Sunday, January 26, 2020
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.
Saturday, January 18, 2020
Advice to Youth Satire
Andy Diaz â€Å"Advice to Youth†The Object of Mark Twain’s article is to point out that the youths of our society are being told to become just like everyone else and that discourages their individuality. He uses sarcasm, so that he can assume the role of an elder in society, the kind of people he attacks, which instructs younger people how to act. Mark Twain does an exemplary job in copying exactly the types of teachings for youth that have been passed down through the ages. The idea of respecting one’s elders has been around for a long time. The article gives expresses of both adult and teen satire. But it mostly, pokes fun at the stereotypical advice and coaching given to youth through the use of firearms. However, the last sentence has a much darker and more of a bitter tone, and attacks the teachings he has just mocked, â€Å"Build your character thoughtfully and painstakingly upon these precepts, and by and by, when you have got it built, you will be surprised and gratified to see how nicely and sharply it resembles everybody else’s. Twain uses satire in describing the story of the boy who almost shoots his grandmother with what he thought was an unloaded gun. Instead of the classic â€Å"life lesson†story where the gun turns out to be loaded and ends up killing someone, he twists it so that the gun isn’t loaded after all and no one gets hurt. The rule at the root of this piece is the idea that individuality should be educated and that rational teachings are normalizing our youth. Mark Twain is great at mocking known teachings because he looks at it from a different angle, not suggesting that they are wrong but rather questioning if they are limiting originality and freedom of thought. Although, he runs into the certain obstacle that people who follow rational wisdom and tell their children the kinds of things he mocks won’t be as open to his argument because he doesn’t preach rational wisdom himself. He uses irrational wisdom to prove his point.
Friday, January 10, 2020
Reluctant Fundamentalist Essay
In The Reluctant Fundamentalist, Mohsin Hamid uses a variety of literary devices to create the tone of confusion and agitation. By doing so, Mohsin Hamid also develops a theme of change. All these literary features combine to make an enthralling page-turner. One literary device used by the author to create restlessness in the reader is diction. The author intentionally uses words to create discomfort in the reader. â€Å"The twin towers of New York’s World Trade center collapsed. And then I smiled. Yes, despicable as it may sound (4-6). †The fact that the protagonist smiles at America’s misfortune only puts the reader at unease. Furthermore, the author himself suggests that the smile was a despicable reaction. However by making the reader anxious, Mohsin Hamid does only but one thing; lure in the reader to continue reading. This quote also denotes Changez’s change towards America. This is because, although Changez was â€Å"the product of an American university; (he) was earning a lucrative American salary; (he) was infatuated with an American woman (28-30)†his initial reaction to this incident was a smile. The repetition of the word â€Å"American†only creates the motif that Changez has supposedly become American. However after the attack, Changez’s smile only suggests otherwise; thereby creating the theme of change. Another literary device used by Mohsin Hamid to create a sense of confusion is the use of ambiguity through dramatic monologue. â€Å"It is hateful to hear another person gloat over one’s country’s misfortune. But surely you cannot be completely innocent of such feelings yourself (22-24). †Because the American does not speak, the reader is forced to create the personality of the American in his or her own mind. Thus, when the protagonist accuses the American of being guilty of such feelings, the protagonist is essentially accusing the reader of those feelings. The ambiguity of this quote further agitates the reader, yet at the same time forces the reader to continue reading in hope of finding more truths about the reader’s self. One more technique the author uses to develop the theme of change as well as compel the reader to continue reading is foreshadowing. After reflecting on his reaction to the attack, Changez still doesn’t know why he smiled; â€Å"So why did part of me desire to see America harmed? I did not know, then (30-31). †Although Changez is not sure why he smiled at the present time, this quote indicates that later on, the reader will find out why he smiled; another reason requiring the reader to finish reading the book. Simultaneously however, the author also foreshadows a lot of change is about to take place since we already know Changez is back in Pakistan and telling this story to an American at a Pakistani cafe. Literary devices and techniques are abundant in this passage, though they all merge to create a sense of perplexity and irony. At the same time however, they all merge to create the theme of change. Though all this contrariness simply obliges the reader to continue reading the book. Not only will the reader finally understand Changez’s reaction to the attack but the reader will also discover new truths about him or herself as he/she continues to read the book.
Thursday, January 2, 2020
Flannery O Connor - 1912 Words
Curiosity about the possibility and conditions of change in identity has been remarkably intense, in fiction and in psychology, during the last century. In talk about literature, this has led to the development of a crude but useful terminological distinction of two sorts of characterization: static and dynamic. A static character, in this vocabulary, is one that does not undergo important change in the course of the story, remaining essentially the same at the end as he or she was at the beginning. A dynamic character, in contrast, is one that does undergo an important change in the course of the story. More specifically, the changes that we are referring to as being undergone here are not changes in circumstances, but†¦show more content†¦The story reveals his feelings about girls throughout the first half of the story. Then finally in the end he has a complicated conversation with his mother. Another difference between the two stories is the setting. The short story, â€Å"Speaking of Courage†took place after the main character, Norman Bowker, returned to his home after the Vietnam War. While in â€Å"Speaking of Courage†, the story took place a few years following World War I. â€Å"Speaking of Courage†took place in Norman Bowker’s van most of the time as he was making revolutions are the lake. In â€Å"Soldier’s Home†the story was mostly on the front porch of Krebs home and inside his kitchen. The dialogue and structure of the two stories are also different. In â€Å"Soldier’s Home†the first half of the story contained no dialogue and in the second half of the story, it contained a whole conversation to the end of the story. In â€Å"Speaking of Courage†, the whole story was mixed up and contained dialogue throughout the whole story. The two stories contained many differences. Both of the stories had a lot of differences but there were more similarities. The main plot of both stories were different, but the theme of the two short stories were similar. The dialogues contained in both stories were also different and the short story â€Å"Speaking of Courage†contained much more dialogue than â€Å"Soldier’s Home†. The point of view, and the characters in the story’s meaning were similar but the structure and dialogueShow MoreRelatedReview Of Flannery O Connor1228 Words  | 5 Pagesauthor of two novels and multiple classic short stories, Flannery O’Connor is widely regarded as one of the greatest fiction writers in American literature. However, as a Southern and devoutly Christian author in the 1950s, O’Connor was often criticized for the religious content and â€Å"grotesque†characters often incorporated into her works. They were considered too â€Å"brutal†, too â€Å"sarcastic.†(The Habit of Being: Letters of Flannery O Connor). O’Connor begged to differ. Through her essay, â€Å"Some AspectsRead MoreFamily, By Flannery O Connor1803 Words  | 8 Pages In every home, there is a different definition of family and how family should treat each other. Two short stories were read by an author named Flannery O’Connor. â€Å"A Good Man is Hard to Find†. It was about a dysfunctional family who encounters a criminal named â€Å"The Misfit†. The grandmother which is the main character is very judgmental towards others and sometimes her own family at times. This story starts off with a disagreement on where to go for a family trip, but they decide on going to FloridaRead MoreThe, By Flannery O Connor1305 Words  | 6 Pagesdisparity between Blacks and their White counterparts can been seen through not only the South, but also throughout America. Flannery O’Connor, often considered one of the great Southern authors of her time, implemented an artistic writing style which gave her writing a unique Southern gothic appeal that previous novels and stories did not possess. Born in Savannah, Georgia, Flannery Oâ€℠¢Connor grew up in a turbulent time regrading race relations. Living most of her life in predominantly white Georgia, itRead MoreFlannery O Connor s Revelation1307 Words  | 6 PagesFlannery O’Connor believed in the power of religion to give new purpose to life. She saw the fall of the old world, felt the force and presence of God, and her allegorical fictions often portray characters who discover themselves transforming to the Catholic mind. Though her literature does not preach, she uses subtle, thematic undertones and it is apparent that as her characters struggle through violence and pain, divine grace is thrown at them. In her story â€Å"Revelation,†the protagonist, Mrs. TurpinRead MoreA Good Man By Flannery O Connor1235 Words  | 5 PagesA good man is hard to find is a short story written by, Flannery O’Connor, in which the she describes the story of a family going out for a trip to Florida. The grandmother in the family seems to be apparently the main character in the story and the main one who tries to convince the family that is dangerous to go out when there’s a man named the Misfit who is ready to attack and kill anyone in his way. No one in the family seemed to believe her, and yet; they all decided to travel. In their wayRead MoreFlannery O Connor And Religion961 Words  | 4 PagesFlannery O’Connor and Religion in Literature In Flannery O’Connor’s short stories, she often relates them to religion or incorporates a religious form in her stories. O’Connor highlights Christian themes throughout her writing. From her stories familiar to the bible, to death and religious views, its simple to see O’Connor’s religious understanding of God. She ties in her Christian views through her writing, which have a distinct and unique quality. O’Connor often expresses her views of God throughoutRead MoreA Good Man By Flannery O Connor Essay1049 Words  | 5 PagesMicah Gonzales Vernon English 1302.03 31 March 2015 A Good Man is Hard to Find In â€Å"A Good Man is Hard to Find, the author, Flannery O’Connor states many points in this short story. It is to be said that O’Connor is a very religious, catholic, in fact (Vol. 2 pg. 97). O’Connor has stated her religious views in most of her fiction stories. O’Connor was also known for her stories on violence. Readers sometimes find O’Connor’s fictions to be weird with such turning points in her stories. â€Å"A GoodRead MoreA Good Man By Flannery O Connor1239 Words  | 5 Pagesis not mentioned directly in the story. Mystery has it, that she has been explaining her story as if she was the good man or who knows what she could be talking about referring someone or telling someone that good men are hard to find. The author Flannery O’connor wrote this story in 1953, where there must different events occurring causing the relations of racism and different inventions throughout the history that had been shaping America. There fore, a lot of important events that must it beenRead MoreA Good Man By Flannery O Connor946 Words  | 4 PagesIn 1955, Flannery O’ Connor published the short story â€Å"A Good Man Is Hard to Find†which became her best-known short story. Although many appreciated her work it received much criticism for its peculiar character, The Misfit. His callous violent behavior made people uncomfortable with her work describing it as consistently distorted and manipulative. The Misfit’s unsentimental and cruel behavior characterizes true psychological disturbance similar to that of Charles Manson and Jeffrey Dahmer. Read MoreA Good Man By Flannery O Connor1795 Words  | 8 Pagesshort story authors in her era, Flannery O Connor wrote many short stories before her death in 1964. A faithful Catholic, religion was a primary theme in her works; she wrote mostly about southern life with religious themes recurring in her work. One of her most famous stories was the 1 955 short story A Good Man Is Hard to Find. The story depicts the heartless execution of a family by a group of escaped inmates led by their leader known as The Misfit. Though O Connor disagreed, many critics associated
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