Saturday, September 7, 2019
The Great Leap Forward Essay Example for Free
The Great Leap Forward Essay The Hundred flowers campaign was followed by a new militant approach to Chinese economics. Shaoqi believed that the PLA and the military complex should be strengthened for several reasons; firstly the rejection of Mao foreign policy (Five Principles of Peaceful Co-existence) in order to prepare for the invasion of Tibet and other island provinces free from mainland influence since the days of the KMT republic. Secondly the war in Korea had created a siege-mentality in China (similar to USSR in the 1930s), China would be ready for invasion. Xiaoping believed that the people could be motivated and ideologically aroused. Mao supported this initial plan believing that this Second Five Year Plan could work better than the first. However he was wary of Shaoqis motives and did not wish to see the people merely exploited and made to work towards unrealistic targets. He initiated the xiafang movement, which took the Leap down to the countryside level. The xiafang movement would have several stages. Primarily it would concentrate on heavy industry and mobilisation of the urban regions. Beijing would begin this with a march to work programme. Next, the increasing of the indoctrinisatation of technology experts and the scientific community. Finally the xiafang movement would move to the rural areas with party cadres and members moving to the people and helping them in agricultural policies. It is at this point that the debate arises critics have suggested that Mao supported the scheme because he was unhappy with the USSRs de-Stalinisation of itself. He was undoubtedly concerned about his countrys over-reliance on Soviet help. The split over the direction occurred in late 1958, by then nearly 750,000 new collectivised farms had been created and agricultural output was at Chinas highest ever, Mao wanted to create forums to discuss problems with the Leap, he also wanted greater self-sufficiency amongst the communes. Shaoqi resisted this idea believing that centralisation was the only means of ensuring success. He introduced the radical mass dormitories with over 5000 people to each one. This new housing was resisted bitterly and Mao argued that it was essential the CCP listened to the people. Zhou Enlai also voiced concerns over plans to release worker from these collectives for overly grand projects such as hydro-plants and irrigation works. Mao quickly seized upon growing disenchantment and distanced himself from the ruling committee. 1959 was a disastrous year for the Chinese economy, in February of that year; Shaoqi admitted that the CCP had exaggerated figures for success. Famine ravaged Maos home province of Hunan and Zhus Jiangxi. Food shortages affected Beijing; raw materials were in short supply for the industrial complex. Xiaoping worsened the situation by creating the Department of Economic Growth ((based upon the Soviet Gosplan model) which centralised directives and set even higher targets. The direct result was the over-production of poor quality goods, a virtual collapse of heavy industry through mismanagement, a malaise and a demoralisation and exhaustion of the peasant population. The intellectual wing of the CCP demanded the plan was scrapped, which led to a vicious purging of the intelligentsia. Mao who personally bore the brunt of blame for the Leap fiasco stepped down from office in April. The following year saw a massive shift in the balance of power; the Second National Congress gave Lui Shaoqi complete control of the CCP and all Maos positions. Defence minister Peng Dehuai openly attacked Maoist policies and firmly placed the blame on Mao. However, Lin Baio a noted Maoist successfully ousted Dehuai out of office and accepted the post of Defence minister. He offers Zhu De the post of C-in-C of the army, who declines. Lin Baio resigns in 1961 after Chinas successful total annexation of Tibet. He is alarmed at the threat to invade Taiwan and the attacks on Jinmen and Mazu. By 1961, the swing to the right was almost complete with Shaoqi in the ascendancy and his fraction most of the positions of power. However with the battle for supremacy gaining momentum by 1962 with Maos spectacular return to power, the political landscape of China would never be the same and the CCP by 1970 would be decimated from top to bottom.
Friday, September 6, 2019
Chinas Managed Float Essay Example for Free
Chinas Managed Float Essay The RMB is unlikely to be floated freely in the near term as the countrys economyfaces internal difficulties during its reform drive and external uncertainties of the globaleconomy, the report quoted Xia Bin, a member of the monetary policy committee of thePeoples Bank of China (PBOC), or the central bank, as saying. To create a relatively stable exchange rate formation environment, the governmenthas to gradually open its capital market, so the RMB can not go global too soon,China moved to shift from a conventional dollar peg system to a managed floatingexchange rate system in 2005, which means the central bank now does not link theyuan only to the U. S. dollar. Chinas RMB go global drive requires totally free exchange of the yuan, which meansthe regulation of capital accounts should be fully opened, and that exchange rates willbe largely determined by the demand and supply in both domestic and global markets. But the country can not handle this at its current stage of economic development, Xiasaid. Xia suggested that the government should well coordinate policies concerning theexchange rate, capital management and reform while matching the reform of itsexchange rate policy with that of capital management during the RMBs regionalizationprocess.
Thursday, September 5, 2019
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‚¬ ’ 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 †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€ ’à ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚ 1 2 à †02 + 1 à ¢Ã‹â€ ’ cos à †Â dx (2.1) where à †0 = dà †dx . For the energy density to be finite this requires à †à ¢Ã¢â‚¬ ’ 2à ₠¬nà ¢Ã‹â€ ’ as x à ¢Ã¢â‚¬ ’ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ and à †à ¢Ã¢â‚¬ ’ 2à ₠¬n+ as x à ¢Ã¢â‚¬ ’ à ¢Ã‹â€ Ã… ¾, where n ± à ¢Ã‹â€ ˆZ. To find the number of twists, N, this is simply N = n+ à ¢Ã‹â€ ’ nà ¢Ã‹â€ ’ = à †(à ¢Ã‹â€ Ã… ¾) à ¢Ã‹â€ ’ à †(à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾) 2à ₠¬ = 1 2à ₠¬ Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ à †0 dx à ¢Ã‹â€ ˆZ This is the equation for the topological charge or the winding number. If we set nà ¢Ã‹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 †(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‚¬ ’ 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€ ’ f(x) = f(x) + ÃŽÂ ´xf0 (x) + à ¢Ã‹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 †0 ÃŽÂ ´Ãƒ †] à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ equates to zero on the boundary because it must satisfy ÃŽÂ ´Ãƒ †( ±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€ ’à ¢Ã‹â€ Ã… ¾ {(à ¢Ã‹â€ ’à †00 + sin à †)ÃŽÂ ´Ãƒ †} dx (2.6) This equation can be minimised minimised further to the second order nonlinear differential equation, à †00 = sin à †(2.7) The solution of this differential equation with the boundary conditions, à †(à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾) = 0 and à †(à ¢Ã‹â€ Ã… ¾) = 2à ₠¬ is the kink. Therefore the kink solution is, à †(x) = 4 tanà ¢Ã‹â€ ’1 e xà ¢Ã‹â€ ’a (2.8) where a is an arbitrary constant. When x = a, this is the position of the kink (à †(a) = à ₠¬). It is clear to see à †= 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€ ’à ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚ à †02 + 4 sin2Â à †2 dx (2.9) By completing the square the equation becomes, E = 1 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚ à †0 à ¢Ã‹â€ ’ 2 sinÂ à †2 2 + 4à †0 sinÂ à †2 dx (2.10) Therefore the energy satisfies the inequality, E > 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ à †0 sinÂ à †2 dx = 2 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ sinÂ à †2 dà †dxdx = 2 Z 2à ₠¬ 0 sinÂ à †2 dà †= à ¢Ã‹â€ ’4 cosÂ à †2 2à ₠¬ 0 = 8 (2.11) In order to obtain the solution which is exactly 8, the term à †0 à ¢Ã‹â€ ’ 2 sin à †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 †0 = 2 sinÂ à †2 (2.12) This is a first order Bogomolny equation. Taking this Bogomolny equation and differentiating with respect to à †0 gives, à †00 = cosÂ à †2Â à †0 = cosÂ à †2 2 sinÂ à †2 = sin à †(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 †02 . From equation (2.8) we have, tan à †4 = e xà ¢Ã‹â€ ’a , therefore 1 4 à †0 sec2Â à †4 = e xà ¢Ã‹â€ ’a This equation gives, à †0 = 4 e xà ¢Ã‹â€ ’a 1 + tan2 à †4 = 4e xà ¢Ã‹â€ ’a 1 + e 2(xà ¢Ã‹â€ ’a) = 2 cosh (x à ¢Ã‹â€ ’ a) = 2 (x à ¢Ã‹â€ ’ a) (2.15) Therefore it can be seen that the energy density is given by ÃŽÂ µ = 42 (x à ¢Ã‹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 †(a) = à ₠¬ [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€ ’à ¢Ã‹â€ Ã… ¾ ÃŽÂ µdx = 4 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ 2 (x à ¢Ã‹â€ ’ a) dx = 4 [tanh (x à ¢Ã‹â€ ’ a)]à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢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€ ’à ¢Ã‹â€ Ã… ¾ 1 2 à †02 dx + Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ (1 à ¢Ã‹â€ ’ cos à †) dx (2.17) Now consider the spatial rescaling x 7à ¢Ã¢â‚¬ ’ x ÃŽÂ » = X, so that à †(x) 7à ¢Ã¢â‚¬ ’ à †(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€ ’à ¢Ã‹â€ Ã… ¾ 1 2 ( 1 ÃŽÂ » dà †dX ) 2ÃŽÂ »dX + Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ (1 à ¢Ã‹â€ ’ cos à †) 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€ ’ 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 †(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 †= (à †1, à †2) be a two-component unit vector, where à †Â · à †= |à †| 2 = 1. By writing à †= (sin à †, cos à †), the energy from (2.1) can be rewritten as E = Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ ( 1 2    dà †dx    2 à ¢Ã‹â€ ’ H  · à †+ |H| ) dx (2.20) where H = (0, 1). [3] In this new formulation à †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 †. There is only one point in which the systems ground state is equal to zero in terms of à †, which is à †= 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 †à ¢Ã¢â‚¬ ’ (0, 1) as x à ¢Ã¢â‚¬ ’  ±Ãƒ ¢Ã‹â€ Ã… ¾. As à †takes the same value at x = à ¢Ã‹â€ ’à ¢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 †: S 1 7à ¢Ã¢â‚¬ ’ S 1 between circles, because à †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 †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 ₠¬ Z à ¢ 蠁 ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚ dà †1 dx à †2 à ¢Ã‹â€ ’ dà †2 dx à †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€ ’à ¢Ã‹â€ Ã… ¾ ( 1 2    dà †dx    2 + A 1 à ¢Ã‹â€ ’ (à †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 †=  ±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 †à ¢Ã¢â‚¬ ’ k as x à ¢Ã¢â‚¬ ’ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾ and à †à ¢Ã¢â‚¬ ’ à ¢Ã‹â€ ’k 15 as x à ¢Ã¢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 †= sin à †2 , cos à †2 , then the energy equation becomes E = 1 4 Z à ¢Ã‹â€ Ã… ¾ à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾Ãƒâ€šÃ‚ 1 2 à †02 + 1 à ¢Ã‹â€ ’ cos à †Â 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 †à ¢Ã¢â‚¬ ’ (0,  ±1) as x à ¢Ã‹â€ â€Å" à ¢Ã‹â€ Ã… ¾ are exactly the kink boundary conditions à †(à ¢Ã‹â€ ’à ¢Ã‹â€ Ã… ¾) = 0 and à †(à ¢Ã‹â€ Ã… ¾) = 2à ₠¬. [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 †1, à †2, à †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€ ‚ µÃƒÅ½Ã‚ ¦)  · (à ¢Ã‹â€ ‚  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€ ‚ 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€ ‚ µÃƒ ¢Ã‹â€ ‚  µÃƒÅ½Ã‚ ¦ + (à ¢Ã‹â€ ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ ‚  µÃƒÅ½Ã‚ ¦) ÃŽÂ ¦ = 0 (3.2) Due to the dot product in à ¢Ã‹â€ ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ ‚  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‚¬ ’ MÃŽÂ ¦ Where M à ¢Ã‹â€ ˆO(3) is a constant matrix. [1] The sigma in the models name represents the fields (à †1, à †2, à ƒ), where à †1 and à †2 are locally unconstrained [1] and à ƒ = p 1 à ¢Ã‹â€ ’ à †2 1 à ¢Ã‹â€ ’ à †2 2 is dependent on à †1 and à †2. The energy for the O(3) sigma model is E = 1 4 Z à ¢Ã‹â€ ‚iÃŽÂ ¦  · à ¢Ã‹â€ ‚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‚¬ ’ (0, 0, 1) as x 2 + y 2 à ¢Ã¢â‚¬ ’ à ¢Ã‹â€ Ã… ¾. Topologically we have R 2 à ¢Ã‹â€  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‚¬ ’ 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 ₠¬ Z ÃŽÂ ¦  · (à ¢Ã‹â€ ‚1ÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ ‚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‚¬ ’ x ÃŽÂ » = X and y 7à ¢Ã¢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€ ‚iÃŽÂ ¦  ± ÃŽÂ µijÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ ‚jÃŽÂ ¦)  · (à ¢Ã‹â€ ‚iÃŽÂ ¦  ± ÃŽÂ µikÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ ‚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 ₠¬ |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€ ‚iÃŽÂ ¦  ± ÃŽÂ µijÃŽÂ ¦ ÃÆ'- à ¢Ã‹â€ ‚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€ ’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 †1 + ià †2) (1 + à †3) (3.8) where à †1 = W + W 1 + |W| 2 , à †2 = i W à ¢Ã‹â€ ’ W 1 + |W| 2 , à †3 = 1 à ¢Ã‹â€ ’ |W| 2 1 + |W| 2 ! (3.9) As ÃŽÂ ¦ tends to the point (0, 0, à ¢Ã‹â€ ’1) then L only intersects X3 = 0 at à ¢Ã‹â€ Ã… ¾, therefore the point (0, 0, à ¢Ã‹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€  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€ ‚ µÃƒ †1) 2 + (à ¢Ã‹â€ ‚ µÃƒ †2) 2 + (à ¢Ã‹â€ ‚ µÃƒ †3) 2 . Firstly we need to partially differentiate à †1, à †2, à †3, giving à ¢Ã‹â€ ‚ µÃƒ †1 = à ¢Ã‹â€ ‚ µW + à ¢Ã‹â€ ‚ µW 1 + |W| 2 à ¢Ã‹â€ ’ (à ¢Ã‹â€ ‚ µW) W + W à ¢Ã‹â€ ‚ µW 1 + |W| 2 2 W + W (3.10) à ¢Ã‹â€ ‚ µÃƒ †2 = i à ¢Ã‹â€ ‚ µW à ¢Ã‹â€ ’ à ¢Ã‹â€ ‚ µW 1 + |W| 2 à ¢Ã‹â€ ’ (à ¢Ã‹â€ ‚ µW) W + W à ¢Ã‹â€ ‚ µW 1 + |W| 2 2 W à ¢Ã‹â€ ’ W Finally, from simplifying (3.37) we get the equation for the topological charge in the new formulation to be N = 1 4à ₠¬ Z 4 1 + |W| 2 2 à ¢Ã‹â€ ‚zW à ¢Ã‹â€ ‚zW à ¢Ã‹â€ ’ à ¢Ã‹â€ ‚zW à ¢Ã‹â€ ‚zW d 2x = 1 à ₠¬ Z |à ¢Ã‹â€ ‚zW| 2 à ¢Ã‹â€ ’ |à ¢Ã‹â€ ‚zW| 2 1 + |W| 2 2 d 2x (3.38) In this formulation it is clear to see E à ¢Ã¢â‚¬ °Ã‚ ¥ 2à ₠¬ |N|, with equality if and only if Bogomolny equation is satisfied à ¢Ã‹â€ ‚W à ¢Ã‹â€ ‚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‚¬ ’ 0 as |z| à ¢Ã¢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€ ’2 ÃŽÂ » 2 + x 2 + y 2 , x 2 + y 2 à ¢Ã‹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 ₠¬ but has N = à ¢Ã‹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€ ’ a (3.41) where ÃŽÂ » is the size of the soliton, ÃŽÂ ¸ is the constant angle of rotation in the (à †1, à †2) plane and a à ¢Ã‹â€ ˆC 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€ ‚ µÃƒÅ½Ã‚ ¦  · à ¢Ã‹â€ ‚  µÃƒÅ½Ã‚ ¦ à ¢Ã‹â€ ’ 1 8 (à ¢Ã‹â€ ‚ µÃƒÅ½Ã‚ ¦ ÃÆ'- à ¢Ã‹â€ ‚Π½ÃƒÅ½Ã‚ ¦)  · (à ¢Ã‹â€ ‚  µÃƒÅ½Ã‚ ¦ ÃÆ'- à ¢Ã‹â€ ‚ ÃŽÂ ½ÃƒÅ½Ã‚ ¦) à ¢Ã‹â€ ’ m2 2 (1 à ¢Ã‹â€ ’ à †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.
Wednesday, September 4, 2019
Cleopatra :: Essays Papers
Cleopatra II. Summary: After the completion of the book, it had let me to believe the book was written for the general audience. Although the author provided many resources, the information was taken from literature that was written during the time. Therefore, some of the quotes were biased either against or favored Cleopatra. For an example of bias against her, the Jewish historian Flavius Josephus called her a â€Å"wicked creature, who was a slave to her lusts, but she still imagined that she wanted everything she could think of, and did her utmost to gain it†¦. As for Antony, he was so entirely overcome by this woman that†¦ he was some way or other bewitched to do whatever she would have him do.†On the contrary, here is an example of bias in favor of Cleopatra: â€Å"a princess well versed in the sciences, disposed to the study of philosophy and counting scholars among her intimate friends. She was the author of works on medicine, charms, and other divisions of the natural scie nces.†This was taken from a tenth-century Arab historian Al Masudi. The author allowed his readers to conclude their own interpretation of Cleopatra by stating both sides of the story. The book was broken down into eight chapters. These chapters spanned the time from 332 B.C. to 30 B.C. In the beginning of the book, it began the story of Cleopatra with Alexander the Great liberating Egypt from Persian control. However, the bulk of the chapters concentrated at 69 B.C. and ended 30 B.C. with the birth and death of Cleopatra. The story of Cleopatra began with her rein over Egypt as queen. This was when she allied and companioned with Caesar in attempt to strengthen her power. It was not long before Caesar was assassinated and his close friend and a powerful general Mark Antony denounced the conspirators. Not long after Caesar’s death, Antony and Cleopatra fell in love and ruled Rome and Egypt together. Together, they had formed an alliance strong enough to take down the most powerful force in the world at the time, Rome. The fall of Antony and Cleopatra began when they were defeated at Actium in Greece against Octavian’s Roman army. Towards the end of the book, the author went into details on the true love that existed between Antony and Cleopatra.
Tuesday, September 3, 2019
Essay on Whartons Ethan Frome: Nature -- Ethan Frome Essays
Nature in Ethan Frome Every winter frigid white bullets, squalling gusts, and icicle shards swaddle the town of Starkfield in a frosty white glaze. It is easy to understand why the people emerge from this six month siege like starved troops capitulating without shelter. Most people evacuate the premises immediately after suffering through a devastating winter, but not Ethan Frome. Circumstances hindered the flight of this man. As one retired stage driver remarked, "Guess he's been in Starkfield too many winters. Most of the smart ones get away." The statement by Harmon Gow, a resident of Starkfield, relates to Ethan Frome, the protagonist of the novel, Ethan Frome. This book pieces together the enigmatic life of a man bound by the shackles of silence and isolation. By deftly heightening suspense and foreshadowing plot, Edith Wharton explores nature's degeneration of human spirit and vitality. Mr. Gow's quote delves into two integral aspects of the book: how the unrelenting blows of nature corrode, yet intertwine with man's spirit, and how the seas... Essay on Wharton's Ethan Frome: Nature -- Ethan Frome Essays Nature in Ethan Frome Every winter frigid white bullets, squalling gusts, and icicle shards swaddle the town of Starkfield in a frosty white glaze. It is easy to understand why the people emerge from this six month siege like starved troops capitulating without shelter. Most people evacuate the premises immediately after suffering through a devastating winter, but not Ethan Frome. Circumstances hindered the flight of this man. As one retired stage driver remarked, "Guess he's been in Starkfield too many winters. Most of the smart ones get away." The statement by Harmon Gow, a resident of Starkfield, relates to Ethan Frome, the protagonist of the novel, Ethan Frome. This book pieces together the enigmatic life of a man bound by the shackles of silence and isolation. By deftly heightening suspense and foreshadowing plot, Edith Wharton explores nature's degeneration of human spirit and vitality. Mr. Gow's quote delves into two integral aspects of the book: how the unrelenting blows of nature corrode, yet intertwine with man's spirit, and how the seas...
Monday, September 2, 2019
Abraham Lincoln Essay -- essays research papers
Abraham Lincoln was the 16th president of the U.S (1861-1865) who brought the Union to victory in the Civil War. Lincoln was born on February 12, 1809 in Kentucky. His father was Thomas Lincoln and his mother was Nancy Hanks, both were pioneer farmers. When Abraham Lincoln was two they moved to nearby Knob Creek, Indiana. The following year his mother died. In 1819 Abraham Lincoln’s father married Sarah Bush Johnston, a kind widow who gained Abraham Lincoln’s friendship. Abraham Lincoln grew up to be a tall, gangling boy who could handle himself. He also showed intellectual promises, even though he had little formal education. In 1831 he moved again to Macon County, Illinois and finally he got a job on a cargo ship sailing down the Mississippi to New Orleans. He then returned to Illinois to settle in New Salem on the Sangamon River, were he became a clerk at a local store. In 1832 he became Captain of a company going to fight in the Black Hawk War. When the war ended he came home and he tried to open a store but that ended in a failure when his partner died. In 1833 he was appoint ed postmaster. But he also had to take up surveying to support himself. In time he was able to pay off his debts and began to study law. In 1834 Lincoln was elected on the Wing ticket to serve in the Lower House until 1841. He emerged as a party leader, so he moved to Springfield the capital of Illinois. At this time he also became a very popular attorney with a partnership of 3 other men. In 1842 ...
Sunday, September 1, 2019
Compare and Contrast: A&P and Everyday Use Essay
A&P is a short story written by American writer John Updike. The story takes place during the summer in a small New England town where everything seems ordinary and gray. The story stars Sammy a nineteen-year-old boy working the checkout line of grocery store by the name of A&P. Sammy is working one day when in walks three teenage girls, wearing only their bathing suits. He is quickly infatuated by the scene for it is not one he comes across to often and begins to watch the girls closely. Sammy pays close attention to the appearance of the girls naming each one according to how he views them, there is â€Å"Plaid†the chunky one who got her name due to her swimsuit, â€Å"Big Tall Goony Goony†who he describes as attractive but falling short to â€Å"Queenie†, the girl who Sammy seems to be most infatuated with, she is the most striking out of the bunch getting her name due to being the leader of the group. Sammy pays close attention to the three girls all the way to the point where they finally come to his register where they are confronted by his manager Lengel a very conservative man who ask the girls to come with their shoulders covered the next time they wish to enter his store. This sparks Sammy to quit and to chase the girls outside where to his surprise are gone. Everyday Use is a short story written by American author Alice Walker. The story takes place in the late 1900’s in the south in a house that was rebuilt after it was burned down by a ferocious fire. The story stars Mama a big-boned woman with hands that are rough from years of physical labor that is poor and uneducated due to never being given the opportunity to break out of her rural life, she is also the story’s narrator. Mama and her daughter Maggie who is shy and self conscience due to her being burned and scarred by the fire that took down their house wait outside their home for the arrival of her older daughter Dee who is the exact opposite of Maggie being that she is educated and quite confident. Dee arrives at the home of her mother with not only a new man, but a new look, she is very colorful and into her â€Å"heritage†. Mama, Maggie, Dee, and Dee’s male friend have dinner in which Dee tells Mama what she’s been doing with herself and ends with Dee wanting to leave with the butter churn. The butter churn isn’t the only item which Dee would like to leave with, she also wants two quilts made by her ancestors, but Mama says no for they are promised to her sister Maggie. Dee leaves the house outraged telling her mother she doesn’t understand her heritage. Two scenes that I feel share a common message is Sammy’s decision to quit A&P and Mama’s decision to not let Dee have her way and give Maggie the quilt. Both characters decided this would be their time to take a stand against something they felt was not right. For Sammy it was a decision driven by his lust for Queenie, for Mama it was finally saying no to the daughter who she feels unappreciated by. Sammy’s decision ends with him having no job and no women, but a better sense of what he wants in life which is a desire to get away from the norm that is his job and town and go out and experience opportunities beyond his limit. Mama’s decision had the opposite effect of what Sammy’s had, she not only grew closer to her daughter Maggie, but she realized that she’s happy being the woman she is, she doesn’t have to change herself . Mama is proud of her life and of being a practical hardworking woman. Both A&P and Everyday Use are similar in the fact that we have characters struggling to find out what it is they want from their lives, but are different in the sense of what the characters actually realize they want.
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