ESSAYS 25, 3 (2012)

Relativistic mechanics in multiple 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 , 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 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 -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 þþ(]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 þþ 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 ¼¼ukq 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, jahj1—and g¼kþ1 velocity U is invariant when the space-axes basis and the 2 vg ¼ 1—that is, jvgj1. 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 ¼¼ak ¼ 1/ k (that is, dt ¼ dt ¼¼dt ¼ dT/ k . 0) and v ¼ v ¼¼ 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 ¼¼Vkg 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 R1. 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 ¼i(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 ¼i(v/c)x1 ¼i(w/c)x2 . 0. Therefore, a more general case, the so-called Lorentz boost in an we have tan u ¼ib.Heretheangleu is an imaginary arbitrary direction (transformations between two inertial number. We have sin u ¼ibf,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 ¼edt 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 ¼bf. 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)] ¼e(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 ¼1 (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, fpffiffiffi ‘ , 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) ¼1, 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 :

Kþ þ1 þ1 þ1 1 (unit matrix) ¼ diag(þ1, þ1, þ1, þ1, þ1) Kþ þ1 þ1 1 PN ¼ diag(þ1, 1, 1, 1, 1) Kþ þ1 1 þ1 PX ¼ diag(1, þ1, 1, 1, 1) Kþ þ1 1 1 C ¼ diag(1, 1, þ1, þ1, þ1) K 1 þ1 þ1 P ¼ diag(þ1, þ1, 1, 1, 1) K 1 þ1 1 N ¼ diag(þ1, 1, þ1, þ1, þ1) K 1 1 þ1 X ¼ diag(1, þ1, þ1, þ1, þ1) K 1 1 1 PC (total inversion) ¼ diag(1, 1, 1, 1, 1)

detðAÞ¼61: ð22Þ K }. The transformations which do not change the sign 2 in front of s (that is, a 2 . 0) we will call orthochronous in In Eq. (21), if we set l ¼ q ¼ 1 and l ¼ q ¼ 2, we relation to s; we will denote them with K ¼ {Kþ ¨ Kþ ¨ accordingly obtain K K }. The transformations which change the signs in X5 front of t and s we will call nonorthochronous in relation 1 2 2 2 g 2 1 ¼ða1Þ þða1Þ ða1Þ ; to the time dimensions; we will denote them with K ¼ g¼3 {Kþ ¨ K }. It is clear that the same transformation can be orthochronous in relation to a given time dimension X5 and nonorthochronous in relation to another time 1 ¼ða1Þ2 þða2Þ2 ðagÞ2; 2 2 2 dimension (for example, the transformation K). The g¼3 þ transformations responsible for the condition det(A) ¼þ1 that is, we will call proper transformations and will denote with vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u Kþ ¼{Kþ ¨ Kþ ¨ Kþ ¨ Kþ }. The transformations for u X5 t g 2 2 which det(A) ¼1 we will call nonproper transformations a1 ¼ 6 1 þ ða Þ ða2Þ ; 1 1 1 and will denote with K ¼ {K ¨ K ¨ K ¨ K }. g¼3 Let us define the following discrete operations which vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l0 l q l0 u present spatial or temporal reflection: x ¼ Pqx , x ¼ u X5 l q l0 l q l0 l q l 2 t g 2 1 2 C qx , x ¼ X qx , x ¼ N qx , where P q ¼ diag(þ1, þ1, a2 ¼ 6 1 þ ða2Þ ða2Þ : l l 1, 1, 1), C q ¼ diag(1, 1, þ1, þ1, þ1), X q ¼ diag(1, g¼3 l þ1, þ1, þ1, þ1), N q ¼ diag(þ1, 1, þ1, þ1, þ1). Since after 1 Therefore, we have nine possible cases: a 1 . 0 and these operations the scalar product remains invariant, 2 1 2 1 2 1 a 2 . 0; a 1 . 0 and a 2 , 0; a 1 , 0 and a 2 . 0; a 1 , 0 they belong also to the full group of transformations. We 2 1 2 1 2 1 and a 2 , 0; a 1 ¼ 0 and a 2 . 0; a 1 ¼ 0anda 2 , 0; a 1 . can connect these operations to the transformations 2 1 2 1 2 0anda 2 ¼ 0; a 1 , 0 and a 2 ¼ 0; and a 1 ¼ 0 and a 2 ¼ 0. obtained previously (see Table II). Adding into consideration Eq. (22), we have altogether 18 The results obtained here will play a very important cases in general. role relative to an antiparticle moving in multidimension- In the case of two-dimensional time it is possible that al time (see Section X). 1 2 1 the equalities a 1 ¼ 0anda 2 ¼ 0 are fulfilled [e.g., if x ¼ Ifwe assume that thestransformations are valid, then the ct,x2 ¼ cs, x3 ¼ x, x4 ¼ y, x5 ¼ z, then they are fulfilled at Lorentz covariance must be violated. The physical laws will the transformation (x1)0 ¼ cs,(x2)0 ¼ct,(x3)0 ¼ x,(x4)0 ¼ not be Lorentz covariant, i.e., they will not be transformed 5 0 1 2 y,(x ) ¼ z]. The transformations where a 1 6¼ 0anda 2 6¼ according to the Lorentz group. The four-dimensional space- 0 we will call nonzero transformations. time interval in this case will not be invariant, while the five- Let us now consider only the nonzero transforma- dimensional interval will be—see Section IV. tions. There are in general eight nonzero transformations In the case of multidimensional time, the CPT symmetry (see Table II). The transformations conserving the will be violated. STR and consequently the Lorentz 1 2 orientation—i.e., for which det(A) ¼ 1, a 1 . 0, a 2 . covariance is placed at the base of CPT symmetry. Indeed, 0—we will call proper orthochronous transformations the even number of reflections of the coordinates in and will denote with Kþ . The transformations which do Minkowski space-time (PT-symmetry) is formally reduced 1 not change the signs in front of t and s (that is, a 1 . 0 to a rotation by an imaginary angle. Due to this fact, the 2 and a 2 . 0) we will call orthochronous in relation to t existing physical theories, which are invariant relative to the and s; we will denote them with K ¼ {Kþ ¨ K }. The Lorentz transforms (i.e., rotations in Minkowski space-time) transformations which do not change the sign in front of t turn out to be automatically CPT invariant. We discuss the 1 (that is, a 1 . 0) we will call orthochronous in relation to problem with CPT symmetry in multidimensional time more t; we will denote them with K ¼ {Kþ ¨ Kþ ¨ K ¨ precisely in Section X. Phys. Essays 25, 3 (2012) 413  D. Waves in two-dimensional time and three- c2 c2 x x0 1 f 1 b2 b2f:cos c0 dimensional space and the Doppler effect t ¼ t þð Þ 2 þ 0 v vut 2 It is clear that in the case of two-dimensional time the 0 c 2 þ xsðf 1Þ b ; d’Alembert operator is not a scalar. Instead, another vw operator must be used, obtained from the scalar product:  2 2 0 c 2 c 2 0  2 x ¼ x 1 þðf 1Þ b þ b f:cos c 2 2 X5 s s w2 wu0 lq ] ] ] ] ] s Ats ” g ¼ þ ; 2 ]xl ]xq c]t c]s ]xg 0 c 2 g¼3 þ x ðf 1Þ b ; t vw where  8 u0 u0 < j ¼ j0 f 1 þ t b2 þ j0 b2f s ; 1 l ¼ q ¼ 1; 2 x x v cos c0 x w cos c0 glq ¼ 1 l ¼ q ¼ 3; 4; 5 : j ¼ j0 ; j ¼ j0: 0 l 6¼ q: y y z z 0 2 2 0 2 0 0 1 2 The expressions x (f 1)(c /vw)b and j b f(u /w cos c ) (Here x ¼ ct,x ¼ cs.) Since the A operator is a scalar, it s x s ts in the formulas for the angular frequency x and the wave follows that t vector j , respectively, can be regarded as corrections in   x 2 2 X5 2 the formulas for the Doppler effect for the case of two- 0 ] ] ] 0 Ats ¼ A ¼ þ : dimensional time. If we set x ¼ 0, we will obtain the ts c]t0 c]s0 ]xg0 s g¼3 formulas for the Doppler effect for the case in which the wave is moving in only one time dimension, t0. Obviously For the case of two-dimensional time and three- in this case, the expressions for x and j differ from the dimensional space, the wave equation takes the form t x   formulas for the relativistic Doppler effect in STR. ]2F ]2F ]2F j 2]2F j ]2F 2 þ 2 þ 2 ¼ 2 þ 2 ; ]x ]y ]z xt ]t xs ]s IV. CAUSAL STRUCTURE OF SPACE-TIME (n, k) ¼ (3, 2) where xt and xs are the angular frequencies of the wave, defined in relation to the time dimensions t and s Let us have a five-dimensional vector Al in space- respectively, and j is the angular wavenumber (ultra- time (n, k) ¼ (3, 2). The scalar product of the vector Al hyperbolic partial differential equations—see Ref. 5). We with itself will bee 3 4 5 have set x ¼ x, x ¼ y,x ¼ z.) Here ut ¼ xt/j and us ¼ xs/j 2 l l q are phase velocities, defined in relation to t and s, ðAÞ ¼ A Al ¼ glqA A X5 respectively. ¼ðA1Þ2 þðA2Þ2 ðAgÞ2; We will examine the Doppler effect in two-dimen- g¼3 sional time and three-dimensional space. In this case it is possible that waves exist which have properties depend- where l, q ¼ 1, 2, 3, 4, 5 and ing only on one or both time dimensions. We will 8 < 1 l ¼ q ¼ 1; 2 consider the general case, when a wave is moving in the g ¼ g ¼ 1 l ¼ q ¼ 3; 4; 5 two time dimensions t, s andinthethreespace lq ql : 0 l 6¼ q: dimensions x, y, z. P 2 1 2 5 g 2 Let us denote with xt and xs the angular frequencies Let us denote (A3,1) ¼ (A ) g¼3ðA Þ . Then we 2 2 2 2 of the wave in the frame K, defined relative to the time will have (A) ¼ (A3,1) þ (A ) . While in Minkowski 2 dimensions t and s, and with j ¼ (jx, jy, jz) the wave space-time [(n, k) ¼ (3, 1)] the value (A3,1) is invariant, in vector of this wave in K. The phase of the wave in K is a space-time (n, k) ¼ (3, 2) it is not invariant—but the 2 2 given by the expression xtt þ xss jR, where R ¼(x, y, z). value (A) is invariant. If, for example, the value (A3,1) is 0 0 spacelike in one frame of reference K [(A )2 , 0], it can Let us denote with xt and xs the angular frequencies of 3,1 0 0 0 0 0 2 0 2 the wave in K , defined relative to t and s , and with j ¼ be 0 [ðA3;1Þ ¼ 0] or timelike [ðA3;1Þ . 0] in another frame 0 0 0 0 of reference K0. jx, jy, jz the wave vector of the wave in K . The phase of the wave in K0 is given by x0t0 þ x0s0 j0R0, where R0 ¼ The causal region of the space-time (n, k) ¼ (3, 2) t s 2 (x0, y0, z0). In order to determine the Doppler effect for the encompasses the region (A) 0, and the causal region of wave under consideration, we have to set to set the wave the space-time (n, k) ¼ (3, 1) encompasses the region 2 phase to be invariant—that is, x t x s – jR x0t0 x0 s0 (A3,1) 0. t þ s ¼ t þ s 1 2 3 4 5 j0R0. Let us set:pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA ¼ cdt, A ¼ cds, A ¼ dx, A ¼ dy, A ¼ 2 2 2 0 0 0 0 dzp,ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq1 ¼ ( dx þ dy þ dz )/(cjdtj) ¼ V/c 0, q2 ¼ We set j ¼jjj jj . 0 (angular wavenumber); ut ¼ xt/ 2 2 2 0 0 0 0 0 q( ffiffiffiffiffiffiffiffiffiffidx þ dy þ dz )/(cjdsj) ¼ W/c 0, ds3.2 ¼jjAjj ¼ j , us ¼ xs/j (phase velocities determined according to t 0 0 0 0 2 and s , respectively); jx/j ¼ cos c . Applying the ðAÞ . transformations in Eq. (12) from K to K0, we define the 0 0 0 0 0 e relation between xt, xs, jx, jy, jz and xt, xsjx, jy, jz: In the formula, the Einstein summation convention is used. 414 Phys. Essays 25, 3 (2012)

to the following equality: 1 1 2 þ 2 ¼ 1: ð24Þ q1 q2

If q1 , 1 (that is, V , c)orq2 , 1 (that is, W , c), then Eq. (24) is impossible. If q1 ¼ 1 (that is, V ¼ c), then Eq. (24) is possible under the condition that q2 ¼ ‘ (that is, W ¼ ‘). This means that dx2 þ dy2 þ dz2 . 0, dt2 . 0, ds2 ¼ 0—i.e., the considered particle moves in space and in the time dimension t but not in the time dimension s. The

same applies for the case q2 ¼ 1 (that is, W ¼ c). From Eq. (24) follows the equality (c2/V2) þ (c2/W2) ¼ 1. Therefore, if the considered particle is moving with velocity c V ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; c2 1 W2

FIG. 1. (Color online) The values of the velocities V and W at which an 2 2 then we have (A) ¼ ds3;2 ¼ 0. According to the results of object is moving in the causal regions of the space-time (n, k) ¼ (3, 2) and Section VIII, if the equality (c2/V2) þ (c2/W2) ¼ 1is (n, k) ¼ (3, 1). fulfilled for the velocities V and W of a particle in the 0 0 2 frame of reference K, then for the velocities V and W of We will consider three cases for the value of (A) . 0 2 2 2 this particle in the frame K , the similar equality First case: (A) . 0—that is, ds3;2 . 0. If A ¼ cds 6¼ 0, 2 2 2 2 then there are three possible cases: (A3,1) . 0,(A3,1) ¼ c c 2 2 2 0,(A ) , 0. If A cds 0, then (A ) . 0. Let us 2 þ 2 ¼ 1 3,1 ¼ ¼ 3,1 ðV0Þ ðW0Þ assume that in the frame of reference K, the following are 2 2 fulfilled: A ¼ cds 6¼ 0 and (A3,1) , 0. It is clear that in the is fulfilled. 0 2 2 system K moving uniformly and rectilinearly in relation Third case: (A) , 0—that is, ds3;2 , 0. We have 0 2 0 2 20 2 2 2 2 to K,wehave(A ) ¼ (A3;1) þ (A ) ¼ (A) . 0. If we (A3,1) , 0. In this case it is not possible that (A3,1) . 0 2/ 0 0 2 2 2 assume that A ¼ cds ¼ 0, then we have (A3;1) . 0. So we or (A3,1) ¼ 0. The condition (A) , 0 is equivalent to the have obtained that in the frame K the value A3,1 is following inequality: 2 0 0 spacelike [(A3,1) , 0] and in K the value A3;1 is timelike [ (A0 )2 . 0]. It is easy to prove that the condition (A)2 . 0 1 1 3;1 2 þ 2 , 1: ð25Þ is equivalent to the following inequality: q1 q2 1 1 The inequality in Eq. (25) is fulfilled only if the þ . 1: ð23Þ 2 2 following inequalities are simultaneously fulfilled: q1 . 1 q1 q2 and q2 . 1 (that is, V . c and W . c). If 0 q1 , 1 (that is, 0 V , c), then the inequality From these considerations we can conclude that the in Eq. (23) will be fulfilled for all values of q2 ‘ (i.e., for causal region of (3 þ 2)-dimensional space-time includes W W q V c all values of , including ¼ ‘). If 1 ¼ 1 (that is, ¼ ), the causal region of (3 þ 1)-dimensional space-time and then the inequality in Eq. (23) will be fulfilled provided presents a larger part of it. that q ,‘(that is, W ,‘). Similar considerations are 2 Figure 1 shows graphs of the functions (c2/V2) þ (c2/ valid for q (and accordingly for the velocity W). 2 W2) ¼ 1andc/V ¼ 1 for non-negative values of V and W. Therefore, if the velocity of a particle defined in relation It is clear that for each point in the darker region and to the one time dimension (as absolute value) is less than on the border of this region [i.e., the region confined by or equal to the speed of light in vacuum, then the velocity the coordinate axes and the graph of the function (c2/V2) of this particle defined in relation to the other time 2 2 dimension can have an arbitrary value without violating þ (c /W ) ¼ 1] corresponds to a combination of values for the velocities V and W where a given particle is moving in the principle. If simultaneously q1 . 1 and q2 . 1, then the inequality in Eq. (23) will be fulfilled for an the causal region of the space-time (n, k) ¼ (3, 2), that is, ds2 0. For each point outside this region there is a appropriate choice of the parameters q1 and q2 (e.g., q1 ¼ 3;2 2 combination of values of the velocities V and W where the 10/9,q2 ¼ 2). It is clear that the condition (A3,1) . 0is 2 given point does not move in the causal region of the equivalent to the inequality V , c, the condition (A3,1) ¼ 2 0 is equivalent to the equality V ¼ c, and the condition space-time (n, k) ¼ (3, 2), that is, ds3;2 , 0. For each point 2 (A3,1) , 0 is equivalent to the inequality V . c. in the region between the abscissa and the straight line c/V 2 2 Second case: (A) ¼ 0—that is, ds3;2 ¼ 0. In this case, ¼ 1 there exists a combination of values for the velocities 2 2 2 2 (A3,1) ¼c ds , and therefore it is not possible that (A3,1) V and W where the particle is moving in the causal region 2 2 2 2 . 0. If A ¼ cds ¼ 0, then (A3,1) ¼ 0. If A ¼ cds 6¼ 0, then of space-time (n, k) ¼ (3, 1); this is the case of STR [(A3,1) 2 2 we have (A3,1) , 0. The condition (A) ¼ 0 is equivalent 0.] Phys. Essays 25, 3 (2012) 415

FIG. 2. (Color online) Causal structure of the space-time (a); causal region in the plane l1 (b) and in the plane l2 (c).

Let us denote with x1, x2 the time dimensions and its base. According to the previous considerations, in with x3, x4, x5 the space dimensions. Let us further for order for a exist causal relation between two events to 3 2 simplicity consider only one space dimension x and the exist, the inequality s3;2 0 must be fulfilled—that is, two time dimensions x1 and x2 (that is, x1 ¼ ct, x2 ¼ cs, x3 (x1)2 þ (x2)2 (x3)2 0. This inequality can be rep- ¼ x, x4 ¼ 0, x5 ¼ 0). Let us assume that at point O, having resented graphically by a double right circular cone. The coordinates x1 ¼ 0, x2 ¼ 0, x3 ¼ 0(x4 ¼ 0, x5 ¼ 0) defined in generatrices of the lateral surface of the cone make an 1 2 the frame K, there has been an E.p Letffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi us further angle of p/4 with the plane x -x . Therefore, we obtain a assume that a given interval of time DT ¼ Dt2 þ Ds2 0 double cone inscribed in the cylinder [Fig. 2(a)]. From the has passed since the event. Our task is to examine the inequalities (x1)2 þ (x2)2 (x3)2 0 and (x1)2 þ (x2)2 causal region attached to the event E. Since the time is c2DT 2 we obtain jxj3 cDT—that is, cDT x3 cDT. two-dimensional, all possible combinations of coordi- The cone and the cylinder have common bases: a circle nates x1, x2 for which the inequality (x1)2 þ (x2)2 c2DT 2 with its center at the point x3 ¼cDT (x1 ¼ 0,x2 ¼ 0) and is valid form a circle in the plane x1-x2 with center at point radius cDT, and a circle with its center at the point x3 ¼ O and radius equal to cDT. If we add the space dimension cDT (x1 ¼ 0, x2 ¼ 0) and radius cDT. 3 3 x in such a way that for an arbitrary value of x the Let us with C1 denote the border region which includes inequality (x1)2 þ (x2)2 c2DT 2 is fulfilled, then we will all points lying on the lateral surface of the cone and the obtain a right circular cylinder with the obtained circle as lateral surface of the cylinder. The region C1 includes also 416 Phys. Essays 25, 3 (2012) the origin O—i.e., it includes all points (x1, x2, x3)for We have assumed that since event E an interval of which the conditions {(x1)2 þ (x2)2 (x3)2 ¼ 0andjxj3 time DT has passed. In this case, the causal region in the 1 2 2 2 2 2 3 cDt}or{(x ) þ (x ) ¼ c DT and jxj cDt} are fulfilled. plane l1 (that is, r2) includes all events which are causally Let us denote with C2 the inner region limited by the connected with event E and happen at a distance jOLj¼ lateral surface of the cone and the lateral surface of the cDT tan w from O (along the axis x3). The causal region in cylinder. The region C2 does not include the origin O or the plane l2 (that is, r4) includes all events which are the lateral surfaces of the cone and the cylinder (i.e., it causally connected with event E and happen in a plane 1 3 does not include the border region C1)—it includes all making an angle a with the plane x -x . points (x1, x2, x3), for which the inequalities (x1)2 þ (x2)2 If a ¼ 0 (i.e., the vector OM is in the plane x1-x3), then 3 2 1 2 2 2 2 2 1 3 (x ) . 0 and (x ) þ (x ) , c DT are fulfilled. From the plane l2 coincides with the plane x -x . (We will have 3 3 these two inequalities we obtain jxj cDT.Ifx ¼ 0, then M1 ” M.) This case is considered in STR. If a ¼ p/2 (i.e., 1 2 2 2 2 2 2 3 0 , (x ) þ (x ) , c DT . the vector OM is in the plane x -x ), then the plane l2 2 3 Let us denote with C3 the region which includes all coincides with the plane x -x . (We will have M2 ” M.) points lying in the inner volume of the cone. The region C3 We can consider two more specific cases depending does not include the origin O or the lateral surface of the on the value of the angle w. cone. It includes the inner parts of the circles that form the First case: w ¼ 0—that is, the plane l1 coincides with bases of the cones (and the cylinder), but does not include the plane x1-x2. In this case the particle M is at rest; i.e., it the points lying on the directrix circumference restricting does not move in the space dimension x3, but only in the 1 2 these circles (bases)—i.e., the region C3 includes all points time dimensions x and x . We have tan w ¼ tan 0 ¼ 0. In 1 2 3 1 2 2 2 3 2 (x , x , x ), for which the inequalities (x ) þ (x ) (x ) , this case the radius of the inner circle r1 in Fig. 2(b) is 3 3 0andjxj cDT are fulfilled. It is clear that if x ¼ 6cDT, equal to 0 (r ¼ cDT tan w ¼ 0), i.e., the causal region r2 of then (x1)2 þ (x2)2 , c2DT2. event E includes all points of the outer circle. Therefore, Let us denote with C4 the region including all points the causal region of event E in the plane l1 (and outside the cylinder. It does not include the lateral surface accordingly in the plane x1-x2) in this case coincides with of the cylinder or the two bases of the cylinder—i.e., it the circle for which (x1)2 þ (x2)2 jjOMjj2 ¼ c2DT 2.In 1 2 3 1 2 includes all points (x , x , x ) for which the inequality (x ) this case we have tan w1 ¼ 0, w1 ¼ 0ifFig. 2(c) shows the 2 2 2 2 3 1 3 þ (x ) . c DT or the inequality jxj . cDT is fulfilled. plane x -x and tan w2 ¼ 0, w2 ¼ 0ifFig. 2(c) shows the 2 3 The causal region includes the regions C1 and C2; the plane x -x . One can say that the two-dimensional time 1 noncausal region includes the regions C3 and C4. ‘‘flows’’ from the origin O in all directions in the plane x - In Fig. 2(a), the vector OM shows the motion of a x2; as a result of this, the described circle is obtained. If we pointlike particle M. The point M1 is the projection of assume that the moments lying on the circumference of 1 3 point M on the plane x -x , and point M2 is the projection this circle are contemporary moments, then the moments of point M on the plane x2-x3. of the inner part of the circle are past moments and the Let us consider the plane l1, which is parallel to the moments outside the circle are future moments (according 1 2 plane x -x and includes point M, and the plane l2, which to the moments defined as contemporary). These consid- is perpendicular to the plane x1-x2 and includes points O erations are valid also for the case of k-dimensional time. qandffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiM. Let us set tan a ¼ x2/x1 and tan w ¼ x3/ It is evident that k-dimensional time will ‘‘flow’’ in the 2 2 ðx1Þ þðx2Þ . If we set x1 ¼pctffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi,x2 ¼ cs,x3 ¼ x . 0, then form of a k-dimensional hypersphere having its center at tan a ¼ v/w and tan w ¼ 1/ ðc2=v2Þþðc2=w2Þ. (In this the origin O. Let us assume that between point O and the case, v and w are the velocities of the particle M defined event E a period of time DT . 0 has passed, which is relative to the time dimensions t,and s—see Subsection determined according to the frame of reference K. In this 1 case all moments inside the k-dimensional hypersphere III.A). Let us denote with w1 the angle between x and 2 with center O and radius DT are in the past and all OM1, and with w2 the angle between x and OM2.Itis moments outside the k-dimensional hypersphere are easy to see that tan w1 ¼ (tan w)/(cos a) [if in Fig. 2(c) one 1 3 future moments. The surface which defines the k- can imagine the plane x -x ] and tan w2 ¼ (tan w)/(sin a) (if one can imagine the plane x2-x3). Let us denote with L dimensional hypersphere is a (k 1)-dimensional hyper- 3 sphere. All moments lying on the mentioned (k 1)- the point where the axis x intersects the plane l1—i.e., point L is the projection of the origin O (x1 ¼ 0, x2 ¼ 0, x3 dimensional hypersphere are present moments. Second case: w ¼ p/4—that is, the upper bases of the ¼ 0) on the plane l1. cylinder and the cone lie in the plane l .Inthiscasethe The causal region r2 in Fig. 2(b) represents the region 1 between the two circumferences having a common center; vector OM lies on one of the generatricesqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi of the conical the outer circumference has a radius R ¼ cDT, and the surface. We have tan w ¼ tan(p/4) ¼ x3/ ðx1Þ2 þðx2Þ2 ¼ 1. 1 2 3 inner one has a radius r ¼ cDT tan w. The causal region r4 If we further set x ¼ ct, x ¼ cs, x ¼ x . 0, then for the in Fig. 2(c) represents two congruent right-angled isosceles velocities v and w of the particle M defined in relation to the 2 2 triangles, namely OAC and OA1C1, which have a common time dimensions t and s, respectively, the equality (c /v ) þ 2 2 apex at point O. It is clear that jACj¼jA1C1j¼2cDT, jOBj (c /wp)¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 will be fulfilled (see Subsectionpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi III.A). Since cos a 2 2 2 2 ¼jABj¼jOBp1ffiffiffij¼jA1B1j¼cDT, jOAj¼jOCj¼jOA1j¼ ¼ 1/v ð1=v Þþð1=w Þ,sina ¼ 1/w ð1=v Þþð1=w Þ ,in jOC1j¼cDT 2. Further, we have jOLj¼cDT tan w. this case we can express the velocities v and w in terms of Phys. Essays 25, 3 (2012) 417

FIG. 3. (Color online) Projection of the causal region (a) on the plane x1-x2, (b) on the plane x1-x3, and (c) on the plane x2-x3, for a change of the angle a (from a to a0). angle a: v ¼ (c/cos a)andw ¼ (c/sin a)—see Subsection Figures 3(a), 3(b), and 3(c) show what happens when III.A. Since tan w ¼ 1, in this case the radius r of the inner the angle a is changed (i.e., the angle a changes to angle 0 0 0 circle r1 from Fig. 2(b) is equal to the radius R of the outer a ). Here tan a1 ¼ (1/cos a), tan a1 ¼ (1/cos a )[Fig. 3(b)] 0 0 circle (r ¼ cDT tan w ¼ cDT ¼ R), i.e., the causal region r2 and tan a2 ¼ (1/sin a), a2 ¼ (1/sin a )[Fig. 3(c)]. includes only the points lying on the outer circumference. In At a ¼ 0ora ¼ p/2 we obtain the well-known case of 1 2 this case we have tan w1 ¼ (1/cos a)ifFig. 2(c) shows the motion along only one axis of time (x or x , respectively). 1 3 plane x -x and tan w2¼(1/sin a)ifFig. 2(c) shows the plane This case (a ¼ 0) is considered in STR. x2-x3. If, for example, a ¼ 0 and x1 ¼ ct, x3 ¼ x, then the 1 3 Let us project the intersection of the plane l1 with the projection of the causal region onto the plane x -x [Fig. cylinder and with the cone [shown in Fig. 2(a)] onto the 3(b)] will coincide with the causal region according to 1 2 plane x -x . Let us do the same with the intersection of STR. If a ¼ 0, then cos a ¼ 1, tan a1 ¼ 1, a1 ¼ p/4, tan a2 ¼ 1 3 the plane l2 with the cylinder and the cone onto x -x and ‘, a2 ¼ p/2. If a ¼ 0, then the projection of the causal x2-x3. Then we obtain Fig. 3(a) and the darker areas region onto the plane x2-x3 [Fig. 3(c)] will be a line 3 shown in Figs. 3(b) and 3(c), respectively. If the vector segment lying on the axis x (a2 ¼ p/2). In this case, if OM presents the motion of a given particle M, then the jjOM2jj ¼ 0, then the velocity of the particle M projections of the motion of this particle onto the plane determined in relation to the two time dimensions x1 1 3 2 3 2 x -x (i.e., the vector OM1) or onto the plane x -x (i.e., and x is equal to 0; but if jjOM2jj 6¼ 0, then the velocity 1 the vector OM2) can lie only in the darker regions of Figs. of the particle M defined in relation to x is greater than 0 3(b) 3(c), respectively. In the opposite case—if the and less than or equal to the speed of light in a vacuum, 2 projections OM1 or OM2 do not lie in the indicated and the velocity of the particle defined in relation to x is regions—then the motion shown by the vector OM is not infinitely large. (See the considerations at the beginning of causally related. In Fig. 3(a), the point M3 is the Section IV.) Likewise, one can apply the same consider- projection of point M onto the plane x1-x2. ations for the case a ¼ p/2. 418 Phys. Essays 25, 3 (2012)

For the angle a0 ¼ 6a þ bp (b ¼ 0, 61, 62, . . .), one ization of the Lorentz boost in a fixed direction in the field can obtain the same projections of the causal region [Figs. of multidimensional space and multidimensional time). 3(b) and 3(c)] as for angle a. Equation (12) can be easily generalized for an arbitrary For an observer in the frame K, each coordinate plane number of space and time dimensions (n 1, k 1). The 1 2 x ¼ const, x ¼ const represents space in the respective time dimensions we will denote with t1, t2, ..., tk, and the instant of time—i.e., these are a set of events which occur space dimensions we will denote with xkþ1, xkþ2, ..., xkþn. for the observer simultaneously. These regions in STR are We assume that the frames K and K0 are in a standard called simultaneous spaces of the given observer. By configuration. Let us denote with analogy with STR, each timelike straight line [or line segment—see, for example, OM in Fig. 2(a)] is parallel to v1 ¼ðv1; 0|fflfflfflffl{zfflfflfflffl}; ...; 0Þ; 0 one of the time axes of a frame of reference K moving n1 uniformly and rectilinearly against K. Therefore, each timelike straight line ‘‘separates’’ the three-dimensional v2 ¼ðv2; 0|fflfflfflffl{zfflfflfflffl}; ...; 0Þ; space into countless simultaneous regions. n1 Let us assume that the line segment OM lies in the plane x1-x3 and not on the axis x1. In this case, the three- . dimensional velocity of K0 in relation to K defined in . relation to the time dimension x2 is infinitely large. However, as noted earlier, the three-dimensional velocity vk ¼ðvk; 0|fflfflfflffl{zfflfflfflffl}; ...; 0Þ 0 of K against K determined in relation to the time n1 1 dimension x must be less than the speed of light in a 0 vacuum. Obviously, in this causal region there exist the vectors of the velocities of K against K, defined infinitely many inertial frames like K0 which posses the respectively in relation to the time dimensions t1, previously mentioned properties. Similar considerations t2, ..., tk. Let us set 2 3 are valid if the line segment OM lies in the plane x -x and 1 2 vffiffiffiffiffiffiffiffiffiffiffiffiffi not on the axis x . b ¼ u uXk In space- with at least two time dimensions, it is t c2 always possible to construct closed timelike curves.f,4 In the v2 q¼1 q space-time (n, k) ¼ (3, 2), we can consider a ‘‘motion’’ in the qffiffiffiffiffiffiffiffiffiffiffiffiffi 2 causal region in the plane t-s which begins at the origin O (t and f ¼ 1/ 1 b . Let us denote with Klq the matrix for ¼ 0,s ¼ 0) and ends at the same point. Indeed, let us set t ¼ transfer between X.sin H, s ¼ X(1 – cos H); x, y, z ¼ const. (Here X ¼ 0 1 const,H [0; 2p].) Then we have ds2 ¼ c2dt2 þ c2ds2 – dx2– t1 3;2 B . C dy2 dz2 ¼ X2dH2 . 0—i.e., the world line is everywhere B . C B . C timelike.4 As pointed out by Foster and Muller,¨ 4 B t C X ¼ B k C B x C The existence of closed time-like curves implies that an B kþ1 C @ . A observer can revisit the past and, if we accept the tenet . of ‘‘,’’ change it in a manner that is incompatible xkþn with the already experienced future. Obviously, any- thing resembling the common notion of causality and 0 1 cannot be maintained under such circumstances. 0 t1 B C It has been argued, however, that such cases arise B . C B . C only due to the misidentification of different space-time B 0 C 0 B tk C points as identical points of the manifold.13 Other authors X ¼ B 0 C; B xkþ1 C have argued that the paradoxes arising from closed B C @ . A timelike curves are real, but can be resolved by an . 14 0 appropriate extension of the space-time manifold (see xkþn Subsection III.B). 0 where l, q ¼ 1,2,..., k þ n. Here we have X ¼ KlqX.Ifl 2 2 k and q k, then Klq ¼ dlq þ (f 1)(c /vlvq)b . Here V. GENERALIZATION OF THE TRANSFORMATION dlq is the Kronecker delta, i.e., FOR n-DIMENSIONAL SPACE AND k-DIMENSIONAL  1 l ¼ q TIME d ¼ : lr 0 l 6¼ q First, we will consider the simplest case, without rotations or translations in space or time (i.e., general- If l ¼ q ¼ k þ 1, then Klq ¼ f.Furthermore,Kr(kþ1) ¼ 2 2 2 (1/vr)b f, K(kþ1)h ¼(c /vh)b f,wherer, h ¼ 1,2,..., k. f This is different from the case of curved manifolds, where closed If l k þ 2orq k þ 2, then Klq ¼ dlq. Therefore, we timelike curves may arise under certain circumstances. have Phys. Essays 25, 3 (2012) 419

 R Xk c2 1 space-time is SO(1, 3, ); together with parity and time- t0 ¼ d t þ b2ðf 1Þt b2fx ; R r rh h v v h v kþ1 reversal it becomes O(1, 3, ). If we introduce an (n þ k)- h¼1 r h r dimensional vector of translation in the space-time, then ð26Þ by analogy with (n, k) ¼ (3, 1)—that is, the inhomoge- neous Lorentz group or Poincare´ group in STR—we can Xk t obtain the most general transformations [see Eq. (18)]. x0 ¼c2b2f h þ fx ; x0 ¼ x ; ð27Þ kþ1 v kþ1 u u h¼1 h VI. VELOCITY-ADDITION LAW where r, h ¼ 1, 2, ..., k, and u ¼ k þ 2, k þ 3, ..., k þ n. The transformations in Eqs. (26) and (27) are equivalent From the transformations obtained in Eqs. (26) and to the Lorentz transformations at t2 0, t3 0, ..., tk (27), it is easy to derive the velocity-addition formulas. 0 0 0 0 and accordingly v2 6‘, v3 6‘, ..., vk 6‘. Let us denote (dxg/dtr) ¼ Vrg and xg/dtr ¼ Vrg, where r ¼ Let us set (dxkþ1/dt1) ¼ v1,(dxkþ1/dt2) ¼ v2,...(dxkþ1/ 1, 2, . . ., k and g ¼ k þ 1, k þ 2,..., k þ n. The velocity- dtk) ¼ vk. Applying Eq. (26), we obtain addition formulas are given as follows: qffiffiffiffiffiffiffiffiffiffiffiffiffi "# Xk 2 dt0 dt 1 b2; 2 c r ¼ r Vrðkþ1Þf 1 b vhV 0 h¼1 hðkþ1Þ that is, V ¼ "#; rðkþ1Þ Xk 2 Vrðkþ1Þ 2 c 0 2 0 2 2 2 2 1 þ b ðf 1Þ f ðdt Þ þþðdt Þ ¼ðdt þþdt Þð1 b Þ: vr v V 1 k 1 k h¼1 h hðkþ1Þ (We used the equalities ð29Þ

dth dth dxkþ1 vr ¼ 3 ¼ ; Vru dt dx dt v V0 ¼ "#; ð30Þ r kþ1 r h ru Xk 2 V c 1 þ rðkþ1 b2 ðf 1Þ f where r, h ¼ 1, 2, . . ., k.) Let us apply rotation in the vr v V 0 0 0 h¼1 h hðkþ1Þ hyperplane of time t1-t2- -tk (see Section VII). We obtain the equality where r, h ¼ 1, 2, . . ., k and u ¼ k þ 2, k þ 3,..., k þ n.We

00 2 00 2 0 2 0 2 have used the equalities ðdt1 Þ þþðdtkÞ ¼ðdt1Þ þþðdtkÞ 2 2 2 dth dxg dth Vrg ¼ðdt1 þþdtkÞð1 b Þ: ¼ 3 ¼ : dt dt dx V ð28Þ r r g hg We note that (V /V ) ¼ (V /V ) ¼ ¼ (V /V ), If we set dt00 ¼ dt00 ¼¼dt00 . 0, then we obtain the 1g 1p 2g 2p kg kp 1 2 k where g, p ¼ k þ 1, k þ 2,..., k þ n. Indeed, (V /V ) ¼ following equalities: rg rp (dxg/dxp), where r ¼ 1, 2, . . ., k. 00 00 00 dt1 ¼ dt2 q¼¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidtk 1 ffiffiffi 2 2 2 2 VII. ROTATIONS IN THE HYPERPLANE OF TIME ¼ p ðdt1 þ dt2 þþdtkÞð1 b Þ: k AND IN THE HYPERPLANE OF SPACE We will use these equations in Section IX.A. Let us assume that a particle under consideration is Let us consider more general transformations, which moving in k time dimensions (t1, t2, ..., tk) and in n space do not include translation in space and time. Let us dimensions (xkþ1, xkþ2, ..., xkþn). According to our introduce the indefinite metric considerations, separately and independently from each other we can apply some rotations of the time axes (t , Ik;n ¼ diagð1|fflfflfflffl{zfflfflfflffl}; ...; 1; |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}1; ...; 1Þ: 1 t2, ..., tk) in the hyperplane of time and some rotations of k n the space axes (xkþ1, xkþ2, ..., xkþn) in the hyperplane of Let us denote with O(k, n, R) the group of all (k þ n) 3 (k space (see Section II and Subsection IX.A). These tr 1 tr þ n) matrices M for which M (Ik,n) M ¼ Ik,n. (Here M transformations are expressed in a change of the basis denotes the transpose of the matrix M and R denotes the of the time axes or of the space axes of the frame of field of real numbers.) The inverse of M is given by M1 ¼ reference under consideration (these are the so-called 1 tr (Ik,n) M Ik,n. The indefinite orthogonal group O(k, n, R) passive linear transformations). The group of all proper conserves the quadratic form defined through the metrics and improper rotations in the hyperplane of time is Ik,n and is isomorphic to the group of all proper and isomorphic to the orthogonal group O(k, R), and the improper rotations in the space-time being considered. group of all proper and improper rotations in the The indefinite special orthogonal group SO(k, n, R) is the hyperplane of space is isomorphic to the orthogonal subgroup of O(k, n, R) consisting of all elements with group O(n, R), where R denotes the field of real numbers. determinant þ1. The group SO(k, n, R) corresponds to the During the simultaneous rotation of the space axes in the group of all proper rotations in the space-time. The hyperplane of the space and the time axes in the proper Lorentz group of four-dimensional Minkowski hyperplane of time, the origin will remain invariant. The 420 Phys. Essays 25, 3 (2012)

2 2 2 2 Xk time interval dT ¼ dt1 þ dt2 þþdtk, the space interval c2 dX 2 dx2 dx2 dx2 dx2 , and the (n k)- ¼ 1: ð31Þ ¼ kþ1 þ kþ2 þþ kþn kþ2 þ V2 2 2 2 2 h¼1 h dimensional interval dsn;k ¼ c dT – dX are invariant during these operations. In Eq. (31), let us set V1 ¼ V2 ¼¼Vk. This can be Let us apply a proper or improper rotation of the done by appropriate rotation in the hyperplane of time so time axes t1, t2, ..., tk in the hyperplane of time. The new that the condition dt1 ¼ dt2 ¼¼dtk is fulfilled (see Sec. time axes obtained after this transformation we will VII). Then we will obtain 0 0 0 denote with t1, t2, ..., tk. Let us denote dT ¼ (dt1, pffiffiffi dt2, ..., dtk). For the vector components of dT before and V1 ¼ V2 ¼¼Vk ¼ c k: ð32Þ 0 2 0 2 after the transformation we have (dt1) þ (dt2) þ þ 0 2 2 2 2 Let us consider the motion of the particle in relation (dtk) ¼ dt1 þ dt2 þþdtk .The transformation under to the reference frame K0, which is moving uniformly and consideration can be presented through the orthogonal rectilinearly in relation to K. The (n þ k)-dimensional matrix A ¼ [a1r]k3k, which belongs to the orthogonal interval in this case is given by the expression (ds0 )2 ¼ R n;k group O(k, ). Here, 1, r ¼ 1, 2, . . ., k. Then we will have c2(dt0 )2 þþc2(dt0 )2 –(dx0 )2 (dx0 )2. Let us tr 1 R 1 k kþ1 kþn A ¼ A , det(A) ¼þ1 [proper rotation, A SO(k, )] or denote det(A) ¼1 (improper rotation). The relation between the vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 0 0 u components dt1, dt2, , dtk and the components dt1, u Xkþn 0 t 0 2 dt2, ..., dtk is given by dT ¼ dT 3 A—that is, ðdxgÞ g¼kþ1 Xk V0 ¼ ; 0 h jdt0 j dtr ¼ dt1a1r: h 1¼1 pffiffiffi (h ¼ 1, 2, . . ., k), and 0 0 0 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Let us set dt1 ¼ dt2 ¼ ¼ dtk ¼ dT/ k . 0. This is u Xkþn possible if the size of the angle made by the vector dT and u u ðdx0 Þ2 each of the time axes t0 , t0 , ..., t0 is equal to p/4. u g 1 2 k ug¼kþ1 u0 ¼ u : We can similarly consider a proper or improper u Xk t 2 rotation of the space axes xkþ1, xkþ2, ..., xkþn in the 0 ðdthÞ hyperplane of space. h¼1 2 Since the (n þ k)-dimensional interval dsn;k is invariant, 2 0 2 VIII. GENERALIZATION OF THE PRINCIPLE OF the equality dsn;k ¼ (dsn;k) ¼ 0 will be fulfilled. From here INVARIANCE OF THE SPEED OF LIGHT IN STR it follows that u0 ¼ c—that is,

In the case of multidimensional time, the principle of Xk c2 invariance of the speed of light defined in STR is violated. ¼ 1: ð33Þ V0 2 This requires a generalization of this principle for the case h¼1 ð hÞ k 1. 0 0 0 In Eq. (33), let us set V1 ¼ V2 ¼¼Vk. This can be Let us assume that the particle under consideration is obtained through an appropriate rotation in the hyper- moving in k time dimensions, denoted with t1, t2, ..., tk, 0 0 0 plane of time so that the condition dt1 ¼ dt2 ¼ ¼ dtk is and in n space dimensions, denoted with xkþ1, fulfilled (see Section VII). Then we will obtain xkþ2, ..., xkþn. We consider the motion of the particle in pffiffiffi 0 0 0 relation to the frame of reference K. The (n þ k)- V1 ¼ V2 ¼¼Vk ¼ c k: ð34Þ 2 dimensional interval in this case is expressed as dsn;k ¼ 2 2 2 2 2 2 Let us summarize: If for the velocities u and Vh of a c dt1 þþc dtk dxkþ1 dxkþn. Let us set 0 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi particle in K Eq. (31) is fulfilled, then for the velocities u u 0 0 u Xkþn and Vh of the particle in K the similar Eq. (33) will be t 2 fulfilled. If for the velocities V of a particle in K Eq. (32) dxg h 0 0 g¼kþ1 is fulfilled, then for the velocities Vh of this particle in K Vh ¼ ; the similar Eq. (34) will be fulfilled. These two statements jdthj are equivalent. For k ¼ 1 we obtain the principle of where h ¼ 1, 2, . . ., k. The total coordinate velocity in this constancy of the speed of light in a vacuum, defined in case (see Section II) is equal to STR: If V ¼ c, then V0 ¼ c. vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u Xkþn u 2 u dxg IX. MOTION OF A PARTICLE IN n-DIMENSIONAL u ug¼kþ1 SPACE AND IN k-DIMENSIONAL TIME u ¼ u . 0: t Xk 2 A. Proper time and generalized velocity dth h¼1 Let us assume that the particle under consideration is 2 Let us now assume that dsn;k ¼ 0. It is easy to prove moving in k temporal dimensions, denoted with t1, that in this case we have u ¼ c—that is, t2, ..., tk, and in n spatial dimensions, denoted with Phys. Essays 25, 3 (2012) 421 xkþ1, xkþ2, ..., xkþn. By analogy with the term ‘‘proper The obtained formula coincides with Eq. (28) (see 2 2 2 time’’ in STR, we will introduce the term ‘‘proper time’’ Section V). We can see that the values dt01, dt02, , dt0k are 00 2 00 2 for the case k 1, which we will denote with T0. transferred according to the same law as (dt1 ) ,(dt2 ) , , 00 2 Since time has k dimensions, the proper time dT0 will (dtk) in Eq. (28)—which had to be expected. The frame K0 be a k-dimensional vector. We have dT0 ¼ (dt01, we will call the proper frame of reference. Taking into dt02, ..., dt0k), where dt01, dt02, ..., dt0k are the projec- consideration Eqs. (28) and (35), we can understand the tions of the proper time on the different time axes. Let us physical meaning of proper time: It is the time measured in set the proper frame of reference K0 where the particle is at rest. vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u The values dt01, dt02, ..., dt0k are projections of the proper uXk t 2 time dT0 on the axes tP01, t02,...,t0k, respectively, of K0. dT0 ¼jjdT0jj ¼ dt0h 0: k 2 2 2 We note that if 1¼1ðc =V1 Þ¼1—that is, dsn;k ¼ 0 h¼1 2 and accordingly dT0 ¼ 0—then the particle will not move Let us consider the motion of a particle in relation to in time (for the observed particle, time will not exist). 2 2 2 2 the frame K. The (n þ k)-dimensional interval, being Indeed, in this case we have dT0 ¼ dt01 þ dt02 þþdt0k ¼ 2 0—that is, dt2 dt2 dt2 0. Thus, for the invariant, is given in this case by the expression dsn;k ¼ 01 ¼ 02 ¼ ¼ 0k ¼ 2 2 2 2 2 2 observed particle the time axes t , t , ..., t do not c dt1 þþc dtk dxkþ1 dxkþn 0. 01 02 0k By analogy with STR, we have to assume that the exist. P k 2 proper time is invariant. If we consider the quotient of the Let us set dt0h ¼ a0hdT0.Wehave a ¼ 1—that qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffih¼1 0h two invariant quantities dsn,k and c, we obtain an kþn 2 is, ja0hj1. Let us set dX ¼ dx . 0, dT ¼ invariant value having the physical dimension of time. qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g¼kþ1 g k 2 We assume that this value is equal to the length of the qrffiffiffiffiffiffiffiffiffiffiffiffiffi¼1 dtr . 0. From Eq. (35) we obtain dT0 ¼ vector dT0. In the general case (k 1), the obtained value 2 dT 1 b . P will be the proper time. We will have dT ¼jds j/c. 2 k 0 n,k If we use the equalities ds2 c dT2 ¼ 0, dT2 ¼ Hence, the length of the vector dT is invariant. n;k 0 0 h¼1 0 dt2 ¼ (dt /a )2, we will have In order to understand the physical meaning of 0h 0h 0h proper time, we will make the following considerations: 2 2 2 2 2 2 2 2 c dt1 þþc dtk dxkþ1 dxkþn c dt01 2 2 2 2 2 2 2 2 We will have dsn;k ¼ c (dt1 þ dt2 þ þ dtk)(1 – b ) ¼ c dt 2 2 0k c dT0, where ¼ 0; ð36Þ 1 b vffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; ¼ u c2dt2 c2dt2 dx2 dx2 c2dT2 uXk 2 1 þþ k kþ1 kþn 0 t c ¼ 0; ð37Þ V2 1¼1 1 c2dt2 c2dt2 dx2 dx2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þþk kþ1 kþn u 2 u Xkþn 2 dt0h t 2 c dxp a0h p¼kþ1 ¼ 0; ð38Þ V1 ¼ . 0: jdt1j 2 2 2 2 2 Let us assume that the (n þ k)-dimensional vector of the c dT dX c dT0 ¼ 0: ð39Þ location of the particle R ¼ (ct1, ..., ctk, xkþ1, ..., xkþn)is According to Eq. (36), instead of (n þ k)-dimensional a function of the proper time T0 ¼ (t01, t02, ..., t0k)—that space-time, we can consider a generalized [(n þ k) þ k)]- is, xl ¼ xl(T0), where l ¼ 1, 2, . . ., k þ n; here xl¼ ctl at l dimensional space-time, where k dimensions are related to ¼ 1, 2, . . ., k. Let us consider the motion of a particle in the projections of the proper time. In the generalized the inertial frame of reference K0, the velocity of which at space-time, k dimensions are timelike (ct1, ct2, ..., ctk) a given moment T0 ¼ (t01, t02, ..., t0k) coincides with the and (n þ k) dimensions are spacelike (xkþ1, xkþ2, ..., xkþn, velocity of the particle. Let the coordinates of the particle ct01, ct02,...,ct0k). In accordance with Eqs. (37) and (38), at this moment be x0l ¼ x0l(T0). Since at the considered instead of (n þ k)-dimensional space-time, we can consider moment the particle is at rest relative to K0,atl ¼ kþ1, a generalized [(n þ 1) þ k)]-dimensional space-time, where kþ2,..., kþn we will have x0l ¼ x0l(T0 þ dT0). So we will the additional dimension is related to the proper time cT0 obtain x01(T0 þ dT0) x01(T0) ¼ dx01, x02(T0 þ dT0) x02 or to the proper time (ct0h/a0h). In this case, k dimensions (T0) ¼ dx02, , x0k(T0 þ dT0)–x0k(T0) ¼ dx0k. are timelike (ct1, ct2, ..., ctk) and (n þ 1) dimensions are Let us set dx0h ¼ cdt0h,whereh ¼ 1, 2, . . ., k.Heret0h ¼ spacelike (xkþ1, xkþ2,...,xkþn, cT0 or xkþ1, xkþ2, ..., xkþn, x0h/c are the time axes of the frameqKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0. Hence we have cdT0¼ ct0h/a0h). According to Eq. (39), instead of (n þ k)- 0 0 0 2 2 2 dsn,k ¼ dsn;k ¼ cdT0,wheredT0 ¼ dt01 þ dt02 þþdt0k. dimensional space-time, we can consider a generalized (2 0 We have dT0 ¼ dT0—that is, þ 1)-dimensional space-time. In this case, one of the 2 2 2 2 2 2 dimensions is timelike (cT) and two of the dimensions are dT0 ¼ dt01 þþdt0k ¼ðdt1 þþdtkÞð1 b Þ: spacelike (X, cT0). Similar considerations are valid in STR ð35Þ (see Section II). 422 Phys. Essays 25, 3 (2012) vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u By analogy with STR, in the case of (n þ k)- uXk uXk dimensional space-time we can define the generalized t 2 t 2 Ut ¼ Utr ¼ Ut0h velocity U. Let us assume that the location of an observed r¼v1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffih¼1 u  particle in the frame K0 is given by the (n þ k)-dimensional uXk Xk 2 cdT t khrvr vector R0 ¼ (ct01, ..., ct0k, x0(kþ1), ..., x0(kþn)); then the ¼ : ð42Þ dT a location of the particle in K is given by the (n þ k)- 0 r¼1 h¼1 0h dimensional vector R ¼ (ct , ..., ct , x , ..., x ). If the 1 k kþ1 kþn Let us set location of the particle in K0 is given by the (n þ k)- vffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u  dimensional vector R0 þ dR0 ¼ [ct01 þ cdt01, ..., ct0k þ uXk uXk 2 t 2 jdxgj t khg cdt0k, x0(kþ1), x0(kþ2), ..., x0(kþn)], then the location of the U ¼ u ¼ ; sg hg dT a particle in K is given by the (n þ k)-dimensional vector R þ h¼1 0 h¼1 0h dR ¼ (ct1 þcdt1, ..., ctk þ cdtk, xkþ1 þ dxkþ1,...,xkþn þ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u dxkþn). For the case of multidimensional time we have u Xkþn u Xkþn dT 3 U ¼ dR, where dT ¼ (dt , dt , ..., dt )anddR ¼ t 2 dX t 2 0 0 01 02 0k Ush ¼ uhg ¼ ðkhgvgÞ ; jdt0hj (cdt1, cdt2, ..., cdtk, dxkþ1, dxkþ2, ..., dxkþn)—see also g¼kþ1 g¼kþ1 Section II. It is easy to prove that the generalized velocity vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U is a k 3 (k þ n) matrix with elements u (h ¼ 1, 2, . . ., k; u hl u Xkþn Xk t l ¼ 1, 2, . . ., kþn)—that is, U ¼ [uhl]k3(kþn). Let us denote U ¼ u2 : 0 1 s hg g¼kþ1 h¼1 u1l B C B u2l C Then for the velocity U we have B C s ul ¼ @ . A: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . u u u Xkþn uXk ukl t 2 t 2 Us ¼ Usg ¼ Ush Then we have dT 3 u ¼ cdt ,dT 3 u ¼ dx (r ¼ 1, gv¼kþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 h¼1 0 r r 0 g g u  2,..., k; g ¼ k þ 1,..., k þ n)—that is, u Xkþn Xk 2 dX t khgvg ¼ : ð43Þ dT a dt01 dt02 dt0k g¼kþ1 h¼1 0h u1r þ u2r þþ ukr ¼ 1; ð40Þ cdtr cdtr cdtr We are going to show the conditions for the velocities

dt01 dt02 dt0k Utr,Ut0h,Ut and Usg,Ush,Us. These conditions are imposed u1g þ u2g þþ ukg ¼ 1: ð41Þ by physical considerations (see also Sections II and VII). dxg dxg dxg Since during a change of the basis of the time axes of It is clear that if for a given d (1 d k) it is true that K0 (the so-called passive linear transformation) the value dt ¼ 0, then the components of the velocity u , 2 2 2 0d d1 of dT0 does not change, the values dX and dT remain ud2, ..., ud(kþn) will be undefined values. If for a given u the same as well [see Eq. (39)]. Likewise, during a change it is true that dtu ¼ 0 (where 1 u k)ordxu ¼ 0 (where of the basis of the spatial or the temporal axes of K, the k k n u u u 2 2 2 þ 1 u þ ), then 1u ¼ 2u ¼¼ ku ¼ 0. values dX , dT ,anddT0 remain constant. It is clear that Let us assume that uhr ¼ khr(cdtr/dt0h),uhg ¼ khg(dxg/ the temporal axes of the frame K0 and the spatial and dt0h), where h, Pr ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n. temporal axes of frame K can be chosen independently k Then we have h¼1 khl ¼ 1, where l ¼ 1, 2, . . ., k þ n. from each other (see Sections II and VII). Therefore, Let us set dtr ¼ vrdT and dxg ¼ vgdX (Pr ¼ 1,2,..., k; g when applying these transformations, some physical k 2 ¼Pk þ 1, k þ 2,..., k þ n). We then have r¼1 vr ¼ 1 and values (in this case velocities and consequently energy or kþn 2 g¼kþ1 vg ¼ 1—that is, jvrj1andjvgj1. momentum—see further) must remain invariant. Let us set In the expression for the velocity U , all time axes of tr qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 u u  K0 are included in the denominator (dT0 ¼ h¼1 dt0h); uXk uXk 2 t cjdtrj t khr we thus assume that the velocity U is invariant during a U ¼ u2 ¼ ; tr tr hr dT a change of the basis of the time axes of K . Therefore, the h¼1 0 h¼1 0h 0 velocity Utr can be presented in the form Utr ¼ ctr(jcdtrj/ vffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u dT0), where ctr . 0 is a parameter that does not depend uXk uXk t 2 cdT t 2 on dt0h or, consequently, on the numbers a0h (h ¼ 1, Ut0h ¼ uhr ¼ ðkhrvrÞ ; k jdt0hj 2,..., ). r¼1 r¼1 In the expression for the velocity U , all time axes of t0hqPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 u K are included in the numerator (dT ¼ r¼1 dtr); we uXk Xk t thus assume that the velocity U is invariant during a U ¼ u2 : t0h t hr change of the basis of the time axes of K. Therefore, the r¼1 h¼1 velocity Ut0h can be presented in the form Ut0h ¼ ct0h(cdT/ Then for the velocity Ut we have jdt0hj), where ct0h . 0 is a parameter that does not depend Phys. Essays 25, 3 (2012) 423 on dtr or, consequently, on the numbers vr (r ¼ 1, Let us set vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2,..., k). u uXkþn Xk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi From here we can conclude that the velocity Ut is t U u2 U2 U2: invariant during a change of the basis of the time axes of ¼ hl ¼ t þ s l¼1 h¼1 K0 and during a change of the basis of time axes of K— i.e., the velocity Ut can be presented in the form Ut ¼ Since Ut ¼ ct(cdT/dT0) and Us ¼ cs(dX/dT0), we obtain ct(cdT/dT0), where ct . 0 is a parameter that does not   cdT 2 dX 2 depend on dt0h and dtr or, consequently, on the numbers 2 2 2 U ¼ ct þcs : a0h and vr (r, h ¼ 1, 2, . . ., k)—see Eq. (42). Indeed, dT0 dT0 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u u In support of these assumptions we can present one uXk uXk uXk t 2 cdT t 2 2 t 2 more argument. In STR, the real space-time is compared Ut ¼ Utr ¼ vrctr ¼ Ut0h dT0 to the dual space of the energy-momentum. These r¼v1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r¼1 h¼1 u considerations must apply also for the case of multidi- uXk cdT t c2 ¼ t0h: mensional time. Therefore, between the total energy and dT a2 0 h¼1 0h total momentum in the multidimensional time there should be a relation similar to the one in Eq. (39).The Let us set velocities U and U participate accordingly in the vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t s u u formulas for the total energy and for the total uXk uXk t t c2 momentum—see Eqs. (61), (71), and (73). If the c ¼ v2c2 ¼ t0h: t r tr a2 equalities U ¼ c (cdT/dT )andU ¼ c (dX/dT )are r¼1 h¼1 0h t t 0 s s 0 valid, then between the total energy and the total Since the values v and c do not depend on the numbers momentum there will be a dependence which is similar r qtrPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 2 to the relation in Eq. (39)—see Eqs. (73) and (74). a0h, the parameter ct ¼ r¼1 vrctr will not either. Since pffiffiffi Let us assume that a a a 1/ k (that is, the values a0h and ct0h do not depend on the numbers vr, 01 ¼ p02ffiffiffi¼¼ 0k ¼ the parameter dt01pffiffiffi¼ dt02 ¼¼dt0k ¼ dT0/ k . 0), vp1 ¼ffiffiffi v2 ¼¼vk ¼ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1/ k (that is, dt ¼ dt ¼¼dt ¼ dT/ k . 0), and v u 1 2 pffiffiffi k kþ1 uXk 2 ¼ vkþ2 ¼¼vkþn ¼ 1/ n, (that is, dxkþ1 ¼ dxkþ2 ¼¼ t ct0h pffiffiffi ct ¼ dx ¼ dX/ n . 0). Let us apply a proper or improper a2 kþn h¼1 0h rotation of the time axes of the frame K0 in the hyperplane of the time. The transformation under will not either. Therefore, the parameter ct does not consideration can be presented through the orthogonal depend on the numbers a0h and vr or, consequently, on dt and dt . matrix A ¼ [ah1]k3k, belonging to the orthogonal group 0h r R R In the expression for the velocity U , all the time axes O(k, ), where denotes the real numbers field (see also sg Section VII). The new time axes obtained after applying of K are included in the denominator (dT ¼ qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 0 0 k this transformationP we willP denote with t0h (h ¼ 1, 2 k 2 k h¼1 dt0h); we thus assume that the velocity Usg is 0 2 2,..., k). Since h¼1ða0hÞ ¼ h¼1 a0h ¼ 1, we will have invariant during a change of the basis of the time axes of 0 1 0 1 0 K0. Therefore, the velocity Usg can be presented in the a01 a01 B a0 C B C form Usg ¼ csg(jdxgj/dT0), where csg . 0 is a parameter B 02 C B a02 C B . C ¼ A 3 B . C; that does not depend on dt0h or, consequently, on the @ . A @ . A numbers a0h (h ¼ 1, 2, . . ., k)—see Eq. (4). 0 a0k a0k In the expression for the velocity Ush, all the space axes of the frame K are included in the numerator (dX ¼ that is, qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kþn 2 dx ); we thus assume that the velocity Ush is Xk Xk g¼kþ1 g 0 1ffiffiffi invariant during a change of the basis of the space axes of a0h ¼ ah1a01 ¼ p ah1: ð44Þ 1¼1 k 1¼1 K. Therefore, the velocity Ush can be presented in the form U c (dX/ dT ), where c . 0 is a parameter sh ¼ sh j 0hj sh Since the velocity Utr is invariant during a change of that does not depend on dx or, consequently, on the g the basis of the time axes of K0, the expression numbers vg (g ¼ k þ 1, k þ 2,..., k þ n)—see Eq. (8).  Xk 2 We can prove by analogy that the velocity Us is khr invariant during a change of the basis of the time axes of a h¼1 0h K0 and during a change of the basis of the space axes of remains constant during a change of the numbers a — K—i.e., the velocity Us can be presented in the form Us ¼ 0h that is, cs(dX/dT0), where cs . 0 is a parameter that does not depend on dt and dx or, consequently, on the numbers   0h g Xk k0 2 Xk k 2 Xk a and v (h ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n)—see hr ¼ hr ¼ k k2 ¼ c2 ; ð45Þ 0h g a0 a hr tr Section II and Eq. (43). h¼1 0h h¼1 0h h¼1 424 Phys. Essays 25, 3 (2012) pffiffiffi where ctr is a parameter that does not depend on the ¼ 1/ k. Since the parameter ct does not dependpffiffiffi on the numbers a0h (h ¼ 1, 2, . . ., k). numbers a0h and vr, we can set a0h ¼ vr ¼ 1/ k. In this Let us consider the orthogonal matrix B ¼ [bhm]k3k, case we have R vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi belonging to the orthogonal group O(k, ). Taking into u u uXk Xk 2 uXk Xk account Eq. (45), we likewise obtain t khrv t c ¼ r ¼ k2 : ð52Þ t a hr k0 Xk b k pffiffiffi Xk r¼1 h¼1 0h r¼1 h¼1 hr ¼ hm mr ¼ k b k : ð46Þ a0 a hm mr 0h m¼1 0m m¼1 [See Eq. (42).] 0 Similar considerations apply for the velocities Usg and From Eqs. (44) and (46) we can define the quantities khr: 00 Ush (see Section II). For example, if the quantities khg are Xk Xk Xk Xk defined at arbitrary values of a0 and v00 ,andthe 0 ah1a01bhmkmr 0h g pffiffiffi khr ¼ ¼ ah1bhmkmr: ð47Þ quantities kmq are defined provided that a0h ¼ 1/ k, vg ¼ a0m pffiffiffi m¼1 1¼1 m¼1 1¼1 1/ n, then we have ! Let us apply a proper or improper rotation of the Xkþn Xk Xk 1 ah1a01bhmkmqv qqg time axes of the frame K in the hyperplane of the time. k00 ¼ q hg Xkþn a The transformation under consideration can be presented q¼kþ1 m¼1 1¼1 0m v h through the orthogonal matrix J [ j ] , belonging to p pg ¼ dr k3k p¼kþ1 the orthogonal group O(k, R). The new time axes Xkþn Xk Xk obtained applying this transformationP weP will denote 00 k 00 2 k 2 ah1bhmkmqqqg with tr (r ¼ 1, 2, . . ., k). Since r¼1ðvrÞ ¼ r¼1 vr ¼ 1, 00 00 00 q¼kþ1 m¼1 1¼1 we will have (v1 , v2 , , v ) ¼ (v1, v2, , vk) 3 J—that is, ¼ : k Xkþn Xk Xk hpg 00 1ffiffiffi v ¼ vdjdr ¼ p jdr: ð48Þ p¼kþ1 r k d¼1 d¼1 ð53Þ

Since the velocity Ut0h is invariant during a change of [See Eq. (10).] Here ah1, bhm, hpg, qqg are elements of the the basis of the temporal axes in K, the expression orthogonal matrices A ¼ [ah1]k3k, B ¼ [bhm]k3k, H ¼ [hpg]n3n, Xk Q ¼ [qqg]n3n, respectively. [The matrices A and B belong to 0 2 R ðkhrvrÞ the orthogonal group O(k, ), and the matrices H and Q r¼1 belong to the orthogonal group P(n, R).] For the remains constant during change of the numbers v —that is, parameter cs we have r vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u rffiffiffiu Xk Xk Xk u Xkþn Xk 2 u Xkþn Xk 2 2 1 2 t khgvg kt ðk00 v00 Þ ¼ ðk0 v Þ ¼ ðk0 Þ ¼ c2 ; ð49Þ c ¼ ¼ k2 : hr r hr r k hr t0h s a n hg r¼1 r¼1 r¼1 g¼kþ1 h¼1 0h g¼kþ1 h¼1 where ct0h is a parameter that does not depend on vr (r ¼ 1, ð54Þ 2, . . ., k). [See Eqs. (11) and (43).] In Eq. (54)p theffiffiffi quantitiesffiffiffi khg are Let us consider the orthogonal matrix L ¼ [lqr]k3k, p determined providedP that a0h ¼ 1= k; vg ¼ 1= n. belonging to the orthogonal group O(k, R). Taking into k Let us set K00 ¼ k00 , l ¼ 1, 2, . . ., k þ n. [See Eqs. account Eq. (49), we likewise obtain l h¼1 hl (51) and (53).] In the general case (i.e., at arbitrary values 0 00 00 Xk Xk of a0h, vr, and vg ), we will have 00 00 0 1 0 k v ¼ k v lqr ¼ pffiffiffi k lqr: ð50Þ hr r hq q hq K 00 q¼1 k q¼1 dT0 3 U ¼ dR; From Eqs. (47), (48), and (50) we can define the where 00 quantities khr: hi 00 00 00 00 00 k v cdT !UK uK ; uK hr r ; Xk Xk Xk ¼ hl hr ¼ 00 0 1 a a b k v l k 3ðkþnÞ Kra0 dT0 k00 ¼ h1 01 hm mq q qr h hr k 00 00 X a0m 00 k v dX q¼1 m¼1 1¼1 uK hg g : vdjdr hg ¼ 00 0 Kg a0hdT0 d¼1 P k Xk Xk Xk [See Eqs.P (40), (41), (44), (48), (51), and (53).] If h¼1 k00 k k (that is, K00 1, l 1, 2, . . ., k n), ah1bhmkmqlqr hl 6¼ h¼1 hl l 6¼ ¼ þ q¼1 m¼1 1¼1 then ¼ ; ð51Þ Xk 00 00 00 00 K 00 k v cdT 00 k v cdT jdr hr r hr r uhr ¼ 00 0 6¼ uhr ¼ 0 d¼1 Kra0hdT0 a0hdT0 where the numbers kmq are defined provided that a0h ¼ vr and Phys. Essays 25, 3 (2012) 425 0 1 00 00 00 00 00 k v dX k v dX e1r uK ¼ hg g 6¼ u00 ¼ hg g : B C hg 00 0 hg 0 B e2r C Kg a0hdT0 a0hdT0 Er ¼ B . C; @ . A Let us set u ¼ (dX/dT ) ¼ (U /c ). We will say that u s 0 s s s e is the total proper velocity of the particle under kr consideration. The relation between the total coordinate where velocity uc ¼ (dX/dT) ¼ (Uc/c)—see Section II—and the k v m c2 total proper velocity us is given through the expression us ¼ hr r 0 qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffi ehr ¼ m0uhrc ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; ð57Þ 2 2 2 uc 1 b .WehaveUs ¼ csUc/(qc ffiffiffiffiffiffiffiffiffiffiffiffiffi1 b )—see Eqs. (2), a0h 1 b 2 (11), (43), and (54). If dt0h ¼ dth 1 b (h ¼ 1, 2, . . ., k), r, h ¼ 1, 2, ..., k [see also Eq. (55)]. The energy of the particle will have k2 total components. If for a given d (1 then c ¼ c (see Sections II and V). s k dt Let us consider the frame K000 , moving uniformly and d ) the condition 0d ¼ 0 is fulfilled, then the velocity components u , u , ..., u will be undefined rectilinearly in relation to K. It is clear that in the frame d1 d2 dk values and thus the energy components e , e , ..., e K000 a particle will have a different generalized velocity U000 , d1 d2 dk 000 000 will also be undefined values. If for a given d (1 d k) given by the matrix U ¼ [uhl]k3(kþn). Let us choose the 000 the condition dtd ¼ 0 is fulfilled—i.e., a given particle is framesffiffiffi K0 and K in suchffiffiffi a way that a0h ¼ dt0h/dT0 ¼ 1/ p p pffiffiffi not moving in the time dimension t —then we have u ¼ k, v000 ¼ dT000 /dT000 ¼ 1/ k, v000 ¼ dx000 /dX000 ¼ 1/ n , where d 1d r r g g u ¼¼u ¼ 0 and therefore e ¼ e ¼¼e ¼0. h, r ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n. In this case 2d kd 1d 2d kd 000 From Eqs. (40) and (57) the following important the quantities k ,defined in the frame K000 —as with k , hl hl equalities are obtained: defined in K—can take arbitrary (real) values.P The only k 000 condition for this is defined with the equality h¼1 khl ¼ dt01e1r þ dt02e2r þþdt0kwkr ¼ cdtr; 1, where l ¼ 1,P 2, . . ., k þ n. (For the quantities khl, the m c k 0 similar equality h¼1 khl ¼ 1 applies.) Due to this fact we 000 000 that is, can assume that the values of khl in the frame K coincide with the corresponding values of khl in the frame K—that a e þ a e þþa e m c2 000 01 1r 02 2r 0k kr ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi0 . 0: ð58Þ is, khl ¼ khl, where h ¼ 1, 2, . . ., k; l ¼ 1, 2, . . ., k þ n. (Thepffiffiffi v 000 000 r 1 b2 values khlffiffiffiare determined provided that a0h ¼ vr ¼ 1/ k, 000 p v ¼ 1/ n, and theffiffiffi values khl are determined provided g p pffiffiffi Let us assume that for a given d (1 d k)wehave that a ¼ v ¼ 1/ k, v ¼ 1/ n.) 0h r g v ,0—that is, the particle under consideration is moving For the velocities u (h ¼ 1, 2, . . ., k; l ¼ 1, 2, . . ., k þ d hl backward in the time dimension t (dt , 0). Let us n)wehave d d simultaneously multiply the numbers vd, a01, a02,...,a0k cdtr khrvrc by 1. This operation corresponds to an inversion of the uhr ¼ khr ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; ð55Þ dt0h 2 time dimension dtd ¼ vddT and an inversion of the proper a0h 1 b time dT0 ¼ (dt01, dt02, ..., dt0k) ¼ (a01dT0, a02dT0, ..., a0kdT0). According to Eq. (58), this operation will not k v cb dxg hg g change the components of the energy e1d, e2d, ..., ekd. (If uhg ¼ khg ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; ð56Þ dt0h 2 r 6¼ d, during this operation the number vr does not a0h 1 b change its sign but the numbers a01, a02, ..., a0k change where r ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n. their signs, and therefore the energy components e1r, e2r, ..., ekr change their signs, as well.) Therefore, if a given particle is moving backward in the time dimension B. Energy and momentum of a particle moving in k- t (that is, dt , 0) and the projections of the proper time dimensional time d d on the time axes are dt01, dt02, ..., dt0k, then we can Let us consider the motion of a particle in relation to assume that the particle is actually moving forward in the the frame K. In the case of the one-dimensional time of time dimension td (that is, dtd . 0) but the projections of STR, a (3 þ 1)-dimensional energy-momentum vector is the proper time on the time axes are dt01, dt02, ..., defined. Likewise, in the case of multidimensional time we dt0k. These results have important consequences for can define a k 3 (k þ n) energy-momentum matrix p ¼ antiparticles in multidimensional timepffiffiffi (see Section X). m0U, where m0 . 0,U ¼ [uhl]k3(kþn). The physical meaning If in Eq. (57)qweffiffiffiffiffiffiffiffiffiffiffiffiffi set a0h ¼ vr ¼ 1/ k, then we will have of the constant m will be clarified later. 2 2 0 ehr ¼ khrm0c / 1 b . From here we can define the Let us denote p ¼ m u and p ¼ m u (h ¼ 1, l 0 l hl 0 hl quantities khr: 2,..., k; l ¼ 1, 2, . . ., k þ n). First, let us consider the qffiffiffiffiffiffiffiffiffiffiffiffiffi components p , where r,h ¼ 1, 2, . . ., k. By analogy with 2 hr ehr 1 b STR, if we multiply by the velocity of light in a vacuum c, khr ¼ 2 : ð59Þ m0c we will obtain the components of the energy ehr. The energy of the particle defined in relation to the time In this case, if khr . 0 then we have ehr . 0, and vice dimension tr has k components: versa. If khr , 0 then we will have ehr , 0, and vice versa. 426 Phys. Essays 25, 3 (2012) qffiffiffiffiffiffiffiffiffiffiffiffiffi 000 2 As pointed out in Subection IX.A, if the frame K is mass m ¼ m0/ 1 b defined in K. Let us set V1 ¼ V2 ¼ moving uniformly and rectilinearly in relation to K, then ¼ Vk ¼ V. (This can be obtained through an 000 for an appropriatepffiffiffi choice of the axes of K0 and K (a0h ¼ appropriate rotation in the hyperplane of time—see 000 vr ¼ 1/ k), the quantities khr will conserve their value Section VII.) Then we have during the transfer from K to K000 . qffiffiffiffiffiffiffiffiffiffiffiffiffim0 qffiffiffiffiffiffiffiffiffiffiffiffiffim0 Let us denote Er ¼jjErjj 0. The aggregate energy, m ¼ ¼ : ð63Þ 2 V2 defined in relation to the time dimension tr, is equal to 1 b 1 kc2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u  uXk 2 uXk 2 Equation (63) gives the relationship between relativ- t m0c jv j t khr E ¼ e2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffir istic mass m, rest mass m , velocity V of a particle, and the r hr a 0 h¼1 1 b2 h¼1 0h number of time dimensions k in which the particle is 2 moving. According to Eq. (63), a given particle has zero mq0cffiffiffiffiffiffiffiffiffiffiffiffiffijvrjctr ¼ ; ð60Þ rest mass if b ¼ 1—that is, 1 b2 Xk c2 where c . 0 is a parameter that does not depend on the ¼ 1 tr V2 numbers a (h ¼ 1, 2, . . ., k)—see Subsection IX.A and 1¼1 1 0h pffiffiffi more precisely Eq. (45). [If we assume that the given (and accordingly V ¼ c k). At k ¼ 1 we obtain the well- particle does not move in the time dimension td (1 d known case of STR. From Eq. (63) it follows that at k), then we will have vd ¼ 0 and accordingly Ed ¼ 0.] constant velocity V (V ¼ const), as the number of time The total energy of the particle which is moving in k dimensions increases, the relativistic mass m decreases. timeqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dimensions will be defined through the expression E P If in Eq. (61) we set V1 ¼ V2 ¼¼Vk ¼ V, then we k 2 ¼ r¼1 Er . 0—that is, obtain vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 2 2 uXk 2 qmffiffiffiffiffiffiffiffiffiffiffiffiffi0c ct 2 m0c t 2 2 m0c ct E ¼ ¼ mc ct: ð64Þ E ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi v c ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; ð61Þ 2 r tr 1 V 1 b2 r¼1 1 b2 kc2 Through Eq. (64) we determine the relation between where ct . 0 is a parameter that does not depend on the total energy E, rest mass m0, relativistic mass m, velocity numbers vr and a0h [see Subsection IX.A and more precisely Eqs. (42) and (52)]. V of a particle, and number of time dimensions k in which Let us assume that the frame K coincides with the the particle is moving. As can be seen from Eq. (64), only for the case k ¼ 1 proper frame of reference K0 (see Subsection IX.A). In do we have following peculiarity: If V c, then E ‘— this case we have dtr ” dt0r (that is, dT ¼ dT0,vr ¼ a0r),dxg ¼ 0, where r ¼ 1,2,..., k; g ¼ k þ 1, k þ 2,..., k þ n. The that is, the energy of the particle is an infinite quantity proper energy or the rest energy of the particle defined in (see Figs. 4pandffiffiffi p5ffiffiffiffiffiffiffiffiffiffiffi). It is clear that if k . 1 and V ¼ c, then E ¼ m c2c k/ k 1. Therefore, if a particle with rest relation to the time dimension tr ” t0r has k components: 0 t 0 1 mass (rest energy) differing from 0 and moving in a e01r number k . 1 of time dimensions with the same B C B e02r C velocity—which is equal to the speed of light in B C E0r ¼ @ . A; . vacuum—then its energy is notp infinite,ffiffiffi but will have a finite value. However, if V c k, then E ‘—that is, e0kr the energy of the particle is an infinite value. (As noted in 2 pffiffiffi where e0hr ¼ khra0rm0c /a0h,r, h ¼ 1, 2, . . ., k.The Section VIII, the velocity V ¼ c k is a constant, invariant aggregate proper energy defined in relation to the time value in relation to all inertial frames of reference.) 2 dimension tr is equal to E0r ¼ m0c jvrjctr. Let us consider two particles L and N, moving The total proper energy of the particle moving in k qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi uniformly and rectilinearly to each other. Let us assume k 2 dimensions is defined by the expression E0 ¼ r¼1 E0r that particle N is moving in only one time dimension (t1) . 0—that is, and that particle L is moving in k time dimensions (t1, 2 t2, ..., tk), where k . 1. We assume that particle N is not E0 ¼ m0c ct: ð62Þ a luxon—i.e., the rest mass of the particle N differs from

It is clear that if k ¼ 1 (and accordingly ct ¼ 1), then E 0. Let us assume, as well, that particle L is a luxon—i.e., 2 ¼ m0c . This is the well-known formula obtained in STR. the velocity of L in relation to N defined according to the Hence, the constant m0 . 0 is equal to the proper mass or different temporal dimensions t1, t2, ..., tk is the same to the rest mass of the particle (at V1 ¼ V2 ¼¼Vk ¼ 0). and is equal to the speed of light in a vacuum. According By analogy with the case k ¼ 1 (i.e., STR), we will to Eq. (64), the total energy of particle L definedffiffiffi ffiffiffiffiffiffiffiffiffiffiffi in 2 p p assume that if a particle has rest mass m0 and is moving relation to particle N is equal to E ¼ m0 c ct k/ k 1 with velocities V1, V2, V3, ..., Vk (defined in relation to (where k . 1). Although particle L is moving in relation the time dimensions t1, t2, ..., tk) in relation to a given to particle N with a velocity equal to the speed of light in a frame of reference K, this particle will have relativistic vacuum, the rest mass of particle L differs from 0: m0 ¼ Phys. Essays 25, 3 (2012) 427

FIG. 4. (Color online) Relation between the quantity e1 and the number of time dimensions k. pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi E k 1/c2c k 6¼ 0. The reason for this result is that 2 t qmffiffiffiffiffiffiffiffiffiffiffiffiffi0c particle L is moving in more than one time dimension (k E ¼ : 1 V2 . 1). Therefore, the particles which are moving with a 2c2 velocity equal to the speed of light in a vacuum (i.e., the If we set V ¼ 0, then according to Eq. (65) we obtain the luxons) can have nonzero rest mass, but on the condition following expression for the proper energy of the particle: that they move in two, three, or more time dimensions. 2 2 2 2 2 m0c E0 m0c k: ð66Þ Let us assume k1l ¼ k2l ¼¼kkl—thatP is, jk1lj¼ k jk2lj¼¼jkklj, l ¼ 1, 2, . . ., k þ n. Since h¼1 khl ¼ 1, From Eq. (64) it follows that if V ¼ const and ct ¼ k (k the minimum value which the quantity jkhlj can take in is an odd number) or ct ¼ k/2 (k is an even number), then this case is 1/k (here h ¼ 1, 2, . . ., k). If k is an odd as the number of time dimensions increases, so does the number, then the maximum value which the quantity jkhlj total energy (see Fig. 4). In this case, the additional time can take is 1; and if k is an even number, then the dimensions ‘‘add’’ additional energy to the energy of the maximum value which jkhlj can take is 1/2. If we take into particle. Let us set ct ¼ k at k ¼ 2k1 þ 1 and ct ¼ k/2 at k ¼ 2 account Eqs. (52) and (54), then for odd values of k we 2k1, where k1 ¼ 0,1,2,3,....Letusdenote e1 ¼ E/m0c . According to Eq. (64) we obtain will obtain 1 ct k and accordingly 1 cs k. For even values of k we will have 1 ct k/2 and accordingly k e1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi 1 cs k/2. We note that if k ¼ 2, then ct ¼ 1 and thus cs 2 1 V ¼ 1. kc2 Since 1 c k, according to Eq. (64) we have t at k ¼ 2k1 þ 1and 2 2 qmffiffiffiffiffiffiffiffiffiffiffiffiffi0c qmffiffiffiffiffiffiffiffiffiffiffiffiffioc k k E : ð65Þ e1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi 1 V2 1 V2 V2 kc2 kc2 2 1 kc2

If k ¼ 2, then ct ¼ 1—that is, at k ¼ 2k1. Figure 4 gives the relation between the quantity

FIG. 5. (Color online) Relation between the quantity e2 and the number of time dimensions k. 428 Phys. Essays 25, 3 (2012) e1 and the number of time dimensions k at different values of these two operations leads to identical results, namely of the ratio V/c (0.000, 0.990, 1.000, 1.900, 2.500, 3.000). changing the signs of all components of the energy and of According to Eq. (64),ifV ¼ const and ct ¼ 1, then as the momentum. A similar statement is valid in STR. the number of time dimensions increases, the total energy These results have important significance in relation to decreases (see Fig. 5). In this case, the additional time antiparticles in multidimensional time (see Section X). pffiffiffi pffiffiffi dimensions ‘‘subtract’’ from the energy of the particle. Let If in Eq. (67) we set a0h ¼ 1/ k, vg ¼ 1/ n, then we 2 us set ct ¼ 1 and e2 ¼ E/m0c . According to Eq. (64) we obtain obtain pffiffiffi m0kqhgffiffiffiffiffiffiffiffiffiffiffiffifficb k 1 phg ¼ ffiffiffi : e2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi p 2 V2 n 1 b 1 kc2 From here we can define the quantities khg: Figure 5 gives the relation between the quantity e2 and the qffiffiffiffiffiffiffiffiffiffiffiffiffi number of time dimensions k at different values of the pffiffiffi p n 1 b2 ratio V/c. hg khg ¼ pffiffiffi : ð69Þ As can be seen from Figs. 4 and 5, only in the case k ¼ m0cb k 1 do we have following peculiarity: If V c, then e 1,2 In this case, if k . 0 then we have p . 0, and vice ‘ and accordingly E ‘. Further, if k ¼ 9 and V 3c, hg hg versa. If k , 0 then we have p , 0, and vice versa. As then e ‘ and accordingly E ‘. hg hg 1,2 already pointed out in Subsection IX.A, if the frame K000 is In the case of multidimensional time, the momentum moving uniformly and rectilinearly in relation to K, then of the particle defined in relation to the space dimension 000 forpffiffiffi appropriateffiffiffi choice of the axes of K0 and K (a0h ¼ 1/ xg has k components: 000 p 0 1 k, vg ¼ 1/ n), the quantities khg conserve their value 000 p1g during transfer from K to K . B C B p2g C Let us denote p ¼jjp jj . 0. The momentum of the B C g g pg ¼ . ; particle defined in relation to the space dimension x is @ . A g p given through the following formula: kg vffiffiffiffiffiffiffiffiffiffiffiffiffiffi u where uXk m jv jc cb t 2 q0 ffiffiffiffiffiffiffiffiffiffiffiffiffig sg pg ¼ phg ¼ : ð70Þ m0khgvgcb h¼1 1 b2 phg ¼ m0uhg ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi : ð67Þ 2 a0h 1 b Here vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi [Here h ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n—see also u  uXk 2 Eq. (56).] The momentum of the particle has nk total t khg c ¼ . 0 components. If for a given d (1 d k) the condition sg a h¼1 0h dt0d ¼ 0 is fulfilled, then the velocity components ud(kþ1), ud(kþ2),...,ud(kþn) will be undefined values and therefore is a parameter which does not depend on the numbers a0h the momentum components pd(kþ1), pd(kþ2),...,pd(kþn) (see Subsection IX.A). will also be undefined. If for a given u (kþ1 u k þ n) The total momentum is defined through the following the condition dxu ¼ 0 is fulfilled—i.e., the particle does formulas: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi not move in the space dimension xu—then we will have u u Xkþn vu ¼ 0 and therefore p1u ¼ p2u ¼¼pku ¼ 0. It is seen t p ¼ p2 0; that if for a given q (1 q k) the condition Vq ¼ 0is s g g¼kþ1 fulfilled, then b ¼ 0 and therefore phg ¼ 0 for all values of h and g. that is, From Eqs. (41) and (67) we obtain the following: m c cb p ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi0 s ¼ mc cb; ð71Þ a01p1g þ a02p2g þþa0kpkg qmffiffiffiffiffiffiffiffiffiffiffiffiffi0cb s s ¼ . 0: ð68Þ 1 b2 vg 2 1 b qPffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kþn 2 2 where cs ¼ v c . 0 is a parameter which does From Eq. (68) it follows that if we simultaneously g¼kþ1 g sg not depend on the numbers vg and a0h [see Subsection multiply the numbers a01, a02, ..., a0k, vkþ1, vkþ2, ..., vkþn IX.A and more precisely Eqs. (43) and (54)]. by the number 1, then in this operation the momentum Let us set V ¼ V ¼ ¼ V ¼ V. (This can be components p will not change. 1 2 k hg obtained through an appropriate rotation in the hyper- According to Eqs. (58) and (68), the multiplication of plane of time—see Section VII.) We have the value m0 by the number 1 is equivalent to the multiplication of all numbers a01, a02, ..., a0k by 1—that m0csV mcsV ps ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi : ð72Þ is, the rest-mass sign inversion is equivalent to the proper V2 k k c2 time inversion dT0 ¼ (dt01, dt02, ..., dt0k). The application Phys. Essays 25, 3 (2012) 429

Equation (72) gives the relation between total (h ¼ 1, 2, . . ., k; l ¼ 1, 2, . . ., k þ n). For the energy and H A B momentum ps, rest mass m0, relativistic mass m, velocity the momentum, respectively, we have E ¼ E þ E þ D H A B D H A B V of a particle, and number of the time dimensions k in þ E —that is, ehr ¼ ehr þ ehr þþehr—and ps ¼ ps þ ps D H A B D which the particle is moving. þ þ ps —that is, phg ¼ phg þ phg þ þ phg (r ¼ 1, As is easy to see, following equalities are fulfilled: 2,..., k; g ¼ k þ 1,..., k þ n). For the total energies of the considered particles, we have cdT c E dX cb ps vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ; ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ : u dT0 2 m0cct dT0 2 m0cs uXk Xk 1 b 1 b H t H 2 E ¼ ðehrÞ ð73Þ r¼1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u [See Eqs. (61) and (71).] Taking into consideration Eqs. uXk Xk t A B D 2 (39) and (73), we obtain the following important equality: ¼ ðehr þ ehr þþehrÞ ; r¼1 h¼1 E2 p2c2 s ¼ m2c4: ð74Þ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 c2 0 u t s uXk Xk A t A 2 Equation (74) gives the relation between total energy E ¼ ðehrÞ ; r¼1 h¼1 E, total momentum ps, and rest mass m0.Ifk ¼ 1, then vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi from Eq. (74) we obtain the well-known equality derived u u in STR: E2 – c2p2 ¼ m2c4. uXk Xk uXk Xk s 0 B t B 2 D t D 2 E ¼ ðehrÞ ; ...; E ¼ ðehrÞ : r¼1 h¼1 r¼1 h¼1 C. Energy-momentum conservation law The total energies EH, EA, EB,...,ED are the In STR, the energy-momentum conservation law is H H A Frobenius norms of the matrices E ¼ [ehr]k3k, E ¼ derived as a consequence of continuous space-time A B B D D [ehr]k3k, E ¼ [ehr]k3k, E ¼ [ehr]k3k, respectively—i.e., symmetry (Noether’s theorem). For example, let us H H A B D A A B E ¼jjE jjF ¼jjE þ E þþE jjF, E ¼jjE jjF, E ¼ consider the process of decay of a particle. Applied to a B D D H A jjE jjF, , E ¼jjE jjF. (The total momentums ps , ps , decay process, energy-momentum conservation states that B D H ps , , ps are the Frobenius norms of the matrices ps ¼ the vector sum of the energy-momentum four-vectors of H A A B B D D [phg]k3n, ps ¼ [phg]k3n, ps ¼ [phg]k3n, , ps ¼ [phg]k3n, the decay products should equal the energy-momentum respectively.) The Frobenius norm possesses the following four-vector of the original particle. Some aspects of the A B D A B property: jjE þ E þþE jjF jjE jjF þjjE þþ problem for conservation of energy and momentum in EDjj jjEAjj þjjEBjj þþjjEDjj (the triangle 3 F F F F multidimensional time are discussed by Dorling. Until inequality property). Therefore, we have now it was accepted that in the case of k-dimensional time, the energy is a k-dimensional vector3–5.But EH EA þ EB þþED: ð76Þ according to the previous considerations, in the case of We obtained the result that in the case of multidi- k-dimensional time and n-dimensional space, the energy is mensional time, the energy conservation law as defined in a k 3 k matrix and the momentum is a k 3 n matrix. In STR will be violated—the magnitude of the total energy this case the energy-momentum conservation law applied of the original particle H is less than the sum of the to the process of decay of a particle will state following: magnitudes of the total energies of the decay products The matrix entrywise sum of the energy-momentum k 3 ( (the particles A, B, ..., D). The same applies also for the k þ n) matrices of the decay products is equal to the momentum of the considered particles—the momentum energy-momentum k 3 (k þ n) matrix of the original conservation law for the case of multidimensional time particle. In particular, the matrix entrywise sum of the differs from the momentum conservation law for the case energy k 3 k matrices of the decay products is equal to the of one-dimensional time (STR). energy k 3 k matrix of the original particle. Moreover, the Let us assume that the particle H is moving only in matrix entrywise sum of the momentum k 3 n matrices of one time dimension t1 and that the remaining particles A, the decay products is equal to the momentum k 3 n B, ..., D are moving in k . 1 time dimensions t1, ..., tk. matrix of the original particle. For example, if we denote According to the energy conservation law, the equality EH H H with p [p ]k3(k n) the energy-momentum matrix of the H A B D hl þ ” e11 ¼ e11 þ e11 þþe11 will be fulfilled. Further, at r . A A B H original particle (particle H) and with p ¼ [phl]k3(kþn), p 1orh . 1 the equality e ¼ 0,will be fulfilled—i.e., B D D hr ¼ [p ]k3(k n), , p ¼ [p ]k3(k n) the energy-momentum A B D hl þ hl þ ehr þ ehr þþehr ¼ 0. matrices of the decay products (of the particles A, Up to now, it has been accepted that particles moving B, ..., D, respectively), then we have in multidimensional time are more unstable and decay more easily than those moving in one-dimensional time.3 pH ¼ pA þ pB þþpD; However, this is only valid if we suppose that in the case that is, of k-dimensional time and n-dimensional space the energy

H A B D is a k-dimensional vector and the momentum is an n- phl ¼ phl þ phl þþphl; ð75Þ dimensional vector. According to our obtained results, in 430 Phys. Essays 25, 3 (2012) the case of k-dimensional time and n-dimensional space H 2 H 2 2 A 2 A 2 2 ðE Þ ðps Þ c H 2 4 ðE Þ ðps Þ c the energy is a k 3 k matrix and the momentum is a k 3 n 2 2 ¼ðm0 Þ c ; 2 2 ct cs ct cs matrix. We are going to prove that in multidimensional ¼ðmAÞ2c4; ...; time, like in one-dimensional time, the sum of the rest 0 masses of the decay products is always less than or equal ðEDÞ2 ðpDÞ2c2 s ¼ðmDÞ2c4: to the rest mass of the original particle. Therefore, c2 c2 0 particles moving in multidimensional time are as stable t s as those moving in one-dimensional time. Indeed, let us It is clear that if the inequality consider the process of decay of the particle H into the sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðEHÞ2 ðpHÞ2c2 ðEAÞ2 ðpAÞ2c2 particles A, B, ..., D. Let us choose the values of khl in s s 2 2 2 2 þ such a way that for a particular choice of time axes in the ct cs scffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit cs frame K, of some space axes in K, and of some time axes ðEDÞ2 ðpDÞ2c2 þ s in K0 (i.e., at a given vr, vg, a0h), the following equalities 2 2 ct cs are fulfilled: is fulfilled, then the inequality mH mA þ mB þþmD H A B D 0 0 0 0 k vH k vA k vB k vD will be valid as well. However, the first inequality is hl l ¼ hl l ¼ hl l ¼¼ hl l : ð77Þ H A B D equivalent to the following inequality: a0h a0h a0h a0h (Here h, r ¼ 1, 2, . . ., k; g ¼ k þ 1, k þ 2,..., k þ n; l ¼ 1, Xk Xk eA eD 2 2,..., k þ n—see Subsection IX.A.) It is obvious that in ð hr þþ hrÞ r¼1 h¼1 this case the parameters c , c will be the same for all t s c2 particles H, A, B, ..., D—i.e., cH ¼ cA ¼ cB ¼¼cD ¼ c t t t t t t Xkþn Xk H A B D . 0, c ¼ c þ c ¼¼c ¼ cs . 0 [see Eqs. (52) and 2 A D 2 s s s s c ðphg þþphgÞ (54)]. For the total energies and the total momentums of g¼kþ1 h¼1 the particles we have c2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s u Xk Xk Xkþn Xk uXk Xk 2 2 A 2 t ðeA Þ c ðp Þ EH ¼ ðeH Þ2 hr hg hr r¼1 h¼1 g¼kþ1 h¼1 r¼1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 c2 u t s uXk Xk t Xkþn Xk ¼ ðeA þ eB þþeD Þ2; Xk Xk hr hr hr D 2 c2 ðpD Þ2 r¼1 h¼1 ðehrÞ hg r¼1 h¼1 g¼kþ1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þþ 2 2 u ct cs uXk Xk A t A 2 E ¼ ðehrÞ ; ...; A B 4 A D 4 þ 2m0 m0 c þþ2m0 m0 c þ: r¼1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi From this inequality we obtain u uXk Xk kþn k t 2 Xk Xk X X D D 2 A B E ¼ ðehrÞ ; A B c p p ehrehr hg hg r¼1 h¼1 g¼kþ1 h¼1 r¼1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 c2 u t s u Xkþn Xk Xk Xk Xkþn Xk t 2 pH ¼ ðpH Þ A D c2 pA pD s hg ehrehr hg hg g¼kþ1 h¼1 g¼kþ1 h¼1 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þþ r¼1 h¼1 þ u c2 c2 u Xkþn Xk t s t A B D 2 A B 4 A D 4 ¼ ðphg þ phg þþphgÞ ; m0 m0 c þþm0 m0 c þ: g¼kþ1 h¼1 In order to find whether the obtained inequality is vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u fulfilled, we have to compare the expressions on each side u Xkþn Xk A t A 2 of the inequality concerning the respective pairs of ps ¼ ðphgÞ ; ...; g¼kþ1 h¼1 particles. For example, let us consider the expressions concerning the pair A and B—these are the expressions vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u Xkþn Xk Xk Xk Xkþn Xk t 2 pD ¼ ðpD Þ : A B c2 pA pB s hg ehrehr hg hg g¼kþ1 h¼1 g¼kþ1 h¼1 r¼1 h¼1 c2 c2 [See Eqs. (60), (61), (70), and (71)]. According to Eq. (74), t s A B 4 we have and m0 m0 c . (The comparison for the remaining pairs of Phys. Essays 25, 3 (2012) 431 particles is done similarly.) Taking into account Eqs. (57), lated by Stuckelberg,¨ Feynman, Sudarshan, and Reca- (67), (52), and (54), we have mi.15–27 According to this principle,

Xk Xk Xkþn Xk positive energy objects traveling backwards in time A B c2 pA pB ehrehr hg hg do not exist; and any negative energy particle r¼1 h¼1 g¼kþ1 h¼1 travelling backwards in time can and must be 2 2 ct cs described as its antiparticle, endowed with positive mAmBc4ð1 bAbBÞ energy and motion forward in time (but going the ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 0 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : opposite way in space).21 1 ðbAÞ2 1 ðbBÞ2 Thus the antiparticle moving forward in time and A A A B B B [We have accepted that (khlvl /a0h) ¼ (khlvl /a0h); therefore possessing positive energy in fact can be regarded as a we have particle moving backward in time and possessing negative 7 2 energy. Xk Xk kA vAkB vB Xk Xk kA vA hr r hr r ¼ hr r This principle can be generalized for the case of k- aA aB aA r¼1 h¼1 0h 0h r¼1 h¼1 0h dimensional time t1, t2, ..., tk. A particle moving back- Xk Xk B B 2 k v ward in a given time dimension td (that is, dtd , 0, where 1 ¼ hr r ¼ c2 aB t d k) and possessing energy r¼1 h¼1 0h 0 1 e1d and B C e ! B 2d C 2 Ed ¼ B . C Xkþn Xk kA vAkB vB Xkþn Xk kA vA @ . A hg g hg g ¼ hg g aA aB aA ekd g¼kþ1 h¼1 0h 0h g¼kþ1 h¼1 0h ! B 2 Xkþn Xk B defined in relation to td can be described as an antiparticle khgvg 2 ¼ ¼ c : t dt aB s moving forward in the time dimension d (that is, d . 0) g¼kþ1 h¼1 0h and possessing energy 0 1 See Eqs. (52), (54), and (77).] We have obtained the e1d expressions B C B e2d C E ¼ B . C A B 4 A B d @ . A qmffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0 m0 c ð1q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib b Þ . e 1 ðbAÞ2 1 ðbBÞ2 kd defined in relation to t . In the case of multidimensional A B 4 A d and m0 m0 c , which have to be compared. Because b and time, the components of energy e (r, h ¼ 1, 2, . . ., k) can B A B 2 hr b are real numbers, we have 0 (b b ) and accept not only positive but also negative values (unlike the case of one-dimensional time). 1 bAbB qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1; First, we will consider the case (n, k) ¼ (3, 1). Some 1 ðbAÞ2 1 ðbBÞ2 authors11,21,23,25,28–33 have shown that the nonorthochro- nous, proper Lorentz transformations (i.e., transforma- respectively. (Both are equivalent.) From here it follows tions which include ‘‘total inversion’’ 1)canbe that connected with the existence of antimatter. The asser- 24 Xk Xk Xkþn Xk tion is reasonable that the term ‘‘antimatter’’ is strictly A B c2 pA pB relativistic, and that on the ground of the double sign in ehrehr hg hg g¼kþ1 h¼1 the formula for energy, the existence of antiparticles could r¼1 h¼1 mAmBc4: 2 2 0 0 be predicted right after 1905, provided that the switching ct cs principle is applied. It is clear that the same inequality will be valid also for the It is important to note that the full group of Lorentz remaining pairs of particles. Therefore, if Eq. (77) is transformations acts as the four-dimensional vector on fulfilled for particles undergoing decay (i.e., the values of the position of a given object, and also on the other four- khl are chosen in an appropriate way), then the inequality dimensional vectors (four-momentum, four-current, etc.) H A B D A B m0 m0 þ m0 þþm0 will be fulfilled. (If b ¼ b ¼ which are connected with the object. We will introduce D _ _ ¼ b then this inequality becomes an equality.) The the new notations P for strong parity and T for strong obtained inequality is similar to that derived in STR. time reversal, in order to denote the inversion of the sign of the first component and of the next three components_ _ X. ANTIPARTICLES IN MULTIDIMENSIONAL TIME of all four-vectors. The operation strong reflection, PT, which changes the sign of the three-vector x and of the In relativistic quantum mechanics, an antiparticle is time t, will change the sign of the three-momentum_ _ p and attached to every particle. The so-called switching of the energy E as well. We can write P ” Ppˆand T ” TEˆ , principle (SP) (or reinterpretation principle) was formu- where pˆand Eˆ are operators respectively changing the 432 Phys. Essays 25, 3 (2012)

signs of the three-momentum p and of the energy E If we denote with Lþ the proper orthochronous (operations of three-momentum and_ energy reversal). In Lorentz transformations (i.e., transformations including this new formalism, the operation P is in fact equivalent_ the unit matrix 1) and with Lþ the proper nonorthochro- to the standard operation P (parity), but the operation T nous Lorentz transformations (i.e., transformations 11,21 is not equivalent to the standard_ operation T (time including total inversion 1), then we will have reversal transformation) because T does not contain the SP operation X, which performs the exchange of emission Lþ ” ð1ÞLþ ” P T Lþ ! CPTLþ: ð82Þ and absorption. [Here the operation (1) means total inversion, as in (1) 29,30,34 _ _ It can be proven that the strong reflection_ _ PT is ” PT ” CPT—see Refs. 11, 29, 30, and 34.i] equivalent to the normal CPT operation: PT!SP CPT. According to the results obtained in Section IX.B, in Indeed, applying the switching principle (SP), we the case of one-dimensional time the formula E ” e11 ¼ 33,34 obtain m0u11c is valid for all free particles, where u11 is the time component of the four-velocity. According to Eq. (58),at ˆ^ SP ” CEp; ð78Þ k ¼ 1 we have where C is the_ conjugation_ of all conserved_ _ additive 2 v1m0c charges. Since P ” Ppˆand T ” TEˆ, we have PT ” PpˆTEˆ . E ” e11 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi : 34 2 Thus we have a01 1 b SPP T ” ðCEˆ p^ÞPp^TEˆ ” CPT: ð79Þ (In the case dT ¼jdt1j . 0, dT01 ¼jdt01j . 0, k11 ¼ 1—see

11 Subsection IX.B.) Let us assume that v1 . 0 and a01 . If we apply the switching principle, then we have 0—that is, the energy of the particle is a positive quantity: E ” e11 . 0. Let us consider the proper nonorthochro- SP ” CCmX ” CX; ð80Þ _ nous Lorentz transformations Lþ, which lead to the where C ” CCm; here Cm is the rest-mass sign inversion exchange of the signs of all time components: and X operates the exchange of absorption and emission. _ 2 The operation C will be called strong conjugation.h One 0 ðv1Þm0c E ¼E ¼ m0ðu11Þc ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi , 0: can easily realize that in the frame of quantum mechanics 2 _ a01 1 b (in the case of states with definite_ _ parity), C ” P5, where 1 5 1 P5 is the chirality operator (C wC ” c w ¼ P5 wP5)—see Let us apply the switching principle, which is expressed in Ref. 11. Let us now consider generalized five-dimensional the inversion of the fifth axis (i.e., inversion of the proper space-time instead of the four-dimensional of space-time time dt01 ¼ a01dT0 or, accordingly, of the rest mass). The of Minkowski, where the fifth dimension corresponds to switching principle will mean multiplying the number a01 proper time and consequently is connected with the rest ¼ dt01/dT0 by the number 1. We obtain the related mass.11,35–46 It turns out that from the geometric point of antiparticle: view, chirality P5 in fact means inversion of the fifth axis 2 2 ðv1Þm0c ðv1Þðm0Þc (i.e., inversion of the proper time or accordingly of the E00 ¼E0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi . 0: 2 2 rest mass). Taking into account Eq. (80), we can write SP ða01Þ 1 b a01 1 b ¼ P X (see Ref. 11). Therefore, SP is a combination of 5 We determined that for the antiparticle (moving in inversion in relation to the fifth axis (proper time and one-dimensional time) we have to attach negative proper accordingly rest mass) and application of the operator X. _ _ time or, accordingly, negative rest mass, but of course Since PT means the sign inversion of all components 11 positive full relativistic mass and energy. _ _ _ of all four-vectors,_ _ for getting such an effect it is enough 11 1 1 Recmi and Ziino formulated so-called strong CPT- to write P T ” PTuˆX , where X ” X has been symmetry: The physical world is symmetric (i.e., physical introduced because ordinary T contains the exchange_ X laws are invariant) during total five-dimensional inversion of emission and absorption exchange (different from T)— of the axes x, y, z, ct, ct01 (where t01 is the proper time). see Ref. 33. With uˆhere is denoted the four-velocity If the number of time dimensions in which a particle inversion. We obtain is moving is greater than one, then the particle will have more than one antiparticle. There will be a violation of SPP T ” CXðPTu^X1Þ ” CPT: ð81Þ Lorentz covariance and therefore of CPT symmetry. In The switching principle changes the sign of the three- the case of multidimensional time, CPT symmetry must momentum but does not change the sign of the three- be exchanged with another generalized symmetry. Now velocity—i.e., it changes the sign of the rest mass. In fact, we are going discuss this issue. the four-velocity inversion is equivalent to the rest-mass Here we will consider the case (n, k) ¼ (3, 2). The time inversion: uˆ” Cm. dimensions we will denote with t, s and the space dimensions with x, y, z. In the case of two-dimensional h We point out that in the formalism used here, the strong conjugation is a unitary operator when acting on the state space, as well as strong i In the formalism used here, CPT is a linear operator in pseudo- time reversal. Euclidean space. Phys. Essays 25, 3 (2012) 433 time the term ‘‘switching principle’’ will mean inversion of By analogy with Eq. (82),inthecaseoftwo the proper time dT0 ¼ (dt01, dt02) ¼ (a01dT0, a02dT0)—that dimensional time we have the following three formulas: is, multiplication of the numbers a01 and a02 by 1. (See SPts ˆ Subsection IX.A.) This is equivalent to rest-mass sign K ” ð1ÞKþ ” PCKþ ! CPtˆsKþ ; ð86Þ inversion. Therefore, every object moving backward in SP the time dimension t (that is, dt , 0) and having energy K ” PXK ts CPtˆK; 87  þ þ ! þ ð Þ e11 E1 ¼ SP e21 ts Kþ ” PNKþ ! CPs ˆ Kþ ; ð88Þ definedinrelationtot can be described as the where the operators tˆ and sˆ correspond to the standard t- corresponding antiobject moving forward in the time reversal transformation and s-reversal transformation dimension t (that is, dt . 0) and having energy operations, and P is the standard operation parity (see  Subsection III.C and more precisely Table II). The e11 E1 ¼ transformations K lead to a change of the signs of e21 the numbers v1 ¼ dt/dT and v2 ¼ ds/dT and therefore to a defined in relation to t. The same considerations apply to change of the signs of all components of the energy ehr (r, the time dimension s as well. h ¼ 1, 2)—see Eq. (58). The transformations Kþ lead to a Let us denote with Aþ, A0, A, A0, Aþ the change of the sign of the number v1 and therefore to a different kinds of antiparticles in two-dimensional time. change of the signs only of the t-components of the energy The antiparticle Aþ is moving backward in the time eh1 (h ¼ 1, 2). The transformations Kþ lead to a change of dimension t and forward in the time dimension s (that is, the sign of the number v2 and therefore to a change of the dt , 0 and ds . 0). The antiparticle A0 is moving signs only of the s-components of the energy eh2 (h ¼ 1, 2). backward in the time dimension t but does not move in The transformations K , Kþ , and Kþ lead to a sign the time dimension s (that is, dt , 0 and ds ¼ 0). The change of the numbers v3 ¼ dx/dX, v4 ¼ dy/dX, v5 ¼ dz/dX antiparticle A moves backward in the time dimensions t and therefore to a sign change of all momentum and s (that is, dt , 0 and ds , 0). The antiparticle A0 components phx, phy, phz (h ¼ 1, 2)—see Eq. (68). The moves backward in the time dimension s but does not same is valid also for the respective components in the move in the time dimension t (that is, dt ¼ 0 and ds , 0). dual spaces. The antiparticle Aþ moves forward in the time dimension Let particle M be moving forward in the time t and backward in the time dimension s (that is, dt . 0 dimensions t and s. Application of the proper orthochro- and ds , 0). nous transformations Kþ in some cases leads to a By analogy with Eq. (78), we have the equality direction change of the time dimensions t or s (see Subsection III.B). Therefore, as a result of the transfor- SPts ” CEˆ p^; ð83Þ mations Kþ applied on the particle M, in these cases we where Eˆ is an operator changing the signs of all obtain some of the listed antiparticles: Aþ, A0, A, A0, or A . components of the energy E [the quantities ehr (r, h ¼ 1, þ 2)], pˆis an operator changing the sign of all components Let us assume, for example, that as a result of the transformations K applied on the particle M, the of the momentum p [the quantities phx, phy, phz (h ¼ 1, 2)], þ and C is the conjugation of all conserved additive charges. antiparticle A (dt , 0,ds , 0) has been created. In this By analogy with Eq. (80), we have the equality case, when applying the operations in Eqs. (86), (87), and (88) we will have the respective antiparticles of the particle

SPts ” CCmX; ð84Þ A. For example, if to the obtained antiparticle A we apply the operations in Eq. (86), then we will obtain the where the operation Cm is the rest-mass sign inversion and the operator X performs the exchange of emission and given particle M. Therefore, with application of the absorption (and vice versa). operations in Eqs. (86), (87), and (88) the directions of t or s are reversed, but it is possible to obtain antiparticles If in Eq. (36) we set (n, k) ¼ (3, 2), then we obtain moving backward in the time dimensions t or s as well as 2 2 2 2 2 2 2 2 2 2 2 c dt þ c ds dx dy dz c dt01 c dt02 ¼ 0: particles moving forward in the time dimensions t or s ð85Þ (depending on the transformations Kþ ). Let us assume that the proper orthochronous By means of Eqs. (84) and (85) we can interpret the transformations Kþ have not changed the directions of meaning of the operation SPts from the geometrical point the time dimensions t and s—that is, dt . 0,ds . 0. of view. According to Eq. (85), we could consider a The transformations K lead to a sign change of all generalized [(3 þ 2) þ 2]-dimensional space-time, where energy components ehr (r, h ¼ 1, 2). Therefore, for the the two additional dimensions are the projections of the energy components (after application of K ) we will have proper time t01 and t02. Therefore, SPts is a combination  e of inversion in relation to the axes ct01 and ct02 and 0 11 Et ¼Et ¼ ; application of the operator X. e21 434 Phys. Essays 25, 3 (2012)

where The transformations Kþ lead to a sign change only 2 of the s-components of the energy eh2 (h ¼ 1, 2). kh1ðv1Þm0c eh1 ¼ m0ðuh1Þc ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; h ¼ 1; 2; Therefore, for the energy components (after application 2 a 1 b of Kþ ) we will have 0h   0 e11 Et ¼ Et ¼ ; 0 e12 e21 Es ¼Es ¼ ; e22 where where 2 kh1qv1ffiffiffiffiffiffiffiffiffiffiffiffiffim0c k ðv Þm c2 eh1 ¼ m0uh1c ¼ ; h ¼ 1; 2; h2 qffiffiffiffiffiffiffiffiffiffiffiffiffi2 0 2 eh2 ¼ m0ðuh2Þc ¼ ; h ¼ 1; 2: a0h 1 b 2 a0h 1 b  00 00 e If we apply SPts we obtain dt ¼dt01 (that is, a ¼a01) 0 12 01 01 Es ¼Es ¼ ; 00 00 e22 and dt02 ¼dt02 (that is, a02 ¼a02. According to Eq. (58), we will have  where 00 0 e11 k ðv Þm c2 Et ¼Et ¼ h2 qffiffiffiffiffiffiffiffiffiffiffiffiffi2 0 e21 eh2 ¼ m0ðuh2Þc ¼ ; h ¼ 1; 2: 2 a0h 1 b and  If we apply SP we obtain dt00 ¼dt (that is, a00 ¼a ) e ts 01 01 01 01 E00 ¼E0 ¼ 12 ; and dt00 ¼dt (that is, a00 ¼a ). According to Eq. s s e 02 02 02 02 22 (58), we will have  where eh1 ¼ (m0)(uh1)c and eh2 ¼ (m0)(uh2)c. As a result 00 0 e11 of these operations we obtain the antiparticle A. Et ¼Et ¼ e21 The transformations Kþ lead to a sign change only of the t-components of the energy eh1 (h ¼ 1, 2). Therefore, and  for the energy components (after application of Kþ )we will have 00 0 e12 Es ¼Es ¼ ;  e22 0 e11 Et ¼Et ¼ ; where e ¼ (m )u c and e ¼ (m )(u )c. As a result e21 h1 0 h1 h2 0 h2 of these operations we obtain the antiparticle Aþ or the where antiparticle A0. 2 According to these considerations, a negative rest kh1ðqvffiffiffiffiffiffiffiffiffiffiffiffiffi1Þm0c eh1 ¼ m0ðuh1Þc ¼ ; h ¼ 1; 2; mass equal to m0 must be attached to the antiparticles 2 a0h 1 b Aþ, A0, A, A0, Aþ. On the ground of the obtained results, it is possible to  e make a generalization of the strong CPT symmetry 0 12 11 Es ¼ Es ¼ ; formulated by Recami and Ziino. In the case of two- e22 dimensional time the physical world is symmetric (i.e., the where physical laws are invariant) for inversion of the axes x, y, 2 z, ct, cs, ct01, ct02 [see Eqs. (84) and (85)]. kh2v2m0c eh2 ¼ m0uh2c ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi ; h ¼ 1; 2: Let us summarize the obtained results: If k ¼ 1, then 2 a0h 1 b the number of antiparticles is 1; if k ¼ 2, then the number of antiparticles is 5. With similar considerations, one can 00 00 If we apply SPts we obtain dt01 ¼dt01 (that is, a01 ¼a01) find that if the number of time dimensions is equal to k, 00 00 and dt02 ¼dt02 (that is, a02 ¼a02). According to Eq. then the number of the different antiparticles is equal to (58), we will have 3k 2k. For example, for the case k ¼ 3 we obtain 33 23 ¼  19 different antiparticles. Indeed, let us denote with t1, 00 0 e11 Et ¼Et ¼ t , ..., t the time dimensions. Since we consider antipar- e21 2 k ticles, it must be true that dtd , 0 for at least one of the and time dimensions td (1 d k). Obviously, for each of the  quantities dt (r ¼ 1, 2, . . ., k) there are three possibilities: e r E00 ¼E0 ¼ 12 ; dt . 0ordt , 0ordt ¼ 0. If we take into consideration s s e r r r 22 these three possibilities and the circumstance that the where eh1 ¼ (m0)(uh1)c and eh2 ¼ (m0)uh2c. As a result number of_ time dimensions is equal to_ k, then we have a k of these operations we obtain the antiparticle Aþ or the total of V (3, k) ¼ 3 cases. (Here V (3, k) denotes the antiparticle A0. respective variations with repetition.) From these cases we Phys. Essays 25, 3 (2012) 435 have to exclude those in which all quantities dt1, inertial reference frames must be such that dt2, ..., dtk are non-negative. These are the cases where ðx1=Þ2 ðx2=Þ2 ðx3=Þ2 ðx4=Þ2 for all values of r (r ¼ 1,_ 2, . . ., k), either dtr . 0ordtr ¼ 0 hi is true. This comes to V(2, k) ¼ 2k cases. Therefore, the ¼ 6 ðx1Þ2 ðx2Þ2 ðx3Þ2 ðx4Þ2 ð89Þ total number of different kinds of antiparticles is equal to 1 2 3 4 3k 2k. for each four-vector x ¼ (x , x , x , x ), which can be a As we pointed out, the antiparticles must move four-dimensional vector on the position of a given backward in at least one of the time dimensions; in the particle, a four-momentum, four-velocity, four-current, remaining k 1 time dimensions they can move forward, etc. In the special case where space-time coordinates are backward, or not at all. It is easy to prove that there exist considered, Eq. (89) takes the following form:  0 2 0 2 0 2 0 2 k k q ðct Þ ðx Þ ðy Þ ðz Þ ð2 1Þ hi q 2 2 2 2 ¼ 6 ðctÞ ðxÞ ðyÞ ðzÞ : ð90Þ different antiparticles which do not move in q of a total k time dimensions. Here 1 q k and The plus sign in the right-hand side of Eqs. (89) and  (90) corresponds to the standard case of subluminal k relative velocities, i.e., concerns bradyons [here (n, k) ¼ (3, q 1)]; the minus sign must be chosen in the case of superluminal relative velocities, i.e., concerns tachyons is the binomial coefficient. [If q ¼ 0, then for each r (r ¼ 1, (see, for example, Refs. 57 and 58). 2,..., k) the condition dt 6¼ 0 is fulfilled.] r The difference between the transformations bradyon Let us have an antiparticle A moving in k-dimen- tachyon and the transformations (n, k) (3, 1) (n, k) sional time. We can prove that in the case of k- ¼ k (1, 3) can be understood best if, instead of four- dimensional time there exist 2 1 different particles M ¼ corresponding to the antiparticle A. [Concerning the dimensional Minkowski space-time, we consider a gener- movement of the particle M in the hyperplane of time, the alized five-dimensional space-time (x, y, z, ct, ct01), where the fifth dimension corresponds to proper time t01 and condition dtr 0 is fulfilled for all values of r (r ¼ 1, k dt therefore is related to the rest mass (see Section X). If we 2,..., ) and the condition r . 0 is fulfilled for at least 2 1 2 2 2 3 2 4 2 0 2 10 2 one value of r.) For example, in the case k 2 we obtain set (s) ¼ (x ) (x ) (x ) (x ) and (s ) ¼ (x ) ¼ 20 2 3 2 40 2 22 1 ¼ 3 different particles M corresponding to the (x ) (x ) (x ) , then according to Eq. (89) we will have (s0)2 ¼ 6(s)2(s0)2 ¼ 6(s)2, where (s)2 . 0. Therefore, antiparticle A: Mþþ (dt . 0, ds . 0), Mþ0 (dt . 0, ds ¼ 0), Eqs. (89) and (90) can be written as M0þ (dt ¼ 0, ds . 0). 10 2 20 2 30 2 40 2 0 2 ðx Þ ðhix Þ ðx Þ ðx Þ ðs Þ ¼ 0 XI. DISTINCTION BETWEEN TACHYONS AND ¼ 6 ðx1Þ2 ðx2Þ2 ðx3Þ2 ðx4Þ2 ðsÞ2 ; ð91Þ PARTICLES MOVING IN MULTIDIMENSIONAL TIME

Recami pointed out that in the course of any 0 2 0 2 0 2 0 2 2 ðct Þ ðhix Þ ðy Þ ðz Þ 7ðct01Þ ¼ 0 generalization of STR in such a way that tachyons are 2 2 2 2 2 included, it will turn out that these particles move in three ¼ 6 ðctÞ ðxÞ ðyÞ ðzÞ ðct01Þ : ð92Þ time dimensions and one space dimension.7 [Here, s ¼ ct —see Eq. (36).] Since in the case of Tegmark5 pointed out that the case (n, k) ¼ (1, 3) is 01 superluminal relative velocities one should choose the mathematically equivalent to the case (n, k) ¼ (3, 1), so minus sign in the right-hand side of Eq. (91), during the that ‘‘all particles are tachyons with imaginary rest transformations bradyon tachyon the signs in the mass.’’47 expressions (x1)2,(x2)2,(x3)2,(x4)2,(s)2 will change. This However, from the physical point of view these two is equivalent to a multiplication of the row matrix S (x1, cases are not equivalent. It is necessary to distinguish ¼ x2, x3, x4, s) by the complex diagonal matrix D diag(i, i, between the transformations bradyon tachyon, where a pffiffiffiffiffiffiffi ¼ i, i, i), where i 1—that is, the product S 3 D. The particle is created moving with a velocity greater than the ¼ speed of light in a vacuum,j and the transformations (n, k) multiplication of the row matrix S by the matrix D corresponds to rotation of all axes x1, x2, x3, x4, s through ¼ (3, 1) (n, k) ¼ (1, 3), where a particle is created moving in three-dimensional time and one-dimensional an angle of p/2 (arg i ¼ p/2). According to Eq. (92), in the case of superluminal space. 0 2 0 2 0 2 0 2 48–56 relative velocities we have (ct ) –(x ) –(y ) –(z ) þ It has been proven that if both postulates of STR 2 are fulfilled (namely the principle of invariance of the (ct01) . Therefore, instead of (3 þ 1)-dimensional space- speed of light and the principle of relativity), then the time, we could consider a generalized [3 þ (1 þ 1)]- transformations describing the transfer between the two dimensional space-time, where the additional dimension corresponds to proper time ct01. In the generalized space- 0 j ct ct We point out that the velocities of the bradyon and of the tachyon time, 1 þ 1 dimensions are timelike ( , 01) and three 0 0 0 aredefinedfromthepointofviewofanobserversituatedin(n, k) ¼ dimensions are spacelike (x , y , z ). These considerations (3, 1) space-time. also concern the dual space: In the generalized dual space, 436 Phys. Essays 25, 3 (2012)

10 0 0 2 0 0 1 þ 1 dimensions are timelike (x ¼ E , s ¼ m0c ) and momentum will have three components phx ¼ m0uhx (h ¼ 1, 20 0 30 0 40 three dimensions are spacelike (x ¼ pxc, x ¼ pyc, x ¼ 2, 3)—see Eq. (67). The following relation will be fulfilled 0 C 2 0 pz ). (Here m0 . 0.) The energy E of the tachyon will [see Eq. (74)]: 0 have only one component, and the momentum ps of the 2 2 0 2 ðE0Þ ðp0 Þ c2 E2 tachyon will have three components. The expression (E ) x ¼ 0 ¼ m2c4 . 0: 0 2 2 2 2 c2 c2 c2 0 –(ps) c ¼m0c , 0 will be valid. Obviously, the t s t 2 2 tachyon will have imaginary rest mass: (im0) ¼m0 , 0. 2 The particle will have real rest mass (m0 . 0). As we know, in the case of (n, k) ¼ (1, 3) the following According to these considerations (see Section IV), in equality is fulfilled for each four-vector x (x1, x2, x3, x4): ¼ the case (n, k) ¼ (1, 3) the particle can have velocity ðx10Þ2 þðx20Þ2 þðx30Þ2 þðx40Þ2 (defined in relation to one of the time dimensions) which is greater than, less than, or equal to the speed of light in a ¼ðx1Þ2 þðx2Þ2 þðx3Þ2 þðx4Þ2: ð93Þ vacuum. Until now it was accepted that if a particle

(Here x1 is spacelike dimension and x2, x3, x4 are timelike moves according to one observer with a velocity which is dimensions.) If we set (s)2 ¼[(x1)2 (x2)2 (x3)2 (x4)2] greater than the speed of light in a vacuum, then the and (s0)2 ¼(x10)2 þ (x20)2 þ (x30)2 þ (x40)2, then in particle has imaginary rest mass (and accordingly it has accordance with Eq. (93) we will have ðs0Þ2 ¼ (s)2 . 0. real mass measured by the observer). However, this is not Therefore, Eq. (93) can be written in the form fulfilled if the particle moves in multidimensional time— in this case it will have real rest mass. 10 2 20 2 30 2 40 2 0 2 ðx Þhiþðx Þ þðx Þ þðx Þ ðs Þ ¼ 0 If the particle moves in the space-time (n, k) ¼ (1, 3) ¼ ðx1Þ2 ðx2Þ2 ðx3Þ2 ðx4Þ2 ðsÞ2: ð94Þ and the condition

X3 It is clear that if (n, k) (3, 1), then we have (s)2 c2 ¼ ¼ , 1 (x1)2 (x2)2 (x3)2 (x4)2 . 0. [See Eq. (91) for the case of 0 2 1¼1 ðV1Þ subluminal relative velocities, i.e., those with the plus sign 0 0 0 in the right-hand side of Eq. (91).] If (n, k) ¼ (1, 3), then is fulfilled, where V1 ¼jdx j/jdt1j (see Sections IV and we have (s)2 ¼[(x1)2 (x2)2 (x3)2 (x4)2] . 0 [see Eq. VIII), then by analogy with Eqs. (91) and (92) we have (94)]. Therefore, in the transformations (n, k) ¼ (3, 1) ðx10Þ2 þðx20Þ2 þðx30Þ2 þðx40Þ2 ðs0Þ2 ¼ 0 (n, k) ¼ (1, 3) the signs are changed in front of the hi 2 2 2 2 2 expressions (x1)2,(x2)2,(x3)2,(x4)2, but unlike in the ¼ ðx1Þ þðx2Þ þðx3Þ þðx4Þ ðsÞ transformations bradyon tachyon, the sign is not changed in front of the expression (s)2. This is equivalent and 1 2 3 4 to multiplication of the row matrix S ¼ (x , x , x , x , s) 0 2 0 2 0 2 0 2 2 2 ðx Þ þðct Þ þðct Þ þðct Þ þðct01Þ þðct02Þ by theffiffiffiffiffiffiffi complex diagonal matrix J ¼ diag(i, i, i, i, 1), where 1 h2 3 p 2 2 2 2 2 i ¼ 1—that is, the product S 3 J. The multiplication of þðct03Þ ¼ 0 ¼ ðxÞ þðct1Þ þðct2Þ þðct3Þ 1 2 3 4 i S by J reflects rotation of the axes x , x , x , x through 2 2 2 angle p/2 around the axis s (the axis s is invariant in ðct01Þ ðct02Þ ðct03Þ : relation to the applied operation). Here (s0)2 ¼(s)2 , 0. According to Eq. (94), in the special case of space- time coordinates we have

0 2 0 2 0 2 0 2 2 2 XII. CONCLUSION ðx Þ þðct1Þ þðct2Þ þðct3Þ ðct01Þ ðct02Þ The simplest way of thinking about and considering ðct Þ2 ¼ 0: ð95Þ 03 multidimensional time is using a branching or train-track

Here t01, t02, t03 are the projections of the proper time T0 model. Meiland proposed a more formal model of [see Eq. (36)]. According to Eq. (95), instead of (1 þ 3)- multidimensional time, where the past can be changed.59 dimensional space-time we can consider a generalized [(1 Despite the changes, however, there is only one past. þ 3) þ 3]-dimensional space-time, where three dimensions According to Meiland, his two-dimensional time model is correspond to the projections of proper time (ct01, ct02, not radically different from our ordinary, one-dimension- ct03). In the generalized space-time, three dimensions are al perception of time. He treats ‘‘the past as a continuant, 0 0 0 59 timelike (ct1, ct2, ct3) and 1 þ 3 dimensions are spacelike as existing at each of several times.’’ 0 (x , ct01, ct02, ct03). In multidimensional time, like in one-dimensional These considerations also concern the dual space: In time, every localized object is moving along a one- the generalized dual space, three dimensions are timelike dimensional timelike world line.5 Therefore, even in two 20 0 30 0 40 0 (x ¼ E1/ct, x ¼ E2/ct, x ¼ E3/ct) and 1 þ 3 dimensions or more dimensions time will look one-dimensional, 10 0 10 20 30 are spacelike (x ¼ pxc/cs, s ¼ E01/ct, s ¼ E02/ct, s ¼ because all physical processes (including thinking) will 0 0 E03/ct)—see Eqs. (62) and (74). (Here ps ” px.) run in a linear sequence, which is characteristic of the According to the considerations in Subsection IX.B, perception of reality. Clocks will work in their usual the total energy of the particle will have nine components manner. Every localized object will have one single 0 0 ehs ¼ m0uhrc (r, h ¼ 1, 2, 3)—see Eq. (57). The particle’s ‘‘history’’ in the multidimensional time. In this sense, Phys. Essays 25, 3 (2012) 437 pffiffiffi the notion an observer will build of multidimensional time The principle of invariance of the speed c k will be will not differ very much from the well-known notion of valid instead of the principle of the invariance of time. However, in the case of multidimensional time there the speed of light c in STR (Section VIII). are problems concerning well-posed causality4,5,13,14 (see The causal region in multidimensional time (which Section IV). is described in Section IV) will differ from the As stated in Section IV, for multidimensional time the causal region in STR. notions of past, present, and future can be defined as well. The relations derived in Subsection IX.B will be If time is one-dimensional, then one point (the present valid between total energy E, total momentum ps, moment) divides the time axis into two separate regions: rest mass m0, relativistic mass m, and velocity V of past and future. If the number of time dimensions is equal a particle. They differ from the relations in the to k, then time can be divided into two k-dimensional STR. regions from one (k 1)-dimensional hypersurface—more The energy-momentum conservation law derived in precisely, from the (k 1)-dimensional hypersphere (see Subsection IX.C will be valid. This law differs from Section IV). The two obtained k-dimensional regions can the energy-momentum conservation law of STR. conditionally be called past and future, and the border A new, different CPT symmetry will be valid region—i.e., the (k 1)-dimensional hypersphere—can be (Section X). called present (see Section IV). In the case of k-dimensional time, there will exist 3k Let us consider two nonrelativistic observers moving 2k different antiparticles (Section X). in different time directions. (In this consideration According to the results obtained in the study: relativistic effects are neglected.) These observers can meet in the space-time and can synchronize their clocks It is proven that in multidimensional time, particles only if their directions of movement are crossing in time. can move in the causal region with a velocity which Let us assume that the observers meet at point O on the is greater than, less than, or equal to the speed of hyperplane of time. Then these observers will separate light in a vacuum (Section IV). For the case of again and will continue to move in their time directions, multidimensional time, it is possible that a particle without any opportunity to meet.5 Let us assume that, can move simultaneously faster that the speed of according to the one of the observers, from the moment of light in a vacuum and forward in the time their meeting (point O) a period DT . 0 has passed. dimensions, which is not true for the case of one- Because in the case of k-dimensional time the present is a dimensional time (STR; Subsection III.B). Thus, if (k 1)-dimensional hypersphere, in this case both the results for the superluminal neutrinos are observers will be at points (moments) which lie on the confirmed,60 this can be explained with the (k 1)-dimensional hypersphere with center O and radius existence of additional time dimensions in which DT (see Section IV). these neutrinos are moving. In a universe of multidimensional time many other In the case of multidimensional time, application of strange things can happen. If two observers are not the proper orthochronous transformations at moving against each other (in space) but they move in certain conditions leads to movement backward different directions in the hyperplane of time, then in the time dimensions (Subsection III.B). according to their opinion the same physical process will It is shown that particles moving faster than the run with different speeds (see, e.g., Subsection III.A). If speed of light in a vacuum can have a real rest mass two observers move in orthogonal directions to each (unlike tachyons), provided that they move in other in the hyperplane of time, then according to each multidimensional time (Subsection IX.B). one of them time for the other observer will ‘‘stand still,’’ Thus the existence of particles moving in multidi- i.e., will not run. If these observers move relative to each mensional time can be proven or rejected through other in space, then according to each one of them the proper experiments. other will move with infinitely high velocity (see Section As a conclusion, we want to point out that if in our IV). universe exist particles moving in two, three, or more time In this study, there are results which can be proven dimensions, then the relations between energy, mass, experimentally. If there exist particles moving in multidi- velocity, and momentum of these particles will be mensional time, the following physical effects can be expressed through the formulas derived in this article. found which are different from the effects derived from STR: ACKNOWLEDGMENTS The transformations derived in Sections III and V will be valid for transfer from one inertial frame of The author is very grateful to Hristo Pavlov for his reference to another. interest in this work and his valuable advice. The law derived in Section VI will be valid for 1I. Bars, J. Terning, and F. Nekoogar (Founding editor), Extra addition of velocities. Dimensions in Space and Time (Springer, Multiversal Journeys, New The Doppler effect derived in Subsection III.D will York, 2007). be valid. 2I. Bars, AIP Conf. Proc. 589, 18–30 (2001). 438 Phys. Essays 25, 3 (2012)

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