APPLICATIONS OF FLUID DYNAMICS BASED ON GAUGE FIELD THEORY APPROACH
|
|
- Ερμόλαος Τρικούπη
- 7 χρόνια πριν
- Προβολές:
Transcript
1 APPLICATIONS OF FLUID DYNAMICS BASED ON GAUGE FIELD THEORY APPROACH A thesis submitted to the Fakultas Pasca-Sarjana Universitas Indonesia in partial fulfillment of the requirements for the degree of Master of Science Graduate Program in Pure and Applied Physics KETUT EKO ARI SAPUTRO Universitas Indonesia 2005
2 Lembar Persetujuan Judul tesis : Applications Of Fluid Dynamics Based On Gauge Field Theory Approach Nama : Ketut Eko Ari Saputro NPM : tesis ini telah diperiksa dan disetujui Depok, 29 Juni 2005 Mengesahkan Pembimbing I Pembimbing II Dr. Terry Mart) Dr. L.T. Handoko) Penguji I Penguji II Penguji III Dr. Agus Salam) Dr. Imam) Dr. Anto Sulaksono) Ketua Program Magister Fisika Program Pasca Sarjana FMIPA UI Dr. Dedi Suyanto)
3 Acknowledgements I began working on my thesis soon as I joined the nonlinear physics group at University of Indonesia on July 2004 under supervision of Dr. L.T. Handoko. He was actually one of the reviewers for my BSc theses at Bogor Institute of Agricultural about the phenomenon on nonlinear optics. After that, I spent my time to study gauge field theory it s something new for me). First of all, I would like to acknowledge my primary advisor, Dr. L.T. Handoko for all guidance, patience and critical comments. I am also indebted to Dr Terry Mart, he taught me about the quantum mechanics. I thank D r.anto Sulaksono and Dr. Agus Salam for his intriguing question during my defense. I would like to appreciate our theoretical group, Albert Sulaiman, Fahd, Jani, Ardi Mustafa, Anton, Freddy, etc for so many valuable discussion. The last but not the least, Tri Apriyani, Dhifa al-hakim and Jauza Nur Azizah for their patience and I always love you all. My study was supported by Nurul Fikri learning center. Jakarta, 2005 Ketut eko Ari Saputro iii
4 Abstract Recently, a new approach to deal with the Navier-Stokes equation has been developed. The equation which governs the fluid dynamics is a a non-linear one and then generally unsolvable. In the new approach, the fluid dynamics is described using the relativistic gauge invariant bosonic lagrangian which could reproduce the Navier-Stokes equation as its equation of motion through the Euler-Lagrange principle. Based on the lagrangian we model the fluid dynamics phenomenon as scattering of either three or four fluid bunches represented by the gauge field A µ = φ, A) d 2 v 2 V, d v ) with v is velocity and d is a parameter to adjust the dimension for any potential V. Further we present all relevant Fynman rules and diagrams, and also provide complete calculations for all vertices induced by three and four fluid fields interactions. References: ) iv
5 Contents Acknowledgements Abstract Contents iii iv v 1 Introduction Background Overview Navier-Stokes Equation From Gauge field theory Introduction Gauge invariant bosonic lagrangian Abelian gauge theory Non-abelian gauge theory The NS equation from the gauge field theory Multi Fluid System Using Gauge Field Theory Approach Feyman Diagram for Fluid System Multi Fluids System Interaction of Three Fluids System Dynamics Interaction of Four Fluids System Dynamics Discussion and conclusion 21 A Massive vector polarization 22 v
6 B Three points amplitude calculation 24 C Four points amplitude calculation 33 References 49 vi
7 Chapter 1 Introduction 1.1 Background The understanding of fluid dynamics like hydrodynamic turbulence is an important problem for nature science, from both, theoretical and experimental point of view, and has been investigated intensively over the last century. However a deep and fully comprehension of the problem remains obscure. Over the last years, the investigation of turbulent hydrodynamics has experienced a revival since turbulence has became a very fruitful research field for theorists who study the analogies between turbulence and field theory, critical phenomena and condensed matter physics, renewing the optimism to solve the turbulence problem. The dynamics of turbulent viscous fluid is expressed by the Navier-Stokes NS) equations of motion, which in a vectorial form is a fluid flow described by, v t + v ) v = 1 ρ P ν 2 v, 1.1) where v is the velocity field, P is pressure, ρ is density and ν is the kinematic viscosity. The equation of continuity reduces to the requirement that the velocity field is divergenceless for incompressible fluids, v = ) In this context, the hydrodynamic turbulence has attracted an enormous interest due to the universal characteristics stressed by an incompressible fluid with high Reynolds numbers in the fully developed turbulent regime. The Reynolds number, R = L/U. where L is the 1
8 integral length-scale of the largest eddies and U is a characteristic large-scale velocity), measures the competition between convective and diffusive processes in an incompressible fluid described by the NS equations. In view of this, the incompressible fluid flow assumes high Reynolds numbers when the velocity increases and, consequently, the solution for Eq. 1.1) becomes unstable and the fluid switches to a new regime of a very complex motion with the velocity varying almost randomly and without any noticeable order. To discover the laws describing what exactly is going on with the fluid in this turbulent regime is very important to both theoretical and applied science. Recently, A. Sulaiman and L.T. Handoko proposed an alternative approach to treat fluid dynamics [1]. In the fluid dynamics which is governed by the NS equation we are mostly interested only in how the forces are mediated, and not in the transition of an initial state to another final state as concerned in particle physics. Based on lagrangian density Navier-stokes gauge field theory, they describes the dynamical of interactions multi fluids system like on particle physics. L NS = 1 4 F µνf a aµν gj aµ A a µ. 1.3) In this thesis, we make a further investigation of the physical contents present in that theory, in order to furnish a better understanding of fluid dynamics. 1.2 Overview This theses is organized as follow. The introduction and background of the problem are given in chapter one. Then the theoretical basic of this thesis, i.e.constructing the Navier-stokes equation from gauge field theory, will be described in chapter two. The main part is presented in chapter three. The discussion will be given in chapter four. The last chapter is devoted for summary. 2
9 Chapter 2 Navier-Stokes Equation From Gauge field theory In this chapter we will construct the Navier-Stokes equation from first principle using relativistic bosonic lagrangian which is invariant under local gauge transformations. We show that by defining the bosonic field to represent the dynamic of fluid in a particular form, a general Navier-Stokes equation with conservative forces can be reproduced exactly. It also induces two new forces, one is relevant for rotational fluid, and the other is due to the fluid s current or density. This approach provides an underlying theory to apply the tools in field theory to the problems in fluid dynamics. 2.1 Introduction The Navier-Stokes NS) equation represents a non-linier system with flow s velocity v vx µ ), where x µ is a 4-dimensional space consists of time and spatial spaces, x µ x 0, x i ) = t, r) = t, x, y, z). Note that throughout the paper we use natural unit, i.e. the light velocity c = 1 such that ct = t and then time and spatial spaces have a same dimension. Also we use the relativistic Minkowski) space, with the metric g µν = 1, 1) = 1, 1, 1, 1) that leads to x 2 = x µ x µ = x µ g µν x ν = x 2 0 x 2 = x 2 0 x 2 1 x 2 2 x 2 3. Since the NS equation is derived from the second Newton s law, in principle it should be derived from analytical mechanics using the principle of at-least action on the hamiltonian as already done in several papers [2]. Some papers also relate it with the Maxwell equation [3]. The relation between NS and Maxwell equations is, however, 3
10 not clear and intuitively understandable claim since both equations represent different systems. Moreover, some authors have also formulated the fluid dynamics in lagrangian with gauge symmetries [4]. However, in those previous works the lagrangian has been constructed from continuity equation. Inspired by those pioneering works, we have tried to construct the NS equation from first principle of analytical mechanics, i.e. starting from lagrangian density. Also concerning that the NS equation is a system with 4-dimensional space as mentioned above, it is natural to borrow the methods in the relativistic field theory which treats time and space equally. Then we start with developing a lagrangian for bosonic field and put a contraint such that it is gauge invariant. Taking the bosonic field to have a particular form representing the dynamics of fluid, we derive the equation of motion which reproduces the NS equation. 2.2 Gauge invariant bosonic lagrangian In the relativistic field theory, the lagrangian density) for a bosonic field A is written as [6], L A = µ A) µ A) + m 2 A A2, 2.1) where m A is a coupling constant with mass dimension, and µ / x µ. The bosonic field has the dimension of [A] = 1 in the unit of mass dimension [m] = 1 [x µ ] = 1). The bosonic particles are, in particle physics, interpreted as the particles which are responsible to mediate the forces between interacting fermions, ψ s. Then, one has to first start from the fermionic lagrangian, L ψ = iψγ µ µ ψ) m ψ ψψ, 2.2) where ψ and ψ are the fermion and anti-fermion fields with the dimension [ψ] = [ψ] = 3/2 then [m ψ ] = 1 as above), while γ µ is the Dirac gamma matrices. In order to expand the theory and incorporate some particular interactions, one should impose some symmetries. 4
11 2.2.1 Abelian gauge theory For simplicity, one might introduce the simplest symmetry called U1) abelian) gauge symmetry. The U1) local transformation 1 is just a phase transformation U exp [ iθx)] of the fermions, that is ψ U ψ U ψ. If one requires that the lagrangian in Eq. 2.2) is invariant under this local transformation, i.e. L L = L, a new term coming from replacing the partial derivative with the covariant one µ D µ µ + iea µ, should be added as, L = L ψ eψγ µ ψ)a µ. 2.3) Here the additional field A µ should be a vector boson since [A µ ] = 1 as shown in Eq. 2.1). This field is known as gauge boson and should be transformed under U1) as, A µ U A µ A µ + 1 e µθ), 2.4) to keep the invariance of Eq. 2.3). Here e is a dimensionless coupling constant interpreted as electric charge later on. The existence of a particle requires that there must be a kinetic term of that particle in the lagrangian. In the case of newly introduced A µ above, it is fulfilled by adding the kinetic term using the standard boson lagrangian in Eq. 2.1). However, it is easy to verify that the kinetic term i.e. the first term) in Eq. 2.1) is not invariant under the transformation of Eq. 2.4). Then one must modify the kinetic term to keep the gauge invariance. This can be done by writing down the kinetic term in the form of anti-symmetric strength tensor F µν [5], L A = 1 4 F µνf µν, 2.5) with F µν µ A ν ν A µ, and the factor of 1/4 is just a normalization factor. On the other hand, the mass term the second term) in Eq. 2.1) is automatically discarded in this theory since the quadratic term of A µ is not invariant and then not allowed) under transformation in Eq. 2.4). In particle physics this result justifies the interpretation of gauge boson A µ as photon which is a massless particle. 1 The terminology local here means that the parameter θ is space dependent, i.e. θ θx). One needs also to put a preassumption that the transformation is infinitesimal, i.e. θ 1. 5
12 Finally, imposing the U1) gauge symmetry, one ends up with the relativistic version of electromagnetic theory, known as the quantum electrodynamics QED), L QED = iψγ µ µ ψ) m ψ ψψ ej µ A µ 1 4 F µνf µν, 2.6) where J µ ψγ µ ψ = ρ, J) = J 0, J) is the 4-vector current of fermion which satisfies the continuity equation, µ J µ = 0, using the Dirac equation governs the fermionic field [6] Non-abelian gauge theory One can furthermore generalize this method by introducing a larger symmetry. This so-called) non-abelian transformation can be written as U exp [ it a θ a x)], where T a s are matrices called generators belong to a particular Lie group and satisfy certain commutation relation like [T a, T b ] = if abc T c, where the anti-symmetric constant f abc is called the structure function of the group [7]. For an example, a special-unitary Lie group SUn) has n 2 1 generators, and the subscripts a, b, c run over 1,, n 2 1. Following exactly the same procedure as Sec , one can construct an invariant lagrangian under this transformation. The differences come only from the noncommutativeness of the generators. This induces µ D µ µ + igt a A a µ, and the non-zero f abc modifies Eq. 2.4) and the strength tensor F µν to, A a µ U A a µ A a µ + 1 g µθ a ) + f abc θ b A c µ, 2.7) F a µν µ A a ν νa a µ gf abc A b µ Ac ν, 2.8) where g is a particular coupling constant as before. One then has the non-abelian NA) gauge invariant lagrangian that is analoguous to Eq. 2.6), L NA = iψγ µ µ ψ) m ψ ψψ gj aµ A a µ 1 4 F a µνf aµν, 2.9) while J aµ ψγ µ T a ψ, and this again satisfies the continuity equation µ J aµ = 0 as before. For instance, in the case of SU3) one knows the quantum chromodynamics QCD) to explain the strong interaction by introducing eight gauge bosons called gluons induced by its eight generators. 6
13 2.3 The NS equation from the gauge field theory In the fluid dynamics which is governed by the NS equation we are mostly interested only in how the forces are mediated, and not in the transition of an initial state to another final state as concerned in particle physics. Within this interest, we need to consider only the bosonic terms in the total lagrangian. Assuming that the lagrangian is invariant under certain gauge symmetry explained in the preceding section, we have, L NS = 1 4 F µνf a aµν gj aµ A a µ. 2.10) We put an attention on the current in second term. It should not be considered as the fermionic current as its original version, since we do not introduce any fermion in our system. For time being we must consider J aµ as just a 4-vector current, and it is induced by different mechanism than the internal interaction in the fluid represented by field A a µ. Actually it is not a big deal to even put J aµ = 0 free field lagrangian), or any arbitrary forms as long as the continuity equation µ J aµ = 0 is kept. According to the principle of at-least action for the action S = d 4 x L NS, i.e. δs = 0, one obtains the Euler-Lagrange equation, µ L NS µ A a ν) L NS A a ν = ) Substituting Eq. 2.10) into Eq. 2.11), this leads to the equation of motion EOM) in term of field A a µ, µ ν A a ν ) 2 A a µ + gj aµ = ) If A µ is considered as a field representing a fluid system for each a, then we have multi fluids system governed by a single form of EOM. Inversely, the current can be derived from Eq. 2.12) to get, J aµ = 1 g ν µ A a ν ν A a µ), 2.13) and one can easily verify that the continuity equation is kept. We note that this equation holds for both abelian and non-abelian cases, since the last term in Eq. 2.8) contributes nothing due to its anti-symmetry. Also, this reproduces the relativistic version of the classical electromagnetic density and current of Maxwell. 7
14 The next task is to rewrite the above EOM to the familiar NS equation. Let us first consider a single field A µ. Then the task can be accomplished by defining the field A µ in term of scalar and vector potentials, A µ = A 0, A i ) = φ, A) ) d 2 v 2 V, d v, 2.14) where d is an arbitrary parameter with the dimension [d] = 1 to keep correct dimension for each element of A µ. V = V r) is any potential induced by conservative forces. The condition for a conservative force F is d r F = 0 with the solution F = φ. This means that the potential V must not contain a derivative of spatial space. We are now going to prove that this choice is correct. From Eq. 2.12) it is straightforward to obtain, µ A a ν νa a µ = g dx ν J aµ. 2.15) First we can perform the calculation for µ = ν where we obtain trivial relation, that is J aµ = 0. Non-trivial relation is obtained for µ ν, 0 A i i A 0 = g dx 0 J i = g dx i J ) Different sign in the right hand side merely reflects the Minkowski metric we use. Now we are ready to derive the NS equation. Substituting the 4-vector potensial in Eq. 2.14) into Eq. 2.16), we obtain d 0 v i + i φ = g J i or, d 0 v + φ = g J, 2.17) where J i dx 0 J i = dx i J 0. Using the scalar potential given in Eq. 2.14), we obtain, d v t + d 2 v 2 V = g J. 2.18) By utilizing the identity 1 2 v 2 = v ) v + v v), we arrive at, v t + v ) v = 1 V d v ω + g J, 2.19) d where ω v is the vorticity. This result reproduces a general NS equation with arbitrary conservative forces V ) and some additional forces. This result justifies our choice for the bosonic field in Eq. 2.14). 8
15 Just to mentioned, the potential could represent the already known ones such as, P r)/ρ r) : pressure V r) = Gm/ r : gravitation. 2.20) ν + η) v) : viscosity Here, P, ρ, G, ν + η denote pressure, density, gravitational constant and viscosity as well. We are able to extract a general force of viscosity, Vviscosity = η ) v + ) ) ν 2 v + ν ω using the identity v) = ω + 2 v. This reproduces two terms relevant for both compressible and incompressible fluids, while the last term contributes to the rotational fluid for non-zero ω. This provides a natural reason for causality relation between viscosity and turbulence as stated in the definition of Reynold number, R ν 1. A general NS equation of multi fluids system can finally be obtained by putting the superscript a back to the equation, v a t + va ) v a = 1 V d a v a ω a + g J a, 2.21) d Here the second term in the right hand side is a new force relevant for rotational fluid, while the last term is due to the current or density of fluid. We would like to note an important issue here. One can take arbitrary current forces in the NS equation Eq. 2.21)), as long as the continuity equation is kept, but should set a small number for g. This is very crucial since we will use the perturbation method of field theory to perform any calculation in fluid dynamics starting from the lagrangian in Eq. 2.10) later on. Taking arbitrary and small enough coupling constant g 1) is needed to ensure that our perturbation works well. 9
16 Chapter 3 Multi Fluid System Using Gauge Field Theory Approach This chapter is the main part of the thesis. We describe multi fluids system by modelling such phenomenon as scatterings of multi buch fluid fields. This can be done by defining the amplitude of such interactions using the Navier-stokes lagrangian developed in the preceeding chapter in a similar manner as in the elementary particle physics. 3.1 Feyman Diagram for Fluid System In the fluid dynamics which is governed by the NS equation we are mostly interested only in how the forces are mediated, and not in the transition of an initial state to another final state as concerned in particle physics. Within this interest, we need to consider only the bosonic terms in the total lagrangian. Assuming that the lagrangian is invariant under certain gauge symmetry, we have from Eq. 2.10), L NS = 1 4 F µν a F aµν. 3.1) Expanding and writing all terms explicitely, L NS = 1 4 F µν a F aµν = 1 4 [ µa a ν νa a µ gf abc A b µ Ac ν ][ µ A aν ν A aµ gf amn A mµ A nν ] = 1 [ µ A a ν ν A a 4 µ) µ A aν ν A aµ ) + 2gf abc A b µa c ν µ A aν ν A aµ ) +g 2 f abc f amn A a µa b νa mµ A nν] 3.2) 10
17 Figure 3.1: Feynman rule for three point interaction. From these terms, we have the quadratic term which gives the propagator for the field, i ] [g µν + ζ 1) kµ k ν δ ab. 3.3) k 2 k 2 Further, the cubic and quartic terms give the interactions for three and four buch fluid fields, and, 2gf abc [g µν k 1 k 2 ) ρ + g νρ k 2 k 3 ) µ + g µρ k 3 k 1 ) ν ], 3.4) g 2 [ f abe f cde g µρ g νλ g µλ g νρ) + f ade f bce g µν g ρλ g µρ g νλ) + f ace f bde g µλ g νρ g µν g ρλ)]. 3.5) Clearly, Eqs. 3.3)??) provide the Feynman rules for all allowed interactions in the theory as depicted in Figs. 3.1 and 3.2. Figure 3.2: Feynman rule for four point interaction. 11
18 3.2 Multi Fluids System In this section, we describe the dynamics of multi fluids system using the NS lagrangian. Using the allowed interactions in obtained in the previous section, we can model the hydrodynamics in term of those interactions. This means, we adopt completely the method which is familiar and used widely in the elementary particle physics. This model can be justified physically by considering that the fluid dynamics is a a result of an interaction of fluid fields which is localized at a particular point and ignoring the fluid states before interacting point. Each fluid is regarded as a separated field. Based on this scenario, we can calculate the amplitude of multi fluids system interaction, and then later on interpret and relate it with a real known observable. As usual, we can rewrite the field in term of its polarization vector as follow, ) d A µ = ɛ µ e ik x with ɛ µ = 2 v 2 V, v, 3.6) where k is four-momentum. Of course, the momentum conservation still works, that is Σk i = ) This decomposition yields the completeness relation for the polarization vector as follow, λ ɛ λ µ ɛλ ν = g µν + k ) ) ) 2 µk ν d M 2 2 v 2 V v 2, 3.8) where the sum is over the three polarization states of massive vector fields. The proof is given in App. A. This is important to deal with the multiplication of external fields in the amplitude as done in the next section Interaction of Three Fluids System Dynamics Having the Feynman rule at hand, we can use Eq. 3.4) to calculate the amplitude of 3 fluid fields. As written in detail in App. B we have found the transition amplitude of 3-points to be, im 3 = 2gf abc [g µν k 1 k 2 ) ρ + g νρ k 2 k 3 ) µ + g µρ k 3 k 1 ) ν ] A µ A ν A ρ. 3.9) 12
19 Then, this result can be immediately calculated further to obtain the squared amplitude using Eq. 3.8), M 3 2 = g 2 f abc ) 2 [ g µν k 1 k 2 ) ρ + g νρ k 2 k 3 ) µ + g µρ k 3 k 1 ) ν ] g µα + k 1µk 1α ) g m 2 ν β + k 2νk 2β ) g 1 m 2 ργ + k 3ρk 3γ ) 2 ] [g αβ k 1 k 2 ) γ + g βγ k 2 k 3 ) α + g αγ k 3 k 1 ) β d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v ) 13
20 Expanding all terms, we have [ M 3 2 = g 2 f abc ) 2 k 1 k 2 ) 2 2k 1 k 2 ) k 2 k 3 ) k 2 k 3 ) 2 2k 1 k 2 ) k 3 k 1 ) k 3 k 1 ) 2 2k 2 k 3 ) k 3 k 1 ) + k2 1 k 1 k 2 ) k 1 k 1 k 2 )k 1 k 2 k 3 ) + k 1 k 2 k 3 )) 2 +2 k 1 k 1 k 2 )k 1 k 3 k 1 ) + k2 2k 1 k 2 ) 2 +2 k 2 k 1 k 2 )k 2 k 3 k 1 ) + k 3 k 1 k 2 )) 2 +2 k 3 k 1 k 2 )k 3 k 3 k 1 ) k 1 k 2 ) 2 k 1 k 2 ) 2 k2 2 k 1 k 2 k 3 )) 2 k2 1k 2 k 3 k 1 )) 2 m2 2 k2 1 k 3 k 1 k 2 )) 2 k2 3 k 1 k 2 k 3 )) 2 k 1 k 3 ) 2 k 3 k 1 ) 2 m2 3 k2 2 k 3 k 1 k 2 )) 2 k 2 k 3 ) 2 k 2 k 3 ) 2 k2 3k 2 k 3 k 1 )) 2 m2 3 + k 1 k 2 ) 2 k 3 k 1 k 2 )) 2 + k 2 k 3 ) 2 k 1 k 2 k 3 )) 2 + k2 1 k 3 k 1 ) k 2 k 1 k 2 )k 2 k 2 k 3 ) + k 2 k 3 k 1 )) k 3 k 1 k 2 )k 3 k 2 k 3 ) + k2 3k 3 k 1 ) k 1 k 2 k 3 )k 1 k 3 k 1 ) + k2 2k 2 k 3 ) k 2 k 2 k 3 )k 2 k 3 k 1 ) + k2 3 k 2 k 3 ) 2 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 1 k 2 ) 2 k 1 k 2 k 1 k 1 k 2 )k 2 k 3 k 1 ) 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 3 k 1 ) m2 2 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 3 k 1 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 3 k 1 ) m2 3 2 k 2 k 3 k 2 k 2 k 3 )k 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 2 k 3 ) m k 3 k 2 k 3 )k 3 k 3 k 1 ) + 2 k 1 k 2 k 2 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) + 2 k 1 k 2 k 1 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) 14
21 + k 1 k 3 ) 2 k 2 k 3 k 1 )) k 1 k 3 k 2 k 3 k 1 k 2 k 3 )k 2 k 3 k 1 ) ] m2 2 m2 3 m2 2 m2 3 d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 Taking into account the momentum conservation, Eq. 3.7), we obtain several kinematic relations, 3.11) k 1 + k 2 + k 3 = 0, 3.12) k i k i = m 2 i = ρ2 i V 2 ; k 1 k 2 k 3 ) 2 = ρ 2 3 ρ2 2 )V 2 ; k 1 k 2 ) 2 = 2ρ ρ2 2 ρ2 3 )V 2 ; k 2 k 3 k 1 ) 2 = ρ 2 1 ρ2 3 )V 2 ; k 2 k 3 ) 2 = 2ρ ρ2 3 ρ2 1 )V 2 ; k 3 k 1 k 2 ) 2 = ρ 2 2 ρ2 1 )V 2 ; k 3 k 1 ) 2 = 2ρ ρ2 3 ρ2 2 )V 2 ; k 1 k 2 = 1 2 ρ2 3 ρ2 1 ρ2 2 )V 2 ; k 1 k 1 k 2 ) 2 = 1 2 3ρ2 1 + ρ2 2 ρ2 3 )V 2 ; k 1 k 3 = 1 2 ρ2 2 ρ2 1 ρ2 3 )V 2 ; k 2 k 2 k 3 ) 2 = 1 2 3ρ2 2 + ρ2 3 ρ2 1 )V 2 ; k 2 k 3 = 1 2 ρ2 1 ρ2 2 ρ2 3 )V 2 ); k 3 k 3 k 1 ) 2 = 1 2 3ρ2 3 + ρ2 1 ρ2 2 )V 2 ; k 1 k 3 k 1 ) 2 = 1 2 ρ2 2 ρ2 3 3ρ2 1 )V 2 ; k 2 k 1 k 2 ) 2 = 1 2 ρ1 3 ρ 2 1 3ρ 2 2)V 2 ; k 3 k 2 k 3 ) 2 = 1 2 ρ2 1 ρ 2 2 3ρ 2 3)V 2 ; Interaction of Four Fluids System Dynamics Following the same procedure as done in Sec , we obtain im 4 = g 2 [ f abe f cde g µρ g νλ g µλ g νρ) ) f ade f bce g µν g ρλ g µρ g νλ) + f ace f bde g µλ g νρ g µν g ρλ)] A µ A ν A ρ A λ, 3.14) 15
22 for 4-point amplitude. Then, the squared amplitude becomes, M 4 2 = g 4 {[ f abe f cde g µρ g νλ g µλ g νρ) + f ade f bce g µν g ρλ g µρ g νλ) + f ace f bde g µλ g νρ g µν g ρλ)] g µα + k ) 1µk 1α g m 2 ν β + k ) 2νk 2β 1 g ργ + k ) 3ρk 3γ g m 2 λ σ + k ) 4λk 4σ 3 m 2 4 [ f abe f cde g αγ g βσ g ασ g βδ) + f ade f bce g αβ g γσ g αγ g βσ) + f ace f bde g ασ g βγ g αβ g γσ)]} d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 d ) 4 2 v V 4 ) 2 v ) 16
23 Expanding the result further, we get M 4 2 = g 4 [ f abe f cde f abe f cde 2 k 1 k 1 k 2 k 2 k 1 k 2 ) 2 + k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m 2 4 m2 2 + k 2 k 2 k 3 k 3 k 2 k 3 ) k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m k 1 k 1 k 3 k 3 k 1 k 3 ) 2 m k 1 k 2 k 1 k 3 k 2 k 3 k 1 k 1 k 2 k 3 ) 2 + k 2 k 2 k 1 k 3 ) 2 ) m2 2 m k 1 k 2 k 1 k 4 k 2 k 4 k 1 k 1 k 2 k 4 ) 2 + k 2 k 2 k 1 k 4 ) 2 ) m2 2 m k 1 k 3 k 1 k 4 k 3 k 4 k 3 k 3 k 1 k 4 ) 2 + k 4 k 4 k 1 k 3 ) 2 ) m k 2 k 3 k 2 k 4 k 3 k 4 k 3 k 3 k 2 k 4 ) 2 + k 4 k 4 k 2 k 3 ) 2 ) m2 3 m2 4 + k ) 1 k 3 k 2 k 4 ) 2 + k 1 k 4 k 2 k 3 ) 2 2k 1 k 3 k 2 k 4 )k 1 k 4 k 2 k 3 ) m2 2 m2 3 m2 4 +f ade f bce f ade f bce k1 k 1 k 2 k 2 k 1 k 2 ) 2 + k 1 k 1 k 3 k 3 k 1 k 3 ) 2 +2 k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m k 2 k 2 k 3 k 3 k 2 k 3 ) 2 + k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m k 1 k 2 k 1 k 3 k 2 k 3 k 2 k 2 k 1 k 3 ) 2 + k 3 k 3 k 1 k 2 ) 2 ) + 2k 1 k 2 k 1 k 4 k 2 k 4 k 1 k 1 k 2 k 4 ) 2 + k 4 k 4 k 1 k 2 ) 2 ) m k 1 k 3 k 1 k 4 k 3 k 4 k 1 k 1 k 3 k 4 ) 2 + k 4 k 4 k 1 k 3 ) 2 ) m2 3 m k 2 k 3 k 2 k 4 k 3 k 4 k 3 k 3 k 2 k 4 ) 2 + k 2 k 2 k 3 k 4 ) 2 ) m k ) 1 k 2 k 3 k 4 ) 2 + k 1 k 3 k 2 k 4 ) 2 2k 1 k 2 k 3 k 4 )k 1 k 2 k 2 k 4 ) m
24 +f ace f bde f ace f bce k1 k 1 k 2 k 2 k 1 k 2 ) k 1 k 1 k 3 k 3 k 1 k 3 ) 2 m2 2 m2 3 + k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m k 2 k 2 k 3 k 3 k 2 k 3 ) 2 + k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m k 1 k 2 k 1 k 3 k 2 k 3 k 1 k 1 k 2 k 3 ) 2 + k 3 k 3 k 1 k 2 ) 2 ) m2 2 m k 1 k 2 k 1 k 4 k 2 k 4 k 2 k 2 k 1 k 4 ) 2 + k 4 k 4 k 1 k 2 ) 2 ) m2 2 m k 1 k 3 k 1 k 4 k 3 k 4 k 3 k 3 k 1 k 4 ) 2 + k 1 k 1 k 3 k 4 ) 2 ) m k 2 k 3 k 2 k 4 k 3 k 4 k 2 k 2 k 2 k 4 ) 2 + k 4 k 4 k 2 k 3 ) 2 ) m2 3 m2 4 + k ) 1 k 2 k 3 k 4 ) 2 + k 1 k 4 k 2 k 3 ) 2 2k 1 k 2 k 3 k 4 )k 1 k 4 k 2 k 3 ) m2 2 m2 3 m2 4 +2f abe f cde f ade f bce k1 k 2 ) 2 k 1 k 1 k 2 k 2 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m k 2 k 3 ) 2 k 2 k 2 k 3 k 3 + k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m k 2 k 2 k 1 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 m2 2 m2 3 + k 1 k 1 k 2 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 4 k 4 k 1 k 3 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 3 k 3 k 2 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m2 3 m2 4 + k ) 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m m
25 +2f abe f cde f ace f bde k1 k 2 ) 2 k 1 k 1 k 2 k 2 + k 1 k 3 ) 2 k 1 k 1 k 3 k 3 m2 2 m2 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m k 1 k 1 k 2 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 + k 2 k 2 k 1 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m2 2 m2 4 + k 3 k 3 k 1 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m2 3 m2 4 + k 4 k 4 k 2 k 3 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m k ) 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 ) m2 2 m2 3 m2 4 +2f ade f bce f ace f bde k1 k 3 ) 2 k 1 k 1 k 3 k 3 m2 3 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m2 4 + k 2 k 3 ) 2 k 2 k 2 k 3 k 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m k 3 k 3 k 1 k 2 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 + k 4 k 4 k 1 k 2 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m2 2 m2 4 + k 1 k 1 k 3 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 2 k 2 k 3 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m k )] 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m2 2 m2 3 m2 4 d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 d ) 4 2 v V 4 ) 2 v ) As a consequence of the momentum conservation, k 1 + k 2 + k 3 + k 4 = 0, 3.17) 19
26 the following relations hold, k i k i = m 2 i = ρ2 i V 2, k 1 k 2 = 1 4 ρ 1v 1 ρ 2 v 2 v 1 v 2 4cosθ) V 2, k 1 k 3 = 1 4 ρ 1v 1 ρ 3 v 3 v 1 v 3 4cosα) V 2, k 1 k 4 = 1 4 4ρ2 1 + ρ 1v 1 ρ 2 v 2 v 1 v 2 4cosθ) + ρ 1 v 1 ρ 3 v 3 v 1 v 3 4cosα))V 2, k 2 k 3 = 1 2ρ ρ ρ2 3 ρ2 4 ) + ρ 1v 1 ρ 2 v 2 v 1 v 2 4cosθ) +ρ 1 v 1 ρ 3 v 3 v 1 v 3 4cosα)) V 2, k 2 k 4 = 1 4 2ρ2 1 + ρ2 3 ρ2 2 ρ2 4 ) + ρ 1v 1 ρ 3 v 3 v 1 v 3 4cosα))V 2, k 3 k 4 = 1 4 2ρ2 1 + ρ2 2 ρ2 3 ρ2 4 ) + ρ 1v 1 ρ 2 v 2 v 1 v 2 4cosθ))V ) 20
27 Chapter 4 Discussion and conclusion In this chapter we will describe an idea about application of the theory, i.e.the interactions of three or four fluids. Application of this theory can be used to show describe the observable in a meeting point of river and sea. This means we can model, for instance the turbuilence, at the point as one fluid coming from the river and the other two fluids fields scattered away to the sea. The application of 4-point scattering is much wider [9]. This can be used to deal with turbulence in any fluid system, nebula for cosmology, dynamics of nano-crystal and so on. In all cases we can model all appropriate observables as a result of scattering of 4 homogeneous fields coming into a point. Finally we have shown that interaction of multi fluids system which is localized on one fixed point could be counted based on gauge field theory. Further, there is a special characteristic in the 3- and 4-point interactions. In the case of 3-point there is no angle dependence at all, while in 4-point case the amplitude depends on two angles of the directions of external fields. 21
28 Appendix A Massive vector polarization In the case of massive vector particle, the wave equation for a particle of mass M for a bosonic field A µ is written as [6], g νµ 2 + M 2 ) ν µ )A µ = 0, A.1) We can determine the inverse of the momentum space operator by solving g νµ k 2 + M 2 ) + k ν k µ ) 1 = δ µ λ Agνλ + Bk λ k ν ), A.2) for A and B. The propagator, which is the quantity in brackets on the right-hand side of A.2) multiplied by i, is found to be, ig νµ + k ν k µ /M 2 ) k 2 M 2. A.3) We can show that the numerator is the sum over the spin states of the massive particle when it is taken on-shell, i.e.k 2 = M 2. We first take the divergence, ν of A.1), which makes cancellation to find M 2 µ A µ = 0 that is, µ A µ = 0, A.4) Thus, for a massive vector particle, we have no choice but to take µ A µ = 0. We should remark that it is not a gauge condition. As a consequence, the equation is reduced to be, with free-particle solutions, A µ = ɛ µ e ik x with ɛ µ = 2 + M 2 )A µ = 0, A.5) 22 ) d 2 v 2 V, v. A.6)
29 The condition in Eq. A.1) demands, k µ ɛ µ = 0, A.7) which reduces the number of independent polarization vector from four to three in a covariant fashion. For a vector particle of mass M, energy E and momentum k along the z axis, the states of helicities λ can be described by polarization vectors ) ɛ λ=±1 0, 1, ±i, 0) d = 2 2 v 2 V, v, ) ɛ λ=0 k, 0, 0, E) d = M 2 v 2 V, v. A.8) Therefore the completeness relation is, λ ɛ λ µ ɛλ ν = g µν + k ) ) ) 2 µk ν d M 2 2 v 2 V v 2 where the sum is over the three polarization states of massive vector particle., A.9) 23
30 Appendix B Three points amplitude calculation M 3 2 = g 2 f abc ) 2 [ g µν k 1 k 2 ) ρ + g νρ k 2 k 3 ) µ + g µρ k 3 k 1 ) ν ] g µα + k 1µk 1α ) g m 2 ν β + k 2νk 2β ) g 1 m 2 ργ + k 3ρk 3γ ) 2 ] [g αβ k 1 k 2 ) γ + g βγ k 2 k 3 ) α + g αγ k 3 k 1 ) β d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 = g 2 f abc ) 2 [left 1) 2) 3) right scalar] = g 2 f abc ) 2 [left 1) 2) 3) A + B + C) scalar] B.1) ] left = [g µν k 1 k 2 ) ρ + g νρ k 2 k 3 ) µ + g µρ k 3 k 1 ) ν B.2) 1) = g µα + k 1µk 1α ); 2) = g νβ + k 2νk 2β ); 3) = g ργ + k 3ρk 3γ ) B.3) ] right = A + B + C) = [g αβ k 1 k 2 ) γ + g βγ k 2 k 3 ) α + g αγ k 3 k 1 ) β B.4) scalar = d 1 2 v 1 2 +V 1 ) 2 v 1 2 ) d 2 2 v 2 2 +V 2 ) 2 v 2 2 ) d 3 2 v 3 2 +V 3 ) 2 v 3 2 ) B.5) 24
31 left 1) = [ gαk ν 1 k 2 ) ρ g νρ k 2 k 3 ) α gαk ρ 3 k 1 ) ν + k 1 ν k 1α k 1 k 2 ) ρ + gνρ k 1 k 2 k 3 )k 1α + kρ 1 k 1αk 3 k 1 ) ν ] B.6) left 1) 2) = [ g αβ k 1 k 2 ) ρ + g ρ β k 2 k 3 ) α + g ρ αk 3 k 1 ) β k 1βk 1α k 1 k 2 ) ρ k 2αk 2β k 1 k 2 ) ρ + k 1 k 2 k 1α k 2β k 1 k 2 ) ρ + k 2 k 3 k 1 )k 1α k 2β k ρ 1 gρ β k 1 k 2 k 3 )k 1α kρ 2k 2β k 2 k 3 ) α kρ 1 k 1αk 3 k 1 ) β gρ αk 2 k 3 k 1 )k 2β + k 1 k 2 k 3 )k 1α k 2β k ρ 2 ] B.7) left 1) 2) 3) = [ g αβ k 1 k 2 ) γ g βγ k 2 k 3 ) α g αγ k 3 k 1 ) β + k 1βk 1α k 1 k 2 ) γ + k 2αk 2β k 1 k 2 ) γ k 1 k 2 k 1α k 2β k 1 k 2 ) γ + k 3 k 1 k 2 )k 3γ g αβ k 3 k 1 k 2 )k 1α k 1β k 3γ k 3 k 1 k 2 )k 2α k 2β k 3γ + g βγk 1 k 2 k 3 )k 1α + k 2βk 2γ k 2 k 3 ) α + k 1 k 2 k 3 k 1 k 2 )k 1α k 2β k 3γ + k 1 k 3 k 2 k 3 k 1 )k 1α k 2β k 3γ m2 2 m2 3 + k 1αk 1γ k 3 k 1 ) β + g αγk 2 k 3 k 1 )k 2β k 1 k 2 k 3 )k 1α k 2β k 2γ + k 3βk 3γ k 2 k 3 ) α k 1 k 2 k 3 )k 1α k 3β k 3γ k 2 k 3 k 2 k 3 ) α k 2β k 3γ k 2 k 3 k 1 )k 1α k 2β k 1γ + k 3αk 3γ k 3 k 1 ) β + k 2 k 3 k 1 k 2 k 3 )k 1α k 2β k 3γ ] 25 k 1 k 3 k 1α k 3 k 1 ) β k 3γ k 2 k 3 k 1 )k 3α k 2β k 3γ B.8)
32 left 1) 2) 3) A) = [ k 1 k 2 ) 2 k 1 k 2 ) k 2 k 3 ) k 1 k 2 ) k 3 k 1 ) + k2 1k 1 k 2 ) 2 + k2 2 k 1 k 2 ) 2 + k 3 k 1 k 2 )) 2 k 1 k 2 ) 2 k 1 k 2 ) 2 + k 1 k 1 k 2 )k 1 k 2 k 3 ) + k 2 k 1 k 2 )k 2 k 2 k 3 ) + k 3 k 1 k 2 )k 3 k 2 k 3 ) k 1 k 2 k 2 k 3 k 1 )k 1 k 1 k 2 ) k 1 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) k2 2k 3 k 1 k 2 ) 2 m2 3 k 2 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) k 1 k 2 k 1 k 2 k 3 )k 2 k 1 k 2 ) + k 1 k 1 k 2 )k 1 k 3 k 1 ) + k 2 k 1 k 2 )k 2 k 3 k 1 ) k2 1 k 3 k 1 k 2 )) 2 + k 3 k 1 k 2 )k 3 k 3 k 1 ) k 1 k 3 k 1 k 3 k 1 )k 3 k 1 k 2 ) k 2 k 3 k 2 k 2 k 3 )k 3 k 1 k 2 ) m2 3 + k 1 k 2 ) 2 k 3 k 1 k 2 )) 2 + k 1 k 2 k 2 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) + k 1 k 2 k 1 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) ] B.9) 26
33 left 1) 2) 3) B) = [ k 1 k 2 ) k 2 k 3 ) k 2 k 3 ) 2 k 3 k 1 ) k 2 k 3 ) + k 1 k 1 k 2 )k 1 k 2 k 3 ) + k 2 k 1 k 2 )k 2 k 2 k 3 ) + k 3 k 1 k 2 )k 3 k 2 k 3 ) k 1 k 2 k 1 k 2 k 3 )k 2 k 1 k 2 ) k 1 k 2 k 1 k 2 k 3 )k 2 k 3 k 1 ) k2 3 k 1 k 2 k 3 )) 2 k 2 k 3 k 2 k 2 k 3 )k 3 k 1 k 2 ) m2 3 k 2 k 3 k 2 k 3 k 1 )k 3 k 2 k 3 ) m2 3 + k 1 k 2 k 3 )) 2 k 2 k 3 ) 2 + k 1 k 2 k 3 )) 2 + k2 2 k 2 k 2 k 3 )) 2 + k2 3 k 2 k 3 ) 2 + k 1 k 2 k 3 )k 1 k 3 k 1 ) + k 2 k 2 k 3 )k 2 k 3 k 1 ) + k 3 k 2 k 3 )k 3 k 3 k 1 ) k2 2k 1 k 2 k 3 )) 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) k 1 k 3 k 1 k 2 k 3 )k 3 k 3 k 1 ) k 2 k 3 ) 2 k 2 k 3 ) 2 m2 3 + k 1 k 2 k 2 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) m2 2 m2 3 + k 1 k 3 k 2 k 3 k 1 k 2 k 3 k 2 k 3 k 1 )) ] B.10) 27
34 left 1) 2) 3) C) = [ k 1 k 2 ) k 3 k 1 ) k 2 k 3 ) k 3 k 1 ) k 3 k 1 ) 2 + k 1 k 1 k 2 )k 1 k 3 k 1 ) + k 2 k 1 k 2 )k 2 k 3 k 1 ) + k 3 k 1 k 2 )k 3 k 3 k 1 ) k 1 k 2 k 1 k 1 k 2 )k 2 k 3 k 1 ) m2 2 k2 1 k 2 k 3 k 1 )) 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 3 k 1 ) k 2 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) k2 3k 2 k 3 k 1 ) 2 m2 3 + k 1 k 2 k 3 )k 1 k 3 k 1 ) + k 2 k 2 k 3 )k 2 k 3 k 1 ) + k2 1k 3 k 1 ) 2 + k 2 k 3 k 1 )) 2 + k 3 k 2 k 3 )k 3 k 3 k 1 ) k 1 k 2 k 1 k 2 k 3 )k 2 k 3 k 1 ) m2 2 k 1 k 3 k 1 k 3 k 1 )k 3 k 1 k 2 ) + k 1 k 3 k 2 k 3 k 1 k 2 k 3 )k 2 k 3 k 1 ) k 1 k 3 ) 2 k 3 k 1 ) 2 + k2 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 2 k 3 ) + k 1 k 2 k 1 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) m2 2 m2 3 + k 1 k 3 ) 2 k 2 k 3 k 1 )) 2 ] B.11) 28
35 left 1) 2) 3) A + B + C) = left 1) 2) 3) right = [ k 1 k 2 ) 2 2k 1 k 2 ) k 2 k 3 ) k 2 k 3 ) 2 2k 1 k 2 ) k 3 k 1 ) k 3 k 1 ) 2 2k 2 k 3 ) k 3 k 1 ) + k2 1 k 1 k 2 ) k 1 k 1 k 2 )k 1 k 2 k 3 ) + k 1 k 2 k 3 )) 2 +2 k 1 k 1 k 2 )k 1 k 3 k 1 ) + k2 2k 1 k 2 ) 2 +2 k 2 k 1 k 2 )k 2 k 3 k 1 ) + k 3 k 1 k 2 )) 2 +2 k 3 k 1 k 2 )k 3 k 3 k 1 ) k 1 k 2 ) 2 k 1 k 2 ) 2 k2 2 k 1 k 2 k 3 )) 2 k2 1k 2 k 3 k 1 )) 2 m2 2 k2 1 k 3 k 1 k 2 )) 2 k2 3 k 1 k 2 k 3 )) 2 k 1 k 3 ) 2 k 3 k 1 ) 2 m2 3 k2 2k 3 k 1 k 2 )) 2 k 2 k 3 ) 2 k 2 k 3 ) 2 k2 3k 2 k 3 k 1 )) 2 m2 3 + k 1 k 2 ) 2 k 3 k 1 k 2 )) 2 + k2 1 k 3 k 1 ) k 2 k 1 k 2 )k 2 k 2 k 3 ) + k 2 k 3 k 1 )) k 3 k 1 k 2 )k 3 k 2 k 3 ) + k2 3k 3 k 1 ) k 1 k 2 k 3 )k 1 k 3 k 1 ) + k2 2k 2 k 3 ) k 2 k 2 k 3 )k 2 k 3 k 1 ) + k2 3 k 2 k 3 ) 2 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 1 k 2 ) 2 k 1 k 2 k 1 k 1 k 2 )k 2 k 3 k 1 ) 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 3 k 1 ) m2 2 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 3 k 1 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 3 k 1 ) m2 3 2 k 2 k 3 k 2 k 2 k 3 )k 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 2 k 3 ) m k 3 k 2 k 3 )k 3 k 3 k 1 ) + 2 k 1 k 2 k 2 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) 29
36 + k 2 k 3 ) 2 k 1 k 2 k 3 )) k 1 k 2 k 1 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) m2 2 m2 3 m2 2 m2 3 + k 1 k 3 ) 2 k 2 k 3 k 1 )) k 1 k 3 k 2 k 3 k 1 k 2 k 3 )k 2 k 3 k 1 ) ] B.12) 30
37 M 3 2 = g 2 f abc ) 2 [left 1) 2) 3) right scalar] = g 2 f abc ) 2 [ k 1 k 2 ) 2 2k 1 k 2 ) k 2 k 3 ) k 2 k 3 ) 2 2k 1 k 2 ) k 3 k 1 ) k 3 k 1 ) 2 2k 2 k 3 ) k 3 k 1 ) + k2 1k 1 k 2 ) k 1 k 1 k 2 )k 1 k 2 k 3 ) + k 1 k 2 k 3 )) 2 +2 k 1 k 1 k 2 )k 1 k 3 k 1 ) + k2 2 k 1 k 2 ) 2 +2 k 2 k 1 k 2 )k 2 k 3 k 1 ) + k 3 k 1 k 2 )) 2 +2 k 3 k 1 k 2 )k 3 k 3 k 1 ) k 1 k 2 ) 2 k 1 k 2 ) 2 k2 2 k 1 k 2 k 3 )) 2 k2 1 k 2 k 3 k 1 )) 2 m2 2 k2 1k 3 k 1 k 2 )) 2 k2 3 k 1 k 2 k 3 )) 2 k 1 k 3 ) 2 k 3 k 1 ) 2 m2 3 k2 2k 3 k 1 k 2 )) 2 m2 3 k 2 k 3 ) 2 k 2 k 3 ) 2 k2 3 k 2 k 3 k 1 )) 2 m2 3 + k 1 k 2 ) 2 k 3 k 1 k 2 )) 2 m2 2 m2 3 + k 2 k 3 ) 2 k 1 k 2 k 3 )) 2 + k2 1 k 3 k 1 ) k 2 k 1 k 2 )k 2 k 2 k 3 ) + k 2 k 3 k 1 )) k 3 k 1 k 2 )k 3 k 2 k 3 ) + k2 3 k 3 k 1 ) k 1 k 2 k 3 )k 1 k 3 k 1 ) + k2 2 k 2 k 3 ) k 2 k 2 k 3 )k 2 k 3 k 1 ) + k2 3 k 2 k 3 ) 2 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 1 k 2 ) 2 k 1 k 2 k 1 k 1 k 2 )k 2 k 3 k 1 ) 2 k 1 k 2 k 1 k 2 k 3 )k 2 k 3 k 1 ) m2 2 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 3 k 1 )k 3 k 1 k 2 ) 2 k 1 k 3 k 1 k 2 k 3 )k 3 k 3 k 1 ) m2 3 2 k 2 k 3 k 2 k 2 k 3 )k 3 k 1 k 2 ) m2 3 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) 2 k 2 k 3 k 2 k 3 k 1 )k 3 k 2 k 3 ) m k 3 k 2 k 3 )k 3 k 3 k 1 ) + 2 k 1 k 2 k 2 k 3 k 1 k 2 k 3 )k 3 k 1 k 2 ) m2 2 m k 1 k 2 k 1 k 3 k 2 k 3 k 1 )k 3 k 1 k 2 ) 31
38 + k 1 k 3 ) 2 k 2 k 3 k 1 )) k 1 k 3 k 2 k 3 k 1 k 2 k 3 )k 2 k 3 k 1 ) ] m2 2 m2 3 m2 2 m2 3 d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 B.13) k 1 + k 2 + k 3 = 0 B.14) k i k i = m 2 i = ρ2 i V 2 ; k 1 k 2 k 3 ) 2 = ρ 2 3 ρ 2 2)V 2 ; k 1 k 2 ) 2 = 2ρ ρ 2 2 ρ 2 3)V 2 ; k 2 k 3 k 1 ) 2 = ρ 2 1 ρ 2 3)V 2 ; k 2 k 3 ) 2 = 2ρ ρ 2 3 ρ 2 1)V 2 ; k 3 k 1 k 2 ) 2 = ρ 2 2 ρ 2 1)V 2 ; k 3 k 1 ) 2 = 2ρ ρ 2 3 ρ 2 2)V 2 ; k 1 k 2 = 1 2 ρ2 3 ρ 2 1 ρ 2 2)V 2 ; k 1 k 1 k 2 ) 2 = 1 2 3ρ2 1 + ρ 2 2 ρ 2 3)V 2 ; k 1 k 3 = 1 2 ρ2 2 ρ 2 1 ρ 2 3)V 2 ; k 2 k 2 k 3 ) 2 = 1 2 3ρ2 2 + ρ 2 3 ρ 2 1)V 2 ; k 2 k 3 = 1 2 ρ2 1 ρ2 2 ρ2 3 )V 2 ); k 3 k 3 k 1 ) 2 = 1 2 3ρ2 3 + ρ2 1 ρ2 2 )V 2 ; k 1 k 3 k 1 ) 2 = 1 2 ρ2 2 ρ2 3 3ρ2 1 )V 2 ; k 2 k 1 k 2 ) 2 = 1 2 ρ1 3 ρ2 1 3ρ2 2 )V 2 ; k 3 k 2 k 3 ) 2 = 1 2 ρ2 1 ρ2 2 3ρ2 3 )V 2 ; B.15) 32
39 Appendix C Four points amplitude calculation M 4 2 = g 4 {[ f abe f cde g µρ g νλ g µλ g νρ) + f ade f bce g µν g ρλ g µρ g νλ) + f ace f bde g µλ g νρ g µν g ρλ)] g µα + k ) 1µk 1α g m 2 νβ + k ) 2νk 2β 1 g ργ + k ) 3ρk 3γ g m 2 λ σ + k ) 4λk 4σ 3 m 2 4 [ f abe f cde g αγ g βσ g ασ g βδ) + f ade f bce g αβ g γσ g αγ g βσ) + f ace f bde g ασ g βγ g αβ g γσ)]} d ) 1 2 v V 1 ) 2 v 1 2 d ) 2 2 v V 2 ) 2 v 2 2 d ) 3 2 v V 3 ) 2 v 3 2 d ) 4 2 v V 4 ) 2 v 4 2 = g 4 [left 1) 2) 3) 4) right scalar] = g 4 [left 1) 2) 3) 4) A + B + C) scalar] C.1) left = [ f abe f cde g µρ g νλ g µλ g νρ) + f ade f bce g µν g ρλ g µρ g νλ) + f ace f bde g µλ g νρ g µν g ρλ)] C.2) 33
40 1) = g µα + k 1µk 1α ); 2) = g m 2 νβ + k 2νk 2β ); 1 3) = g ργ + k 3ρk 3γ ); 4) = g λ σ + k 4λk 4σ ); C.3) m 2 4 right = A + B + C) = [ f abe f cde g αβ g γσ g αγ g βσ) + f ade f bce g αβ g γσ g αγ g βσ) +f ace f bde g ασ g βγ g αβ g γσ)] C.4) scalar = d 1 ) 2 v V 1 ) 2 v 1 2 d 3 2 v V 3 ) 2 v 3 2 d 2 ) ) 2 v V 2 ) 2 v 2 2 d 4 2 v V 4 ) 2 v 4 2 ) C.5) left 1) = [ f abe f cde g νλ gα ρ + gνρ gα λ + kρ 1k 1α gνλ g νρ kλ 1 k ) 1α ) +f ade f bce g ρλ g ν α + g νλ g ρ α + g ρλ kν 1 k 1α g νλ kρ 1 k 1α +f ace f bde g νρ gα λ + gρλ gα ν + kλ gνρ 1 k 1α g ρλ kν 1k 1α left 1) 2) = [f abe f cde gα ρ gλ β gρ β gλ α k1k ρ 1α gλ β + g ρ k1 λ k 1α m 2 β 1 gα ρ k2 λk 2β + gα λ k ρ 2 k 2β + kρ 1 kλ 2 k 1αk 2β m2 2 +f ade f bce g αβ g ρλ gα ρ gλ β k 1βk 1α gρλ + g m 2 β λ 1 g ρλ k 2αk 2β + gα ρ k2 λk 2β + g ρλ k 1 k 2 k 1α k 2β kλ 1 kρ 2 k ) 1αk 2β m2 2 k ρ 1k 1α f ace f g bde ρ β gλ α g αβg ρλ g ρ k1 λ k 1α β + g ρλ k 1βk 1α gα λ k ρ 2 k 2β + g ρλ k 2αk 2β + kλ 1 kρ 2 k 1αk 2β )] kρ 1 kλ 2 k 1αk 2β g ρλ k 1 k 2 k 1α k 2β ) )] C.6) C.7) 34
41 left 1) 2) 3) = [f abe f cde gβg λ αγ + g βγ gα λ + gβ λ k 1α k 1γ g βγ k λ 1 k 1α +g αγ k λ 2 k 2β gα λ k 2β k 2γ kλ 2 k 1αk 2β k 1γ + kλ 1 k 1αk 2β k 2γ +gβ λ k 3α k 3γ kλ 2 k 3αk 2β k 3γ gα λ k 3β k 3γ gβ λ k 1 k 3 k 1α k 3γ m2 3 + gα λ k 2 k 3 k 2β k 3γ + kλ 1 k 1α k 3β k 3γ m2 3 k 2 k 3 k1 λk ) 1αk 2β k 3γ + k 1 k 3 k2 λk 1αk 2β k 3γ +f ade f bce g λ γ g αβ + g λ β g αγ + g λ γ k 1α k 1β g λ β k 1α k 1γ +g λ γ k 2α k 2β g αγ k λ 2 k 2β g λ γ k 1 k 2 k 1α k 2β + kλ 2 k 1αk 2β k 1γ +g αβ k λ 3 k 3γ kλ 3 k 2αk 2β k 3γ m2 3 gβ λ k 3α k 3γ + kλ 2 k 3αk 2β k 3γ m2 3 kλ 3 k 1α k 1β k 3γ m2 3 + gβ λ k 1 k 3 k 1α k 3γ m2 3 + k 1 k 2 k λ 3 k 1αk 2β k 3γ m2 2 m2 3 +f ace f bde g βγ gα λ + gλ γ g k1 λ αβ + g k 1α βγ g λ k 1α k 1β m 2 γ 1 k 1 k 3 k λ 2 k 1αk 2β k 3γ m2 2 m2 3 ) +gα λ k 2β k 2γ g λ γ k 2α k 2β kλ 1 k 1αk 2β k 2γ m2 2 + g λ γ k 1 k 2 k 1α k 2β m2 2 +gα λ k 3β k 3γ gα λ k 2 k 3 k 2β k 3γ m2 3 g αβ k λ 3 k 3γ kλ 1 k 1αk 3β k 3γ + kλ 3 k 2α k 2β k 3γ m2 3 + kλ 3 k 1αk 1β k 3γ + k 2 k 3 k λ 1 k 1α k 2β k 3γ m2 2 m2 3 k )] 1 k 2 k3 λ k 1α k 2β k 3γ m2 2 m2 3 C.8) 35
42 left 1) 2) 3) 4) = [f abe f cde g αγ g βσ g ασ g βγ g βσ k 1α k 1γ g αγ k 2β k 2σ g βσ k 3α k 3γ + k 3αk 2β k 3γ k 2σ m2 3 g αγ k 4β k 4σ m 2 4 +g αγ k 2 k 4 k 2β k 4σ m2 4 + k 3αk 4β k 3γ k 4σ m g ασ k 2β k 2γ + g ασ k 3β k 3γ + k 1αk 2β k 1γ k 2σ m2 2 g ασ k 2 k 3 k 2β k 3γ m2 3 + g βγ k 4α k 4σ m 2 4 k 2 k 4 k 3α k 2β k 3γ k 4σ m g βσ k 1 k 3 k 1α k 3γ + k 1αk 4β k 1γ k 4σ m 2 4 k 4αk 2β k 2γ k 4σ m2 4 k 4αk 3β k 3γ k 4σ m k 1 k 3 k 2 k 4 k 1α k 2β k 3γ k 4σ m g βγ k 1α k 1σ k 1αk 2β k 2γ k 1σ m2 2 k 1αk 3β k 3γ k 1σ k 1 k 3 k 1α k 2β k 3γ k 2σ m2 2 m2 3 g βγ k 1 k 4 k 1α k 4σ m 2 4 k 2 k 4 k 1α k 2β k 1γ k 4σ m2 2 m2 4 k 1 k 3 k 1α k 4β k 3γ k 4σ m k 2 k 3 k 4α k 2β k 3γ k 4σ m 2 4 +f ade f bce g αβ g γσ g ασ g βγ g γσ k 1α k 1β g γσ k 2α k 2β g αβ k 3γ k 3σ + k 2αk 2β k 3γ k 3σ g αβ k 4γ k 4σ m k 2αk 2β k 4γ k 4σ m 2 4 +g αβ k 3 k 4 k 3 γk 4σ m g αγ k 2β k 2σ + g βσ k 3α k 3γ k 3αk 2β k 3γ k 2σ + g αγ k 4β k 4σ m 2 4 k 3 k 4 k 2α k 2β k 3γ k 4σ m2 3 m2 4 k 1 k 4 k 2 k 3 k 1α k 2β k 3γ k 4σ m g γ σ k 1 k 2 k 1α k 2β + k 1αk 1β k 3γ k 3σ m2 3 + g βσ k 1α k 1γ + k 2 k 3 k 1α k 2β k 3γ k 1σ m2 2 m2 3 + k 1 k 4 k 1α k 2β k 2γ k 4σ m2 2 m2 4 + k 1 k 4 k 1α k 3β k 3γ k 4σ m 2 4 ) k 1αk 2β k 1γ k 2σ g βσ k 1 k 3 k 1α k 3γ m2 3 k 1 k 2 k 1α k 2β k 3γ k 3σ + k 1αk 1β k 4γ k 4σ m2 4 g αγ k 2 k 4 k 2β k 4σ m 2 4 k 3αk 4β k 3γ k 4σ m k 1 k 2 k 3 k 4 k 1α k 2β k 3γ k 4σ m k 2 k 4 k 3α k 2β k 3γ k 4σ m2 3 m2 4 f ace f bde g ασ g βγ g αβ g γσ g βγ k 1α k 1σ k 1αk 4β k 1γ k 4σ m2 4 k 1 k 2 k 1α k 2β k 4γ k 4σ m 2 4 k 3 k 4 k 1α k 1β k 3γ k 4σ m 2 4 k 1 k 3 k 2 k 4 k 1α k 2β k 3γ k 4σ m g γσ k 1α k 1β + k 1 k 3 k 1α k 2β k 3γ k 2σ ) + k 2 k 4 k 1α k 2β k 1γ k 4σ m k 1 k 3 k 1α k 4β k 3γ k 4σ m 2 4 C.9)
43 g ασ k 2β k 2γ g ασ k 3β k 3γ +g ασ k 2 k 3 k 2β k 3γ m2 3 g βγ k 4α k 4σ m k 4αk 2β k 2γ k 4σ m k 4αk 3β k 3γ k 4σ m2 4 + g γσ k 2α k 2β + g αβ k 3γ k 3σ + k 1αk 2β k 2γ k 1σ m2 2 + k 1αk 3β k 3γ k 1σ k 2αk 2β k 3γ k 3σ m2 3 + g αβ k 4γ k 4σ m 2 4 k 2 k 3 k 4α k 2β k 3γ k 4σ m 2 4 k 2αk 2β k 4γ k 4σ m 2 4 g αβ k 3 k 4 k 3γ k 4σ m2 4 + k 1 k 4 k 2 k 3 k 1α k 2β k 3γ k 4σ m g βγ k 1 k 4 k 1α k 4σ m 2 4 g γσ k 1 k 2 k 1α k 2β m2 2 k 1αk 1β k 3γ k 3σ k 2 k 3 k 1α k 2β k 3γ k 1σ m2 2 m2 3 k 1αk 1β k 4γ k 4σ m 2 4 k 1 k 4 k 1α k 2β k 2γ k 4σ m k 3 k 4 k 2α k 2β k 3γ k 4σ m 2 4 k 1 k 4 k 1α k 3β k 3γ k 4σ m2 3 m2 4 k 1 k 2 k 3 k 4 k 1α k 2β k 3γ k 4σ m k 1 k 2 k 1α k 2β k 3γ k 3σ m2 2 m2 3 + k 1 k 2 k 1α k 2β k 4γ k 4σ m 2 4 )] + k 3 k 4 k 1α k 1β k 3γ k 4σ m2 3 m2 4 C.10) 37
44 left 1) 2) 3) 4) A) = [f abe f cde f abe f cde 2 k 1 k 1 k 2 k 2 k 1 k 2 ) 2 + k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m2 4 + k 2 k 2 k 3 k 3 k 2 k 3 ) k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m2 4 + k 1 k 1 k 3 k 3 k 1 k 3 ) 2 + 2k 1 k 2 k 1 k 3 k 2 k 3 k 1 k 1 k 2 k 3 ) 2 + k 2 k 2 k 1 k 3 ) 2 ) + 2k 1 k 2 k 1 k 4 k 2 k 4 k 1 k 1 k 2 k 4 ) 2 + k 2 k 2 k 1 k 4 ) 2 ) m k 1 k 3 k 1 k 4 k 3 k 4 k 3 k 3 k 1 k 4 ) 2 + k 4 k 4 k 1 k 3 ) 2 ) m2 3 m k 2 k 3 k 2 k 4 k 3 k 4 k 3 k 3 k 2 k 4 ) 2 + k 4 k 4 k 2 k 3 ) 2 ) m k ) 1 k 3 k 2 k 4 ) 2 + k 1 k 4 k 2 k 3 ) 2 2k 1 k 3 k 2 k 4 )k 1 k 4 k 2 k 3 ) m 2 4 +f abe f cde f ade f bce k1 k 2 ) 2 k 1 k 1 k 2 k 2 m2 2 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m2 4 + k 2 k 3 ) 2 k 2 k 2 k 3 k 3 m2 3 + k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m2 4 + k 2 k 2 k 1 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 + k 1 k 1 k 2 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 4 k 4 k 1 k 3 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m2 3 m2 4 + k 3 k 3 k 2 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m k ) 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m m2 3 m2 4 +f abe f cde f ace f bde k1 k 2 ) 2 k 1 k 1 k 2 k 2 m2 2 + k 1 k 3 ) 2 k 1 k 1 k 3 k 3 m2 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m
45 + k 1 k 1 k 2 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 m2 2 m2 3 + k 2 k 2 k 1 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 3 k 3 k 1 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 4 k 4 k 2 k 3 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m2 3 m2 4 + k )] 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 ) m 2 4 C.11) 39
46 left 1) 2) 3) 4) B) = [ f abe f cde f ade f bce k1 k 2 ) 2 k 1 k 1 k 2 k 2 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m k 2 k 3 ) 2 k 2 k 2 k 3 k 3 + k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m k 2 k 2 k 1 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 m2 2 m2 3 + k 1 k 1 k 2 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 4 k 4 k 1 k 3 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 3 k 3 k 2 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m2 3 m2 4 + k ) 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m m 2 4 +f ade f bce f ade f bce k1 k 1 k 2 k 2 k 1 k 2 ) 2 +2 k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m2 4 + k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m k 2 k 2 k 3 k 3 k 2 k 3 ) 2 m2 3 + k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m k 1 k 1 k 3 k 3 k 1 k 3 ) 2 + 2k 1 k 2 k 1 k 3 k 2 k 3 k 2 k 2 k 1 k 3 ) 2 + k 3 k 3 k 1 k 2 ) 2 ) + 2k 1 k 2 k 1 k 4 k 2 k 4 k 1 k 1 k 2 k 4 ) 2 + k 4 k 4 k 1 k 2 ) 2 ) m2 2 m k 1 k 3 k 1 k 4 k 3 k 4 k 1 k 1 k 3 k 4 ) 2 + k 4 k 4 k 1 k 3 ) 2 ) m k 2 k 3 k 2 k 4 k 3 k 4 k 3 k 3 k 2 k 4 ) 2 + k 2 k 2 k 3 k 4 ) 2 ) m k ) 1 k 2 k 3 k 4 ) 2 + k 1 k 3 k 2 k 4 ) 2 2k 1 k 2 k 3 k 4 )k 1 k 2 k 2 k 4 ) m2 2 m2 3 m2 4 +f ade f bce f ace f bde k1 k 3 ) 2 k 1 k 1 k 3 k 3 m2 3 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m2 4 + k 2 k 3 ) 2 k 2 k 2 k 3 k 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m k 3 k 3 k 1 k 2 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 40
47 + k 4 k 4 k 1 k 2 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m2 2 m2 4 + k 1 k 1 k 3 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 2 k 2 k 3 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m k )] 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m2 2 m2 3 m2 4 C.12) 41
48 left 1) 2) 3) 4) C) = [ f abe f cde f ace f bde k1 k 2 ) 2 k 1 k 1 k 2 k 2 + k 1 k 3 ) 2 k 1 k 1 k 3 k 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m k 3 k 4 ) 2 k 3 k 3 k 4 k 4 m k 1 k 1 k 2 k 3 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 m2 2 m2 3 + k 2 k 2 k 1 k 4 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 3 k 3 k 1 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m k 4 k 4 k 2 k 3 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m2 3 m2 4 + k ) 1 k 3 k 2 k 4 k 1 k 4 k 2 k 3 )k 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 ) m 2 4 +f ade f bce f ace f bde k1 k 3 ) 2 k 1 k 1 k 3 k 3 + k 1 k 4 ) 2 k 1 k 1 k 4 k 4 m k 2 k 3 ) 2 k 2 k 2 k 3 k 3 m2 3 + k 2 k 4 ) 2 k 2 k 2 k 4 k 4 m2 4 + k 3 k 3 k 1 k 2 ) 2 k 1 k 2 k 1 k 3 k 2 k 3 + k 4 k 4 k 1 k 2 ) 2 k 1 k 2 k 1 k 4 k 2 k 4 m k 1 k 1 k 3 k 4 ) 2 k 1 k 3 k 1 k 4 k 3 k 4 m2 3 m2 4 + k 2 k 2 k 3 k 4 ) 2 k 2 k 3 k 2 k 4 k 3 k 4 m k ) 1 k 4 k 2 k 3 k 1 k 2 k 3 k 4 )k 1 k 2 k 3 k 4 k 1 k 3 k 2 k 4 ) m 2 4 +f ace f bde f ace f bce k1 k 1 k 2 k 2 k 1 k 2 ) 2 m2 2 + k 1 k 1 k 4 k 4 k 1 k 4 ) 2 m k 2 k 2 k 4 k 4 k 2 k 4 ) 2 m k 2 k 2 k 3 k 3 k 2 k 3 ) 2 m2 3 + k 3 k 3 k 4 k 4 k 3 k 4 ) 2 m k 1 k 1 k 3 k 3 k 1 k 3 ) 2 m2 3 42
Space-Time Symmetries
Chapter Space-Time Symmetries In classical fiel theory any continuous symmetry of the action generates a conserve current by Noether's proceure. If the Lagrangian is not invariant but only shifts by a
6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
derivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Section 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Matrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)
1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations
Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
EE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Concrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
C.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Symmetric Stress-Energy Tensor
Chapter 3 Symmetric Stress-Energy ensor We noticed that Noether s conserved currents are arbitrary up to the addition of a divergence-less field. Exploiting this freedom the canonical stress-energy tensor
Homework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Partial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Example Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Section 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Higher Derivative Gravity Theories
Higher Derivative Gravity Theories Black Holes in AdS space-times James Mashiyane Supervisor: Prof Kevin Goldstein University of the Witwatersrand Second Mandelstam, 20 January 2018 James Mashiyane WITS)
= {{D α, D α }, D α }. = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ.
PHY 396 T: SUSY Solutions for problem set #1. Problem 2(a): First of all, [D α, D 2 D α D α ] = {D α, D α }D α D α {D α, D α } = {D α, D α }D α + D α {D α, D α } (S.1) = {{D α, D α }, D α }. Second, {D
Chapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
1 String with massive end-points
1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε
6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Congruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and
( ) 2 and compare to M.
Problems and Solutions for Section 4.2 4.9 through 4.33) 4.9 Calculate the square root of the matrix 3!0 M!0 8 Hint: Let M / 2 a!b ; calculate M / 2!b c ) 2 and compare to M. Solution: Given: 3!0 M!0 8
Srednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Second Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
the total number of electrons passing through the lamp.
1. A 12 V 36 W lamp is lit to normal brightness using a 12 V car battery of negligible internal resistance. The lamp is switched on for one hour (3600 s). For the time of 1 hour, calculate (i) the energy
Lecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C By Tom Irvine Email: tomirvine@aol.com August 6, 8 Introduction The obective is to derive a Miles equation which gives the overall response
ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
CRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
[1] P Q. Fig. 3.1
1 (a) Define resistance....... [1] (b) The smallest conductor within a computer processing chip can be represented as a rectangular block that is one atom high, four atoms wide and twenty atoms long. One
Tutorial problem set 6,
GENERAL RELATIVITY Tutorial problem set 6, 01.11.2013. SOLUTIONS PROBLEM 1 Killing vectors. a Show that the commutator of two Killing vectors is a Killing vector. Show that a linear combination with constant
Homework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Variational Wavefunction for the Helium Atom
Technische Universität Graz Institut für Festkörperphysik Student project Variational Wavefunction for the Helium Atom Molecular and Solid State Physics 53. submitted on: 3. November 9 by: Markus Krammer
The Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
PARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
Assalamu `alaikum wr. wb.
LUMP SUM Assalamu `alaikum wr. wb. LUMP SUM Wassalamu alaikum wr. wb. Assalamu `alaikum wr. wb. LUMP SUM Wassalamu alaikum wr. wb. LUMP SUM Lump sum lump sum lump sum. lump sum fixed price lump sum lump
Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Approximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Second Order RLC Filters
ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor
Every set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Forced Pendulum Numerical approach
Numerical approach UiO April 8, 2014 Physical problem and equation We have a pendulum of length l, with mass m. The pendulum is subject to gravitation as well as both a forcing and linear resistance force.
F-TF Sum and Difference angle
F-TF Sum and Difference angle formulas Alignments to Content Standards: F-TF.C.9 Task In this task, you will show how all of the sum and difference angle formulas can be derived from a single formula when
Non-Abelian Gauge Fields
Chapter 5 Non-Abelian Gauge Fields The simplest example starts with two Fermions Dirac particles) ψ 1, ψ 2, degenerate in mass, and hence satisfying in the absence of interactions γ 1 i + m)ψ 1 = 0, γ
The kinetic and potential energies as T = 1 2. (m i η2 i k(η i+1 η i ) 2 ). (3) The Hooke s law F = Y ξ, (6) with a discrete analog
Lecture 12: Introduction to Analytical Mechanics of Continuous Systems Lagrangian Density for Continuous Systems The kinetic and potential energies as T = 1 2 i η2 i (1 and V = 1 2 i+1 η i 2, i (2 where
On a four-dimensional hyperbolic manifold with finite volume
BULETINUL ACADEMIEI DE ŞTIINŢE A REPUBLICII MOLDOVA. MATEMATICA Numbers 2(72) 3(73), 2013, Pages 80 89 ISSN 1024 7696 On a four-dimensional hyperbolic manifold with finite volume I.S.Gutsul Abstract. In
Potential Dividers. 46 minutes. 46 marks. Page 1 of 11
Potential Dividers 46 minutes 46 marks Page 1 of 11 Q1. In the circuit shown in the figure below, the battery, of negligible internal resistance, has an emf of 30 V. The pd across the lamp is 6.0 V and
Reminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Math221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Strain gauge and rosettes
Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
( y) Partial Differential Equations
Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate
Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
D Alembert s Solution to the Wave Equation
D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique
상대론적고에너지중이온충돌에서 제트입자와관련된제동복사 박가영 인하대학교 윤진희교수님, 권민정교수님
상대론적고에너지중이온충돌에서 제트입자와관련된제동복사 박가영 인하대학교 윤진희교수님, 권민정교수님 Motivation Bremsstrahlung is a major rocess losing energies while jet articles get through the medium. BUT it should be quite different from low energy
Problem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
Statistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
Solutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!
MTH U341 urface Integrals, tokes theorem, the divergence theorem To be turned in Wed., Dec. 1. 1. Let be the sphere of radius a, x 2 + y 2 + z 2 a 2. a. Use spherical coordinates (with ρ a) to parametrize.
9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr
9.9 #. Area inside the oval limaçon r = + cos. To graph, start with = so r =. Compute d = sin. Interesting points are where d vanishes, or at =,,, etc. For these values of we compute r:,,, and the values
Relativistic particle dynamics and deformed symmetry
Relativistic particle dynamics and deformed Poincare symmetry Department for Theoretical Physics, Ivan Franko Lviv National University XXXIII Max Born Symposium, Wroclaw Outline Lorentz-covariant deformed
Chapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
TMA4115 Matematikk 3
TMA4115 Matematikk 3 Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet Trondheim Spring 2010 Lecture 12: Mathematics Marvellous Matrices Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet
Numerical Analysis FMN011
Numerical Analysis FMN011 Carmen Arévalo Lund University carmen@maths.lth.se Lecture 12 Periodic data A function g has period P if g(x + P ) = g(x) Model: Trigonometric polynomial of order M T M (x) =
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Written Examination. Antennas and Propagation (AA ) April 26, 2017.
Written Examination Antennas and Propagation (AA. 6-7) April 6, 7. Problem ( points) Let us consider a wire antenna as in Fig. characterized by a z-oriented linear filamentary current I(z) = I cos(kz)ẑ
Μονοβάθμια Συστήματα: Εξίσωση Κίνησης, Διατύπωση του Προβλήματος και Μέθοδοι Επίλυσης. Απόστολος Σ. Παπαγεωργίου
Μονοβάθμια Συστήματα: Εξίσωση Κίνησης, Διατύπωση του Προβλήματος και Μέθοδοι Επίλυσης VISCOUSLY DAMPED 1-DOF SYSTEM Μονοβάθμια Συστήματα με Ιξώδη Απόσβεση Equation of Motion (Εξίσωση Κίνησης): Complete
Derivation of Optical-Bloch Equations
Appendix C Derivation of Optical-Bloch Equations In this appendix the optical-bloch equations that give the populations and coherences for an idealized three-level Λ system, Fig. 3. on page 47, will be
Finite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Math 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
Dr. D. Dinev, Department of Structural Mechanics, UACEG
Lecture 4 Material behavior: Constitutive equations Field of the game Print version Lecture on Theory of lasticity and Plasticity of Dr. D. Dinev, Department of Structural Mechanics, UACG 4.1 Contents
Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
Other Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
ΠΑΝΕΠΙΣΤΗΜΙΟ ΠΕΙΡΑΙΑ ΤΜΗΜΑ ΝΑΥΤΙΛΙΑΚΩΝ ΣΠΟΥΔΩΝ ΠΡΟΓΡΑΜΜΑ ΜΕΤΑΠΤΥΧΙΑΚΩΝ ΣΠΟΥΔΩΝ ΣΤΗΝ ΝΑΥΤΙΛΙΑ
ΠΑΝΕΠΙΣΤΗΜΙΟ ΠΕΙΡΑΙΑ ΤΜΗΜΑ ΝΑΥΤΙΛΙΑΚΩΝ ΣΠΟΥΔΩΝ ΠΡΟΓΡΑΜΜΑ ΜΕΤΑΠΤΥΧΙΑΚΩΝ ΣΠΟΥΔΩΝ ΣΤΗΝ ΝΑΥΤΙΛΙΑ ΝΟΜΙΚΟ ΚΑΙ ΘΕΣΜΙΚΟ ΦΟΡΟΛΟΓΙΚΟ ΠΛΑΙΣΙΟ ΚΤΗΣΗΣ ΚΑΙ ΕΚΜΕΤΑΛΛΕΥΣΗΣ ΠΛΟΙΟΥ ΔΙΠΛΩΜΑΤΙΚΗ ΕΡΓΑΣΙΑ που υποβλήθηκε στο
SOLVING CUBICS AND QUARTICS BY RADICALS
SOLVING CUBICS AND QUARTICS BY RADICALS The purpose of this handout is to record the classical formulas expressing the roots of degree three and degree four polynomials in terms of radicals. We begin with
Physics 582, Problem Set 2 Solutions
Physics 582, Problem Set 2 Solutions TAs: Hart Goldman and Ramanjit Sohal Fall 2018 Symmetries and Conservation Laws In this problem set we return to a study of scalar electrodynamics which has the Lagrangian
Exercises to Statistics of Material Fatigue No. 5
Prof. Dr. Christine Müller Dipl.-Math. Christoph Kustosz Eercises to Statistics of Material Fatigue No. 5 E. 9 (5 a Show, that a Fisher information matri for a two dimensional parameter θ (θ,θ 2 R 2, can
Dirac Matrices and Lorentz Spinors
Dirac Matrices and Lorentz Spinors Background: In 3D, the spinor j = 1 representation of the Spin3) rotation group is constructed from the Pauli matrices σ x, σ y, and σ k, which obey both commutation
Parametrized Surfaces
Parametrized Surfaces Recall from our unit on vector-valued functions at the beginning of the semester that an R 3 -valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some
Bounding Nonsplitting Enumeration Degrees
Bounding Nonsplitting Enumeration Degrees Thomas F. Kent Andrea Sorbi Università degli Studi di Siena Italia July 18, 2007 Goal: Introduce a form of Σ 0 2-permitting for the enumeration degrees. Till now,
(1) Describe the process by which mercury atoms become excited in a fluorescent tube (3)
Q1. (a) A fluorescent tube is filled with mercury vapour at low pressure. In order to emit electromagnetic radiation the mercury atoms must first be excited. (i) What is meant by an excited atom? (1) (ii)
ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
Uniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula