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Gravitational back-reaction from colliding travelling waves on a cosmic string Wells, R. Glenn 1994

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GRAVITATIONAL BACK-REACTION FROM COLLIDING TRAVELLING WAVESON A COSMIC STRINGByR. Glenn WellsB.Sc.(Hon), The University of Calgary, 1991A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER’S OF SCIENCEinTHE FACULTY OF GRADUATE STUDIESDEPARTMENT OF PHYSICSWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAJanuary 1994© R. Glenn Wells, 1994In presenting this thesis in partial fulfilment of the requirements for an advanced degree atthe University of British Columbia, I agree that the Library shall make it freely availablefor reference and study. I further agree that permission for extensive copying of thisthesis for scholarly purposes may be granted by the head of my department or by hisor her representatives. It is understood that copying or publication of this thesis forfinancial gain shall not be allowed without my written permission.Department of PhysicsThe University of British Columbia6224 Agricultural RoadVancouver, B.C., CanadaV6T 1Z1ate:,( /9y.AbstractIn this thesis, I investigate the effects of the gravitational back-reaction force on twocolliding triangular pulses travelling along an infinite cosmic string. The force is determined using a perturbative approach in the Lorentz gauge. This gauge choice is foundto be reasonable in several aspects. Firstly, there is no gravitational back-reaction forceon a single travelling wave. Secondly, the energy emitted by the string as gravitationalradiation is the same as that lost by the string due to the back-reaction. Finally, the forcecan be made perpendicular to the string such that the perturbed string trajectory continues to satisfy the equations of motion. Unfortunately, the force equations contain anarbitrary gauge freedom making it difficult to form definite statements about the effectsof the back-reaction force. However, choosing a reasonable value for the gauge parameterresults in the two wave pulses being flattened out and having their sharp kinks roundedoff.11Table of ContentsAbstract iiTable of Contents iiiList of Figures vAcknowledgements ViiiIntroduction 11 Cosmic Strings 51.1 Notation 61.2 The Stress-energy Tensor 71.3 The Single Travelling Wave 91.4 Waves Travelling in Opposite Directions 111.5 The Emission of Radiation 132 The Formalism 212.1 The Force Equation 242.2 The Back-Reaction on a Single Wave 262.3 The Energy Lost by the String 342.4 Evaluating the Force for Triangular Waves 412.5 Other Features of the Force Equations 451113 The Composite Metric 493.1 Failures of the Direct Approach 493.2 The Composite Metric 533.3 Force Equations for Single Travelling Waves 564 The Results 614.1 The Force Components 624.2 Time Slices 654.3 U-slices 794.4 V-slices 885 The Perpendicular Force 935.1 Removing the Parallel Components 945.2 Force on a Single Wave Revisited 975.3 Energy Equivalence Revisited 985.4 Effect of the Perpendicular Force 996 Conclusions 104Bibliography 110A Comparing Single Wave Metrics 112B Gauge Condition for the Residual Wave 115C Gauge Independence of the Left-Moving Delta Functions 118ivList of Figures1.1 A small travelling wave on a cosmic string moving up the side of a largerone 141.2 The distorted appearance of a wave when seen from the reference frameof a sloped string 202.1 Degeneracies in ii and i 433.1 The two travelling waves before the collision 503.2 Infinite force spikes at time t = —115 523.3 Decomposition of the two-wave string 543.4 The residual component of the string metric 553.5 Amplitudes of the delta function generated by the right-moving wave 583.6 Amplitudes of the delta function generated by the left-moving wave . 583.7 The effects of the scale factor on the amplitude of the delta function contribution to the force 594.1 The decomposition of the residual metric’s contribution to the force att=—90 634.2 Force contributions at t = —90 from the two single travelling waves andthe straight string using Vachaspati’s metric 644.3 Force at time t = —119 684.4 Force at time t = —115 684.5 Force at time t = —110 69v4.6 Force at time t = —101 69714.7 Force at time t = —994.8 Force at time t = —90 714.9 Force at time t = —85 724.10 Force at time t = —81 724.11 Force at time t = —79 744.12 Force at time t = —50 744.13 Force at time t = —21 754.14 Force at time t = —18 754.15 Force at time t = —15 764.16 Force at time t —1 764.17 Force at time t = 1 774.18 Force at time t = 15 774.19 Force at time t = 21 784.20 Force at time t = 150 784.21 Force at u = —105 804.22 Force at u = —101 804.23 Force at u = —100.1 804.24 Force at u = —99.9 804.25 Force at u = —99 824.26 Force at u = —80 824.27 Force at u = —5 824.28 Force at u = —1 824.29 Force at u = —0.1 834.30 Force at u = 1 834.31 Force at u = 20 83vi4.32 Force at u = 95 834.33 Force at u = 99 844.34 Force at u = 99.9 844.35 Force at ‘a = 101 844.36 Force at ‘a = 110 844.37 Force at ‘a = 150 854.38 Force at ‘a = 500 854.39 Integrated finite forces before ‘a = —100 874.40 Delta function contribution to the integrated force before ‘a = —100 . . 874.41 Integrated finite forces after ‘a = —100 894.42 Delta function contribution to the integrated force after ‘a = —100 . . 894.43 Integrated force after ‘a = 100. . 904.44 Force at v = —25 924.45 Force at v = —21 924.46 Force at v = —20.1. . 924.47 Force at v = —19 925.1 Amplitudes of the delta function contribution to the perpendicular forceat a scale of 1 1015.2 Amplitudes of the delta function contribution to the perpendicular forceat a scale of 0.01 102viiAcknowledgementsI would like to thank my supervisor, Bill Unruh for all of his help and guidance throughoutthis project. My special thanks also to Kristin Schleich for her critical and expeditiousreading of this thesis.Finally, I would like to thank my family and friends for their continual encouragementand support.vii’IntroductionGravitational radiation was first predicted by Einstein in 1918 [1]. However, it was notuntil recently that there was any experimental evidence for its existence. During the lasttwo decades, J. Taylor and his collaborators have been studying the binary pulsar systemPSR 1913+16 with the Aricebo telescope in Peurto Rico [2]. The binary system emitsgravitational radiation and this loss of energy causes the system to slowly spiral inwards.Because one of the binary stars is a millisecond pulsar, Talyor was able to use the pulsesto accurately measure this inward decay. The results they found perfectly match withthe predictions of General Relativity for the back-reaction of the gravitational radiationon the system thus providing the first indirect evidence of gravitational radiation andadditional verification of Einstein’s theory. Any system which emits gravity waves shouldexperience back-reaction effects. One such system is the collision of travelling waves ona cosmic string. It is the gravitational back-reaction on cosmic strings which forms thebasis of this thesis.Cosmic strings are formed as the universe cools below a critical temperature and thesymmetry of the grand unified theories (GUT’s) is spontaneously broken [3, 4]. As asimplistic model of a GUT, consider a field theory containing a complex scalar Higgsfield and a U(1) symmetry:—* When this symmetry is broken, the Higgsfield acquires a vacuum expectation value of < >= e0 where j is related to theself-interaction potential of the theory. The parameter 9 is arbitrary which means thatthe vacuum state is degenerate; solutions form a circle of radius ij in the complex qspace. The angle which 0 acquires as the universe cools is random and different parts ofthe universe can acquire different 0 values. As a result the vacuum expectation value is1uncorrelated at distances of separation greater than which is of the same order as theage of the universe at the time of the phase transition. In other words, 0 can be differenton length scales greater than e.It is energetically favorable for the universe to have a uniform 0 value. However, theuniverse’s topology may be such that it is not possible to achieve this uniformity througha continuous evolution of 0. Because < > is single valued, along any closed ioop inspace, z0 = 2irn where n is an integer. If n is non-zero, say n = 1, then the loop cannotbe continuously shrunk down to zero radius because this would involve a discontinuouschange in n from one to zero. As a result, one has a tube-like volume of trapped falsevacuum. The tube is either closed or infinitely long because otherwise one could movethe loop around the end of the tube and continuously deform it to zero radius. The tubeof topologically trapped false vacuum is called a cosmic string.Cosmic strings are cosmologically important as they provide a possible explanationfor the creation of structure in the universe (for example see references [3, 5, 6, 7] amongothers). Cosmic string loops, which form when strings collide and intercommute, canact as seeds for the formation of galaxies, accreting matter and radiating energy as theyoscillate, decay, and finally disappear. They can also generate larger structures because ofthe wake they create while moving through space. The spacetime around cosmic stringsis quite simple. It is exactly like fiat space locally, but globally there exists a conicaldeficit. This means that the circumference of a circle surrounding a cosmic string is not2irr as it is in Euclidian space but rather 27rr(1 — 4G1i) where G is a very small quantity.As the string moves through space, matter effectively falls toward the deficit creating awake behind the string. This wake could be the cause of some of the large scale structurein the universe.Besides their possible cosmological significance, cosmic strings are also interesting intheir own right. They have some unique properties that make them very different from2ordinary matter. These properties include the fact that stretching a string does not giveyou a longer thinner string, such as one gets upon stretching a rubber band, but insteadcauses an increase in the quantity of cosmic string. Also, unlike regular travelling waveson a normal string, two waves moving in opposite directions on a cosmic string cannotbe expressed as a superposition of two single travelling waves, f(z + t) + g(z — t) wherez and t correspond to a standard Cartesian coordinate and coordinate time respectively.Another interesting feature of cosmic strings is that they could cause astronomicallyobservable effects. The conical deficit in the spacetime around cosmic strings causes thebending of light rays which might lead to the double imaging of stars or galaxies [8]. Alsowhen two oppositely moving travelling waves collide on a long cosmic string, they emitgravitational radiation [9, 10]. This radiation, though very small in amplitude, mightalso eventually be observable.The cosmological aspects of strings are perhaps their most important feature. Tomore completely understand the role they may have played in the early universe, it isnecessary to comprehend the mechanisms by which string loops decay and kinky stringsare damped or straightened out. To this end, it is important to understand the effects ofgravitational back-reaction on the cosmic string. A valuable step towards this goal is todetermine how the radiation from two colliding wave pulses on a string affects the shapeand motion of the two waves. This thesis investigates the effects of the back-reactionforces on the collision of two triangular wave pulses travelling along a cosmic string.The thesis is divided into six chapters. The first chapter provides the basic propertiesof a cosmic string required for a study of the back-reaction forces: the stress-energytensor and the metric. It also demonstrates that the radiation from the collision of twotriangular pulses is emitted when two oppositely moving kinks in the string intersect.Chapter two develops the equation for the back-reaction force. This equation, in theLorentz gauge, is shown to have reasonable properties: the perturbative solution does not3violate the equations of motion in flat spacetime, the single travelling wave feels no back-reaction force, and the energy lost by the string is the same as that emitted in the form ofgravitational radiation. Some other features of the equations are also discussed. The thirdchapter discusses the problems that arise through direct application of the back-reactionequation found in chapter two. Chapter three also develops a solution to this problem,the composite metric. The construction of the composite metric is explained and theequations for the back-reaction force due to the Vachispati components of the compositemetric on the actual cosmic string with two travelling waves are stated. Chapter fourshows what the back-reaction force on the string looks like on various different time slicesthroughout the collision. In addition, u-slices of the collision and the effects of the forceon the right-moving wave are shown. The symmetry of the results with regards to theleft- and right-moving waves is also demonstrated. Chapter five discusses a failing of theforce as determined in chapters two and three. This failing is the presence of componentsof the force which are parallel to the string. These components cause a coordinate changewhich prevents the perturbed string trajectory from satisfying the coordinate conditionsin curved spacetime. Finally, in chapter six, I summarize my conclusions regarding theeffects of the back-reaction force on the two colliding waves.4Chapter 1Cosmic StringsCosmic strings are characterized by a single parameter, the energy of the symmetrybreaking, 7]. Based on dimensional arguments, the radius of the string is 6 - 77_i andthe mass per unit length of the string is j2 For strings creating at the GrandUnifying scales, this corresponds to 6 1032m and a mass density of t 1021kg/m[8]. To put these numbers in some perspective, the radius of the Hydrogen nucleus isonly 105m. The volume mass density of the string is 1085kg/m3whereas the density ofa neutron star is only 105kg/m3.The extremely thin nature of the cosmic string makesit reasonable to consider the string as an object with only one spatial dimension.1 Thiswill be assumed throughout this thesis.Despite the high mass of cosmic strings, because of their extremely thin nature, theirgravitational effects still lie within the weak gravity regime. Using a system of unitsin which G=c=1, the relation between mass and length is lg= 1028cm and so G[L isunitless with a value of only 1O_6. Because this parameter is quite small, the right-handside of Einstein’s equations is close to zero and one is justified in using the weak fieldapproximation to the theory.Using linearized gravity and the approximation that a string is only a one dimensionalobject, it is quite straightforward to find the stress-energy tensor and the metric whichdescribes the string and its equations of motion.‘Garfinkle shows [11] that ignoring the internal structure of the string and treating the string as aninfinitely thin object is a valid approximation for determining its gravitational field.5Chapter 1. Cosmic Strings 61.1 NotationIn this thesis I investigate the effects of gravitational back-reaction on colliding travellingwaves moving along a cosmic string. This investigation assumes that only gravitationalradiation is present to cause a back-reaction effect, that is, I assume the string is uncharged and no electromagnetic radiation is being emitted. As the gravitational field ofcosmic strings is weak, I will be using linearized gravity. The purpose of this section isto clarify the notation that I will be using throughout the thesis.In linearized gravity, or the weak gravity limit, one assumes that the full spacetimemetric g. can be broken into two components r, and The first is just the metricfor fiat Minkowskian spacetime which can be expressed as = diag(—1, 1, 1, 1) whilethe second represents a very small perturbation to that spacetime.In linearized gravity, one works only to first order in h, and so raises and lowersthe indices on using just the fiat spacetime metric i. Therefore, one can write theChristoffel, or connection, symbols for the spacetime as1P1L —— \gv,3,a —i-—= fr’(h,,a + hva,— ha,v) (1.1)In linearized gravity Einstein’s field equations are [12]7, aLt, a j, a a‘va,Ji—‘‘fw,a —— = 16KT1z, (1.2)These equations may be simplified by writing them in terms of“ = — andby specifying a gauge.The gauge I will be working in is the Lorentz gauge. For a metric to be in the Lorentzgauge, it must satisfy the requirement thatV= o (1.3)Chapter 1. Cosmic Strings 7This requirement does not, however, completely fix the gauge. Any transformation ofthe form—— (1.4)with satisfying = 0 is still permissible. Again, indices are raisedand lowered using the metric of fiat spacetime, i. This gauge freedom is critical indetermining the back-reaction force as shall be seen in chapters two and three.In the Lorentz gauge, Einstein’s equations reduce down toDh = —167rGT, (1.5)where G is Newton’s universal gravitational constant.1.2 The Stress-energy TensorA first step towards finding the back-reaction forces is to describe the stress-energy tensor,and thereby the metric, of the cosmic string. An infinite straight string lying along thez-axis is invariant with respect to Lorentz boosts in the z-direction and only motionin the transverse direction can have physical significance. These conditions restrict thestress-energy tensor to have the formT = T = fL5(x)ö(y) T’ = 0 otherwise (1.6)To determine the stress-energy tensor for a string which is not straight, it is easiestto work from the action for the string. The Nambu action for the string can be derivedfrom first principles [13, 14, 15] and, heuristically, the argument is as follows [16]. Onebegins by observing that the action for a relativistic particle is just given by the integralover the world line of the particle.I dx dx5particle = _mfds = _mfdt (1.7)Chapter 1. Cosmic Strings 8where m is the mass, x is the position, g is the metric of the background spacetime,and r is timelike parameter characterizing the world line of the particle. Like the freeparticle, the cosmic string also has no external forces acting on it but instead of a mass,m, the string is characterized by a mass density . In addition the string can be describedby Lorentz-invariant field equations that are independent in ‘i- and 1, a spacelike parameter directed along the length of the string and, therefore, the string is invariant underLorentz boosts parallel to the string. This means that only velocities perpendicular tothe string have any physical meaning and so the action should only be a function of theperpendicular velocities, and not of the full velocity . Using 1 to parameterizethe elemental length of the string so thatrn=ifdlone gets that the action of the string is [13, 14, 15]ISstring = —ii] dldr= —fdldr/1— (1.8)where- a a Ia oaN a a=and = 1 (1.9)To put this action into the explicitly covariant form of the Nambu action, one reparametrizes using the timelike parameter ° and the spacelike parameter ‘ [3].SNambu _f7d2 (1.10)where y is given byOx!’ Ox7jk = g,v---- (1.11)7 = 73.—(± . x’)2 — (1.12)Chapter 1. Cosmic Strings 9and where the dot and the prime refer to derivation with respect to and respectively.The indices j and k are two-dimensional indices of the string’s internal parameter .The equations of motion are obtained by varying the action with respect to[17] 2•— x’2’9}+{7[(± —±2x’]} = 0 (1.13)The stress-energy tensor is found by varying the action with respect to the fourdimensional metric grn,.T(z) = fd27_4)(z—X[X!212±’ +X2!12X!h) — (th • x’)(bx” + x’i)] (1.14)As a check on the Nambu action, one can choose ° = t, ‘ = z, x = 0, and y = 0 in(1.14) and get back the stress-energy tensor for a long straight string, (1.6).1.3 The Single Travelling WaveAs the action is invariant under reparametrization of , it is possible to choose a whichsimplifies (1.14). Again choosing ° = t and = z hut this time allowing x = f(z ± t)and y = g(z ± t), one gets the stress-energy tensor of a single travelling wave pulse onthe string [17]. For instance, taking f and g to be function of z — t, the stress-energytensor is:2These equations of motion are found using the flat Minkowski spacetime metric. More generalequations of motion are given in section 2.1.Chapter 1. Cosmic Strings 101+f’2+g—f’ —g’ f’2+g’—f’ 0 0 —f’T’ = —— g) (1.15)—g’ 0 0 —g’f’2+g’—f’ —g’ —1+f’2+g’Making the linearized gravity approximations, it is possible to integrate (1.5) usingthe retarded Green’s function 5((z—x())2)6(z°—x°) and obtain the perturbation metricof a travelling wave on the string [17].f’2+g’—f’, —g’ f’2+g’—f’ 1 0= h (1.16)—g’ 0 1 —g’f’2+g’—f’ —g’ f’2+g’where h = (—4G[t) ln(22[(x— f)2 + (y — g)2] and 2 is just an integration constant.3It is quite straightforward to see that the single travelling wave gives off no radiation.Weinberg ‘s equation for the energy emitted by a source of gravity waves [18] (2.48) saysthat the energy depends onT*(k)T(k)— Ta(k)l2 (1.17)This is equal tof fdxdx’exp(ik. (x — x’)){T(x’)T(x) — T(x)T(x’)} (1.18)The symmetries of the stress-energy tensor in the z and t coordinates make it easy to seethat the integrand in (1.18) vanishes. Thus there is no radiation given off by the single3This solution is exactly the same as that obtained by using an advanced Green’s function instead ofa retarded one. The equality of the advanced and retarded solutions is another way of seeing that thethere is no radiation given off by the single travelling wave; the radiation metric is just the differencebetween the advanced and retarded metrics.Chapter 1. Cosmic Strings 11travelling wave. This will serve as a useful check on the equations for the back-reactionforce. Because there is no radiation emitted, there should be no back-reaction force fora single travelling wave.Unfortunately, it is not possible to use this coordinate choice to study two wavesmoving in opposite directions on the same string. Although the equations of motion(1.13) are satisfied for f and g being functions of either z — t or z + t, a combination ofright- and left-moving waves does not satisfy the equations. The two-wave solution is nota simple superposition of two single travelling wave solutions in this parameterization.To deal with colliding waves on a cosmic string, it is necessary to choose a differentparametrization of the string trajectory.1.4 Waves Travelling in Opposite DirectionsIt is always possible to choose the string parameters in such a way as to satisfy thefollowing conditions [8]:and x2+x’=O (1.19)This causes the equation of motion (1.13) to reduce down to the form= 0 (1.20)The two conditions in (1.19) do not, however, completely fix the parametrization of thestring. It is still possible to choose the timelike parameter, to be t, the coordinatetime. Under this specification, the conditions (1.19) reduce to= 0 = 1 (1.21)with the equations of motion becoming(1.22)Chapter 1. Cosmic Strings 12As this form of the equations of motion is just the wave equation, it is easy to see that theparametrization specified by the conditions (1.19) does permit the superposition of wavestravelling in both directions. The timelike parameter 4° = r = t can be identified withthe coordinate time. The spacelike parameter is usually denoted a and is proportionalto the total energy or mass of the string.mass =where i is the mass density. So when the string is straight and fiat, a corresponds to thelength of the string.The general solution of (1.22) is a linear combination of left and right moving waves.(a,t) = [(t - a) + (t + a)] (1.23)with the only constraint on and b being(1.24)With this parametrization, the stress-energy tensor becomesT(,t) = fda(1’— (a,t)) (1.25)It is often useful to relabel the parameters in terms of the null coordinates u = andv = With these variables, the stress-energy tensor (1.25) can be written asT(z) = + OvxOxv)(4)(z—x(u,v)) (1.26)In this choice of parameters, the conditions on x are reformulated as= gaavx9vx = 0 (1.27)= = 1 (1.28)Chapter 1. Cosmic Strings 131.5 The Emission of RadiationIt is useful to examine where the gravitational radiation is emitted from when two wavescollide on a string. This will help to explain the general trend of the back-reaction forcesproduced.The waves which I chose to study in this thesis were two oppositely moving triangularpulses. This pulse shape was chosen in order to make the slope of the string piecewiseconstant. A piecewise constant slope makes it much easier to analytically evaluate theback-reaction force. As is shown in this section, this choice also means that the source ofthe gravitational radiation is restricted to specific points, those points at which oppositelymoving kinks on the string collide. It was hoped that this later aspect of the waveswould result in a simpler picture of the back-reaction forces and consequently a clearerunderstanding of what was happening throughout the collision.If the two waves which are colliding on the string are triangular pulses, then at somepoint during the collision, one of the waves will be travelling along a portion of the secondwave which is flat but at an angle to the main part of the string. If the first wave is muchsmaller than the second, then the entire first wave could be on this sloped but straightsection of string at once, figure (1.1). As the physical results of General Relativity areindependent of the frame of reference chosen, it should be possible to rotate our frameof reference and view the first wave as just moving along a flat string. In this case, themetric perturbation would be that of a single travelling wave and there would be noradiation emitted. This can be shown to be the case.For simplicity, consider a triangular right-moving wave restricted to the yz-plane.This wave can be expressed byChapter 1. Cosmic Strings 14Two Waves CollidingI I I I I I I I I I I I I I I I I I I I I I I I60 -0.—4)C) 40 -ci)-20-0 I I I I I I I I I I I—300 —200 —100 0 100 200z—directionFigure 1.1: A small left-moving wave on a cosmic string travelling along the fiat sectionof a larger right-moving triangular wave.x°(u) = ux’(u) = 0x2Qu) = /3(u + l)6(u + 1) — 2/3u0(u) + /3(u — l)O(u — 1) (1.29)x(u) = -u-(- 1)(u+l)0(u+l) -(1 -)(u-l)0(u-l)where /3 is a positive value between 0 and 1, = /1 — /32, 1 is some positive constant,and 0(u) is the Heaviside function. For now, I consider the left-moving wave to be flat,that isx(v) = (v,0,0,v) (1.30)Consider a point at which —l <u <0. The metric perturbation is found using (1.5), theGreen’s function for the wave equation, and the stress-energy tensor.v(z) = 8GfT(x)((x_z)2) (1.31)Chapter 1. Cosmic Strings 15Using the form of T’ given in (1.26), one obtains2 0/31—13- 0 00 0h(z) = 8Gtf dudvb((z —x(u,v))2) (1.32)/3 00 /31—13 0 ,8 —2/3If A is the Lorentz transformation between the frames of reference, then h(z) in thenew reference frame is h’(z) = AhAT(A_lz). A general Lorentz transformation can beexpressed as the product of a boost and a rotation. It is expected that A will involvefirst rotating the string such that it lies along the z-axis and then boosting it so that itstransverse velocity is zero. Let the Lorentz transformation which takes the sloped stringinto a straight flat string be7 0—v’yO 10 0 00 1 0 0 01 0 0(1.33)—7)7 0 0 0 0 cos() sin()0 0 0 1 0 0 — sin() cos()where is the angle of the rotation, v is the velocity between the old and the new frames,and = — v21• A boost transverse to the z-axis does not mix the z-componentof the tensor with any of the others, therefore, the rotation must cause the and 23components of the tensor to vanish. This fixes the value of phi as = — arcsiri(Because the perturbation metric of a straight string is diagonal with and -22 equalto zero, the boost should cause the h22 component to vanish. This fixes the value ofthe velocity at v = /3//2(1 + /3). The A’ operating on z has no effect because thecoordinates only appear in the argument of the delta function in the form of a LorentzChapter 1. Cosmic Strings 16invariant. Therefore,(i+,ã) 0 0 0= 8Gf dudv((z —x(u,v))2) 0 0 0 0 (1.34)0 0 0 —(1+)The only difference between this metric and the metric of a straight string is a constantfactor of (1 + )/2. This factor is merely a question of rescaling the u and v parameters.Because of the presence of ‘dudv 9 9’ in the integrand of the stress-energy tensor (1.26),transformations of the form v —÷ av + and u —* !3u + cause no change in the integral(1.34). The Jacobian from the change of variables in the integration measure cancelsthe extra terms arising from the partial derivatives. However, when the integral over thedelta function is performed a term of the form 2ja (x— z)I or 2IOx. (x — z)j appearsin the denominator. Consider, for example, what occurs if one integrates over v first,resulting in the second term appearing in the denominator. If this is done, then for astraight string,9x• (x — z) = —2u + Ox• z (1.35)whereas for the sloped case,Ox.(x—z) = —u—$(u-i-l)+l-i-Ox.z (1.36)A change in u and v does not affect affect the coordinates of the string trajectory, x, andso it is possible to make the determinants equal by the repararnetrization1— 9x•z ) (1.37)2Making this change in variables, causesôx.(x—z) ‘,1(—2u+Ox.z) (1.38)Chapter 1. Cosmic Strings 17This results in exactly the factor required to cancel out the difference in scale that arosein the Lorentz transformation. Thus, the metric for the sloped piece of string is exactlythe same in the new reference frame as that of a straight flat piece of string.The integral could also have been done by integrating first in u instead of v. Doingthis and following similar arguments results in a similar reparametrization of v. As theorder of the integration should commute, it is actually necessary to reparametrize bothof the variables u and v in both cases. Fortunately, this does not change the result asthe denominator depends only on one of the two parameters (and the trajectory of thestring, x, is independent of the parametrization).Throughout this argument, x(v) was always fiat, so the string trajectory is reallyjust that of a single travelling wave. The metric in this case was solved for exactly byVachispati [17] (1.16). One can also, therefore, demonstrate this same result in terms ofhis metric. Again, for simplicity, I assume that g = 0 restricting the wave to the yz-plane.The metric is thenf’2 0 —f’ f’2010 0 (1.39)—f’ 0 1 —f’f’2 0 f1 f12Using the Lorentz transformation (1.33), and following the same reasoning as before, onefinds that g = tan’(f’) and v = —f’//1 + f’2. In this case, however, the coordinatesare not in the form of a Lorentz scalar and so do transform under A—’.= t1+f’2+f’xf’(t-z)xx=_____f’2t+zzz= yl+f,+fx (1.40)Chapter 1. Cosmic Strings 18However, the expression f(t — z) is just that of a straight, but sloped line, so by choosingthe origin correctly, f(t — z) = [t— z]f’. Therefore, under the inverse Lorentz transformation,ln(2{(x- f(t - z))2 + y2}) -1n(2{( f))22}) 1n(22{x+y}) (1.41)This is exactly the factor for a straight fiat string. As a result, the metric transforms as0000h(z) = AhAT(A_lz) = —4Gln(2{x+y})0 1 0 0 (1.42)00100000As before, the metric of a straight but sloped string is the same as that of a straight flatstring.Because the metric of a straight sloped string is the same as that of a straight flatstring, one might expect that the metric from a wave travelling on a straight sloped stringwould be the same as that of the wave travelling on a straight fiat string. This is almost,but not quite, correct. What I found is that a wave on the sloped string, when viewed inthe frame of reference of that sloped string appears as a single travelling wave, but onewhich is slightly distorted from how it would have appeared on a fiat string.To see this, one considers a left-moving triangular wave pulse of similar form to theright-moving pulse considered earlier.Chapter 1. Cosmic Strings 19x°(v) = vx’(v) = 0x2(v) = (v +12)8(v + 12) — 2cvO(v) + c(v —l2)O(v — 12) (1.43)x3(v) = v+(—l)(v+l)O(v+l(1—&)(v—1)9(v—where o is a positive value between 0 and 1, & = — , 12 is some positive constant.Considering a point at which —l <u < 0 and—12 < v < 0, one finds:2 0 a+3 ti—/3v(z) =8GLfdudv((z_x)2) 0 0 0 0 (1.44)/+a 0 0—3 0 /3i—o3 —23öApplying the same type of Lorentz transformation as was used to go into the frame ofreference of the sloped right-moving string, one finds that /i is proportional to2 0 ii0 0 0 0(145)0 0/1_2_1 0 i _2/1_,2where ij is a parameter between 0 and 1. The factor in front of the matrix is cancelled inthe same manner as before by reparametrizing the string. This results in the metric fora left-moving wave on a fiat string, but one which is parametrized by ‘ij and not i. Theparameter i can be implicitly expressed in terms of the v and through the followingexpression.2(1+3i—c6) (146)(1+&—c49+1—)If—l< u <0 then i> a if—i2 <v <0, i a ifO <V <12, and i=a =0 otherwise.This results in a distorted image of the original left-moving wave, figure (1.2).Chapter 1. Cosmic Strings 20Figure 1.2: A wave’s appearance is distorted when viewed from the frame of reference ofa sloped piece of string (b) as compared to how it would have appeared on a straight flatpiece of string (a).However, as the same transformation may be applied to all sections of the left-movingwave (if the wave is entirely on the section of the right-moving wave for which —l <u <0), the result is just the metric of a single left-moving travelling wave. Thus, there is noradiation produced by the left-moving wave travelling along a straight but sloped sectionof string. Making piece-wise transformations of this sort, it is possible to see that theonly time at which it is not possible to view the metric produced by the string as thatof a single travelling wave emitting no radiation occurs when the kinks of oppositelymoving waves collide. Here there is no reference frame in which the component of theright-moving wave is flat across the kink of the left-moving wave or vice-versa. Therefore,the gravitational radiation emitted from a collision of two triangular waves on a cosmicstring must be produced when the kinks of two oppositely moving waves collide.(a) (b)Chapter 2The FormalismAlthough determining the effects of back-reaction on a source emitting gravitationalradiation is, in general, fairly difficult, in the case of cosmic strings it is much easierbecause one can use a perturbative approach to the problem. If the gravity waves beingemitted by a source contain a large amount of energy, this energy will cause significantdistortions to the background metric of spacetime. These distortions would affect themotion of the source generating the waves and alter the characteristics of the wavesactually produced. However, this means that the gravity waves would have originallycaused different distortions in spacetime, and so on. Because the answer affects theformulation of the problem, these types of problems are quite difficult to solve.Another difficulty with this problem is that the back-reaction ‘forces’ are entirelygauge dependent. It has been shown [19] that a cosmic string can always be viewedas being a geodesic two-surface. Although the internal metric can change, there are noinvariant ‘forces’ on the string. In this view all of the effects of the radiation appear asdistortions of the background metric; the string stays perfectly straight. This picture isnot, however, a very useful way of thinking about the back-reaction. One would ratherconsider the string as moving in some kind of fixed almost flat background and have thestring’s motion altered by the back-reaction. This is closer to what a person floating inspace would observe as happening to the string. Gravitational waves are emitted fromthe string and the wave on the string is subsequently damped out. One way to fix thebackground and thereby force the back-reaction effects onto the string itself is to choose a21Chapter 2. The Formalism 22specific gauge. Unfortunately, the choice of gauge is not an obvious one and so whicheverone is chosen, it must be shown that the gauge is in some sense reasonable.With cosmic strings, the first difficulty can be overcome because the disturbanceof fiat spacetime caused by the emitted radiation is very small, G 10—6. Thisfeature makes it possible to use the weak field approximations of linear gravity and aperturbative approach to the back-reaction problem. Choosing a specific gauge, one caninitially assume that the effects of the changes to the background spacetime caused by theradiation do not significantly alter the motion of the string. That is, one assumes thereare no back-reaction effects on the string in order to obtain a first approximation for theradiation produced. Once one has the radiation emitted, it is then possible to determinewhat its effects on the string would be. Taking into account this slightly revised motionof the string, the radiation produced can be modified. If the changes caused at eachstep are small, then this sort of iterative approach will converge on the real solution forthe radiation and its corresponding effects on the string. This is the technique used byQuashnock and Spergel [20] in determining the effects of gravitational back-reaction onoscillating cosmic string loops. Their approach can also be made to apply to collidingtravelling waves on long cosmic strings. In my case I will just be determining the firstcorrection to the motion of the string.The second difficulty is the gauge dependence of the force. To remove this dependence,I chose to work in the Lorentz gauge (1.3). Unfortunately, this gauge still contains somegauge freedom in the form of any gauge transformation satisfying D = 0.An analogous gauge freedom arises in the simpler case of Maxwell’s equations forelectromagnetism. There, the Lorentz gauge choice permits the freedom Ac, Ac, + Oc,Xwhere Ac, is the four vector potential of the fields and X is a scalar satisfying DX 0.The gauge can be completely fixed by specifying a condition on Ac, such as A0 = 0 [21].Similarly, with gravitational radiation, one often fixes the gauge by choosing to workChapter 2. The Formalism 23in the transverse traceless or TT gauge. With this choice the freedom of the Lorentzgauge is removed by forcing the metric to obey the following conditions [12].h=0 h03 =0Unfortunately, only pure gravitational waves can be put into the TT gauge. Because Iam using the metric generated by the string, it is not a sourceless metric and cannot beput into the TT gauge.The best way in which to restrict the Lorentz gauge for this problem is not immediately obvious. Therefore, I decided to explore the force equations using the Lorentzgauge with its associated freedom and see what restrictions were required in order forthe force to satisfy some reasonable conditions (sections 2.1 to 2.3). It was hoped thatrequiring some reasonable properties of the force would sufficiently restrict the gaugesuch that any remaining freedoms would not influence the back-reaction effects on thestring.This approach was partially successful. In section 2.2, it becomes necessary to restrictthe metric to Vachaspati’s gauge. This reduces the gauge freedom down to a singleparameter, Vachaspati’s integration constant Q2. Unfortunately, although the influenceof this remaining freedom cancels out to some extent, (section 4.3 and Appendix C), itcontinues to have an effect on the back-reaction forces as will be seen in chapter four.It is necessary, therefore, to completely fix the gauge. This can be done by selectinga value for the integration constant 2, section 3.3, or by fixing the gauge from thebeginning using conditions similar to those which restrict gravitational wave metrics tothe TT gauge. For example, one might try imposing just the h03 = 0 condition. Thislater approach was not investigated in this thesis.Chapter 2. The Formalism 242.1 The Force EquationThe first step I take in determining the effects of the emission of radiation from cosmicstrings is to develop the equations of motion in terms of a metric more general than thefiat Minkowski metric which was used in section 1.2. Denoting the metric as g, this isdone by minimizing the Nambu action with respect to the coordinate x.S _zfddr7=— f — (2.1)With the standard co-ordinate choice ofOrXOyX = 0 O,-XOrX+OcyX6oX = 0 (2.2)one obtains the equations of motion [20](O — O) =—r(oo — xOx) (2.3)Switching to the null co-ordinates a and v, the conditions (2.2) becomeöxOx = 0 = 0 (2.4)and the equation of motion is0uvZ (2.5)In the case of the back-reaction problem for the cosmic string, the string is itself thesource of the perturbation metric h. Therefore, using (1.1) and making the weak fieldapproximation in determining the Christoffel symbol, one can write (2.5) as8uOvZ (ha7, + h7,a — (2.6)Chapter 2. The Formalism 25where= 8G 8 f dudv Fa ((x — z)2) 6(z° — x°) (2.7)andF = Ox&x + OvxOux3 — (2.8)Equation (2.7) is found by starting with equation (1.5). That is, I have chosen to operate in the Lorentz gauge. The Heaviside function in (2.7) is inserted to insure causality.Henceforth, I shall use x to denote the point on the string generating the metric perturbation and z to denote the point on the string being affected by the perturbation.Equation (2.6) determines the force on the string at a specific point z(u, v). To findthe effect of this force on the string, one integrates with respect to either u or v. Forexample, one finds the change in z, L(Oz), due to the metric perturbation h byintegrating with respect to the variable v.= fdvOãvz (2.9)To find the first order change in the trajectory, one starts with a solution to theequations of motion in flat Minkowski spacetime so that conditions (2.4) are satisfiedwith respect to ij,,. A reasonable requirement of the force is that its effects shouldnot result in a perturbed trajectory which violates this initial gauge condition. The forcegiven by equation (2.6) meets this requirement. The change, (t9z’), can be safely addedto because this combination still satisfies the conditions (2.4) in flat spacetime.’‘The equations of motion (2.5) require that the gauge conditions be satisfied. Therefore, as was laterpointed out to me, a better requirement of the force is that the perturbed trajectory satisfy not oniythe gauge conditions in flat spacetime, but also those in the perturbed spacetime. The force (2.6) doesnot satisfy this condition because azz”h, 0. This is discussed further in section 2.2 and inChapter 5.Chapter 2. The Formalism 26== — dv Uz1,1fY ha7,9uZ OvZ= _0uZ7OuZfdVha,Ovz (2.10)Because h h(z), the integrand in (2.10) is just a total derivative of h with respect tov and the integral can be expressed as:Jdvk7,Oz = JdVha7,v h7It (2.11)Because of the causality requirement, h7 = 0 at the lower bound of v. One needs to thenconsider the upper limit. For clarity, I shall use ü and 3 to indicate the null coordinatesof x in order to distinguish them from u and v, the null coordinates of z.= 8Gt1imJdüdPa7b((x(ü,3) — z(u,v))2)8(z° — x°)= 8Glimfdu F O(z°—x°) (2.12)2IOx. (x — z)IBut, in the limit as v goes to infinity, IOx. (x — z)I = I(x Oz)v + fQü, €, also goesto infinity. Therefore, the integrand goes to zero and the entire integral in (2.12) mustbe zero. This means that ãz• (ãz + zaz) = 0. Thus the change in the motion of thestring satisfies the co-ordinate conditions (2.4) for the string in flat spacetime.It should be noted in passing that by symmetry h1,,7°°= 0 also. This fact will beuseful later on.2.2 The Back-Reaction on a Single WaveAs was seen in the previous chapter, a single travelling wave on a string emits no radiation.Therefore, for the Lorentz gauge to be a reasonable gauge choice, one would expect thatthe single wave experience no back-reaction force. All trajectories, z, of the string mustChapter 2. The Formalism 27satisfy the wave equation and so can always be written as the composition of left andright moving components, z(v) and z(u) respectively. A single right-moving travellingwave can be created by requiring that z(v) = az v where z is some constant four-vectorand by letting z(u) be an arbitrary function with the restriction that Oz and z bothsatisfy the coordinate conditions (2.4). It should also be noted that as ãz is constant,it will be the same at both the point on the string which generates the back-reactionforce and the point on the string which is acted upon by the back-reaction force. Thatis övzIt(v) = 8x’(i3) = k” where ki’ is some constant value for all v and ii.The back-reaction force, 8Oz”, is given by equation (2.6). The first term in (2.6)involves which is proportional tof + OxOx7— Ox) 6((x — z)2)O(z° — x°)— 0 [Umaxd (070x + 0jx.0x — 0x) 0 za0z° J— 2(0x. (x— z))1 JUmax dü +x130x —i70x . ãx 0z2— (0x.(x—z))= 0 (2.13)because05Xcx0vZa= avav = oThe boundary Umax is the solution of (x(i, 3) — z)2 = 0 for ii in the limit as i3 goes tonegative infinity and which takes into account the causality condition that z° > x°.Equation (2.13) shows that the first term contributes nothing to the force for a singleright-moving wave.2The result in equation (2.13) assumes that there is no contribution from actingon the boundaries of the integration. To see that this is the case, one inserts a past-time2The cancellation in (2.13) seems to depend on doing the i integration first. This is not actually thecase; the order of integration does commute. This can be seen most easily with the specific example ofthe colliding triangular pulses as will be shown later in section 2.4.Chapter 2. The Formalism 28cut-off, O(ü + i3 + r0), into the integral.3 Taking T0 to +oo at the end of the calculationwill then remove the cut-off. This technique changes the bounds of the integration. Thenew integration limits are found by solvingü = —r0i(ü,z)— —Tt2—2tu—2uaf.(b—F)—(b—F) (214)2[Ox.Oxü+t+O,f.(b—)]where t = u + v and b = x(ü, i3) — — O,xi3. This expression is just a quadratic in ii.Assuming T0 is very large yields the following implicit values for iiDjx.(b—z)U——T4, +aX .— 2(Ox.z—Of.b)U— —r0(Dx.Ox)— Umax ——2r08xwherep = [X(Umax) — = [X(Umax) z]? + [X(Umax) — (2.16)and the subscripts 1 and 2 refer to the components of the four-vector (x — z)As m becomes large and negative, at the lower boundary,O 1’________—r0 + j = bounded terms (2.17)Ozc \ OXOX jHowever, the integrand involves a term 6x (x — z) andlirn Ox• (x — z) = urn Ox• (&uxü + b — z) = (2.18)Therefore, the lower boundary will make no contribution in the limit as T0 goes to infinity.To examine the contribution from the upper boundary one approximates Ox7 asOUXIU=Um This approximation becomes exact in the limit as T0 goes to infinity and Ii3A problem with using a past-time cut-off is that it destroys the Lorentz gauge condition for anyfinite choice of r0. However, the gauge is restored if r0 is taken to infinity.Chapter 2. The Formalism 29goes to Umax. Therefore, at the upper boundary,Of fJ 9Xchm Umax— ) = (2.19)UUmax OZ’ 2T0 OX OjX O3x OxBecause the derivative is proportional to the contribution from this boundary termcontains the contraction zOxa which vanishes. Therefore, the upper boundary termalso gives no contribution to the integral.The second term in (2.6) has the form Using (2.7), one can expressas=— z)2) (2.20)The O1,z term can be brought inside the integral and so the integrand is proportional toFai3OvZ == (OxOx +—= O28x.Ox —= 0 (2.21)The integrand is zero and, therefore,h7,Oz13Ovza is also zero. The third term in (2.6)also involves a contraction of Oz’ with F and so it too will cancel to zero. Therefore= 0 for a single right-moving wave and, by symmetry, for a left-moving singlewave as well.Because the back-reaction force on the single travelling wave is zero, one expects thatthe solutions for the trajectories of single travelling waves should satisfy not only thecoordinate condition in flat space,= 1]IwOvZ1IOvZi’ = 0 (2.22)but also the coordinate condition in curved spacetime. That is, one expects that for asingle travelling wave= = 0 (2.23)Chapter 2. The Formalism 30Considering once again the example of the right-moving wave, the first of the twoconditions in (2.23) is easily seen to be true. The metric h,2 expressed as an integralinvolves the factor F. Pulling the Oz’’ term inside the integral, one gets the contractionFOz’-”. As this factor equates to zero, one obtains the desired result: kpvãvz1Lövzi1 = 0.The second condition in (2.23) is not true for the full integral expression of the metric.In order to satisfy (2.23), it will be necessary to change the gauge. Consider h, for thesingle travelling wave which is calculated using Einstein’s equations (1.5) and a retardedGreen’s function.fUm Fh(z) = 8Gt / dü (2.24)J—oo 2Ox. (x — z) i(it,z)Equation (2.24) has some problems with infinities. At ii = —, Ox. (x — z) = q, = ccwhile at ü = umax, q = 0. As before, one resolves these problems by inserting a pasttime cut-off, 8(ü + i3 + r0). Installing this cut-off changes the bounds of the integrationover I. The new bounds are given by (2.15). With hindsight, it helps if the integral isalso split in two at the pointU = Umax— Po (2.25)—2r0Ox Of1X UUmwhere p0 is some arbitrary constant. Therefore,I [Umaz_2°h(z) = 4G[L i duJ—T0 + 3x. (x — z)+JUmx__2ro4z.aadü (2.26)Um—2ro . (x — z) jI now consider each of the two integrals separately. To evaluate the lower integral,one makes the approximation that Oj,x’ = OXjUm. Rewriting ãx (x — z) asOj5x OüX(Umax— ), the first integral in equation (2.26) becomes4GIf_20 d(Umax) = 4G1t ln() (2.27)—2i&z OüX(Umax — U) Ox UUmChapter 2. The Formalism 31Associating 1/pr, with Vachaspati’s integration constant 22, (2.27) can be shown tobe exactly Vachaspati’s solution for the metric of a right-moving single travelling wave(Appendix A).= 4GiOxln(2p) (2.28)U—Um,For a single travelling wave, if z is a point on the string, Umax = u. The equation forthe maximum value of ii isz— . bU = ‘Umax = (2.29)a•where b = x(u) — iiOx. But, for the back-reaction on a single left-moving wave, z is apoint on the string and so Ox = Oz. Therefore, (2.29) can be rewritten as— az.z(u)—az.bUmax—avz. ouxA solution to this equation is that OXUmx = 9z which implies that Umax = u. Tosee this, rewrite (2.30) asOvZ X(U)Iü=u = 9z• z(U) (2.31)Expanding xü) about i = u in a Taylor series givesavz öXI(Uma — u) + OXIu(Umax — u)2 +... = 0 (2.32)The solution to this equation is Umax = u. The Taylor series expansion of x(u) maynot, however, be justified in the case of a piecewise continuous function such as (1.30).Nevertheless, Umax = U 1S the only solution in this case also. Suppose, for example, thatU = —l— E with E > 0 and ii> —l so that Ox 8z. Then-(-1 - 1)U-(_l(/fZ /32-1))‘Umax— —1— i— _l(1+\/1_/32)_2- (i+i_/32)= -l-- /32) < -l (2.33)Chapter 2. The Formalism 32Other cases also yield contradictions leading one to the conclusion that UmaT = u for thesingle travelling wave.Because UmaT = u, one can rewrite (2.28) as_________2= 4G ln(2 p) (2.34)(JuzFrom this it is easy to see thath16zh9z’= 0 because FDz = 0. Therefore, thissection of the integral, (2.27), which is equivalent to Vachaspati’s metric for a right-moving wave, satisfies the second of the conditions in (2.23). Using this form of themetric, it is easy to see that Vachaspati’s metric also satisfies the first condition in(2.23). F1az’2 equals zero which implies that haz’1Oz” also equals zero.4I now return to the second integral in equation (2.26). It is this integral whichcauses problems for if one actually evaluates the integral, one finds that isnon-zero. However, this integral just corresponds to a gauge transformation within theLorentz gauge. Because the integral is symmetric with respect to the cr and indices,one can associate it with a gauge transformation of the forma3Clmaz 2 8 8- P= 2Gj I - ° dü (2.35)i_To + ãx . (x — z)Taking the divergence of one finds that it is non-zero for any finite value of r0 butzero in the limit as r0 goes to infinity. The non-zero nature of D at finite values of r0 isbecause the past-time cut-off destroys the Lorentz gauge condition. Taking r0 to infinityremoves the cut-off and consequently restores the gauge. One can show this directly bytaking the divergence of (2.35).As has been shown previously, a derivative acting on the upper boundary gives nocontribution to the integral. The derivative, , acting on the lower boundary results in“It can be shown that Vachaspati’s solution for the metric also satisfies the criterion that the backreaction force on a single wave be zero.Chapter 2. The Formalism 33the term_frc32 36O3x. (x — z) Ox xUUmzin the limit as r0 goes to infinity. However, the contraction F°Ox is zero and so thelower boundary also makes no contribution to the integral. Acting on the integrand, thederivative givesö fUma— —2r0 Oz,ôj2Gp— i duOZ J—,- + (x — z)(Urn 2r88z=2GL1 dui—To + (Ox (x — z))2= 0 (2.37)Therefore, J° = 0 and the gauge transformation given by (2.35) is a transformationwithin the Lorentz gauge.The gauge freedom within the Lorentz gauge has a significant impact on how reasonable the gauge choice is. While Vachaspati’s gauge satisfies the conditions (2.23), thefull integral expression of the metric does not. One implication of this is that the metricchosen for investigating the collision of two waves should reduce to Vachaspati’s gaugechoice at least in those regions before the waves collide in which the two waves may beconsidered as just two single travelling waves on a string. Using the integral expressionfor the metric does not satisfy this criterion and, as shall be seen in chapter three, leadsto unreasonable results. The back-reaction forces turn out to be non-integrably infinite.This is the first indication one has that a better metric choice is the composite metricdiscussed in chapter three.Unfortunately, this is not the whole solution to the problem. Another difficulty ariseswhich is related to satisfying ôuzuzvhv = 0 in the case of two colliding waves. Towardsthe end of working on this thesis, it was discovered that the force contains a componentChapter 2. The Formalism 34parallel to the string. This parallel component causes a shift in the internal null coordinates of the string which causes the perturbed solution of the string to no longer satisfythe null coordinate conditions (2.4). That is o.This means that what I have called the force, equation (2.6) is not what one would liketo define as the physically reasonable force. Because only motion perpendicular to thestring has any physical meaning, one would like for the force to also be perpendicular.Although the force (2.6) does give a first order correction in flat spacetime, it would bemore reasonable to remove the contribution from the parallel components. Doing this,however, does not solve what turns out to be the major drawback of using the Lorentzgauge to study the back-reaction on a cosmic string, the gauge freedom of the solution.As discussed in Chapter 5, even the perpendicular force still contains the gauge dependence expressed by the arbitrary integration constant, 22, in Vachaspati’s single wavemetric. To see where and how this gauge freedom influences the back-reaction and alsoto get a rough idea of the effects of the gravitational back-reaction on the collision oftwo travelling waves it is sufficient to use the force definition given by equation (2.6).This force also has the advantage of being much simpler to work with than the correctedperpendicular force, equation (5.13).2.3 The Energy Lost by the StringAnother reasonable requirement of the gauge choice is that the change in energy of thestring produced by the back-reaction forces should equal the energy radiated away fromthe string in the form of gravitational radiation. To find the change in the energy of thestring, one first determines the derivative of the stress-energy tensor,Chapter 2. The Formalism 35T,(y) = pf du dv64)(y — z(u, v)) {Ouz3vzv + auzvOvzP}=— f du dv { [ö()(y — z(u, v))] Oz + — z(u, v))]U=+oo V=+oo=— f dv4)( — z) — p fdu64)(y — z)+ 2p J du dv64(y — z) Oãz (2.38)For finite values of y the boundary terms disappear and one is left with5T”’,(y) = 2pJdudvOuOvz5(4)(y— z) (2.39)The four-momentum can be written as= fd3l7T10 (2.40)Unfortunately, this equation assumes that the stress-energy tensor goes to zero outsideof some finite spatial region. This assumption is not satisfied in my case because thecosmic string is infinite in the y3- or z-direction. However, I am only interested in theamount by which the back-reaction force changes the momentum of the string. Betweenany two instants in time, the back-reaction force is confined to a finite length of thestring as are the two colliding wave pulses. Therefore, one can always choose a constant,L, large enough such that the region containing the wave pulses and the back-reactionforces is contained between —L < z < +L. The string outside of this region is straight.A straight string does not experience back-reaction forces and so one can subtract thefour-momentum of the straight string from that of the string with two colliding waves.The remaining momentum will still include the effects of the back-reaction forces butwill only involve an integral over a finite region, a sphere of radius L. This allows one to5The parallel component of the force contributes nothing to as discussed in section 5.3 so thisenergy equivalence will hold for both the full force (2.6) and the perpendicular force (5.13).Chapter 2. The Formalism 36use equation (2.40) to determine the change in the momentum of the string due to theback-reaction forces.The change in momentum between times t1 and t2 is=dy°P,o=dy° fd3T°,0=£2 dy° jd3(T, — (2.41)where i is an integer ranging from 1 to 3 and V is a spherical volume of radius L. Onecan also consider the bounds on the integration over dy1 and dy2 to be —L to +L becausethe string is finite in those directions and so the stress-energy tensor is zero beyond thisrange.The second integral in equation (2.41) can be rewritten as two terms representingthe stress-energy tensor containing the back-reaction forces, TB and the tensor of thestraight flat string, TF’”. These integrals are then simplified using Gauss’ theorem.£2 dy° f d3=dy° fd3(TB,— TF,)= f dyof= f2dy° fd2S n (T8— TFt) (2.42)where nt is the normal to the surface S which is a sphere of radius L. On this surfacethe stress-energy tensor of the string with two waves is the same as that of the straightstring and so the second term in equation (2.41) is zero. Taking L to infinity does notchange this result.The first integral in equation (2.41) can be evaluated using equation (2.39).Chapter 2. The Formalism 37ft2 f ft2 fJ dy° J d3T’, = ] dy°] d3(TB, — TF’”,U)= 2t f dudv f2 dy° Ld3il( [O3zö(4)(y — z)IB_[ôaz5(4)(y— z)]F) (2.43)but, 8,Uz” = 0 for the straight string, and so equation (2.43) equals2tfdudv [OOVz]BO(t2— y°)O(y° — t’)e(L —y3)O(y + L) (2.44)Taking the limit as L goes to infinity removes two of the Heaviside functions in equation(2.44) and taking t2 and t1 to positive and negative infinity respectively removes theother two.Dropping the subscript B, one finds that the change in the four-momentum due tothe back-reaction forces on the string is given byzP = 2iifdudvOuavz(u,v) (2.45)The change in the energy of the string is then2iLtfdudvauavzo=— J dudv OzOz + h7,a — h,7) (2.46)Because the first two terms in this integral are just total derivatives, they equate to zerowhen evaluated (2.12). The third term can be expressed, using (2.7) and (2.8) as=— z)2)9(z°— x°)] (2.47)The actual energy radiated away from the string can be found using Weinberg’s formula for the total gravitational wave energy{18, 201. Unfortunately, there are difficultiesassociated with applying this formula to my problem. Weinberg’s formula assumes thatChapter 2. The Formalism 38the metric, g1jv, may be written as ij + h, where ij, is the metric for flat Minkowskispacetime and is a metric which goes to zero as the distance from the source of thegravitational radiation goes to infinity. However, the spacetime of a cosmic string contains a conical deficit and so when the metric for the string is split into ij + theh will not go to zero at large distances from the string.‘When the mass density, u, of the string is small, the conical deficit is also very smalland so the metric for the string does not deviate very much from Minkowski spacetime.The associated although not zero at large distances from the string, is small andone might hope that the error associated with using Weinberg’s formula would not besignificant. In the limit where t goes to zero, this error vanishes but in the limit wherebecomes very large the error becomes more and more significant. However, for largevalues of , the entire approach followed in this thesis becomes invalid. Neither the weakgravity limit nor the perturbation approach to finding the corrections to the trajectoryof the string are applicable any longer. Fortunately, the only situation which I aminvestigating in this thesis is the one in which is very small.The reason for which Weinberg assumes that the perturbation metric h,LV goes tozero as the distance from the source goes to infinity is that he requires convergence of theintegral for the total energy and momentum of the source, fd3fl’°”. One way in whichto achieve this is for the source to be confined to a finite region. In this case, at large r,T = O() and the integral converges. Although h1 for my problem does not approachzero at spatial infinity, the source of the radiation is still finite. The radiation is onlyemitted during the collision of the two oppositely moving wave pulses. As both wavepulses are of finite size, the radiation is emitted from a bounded region in spacetime.Another way to look at this is to see that the residual metric of the two colliding wavepulses (Chapter 3) is a finite metric that does go to zero at large distances. This is theonly component of the metric for the two colliding wave pulses that emits radiation [5].Chapter 2. The Formalism 39The radiating portion of the metric does satisfy Weinberg’s criterion.Unfortunately, this is not a solution to the problem. Weinberg’s formula involves thestress-energy tensor, or the metric, in the form of a quadratic. The full metric for thecollision of the two waves is a sum of the residual metric plus other non-radiating metriccomponents which are not finite in size and may not satisfy Weinberg’s requirements.One must, therefore, be concerned about the cross terms of these metric components withthe residual metric. If the cross terms make any contribution to the radiated energy oforder h2 then Weinberg’s formula cannot be used unless the integral over all space of thestress-energy tensors for the other metric components also converge. One might suspectthat the cross terms to not contribute to the energy equation because the radiationemitted from the colliding waves is polarized whereas the straight string component ofthe metric, for instance, is cylindrically symmetric. However, I was unable to show thatthe cross terms do not contribute significantly nor that the integrals of their stress-energytensors converge and so the results of this section must be viewed with caution.Henceforth, I shall assume that Weinberg’s formula can be applied to my problem.Under this assumption, the energy per unit solid angle emitted in a direction is= 2Gj w2 — IT(,w)2] dw (2.48)where T1w(k) is the Fourier transform of the stress-energy tensor and the * indicatescomplex conjugation. The Fourier transform of the stress-energy tensor is given by,T(k) = fd4zT(z)exp(ik.z) (2.49)where k is the wave-vector of the gravitational wave emitted by the string. Usingfw2dwd = fd4k6(k2)k0c(k0) (2.50)one finds after some algebra that (2.48) gives= 4 J dudvOuzAOvzv f düd fr Jd4k6(k2)c(kO)kOe_x) (2.51)Chapter 2. The Formalism 40Itzykson and Zuber [22] give the radiation Green’s function D(z-x) and its Fourier decomposition asD(z—x) = _€(z0_x0)((x_z)2)= (2)3fd4ke_(k2)f(k0) (2.52)Because E(x) 28(x) — 1, and as antisymmetry makes the —1 component vanish, (2.51)can be rewritten as= 8G[L2fdudvfdd aUzaoVzaO[o(z0— x°)((z— x)2)] (2.53)Finally, by comparing equations (2.47) and (2.53), one can see that the energy emittedby the string according to Weinberg’s equation is just the energy lost by the string due tothe back-reaction force. There is no gravitational wave energy unaccounted for and so allof the effects of the back-reaction are occurring on the string and not in the backgroundmetric.The equations for the back-reaction force in the Lorentz gauge do seem to satisfysome reasonable requirements. First, the net effect of the force is a change in the stringtrajectory which maintains the coordinate conditions in flat spacetime. Secondly, thesingle travelling wave experiences no back-reaction force. Lastly, the energy lost by thestring is equal to that emitted as gravitational radiation. For these reasons, the Lorentzgauge seems to be a reasonable gauge choice in which to investigate the back-reactionforce.Unfortunately, despite the aforementioned attributes, the Lorentz gauge choice alsohas some unreasonable features. Chief among these is the fact that this choice of gaugeleads to equations for the force which contain an arbitrary parameter, section 3.3. Thisparameter has a significant influence on the effect of the back-reaction forces. Anotherunsatisfactory feature is that the single travelling wave components of the compositeChapter 2. The Formalism 41metric (section 3.2) make a contribution to the back-reaction force. This is somewhatdisturbing because the single travelling waves do not emit any gravitational radiation andso one does not expect them to contribute to the force. There are also problems withthe perturbed string trajectory not satisfying the coordinate conditions and the forcenot being perpendicular to the string.6 Therefore, although the Lorentz gauge choicedoes seem very reasonable in some regards, it also retains a few flaws indicating thatperhaps a better gauge choice is still possible. However, what this gauge might be is farfrom evident and so I shall continue to investigate the back-reaction force in the Lorentzgauge.2.4 Evaluating the Force for Triangular WavesAs a useful first step towards determining the back-reaction on two colliding wave pulses,one can find the force due to equation (2.6) using the full integral expression of the metric.Although there are difficulties associated with doing this (section 2.2), it will provide auseful illustration of the problems involved with the gauge choice and the freedom ofthe Lorentz gauge itself. Additionally, the equations developed here are required for thecomposite metric. The composite metric which is formulated in chapter three is a metricwhich uses Vachaspati’s gauge for those parts of the metric which behave like singletravelling waves. It solves many but not all of the problems one encounters with the fullintegral expression for the metric.My goal then is to evaluate the back-reaction force for the collision of two travellingwaves on a cosmic string. Although for complicated waves this could be difficult, forthe simplified situation which I chose, the case of two triangular pulses, it becomesstraightforward. Because derivatives of the wave are piecewise constant, one can divide6This problem can actually be corrected by removing the parallel components of the force as is shownin Chapter 5.Chapter 2. The Formalism 42up the integration in the terms into regions over which F is a constant. Thispermits the analytical evaluation of the force. The force from (2.6) will involve threeterms of the type= 8G[L07f düdF((x — z)2)O(z° — x°) (2.54)= 4Gf1fd-- (2.55)Ure u(v,z)where Üret refers to the retarded solution of (x(il, ) — z)2 = 0. Because Ox• Ox = 0, thedeterminant in the integrand does not involve u. As the slopes Ox and Ox are constantsover the integration region, the integral can be evaluated.=v1r= 4GO7Ox . axln(Ox. (x — z)) - -Uret VV2(O-x . Ox)(Ojx . (x — z))>< { + • — 3x } (2.56)Ure U,.ewhereav 0 if v is a constant= (2.57)0z7 (a—z)7 otherwiseV.This evaluation was done by finding the integral over u first. It is also possible to determine the force by integrating over v first. Although superficially the formulae aredifferent, the two approaches give the same value for the integral unless the first integration is being done over a region which is degenerate in either ü or i. In other words,there is no difficulty commuting the two integrals unless one is in a region for which asingle value of i in the equation (x(ñ, i3)— z)2 = 0 results in multiple solutions for il orfor which a single value of Ii gives multiple values of i.If z is on the string, then evaluating the path in iii-space over which (x— z)2 = 0results in some degeneracies. An example of a path with degeneracies is given in figureChapter 2. The Formalism 43Degenerate ii’—p1ot Non—degenerate —p1ot100 10050 50U) U)0-—50 —50—100 —100—40—20 0 20 40 —40—20 0 20 40c’—vaiues ‘v—values(a) (b)Figure 2.1: Figure (a) shows the degeneracies which arise in üi3-space along the path(x(ii, i3)— z)2 = 0 when z lies on the string. Figure (b) shows a similar path when z is apoint not on the string.(2.la). The z-values in figure (2.la) correspond to a choice of u = 80 and v = 15 whichit will also be noted are the coordinates of the corner point. The values of i1 are seento be degenerate over a range of ü = 0 to ü = 100. The degeneracies in fi occur where0 < < 20. Although the curve also appears flat in the region where i3 < —20, this isjust a feature of the scale of the plot. If enlarged, this region would have a slight slopeto it. Similarly for the ii < —100, the curve is very steep but not quite vertical. Thedegeneracies occur over ranges of ü and i values for which 8,z = ôx and Oz = Ox. Itis therefore best to integrate first over i3 and then over for values of ii u and thenreverse the order of integration at the point (ii, i) = (u, v). Figure (2.lb) shows that fora value of z off of the string, there are no degenerate regions and so the integrals wouldcommute.By the same reasoning as in section 2.2, there is no contribution to the force fromChapter 2. The Formalism 44the boundary terms of equation (2.55). Once more installing a past-time cut-off, theintegration bounds becomeOjx.(b—z)V— —T0+V = Vmax— (2.58)—2r0OxThe derivative acting on the lower bound, near = —r0 results in terms which arebounded. However, the integrand involves the term Ojx. (x — z) in the denominator. Asr0 goes to infinity, Ox (x — z) goes to infinity and the lower boundary term goes to zero.At the other boundary, for large T0,07V = O,X (2.59)This term exactly cancels the other Ox7 term in equation (2.56) making the contributionto the force zero at this boundary also.As the contributions to the force from the boundaries at i Vmaz (u ‘. 00) and—*—oc are zero, one is able to obtain a specific value for the force at a given point z.Integrating the force over the internal co-ordinate v of z(u, v) will give the effect of theforce on the trajectory Oz’- of the string whereas integrating over u will yield the changein ôvzIA. The force as determined by equation (2.56) for the collision of two triangularwaves is discussed in Chapter 3.Before examining the force on two colliding waves, I return briefly to the force on asingle travelling wave. For the specific case of a triangular moving pulse, one can seeexplicitly that the force is zero regardless of whether one integrates first over ii or over 3.For instance, if one considers a left-moving wave. The kinks in the string, those placeswhere the integral has to be broken up to insure that the slopes Ox and Ox are constantthroughout the integration region, only occur at kinks in €. If one integrates first over i,Chapter 2. The Formalism 45then the integrand will include terms proportional toF[(x_z)7Ox.Ox (2.60)If Oz = Ox contracts on either the a or the index, then as has been shown before(2.21), the integrand would equate to zero. If the contraction occurs on the y index, thenthe integrand takes the form= 0 (2.61)If the integration is done over the ü index first, then the integrand is simply of the formFöx7 and the contraction of Oz on any of the three indices gives zero due to thecoordinate conditions (2.4). The force on a single travelling wave is zero regardless of theorder of integration.2.5 Other Features of the Force EquationsAnother feature of the force equations is that when ü < u then > v and vice-versa.First define q and q as= (x — z) (2.62)q = Oj,x. (x — z) (2.63)When solving the argument of the delta function, (x — z)2, for ü or i one finds that thecausality condition forces the solution to be such that= z°—x°+ã.(f—F)= ( — i) + ( — ) (2.64)As Oj has a magnitude of one, it follows that q, and similarly qv, are always greaterthan or equal to zero.Chapter 2. The Formalism 46The argument of the delta function requires that= 0 (2.65)If one considers ü to be a function of i and z, then one finds that=- (2.66)Ov qAs q and q are both always positive, this means that if is increasing then i is decreasingand vice versa.Now one finds the value of ü as —* —oo. In the case of triangular waves, x(ü, )can be expressed as x(i) + x((v)) for which both x(il) and x(i3) have piecewise constantslopes. Assuming x0 = i + i and writing x() as Ox ‘ü + b where= x(i) — Oxü (2.67)and b is also piecewise constant over the same ranges as Ox, this allows one to solve(2.65) for ü.— (z°—)2—()+bn—) p268— 2(z°—+O•(f(i)+bji—)Equation (2.68) actually gives four possible solutions for ü based on the four differentpossible ranges over which ôux and b are constant. However, only at most one of thesevalues is consistent with the Ox, pair used to find it. This permits equation (2.68) tobe used for a unique determination of ii.Taking the limit of large negative in (2.68), yields the following equation for ü(actually for umax):—z°+z314___________u = = (2.69)—i+ajx Ox•OxEquation (2.69) can then be checked for all possible ranges of u and v and it can beconfirmed that l u. For example, consider the case where—12 <v <0 and u < —1. IfChapter 2. The Formalism 47fl is to be less than u, then < —l also. Therefore, equation (2.69) becomes— —2u—v+äv+(—1)l—21= u+(v+l2)(1—c)>uwith the equality only occurring where o = 0 (& = 1). Checking all other cases, one findsthat this holds true regardless of the value of v or u. By similar arguments, one findsthat if i = v — where is small but positive, then €i — u < . As ii is monotonicallydecreasing while is increasing, it follows that i < v ii > u and by symmetry,i < U = 3 > v.In the special case that i3 = v then one finds that Ox’ = From equation (2.65),one hasu—u = (u)—F(u)I (2.70)This implies that there can be no kinks in the string between (u) and 1(u). Thereforex(u) is at a point which has the same slope as z(u) or is at the kink which adjoins thesection of string including z(u). For the instant where 3 = v, the value of ii is not uniquelydetermined by (2.65). The value for ii is degenerate as was seen in figure (2.la).7 It isworth noting, however, that if Ojx)L = Oz’, then the contribution to the force is zerobecause of the contractions with Oz and t9z’3 in (2.5) and the coordinate condition(1.28). This implies that a piece of string does not contribute to the back-reaction forceon itself. All of the back-reaction forces on the fiat pieces of the string are non-local.The forces are all due to perturbations in spacetime generated by the string at an earliertime. Although this is true for all fiat pieces of the string, it may not be true at the kinksthemselves. At those points, Oz and ôz are not well defined and the use of equation7Because of the symmetry of the equations, there are also degeneracies in 3 when = u and O z = x.Chapter 2. The Formalism 48(2.6) for the force is questionable. The non-locality of the force will become more evidentwhen I look at the force generated by two colliding waves in greater detail in Chapter 4.Chapter 3The Composite MetricIn the previous chapter, it was seen that the Lorentz gauge choice leads to a force whichis reasonable in several aspects but also contains some disturbing features. Directlyapplying the equations developed by Quashnock and Spergel [20], one obtains the forceequation (2.6) and the full integral expression for the metric and its derivatives, equation(2.7). However, in examining the effects of this force on the single travelling wave,section 2.2, it was found that there were some difficulties with this particular gauge choice.In order to satisfy the condition that h8z’ zt = 0 for a single wave, it was necessaryto make a gauge transformation into the Vachaspati gauge. This suggests that the fullintegral expression for the metric (2.7) is not the best gauge choice. However, ignoringthis problem and blindly evaluating this force anyways provides a useful illustrationof the dependence of the force on the chosen gauge. Evaluating (2.6) for two collidingtriangular pulses results in back-reaction forces on the string which contain non-integrableinfinities. This problem is corrected by constructing a composite metric which is explicitlyin Vachaspati’s gauge for those times when the two waves on the string may treated astwo completely separate single travelling waves.3.1 Failures of the Direct ApproachUsing equations (2.6) and (2.56) one can investigate the effect of the back-reaction forceon two colliding travelling waves. To attempt this, I used two triangular pulses whichwere for simplicity restricted to lie in the yz-plane. The pulses had the form of (1.30) for49Chapter 3. The Composite Metric 50the right-moving wave with /3 = 1/ and 1 100 and (1.44) for the left-moving wavewith c = 0.5 and 12 = 20. This simplification means that all forces in the x-direction arezero. The magnitude shown for the force has been scaled so that 4Gii = 1.In order to provide a clearer picture of the two wave pulses that I am colliding, animage of the waves before they collide is shown in figure (3.1). At time t = —120, thetwo wave pulses collide and figure (3.2) shows the force on the string in the y-directionshortly after the collision at time t = —115. Also shown beside the graph of the force isthe position of the waves at the same time. The wave image shows not only the collisionof the two wave pulses (as a solid line) but also the location of the single wave pulses hadthey not collided (shown as dotted and dashed lines).The two force spikes seen in figure (3.2) are caused by the kinks at ü = —100 (theforce near z = 85) and i = —20 (the force near z = 75). These forces seem to propagatealong the string keeping pace with the non-colliding wave pulses. The forces are due tothe change in the derivative of the perturbation metric at the kink acting on the samewave as it moves along a piece of the second wave having a different slope. For exampleTravelling Wave Pulses Before Collision0.4-)C.’)06040200—400 —200 0 200z — positionFigure 3.1: The two travelling wave pulses at time t = -150.Chapter 3. The Composite Metric 51the force at position z = 75 is due to the effect of the change in the derivative of themetric at the i3 = —20 kink of the left-moving travelling wave as it travelled along theflat portion of the string acting on the left-moving wave as it moves up the front face ofthe right-moving triangular pulse.Unfortunately, neither of the force spikes shown in figure (3.2) are bounded and ifone tries to integrate the force, thereby obtaining the effect of this force on the string,one obtains an infinite result. This is not reasonable and illustrates the difficulties whichcan arise from an inappropriate gauge choice.Closer examination of the infinite spikes in figure (3.2) indicates a route by which onecan get around this problem. The spikes are generated by the derivatives of the metricof the string at times before the two waves have collided. At these times, the metric ofthe string should be identical to that of a single travelling wave because the two wavesare not causally connected. The metric for a single travelling wave was solved for byVachispati [17] and also obeys the Lorentz gauge condition. In addition, the infinitieswhich arise in the force expression are found to be of the same form as those which arisewhen solving (1.5) for the metric of the single wave (section 2.2). Both of these pointsindicate that, as was suspected, one should be using Vachispati’s solution for the metricin order to determine the force generated by the waves at times before they have collided.One of the other problems that arises with the full integral expression for the metricwas that in order to evaluate it, it was necessary to use a past-time cut-off. This alsopresents difficulties because for all finite values of r12, this cut-off causes a violation of theLorentz gauge condition. The full integral expression is still required for those times afterthe waves start colliding and the Vachaspati solution can no longer completely describethe string. Therefore, it is also necessary to find a way to remove this cut-off dependencein the metric and satisfy the Lorentz gauge for finite values of r0.A metric which satisfies these conditions is the composite metric developed by UnruhChapter 3. The Composite Metric 52Infinite Force Spikes at t = —11510 I I I I I I - I I I I I I8—6—c)0--:2-0I I I I I I40 60 80 100 120z —position(a)Colliding Waves at t = —115IE.40 60 80 100 120z —position(b)Figure 3.2: Shown here in (a) are the infinite force spikes that are obtained if the two-wavemetric is used directly in the force equations. Also shown in (b) is the position of thecolliding waves at this time.Chapter 3. The Composite Metric 53and Vollick [5].3.2 The Composite MetricBecause I am operating in the weak gravity regime, it is possible to linearly decomposethe two-wave metric into several separate metrics [5]. First one breaks down the metricinto the metric for a straight string, two single travelling waves, and a residual piece.The decomposition of the two-wave string is shown diagrammatically in figure (3.3).The residual piece of string can also be broken up as shown in figure (3.4). The twosingle wave strings are aligned with the two-wave string so as to exactly cancel eachother out (once the straight string is also included) making the stress-energy tensor forthe residual component of the two-wave string vanish up until the time of collision.The advantage of doing this is that now, each of the individual components of theentire composite metric can be evaluated individually. The two single travelling wavesand the straight string from the decomposition of the string with two waves, figure (3.3),can be represented by Vachispati’s metric. It is known that this metric obeys the Lorentzgauge conditions and, from section (3.1), using this metric for the single travelling wavesshould remove the non-integrable infinities that were being found initially.As shown in Appendix B, the metric for the residual component of the string alsoobeys the gauge conditions as long as one chooses a past-time cut-off at some time priorto the collision of the two waves. At this cut-off time the components of the residualstring all cancel making the stress-energy tensor for the residual string, and consequentlythe residual’s metric, vanish. The problems with satisfying the gauge conditions alsodisappear. The metric for the residual component, and the contribution of its derivativeto the force, can be evaluated using some finite value for the past-time cut-off and equation (2.56). Because the stress-energy tensor is zero for any cut-off before the time ofChapter 3. The Composite Metric 54Decomposition of Two Colliding Waves(a)± (b)(a)± (d)Figure 3.3: The string with two travelling waves can be broken up into four components,the left- and right-moving travelling waves, (a) and (b), the straight string (c), and aresidual piece of string (d) in which the dotted line segments correspond to negativestring components.Chapter 3. The Composite Metric 55Decomposition of Residual ComponentFigure 3.4: The residual component of the initial decomposition can also be representedas the sum of four different pieces: a two-wave component (a), a straight piece of string(b), and a left- and right-moving single wave component (c) and (d).±()(b)(c)(d)Chapter 3. The Composite Metric 56the collision, the metric and it’s derivatives are independent of the value chosen for thecut-off, although each individual component is affected by the choice.This illustrates some of the difficulties associated with the gauge dependence of theforce. The previous solution involved only one of the four pieces of the residual metric,namely the two-wave component. But, in order to deal with the infinite limits on theintegral for this case, it was necessary to install a past-time cut-off. Unfortunately, themetric in this case strongly depends on the value of the cut-off. If one effectively removesthe cut-off by taking r0 to infinity, one obtains infinite, non-integrable forces, a very non-physical result. The difference in gauge between this situation and that of the compositemetric is the difference between a physically reasonable result and a very unreasonableone.Through the use of the composite metric, it is possible to remove the non-integrableinfinities through the use of the Vachispati solution for the single travelling waves. Thecomposite metric also satisfies the Lorentz gauge conditions at all times, a feature absentfrom the previous solution.3.3 Force Equations for Single Travelling WavesEquations for the contribution to the force due to the residual component of the stringhave already been found. One can simply use equation (2.56) in equation (2.6) with afinite cut-off for each piece of the residual metric. In this case the boundary terms at theupper and lower ends of the integral should also be included although they will cancel outwhen components are summed together to give the contribution from the entire residualmetric.Determining the contribution to the back-reaction force from the single travellingwaves and the straight string is quite straightforward. Equation (2.6) is used insertingChapter 3. The Composite Metric 57derivatives of Vachaspati’s metric for the terms. Starting with equation (2.28) forthe metric’, the derivative for the right-moving wave is found to be= 4G{[(axX)2] Ox7 ln(2(x — Z)2)(Umax — Ukink)kinksFaj 2________+ 2 UfXX—Z) —X—Z)O-x.O--x(x—z) OuxOxwhere Ii is evaluated at ü = umax andq3 = ‘c8 —9 (3.2)(Oix . Oix)2 (Ox. Oix)2i=ikk+e(ôx . O)-kinkThe magnitude of the delta force portion of equation (3.1) is plotted in figures (3.6)and (3.5) for the six different kinks found in my collision of two triangular pulses. Themagnitudes have been scaled so that 4G1i = 1. The delta functions in equation (3.1)lead to an infinite contribution to the force, but this contribution is integratable. Whendetermining the effects of the back-reaction force on the string itself, one must integratethe force. The contribution that the delta function component of the force, or delta force,makes to this integral is just the coefficient in front of the delta function, the magnitudeof the delta function. For this reason, the delta forces are represented as single linesextending out of the graphs in chapter four whereas in figures (3.6) to (3.7) they arerepresented in a more instructive manner by their magnitudes.It is worth noting that the integration constant 2 from Vachaspati’s solution does notcompletely disappear from the derivative which reflects the remaining gauge freedom stillleft in the composite metric solution. The constant is still present in the delta functionportion of equation (3.1). This constant is an. arbitrary parameter that one is free tovary in the equations. The effects of varying this parameter are shown in figure (3.7).The presence of the arbitrary integration constant in the expressions for the back-reaction force is the least satisfactory aspect of my gauge choice. Changing the integration‘This expression for the metric is not in the same form as given by Vachispati, however as shown inAppendix A, the two forms are equivalent.ci)ci)SFigure 3.6: The amplitude to the delta function contribution by the left-moving wave tothe force is shown. The three different curves, (a), (b), and (c), indicate the effect of theforce due to the delta function at the three kinks in the left-moving wave at i3 = —20, 0, 20respectively.Chapter 3. The Composite Metric 58Delta Function Amplitude from Right—Moving Wave100—10—20TimeFigure 3.5: The amplitude of the delta function contribution to the force by theright-moving wave is shown. The three different curves, (a), (b), and (c), indicate the effect of the force due to the delta function generated by the three kinks in the right-movingwave at ii = —100,0, 100 respectively.Delta Function Amplitude from Left—Moving Wave—100 0 100420—2—4—61 I I I I(a)I I I I I(b) (c)—100 0 100 —100 0 100 —100 0 100TimeChapter 3. The Composite Metric 59Scale = 10 Scale = 1:: 1010—-•- (1)02—’I’’’ ‘‘‘‘I• •0--• S—b—10 --—20--—I I I I I I—100 0 100 —100 0 100Time Time(a) (b)Scale = 0.1 Scale = 0.0120I I I I I I I I I I I I I IJ•20Q) -0_ 0-.4-)H JJJ-- S—20—20—-—30--I I I I—40 - I I I I I—100 0 100 —100 0 100Time Time(c) (d)Figure 3.7: The effect of different values of the integration constant is demonstratedfor the right-moving wave. Graphs (a),(b),(c),and(d) show the amplitude of the deltafunction for constant values of 1,10,0.1,0.01 respectively.Chapter 3. The Composite Metric 60constant in Vachaspati’s solution corresponds to moving about in the Lorentz gauge.That is, it corresponds to making a gauge transformation of the form D = 0. Aswas seen in section 3.1, this small freedom can have a strong effect on the resultingforce. As is demonstrated in figure (3.7) and as will be shown explicitly in the nextchapter, changing the integration constant in the force equations has a significant effecton the back-reaction force. Although the ‘spikes’ in the delta function amplitudes alwaysmaintain the same sign regardless of the choice of integration constant, just a shortdistance away, the amplitude can vary from strongly negative values to strongly positiveones. Also, even though the spikes maintain their sign, their magnitude is still affectedby changes in the scale factor. No obvious choices or physical reasons have been foundto select one integration constant, or gauge, over another and for this reason the forcesthat are obtained using my gauge choice are not very well defined. Only in regions suchas u> 100 where all of the delta contributions are zero, can one be confident about theshape of the back-reaction force.For consistency and simplicity, I chose the constant to be such that the value of thedelta function force is the same sign for each kink. For example, in the case of theright-moving wave’s metric, I chose(x—z)2Io,iioo (3.3)This choice makes all of the delta force contributions from the kinks at ii = +100 positiveand those from the kink at i = 0 all negative. A similar choice for the left-moving deltafunctions would be the value of (x— z)2 at u = 0 and i = —20 which is 1.62 x iO.One is free to choose a different constant for the left and the right-moving waves becausethe integration constants for the two metrics are independent of each other.Chapter 4The ResultsIn the previous chapter, it was seen that determining the force by directly evaluating thederivative of the metric as given in equation (2.7), leads to questionable results. Not onlydid the results have problems with the validity of the gauge conditions at the boundariesof the integral, but this approach also led to non-integrable infinite forces. To solve theseproblems, a composite metric for the string with two colliding waves was developed. Thislater approach results in forces which have more reasonable properties.1 The integratedforce on the string increases in magnitude as one nears the point of collision of twooppositely moving kinks. This is expected because one would like to think that thegravitational back-reaction would be strongest closest to those points where the radiationis emitted. The forces also tend to smooth out the sharp kinks of the triangular pulsesused. This too seems reasonable as physical intuition leads one to expect that the lossof energy by the waves in terms of gravitational radiation would lead to a damping outof the waves themselves. One unfortunate aspect of the composite metric is that it leadsto a much more complicated form for the force. Although each individual component isfairly simple, the resultant sum is not.1This force is not, however, perpendicular to the string. The effect of the parallel force componentsis discussed in Chapter 5.61Chapter 4. The Results 624.1 The Force ComponentsBy breaking up the metric for two waves on a cosmic string in the manner shown in figures(3.3) and (3.4) of the previous chapter, one obtains seven different metrics which must beevaluated separately for each different point on the string. These seven components arethen summed appropriately to give the final force on the string. It is informative to firstexamine the individual components in order to see how the infinities of the more directapproach are removed. Choosing a time-slice of the force at t = —90, the forces fromeach piece of the composite metric are shown in figures (4.1) and (4.2). The appropriatesigns on the forces have been included so that the total force, figure (4.2[d]), is just thesum of the other seven components.Figure (4.1) shows the four components which make up the residual metric. The firstgraph shows the force due to the two-wave component. Here one can see that, initially,the non-integrable infinities are still present. In fact this metric is exactly that whichwas first computed and found unacceptable in chapter three. The next two graphs offigure (4.1) show how the infinity problem is solved. They show that the infinite forcecontributions from the two single travelling waves are exactly the negative of those inthe first graph. Therefore, when the components are added together, the infinities cancelout. The fourth graph shows the contribution from the straight piece of string.Figure (4.2) displays the forces generated by the two single travelling waves and thestraight string with metrics having the form of Vachaspati’s solution. Also shown is theresultant force when all seven components of figures (4.1) and (4.2) are added together.The two single waves generate the only surviving infinite contributions to the force inthe form of delta function forces. However, these are not a problem as, once integrated,the delta function gives a finite quantity. The delta function forces are shown as singlelines extending off of the graph. It is interesting to note that the single travelling wavesChapter 4. The ResultsC)020ci)C.)0r.ioci)063Figure 4.1: Shown here are the contributions to the back-reaction force from the fourcomponents of the residual metric. Graph (a) gives the contribution from the two-wavemetric while (b) and (c) give the contribution from the left- and right-moving single wavecomponents. Graph (d) shows the effect of the straight piece of string.Two Waves0•II III 11111111JI 1111111111111200—2030Right—Moving Wave1111111 11111ii I 11111 -40 60 80 100 120z—position(b)—20—3040 60 80 100 120z—position(a)Left—Moving Wave Straight String1 11111 II I . 0.05-0ci)- C.)—0.05--—0.1I040 60 80 100 120z — position(c)40 60 80 100 120z—position(d)Chapter 4. The Results1)C)0C)00C)064Figure 4.2: The force contributions at t = —90 from the two single travelling wavesand the straight string which were found by using Vachaspati’s metric. Graphs (a) and(b) show the left- and right-moving waves respectively. Graph (c) presents the straightstring’s effect while figure (d) shows the net force including the contribution from theresidual metric. The single lines extending out of the graphs represent the delta functioncomponents of the force.Right—Moving WaveI I I I I I I I I I I ILeft—Moving WaveI I I I i i I i i i I0.30.20.10—0.10.200.30.20.10—0.10.10.050—0.054040 60 80 100 120z —position(a)Straight Stringci)C)040 60 80 100 120z—positiori(b)Total ForceI I I p I II I I I I I I I I i—0.460 80 100 120z —position(c)40 60 80 100 120z—position(d)Chapter 4. The Results 65while having no back-reaction on themselves, do contribute to the back-reaction force onthe string with the colliding waves. The back-reaction force does not come entirely fromthe residual component of the string, the piece that in some sense generates all of thegravitational radiation [5]. Additionally, the force from the straight string can be seen tobe exactly the same as that from the straight string part of the residual metric. Becauseof the difference in sign between these two components, their contributions will exactlycancel in the final sum. It is worthwhile to notice that the finite part of the single waveforces do not cancel with their residual counterparts. This is the main source of much ofthe complexity seen in the resultant back-reaction force on the string.4.2 Time SlicesAs an initial investigation into the back-reaction force on the two colliding waves, Iconsidered time-like slices of the force. For my colliding waves, I once again chose toexamine two triangular pulses. The two pulses used are of the form of (1.30) for theright-moving wave with /3 i/\/ and 1 = 100 and (1.44) for the left-moving componentwith a = 0.5 and 12 = 20. Waves of two different sizes and shapes were selected in orderto reduce the symmetry of the system somewhat and to make it easier to distinguish theforces from one wave as compared to the other. Because the waves are restricted to theyz-plane for the purposes of simplification, the force in the x-direction is always zero.The forces represented here are the forces in the y-direction. This direction was chosenbecause only motion perpendicular to the cosmic string has any physical meaning. Asmentioned in section 1.2, the field equations representing the cosmic string are invariantto boosts and translations along the length of the string. The force in the y-directionalways has at least a component in the direction perpendicular to the string. As discussedin section 3.3, the delta function contribution to the force from the two single travellingChapter 4. The Results 66waves has been scaled so that the force from each kink is always of the same sign. Thedelta functions will be represented in the figures of the force as single lines extending outof the graph. The magnitude shown for the finite force contributions has been scaled sothat 4G[L = 1.As expected, before the time at which the two waves first collide there is no forceon the string. Before the collision, there is no gravitational radiation emitted from thestring and each of the waves can be viewed as just a separate single travelling wavewhich has been shown to experience no gravitational back-reaction. Therefore, the firstfigure presented, figure (4.3), shows the forces just after the start of the collision at timet = —119. Below this is an image of the string at the same time which also indicateswhat the position of the two single travelling waves would have been had they not collided(the dotted and dashed lines). The delta function forces appear here and as can be seenfrom later figures, travel along the string keeping pace with the generating kinks of theundisturbed single waves. The force also contains two bounded spikes which shrink andspread out as time goes on.In some sense the effects of the force spreading out and moving further apart canbe thought of as the effects of the gravitational wave pulse as it expands out from thekink collision where it was created. However, because the radiation was generated at onepoint in spacetime, one would expect the effects of this radiation at a later instant intime to be confined to very small sections of the string. The effects instead are spreadout over a broad region. This exemplifies the non-local nature of the back-reaction force.The forces affecting the string at any given time were generated by perturbations in themetric of the string at some earlier time. The perturbations propagate at the speed oflight and travel a variety of different distances catching up to the string in many differentlocations. This causes the forces of the string back on itself to be spread out over largerlengths of string than might at first be expected. Not only the finite forces, but also theChapter 4. The Results 67delta function contributions were generated by the kinks of the travelling wave at muchearlier times than that at which they recollide with the string, times in fact before thecollision even began.This brings up another difficulty with this choice of the Lorentz gauge. The back-reaction forces, most notably the delta function contributions, are not always generatedafter the collision of the waves. Some force contributions are caused by perturbationsin the metric of the string at times long before the collision. At these times, there hadbeen no gravitational radiation produced and one would have expected no gravitationalback-reaction. Although none of these forces affect the string until after the collision hasbegun, it is still somewhat disturbing that they should influence the back-reaction at all.It would have been nicer if the source of the back-reaction forces was confined to thecollision of the oppositely moving kinks, the source of the gravitational radiation, but inthe Lorentz gauge this does not appear to be the case.Returning to the time-slice graphs, figures (4.4) and (4.5) show the shape of the forceat two later times demonstrating the propagation of the force away from its point oforigin. At time t = —101, one sees a significant change in the shape of the force asits leading edge flips sign. This is just caused by the right-moving undisturbed wavecomponent of the metric crossing the position of the next kink in the left-moving wave atv = 0. When this occurs the slope Oz” changes causing the switch in sign of the force.After the collision with the next kink in the left-moving wave at v = 0 and timet = —100, another double spike of force appears similar to that seen at time t = —119,figure (4.7). Another delta force line has also appeared, generated by the kink whichwas just crossed. As one proceeds to later and later times, one passes the force at timet = —90 shown in figure (4.8) then figures (4.9) and (4.10) showing the force just beforethe collision with the last kink of the left-moving wave. One interesting feature of figure(4.9) is the spike located at z = 115. The feature is occurring away from any kinkChapter 4. The Results000C) —0.10—0.2—0.30C)068t = —119 t = —1150.1.0II jill iiI I I i I i i i I111111 1 11111•lI I II 11111110.50-0.5—1—1.52015105040 60 80 100 120z —position(a)Waves at t=—119.111111111111111.I I I I I I I40 60 80 100 120z—position(b)Figure 4.3: (a) Back-reaction force on thestring in the y-direction at time t = —119.Delta forces are shown is single lines extending out of the graph. (b) The positionof the waves on the string at the same time.40 60 80 100 120z—position(a)Waves at t=—11525 IIIIIII1I0—, 1TTTE40 60 80 100 120z—position(b)Figure 4.4: Force at time t = —115Chapter 4. The Resultsci)C.)0ci)C)0ci)C)0ci)C)0t = —10169t = —1100.1 III II liii0-0.1—0.2 I I400.150.10.050—0.05—0.1 I I I i i i I i i I i60 80 100 120 40 60 80 100 120z—position z—position(a) (a)Waves at t=—110 Waves at t=—101II 111111111 I_ 111111 1111111E:NN\I I 0 I I40 60 80 100 120 40 60 80 100 120z—position z—position2010020(b) (b)Figure 4.5: Force at time t = —110 Figure 4.6: Force at time t = —101Chapter 4. The Results 70collision and becomes less significant at later times when the string is closer to the nextkink collision. However, one also notes that it is near this value that the undisturbedtravelling wave ‘passes through’ the two-wave solution. The force is proportional to 1/)where \ is a measure of the separation between the position of the string generating theforce x and the point being affected by the force z. As this separation becomes small,the force becomes correspondingly large. The spike contribution to the force comes fromthe residual component of the metric. As such, the kink of the single travelling wave iswhat generates the force. The kink, however, never intersects the string before v > 20where the left-moving component of the string is flat. In this region the force is zero.Therefore, because the separation distance is always non-zero, the spike seen in figure(4.9) remains bounded. An interesting feature of figure (4.10) is that the delta functionforce has disappeared. The kink which was generating this force has crossed the last kinkof the oppositely moving wave and is once more on a flat piece of string. It has returnedto the configuration of a single travelling wave and so a large portion of the back-reactionforce including the delta function contribution is reduced to zero. However, the force doesnot completely disappear. Because the wave was delayed with respect to its counterpartwhich did not collide, and because the two travelling waves are now causally connectedto each other, a small back-reaction force still remains as will be seen in the next figures.Crossing the final kink of the left-moving wave again results in the appearance ofspike-like forces, figure (4.11), which decrease in amplitude as one moves away from thepoint of collision, figures (4.12) and (4.13). An interesting feature first appears in figure(4.11) at z = 118 and is seen more clearly in figure (4.12) at z = 145. This small spikeis the remaining effects of the kink at ü = —100 as it moves away from the collision. Itis the back-reaction forces caused by the displacement of the right-moving wave whichcollided with respect to the one which did not. This effect decreases in amplitude as onemoves to later and later times and is the first indication that the back-reaction effects doChapter 4. The ResultsC)0C)F.1071t= —991086420ci)C)0El 11111 III II: i II IIci)000.20—0.2—0.4—240t= —90111111 II I11111111-40 60 80 100 120z—position(a)12060 80 100z —position(a)3020100Waves at t=—110 Waves at t=—101I I. ) I III: 40-—1;—1 I i 0—“ I I i40 60 80 100 120 40 60 80 100 120z —position(b)z— position(b)Figure 4.7: Force at time t = —99 Figure 4.8: Force at time t = —90Chapter 4. The Resultsa)C.)00072t= —85 t = —81a)C.)0:1I I I I I I I I I I I I I I I IIII 1111111 III 111111.a)000.150.10.050—0.05—0.10—51040302010040 60 80 100 120 140z—position(a)Waves at t=—8540 60 80 100 120 140z—position(a)Waves at t=—811 11111 11111 111111111 I 50 I II 111111 III III I—40 60 80 100 120 140 40 60 80 100 120 140z —position(b)z—position(b)Figure 4.9: Force at time t = —85 Figure 4.10: Force at time t = —81Chapter 4. The Results 73not stop immediately after the collision is ‘finished’.As the u = 0 kink of the right-moving wave begins to pass through the left-movingwave, the sequence of events observed for the kink at u = —100 repeats itself. Sets ofdouble spikes appear just following the collisions, figures (4.14) and (4.17), and spreadapart decreasing in size, figures (4.15) and (4.16). The sign of the force flips just beforethe collision with the middle kink, figure(4.16), due to the undisturbed wave effectivelycrossing the kink ahead of the actual collision. Even the spike due to small separationvalues of \ reappears, figure (4.18), as the strings once again ‘pass through’ each other.One important difference is that the forces, both the finite contributions and the deltaforce contribution from the right moving wave, have changed sign. This feature shall bediscussed further in the next section when I consider u-slices of the force.The collision of the right-moving kink at t = 100 proceeds in a manner very similarto that of the previous two kinks. The signs of the forces are the same as those for thecollision of the first kink.Finally, once the waves have finished passing through one another, one might expectthat the force would drop entirely back to zero again. The string is once more in aposition where it can be viewed as simply two single travelling waves. However, unlikebefore the collision, the two waves are now causally connected and can affect each otherresulting in a small but non-zero force as shown in figure (4.20). Also, the situationis further changed because the two travelling waves were each delayed by the collision.As one moves to later times, the shape of the force stays essentially the same but theamplitude decreases slowly towards zero. The three spikes seen at z = —140, —100, and—40 are caused by the displacement of the colliding left-moving wave with respect to itsundisturbed counterpart. Similar spikes are present due to the displacement of the rightmoving wave seen in this figure at z = 0 and also seen in figure (4.12). The effects dueto the right-moving wave appear further apart because the right-moving wave is muchChapter 4. The Resultsci)C)040 60 80 100 120 140z—positionci)C-)074t= —79ci)C.)0I IlIllIllIllIll 111111.IZLiIIII IIIci)C.)0r.t= —50111111 III IIIliii liii3210—1504030201000.060.040.020—0.02—0.04—0.066040200(a)Waves at t=—79-I I I I i ii i I40 60 80 100 120 140z—position(b)0 50 100 150z—position(a)Waves at t=—50.11111 11111111111111111111 II I1 IIo 50 100 150z—position(b)Figure 4.11: Force at time t = —79 Figure 4.12: Force at time t = —50Chapter 4. The Resultsci)C)0ci)C)075t= —21 t = —18a)C)0ci)C)06420—2—50liii[ I I I I0.030.02 -0.01 - -—0.02 --iIiiiIiiiiIiiIii-t—50 0 50 100 150 200z —position(a)Waves at t=—21I I I I I I I I I I I I I I I I I I I0 ——---—50 0 50 100 150 200z —position(b)0 50 100z—position(a)Waves at t=—1811111110 --- I -—50 0 50 100z—position(b)Figure 4.13: Force at time t = —21 Figure 4.14: Force at time t = —18Chapter 4. The Results 76a) a)o 05-I s-IOflR 0t = —151t=—10—50 0 50 100z —position(a)Waves at t=—150.10—0.1—50 0 50 100z—position(a)Waves at t=—18060j4020060200—50 0 50 100 —50 0 50 100z—position z—position(b) (b)Figure 4.15: Force at time t = —15 Figure 4.16: Force at time t = —1Chapter 4. The Results0.)C.)00)C.)077—50 0 50 100z—positiont=1 t=15I I I I I I I I I I I I I-:I I—50 0 50 10020—2—4—6—8—1080600)C)40200z —position(a)1086420—2600)400200Waves at t=1(a)Waves at t=15—50 0 50 100z—position(b)—50 0 50 100z—position(b)Figure 4.17: Force at time t = 1 Figure 4.18: Force at time t = 15Chapter 4. The Resultsa)C-)0rz78t=21 t=150302010000I I I I I I I I I4 I II 11111 I II-a)C.)0—50 0 50 100z —position(a)Waves at t=21080600.050—0.05—200 —100z — position(a)Waves at t=150I I I I I I60-20-0 I I I—200 —100 0z—position(b)200—50 0 50 100z —position(b)Figure 4.19: Force at time t = 21 Figure 4.20: Force at time t = 150Chapter 4. The Results 79larger than the left-moving one.4.3 U-slicesAlthough the time slices of the force provide a good visual picture of where the forces areat a given point during the collision and are also easy to reference back to the positionof the string at that time, for the purposes of determining the effect of the collision onsay the right-moving travelling wave, it is more convenient to use u-slices of the forces.This has been done in figures (4.21) to (4.38) where the force in the y-direction has beenplotted against v for various different constant u values.The first set of figures shows the force as one approaches the first kink at u = —100.Shown in figure (4.21), there is no force on the string up until one is fairly close to thekink. By u = —101, figure (4.22), the force has appeared and by u = —100.1, figure(4.23), it has grown to a significant size. The major feature of the force is that it formstwo spikes which approach v = 0 and v = 20 as u tends to —100 and that there is alsoa delta force contribution from the right-moving single wave kink at ü = —100 whichapproaches v = —20. That is, as one approaches the kink in u, one finds that the majorcomponents of the force are coming from points approaching the kinks in v. The force isstrongest near the collisions of the oppositely moving kinks as was seen in the time slices.There are no contributions from the delta function forces generated by the left-movingwave at values of u < —100 or u > 100. For intermediate values of u, the left-movingdelta forces tend to act against each other, reducing their importance. What one is leftwith is a force distribution which is predominantly positive and which is increasing inmagnitude as one nears the kink.Just past the kink, at u = —99, the force quickly drops in magnitude, figure (4.25).The force then stays relatively small until one begin to get near the next kink at u = 0.Chapter 4. The Results 80u=—105 u=—1012 4IIIIIIII IIIIIl4-.4---2- = 0ci)- 0)C)- C)0- 0•---—2--- —4--—4---I I I I I— I I I I—40 —20 0 20 40 —40 —20 0 20 40v—value v—valueFigure 4.21: Force at u = —105 Figure 4.22: Force at u = —101u = —100.1 u = —99.910 II1IIrrm 1111111111 10 111111 11111115-- 5--ci) •- 0)C) •-C)0 •- 0•.0—- 0—5 IIIIIIIIIIIIIIIIIIIII—5 1111111.,. 111111111—40 —20 0 20 40 —40 —20 0 20 40v—value v—valueFigure 4.23: Force at u = —100.1 Figure 4.24: Force at u = —99.9Chapter 4. The Results 81This is shown in figures (4.26), (4.27), (4.28), and (4.29). By the u = —1 slice the forceis once again definitely beginning to grow and is taking much the same form as was seenin the approach to the kink at u = —100. The main difference is that the forces are nowpredominantly negative in direction.Past the kink at u = 0, the force once more decreases again. Approaching the finalkink at u = 100, figures (4.33) and (4.34), the force is growing again as it did before andis oriented primarily in the positive direction.After the last collision, at values of u greater than u = 100, the forces while not zero,are steadily decreasing. Also, for these values of u, there are no delta function forcecontributions from either the left- or right-moving waves. These forces can be seen infigures (4.37) and (4.38).One of the main goals in determining the back-reaction forces on a cosmic stringis to, by integrating these forces, find out how the motion of the string is affected. Byintegrating the force with respect to the variable v, one obtains (Oz), that is one findsout by what amount the slope of the right-moving component of the string is changed bythe back-reaction force. Similarly, by integrating over the u variable, one could find theeffect on the left-moving component. To get some idea of how the triangular pulses wereaffected by the back-reaction, I integrated the u-slice forces for a few different values ofU.The main effect of the forces on the right-moving wave occurs near the kinks. As oneapproaches the values of the kinks at u ±100,0, the magnitude of the integrated forceincreases dramatically. This is shown in figure (4.39) and (4.40) which give the value ofthe integrated force as one approaches the kink at u = —100. Figure (4.39) shows thecontribution from the finite components of the force. Here it is quite easy to see that theforce is positive and increasing as one nears the kink. A positive force means that theslope is being increased. The slope on the other side of the kink was originally larger thanChapter 4. The Resultsa)C)0Figure 4.27: Force at u = —5C)C)082u= —99 u=—80111111 1111111 111111liii 11111 111111 IIa)C)042C)C)00—20.040.020—0.02—0.04—40 —20 0 20 40v—valueFigure 4.25: Force at ‘u = —99u=—5—40 —20 0 20 40v—value10.500.5—1420—2—4Figure 4.26: Force at u = —80u=—1—40 —20 0 20 40v—value_r 111111111__iiIiiiI ..LIIIIIIIIIIII—40 —20 0 20 40v—valueFigure 4.28: Force at zt = —1Chapter 4. The Results 83= —0.111111111u=1:‘‘‘‘. rir...l. 11111I1IIIIl1111HI I I Iliii 1111111 liiia)C)0000r.0—20—400.20—0.2—40 —20 0 20v—value406040a)C)00—202a)C)00—2Figure 4.30: Force at u = 1u=95—40 —20 0 20 40v—valueFigure 4.29: Force at u = —0.1u=20111 IIIIII 11111111.-II III 111111 111111——40 —20 0 20 40 —40 —20 0 20 40v—value v—valueFigure 4.31: Force at u = 20 Figure 4.32: Force at z = 95Chapter 4. The Results0C)0C)0u = 110Figure 4.36: Force at u = 11084u=99 u=99.9a)C-)0C-)C)0- 11111 Tir iii1 liiiLIIII I IIII I 11111111111111—40 —20 0 20 40v—valueFigure 4.33: Force at u = 99u= 10120—26420—2—4—40 —20 0 20 40v—valueFigure 4.34: Force at u = 99940200—200—1—2.1111111 1111111 11111..1 I I i i I i i i i I i III liii l III II I liIi•IIII 11111111 I I—60 —40 —20 0 20 40v—value—60 —40 —20 0 20 40v—valueFigure 4.35: Force at u = 101Chapter 4. The Results 85u=150 u=5000.5—60 —40 —20 0 20 —60 —40 —20 0 20v—value v—valueFigure 4.37: Force at tt = 150 Figure 4.38: Force at u = 500that at u < —100 and so the back-reaction force is acting to reduce the difference betweenthe slopes on either side of the kink, to smooth out the kink. Because the integrated forceis increasing, the effect is greater nearer the kink than farther away which indicates thatthe force is also rounding off the sharp corners of the triangular pulse. This correspondsto a removal of some of the higher frequency components of the pulse, a result which wasfound in an entirely different manner by Hindmarsh [10] for the special case of very smallperturbations to the string. My results indicate that it is true for larger waves as well.The effect of the delta force contributions is also increasing in a positive direction asone approaches the kink. However, because of the arbitrary scaling factor, these resultsare more ambiguous. Shown in figure (4.40) are two different choices for the scale. Thesolid line indicates a scale of 1 while the dashed line is for a scale of 0.008 or 1/125which is the scale which just causes all of the right-moving delta forces to always havethe same sign, section 3.3. Regardless of scale, the delta force always becomes positive40 40Chapter 4. The Results 86just before the kink, however, the scale does affect its sign a short distance away fromthe kink. If the force is positive, as with a scale of 0.008, then its effects serve to enhancethose of the finite forces. A negative force, on the other hand, serves to reduce theeffect or perhaps even cause the slope to go first negative before becoming positive. Thiswould generate a dip in the string before the kink was reached. The arbitrariness of theintegration constant makes it impossible to say anything definite about the shape of theforce. Nevertheless, in all cases the difference between the slopes immediately on eitherside of the kink would be reduced and the kink would in some sense be smoothed out.The rounding off of the kink at u = —100 is further enhanced by the effects of theforce at u > —100. Shown in figure (4.41), the effect of the force on the other side ofthe kink is also decreasing in magnitude as one moves away from the kink. The finiteforces are negative in direction which acts to further reduce the difference between thetwo slopes.On this side of the kink, the right-moving delta function is zero but the left-movingdelta functions are now non-zero. It is worth noting, however, that the delta functionsserve to exactly cancel each other except for a small constant negative factor. Thisfactor is independent of the value of the integration constant. The effect of the left-moving delta functions on the right-moving wave is independent of the gauge freedom inVachaspati’s solution. This result can be shown analytically, Appendix C, and holds forall values of u. This means that the only parts of the force which continue to depend onthe gauge choice are those parts which contain a non-zero contribution from the rightmoving delta functions. These are three small regions just to the left of each of the kinks:approximately those regions where —103 < u < —100, —3 < ti < 0, and 97 < zi < 100.Unfortunately, it is at these locations that some of the most significant damping androunding off of the wave seems to be occurring.The effect of the left-moving delta functions to the right of u = —100 is not onlyChapter 4. The Results 87Integrated Finite Force Components for u < —100I I I I I I I I I I I I I I I IC.)-V-0-ctci)-I I I I I I I I I I I I ——101 —100.8 —100.6 —100.4 —100.2u—valueFigure 4.39: The integrated force on the right-moving wave before the kink at u = —100.Magnitude of the Delta Force Component, u < —100—I I I I I I I IC1. 10—10—I I I——101 —100.8 —100.6 —100.4 —100.2u—valueFigure 4.40: The integrated force from the delta function contribution at values of u lessthat —100. The solid line represents a scale of 1 while the dashed line represents a scaleof 0.008.Chapter 4. The Results 88unambiguous but is also negative. This serves to further reduce the difference betweenthe slopes on either side of the kink. However, the forces at this kink are constant sothey do not contribute to the rounding off of the kink as do the finite force components.Similar results can be obtained for the integration of the force as one nears the kinkat u = 0 where the force is increasing but negative for values of u below zero and positiveand decreasing above zero. Approaching u = 100 from below, the force is again positiveand increasing. Past the last kink, the force is negative and decreasing in amplitude asis shown in figure (4.43). Here there is no ambiguity due to the scale factor becausethe amplitudes of all of the delta force contributions in this region are zero. Here, aswith the forces on either side of u = —100, a negative force serves to decrease thedifference between the slope on one side of the kink with respect to the other. Becausethe magnitude increases as one nears the kink, the forces will induce a curving in thestring which will smooth out the kink and reduce the high frequency components of thewave.Because of the arbitrary nature of the scale factor, although the kink itself is bothrounded off and reduced, at short distances to the left of the kink it is not possible tosay what exactly the effects of the back-reaction force will be. Only along sections of thestring, such as the region at u > 100 where there are no delta forces, or the region justbeyond u = —100 where the effects of the integration constant cancel out, can one makeany sort of true description of the shape of the force. This is perhaps the most seriousfailing of this approach to determining the back-reaction forces.4.4 V-slicesAlthough there is certainly some indication in the time slices of the force, it is far fromobvious in the u-slices that the force on the left-moving wave is very similar to that onChapter 4. The Results 89Integrated Finite Force Components for u > —100I I I I I I I I I I I I I I I I—2 -VC)0a)o —4 -a)I I I I I I I I I I—99.8 —99.6 —99.4 —99.2 —99u—valueFigure 4.41: The integrated finite force on the right-moving wave just past the kink atu = —100.Magnitude of the Delta Force Component, u > —100I I I I I I I I I I IVC)C)I I I I I I I I—99.8 —99.6 —99.4 —99.2 —99u—valueFigure 4.42: The integrated force from the delta function contribution at values of zigreater than —100. The solid line represents a scale of 1 while the dashed line representsa scale of 1.62 x iO’. The two lines exactly overlap indicating that there is no gaugedependence in this part of the force.Chapter 4. The Results 90C)0ci)ctsbEci)Integrated Force After u= 100100 101 102 103 104 105u—value—2—4—6—8Figure 4.43: The integrated force on the right-moving wave after the kink at u = 100.the right-moving one. There is no intrinsic difference between the two waves and so oneexpects that the forces should be in some sense symmetric. This is shown to be the caseby examining briefly some v-slices of the force.Here one sees that the force appears a greater distance away from the kink, at v = —25,figure (4.44), than occurred with the u-slicings. However, as in the case of the right-moving wave the force spikes appear and approach the kink values of u = +100,0 asone nears the kink at v = —20. These forces are very similar in shape to those seen asone neared the kink at u = —100 with the u-slices. The appearance of the delta forcespike caused by the kink at v = 0 and seen at position u = 80 in figure (4.45) is justthe leading spike of the forces that appear as one nears the next kink at v = 0. Theforces are overlapping and also appearing at greater distances from the kink than did theirright-moving counterparts simply because of the difference in size between the two waves.Although the delta force spike from v = 0 is negative, its magnitude is much smaller thanthe delta force contribution from the kink at v = —20 and so the resultant integratedChapter 4. The Results 91force on the left-moving wave is positive as was the case for the right-moving wave. Justpast the kink at v = —19, figure(4.47), the force once again drops in magnitude and hasa form similar to that of the u-slices of the force just past the first kink at u = —99.Chapter 4. The Results 92a)0Ca) a)o cio 0v= —25 v= —21.1 I I I I I I I I I I I I I I I—1111111 liii IIi1: I I I I I I I I3- -:—100 0 100 200u—valueFigure 4.44: Force at v = —25v = —20.110.501086420—2—100 0 100 200u—valuea)00—110.50—0.5liii liii 11111Figure 4.45: Force at v = —21v = —19liii liii liii—-II I — 11111111 I——100 0 100 200u—value—100 0 100 200u—valueFigure 4.46: Force at v = —20.1 Figure 4.47: Force at v = —19Chapter 5The Perpendicular ForceIt was noticed in chapter two that the force given by equation (2.6) yields a perturbationmetric for which (OzI2 + OuzP)(Ouzv + + h) 0. This was discoveredwhen it was found that OUz’öUz11h.LV 0 even though = 0. This indicatesthat there is a problem with the force expression because the coordinate conditions (2.4)are not satisfied for the perturbed metric. The perturbation caused by this force resultsin a trajectory of the string which no longer satisfies the equations of motion. If one stopsat the first order corrections to the string trajectory, one could still use this definition ofthe force to get a rough idea of the perturbations caused by the back-reaction. However,this is not a very satisfactory solution.The force fails in this regard because it contains components which are not perpendicular to the string. Because the string is invariant to boosts parallel to its axis, onlyquantities perpendicular to the string have any real meaning. Therefore, a more reasonable definition of the force would be one which contained no parallel components. Onecan obtain a force of this form be subtracting out the parallel components from the forceequation (2.6). This is permissible because it turns out that the parallel force just causesa change in the internal coordinate system.It is a good idea to correct for this coordinate change because the derivation of theforce equation is based upon the satisfaction of the null coordinate conditions. A changein coordinates could shift one out of a null coordinate system, invalidating the forceequation. By redefining the force as only that component which is perpendicular to the93Chapter 5. The Perpendicular Force 94string, one maintains the initial choice of null coordinates. There would be no ambiguitywith respect to the continued satisfaction of the null coordinate conditions and the forcewould no longer contain any ‘unreal’ components. The perturbed metric would stay inthe solution space of the equations of motion.5.1 Removing the Parallel ComponentsIt is desired that the force be perpendicular to the string. Not only is this more physicallyreasonable, but this would also cause the null coordinate condition to be true not just inthe flat background metric but also under the perturbed metric. That is, there would beno coordinate change due to a force acting parallel to the string.If the perturbed string trajectory is to be parametrized by the same null coordinatesas the initial solution, then the following condition must hold true.= 0 (5.1)The subscript o refers to the original or zero order approximation to the string trajectorywhile the 1 indicates the first order approximation. Therefore, to first order in theperturbation,1= (5.2)This relation can also be found by starting with the force expression= (5.3)Contracting both sides with one getsOUZOUOVZT1/.Ll) = —r a=— ôz= (5.4)Chapter 5. The Perpendicular Force 95But,zvzT),w = (5.5)Therefore,OuZOuZ?7w, = _Ouzãuzvhiiv (5.6)For the force to be perpendicular to the string, it is necessary that zvO vz71pz, = 0.The perturbed component of the trajectory, z, is generated by a force which hascomponents both perpendicular and parallel to the string. Therefore, one can also expressz in terms of the components I, which is generated by the perpendicular force, and, which is generated by the parallel force. The force also causes a change in thecoordinates, ) —+ ) + where the index i runs over the two internal coordinates u andv, \ corresponding to the original null coordinates and to the first order correction tothese coordinates [23]. As a result, zi can be expressed to first order in the perturbationas= = z+eä (5.7)Because I is the perturbation caused by the perpendicular component of the force,it is that part of z which satisfies= 0 (5.8)This is equivalent to requiring that= = k (5.9)where k is a constant which I choose to be equal to zero for convenience. Therefore, Isatisfies the conditions= av = 0 (5.10)Chapter 5. The Perpendicular Force 96Using these relations in equation (5.7), one obtains=aZ”== (5.11)Using equation (5.6), one can rearrange equation (5.11) to solve for the derivative of thecoordinate shift, 9. 9,,c1i is found with the analogous equation for Theresults area5V—a 2a “-‘hS — U0 U0VLJZQ ‘vZo 17li”—aVzaVzhliV (5.12)V 11Z0 vZ0 77Substituting equation (5.7) into the equation for the force and rearranging, it ispossible to find an equation for the perpendicular force.8z1A==— aUaV— aa’a2zli a2zli—a’0 1VSan2‘-‘uSav2where and aeu are given in equation (5.12) and are equal toa au h a — a u az aVz llivUV’,— /3 l-”’7 Uo U’, /32aza0 a0—a P9 v a2liaa aL-v — UZQ UZ0 h— av V Z hw‘U VS /3 l-7 .VZo uS /32aa0 aVZO 77The force given by equation (5.13) is now explicitly perpendicular to the string. To seethis, consider the contraction of the force with aUz’r,liV. The first term in (5.13) canexpanded using equation (2.6)._Plia/3 aVZOaUZOliV = — li7 (h7,/3 + h87, — ha/3,7) aUZOliV= (5.15)Chapter 5. The Perpendicular Force 97Performing the contraction with the rest of the terms in equation (5.13), one findsO2z{aa Oz — 8UOV—— } 8Uz”1 Oz°8z”— 3 Uo U0 h a a7p,7 V0 VoIS 2ãz ô,z0 i]ZPi 82z- a v U I°P 8 — voI U’, V0 U, 91 U 4thOUZ ‘9Zo ?] uV j= (5.16)It is easy to see that equation (5.16) cancels with (5.15) and so = 0.Similarly, OVz OVzlrl/JJI = 0. The force given by equation (5.13) is perpendicular to thehypersurface of the string defined by the two null vectors OUz and Additionally,the string trajectory, corrected to first order in the perturbation, still satisfies the nullcoordinate conditions. That is, the shift in coordinates caused by the force parallel tothe string has been cancelled out. The new trajectory is still a solution to the equationsof motion and there is no doubt as to the validity of the force equation.5.2 Force on a Single Wave RevisitedIn chapter two, it was found that the force obeyed two other reasonable conditions. Onecondition was that the force on a single travelling wave be explicitly zero. This conditionalso holds true for the perpendicular force of equation (5.13). It has already been shownin section 2.2 that the first term in (5.13) vanishes for the single wave pulse. It remainsto check that the parallel force terms also make no contribution.Consider the right-moving wave. For this case, ãz = Ox =constant, for example,x(3)=(i, 0,0, i1) and so 0x = (1,0,0, 1). Using the composite metric for h, the singlewave is represented by Vachaspati’s metric.—I” 2= —4Gp ln( p) (5.17)0VUUUmazChapter 5. The Perpendicular Force 98where p (X(Umax) — z)2 = (X2Umax) — z2) + (X’(Umaz) — z1)2. The derivatives of thecoordinate transformation, and 8U, involve contractions of either 9z or ãz withthe metric. Both of these contractions, as shown in section 2.2, equate to zero.The remaining terms of the parallel force contain a contraction of either Oz orwith the derivative of the metric. As given in section 3.3, the derivative of the Vachaspatimetric is= 4G{[(ãxx)2] Ox7 ln(2(x — Z)2)(Umax — Ukk)kinksF 2 Ox+ 2 (x — z) — (x — z)7 (5.18)O-x.O-x(x—z)where ü is evaluated at u Umax and___________— 13 — Fcx13(9x Oux)2 — (Ox. OUx)UUkjnk+E (x. 9—x)2 UkjnkETherefore, in the contraction 8z”h,7 corresponds to a contraction of zIAF,,which always vanishes for the right-moving wave. The last term, O8’, contains thecontraction Oz”h11,7 which again involves terms of the form z’’F. Because thismetric is evaluated at€ = Umax and also because Umax = u for a point z on the string,= 0.All of the terms from the parallel force are zero for the case of a single right-movingwave, as is the contribution from the first term. Consequently, the perpendicular back-reaction force on a single right-moving pulse is also zero. An exactly analogous argumentfollows for the left-moving wave and so, as before, the force equation developed in thischapter yields the expected result of no back-reaction for single travelling waves.5.3 Energy Equivalence RevisitedThe other reasonable condition on the force in chapter two related to the energy of thestring. In section 2.3 it was seen that the full force gave an equivalence between the energyChapter 5. The Perpendicular Force 99emitted by the string as gravitational radiation and the energy lost by the string due tothe back-reaction. However, because the parallel component of the force seems to justperform a coordinate transformation on the string and as the energy is the integral overthe entire world sheet of the string, one might expect that there would be no differencebetween using the full force expression or just the perpendicular component of it. Thatis, one expects that the parallel component of the force does not contribute to the energyloss of the string.As was demonstrated in section 2.1, the integral over the force is orthogonal to c9z.OuZ]wfdV8uDvZ 0 (5.19)Therefore, the component of the force parallel to O,z, makes no net contribution to theintegral. Similarly, the integral over u of the force is orthogonal to O,z0 and so thecomponent of the force parallel to Oz also makes no contribution. Because the equationfor the energy of the string involves an integral over both u and v, the only part of theforce which makes a contribution is the perpendicular component. This means that theenergy equivalence relation holds for both the force developed at the start of this chapterand the one used in chapter two.5.4 Effect of the Perpendicular ForceBecause the force given by equation (5.13) does not cause a shift in the internal coordinates, one might hope that this force would not have the gauge problems that wereencountered with the full force, equation (2.6). Unfortunately, this does not occur. Themajor gauge problem with the full force is that there is an arbitrary integration constantpresent in the delta function component of the force. This freedom means that the shapeof the force was very ambiguous and although the integrated force did seem to havesome reasonable behavior in the finite portion, the results are highly dependent on anChapter 5. The Perpendicular Force 100appropriate choice of the integration constant or scale of the delta function. The twocorrection terms both involve derivatives of the metric and so will involve deltafunctions containing the same scale factor. Unfortunately, these delta function contributions do not cancel out the previous ones. This can be seen in figures (5.4) and (5.4). Inaddition, the correction terms do have a scale dependence in the region ti> —100 wherethe full force did not (Appendix C). The perpendicular force increases the ambiguity inthe back-reaction effects instead of decreasing it.Figure (5.4) shows the amplitudes of the delta function contributions from the fullforce, the parallel component of the force, and the perpendicular component which isjust the full force minus the part parallel to the string. One quickly sees that not onlyare the magnitudes of the force different but also the qualitative shape of the force isdifferent. The perpendicular component of the delta function amplitudes is evidentlynot zero and so the ambiguity associated with the arbitrary integration constant is stillpresent. Switching to the perpendicular force does not solve the problems of the gaugefreedom.Figure (5.4) shows the delta function amplitudes with a scale of 0.01 instead of 1.Again, the perpendicular delta function contribution is non-zero and in fact qualitativelynot very different from the full force.The perpendicular force provides a nicer view of what one means by the back-reactionon a cosmic string. However, it still exhibits many of the problematic features of the fullforce. It still contains the difficulties associated with the gauge freedom of Vachaspati’ssolution. In fact, in some regions, it contains more difficulties than the full force. Thegauge dependence in the left-moving delta function cancelled out for the full force butdoes not do so for the perpendicular force. Additionally, the perpendicular force containsambiguities in locations other than just those of the full force. The full force deltafunctions occur at the position of the kinks of the undisturbed waves. The perpendicularChapter 5. The Perpendicular Force 101Full Force Parallel ForceI I I I I I I I I. I I I I Ia.) 5-.I I I— I I I I—100 0 100 —100 0 100Time Time(a) (b)Perpendicular Force20 I I I II I I—100 0 100Time(c)Figure 5.1: These figures show the amplitude of the delta function contributionto the full force (a), the parallel force (b), and the perpendicular force (c) whichis just the full force minus the parallel component. The integration constantfor these graphs is 1.Chapter 5. The Perpendicular Force 102Full Force Parallel ForceI I I I I I I I - r I -20- -10 -- ci)—i I I I _._. I I I I I I—100 0 100 —100 0 100Time Time(a) (b)Perpendicular ForceI I I I I I I I I I:1 I I I—100 0 100Time(c)Figure 5.2: These figures show the amplitude of the delta function contributionto the full force (a), the parallel force (b), and the perpendicular force (c). Theintegration constant for these graphs is 0.01.Chapter 5. The Perpendicular Force 103force also contains delta functions at the kinks of the colliding waves on the string dueto terms like These terms are multiplied by the metric which also depends on thegauge choice, the integration constant.The continued presence of the gauge freedom indicates that before one can have anyconfidence in the effects of either the full force or the perpendicular force, some meanswill have to be found to completely fix the gauge. This requires either finding a physicallyacceptable reason for choosing one specific integration constant, or deciding on a gaugefixing condition before developing the force expression.Based on the minimal effect that the correction terms had upon the right-movingdelta force contribution to the force, there is also some indication that the perpendicularforce would behave in a manner qualitatively similar to the full force. For this reasonand because the perpendicular force does not solve the gauge problems of the full force,I have not explored this possibility in any depth.Chapter 6ConclusionsIn this paper, I have investigated the gravitational back-reaction of two colliding travelling waves on a cosmic string, more specifically the back-reaction force generated in theLorentz gauge by two oppositely moving triangular pulses as they pass through one another. It was found that the approach used by Quashnock and Spergel could be modifiedin order to be applicable to this situation [20]. The result was that the back-reactionforce, though somewhat complicated in appearance, leads to a damping out of the twowaves. The damping effects are greatest near the collisions of the kinks in the waves,those points from which the gravitational radiation is emitted.Quashnock and Spergel’s technique was developed for the case of oscillating ioops ofstring [20]. Their method cannot be directly applied to the case of colliding waves oninfinitely long strings. If one attempts to do this, it was found that the resulting back-reaction force becomes non-integrably infinite. The reason for this is that the form ofthe metric that they used is in a gauge which is inappropriate for this problem. Makinga gauge transformation within the Lorentz gauge, the original metric was converted intothe form of a composite metric. This gauge choice proved more successful for studyingthe back-reaction forces. The composite metric reduces to Vachaspati’s gauge choice forsingle travelling waves. The composite metric also solves the difficulties associated withthe infinite boundaries of the metric integral. Because of how the composite metric isconstructed, the dependence on the value of the past-time cut-off is removed. As a result,the composite metric satisfies the Lorentz gauge conditions at finite values of the cut-off.104Chapter 6. Conclusions 105Using the composite metric a.s the source of the gravitational back-reaction force, it wasfound that the infinities encountered with the original metric had been reduced to deltafunctions.The shape and integrated effect of the force was determined for several different pointson the string. The time slices of the force indicate that the force is generated by thecollision of oppositely moving kinks. The forces first appear just after the kinks collideand then propagate away from their point of origin and decrease in amplitude. Thepoint of origin of the back-reaction forces is the collision of oppositely moving kinkson the string. This result is quite reasonable because it is at these locations that thegravitational radiation is produced.The effects of the gravitational back-reaction are entirely non-local. The back-reactionequations indicate that there is no effect on a point of the string caused by a forcegenerated at that same point. The delta function forces, for instance, are off-set fromthe kinks on the two-wave string. They align instead with the kinks of the undisturbedtravelling wave which is at a different spatial position. The source of the perturbationwhich generated the force must be on the past light cone of the point on the stringbeing effected. Therefore, in order to have travelled the necessary spatial distance, theperturbation must have come from some time in the past of and some distance away fromthe affected point on the string. This implies that the force effects are very non-local. Thenon-locality is also evident in the width of the back-reaction forces. The gravitationalradiation is only emitted where two kinks collide, a single point in spacetime. Therefore,one would expect that the forces caused by the gravitational radiation would also affectonly a small portion of the string at once. This is certainly true of the delta functionforces, but the finite contributions to the back-reaction are drawn out over much largerregions. This indicates that the metric deviations causing the force occur over a varietyof times and travel various distances before catching up with the string again. The timeChapter 6. Conclusions 106slices also indicate that back-reaction effects are not confined to the period during whichthe waves are actually passing through one another. A small decreasing force, caused bythe delay suffered by one travelling wave as it passed over the other, continues to affectthe wave pulses long after they have separated.The effect of these forces on the right-moving component of the string was examinedusing u-slices of the force. It was seen that the integrated force increases in magnitude asone approaches the kink. The integral over the finite components of the force is positiveas one approaches the kinks at u = ±100 from below and negative as one nears thekink at u = 0. As one approaches the kinks from above, the forces are of the oppositesign. The sign of the integrated force is important as it determines how the slope ofthe string components will be affected. The integrated force as one approaches the kinkat u = —100 from below is positive indicating that the slope is being increased. Thenegative force for u > —100 reduces the slope on the other side of the kink. This makesthe slope on the left of the kink closer to that on the right. The forces as one nears u = 0and u = 100 also serve to reduce the the difference between the slopes on either side ofthe kinks.Unfortunately, there is some ambiguity as to the sign of the right-moving delta function contribution. Because of the arbitrary scale factor it is possible to shift the positionof the delta function component of the force. Although, regardless of the scale, the forcealways has the same sign as the integrated finite forces in the immediate neighborhood ofthe kink, at an arbitrarily small distance away from the kink, it can have a value whichis very different. The effects of the delta functions can completely dominate those of thefinite force contributions resulting in significantly different effects on the string. Insteadof the kinks just being smoothly rounded off, sharp dips can be produced in the string.The arbitrary scale factor in the force equations represents the major problem with thisapproach to finding the back-reaction force.Chapter 6. Conclusions 107Another disturbing aspect of my choice of gauge is that the back-reaction effects arecaused not just by the residual component of the metric, that part which generates thegravitational radiation [5], but also by the single travelling wave components. A singletravelling wave emits no gravitational radiation and yet it does make a contribution tothe back-reaction force. Furthermore, the contribution does not cancel out with somepart of the residual metric as did the straight string contribution. Additionally, the deltafunction forces seem to come from times before the collision had even occurred, timesat which there was no gravitational radiation produced by the system at all. Thesedifficulties together with the problem of the arbitrary scale factor indicate that despiteits reasonable features, the force developed in this thesis still contains too much gaugefreedom. To study gravitational back-reaction on cosmic strings, therefore, one needs tofind a better method of restricting the Lorentz gauge freedom or one must find a differentgauge entirely.However, assuming scale factors which do not make the behavior of the delta functionforces radically different than the finite forces, for example, the scales which I chose, itis still possible to determine the effects of the back-reaction force in the Lorentz gauge.Also there are some regions of the wave which are not affected by the arbitrary natureof the scale factor. Regions such as the one at u > —100 where the dependence of thescale factor cancels out or the region at u > 100 where the delta function contribution iszero. In these cases as well as those for which I have chosen an appropriate gauge, themagnitude of the integrated force increases as one examines points closer and closer tothe kinks. This increase, combined with the reduction in the slope differential across thekinks, results in the waves being damped out. The pulse is slightly flattened out and thesharp corners are rounded off. These results are in agreement with other work done onthe back-reaction of cosmic strings [10, 20].As expected, the forces have a high degree of symmetry between the left-moving andChapter 6. Conclusions 108right-moving waves. This was shown briefly in section 4.4 and indicates that the effectsof the force, though only evaluated for the right-moving wave, are also applicable to theleft-moving wave. The left-moving pulse is also damped down and rounded off by itsinteraction with the right-moving wave.Another difficulty with the force used is that it contains components parallel to theaxis of the string. This results in the perturbed trajectory of the string no longer satisfyingthe null coordinate conditions in the curved spacetime. A better definition of the ‘real’force is just the perpendicular component of the force. The perpendicular force satisfiesnot only all of the conditions that the original force does but also the null coordinatecondition for the perturbed metric. However, although the perpendicular definition ofthe force is a better definition, it still exhibits the major failings of the original force. Itstill contains the arbitrary integration constant and still has contributions to the forcefrom the single travelling wave components at times before the collision had occurred.Quantitatively, the two forces give different results but qualitatively, there are indicationsthat the two forces are very similar. Nevertheless, the perpendicular force is still apossibility that could be explored.One can conclude, therefore, that the choice of the Lorentz gauge results in a back-reaction force which is reasonable in many respects. The energy emitted by the stringin the form of gravitational radiation is the same as that lost by the string due to theback-reaction forces. Also, in this gauge, the single travelling wave has no back-reactionforce as is expected because it emits no gravitational radiation. Unfortunately, thisgauge is not perfect being flawed most noticeably by the presence of an arbitrary scalefactor in the force equations. The gauge freedom of this arbitrary constant means thatit is difficult to make any definite statements about the effects of the gravitational backreaction indicating that fixing the gauge in a different manner might be more informative.Chapter 6. Conclusions 109However, a reasonable choice for the scale factor does permit a determination of the backreaction forces on two colliding travelling waves. In this case the back-reaction servesto damp out the waves, rounding off the corners of the wave and straightening out thestring. These results are in agreement with previous work done on the back-reaction ofcosmic strings [20, 10] and in accord with one’s physical intuition about how the stringsshould behave.Bibliography[1] Einstein, A., Ber. Preuss. Akad. Wiss. Berlin, 8, 154 (1918).[2] Taylor, J. Binary Pulsar, Gravitational Waves A colloquium given at the Universityof British Columbia on November 18, 1993.[3] Vilenkin, A., Phys. Rep. 121, 263 (1985).[4] Kibble, T.W.B., Phys. Rep. 67, 183 (1980).[5] Unruh, W.G. and D.N. Vollick, Cosmic Strings: Structure Formation and RadiationUniversity of British Columbia preprint, (1993).[6] Allen, B. and E.P.S. Shellard, Phys. Rev. D 45, 1898 (1992).[7] Vollick, D.N., Phys. Rev. D 45, 1884 (1992).[8] Vilenkin, A., Gravitational Interactions of Cosmic Strings 300 Years of Gravitation(New York, Cambridge University Press, 1990).[9] Sakellariadou, M., Phys. Rev. D 42, 354 (1990).[10] Hindmarsh, M., Phys. Lett. B 251, 28 (1990).[11] Garfinkle, D., Phys. Rev. D 32, 1323 (1985).[12] Misner, C.W., K.S. Thorne and J.A. Wheeler, Gravitation (W.H. Freeman andCompany, New York, 1973).[13] Nambu, Y., Lectures at the Copenhagen Summer Symposium (1970).[14] Goddard, P. et. al., Nuc. Phys. B56, 109 (1973).[15] Nielsen, H.B. and P. Olesen, Nuc. Phys. B61, 45 (1973).[16] Unruh, W.G., private communication (October, 1993).[17] Vachaspati, T., Nuc. Phys. B 277, 593 (1986).[18] Weinberg, S., Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity (John Wiley & Sons, Toronto, 1972).110Bibliography 111[19] Unruh, W.G., G. Hayward, W. Israel, and D. McManus, Phys. Rev. Lett. 62, 2897(1989).[20] Quashnock, J.M. and D.N. Spergel, Phys. Rev. D 42, 2505 (1990).[21] Wald, R. M., General Relativity (University of Chicago Press, Chicago, 1984).[22] Itzykson, C. and J. Zuber, Quantum Field Theory (McGraw-Hill, New York, 1980),p. 34.[23] Unruh, W.G., private communication (December, 1993).Appendix AComparing Single Wave MetricsAlthough on the surface the form that I have used in this paper for a single travellingwave on a string appears different Vachaspati’s solution [17], they are actually identical.Vachaspati’s solution for a right-moving travelling wave is (1.16):f’2+g’—f’ —g’ f’2+g’—f’ 1 0 —f’= h (A.1)—g’ 0 1 —g’f’2+g’—f’ —g’ f’2+g’where h = (—4Gt) ln(2[(x— f)2 + (y — g)2]), is an integration constant, and f and gare functions of z3 — z0 where z is the jth component of the four-vector z. From (2.28)one has= —4GiO.axln(2p) (A.2)UUm axwhere p = (X(Uma) — z)2 = (X2Umax) — z2) + (X’(Umax) — z’)2.Recalling that for a right-moving travelling wave, x = (1, 0, 0, 1), one first considerthe arguments of the logarithm. At Umax,Ox.x(u) =— = — z0 (A.3)As f measures the displacement of the string from the z-axis at the position z3 — z0 inthe 1-direction. Because of (A.3), f is equivalent to X’(Umax). The argument is the same112Appendix A. Comparing Single Wave Metrics 113for g and, therefore, the arguments of the logarithms are the same, up to an integrationconstant.Next, consider the components of the tensor.1 3 0(Jux— 0O(x —x) Ou Ou= f’8x.ôxaux’= f = (A.4)Ox t9xHowever, from (A.2) the same component of the tensor is—F°’— —9x1— —The two tensor components are the same. Similarly, one finds that the F°2 componentsare also identical. The F component from (A.2) is just—F”——Ox.because Ox’ = 0. Similarly for the F22 and F’2 components.All that remains to check is that the F°3 components are the same. Consider firstthe f’ 2 + term.= ( O,x’ )2 ( Ox2 2 (A.7)\OuX OvX \.0ux OvXJbut(Ox)2 = (Ox°)2+ (Ox’)2 + (ôx2)+ (ax3)2 (A.8)therefore,f 02 ( 32I 2 ! 2 — J — -‘u’- )(Ox.Ox)— (ax° — Ox3)(O ° + 8x3)— (Ox .——(6x° + Ox3)—-F°3= ax ax(A.9)Appendix A. Comparing Single Wave Metrics 114Observing thatF°° = 2ôx°8x°+ãx•Ox= 2Ox° + (Ox3 — a°)= F°3 (A.1O)= F33 (A.11)it is easy to see that the remaining components of the tensors in (A.1) and (A.2) areequivalent.Therefore one concludes that (A.1) and (A.2) are identical up to the arbitrary integration constant.Appendix BGauge Condition for the Residual WaveTo find the metric for the string, either for single travelling waves, or in the case wherethere are two waves on the string, one evaluates (1.5) using the causal Green’s functionof the wave equation. To insert a past-time cut-off, a Heaviside function of the formO(u + v + T) is inserted into the integral. Taking the value of r to positive infinity theneffectively removes the cut-off. If this is done, however, for any finite value of r, one findsthat the metric does not obey the Lorentz gauge condition. Physically, this is due to theabrupt change that one has effectively inserted into the stress-energy tensor. At time ---rthe string’s energy abruptly goes from zero to some non-zero value. The conservationof stress-energy is not obeyed across this boundary. Mathematically, this appears in theintegral as a dependence on z of the boundary condition. Doing the integration overthe delta function, 6((x(u, v) — z)2), causes a change in the Heaviside function for thepast-time cut-off of say,8(u+v+r)—O(u+v(u,z)+r) (B.12)As the derivative of vQu, z) with respect to z is now non-zero, the Lorentz gauge conditionno longer holds.In building up the metric that was used to calculate the back-reaction force, I usedthe metric of two Vachaspati-like single travelling waves, a straight string, and a residualcontribution. The metric for the straight string obeys the Lorentz gauge conditions asdoes the metric for Vachaspati’s single travelling waves [17]. It is necessary, however, toconfirm that the residual metric also obeys the gauge conditions.115Appendix B. Gauge Condition for the Residual Wave 116The residual metric is build up from the two-wave solution plus a straight stringminus two oppositely moving single travelling wave solutions. This is represented diagrammatically in figure (3.4). The metric, , can therefore be written ash=h1—2k+ (B.13)where subscript 1 refers to the two-wave solution, subscripts 2 and 3 refer to the left-and right-moving single wave solutions respectively, and subscript 4 refers to the straightstring solution. Setting 8Gt = 1, one can then write= f dudv ((x — z)2)O(z° — u — v) (u + v + r) (B.14)where T is some constant past-time cut-off. Consider now, the Lorentz gauge condition.= f dudv [4’((x — z)2) (x — z) x6,xTh e(z° — u — v) 8(n + V + T)+26((x — z)2)OxOx 6(z° — u — v) 8(u + v + r)]v+oo= J du O(z° — u — v) O(u + V + T) 6((x —— fdvo((x— z)2) [f duax(z0— u — v)8(u + v + r)]+ f dv Ox O(z° — u — v)O(u + v + T) 6((x — z)2)— fduo((x _z)2) [fdvaxo(z0 uv)O(u+v+r)]+ f dudv (Ox + ö((x — z)2) ö(z° — u — v) 8(u + V + T)= f dudv (Ox + Ox) ö((x — z)2)x[O(z°— u— v)(u + v + T) — b(z° — — v)8(u +v +r)]+ f dudv (Ox + Oxfl 6((x — z)2) 6(z° — — v) 8(n + v + r)= f dudv (Ox + Ox) ö((x — z)2) 6(u + v + r)aux OXcr= + (B.15)2Ox . (x — z)ZL=u(v,z),v=—T—u 2Ox (x — z) v=v(u,z),u—r—vAppendix B. Gauge Condition for the Residual Wave 117Individually, h does not obey the Lorentz gauge condition. However, (B.15) is evaluated at the time —T, = u+v = —r. This is a time before any collision of the two waveshas occurred and by construction, the single travelling wave components of the residualwave exactly line up with the two-wave component. Therefore, when the summation isdone to obtain Ii for the whole residual wave, the contributions from the cut-off, those of(B.15), will exactly cancel and the residual wave will obey the Lorentz gauge condition.Appendix CGauge Independence of the Left-Moving Delta FunctionsOne of the major difficulties associated with the force which was developed in this thesisis that the delta function contributions to the force contain an arbitrary factor. Thesedelta functions arise in the Vachaspati form of the metric for a single travelling wave. Thearbitrary integration constant, 2, means that the back-reaction forces obtained for thetwo colliding triangular wave pulses are not well defined. It would have been better if thepart of the force which depends on the integration constant had cancelled out. For theright-moving delta function components this does not occur. However, when studyingthe effect of the force on the right-moving wave, the dependence of the left-moving deltafunction force on 122 does disappear. The cancellation of this gauge freedom can be shownanalytically.The back-reaction force, as given by equation (2.6) involves three terms each of whichcontains a derivative of the metric.vz = +—(C.16)For the single travelling waves, the derivatives of the metric take a form such as thatgiven in equation (3.1).= 4Gi[(Oxx)2] ln(122(x — Z)2)6(Vma — Vkk) (c.17)kinks V ‘Uwhere €3 is evaluated at V = Vmax and________—________C18(8x. Ox)2— (0x . O-x)2=kk+E— (Ox Ux)2kkink6( .118Appendix C. Gauge Independence of the Left-Moving Delta Functions 119Splitting the logarithm in (C.17) into two pieces,1n(2(x — z)2) = ln(2) + ln((x — z)2)one can consider just that part of the logarithm which is dependent on the constantWhen integrating over v, the delta function, i5(Vmax— vkk), will introduce a factorof (Okx . . Ox) into the denominator of (C.17). The absolute value signs maybe dropped as this factor is always positive.I consider first the contributions to the force from the delta functions in the lasttwo terms of equation (C.16). For the purpose of concreteness, I will demonstrate theircancellation at the point u = —80. At this point the delta functions each lie on sectionsof the string for which Oz takes on a different value. The same result, however, can beshown for all points along the string. As I am oniy interested here in the gauge dependentparts of the force, I shall only consider those terms containing 1n(2). Because all of theseterms are proportional to a delta function, the integration over v is straightforward.(h — 1la,7) Ouza Oz = 4Gt1n(2) { a .kinks ( vX x z) VVkink+E— Fa0Z0vZ0uX7 (C 19)(x. Ox)(ã-x. Oz) kink+ewhere àx is the slope of the right-moving component of the string as ii tends towardsnegative infinity and v is evaluated at the value of v for which Vma = Vkink.Using the simplification that O,z . ôjx = 0 if Ovzv = ax”, one finds that three ofthe six terms in (C.19) are zero. There are three different values that 8kx and Oz cantake on corresponding to the three different slopes of the left-moving triangular wave.Labelling these three slopes as Ox where i is a value from 1 to 3, equation (C.19) canbe rewritten asAppendix C. Gauge Independence of the Left-Moving Delta Functions 120—ha,7)zOvz= 4Gbtln(12)(a ôxi)(Ox .Ox2)(O . 0x3) X{(Ojx.Oxi aX.auZavX27 + X•OvX2OvX1 •3uZOIX78jX ãvXl OvX2 Z X7 + 9x ôvX2 Ojx 3z 3xi7)(9x . 9x3)+(6jx 8vX2DiX •Ou2OvX37 + Dx ôvx3Dvx2—Ox OvX2 OvX3 OuZ OüXy + OjX OvX3 c9x Ozax27)(ox .• O-jX OuZ OvXl7 + 0X avxi av3 3uZ Oi1X7jX OvX3 0vXl OZ OjX7 + OjX . 0vXl OX . Ox2)C.20)All of the terms in equation (C.20) cancel indicating that the last two terms in the forceequation (C.16) have no dependence on the integration constant.Consider now the first term in equation (C.16). Expanding ha7, uzaOvz in a similarfashion to that done for equation (C.20), one finds that3ha7,Ouz’vz = 4Gtln(f2)f Ox . Ox+i7+ Ovxi+i . Ouz Oiix7 — Ox+i . a2oz7Ox— aix. OZ X+8X OU29U27— (C 21)ojx•ovxj Jwhere i = 4 is equivalent to i = 1. In this equation, all of the terms correspondto Ox terms. The terms containing Oz are eliminated. Because all of the terms in(C.21) are independent of Oz and because Ox and Oz are constant for the terms beingconsidered, the summation equates exactly to zero. Each of the three different slopes,occurs twice, once being added to the sum and once being subtracted. Therefore,the first term of the force equation (C.16) also vanishes. There is no contribution tothe back-reaction from the terms in the force equation which depend on the integrationAppendix C. Gauge Independence of the Left-Moving Delta Functions 121constant.Although the independence of equation (C.16) has oniy been shown for the force atu = —80, it also exactly cancels for all values of u between —100 and 100.1 That is,the left-moving delta function contributions to the back-reaction do not depend uponthe arbitrary integration constants, only the right-moving ones do. This means that inthose regions where there is no right-moving delta force, such as at —100 < u < —3 andu> 100, there is no ambiguity as to the shape of the back-reaction.Unfortunately, the perpendicular force found in Chapter 5 does not exhibit this feature. The two terms, O’ and OO’, which contain contributions to the delta function, cannot cancel out in this manner. The cancellations no longer work because of thepresence of extra 9z and 9z terms and also because of the lack of a term equivalent tothe third term in equation (C.16). Therefore, in this respect, the force which still includescomponents parallel to the string contains less ambiguity than the force in which theyhave been removed. This is yet another indication that the perpendicular force will notbe a great improvement over the full force.‘For values of u > 100 and u < —100, the delta functions from the left-moving wave have zeroamplitude and so do not contribute to the force at all.

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