Jane Austen s Pride And Prejudice - 2530 Words

Just as one cannot love anyone else until they love themselves, one cannot be true to anyone else until they are true to themselves. The novel Pride and Prejudice illustrates a powerful message of being true to oneself through the powers of the author s own outlook. As an opinionated women of her time, Jane Austen voiced her views through her own writing. Even though the objective of the book was to entertain readers, Austen never failed to wittily use plot and characters to express her own stance. Her distinctive own style of writing has left future generations something to scrutinize over. In the play, â€Å"Hamlet†, Shakespeare wrote â€Å"to thine own self be true,/And it must follow, as the night the day,/Thou canst not then be false to any†¦show more content†¦The path to finding oneself relies on the on being true to oneself. Many cruise by life following societal standards without questioning oneself. This allows one to be average and normal. While many are afraid of being an outcast, some are willing to take the risks and challenge society; in Pride and Prejudice, Elizabeth and Mr. Darcy show examples of defying the standards of society to fight for their own beliefs. During the beginning of the novel, the Bingley sisters criticize Elizabeth for walking in the rain to see her sick sister, Jane. One of the sisters exclaim, â€Å" To walk three miles, or four miles, or five miles, or whatever it is, above her ankles in dirt, and alone, quite alone! † (8.10). For them as â€Å"very fine ladies† (4.11), it was considered inappropriate for a lady to walk alone in mud although it was for the sake of her sister. However, Elizabeth allowed herself to be the subject of judgments if it meant helping her family. Later in the book, she makes another decision that is frowned upon by people around her. She rejects Mr. Collins who is a cousin of her father. A sole objective for a women in the eighteenth century was to get married t o a respectable man. Elizabeth, despite a good marriage opportunity, rejects Mr. Collins. She believes that marriage without love is doomed to misery. Through Elizabeth, Austen successfully conveys feminine individuality. Austen deliberately

Essay on Three Recomendations for the Democratic Republic...

Three Recommendations for the DRC The Democratic Republic of Congo (DRC) has undergone two recent wars in 1998 and 1999-2008 with Ugandan rebels and Uganda itself. In the wake of these wars, an estimated 6 million people lost their lives due to war-induced causes. Horrific human rights abuses such as systematic rape and murder have created a horrific humanitarian crisis in the DRC (The International Coalition for the Responsibility to Protect website, n.d.). Currently, this violence is associated with conflict materials; in the area of the DRC, this is not related just to diamonds, but it is related to this and the mining of gold, copper, cobalt, gold, tantalum and tin. Neither the military nor the government appears to have concern†¦show more content†¦The final recommendation for the DRC is a focus on governmental corruption. Many times the different governmental organizations, municipalities and departments fail to follow the budget passed by the national legislature, instead using the money for the po liticians and bureaucrats own purposes. By increasing transparency within these ministries, strengthening financial reporting and cash management procedures, as well as auditing the results, the government could deal with the high level of corruption and instead spend the money on the citizens and their needs as was expected with the passage of the budgets. In doing so, the country’s poorest citizens will have access to social services designed to improve health, education and agriculture (Congo-Kinshasa: World Bank supports strong financial management in Democratic Republic of Congo, 2014). With financial integrity the funds can be utilized for the populations for which they were intended, not to line the pockets of the bureaucrats. Conclusion With reform the DRC’s government can offer transparency and auditing tools to ensure that the budgets that are passed are implemented and the monies go to the people for whom they were intended. This can ensure access to resources such as access to services intended to improve health outcomes, education and agriculture. Access to health care can

Attacking Wifi Nets with Traffic Injection Free Essays

string(68) " or with fine which may extend up to two lakh rupees, or with both\." I am very much thankful to him. I benefited a lot discussing with him. I am also thankful to my parents who encouraged me and provided such a motivation, so I became able to perform this. We will write a custom essay sample on Attacking Wifi Nets with Traffic Injection or any similar topic only for you Order Now I am also thankful to all my friends and those who helped me directly or indirectly in completion of my project. CONTENTS †¢Introduction †¢Crime Definition †¢Laws that have been violated †¢Possible Punishments (IT ACT + INTERNATIONAL LAWS) †¢Unlawful Losses and Gains †¢Working of Attacks †¢Description of Tools INTRODUCTION This term paper is based on â€Å"attacking wifi nets with traffic injection† also nown as packet injection which simply means the hacking of wireless networks with different techniques to send extra amount of traffic (packets, frames, duplicate copies) on a network by which a hacker can able to access the information and identity that a client is using. Some techniques are wireless network sniffing, DOS (denial of service attack), Man in the middle attack etc. Attacks on wireless LANs (WLAN’s) and wireless-enabled laptops are a quick and easy way for hackers to steal data and enter the corporate network. Many types of tools are used to perform hacking. Some of them are named as aircrack-ng, airjack etc. thts paper will later give brief information on tools used , working of tools ,losses and gains with hacking etc. These type of attacks are known as INTEGRITY attacks. Wireless networks broadcast their packets using radio frequency or optical wavelengths. A modern laptop computer can listen in. Worse, an attacker can manufacture new packets on the fly and persuade wireless stations to accept his packets as legitimate. We already know 802. 11 networks are weak. Open networks are prone to any well-known LAN perimeter attack WEP is vulnerable. Traffic injection has changed things like †¢Increased DoS (denial of service) capabilities †¢Dramatically decreased WEP cracking achievement time †¢Allows traffic tampering †¢Allows stations attacks CRIME DEFINITION Cyber Crime –A crime where the computer is used as a tool or target. Cyber crime encompasses any criminal act dealing with computers and networks (called hacking). Additionally, cyber crime also includes traditional crimes conducted through the Internet. For example; hate crimes, telemarketing and Internet fraud, identity theft, and credit card account thefts are considered to be cyber crimes when the illegal activities are committed through the use of a computer and the Internet. Hacking – Traffic injection attacks comes under hacking. It is defined as whomever with the intent to cause or knowing that he is likely to cause wrongful loss or damage to the public or any person destroys or deletes or alters any information residing in a computer resource or diminishes its value or utility or affects it injuriously by any means, commits hacking. Hacking may also occur when a person willfully, knowingly, and without authorization or without reasonable grounds to believe that he or she has such authorization, destroys data, computer programs, or supporting documentation residing or existing internal or external to a computer, computer system, or computer network. Besides the destruction of such data, hacking may also be defined to include the disclosure, use or taking of the data commits an offense against intellectual property. This paper is a survey of wireless attack tools focusing on 802. 11 and Bluetooth. It includes attack tools for three major categories: confidentiality, integrity, and availability. Confidentiality attack tools focus on the content of the data and are best known for encryption cracking. Integrity attacks tools focus on the data in transmission and include frame insertion, man in the middle, and replay attacks. Finally, availability attack tools focus on Denial of Service (DoS) attacks. Law That Have Been Violated The laws that have been violated are section 43,65 and 66 of IT ACT 2000. Section 43 of IT ACT 2000, defines as If any person without permission of the owner or any other person who is in charge of a computer, computer system or computer network, — (a) Accesses or secures access to such computer, computer system or computer Network; (b) Downloads, copies or extracts any data, computer data base or information from such computer, computer system or computer network including information or data held or stored in any removable storage medium; c) Introduces or causes to be introduced any computer contaminant or computer virus into any computer, computer system or computer network; (d) Damages or causes to be damaged any computer, computer system or computer network, data, computer data base or any other programs residing in such computer, computer system or computer network; (e) Disrupts or causes disruption of any computer, computer system or computer network; (f) Denies or caus es the denial of access to any person authorized to access any computer, computer system or computer network by any means; g) Provides any assistance to any person to facilitate access to a computer, computer system or computer network in contravention of the provisions of this Act, rules or regulations made there under; (h) Charges the services availed of by a person to the account of another person by tampering with or manipulating any computer, computer system, or computer network, Section 65 of IT ACT 2000, defines as, Tampering with computer source documents Whoever knowingly or intentionally conceals, destroys or alters or intentionally or knowingly causes another to conceal, destroy or alter any computer source code used for a computer, computer programs, computer system or computer network, when the computer source code is required to be kept or maintained by law for the time being in force, shall be punishable with imprisonment up to three years, or with fine which may extend up to two lakh rupees, or with both. You read "Attacking Wifi Nets with Traffic Injection" in category "Papers" Section 66 of IT ACT 2000, defines as, (1) Whoever with the intent to cause or knowing that he is likely to cause rongful loss or damage to the public or any person destroys or deletes or alters any information residing in a computer resource or diminishes its value or utility or affects it injuriously by any means, commits hacking. (2) Whoever commits hacking shall be punished with imprisonment up to three years, or with fine which may extend up to tw o lakh rupees, or with both. POSSIBLE PUNISHMENTS (IT ACT + International laws) Cyber crime is a type of crime that not only destroys the security system of a country but also its financial system. One supporter of legislation against cyber crime, Rep. Lamar Smith (R-Texas), stated, â€Å"Our mouse can be just as dangerous as a bullet or a bomb. † Cyber attackers should be penalized and punished severely and most cyber crimes have penalties reflecting the severity of the crime committed. Although in the past many laws against cyber crimes were insufficient, law enforcement agencies and governments have recently proposed many innovative plans for fighting cyber crimes. Punishment Cybercrime must be dealt with very seriously because it causes a lot of damage to businesses and the actual punishment should depend on the type of fraud used. The penalty for illegally accessing a computer system ranges from 6 months to 5 years. The penalty for the unofficial modification on a computer ranges from 5 to 10 years. Other penalties are listed below: Telecommunication service theft: The theft of telecommunication services is a very common theft and is punished with a heavy fine and imprisonment. Communications intercept crime: This is a Class-D crime which is followed by a severe punishment of 1 to 5 years of imprisonment with a fine. Other cyber crimes like telecommunication piracy, offensive material dissemination, and other cyber frauds also belong to this category. Information Technology Act-2000: According to this act, different penalties are available for different crimes. Some of the penalties are as follows: Computer source document tampering: The person who changes the source code on the website or any computer program will get a punishment up to 3 years of imprisonment or fine. Computer hacking: The individual who hacks the computer or computer devices will get an imprisonment up to 3 years or a fine. Government protected system: An act of trying to gain access to a system which is a protected system by the government, will result in imprisonment for 10 years and a heavy fine. The introduction of such penalties have lead to a drastic reduction in the cyber crime rates as more and more criminals are becoming aware of the penalties related to them. Spreading the word about the penalties of cyber crime can serve as a deterrent against such crime. Penalties relating to cyber crime will vary depending on the country and legislation in place. Punishments according to IT ACT 2000 The person who commits the crime shall be liable to pay damages by way of compensation not exceeding one crore rupees to the person so affected according to section 43 of IT ACT. The person who commits the crime shall be punishable with imprisonment up to three years, or with fine which may extend up to two lakh rupees, or with both according to section 65 of IT ACT. Whoever commits hacking shall be punished with imprisonment up to three years, or with fine which may extend up to two lakh rupees, or with both according to section 66 of IT ACT 2000 INTERNATIONAL LAWS In USA section 18 U. S. C.  § 1030 A) a fine under this title or imprisonment for not more than ten years, or both, in the case of an offense under subsection (a)(1) of this section which does not occur after a conviction for another offense under this section, or an attempt to commit an offense punishable under this subparagraph; and (B) a fine under this title or imprisonment for not more than twenty years, or both, i n the case of an offense under subsection (a)(1) of this section which occurs after a conviction for another offense under this section, or an attempt to commit an offense punishable under this subparagraph; In Canada The person who commits the crime is guilty of an indictable offence and liable to imprisonment for a term not exceeding ten years, or is guilty of an offence punishable on summary conviction. UNLAWFUL LOSSES AND GAINS Losses due to hacking Hackers targeted major companies including Sony, RSA Security, and Citigroup, but also governmental websites and smaller firms. Many companies could have prevented the attacks. Because of their vulnerabilities, they not only lost money, but also risked losing clients, prestige and market share. Multitudes of people were affected by their security breaches Recent reports showed hackers earned $12. billion in 2011, mainly by spamming, phishing, and online frauds. Some companies have made their financial losses public, while others chose not to disclose them. Here’s a top 5 of the declared losses caused by hackings from last year until present. Undeclared losses may even exceed these ones. 1. $171 million – Sony Hacked in April to June 2 011, Sony is by far the most famous recent security attack. After its Playstation network was shut down by LulzSec, Sony reportedly lost almost $171 million. The hack affected 77 million accounts and is still considered the worst gaming community data breach ever. Attackers stole valuable information: full names, logins, passwords, e-mails, home addresses, purchase history, and credit card numbers. 2. $2. 7 million – Citigroup Hacked in June 2011, Citigroup was not a difficult target for hackers. They exploited a basic online vulnerability and stole account information from 200,000 clients. Because of the hacking, Citigroup said it lost $2. 7 million. Just a few months before the attack, the company was affected by another security breach. It started at Epsilon, an email marketing provider for 2,500 large companies including Citigroup. Specialists estimated that the Epsilon breach affected millions of people and produced an overall $4 billion loss. 3. $2 million – Stratfor Last Christmas wasn’t so joyful for Stratfor Global Intelligence. Anonymous members hacked the US research group and published confidential information from 4,000 clients, threatening they could also give details about 90,000 credit card accounts. The hackers stated that Stratfor was â€Å"clueless†¦when it comes to database security†. According to the criminal complaint, the hack cost Stratfor $2 million. 4. $2 million – ATamp;T The US carrier was hacked last year, but said no account information was exposed. They said they warned one million customers about the security breach. Money stolen from the hacked business accounts was used by a group related to Al Qaeda to fund terrorist attacks in Asia. According to reports, refunding costumers cost ATamp;T almost $2 million. 5. $1 million – Fidelity Investments, Scottrade, E*Trade, Charles Schwab The most recent declared losses were in a brokerage scam. A Russian national was charged in the US with $1. 4 million in computer and hacking crimes. $1 million was stolen from stock brokerages Fidelity Investments, Scottrade, E*Trade, and Charles Schwab. The rest of the money was taken from fraudulent tax refunds, with the stolen identities of more than 300 people. Gains To Hackers * To use your computer: * as an Internet Relay Chat (IRC) server – hackers wouldn’t want to discuss openly about their activities on their ‘own’ servers * as Storage for Illicit Material (ex. pirated software, pirated music, pornography, hacking tools etc) * as part of a DDoS Attack – where many computers are controlled by hackers in an attempt to cause resource starvation on a victim’s computers or networks * To steal services and/or valuable files For thrill and excitement * To get even – maybe an IT staff who was terminated, or other parties you’ve ‘wronged’ * As a publicity stunt – an example of which was reported in 1998 by Jim Hu in MTV â€Å"hack† backfires * Knowledge/Experiment/Ethical – some hackers probe a computer system to find its security vulnerabilities and then inform the system administrator to help improve their security * Another possible reason is that the hackers might suffer from a disease called Asperger syndrome (AS). They are people who are very good with numbers and at focusing on a problem for a very long period of time, but are not good in social relationships. How AS can possibly be linked to hacking behavior was discussed more thoroughly by M. J. Zuckerman in his ‘USA Today’ article, What fuels the mind of a hacker? * Curiosity * To spy on friends, family members or even business rivals * Prestige – bragging rights in their social circle (particularly if they’ve hacked high-profile sites or systems) * Intellectual Challenge Money – although most hackers are not motivated by financial gain; many professional criminals make money by using hacking techniques either to * set up fake e-commerce sites to collect credit card details * gain entry to servers that contain credit cards details * engage in other forms of credit card fraud WORKING OF ATTACKS Before studying about how traffic injection attacks works there are some basic terms we shoud have to know WEP Wir ed Equivalent Privacy (WEP) is a shared-secret key encryption system used to encrypt packets transmitted between a station and an AP. The WEP algorithm is intended to protect wireless communication from eavesdropping. A secondary function of WEP is to prevent unauthorized access to a wireless network. WEP encrypts the payload of data packets. Management and control frames are always transmitted in the clear. WEP uses the RC4 encryption algorithm. The shared-secret key is either 40 or 104 bits long. The key is chosen by the system administrator. This key must be shared among all the stations and the AP using mechanisms that are not specified in the IEEE 802. 11. FRAMES Both the station and AP radiate and gather 802. 1 frames as needed. The format of frames is illustrated below. Most of the frames contain IP packets. The other frames are for the management and control of the wireless connection. There are three classes of frames. The management frames establish and maintain communications. These are of Association request, Association response, Reassociation request, Reassociation response, Probe request, Probe respo nse, Beacon, Announcement traffic indication message, Disassociation, Authentication, Deauthentication types. The SSID is part of several of the management frames. Management messages are always sent in the clear, even when link encryption (WEP or WPA) is used, so the SSID is visible to anyone who can intercept these frames. Authentication Authentication is the process of proving identity of a station to another station or AP. In the open system authentication, all stations are authenticated without any checking. A station A sends an Authentication management frame that contains the identity of A, to station B. Station B replies with a frame that indicates recognition, addressed to A. In the closed network architecture, the stations must know the SSID of the AP in order to connect to the AP. The shared key authentication uses a standard challenge and response along with a shared secret key. Traffic injection quick HOWTO †¢1 Insert adapter †¢2 Load driver and activate adapter †¢3 Set driver into monitor mode (real 802. 11 mode) †¢4 Set appropriate channel †¢5 Open PF PACKET/RAW socket on interface (Linux only) †¢6 Use your socket and play †¢ Still, you need a 802. 11 stack over your socket and/or good libs †¢ and tools so you can communicate WORKING – This phase of term paper describes the working of attack by using one tool called INJECTION WIZARD Injection Wizard is an application for injecting traffic into WEP-protected Wi-Fi networks, like aireplay-ng, but it’s much more easy to use and it can work with worse conditions (for example, more interferences, weaker transmitted/received signals, more restricted access points, etc). The higher the traffic of the network, the earlier we will be able to crack a WEP key with tools like aircrack-ng, airsnort, dwepcrack, weplab, WEPAttack, WEPCrack, etc. However, injecting traffic is not easy because you must build or capture a frame that causes a response frame in any other station (that is, a wireless node). This application carries out automatically all the needed actions to build a frame that causes a response in other station. These actions can be summarized in the following sequence of steps: 1. The application scans Wi-Fi networks and shows a list of WEP-protected networks, then it allows the user to select one of them. 2. It joins the selected network and monitors that network in order to find a data frame. 3. It tries to extract a keystream prefix from the captured frame and then it tries to extend the keystream up to 40 bytes by means of the W. A. Arbaugh’s inductive chosen plaintext attack. 4. It tries to find a host (for example, a connected computer, a network device, etc), which has an IP address belonging to a predefined range, by injecting forged ARP packets. 5. After finding an active host, it injects ARP packets targeted at that host. Some of the benefits of this application are easiness of use (due to its graphical interface, automatic operation, etc) androbustness (detection/management of network disconnections, repetition of failed actions, etc). Moreover, the Arbaugh’s inductive attack can be performed by any Wi-Fi interface supporting injection in monitor mode, because the interface driver doesn’t need any additional patch as it’s usual to happen with the Bittau’s fragmentation attack. Besides its higher applicability, this attack is generally more reliable than Chop-Chop attack for recovering a keystream of a given size, because it doesn’t have to inject any frame larger than needed. This application is distributed under the terms of the GNU General Public License version 2 (read the license. tm file for more details) and comes with absolutely no warranty. The author assumes no responsibility derived from the use or the distribution of this program. The copyright of this application is owned by Fernando Pablo Romero Navarro (May 2010). Injection Wizard has made use of (with convenient modifications) the following free software applications: * scapy (version 2. 0. 1), distributed under the license: GNU GPL version 2. Copyright: Philippe Biondi,2009 (http://www. secdev. org/projects/scapy). * python-wifi (version 0. 3. 1), distributed under the license: GNU LGPL version 2. 1. Copyright: Roman Joost, 2004-2008 Software Requirements For the client application (graphical interface): †¢Any system with a recent Java virtual machine: JRE version 1. 6 or later. For the server application: * A Linux box with a recent kernel, so it should support Wireless Extensions version 22 or later (since kernel version 2. 6. 21) and the mac80211 stack for Wi-Fi interfaces (since kernel version 2. 6. 24, it is supported by many Wi-Fi adapter drivers). * A Wi-Fi network interface driver supporting injection in monitor mode (sometimes it’s required to patch the driver for supporting this feature). The iw system command, if it’s not provided by your Linux distribution you can get it by installing the aircrack-ng package or by compiling the source code that can be downloaded from: http://wireless. kernel. org/download/iw. * A Python interpreter with version 2. 5, later versions might also work. Instructions 1. Uncompress the injwiz. zip file. 2. Copy the client directory on a system with a Java virtual machine accessible from the command path (for example, launch a shell, enter the client directory, execute the command: java -version and check the command outputs the JRE version number). . Copy the server directory on a Linux box. If the client and server directories weren’t copied on the same machine, you should edit the runserver. sh script (in the server directory) and replace the IP address: 127. 0. 0. 1 with the IP address of the Linux box’s network interface that is attached to the same network that the client machine (i. e. the computer that hosts the clientdirectory). 4. Enter the server directory and run the script: . /runserver. sh (the Python interpreter should be accessible from the command path. You can check this by running: ython -V from the command line and verifying that the interpreter version is showed). 5. On the client machine, enter the client directory and run either the script: . /runclient. sh (for Linux or Unix-like operating systems providing a shell compatible with the Bourne shell and whose path for the executable file is: /bin/sh) orrunclient. bat (for Windows). DESCRIPTION OF TOOLS The tools used for packet injection purpose s are divided into two categories Hardware and software 1. Software Serious hackers usually use Linux-based open source penetration test tools from which to launch their attacks. This section details some of the more popular tools that can be used to search out and hack wifi networks. †¢Aircrack-ng: This suite of tools includes 802. 11 WEP and WPA-PSK key cracking programs that can capture wireless packets and recover keys once enough information been captured. Aircrack-ng supports newer techniques that make WEP cracking much faster. This software has been downloaded over 20,000 times. †¢Airjack: An 802. 11 packet injection tool, Airjack was originally used as a development tool to capture and inject or replay packets. In particular, Airjack can be used to inject forged deauthentication packets, a fundamental technique used in many denial-of-service and Man-in-the-Middle attacks. Repeatedly injecting deauthentication packets into a network wreaks havoc on the connections between wireless clients and access points. †¢AirSnort: AirSnort is wireless LAN (CLAN) tool which recovers WEP encryption keys. AirSnort works by passively monitoring transmissions, and then computing the encryption key when enough packets have been gathered. After that point, all data sent over the network can be decrypted into plain text using the cracked WEP key. †¢Cain ;amp; Able: This is a multi-purpose tool that can intercept network traffic, using information contained in those packets to crack encrypted passwords using dictionary, brute-force and cryptanalysis attack methods, record VoIP conversations, recover wireless network keys, and analyze routing protocols. Its main purpose is the simplified recovery of passwords and credentials. This software has been downloaded over 400,000 times. CommView for WiFi: This commercial product is designed for capturing and analyzing wifi network packets. CommView for WiFi uses a wireless adapter to capture, decode, and analyze packets sent over a single channel. It allows hackers to view the list of network connections and vital IP statistics and examine individual packets. †¢ElcomSoft Wireless Security Auditor: This is an all-in-one cracking solution that automatically locates wirel ess networks, intercepts data packets, and uses cryptanalysis techniques to crack WPA/WPA2 PSKs. This software displays all available wireless networks, identified by channel number, AP MAC address, SSID, speed, load, and encryption parameters. While these capabilities can be found in open source tools, ElcomSoft provides a more polished product for professional use by wireless security auditors. †¢Ettercap: Ettercap can be used to perform man-in-the-middle attacks, sniff live connections, and filter intercepted packets on the fly. It includes many features for network and host analysis. This shareware has been downloaded nearly 800,000 times. Firesheep: This is a plug-in to the Firefox browser that allows the hacker to capture SSL session cookies sent over any unencrypted network (like an open wifi network) and use them to possibly steal their owner’s identities. It is extremely common for websites to protect user passwords by encrypting the initial login with SSL, but then never encrypt anything else sent after login, which leaves the cookie (and the user) vulnerab le to â€Å"sidejacking. † When a hacker uses Firesheep to grab these cookies, he may then use the SSL-authenticated session to access the user’s account. Hotspotter: Like KARMA, Hotspotter is another wireless attack tool that mimics any access point being searched for by nearby clients, and then dupes users into connecting to it instead. †¢IKECrack: This is an open source IPsec VPN authentication cracking tool which uses brute force attack methods to analyze captured Internet Key Exchange (IKE) packets to find valid VPN user identity and secret key combinations. Once cracked, these credentials can be used to gain unauthorized access to an IPsec VPN. KARMA: This evil twin attack listens to nearby wireless clients to determine the name of the network they are searching for and then pretends to be that access point. Once a victim connects to a KARMA evil twin, this tool can be used to redirect web, FTP, and email requests to phone sites in order to steal logins and passwords. †¢Kismet: K ismet takes an intrusion detection approach to wireless security, and can be used to detect and analyze access points within radio range of the computer on which it is installed. This software reports SSIDs (Service Set Identifiers – used to distinguish one wireless network from another) advertised by nearby access points, whether or not the access point is using WEP, and the range of IP addresses being used by connected clients. †¢NetStumbler: This tool turns any WiFi-enabled Windows laptop into an 802. 11 network detector. NetStumbler and dozens of similar â€Å"war driving† programs can be used with other attack tools to find and hack into discovered wifi networks. †¢WireShark: WireShark is a freeware LAN analyzer that can be used to passively capture 802. 11 packets being transmitted over a wifi network. This software has been downloaded millions of times. 2. Hardware †¢For hackers that prefer a turn-key package, there are also hardware wireless hacking tools available. We’ve highlighted one called WiFi Pineapple. It’s a simple, small, portable device that can be carried into any hotspot and used to attract any laptop trying to find a wifi access point. The Pineapple uses a technique called an Evil Twin attack. Hackers have used tools like KARMA to do the same thing for years, but with Pineapple, now you can buy a piece of hardware for only $100 that allows you to become a hacker without downloading or installing any software. Here’s what their website says: â€Å"Of course all of the Internet traffic flowing through the pineapple such as e-mail, instant messages and browser sessions are easily viewed or even modified by the pineapple holder. † REFERENCES http://www. cse. wustl. edu/~jain//cse571-07/ftp/wireless_hacking/index. html http://www. cs. wright. edu/~pmateti/InternetSecurity/Lectures/WirelessHacks/Mateti-WirelessHacks. htm#_Toc77524642 http://www. webopedia. com/TERM/C/cyber_crime. html http://www. wi-fiplanet. com/tutorials/article. php/3568066 file:///C:/Users/jsk/Desktop/Wireless%20Hackers%20101. htm How to cite Attacking Wifi Nets with Traffic Injection, Papers

Virtual Machines Virtualization

Question: Discuss about the Virtual Machinesfor Virtualization. Answer: Introduction Virtualization is a very important part of the computer technology that has evolved in recent years. It has evolved into the most efficient form of storage of data. This report consists of the description of different features of the virtualization.The report contains information about the role and importance of the virtualization and also information about the virtual appliances and the vApp. The Role of Virtualization in Data Center are: Firstly, it provides disaster recovery in the data centers (Khajehei, 2014). The data that are lost due to some system failure or even due to some external effects can be recovered very easily via virtualization. The data that is back up on the cloud is available throughout the cloud and can be back up at any time. Secondly, the virtualization technique provides hardware independence to the data centers. Usually earlier huge storage spaces were required in the machines to store large amounts of data. But, with virtualization a large amount of data can be stored in the clouds virtually. This would decrease the amount of hardware that would be required for storing the data. Sometimes even no hardware would be required. Since, all the data can be stored in the virtual memories. The Importance of Virtualization are: Firstly, the virtualization process is very easy to implement (Thakral Kadam, 2014). It just requires the user to register on the cloud and hire a storage space for themselves in which they can store all their data very easily. In addition to this the user does not require to upgrade any of the system or soft wares. Hence the up gradation is also very for the user. Secondly, the virtualization method is very cost effective. The cost of a hard ware device is very high in todays world. And the price increase with the need requirement of more space by the user. But, it is very cheap for the user to purchase a virtual space on the cloud. Hence, this provides the cost effectiveness for the user. Thirdly, the system is very fast. The virtualization allows the users to upload and download the data very easily. In addition to this upload and the download processes are very fast depending upon the speed of the network. The safety of the data of the user is also provided very efficiently for the user. The Technology is used by IBM. The IBM cloud data virtualization is the organization that has implemented this technology successfully (Zheng,Wang Wang, 2014). The IBM data centers are online. So, the data import and export of the company takes place online. The company also provides this facility to the users and other companies. The company also provides their visualization tool for purposes other than storage facilities. They also provide storage facilities to the private user free of cost up to a certain level. The Key Attributes of a Virtual Machine are: Simple to install and configure: The installation and the configuration procedures of the virtual machine are very simple (Peng et al., 2015). The installation tools for the virtual machine are easily available on the internet. In addition to this the configuration processes are also available very easily on the internet. For the configuration process the machines are to integrated with the APIS that are easily available in the VMware vSphere and the vCenter Server. In addition to the configuration procedure the backup and the restore process are also very easy to implement. VMware certified: The Virtual Machines are certified by the VMware certification. Hence, the systems are very reliable and authentic. The VMware vSphere contains the data protection APIs that are very efficient. In addition to this the APIs are updated automatically s soon as they are released. Application Aware: The Virtual Machines are very aware of the application that they are required to run. They automatically develop the require applications and also update them when require, the backup processes are not hampered. Scalable and efficient: The Virtual Machine are very efficient in providing backup. The backup software have to evolve so the IT efficiency is increased. IN addition to this the system are easily scalable. It provides the users the option of scalability. So, that they can increase or decrease their requirements very easily. Affordable: The Virtual Machines are very affordable. The cost of the Virtual Machine are very less in comparison to the various types of hardware components. So, they can be considered as very cheap and affordable of the users and also the organizations. The Virtual machine can also be used as a Server. Many of the Organizations that provide the virtual machines as a service provide the server option of the Virtual machines (Gao et al., 2013). The main that is to be considered while adopting the virtual machine as the server is the nature of the system. It is to be identified weather the system is a fully virtual system or it is partially virtual or the operating system of the system is virtual. In the fully virtual system the VM as a server is to be used. In the partially virtual system the proportion of the virtual system and the hardware components are to be considered. And in the Software virtualization the virtual server should not be used. This is because of the system is composed of the hardware components. Virtual Appliance: The virtual appliances are the virtual machine that are already configured, and can readily run on the hypervisors (Kecskemeti et al., 2014). It is basically derives from the software appliances. Basically, if a virtual software appliance is installed in the on the virtual machine it could be termed as the virtual appliance. The virtual appliance is the imaging instance of the software appliance that is installed into the virtual machine. vApp: The vApp is the container and it can contain one or more than one virtual machines in them (Bigelow, 2014). They are like the resource pool. They also share functionalities of the virtual machines. The main features of the vApp are that they can easily be power on and offed and can also be cloned very easily. The main Difference Between the Virtual Appliance and the vApp are: The vApps can be run on only the hypervisor environment, while the virtual appliance can run on all the environments such as hypervisor as well as the hosted environments. The Virtual Appliance are Required During the Dual Boot of a machine. On several occasions the machines are slowed down if dual boot is performed on the machine (Chih-Liang et al., 2015). Hence, the virtual appliance can be used for dual booting the system. It would provide a virtual instance of a system on the machine. This would result in increased efficiency of the system. The virtual appliances [provide the user the opportunity to use multiple instances of different systems without having to boot the systems into their machines. This would allow the user to switching between different operating system without reducing the efficiency of the system. In addition to this, the users that require a particular software for only a limited period of time can use the virtual appliances do that they would not require to install the software into their machine. This would be saving the space on their machine and would also save the cost of buying the application. The Three main Virtual Appliances Provider are: IBM, Flexera Software and Fortinet.The IBM Partner World provides the virtual appliance of various sectors (Kandan et al., 2014). They provide the services in the fields of storage as well as the different other services. They are one of the main providers of the virtual appliances as a service. The Flexera Software provides the virtual appliances that can help in the business sectors. They perform the software licensing facilities and the entitlement management solutions. The Fortinet provides the security oriented virtual appliances. They are basically, concerned with the development of the firewall services. They provide the security of the application with their virtual appliances. Conclusion For conclusion, it can be said that the features of virtualization has been described very aptly in this report. The role and importance of virtualization has been successfully demonstrated in this report. In addition to this the virtualization as a server can also be used. In addition to this, the virtual appliances and vApp and the difference in between them have also been described in the report. References Bigelow, S. J. (2014). How can an enterprise benefit from using a VMware vApp?. Chih-Liang, H. S. U., CHENG, W. L., HUANG, Y. C. (2015).U.S. Patent Application No. 14/825,864. Gao, Y., Guan, H., Qi, Z., Hou, Y., Liu, L. (2013). A multi-objective ant colony system algorithm for virtual machine placement in cloud computing.Journal of Computer and System Sciences,79(8), 1230-1242. Kandan, R., Alli, M. Z., Ong, H. (2014, January). DiAF: A Dynamic virtual Appliance provision and management Framework for cloud Computing. InProceedings of the International Conference on Grid Computing and Applications (GCA)(p. 1). The Steering Committee of The World Congress in Computer Science, Computer Engineering and Applied Computing (WorldComp). Kecskemeti, G., Terstyanszky, G., Kacsuk, P., Nemeth, Z. (2014). Towards efficient virtual appliance delivery with minimal manageable virtual appliances.IEEE Transactions on Services Computing,7(2), 279-292. Khajehei, K. (2014). Role of virtualization in cloud computing.International Journal of Advance Research in Computer Science and Management Studies,2(4). Peng, Z., Xu, B., Gates, A. M., Cui, D., Lin, W. (2015). The Deasibility and Properties of Dividing Virtual Machine Resources using the Virtual Machine Cluster as the Unit in Cloud Computing.TIIS,9(7), 2649-2666. Thakral, M. A., Kadam, S. A. (2014). Effective use of Virtualization in Computer Laboratory Management. InNational Conference on Innovations in IT and Management. Pune, Maharashtra, India. Zheng, K., Wang, X., Li, L., Wang, X. (2014, April). Joint power optimization of data center network and servers with correlation analysis. InINFOCOM, 2014 Proceedings IEEE(pp. 2598-2606). IEEE.

25 Quotes to Inspire Thoughtful Written Sentiments

25 Quotes to Inspire Thoughtful Written Sentiments Sometimes its easy to take friends and family for granted, which is why showing appreciation is so important. As philosopher Voltaire said, Appreciation is a wonderful thing: It makes what is excellent in others belong to us as well. When you take the time to express thanks and gratitude, you help build and strengthen bonds of trust and love. It doesnt matter whether you send a card or make a phone call. Appreciation, however you express it, builds bridges and fosters healthy relationships. Of course, appreciation should always be sincere. For example, when you praise a family member for their cooking, mention what you specifically liked about the dish, and thank them for preparing it so well. If a friend has thrown you a surprise birthday party, offer your sincere thanks. Remember to say what you enjoyed most about the celebration. Everyone loves a thoughtful thank-you card, but finding the right words to show your appreciation is not always easy. The following is a list of quotes on the subject of appreciation and gratitude from well-known artists, writers, world leaders, and others to help you create your own special sentiments. You could also include the entire attributed quote if it makes sense. Maya Angelou: When we give cheerfully and accept gratefully, everyone is blessed. Guillaume Apollinaire: Now and then it’s good to pause in our pursuit of happiness and just be happy. Thomas Aquinas: There is nothing on this earth more to be prized than true friendship. Marcus Aurelius: Dwell on the beauty of life. Watch the stars, and see yourself running with them. Leo Buscaglia aka Dr. Love: Too often we underestimate the power of a touch, a smile, a kind word, a listening ear, an honest compliment, or the smallest act of caring, all of which have the potential to turn a life around. Henry Clay: Courtesies of a small and trivial character are the ones which strike deepest in the gratefully and appreciating heart. Ralph Waldo Emerson: A friend may well be reckoned the masterpiece of nature. Helen Keller: Words are never warm and tender enough to express ones appreciation of a great kindness. Dalai Lama aka Tenzin Gyatso: The roots of all goodness lie in the soil of appreciation for goodness. Washington Irving: Sweet is the memory of distant friends! Like the mellow rays of the departing sun, it falls tenderly, yet sadly, on the heart. President John F. Kennedy: As we express our gratitude, we must never forget that the highest appreciation is not to utter words, but to live by them. Steve Maraboli: Forget yesterday - it has already forgotten you. Dont sweat tomorrow - you havent even met. Instead, open your eyes and your heart to a truly precious gift - today. Willie Nelson: When I started counting my blessings, my whole life turned around. Marcel Proust: Let us be grateful to the people who make us happy; they are the charming gardeners who make our souls blossom. Albert Schweitzer: At times our own light goes out and is rekindled by a spark from another person. Each of us has cause to think with deep gratitude of those who have lighted the flame within us. Mark Twain aka Samuel Langhorne Clemens: To get the full value of joy you must have someone to divide it with. Kindness is a language which the deaf can hear and the blind can see. Voltaire: Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well. William Arthur Ward: Flatter me, and I may not believe you. Criticize me, and I may not like you. Ignore me, and I may not forgive you. Encourage me, and I may not forget you. Booker T. Washington: Any mans life will be filled with constant and unexpected encouragement if he makes up his mind to do his level best each day. Mae West aka Mary Jane West: Too much of a good thing can be wonderful! Walt Whitman: I have learned that to be with those I like is enough. Oscar Wilde: The smallest act of kindness is worth more than the grandest intention. Thornton Wilder: We can only be said to be alive in those moments when our hearts are conscious of our treasures. Oprah Winfrey: Be thankful for what you have; youll end up having more. If you concentrate on what you dont have, you will never, ever have enough.

Calculating the Mean Absolute Deviation

Calculating the Mean Absolute Deviation There are many measurements of spread or dispersion in statistics. Although the range and standard deviation are most commonly used, there are other ways to quantify dispersion.  We will look at how to calculate the mean absolute deviation for a data set.   Definition We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. The formula displayed with this article is the formal definition of the mean absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic. We start with an average, or measurement of the center, of a data set, which we will denote by m.  Next, we find how much each of the data values deviates from m.  This means that we take the difference between each of the data values and m.  After this, we take the absolute value of each of the difference from the previous step. In other words, we drop any negative signs for any of the differences.  The reason for doing this is that there are positive and negative deviations from m.  If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.Now we add together all of these absolute values.Finally, we divide this sum by n, which is the total number of data values.  The result is the mean absolute deviation. Variations There are several variations for the above process.  Note that we did not specify exactly what m is. The reason for this is that we could use a variety of statistics for m.  Typically this is the center of our data set, and so any of the measurements of central tendency can be used. The most common statistical measurements of the center of a data set are the mean, median and the mode.  Thus any of these could be used as m in the calculation of the mean absolute deviation. This is why it is common to refer to the mean absolute deviation about the mean or the mean absolute deviation about the median. We will see several examples of this. Example:  Mean Absolute Deviation About the Mean Suppose that we start with the following data set: 1, 2, 2, 3, 5, 7, 7, 7, 7, 9. The mean of this data set is 5.  The following table will organize our work in calculating the mean absolute deviation about the mean.   Data Value Deviation from mean Absolute Value of Deviation 1 1 - 5 = -4 |-4| = 4 2 2 - 5 = -3 |-3| = 3 2 2 - 5 = -3 |-3| = 3 3 3 - 5 = -2 |-2| = 2 5 5 - 5 = 0 |0| = 0 7 7 - 5 = 2 |2| = 2 7 7 - 5 = 2 |2| = 2 7 7 - 5 = 2 |2| = 2 7 7 - 5 = 2 |2| = 2 9 9 - 5 = 4 |4| = 4 Total of Absolute Deviations: 24 We now divide this sum by 10, since there are a total of ten data values.  The mean absolute deviation about the mean is 24/10 2.4. Example:  Mean Absolute Deviation About the Mean Now we start with a different data set: 1, 1, 4, 5, 5, 5, 5, 7, 7, 10. Just like the previous data set, the mean of this data set is 5.   Data Value Deviation from mean Absolute Value of Deviation 1 1 - 5 = -4 |-4| = 4 1 1 - 5 = -4 |-4| = 4 4 4 - 5 = -1 |-1| = 1 5 5 - 5 = 0 |0| = 0 5 5 - 5 = 0 |0| = 0 5 5 - 5 = 0 |0| = 0 5 5 - 5 = 0 |0| = 0 7 7 - 5 = 2 |2| = 2 7 7 - 5 = 2 |2| = 2 10 10 - 5 = 5 |5| = 5 Total of Absolute Deviations: 18 Thus the mean absolute deviation about the mean is 18/10 1.8.  We compare this result to the first example.  Although the mean was identical for each of these examples, the data in the first example was more spread out. We see from these two examples that the mean absolute deviation from the first example is greater than the mean absolute deviation from the second example. The greater the mean absolute deviation, the greater the dispersion of our data. Example:  Mean Absolute Deviation About the Median Start with the same data set as the first example: 1, 2, 2, 3, 5, 7, 7, 7, 7, 9. The median of the data set is 6.  In the following table,  we show the details of the calculation of the mean absolute deviation about the median. Data Value Deviation from median Absolute Value of Deviation 1 1 - 6 = -5 |-5| = 5 2 2 - 6 = -4 |-4| = 4 2 2 - 6 = -4 |-4| = 4 3 3 - 6 = -3 |-3| = 3 5 5 - 6 = -1 |-1| = 1 7 7 - 6 = 1 |1| = 1 7 7 - 6 = 1 |1| = 1 7 7 - 6 = 1 |1| = 1 7 7 - 6 = 1 |1| = 1 9 9 - 6 = 3 |3| = 3 Total of Absolute Deviations: 24 Again we divide the total by 10 and obtain a mean average deviation about the median as 24/10 2.4. Example:  Mean Absolute Deviation About the Median Start with the same data set as before: 1, 2, 2, 3, 5, 7, 7, 7, 7, 9. This time we find the mode of this data set to be 7.  In the following table,  we show the details of the calculation of the mean absolute deviation about the mode. Data Deviation from mode Absolute Value of Deviation 1 1 - 7 = -6 |-5| = 6 2 2 - 7 = -5 |-5| = 5 2 2 - 7 = -5 |-5| = 5 3 3 - 7 = -4 |-4| = 4 5 5 - 7 = -2 |-2| = 2 7 7 - 7 = 0 |0| = 0 7 7 - 7 = 0 |0| = 0 7 7 - 7 = 0 |0| = 0 7 7 - 7 = 0 |0| = 0 9 9 - 7 = 2 |2| = 2 Total of Absolute Deviations: 22 We divide the sum of the absolute deviations and see that we have a mean absolute deviation about the mode of 22/10 2.2. Fast Facts There are a few basic properties concerning mean absolute deviations The mean absolute deviation about the median is always less than or equal to the mean absolute deviation about the mean.The standard deviation is greater than or equal to the mean absolute deviation about the mean.The mean absolute deviation is sometimes abbreviated by MAD.  Unfortunately, this can be ambiguous as MAD may alternately refer to the median absolute deviation.The mean absolute deviation for a normal distribution is approximately 0.8 times the size of the standard deviation. Common Uses The mean absolute deviation has a few applications.  The first application is that this statistic may be used to teach some of the ideas behind the standard deviation. The mean absolute deviation about the mean is much easier to calculate than the standard deviation. It does not require us to square the deviations, and we do not need to find a square root at the end of our calculation. Furthermore, the mean absolute deviation is more intuitively connected to the spread of the data set than what the standard deviation is. This is why the mean absolute deviation is sometimes taught first, before introducing the standard deviation. Some have gone so far as to argue that the standard deviation should be replaced by the mean absolute deviation.  Although the standard deviation is important for scientific and mathematical applications, it is not as intuitive as the mean absolute deviation. For day-to-day applications, the mean absolute deviation is a more tangible way to measure how spread out data are.

Assignment One Essay Example | Topics and Well Written Essays - 1500 words

Assignment One - Essay Example As the utilization of ICT increases with the creation of new applications, the governments continue to adopt these services. The implementation of the ICT technologies by governments matches the millennium development goals (MDG’s) that focus on digitizing government services. Although embracing some of the ICT technologies strains the financial budgets of countries, all governmental stakeholders agree that it is immensely critical in facilitating public services. Evidently, there has been an evolution of modern communication methods. The methods through which citizens interact with the government have changed remarkably from counter transactions to digital channels, for example, government websites. Rwanda is one of the countries that are currently incorporating ICT technologies in their governance structures (Watkins, 2008). This entails public sector services such as healthcare, license acquisition, taxation and education. Due to this, the government will be able to provide better services. The adoption of ICT solutions by various countries is in coherence with the globalization objective of computerizing administration procedures. The digitization processes are within the millennium development goals (MDG’s) of developing counties. ICT enhancement addresses Rwanda’s MDGs. The Rwandan government acknowledges the rising utilization of ICT in daily lives of its citizens. Consequently, the enhancement of government services through the ICT will be highly beneficial. It will assist the governance process in achieving exemplary echelons of performance. Furthermore, ICT adoption will increase the interaction of the authorities and citizens, further enhancing responsiveness of Rwanda’s government to their citizens’ concerns. For example, the Rwandan health ministry can create websites that focus on the needs of the citizens (Watkins, 2008). This means the citizens can apply health services online and share information. Additionally, the

Bush's War (PBS) Analasys Essay Example | Topics and Well Written Essays - 750 words

Bush's War (PBS) Analasys - Essay Example Cheney advocated the use of controversial interrogation techniques to obtain information that would link Saddam Hussein to the 9/11 attacks. In the documentary, Cheney and Rumsfeld supported a pentagon (military) led alternative that led to the decision to invade Iraq. Cheney was also involved in securing controversial secret legal opinions from the Justice Department that would grant President Bush unrestricted broad authority to wage 'war' without the consent of the U.S congress. Cheney also supported the use of 'enhanced combat and interrogation techniques' against captured combatants. President Bush's fixation on invading Iraq was borne out of his distrust of Saddam Hussein. Bush stated in the documentary that Saddam was "an evil man who gassed his own people" In reference to Saddam Hussein, Bush declared after the 9/11 attacks, that his administration would hunt down the Islamic fundamentalists and "those who harbour them." Bush believed that Saddam Hussein was providing support for Al-Qaeda. The intelligence information that was used as a predicate to the invasion was manipulated in order to justify the war. This invasion 'policy' was hatched and promoted chiefly by Cheney and Rumsfeld although there was a lack of substantial evidence that linked Saddam to Al-Qaeda. The Central Intelligence Agency (CIA) Director, George Tenet did not initially support the Iraq invasion on account of the lack of credible intelligence evidence available. Instead of relying and heeding to CIA intelligence reports, Cheney and Rumsfeld formed a parallel and secretive intelligence unit in the Pentagon to analyse evidence that would hitherto link Saddam Hussein to Al Qaeda (Chapter 12). Cheney also pressured CIA analysts who were preparing a National Intelligence Estimate, to include language that would support the invasion policy. The CIA analysts have since reported that Cheney and his staff wanted the report to indicate that Saddam had or was seeking to acquire Weapons of Mass Destruction ( WMD). This attempts led the administration to use "highly dubious" and un-corroborated evidence that stated that Saddam Hussein had attempted to purchase 'yellow cake' Uranium (a key component for producing a nuclear weapon) from Niger (Chapter 12). Why was the press unable to bring this story to light earlier Although some sections of the press were critical of the plans, President Bush had a 90% popularity rate. The national press were therefore weary and feared a public backlash if they did not seem to be supporting the President in war time. How would you assess Rumsfeld's role in this issue Donald Rumsfeld was the one of the Architects of the invasion of Iraq. Rumsfeld first succeeded in taking the lead role in the 'war on terror' from the CIA in Afghanistan and subsequently in the Iraq invasion plans. He wanted to be the solely in charge, "100% responsible" and determined to go to war with Saddam at all cost. He continued to claim that Saddam Hussein had WMD (Chapter 13). Rumsfeld also withheld critical information form the White House and undermined both the State Department and the CIA all in a bid to ensure that the invasion took

Audit of Panera Bread Company Quality Service and Market Share Process Research Paper

Audit of Panera Bread Company Quality Service and Market Share Process - Research Paper Example This is because, few customers subscribe to this company and, therefore, the expenses incurred by the company compared to the revenue generated from the services rendered is relatively high. In order to decrease its liabilities, and hence portray a positive image of the company, managers may attempt to take the losses into a different account especially the expenses accounts. In addition, when customers are disappointed in the orders they make, there is a higher possibility of them demanding that their order be remade or refunded. This environment provides the possibility of pilfering by employees as they can always say they had to remake an order with no hint of plausibility. Increased customer returns and higher rates of pilfering by the employees increases the costs of goods sold by the company. Another risk is that Panera Bread is in the provision of services it offers on a national, regional, and local level, and this reduces its revenues and market share. When customers get poo r services, it increases their chances of shifting to another competing company offering the same services. This reduces the amount of revenue and the company incurs expenses such as, in advertising and improving its services to be better than those offered by their competitors. However, the company must make sure not to increase its prices since the customers will again shift to the competing companies. Since advertising is necessary, the managers then tend to hide the expenses behind the advertising expense in the statement of account. Controls To alleviate the risks coming from poor services, Panera Bread should put into practice several controls. The company, to alleviate the risk of accounts payable being understated, as a result of increased expenses and, therefore, managers resulting to understating the expenses, should require proper authorisation of orders and entry of these orders in its ledger. The company should establish regular receipt book control, and ledger book to help reduce the irregularities. In a bid to entice customers, advertising is done. A control on this should also be enacted where the company should ensure that there are supporting documents any time a manager books an advertising expense. It should also provide an appropriate division of duties such that a manager does not fake any advertising expenses. The company, to deal with the risks of losing its customers, can offer differentiated services from those that it normally offers. This can be done through, introducing offers, for example, ‘buy one get two’. Also, the maintenance of a serene environment can also be an advantage. This increases the influx of customers thereby solving the problem of managers faking expenses to reduce the liabilities since the revenue will increase in the long run. Audit Objectives During this audit, I plan to test the accuracy and the valuation of the contingent liability for losses associated with poor services, the completeness of pur chases and accounts payable and the existence of advertising expenses. Risk Assessment As a result, many emerging enterprises, especially in the food service industry, restaurants are always searching for ways of increasing their profits: Therefore, I assessed natural risk throughout all financial assertions. Moreover, the contingent liability is an estimation and, therefore, at a high risk of manipulation by the managers. The controls around the accounts payable

Basics of Topological Solutons

Basics of Topological Solutons Research into topological solitons began in the 1960s, when the fully nonlinear form of the classical field equations, were being thoroughly explored by mathematicians and theoretical physicists. Topological solitons were first examined when the solutions to these equations were interpreted as candidates for particles of the theory [1]. The particles that were observed from the results were different from the usual elementary particles. Topological solitons appeared to behave like normal particles in the sense that they were found to be localised and have finite energy [4]. However, the solitons topological structure distinguished them from the other particles. Topological solitons carry a topological charge (also known as the winding number), which results in these particlelike objects being stable. The topological charge is usually denoted by a single integer, N; it is a conserved quantity, i.e. it is constant unless a collision occurs, and it is equal to the total number of partic les, which means as |N| increases, the energy also increases. The conservation of the topological charge is due to the topological structure of the target space in which the soliton is defined. The most basic example of soliton has topological charge, N = 1, which is a stable solution, due to the fact a single soliton is unable to decay. 3 If the solution to a nonlinear classical field equation has the properties of being particle-like, stable, have finite mass; and the energy density is localised to a finite region of space, with a smooth structure; then this solution is a topological soliton. In addition to solitons existing with topological charge, N, there also exist antisolitons with -N. In the event of a collision between a soliton and an antisoliton, it is possible for them to annihilate each other or be pair-produced [1]. It is also possible for multi-soliton states to exist. Any field composition where N > 1, is known as a multi-soliton state. Likewise, multi-solitons also carry a topological charge which again means they are stable. Multi-state solitons either decay into N well separated charge 1 solitons or they can relax to a classical bound state of N solitons [1]. The energy and length scale [1] (a particular length which is determined to one order of magnitude.) the constant in the Lagrangian and field equations which represents the strength of the interaction between the particle and the field, also known as the coupling constant. The energy of a topological soliton is equal to its rest mass in a Lorentz invariant theory. [5] [6] Lorentz invariant: A quantity that does not change due to a transformation relating the space-time coordinates of one frame of reference to another in special relativity; a quantity that is independent of the inertial frame. In contrast to the topological soliton, the elementary particles mass is proportional to Plancks constant, ~. In the limit ~ à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 0, the elementary particles mass goes to zero where as the topological solitons mass is finite. The quantization of the wave-like fields which satisfy the linearized field equations [1] determines the elementary particle states, where the interactions between the particles are determined by the nonlinear terms A fundamental discovery in supporting the research of topological solitons is that, given the coupling constants take special values, then the field equations can be reduced from second order to first order partial differential equations.[1] In general, the resulting first order equations are known as Bogomolny equations. These equations do not involve any time derivatives, and their solutions are either static soliton or multi-soliton configurations. [1] In these given field theories, if the field satisfies the Bogomolny equation then the energy is bounded below by a numerical multiple of the modulus of the topological charge, N, so the solutions of a Bogomolny equation with a certain 4 charge will all have the same energy value. [1] The solutions of the Bogomolny equations are automatically stable [1] because the fields minimize the energy [1]. As well as this they naturally satisfy the Euler-Lagrange equations of motion, which implies the static solutions are a stationary point of the energy. [1] Kinks are solutions to the first-order Bogomolny equation which we shall see in the following chapter Figure 2.2 shows a model of an infinite pendulum strip, with the angle à Ã¢â‚¬   being the angle to the downward vertical [3]. The energy (with all constraints set to 1) is E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚   1 2 à Ã¢â‚¬   02 + 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ cos à Ã¢â‚¬  Ãƒâ€šÃ‚   dx (2.1) where à Ã¢â‚¬   0 = dà Ã¢â‚¬   dx . For the energy density to be finite this requires à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 2à Ã¢â€š ¬nà ¢Ã‹â€ Ã¢â‚¬â„¢ as x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ and à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 2à Ã¢â€š ¬n+ as x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã… ¾, where n ± à ¢Ã‹â€ Ã‹â€  Z. To find the number of twists, N, this is simply N = n+ à ¢Ã‹â€ Ã¢â‚¬â„¢ nà ¢Ã‹â€ Ã¢â‚¬â„¢ = à Ã¢â‚¬   (à ¢Ã‹â€ Ã… ¾) à ¢Ã‹â€ Ã¢â‚¬â„¢ à Ã¢â‚¬   (à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾) 2à Ã¢â€š ¬ = 1 2à Ã¢â€š ¬ Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ à Ã¢â‚¬   0 dx à ¢Ã‹â€ Ã‹â€  Z This is the equation for the topological charge or the winding number. If we set nà ¢Ã‹â€ Ã¢â‚¬â„¢ = 0 and n+ = 1 then N = 1, this gives the lowest possible energy for a topological soliton. This is called a kink, and it is the term we use for the one spatial dimension soliton with a single scalar field. The name kink is due to the shape of the scalar field when plotted as a function of x [1]. Knowing that a kink gives the minimum of the energy, it is possible to apply the calculus of variations to derive a differential equation à Ã¢â‚¬  (x) and then solve it[3] to give the shape of the kink. Given a differentiable function on the real line, f(x), it is possible to find the minimum of f(x) by finding the solutions of f 0 (x) = 0, i.e. by finding the stationary points of f(x) [3]. It is achievable to derive this differential equation, f(x), by making a small change to x, i.e. x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ x + ÃŽÂ ´x, and from this calculate the change in the value of the function to lea ding order in the variaton ÃŽÂ ´x [3]. ÃŽÂ ´f(x) = f(x + ÃŽÂ ´x) à ¢Ã‹â€ Ã¢â‚¬â„¢ f(x) = f(x) + ÃŽÂ ´xf0 (x) + à ¢Ã‹â€ Ã¢â‚¬â„¢ f(x) = f 0 (x)ÃŽÂ ´x + If f 0 (x) 0. If f 0 (x) > 0 then we can make ÃŽÂ ´f(x) The term [à Ã¢â‚¬   0 ÃŽÂ ´Ãƒ Ã¢â‚¬  ] à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ equates to zero on the boundary because it must satisfy ÃŽÂ ´Ãƒ Ã¢â‚¬  ( ±Ãƒ ¢Ã‹â€ Ã… ¾) = 0 as we cannot change the boundary conditions, so ÃŽÂ ´E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ {(à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ Ã¢â‚¬   00 + sin à Ã¢â‚¬  )ÃŽÂ ´Ãƒ Ã¢â‚¬  } dx (2.6) This equation can be minimised minimised further to the second order nonlinear differential equation, à Ã¢â‚¬   00 = sin à Ã¢â‚¬   (2.7) The solution of this differential equation with the boundary conditions, à Ã¢â‚¬  (à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾) = 0 and à Ã¢â‚¬  (à ¢Ã‹â€ Ã… ¾) = 2à Ã¢â€š ¬ is the kink. Therefore the kink solution is, à Ã¢â‚¬  (x) = 4 tanà ¢Ã‹â€ Ã¢â‚¬â„¢1 e xà ¢Ã‹â€ Ã¢â‚¬â„¢a (2.8) where a is an arbitrary constant. When x = a, this is the position of the kink (à Ã¢â‚¬  (a) = à Ã¢â€š ¬). It is clear to see à Ã¢â‚¬   = 0 is also a solution to the differential equation , however, it does not satisfy the boundary conditions. It is possible to find a lower bound on the kink energy without solving a differential equation [3]. First of all we need to rewrite the energy equation (2.1), using the double angle formula the equation becomes, E = 1 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚   à Ã¢â‚¬   02 + 4 sin2   à Ã¢â‚¬   2   dx (2.9) By completing the square the equation becomes, E = 1 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚   à Ã¢â‚¬   0 à ¢Ã‹â€ Ã¢â‚¬â„¢ 2 sin   à Ã¢â‚¬   2 2 + 4à Ã¢â‚¬   0 sin   à Ã¢â‚¬   2 dx (2.10) Therefore the energy satisfies the inequality, E > 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ à Ã¢â‚¬   0 sin   à Ã¢â‚¬   2   dx = 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ sin   à Ã¢â‚¬   2   dà Ã¢â‚¬   dxdx = 2 Z 2à Ã¢â€š ¬ 0 sin   à Ã¢â‚¬   2   dà Ã¢â‚¬   = à ¢Ã‹â€ Ã¢â‚¬â„¢4   cos   à Ã¢â‚¬   2 2à Ã¢â€š ¬ 0 = 8 (2.11) In order to obtain the solution which is exactly 8, the term à Ã¢â‚¬   0 à ¢Ã‹â€ Ã¢â‚¬â„¢ 2 sin à Ã¢â‚¬   2 2 would have to be exactly 0. Therefore the lower bound on the kink energy is calculated by the solution to the equation, à Ã¢â‚¬   0 = 2 sin   à Ã¢â‚¬   2   (2.12) This is a first order Bogomolny equation. Taking this Bogomolny equation and differentiating with respect to à Ã¢â‚¬   0 gives, à Ã¢â‚¬   00 = cos   à Ã¢â‚¬   2   à Ã¢â‚¬   0 = cos   à Ã¢â‚¬   2   2 sin   à Ã¢â‚¬   2   = sin à Ã¢â‚¬   (2.13) This shows that a solution of the Bogomo lny equation (2.12) gives the output of the kink solution (2.7). To calculate the energy density ÃŽÂ µ, equation (2.1), we need to use the fact that the Bogomolny equation shows that ÃŽÂ µ = à Ã¢â‚¬   02 . From equation (2.8) we have, tan à Ã¢â‚¬   4   = e xà ¢Ã‹â€ Ã¢â‚¬â„¢a , therefore 1 4 à Ã¢â‚¬   0 sec2   à Ã¢â‚¬   4 = e xà ¢Ã‹â€ Ã¢â‚¬â„¢a This equation gives, à Ã¢â‚¬   0 = 4 e xà ¢Ã‹â€ Ã¢â‚¬â„¢a 1 + tan2 à Ã¢â‚¬   4   = 4e xà ¢Ã‹â€ Ã¢â‚¬â„¢a 1 + e 2(xà ¢Ã‹â€ Ã¢â‚¬â„¢a) = 2 cosh (x à ¢Ã‹â€ Ã¢â‚¬â„¢ a) = 2 (x à ¢Ã‹â€ Ã¢â‚¬â„¢ a) (2.15) Therefore it can be seen that the energy density is given by ÃŽÂ µ = 42 (x à ¢Ã‹â€ Ã¢â‚¬â„¢ a) From this we get the solution of a lump with a maximal value of 4 when x = a. This maximal value is the position of the kink. The position of the kink is also the position of the pendulum strip when it is exactly upside down, this is due to the fact à Ã¢â‚¬  (a) = à Ã¢â€š ¬ [3]. Using this interpretation for the energy density, it can be verified that the energy is equal to the lower bound E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ ÃŽÂ µdx = 4 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ 2 (x à ¢Ã‹â€ Ã¢â‚¬â„¢ a) dx = 4 [tanh (x à ¢Ã‹â€ Ã¢â‚¬â„¢ a)]à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ = 8 (2.16) For N > 1 i.e. more than one kink, E > 8|N|. In order t o obtain the lower bound of N > 1 kinks, the kinks must be infinitely apart to create N infinitely separated kinks. This means there must be a repulsive force between kinks. We shall now look at applying Derricks theorem [3] to kinks to show that it does not rule out the existence of topological solitons. Derricks Theorem: If the energy E has no stationary points with respect to spatial rescaling then it has no solutions with 0 Derricks theorem can only be applied to an infinite domain. Firstly, the energy terms need to be split according to the powers of the derivative, E = E2 + E0 = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ 1 2 à Ã¢â‚¬   02 dx + Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ (1 à ¢Ã‹â€ Ã¢â‚¬â„¢ cos à Ã¢â‚¬  ) dx (2.17) Now consider the spatial rescaling x 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ x ÃŽÂ » = X, so that à Ã¢â‚¬   (x) 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à Ã¢â‚¬   (X), with dx = ÃŽÂ »dX, d dx = 1 ÃŽÂ » d dX . Under this rescaling the energy becomes E (ÃŽÂ »), E(ÃŽÂ ») = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ 1 2 ( 1 ÃŽÂ » dà Ã¢â‚¬   dX ) 2ÃŽÂ »dX + Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ (1 à ¢Ã‹â€ Ã¢â‚¬â„¢ cos à Ã¢â‚¬  ) ÃŽÂ »dX = 1 ÃŽÂ » E2 + ÃŽÂ »E0 (2.18) It is now important to see whether E(ÃŽÂ ») has a stationary point with respect to ÃŽÂ », dE (ÃŽÂ ») dÃŽÂ » = à ¢Ã‹â€ Ã¢â‚¬â„¢ 1 ÃŽÂ » 2 E2 + E0 = 0 (2.19) if ÃŽÂ » = qE2 E0 , where ÃŽÂ » equals the size of the soliton. From this we can see a stationary point exists, so by Derricks theorem we cannot rule out the possibility of a topological soliton solution existing. We already know this is the case due to already finding the kink solution earlier. If it is found that à Ã¢â‚¬  (x) is a solution then the stationary point corresponds to no rescaling [3], so ÃŽÂ » = 1, meaning E2 = E0. This is known as a virial relation. In order to extend the kink example to higher spatial dimensions, we will rewrite it using different variables. If we let à Ã¢â‚¬   = (à Ã¢â‚¬  1, à Ã¢â‚¬  2) be a two-component unit vector, where à Ã¢â‚¬    · à Ã¢â‚¬   = |à Ã¢â‚¬  | 2 = 1. By writing à Ã¢â‚¬   = (sin à Ã¢â‚¬  , cos à Ã¢â‚¬  ), the energy from (2.1) can be rewritten as E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ ( 1 2  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚   dà Ã¢â‚¬   dx  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚   2 à ¢Ã‹â€ Ã¢â‚¬â„¢ H  · à Ã¢â‚¬   + |H| ) dx (2.20) where H = (0, 1). [3] In this new formulation à Ã¢â‚¬   represents the direction of the local magnetization (restricted to the plane) in a ferromagnetic medium [3] and H represents the constant background magnetic field which is also restricted to lie within the same plane as à Ã¢â‚¬  . There is only one point in which the systems ground state is equal to zero in terms of à Ã¢â‚¬  , which is à Ã¢â‚¬   = H |H| = (0, 1 ). Any structure with finite energy has to approach this zero energy ground state at spatial infinity, therefore the boundary conditions are à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ (0, 1) as x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢  ±Ãƒ ¢Ã‹â€ Ã… ¾. As à Ã¢â‚¬   takes the same value at x = à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ and x = +à ¢Ã‹â€ Ã… ¾, then these points can be identified so the target space, which is the real line R, topologically becomes a circle, S 1 of infinite radius. Therefore we have the mapping à Ã¢â‚¬   : S 1 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ S 1 between circles, because à Ã¢â‚¬   is a two-component vector so it also lies on a circle of unit radius. [3] The mapping between circles has a topological charge (winding number), N, which counts the number of times à Ã¢â‚¬   winds around the unit circle as x varies over the whole real line. [3] The topological charge is equal to the equation defined earlier in (2.2), but using the new variables it is given by the expression N = 1 2à Ã¢â€š ¬ Z à ¢ 蠁 ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚   dà Ã¢â‚¬  1 dx à Ã¢â‚¬  2 à ¢Ã‹â€ Ã¢â‚¬â„¢ dà Ã¢â‚¬  2 dx à Ã¢â‚¬  1   dx (2.21) If we consider a restricted ferromagnetic system in which there is the absence of a background magnetic field (H = 0); it is still possible for a topological soliton to exist if there is an easy axis anisotropy. [3] Magnetic anisotropy is the directional dependence of a materials magnetic property, and the easy axis is a energetically favorable direction if spontaneous magnetization occurs.[7] The energy for this system is E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ ( 1 2  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚   dà Ã¢â‚¬   dx  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚  Ãƒâ€šÃ‚   2 + A 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ (à Ã¢â‚¬    · k) 2   ) dx (2.22) where A > 0 is the anisotropy constant and k is the unit vector which specifies the easy axis. [3] For this type of system there are two zero energy ground states, à Ã¢â‚¬   =  ±k. The kink in t his system, also called a domain wall, interpolates between the two zero energy ground states and has boundary conditions à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ k as x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾ and à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã¢â‚¬â„¢k 15 as x à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ +à ¢Ã‹â€ Ã… ¾. Therefore the domain wall does not have a full twist of a kink and only has a half-twist. It is possible to map this system to our original kink example by a change of variables. If we set k = (0, 1) for convenience, and choose A = 1 2 . Setting à Ã¢â‚¬   = sin à Ã¢â‚¬   2   , cos à Ã¢â‚¬   2 , then the energy equation becomes E = 1 4 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚   1 2 à Ã¢â‚¬   02 + 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ cos à Ã¢â‚¬  Ãƒâ€šÃ‚   dx (2.23) which is equal to the energy equation (2.1) but with a normalization factor of 1 4 . The domain wall boundaries are à Ã¢â‚¬   à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ (0,  ±1) as x à ¢Ã‹â€ Ã¢â‚¬Å" à ¢Ã‹â€ Ã… ¾ are exactly the kink boundary conditions à Ã¢â‚¬   (à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¾) = 0 and à Ã¢â‚¬   (à ¢Ã‹â€ Ã… ¾) = 2à Ã¢â€š ¬. [1] This chapter will focus on topological solitons in (2+1) spatial dimensions. It would be incorrect to use the term soliton for these solutions due to their lack of stability, instead they are often referred to as lumps. The solutions for these lumps are given explicitly by rational maps between Riemann spheres. [1] For this chapter we shall be looking at one of the simplest Lorentz invariant sigma models in (2+1) spatial dimensions which renders static topological soliton solutions; the O(3) sigma model in the plane. [1] A sigma model is a nonlinear scalar field theory, where the field takes values in a target space which is a curved Riemannian manifold, usually with large symmetry. [1] For the O(3) sigma model the target space is the unit 2-sphere, S 2 . This model uses three real scalar fields, ÃŽÂ ¦ = (à Ã¢â‚¬  1, à Ã¢â‚¬  2, à Ã¢â‚¬  3), which are functions of the space-time coordinates (t, x, y) in (2+1) spatial dimensions. [2] The O(3) model is defined by the Lagrangia n density L = 1 4 (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒÅ½Ã‚ ¦)  · (à ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦)  with the constraint ÃŽÂ ¦  · ÃŽÂ ¦ = 1. For this equation the indices represent the space-time coordinates and take the values 0, 1, 2, and à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µ is partial differentiation with respect to X µ . [2] From (3.1), the Euler-Lagrange equation can be derived, which is à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦ + (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦) ÃŽÂ ¦ = 0 (3.2) Due to the dot product in à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦, this shows that the Euclidean metric of R 3 is being used, and this becomes the standard metric on the target space S 2 when the constraint ÃŽÂ ¦  · ÃŽÂ ¦ = 1 is being imposed. [1] For the sigma model we are exploring, the O(3) represents the global symmetry in the target space corresponding to the rotation s: ÃŽÂ ¦ 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ MÃŽÂ ¦ Where M à ¢Ã‹â€ Ã‹â€  O(3) is a constant matrix. [1] The sigma in the models name represents the fields (à Ã¢â‚¬  1, à Ã¢â‚¬  2, à Ã†â€™), where à Ã¢â‚¬  1 and à Ã¢â‚¬  2 are locally unconstrained [1] and à Ã†â€™ = p 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ à Ã¢â‚¬   2 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ à Ã¢â‚¬   2 2 is dependent on à Ã¢â‚¬  1 and à Ã¢â‚¬  2. The energy for the O(3) sigma model is E = 1 4 Z à ¢Ã‹â€ Ã¢â‚¬Å¡iÃŽÂ ¦  · à ¢Ã‹â€ Ã¢â‚¬Å¡iÃŽÂ ¦d 2x (3.3) where i = 1, 2 runs over the spatial indices. In order for the energy to be finite, ÃŽÂ ¦ has to tend to a constant vector at spatial infinity, so without loss of generality we are able to set the boundary condition ÃŽÂ ¦ à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ (0, 0, 1) as x 2 + y 2 à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã… ¾. Topologically we have R 2 à ¢Ã‹â€ Ã‚ ª {à ¢Ã‹â€ Ã… ¾}, which is interpreted as a sphere S 2 via the stereographic projection. (The sphere itself has finite radius.) Therefore we are considering mapping between spheres ÃŽÂ ¦ : S 2 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ S 2 . Just like in our kink example, mapping between spheres means there exists a topological charge, which can be found using N = 1 4à Ã¢â€š ¬ Z ÃŽÂ ¦  · (à ¢Ã‹â€ Ã¢â‚¬Å¡1ÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡2ÃŽÂ ¦) d 2x (3.4) The topological charge represents the number of lumps in the field configuration [1], since generally there are N well-separated, localized areas where the energy density is concentrated and each area has one unit of charge. However, as the lumps approach each other this is no longer the case. In order to apply Derricks theorem to the energy (3.3), we would need to consider the scaling x 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ x ÃŽÂ » = X and y 7à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ y ÃŽÂ » = Y which would give E (ÃŽÂ ») = E. The energy is independent of ÃŽÂ », therefore any value of ÃŽÂ » is a stationary point since the energy does not change from spatial rescaling. If we integrate the inequality  (à ¢Ã‹â€ Ã¢â‚¬Å¡iÃŽÂ ¦  ± ÃŽÂ µijÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡jÃŽÂ ¦)  · (à ¢Ã‹â€ Ã¢â‚¬Å¡iÃŽÂ ¦  ± ÃŽÂ µikÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡kÃŽÂ ¦) à ¢Ã¢â‚¬ °Ã‚ ¥ 0 (3.5) over the plane and use the equations (3.3) and (3.4) for the energy density and the topological charge respectively [1], then we get the Bogomolny bound E à ¢Ã¢â‚¬ °Ã‚ ¥ 2à Ã¢â€š ¬ |N| (3.6) This Bogomolny bound is the lower bound of the energy in terms of lumps. [1] If the field is a solution to one of the first-order Bogomolny equations à ¢Ã‹â€ Ã¢â‚¬Å¡iÃŽÂ ¦  ± ÃŽÂ µijÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡jÃŽÂ ¦ = 0 (3.7) then the energy is equal to the Bogomolny bound. In order to analyse the Bogomolny equations it is best to make the following changes of variables. For the first change in variable let X = (X1, X2, X3) denote the Cartesian coordinates in R 3 and take X = ÃŽÂ ¦ to be a point on the unit sphere, (X2 1 , X2 2 , X2 3 ) = 1. Let L be the line going through X = (0, 0, à ¢Ã‹â€ Ã¢â‚¬â„¢1) and ÃŽÂ ¦ and set W = X1 + iX2 to be the complex coordinate of the point where L intersects the plane at X3 = 0. We then get W = (à Ã¢â‚¬  1 + ià Ã¢â‚¬  2) (1 + à Ã¢â‚¬  3) (3.8) where à Ã¢â‚¬  1 =   W + W 1 + |W| 2   , à Ã¢â‚¬  2 = i   W à ¢Ã‹â€ Ã¢â‚¬â„¢ W 1 + |W| 2   , à Ã¢â‚¬  3 = 1 à ¢Ã‹â€ Ã¢â‚¬â„¢ |W| 2 1 + |W| 2 ! (3.9) As ÃŽÂ ¦ tends to the point (0, 0, à ¢Ã‹â€ Ã¢â‚¬â„¢1) then L only intersects X3 = 0 at à ¢Ã‹â€ Ã… ¾, therefore the point (0, 0, à ¢Ã‹â€ Ã¢â‚¬â„¢1) maps to the point W = à ¢Ã‹â€ Ã… ¾. This method of assigning each point on the sphere to a point in C à ¢Ã‹â€ Ã‚ ª {à ¢Ã‹â€ Ã… ¾} is called stereographic projection as seen in Figure 3.1.[3] The next change in variable comes from using a complex coordinate in the (x, y) plane by letting z = x + iy. Using this formation it is possible to rewrite the Lagrangian density, from (3.1) L = 1 4 ( à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ Ã¢â‚¬  1) 2 + (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ Ã¢â‚¬  2) 2 + (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ Ã¢â‚¬  3) 2   . Firstly we need to partially differentiate à Ã¢â‚¬  1, à Ã¢â‚¬  2, à Ã¢â‚¬  3, giving à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ Ã¢â‚¬  1 = à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW + à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW 1 + |W| 2 à ¢Ã‹â€ Ã¢â‚¬â„¢ (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW) W + W à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW   1 + |W| 2 2 W + W   (3.10) à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒ Ã¢â‚¬  2 = i à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW à ¢Ã‹â€ Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW 1 + |W| 2 à ¢Ã‹â€ Ã¢â‚¬â„¢ (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW) W + W à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µW   1 + |W| 2 2 W à ¢Ã‹â€ Ã¢â‚¬â„¢ W Finally, from simplifying (3.37) we get the equation for the topological charge in the new formulation to be N = 1 4à Ã¢â€š ¬ Z 4 1 + |W| 2 2 à ¢Ã‹â€ Ã¢â‚¬Å¡zW à ¢Ã‹â€ Ã¢â‚¬Å¡zW à ¢Ã‹â€ Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã¢â‚¬Å¡zW à ¢Ã‹â€ Ã¢â‚¬Å¡zW   d 2x = 1 à Ã¢â€š ¬ Z |à ¢Ã‹â€ Ã¢â‚¬Å¡zW| 2 à ¢Ã‹â€ Ã¢â‚¬â„¢ |à ¢Ã‹â€ Ã¢â‚¬Å¡zW| 2   1 + |W| 2 2 d 2x (3.38) In this formulation it is clear to see E à ¢Ã¢â‚¬ °Ã‚ ¥ 2à Ã¢â€š ¬ |N|, with equality if and only if Bogomolny equation is satisfied à ¢Ã‹â€ Ã¢â‚¬Å¡W à ¢Ã‹â€ Ã¢â‚¬Å¡z = 0 (3.39) This equation shows that W is a holomorphic function of z only. [4] Due to the requirement that the total energy is finite, together with the boundary condition [4] W à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ 0 as |z| à ¢Ã¢â‚¬  Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã… ¾, this means that N is finite. [3] The simplest solution for the Bogomolny equation would be W = ÃŽÂ » z , where ÃŽÂ » is a real and positive constant. Applying this to the equation (3.9) yields the solution for t he N = 1 solution ÃŽÂ ¦ =   2 ÃŽÂ » 2 + x 2 + y 2 , à ¢Ã‹â€ Ã¢â‚¬â„¢2 ÃŽÂ » 2 + x 2 + y 2 , x 2 + y 2 à ¢Ã‹â€ Ã¢â‚¬â„¢ ÃŽÂ » 2 ÃŽÂ » 2 + x 2 + y 2 (3.40) If we change the negative sign in the second component to a positive sign then we get the solution of the anti-Bogomolny equation (3.7) (with the minus sign), which also has E = 2à Ã¢â€š ¬ but has N = à ¢Ã‹â€ Ã¢â‚¬â„¢1. This soliton is located at thee origin because W(0) = à ¢Ã‹â€ Ã… ¾. [3] The N = 1 general solution has 4 real parameters and is given by the Bogomolny solution W = ÃŽÂ »eiÃŽÂ ¸ z à ¢Ã‹â€ Ã¢â‚¬â„¢ a (3.41) where ÃŽÂ » is the size of the soliton, ÃŽÂ ¸ is the constant angle of rotation in the (à Ã¢â‚¬  1, à Ã¢â‚¬  2) plane and a à ¢Ã‹â€ Ã‹â€  C is the position of the soliton in the complex plane, z = x + iy. The O(3) sigma model can be modified to stabilise a lump, and the simplest way in doing this is by introducing extra terms into the Lagrangian which break the conformal invariance of the static energy. [1] These new terms must scale as negative and positive powers of a spatial dilation factor. [1] An example of this is the Baby Skyrme model which is given by the Lagrangian L = 1 4 à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦ à ¢Ã‹â€ Ã¢â‚¬â„¢ 1 8 (à ¢Ã‹â€ Ã¢â‚¬Å¡Ãƒâ€šÃ‚ µÃƒÅ½Ã‚ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡ÃƒÅ½Ã‚ ½ÃƒÅ½Ã‚ ¦)  · (à ¢Ã‹â€ Ã¢â‚¬Å¡  µÃƒÅ½Ã‚ ¦ ÃÆ'- à ¢Ã‹â€ Ã¢â‚¬Å¡ ÃŽÂ ½ÃƒÅ½Ã‚ ¦) à ¢Ã‹â€ Ã¢â‚¬â„¢ m2 2 (1 à ¢Ã‹â€ Ã¢â‚¬â„¢ à Ã¢â‚¬  3) (3.42) where the constraint ÃŽÂ ¦  · ÃŽÂ ¦ = 1 is implied. As we can see the first term in this Lagrangian is simply that of the O(3) sigma model. The second term in (3.42), is known as the Skyrme term and the final term in this Lagrangian is the mass term. The complete understanding of topological solitons is unknown and there are very limited experimental tests of many of the theories of topological solitons and their mathematical results. However, there is evidence of topological solitons existing in some physical systems, for example in one-dimensional systems they exist in optical fibres and narrow water channels. [1] Topological solitons can be applied to a range of different areas including particle physics, condensed matter physics, nuclear physics and cosmology. They also can be applied within technology, which involves using topological solitons in the design for the next generation of data storage devices. [3] In August 2016, a 7 million pound research programme, being led by Durham University, was announced into looking at how magnetic skyrmions can be used in creating efficient ways to store data. [10] Magnetic skyrmions are a theoretical particle in three spatial dimensions which have been observed experimentally in condensed matter systems. [11] This type of soliton was first predicted by scientists back in 1962, but was first observed experimentally in 2009. [10] In certain types of magnetic material it is possible for these magnetic skyrmions to be created,manipulated and controlled[10], and because of their size and structure it is possible for them to be tightly packed together. The structure inside the skyrmions [10] Due to this and the force which locks the magnetic field into the skyrmion arrangement, any magnetic information which is encoded by skyrmions is very robust. [10] It is thought that it will be possible to move these magnetic skyrmions with a lot less energy than the ferromagnetic domain being used in current data storage devices of smartphones and computers. Therefore, these magnetic skyrmions could revolutionise data storage devices, as the devices could be created on a smaller scale and use a lot less energy, meaning they would be more cost effective and would generate less heat. This project has given an insight into the very basics of topological solutons by analysing the energy and topological charge equations for kinks in one spatial dimension and lumps in (2+1) spatial dimensions. From the energy equation for a kink, we could derive the solution of a kink and find the lower energy bound. From the lump model, we successfully changed the variables for the energy, topological charge and the Lagrange equation for a lump to be able to analyse the Bogomolny equation. From this change of variables of the Lagrange equation we successfully solved the Euler-Lagrange equations of motion for the lump model. This research project has been captivating and has given me an insight into how the complex mathematics we learn is applied to real world situations. I first became interested in this topic after attending the London Mathematical Societys summer 33 school in 2016, where I had the privilege of attending a few lectures given by Dr Paul Sutcliffe, one of the authors of the book on Topological Solitons. It was in these few lectures where I first learnt about topological solitons and some of their applications, and this inspired me for my research project as I wanted to study the topic further. Although this project has been thoroughly enjoyable, it came with challenging aspects, due to its complex mathematics in such a specialised subject. As a result of this topic being so specific, I was very limited in the resources I had for my research, my main resource being the book on topological solitons by Dr Paul Sutcliffe and Dr Nicholas Manton. I have gained a lot of new skills from this research project and it has given me an opportunity to apply my current mathematical knowledge. There is an endless amount of research that can be continued within this subject. I, for example, would have liked to do some further research into the (2+1) spatial dimension model of the Baby Skyrmion and, like the lump example, solve the EulerLagrange equations motion . As well as this, I would have liked to input the equations of motion I solved for the lump model in Maple, so it was possible to simulate two lumps colliding and from this graph the energy density. It would have been really interesting to research further into topological solitons in three spatial dimensions, specifically Skyrmions, to learn further about their technological applications. However, the mathematics used for this model is very challenging and specialised, and goes beyond my understanding and knowledge.

Capote/Krakauer Comparison :: essays papers

Capote/Krakauer Comparison Essay The most important thing any writer can do is to give their characters a feel of dimension to make them seem real. Although Capote and Krakauer do that in very different ways in In Cold Blood and Into Thin Air, they both reached the same end result: characters you believe. They give them thoughts, faces and personalities. They don’t portray everyone as flawless, they display the faults and the little quirks. They give them life through words, making these stories believable. Despite the fact both incidents happened years before each book was written, the use of detailed facts and personality profiles make each story seem incredibly realistic. But while Capote chooses to write an entirely objective piece, Krakauer relies heavily on personal opinion and experience, creating two very distinct frames of mind and causing the reader too see the characters in each book very differently. In 1959 the Clutter family was murdered in a tiny Kansas town called Holcomb. Six years later Truman Capote wrote a very detailed book about the whole case, from the day of the murder to the court case prosecuting the two murderers, Dick and Perry. Although he wasn’t there when the four murders happened, through word choice, description and characterization he creates an accurate portrait of the many intense events surrounding such a tragic story. In comparison, in 1996 esteemed climber Rob Hall led an expedition of moderately experienced climbers attempting to climb Mt. Everest, only to result in disaster and the loss of nine people’s lives. Jon Krakauer was a member of that expedition, and wrote a piece about the misadventure for Outside magazine. Feeling there was more to be said, soon after he wrote a book. Krakauer takes a similar approach as Capote, yet inserting more opinions and less of a feeling of objectiveness to his characters. This is most likely since Krakauer was living Everest first hand, as opposed to Capote who put himself into the environment years later, picking up details here and there instead of relying solely on memory and friends. One of Capote’s greatest strengths is to create thought for his characters, making it almost appear as if he knows what they are thinking. All summer Perry undulated between half-awake stupors and stickly, sweat-drenched sleep. Voices roared through his head; one voice persistently asked him, â€Å"Where is Jesus? Where?† And once he woke up shouting, â€Å"The bird is Jesus! The Bird is Jesus!† (381) This selection almost creates a feeling that Capote is talking about himself as opposed to a man he never met. Capote/Krakauer Comparison :: essays papers Capote/Krakauer Comparison Essay The most important thing any writer can do is to give their characters a feel of dimension to make them seem real. Although Capote and Krakauer do that in very different ways in In Cold Blood and Into Thin Air, they both reached the same end result: characters you believe. They give them thoughts, faces and personalities. They don’t portray everyone as flawless, they display the faults and the little quirks. They give them life through words, making these stories believable. Despite the fact both incidents happened years before each book was written, the use of detailed facts and personality profiles make each story seem incredibly realistic. But while Capote chooses to write an entirely objective piece, Krakauer relies heavily on personal opinion and experience, creating two very distinct frames of mind and causing the reader too see the characters in each book very differently. In 1959 the Clutter family was murdered in a tiny Kansas town called Holcomb. Six years later Truman Capote wrote a very detailed book about the whole case, from the day of the murder to the court case prosecuting the two murderers, Dick and Perry. Although he wasn’t there when the four murders happened, through word choice, description and characterization he creates an accurate portrait of the many intense events surrounding such a tragic story. In comparison, in 1996 esteemed climber Rob Hall led an expedition of moderately experienced climbers attempting to climb Mt. Everest, only to result in disaster and the loss of nine people’s lives. Jon Krakauer was a member of that expedition, and wrote a piece about the misadventure for Outside magazine. Feeling there was more to be said, soon after he wrote a book. Krakauer takes a similar approach as Capote, yet inserting more opinions and less of a feeling of objectiveness to his characters. This is most likely since Krakauer was living Everest first hand, as opposed to Capote who put himself into the environment years later, picking up details here and there instead of relying solely on memory and friends. One of Capote’s greatest strengths is to create thought for his characters, making it almost appear as if he knows what they are thinking. All summer Perry undulated between half-awake stupors and stickly, sweat-drenched sleep. Voices roared through his head; one voice persistently asked him, â€Å"Where is Jesus? Where?† And once he woke up shouting, â€Å"The bird is Jesus! The Bird is Jesus!† (381) This selection almost creates a feeling that Capote is talking about himself as opposed to a man he never met.