Visit Eldercare Columns and select five of the topics listed. Then email your choices to familyeldercare@juno.com and receive those columns by return email. Do your aging parents resent, resist or refuse your help?
Have they shut you out?
Does worry control your life?
When the Old Block Chips comes to the rescue.
"Emotional, legal and financial issues can overwhelm adult children when their challenging elderly parents need but resist care. Carol's survival guide presents creative solutions for managing difficult elders, and reveals a realstic look at a faulty eldercare system." ~ Jacqueline Marcell, author Elder Rage.
After my mother developed dementia and my father booted me from his life, I felt like a geriatric orphan, abandoned by both parents in my own very adult years. For months shock, devastation and pain put my life in freeze-frame. I suffered financial, emotional and social losses. When the daze finally cleared, part of my personal therapy took the form of researching what I thought was my bizarre situation.
Parental rejection of adult children is not as rare as I thought. Because many children who are rejected or tormented by ornery elderly parents keep it to themselves, this is an under-reported facet of eldercare, one that will escalate as the population ages. I discovered another under-reported phenomenon, that of the toxic ager. Some of our parents' late life rejection and difficult behavior results from toxic aging.
I found plenty of information on eldercare but none of it addressed the problems faced by estranged adult children and little of it applied to children with supremely difficult parents. Dealing with eldercare while helping ourselves through the trauma of parental rejection as we approach our own senior moments poses special challenges (I love a good euphemism).
I am a writer who knows the healing value of writing, so producing a book was a natural response. Writing about all I learned helped me deal with this tragic situation. As I met an increasing number of geriatric orphans, I realized the information I gathered could benefit many others.
Sometimes we need a nudge to do what we know will improve our lives. Reading about techniques and strategies helps, but how many of us practice what we read in self-help books? There wouldn't be so many of them if we did. So, I revised the content, adding exercises at the end of each chapter to get you started on your personalized plan for better caregiving, better health and a better life - yours!
Print books are costly and take a long time to publish and distribute. I did try that route. A major publisher told me that, while it was an important topic, eldercare books don't fare well commercially.
So many geriatric orphans I met in my research and self-help journey asked me when the book would be available that I didn't want to wait for a publisher to recognize the market potential of the growing geriatric orphan population. I decided to present it economically and quickly. Only the ebook format can deliver those benefits.
*How to cope with late-life parental rejection.
*How to care for yourself and why it's important that you do.
*Legal rights and recourses for you and your parents.
*Potential financial implications of estrangement.
*How to deal with your parents' doctors.
*Ways to communicate in a strained or broken relationship.
*How to avoid being a difficult ager to your children.
*What emotional and fiscal vulnerabilities to watch for.
- Toxicity Assessment Check List.
- Dozens of resources to help you and your parents.
"Emotional, legal and financial issues can overwhelm adult children when their challenging elderly parents need but resist care. Carol's survival guide presents creative solutions for managing difficult elders, and reveals a realistic look at a faulty eldercare system."- Jacqueline Marcell, author of Elder Rage and host of "Coping with Caregiving" Internet radio program.
"As our society ages, care of an elderly parent is rapidly becoming the most trying issue for adult children for many reasons. . . .Carol Celeste has addressed one aspect of this aging problem—the estranged adult child trying to care for a parent who is both in denial about needing care and openly resentful of the adult child's efforts and offers a humorous, yet insightful discussion of the issues and problems confronted by all."- Edward J. Carnot, Attorney and author of Is Your Parent in Good Hands?
"We all want to believe that aging parents are sweet and loveable. Consequently, when an incessant pattern of verbal and emotional abuse occurs, it becomes toxic—a hidden phenomenon—devastating to families. I am deeply grateful to Carol for using her personal experience and knowledge to help bring this concealed time bomb into public awareness." - Gloria Davenport, Ph.D. and author of Working with Toxic Older Adults.
"I just read the entire ebook and it is worth twice the price!" Reader S.B.
"I cannot thank you enough for this book...sending heartfelt appreciation!" Reader J.W.
"Your book meant a lot to me. There were so many places where the description was so on target." Reader M.W.
Carol Celeste is a California-based award-winning freelance writer and editor, syndicated eldercare columnist and recovering geriatric orphan. Her work covers a range of topics, including travel and business. She branched into eldercare and caregiver well-being when she became a caregiver and geriatric orphan several years ago. She operates Writing to Heal, Writing to Growsm online personal and therapeutic writing courses and licenses a variety of therapeutic writing programs, including a program for personal caregivers. Reach Carol at familyeldercare@juno.com or carol@writingtoheal.com.
This site is intended to be a self-help site. The information contained in this site is not medical or legal advice and does not substitute for medical or legal advice. Always consult an appropriate professional about these issues.


text:
IJshockey is een van de sporten die beoefend werden tijdens de Olympische Winterspelen 1928 in St. Moritz.
Dit ijshockeytoernooi was tevens het derde wereldkampioenschap en het 13e Europees kampioenschap.

Historiek
Er namen 11 teams deel. De Verenigde Staten, 4 jaar eerder nog tweede, ontbraken omdat het land op het moment van het Olympisch toernooi niet aangesloten was bij de internationale ijshockey federatie LIHG. Canada was rechtstreeks geplaatst voor de finaleronde, terwijl de overige 10 landen, verdeeld over 3 poules, de overige finalerondeplaatsen betwistten.

Heren

Voorronde

Groep A

Groep B

Groep C

Finaleronde

Eindrangschikking

IJshockey
IJshockey op de Olympische Spelen
1928
IJshockey in 1928
IJshockey in Zwitserland

redpajama_set_name: RedPajamaWikipedia


The New Zealand Company was formed to develop settlements in New Zealand for British migrants. William Wakefield led an advance expedition in the Tory to Port Nicholson (Wellington) in 1839. On board was German scientist Ernst Dieffenbach, who was to report on the plants, animals and resources of the new land. When the Tory reached New Plymouth, Dieffenbach stayed ashore to explore. He decided that the best view would be from the top of Mt Taranaki (Mt Egmont). There were warnings that the mountain was tapu (under spiritual restriction), but Dieffenbach set off to climb it with the black American cook of the company's land agent Dicky Barrett. They took a tohunga (Māori priest) as guide. The first attempt failed, but in the company of the whaler 'Worser' Heberley, Dieffenbach succeeded, while his Māori companions remained below praying.
Over the next two years Dieffenbach explored Northland and the centre of the North Island, and visited the Chatham Islands.
The New Zealand Company urgently needed flat, fertile, habitable land for the Port Nicholson settlement, and the country west towards Taranaki was examined. Edward Jerningham Wakefield went from Whanganui to Pātea in March 1840. In August, Robert Park, Robert Stokes and Charles Heaphy surveyed the land from Porirua to Taranaki.
They took longer to discover the country east of Wellington. In 1841 the company's chief surveyor, William Mein Smith, sent Stokes over the Remutaka Range to Lake Wairarapa. The following year the company sent Charles Kettle and Alfred Wills with seven Māori through the Manawatū Gorge. Led by the Māori guide Eahu, they sidled southwards along the Tararua Range before finding a good route back over the Remutakas. They reported that the Wairarapa was like an English park, which helped encourage settlement of the area over the next few years.
By the mid-1840s, through the efforts of missionaries and the New Zealand Company, European exploration of the North Island was largely complete (except for remote areas of the Tararua, Ruahine and Urewera ranges).


text:
Q: Removing tuple from nested list with tuple and lists I have a nested output in the form of this:
[(3, [['Avg_Order_Insert_Size', '572.9964086553654'], ['Avg_Order_Insert_Size', '34.858670832934195'],
['Avg_Order_Insert_Size', '22.09531308137768']])]

And I want to get rid of the int (3) which is in a tuple, and save the lists containing strings and ints.
How do I accomplish this in the best way?
My goal is to use the lists within the tuple for creating a dictionary later on, but while these ints are there within the tuple I don't know what to do. Basically I think I want an output such as:
[([['Avg_Order_Insert_Size', '572.9964086553654'], ['Avg_Order_Insert_Size', '34.858670832934195'],
['Avg_Order_Insert_Size', '22.09531308137768']])]

A: I think this is enough:
>>> value = [(3, [['Avg_Order_Insert_Size', ...
>>> value[0] = value[0][1:]
>>> value
    [([['Avg_Order_Insert_Size', '572.9964086553654'], ['Avg_Order_Insert_Size', '34.858670832934195'], ['Avg_Order_Insert_Size', '22.09531308137768']],)]

redpajama_set_name: RedPajamaStackExchange


Welcome to The Church Network's member and guest interface. From here you can manage your relationship with the network. National participation in the network has many benefits as you will discover during your participation. After logging in to your account please use the menu above to access information and resources. After logging in or creating an account you will be redirected to My Account to see your personal profile, renew or join or check your CEU status.
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VERED organic botanicals is an organic skin and body care line integrating therapeutic botanical essential oils that are 100% chemical and preservative free. Creator Vered Back is a master esthetician and herbalist who brings 23 years of combined experience to VERED organic botanicals - integrating holistic modalities such as phytotherapy, medicinal plant research, skilled esthetician expertise, extensive bodywork, and aromatherapy - to develop and share these high quality, 85%-100% organic products for face and body.
No Vered Organic Botanicals products found for these filters.


Statistics Netherlands publishes cause-of-death statistics since 1901. By reclassifications (International Classification of Diseases, now version ICD-10) and by changes in coding rules several method breaks in these statistics have arisen. In the past coding rules, deviating from ICD classifications, have not been documented consistently. The aim of this study is to examine cause-of-death series from 1950 to 2010 for breaks and for possibilities to mend these.


text:
Q: Why does CBCentralManager report "Peer removed pairing information" with 1st phone after 2nd phone is paired with device? Background
I'm using a couple of different hobbyist BLE devices (HM-10).  This particular one is an example of one of them.
It's a simple 4-pin device and I've set all the parameters on it using an Arduino Nano and I can see the data coming into it live since it is connected to USB of my computer and I'm monitoring serial data using Arduino Serial Monitor.

The device works great.  I have it set to Type 3 which forces it to pair/bond (so iphone or ipad will force you to enter the 6-digit PASS code, which I've set on the device.
iPhone App Works
I have written a simple Single-View iOS app which runs on my iphone / ipad and finds local BLE peripherals.
Everything works great.
When I select the item (HM-10) shown above by name I see the iOS pairing dialog & successfully pair the item.
I can send data and everything works fine.  If I go out of distance or turn off bluetooth on my device then come back into proximity or turn the device back on, then the iphone can send data again and I see it in the Serial Monitor.  Works great.
** The Problem **
The problem occurs if I then pair this device with any other iphone or ipad.
After that, the new iphone / ipad can send data to the device with no problem.
However, if I then go back to my original iphone and attempt to send data again, I see the following error raised by the CBCentralManager:
"Peer removed pairing information."
After that, I cannot send data from the original iphone to the device again, unless I go to Bluetooth settings on the phone and say, "Forget this device".
Android & HC-06 Bluetooth Devices
I've developed simple Android apps which use a similar Bluetooth classic device (HC-06) and I could pair multiple devices.
Questions
Does BLE 4.0 Allow multiple devices to be paired & bonded?
Do you know why the Peripheral would remove the pairing info?  (again, this only occurs if I pair the device with a new iphone / ipad).
Is there a way to send something via iOS app that would tell it to renew the pairing information?
Do you have any suggestions (code to try or other) that would allow me to definitively know that this device does or does not allow multiple phones to pair with it? (I've contacted the manufacturer also.)
Note: I also have a another BLE device here and it seems to exhibit the same behavior.  This seems odd though, because I have a BLE speaker that will pair with multiple devices so it must be possible.  And I understand that could depend upon the BLE hardware.

A:
Does BLE 4.0 Allow multiple devices to be paired & bonded?

Sure. But that doesn't mean that the device does. It needs to allocate memory for it, and smaller devices often don't.

Do you know why the Peripheral would remove the pairing info? (again, this only occurs if I pair the device with a new iphone / ipad).

It doesn't remove the pairing info. It's telling you the device has removed the pairing info. Pairing info includes a shared secret, so if one side throws away their secret, there's no point in you keeping yours. You have to create a new one.

Is there a way to send something via iOS app that would tell it to renew the pairing information?

Unfortunately no. Apple doesn't give any access to the pairing experience.
There isn't any standardized way to ask devices this kind of information. Some devices support a single pairing Some support two. Some support "a few." But it's really up to the device.
It is very possible that you will get a better overall experience with a Bluetooth 4.1 or 4.2 device, because they've added better security protocols. With iOS 13, the phone also supports cross-transport key derivation, which possibly will help here (if the device also supports it). But I don't think you can improve things with a Bluetooth 4.0 device.


2021 Chevrolet Avalanche Price, Specs, Release Date – The particular 2021 Chevrolet Avalanche comes from General Motors firm and also is also a whole take to all of the company-new brand lorries in addition to the company-new assortment. There may be a couple of combination relating to complete strategy with design in addition to the engine. Much more changes will probably be accessible in many starts expected days that happen to be helped by brand name-new modern-day technological innovation and far far better productivity. Fancy seem together with the 5-doorway combination will undoubtedly function as the stress from the new format, which may furthermore get enhancements with lots of improvements if necessary. Charge and convenience will inevitably rely on capabilities upfront along with not for foundation versions. It can be supposed to gain a relatively significant difference in USA marketplace and bigger.
The newly designed outside areas could make this automobile refreshing and trendy. The modifications are used virtually in all the methods. The back verdict in the auto you can find steel shades that happen to be lovely with a brand new design. Next, to, its part may have new Well guided taillights plus exhaust method. The most up-to-date lamps method is likely to make the looks more attractive.
In the entrance portion, you will realize the newest grille and front lights. Instructed modern day technological innovation strategy will comprehensive the style of the leading aspect automobile. The fender is made to become sportier that also secure the full visual appeal in the 2021 Chevrolet Avalanche. In that case, your current body in the vehicle is made out of the most effective product which is actually lightweight aluminum which will help the car to have much better stableness particularly in rapid speed driving a vehicle. Right after interested in the exterior design, the inner structure will even amaze individuals.
In this particular Chevrolet, you will observe new capabilities that are incorporated with increased new technological know-how. The cabin will probably be embellished while using leather-based substance chairs combined with the regulates. You will find numerous doorways also to enable you to efficiency the user and travelers ahead of time into this vehicle.
The amusement features will handle the actual infotainment prepare, music system; on the web online link, touch screen display, loudspeakers, together with environment control capabilities. Also, a safety method will almost certainly protect the newest chair straps procedure, automobile vehicle parking sensors, back again finish-see camera, in addition to safety bags. Every day, the alterations, as well as changes provided to this particular car, will comfort and ease each vehicle operator and vacationers in having driving a car contact with this specific 2021 Chevrolet Avalanche.
For the first time, 2021 Chevrolet Avalanche engine will definitely be considered a handful of choices supplied. From many decision integrated, you will find a 5.3-Liters V8 vehicle together with 355 hp utilizing get gasoline use that is more than famous. Potential is definitely an effect for the engine to acquire 60 km/h when it comes to 6.9 moments. Another blend is hybrid using a diesel engine plus generally productive and probable during American industry. It might get a lot better performance throughout the city and even severe discipline also.
In the past year or 2020, we are going to have additional common information regarding closing presenting this specific entirely new 2021 Chevrolet Avalanche. The conclusion item is being offered within the last several weeks of the coming year nevertheless a lot of several other specs continue being unfamiliar. A number of the pundits believe and also forecast, sted that about the price can provide some advice. But still, without clear information, we cannot think possible.


text:
Leigh Curran
Leigh Curran is the author of four full-length plays - The Lunch Girls, Alterations (both published by Samuel French), Walking the Blonde and Body Beautiful (both on newplayexchange.org) Leigh's one-acts have been produced nationally and internationally. The Lunch Girls had its world premiere at The Long Wharf Theatre under the direction of Arvin Brown and was a finalist for the Susan Smith Blackburn Award before being directed off-Broadway by Stuart Ross. Alterations was also produced Off-Broadway under the direction of Austin Pendleton, and Walking the Blonde was produced at Circle Rep under the direction of Paul Benedict; at La Mama under the direction of Leonard Foglia and at Theatre Geo in Hollywood under the direction of Linda Carlson. Body Beautiful was developed through a series of readings and a workshop in 2019 at the Fountain Theatre in Hollywood. Ms. Curran is a member of the LA Poets and Writers Collective and the author of four chapbooks. Her poetry has been published in Slant, Rattle and Onthebus. Her first novel, Going Nowhere Sideways, was published by Fithian Press – it is an underground favorite and is available in on-line bookstores. As an actress Ms. Curran has appeared on, off, and way off Broadway and was an original member of the Los Angeles Women's Shakespeare Company. For 22 years Ms. Curran was the Founder/Artistic Director of the Virginia Avenue Project - a non-profit using the Arts and long-term, one-on-one mentoring to give underserved children the skills to think creatively, critically, and courageously about life goals and choices. Leigh's solo show, Why Water Falls, was produced at the Hollywood Fringe Festival where it received the Producers Encore Award and subsequently ran Off-Off Broadway in 2016. Why Water Falls is available on newplayexchange.org – it can be performed by one or more actors. Website: www.leighcurran.net
https://leighcurran.net/
FB: Leigh Curran
I feel most complete as an artist when I'm performing in something I've written. A few years ago I wrote myself a solo show but my real thrill comes from working off really good actors who are excited to join me in bringing my stories to life. My hope is the stigma surrounding writer/performers can be undone along with the stigma surrounding plays about older people - often written by older writers - many of them women. In a perfect world, the life journey would be explored and celebrated from all vantage points. My first play, The Lunch Girls, made a big splash when it was first produced - I was in my mid 30s and didn't really know what I was doing. Now I I like to think I do ... in any event, I have a whole lot more to write about! And I plan to keep writing about it because one day, when we can finally throw our masks to the wind, theatre will return and people will once again come together to share deep, funny, sad, complex, surprising, thought provoking, live experiences.
Another play and another novel.
Lesbian, Gay, Transgender, Buddhist, People of Color, Equality, Love, Aging, Loyalty, Mothers and Daughters, Sexual Identity, Transformation, Dogs, New York City, Abortion, Comedy, Drama, Therapy, Waitresses, Spirituality

redpajama_set_name: RedPajamaCommonCrawl


text:
package com.compomics.pride_asa_pipeline.core.gui.view;

import javax.swing.JPanel;

/**
 *
 * @author Niels Hulstaert
 * @author Harald Barsnes
 */
public class SummaryPanel extends javax.swing.JPanel {

    /**
     * Creates new form SummaryPanel
     */
    public SummaryPanel() {
        initComponents();
    }

    public JPanel getIdentificationsChartParentPanel() {
        return identificationsChartParentPanel;
    }

    public JPanel getModificationsChartParentPanel() {
        return modificationsChartParentPanel;
    }

    public JPanel getScoresChartParentPanel() {
        return scoresChartParentPanel;
    }

    public JPanel getPrecursorMassDeltaChartParentPanel() {
        return precursorMassDeltaChartParentPanel;
    }

    public JPanel getFragmentIonMassDeltaChartParentPanel() {
        return fragmentIonMassDeltaChartParentPanel;
    }

    public JPanel getbIonCoverageChartParentPanel() {
        return bIonCoverageChartParentPanel;
    }

    public JPanel getyIonCoverageChartParentPanel() {
        return yIonCoverageChartParentPanel;
    }

    /**
     * This method is called from within the constructor to initialize the form.
     * WARNING: Do NOT modify this code. The content of this method is always
     * regenerated by the Form Editor.
     */
    @SuppressWarnings("unchecked")
    // <editor-fold defaultstate="collapsed" desc="Generated Code">//GEN-BEGIN:initComponents
    private void initComponents() {
        java.awt.GridBagConstraints gridBagConstraints;

        identificationsChartParentPanel = new javax.swing.JPanel();
        modificationsChartParentPanel = new javax.swing.JPanel();
        generalSummaryPanel = new javax.swing.JPanel();
        precursorMassDeltaChartParentPanel = new javax.swing.JPanel();
        scoresChartParentPanel = new javax.swing.JPanel();
        fragmentIonMassDeltaChartParentPanel = new javax.swing.JPanel();
        ionCoveragesChartPanel = new javax.swing.JPanel();
        bIonCoverageChartParentPanel = new javax.swing.JPanel();
        yIonCoverageChartParentPanel = new javax.swing.JPanel();

        setBackground(new java.awt.Color(255, 255, 255));
        setOpaque(false);
        setPreferredSize(new java.awt.Dimension(400, 200));
        java.awt.GridBagLayout layout = new java.awt.GridBagLayout();
        layout.columnWidths = new int[] {0, 5, 0};
        layout.rowHeights = new int[] {0, 5, 0};
        setLayout(layout);

        identificationsChartParentPanel.setBackground(new java.awt.Color(255, 255, 255));
        identificationsChartParentPanel.setBorder(javax.swing.BorderFactory.createTitledBorder("Identifications"));
        identificationsChartParentPanel.setOpaque(false);
        identificationsChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 0;
        gridBagConstraints.gridy = 0;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.33;
        gridBagConstraints.weighty = 0.5;
        gridBagConstraints.insets = new java.awt.Insets(6, 6, 0, 0);
        add(identificationsChartParentPanel, gridBagConstraints);

        modificationsChartParentPanel.setBackground(new java.awt.Color(255, 255, 255));
        modificationsChartParentPanel.setBorder(javax.swing.BorderFactory.createTitledBorder("Modifications"));
        modificationsChartParentPanel.setOpaque(false);
        modificationsChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 0;
        gridBagConstraints.gridy = 2;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.33;
        gridBagConstraints.weighty = 0.5;
        gridBagConstraints.insets = new java.awt.Insets(0, 6, 6, 0);
        add(modificationsChartParentPanel, gridBagConstraints);

        generalSummaryPanel.setBorder(javax.swing.BorderFactory.createTitledBorder("General"));
        generalSummaryPanel.setOpaque(false);
        generalSummaryPanel.setPreferredSize(new java.awt.Dimension(12, 23));
        generalSummaryPanel.setLayout(new java.awt.GridBagLayout());

        precursorMassDeltaChartParentPanel.setOpaque(false);
        precursorMassDeltaChartParentPanel.setPreferredSize(new java.awt.Dimension(20, 20));
        precursorMassDeltaChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 0;
        gridBagConstraints.gridy = 0;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.5;
        gridBagConstraints.weighty = 0.5;
        generalSummaryPanel.add(precursorMassDeltaChartParentPanel, gridBagConstraints);

        scoresChartParentPanel.setOpaque(false);
        scoresChartParentPanel.setPreferredSize(new java.awt.Dimension(20, 20));
        scoresChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 1;
        gridBagConstraints.gridy = 1;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.5;
        gridBagConstraints.weighty = 0.5;
        generalSummaryPanel.add(scoresChartParentPanel, gridBagConstraints);

        fragmentIonMassDeltaChartParentPanel.setOpaque(false);
        fragmentIonMassDeltaChartParentPanel.setPreferredSize(new java.awt.Dimension(20, 20));
        fragmentIonMassDeltaChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 0;
        gridBagConstraints.gridy = 1;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.5;
        gridBagConstraints.weighty = 0.5;
        generalSummaryPanel.add(fragmentIonMassDeltaChartParentPanel, gridBagConstraints);

        ionCoveragesChartPanel.setOpaque(false);
        ionCoveragesChartPanel.setPreferredSize(new java.awt.Dimension(20, 20));
        ionCoveragesChartPanel.setLayout(new java.awt.GridBagLayout());

        bIonCoverageChartParentPanel.setOpaque(false);
        bIonCoverageChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 1.0;
        gridBagConstraints.weighty = 0.5;
        ionCoveragesChartPanel.add(bIonCoverageChartParentPanel, gridBagConstraints);

        yIonCoverageChartParentPanel.setOpaque(false);
        yIonCoverageChartParentPanel.setLayout(new java.awt.GridBagLayout());
        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 0;
        gridBagConstraints.gridy = 1;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 1.0;
        gridBagConstraints.weighty = 0.5;
        ionCoveragesChartPanel.add(yIonCoverageChartParentPanel, gridBagConstraints);

        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 1;
        gridBagConstraints.gridy = 0;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.5;
        gridBagConstraints.weighty = 0.5;
        generalSummaryPanel.add(ionCoveragesChartPanel, gridBagConstraints);

        gridBagConstraints = new java.awt.GridBagConstraints();
        gridBagConstraints.gridx = 2;
        gridBagConstraints.gridy = 0;
        gridBagConstraints.gridheight = 3;
        gridBagConstraints.fill = java.awt.GridBagConstraints.BOTH;
        gridBagConstraints.weightx = 0.66;
        gridBagConstraints.weighty = 1.0;
        gridBagConstraints.insets = new java.awt.Insets(6, 0, 6, 6);
        add(generalSummaryPanel, gridBagConstraints);
    }// </editor-fold>//GEN-END:initComponents
    // Variables declaration - do not modify//GEN-BEGIN:variables
    private javax.swing.JPanel bIonCoverageChartParentPanel;
    private javax.swing.JPanel fragmentIonMassDeltaChartParentPanel;
    private javax.swing.JPanel generalSummaryPanel;
    private javax.swing.JPanel identificationsChartParentPanel;
    private javax.swing.JPanel ionCoveragesChartPanel;
    private javax.swing.JPanel modificationsChartParentPanel;
    private javax.swing.JPanel precursorMassDeltaChartParentPanel;
    private javax.swing.JPanel scoresChartParentPanel;
    private javax.swing.JPanel yIonCoverageChartParentPanel;
    // End of variables declaration//GEN-END:variables
}

redpajama_set_name: RedPajamaGithub


It's just that on days I'm working, I can't do that.
Make sure she knows the plan, so you both can dress appropriately. If I had one thing to add, mixed race marriages are quite similar.
Because she already lives the covenants in most of the meaningful ways. Things have worked out pretty well with us so far.


The training of a new teacher at a Bradford school is being funded by a grant from the Yorkshire Young Achievers Foundation.
Fiona Bradford has joined Buttershaw Business and Enterprise College in Bradford as part of the Teach First programme and has been part funded by the Foundation which raises money for young people and for projects benefiting them across the region.
Teach First is an innovative charity that is working to break the link between low family income and poor educational attainment, which is greater in the UK than in almost any other developed country.
Teach First recruits, trains and supports teachers working in schools in low income communities to raise the achievement, aspirations and access to opportunity of children from low socio-economic backgrounds, whilst developing a network of leaders with a commitment to ending inequality in education from both inside and outside the classroom.
Yorkshire Young Achievers Foundation Vice Chairman Richard Stroud said: "The fantastic work done by Teach First was highlighted last year by another Bradford teacher, Liam Gilbert, who won the Achievement in Education Award at the Yorkshire Young Achievers Awards.
Fiona Bradford said: "We believe the partnership between the Yorkshire Young Achievers Foundation and Teach First represents a common drive, both locally and nationally, towards providing opportunities for young people; the Teach First values of access, aspiration and achievement can be applied to more than just education - these human rights should underpin individual and community visions. Combined with the Foundation's recognition of those who overcome adversity to attain goals, we, at Buttershaw Business and Enterprise College, are delighted to be involved in the partnership, and it is exciting to think about the opportunities such a collaboration may bring for young people today and in the future, whether in Buttershaw or further afield."


text:
[Review] Mario Party 10
By Editorial StaffApril 7, 2015One Comment
Platform: Nintendo Wii U
Mario Party 10, the latest installment in the Mario Party series from Nintendo, gives one to four players the chance to go head to head in a virtual board game. Just as in other board games, players have to roll dice and advance along a game board in order to win. In Mario Party 10, the three main game modes are Amiibo Party, Bowser Party, and Mario Party. Each has slightly different rules and features.
This game is especially great for groups of people, so the Pixelkin staff decided to play the game together. Here's what we have to say.
Courtney: My favorite part of Mario Party 10 is definitely the minigames. These short and frantic levels between turns definitely kept things exciting. Since each one has different rules, they keep the gameplay fresh and diverse. Each of the three game modes features minigames, and you can play them by themselves without using a game board at all.
If you're going to do a whole game board, though, I definitely liked the standard Mario Party mode best. Even though we were competing with each other, I had a fun time laughing and supporting Simone, Nicole, and Linda as we worked our way through the board.
Simone: Mario Party is Mario Party. If you like getting competitive with your friends, you'll have a pretty good time. I love that the minigames do their best to keep people on an even playing field. Even if you might destroy your competitors at one minigame, there's no guarantee you'll be as good at the next one—and some minigames are just plain impossible for anyone. Is that good or bad? It depends on what you're looking for. It's never a good idea to get too invested in winning Mario Party. You'll only be disappointed.
But if you can set aside your dignity and just enjoy a healthy rivalry, Mario Party is a fun use of your time.
Nicole: I haven't played a Mario Party game since Mario Party 2, so some of the differences were a bit jarring to me. However, I like how the characters all move together rather than one-by-one. It makes for a quicker game and an overall cooperative feel. I also love the minigames. They've always been wacky and funny and they continue to be that in Mario Party 10. I think the minigames are what made us laugh the most during our play session. I especially liked the "soccer" minigame where are characters were rolling around on large beach balls while trying to play soccer. The fact that it was difficult to control didn't make it frustrating like in other games. In Mario Party, it just adds to fun because everyone is basically as bad as you are.
Linda: I loved playing this game because I was playing with people I can have a good time with. That's the same for me whether I'm playing cards or a board game or Mario Party. Mario Party is extra fun because unlike most other video games it lets up to four people play at once. You can laugh at your own mistakes, taunt others, and really have fun.
What Doesn't Work
Courtney: The Bowser Party mode was…okay. Playing as the bad guy made me feel alienated from Simone and Nicole. I definitely prefer playing on a team. I guess messing with other players just isn't my cup of tea. Plus, as Bowser, I felt more aware of the time I spent sitting around waiting.
Bowser Party mode is just…okay.
Mario Party 10 didn't have a good grasp of how to use the Gamepad effectively. It was hardly used at all for the Mario Party mode and used way too much for the Amiibo Party mode.
Simone: Previous Mario Party games would let you choose how long you wanted your game to be. I really missed this feature, even though the games are a relatively short 30 minutes on average. Similarly, it was unfortunate that there is no drop-in/drop-out for players. If someone has to leave in the middle of the game, you either have to start over, wait for them to get back…or hand off their controller to another player, like we did. It was fun for us but certainly made things unfair.
Sure, generally you don't start a multiplayer game unless you're sure you can finish it. But games like Mario Party, which aim to please families playing together, have a bigger obligation to consider that players might be on different schedules.
Like Courtney said, Bowser Party was also a disappointment for me too. Not because I didn't enjoy Courtney beating up on me—that was actually pretty fun! But towards the end of the game, my companions all got knocked out. It was sad to see Nicole and our computer-controlled friends sitting on the sidelines while Courtney and I were the only ones left in the game. Our rivalry is great, but I wanted a Mario Party, not a Mario Duel.
Nicole: The Bowser Party mode seemed really cool to me until we played it. It did have some neat aspects, like Courtney being able to draw on the screen to try to lead us to pick the wrong path or treasure chest. But the thing that bothered me the most is that players get kicked out one by one. I understand that's the whole point of playing as Bowser, but it was very frustrating to just sit and watch Simone battle Bowser all by herself. The normal mode makes for a better experience because everyone gets to play to the very end.
Courtney: Mario Party 10 is worth checking out, especially if you like to game together as a family. The Mario Party games have long been a quintessential addition to family game night, and at the very least this game tried to add something new. If you already have an older installment of Mario Party that you're not tired of just yet, there's really no hurry to rush out and buy this game. Also, unless you know you'll be playing Mario Party 10 a lot, I wouldn't bother with the amiibo.
Simone: Our Mario Party 10 experience was pretty fun. The variety of minigames was decent, though we found it kind of odd that so many of them were casino-themed. I guess Bowser is a gambler. I wouldn't call it a must-have, it's a great addition to your collection of multiplayer games. People of all ages can play it together, and be equally bad at it. If that's what you're after, this is a game you can have fun with.
Nicole: Mario Party is still one the best if not the best party game out there. Playing with my Pixelkin cohorts brought back fond memories of playing with my husband and friends years ago. I love that it makes things fun for everyone, regardless of your gaming skill. The minigames are a blast and our playtime was a lot of fun. Unfortunately Bowser mode isn't as cool as it seems and the amiibo integration is just laughable. I would recommend the game on its basic party mode especially if you haven't played a Mario Party in a while or if you've never played one before.
Linda: It's a big deal to me to find a game that's really fun for a big group and easy enough to play that people of different skill levels can understand the game and enjoy it. Mario Party definitely fits the bill for multi-generational family gatherings.
amiibofamily game nightgame reviewMario Partymario party 10NintendoNintendo Wii U
This article was written by Editorial Staff
You can read about Linda Breneman, Nicole Tanner, Keezy Young, Courtney Holmes and Simone de Rochefort right here.
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Worried about how anxious you feel? Wondering 'what is anxiety'? Learn how to tell the difference between 'normal' anxiety and an anxiety disorder. Find out about potential causes of anxiety disorders, anxiety symptoms and what to do if you feel that anxiety is affecting your life.
you avoid doing things because of anxiety.
Everyone feels anxious at times, and a certain level of anxiety is both normal and even helpful in some situations. Anxiety is our body's way of keeping us safe. For instance, imagine you're walking home, and you're dragging your feet because you're tired. Out of the corner of your eye, you think you see a snake. Suddenly, you forget how tired you are and have a burst of energy that helps you to get away.
Anxiety can also motivate us. If we feel a bit anxious about an assignment that's due, it can help us to get it done on time. However, feeling too much anxiety about something usually gets in the way and can be unhealthy.
Anxiety disorder occurs when anxiety starts to severely impact on a person's life. Rather than feeling anxious in response to actual danger, someone with an anxiety disorder will experience the same symptoms in situations they perceive as dangerous (e.g. meeting new people or taking public transportation).
Generalised anxiety disorder: excessive worry about anything and everything, including worrying about worrying.
Social anxiety disorder: anxiety in social situations, often rooted in the fear of doing something wrong and being judged by others.
Panic disorder: repeated panic attacks and worry about future panic attacks.
Agoraphobia: anxiety about having a panic attack in certain situations and not being able to escape or to get help.
Specific phobias: intense fear of objects or situations (e.g. dogs).
Anxiety disorders aren't caused by a single factor, but rather by a combination of things. A family history of anxiety, a person's physical health, personality traits and stressful life experiences can all be contributing factors. Anxiety disorders are likely to develop over time and are usually sustained by unhelpful thinking patterns.
Learn some different relaxation strategies that can help with anxious thoughts and feelings.
Use problem solving to work through challenges that seem overwhelming.
Download ReachOut WorryTime and learn how to control your worrying.


text:
Hearts of Gold
Posted on December 22, 2020 July 28, 2021 by John Chilton
A Word for the Season of Giving
Too often, we are fed all of the bad news and very little of the good. But it's the good news that makes us smile. The good news is what makes life worth living. Renewed faith in humanity and restored purity in a society that feels overwhelmingly dark. So here are a few short stories of people across our country who acted in kindness from Hearts of pure Gold!
A Wish Spent For Others
Javier Amos was nine years old when he was diagnosed with lymphoblastic leukemia. As they do, the "Make A Wish" foundation reached out to grant Javier his. Most children want to take a trip or be greeted and hang out with their favorite celebrity or Superhero. Javier, did not ask for that. Instead, he wished for a Pizza Party for his entire Elementary School!
The whole school, one Friday afternoon, gathered to form a tunnel of appreciation to Javier's gesture of kindness. Jo Evenson, who heads "Make A Wish" in South Dakota, attended the event, and he said he had never seen anything like it before. Javier had his own honorary procession through a tunnel of his peers and teachers shouting happy words of thanks and encouragement. The school ate pizza and played on jump houses and jungle gyms for the last two-hours of their educational day, before going home in their school buses.
Most children would have asked for something for themselves, but it takes a true "Heart of Gold" to ask for something that benefits everyone else. Javier Amos is now eleven years old and in remission.
For More click Here.
Christmas in September
The Allen family was given the worst kind of news imaginable. Brody, their two year old son, was given mere weeks to live due to brain cancer. This was September, and his parents knew he would never make it to his favorite holiday—Christmas. So the couple decided to begin decorating their yard and home like they would for Christmas, giving Brody one last 'winter wonderland'. Brody helped his mother decorate the tree and they supervised his father as he strung the lights on the porch and overhangs, hung the wreaths and set up the light-up decorations in the yard. Heartbroken, his family choked back their own emotion to make sure this was the best 'Christmas' their son could ever ask for.
Soon, to the Allens' surprise, the neighbors began to decorate as well. (Revealing their true Hearts of Gold) Every home got involved stringing lights and setting up trees and other outdoor decorations. Eventually the entire neighborhood was decorated. From rooftop inflatables to fake snow sprinkled across the lawns, Christmas had come in September for Brody. As parents, you do what you can, but the neighbors going over and beyond made this boy's last holiday a special one.
Brody Allen lost his fight with brain cancer less than a month later, 2 weeks before Halloween. But his story and the love that his neighbors showed will live on in us. Christmas doesn't have to be one month out of the year.
An Angel. An Icon.
An icon. A legend. A performer. An Angel. Dolly Parton is all of those things and more. She has done good for those around her out of love, more so than most of the world. Her name should be written in the stars with other 'Angels' like Mother Theresa and Lottie Moon and all of the Saints. She gives and gives and finds ways to make that giving go further!
In 2016, a wildfire ravaged the Great Smoky Mountains in Tennessee. Hundreds of citizens were left homeless, displaced and without the essential needs to survive. Dolly immediately launched the 'My People Fund', promising to give families like these $1,000 per month stipend until the catastrophe could be sorted out. Dolly called on other artists to help raise money by doing concerts and events. Through these efforts they raised nearly $10 million caring for the needs of 1,500 families and nearly 5,000 individuals. The foundation also partnered with the Mountain Tough Recovery Team to give another $3 million in additional support through the months that followed.
Lives were saved and hearts were warmed through that dark, cold winter, because Dolly Parton exemplifies those with Hearts of Gold. And she continues to do God's Work in this world…
A Real Life Super Hero
A real life Superhero, Chris Evans played Captain America on the big screen for the Marvel Cinematic Universe. But in his day to day life he is just as much a Superhero for the people. From rescuing a dog on a movie set to making virtual calls to young people to encourage their own acts of kindness, Chris Evans goes far beyond his call of duty.
Most memorable, however, was a few years ago when Evans teamed with co-star Chris Pratt who played Starlord in Guardians of the Galaxy, to visit a children's hospital in Seattle. They went in full Superhero costumes, Chris Evans even sported his shield. Needless to say, the kids were over-the-moon excited for the visitors who certainly sparked their spirits that day.
On Twitter, Chris posted a picture of him with one of the children he visited. Writing, "AMAZING day at @seattlechildren w @prattprattpratt. Met some TRUE super heroes! Feeling inspired, blessed, touched. Thank you w all my heart." His Ex-girlfriend, Jenny Slate, said it best, "… if you could paint his heart, it would probably be golden." And that's something coming from an ex. Yet another example of celebrities with hearts of gold!
Cameron Lyle, college sophomore at the University of New Hampshire, was hanging out with his friends in the cafeteria when a few players from the football team playfully jested with him to sign up at the 'Be The Match' station for Bone Marrow Donorship. As most Sophomores would, he was easily pressured into the simple task and with a cheek swab and a signature, he was an official donor.
Throughout his next two years of college, Lyle made a name for himself in Track and Field taking home silver medal after silver medal and positioning himself to really 'go for the gold' his Senior year. Poised to take the gold in shot put at the conference championships in May, Lyle readied himself for victory. And then he got the phone call. He was a match.
"When I got the phone call, I was so shocked," said Lyle. "I totally forgot I even signed up a couple of years before. Even then, I did not think I would be a match. The odds are so ridiculous."
A Single Moment Shows a Man's True Colors
Additional testing took place to confirm he was a definite match, and when it was discovered that he was, they immediately set him up for surgery. But there was an issue. The surgery was set for April 25th, just one short month before the conference championships.
Bone marrow transplants are no walk in the park. They are intense, especially the one he was having done, and if he went through with it, he would have to sit out of the championships as well as the rest of his Senior season of sports. Terrified to tell his coach he took a breath and just spat it out. He found that his coach was much more supportive than Lyle could have imagined. His coach told him, "If you have the chance to save a life, do whatever it takes."
Lyle's choice did just that. He saved the life of a 28 year old man who had been diagnosed with lymphoblastic leukemia a few days before Christmas in 2012. The man was a father of nineteen-month-old daughter and was set to be married. But it was not only one life that was changed. After the donation, he was given the NCAA Award of Valor which brought streams of news media to the cause. Registration to the National Marrow Donor program tripled so that those who need Bone Marrow transplants might find their match too.
Cameron is proof that we can all be donors, whether of blood or bone or simple kindness. We can all have hearts of gold!
A Call to Activate your Heart of Gold
So what is the difference between these people and us? Is it that, within their chest, there actually IS a heart made of gold? No. We all have the potential to be good, even GREAT. The only defining difference is that these people DID. They chose to act in kindness. They did things that were within their power to accomplish, but that would also make the highest impact. So don't forget, this Christmas especially, to look around at the needs of others and find the ways to best 'activate' your own heart of gold. We are all in this together, as we look forward to a brighter tomorrow!
For More Blogs like "Hearts of Gold" Click here!
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text:
\section{ The domain wall. }

{}~~~~~A first-order phase transitions which take place in the most of
cosmological models proceed through the nucleation of the new phase bubbles
(see e.g. \cite{B88,HKLLM92,L77}). Such processes take place at the very
beginning of the phase transition. At the final stage of transition the old
phase fragments which are left up to that moment also take the spherical
form (for example due to surface tension effects or dissipation). The
surface of phase separation in the case of the first-order phase transition
is represented by the so-called domain wall. Such objects are created when
the vacuum manifold for the order parameter or scalar field $\varphi$
driving the symmetry breaking has a discrete symmetry  \cite{V85,V91}. At
the time of transition there can be both infinite and closed surface walls.

To model the essential features of a domain wall, we will  consider a
scalar field with the Lagrangian

$$
L = \frac{1}{2} g^{\mu\nu}\partial _\mu\varphi\partial _\nu\varphi -
V(\varphi ) ,
$$
where $ V(\varphi )$ has two minima at nonzero $\varphi$ . Here the greek
indices take the values 0,1,2,3 and the metric has the signature
$(+,-,-,-)$. The example of such potential with discrete symmetry
$\varphi \rightarrow -\varphi$ is

$$
V(\varphi ) = -\frac{M_G^2}{2}\varphi^2 + \frac{\lambda}{4}\varphi^4 ,
$$
$M_G^2/\lambda$   representing the symmetry breaking scale and
$ \lambda $ --- the  self coupling constant.

  In the case of a flat wall in Minkowski space the  equations  of
motion

$$
\partial _\mu\partial_ \nu\varphi - \frac{\partial V(\varphi )}{\partial
\varphi} = 0
$$
admit the classical "domain wall" or "membrane"  solution,  depending on one
coordinate:

$$
\varphi (x) = \frac{M_G}{\sqrt{\lambda}}\cdot \tanh\left[
\frac{M_G}{\sqrt{2}} ( x - x_0 ) \right] ,
$$
where $x$ is the normal direction to the wall and $x_0$ is the wall position.

In this case the VEV of scalar field
$<\varphi > \rightarrow \pm <\varphi_0  > = \pm M_G/\sqrt{\lambda}$ on
either side of the  wall and thus there is defined the wall surface $\Sigma$
--- the  three  dimensional surface on which $\varphi = 0$.

The energy-momentum tensor

$$
T_{\mu\nu} = \partial_\mu\varphi\cdot\partial_\nu\varphi - g_{\mu\nu}L
$$
for this solution takes the form :

\begin{eqnarray}
T_{00} = \left( \frac{\partial\varphi}{\partial x} \right)^2 =
\frac{M_G^4}{2\lambda}\cdot \cosh^{-4}\left[ \frac{M_G}{2} ( x - x_0 )
\right] ;
\nonumber \\
T_{xx} = 0 ; \nonumber \\
T_{yy} = T_{zz} = - T_{00}.
\label{2.6}
\end{eqnarray}

{}From the first equation of (\ref{2.6}) it is easy to see, that the
thickness and energy per unit area of this wall are respectively:

$$
\Delta \sim \frac{1}{M_G} ;
$$
$$
\sigma = \int\limits_{-\infty}^{+\infty} dx T_{00} \sim
\frac{M_G^3}{\lambda} .
$$

{}From the third equation of (\ref{2.6}) one can see that the wall solution
has a high pressure along the wall, which plays an essential role in
gravitational behavior of such objects (see Tolmans formula (\ref{0.2})).

\section{ Thin wall approximation. }

{}~~~~~In the limit of vanishing thickness of  the  domain  wall it can be
considered as an infinitely thin three-dimensional hypersurface $\Sigma$ with
the energy-momentum tensor $T^\mu_\nu$ having, in general, singularities on
it. This hypersurface divides the Riemannian manifold $^{(4)}V$ into two
parts $V^+$ and $V^-$. Each of them contains $\Sigma$  as a part of its
boundary.

    We intend to apply the Einstein equations

\begin{equation}
G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}\cdot g_{\mu\nu}R = 8\pi G\cdot
T_{\mu\nu}
\label{3.1}
\end{equation}
to such thin boundaries of phase separation and investigate their dynamics and
gravitational behavior using  the  thin-wall approximation. In (\ref{3.1})
$R_{\mu\nu}$ is the Ricci tensor, expressed in terms of Christoffel
connections $\Gamma^\lambda_{\mu\nu}$ in the usual way:

$$
R_{\mu\nu} = \partial_{[\rho}\Gamma_{\mu ]\nu}^\rho +
\Gamma^\sigma_{\rho [\sigma}\Gamma_{\mu ]\nu}^\rho
$$
(here the square brackets denote the antisymmetrization)
and
$$
\Gamma_{\mu\nu}^\lambda = \frac{1}{2}\cdot g^{\lambda\sigma}
( \partial_\nu g_{\sigma\mu} + \partial_\mu g_{\sigma\nu} -
\partial_\sigma g_{\mu\nu} ) ,
$$
$G = M_{Pl}^{-2}$ is the gravitational constant,
$M_{Pl} = 1,2\cdot 10^{19} GeV$ is the Planck mass, $\partial_\mu$ is usual
derivative.  (One must realize, that when speaking  about  a  thin shell we
always keep in mind that the thickness of the shell must not be smaller then
$1/M_{Pl}$, provided gravity is described just by the classical Einstein
equations).

We shall restrict ourselves to the case of time-like hypersurfaces $\Sigma$
with the energy-momentum tensor having singularities no stronger, then those
given by $\delta$-functions. Then the first derivatives of $g_{\mu\nu}$ are
discontinuous on $\Sigma$, while metric itself is still continuous on
$\Sigma$.  Gravitational formalism for such boundaries is considered in
details in papers \cite{BKT87,C70,CI67, I66,K68,MTW73}.

Let $\{ y^\mu \}^+\quad  (\{ y^\mu \}^- )$ be an arbitrary coordinate
system  in  $V^+\; (V^-)$ region, and $\{\xi^i\}$ be an arbitrary
coordinate system on $\Sigma$ (here the latin indices run through 0,1,2).
The coordinate charts $\{ y^\mu\}^+$ and $\{ y^\mu\}^-$ need not join
smoothly on $\Sigma$ .

    Let us assume, that equation of hypersurface $\Sigma$ in the  chosen
coordinates $y^\mu$ has the form

\begin{equation}
F(y^\mu ) = 0
\label{4.1}
\end{equation}
(Since all equations in the regions $V^+$ and $V^-$ differ only by the
indices "$+$" and "$-$", we, for simplicity, will not write them separately
if not necessary).

We can introduce the new function
\begin{equation}
n(y^\mu ) = \varepsilon \frac{F(y^\mu
)}{\sqrt{g_{\lambda\nu}\partial^\lambda F\cdot\partial^\nu F}}
\label{4.2}
\end{equation}
which describes the surface $\Sigma$. Here the sign function $\varepsilon$
depends on the orientation of the (1+2)-surface $\Sigma$.
If a displacement vector $dy^\mu$ lies on the hypersurface $n
= const$, then

$$
dn = \partial_\mu n \cdot dy^\mu = 0  ,
$$
$\partial_\mu$ being the usual derivative. Therefore a vector

\begin{equation}
N_\mu \equiv \partial_\mu n |_\Sigma
\label{4.3}
\end{equation}
is the unit normal vector to the hypersurface $\Sigma$ :

$$
N_\mu N^\mu = - 1
$$
(recall that we consider the timelike $\Sigma$ ).

We can also introduce the unit vectors, tangential to $\Sigma$ :

$$
e^\mu_i = \frac{\partial y^\mu}{\partial \xi^i}.
$$
The tetrad field $(N^\mu, e^\mu_i)$ is thereby defined on $\Sigma$ .

For our further calculations it is convenient  to  split  the field
equations into their components orthogonal and tangential to the wall
surface $\Sigma$ \cite{BKT87,C70,CI67,I66,K68,MTW73}.

For this purpose the interval, which in $y^\mu$ chart has the form

$$
ds^2 = g_{\mu\nu}dy^\mu dy^\nu
$$
($g_{\mu\nu}$ is the metric tensor, which determines the geometry in $V$
region), can be written in Gaussian normal coordinates in the form

\begin{equation}
ds^2 = -dn^2 + \gamma_{ij}(\xi^k, n)d\xi^id\xi^j .
\label{4.7}
\end{equation}
{}From (\ref{4.1}) and (\ref{4.2}) it is clear, that $n = 0$ is the equation
of
hypersurface $\Sigma$ and thus the interval

$$
ds^2 = \gamma_{ij}(\xi^k)d\xi^id\xi^j
$$
determines the 3-geometry on $\Sigma$.

Any  vector  and  tensor  naturally  is  splitted   into   its components
orthogonal and tangential to $\Sigma$ :

$$
A^\nu = A^nN^\nu + A^ie^\nu_i,
$$
$$
Q^{\mu\nu} = Q^{nn}N^\mu N^\nu + Q^{ni}e^\mu_iN^\nu + Q^{jn}N^\mu e^\nu_j
+ Q^{ij}e^\mu_ie^\nu_j .
$$
Using the Gaussian coordinates the Einstein equations (\ref{3.1}) also can be
decomposed into scalar, 3-vector  and  3-tensor  parts  in  respect  to  the
coordinate transformations on Hypersurface $\Sigma$ :

\begin{eqnarray}
G^n_n = - \frac{1}{2}\cdot\left( ^{(3)}R + (K^i_i)^2 - K^i_lK^l_i \right) =
8\pi G\cdot T^n_n  ; \nonumber \\
G^n_i = D_mK^m_i - D_iK^l_l = 8\pi G\cdot T^n_i ; \nonumber \\
G^i_j = ^{(3)}G^i_j - N^\mu D_\mu (K^i_j - \delta^i_jK^l_l) +
K^l_lK^i_j - \nonumber \\
\frac{1}{2}\cdot \delta^i_j \left( (K^l_l)^2 + K^l_mK^m_l
\right) = 8\pi G\cdot T^i_j .
\label{4.11}
\end{eqnarray}
Here the $D_\mu$ denotes covariant differentiation with  respect of the
connection in $V$, while $D_i$ denotes  3-dimensional covariant
differentiation with respect to  the  connection  on $\Sigma$ , $ ^{(3)}R$
is three dimensional scalar curvature, $ ^{(3)}G^i_j$ is three dimensional
Einstein tensor and $K^i_j$ is the extrinsic curvature tensor of the
hypersurface $\Sigma$ , which is defined in the following way:

\begin{equation}
K_{ij} = - e^\mu_ie^\nu_j D_\nu N_\mu  .
\label{4.12}
\end{equation}
In  the  equations  (\ref{4.11})  we  have  used   the   well-known
Gauss-Kodazzi equations \cite{BKT87,C70,CI67,I66,K68,MTW73}.

    Note, that in the Gaussian coordinates (\ref{4.7}) the  tensor  of
extrinsic curvature (\ref{4.12}) acquires the simple form:

\begin{equation}
K_{ij} = -\Gamma^n_{ij} = -\frac{1}{2}\cdot \partial_n\gamma_{ij}  .
\label{4.13}
\end{equation}

Since the 3-geometry on the surface $\Sigma$ is, by assumption, well
defined, the components of Christoffel symbols (in  the  intrinsic
coordinates $\xi^i$), not containing indices $n$ are regular.  Components
with two or three indices $n$ are equal to zero, while with one $n$ --- are
discontinuous and have the step-function behavior when crossing $\Sigma$.
However $g_{\mu\nu}$ is assumed to be regular on $\Sigma$,  and
$g^+_{\mu\nu}$ and $g^-_{\mu\nu}$ have to match continuously on the shell.
Three-curvature tensor $ ^{(3)}R_{ij}$  of hypersurface $\Sigma$ also does not
contain singularities and is expressed  in terms of 3-metric tensor
$\gamma_{ij}$ in the usual way just as $R_{\mu\nu}$ is  expressed  by
$g_{\mu\nu}$.

Integrating the $(ij)$ component of equation (\ref{4.11}) in the normal
direction by the proper distance $dn$ through $\Sigma$ we can get the
so-called Lanczos equation (see \cite{BKT87,C70,CI67,I66,K68,MTW73}).

\begin{equation}
\left[ K^i_j \right] - \delta^i_j \left[ K^l_l \right] =
8\pi G\cdot S^i_j ,
\label{4.15}
\end{equation}
where

$$
\left[ K^i_j \right] \equiv lim_{\epsilon \rightarrow 0 }
\left( K^i_j (n=+\epsilon ) - K^i_j (n=-\epsilon ) \right)
$$
is the discontinuity of the outer curvature tensor and
$$
S^i_j = lim_{\epsilon \rightarrow 0 } \int\limits_{-\epsilon}^{+\epsilon}
T^i_j dn
$$
is the intrinsic surface energy tensor  on  the  $\Sigma$.

Noting (\ref{4.13}), writing the $(nn)$ and $(ni)$ components of Einstein
equations in the regions $V^+$ and $V^-$ and subtracting the corresponding
equations for the $V^+$ region from those for the $V^-$ region one yields
(using (\ref{4.15})):

\begin{eqnarray}
 D_jS^j_i + \left[ T^n_i \right] = 0 \nonumber \\
\{ K^i_j \} S_i^j + \left[ T^n_n \right] = 0,
\label{4.18}
\end{eqnarray}
where
 $\{ K_j^i \} = \frac{1}{2} \cdot lim_{\epsilon \rightarrow 0 }
\left( K^i_j (n=+\epsilon ) + K^i_j (n=-\epsilon )\right) $.

Now to describe completely the gravitational behavior and dynamics of bubbles
one needs to investigate the outer curvature and energy-momentum tensors of
these objects, insert these quantities into motion equations (\ref{4.15})
and (\ref{4.18}) and solve them.

\section{ The surface stress-energy tensor.}

{}~~~~~We shall restrict ourselves to a pure vacuum case, i.e.  when
there are no particles at the either sides off the shell.

    An observer, which moves with an element of the  shell  finds that the
momentum of the matter all the time lies on  the  surface of the shell i.e.
$$
T_{\mu\nu}\cdot N^\mu = 0
$$
on $\Sigma$  and

$$
T_{\mu\nu} = 0
$$
outside $\Sigma$ . Thus in the intrinsic coordinates

\begin{equation}
T_n^n = T_n^i = 0
\label{5.3}
\end{equation}
and the surface energy-momentum tensor for an  observer  on  $\Sigma$   is
represented by the tensor $S^i_j$.

It is easy to see that with the condition (\ref{5.3}) the motion equations
(\ref{4.18}) are satisfied automatically (they become identities) and one is
left only with Lanczos equations (\ref{4.15}).

Now let us suppose, that the surface energy-momentum tensor has the
structure of an ideal fluid

\begin{equation}
S^{ij} = (\sigma + p )u^iu^j - p\gamma^{ij} ,
\label{5.4}
\end{equation}
where $u^i$ is the 3-velocity of an element of hypersurface $\Sigma$ ,  while
$\sigma$  and  $p$ are the surface energy density (energy per unit area) and
surface tension respectively.

    The energy-momentum conservation in the intrinsic coordinates
implies that equation

\begin{equation}
D_iS^{ij} = 0
\label{5.5}
\end{equation}
expresses the energy-momentum balance of matter on the shell.  For
tensor (\ref{5.4}) this conservation equation becomes

\begin{equation}
u^j\cdot D_i\left[ (\sigma + p )u^i \right] + (\sigma + p)u^iD_iu^j -
\gamma^{ij}\cdot D_ip = 0  .
\label{5.6}
\end{equation}
Multiplying (\ref{5.6}) by $u_j$ we yield

\begin{equation}
D_i\left[ (\sigma + p)u^i \right] - u^i\cdot D_ip = 0.
\label{5.7}
\end{equation}
Feeding the last relation back into (\ref{5.6}) we obtain

\begin{equation}
(\gamma^{ij} - u^iu^j)\cdot D_ip - (\sigma + p)u^iD_iu^j = 0 .
\label{5.8}
\end{equation}

    The conservation equations (\ref{5.7}) and (\ref{5.8}) are easily
solved in two cases.

    (a). For the dust wall

$$
p = 0
$$
and we have

$$
D_i(\sigma u^i) = 0 ,
$$
that states that the total amount of dust is conserved.

    (b). For the domain wall

$$
p = -\sigma
$$
and from (\ref{5.7}) it follows immediately that

$$
\sigma = const  .
$$

This two examples are considered in section 7 in spherically symmetrical
case.

\section{ The geometry of spherical vacuum shells.}

{}~~~~~Here we consider the bubbles of spherical form. For the spherical
shells the Lanczos equations (\ref{4.15}) (the only nontrivial equation
of motion for the vacuum case) takes the form (see \cite{BKT87}):

\begin{eqnarray}
 \left[ K^2_2 \right] = 4\pi GS^0_0 ,    \nonumber  \\
 \left[ K_0^0 \right] + \left[ K^2_2 \right] = 8\pi GS^2_2 .
\label{7.6}
\end{eqnarray}

For the spherical shell the most convenient  choice  for  the outer region
coordinates $y^\mu$ is the ordinary  spherical  coordinate system. The
symmetry features of the problem allows us to  choose the coordinates
$\vartheta$ and $\varphi$ to be continuous across $\Sigma$ , i.e.

$$
\vartheta^+ = \vartheta^- = \vartheta ;
$$
$$
\varphi^+ = \varphi^- = \varphi .
$$

    The metric off the shell in the $V^\pm$ regions must be the solution
of spherically symmetrical Einstein equations. According to Birkhoff's
theorem such metric has the form \cite{LL75,MTW73,T69}

\begin{equation}
(ds^\pm )^2 = f^\pm \cdot (dt^\pm )^2 - \frac{1}{f^\pm }\cdot (dr^\pm )^2 -
(r^\pm )^2\cdot d\Omega^2 ,
\label{6.2}
\end{equation}
where

$$
d\Omega^2 = d\vartheta^2 + \sin^2\vartheta\cdot d\varphi^2.
$$

    For the intrinsic coordinates $\xi^i$  of the shell  we  shall  use
the proper time $\tau$  and the spherical angles $\vartheta$ ,$\varphi$ .
Then the metric on (1+2)-surface $\Sigma$ , induced both by exterior
$(ds^+)^2$ and interior $(ds^-)^2$ metrics (\ref{6.2}), can be expressed as

\begin{equation}
ds^2 = d\tau^2 - R^2(\tau )d\Omega^2 ,
\label{6.6}
\end{equation}
where $R(\tau )$ is the shell radius.

    Generally speaking, the time and radial  coordinates  are  not
continuous on the shell, but some restrictions on these coordinates is
obtained from the junction of metrics (\ref{6.2}), (\ref{6.6}):

\begin{equation}
(ds^+)^2 = (ds^-)^2 = ds^2_{|\Sigma} .
\label{6.7}
\end{equation}
The radius of the shell $R(\tau )$ can be described  in  the  coordinate
invariant way, so the (\ref{6.7}) gives only two conditions.  Identification
of the two-spheres $(r,t = const)$ on $\Sigma$  yields

\begin{equation}
r^+ = r^- = R(\tau )_{|\Sigma } ,
\label{6.8}
\end{equation}
while the time  coordinate  can  be  discontinuous  on  the  shell:
$(t^+ \neq t^-)_{|\Sigma}$. Comparison of timelike lines
$(\varphi ,\vartheta  = const)$ gives

\begin{equation}
d\tau^2 = f^+\cdot (dt^+)^2 - \frac{1}{f^+}\cdot dR^2 = f^-\cdot (dt^-)^2 -
\frac{1}{f^-}\cdot dR^2 .
\label{6.9}
\end{equation}
{}From these relations it is easy to find that

$$
\left( f^\pm \dot t^\pm \right)^2 = f^\pm +  \dot R^2 ,
$$
\begin{equation}
\frac{1}{f^\pm }\cdot \left( \frac{dR}{dt^\pm } \right) ^2 = f^\pm -
\frac{1}{(\dot t^\pm )^2}
= \frac{f^\pm \cdot \dot R^2}{f^\pm + \dot R^2},
\label{6.11}
\end{equation}
where the overdot denotes the derivation with respect of  proper  time
$\tau$ . Thus, after the procedure of junction (\ref{6.7}), there remains
only one unknown function $R(\tau )$ which must obey the  Einstein
equations on the shell.  It means that from the two equations of motion
(\ref{7.6}) only one is independent. We shall choose and solve the first
one (then the second one will be satisfied automatically).  Now we have to
find the outer curvature tensor.

It is easy to see, that equation (\ref{6.8}) is the equation of motion of
the shell (compare with (\ref{4.1})):

$$
F = r - R(\tau ) = 0 .
$$
Noting, that

$$
dF/dt = - dR/dt;
$$
$$
\quad dF/dr = 1
$$
and using equations (\ref{6.2}),(\ref{6.11}) we can compute the outer
normal to $\Sigma$ (see (\ref{4.2}) and (\ref{4.3})):

$$
N_0 = - \left( \frac{dR}{dt} \right) \cdot N_1 ,
$$
$$
N_1 = \varepsilon \cdot |\dot t| ,
$$
$$
N_2 = N_3 = 0 .
$$
Here $\varepsilon$ is a sign function, which depends on the direction of an
outer normal $N_\mu$, i.e. on the orientation of (1+2)-surface $\Sigma$:

\begin{equation}
\varepsilon = sign\left( \dot t, \frac{\partial r}{\partial q} \right),
\label{e}
\end{equation}
where $q$ is any coordinate increasing along the direction from the
bubble-center.

  Using formulae (\ref{4.12}) one can find the components of  extrinsic
curvature tensor:

\begin{eqnarray}
 K_0^0 = K^2_2 + \frac{R}{\dot R} \cdot \dot K^2_2 , \nonumber \\
K^2_2 = - \varepsilon \frac{|f\dot t|}{R} .
\label{7.4}
\end{eqnarray}
Note that the first equation of (\ref{6.11}) admits both signs for $f\dot
t$. The sign of $\varepsilon$ is equivalent to tha sign of $f\dot t$, thus
instead of the second equation (\ref{7.4}) one can write
\begin{equation}
K^2_2 = - \frac{f\dot t}{R} .
\label{7.41}
\end{equation}

Using (\ref{6.11}) one can yield from (\ref{7.4}):

$$
(K_\pm )^0_0 = -\varepsilon_\pm \frac{\ddot R + \frac{1}{2}\cdot (f^\pm)'
}{\sqrt{f^\pm + \dot R^2}} ,
$$
\begin{equation}
(K_\pm )^2_2 = - \frac{\varepsilon_\pm }{R}\cdot \sqrt{f^\pm + \dot R^2} ,
\label{7.5}
\end{equation}
where $f'$ denotes the derivation by $R$.

Thus we have shown that the dynamics of  spherically  symmetrical shell is
completely described by the first equation (\ref{7.6})

\begin{equation}
K_+ - K_- = 4\pi G \sigma ,
\label{8.1}
\end{equation}
where $\sigma = S_0^0$ is the energy density of the shell and $K$ is the only
independent component $K_2^2$ of the extrinsic curvature tensor described by
(\ref{7.41}) or (\ref{7.5}).

The fact that the first equation of (\ref{6.11}) admits both signs for
$f\dot t$ leads to the new class of solutions of Einstein equations for
bubbles with outer $\varepsilon_+ = -1$. This was missed in earlier works
(for example \cite{BKT87}), where it was considered that for bubbles, lying
above the Schwarzschild horizon, in outer space always $\varepsilon_+ = 1$.
The correct sign must be chosen from boundary conditions. We shall discuss
this subject below.

\section{ The sign ambiguity of extrinsic curvature. }

{}~~~~~Investigating the dynamics of vacuum spherical shells by means of
the motion equation (\ref{8.1}) one has to be careful when choosing the
sign of $\varepsilon$ and thus of $f\dot t$. For the given the inner and the
outer metrics $(ds^\pm)^2$ the $\varepsilon_\pm$ determines the global
geometry, i.e. how the inner geometry is stuck with the outer one. In
some cases the junction is impossible.

 As we have mentioned above, the sign of $\varepsilon$ depends on the
direction of an outer normal $N_\mu$, i.e. on the orientation of
(1+2)-surface $\Sigma$:

\begin{equation}
\varepsilon = sign\left( \dot t, \frac{\partial r}{\partial q} \right),
\label{e}
\end{equation}
where $q$ is any coordinate increasing along the direction from the
bubble-center.

Generally speaking, the sign of $\varepsilon$ may change at the shell motion,
for example, when the shell passes the horizon. If the initial conditions
allow the shell to collapse, then in the case $\varepsilon_+ > 0$ the shell
forms the black hole, while in the case $\varepsilon_+ < 0 $ --- the
warmhole.

In paper \cite{BKT87} the classification of possible geometries for different
signs of $\varepsilon_+$ and $\varepsilon_-$ was done. However in this paper
the dependence of the sign-function (\ref{e}) on the mutual orientation of
time coordinates on and off the shell, i.e. on the sign of $\dot t$ was
overlooked.  The bubble is the (1+2)-surface embedded in the 4-dimensional
space-time $^{(4)}V$ and not only the 2-sphere as it appears if we do not
consider the direction of time coordinate. Thus we get the more general
classification of bubble geometries compared to those presented in previous
papers.

  Generally speaking, we have four different cases of equation
(\ref{8.1}), depending on the signature of $\varepsilon_\pm$ or $(f \dot
t)^\pm$ and, therefore, on the signature of $K_\pm$.
The sign of $\varepsilon$ must be chosen from the motion equation
(\ref{8.1}) and the boundary conditions mentioned in the Introduction. One
has to take into account the positiveness of the surface energy density
$\sigma$ and the fact that there must not be the contradiction between the
Newton's limit of Einstein equations (\ref{8.1}) and Tolman's formula
(\ref{0.2}).

Let us discuss this problem and examine the signatures of $\varepsilon$
or $f\dot t$.

  a) The signature of $f^\pm$ ($\dot t_\pm > 0$).

Let us assume for a while that $\dot t_\pm > 0$. The the signature of $f\dot
t$ depends on the sign of $f$.
The general form of $f$ for the spherically symmetrical source is

\begin{equation}
f = 1 - \frac{2Gm}{r} + \frac{Ge^2}{r^2} - G\Lambda r^2 ,
\label{8.5}
\end{equation}
where $m$ is the mass of the source, $e$ is its charge,
$\Lambda = (8\pi /3) \cdot \rho $ and $\rho $  is
the vacuum energy density. In the simplest case

$$
e = \Lambda = 0
$$
and
\begin{eqnarray}
f_+ = 1 - \frac{2Gm}{R} ; \nonumber \\
f_- = 1 .
\label{8.7}
\end{eqnarray}
The signature of $f_+$ is positive if the radius of the shell lies above
the Schwarzschild horizon, i.e. when $R > 2m$ and is  negative  beyond
it. So, $f\dot t$ changes its signature, when the radius of  shell  passes
the horizon. The standart metric (\ref{8.5}) has an unphysical singularity
at the horizon caused by a poor  choice  of  the  coordinate system. To be
more correct in correspondence between the signature of $K$ and the radius
of shell, one can use another coordinate  system, for example the isotropic
coordinates (see \cite{BGG87}), which  have no singularities at the horizon.
In ref. \cite{BGG87} it was investigated (in Kruscal-Szekares coordinates)
the case when the value $f\dot t$ becomes negative for the bubbles crossing
the horizon, the systematical catalog of possible
solutions of the equation of motion was obtained and was shown, that two
signatures of $f$ correspond to the  two different, but equally acceptable
trajectories of shell.

  b). The signature of $\dot t_\pm$ ($f^{\pm} > 0$).

Here we shall examine the mostly interesting case of macroscopic bubbles
we shall assume, that the shell  always  lies  outside  the horizon  and
thus the unusual features of  the  space geometry are avoided. In this case
$f^\pm > 0$ and the signature of $f\dot t$ is determined by the sign of
$\dot t$.

Note that the example, when time flows in  opposite directions is the case
of Reissner-Nordstrom metric, when in the  region between the upper and
Cauchy radii the time coordinate changes its direction to the opposite. In
this case (in difference  to  the ours) the metric covers the whole
space-time $(r > 0)$ and such  feature was easy to notice.  The problem of
sign ambiguity entered due to taking a square root in equation (\ref{6.11}).
Similar situation is in Dirac theory when one has states with negative
energy that correspond to antiparticles.

We shall choose the positive direction for the 2-spheres in the way that the
radii increase in the direction of the outer normal. Then orientation of
$\Sigma$ is determined by the direction of time flow on the shell and

\begin{equation}
\varepsilon = sign \dot t
\label{eps}
\end{equation}

  It is clear, that the signature of $\dot t_\pm$ depends on the  direction
of time flowing in $V^\pm$ regions and on the shell. It is  natural  to
assume, that in $V^\pm$ regions time flows in the same direction i.e.

\begin{equation}
sign (t_+) = sign (t_-)
\label{8.8}
\end{equation}
while the direction of intrinsic time $\tau$  has to be found from the
equation of motion. From (\ref{eps}) and (\ref{8.8}) it is clear that
$\varepsilon_+ = \varepsilon_- \equiv \varepsilon = 1$ when $t_\pm$ and
$\tau$ flow in the same direction and $\varepsilon = -1$ in the opposite
case. Assumption (\ref{8.8}) leaves only two cases of equation (\ref{8.1}):
\begin{equation}
\sqrt{f_+ + \dot R^2} - \sqrt{f_- + \dot R^2} = \pm 4\pi\sigma GR
\label{em}
\end{equation}
which depend on the values of $f_\pm$.

  If $f_- > f_+$, than $| K_+ | < | K_- |$ and we have to take the lower
sign $(-)$ in (\ref{em}), because the energy density of the shell  is
positive (see (\ref{8.1})). According to (\ref{7.41}) and (\ref{8.8}) this
means  that $\dot t_\pm > 0$  and thus the time off and on the shell flows
in the same direction.

  Similarly if $f_- < f_+$ we have $\dot t_\pm < 0$, that means that the time
on the shall flows in the opposite direction to the  time  in  $V^\pm$
regions.

  Note, that for the simplest case (\ref{8.7}) the signature of $\dot t$
is connected with the sign of constant $m$, which is the integration
constant of solution of Einstein equations in $V^+$ region. This  constant
is fixed by matching of $f_+$ to the solution of Einstein  equations  on the
shell in the same coordinates. In the Newton approximation

$$
f_+ = 1 - 2G\cdot \int \left( T_0^0 - \frac{1}{2}\cdot T \right) \cdot
\frac{1}{R}\cdot dV  .
$$
The value and the signature of $m$ is determined by this formula and it can
be positive as well as negative dependently on boundary conditions.
In the next section we shall see that the correct choice of sign in
(\ref{em}) cancels the contradictions between Tolman's formula and Newton's
limit of (\ref{em}).

\section{Different kinds of shells}

{}~~~~~Here we shall consider two cases of macroscopic bubbles and show how
one has to choose the different signs of $\dot t$.

\medskip
{\large \bf a. The dust walls in vacuum.  }
\medskip

As we have mentioned above, for the shell of dust

\begin{equation}
p=0
\label{9.1}
\end{equation}
and the surface energy tensor of layer (\ref{5.4}) is equal to

$$
S^{ij} = \sigma u^iu^j  .
$$
Conservation law (\ref{5.7}) now has the form

\begin{equation}
D_i(\sigma u^i) = 0
\label{9.3}
\end{equation}
stating, that the total amount of dust is conserved. For the spherical shells
in the intrinsic coordinates $\xi^i$ with  the  interval
(\ref{6.6}) we have

$$
u^0 = 1 ;
$$
$$
u^1 = u^2 = 0 .
$$
Therefore

$$
D_iu^i = \Gamma^i_{0i} = \frac{2\dot R}{R} .
$$
So that equation (\ref{9.3}) reduces simply to

$$
\dot \sigma + 2\sigma \dot R/R = 0  .
$$
This equation is easily solved and yields

\begin{equation}
\sigma = \frac{const}{R^2} .
\label{9.7}
\end{equation}
To fix the constant in (\ref{9.7}) one can use  the  Tolman's  formula for
the proper mass of  the  shell  in  the  state  of  rest  (see (\ref{0.2})).
For the ideal fluid

\begin{equation}
M = 8\pi \cdot \int \left[ \sigma - \frac{1}{2}\cdot (\sigma - 2p) \right]
\cdot \delta (r-R)\cdot r^2dr = 4\pi R^2 \sigma = 4\pi \cdot const  .
\label{9.8}
\end{equation}
Using (\ref{9.1}) and (\ref{9.7}) for the $const$ from (\ref{9.8}) we obtain

$$
const = \frac{M}{4\pi} .
$$

In the simplest case of uncharged shells of dust in vacuum when the metric
is given by the formulae (\ref{8.7}) the motion equation (\ref{em}) takes
the form

\begin{equation}
\sqrt{1 - 2Gm/R + \dot R^2} - \sqrt{1 + \dot R^2} = \pm 4\pi \sigma G R =
\pm G\frac{M}{R} .
\label{9.10}
\end{equation}
We can rewrite (\ref{9.10}) in the form

\begin{equation}
m = \mp M\cdot \sqrt{1+\dot R^2} - G\frac{M^2}{2R} .
\label{9.12}
\end{equation}
The term $-GM^2/2R$ represents the gravitational interaction energy
according to the Newton's law, and $M\cdot \sqrt{1+\dot R^2}$ represents the
internal mass of the shell (the root is the analog of the Lorentz factor and
is equal to 1 in equilibrium, when $R = 0$ ).

{}From this relation it is easy to find that one has to choose in (\ref{9.10})
the {\it lower} sign "$-$" corresponding to $f\dot t > 0$ ($\varepsilon =
1$). Certainly, in this case $m > 0$, $f_+ < f_-$ and the condition of
surface energy density positiveness is satisfied. Besides, it is easy to see
that with this choice of sign the static Newton's limit of (\ref{9.12})
coinsides with Tolman's formula (\ref{9.8}) (though Tolman's formula is
valid only for static states, which, as we'll see, do not occur in this
simplest model, the term corresponding to gravitational energy of the shell
must enter the expression for the shell mass with the correct sign. Then in
static Newton's limit ($\dot R = 0; \quad G = 0$) the mass of the shell will
be described by Tolman's formula).

Note that if one would choose the {\it upper} sign "$+$" in (\ref{9.10})
then he should receive the contradiction between the static Newtons limit of
(\ref{9.12}) and Tolman's formula (\ref{9.8}).

Now let us show that in this simplest model there is no equilibrium state
for the dust shell. The conditions

\begin{equation}
\dot R = 0 , \quad \ddot R = 0
\label{9.13}
\end{equation}
must be satisfied for the spherical shell  to  be  in  equilibrium state.
The first of them leads from (\ref{9.12}) to the equation

$$
m = M - G\frac{M^2}{2R} .
$$
Taking $d/d\tau$ of (\ref{9.12}) we obtain

$$
\ddot R = \frac{1}{M^2}\cdot \left( m + G\frac{M^2}{2R} \right) \cdot
\frac{dm}{dR} ,
$$
so the second condition (\ref{9.13}) gives the stipulation

\begin{equation}
\frac{dm}{dR} = 0 .
\label{9.16}
\end{equation}
Using (\ref{9.12}) and (\ref{9.16}) one can see, that the radius of stability
of the shell $R_{stab} \rightarrow \infty$, i.e. there is no equilibrium
configuration for the dust wall in vacuum.

\medskip
{\large \bf b. Domain wall in vacuum.}
\medskip

In the case of domain wall one has the picture opposite to the previous
case.  For the domain wall $$p =\sigma$$ and the surface energy tensor of
layer (\ref{5.4}) is

$$
S^{ij} = \sigma\gamma^{ij} .
$$
{}From the conservation law (\ref{5.7})

$$
D_i\sigma = 0
$$
we obtain

$$
\sigma = const .
$$
{}From the Tolman's formula (\ref{0.2}) we have:

\begin{equation}
M = 8\pi \cdot \int \left[ \sigma - \frac{1}{2}\cdot (\sigma - 2p) \right]
\cdot \delta (r-R)\cdot r^2dr = -4\pi\sigma R^2  .
\label{10.4}
\end{equation}
One can see that the proper mass of layer is negative, if the energy density
of domain wall $\sigma$  is positive.

Considering again the simplest case (\ref{8.7}) one obtains the motion
equation
\begin{equation}
\sqrt{1- \frac{2Gm}{R} +\dot R^2} - \sqrt{1 + \dot R^2} = \pm 4\pi \sigma GR
= \mp G\frac{M}{R},
\label{10.6}
\end{equation}
where $M$ is described by (\ref{10.4}).

Again let us rewrite (\ref{10.6}) in the form
\begin{equation}
m = \pm M\cdot \sqrt{1+\dot R^2} - G\frac{M^2}{2R} .
\label{10.7}
\end{equation}
This formula as well as (\ref{9.12}) corresponds to the energy balance
equation and must contain the Newton's gravitational energy of the shell.

In opposite to dust-wall case, now one has to choose in (\ref{10.6}) the
{\it upper} sign "$-$" corresponding to $f\dot t < 0$ ($\varepsilon = -1$).
Certainly, in this case $m < 0$, $f_+ > 1$ and the surface energy density
positiveness condition is satisfied. Besides, for this choice of sign the
static Newton's limit of (\ref{10.7}) coinsides with Tolman's formula
(\ref{10.4}).

We emphasize once more that though Tolman's formula is valid only for static
states, the term corresponding to gravitational energy of bubble must enter
the expression for the bubble mass with the correct sign. Then in static
Newton' limit the mass of the bubble will be described by Tolman's formula.

One can see, that when one chooses the upper sign in (\ref{10.6}) then

$$
m = - \left( 4\pi\sigma R^2 \sqrt{1+\dot R^2} +
8\pi^2\sigma^2GR^3 \right).
$$
In this expression the mass is negative and we can  believe,  that distant
observer will be repelled by the domain wall.

  If one considers the equation (\ref{10.6}) with different sign in the
right side (see \cite{BKT87,IS84,L88}), one will obtain the positive value
for mass

$$
m =  \kappa R^2 \sqrt{1+\dot R^2} - \frac{G\kappa^2R^3}{2}.
$$
This expression contains the rest mass (\ref{10.4}) with the  wrong  sign
(the first term with $\dot R  = 0$) and contradicts with Tolman's formula
(\ref{10.4}). Thus Tolman's formula helps us to choose the correct sign in a
motion equation (\ref{10.6}).

   Thus the correct choice of the sign in the motion equation (\ref{10.6})
leads to the new solution of Einstein equations for bubbles (Schwarzschild
solution with negative mass) which shows that spherical  domain  walls are
repulsive. This fact requires a further investigation of bubble-dynamics and
their creation in cosmological models.

\section{ Schwarzschild space with negative mass.}

{}~~~~~Let us investigate the properties of Schwarzschild metric

\begin{equation}
ds^2 = \left( 1 - \frac{b}{r} \right) \cdot dt^2 - \frac{1}{1-b/r}
\cdot dr^2 - r^2 \cdot (d\theta^2 + \sin^2\theta \cdot d\varphi ^2 )
\label{11.1}
\end{equation}
in the case, when constant $b$ (which is related to the active  gravitational
mass of the source), can be negative.

First of all, for such spaces $g_{00} > 1$ and the velocity of light exceeds
its velocity measured in Minkowski space. However this fact does not cause
appearance of takhions.

For the metric (\ref{11.1}) the nonzero components of curvature tensor are

$$
R_{trtr} = \frac{b}{r^3} ,
$$
$$
R_{t\theta t\theta} = \frac{R_{t\varphi t\varphi}}{\sin^2\theta} =
-\frac{b(r-b)}{2r^2} ,
$$
$$
R_{r\theta r \theta} = \frac{R_{r\varphi r\varphi}}{\sin^2\theta} =
-\frac{b}{2(r-b)} ,
$$
$$
R_{\theta\varphi\theta\varphi} = - br\cdot \sin^2\theta .
$$
In this case the complex invariants of gravitational field have the form

$$
I_1 = \frac{1}{48} \cdot R_{\alpha\beta\gamma\delta}\cdot
\left( R^{\alpha\beta\gamma\delta} - \frac{i}{2}\cdot
\varepsilon^{\alpha\beta\mu\nu} \cdot R^{\gamma\delta}_{\mu\nu} \right) =
\left( \frac{b}{2r^3} \right) ^2 ,
$$
$$
I_2 = \frac{1}{96} \cdot R_{\alpha\beta\mu\nu}R^{\mu\nu\rho\sigma} \cdot
\left( R^{\alpha\beta}_{\sigma\rho} - \frac{i}{2}\cdot
\varepsilon_{\sigma\rho\lambda\kappa}\cdot R^{\lambda\kappa\alpha\beta}
\right) = -\left( \frac{b}{2r^3} \right) ^3 .
$$
One can see, that for the negative $b$ the second invariant changes its
sign. The result of this fact is that the  3-space  $(t = const )$ Gauss
curvature for the plains, normal to radius also changes  its sign:

\begin{equation}
k = \frac{P_{\theta\varphi\theta\varphi}}{\gamma_{\theta\theta}\cdot
\gamma_{\varphi\varphi}} = - P^r_r = \frac{b}{r^3} .
\label{11.4}
\end{equation}
Here $P_{\theta\varphi\theta\varphi}$, $P^r_r$, $\gamma_{\theta\theta}$,
$\gamma_{\varphi\varphi}$ are 3-curvatures and 3-metric  tensors of space
(\ref{11.1}) in the case $t = const$. Equation (\ref{11.4})  means, that for
the different signs of $b$ this subspaces belong to  the different types of
surfaces. To show this let us  embed  metric (\ref{11.1}) into Euclidean
space with more then 4 dimensions.

  An embedding of the linear element (\ref{11.1})  in  5  dimensions  is
impossible (see \cite{F59,R65}). Embedding of (\ref{11.1}) with positive
$b$ in  6 dimensions with signature 2+4

$$
ds^2 = dz_1^2 + dz_2^2 - dz_3^2 - dz_4^2 - dz_5^2 - dz_6^2
$$
is given by

\begin{eqnarray}
z_1 = \left( 1 - \frac{b}{r} \right) \cdot \cos t , \nonumber \\
z_2 = \left( 1 - \frac{b}{r} \right) \cdot \sin t , \nonumber \\
z_3 = \int \left( \frac{b(b+4r^3)}{4r^3(r-b)} \right) ^{1/2} \cdot dr ,
\nonumber \\
z_4 = r \cdot \cos \theta , \nonumber \\
z_5 = r \cdot \sin \theta \cdot \cos \varphi, \nonumber \\
z_6 = r \cdot \sin \theta \cdot \sin \varphi .
\label{11.6}
\end{eqnarray}
It is possible to eliminate the coordinates

$$
z_1^2 + z_2^2 = 1 - \frac{b}{r} ,
$$
$$
z_3 = \int \left( \frac{b(b+4r^3)}{4r^3(r-b)} \right) ^{1/2} \cdot dr ,
$$
$$
z_4^2 + z_5^2 + z_6^2 = r^2 .
$$
We note, that this surface in $z_1z_2$ plane is the 1-sphere,  in  the
$z_4z_5z_6$ is the 2-sphere and $z_3$ is space-like. Time-like  coordinates
$z_1$ and $z_2$ are periodic functions of $t$ so that embedding (\ref{11.6})
identifies distinct points of the original manifold. This suggests replacing
the trigonometrical functions by the hyperbolic functions and embedding of
(\ref{11.1}) for the positive $b$ is possible for the signature 1+5:

$$
ds^2 = dz_1^2 - dz_2^2 - dz_3^2 - dz_4^2 - dz_5^2 - dz_6^2 ,
$$
where
\begin{eqnarray}
z_1 = \left( 1 - \frac{b}{r} \right) \cdot \sinh t , \nonumber \\
z_2 = \left( 1 - \frac{b}{r} \right) \cdot \cosh t , \nonumber \\
z_3 = \int \left( \frac{b(4r^3-b)}{4r^3(r-b)} \right) ^{1/2} \cdot dr ,
\nonumber \\
z_4 = r \cdot \cos \theta , \nonumber \\
z_5 = r \cdot \sin \theta \cdot \cos \varphi, \nonumber \\
z_6 = r \cdot \sin \theta \cdot \sin \varphi .
\label{11.9}
\end{eqnarray}
We note that in $z_1z_2$ plane this surface now is hyperbola

$$
z_2^2 - z_1^2 = \left( 1 - \frac{b}{r} \right) .
$$

  For the negative mass
$$
   b  < 0
$$
we can see from (\ref{11.6}) and (\ref{11.9}), that coordinate $z_3$ became
complex and embedding (\ref{11.6}) now takes place in the space with
signature 3+3

$$
ds^2 = dz_1^2 + dz_2^2 + dz_3^2 - dz_4^2 - dz_5^2 - dz_6^2
$$
where

$$
z_1 = \left( 1 - \frac{b}{r} \right) \cdot \cos t ,
$$
$$
z_2 = \left( 1 - \frac{b}{r} \right) \cdot \sin t ,
$$
$$
z_3 = \int \left( \frac{b(b-4r^3)}{4r^3(r-b)} \right) ^{1/2} \cdot dr ,
$$
$$
z_4 = r \cdot \cos \theta ,
$$
$$
z_5 = r \cdot \sin \theta \cdot \cos \varphi,
$$
$$
z_6 = r \cdot \sin \theta \cdot \sin \varphi .
$$
Now the coordinate $z_3$ became time-like. Embedding (\ref{11.9}) is
possible for the signature 2+4

$$
ds^2 = - dz_1^2 + dz_2^2 + dz_3^2 - dz_4^2 - dz_5^2 - dz_6^2
$$
where

$$
z_1 = \left( 1 - \frac{b}{r} \right) \cdot \sinh t ,
$$
$$
z_2 = \left( 1 - \frac{b}{r} \right) \cdot \cosh t ,
$$
$$
z_3 = \int \left( \frac{b(b-4r^3)}{4r^3(r-b)} \right) ^{1/2} \cdot dr ,
$$
$$
z_4 = r \cdot \cos \theta ,
$$
$$
z_5 = r \cdot \sin \theta \cdot \cos \varphi,
$$
$$
z_6 = r \cdot \sin \theta \cdot \sin \varphi .
$$
In this case the coordinate $z_1$ is space-like and we can not  identify the
surface in $z_1z_2z_3$ space. $z_4z_5z_6$ surface is  the  2-sphere $r = R$
in all the cases and is identified in our work with the surface of domain
bubble.

To conclud this section we can say that the space (\ref{11.1}) with $b < 0$
can be embedded into a 6-dimensional space with signature (2+4) or (3+3),
while the usual Schwarzschild space $(b > 0)$ --- into the space with
signature (1+5) or (2+4) respectively.

\section{ The model.}

{}~~~~After we have formulated our idea \cite{BG94} how to avoid the
disagreements mentioned at the beginning of this paper let  us  investigate
a more general case of a spherically symmetrical charged bubble in vacuum,
when the metric outside the bubble is

$$
f_+ = 1 - \frac{2Gm}{r} + \frac{Ge^2}{r^2} - G\Lambda_+r^2 ,
$$
while inside ---

$$
f_- = 1 - G\Lambda_-r^2 ,
$$
where $\Lambda_\pm \equiv \frac{8\pi}{3}\cdot\rho$ , $\rho$
being the vacuum energy density in $V^\pm$ regions, and $e$ is the charge on
the shell.

  Now the motion equation (\ref{8.1}) takes the form
\begin{equation}
\sqrt{\dot R^2 + 1 - \Lambda_+GR^2 - \frac{2Gm}{R} + \frac{Ge^2}{R}} -
\sqrt{\dot R^2 + 1 - \Lambda_-GR^2} = \pm G\kappa R ,
\label{26}
\end{equation}
where

$$
\kappa \equiv 4\pi \sigma .
$$
Finding $m$ from this equation we yield:

\begin{equation}
m = \frac{\Lambda_- - \Lambda_+}{2}\cdot R^3 - \frac{G\kappa^2}{2}\cdot R^3
+ \frac{e^2}{2R} \mp \kappa R^2\cdot \sqrt{\dot R^2 + 1 - \Lambda_-GR^2}
\label{27}
\end{equation}
It is easy to understand the meaning of the terms in (\ref{27}). The first
term is the volume energy of the bubble (a difference between the old and
new vacuum energy densities). The second term  represents an energy of
gravitational self-interaction of  the  shell (the surface-surface binding
energy). The third term is the  electrostatic energy lying in the
three-space outside the bubble. The last term contains the kinetic energy of
the shell  and  surface-volume binding energy. As we have mentioned above,
we have to choose  the upper sign in equations (\ref{26}) and (\ref{27}) to
avoid the disagreements with Tolman's formula (\ref{10.4}).

  To examine dynamics of the bubble let us rewrite  the  equation
(\ref{27}) in the following way:

\begin{equation}
\dot R^2 - \Biggl[ \frac{1}{\kappa^2}\cdot\biggl( -\frac{m}{R^2} -
\frac{a}{2}\cdot R + \frac{e^2}{2R^3}\biggr)^2 + \Lambda_-GR^2 \Biggr] = - 1 ,
\label{28}
\end{equation}
where $a \equiv \Lambda_+ - \Lambda_- + G\kappa^2$. It is worth to note that
in this  equation the sign ambiguity disappears due to squaring. Introducing
new  dimensionless variables

\begin{eqnarray}
z \equiv \frac{R}{(-2m)^{1/3}} \cdot
\bigl( a^2 + 4\kappa^2\Lambda_-G \bigr)^{1/6} ,   \nonumber\\
\tau' \equiv \frac{\tau}{2\kappa}\cdot\bigl( a^2 + 4\kappa^2\Lambda_-G
\bigr)^{1/2}
\label{29}
\end{eqnarray}
and dimensionless parameters

$$
A \equiv a\cdot
\bigl( a^2 + 4\kappa^2\Lambda_-G \bigr)^{-1/2}  ,
$$
$$
E \equiv - \frac{4\kappa^2}{(-2m)^{2/3}} \cdot
\bigl( a^2 + 4\kappa^2\Lambda_-G \bigr)^{-2/3}  ,
$$
$$
Q^2 \equiv \frac{e^2}{(-2m)^{4/3}} \cdot
\bigl( a^2 + 4\kappa^2\Lambda_-G \bigr)^{1/6}  ,
$$
we can represent the motion equation (\ref{28}) as

$$
\biggl( \frac{dz}{d\tau'} \biggr)^2 + U(z) = E ,
$$
which is identical to that of the point like particle  with  energy $E$ ,
moving in one dimension under the influence of the potential

\begin{equation}
U(z) = - \Biggl[ z^2 - \frac{2A}{z}\cdot\biggl(1+\frac{Q^2}{z}\biggr) +
\frac{1}{z^4}\cdot\biggl(1+\frac{Q^2}{z}\biggr)^2\Biggr] ,
\label{32}
\end{equation}
For real trajectories $U$ must be negative since $E < 0$. Such potentials
(but for the case of uncharged shells $(Q = 0)$ with $ m > 0$) were
discussed in \cite{AKMS87,BGG87}.

  In the equilibrium state $\dot z_{|z=z_0} = 0$, where $z_0$ is the
equilibrium point,  $U(z_0) = E$ and one can find the critical mass of the
bubble:

\begin{equation}
m_0 = - \frac{4\kappa^3}{(a^2 + 4\kappa^2G\Lambda_-)\cdot {U_0}^{3/2}}
\label{33}
\end{equation}
where $U = |U(z_0)| > 0$. From (\ref{29}) for the equilibrium radius we have

\begin{equation}
R_0 = \frac{2\kappa z_0}{{U_0}^{1/2}\cdot (a^2 + 4\kappa^2G\Lambda_-)^{1/2}} .
\label{34}
\end{equation}
{}From this formula it is easy to notice that for the de Sitter spaces
$(\Lambda >0)$ the radius is maximal, when \it a \rm is minimal. So to get
the macroscopic bubbles it is worth to set $\Lambda_+ = \Lambda_- = \Lambda$.
In  this case the domain wall separates the  phases  with  the  same  vacuum
energy. Then $a \sim G^2\kappa^4$  which is negligible compared to the second
term in brackets. The charge enters the formulae  (\ref{33}), (\ref{34}) only
through $U_0$ and has very small influence on the values $m_0$ and $R_0$. So
we can set $Q = 0$. With this assumptions one finds  that  \it U  \rm reaches
its maximum $U_0 \sim 1$ at the point $z_0 \sim 1$. This implies that

$$
 m_0 \sim - \frac{\kappa}{G\Lambda} , \\
 R_0 \sim \frac{1}{\sqrt{G\Lambda}} .
$$

Considering a scenario for domain wall production in models with spontaneous
breaking of some gauge symmetry group, it is easy to see that $\Lambda \sim
\alpha^{-1}{M_G}^4$, $\kappa \sim \alpha^{-1}{M_G}^3$ and the thickness of
the wall $d \sim \frac{1}{M_G}$, where $M_G$ is the symmetry breaking scale
and $\alpha \sim 10^{-2}$ is  the coupling constant \cite{BKT87,V85,V91}.

  For the different scales of particle physics we yield:

  a). The  Electro-Weak  scale:  $M_G \sim 10^2GeV$.  Then $\Lambda \sim
10^{10}GeV^4$,
$\kappa \sim 10^8GeV^3$  and $R_0 \sim 10^{14}GeV^{-1}$,
$m_0 \sim -10^{36}GeV$. The  radius  of  such
bubble is about 1 cm, and its negative mass is about million tons.

  b). The scale of family symmetry \cite{C91}: $M_G \sim 10^4 \div 10^{10}GeV$.
Then $\Lambda \sim 10^{18} \div 10^{42}GeV^4$, $\kappa \sim 10^{14} \div
10^{32}GeV^3$  and  $R_0 \sim 10^{10} \div 10^{-2}GeV^{-1} \sim
10^{-4} \div 10^{-16}cm$
, $m_0 \sim - (10^{34} \div 10^{28})GeV \sim - (10^{10} \div 10^4)g$.

  c).  The  scale  of  Grand   Unification:  $M_G \sim 10^{15}GeV$.   Then
$\Lambda \sim 10^{62}GeV^4$, $\kappa \sim 10^{47}GeV^3$ and thus
$R_0 \sim 10^{-12}GeV^{-1}$, $m_0 \sim -10^{23}GeV \sim 0,1g$. This is a very
small radius and  there  arises  the  question
of the validity of the thin-wall approximation ($ d \sim 10^{-15}GeV^{-1}$).

  We see that in all the cases the mass of the bubble is negative and  it
exhibits a strong gravitational repulsion.

\section{The problem of stability.}

  ~~~~Unfortunately this equilibrium state seems to be  unstable.  Potential
(\ref{32}) has the single maximum. In contrary to the  case  with positive
$m$ \cite{BKT87}, the charge does not stabilize the  bubble.  Perhaps one
could find the stable equilibrium states of bubble with negative mass in
some nonvacuum models or in models with rotating bubble.

  In \cite{GKHK78} it is suggested that the inclusion of a nonvanishing
angular momentum might stabilize the shell.  The  surface  energy-momentum
density tensor of a charged rotating shell is given by \cite{I70,L84}

\begin{equation}
T^i_j = - \sigma u^i u_j + \delta^i_j \sigma ,\qquad i,j = 0, 2, 3
\label{36}
\end{equation}
It consists of a mixture of two perfect fluids. The first term can be
interpreted as "dust" particles with negative  energy  density.  The second
one represents a domain wall. The negative mass  "dust" increases the
repulsive character of the bubble and  can  keep  it static. Exterior for of
a spinning charged shell is the  Kerr-Newman metric. Our simple analysis
based on spherical symmetrical equations is not  valid  in  this  case  and
model  (\ref{36})  requires  further investigations.

If the problem of stability will be solved there will be an interesting
possibility of creation of the static singular shell of macroscopic size and
with repulsive gravitational field. Radius and repulsive features of such
objects depend on the scale of symmetry violation and can be varied slowly
by changing the forces stabilizing the bubble.

\bigskip
{\large \bf Acknowledgment . }
\medskip

The research described in this publication was made possible in part by
Grant MXL000 from the International Science Foundation.

redpajama_set_name: RedPajamaArXiv