Open Collections

UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

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

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata

Download

Media
831-ubc_1994-0109.pdf [ 2.11MB ]
Metadata
JSON: 831-1.0085445.json
JSON-LD: 831-1.0085445-ld.json
RDF/XML (Pretty): 831-1.0085445-rdf.xml
RDF/JSON: 831-1.0085445-rdf.json
Turtle: 831-1.0085445-turtle.txt
N-Triples: 831-1.0085445-rdf-ntriples.txt
Original Record: 831-1.0085445-source.json
Full Text
831-1.0085445-fulltext.txt
Citation
831-1.0085445.ris

Full Text

GRAVITATIONAL BACK-REACTION FROM COLLIDING TRAVELLING WAVES ON A COSMIC STRING By R. Glenn Wells B.Sc.(Hon), The University of Calgary, 1991  A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER’S OF SCIENCE  in THE FACULTY OF GRADUATE STUDIES DEPARTMENT OF PHYSICS  We accept this thesis as conforming to the required standard  THE UNIVERSITY OF BRITISH COLUMBIA  January 1994  ©  R. Glenn Wells, 1994  In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my department or by his or her representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission.  Department of Physics The University of British Columbia 6224 Agricultural Road Vancouver, B.C., Canada V6T 1Z1  ate:  ,( /9y.  Abstract  In this thesis, I investigate the effects of the gravitational back-reaction force on two colliding triangular pulses travelling along an infinite cosmic string. The force is deter mined using a perturbative approach in the Lorentz gauge. This gauge choice is found to be reasonable in several aspects. Firstly, there is no gravitational back-reaction force on a single travelling wave. Secondly, the energy emitted by the string as gravitational radiation is the same as that lost by the string due to the back-reaction. Finally, the force can be made perpendicular to the string such that the perturbed string trajectory con tinues to satisfy the equations of motion. Unfortunately, the force equations contain an arbitrary gauge freedom making it difficult to form definite statements about the effects of the back-reaction force. However, choosing a reasonable value for the gauge parameter results in the two wave pulses being flattened out and having their sharp kinks rounded off.  11  Table of Contents  Abstract  ii  Table of Contents  iii  List of Figures  v  Acknowledgements  Viii  Introduction 1  2  1  Cosmic Strings  5  1.1  Notation  6  1.2  The Stress-energy Tensor  7  1.3  The Single Travelling Wave  9  1.4  Waves Travelling in Opposite Directions  11  1.5  The Emission of Radiation  13  The Formalism  21  2.1  The Force Equation  24  2.2  The Back-Reaction on a Single Wave  26  2.3  The Energy Lost by the String  34  2.4  Evaluating the Force for Triangular Waves  41  2.5  Other Features of the Force Equations  45  111  3  4  5  6  The Composite Metric  49  3.1  Failures of the Direct Approach  49  3.2  The Composite Metric  53  3.3  Force Equations for Single Travelling Waves  56  The Results  61  4.1  The Force Components  62  4.2  Time Slices  65  4.3  U-slices  79  4.4  V-slices  88  The Perpendicular Force  93  5.1  Removing the Parallel Components  94  5.2  Force on a Single Wave Revisited  97  5.3  Energy Equivalence Revisited  98  5.4  Effect of the Perpendicular Force  99  Conclusions  104  Bibliography  110  A Comparing Single Wave Metrics  112  B Gauge Condition for the Residual Wave  115  C Gauge Independence of the Left-Moving Delta Functions  118  iv  List of Figures  1.1  A small travelling wave on a cosmic string moving up the side of a larger one  1.2  14  The distorted appearance of a wave when seen from the reference frame of a sloped string  20  2.1  Degeneracies in ii and i  43  3.1  The two travelling waves before the collision  50  3.2  Infinite force spikes at time t  —115  52  3.3  Decomposition of the two-wave string  54  3.4  The residual component of the string metric  55  3.5  Amplitudes of the delta function generated by the right-moving wave  58  3.6  Amplitudes of the delta function generated by the left-moving wave  3.7  The effects of the scale factor on the amplitude of the delta function con  =  .  tribution to the force 4.1  59  The decomposition of the residual metric’s contribution to the force at t=—90  4.2  58  63  Force contributions at t  =  —90 from the two single travelling waves and  the straight string using Vachaspati’s metric  64  4.3  Force at time t  =  —119  68  4.4  Force at time t  =  —115  68  4.5  Force at time t  =  —110  69  v  4.6  Force at time t  =  —101  69  4.7  Force at time t  =  —99  71  4.8  Force at time t  =  —90  71  4.9  Force at time t  =  —85  72  4.10 Force at time t  =  —81  72  4.11 Force at time t  =  —79  74  4.12 Force at time t  =  —50  74  4.13 Force at time t  =  —21  75  4.14 Force at time t  =  —18  75  4.15 Force at time t  =  —15  76  —1  76  4.16 Force at time t 4.17 Force at time t  =  1  77  4.18 Force at time t  =  15  77  4.19 Force at time t  =  21  78  4.20 Force at time t  =  150  78  4.21 Force at u  =  —105  80  4.22 Force at u  =  —101  80  4.23 Force at u  =  —100.1  80  4.24 Force at u  =  —99.9  80  4.25 Force at u  =  —99  82  4.26 Force at u  =  —80  82  4.27 Force at u  =  —5  82  4.28 Force at u  =  —1  82  4.29 Force at u  =  —0.1  83  4.30 Force at u  =  1  83  4.31 Force at u  =  20  83 vi  4.32 Force at u  =  95  83  4.33 Force at u  =  99  84  4.34 Force at u  =  99.9  84  4.35 Force at ‘a  =  101  84  ‘a =  110  84  4.36 Force at  4.37 Force at ‘a  =  150  85  4.38 Force at ‘a  =  500  85  4.39 Integrated finite forces before  ‘a =  —100  87  4.40 Delta function contribution to the integrated force before 4.41 Integrated finite forces after  ‘a =  ‘a =  —100  .  .  —100  4.42 Delta function contribution to the integrated force after 4.43 Integrated force after  ‘a =  100  87 89  ‘a =  —100  .  .  .  .  89 90  4.44 Force at v  =  —25  92  4.45 Force at v  =  —21  92  4.46 Force at v  =  —20.1  4.47 Force at v  =  —19  5.1  .  .  92  Amplitudes of the delta function contribution to the perpendicular force at a scale of 1  5.2  92  101  Amplitudes of the delta function contribution to the perpendicular force at a scale of 0.01  102  vii  Acknowledgements  I would like to thank my supervisor, Bill Unruh for all of his help and guidance throughout this project. My special thanks also to Kristin Schleich for her critical and expeditious reading of this thesis. Finally, I would like to thank my family and friends for their continual encouragement and support.  vii’  Introduction  Gravitational radiation was first predicted by Einstein in 1918 [1]. However, it was not until recently that there was any experimental evidence for its existence. During the last two decades, J. Taylor and his collaborators have been studying the binary pulsar system  PSR 1913+16 with the Aricebo telescope in Peurto Rico [2]. The binary system emits gravitational 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 pulses to accurately measure this inward decay. The results they found perfectly match with the predictions of General Relativity for the back-reaction of the gravitational radiation on the system thus providing the first indirect evidence of gravitational radiation and additional verification of Einstein’s theory. Any system which emits gravity waves should experience back-reaction effects. One such system is the collision of travelling waves on a cosmic string. It is the gravitational back-reaction on cosmic strings which forms the basis of this thesis. Cosmic strings are formed as the universe cools below a critical temperature and the symmetry of the grand unified theories (GUT’s) is spontaneously broken [3, 4]. As a simplistic model of a GUT, consider a field theory containing a complex scalar Higgs field  and a U(1) symmetry:  When this symmetry is broken, the Higgs  —*  field acquires a vacuum expectation value of <  >=  0 where e  j  is related to the  self-interaction potential of the theory. The parameter 9 is arbitrary which means that the vacuum state is degenerate; solutions form a circle of radius  ij  in the complex q  space. The angle which 0 acquires as the universe cools is random and different parts of the universe can acquire different 0 values. As a result the vacuum expectation value is 1  uncorrelated at distances of separation greater than  which is of the same order as the  age of the universe at the time of the phase transition. In other words, 0 can be different on length scales greater than  e.  It is energetically favorable for the universe to have a uniform 0 value. However, the universe’s topology may be such that it is not possible to achieve this uniformity through a continuous evolution of 0. Because < space, z0  =  >  is single valued, along any closed ioop in  2irn where n is an integer. If n is non-zero, say n  =  1, then the loop cannot  be continuously shrunk down to zero radius because this would involve a discontinuous change in n from one to zero. As a result, one has a tube-like volume of trapped false vacuum. The tube is either closed or infinitely long because otherwise one could move the loop around the end of the tube and continuously deform it to zero radius. The tube of topologically trapped false vacuum is called a cosmic string. Cosmic strings are cosmologically important as they provide a possible explanation for the creation of structure in the universe (for example see references [3, 5, 6, 7] among others). Cosmic string loops, which form when strings collide and intercommute, can act as seeds for the formation of galaxies, accreting matter and radiating energy as they oscillate, decay, and finally disappear. They can also generate larger structures because of the wake they create while moving through space. The spacetime around cosmic strings is quite simple. It is exactly like fiat space locally, but globally there exists a conical deficit. This means that the circumference of a circle surrounding a cosmic string is not 2irr as it is in Euclidian space but rather 27rr(1  —  4G i 1 ) where G is a very small quantity.  As the string moves through space, matter effectively falls toward the deficit creating a wake behind the string. This wake could be the cause of some of the large scale structure in the universe. Besides their possible cosmological significance, cosmic strings are also interesting in their own right. They have some unique properties that make them very different from 2  ordinary matter. These properties include the fact that stretching a string does not give you a longer thinner string, such as one gets upon stretching a rubber band, but instead causes an increase in the quantity of cosmic string. Also, unlike regular travelling waves on a normal string, two waves moving in opposite directions on a cosmic string cannot be expressed as a superposition of two single travelling waves, f(z + t) + g(z  —  t) where  z and t correspond to a standard Cartesian coordinate and coordinate time respectively.  Another interesting feature of cosmic strings is that they could cause astronomically observable effects. The conical deficit in the spacetime around cosmic strings causes the bending of light rays which might lead to the double imaging of stars or galaxies [8]. Also when two oppositely moving travelling waves collide on a long cosmic string, they emit gravitational radiation [9, 10]. This radiation, though very small in amplitude, might also eventually be observable. The cosmological aspects of strings are perhaps their most important feature. To more completely understand the role they may have played in the early universe, it is necessary to comprehend the mechanisms by which string loops decay and kinky strings are damped or straightened out. To this end, it is important to understand the effects of gravitational back-reaction on the cosmic string. A valuable step towards this goal is to determine how the radiation from two colliding wave pulses on a string affects the shape and motion of the two waves. This thesis investigates the effects of the back-reaction forces 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 properties of a cosmic string required for a study of the back-reaction forces: the stress-energy tensor and the metric. It also demonstrates that the radiation from the collision of two triangular 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 the Lorentz gauge, is shown to have reasonable properties: the perturbative solution does not 3  violate the equations of motion in flat spacetime, the single travelling wave feels no backreaction force, and the energy lost by the string is the same as that emitted in the form of gravitational radiation. Some other features of the equations are also discussed. The third chapter discusses the problems that arise through direct application of the back-reaction equation 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 the equations for the back-reaction force due to the Vachispati components of the composite metric on the actual cosmic string with two travelling waves are stated. Chapter four shows what the back-reaction force on the string looks like on various different time slices throughout the collision. In addition, u-slices of the collision and the effects of the force on the right-moving wave are shown. The symmetry of the results with regards to the left- and right-moving waves is also demonstrated. Chapter five discusses a failing of the force as determined in chapters two and three. This failing is the presence of components of the force which are parallel to the string. These components cause a coordinate change which prevents the perturbed string trajectory from satisfying the coordinate conditions in curved spacetime. Finally, in chapter six, I summarize my conclusions regarding the effects of the back-reaction force on the two colliding waves.  4  Chapter 1  Cosmic Strings  Cosmic strings are characterized by a single parameter, the energy of the symmetry breaking,  7].  Based on dimensional arguments, the radius of the string is 6  the mass per unit length of the string is Unifying scales, this corresponds to 6  2 j  -  77_i  and  For strings creating at the Grand  m and a mass density of 32 10  t  kg/m 21 10  [8]. To put these numbers in some perspective, the radius of the Hydrogen nucleus is only 5 10 m . The volume mass density of the string is 10 3 whereas the density of kg/m 85 a neutron star is only 10 . The extremely thin nature of the cosmic string makes 3 kg/m 15 it reasonable to consider the string as an object with only one spatial dimension. 1 This will be assumed throughout this thesis. Despite the high mass of cosmic strings, because of their extremely thin nature, their gravitational effects still lie within the weak gravity regime. Using a system of units in which G=c=1, the relation between mass and length is lg= 10 cm and so G[L is 28 unitless with a value of only 1O_6. Because this parameter is quite small, the right-hand side of Einstein’s equations is close to zero and one is justified in using the weak field approximation to the theory. Using linearized gravity and the approximation that a string is only a one dimensional object, it is quite straightforward to find the stress-energy tensor and the metric which describes the string and its equations of motion. ‘Garfinkle shows [11] that ignoring the internal structure of the string and treating the string as an infinitely thin object is a valid approximation for determining its gravitational field.  5  Chapter 1. Cosmic Strings  1.1  6  Notation  In this thesis I investigate the effects of gravitational back-reaction on colliding travelling waves moving along a cosmic string. This investigation assumes that only gravitational radiation is present to cause a back-reaction effect, that is, I assume the string is un charged and no electromagnetic radiation is being emitted. As the gravitational field of cosmic strings is weak, I will be using linearized gravity. The purpose of this section is to clarify the notation that I will be using throughout the thesis. In linearized gravity, or the weak gravity limit, one assumes that the full spacetime metric g. can be broken into two components r, and  The first is just the metric  for fiat Minkowskian spacetime which can be expressed as  = diag(—1, 1, 1, 1) while  the second represents a very small perturbation to that spacetime. In linearized gravity, one works only to first order in h, and so raises and lowers the indices on  using just the fiat spacetime metric  i.  Therefore, one can write the  Christoffel, or connection, symbols for the spacetime as P1L  —  1  \gv,3,a  —  =  —i-  —  fr’(h,,a + hva,  ha,v)  (1.1)  = 16KT z, 11  (1.2)  —  In linearized gravity Einstein’s field equations are [12] 7,  aLt,  ‘va,Ji  a  j,  —  ‘‘fw,a  a  a —  —  These equations may be simplified by writing them in terms of  “  =  —  and  by specifying a gauge. The gauge I will be working in is the Lorentz gauge. For a metric to be in the Lorentz gauge, it must satisfy the requirement that V  =  o  (1.3)  Chapter 1. Cosmic Strings  7  This requirement does not, however, completely fix the gauge. Any transformation of the form —  with  satisfying  =  (1.4)  —  0 is still permissible. Again, indices are raised  and lowered using the metric of fiat spacetime,  i.  This gauge freedom is critical in  determining the back-reaction force as shall be seen in chapters two and three. In the Lorentz gauge, Einstein’s equations reduce down to Dh  =  (1.5)  —167rGT,  where G is Newton’s universal gravitational constant. 1.2  The Stress-energy Tensor  A 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 the z-axis is invariant with respect to Lorentz boosts in the z-direction and only motion in the transverse direction can have physical significance. These conditions restrict the stress-energy tensor to have the form T  =  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 easiest to work from the action for the string. The Nambu action for the string can be derived from first principles [13, 14, 15] and, heuristically, the argument is as follows [16]. One begins by observing that the action for a relativistic particle is just given by the integral over the world line of the particle. particle 5  =  _mfds  =  _mfdt  I  dx dx  (1.7)  Chapter 1. Cosmic Strings  8  where 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 free particle, 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 described  by Lorentz-invariant field equations that are independent in  ‘i-  and 1, a spacelike param  eter directed along the length of the string and, therefore, the string is invariant under Lorentz boosts parallel to the string. This means that only velocities perpendicular to the string have any physical meaning and so the action should only be a function of the perpendicular velocities,  and not of the full velocity  .  Using 1 to parameterize  the elemental length of the string so that rn=ifdl one gets that the action of the string is [13, 14, 15] Sstring  I  dldr  =  —ii]  =  —fdldr/1—  (1.8)  where -  a  a Ia oaN  and  a a  = 1 (1.9) = To put this action into the explicitly covariant form of the Nambu action, one repara metrizes using the timelike parameter  °  and the spacelike parameter  _f d 7 2  SNambu  ‘  [3].  (1.10)  where y is given by 7jk  Ox!’ Ox = g,v----  7 =  73.  (± x’) 2 .  —  (1.11) —  (1.12)  Chapter 1. Cosmic Strings  9  and where the dot and the prime refer to derivation with respect to The indices  j  and  respectively.  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[(±  —  ± x 2 ’]}  =  0  (1.13)  The stress-energy tensor is found by varying the action with respect to the four dimensional metric grn,. T(z)  =  fd2 _ 7 4)(z  —  X[X! 1 2 2±’  +  X!12X!h) 2 X  As a check on the Nambu action, one can choose  —  (th  °  •  =  x’)(bx”  t,  ‘  =  +  x’i)]  z, x  =  (1.14)  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 Wave  As the action is invariant under reparametrization of , it is possible to choose a simplifies (1.14). Again choosing  and y  =  ° =  t and  =  z hut this time allowing x  =  which f(z ± t)  g(z ± t), one gets the stress-energy tensor of a single travelling wave pulse on  the string [17]. For instance, taking  f  and g to be function of z  —  t, the stress-energy  tensor is: These equations of motion are found using the flat Minkowski spacetime metric. More general 2 equations of motion are given in section 2.1.  Chapter 1. Cosmic Strings  T’  =  —  —  10  +g’ 2 1+f’  —f’  —g’  +g’ 2 f’  —f’  0  0  —f’  —g’  0  0  —g’  +g’ 2 f’  —f’  g)  (1.15)  +g’ 2 —g’ —1+f’  Making the linearized gravity approximations, it is possible to integrate (1.5) using the retarded Green’s function 2 5((z—x()) ) 6(z° —x°) and obtain the perturbation metric of a travelling wave on the string [17].  =  where h  =  (—4G[t) ln(2 [(x 2  +g’ 2 f’  —f’,  —f’  1  0  —g’  0  1  +g’ 2 f’  —f’  h  —  2 f)  + (y  —g’ f’ +g’ 2 (1.16)  —  —g’  —g’ f’ +g’ 2  g) ] 2 ) and  2  is just an integration constant. 3  It 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) says that the energy depends on T*(k)T(k)  —  Ta(k)l2  (1.17)  This is equal to  f  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 see that the integrand in (1.18) vanishes. Thus there is no radiation given off by the single This solution is exactly the same as that obtained by using an advanced Green’s function instead of 3 a retarded one. The equality of the advanced and retarded solutions is another way of seeing that the there is no radiation given off by the single travelling wave; the radiation metric is just the difference between the advanced and retarded metrics.  Chapter 1. Cosmic Strings  11  travelling wave. This will serve as a useful check on the equations for the back-reaction force. Because there is no radiation emitted, there should be no back-reaction force for a single travelling wave. Unfortunately, it is not possible to use this coordinate choice to study two waves moving 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 of  right- and left-moving waves does not satisfy the equations. The two-wave solution is not a 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 different parametrization of the string trajectory. 1.4  Waves Travelling in Opposite Directions  It is always possible to choose the string parameters  in such a way as to satisfy the  following conditions [8]: and  +x’ 2 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 the string. It is still possible to choose the timelike parameter,  to be t, the coordinate  time. 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  12  As this form of the equations of motion is just the wave equation, it is easy to see that the parametrization specified by the conditions (1.19) does permit the superposition of waves travelling in both directions. The timelike parameter the coordinate time. The spacelike parameter  4°  =  r  =  t can be identified with  is usually denoted a and is proportional  to 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 the  length of the string. The general solution of (1.22) is a linear combination of left and right moving waves. (a,t) with the only constraint on  =  [(t  -  a) + (t + a)]  (1.23)  and b being (1.24)  With this parametrization, the stress-energy tensor becomes T(,t)  =  fda(1’  (a,t))  —  It is often useful to relabel the parameters in terms of the null coordinates u v  =  (1.25) =  and  With these variables, the stress-energy tensor (1.25) can be written as T(z)  )(z 4 + OvxOxv)(  =  —  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  1.5  13  The Emission of Radiation  It is useful to examine where the gravitational radiation is emitted from when two waves collide on a string. This will help to explain the general trend of the back-reaction forces produced. The waves which I chose to study in this thesis were two oppositely moving triangular pulses. This pulse shape was chosen in order to make the slope of the string piecewise constant. A piecewise constant slope makes it much easier to analytically evaluate the back-reaction force. As is shown in this section, this choice also means that the source of the gravitational radiation is restricted to specific points, those points at which oppositely moving kinks on the string collide. It was hoped that this later aspect of the waves would result in a simpler picture of the back-reaction forces and consequently a clearer understanding of what was happening throughout the collision. If the two waves which are colliding on the string are triangular pulses, then at some point during the collision, one of the waves will be travelling along a portion of the second wave which is flat but at an angle to the main part of the string. If the first wave is much smaller than the second, then the entire first wave could be on this sloped but straight section of string at once, figure (1.1). As the physical results of General Relativity are independent of the frame of reference chosen, it should be possible to rotate our frame of reference and view the first wave as just moving along a flat string. In this case, the metric perturbation would be that of a single travelling wave and there would be no radiation 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 by  Chapter 1. Cosmic Strings  14  Two Waves Colliding I  60  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  -  0  .—  4)  C)  ci)  40  -  -  20  -  0  I  —300  I  I  I  I  I  —200  I  I  —100 0 z—direction  I  100  I  I  200  Figure 1.1: A small left-moving wave on a cosmic string travelling along the fiat section of a larger right-moving triangular wave.  x°(u)  =  u  x’(u)  =  0  x Q 2 u)  =  /3(u + l)6(u + 1)  x(u)  =  -u-(- 1)(u+l)0(u+l) -(1 -)(u-l)0(u-l)  —  2/3u0(u) + /3(u  where /3 is a positive value between 0 and 1,  =  /1  —  —  /32,  l)O(u  —  1)  (1.29)  1 is some positive constant,  and 0(u) is the Heaviside function. For now, I consider the left-moving wave to be flat, that is x(v)  =  (v,0,0,v)  (1.30)  Consider a point at which —l <u <0. The metric perturbation is found using (1.5), the Green’s function for the wave equation, and the stress-energy tensor. v(z)  =  8GfT(x)((x_z)2)  (1.31)  Chapter 1. Cosmic Strings  15  Using the form of T’ given in (1.26), one obtains  -  h(z)  =  8Gtf dudvb((z —x(u,v)) ) 2  2  0/31—13  0  00  0  /3  00  /3  1—13 0 ,8  (1.32)  —2/3  If A is the Lorentz transformation between the frames of reference, then h(z) in the new reference frame is h’(z)  =  AhAT(A_lz). A general Lorentz transformation can be  expressed as the product of a boost and a rotation. It is expected that A will involve first rotating the string such that it lies along the z-axis and then boosting it so that its transverse velocity is zero. Let the Lorentz transformation which takes the sloped string into a straight flat string be 7  0—v’yO  10  0  0  0  1  0  01  0  0  0  0 0  cos()  sin()  1  0 0  0  —7)7 0 0 where and  0  0  —  (1.33)  sin() cos()  is the angle of the rotation, v is the velocity between the old and the new frames, =  v21• —  A boost transverse to the z-axis does not mix the z-component  of the tensor with any of the others, therefore, the rotation must cause the components of the tensor to vanish. This fixes the value of phi as  =  Because the perturbation metric of a straight string is diagonal with  —  and  23  arcsiri( and  -22  equal  to zero, the boost should cause the h 22 component to vanish. This fixes the value of the velocity at v  =  /3//2(1 + /3). The A’ operating on z has no effect because the  coordinates only appear in the argument of the delta function in the form of a Lorentz  Chapter 1. Cosmic Strings  16  invariant. Therefore, (i+,ã) 0 0 =  8Gf dudv((z —x(u,v)) ) 2  0  0  0 0  0  0 0 —(1+)  0  (1.34)  The only difference between this metric and the metric of a straight string is a constant factor 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  —*  cause no change in the integral  !3u +  (1.34). The Jacobian from the change of variables in the integration measure cancels the extra terms arising from the partial derivatives. However, when the integral over the delta function is performed a term of the form 2ja (x  —  z)I or 2IOx. (x  —  z)j  appears  in 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 a straight string, 9x•  (x  —  —2u + Ox• z  (1.35)  —u—$(u-i-l)+l-i-Ox.z  (1.36)  z)  =  whereas for the sloped case, Ox.(x—z)  =  A change in u and v does not affect affect the coordinates of the string trajectory, x, and so it is possible to make the determinants equal by the repararnetrization 1—  9x•z  2  )  (1.37)  Making this change in variables, causes ôx.(x—z)  ‘,  1 ( —2u+Ox.z)  (1.38)  Chapter 1. Cosmic Strings  17  This results in exactly the factor required to cancel out the difference in scale that arose in the Lorentz transformation. Thus, the metric for the sloped piece of string is exactly the 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. Doing this and following similar arguments results in a similar reparametrization of v. As the order of the integration should commute, it is actually necessary to reparametrize both of the variables u and v in both cases. Fortunately, this does not change the result as the denominator depends only on one of the two parameters (and the trajectory of the string, x, is independent of the parametrization). Throughout this argument, x(v) was always fiat, so the string trajectory is really just that of a single travelling wave. The metric in this case was solved for exactly by Vachispati [17] (1.16). One can also, therefore, demonstrate this same result in terms of his metric. Again, for simplicity, I assume that g  0 restricting the wave to the yz-plane.  =  The metric is then  2 f’  0  —f’ f’ 2  010  —f’ 2 f’  0 0  0  (1.39)  —f’ 1 f f 12  1  Using the Lorentz transformation (1.33), and following the same reasoning as before, one finds that g  =  tan’(f’) and v  =  —f’//1 +  . 2 f’  In this case, however, the coordinates  are not in the form of a Lorentz scalar and so do transform under A—’. =  xx  zz  t1+f’2+f’x  f’(t-z) =  =  t+z 2 f’ yl+f, + 2 fx  (1.40)  Chapter 1. Cosmic Strings  However, the expression f(t the origin correctly, f(t  —  18  —  z)  z) is just that of a straight, but sloped line, so by choosing  =  [t  —  z]f’. Therefore, under the inverse Lorentz transfor  mation, ln(2{(x  -  f(t  -  z))2  1n(2{(  }) 2 +y  -  2 f))  2})  {x 1n(2 + 2 } y )  (1.41)  This is exactly the factor for a straight fiat string. As a result, the metric transforms as 0000  h(z)  =  AhAT(A_lz)  =  {x —4Gln( + 2 } y )  0 1  0 0  (1.42)  0010 0000 As before, the metric of a straight but sloped string is the same as that of a straight flat string. Because the metric of a straight sloped string is the same as that of a straight flat string, one might expect that the metric from a wave travelling on a straight sloped string would 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 in the frame of reference of that sloped string appears as a single travelling wave, but one which 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 the right-moving pulse considered earlier.  Chapter 1. Cosmic Strings  19  x°(v)  =  v  x’(v)  =  0  x ( 2 v)  =  (v + 1 )8(v + 12) 2  x ( 3 v)  =  v+(—l ) 2 ) )9(v—1 —&)(v—1 ) O(v +( )(v+l 1 2 +l  —  2cvO(v) + c(v  where o is a positive value between 0 and 1, &  =  —  —  ,  )l 2 O(v  —  (1.43)  12)  12 is some positive constant.  Considering a point at which —l <u < 0 and —12 < v < 0, one finds:  v(z) =8GLfdudv((z_x)2)  2  0  a+3  ti—/3  0  0  0  0  /+a 0  0  —3 0 /3i—o3  (1.44)  —23ö  Applying the same type of Lorentz transformation as was used to go into the frame of reference of the sloped right-moving string, one finds that /i is proportional to 2  0  0  0 0  ii  0  (145)  0 0 /1_2_1  where  ij  0  i  _2/1_,2  is a parameter between 0 and 1. The factor in front of the matrix is cancelled in  the same manner as before by reparametrizing the string. This results in the metric for a left-moving wave on a fiat string, but one which is parametrized by parameter  i  can be implicitly expressed in terms of the v and  ‘ij  and not i. The  through the following  expression.  If—l< u <0 then  i>  2(1+3i—c6) (1+&—c49+1—) 2 <v <0, i a ifO <V <12, and a if—i  (146) i=a  =0 otherwise.  This results in a distorted image of the original left-moving wave, figure (1.2).  Chapter 1. Cosmic Strings  20  (a)  (b)  Figure 1.2: A wave’s appearance is distorted when viewed from the frame of reference of a sloped piece of string (b) as compared to how it would have appeared on a straight flat piece of string (a). However, as the same transformation may be applied to all sections of the left-moving wave (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 no radiation produced by the left-moving wave travelling along a straight but sloped section of string. Making  piece-wise  transformations  of  this  sort,  it is possible to see that the  only time at which it is not possible to view the metric produced by the string as that of a single travelling wave emitting no radiation occurs when the kinks of oppositely moving waves collide. Here there is no reference frame in which the component of the right-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 cosmic string must be produced when the kinks of two oppositely moving waves collide.  Chapter 2  The Formalism  Although determining the effects of back-reaction on a source emitting gravitational radiation is, in general, fairly difficult, in the case of cosmic strings it is much easier because one can use a perturbative approach to the problem. If the gravity waves being emitted by a source contain a large amount of energy, this energy will cause significant distortions to the background metric of spacetime. These distortions would affect the motion of the source generating the waves and alter the characteristics of the waves actually produced. However, this means that the gravity waves would have originally caused different distortions in spacetime, and so on. Because the answer affects the formulation of the problem, these types of problems are quite difficult to solve. Another difficulty with this problem is that the back-reaction ‘forces’ are entirely gauge dependent. It has been shown [19] that a cosmic string can always be viewed as being a geodesic two-surface. Although the internal metric can change, there are no invariant ‘forces’ on the string. In this view all of the effects of the radiation appear as distortions of the background metric; the string stays perfectly straight. This picture is not, however, a very useful way of thinking about the back-reaction. One would rather consider the string as moving in some kind of fixed almost flat background and have the string’s motion altered by the back-reaction. This is closer to what a person floating in space would observe as happening to the string. Gravitational waves are emitted from the string and the wave on the string is subsequently damped out. One way to fix the background and thereby force the back-reaction effects onto the string itself is to choose a 21  Chapter 2. The Formalism  22  specific gauge. Unfortunately, the choice of gauge is not an obvious one and so whichever one 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 disturbance of fiat spacetime caused by the emitted radiation is very small, G  10—6.  This  feature makes it possible to use the weak field approximations of linear gravity and a perturbative approach to the back-reaction problem. Choosing a specific gauge, one can initially assume that the effects of the changes to the background spacetime caused by the radiation do not significantly alter the motion of the string. That is, one assumes there are no back-reaction effects on the string in order to obtain a first approximation for the radiation produced. Once one has the radiation emitted, it is then possible to determine what its effects on the string would be. Taking into account this slightly revised motion of the string, the radiation produced can be modified. If the changes caused at each step are small, then this sort of iterative approach will converge on the real solution for the radiation and its corresponding effects on the string. This is the technique used by Quashnock and Spergel [20] in determining the effects of gravitational back-reaction on oscillating cosmic string loops. Their approach can also be made to apply to colliding travelling waves on long cosmic strings. In my case I will just be determining the first correction 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 some gauge freedom in the form of any gauge transformation satisfying D  =  0.  An analogous gauge freedom arises in the simpler case of Maxwell’s equations for electromagnetism. There, the Lorentz gauge choice permits the freedom Ac,  Ac, + Oc,X  where Ac, is the four vector potential of the fields and X is a scalar satisfying DX The gauge can be completely fixed by specifying a condition on Ac, such as A 0  =  0. 0 [21].  Similarly, with gravitational radiation, one often fixes the gauge by choosing to work  Chapter 2. The Formalism  23  in the transverse traceless or TT gauge. With this choice the freedom of the Lorentz gauge is removed by forcing the metric to obey the following conditions [12]. h=0  h 0 3 =0  Unfortunately, only pure gravitational waves can be put into the TT gauge. Because I am using the metric generated by the string, it is not a sourceless metric and cannot be put into the TT gauge. The best way in which to restrict the Lorentz gauge for this problem is not imme diately obvious. Therefore, I decided to explore the force equations using the Lorentz gauge with its associated freedom and see what restrictions were required in order for the force to satisfy some reasonable conditions (sections 2.1 to 2.3). It was hoped that requiring some reasonable properties of the force would sufficiently restrict the gauge such that any remaining freedoms would not influence the back-reaction effects on the string. This approach was partially successful. In section 2.2, it becomes necessary to restrict the metric to Vachaspati’s gauge. This reduces the gauge freedom down to a single parameter, Vachaspati’s integration constant Q . Unfortunately, although the influence 2 of this remaining freedom cancels out to some extent, (section 4.3 and Appendix C), it continues 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 selecting a value for the integration constant  2,  section 3.3, or by fixing the gauge from the  beginning using conditions similar to those which restrict gravitational wave metrics to the TT gauge. For example, one might try imposing just the 0 h 3 later approach was not investigated in this thesis.  =  0 condition. This  Chapter 2. The Formalism  2.1  24  The Force Equation  The first step I take in determining the effects of the emission of radiation from cosmic strings is to develop the equations of motion in terms of a metric more general than the fiat Minkowski metric which was used in section 1.2. Denoting the metric as g, this is done by minimizing the Nambu action with respect to the coordinate x. 7 _zfddr  S =  —  f  (2.1)  —  With the standard co-ordinate choice of OrXOyX  =  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) become öxOx = 0  = 0  (2.4)  and the equation of motion is (2.5)  vZ 0 u  In the case of the back-reaction problem for the cosmic string, the string is itself the source of the perturbation metric h. Therefore, using (1.1) and making the weak field approximation in determining the Christoffel symbol, one can write (2.5) as 8uOvZ  , + 7 7 (ha h , a  —  (2.6)  Chapter 2. The Formalism  25  where  f  dudv Fa ((x  =  8G  =  Ox&x + OvxOux 3  8  —  ) 6(z° 2 z)  —  x°)  (2.7)  and F  —  (2.8)  Equation (2.7) is found by starting with equation (1.5). That is, I have chosen to oper ate 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 pertur bation 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 find the effect of this force on the string, one integrates with respect to either u or v. For example, one finds the change in z, L(Oz), due to the metric perturbation h by integrating with respect to the variable v. =  fdvOãvz  (2.9)  To find the first order change in the trajectory, one starts with a solution to the equations of motion in flat Minkowski spacetime so that conditions (2.4) are satisfied with respect to  ij,,.  A reasonable requirement of the force is that its effects should  not result in a perturbed trajectory which violates this initial gauge condition. The force given by equation (2.6) meets this requirement. The change, (t9z’), can be safely added to  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 later pointed out to me, a better requirement of the force is that the perturbed trajectory satisfy not oniy the gauge conditions in flat spacetime, but also those in the perturbed spacetime. The force (2.6) does not satisfy this condition because azz”h, 0. This is discussed further in section 2.2 and in Chapter 5.  Chapter 2. The Formalism  26  =  =  9 u Z OvZ  7 _0uZ7OuZfd , Ovz Vha  =  Because h  Y7 dv Uz , f1 1 ha ,  —  (2.10)  h(z), the integrand in (2.10) is just a total derivative of h with respect to  v and the integral can be expressed as:  J  dvk , 7 Oz  =  Because of the causality requirement, h 7  h I 7 t  JdVha , 7 v =  (2.11)  0 at the lower bound of v. One needs to then  consider the upper limit. For clarity, I shall use ü and 3 to indicate the null coordinates of x in order to distinguish them from u and v, the null coordinates of z. =  7 8Gt1imJdüdP b ((x(ü,3) a  =  8Glimfdu  But, in the limit as v goes to infinity,  F  2IOx. (x  —  IOx. (x z)I —  z(u,v)) ) 2 8(z°  —  x°)  O(z°—x°)  z)I =  —  I(x  Oz)v + fQü, €,  (2.12) also goes  to infinity. Therefore, the integrand goes to zero and the entire integral in (2.12) must be zero. This means that ãz• (ãz + zaz)  =  0. Thus the change in the motion of the  string satisfies the co-ordinate conditions (2.4) for the string in flat spacetime. It should be noted in passing that by symmetry h 7 , 1 ,  °°  =  0 also. This fact will be  useful later on. 2.2  The Back-Reaction on a Single Wave  As 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 that the single wave experience no back-reaction force. All trajectories, z, of the string must  Chapter 2. The Formalism  27  satisfy the wave equation and so can always be written as the composition of left and right moving components, z(v) and z(u) respectively. A single right-moving travelling wave can be created by requiring that z(v)  =  az v where z is some constant four-vector  and by letting z(u) be an arbitrary function with the restriction that Oz and z both satisfy 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-reaction force and the point on the string which is acted upon by the back-reaction force. That is 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 to  f  7 + OxOx  0  —  [Umax  0z° J— 1 2 =  d  (0 0 7 x + 0jx.0x 2(0x. (x z))  —  ) O(z° 2 z)  0x)  —  —  x°)  0 za  —  JUmax  dü —  Ox) 6((x  —  13 i x 0x 7 + 0x (0x.(x—z)) —  .  ãx 0z  0  (2.13)  because Xcx0vZa 5 0  The boundary  Umax  =  avav  is the solution of (x(i, 3)  —  2 z)  =  =  o  0 for ii in the limit as i3 goes to  negative 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 single right-moving wave. 2 The result in equation (2.13) assumes that there is no contribution from  acting  on the boundaries of the integration. To see that this is the case, one inserts a past-time The cancellation in (2.13) seems to depend on doing the i integration first. This is not actually the 2 case; the order of integration does commute. This can be seen most easily with the specific example of the colliding triangular pulses as will be shown later in section 2.4.  Chapter 2. The Formalism  28  cut-off, O(ü + i3 + r 3 Taking ), into the integral. 0  0 T  to +oo at the end of the calculation  will then remove the cut-off. This technique changes the bounds of the integration. The new integration limits are found by solving ü  =  — —  where t  =  Assuming  u + v and b 0 T  =  —r — 0 i(ü,z) —2tu—2uaf.(b—F)—(b—F) t 2 —T 2[Ox.Oxü+t+O,f.(b—)] x(ü, i3)  —  —  (214)  O,xi3. This expression is just a quadratic in ii.  is very large yields the following implicit values for ii U  —  —  U —  —  Djx.(b—z)  +  , 4 —T  aX  .  2(Ox.z—Of.b) (Dx.Ox) 0 —r Umax  —  0 8x —2r  where p  =  [X(Umax)  =  —  z]? +  [X(Umax)  [X(Umax)  (2.16)  —  and the subscripts 1 and 2 refer to the components of the four-vector (x As  m  —  z)  becomes large and negative, at the lower boundary, O 1’ —r + 0 Ozc \  OXOX  However, the integrand involves a term 6x (x lirn Ox• (x  —  z)  =  j  =  j  —  bounded terms  (2.17)  z) and  urn Ox• (&uxü + b  —  z)  (2.18)  =  Therefore, the lower boundary will make no contribution in the limit as  0 T  goes to infinity.  To examine the contribution from the upper boundary one approximates Ox 7 as OUXIU=Um  This approximation becomes exact in the limit as  0 T  goes to infinity and Ii  A problem with using a past-time cut-off is that it destroys the Lorentz gauge condition for any 3 finite choice of r . However, the gauge is restored if r 0 0 is taken to infinity.  Chapter 2. The Formalism  goes to  Umax.  29  Therefore, at the upper boundary, Of  hm UUmax  OZ’  fJ  Umax  0 OX 2T  —  Because the derivative is proportional to contains the contraction  OjX  )  =  9Xc  x Ox 3 O  (2.19)  the contribution from this boundary term  zOxa which vanishes. Therefore, the upper boundary term  also gives no contribution to the integral. The second term in (2.6) has the form  Using (2.7), one can express  as =  —  ) 2 z)  (2.20)  The 1 O , z term can be brought inside the integral and so the integrand is proportional to Fai O 3 vZ  = =  (OxOx +  =  O2 O 8 x.Ox  =  0  —  —  (2.21)  ,Oz1 7 h O vza is also zero. The third term in (2.6) The integrand is zero and, therefore, 3 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 single  wave as well. Because the back-reaction force on the single travelling wave is zero, one expects that the solutions for the trajectories of single travelling waves should satisfy not only the coordinate condition in flat space, =  1]IwOvZ1IOvZi’ =  0  (2.22)  but also the coordinate condition in curved spacetime. That is, one expects that for a single travelling wave =  =  0  (2.23)  Chapter 2. The Formalism  30  Considering once again the example of the right-moving wave, the first of the two conditions in (2.23) is easily seen to be true. The metric h, 2 expressed as an integral involves the factor F. Pulling the Oz’’ term inside the integral, one gets the contraction 1 FOz’-”. As this factor equates to zero, one obtains the desired result: kpvãvz1Lövzi  =  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 the single travelling wave which is calculated using Einstein’s equations (1.5) and a retarded Green’s function. fUm  h(z)  =  / J—oo  8Gt  dü  F 2Ox. (x  Equation (2.24) has some problems with infinities. At ii while at ü  = umax,  q  =  (2.24)  z)  —  =  i(it,z) —,  Ox. (x  z)  —  =  q,  =  cc  0. As before, one resolves these problems by inserting a past  time cut-off, 8(ü + i3 + r ). Installing this cut-off changes the bounds of the integration 0 over I. The new bounds are given by (2.15). With hindsight, it helps if the integral is also split in two at the point Po  —  U  =  Umax  0 Ox —2r  X 1 Of  (2.25) UUm  where p 0 is some arbitrary constant. Therefore, h(z)  =  4G[L  Ii  [Umaz_ ° 2 a  du  0 J—T  +  +  3. (x x  JUmx__2ro z 4 .aa Um  —  z)  dü .  —2ro  (x  —  z)  (2.26)  j  I now consider each of the two integrals separately. To evaluate the lower integral, one makes the approximation that Oj,x’ Oj5x  OüX(Umax  —  4GIf_20 —2i&z  =  OXjUm.  Rewriting ãx  (x  —  z) as  ), the first integral in equation (2.26) becomes d(Umax)  = OüX(Umax  —  U)  t 1 4G  Ox  ln() (2.27) UUm  Chapter 2. The Formalism  31  Associating 1/pr, with Vachaspati’s integration constant  22,  (2.27) can be shown to  be exactly Vachaspati’s solution for the metric of a right-moving single travelling wave (Appendix A). = 4Gi  ln(2p) Ox  (2.28) U—Um,  For a single travelling wave, if z is a point on the string,  Umax =  u. The equation for  the maximum value of ii is z U  where b = x(u)  —  =  ‘Umax  =  b  .  —  (2.29)  a•  iiOx. But, for the back-reaction on a single left-moving wave, z is a  point on the string and so Ox =  Therefore, (2.29) can be rewritten as  Oz.  az.z(u)—az.b  —  Umax  —  avz. oux  A solution to this equation is that  = 9z which implies that  OXUmx  Umax  =  u. To  see this, rewrite (2.30) as OvZ  X(U)Iü=u  (2.31)  = 9z• z(U)  Expanding xü) about i = u in a Taylor series gives avz  öXI(Uma  —  The solution to this equation is  u) +  OXIu(Umax  Umax =  —  2 +... = 0 u)  (2.32)  u. The Taylor series expansion of x(u) may  not, however, be justified in the case of a piecewise continuous function such as (1.30). Nevertheless, U =  —l  —  E  Umax = U 1S  with  E  >  the only solution in this case also. Suppose, for example, that  0 and ii> —l so that Ox -  ‘Umax —  —  (-1  -  1)U  -  8z. Then  (_l(/fZ  /32 -  1))  —1— i _l(1+\/1_/32)_2  -  (i+i_/32)  = -l -  /32)  <  -l  (2.33)  Chapter 2. The Formalism  32  Other cases also yield contradictions leading one to the conclusion that  UmaT =  u for the  single travelling wave. Because  UmaT =  u, one can rewrite (2.28) as =  4G (Juz  From this it is easy to see that ’ 6zh9z 1 h  =  ln(2 2 p)  (2.34)  0 because FDz  =  0. Therefore, this  section of the integral, (2.27), which is equivalent to Vachaspati’s metric for a rightmoving wave, satisfies the second of the conditions in (2.23). Using this form of the metric, it is easy to see that Vachaspati’s metric also satisfies the first condition in (2.23). F 2 a 1 z’ equals zero which implies that 1 4 haz’ O z” also equals zero. I now return to the second integral in equation (2.26).  It is this integral which  causes problems for if one actually evaluates the integral, one finds that  is  non-zero. However, this integral just corresponds to a gauge transformation within the Lorentz gauge. Because the integral is symmetric with respect to the cr and  indices,  one can associate it with a gauge transformation of the form Clmaz  = 2Gj I  i_To  Taking the divergence of  -  2  °  8  +  8-  dü  P  a3  ãx (x .  —  z)  (2.35)  one finds that it is non-zero for any finite value of r 0 but  zero in the limit as r 0 goes to infinity. The non-zero nature of D at finite values of r 0 is because the past-time cut-off destroys the Lorentz gauge condition. Taking r 0 to infinity removes the cut-off and consequently restores the gauge. One can show this directly by taking the divergence of (2.35). As has been shown previously, a derivative acting on the upper boundary gives no contribution 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 back reaction force on a single wave be zero.  Chapter 2. The Formalism  33  the term _frc3  O x 3 . (x  —  z) Ox  x  2 36 UUmz  0 goes to infinity. However, the contraction F°Ox is zero and so the in the limit as r lower boundary also makes no contribution to the integral. Acting on the integrand, the derivative gives ö 2Gp— OZ  fUma —  0 Oz,ôj —2r  i  J—,-  du  (x  +  —  z) (Urn  =2GL1 i—To + =  Therefore, J°  =  0  2r88z  du  (Ox (x  —  2 z))  (2.37)  0 and the gauge transformation given by (2.35) is a transformation  within the Lorentz gauge. The gauge freedom within the Lorentz gauge has a significant impact on how reason able the gauge choice is. While Vachaspati’s gauge satisfies the conditions (2.23), the full integral expression of the metric does not. One implication of this is that the metric chosen for investigating the collision of two waves should reduce to Vachaspati’s gauge choice at least in those regions before the waves collide in which the two waves may be considered as just two single travelling waves on a string. Using the integral expression for the metric does not satisfy this criterion and, as shall be seen in chapter three, leads to 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 metric discussed in chapter three. Unfortunately, this is not the whole solution to the problem. Another difficulty arises which is related to satisfying ôuzuzvhv  =  0 in the case of two colliding waves. Towards  the end of working on this thesis, it was discovered that the force contains a component  Chapter 2. The Formalism  34  parallel to the string. This parallel component causes a shift in the internal null coordi nates of the string which causes the perturbed solution of the string to no longer satisfy  o.  the null coordinate conditions (2.4). That is  This means that what I have called the force, equation (2.6) is not what one would like to define as the physically reasonable force. Because only motion perpendicular to the string 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 be more 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 Lorentz gauge 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 depen dence expressed by the arbitrary integration constant,  22,  in Vachaspati’s single wave  metric. To see where and how this gauge freedom influences the back-reaction and also to get a rough idea of the effects of the gravitational back-reaction on the collision of two 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 corrected perpendicular force, equation (5.13).  2.3  The Energy Lost by the String  Another reasonable requirement of the gauge choice is that the change in energy of the string produced by the back-reaction forces should equal the energy radiated away from the string in the form of gravitational radiation. To find the change in the energy of the string, one first determines the derivative of the stress-energy tensor,  Chapter 2. The Formalism  T,(y)  =  p  f  f f J  du dv 4 6 ) (y  du dv  = —  35  {  —  + 2p  z(u, v)) {Ouz3vzv + auzvOvzP}  [ö()(y  )( dv 4  =  —  z(u, v))] Oz +  —  —  z(u, v))]  U=+oo —  z)  du dv 6 (y 4  V=+oo —  —  )(y 4 p fdu 6  —  z)  z) Oãz  (2.38)  For finite values of y the boundary terms disappear and one is left with 5 T”’,(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 outside of some finite spatial region. This assumption is not satisfied in my case because the cosmic string is infinite in the y - or z-direction. However, I am only interested in the 3 amount by which the back-reaction force changes the momentum of the string. Between any two instants in time, the back-reaction force is confined to a finite length of the string 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-reaction forces 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 the four-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 but will only involve an integral over a finite region, a sphere of radius L. This allows one to The parallel component of the force contributes nothing to 5 as discussed in section 5.3 so this energy equivalence will hold for both the full force (2.6) and the perpendicular force (5.13).  Chapter 2. The Formalism  36  use equation (2.40) to determine the change in the momentum of the string due to the back-reaction forces. The change in momentum between times t 1 and t 2 is dy°P,o =  dy°  f  dy°  (T, 3 jd  = £2  =  0 T 3 d °,  (2.41)  —  where i is an integer ranging from 1 to 3 and V is a spherical volume of radius L. One can also consider the bounds on the integration over dy 1 and dy 2 to be —L to +L because the string is finite in those directions and so the stress-energy tensor is zero beyond this range. The second integral in equation (2.41) can be rewritten as two terms representing the stress-energy tensor containing the back-reaction forces, TB and the tensor of the straight flat string, TF’”. These integrals are then simplified using Gauss’ theorem. £2  dy°  f  3 d  dy° = = =  ff2  f  d ( 3 TB,  —  TF,)  dyof dy° fd2S n (T 8  —  ) t TF  (2.42)  where nt is the normal to the surface S which is a sphere of radius L. On this surface the stress-energy tensor of the string with two waves is the same as that of the straight string and so the second term in equation (2.41) is zero. Taking L to infinity does not change this result. The first integral in equation (2.41) can be evaluated using equation (2.39).  Chapter 2. The Formalism  ft2  J  37  ft2  f  dy°  J  d T 3 ’,  =  ]  =  2t  f  dy°] 3 d ( TB,  f  dudv  —  f2 L dy°  TF’”,U) [O3zö( ) 4 (y  d i 3 l(  —  _[ôaz5( ) 4 (y —  z)IB (2.43)  z)]F)  but, 8,Uz” = 0 for the straight string, and so equation (2.43) equals 2tfdudv [OOVz]BO(t 2  —  y°)O(y°  —  t’)e(L  —  )O(y + L) 3 y  (2.44)  Taking the limit as L goes to infinity removes two of the Heaviside functions in equation (2.44) and taking t 1 to positive and negative infinity respectively removes the 2 and t other two. Dropping the subscript B, one finds that the change in the four-momentum due to the back-reaction forces on the string is given by zP = 2iifdudvOuavz(u,v)  (2.45)  The change in the energy of the string is then 2iLtfdudvauavzo = —  J  dudv OzOz  +7 h , a  —  ) 7 h,  (2.46)  Because the first two terms in this integral are just total derivatives, they equate to zero when 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 for mula for the total gravitational wave energy{18, 201. Unfortunately, there are difficulties associated with applying this formula to my problem. Weinberg’s formula assumes that  Chapter 2. The Formalism  38  the metric, 1 jg v, may be written as spacetime and  ij  + h, where  ij,  is the metric for flat Minkowski  is a metric which goes to zero as the distance from the source of the  gravitational radiation goes to infinity. However, the spacetime of a cosmic string con tains a conical deficit and so when the metric for the string is split into  ij  +  the  h 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 small and 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 and  one might hope that the error associated with using Weinberg’s formula would not be significant. In the limit where  t  goes to zero, this error vanishes but in the limit where  becomes very large the error becomes more and more significant. However, for large values of , the entire approach followed in this thesis becomes invalid. Neither the weak gravity limit nor the perturbation approach to finding the corrections to the trajectory of the string are applicable any longer.  Fortunately, the only situation which I am  investigating in this thesis is the one in which  is very small.  The reason for which Weinberg assumes that the perturbation metric h,LV goes to zero as the distance from the source goes to infinity is that he requires convergence of the integral for the total energy and momentum of the source,  fl’°”. 3 fd  One way in which  to 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 h 1 for my problem does not approach  zero at spatial infinity, the source of the radiation is still finite. The radiation is only emitted during the collision of the two oppositely moving wave pulses. As both wave pulses 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 wave pulses (Chapter 3) is a finite metric that does go to zero at large distances. This is the only component of the metric for the two colliding wave pulses that emits radiation [5].  Chapter 2. The Formalism  39  The radiating portion of the metric does satisfy Weinberg’s criterion. Unfortunately, this is not a solution to the problem. Weinberg’s formula involves the stress-energy tensor, or the metric, in the form of a quadratic. The full metric for the collision of the two waves is a sum of the residual metric plus other non-radiating metric components 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 with the residual metric. If the cross terms make any contribution to the radiated energy of order h 2 then Weinberg’s formula cannot be used unless the integral over all space of the stress-energy tensors for the other metric components also converge. One might suspect that the cross terms to not contribute to the energy equation because the radiation emitted from the colliding waves is polarized whereas the straight string component of the metric, for instance, is cylindrically symmetric. However, I was unable to show that the cross terms do not contribute significantly nor that the integrals of their stress-energy tensors 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 =  2Gj w 2  —  is  IT(,w)2] dw  where T1w(k) is the Fourier transform of the stress-energy tensor and the  (2.48) *  indicates  complex 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. Using fw2dwd  =  fd4k6(k2)k0c(k0)  (2.50)  one finds after some algebra that (2.48) gives =  4  J  dudvOuzAOvzv  f  düd fr  J  )c(kO)kOe_x) 2 k 6(k 4 d  (2.51)  Chapter 2. The Formalism  40  Itzykson and Zuber [22] give the radiation Green’s function D(z-x) and its Fourier de composition as D(z—x)  Because E(x)  28(x)  —  =  _€(z0_x0)((x_z)2)  =  (2)3  fd4ke_(k2)f(k0)  (2.52)  1, and as antisymmetry makes the —1 component vanish, (2.51)  can be rewritten as =  8G[L2fdudvfdd  0 aUzaoVzaO[ o(z  —  x°)((z  —  x) ) 2 ]  (2.53)  Finally, by comparing equations (2.47) and (2.53), one can see that the energy emitted by the string according to Weinberg’s equation is just the energy lost by the string due to the back-reaction force. There is no gravitational wave energy unaccounted for and so all of the effects of the back-reaction are occurring on the string and not in the background metric. The equations for the back-reaction force in the Lorentz gauge do seem to satisfy some reasonable requirements. First, the net effect of the force is a change in the string trajectory which maintains the coordinate conditions in flat spacetime. Secondly, the single travelling wave experiences no back-reaction force. Lastly, the energy lost by the string is equal to that emitted as gravitational radiation. For these reasons, the Lorentz gauge seems to be a reasonable gauge choice in which to investigate the back-reaction force. Unfortunately, despite the aforementioned attributes, the Lorentz gauge choice also has some unreasonable features. Chief among these is the fact that this choice of gauge leads to equations for the force which contain an arbitrary parameter, section 3.3. This parameter has a significant influence on the effect of the back-reaction forces. Another unsatisfactory feature is that the single travelling wave components of the composite  Chapter 2. The Formalism  41  metric (section 3.2) make a contribution to the back-reaction force. This is somewhat disturbing because the single travelling waves do not emit any gravitational radiation and so one does not expect them to contribute to the force. There are also problems with the perturbed string trajectory not satisfying the coordinate conditions and the force 6 Therefore, although the Lorentz gauge choice not being perpendicular to the string. does seem very reasonable in some regards, it also retains a few flaws indicating that perhaps a better gauge choice is still possible. However, what this gauge might be is far from evident and so I shall continue to investigate the back-reaction force in the Lorentz gauge.  2.4  Evaluating the Force for Triangular Waves  As 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 a useful illustration of the problems involved with the gauge choice and the freedom of the Lorentz gauge itself. Additionally, the equations developed here are required for the composite metric. The composite metric which is formulated in chapter three is a metric which uses Vachaspati’s gauge for those parts of the metric which behave like single travelling waves. It solves many but not all of the problems one encounters with the full integral expression for the metric. My goal then is to evaluate the back-reaction force for the collision of two travelling waves on a cosmic string. Although for complicated waves this could be difficult, for the simplified situation which I chose, the case of two triangular pulses, it becomes straightforward. Because derivatives of the wave are piecewise constant, one can divide This problem can actually be corrected by removing the parallel components of the force as is shown 6 in Chapter 5.  Chapter 2. The Formalism  42  up the integration in the  terms into regions over which F is a constant. This  permits the analytical evaluation of the force. The force from (2.6) will involve three terms of the type  f  =  7 8G[L0  =  4Gf f 1 d  düdF((x  )z) 2 O(z°  —  —  x°)  (2.54) (2.55)  -  -  Ure  where  Üret  refers to the retarded solution of (x(il, )  —  2 z)  u(v,z)  =  0. Because Ox• Ox  =  0, the  determinant in the integrand does not involve u. As the slopes Ox and Ox are constants over the integration region, the integral can be evaluated. =v1  = 4GO 7  r  Ox ax  ln(Ox. (x  —  z))  -  -  .  Uret  (O-x Ox)(Ojx (x .  ><  .  {  —  VV2  z))  +  •  —  Ure  3x U,.e  }  (2.56)  where av 0z7  0  if v is a constant (2.57)  = 7 (a—z)  otherwise V.  This evaluation was done by finding the integral over u first. It is also possible to de termine the force by integrating over v first. Although superficially the formulae are different, the two approaches give the same value for the integral unless the first inte gration 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 a single value of i in the equation (x(ñ, i3)  —  2 z)  =  0 results in multiple solutions for il or  for 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  —  2 z)  =  0  results in some degeneracies. An example of a path with degeneracies is given in figure  43  Chapter 2. The Formalism  Non—degenerate —p1ot  Degenerate ii’—p1ot  100  100  50  50 U)  U)  0  -  —50  —50  —100  —100  —40—20 0  20  —40—20 0  40  20  40  ‘v—values  c’—vaiues (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 a point not on the string. —  (2.la). The z-values in figure (2.la) correspond to a choice of u  =  80 and v  =  15 which  it will also be noted are the coordinates of the corner point. The values of i1 are seen to be degenerate over a range of ü 0 <  =  0 to ü  =  100. The degeneracies in fi occur where  < 20. Although the curve also appears flat in the region where i3 < —20, this is  just a feature of the scale of the plot. If enlarged, this region would have a slight slope to it. Similarly for the ii < —100, the curve is very steep but not quite vertical. The degeneracies occur over ranges of ü and i values for which 8,z is therefore best to integrate first over i3 and then over reverse the order of integration at the point (ii, i)  =  =  ôx and Oz  for values of ii  =  Ox. It  u and then  (u, v). Figure (2.lb) shows that for  a value of z off of the string, there are no degenerate regions and so the integrals would commute. By the same reasoning as in section 2.2, there is no contribution to the force from  Chapter 2. The Formalism  44  the boundary terms of equation (2.55). Once more installing a past-time cut-off, the integration bounds become Ojx.(b—z) V  —  + 0 —T —  V  =  Vmax  (2.58)  —2r O 0 x  The derivative acting on the lower bound, near  =  0 results in terms which are —r  bounded. However, the integrand involves the term Ojx. (x  —  z) in the denominator. As  0 goes to infinity, Ox (x — z) goes to infinity and the lower boundary term goes to zero. r At the other boundary, for large V 7 0  , 0 T  =  (2.59)  O,X  This term exactly cancels the other Ox 7 term in equation (2.56) making the contribution to 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 the force on the trajectory Oz’- of the string whereas integrating over u will yield the change in ôvzIA. The force as determined by equation (2.56) for the collision of two triangular waves is discussed in Chapter 3. Before examining the force on two colliding waves, I return briefly to the force on a single travelling wave. For the specific case of a triangular moving pulse, one can see explicitly 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 places where the integral has to be broken up to insure that the slopes Ox and Ox are constant throughout the integration region, only occur at kinks in €. If one integrates first over i,  Chapter 2. The Formalism  45  then the integrand will include terms proportional to F If Oz  =  [(x_z)7Ox.Ox  Ox contracts on either the a or the  (2.60)  index, then as has been shown before  (2.21), the integrand would equate to zero. If the contraction occurs on the y index, then the integrand takes the form =  0  (2.61)  If the integration is done over the ü index first, then the integrand is simply of the form 7 and the contraction of Oz on any of the three indices gives zero due to the Föx coordinate conditions (2.4). The force on a single travelling wave is zero regardless of the order of integration.  2.5  Other Features of the Force Equations  Another feature of the force equations is that when ü < u then  >  v and vice-versa.  First define q and q as  q  =  (x  =  Oj,x. (x  —  —  z)  (2.62)  z)  (2.63)  When solving the argument of the delta function, (x  —  , for ü or i one finds that the 2 z)  causality condition forces the solution to be such that  =  =  z°—x°+ã.(f—F)  (  —  As Oj has a magnitude of one, it follows that than or equal to zero.  ( )  i) +  (2.64)  —  q,  and similarly  qv,  are always greater  Chapter 2. The Formalism  46  The 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 =  Ov  (2.66)  -  q  As q and q are both always positive, this means that if is increasing then i is decreasing and 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 constant slopes. Assuming x 0  =  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. — —  —()+bn—) 2 (z°—) 2(z°—+O•(f(i)+bji—)  p268  Equation (2.68) actually gives four possible solutions for ü based on the four different possible ranges over which ôux and b are constant. However, only at most one of these values is consistent with the Ox,  pair used to find it. This permits equation (2.68) to  be used for a unique determination of ii. Taking the limit of large negative (actually for  in (2.68), yields the following equation for ü  umax):  u  =  —z°+z — 3 14 3 —i+ajx  =  Ox•Ox  (2.69)  Equation (2.69) can then be checked for all possible ranges of u and v and it can be confirmed that l  u. For example, consider the case where —12 <v <0 and u < —1. If  47  Chapter 2. The Formalism  fl is to be less than u, then  < —l  —  =  also. Therefore, equation (2.69) becomes —2u—v+äv+(—1)l 2 —2 1 )(1—c) 2 u+(v+l  >u  with the equality only occurring where o  =  0  (&  =  1). Checking all other cases, one finds  that this holds true regardless of the value of v or u. By similar arguments, one finds that if i  =  v  —  decreasing while i <  U  =  where  is small but positive, then €i  —  u <  is increasing, it follows that i < v  .  As ii is monotonically  ii > u and by symmetry,  3 > v.  In the special case that i3  =  v then one finds that Ox’  =  From equation (2.65),  one has u—u  =  (u)—F(u)I  (2.70)  This implies that there can be no kinks in the string between  (u)  and 1(u). Therefore  x(u) is at a point which has the same slope as z(u) or is at the kink which adjoins the section of string including z(u). For the instant where 3  =  v, the value of ii is not uniquely  determined by (2.65). The value for ii is degenerate as was seen in figure (2.la). 7 It is worth noting, however, that if Ojx)L  =  Oz’, then the contribution to the force is zero  because 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 force on 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 earlier time. Although this is true for all fiat pieces of the string, it may not be true at the kinks themselves. At those points, Oz and ôz are not well defined and the use of equation Because of the symmetry of the equations, there are also degeneracies in 3 when 7  =  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 evident when I look at the force generated by two colliding waves in greater detail in Chapter 4.  Chapter 3  The Composite Metric  In the previous chapter, it was seen that the Lorentz gauge choice leads to a force which is reasonable in several aspects but also contains some disturbing features.  Directly  applying the equations developed by Quashnock and Spergel [20], one obtains the force equation (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 necessary  to make a gauge transformation into the Vachaspati gauge. This suggests that the full integral expression for the metric (2.7) is not the best gauge choice. However, ignoring this problem and blindly evaluating this force anyways provides a useful illustration of the dependence of the force on the chosen gauge. Evaluating (2.6) for two colliding triangular pulses results in back-reaction forces on the string which contain non-integrable infinities. This problem is corrected by constructing a composite metric which is explicitly in Vachaspati’s gauge for those times when the two waves on the string may treated as two completely separate single travelling waves. 3.1  Failures of the Direct Approach  Using equations (2.6) and (2.56) one can investigate the effect of the back-reaction force on two colliding travelling waves. To attempt this, I used two triangular pulses which were for simplicity restricted to lie in the yz-plane. The pulses had the form of (1.30) for 49  Chapter 3. The Composite Metric  50  Travelling Wave Pulses Before Collision 60 0 .4-)  C.’)  40  0  20 0 —400  —200  0  z position Figure 3.1: The two travelling wave pulses at time t  200  —  the right-moving wave with /3 with c  =  0.5 and 12  =  =  1/ and 1  =  -150.  100 and (1.44) for the left-moving wave  20. This simplification means that all forces in the x-direction are  zero. 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, an image of the waves before they collide is shown in figure (3.1). At time t  =  —120, the  two wave pulses collide and figure (3.2) shows the force on the string in the y-direction shortly after the collision at time t  =  —115. Also shown beside the graph of the force is  the position of the waves at the same time. The wave image shows not only the collision of the two wave pulses (as a solid line) but also the location of the single wave pulses had they not collided (shown as dotted and dashed lines). The two force spikes seen in figure (3.2) are caused by the kinks at ü force near z  =  85) and i  =  —20 (the force near z  =  =  —100 (the  75). These forces seem to propagate  along the string keeping pace with the non-colliding wave pulses. The forces are due to the change in the derivative of the perturbation metric at the kink acting on the same wave as it moves along a piece of the second wave having a different slope. For example  Chapter 3. The Composite Metric  the force at position z metric at the i3  =  =  51  75 is due to the effect of the change in the derivative of the  —20 kink of the left-moving travelling wave as it travelled along the  flat portion of the string acting on the left-moving wave as it moves up the front face of the right-moving triangular pulse. Unfortunately, neither of the force spikes shown in figure (3.2) are bounded and if one 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 which can arise from an inappropriate gauge choice. Closer examination of the infinite spikes in figure (3.2) indicates a route by which one can get around this problem. The spikes are generated by the derivatives of the metric of the string at times before the two waves have collided. At these times, the metric of the string should be identical to that of a single travelling wave because the two waves are not causally connected. The metric for a single travelling wave was solved for by Vachispati [17] and also obeys the Lorentz gauge condition. In addition, the infinities which arise in the force expression are found to be of the same form as those which arise when solving (1.5) for the metric of the single wave (section 2.2). Both of these points indicate that, as was suspected, one should be using Vachispati’s solution for the metric in 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 metric was that in order to evaluate it, it was necessary to use a past-time cut-off. This also presents difficulties because for all finite values of , 12 this cut-off causes a violation of the r Lorentz gauge condition. The full integral expression is still required for those times after the waves start colliding and the Vachaspati solution can no longer completely describe the string. Therefore, it is also necessary to find a way to remove this cut-off dependence in the metric and satisfy the Lorentz gauge for finite values of r . 0 A metric which satisfies these conditions is the composite metric developed by Unruh  Chapter 3. The Composite Metric  52  Infinite Force Spikes at t 10  I  I  I  I  I  I  -  =  I  —115 I  I  I  I  8 c)  I  —  6  —  0  -  -:  2  -  0 I  I  I  40  60  I  I  I  80  100  120  z —position (a)  Colliding Waves at t  =  —115  I E. 40  60  80  100  120  z —position  (b) Figure 3.2: Shown here in (a) are the infinite force spikes that are obtained if the two-wave metric is used directly in the force equations. Also shown in (b) is the position of the colliding waves at this time.  Chapter 3. The Composite Metric  53  and Vollick [5].  3.2  The Composite Metric  Because I am operating in the weak gravity regime, it is possible to linearly decompose the two-wave metric into several separate metrics [5]. First one breaks down the metric into 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 two single wave strings are aligned with the two-wave string so as to exactly cancel each other out (once the straight string is also included) making the stress-energy tensor for the 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 the entire composite metric can be evaluated individually. The two single travelling waves and 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 Lorentz gauge conditions and, from section (3.1), using this metric for the single travelling waves should remove the non-integrable infinities that were being found initially. As shown in Appendix B, the metric for the residual component of the string also obeys the gauge conditions as long as one chooses a past-time cut-off at some time prior to the collision of the two waves. At this cut-off time the components of the residual string all cancel making the stress-energy tensor for the residual string, and consequently the residual’s metric, vanish. The problems with satisfying the gauge conditions also disappear. The metric for the residual component, and the contribution of its derivative to the force, can be evaluated using some finite value for the past-time cut-off and equa tion (2.56). Because the stress-energy tensor is zero for any cut-off before the time of  Chapter 3. The Composite Metric  54  Decomposition 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 a residual piece of string (d) in which the dotted line segments correspond to negative string components.  Chapter 3. The Composite Metric  55  Decomposition of Residual Component  () ±  (b)  (c)  (d)  Figure 3.4: The residual component of the initial decomposition can also be represented as 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).  Chapter 3. The Composite Metric  56  the collision, the metric and it’s derivatives are independent of the value chosen for the cut-off, although each individual component is affected by the choice. This illustrates some of the difficulties associated with the gauge dependence of the force. 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 the integral for this case, it was necessary to install a past-time cut-off. Unfortunately, the metric in this case strongly depends on the value of the cut-off. If one effectively removes 0 to infinity, one obtains infinite, non-integrable forces, a very nonthe cut-off by taking r physical result. The difference in gauge between this situation and that of the composite metric is the difference between a physically reasonable result and a very unreasonable one. Through the use of the composite metric, it is possible to remove the non-integrable infinities through the use of the Vachispati solution for the single travelling waves. The composite metric also satisfies the Lorentz gauge conditions at all times, a feature absent from the previous solution.  3.3  Force Equations for Single Travelling Waves  Equations for the contribution to the force due to the residual component of the string have already been found. One can simply use equation (2.56) in equation (2.6) with a finite cut-off for each piece of the residual metric. In this case the boundary terms at the upper and lower ends of the integral should also be included although they will cancel out when components are summed together to give the contribution from the entire residual metric. Determining the contribution to the back-reaction force from the single travelling waves and the straight string is quite straightforward. Equation (2.6) is used inserting  Chapter 3. The Composite Metric  57  derivatives of Vachaspati’s metric for the  terms. Starting with equation (2.28) for  the metric’, the derivative for the right-moving wave is found to be =  4G {  +  kinks  2 Faj O-x.O--x(x—z) 2  where Ii is evaluated at ü q3  (Oix Oix) 2 .  [(axX)2]  = umax =  7 2 Ox ln( ( x  UfXX—Z)  —  Z) ) 2 (Umax  —  Ukink)  —X—Z)  OuxOx  and ‘c8  (Ox. Oix) 2  —  (ôx O)-9  (3.2)  .  i=ikk+e  kink  The 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. The magnitudes have been scaled so that 4G i 1  =  1. The delta functions in equation (3.1)  lead to an infinite contribution to the force, but this contribution is integratable. When determining the effects of the back-reaction force on the string itself, one must integrate the 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 magnitude of the delta function. For this reason, the delta forces are represented as single lines extending out of the graphs in chapter four whereas in figures (3.6) to (3.7) they are represented in a more instructive manner by their magnitudes. It is worth noting that the integration constant  2  from Vachaspati’s solution does not  completely disappear from the derivative which reflects the remaining gauge freedom still left in the composite metric solution. The constant is still present in the delta function portion of equation (3.1). This constant is an. arbitrary parameter that one is free to vary 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 backreaction 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 in Appendix A, the two forms are equivalent.  Chapter 3. The Composite Metric  58  Delta Function Amplitude from Right—Moving Wave 10 ci)  0 —10 —20 —100  0 Time  100  Figure 3.5: The amplitude of the delta function contribution to the force by the right-moving wave is shown. The three different curves, (a), (b), and (c), indicate the ef fect of the force due to the delta function generated by the three kinks in the right-moving wave at ii = —100,0, 100 respectively. Delta Function Amplitude from Left—Moving Wave 4  1  I  I  I  I  2 ci)  0  S  —2 —4 —6  (a) I  I  —100  (b) I  I  0  (c)  I  100  —100  0 Time  100  —100  0  100  Figure 3.6: The amplitude to the delta function contribution by the left-moving wave to the force is shown. The three different curves, (a), (b), and (c), indicate the effect of the force due to the delta function at the three kinks in the left-moving wave at i3 = —20, 0, 20 respectively.  Chapter 3. The Composite Metric  Scale 2—’I’’’  59  Scale  10  =  1  =  ‘‘‘‘I  :: 10—•  10  -  -  (1)  0 •  •  -  -  0  • —10  S—b  -  -  —20--  —  I  I  I  I  I  I  —100  0 Time  100  —100  0 Time  100  (a)  Scale I  20  Q)  I  (b)  0.1  =  I  I  I  I  Scale I  I  I  H J•20 -  0_  0  -  -  .4-)  -  I  I  I  I  I  JJJ  S  —20  —20— —30-  0.01  =  -  I  I  I  —100  0 Time  100  I  (c)  -  —40  -  I  —100  I  I  I  0 Time  100  I  (d)  Figure 3.7: The effect of different values of the integration constant is demonstrated for the right-moving wave. Graphs (a),(b),(c),and(d) show the amplitude of the delta function for constant values of 1,10,0.1,0.01 respectively.  Chapter 3. The Composite Metric  60  constant 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. As  was seen in section 3.1, this small freedom can have a strong effect on the resulting force. As is demonstrated in figure (3.7) and as will be shown explicitly in the next chapter, changing the integration constant in the force equations has a significant effect on the back-reaction force. Although the ‘spikes’ in the delta function amplitudes always maintain the same sign regardless of the choice of integration constant, just a short distance away, the amplitude can vary from strongly negative values to strongly positive ones. Also, even though the spikes maintain their sign, their magnitude is still affected by changes in the scale factor. No obvious choices or physical reasons have been found to select one integration constant, or gauge, over another and for this reason the forces that are obtained using my gauge choice are not very well defined. Only in regions such as u> 100 where all of the delta contributions are zero, can one be confident about the shape of the back-reaction force. For consistency and simplicity, I chose the constant to be such that the value of the delta function force is the same sign for each kink. For example, in the case of the right-moving wave’s metric, I chose  (x z) Io,iioo 2  (3.3)  —  This choice makes all of the delta force contributions from the kinks at ii and those from the kink at i  =  =  +100 positive  0 all negative. A similar choice for the left-moving delta  functions would be the value of (x  —  2 at u z)  =  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 because the integration constants for the two metrics are independent of each other.  Chapter 4  The Results  In the previous chapter, it was seen that determining the force by directly evaluating the derivative of the metric as given in equation (2.7), leads to questionable results. Not only did the results have problems with the validity of the gauge conditions at the boundaries of the integral, but this approach also led to non-integrable infinite forces. To solve these problems, a composite metric for the string with two colliding waves was developed. This later approach results in forces which have more reasonable properties. 1 The integrated force on the string increases in magnitude as one nears the point of collision of two oppositely moving kinks. This is expected because one would like to think that the gravitational back-reaction would be strongest closest to those points where the radiation is emitted. The forces also tend to smooth out the sharp kinks of the triangular pulses used. This too seems reasonable as physical intuition leads one to expect that the loss of energy by the waves in terms of gravitational radiation would lead to a damping out of the waves themselves. One unfortunate aspect of the composite metric is that it leads to a much more complicated form for the force. Although each individual component is fairly simple, the resultant sum is not. This force is not, however, perpendicular to the string. The effect of the parallel force components 1 is discussed in Chapter 5.  61  Chapter 4. The Results  4.1  62  The Force Components  By 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 be evaluated separately for each different point on the string. These seven components are then summed appropriately to give the final force on the string. It is informative to first examine the individual components in order to see how the infinities of the more direct approach are removed. Choosing a time-slice of the force at t  =  —90, the forces from  each piece of the composite metric are shown in figures (4.1) and (4.2). The appropriate signs on the forces have been included so that the total force, figure (4.2[d]), is just the sum of the other seven components. Figure (4.1) shows the four components which make up the residual metric. The first graph 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 which was first computed and found unacceptable in chapter three. The next two graphs of  figure (4.1) show how the infinity problem is solved. They show that the infinite force contributions from the two single travelling waves are exactly the negative of those in the first graph. Therefore, when the components are added together, the infinities cancel out. 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 the straight string with metrics having the form of Vachaspati’s solution. Also shown is the resultant 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 in the 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 single lines extending off of the graph. It is interesting to note that the single travelling waves  Chapter 4. The Results  63  Two Waves •II  III  Right—Moving Wave  11111111  1111111  11111  0  20 ci)  C)  J  0  0  0  —20 —20 I  40  —30  1111111111111  60 80 100 z—position  120  ii  40  (a) II  11111  11111  60 80 100 z—position  -  120  (b)  Left—Moving Wave 1  I  I  Straight String 0.05  .  30 -  ci)  0  20  C.)  ci) C.)  -  —0.05  0  r.io  -  -  —0.1 0 I  40  60  80  100  z position —  (c)  120  40  60 80 100 z—position  120  (d)  Figure 4.1: Shown here are the contributions to the back-reaction force from the four components of the residual metric. Graph (a) gives the contribution from the two-wave metric while (b) and (c) give the contribution from the left- and right-moving single wave components. Graph (d) shows the effect of the straight piece of string.  64  Chapter 4. The Results  Left—Moving Wave  Right—Moving Wave 0.3  I  I  I  I  I  I  I  I  I  I  I  0.3  I  0.2 1) C) 0  0.2 C)  0.1  0  0 —0.1  0.1  0 I  40  I  I  I  i  I  i  i  i  i  I  60 80 100 z —position  —0.1  120  40  100 60 80 z—positiori  (a)  120  (b)  Straight String  Total Force I I I p I  I  0.2 0.1 ci) C)  0  0 C)  0.05  0  0  0  —0.4  —0.05  I  40  100 60 80 z —position (c)  120  40  I  I  I  I  I  I  I  I  i  60 80 100 z—position  120  (d)  Figure 4.2: The force contributions at t = —90 from the two single travelling waves and 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 straight string’s effect while figure (d) shows the net force including the contribution from the residual metric. The single lines extending out of the graphs represent the delta function components of the force.  Chapter 4. The Results  65  while having no back-reaction on themselves, do contribute to the back-reaction force on the string with the colliding waves. The back-reaction force does not come entirely from the residual component of the string, the piece that in some sense generates all of the gravitational radiation [5]. Additionally, the force from the straight string can be seen to be exactly the same as that from the straight string part of the residual metric. Because of the difference in sign between these two components, their contributions will exactly cancel in the final sum. It is worthwhile to notice that the finite part of the single wave forces do not cancel with their residual counterparts. This is the main source of much of the complexity seen in the resultant back-reaction force on the string. 4.2  Time Slices  As an initial investigation into the back-reaction force on the two colliding waves, I considered time-like slices of the force. For my colliding waves, I once again chose to examine two triangular pulses. The two pulses used are of the form of (1.30) for the right-moving wave with /3 with a  =  0.5 and 12  =  i/\/ and 1  =  100 and (1.44) for the left-moving component  20. Waves of two different sizes and shapes were selected in order  to reduce the symmetry of the system somewhat and to make it easier to distinguish the forces from one wave as compared to the other. Because the waves are restricted to the yz-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 chosen because only motion perpendicular to the cosmic string has any physical meaning. As mentioned in section 1.2, the field equations representing the cosmic string are invariant to boosts and translations along the length of the string. The force in the y-direction always has at least a component in the direction perpendicular to the string. As discussed in section 3.3, the delta function contribution to the force from the two single travelling  Chapter 4. The Results  66  waves has been scaled so that the force from each kink is always of the same sign. The delta functions will be represented in the figures of the force as single lines extending out of the graph. The magnitude shown for the finite force contributions has been scaled so that 4G[L  =  1.  As expected, before the time at which the two waves first collide there is no force on the string. Before the collision, there is no gravitational radiation emitted from the string and each of the waves can be viewed as just a separate single travelling wave which has been shown to experience no gravitational back-reaction. Therefore, the first figure presented, figure (4.3), shows the forces just after the start of the collision at time t  =  —119. Below this is an image of the string at the same time which also indicates  what 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 seen from later figures, travel along the string keeping pace with the generating kinks of the undisturbed single waves. The force also contains two bounded spikes which shrink and spread out as time goes on. In some sense the effects of the force spreading out and moving further apart can be thought of as the effects of the gravitational wave pulse as it expands out from the kink collision where it was created. However, because the radiation was generated at one point in spacetime, one would expect the effects of this radiation at a later instant in time to be confined to very small sections of the string. The effects instead are spread out 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 the metric of the string at some earlier time. The perturbations propagate at the speed of light and travel a variety of different distances catching up to the string in many different locations. This causes the forces of the string back on itself to be spread out over larger lengths of string than might at first be expected. Not only the finite forces, but also the  Chapter 4. The Results  67  delta function contributions were generated by the kinks of the travelling wave at much earlier times than that at which they recollide with the string, times in fact before the collision even began. This brings up another difficulty with this choice of the Lorentz gauge. The backreaction forces, most notably the delta function contributions, are not always generated after the collision of the waves. Some force contributions are caused by perturbations in the metric of the string at times long before the collision. At these times, there had been no gravitational radiation produced and one would have expected no gravitational back-reaction. Although none of these forces affect the string until after the collision has begun, 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 the collision of the oppositely moving kinks, the source of the gravitational radiation, but in the 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 force at two later times demonstrating the propagation of the force away from its point of origin. At time t  =  —101, one sees a significant change in the shape of the force as  its leading edge flips sign. This is just caused by the right-moving undisturbed wave component of the metric crossing the position of the next kink in the left-moving wave at v  =  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  t  =  =  0 and time  —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 which was just crossed. As one proceeds to later and later times, one passes the force at time t  =  —90 shown in figure (4.8) then figures (4.9) and (4.10) showing the force just before  the 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 kink  Chapter 4. The Results  t II  0.5  68  —119  =  jill  t 0.1.  ii  =  111111  —115 1  11111  0  0 0  C) —0.1  -0.5  0  —0.2  —1  —0.3  —1.5  I  I  40  I  I  i  i  i  i  I  60 80 100 z —position  •lI  120  40  (a)  I  II  1111111  60 80 100 z—position  120  (a)  Waves at t=—119  Waves at t=—115  .111111111111111.  25  IIIIIII1I  20 15 0 C)  0  0  10  0  5  0  I  40  I  I  I  I  I  I  60 80 100 z—position  120  (b) Figure 4.3: (a) Back-reaction force on the string in the y-direction at time t = —119. Delta forces are shown is single lines ex tending out of the graph. (b) The position of the waves on the string at the same time.  0—, 40  1TTTE  60 80 100 z—position  120  (b)  Figure 4.4: Force at time t  =  —115  Chapter 4. The Results  t  69  —110  =  III  0.1  II  t  —101  =  0.15  liii  0.1 0 ci) C.)  ci) C)  0  0  0.05 0  -0.1  —0.05 —0.2  I  40  —0.1  I  60 80 100 z—position  120  I  40  I  I  i  i  i  i  I  i  60 80 100 z—position  (a)  120  (a)  Waves at t=—110 II  I  i  111111111  Waves at t=—101 I_  1111111  111111  20 ci)  C)  ci) 20 C)  0  0  10  E:NN\  0  I  40  0  I  60 80 100 z—position  120  (b) Figure 4.5: Force at time t  I  40  I  60 80 100 z—position  120  (b) =  —110  Figure 4.6: Force at time t  =  —101  70  Chapter 4. The Results  collision and becomes less significant at later times when the string is closer to the next kink collision. However, one also notes that it is near this value that the undisturbed travelling 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 the force 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 from the residual component of the metric. As such, the kink of the single travelling wave is what generates the force. The kink, however, never intersects the string before v > 20 where 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 function force has disappeared. The kink which was generating this force has crossed the last kink of the oppositely moving wave and is once more on a flat piece of string. It has returned to the configuration of a single travelling wave and so a large portion of the back-reaction force including the delta function contribution is reduced to zero. However, the force does not completely disappear. Because the wave was delayed with respect to its counterpart which did not collide, and because the two travelling waves are now causally connected to 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 of spike-like forces, figure (4.11), which decrease in amplitude as one moves away from the point 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  is the remaining effects of the kink at ü  =  =  145. This small spike  —100 as it moves away from the collision. It  is the back-reaction forces caused by the displacement of the right-moving wave which collided with respect to the one which did not. This effect decreases in amplitude as one moves to later and later times and is the first indication that the back-reaction effects do  Chapter 4. The Results  71  t= —99 10  El  t= —90  III  11111  II  6  0  4  0  —0.2  2 0 —2 : 40  0  ci)  ci) 0  —0.4 i  II  11111111  II  60 80 100 z —position  120  -  60 80 100 z—position  40  (a)  120  (a)  Waves at t=—110  Waves at t=—101 I.  I  )  20  I  III  40-  :  30  C)  I  0.2  8  C)  II  111111  C) 1 F.  0  0  10 0 —1;—1  40  I  60  80  0  i  100  120  z —position  40  60  I  I  80  100  i  120  z position —  (b) Figure 4.7: Force at time t  —“  (b) =  —99  Figure 4.8: Force at time t  =  —90  Chapter 4. The Results  72  t= —85 :1  I  I  I  I  I  I  I  I  I  I  I  t  I  I  I  I  —81  0.15  I  0  0.1  a)  a)  C.)  0  0  =  —5  0.05  0  0 —0.05  10 III  40  1111111  III  111111.  —0.1  60 80 100 120 140 z—position  40  60 80 100 120 140 z—position  (a)  (a)  Waves at t=—85 1  11111  11111  Waves at t=—81  111111111  50 I  I  II  111111  III  III  I—  40 a)  30  C.)  0  0  0  20 10 0 40  60 80 100 120 140 z —position (b)  Figure 4.9: Force at time t  40  60 80 100 120 140 z—position (b)  =  —85  Figure 4.10: Force at time t  =  —81  Chapter 4. The Results  73  not stop immediately after the collision is ‘finished’. As the u  =  0 kink of the right-moving wave begins to pass through the left-moving  wave, the sequence of events observed for the kink at u  =  —100 repeats itself. Sets of  double spikes appear just following the collisions, figures (4.14) and (4.17), and spread apart decreasing in size, figures (4.15) and (4.16). The sign of the force flips just before the collision with the middle kink, figure(4.16), due to the undisturbed wave effectively crossing the kink ahead of the actual collision. Even the spike due to small separation values 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 delta force contribution from the right moving wave, have changed sign. This feature shall be discussed 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 similar  to that of the previous two kinks. The signs of the forces are the same as those for the collision of the first kink. Finally, once the waves have finished passing through one another, one might expect that the force would drop entirely back to zero again. The string is once more in a position where it can be viewed as simply two single travelling waves. However, unlike before the collision, the two waves are now causally connected and can affect each other resulting in a small but non-zero force as shown in figure (4.20). Also, the situation is 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 the amplitude 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 its undisturbed counterpart. Similar spikes are present due to the displacement of the right moving wave seen in this figure at z  =  0 and also seen in figure (4.12). The effects due  to the right-moving wave appear further apart because the right-moving wave is much  Chapter 4. The Results  74  t= —79 I  IlIllIllIllIll  t= —50 0.06  111111.  3  III  liii  liii  0.02  ci)  0  III  0.04  2 C.)  111111  ci) C.)  1  0  0  r.  —0.02  0  —0.04  —1 IZLiIIII  40  III  60 80 100 120 140 z—position  —0.06  0  100 50 150 z—position  (a)  (a)  Waves at t=—79  Waves at t=—50 .11111  50  1111111111  60  40 ci) 30 C)  ci) C-)  0  0  20  40 20  10  0  I 40  -I  I i ii i I I 60 80 100 120 140 z—position (b)  Figure 4.11: Force at time t  0  1111111111  o  50  II  100  I1  II  150  z—position (b)  =  —79  Figure 4.12: Force at time t  =  —50  75  Chapter 4. The Results  t  t= —21  —18  =  liii  0.03 6  0.02 a)  C)  0.01  -  -  -  ci)  4  C)  0  0  2 0  —0.02 --iIiiiIiiiiIiiIii-t 50 100 150 200 —50 0 z —position  —2  [  I  I  I  I  0 50 z—position  —50  (a)  (a)  Waves at t=—18  Waves at t=—21 I I  I I I  I  I  I  I  100  I  I  I I  I  I  I  I  I I  1111111  ci)  ci)  0  0  C)  C)  0  ——---  —50  0  50  100 150 200  z —position (b) Figure 4.13: Force at time t  0 —50  -  -  I  -  0 50 z—position  100  (b) =  —21  Figure 4.14: Force at time t  =  —18  76  Chapter 4. The Results  t  =  t=—1  —15  1  0.1  a) o  a) 0 s-I  5-I  0  OflR  —0.1  0 —50  0  50 0 z —position  100  —50  (a)  0 50 z—position  100  (a)  Waves at t=—15  Waves at t=—1 80  60  60  j40 20  20  0  0  —50  50 0 z—position  100  (b) Figure 4.15: Force at time t  —50  0 50 z—position  100  (b) =  —15  Figure 4.16: Force at time t  =  —1  Chapter 4. The Results  77  t=1  t=15  2  10  0  8  —2 0.) C.)  I  I  I  I  I  I  I  I  I  I  I  I-:  6 0) C.)  —4  0  I  0  4  —6  2  —8  0  —10  —2 —50  0 50 z —position  100  I —50  (a)  Waves  I  0 50 z—position  100  (a)  at t=1  Waves at t=15  80 60  60 0) C)  0)  40  40  0  20  20  0  0 —50  0 50 z—position  100  (b) Figure 4.17: Force at time t  —50  0 50 z—position  100  (b) =  1  Figure 4.18: Force at time t  =  15  Chapter 4.  The Results  78  t=150  t=21  30  I  I  I  I  I  I  I  I  I  0.05  20  a)  0  0  0  C.)  10  0  —0.05  0 4  I  II  —50  11111  I  II-  0 50 z —position  100  —200  —100 0 z position —  (a)  (a)  Waves at t=21  Waves at t=150 I  80  I  I  I  I  I  6060  a)  C-) 0  rz 20-  20 0 —50  0 50 z —position  100  (b) Figure 4.19: Force at time t  0 —200  I  I  I  —100 0 z—position (b)  =  21  Figure 4.20: Force at time t  =  150  79  Chapter 4. The Results  larger than the left-moving one.  4.3  U-slices  Although the time slices of the force provide a good visual picture of where the forces are at a given point during the collision and are also easy to reference back to the position of the string at that time, for the purposes of determining the effect of the collision on say 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 been plotted 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 the kink. 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 forms two spikes which approach v  =  0 and v  =  20 as u tends to —100 and that there is also  a delta force contribution from the right-moving single wave kink at ü approaches v  =  =  —100 which  —20. That is, as one approaches the kink in u, one finds that the major  components of the force are coming from points approaching the kinks in v. The force is strongest 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-moving wave at values of u < —100 or u > 100. For intermediate values of u, the left-moving delta forces tend to act against each other, reducing their importance. What one is left with is a force distribution which is predominantly positive and which is increasing in magnitude 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  80  u=—105  u=—101 2  4-  -  ci) C)  -  -  0  -  •  I  I I I I 20 —40 —20 0 v—value  -  —4-  =  —  =  I I 20 —40 —20 0 v—value  Figure 4.22: Force at u  —105  u  —100.1  I 1 II r rm II  -  -  40  10  1111111111  5-  -  •  -  •  -  •  111111  =  I  =  I 40  —101  —99.9  1111111  5-  -  •  -  0) C) 0  .  0  —5  -  -  u  0  0  -  Figure 4.21: Force at u  ci) C)  0) C)  -  —4-  10  0  -  —2-  IIIIIl4-.  -  =  2-  4IIIIIIII  —-  IIIIIIIIIIIIIIIIIIIII  —40 —20 0 20 v—value Figure 4.23: Force at u  =  40  —100.1  0  —5  1111111.,.  111111111  —40 —20 0 20 v—value Figure 4.24: Force at u  =  40  —99.9  Chapter 4. The Results  81  This is shown in figures (4.26), (4.27), (4.28), and (4.29). By the u  =  —1 slice the force  is once again definitely beginning to grow and is taking much the same form as was seen in the approach to the kink at u  =  —100. The main difference is that the forces are now  predominantly negative in direction. Past the kink at u kink at u  =  =  0, the force once more decreases again. Approaching the final  100, figures (4.33) and (4.34), the force is growing again as it did before and  is 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 force contributions from either the left- or right-moving waves. These forces can be seen in figures (4.37) and (4.38). One of the main goals in determining the back-reaction forces on a cosmic string is to, by integrating these forces, find out how the motion of the string is affected. By integrating the force with respect to the variable v, one obtains (Oz), that is one finds out by what amount the slope of the right-moving component of the string is changed by the back-reaction force. Similarly, by integrating over the u variable, one could find the effect on the left-moving component. To get some idea of how the triangular pulses were affected by the back-reaction, I integrated the u-slice forces for a few different values of U.  The main effect of the forces on the right-moving wave occurs near the kinks. As one approaches the values of the kinks at u  ±100,0, the magnitude of the integrated force  increases dramatically. This is shown in figure (4.39) and (4.40) which give the value of the integrated force as one approaches the kink at u  =  —100. Figure (4.39) shows the  contribution from the finite components of the force. Here it is quite easy to see that the force is positive and increasing as one nears the kink. A positive force means that the slope is being increased. The slope on the other side of the kink was originally larger than  82  Chapter 4. The Results  u=—80  u= —99 4  111111  1111111  1  111111  0.5 2 C) C)  a)  0  0  C)  0  0 0.5  —2  liii  11111  111111  —40 —20 0 20 v—value Figure 4.25: Force at ‘u  II  —1  40  =  —99  —40 —20 0 20 v—value Figure 4.26: Force at u  u=—5 0.04  —80  _r  111111111  2  0.02 a)  0  =  u=—1 4  C)  40  C) C)  0  0  —0.02  __  0  —2  —0.04 —40 —20 0 20 v—value Figure 4.27: Force at u  40  =  —5  —4 iiIiiiI ..LIIIIIIIIIIII —40 —20 0 20 40 v—value Figure 4.28: Force at  zt =  —1  83  Chapter 4. The Results  = ‘‘‘‘  :  u=1  —0.1 ir...l.  11111111  60  0  a)  C)  0  —20  40  .  r  —40  a) C)  0  0 —20  liii  1111111  liii  —40 —20 0 20 v—value  I  =  I  H  IIIIII  I  Figure 4.30: Force at u  —0.1  u=20 111  I  20 —40 —20 0 v—value  40  Figure 4.29: Force at u  40  =  1  u=95  11111111.  2  0.2 a)  0  C)  0  00  0  r.  11111I1IIIIl1111  0  —  -  —2  —0.2 II  III  111111  111111  —40 —20 0 20 v—value Figure 4.31: Force at u  40  =  20  —40 —20 0 20 v—value Figure 4.32: Force at z  40  =  95  84  Chapter 4. The Results  u=99 I  I  u=99.9  I -  2  a)  C-)  0  C-) C)  liii  20  0  —2  0  III  11111111111111  I  —40 —20 0 20 v—value  —20  40  Figure 4.33: Force at u  =  99  LIII —40 —20 0 20 40 v—value  Figure 4.34: Force at u  u= 101 .1111111  u  1111111  II  11111.  6  liii  l  = III  =  999  110 II  I  liIi  I  I  0  4  0  ii 1 iii  40  0  0 C)  Tir  11111  2  C) 0  0  —1  —2 —2  —4  .1  I  I  i  i  I  i  i  i  i  I  I  i  —60 —40 —20 0 20 v—value Figure 4.35: Force at u  =  •IIII  11111111  40  —60 —40 —20 0 20 v—value  40  101  Figure 4.36: Force at u  110  =  Chapter 4. The Results  85  u=150  u=500 0.5  20 —60 —40 —20 0 v—value  40  —60 —40 —20 0 20 v—value  40  Figure 4.37: Force at  150  Figure 4.38: Force at u  500  tt =  =  that at u < —100 and so the back-reaction force is acting to reduce the difference between the slopes on either side of the kink, to smooth out the kink. Because the integrated force is increasing, the effect is greater nearer the kink than farther away which indicates that the force is also rounding off the sharp corners of the triangular pulse. This corresponds to a removal of some of the higher frequency components of the pulse, a result which was found in an entirely different manner by Hindmarsh [10] for the special case of very small perturbations 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 as one approaches the kink. However, because of the arbitrary scaling factor, these results are more ambiguous. Shown in figure (4.40) are two different choices for the scale. The solid line indicates a scale of 1 while the dashed line is for a scale of 0.008 or 1/125 which is the scale which just causes all of the right-moving delta forces to always have the same sign, section 3.3. Regardless of scale, the delta force always becomes positive  Chapter 4. The Results  86  just before the kink, however, the scale does affect its sign a short distance away from the kink. If the force is positive, as with a scale of 0.008, then its effects serve to enhance those of the finite forces. A negative force, on the other hand, serves to reduce the effect or perhaps even cause the slope to go first negative before becoming positive. This would generate a dip in the string before the kink  was  reached. The arbitrariness of the  integration constant makes it impossible to say anything definite about the shape of the force. Nevertheless, in all cases the difference between the slopes immediately on either side 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 the  force at u > —100. Shown in figure (4.41), the effect of the force on the other side of the kink is also decreasing in magnitude as one moves away from the kink. The finite forces are negative in direction which acts to further reduce the difference between the two slopes. On this side of the kink, the right-moving delta function is zero but the left-moving delta functions are now non-zero. It is worth noting, however, that the delta functions serve to exactly cancel each other except for a small constant negative factor. This factor is independent of the value of the integration constant. The effect of the leftmoving delta functions on the right-moving wave is independent of the gauge freedom in Vachaspati’s solution. This result can be shown analytically, Appendix C, and holds for all values of u. This means that the only parts of the force which continue to depend on the gauge choice are those parts which contain a non-zero contribution from the right moving 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 and rounding off of the wave seems to be occurring. The effect of the left-moving delta functions to the right of u  =  —100 is not only  Chapter 4. The Results  87  Integrated Finite Force Components for u < —100 I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  C.)  -  V  -  0  -  ct ci) -  I  I  I  I  —100.8  —101  I  I  —100.6 u—value  I  I  I  I  I  I  —  —100.2  —100.4  Figure 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  —  C1.  10  —10  —  —101  I  I  I  I  I  —100.8  —100.6  I  I  I  I  —100.4  —  —100.2  u—value Figure 4.40: The integrated force from the delta function contribution at values of u less that —100. The solid line represents a scale of 1 while the dashed line represents a scale of 0.008.  Chapter 4. The Results  88  unambiguous but is also negative. This serves to further reduce the difference between the slopes on either side of the kink. However, the forces at this kink are constant so they 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 kink at u  =  0 where the force is increasing but negative for values of u below zero and positive  and decreasing above zero. Approaching u  =  100 from below, the force is again positive  and increasing. Past the last kink, the force is negative and decreasing in amplitude as is shown in figure (4.43). Here there is no ambiguity due to the scale factor because the amplitudes of all of the delta force contributions in this region are zero. Here, as with the forces on either side of u  =  —100, a negative force serves to decrease the  difference between the slope on one side of the kink with respect to the other. Because the magnitude increases as one nears the kink, the forces will induce a curving in the string which will smooth out the kink and reduce the high frequency components of the wave. Because of the arbitrary nature of the scale factor, although the kink itself is both rounded off and reduced, at short distances to the left of the kink it is not possible to say what exactly the effects of the back-reaction force will be. Only along sections of the string, such as the region at u > 100 where there are no delta forces, or the region just beyond u  =  —100 where the effects of the integration constant cancel out, can one make  any sort of true description of the shape of the force. This is perhaps the most serious failing of this approach to determining the back-reaction forces.  4.4  V-slices  Although there is certainly some indication in the time slices of the force, it is far from obvious in the u-slices that the force on the left-moving wave is very similar to that on  Chapter 4. The Results  89  Integrated Finite Force Components for u > —100 —2  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  I  -  V C) 0  a) o —4 a)  -  I —99.8  I  —99.6  —99.4 u—value  —99.2  —99  Figure 4.41: The integrated finite force on the right-moving wave just past the kink at u = —100.  Magnitude of the Delta Force Component, u > —100 I  I  I  I  I  I  I  I  I  I  I  V  C) C)  I  I  —99.8  I  I  I  —99.6  I  I  —99.4 u—value  I  —99.2  —99  Figure 4.42: The integrated force from the delta function contribution at values of zi greater than —100. The solid line represents a scale of 1 while the dashed line represents a scale of 1.62 x iO’. The two lines exactly overlap indicating that there is no gauge dependence in this part of the force.  Chapter 4. The Results  90  Integrated Force After u= 100  —2  C) 0  ci) —4 cts bE  ci)  —6 —8 100  102  101  104  103  105  u—value Figure 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 one expects that the forces should be in some sense symmetric. This is shown to be the case by 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 rightmoving wave the force spikes appear and approach the kink values of u one nears the kink at v  =  one neared the kink at u  =  +100,0 as  —20. These forces are very similar in shape to those seen as =  —100 with the u-slices. The appearance of the delta force  spike caused by the kink at v  =  0 and seen at position u  =  80 in figure (4.45) is just  the leading spike of the forces that appear as one nears the next kink at v  =  0. The  forces are overlapping and also appearing at greater distances from the kink than did their right-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 than  the delta force contribution from the kink at v  =  —20 and so the resultant integrated  91  Chapter 4. The Results  force on the left-moving wave is positive as was the case for the right-moving wave. Just past the kink at v  =  —19, figure(4.47), the force once again drops in magnitude and has  a form similar to that of the u-slices of the force just past the first kink at u  =  —99.  92  Chapter 4. The Results  —25  v= .1  I  I  I  I  I  I  I  I  I  I  v= I  I  I  I  i1:  I—  —21 I  I  I  I  I  I  I  3-  1 a)  a)  0 C  0 0  0.5  0 1111111  —100  liii  0 100 u—value  v liii  =  —1  200  Figure 4.44: Force at v  10  II  =  —25  —100  0 100 u—value  v  —20.1 liii  =  liii  11111  200  =  —21  —19 liii  liii—  1  6 a)  0.5  ci  4  0  0  2  -  —0.5  0 —2  -  Figure 4.45: Force at v  8 a) o o  :  I  II  —100  0 100 u—value  Figure 4.46: Force at v  200  =  —20.1  —100  I  —  11111111  0 100 u—value  Figure 4.47: Force at v  I—  200  =  —19  Chapter 5  The Perpendicular Force  It was noticed in chapter two that the force given by equation (2.6) yields a perturbation metric  for which (OzI2 + OuzP)(Ouzv +  when it was found that 11 OUz’öUz . 1 h LV  0 even though  + h)  0. This was discovered =  0. This indicates  that 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 results in a trajectory of the string which no longer satisfies the equations of motion. If one stops at the first order corrections to the string trajectory, one could still use this definition of the 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 perpen dicular to the string. Because the string is invariant to boosts parallel to its axis, only quantities perpendicular to the string have any real meaning. Therefore, a more reason able definition of the force would be one which contained no parallel components. One can obtain a force of this form be subtracting out the parallel components from the force equation (2.6). This is permissible because it turns out that the parallel force just causes a change in the internal coordinate system. It is a good idea to correct for this coordinate change because the derivation of the force equation is based upon the satisfaction of the null coordinate conditions. A change in coordinates could shift one out of a null coordinate system, invalidating the force equation. By redefining the force as only that component which is perpendicular to the 93  Chapter 5. The Perpendicular Force  94  string, one maintains the initial choice of null coordinates. There would be no ambiguity with respect to the continued satisfaction of the null coordinate conditions and the force would no longer contain any ‘unreal’ components. The perturbed metric would stay in the solution space of the equations of motion.  5.1  Removing the Parallel Components  It is desired that the force be perpendicular to the string. Not only is this more physically reasonable, but this would also cause the null coordinate condition to be true not just in the flat background metric but also under the perturbed metric. That is, there would be no coordinate change due to a force acting parallel to the string. If the perturbed string trajectory is to be parametrized by the same null coordinates as 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 trajectory while the 1 indicates the first order approximation. Therefore, to first order in the perturbation, 1  (5.2)  =  This relation can also be found by starting with the force expression (5.3)  =  Contracting both sides with  one gets  OUZOUOVZT1/.Ll)  =  =  =  —r —  a ôz (5.4)  95  Chapter 5. The Perpendicular Force  But, 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 has components both perpendicular and parallel to the string. Therefore, one can also express z in terms of the components I, which is generated by the perpendicular force, and ,  which is generated by the parallel force.  coordinates, )  —+  )+  The force also causes a change in the  where the index i runs over the two internal coordinates u and  v, \ corresponding to the original null coordinates and  to the first order correction to  these coordinates [23]. As a result, zi can be expressed to first order in the perturbation as =  =  (5.7)  z+eä  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, I satisfies the conditions  =  av  =  0  (5.10)  Chapter 5. The Perpendicular Force  96  Using 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 the coordinate shift, 9.  is found with the analogous equation for  9,,c1i  The  results are “-‘h  a5V  S  —aU aU 2 0 0  —  LJZQ  ‘vZoV  li” 17  —aVzaVzhliV V  0 Z 11  (5.12)  0 77 vZ  Substituting equation (5.7) into the equation for the force and rearranging, it is possible to find an equation for the perpendicular force. A 1 8z = = —  zli 2 a 2 an  and  aeu  —  0  1  2 av  are given in equation (5.12) and  u  aUV’, a  /3  —  are equal to  u az  h l-”’7  aUo  —  h l-7  .VZo  —  aU’,  2az a 0  L-v a‘U aVS  v  —aUZQP9UZ 0  —  aa’  zli 2 a  ‘-‘uS  —a’ VS  where  aUaV  2a a 0/3  a  v  uS  aVz  lliv  /3 0 a lV 2 a iaZ  hw  /3 aVZO 77  The force given by equation (5.13) is now explicitly perpendicular to the string. To see this, consider the contraction of the force with  aUz’r,liV.  The first term in (5.13) can  expanded using equation (2.6). _Plia/3  aVZOaUZOliV  =  =  li7 —  ,/3 + , 7 (h 87 h  —  ) 7 ha/3,  aUZOliV  (5.15)  Chapter 5. The Perpendicular Force  97  Performing the contraction with the rest of the terms in equation (5.13), one finds —  1  — —  }  z 2 O  {aa Oz 3  8UOV  — —  Oz°8z” 0 U Uo  h 7p,7 aV aVo 0  ô,z i] IS 2ãz 0 -  I  8Uz”  a  v  U  ZPi  I°P  U’,  OUZ  8V 0  z 2 8  —  Zo ?] 9 ‘  vo 91 uV j  U,  It is easy to see that equation (5.16) cancels with (5.15) and so OVz  4th  (5.16)  =  Similarly,  U  OVzlrl/JJI  =  =  0.  0. The force given by equation (5.13) is perpendicular to the  hypersurface 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 null coordinate conditions. That is, the shift in coordinates caused by the force parallel to the string has been cancelled out. The new trajectory is still a solution to the equations of motion and there is no doubt as to the validity of the force equation. 5.2  Force on a Single Wave Revisited  In chapter two, it was found that the force obeyed two other reasonable conditions. One condition was that the force on a single travelling wave be explicitly zero. This condition also holds true for the perpendicular force of equation (5.13). It has already been shown in section 2.2 that the first term in (5.13) vanishes for the single wave pulse. It remains to check that the parallel force terms also make no contribution. Consider the right-moving wave. For this case, x(3)  =  (i, 0,0, i1) and so 0x  =  ãz  =  Ox =constant, for example,  (1,0,0, 1). Using the composite metric for h, the single  wave is represented by Vachaspati’s metric. —  =  —4Gp  I” U V 0  ln( 2 p)  (5.17) UUmaz  Chapter 5. The Perpendicular Force  where  p  (X(Umax)  2 z)  —  98  = (X (Umax) 2  coordinate transformation,  and  —  )+ 2 z  U, 8  (X’(Umaz)  —  . The derivatives of the 2 ) 1 z  involve contractions of either  z 9  or  ãz  with  the 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  or  with the derivative of the metric. As given in section 3.3, the derivative of the Vachaspati metric is =  4G {  +  [(ãxx)2]  kinks  F 2 O-x.O-x(x—z) 2  where ü is evaluated at  u  Umax  Therefore, in  —  (x  —  —  )(Umax 2 Z)  Ox  z)  —  (x  —  —  Ukk)  (5.18)  7 z)  and 13  —  (9x Oux) 2  7 2 Ox ln( ( x  (Ox. OUx)  13 Fcx  —  UUkjnk+E  (x.  2 9—x)  UkjnkE  the contraction 8z”h, 7 corresponds to a contraction of  zIAF,,  which always vanishes for the right-moving wave. The last term, O8’, contains the contraction Oz”h 7 which again involves terms of the form z’’F. , 11 metric is evaluated at € =  = Umax  and also because  Umax =  u for a point  z  Because this on the string,  0.  All of the terms from the parallel force are zero for the case of a single right-moving wave, as is the contribution from the first term. Consequently, the perpendicular backreaction force on a single right-moving pulse is also zero. An exactly analogous argument follows for the left-moving wave and so, as before, the force equation developed in this chapter yields the expected result of no back-reaction for single travelling waves. 5.3  Energy Equivalence Revisited  The other reasonable condition on the force in chapter two related to the energy of the string. In section 2.3 it was seen that the full force gave an equivalence between the energy  99  Chapter 5. The Perpendicular Force  emitted by the string as gravitational radiation and the energy lost by the string due to the back-reaction. However, because the parallel component of the force seems to just perform a coordinate transformation on the string and as the energy is the integral over the entire world sheet of the string, one might expect that there would be no difference between using the full force expression or just the perpendicular component of it. That is, one expects that the parallel component of the force does not contribute to the energy loss of the string. As was demonstrated in section 2.1, the integral over the force is orthogonal to c z. 9 OuZ]wfdV8uDvZ  0  (5.19)  Therefore, the component of the force parallel to O,z, makes no net contribution to the 0 and so the integral. Similarly, the integral over u of the force is orthogonal to O,z  component of the force parallel to Oz also makes no contribution. Because the equation for the energy of the string involves an integral over both u and v, the only part of the force which makes a contribution is the perpendicular component. This means that the energy equivalence relation holds for both the force developed at the start of this chapter and the one used in chapter two.  5.4  Effect of the Perpendicular Force  Because the force given by equation (5.13) does not cause a shift in the internal coor dinates, one might hope that this force would not have the gauge problems that were encountered with the full force, equation (2.6). Unfortunately, this does not occur. The major gauge problem with the full force is that there is an arbitrary integration constant present in the delta function component of the force. This freedom means that the shape of the force was very ambiguous and although the integrated force did seem to have some reasonable behavior in the finite portion, the results are highly dependent on an  100  Chapter 5. The Perpendicular Force  appropriate choice of the integration constant or scale of the delta function. The two correction terms  both involve derivatives of the metric and so will involve delta  functions containing the same scale factor. Unfortunately, these delta function contribu tions do not cancel out the previous ones. This can be seen in figures (5.4) and (5.4). In addition, the correction terms do have a scale dependence in the region  ti>  —100 where  the full force did not (Appendix C). The perpendicular force increases the ambiguity in the back-reaction effects instead of decreasing it. Figure (5.4) shows the amplitudes of the delta function contributions from the full force, the parallel component of the force, and the perpendicular component which is just the full force minus the part parallel to the string. One quickly sees that not only are the magnitudes of the force different but also the qualitative shape of the force is different. The perpendicular component of the delta function amplitudes is evidently not zero and so the ambiguity associated with the arbitrary integration constant is still present. Switching to the perpendicular force does not solve the problems of the gauge freedom. 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 qualitatively not very different from the full force. The perpendicular force provides a nicer view of what one means by the back-reaction on a cosmic string. However, it still exhibits many of the problematic features of the full force. It still contains the difficulties associated with the gauge freedom of Vachaspati’s solution. In fact, in some regions, it contains more difficulties than the full force. The gauge dependence in the left-moving delta function cancelled out for the full force but does not do so for the perpendicular force. Additionally, the perpendicular force contains ambiguities in locations other than just those of the full force. The full force delta functions occur at the position of the kinks of the undisturbed waves. The perpendicular  Chapter 5. The Perpendicular Force  101  Full Force I  I  I  I  I  I  I  Parallel Force I.  I  I  I  I  I  I  5-.  a.)  I I —100  I 0  —  100  I —100 I  Time  I  I  0  100  Time  (a)  (b)  Perpendicular Force I  20  I  —100  I  I  I  I  I  0 Time  100  (c) Figure 5.1: These figures show the amplitude of the delta function contribution to the full force (a), the parallel force (b), and the perpendicular force (c) which is just the full force minus the parallel component. The integration constant for these graphs is 1.  Chapter 5. The Perpendicular Force  102  Full Force I  20-  I  I  I  I  Parallel Force I  I  I  r  -  I  -  10  I —100  —i  I 0  I 100  -  ci)  -  I I —100  _._.  Time  I  0  I  I  I 100  Time  (a)  (b)  Perpendicular Force I  I  :1 I  —100  I  I  I  I  I  I  I  I  I  I  0  100  Time (c) Figure 5.2: These figures show the amplitude of the delta function contribution to the full force (a), the parallel force (b), and the perpendicular force (c). The integration constant for these graphs is 0.01.  -  Chapter 5. The Perpendicular Force  103  force also contains delta functions at the kinks of the colliding waves on the string due to terms like  These terms are multiplied by the metric which also depends on the  gauge choice, the integration constant. The continued presence of the gauge freedom indicates that before one can have any confidence in the effects of either the full force or the perpendicular force, some means will have to be found to completely fix the gauge. This requires either finding a physically acceptable reason for choosing one specific integration constant, or deciding on a gauge fixing condition before developing the force expression. Based on the minimal effect that the correction terms had upon the right-moving delta force contribution to the force, there is also some indication that the perpendicular force would behave in a manner qualitatively similar to the full force. For this reason and because the perpendicular force does not solve the gauge problems of the full force, I have not explored this possibility in any depth.  Chapter 6  Conclusions  In this paper, I have investigated the gravitational back-reaction of two colliding travel ling waves on a cosmic string, more specifically the back-reaction force generated in the Lorentz gauge by two oppositely moving triangular pulses as they pass through one an other. It was found that the approach used by Quashnock and Spergel could be modified in order to be applicable to this situation [20]. The result was that the back-reaction force, though somewhat complicated in appearance, leads to a damping out of the two waves. The damping effects are greatest near the collisions of the kinks in the waves, those points from which the gravitational radiation is emitted.  Q uashnock  and Spergel’s technique was developed for the case of oscillating ioops of  string [20]. Their method cannot be directly applied to the case of colliding waves on infinitely long strings. If one attempts to do this, it was found that the resulting backreaction force becomes non-integrably infinite. The reason for this is that the form of the metric that they used is in a gauge which is inappropriate for this problem. Making a gauge transformation within the Lorentz gauge, the original metric was converted into the form of a composite metric. This gauge choice proved more successful for studying the back-reaction forces. The composite metric reduces to Vachaspati’s gauge choice for single travelling waves. The composite metric also solves the difficulties associated with the infinite boundaries of the metric integral. Because of how the composite metric is constructed, 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. 104  Chapter 6. Conclusions  105  Using the composite metric a.s the source of the gravitational back-reaction force, it was found that the infinities encountered with the original metric had been reduced to delta functions. The shape and integrated effect of the force was determined for several different points on the string. The time slices of the force indicate that the force is generated by the collision of oppositely moving kinks. The forces first appear just after the kinks collide and then propagate away from their point of origin and decrease in amplitude. The point of origin of the back-reaction forces is the collision of oppositely moving kinks on the string. This result is quite reasonable because it is at these locations that the gravitational radiation is produced. The effects of the gravitational back-reaction are entirely non-local. The back-reaction equations indicate that there is no effect on a point of the string caused by a force generated at that same point. The delta function forces, for instance, are off-set from the kinks on the two-wave string. They align instead with the kinks of the undisturbed travelling wave which is at a different spatial position. The source of the perturbation which generated the force must be on the past light cone of the point on the string being effected. Therefore, in order to have travelled the necessary spatial distance, the perturbation must have come from some time in the past of and some distance away from the affected point on the string. This implies that the force effects are very non-local. The non-locality is also evident in the width of the back-reaction forces. The gravitational radiation 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 affect only a small portion of the string at once. This is certainly true of the delta function forces, but the finite contributions to the back-reaction are drawn out over much larger regions. This indicates that the metric deviations causing the force occur over a variety of times and travel various distances before catching up with the string again. The time  Chapter 6. Conclusions  106  slices also indicate that back-reaction effects are not confined to the period during which the waves are actually passing through one another. A small decreasing force, caused by the delay suffered by one travelling wave as it passed over the other, continues to affect the wave pulses long after they have separated. The effect of these forces on the right-moving component of the string was examined using u-slices of the force. It was seen that the integrated force increases in magnitude as one approaches the kink. The integral over the finite components of the force is positive as one approaches the kinks at u kink at u  =  =  ±100 from below and negative as one nears the  0. As one approaches the kinks from above, the forces are of the opposite  sign. The sign of the integrated force is important as it determines how the slope of the string components will be affected. The integrated force as one approaches the kink at u  =  —100 from below is positive indicating that the slope is being increased. The  negative force for u > —100 reduces the slope on the other side of the kink. This makes the slope on the left of the kink closer to that on the right. The forces as one nears u and u  =  =  0  100 also serve to reduce the the difference between the slopes on either side of  the kinks. Unfortunately, there is some ambiguity as to the sign of the right-moving delta func tion contribution. Because of the arbitrary scale factor it is possible to shift the position of the delta function component of the force. Although, regardless of the scale, the force always has the same sign as the integrated finite forces in the immediate neighborhood of the kink, at an arbitrarily small distance away from the kink, it can have a value which is very different. The effects of the delta functions can completely dominate those of the finite force contributions resulting in significantly different effects on the string. Instead of 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 this approach to finding the back-reaction force.  Chapter 6. Conclusions  107  Another disturbing aspect of my choice of gauge is that the back-reaction effects are caused not just by the residual component of the metric, that part which generates the gravitational radiation [5], but also by the single travelling wave components. A single travelling wave emits no gravitational radiation and yet it does make a contribution to the back-reaction force. Furthermore, the contribution does not cancel out with some part of the residual metric as did the straight string contribution. Additionally, the delta function forces seem to come from times before the collision had even occurred, times at which there was no gravitational radiation produced by the system at all. These difficulties together with the problem of the arbitrary scale factor indicate that despite its reasonable features, the force developed in this thesis still contains too much gauge freedom. To study gravitational back-reaction on cosmic strings, therefore, one needs to find a better method of restricting the Lorentz gauge freedom or one must find a different gauge entirely. However, assuming scale factors which do not make the behavior of the delta function forces radically different than the finite forces, for example, the scales which I chose, it is 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 nature of the scale factor. Regions such as the one at u > —100 where the dependence of the scale factor cancels out or the region at u > 100 where the delta function contribution is zero. In these cases as well as those for which I have chosen an appropriate gauge, the magnitude of the integrated force increases as one examines points closer and closer to the kinks. This increase, combined with the reduction in the slope differential across the kinks, results in the waves being damped out. The pulse is slightly flattened out and the sharp corners are rounded off. These results are in agreement with other work done on the back-reaction of cosmic strings [10, 20]. As expected, the forces have a high degree of symmetry between the left-moving and  Chapter 6. Conclusions  108  right-moving waves. This was shown briefly in section 4.4 and indicates that the effects of the force, though only evaluated for the right-moving wave, are also applicable to the left-moving wave. The left-moving pulse is also damped down and rounded off by its interaction with the right-moving wave. Another difficulty with the force used is that it contains components parallel to the axis of the string. This results in the perturbed trajectory of the string no longer satisfying the 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 satisfies not only all of the conditions that the original force does but also the null coordinate condition for the perturbed metric. However, although the perpendicular definition of the force is a better definition, it still exhibits the major failings of the original force. It still contains the arbitrary integration constant and still has contributions to the force from the single travelling wave components at times before the collision had occurred. Quantitatively, the two forces give different results but qualitatively, there are indications that the two forces are very similar. Nevertheless, the perpendicular force is still a possibility that could be explored. One can conclude, therefore, that the choice of the Lorentz gauge results in a backreaction force which is reasonable in many respects. The energy emitted by the string in the form of gravitational radiation is the same as that lost by the string due to the back-reaction forces. Also, in this gauge, the single travelling wave has no back-reaction force as is expected because it emits no gravitational radiation. Unfortunately, this gauge is not perfect being flawed most noticeably by the presence of an arbitrary scale factor in the force equations. The gauge freedom of this arbitrary constant means that it is difficult to make any definite statements about the effects of the gravitational back reaction indicating that fixing the gauge in a different manner might be more informative.  Chapter 6. Conclusions  109  However, a reasonable choice for the scale factor does permit a determination of the back reaction forces on two colliding travelling waves. In this case the back-reaction serves to damp out the waves, rounding off the corners of the wave and straightening out the string. These results are in agreement with previous work done on the back-reaction of cosmic strings [20, 10] and in accord with one’s physical intuition about how the strings should 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 University of 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 Radiation University 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 and Company, 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 Gen eral Theory of Relativity (John Wiley & Sons, Toronto, 1972).  110  Bibliography  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 A  Comparing Single Wave Metrics  Although on the surface the form that I have used in this paper for a single travelling wave 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):  =  (—4Gt) ln( [(x 2  3 are functions of z  —  —f’  —f’  1  0  —f’  —g’  0  1  —g’  +g’ f’ 2  —f’  h  =  where h  +g’ f’ 2  —  2 + (y f)  —  0 where z is the z  —g’ 2 +g’ f’  —g’ 2 +g’ f’ is an integration constant, and  ]), 2 g)  jth  (A.1)  f  and g  component of the four-vector z. From (2.28)  one has =  where p  = (X(Uma)  —  2 z)  ln(2p)  —4Gi  = (X (Umax) 2  O.ax —  )+ 2 z  (X’(Umax)  Recalling that for a right-moving travelling wave, x the arguments of the logarithm. At  —  f  —  =  . 2 z’)  (1, 0, 0, 1), one first consider  Umax,  Ox.x(u)  As  (A.2) UUm ax  =  =  —  (A.3)  0 z  measures the displacement of the string from the z-axis at the position z 3  the 1-direction. Because of (A.3),  f  is equivalent to 112  X’(Umax).  —  0 in z  The argument is the same  Appendix A. Comparing Single Wave Metrics  113  for g and, therefore, the arguments of the logarithms are the same, up to an integration constant. Next, consider the components of the tensor. (Jux —  = =  f  1  3  0  0 O(x —x)  Ou  Ou  f’8x.ôx aux’  =  (A.4)  Ox t9x  However, from (A.2) the same component of the tensor is —F°’  1 —9x  —  —  —  The two tensor components are the same. Similarly, one finds that the F° 2 components are also identical. The F  component from (A.2) is just  —F”  —  —  Ox because Ox’  =  .  2 components. 0. Similarly for the F 22 and F’  All that remains to check is that the F° 3 components are the same. Consider first the  f’ 2 +  term. =  (  )2  O,x’  \OuX  (  2  Ox 2 \.0ux OvXJ  OvX  (A.7)  but 2 (Ox)  2 + (Ox’) (Ox°) 2 + (ôx ) + (ax 2 2 ) 3  =  (A.8)  therefore, I  2  !  2  f —  02 J  —  ( 32 -‘u’- )  (Ox Ox) (ax° Ox )(Ox° + 8x 3 ) 3 (Ox —(6x° + Ox ) 3 .  —  —  —  —  .  —  3 -F° =  ax ax  (A.9)  114  Appendix A. Comparing Single Wave Metrics  Observing that F°°  =  2ôx°8x°+ãx•Ox  =  3 2Ox° + (Ox  =  3 F°  (A.1O)  =  33 F  (A.11)  —  a°)  it is easy to see that the remaining components of the tensors in (A.1) and (A.2) are equivalent. Therefore one concludes that (A.1) and (A.2) are identical up to the arbitrary inte gration constant.  Appendix B  Gauge Condition for the Residual Wave  To find the metric for the string, either for single travelling waves, or in the case where there are two waves on the string, one evaluates (1.5) using the causal Green’s function of the wave equation. To insert a past-time cut-off, a Heaviside function of the form O(u + v +  T)  is inserted into the integral. Taking the value of r to positive infinity then  effectively removes the cut-off. If this is done, however, for any finite value of r, one finds that the metric does not obey the Lorentz gauge condition. Physically, this is due to the abrupt change that one has effectively inserted into the stress-energy tensor. At time ---r the string’s energy abruptly goes from zero to some non-zero value. The conservation of stress-energy is not obeyed across this boundary. Mathematically, this appears in the integral as a dependence on z of the boundary condition. Doing the integration over the delta function, 6((x(u, v)  —  ), causes a change in the Heaviside function for the 2 z)  past-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 condition no longer holds. In building up the metric that was used to calculate the back-reaction force, I used the metric of two Vachaspati-like single travelling waves, a straight string, and a residual contribution. The metric for the straight string obeys the Lorentz gauge conditions as does the metric for Vachaspati’s single travelling waves [17]. It is necessary, however, to confirm that the residual metric also obeys the gauge conditions. 115  116  Appendix B. Gauge Condition for the Residual Wave  The residual metric is build up from the two-wave solution plus a straight string minus two oppositely moving single travelling wave solutions. This is represented dia grammatically in figure (3.4). The metric,  can therefore be written as  ,  4 1 h=h — 2 + 3 h h k  (B.13)  where subscript 1 refers to the two-wave solution, subscripts 2 and 3 refer to the leftand right-moving single wave solutions respectively, and subscript 4 refers to the straight string solution. Setting 8Gt  =  where  T  f  =  1, one can then write  dudv  ((x  —  ) O(z° 2 z)  —  u  —  v) (u + v + r)  (B.14)  is some constant past-time cut-off. Consider now, the Lorentz gauge condition.  =  f  dudv [4’((x +26((x  =  —  J  du  +  —  +  O(z°  f  f  —  u  z) x6,xTh e(z°  v) O(u +  —  [f  ) 2 z) —  u  —  6(z°  —  _z)2)  dudv (Ox +  f  —  u  —  —  —  u  —  —  v) 8(n +  V  +  T)  8(u + v + r)]  v)  (x  6((x —  —  —  u  —  T) 6((x  v)8(u + v + r)] z)2)  —  uv)O(u+v+r)]  ) ö(z° 2 z)  —  u  —  v) 8(u +  V  +  T)  ) 2 z)  dudv (Ox + Oxfl 6((x  .  —  v)(u + v + —  T)  —  ) 6(z° 2 z)  b(z° —  —  —  —  v)8(u +v +r)]  v) 8(n + v + r)  ) 6(u + v + r) 2 z) OXcr  aux 2Ox  T)  0 duax(z  ö((x  u  +  [fdvaxo(z0  dudv (Ox + Ox) ö((x  =  V  v)O(u + v +  dudv (Ox + Ox) ö((x  +  f  —  ) OxOx 2 z)  dv Ox O(z°  fduo((x  f  —  x[O(z°  =  ) (x 2 z)  v+oo  — fdvo((x  =  —  —  z)  + ZL=u(v,z),v=—T—u  2Ox (x  —  z)  (B.15) v=v(u,z),u—r—v  Appendix B. Gauge Condition for the Residual Wave  117  Individually, h does not obey the Lorentz gauge condition. However, (B.15) is evalu ated at the time  —T,  =  u+v  =  —r. This is a time before any collision of the two waves  has occurred and by construction, the single travelling wave components of the residual wave exactly line up with the two-wave component. Therefore, when the summation is done 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 C  Gauge Independence of the Left-Moving Delta Functions  One of the major difficulties associated with the force which was developed in this thesis is that the delta function contributions to the force contain an arbitrary factor. These delta functions arise in the Vachaspati form of the metric for a single travelling wave. The arbitrary integration constant,  2,  means that the back-reaction forces obtained for the  two colliding triangular wave pulses are not well defined. It would have been better if the part of the force which depends on the integration constant had cancelled out. For the right-moving delta function components this does not occur. However, when studying the effect of the force on the right-moving wave, the dependence of the left-moving delta function force on  122  does disappear. The cancellation of this gauge freedom can be shown  analytically. The back-reaction force, as given by equation (2.6) involves three terms each of which contains a derivative of the metric. +  vz =  (C.16)  —  For the single travelling waves, the derivatives of the metric take a form such as that given in equation (3.1). =  4Gi  where €3 is evaluated at  (x 2 ln(12  [(Oxx)2] V ‘U  kinks  V = Vmax  —  )6(Vma 2 Z)  —  Vkk)  and  ( C18  —  (8x. Ox) 2  —  (c.17)  (0x O-x) 2 .  —  =kk+E  118  (Ox  2 Ux)  .  kkink6  Appendix C. Gauge Independence of the Left-Moving Delta Functions  119  Splitting the logarithm in (C.17) into two pieces, (x 2 1n(  —  ) 2 z)  =  ) + ln((x 2 ln(  —  ) 2 z)  one can consider just that part of the logarithm which is dependent on the constant When integrating over v, the delta function, of (Okx  .  .  i5(Vmax  —  vkk),  will introduce a factor  Ox) into the denominator of (C.17). The absolute value signs may  be dropped as this factor is always positive. I consider first the contributions to the force from the delta functions in the last two terms of equation (C.16). For the purpose of concreteness, I will demonstrate their cancellation at the point u  =  —80. At this point the delta functions each lie on sections  of the string for which Oz takes on a different value. The same result, however, can be shown for all points along the string. As I am oniy interested here in the gauge dependent parts of the force, I shall only consider those terms containing 1n( ). Because all of these 2 terms are proportional to a delta function, the integration over v is straightforward. (h  —  1la,7) Ouza Oz  =  ) 2 4Gt1n( kinks  {  a (  .  x  vX  z)  VVkink+E  7 Fa0Z0vZ0uX  —  (x. Ox)(ã-x. Oz)  (C 19) kink+e  where àx is the slope of the right-moving component of the string as ii tends towards negative infinity and v is evaluated at the value of v for which Using the simplification that O,z ôjx .  =  0 if  Vma = Vkink.  Ovzv = ax”,  one finds that three of  the six terms in (C.19) are zero. There are three different values that 8 kx and Oz can take 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) can be rewritten as  Appendix C. Gauge Independence of the Left-Moving Delta Functions  —  ha, ) 7 zOvz  =  4Gbtln(1 ) 2 (a  120  ôxi)(Ox 2 ) 3 .Ox ) (Ox 0x  X  .  {(Ojx.Oxi 8 j X  7 ãvXl OvX2 Z X  +(6jx  vX2DiX •Ou2OvX3 8 7  —Ox  OvX2 OvX3  • jX  7 aX.auZavX2  OuZ  O-jX  OüXy  OuZ OvXl 7 OZ OjX 7  OvX3 0 vXl  +  X•OvX2OvX1 •3uZOI 7 X 1  + 9x  2 ôvX  Ojx 3z 7 ) 3 3xi ) (9x 9x .  + Dx ôvx3Dvx2 +  OjX  OvX3  + 0X +  OjX  .  c9x Oz ax )(ox 27  avxi av 3  .  7 X 1 uZ Oi 3  vXl OX 0  .  )C.20) 2 Ox  All of the terms in equation (C.20) cancel indicating that the last two terms in the force equation (C.16) have no dependence on the integration constant. Consider now the first term in equation (C.16). Expanding ha , 7  uzaOvz in a similar  fashion to that done for equation (C.20), one finds that 3  ,Ouz’vz 7 ha  =  ) 2 4Gtln(f  f Ox —  7+ Ox+i  .  Ovxi+i  .  Ouz  7 Oiix  —  =  aix. OZ  4 is equivalent to i  .  7 2 oz a  Ox X+8X  U27 9 OU2  —  ojx•ovxj where i  Ox+i  =  1. In this equation, all of the  J  (C 21)  terms correspond  to 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 being considered, 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 to the back-reaction from the terms in the force equation which depend on the integration  Appendix C. Gauge Independence of the Left-Moving Delta Functions  121  constant. Although the independence of equation (C.16) has oniy been shown for the force at u  =  —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 upon the arbitrary integration constants, only the right-moving ones do. This means that in those regions where there is no right-moving delta force, such as at —100 < u < —3 and u> 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 fea ture. The two terms, O’ and OO’, which contain contributions to the delta func tion, cannot cancel out in this manner. The cancellations no longer work because of the presence of extra 9 z and 9z terms and also because of the lack of a term equivalent to the third term in equation (C.16). Therefore, in this respect, the force which still includes components parallel to the string contains less ambiguity than the force in which they have been removed. This is yet another indication that the perpendicular force will not be a great improvement over the full force.  ‘For values of u > 100 and u < —100, the delta functions from the left-moving wave have zero amplitude and so do not contribute to the force at all.  

Cite

Citation Scheme:

        

Citations by CSL (citeproc-js)

Usage Statistics

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.831.1-0085445/manifest

Comment

Related Items