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The Geometry of Minkowski Space in Terms of Hyperbolic Angles Journal of the Korean Physical Society, Vol. 55, No. 6, December 2009, pp. 2323∼2327 The Geometry of Minkowski Space in Terms of Hyperbolic Angles Jean S. Chung and Won Sik L’Yi∗ Department of Physics, Chungbuk National University, Cheongju 360-763 Jin Hyun Chung Department of Liberal Arts, Chosun College of Science & Technology, Kwangju 501-744 (Received 13 November 2009) The geometry of Minkowski space is investigated by using concepts such as hyperbolic angles, hyperbolic curves, and hyperbolic arc lengths. The hyperbolic angle between two inertial observers 1 is given by ϑ = 2 log{(1 + v)/(1 − v)}. The usual scalar product between any two Lorentz vectors can be written in terms of the hyperbolic angle between them. The scalar product of two timelike vectors A and B, for example, can be written as A · B = |A||B| cosh ϑ, where |A| and |B| are their Lorentz invariant lengths. This is a natural generalizations of Euclidean geometry. PACS numbers: 03.30 Keywords: Special Relativity DOI: 10.3938/jkps.55.2323 I. INTRODUCTION circumference of a hyperbolic curve and ρ is the invariant length of the curve. It is also shown that the coordinate The geometry of general relativity is locally transformation rules can be obtained from the hyperbolic Minkowskian. To investigate the geometry of special rel- angles by using hyperbolic cosine and sine functions. ativity, one may use concepts such as angles and lengths In Section III, the scalar product between vectors is obtained from Euclidean geometry [1]. However even defined in terms of the hyperbolic angle and the invariant though they are intuitive, these Euclidean concepts do lengths. This definition is shown to reduce to the usual not have geometric roots in the Minkowski space. scalar product in the component form. The conclusion Another way of understanding the geometry of the spe- is given in Section IV. cial relativity is by using the affine geometry [2]. It is quite simple and intuitive, thus providing a convenient way of starting a geometric approach to the relativity. II. AFFINE GEOMETRY AND THE The only weak points are that it is a little bit abstract HYPERBOLIC ANGLES and that it does not use coordinate systems and the re- lated angles, which are strong points of Euclidean geom- Consider an inertial observer O who records the time etry. and space coordinates of events.1 The coordinates In this paper, we consider the geometry of special rela- (t , x ) of an event E can be determined by using light tivity by using concepts such as hyperbolic curves, hyper- E E signals, whose velocity is independent of the inertial ob- bolic angles, hyperbolic sine and cosine functions, which server. are projection functions along coordinate axes, scalar Suppose that O sends a light signal at time t along products of vectors in terms of the hyperbolic angles, 1 the +x direction, which reaches x at time t and is then and so on. We show that when one defines basic con- E E reflected back to O. When the time of reception by O is cepts of special relativity by using these terms, one is t , we have able to interpret the geometry of Minkowski space in a 2 t + t Euclidean way. t = 2 1 , (1) This paper is organized in the following way: In section E 2 II, the hyperbolic angle between two inertial observers is t2 − t1 x = . (2) introduced. This angle, which is related to the Bondi E 2 factor K by K = eϑ, is shown to be equal to the hyper- bolic angle defined by ϑ = s/ρ, where s is the hyperbolic 1 For simplicity, we assume that the space has just one dimension ∗E-mail: [email protected] and that the velocity of light is unity. -2323- -2324- Journal of the Korean Physical Society, Vol. 55, No. 6, December 2009 Fig. 1. When an inertial observer sends two light signals of time duration ∆t to another inertial observer, the time duration received by the second observer is K∆t, where K is the so called Bondi factor. Fig. 3. Two light signals, whose reflections are denoted by events A and B, are received by an observer. To determine the hyperbolic angle between two spacelike coordinate axes, 0 we assume that tA = tB = 0 and xA = xB. Then, it can be 0 shown that xA = xB cosh ϑ, where ϑ is the hyperbolic angle between two observers. This shows that the angle between the two spacial coordinate axes x and x0 is equal to the angle between the two temporal coordinate axes t and t0, that is, ϑ. Here, the hyperbolic angle is denoted by an arrowed segment of a hyperbolic curve. of E are K2 + 1 t = t , (4) E 2 0 Fig. 2. When an observer sends a light signal at time K2 − 1 t0, which is reflected by another observer, this signal returns xE = t0. (5) 2 2 to the original observer at time K t0. From this, the original 1 2 observer concludes that the signal was bounced at 2 (K +1)t0 When one uses the parametrization 1 2 when the other observer was at a distance of 2 (K −1)t0 from him. K = eϑ, (6) 0 one finds that tE and xE are given in terms of tE by Consider another observer O0, who moves along the 0 x-axis with a relative velocity v. We assume that O and tE = cosh ϑ tE, (7) 0 0 0 0 O are at x = x = 0 when t = t = 0. When O sends two xE = sinh ϑ tE. (8) light signals of time duration ∆t to O0, the time interval ∆t0 of the signal that O0 receives is given by It is quite interesting that (7) and (8) can be inter- preted as projections of a timelike length t0 along t and 0 E ∆t = K∆t, (3) x coordinates respectively. Compared to the usual pro- jection in Euclidean space, this projection in Minkowski where K, which is known as the Bondi factor [2,3], de- space is performed with hyperbolic cosine and sine func- pends only on v (See Fig. 1). tions rather than the usual trigonometric functions. In To investigate the physical meaning of ϑ, we consider addition to this, (7) means the t and t0 axes are separated the following light signal propagations. At time t , O 0 by the hyperbolic angle ϑ. sends a signal to O0, who receives it at Kt (See Fig. 2). 0 Furthermore, the velocity of O0, that is, x /t , is given We use the letter E to denote this event of reception of E E 0 0 by v = tanh ϑ, or the signal by O . That is, tE = Kt0. When this signal is bounced back to O, the time of reception measured by 1 1 + v 2 ϑ = log . (9) O is K t0. This means that the t and the x coordinates 2 1 − v The Geometry of Minkowski Space in Terms of Hyperbolic Angles – Jean S. Chung et al. -2325- Fig. 4. The Lorentz transformation can be read off from Fig. 5. A spacelike vector can be defined in terms of the this diagram. The dotted arrows denote various projections invariant length ρ and the hyperbolic angle from a spacelike 0 along the coordinates axes. For example, xE = xE cosh ϑ + coordinate axis. If the vector is timelike, the hyperbolic angle 0 tE sinh ϑ. should be measured from a timelike coordinate axis. When one uses the fact that velocity of light is indepen- fined by ρ2 = t2 − x2, is Lorentz invariant. The ρ- dent of the inertial observer, it is quite easy to prove that constant curves are hyperbolic curves. the Euclidean separation angle between x and x0 is equal To understand the geometric meaning of hyperbolic to the angle between t and t0. However, we are dealing angles, we consider the infinitesimal invariant length with a non-Euclidean geometry, so we need to prove the ds2 = dx2 − dt2. When one uses the parametrization same fact for hyperbolic angles in Minkowski space. This t = ρ cosh ϑ, x = ρ sinh ϑ, this can be written as can be proven in the following way. 2 2 2 2 Suppose that an observer O at x = 0 sends a light ds = ρ dϑ − dρ . (13) signal along the +x direction at time −t1, which is then This means that for a hyperbolic curve of constant ρ, the reflected by an event A (See Fig. 3). We assume that arc length s is given by this signal returns to O at t1. It is clear that tA = 0 and xA = t1. s = ρϑ. (14) Observer O sends another signal at t = −t2, which is In this sense, ϑ is the hyperbolic angle given by s/ρ. Note reflected by an event B, where xB = xA. That is, for O, the two signals are reflected at the same position. This that when ρ is timelike, s is spacelike, and vise versa. means that the duration of the two sent signals is the Furthermore, when one assumes that an event E is a 0 same as that of the received ones. Therefore, the time of distance of ρ away from the origin and makes an ϑ angle with the x0-axis, one has arrival of the second signal is 2t1 − t2. 0 Now, consider another observer O , who moves with x = ρ cosh(ϑ0 + ϑ) relative velocity v. When one extrapolates the second 0 0 0 = x cosh ϑ + t sinh ϑ, (15) signal back to the past, it seems to originate at ti = 0 0 −1 −Kt2.
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