Basics of Topological Solutons

Basics of Topological SolutonsResearch into topological solitary waves began in the 1960s, when the fully non delimitatear form of the classical field of study equivalences, were universe thoroughly explored by mathematicians and theoretical physicists. Topological solitons were first examined when the events to these equatings 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 local anestheticised and have finite capacity 4. However, the solitons topological organize 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 de n onenessd by a single integer, N it is a conserved quantity, i.e. it is consta nt unless a collision occurs, and it is cosinet to the total number of particles, which delegacy as N increases, the energy also increases. The conservation of the topological charge is collectable to the topological structure of the tar liquidate 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 equating has the properties of being particle-like, stable, have finite mass and the energy closeness is localised to a finite office of space, with a smooth structure thence this solution is a topological soliton. In addition to solitons existing with topological charge, N, thither 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 fundamental law 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(p) separated charge 1 solitons or they can relax to a classical curtail state of N solitons 1. The energy and length scale 1 (a extra length which is de full termined 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 mate constant. The energy of a topological soliton is able 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 0, the elementary particles mass goes to zero where as the topological solitons mass is finite. The quantization of the wave-like fields which conform to the linearized field equations 1 determines the elementary particle states, where the interactions between the particles are determined by the nonlinear termsA fundamental disco actually in supporting the research of topological solitons is that, apt(p) 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 effrontery 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 s olutions of a Bogomolny equation with a certain 4 charge will all have the corresponding energy value. 1 The solutions of the Bogomolny equations are automatically stable 1 because the fields slander 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 chapterFigure 2.2 appearings a model of an infinite pendulum strip, with the angle being the angle to the downward vertical 3. The energy (with all constraints site to 1) is E = Z 1 2 02 + 1 cos dx (2.1) where 0 = d dx . For the energy density to be finite this requires 2n as x and 2n+ as x , where n Z. To take a chance the number of twists, N, this is simply N = n+ n = () () 2 = 1 2 Z 0 dx ZThis is the equation for the topological charge or the winding number. If we set n = 0 and n+ = 1 then N = 1, th is gives the lowest possible energy for a topological soliton. This is called a crook, and it is the term we use for the one spacial dimension soliton with a single scalar field. The name kink is due to the constrain 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 fall in the calculus of variations to earn a differential equation (x) and then solve it3 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 smaller change to x, i.e. x x + x, and from this calculate the change in the value of the function to leading order in the variaton x 3. f(x) = f(x + x) f(x) = f(x) + xf0 (x) + f(x) = f 0 (x)x + If f 0 (x) 0. If f 0 (x) 0 then we can make f(x) The term 0 equates to zero on the limit point because it must satisfy () = 0 as we cannot change the boundary conditions, so E = Z ( 00 + sin ) dx (2.6) This equation can be minimised minimised foster to the second order nonlinear differential equation, 00 = sin (2.7) The solution of this differential equation with the boundary conditions, () = 0 and () = 2 is the kink. wherefore the kink solution is, (x) = 4 tan1 e xa (2.8) where a is an arbitrary constant. When x = a, this is the military posture of the kink ((a) = ). It is clear to see = 0 is also a solution to the differential equation, however, it does not satisfy the boundary conditions. It is possible to find a set about bound on the kink energy without solving a differential equation 3. prime(prenominal) of all we extremity to rewrite the energy equation (2.1), using the double angle formula the equation becomes, E = 1 2 Z 02 + 4 sin2 2 dx (2.9) By completing the square the equation becomes, E = 1 2 Z 0 2 sin 2 2 + 4 0 sin 2 dx (2.10) Therefore the energy satisfies the inequality,E 2 Z 0 sin 2 dx = 2 Z sin 2 d dxdx = 2 Z 2 0 sin 2 d = 4 cos 2 2 0 = 8 (2.11) In order to obtain the solution which is exactly 8, the term 0 2 sin 2 2 would have to be exactly 0. Therefore the lower bound on the kink energy is calculated by the solution to the equation, 0 = 2 sin 2 (2.12) This is a first order Bogomolny equation. winning this Bogomolny equation and differentiating with compliments to 0 gives, 00 = cos 2 0 = cos 2 2 sin 2 = sin (2.13) This shows that a solution of the Bogomolny equation (2.12) gives the output of the kink solution (2.7). To calculate the energy density , equation (2.1), we need to use the fact that the Bogomolny equation shows that = 02 . From equation (2.8) we have, tan 4 = e xa , whence 1 4 0 sec2 4 = e xaThis equation gives, 0 = 4 e xa 1 + tan2 4 = 4e xa 1 + e 2(xa) = 2 cosh (x a) = 2 (x a) (2.15) Therefore it can be seen that the energy density is given by = 42 (x 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 (a) = 3. Using this interpretation for the energy density, it can be confirm that the energy is equal to the lower bound E = Z dx = 4 Z 2 (x a) dx = 4 tanh (x a) = 8 (2.16) For N 1 i.e. more than one kink, E 8N. In order to obtain the lower bound of N 1 kinks, the kinks must be infinitely apart to hold 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 simply be applied to an infinite domai n. Firstly, the energy terms need to be split according to the powers of the derivative, E = E2 + E0 = Z 1 2 02 dx + Z (1 cos ) dx (2.17) Now consider the spatial rescaling x 7 x = X, so that (x) 7 (X), with dx = dX, d dx = 1 d dX . Under this rescaling the energy becomes E (), E() = Z 1 2 ( 1 d dX ) 2dX + Z (1 cos ) dX = 1 E2 + E0 (2.18) It is now important to see whether E() has a stationary point with respect to , dE () d = 1 2 E2 + E0 = 0 (2.19) if = qE2 E0 , where equals the size of the soliton. From this we can see a stationary point exists, so by Derricks theorem we cannot rule out the happening 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 (x) is a solution then the stationary point corresponds to no rescaling 3, so = 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 = (1, 2) be a two-component unit transmitter, where = 2 = 1. By writing = (sin , cos ), the energy from (2.1) can be rewritten as E = Z ( 1 2 d dx 2 H + H ) dx (2.20) where H = (0, 1). 3 In this new formulation represents the direction of the local magnetic field strength ( restricted to the plane) in a ferromagnetic medium 3 and H represents the constant spikeletground magnetic field which is also restricted to lie within the same plane as . There is only one point in which the systems ground state is equal to zero in terms of , which is = 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 (0, 1) as x . As takes the same value at x = and x = +, then these points can be identified so the target space, which is the real line R, topologically becomes a rophy, S 1 of infinite radius. Therefore we have the mapping S 1 7 S 1 b etween circles, because 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 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 2 Z d1 dx 2 d2 dx 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 ( 1 2 d dx 2 + A 1 ( 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, = k. The kink in this system, also called a domain wall, interpolates between the two zero energy ground states and has boundary conditions k as x and k 15 as x +. 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 authoritative kink example by a change of variables. If we set k = (0, 1) for convenience, and choose A = 1 2 . Setting = sin 2 , cos 2 , then the energy equation becomes E = 1 4 Z 1 2 02 + 1 cos 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 (0, 1) as x are exactly the kink boundary conditions () = 0 and () = 2. 1This 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, quite 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 iii real scalar fields, = (1, 2, 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 Lagrangian density L = 1 4 () ( )with the constraint = 1. For this equation the indices represent the space-time coordinates and take the values 0, 1, 2, and is partial differentiation with respect to X . 2 From (3.1), the Euler-Lagrange equation can be derive d, which is + ( ) = 0 (3.2) Due to the dot reaping in , 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 = 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 revolutions 7 M Where M O(3) is a constant matrix. 1 The sigma in the models name represents the fields (1, 2, ), where 1 and 2 are locally unconstrained 1 and = p 1 2 1 2 2 is dependent on 1 and 2. The energy for the O(3) sigma model is E = 1 4 Z i id 2x (3.3) where i = 1, 2 runs over the spatial indices. In order for the energy to be finite, has to tend to a constant vector at spatial infinity, so without loss of generality we are able to set the boundary condition (0, 0, 1) as x 2 + y 2 . Topologically we have R 2 , which is interpreted as a sphere S 2 via the stereographic riddance. (The sphere itself has finite radius.) Therefore we are considering mapping between spheres S 2 7 S 2 . Just like in our kink example, mapping between spheres means there exists a topological charge, which can be found using N = 1 4 Z (1 - 2) 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 turn 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 7 x = X and y 7 y = Y which would give E () = E. The energy is independent of , therefore any value of is a stationary point since the energy does not change from spatial rescaling.If we integrate the inequality(i ij - j) (i ik - k) 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 bo und E 2 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 i ij - j = 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 = 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, 1) and 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 = (1 + i2) (1 + 3) (3.8) where 1 = W + W 1 + W 2 , 2 = i W W 1 + W 2 , 3 = 1 W 2 1 + W 2 (3.9) As tends to the point (0, 0, 1) then L only intersects X3 = 0 at , therefore the point (0, 0, 1) maps to the point W = . This method of as mansioning each point on the sphere to a point in C 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 let z = x + iy. Using this formation it is possible to rewrite the Lagrangian density, from (3.1) L = 1 4 (1) 2 + (2) 2 + (3) 2 . Firstly we need to partially differentiate 1, 2, 3, giving 1 = W + W 1 + W 2 (W) W + W W 1 + W 2 2 W + W (3.10) 2 = i W W 1 + W 2 (W) W + W W 1 + W 2 2 W WFinally, from simplifying (3.37) we get the equation for the topological charge in the new formulation to be N = 1 4 Z 4 1 + W 2 2 zW zW zW zW d 2x = 1 Z zW 2 zW 2 1 + W 2 2 d 2x (3.38) In this formulation it is clear to see E 2 N, with equality if and only if Bogomolny equation is satisfied W 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 0 as z , this means that N is finite. 3 The simplest solution for the Bogomolny equation would be W = z , where is a real and positive constant. Applying this to the equation (3.9) yields the solution for the N = 1 solution = 2 2 + x 2 + y 2 , 2 2 + x 2 + y 2 , x 2 + y 2 2 2 + x 2 + y 2 (3.40) If we change the blackball 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 = 2 but has N = 1. This soliton is located at thee origin because W(0) = . 3 The N = 1 general solution has 4 real parameters and is given by the Bogomolny solution W = ei z a (3.41) where is the size of the soliton, is the constant angle of rotation in the (1, 2) plane and a C is the position of the soliton in the complex plane, z = x + iy.The O(3) sigma model can be modified to brace 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 facto r. 1 An example of this is the Baby Skyrme model which is given by the Lagrangian L = 1 4 1 8 ( - ) ( - ) m2 2 (1 3) (3.42) where the constraint = 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. 1Topological 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 controlled10, 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 skyrmi ons 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 re search 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 attention 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 godly me for my research project as I wanted to study the topic further. Although this project has been thoroughly enjoyable, it came with contend aspects, due to its complex mathematics in much(prenominal) a specialised subject. As a result of this topic being so specific, I was very limited in the mental imagerys 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 fro m 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 stimulant drug 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 elicit 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.