PHYSICS ESSAYS 25, 3 (2012)
Relativistic mechanics in multiple time dimensions Milen V. Veleva) Ivailo Street, No. 68A, 8000 Burgas, Bulgaria (Received 19 March 2012; accepted 28 June 2012; published online 21 September 2012) Abstract: This article discusses the motion of particles in multiple time dimensions and in multiple space dimensions. Transformations are presented for the transfer from one inertial frame of reference to another inertial frame of reference for the case of multidimensional time. The implications are indicated of the existence of a large number of time dimensions on physical laws like the Lorentz covariance, CPT symmetry, the principle of invariance of the speed of light, the law of addition of velocities, the energy-momentum conservation law, etc. The Doppler effect is obtained for the case of multidimensional time. Relations are derived between energy, mass, and momentum of a particle and the number of time dimensions in which the particle is moving. The energy-momentum conservation law is formulated for the case of multidimensional time. It is proven that if certain conditions are met, then particles moving in multidimensional time are as stable as particles moving in one-dimensional time. This result differs from the view generally accepted until now [J. Dorling, Am. J. Phys. 38, 539 (1970)]. It is proven that luxons may have nonzero rest mass, but only provided that they move in multidimensional time. The causal structure of space-time is examined. It is shown that in multidimensional time, under certain circumstances, a particle can move in the causal region faster than the speed of light in vacuum. In the case of multidimensional time, the application of the proper orthochronous transformations at certain conditions leads to movement backwards in the time dimensions. It is concluded that the number of different antiparticles in the k-dimensional time is equal to 3k 2k. Differences between tachyons and particles moving in multidimensional time are indicated. It is shown that particles moving faster than the speed of light in vacuum can have a real rest mass (unlike tachyons), provided that they move in multidimensional time. Ó 2012 Physics Essays Publication. [DOI: 10.4006/0836-1398-25.3.403]
Resum´ e´: L’article traite du mouvement des particules dans un temps et espace multi- dimensionnels. Les transformations de ref´erentiels´ inertiels d’un systeme` a` l’autre sont deduites´ dans un temps multidimensionnel. Les consequences´ de l’existence d’un plus grand nombre de dimensions temporelles sur les lois physiques sont demontr´ ees:´ sur l’invariance de Lorentz, sur la symetrie´ CPT, sur le principe de l’invariance de la vitesse de la lumiere,` sur la loi d’accumulation des vitesses, sur la loi de conservation de l’energie-impulsion´ etc. L’effet Doppler est obtenu dans un temps multidimensionnel. Les correlations´ entre l’energie,´ la masse, l’impulsion d’une particule donnee´ sont deduites,´ ainsi que le nombre des dimensions temporelles dans lesquelles cette particule se meut. Formulee´ a` et´e´ la loi de conservation de l’energie-impulsion´ en cas de temps multidimensionnel. Il est demontr´ e´ que si ont et´e´ satisfaites certaines conditions, les particules qui se deplacent´ dans un temps multidimensionnel sont tout aussi stables que les particules se depla´ cant¸ dans un temps unidimensionnel. Se resultat´ tranche avec le point de vue adopte´ jusqu’apr` esent´ [J. Dorling, Am. J. Phys. 38, 539–540 (1970)]. Il est demontr´ e´ que les luxons peuvent avoir en repos une masse non egale´ az` ero,´ mais ala` condition qu’ils se meuvent dans un temps multidimensionnel. La structure causale de l’espace-temps est etudi´ ee.´ Il est demontr´ e´ que dans un temps multidimensionnel, dans certaines conditions, une particule peut se mouvoir dans le champ causal plus rapidement que la vitesse de la lumiere` dans du vide. En cas de temps multidimensionnel, l’application des propres transformations orthochrones mene,` dans certaines conditions, a` une marche en arriere` dans la mesure du temps. Nous atteignons la conclusion que le nombre des differentes´ antiparticules dans un temps k-dimensionnel est egal´ a(3` k 2k). Les differences´ entre les tachyons et les particules se mouvant dans un temps multidimensionnel sont montrees.´ Il est demontr´ e´ que les particules qui se meuvent plus rapidement que la vitesse de la lumiere` dans du vide peuvent avoir une masse reelle´ en repos (a` la difference´ des tachyons), mais ala` condition qu’elles se meuvent dans un temps multidimensionnel.
Key words: Multidimensional Time; Special Relativity; Mass-Energy Equivalence; Energy-Momentum Conservation Law; Antiparticles; Tachyons; Lorentz Transformations; Invariance of the Speed of Light. a)[email protected], [email protected]
0836-1398/2012/25(3)/403/36/$25.00 403 Q 2012 Physics Essays Publication 404 Phys. Essays 25, 3 (2012)
I. INTRODUCTION exist if we examined an object moving in multidimen- sional time?’’ The concept of multidimensional time has been In order to find experimental evidence for the introduced and has become more important in contem- existence of particles moving in multidimensional time, porary physical theories.1–5 According to the inflation it is necessary to know their physical properties. As noted theory of the big bang, the visible universe is only a small by Recami in another context—the experimental search part of the multiverse, and it is possible that many other for the hypothetical particles named tachyons—‘‘it is not universes have emerged in which conditions are entirely possible to make a meaningful experiment without a good different from the conditions of our universe.6 Up to now, theory.’’7 no physical principle or law has been found that The main objective of this article is to generalize the determines the possible number of spatial dimensions special theory of relativity (STR) for the cases of and of temporal dimensions (or limits the number of multidimensional time and multidimensional space. There spatial and temporal dimensions to a value which differs is a need to clarify not only the mathematical but also the from the observed number in our universe). Due to this physical meaning of multidimensional time. fact, the number of spatial and temporal dimensions in In this respect, the study raises several basic tasks: our universe is more probably a result of chance than a result of unknown processes acting during the initial deriving transformations for the transition between development phases of the universe. Through the inertial frames of reference for the case where the anthropic principle it is explained that we live in a number of time dimensions is greater than one; universe with more than three dimensions of space (or 10 establishing the implications arising from the dimensions, as predicted by M-theory) but only one existence of a large number of dimensions of time dimension of time. Therefore, in the other universes that on physical laws—the Lorentz covariance, CPT are part of the multiverse it is quite possible that space symmetry, the constancy of the speed of light, the and time have entirely different numbers of dimensions law of addition of velocities, the energy-momentum than the dimensions in our universe. We can assume the conservation law, etc.; existence of universes having two, three, four, or more deriving the Doppler effect for the case of temporal dimensions. The relation between the anthropic multidimensional time; principle and the number of spatial and temporal examining the causal structure of space-time; dimensions is considered by Tegmark.5 Here we do not deriving formulas for momentum and energy for discuss this matter. the case of more than one time dimension; Asshowninsomestudies,1,2 it is possible to establishing the exact relationship between the formulated physically meaningful theories with two time energy of a particle and the number of time dimensions. Bars noted that ‘‘two-time physics could be dimensions in which the particle is moving; viewed as a device for gaining a better understanding of formulating the energy-momentum conservation one-time physics, but beyond this, two- time physics law; offers new vistas in the search of the unified theory while considering antiparticles in multidimensional time; raising deep questions about the meaning of spacetime.’’2 and For systems that are not yet understood or even distinguishing between tachyons and particles constructed, such as M-theory, two-time physics points moving in multidimensional time. to a possible approach for a more symmetric and more The problem with the generalization of STR for the revealing formulation in 11 þ 2 dimensionsa that can lead case of multidimensional time is still not sufficiently to deeper insights, including a better understanding of studied and is only briefly mentioned in different studies space and time. The two-time physics approach could be concerning the topic. The consequences on physical laws one of the possible avenues to construct the most of the existence of multidimensional time have also not symmetric version of the fundamental theory.1,2 been well studied. Up to now there has been no As noted by Tegmark, ‘‘Even when m . 1, there is no distinction between tachyons and particles moving in obvious reason why an observer could not, none the less, multidimensional time. perceive time as being one-dimensional, thereby main- taining the pattern of having ‘thoughts’ in a one- dimensional succession that characterizes our own reality II. GENERAL CONSIDERATIONS perception. If the observer is a localized object, it will Important for this study is following question: Are travel along an essentially one-dimensional (timelike) there physical arguments and grounds allowing general world line through the (n þ m)-dimensional space-time conclusions concerning the dimension of time? Related to manifold.’’5,b Thus it is fully reasonable to ask the this question is another: Is Minkowski space-time real question ‘‘What relations, effects, and features would and should we accept time as the fourth dimension, given the fact that STR can be equally formulated in a three- a Two-time physics introduces one additional space dimension and one additional time dimension. dimensional or a four-dimensional language? As noted by b Here m is the number of time dimensions and n is the number of Petkov, of course, we have to solve this issue before space dimensions. seriously discussing a theory involving a large number of Phys. Essays 25, 3 (2012) 405
8 dimensions (of space or of time). This question is accordingly, dr ¼ (]r/]t1)dt1 þ (]r/]t2)dt2 þ þ(]r/]tk)dtk, important because later in this article, additional time where r ¼ (xkþ1, xkþ2, ..., xkþn) is the radius vector, dimensions are introduced. The arguments applying to defining the position of the particle. Then we have xg ¼ one time dimension will be valid for two, three, or more xg(t1, t2, ..., tk) and dxg ¼ (]xg/]t1)dt1 þ (]xg/]t2)dt2 þ time dimensions—i.e., they will be applicable also to these þ (]xg/]tk), where g ¼ k þ 1, k þ 2,..., k þ n. In the general cases. It has been shown that the block universe view, case, each one of the functions xg ¼ xg(t1, t2, ..., tk) must regarding the universe as a timelessly existing four- be represented by a k-dimensional hypersurface in the (k dimensional world, is the only one that is consistent with þ 1)-dimensional space-time t1, t2, ..., tk, xg—that is, it special relativity.9 Some arguments have been made in will not be presented as a one-dimensional world line. favor of the statement that special relativity alone can Therefore, the functions xg ¼ xg(t1, t2, ..., tk) could not resolve the debate on whether the world is three- describe the movement of a pointlike particle in the space- dimensional or four-dimensional. If the world were time. Each one-dimensional world line in the (k þ 1)- three-dimensional, the kinematic consequences of special dimensional space-time t1, t2, ..., tk, xg is defined through relativity—and, more importantly, the experiments con- a system of k equations: F1g(t1, t2, ..., tk, xg) ¼ 0, F2g(t1, 9,10 firming them—would be impossible. Therefore, time is t2, ..., tk, xg) ¼ 0,..., Fkg(t1, t2, ..., tk, xg) ¼ 0. (During indeed an extra dimension and it is fully justified and the uniform and rectilinear movement of the particle, reasonable to set and examine the issue of the dimen- which can be presented through a straight world line, the sionality of time. functions F1g, F2g, ..., Fkg are linear.) From here we can The interval in Minkowski space-time is an invariant derive the equalities xg ¼ f1g(t1) ¼ f2g(t2) ¼ ¼ fkg(tk), 2 2 2 2 2 2 and is given by the expression ds ¼ c dt dx dy dz . where f1g, f2g, ..., fkg are different (linear) functions of the But instead of the four-dimensional space-time of Min- variables t1, t2, ..., tk, respectively. Thus, we have dxg ¼ 0 kowski, we can consider a general five-dimensional space- fhg(th)dth, where h ¼ 1, 2, . . ., k. For the case where the time (dx, dy, dz, cdt, dn). Here the fifth dimension dn reflects particle moves along a straight world line, the derivatives f 0 (t ), f 0 (t ), , f 0 (t ) are constants which define the the proper time cdt0 or the proper length dl0 or is equal to 0. 1g 1 2g 2 kg k (In Ref. 11 a similar model is considered, but the fifth velocities of the particle in relation to t1, t2, ..., tk, 0 respectively. (We will find that f (th) ¼ dxg/dth ¼ Vhg; see dimension correspondspffiffiffiffiffiffiffi to the proper timepffiffiffiffiffiffiffiffiffifficdt0.) By hg 2 2 2 2 the considerations at the end of Section II.) definition, cdt0 ¼ ds if ds 0anddl0 ¼ ds if ds 2 In order to determine the velocities of a given particle 0. Therefore, if ds . 0, the fifth dimension dn ¼ cdt0 is 2 2 for the case of multidimensional time, we are going to use spacelike; and if ds , 0, then dn ¼ dl0 is timelike. If ds ¼ 0, then d ¼ 0islightlike.Ifds2 , 0, then dn (or dn/c)canbe following considerations. Let us consider a particle regarded as a second, additional time dimension. It should moving uniformly and rectilinearly with velocity U in be noted that the fifth dimension is invariant. If the quantity relation to the frame of reference K. Let us assume that at dn is not invariant, then the time is not one-dimensional in the moment T ¼ (t1, t2,...,tk), the location of this particle the usual sense (see Section IV). (A five-dimensional model is defined by the radius vector r ¼ (xkþ1, xkþ2, ..., xkþn), of space-time is used in Sections X and XI.) and at the moment T þ dT ¼ (t1 þ dt1, t2 þ dt2, ..., tk þ The number of time dimensions we will denote with dtk), the location of the particle is defined by the radius k, and that of space dimensions with n. The time vector r þ dr ¼ (xkþ1 þ dxkþ1, xkþ2 þ dxkþ2, ..., xkþn þ dx ). In the case of multidimensional time we will have dimensions themselves we will denote with x ¼ ct , x ¼ kþn 1 1 2 dT 3 U ¼ dr, where dT ¼ (dt , dt , ..., dt )anddr ¼ ct , ..., x ¼ ct , and the space dimensions with x , 1 2 k 2 k k kþ1 (dx , dx , ..., dx ). As can be easily seen, the x , ..., x . kþ1 kþ2 kþn kþ2 kþn velocity U is a k 3 n matrix having elements u (h 1, The metric signature in the case of k-dimensional hg ¼ 2,..., k; g ¼ k þ 1, k þ 2,..., k þ n), i.e., U ¼ [u ] . Let time and n -dimensional space will be hg k3n c us denote ðþ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}; þ; ...; þ ; |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}; ; ...; Þ. Therefore, the (n þ k)- 0 1 k n 2 u1g dimensional interval dsn,k is given by the expression dsn;k B C 2 2 2 2 2 B u2g C ¼ c dt1 þ þ c dt dx . It is clear that the (n þ k)- u B C; k kþn g ¼ @ . A dimensional interval dsn,k is invariant. . In the case of multidimensional time, the velocities of ukg a particle according to different time dimensions cannot where g ¼ k þ 1,kþ 2,..., k þ n. Then we have dT 3 u ¼ be defined as a set of partial derivatives of the g dxg, that is: independent variables t1, t2, ..., tk. Indeed, the movement of a pointlike particle in the general case can be presented dt1 dt2 dtk u1g þ u2g þ þ ukg ¼ 1: ð1Þ as a one-dimensional (timelike) world line in (k þ n)- dxg dxg dxg dimensional space-time. Let us set r ¼ r(t1, t2, ..., tk); It is clear that if for a given d (1 d k) we have dtd ¼ 0, then the components of the velocity ud(k 1), c For the purpose of this article, we will use the timelike convention þ for the metric signature (i.e., we choose positive signs for the squares ud(kþ2),...,ud(kþn) will be undefined quantities. If for a of timelike dimensions and negative signs for the squares of given q (k þ 1 q k þ n) we have dxq ¼ 0, then it spacelike dimensions). follows that u1q ¼ u2q ¼ ¼ukq ¼ 0. 406 Phys. Essays 25, 3 (2012) P k Let us set uhg ¼ khg(dxg/dth), where h¼1 khg ¼ 1 and g invariant when a change of the time-axes basis of the ¼ k þ 1, k þ 2,..., k þ n. Let us further denote frame K is made. Thus, the velocity Ug can be presented vffiffiffiffiffiffiffiffiffiffiffiffiffiffi u as Ug ¼ cg(jdxgj/dT). Here cg . 0 is a parameter not uXk t depending on the quantities dth, respectively on the dT ¼ dt2 . 0; h numbers ah (h ¼ 1, 2, . . ., k). h¼1 Since in the expression for the velocity Uh all the vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u space axes of theq frameffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK are represented in the k n P u Xþ kþn 2 t 2 numerator (dX ¼ g¼kþ1 dxg), we accept that the dX ¼ dxg . 0; g¼kþ1 velocity Uh remains invariant when the space-axes basis of the frame K is changed. Thus, the velocity Uh can be presented as Uh ¼ ch(dX/jdthj). Here ch . 0 is a parameter dth ¼ ahdT; not depending on the quantities dxg and consequently on the numbers vg (g ¼ k þ 1, k þ 2,..., k þ n). dxg ¼ vgdX: P P From these considerations we can conclude that the k 2 kþn Then we have h¼1 ah ¼ 1—that is, jahj 1—and g¼kþ1 velocity U is invariant when the space-axes basis and the 2 vg ¼ 1—that is, jvgj 1. time-axes basis of the frame K are changed—i.e., the Let us set velocity U can be presented as U ¼ c(dX/dT). Here c . 0 vffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi is a parameter not depending on the values dt and dx u u h g uXk uXk 2 and consequently on the numbers a and v [h 1, t jdxgj t khg h g ¼ U ¼ u2 ¼ ; g hg dT a 2, . . ., k; g ¼ k þ 1, k þ 2, . . ., k þ n;seeEq.(2)]. Indeed, h¼1 h¼1 h vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u u vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u Xkþn u Xkþn uXk u u t 2 dX t 2 2 t 2 u Xkþn u Xkþn U ¼ Ug ¼ vgcg ¼ Uh t dX t 2 dT U ¼ u2 ¼ ðk v Þ ; g¼kþ1 g¼kþ1 h¼1 h hg jdt j hg g g¼kþ1 h g¼kþ1 vffiffiffiffiffiffiffiffiffiffiffiffiffi u uXk vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dX t c2 u h: u Xkþn Xk ¼ 2 t dT ah U ¼ u2 : h¼1 hg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g¼kþ1 h¼1 P P kþn 2 2 k 2 2 Let us set c ¼ g¼kþ1 vgcg ¼ h¼1ðch=ahÞ.Sincethe For the velocity U we have values v and c do not depend on the numbers a , vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffi g g qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h u u kþn 2 2 u Xkþn uXk neither does the parameter c ¼ v c .Sincethe t t g¼kþ1 g g U ¼ U2 ¼ U2 g h values ah and ch do not dependqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi on the numbers vg, g¼kþ1 h¼1 P vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi neither does the parameter c ¼ k ðc2=a2Þ. Thus, the u h¼1 h h u Xkþn Xk 2 dX t khgvg parameter c does not depend on ah or vg,and ¼ : ð2Þ dT a consequently does not depend on the values dth or dxg. g¼kþ1 h¼1 h pffiffiffi Let us assume that a1p¼ffiffiffia2 ¼ ¼ak ¼ 1/ k (that is, dt ¼ dt ¼ ¼dt ¼ dT/ k . 0) and v ¼ v ¼ ¼ We will show the conditions for the velocities Ug, Uh, 1 2pffiffiffi k kþ1 kþ2 pffiffiffi and U. These conditions are imposed due to physical vkþn ¼ 1/ n (that is, dxkþ1 ¼ dxkþ2 ¼ ¼ dxkþn ¼ dX/ n . considerations (see also Subsection IX.A). When the time- 0). Let us apply a proper or improper rotation of the time axes basis of the frame K is changed (these are the so- axes of the frame K in the hyperplane of time. The called passive linear transformations) and the value of transformation under consideration can be presented 2 2 2 2 2 through the orthogonal matrix A [a ] , belonging dT is still the same, then the quantity dX ¼ c dT dsn;k ¼ h1 k3k remains the same as well. Likewise, when the basis of the to the orthogonal group O(k, R), where R denotes the real tr 1 space axes of the frame K is changed, then the values dX2 numbers field (see also Section VII). (We have A ¼ A , 2 2 2 2 det(A) ¼ 61. Here Atr denotes the transpose of the matrix and c dT ¼ dsn;k þ dX remain the same. It is clear that the choice of time axes of the frame K can be made A.) The new time axes, obtained after applying this 0 independently from the choice of space axes of the frame transformation,P weP will denote with th,(h ¼ 1, 2, . . ., k). k 0 2 k 2 K, and vice versa (see also Section VII and Subsection Since h¼1ðahÞ ¼ h¼1 ah ¼ 1, we have 0 1 0 1 IX.A). Thus, when these transformations are applied, 0 a1 a1 some physical quantities (in our case the velocities) must B 0 C B C B a2 C B a2 C remain invariant. B . C ¼ A 3 B . C; @ . A @ . A Since in the expression for the velocity Ug all the time 0 axes of the frame K are represented in the denominator ak ak qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 (dT ¼ h¼1 dth), we accept that the velocity Ug is that is, Phys. Essays 25, 3 (2012) 407 ! Xk Xk Xkþn Xk Xk 0 1 00 1 ah1a1bhmkmqvqqqg a ¼ ah1a1 ¼ pffiffiffi ah1: ð3Þ k ¼ h hg Xkþn a 1¼1 k 1¼1 q¼kþ1 m¼1 1¼1 m vphpg p¼kþ1 Since the velocity Ug remains invariant when the Ptime-axes basis is changed for the frame K, the expression Xkþn Xk Xk k 2 ah1bhmkmqqqg h¼1ðkhg=ahÞ is constant when there is a change in the q¼kþ1 m¼1 1¼1 values of ah—that is, ¼ : ð10Þ Xkþn Xk 0 2 Xk 2 Xk hpg khg khg 2 2 p k 1 0 ¼ ¼ k khg ¼ cg; ð4Þ ¼ þ ah ah pffiffiffi h¼1 h¼1 h¼1 Here the values k are defined provided that a 1/ k, pffiffiffi mq h ¼ where cg is a parameter not depending on the numbers ah vg ¼ 1/ n are fulfilled. Since the parameter c doespffiffiffi not depend of the numbers a or v , we can set a 1/ k, v (h ¼ 1, 2, . . ., k). pffiffiffi h g h ¼ g ¼ Let us consider the orthogonal matrix B ¼ [bhm]k3k, 1/ n. In this case, we have vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi belonging to the orthogonal group O(k, R). Taking into u rffiffiffiu u Xkþn Xk 2 u Xkþn Xk account Eq. (4), we likewise have t khgvg kt c ¼ ¼ k2 : ð11Þ a n hg 0 g k 1 h 1 h g k 1 h 1 k Xk b k pffiffiffi Xk ¼ þ ¼ ¼ þ ¼ hg ¼ hm mg ¼ k b k : ð5Þ 0 hm mg [See Eq. (2).] ah m 1 am m 1 P ¼ ¼ 00 k 00 Let us set Kg ¼ h¼1 khg, g ¼ k þ1,kþ2, kþ n [see Eq. From Eqs. (3) and (5) we can define the values k0 : 0 hg (10)]. In the general case (i.e., at arbitrary values of ah and 00 Xk Xk Xk Xk vg ), we will have 0 ah1a1bhmkmg k ¼ ¼ a b k : ð6Þ 00 hg a h1 hm mg dT 3 UK ¼ dr; m¼1 1¼1 m m¼1 1¼1 Let us apply a proper or improper rotation of the space where axes of the frame K in the hyperplane of space. The hi 00 00 K 00 K 00 K 00 khgvg dX transformation under consideration can be presented U ¼ uhg ; uhg ¼ 00 0 : k 3 n Kg ahdT through the orthogonal matrix H ¼ [hpg]n3n, belonging P P k 00 k to the orthogonal group O(n, R). Here g ¼ k þ 1, k þ [See Eqs. (1), (3), (7), and (10).] If h¼1 khg 6¼ h¼1 khg 00 2,..., k þ n; p ¼ k þ 1, k þ 2,..., k þ n. The new space (that is, Kg 6¼ 1, g ¼ k þ 1, k þ 2, , k þ n), then axes, obtained after applying this transformation, we will 00 00 00 00 P 00 k v dX k v dX 00 kþn uK ¼ hg g 6¼ u00 ¼ hg g : denote withP xg (g ¼ k þ 1, k þ 2,..., k þ n). Since g¼kþ1 hg 00 0 hg 0 00 2 kþn 2 00 00 00 Kg ahdT ahdT ðvg Þ ¼ g¼kþ1 vg ¼ 1, we have (vkþ1, vkþ2, vkþn ¼ vkþ1, vkþ2, , vkþn) 3 H—that is, Let us denote dxg/dth ¼ uhg/khg ¼ Vhg. It is clear that Xkþn 1 Xkþn v00 ¼ v h ¼ pffiffiffi h : ð7Þ dxg ¼ V1g dt1 ¼ V2g dt2 ¼ ¼Vkg dtk. Let us denote Vh ¼ g p pg n pg p¼kþ1 p¼kþ1 (dr/dth), where dr ¼ (dxkþ1, dxkþ2, ..., dxkþn). Then we have Vh ¼ [Vh(kþ1), Vh(kþ2), ..., Vh(kþn)]. We will say that Since the velocity U is invariant when the space-axes h P Vh is the velocity of the particle under consideration in kþn 0 2 basis of K is changed, the expression g¼kþ1ðkhgvgÞ relation to the frame K, defined in relation to the time remains constant when the values of vg are changed—that dimension th. Let us set Vh ¼jjVhjj ¼ dX/jdthj¼Uh/ch . 0. is, Then the equation jdthj/jdt1j¼V1j/Vh is fulfilled, where h, 1 ¼ 1, 2, . . ., k and dX 6¼ 0. Xkþh Xkþn 1 Xkþn ðk00 v00 Þ2 ¼ ðk0 v Þ2 ¼ k0 ¼ c2: ð8Þ Let us set u ¼ dX/dT ¼ U/c. We will say that u is the hg g hg g n hg h g¼kþ1 g¼kþ1 g¼kþ1 total coordinate velocity of the considered particle in relation to K. It is easy to see that Here ch is a parameter not depending on vg (g ¼ k þ 1, k þ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2,..., k þ n). u ¼ u 0 1 ¼ u : Let us consider the orthogonal matrix Q ¼ [q ] , u uXk qg n3n u B C t 1 belonging to the orthogonal group O(n, R). Taking into uXk B C 2 u B 1 C Vh account Eq. (8), we likewise have u B C h¼1 u @Xn A t h¼1 V2 Xkþn 1 Xkþn hg k00 v00 ¼ k0 v q ¼ pffiffiffi k0 q : ð9Þ g¼1 hg g hq q qg n hq qg q¼kþ1 q¼kþ1 We will say thatP a particle is at rest relative to the kþn 2 From Eqs. (6), (7), and (9) we can define the values frame of reference if g¼kþ1 dxg ¼ 0andVh ¼ 0(h ¼ 1, 00 khg: 2,..., k). 408 Phys. Essays 25, 3 (2012)
We assume that all time dimensions we are going to mations with no rotation)—see, for example, Ref. 12. consider—t1, t2, ..., tk (or t, s, . . .)—are homogeneous First we will consider a proper rotation in the plane x1-x2 (i.e., all moments of a given time dimension are equal), through angle a, where the other three dimensions (x3, x4, and that space is homogeneous and isotropic (i.e., all x5) remain invariant. This transformation is described by points and all directions of the space are equal). the matrix 0 1 cos a sin a 000 III. TRANSFORMATIONS FROM ONE INERTIAL B C B C B sin a cos a 000C FRAME OF REFERENCE TO ANOTHER FOR THE B C R ¼ B 0 0 100C: CASE OF (n, k) ¼ (3, 2) B C @ 0 0 010A A. Derivation of the transformations 0 0 001 First, we will consider case with (n, k) ¼ (3, 2), i.e., three space dimensions and two time dimensions. The Let us denote with x1R and x2R the new axes which three space dimensions we will denote with x, y, z. The arise from the rotation R of the axes x1 and x2, first time dimension we will denote with t, and the second respectively. Then we will consider a proper rotation in with s. the plane x1R-x3 through angle u, where the other three The interval in the five-dimensional space-time under dimensions (x2R, x4, x5) remain invariant. This transfor- 2 2 2 consideration is given by the expression ds3;2 ¼ c dt þ mation is described by the matrix 2 2 2 2 2 c ds – dx dy dz . The interval ds3,2 is invariant. 0 1 0 cos u 0 sin u 00 Let us consider the two inertial frames K and K B C B C (moving uniformly and rectilinearly to each other). We B 0 1000C 0 B C assume that the velocity of the frame K against K, defined L ¼ B sin u 0 cos u 00C: in relation to the first time dimension t, is equal to the B C @ 0 0010A vector v, and that defined in relation to the second time dimension s it is equal to the vector w (see Section II). 0 0001 Let us denote with x, y, and z the axes of the frame K, In order to derive the transformations between K and 0 0 0 0 and with x , y , and z the axes of the frame K . The two K0, we will consecutively apply the operations R, L,and time dimensions defined in the frame K we will denote R 1. 0 0 0 with t and s; in the frame K , with t and s . Let us denote First, we will apply the R transformation. As can be with point Q the origin of the spatial frame of reference K easily seen, tan a ¼ x /x .Ifx0 ¼ 0, then x ¼ i(v/c)x ¼ (that is, x ¼ 0, y ¼ 0, z ¼ 0), and with point Q0 the origin of 2 1 3 3 1 i(w/c)x2 . 0 and thus tan a ¼pv/ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiw. Here the angle a is a the spatial frame of reference K0 (that is, x0 ¼ 0, y0 ¼ 0, z0 ¼ real number. Let us set b ¼ 1/ ðc2=v2Þþðc2=w2Þ, f ¼ 1/ 0). We choose the frames K and K0 in such a way that pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 b2. (We will have 0 b 1 and f 1; see also point Q0 is moving along the axis x in the direction of Section IV.) We have cos a cb/v, sin a cb/w. increasing values of x. Further, we can choose the axes of ¼ ¼ the frames K and K0 in such a way that for an observer It is clear that if v ¼ 0, then w ¼ 0, and vice versa. We connected to K, the axis x coincides with the axis x0; the will consider more specific cases in relation to the value of axes y and z are parallel to the axes y0 and z0, respectively; a. Let us assume that a ¼ bp, where b ¼ 0, 61, 62,... and the homonymous axes have the same direction. As (i.e., the motion occurs only along the axis x1 ¼ ict). It is the initial moment we accept (t, t0), (s, s0), where point Q0 clear that if v 6¼ 0, then w ¼ 6‘. If we assume that a ¼ p/2 coincides with point Q (i.e., at the moments t ¼ t0 ¼ 0 and þ bp (i.e., the motion occurs only along the axis x2 ¼ ics), s ¼ s0 ¼ 0, point Q0 ” point Q). Having all these it is clear that if w 6¼ 0, then v ¼ 6‘. conditions fulfilled, we can say for K and K0 that they are We have accepted that point Q0 is moving in the in a standard configuration. Let us set v ¼ (v, 0, 0) and w ¼ direction of increasing values of x, which means that if x0 (w, 0, 0), where (v, 0, 0) are the respective projections of ¼ 0, then x ¼ vt ¼ ws . 0. We have the following: If a [0; the velocity vector v on the axes x, y, z of the frame K and p/2], then t 0ands 0, and therefore v . 0andw . 0; (w, 0, 0) are the respective projections of the velocity if a (p/2; p], then t , 0 and s 0, and therefore v , 0 vector w on the axes x, y, z of K. Let us assume that a and w . 0; if a (p;3p/2), then t , 0 and s , 0, and particle has coordinates (t, s, x, y, z)inK and (t0, s0, x0, y0, therefore v , 0 and w , 0; if a [3p/2; 2p), then t 0and z0)inK0. s , 0, and therefore v . 0 and w , 0. x ict x ic x x x y x z Let us denote 1 ¼ , 2 ¼ s, 3 ¼ , 4 ¼ , 5 ¼ Let us now apply the transformationqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiL.Itcanbe and x0 ¼ ict0, x0 ¼ ics0, x0 ¼ x0, x0 ¼ y0 , x0 ¼ z0. In order to 1 2 3 4 5 easily seen that tan u ¼ x /x ¼ x / x2 þ x2.Further, derive the transformations between K and K0, we will use 3 1R 3 1 2 0 the same approach as for the Lorentz transformations in if x3 ¼ 0, then x3 ¼ i(v/c)x1 ¼ i(w/c)x2 . 0. Therefore, a more general case, the so-called Lorentz boost in an we have tan u ¼ ib.Heretheangleu is an imaginary arbitrary direction (transformations between two inertial number. We have sin u ¼ ibf,cosu ¼ f.Thesignsin frames of reference whose x, y, z axes are parallel and these expressions are chosen so that when v 0andw whose space-time origins coincide, i.e., Lorentz transfor- 0, we have t0 t, s0 s,x0 x, y0 y, z0 z. Phys. Essays 25, 3 (2012) 409
Finally, let us apply the transformation 0 1 cos a sin a 000 B C B sin a cos a 000C 1 B C R ¼ B 0 0 100C; @ 0 0 010A 0 0 001 which is the inverse of the first transformation R. The matrix for the transfer from the coordinates x1, x2, x3, x4, x5 to 0 0 0 0 0 x1,x2, x3, x4, x5 is equal to the product 0 1 1 þðcos u 1Þcos2 a ðcos u 1Þsin a cos a cos a sin u 00 B 2 C B ðcos u 1Þsin a cos a 1 þðcos u 1Þsin a sin a sin u 00C 1 B C R 3 L 3 R ¼ B cos a sin u sin a sin u cos u 00C: @ A 00010 00001 Let us express sin a, cos a, sin u, cos u through v/c and w/c and go back to the old coordinates t, s, x, y, z, t0, s0, x0, y0, z0. Then finally we obtain the following expression: 0 1 c2 c2 1 B 1 þðf 1Þ b2 ðf 1Þ b2 b2f 00C B v2 vw v C 0 1 B C0 1 0 B C t B c2 c2 1 C t B 0 C B ðf 1Þ b2 1 þðf 1Þ b2 b2f 00CB C B s C B 2 CB s C B 0 C B vw w w CB C B x C ¼ B CB x C: ð12Þ @ y0 A B c2 c2 C@ y A B b2f b2ff00C 0 B C z B v w C z @ 00010A 00001
0 00 00 00 00 We can easily prove that if v ¼ 0 and w ¼ 0, then t ¼ t, transformation D between K(t, s, x, y, z)andK (t1 , t2 , x , s0 ¼ s, x0 ¼ x, y0 ¼ y, z0 ¼ z. The transformations in Eq. (12) y00 , z00 ) is obtained by multiplying the matrix K from Eq. belong to the group of proper orthochronous transfor- (12) for the transfer from K to K0 by the matrix M; that is, mations, which we will denote with Kþ (see Section D ¼ M 3 K. III.C). The transformations in Eq. (12) are equivalent to Let us apply a (proper or improper) rotation in the 0 0 00 2 00 2 0 2 the Lorentz transformations as s 0 and accordingly as plane t -s . We obtain the equation (dt1 ) þ (dt2 ) ¼ (dt ) 0 2 00 0 0 00 0 w 6‘. þ (ds ) . Here dt1 ¼ dt cos r þ ds sin r, dt2 ¼ edt sin r þ 0 00 Let us set u ¼ iU. Taking into account the fact that eds cos r, where e ¼ 61. It is easy to prove that if dt1 ¼ 00 0 0 0 0 cos u ¼ cos(iU) ¼ cosh U and sin u ¼ sin(iU) ¼ i sinh U,we dt2 . 0, then r ¼ a e(p/4) þ 2bp, where tan a ¼ ds /dt , b obtain cosh U ¼ f and sinh U ¼ bf. ¼ 0, 61, 62, . . . Since the vector dT0 ¼ (dt0, ds0) makes an 0 0 0 0 00 In the plane of time t -s , one can apply a proper or angle a with the axis t , the axis t1 makes an angle r with 0 00 an improper rotation. These transformations present the axis t , and the axis t2 makes an angle r þ e(p/2) with change of the time-axes basis of the frame K (the so- the axis t0, the vector dT0 makes an angle a0 r ¼ e(p/4) 00 0 called passive linear transformation). The orthogonal 2bp with the axis t1 and an angle a [r þ e(p/2)] ¼ e(p/4) 0 0 00 matrix M for the transfer from the old coordinates t , s , 2bp with the axis t2 . The size of the angle between the 0 0 0 00 00 00 00 00 0 00 00 x , y , z to the new coordinates t1 , t2 , x , y , z ,is vector dT and each of the axes t1 and t2 is equal to p/4 0 1 (see Section VII). cos r sin r 000 B C Let us consider two events which are causally e sin recos r 000 2 2 2 2 2 B C connected, i.e., for which Ds2 ¼ c Dt þ c Ds – Dx M ¼ B 0 0 100C; 3;2 B C Dy2 Dz2 0 (see Section IV). From Eq. (12) it is seen @ 0 0 010A that if Dt ¼ 0,Dx ¼ 0,Dy ¼ 0,Dz ¼ 0, then Dt0 ¼ (f 1)(c2/ 0 0 001 vw)b2Ds and Dx0 ¼ (c2/w)b2fDs. Thus, if two events are where e ¼þ1 (proper rotation) or e ¼ 1 (improper causally connected and if in an inertial reference frame K rotation, rotary reflection). The following equations are the coordinates of these events, defined according to x, y, 00 2 00 2 0 2 0 2 00 0 00 0 00 0 0 fulfilled: (t1 ) þ (t2 ) ¼ (t ) þ (s ) , x ¼ x , y ¼ y , z ¼ z . z, t, coincide, then in another inertial frame K the It is clear that for these transformations, the five- coordinates of the events according to x0, y0, z0, t0 cannot 2 0 dimensional interval ds3;2 is invariant. If a proper or coincide. In this case it is possible that Dt 6¼ 0 and thus improper rotation is applied in the plane of time t0-s0, the Dx0 6¼ 0 (provided that Ds 6¼ 0 and w 6¼ 6‘). 410 Phys. Essays 25, 3 (2012)
B. Backwards motion in the two-dimensional time b2f2 þ 4f 4 , 0: ð15Þ 2 2 2 2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi In STR, the condition s ¼ c t x y z . 0, t . Taking into consideration that f ¼ 1/ 1 b2, we find 0 cannot be changed through proper orthochronous that at b 1, the expression in Eq. (15) tends toward ‘. Lorentz transformations L into the condition s2 . 0, t þ Therefore, if r ¼ c/v ¼ f/2(f 1) and b 1 (that is, fp ffiffiffi ‘ , 0. If it could be done, then on account of continuity one and r 1/2), then Dt0/Dt ‘. Further, if b ¼ (2 2)/3 could also have t ¼ 0—which is impossible if s2 . 0. For 2 (that is, f ¼ 3andr ¼ 3/4), then the expression in Eq. (15) this reason, the region s . 0, t . 0 (inside of the positive becomes equal to 0. It is easy to find that if r [1/2; 3/4), light cone) is called the absolute future. Similarly, the then Eq. (14) is fulfilled and therefore Dt0/Dt , 0, which region s2 . 0, t , 0 (inside of the negative light cone) is we wanted to prove. If r ¼ 3/4, then Dt0/Dt ¼ 0—that is, called the absolute past relative to t ¼ 0. These Dt0 ¼ 0—and if r 1/2, then Dt0/Dt ‘—that is, Dt0 considerations are not valid for the case of multidimen- ‘. sional time. For example, for the case k ¼ 2, the condition According to Eq. (12), the following equality will be s2 ¼ c2t2 þ c2s2 – x2 – y2 – z2 . 0, t . 0, s . 0 can be 3;2 valid: changed through proper orthochronous transformations K (see Subsection III.C) into the condition s2 . 0, t 2 2 þ 3;2 0 c 2 c 2 2 Ds ¼ðf 1Þ b Dt þ 1 þðf 1Þ b Ds 0, s 0, the condition s3;2 . 0, t 0, s 0, or the vw w2 2 condition s3;2 . 0, t 0, s 0. Indeed, since t and s are 1 b2fDx: ð16Þ independent variables, it is possible that following w conditions are simultaneously fulfilled: t ¼ 0, s 6¼ 0, c2s2 x2 y2 z2 . 0. Thus the equality t ¼ 0 does not Since Ds ¼ 0 and Dx ¼ cDt,wehave contradict the inequality s2 . 0. Due to the same 2 3;2 0 c 2 c 2 considerations, the equality s ¼ 0 also does not contradict Ds ¼ðf 1Þ b b f Dt: 2 vw w the inequality s3;2 . 0. The application of the transformations in Eq. (12), Let us assume that w . 0. It is easy to prove that if c/v ¼ f/ which belong to the group of the proper orthochronous 2(f 1), then (f 1)(c2/vw)b2 (c/w)b2f , 0. Therefore, in 0 transformations Kþ , at certain conditions leads to this case we have Ds , 0. If b 1 and accordingly f movement backward in the time dimensions t and s. Let ‘, r 1/2, then Ds0/Dt ‘—that is, Ds0 ‘. 2 2 2 2 2 2 2 2 0 us assume s3;2 ¼ c Dt þ c Ds – Dx – Dy – Dz 0. We Let us now assume that Ds . 0. In this case Dt is accept that Dx . 0, Dy ¼ 0,Dz ¼ 0, Dt . 0,Ds 0. given by Eq. (13) and Ds0 is given by Eq. (16). Further, According to Eq. (12), following equality will be fulfilled: 2 from the condition Ds3;p2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 (and Dy ¼ 0, Dz ¼ 0) it follows that Dx c Dt2 þ Ds2.LetussetDx ¼ c2 c2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 2 2 vc Dt2 þ Ds2, where 0 v 1. We are going to examine Dt ¼ 1 þðf 1Þ 2 b Dt þðf 1Þ b Ds v vw for which values of the velocities v and w the conditions 1 b2fDx: ð13Þ Dt0 , 0 and Ds0 , 0 are fulfilled. We accept that r ¼ c/v ¼ v f/2(f 1) and r [1/2; 3/4). In this case the following We assume that v . 0. Further, we will examine for inequalities are fulfilled: 1 þ (f 1)(c2/v2)b2 (c/v)b2f , 0 0 which values of the velocities v and w the condition Dt , and (f 1)(c2/vw)b2 (c/w)b2f , 0. These two expressions 0 is fulfilled. tend toward ‘ at b 1, and accordingly f ‘, r 1/2. 2 First, we will assume, that Ds ¼ 0. Since Ds3;2 0, we Since Dt, Ds, and Dx are independent variables, we can will have Dx cDt. Let us set Dx ¼ cDt. In this case the choose Dt large enough, Ds small enough, and v close 0 inequality Dt , 0 is equivalent to the inequality enough to 1 that the following expressions take arbitrarily Dt0 c2 c small values: ¼ 1 þðf 1Þ b2 b2f , 0: ð14Þ Dt v2 v c2 Ds ðf 1Þ b2 ; If we set r ¼ c/v, then we obtain a quadratic inequality vw Dt in relation to the parameter r. Since in the expressions for rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 2 b and f there are two independent variables c/v and c/w, c 2 Ds 2 2 b f 1 v 1 þ ; for which the only restriction imposed is (c/v) þ (c/w) v Dt2 1 (see Section IV), we can set (c/v)2 þ (c/w)2 ¼ const 1 and therefore b ¼ const 1 and f ¼ const 1. Further, 2 we can find the first and second derivatives of the function c 2 Ds 1 þðf 1Þ 2 b ; f(r) ¼ (f 1)b2r2 b2fr þ 1. Then we discover that the w Dt function f(r) has a minimum at r ¼ f/2(f 1) and f . 1. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! This means that at a fixed value of 1 b . 0 (and c Ds2 b2f 1 v 1 þ : accordingly of f . 1), the function f(r) reaches its w Dt2 minimum at r ¼ f/2(f 1). Let us set r ¼ c/v ¼ f/2(f 1) and f . 1. According to the inequality in Eq. (14),we Therefore, if r [1/2; 3/4), then for appropriate values of have Dt, Ds, and Dx the following inequalities will be fulfilled: Phys. Essays 25, 3 (2012) 411
TABLE I. Some values of Dt0 and Ds0, provided that Dt ¼ 1, Ds ¼ 0.3, v ¼ 0.999. c c bfr ¼ Dt0 Ds0 v w 0.799 1.663 1.254 Imaginary number Complex number Complex number 0.800 1.667 1.250 0.000 0.276 0.300 0.840 1.843 1.093 0.472 0.320 0.007 0.841649 1.852 1.087 0.480 0.320 0.000 0.950 3.203 0.727 0.761 0.189 0.549 0.960 3.571 0.694 0.776 0.142 0.659 0.970 4.113 0.661 0.791 0.071 0.813 0.976556 4.645 0.637 0.802 0.000 -0.958 0.980 5.025 0.624 0.807 0.051 1.060 0.990 7.089 0.582 0.825 0.336 1.594 0.999 22.366 0.523 0.853 2.487 5.384 0.999999 707.107 0.501 0.866 99.434 173.329
2 2 X5 c 2 c 2 c 2 Ds 1 þðf 1Þ b b f þðf 1Þ b ðx1Þ2 þðx2Þ2 ðxgÞ2 ¼ðx10Þ2 þðx20Þ2 v2 v !vw Dt rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g¼3 2 c 2 Ds X5 þ b f 1 v 1 þ , 0; 2 v Dt2 ðxg0Þ : ð17Þ g¼3 c2 c c2 Ds The general relations which fulfill this condition must f 1 b2 b2f 1 f 1 b2 ð Þ þ þð Þ 2 have the form vw wrffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!w Dt 2 l l q l c 2 Ds x = ¼ a x þ b ; ð18Þ þ b f 1 v 1 þ , 0; q w Dt2 where l, q ¼ 1, 2, 3, 4, 5.d Here bl are five constant values, l l that is, which are equal to the values of x 0 for the case when x ¼ 0(l ¼ 1, 2, 3, 4, 5)—that is, bl is a five-dimensional vector 2 2 0 c 2 c 2 of translation in the space-time. If the origins of both Dt ¼ 1 þðf 1Þ b Dt þðf 1Þ b Ds l pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv2 vw reference systems are the same, then b ¼ 0(l ¼ 1, 2, 3, 4, c b2fv Dt2 þ Ds2 , 0; 5). We will further examine the transformations that do v not include translations in space and time, i.e., c2 c2 xl0 ¼ alxq: ð19Þ Ds0 ¼ðf 1Þ b2Dt þ 1 þðf 1Þ b2 Ds q vwpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi w2 c Let us introduce the notation b2fv Dt2 þ Ds2 , 0: 8 w < 1 l ¼ q ¼ 1; 2 An example of this is shown in Table I. From Table I glq ¼ gql ¼ : 1 l ¼ q ¼ 3; 4; 5 one can see that if b , 0.800, then the value c/w is an 0 l 6¼ q: 0 0 imaginary number and Dt and Ds are complex numbers; In this way, Eq. (17) can be presented in the form if b ¼ 0.800, then Dt0 . 0,Ds0 . 0; if b ’ 0.841649, then l q l0 q0 Dt0 . 0,Ds0 ¼ 0; if b ¼ 0.950, then Dt0 . 0,Ds0 , 0; if b ’ glqx x ¼ glqx x : ð20Þ 0 0 0 0.976556, then Dt ¼ 0,Ds , 0; if b ¼ 0.999, then Dt , If we substitute Eq. (19) into the right-hand side of Eq. 0 0,Ds , 0. According to Table I, for the case of two- (20) and compare the coefficients in front of x,wehave dimensional time, it is possible that the conditions v . c k r and Dt0 . 0 can be simultaneously fulfilled (unlike the glq ¼ gkralaq: ð21Þ case of one-dimensional time in STR). For example, if r ¼ (Here k, r ¼ 1, 2, 3, 4, 5). Let us define a 5 3 5matrix(A)lq, c/v ¼ 0.694, then Dt0 ¼ 0.142 . 0. l l with elements a q:(A)lq ” a q. Similarly, let us define the elements of the matrix (G)lq:(G)lq ” glq.Inmatrix tr C. General properties of the transformations presentation, Eq. (21) can be written as (G)lq ¼ (A GA)lq; therefore det(G) ¼ det(AtrGA) ¼ det(Atr)det(G)det(A). We are going to examine some of the properties of the Taking into consideration the fact that det(Atr) ¼ det(A) 0 general transformations between K and K . Let us denote and det(G) ¼ 1, we obtain with x1, x2 the time dimensions and with x3, x4, x5 the 1 2 3 4 space dimensions; for example x ¼ ct, x ¼ cs, x ¼ x, x ¼ d In this and the following formulas, the Einstein summation 5 y, x ¼ z. The following equality is fulfilled: convention for repeating indices is used. 412 Phys. Essays 25, 3 (2012)
TABLE II. Decomposition of the group of nonzero transformations.
0 1 2 Transformations between K and K det(A) sign of a 1 sign of a 2 Operations applied on Kþ :