The calculations were done with standard software and the results for representative values of the parameters are shown in Figs. \ref{fcc1}, \ref{fcc2} and \ref{fcc3}. When $0<\alpha <1$ we can summarize the results of Figs. \ref{fcc1} and \ref{fcc2} as follows:
\begin{enumerate}
\item If $\Lambda > 0$, for any value of $AM^{\alpha +1}$ there is always one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and tends to the horizon radius $r_{h}^{dS}/M$ of the original manifold for large values of $AM^{\alpha +1}$. Except for the presence of the cosmological horizon, the results are very similar to the ones obtained for the Schwarzschild wormholes ($\Lambda =0$), which were studied in the previous section (uncharged case).
\item When $\Lambda <0$ and $|\Lambda|M^{2}$ is not large, for any value of $AM^{\alpha +1}$ there is one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and tends to the horizon radius $r_{h}^{AdS}/M$ of the original manifold for large values of $AM^{\alpha +1}$.
\item If $\Lambda <0$ and $|\Lambda|M^{2}$ is large, for small values of $AM^{\alpha +1}$ there is one unstable solution. For intermediate values of $AM^{\alpha +1}$ there are three solutions: the larger and the smaller ones are unstable and the middle one stable. For large values of $AM^{\alpha +1}$ there is one unstable solution, which is very close to the horizon $r_{h}^{AdS}/M$ of the original manifold.
\end{enumerate}
The case $\alpha =1$ corresponds to the Chaplygin gas wormholes previously studied in Ref. \cite{chaply}. The results are shown in Fig. \ref{fcc3} for comparison. We have that
\begin{enumerate}
\item When $\Lambda > 0$, for small values of $AM^{\alpha +1}$ there is always one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and it cuts to the horizon radius $r_{h}^{dS}/M$ of the original manifold for a finite value of $AM^{\alpha +1}$, so there are no solutions for large values of $AM^{\alpha +1}$. Again, except for the presence of the cosmological horizon, the results are very similar to the ones obtained for the Schwarzschild wormholes ($\Lambda =0$), which were analyzed in the previous section (uncharged case).
\item If $\Lambda < 0$ and $|\Lambda| M^{2}$ is not large, for small values of $AM^{\alpha +1}$ there is always one unstable solution. The throat radius $a_{0}/M$ decreases with $AM^{\alpha +1}$ and it cuts to the horizon radius $r_{h}^{AdS}/M$ of the original manifold for a finite value of $AM^{\alpha +1}$, so there are no solutions for large values of $AM^{\alpha +1}$.
\item  When $\Lambda <0$ and $|\Lambda| M^{2}$ is large, for small values of $AM^{\alpha +1}$ there is one unstable solution. For intermediate values of $AM^{\alpha +1}$ there are two solutions: the larger one is unstable and the smaller one is stable. For large values of $AM^{\alpha +1}$ there are no solutions.
\end{enumerate}
We can see that the main difference between the cases $0<\alpha <1$ and $\alpha =1$ is that in the former case there are values of the parameters for which exist one unstable additional solution for anti--de Sitter wormholes.

\section{Conclusions}\label{conclu}

In this paper, spherically symmetric thin--shell wormholes supported by a generalized Chaplygin gas were theoretically constructed by using the usual cut and paste procedure for a general class of metrics. Such kind of fluid has received great attention in cosmology in the last few years, because it provides a possible explanation for the accelerated expansion of the Universe, and it has also been considered in previous wormhole studies. For a general class of metrics, the equation that determines the possible radii of the throat for static wormholes was obtained and the stability of the static configurations under radial perturbations was analyzed using the standard potential method. The energy density and the pressure at the throat were obtained as functions of the throat radius. Examples of wormholes made from Reissner--Nordstr\"om and Schwarzschild with a cosmological constant metrics were analyzed in detail and the results were compared with those obtained in a previous work \cite{chaply} for the original Chaplygin gas. It was found that for properly chosen values of the parameters, stable solutions are also possible when the gas exponent $\alpha $ is smaller than $1$. The main difference when $0<\alpha <1$ is the presence of an extra unstable solution which appears for some values of the parameters in both charged and anti--de Sitter wormholes. In the charged case it happens when the charge $|Q|$ is slightly smaller than the mass $M$, while in the anti--de Sitter case for large values of $|\Lambda| M^2$.

\section*{Acknowledgments}

This work has been supported by Universidad de Buenos Aires and CONICET.

redpajama_set_name: RedPajamaArXiv


text:
Q: Can't connect to local MySQL server through socket '/cloudsql/my-gcp-project:europe-west2:instance' I'm getting an error when trying to connect to a MySQL server using a service in Cloud Run. This is the method I'm using to connect:
def get_connection(self) -> mysql.connector.connection.MySQLConnection:
    """ Get a connection to the database

    Returns:
        mysql.connector.connection.MySQLConnection: Connection to the
            database"""
    db_socket_dir = '/cloudsql'

    config = {
        'user': self.user,
        'password': self.password,
        'host': self.host,
        'client_flags': [ClientFlag.SSL],
        'ssl_ca': './ssl/server-ca.pem',
        'ssl_cert': './ssl/client-cert.pem',
        'ssl_key': './ssl/client-key.pem',
        'database': self.database
    }

    if os.environ['LOCAL_RUN'] == "no":
        config['unix_socket'] = '{}/{}'.format(
            db_socket_dir,
            self.instance_connection_name
        )
    else:
        config['port'] = 3307

    return mysql.connector.connect(**config)

Notably, this works fine when I run everything locally. I.e. when I have a proxy server running and my service running in terminal. I have also used this method in another service and it worked. My feeling is that it is linked to the error message saying "Can't connect to local MySQL server" because I am not intentionally trying to connect to a local server, but I can't figure why the method would think I am.
Other context, the value of 'host' is the Public IP address for my instance.
Any ideas?
I have looked at a lot of other similar questions on here but nothing seems to provide an answer to my exact problem.
The full error message is:
Traceback (most recent call last):
  File "/usr/local/lib/python3.9/site-packages/flask/app.py", line 2073, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/lib/python3.9/site-packages/flask/app.py", line 1518, in
    full_dispatch_request
      rv = self.handle_user_exception(e)
  File "/usr/local/lib/python3.9/site-packages/flask/app.py", line 1516, in
    full_dispatch_request
    rv = self.dispatch_request()
  File "/usr/local/lib/python3.9/site-packages/flask/app.py", line 1502, in
    dispatch_request
    return self.ensure_sync(self.view_functions[rule.endpoint])(**req.view_args)
  File "/app/main.py", line 28, in run_query_ep
    sql = SQL()
  File "/app/SQL.py", line 29, in __init__
    self.connection = self.get_connection()
  File "/app/SQL.py", line 58, in get_connection
    return mysql.connector.connect(**config)
  File "/usr/local/lib/python3.9/site-packages/mysql/connector/__init__.py", line 272, in connect
    return CMySQLConnection(*args, **kwargs)
  File "/usr/local/lib/python3.9/site-packages/mysql/connector/connection_cext.py", line 85, in __init__
    self.connect(**kwargs)
  File "/usr/local/lib/python3.9/site-packages/mysql/connector/abstracts.py", line 1028, in connect
    self._open_connection()
  File "/usr/local/lib/python3.9/site-packages/mysql/connector/connection_cext.py", line 241, in _open_connection
    raise errors.get_mysql_exception(msg=exc.msg, errno=exc.errno,
  mysql.connector.errors.InterfaceError: 2002 (HY000): Can't connect to local MySQL server through socket '/cloudsql/my-gcp-project:europe-west2:instance' (2)

EDIT:
I'm using a cloudbuild.yaml file to deploy the service:
steps:
  - name: "gcr.io/cloud-builders/docker"
    args:
      - build
      - "--no-cache"
      - "-f"
      - "backend/SQL/Dockerfile"
      - "-t"
      - "gcr.io/$PROJECT_ID/sql:${SHORT_SHA}"
      - "./backend/SQL"
    id: Build
  - name: gcr.io/cloud-builders/docker
    args:
      - push
      - "gcr.io/$PROJECT_ID/sql"
    id: Push
  - name: "gcr.io/google.com/cloudsdktool/cloud-sdk:slim"
    args:
      - run
      - deploy
      - sql
      - "--image=gcr.io/$PROJECT_ID/sql:${SHORT_SHA}"
      - "--add-cloudsql-instances=${_CLOUD_SQL_ADDRESS}"
      - "--service-account=sql-service@my-gcp-project.iam.gserviceaccount.com"
      - "--region=europe-west2"
      - "--platform=managed"
      - "--allow-unauthenticated"
      id: Deploy
      entrypoint: gcloud

tags:
  - gcp-cloud-build-deploy-cloud-run
  - gcp-cloud-build-deploy-cloud-run-managed
  - sql


Previous post: Celebrating four years of weaving!
This will have to be cut down substantially. I count ten slots for non-bonbons (the candied peels can all go in a single candy cup) and 26 kinds of bonbons (ganache-based chocolates). That's 35 kinds of chocolates, and I know from experience that only 25-28 will fit into a box. (Last year had 28 flavors and we could only fit 26 into a box.) So I may have to do some consolidation, and some elimination.
At any rate, this year's chocolate order is just about ready, and starting in November I will begin devoting my weekends to candy-making. Nonperishables first: chocolate covered dried fruits, caramels. Then the less-perishable fudges. The weekend before Thanksgiving, I start in on chocolate shells for the bonbons. And, finally, on Thanksgiving weekend, a frenzy of bonbon-making (they are very perishable), culminating in the annual chocolate-packing party on Sunday. Four or five friends come over and help pack about 2500 chocolates/candies into candy cups, then into boxes, then into shipping boxes. Then I put on mailing labels and make the annual trip to the post office with a car stuffed to the gills with boxes. The candy-packers go home with all the leftover chocolates they want to take. (It's a popular job!) And chocolate season is over for the year.
Woven shibori update: I dyed the most recent test, with disappointing results. First, I don't think I used enough soda ash, so the gold mostly didn't "take". Second, the black dye bled badly (perhaps the soda ash contributed?) and produced a rather saddened effect. Still contemplating what to do. I have some WashFast acid dyes which I think (but am not sure) have a smaller dye molecule so should penetrate the resist better, so perhaps I will try that in an acid dyebath for next time. I may also switch to nylon monofilament in an attempt to get something that removes more smoothly. Any idea where I can find monofilament that's about the weight of 60/2 silk? The fishing section of a sport store, perhaps?
Places like Hancock fabrics and JoAnns usually sell spools of nylon monofilament. The "invisible" sewing thread. You might also consider the nylon thread for upholstery – it is stronger, but perhaps thicker.
Tien, every year your chocolate making fascinates me! It sounds heavenly and like so much fun. This year I swear I'm going to make at least a few kinds of candy — although they'll be more of the Homemade Candy by Farm Journal type than the exquisite chocolates you're describing here.


text:
Sustainability is a key driver for 3M and its global reach
The global, science-based company has recently released its 2020 Sustainability Report, reflecting its core purpose of growing 3M's business by collaborating to solve some of humanity's greatest challenges.
The report outlines 3M's vision for a sustainable future, alongside its key commitments and achievements, including its use of recycled materials in its products.
A central focus for 3M has been to embed sustainability in every new product it brings to market.
3M launches hundreds of new products each year, spanning nearly every industry. Beginning in 2019, each 3M product entering the new product commercialisation process included a Sustainability Value Commitment that demonstrates how it impacts the company's sustainable goals.
3M's close working relationship with its supply chain partners aims to source raw materials purchased from sustainable and or recycled resources.
Examples of 3M's Sustainability Value Commitment for new products include reusability, recyclability, energy savings, waste reduction, water savings, responsible sourcing, and the use of renewable materials appropriate to the specific product, from the beginning to the end of each product's life cycle.
3M packaging materials are reviewed annually and importantly PVC is being phased out and replaced by PET or recycled paperboard all contributing to the reduction of waste to the litter stream.
3M Australia is a member of Chemistry Australia's initiative, Plastics Stewardship Australia.
Plastics Stewardship Australia continues to work with communities, industry and government to develop solutions for the complex challenges associated with managing Australia's waste.
3M Australia is also a long-standing member of the Australian Packaging Covenant as well as being a member of the Close the Loop (Previously Planet Ark) organization.
Refer to 3M Sustainability goals and policies for more information.


Combined Storage Weight. Closely relevant to the number of dumbbells you need this to hold is the weight the fact that the rack needs to hold. Assuming you have a lot of heavier hand weights then you will need a very good rack. Before buying any dumbbell rack, ensure you find out its support capacity to check that you aren't overloading it. It is very easy to underestimate how much all of your hand weights weigh- ten pairs starting from 20 kg to 52 kg might weigh about 600 kg, so make sure that you find a rack that is lots of strong enough for your needs. Overloading a rack is dangerous, mainly because it could potentially collapse which could experience very nasty consequences taking into consideration how heavy a filled rack is.
Victoriajacksonshow - 25 cool graduation party ideas hative. Make your graduation party the focus of all your guests for the great celebration of a job well done as well as your bright future you deserved such a great celebration of your achievements personalize your graduation party by following these 25 useful tips to catch everyone's attention. 25 creative graduation announcement ideas hative. It celebrates not only how far the graduate has arrived, but the excitement of the new journey to start with it's very important to mark your graduation status using unique graduation announcements for your gorgeous graduation celebration here are some great ideas for creative graduation announcements to inspire you. 25 cool graduation cake ideas hative. Personalize the cake with shapes like graduation scroll, books or even graduation cakes you can also sign the name of the person, graduation date and even the name of the university to memorize the great day here are some cool ideas to incorporate to whip up a stunning graduation cake. Beautiful graduation decoration ideas homemade for 25 diy. It is, beautiful graduation decoration ideas homemade for 25 diy graduation party decoration ideas hative, a picture of graduation decoration ideas homemade that might be your ideas choosing the best decorating room there are countless ideas and inspirations which might be useful for your work on setting up your room interior design. Newest graduation decoration ideas homemade for 25 cool. It is, newest graduation decoration ideas homemade for 25 cool diy graduation cap ideas hative, an image of graduation decoration ideas homemade which could be your insight choosing the best decorating room there are many ideas and inspirations which might be useful for your work on building your room house interior design. 25 cool diy graduation cap ideas hative life. 25 cool diy graduation cap ideas hative visit discover ideas about diy graduation caps hotter by one degree: glittering graduation cap create this stunning graduation cap with gold glittering cap board and black glittering letters you can have a great party even on a budget with these diy graduation party ideas long time coming but. Attractive graduation decoration ideas homemade for 25 diy. You can browse another photos similar to this from the designs gallery below including: 33 graduation party ideas for high school for 2017, 558 best graduation party ideas images on pinterest, 25 cool diy graduation cap ideas hative, graduation centerpieces, 3 simple and inexpensive graduation decorations, 25 diy graduation party ideas a little. 25 cool diy graduation cap ideas hative furniture. 25 cool diy graduation cap ideas hative visit discover ideas about graduation hats no bow graduation cap congratulation to the graduate success with this elegant graduation cap with all its sparkling beadings as well as smooth pink ribbon bow to add up more beauty to this art piece graduation party ideas see more what others are. 25 diy graduation party decoration ideas hative. 25 diy graduation party decoration ideas hative 25 diy graduation party decoration ideas hative visit discover ideas about graduation 2016 cut paper into triangles, mark up the sign of your school thread them through to create this graduation pennant garland to reflect the spirit of your school and ornament your graduation party. 25 killer ideas to throw an amazing graduation party. To make your party planning just a tad easier and a lot less stressful , i've rounded up the best of the best grad party inspiration and ideas so you can host the best bash around! cool themes, amazing dessert table ideas, diy ideas galore and unique decorations, it's all here! check out these 25 killer graduation party ideas!.


text:
Twitter Amplifying Women's Sports With #GoldenTweet Awards
Media / June 21st, 2019
By Ian Thomas
Photo Credit: Michael Chow-USA TODAY Sports
Twitter has rolled out several initiatives alongside the 2019 FIFA Women's World Cup, aiming to drive further awareness to women's sports.
The social networking service is now launching its latest effort, which will celebrate the best tweets about the best female soccer players in the world.
That will come via the #GoldenTweet Awards, which will see members of Twitter's analytics team identify the best tweets about the tournament by users in a variety of countries, including Argentina, Australia, Brazil, Canada, Chile, France, Germany, Japan, Mexico, Spain, South Korea, and the UK. Following the final, the most popular tweets across multiple categories will be crowned #GoldenTweet winners. Each of those winners will be presented a limited-edition trophy that features the Twitter bird.
"Women's sports lack the visibility they deserve and we hope the #GoldenTweet Awards are one way to change that," said Jay Bavishi, senior manager and global lead on quadrennial events at Twitter. "By amplifying the conversation around the #FIFAWWC on Twitter, we're hoping to bring more visibility and awareness to these incredible athletes."
In the U.S., Twitter will hand out awards across four separate categories: athlete, brand, fan and journalist.
To find those best U.S. Tweets, Twitter has put together a panel of judges who have knowledge of both social media and soccer. The group includes Twitter Head of US Content Partnerships Laura Froelich, ESPN Senior Vice President of Marketing Laura Gentile, Colorado Rapids Goalkeeper and former U.S. Men's National Team Player Tim Howard, and Karina LeBlanc, head of women's football for Concacaf and a FOX Sports' FIFA Women's World Cup analyst.
"We're witnessing greatness on the pitch, and the world is coming together to talk about it on Twitter," Froelich said. "I'm looking forward to seeing the best of the best compete for the FIFA Women's World Cup Winner's trophy, as well as our #GoldenTweet honors!"
The idea was spawned following the success of a similar 'Golden Tweet' award that the Twitter UK office handed out following the 2018 FIFA World Cup last summer. Five awards were handed out to the tweets that had the most traction and retweets within the U.K. For example, The player award was given to Kyle Walker, whose usage of a meme that featured fellow English National Team Defender Harry Maguire has garnered more than 70,000 retweets and 270,000 likes to-date.
Yeah so a good header doesn't hurt. I mean the moment you head it proper, you feel it's a good one. Know what I mean love? pic.twitter.com/a5b8UqDjv2
— Kyle Walker (@kylewalker2) July 7, 2018
After the effort gained traction – even Walker's Manchester City teammates joked about the award with him on Twitter – the company decided to expand it to even more countries during the Women's World Cup this year.
It fits into Twitter's larger strategy to elevate the level of conversation around women's sports. For the Women's World Cup in particular, Twitter has a wide-ranging partnership with Fox in the U.S. to post instant goal highlights as well as livestream a nightly recap show from Paris. Twitter is also creating individual pages for each match, curating the best content from publishers.
"I'm proud to be a part of this FIFA Women's World Cup, as it is showing that women can play sports at the highest level and our game continues to grow and develop globally," LeBlanc said.
While the Women's World Cup has proven to be a boon for Twitter – during the final match of the 2015 Women's World Cup between the United States and Japan, tweets about women's soccer spiked 2,000% – Twitter is also hoping that these awards will encourage users to not only tweet about the on-field successes of the players, but issues that matter to the players off it as well, such as equality.
"Sports are a unifier, and supporting women's sports – their growth and vital role in our society – is a must for girls everywhere," Gentile said. "The #GoldenTweet campaign will help spotlight incredible athletes and drive change."
Winners of the awards will be determined in the week or so following the Women's World Cup Final, set to be played on July 7.
"Having always been a fan and a supporter of the USWNT, I'm proud to help amplify their already passionate fanbase with the #GoldenTweet campaign and celebrate these incredible women during their time in France as they compete for the FIFA Women's World Cup," Howard said.
ESPN Being More 'Strategic' With New Podcasts
How Black Panther And Iron Man Found A Home With The Brooklyn Nets
by Ed Moran / January 20, 2020
Tokyo Olympics Kicking Off NBC's Peacock Streaming Service Sports Strategy


text:
Reformist Newspaper in Iran Shut Down Over Imam Ali Article
While the first newspaper under President Rouhani's tenure has been shut down for covering a sensitive religious topic, there are signs that the days of mass newspaper closures have not returned.
The staff of Bahar newspaper poses with a copy of their paper after it was banned, in Tehran, Oct. 30, 2013. - Twitter/@pedestrian
A correspondent in Tehran
TEHRAN — On Monday, Oct. 28, Iran's Press Supervisory Board banned the reformist newspaper Bahar after it published a controversial article on Imam Ali. Alaedin Zohourian, chairman of the paper's board, told the Mehr news agency on Oct. 28 that the case has been referred to the courts. The government of President Hassan Rouhani has thus far gone along with the decision, which was not necessarily seen as part of a broader media crackdown.
Bahar published the article that sealed its fate on Oct. 28. In "Imam Ali, a Political Leader or a Religious Model?" the writer, Ali Asghar Gharavi, claimed that the religious role of the first Shiite imam was more important than his political role. He wrote, "He [Ali] is not just the political ruler for a few days of passing power in this world. Ali, more than being the commander of the faithful, is the imam and the role model for humanity."
The article appeared one day before the Eid al-Ghadeer in Iran, an event and major point of contestation between the Shiites of Iran and the Sunni Muslims of the Arab world. Shiite Muslims believe that on this day, Muhammad, the prophet of Islam, appointed Ali as his successor. Historically, however, the first three caliphs ruled before Ali. For Sunnis, Ali is simply the fourth caliph, but Shiites consider him to be the first (of twelve) imams.
A scholar of religion, who spoke to Al-Monitor on condition of anonymity, asserted, "Look, this article has crossed two red lines. One is that its author is a well-known member of the Freedom Movement of Iran [Nehzat-e Azadi-e Iran]," the opposition party that has been banned from political activities for more than two decades. During the past three decades, especially after the unrest following the 2009 presidential elections, a number of party members, including its leader and other prominent figures, were arrested. "The second problem," the scholar continued, "is the content of the article."
"The establishment in Iran is stressing the point that the velayat-e faqih [guardianship of the jurist] is in fact the continuation of the rulership [velayat] of Imam Ali. Knowing this fact, Gharavi still emphasized the point that Imam Ali was more of a spiritual leader than a political one. In other words, he has questioned the position of Ayatollah [Ali] Khamenei as the political leader." In one part of the article Gharavi wrote, "Imam Ali repeatedly mentions in Nahj al-BalaghaI [a book of quotations and sermons attributed to him] that political rule can only be achieved by people voting and paying allegiance."
Imposing a ban on Bahar, a strong supporter of the reformists during election campaigns, has angered some journalists belonging to the reformist faction. Many of them have asked Rouhani, via social networks, to block any return to the practice of closing newspapers.
A journalist working for the reformist Etemaad, who also asked to remain anonymous, observed, "They are starting to put pressure on the press. Conservatives are clearly showing that they could only tolerate this situation for three months. This is exactly what happened in 1997 when reformist President Mohammad Khatami took office. I am surprised at Rouhani's silence. If he starts his retreat this early, the conservatives will become bolder and bolder. He needs to learn a lesson from what happened during Khatami's term."
The first newspaper shuttered under Khatami's presidency was Jame'e. It began publishing in February 1998 and was closed in June that same year. During the spring of 2000, numerous reformist publications were simultaneously banned by order of Ayatollah Khamenei.
Unfortunately for reformist journalists, Rouhani's administration has not criticized the decision of the Press Supervisory Board, and, in fact, Minister of Culture and Islamic Guidance Ali Jannati has called it a "legitimate move." According to the Islamic Republic News Agency, during an interview on Oct. 28, Jannati said, "This article is falsifying the Islamic history, and it has also played a role in creating religious conflict in the country."
A journalist who had previously interviewed Jannati told Al-Monitor, "He [Jannati] is, after all, the son of a cleric. He also retreats quite quickly. This is exactly what I expected of him. I have heard his answers to the questions I have asked him privately and I have seen how those answers change as soon as he is in an official situation." He claimed, "He obviously wants to keep his job."
The editorial board of Bahar, which has been repeatedly closed down during the past few years, has worked hard at damage control. It immediately apologized and announced that the article's publication had been the result of a technical error. Yet, the pressure from the conservatives was so strong that even the voluntarily suspension proposed by the chief editor failed to stop the Press Supervisory Board from shuttering the paper. Following publication of the article, 26 members of the parliament interpellated Jannati and labeled the article "obscene."
One journalist working for Bahar spoke to Al-Monitor, asserting, "It is quite obvious that they were looking for an excuse. We have officially apologized. We have accepted that the article was unsuitable for publication. What else could we have done? We don't want complete freedom of speech, but if they wanted to show a little tolerance, they would accept our letter of apology. The devastating thing is that numerous journalists, whose source of income was coming from this newspaper, are left unemployed. They can no longer pay their rents or their mortgages. Unemployment is very depressing. Today I was about to go to the office, around noon, when I just remembered that the newspaper was shut down. It almost made me cry."
A reformist editor of another paper in Iran believes that what happened to Bahar does not necessarily forebode a new crackdown on the press similar to those under Khatami or even President Mahmoud Ahmadinejad. He told Al-Monitor, "When a newspaper does not have an editor in chief and is being managed in an unorganized manner, these things can happen. Anyone who has worked as a journalist for even one day in Iran knows that publishing such an article will result in the newspaper being closed down. Rouhani promised to expand the freedom of press, and he has, but he never promised that newspapers could question the position of Khamenei and still operate. This is beyond his authority."
The editor further observed, "Rouhani said, from day one, that he is looking for moderation. It is fine if a newspaper wants to be a little more radical, but it is not logical to ask for the full support of the government. The authority of the president in Iran is limited."
Despite the closure of Bahar, it is undeniable that in the last three months there has been more press freedom in Iran, with reformist newspapers having increasingly challenged what were previously red lines. In September, the reformist Shargh Daily published an interview with Alan Eyre, the US Department of State Persian-language spokesperson, and many reformists journalists who had essentially been silenced for four years are no longer afraid to take a critical position when interviewing conservative or even hard-line politicians. What's more, some publications that had been closed down have recently been issued permits to once again operate.
Alireza, a 40-year-old civil engineer reading the headlines at a newsstand on Valiasr Street in Tehran, offered, "I think the situation has improved a lot for the press in the past three months. I did not know that a ban had been imposed on Bahar, but I know that the newspapers Hammihan and Neshat received publishing permits. Other newspapers also seem to be publishing more freely these days. I think, regardless of what happened with Bahar, I have reasons to be hopeful."
At times, Al-Monitor withholds the bylines of our correspondents for the protection of our authors. Different authors may have written the individual stories identified on this page.