In order to show a relativistic version of the Gauss-Bonnet theorem, an oriented pseudo angle between any two units or null vectors on the Minkowski plane was presented by Helzer [4]. Pseudo-angles were introduced as a generalization of the oriented hyperbolic angles between the unit vectors which were determined in [1], and [7]. Thus, it can be shown that the oriented hyperbolic angles between the unit vectors in the Minkowski spaces are equal to the oriented pseudo-angles between those vectors [6].

E. Nesovic [6] investigated if the measure of the unoriented pseudo-angles can be represented with respect to the hyperbolic arcs of finite hyperbolic lengths. So, she defined the pseudo-perpendicular vectors in the Minkowski spaces. According to any unit or null vectors, she showed that she could associate completely eight vectors being pseudo-perpendicular on the Minkowski plane. Using the pseudo-perpendicular vectors, the geometric meaning of the oriented pseudo-angles was presented with regard to the hyperbolic arcs of finite hyperbolic lengths.

In [3], Chung et al. introduced the hyperbolic angle between two inertial observers. By using the hyperbolic cosine and sine functions, it was shown that the coordinate transformation rules may be obtained from the hyperbolic angles. Also, they presented that the angle connecting with the Bondi factor K by K = e^v is equal to the hyperbolic angle defined by v=sρ, where s is the hyperbolic circumference of a hyperbolic curve and ρ is the invariant length of the curve.

In this study, by following the reference [3], we present some possible cases of the hyperbolic angle between two unit spacelike or timelike vectors in terms of the causal characters of these vectors.

The Minkowski plane E12 is an affine plane endowed with the standard flat metric given by

where (x₁, x₂) is a rectangular coordinate system of E12. A vector v ≠ 0 in E12can have one of the three causal characters: it can be spacelike, timelike or null (lightlike) if g (v, v) > 0, g(v, v) < 0 or g (v, v) = 0 and v ≠ 0, respectively. In particular, the norm of a vector v is given by ‖v‖L=∣g(v,v)∣. Two vectors v and w in E12 are said to be orthogonal if g (v,w) = 0. A curve α:I→E12 is defined as spacelike (resp. timelike, null) at t if α′ (t) is a spacelike (resp. timelike, null) vector. If e₂ = (0, 1) is a unit timelike vector, an arbitrary timelike vector v in E12 is said to be future-pointing if g (v, e₂) < 0, or past-pointing if g (v, e₂) > 0. Two timelike vectors v and w have the same time-orientation if they are either future-pointing or past-pointing vectors. Consider that the ordered basis {e₂, e₁} gives the positive orientation of the Minkowski plane E12 with the above mentioned metric.

Let ℝ22 be the set of matrices of two rows and two columns. Let A = [a_ij ], B=[bjk]∈ℝ22. A Lorentzian matrix multiplication denoted by “·_L” is defined as

The 2x2 L-identity matrix corresponding to the Lorentzian matrix multiplication denoted by I₂ is as follows:

Now, the geometry of the Minkowski spaces will be studied in terms of the hyperbolic angles. Consider an inertial observer φ that presents the time and space coordinates of the events. By using a light signal, the coordinates (x_E, t_E) of an event E can be specified and the velocity of the light signal is independent of the inertial observer.

For the sake of the argument, the light signal is sent at time t₁ towards the +x direction by φ and this light signal reaches x_E at time t_E. Then, the signal is mirrored back to φ. If t₂ is the time of reception by φ, the below mentioned equalities are obtained:

and

Once two light signals with the time duration Δt are sent from the inertial observer to another inertial observer, the second one gets the time duration KΔt, where K is called the Bondi factor. So, when the signal is sent from φ to φ′ at time t₀, the second observer φ′ receives it at time Kt₀. The occurrence of signal reception by φ′ is symbolized as the letter E and then, this signal returns back to φ at time K²t₀. If the parametrization K = e^v is used to determine the hyperbolic angle between two temporal coordinate axes t and t′, x_E and t_E are found in terms of tE′ as the below equalities:

where tE′=Kt0. Also, the angle between two spatial coordinate axes x and x′ is equal to v. So, x_B and t_B are obtained in terms of xB′:

[3]. Furthermore, since the velocity of the inertial observer φ is xEtE, any one of two following equalities is easily obtained:

or

For more details, we refer the readers to [2, 3].

3.1. Hyperbolic Angles Between Two Timelike Unit Vectors

The hyperbolic circle is given by the set:

The set H01 of the unit vectors has two components, i.e.,

The vectors in H+1∪H-1 are timelike [5]. If two unit timelike (spacelike) vectors do not belong to the same component of H01(S11), the angle between these vectors is not determined. The angle between the unit timelike vectors on H+1 will be discussed.

Suppose that the letter E is a point on H+1, that is, a vector OE→=e→ is the radius of the circle. Now, consider that a light signal is sent to t = t₀. This signal reflects from the point E and returns to the t axis. If the equalities (2.1), (2.2) and K = e^v₀are used, the coordinates of this point are defined as follows:

where tE′=Kt0 [3]. Since tE′=‖e→‖L, it is possible to write the below expression

(See Fig. 2). It is shown that the hyperbolic angle depends on t₀, that is, 0 < t₀ ≤ 1. If t₀ = 1 is taken, the hyperbolic angle is equal to zero. In this case, the unit timelike vector is located on the time axis t. Besides, if the graphic of a logarithmic function is used,

is obtained. Thus, it is clear that the hyperbolic angle v₀ increases, reaching the infinite value while t₀ approaches to zero. Moreover, since the velocity of the vector ⥱ is xEtE, it is possible to write

(See Fig. 2). By virtue of v0=ln|1t0|, it is easy to see that the velocity is equal to

Hence, the velocity increases while the time decreases. If we use the equation (3.1),

is calculated. Here, because 0 < t₀ ≤ 1, the velocity ϑ can be written as the following inequality

In this case, if these conventions are taken into consideration, it is certain that the below mentioned equalities are seen

Now, at the time t₁< t₀, let us send one more light signal to this H+1. This signal reaches the point A and returns to the t axis. Here, the coordinates of this point are similar to the coordinates of the point E and the vector 0A→=a→ is the radius of the hyperbolic circle again (See Fig. 2). So, the hyperbolic angle between the axis t and the unit timelike vector a⃗ is equal to

where tA′=Kt1 and K = e^v₁. Similarly, the hyperbolic angle v₁ ∈ ℝ⁺ ∪ {0} is available. It is easily seen that v₁> v₀. Thus, the hyperbolic angle between the unit timelike vectors from ⥱ to a⃗ is defined as

where v′ ∈ ℝ⁺ ∪ {0}.

The Lorentz circle is expressed by

Two components

belong to the set of S11. The vectors in S11+∪S11- are spacelike. The angle between unit spacelike vectors on S11+ will be discussed. Consider that a light signal is sent at time t = −1. This signal is reflected from a point F and then, returns back to t = 1. It is easily seen that

On the other hand, since ta′=K-1,tb′=-K, the following equalities are obtained

and

Because the expression tF′=0, v = 0 is easily obtained. Moreover, due to the equation xF′=cosh v,xF′=1 is calculated. Now, let us send the light signal at time -∞<-t1′<-1, i.e., 1<t1′<∞. This signal reflects from a point B. After that, another signal is sent at time -1<-t2′<0, i.e., 0<t2′<1 and it reflects from a point C, where x_C = x_B (See Fig. 3).

If this point C is on S11+, the vector OC→=c→ is the radius of this Lorentz circle. If the equalities (2.1), (2.2) and K=ev2′ are used, there exist the below mentioned expressions

[3]. Because xC′=‖c→‖L,,

is easily obtained. If the calculations and geometric comments mentioned above are taken into consideration, it becomes obvious that the hyperbolic angle depends on time parameters. Chung et al. [3] show that t1′=K cosh (v2′)t2′. In this case, the following expressions can be written

If we think about the equalities (3.3) and (3.4), it is clear that the below expression can be easily obtained

In this case, when the above mentioned equalities are taken into consideration, it is easily seen that

The equality (3.6) shows the relationship between t1′ and t2′. Here, while t2′=1,   t1′=1, is obtained. Moreover, the equation (3.5) indicates that while t2′ approaches zero, t1′ approaches the infinite value.

At the present time, let us send two light signals at times t1″>t1′ and t2″>t2′. These signals reflect from two points x_D and D, respectively (See Fig. 3). Here, the vector OD→=d→ is the radius of the Lorentz circle, and the coordinates of the point D are obtained according to the coordinates of the point C. In this case, it is easily seen that

where K=ev2″. Similarly, the equality t1″=cosh(ln (t2″)) can be available. Thus, it is presented that the hyperbolic angle between the unit spacelike vectors from c→tod→ is given by

where v″, v″,v2′,v2″∈ℝ+∪{0}..

Theorem 3.2

Suppose that two points ‘C and D’ are located on S11+and the coordinates of these points are y=(cosh v2′,sinh v2′)and Y′=(cosh v2″,sinh v2″), respectively. Here, the hyperbolic angles are v2″=ln|1t2′|and v2″=ln|1t2″|, where t2″<t2′. When the point C is rotated to the hyperbolic angle v″=ln|t2′t2″|, it reaches the point D.

Suppose that the points A and C are located on H+1 and S11+, respectively. So, OA→=a→ is the unit timelike vector and OC→=c→ is the unit spacelike vector. In sections (3.1) and (3.2), it was shown that the angle between the vector a⃗ and the axis t is defined as v1=ln|1t1| and similarly, the angle between the vector c⃗ and the axis x is determined by v2′=ln|1t2′|, where ∣t2′∣>∣t1∣ (See Fig. 4).

Let us choose a point S (A) which is reflection with respect to the line x = t of the point A. The point S (A) is located on the S11+ circle and OS(A)→=s→ is the unit spacelike vector. Thus, the angle between the vector a⃗ and the axis t is equal to the angle between the vector →︀ and the axis x. Consequently, the hyperbolic angle between two unit spacelike vectors c⃗ and →︀ is shown by

This paper points out that the angle between the unit timelike vector ⥱ and the time axis t is v0=ln|1t0|, where 0 < t₀ ≤ 1, and the angle between the unit spacelike vector c→ and the space axis x is v2′=ln|1t2′|, where 0<t2′<1. It is seen that the hyperbolic angle between the unit timelike or spacelike vectors changes according to the time axis t. If the time parameter t decreases, then both the hyperbolic angle v and the velocity ϑ increase.