The Symmetry of Relative Motion. P.R. Asquith

Transcription

1 The Symmetry of Relative Motion P.R. Asquith Abstract A symmetrical spacetime model of relative motion is developed in relation to the hyperbola, t² x² = 1. The model shows the Worldline of P (Inertial Frame coordinates x P, t P ) moving symmetrically away from that of Q. If a ray of light leaves P at x P = 0, t P = a-b, is reflected from an event H on Q (x Q = 0, x P = b) and returns to P at x P = 0, t P = a+b, the value t P =a is an overestimate of the time on Ps clock as H occurs. The time overestimate results in an underestimate by P of the velocity of Q relative to P. There is therefore a velocity v = x P /t P = b/a, which is less than the velocity w = x P /(time on Ps clock as H occurs) = b/(<a) derived from a symmetrical model. The former, v, the usual definition, is limited by the equations to less than the speed of light; the latter, w, is not limited. The "twin paradox" is solved. Slow Relative Motion Relative motion is symmetrical. Initially, slow relative motion is discussed, that is, a motion that is insignificant in relation to the speed of light. The reason for this is that the symmetry can be checked using any two sticks, so long as each has at least one identifying mark to act as a point of reference. The sticks (or other objects) can be thought of as reference frames P and Q moving relative to each other. P and Q are defined as being in Standard Configuration when their axes are parallel and Q is moving along P s positive x P axis (Figure 1). Figure 1 When P and Q are moving relative to each other in Standard Configuration, the movement has two components to it. One is the movement of Q along P s positive x P axis. The other is the movement of P along Q s negative x Q axis. The relative motion always has these two aspects. The two are equal and opposite. Relative motion is only these two movements. Movement of Q along P s positive x P axis cannot occur without movement of P along Q s negative x Q axis. If there is no movement of P along Q and Q along P, there is no relative motion. These two symmetrical movements essentially define relative motion. There is no way to give more importance to one of the movements as they are interlinked and interdependent.

2 Inertial Frames Newton s first law is " A body at rest will remain at rest and a body in motion will continue in motion at a constant speed in a straight line unless acted upon by some outside force". Such a body not acted upon by an external force is called an Inertial Body. An Inertial Body can be allocated a Frame of Reference. It may then be called an Inertial Frame. If P and Q are Inertial Frames, it is logical to represent the relative motion of P and Q on a graph as equal and opposite about an axis of symmetry (Figure 2). Figure 2 The Spacetime Diagram Figure 3

3 A spacetime diagram (Figure 3) is a useful way of representing a ray of light in relation to an Inertial Frame. The unit of distance is one meter. The unit of time is defined as the time it takes light to travel one meter. Thus the speed of light, c, is defined as 1. A light ray bisects the angle between the x and t axes, because distance equals time (Ref 1). Since Minkowski first introduced the concept, the spacetime diagram has been represented by Figure 4. Orthogonal x p and t p axes are the reference frame P of an observer. There are oblique x Q and t Q axes, set at an angle θ to the xp and t p axes respectively, representing the Inertial Frame Q moving in relation to P.(Figure 4) Figure 4 This diagram is clearly asymmetrical between reference frames P and Q. Mathematical Derivation of The Special Theory of Relativity The special theory of relativity examines the relationship between distance and time for Inertial Frames when their relative motion is significant in relation to the speed of light. The mathematics of the Special Theory of Relativity has been derived on the basis of an observer in a "stationary" reference frame P and a second reference frame, Q, "moving with velocity v" in relation to P (Ref 4). The concept of P being stationary with Q moving is asymmetrical, consistent with the asymmetry of the spacetime diagram of Figure 4. As velocity has been defined purely in terms of measures from reference frame P (x P /t P ) or reference frame Q (x Q /t Q ) the definition of velocity itself is asymmetrical in relation to frames P and Q. It will be shown that deriving the transformation equations between P and Q without reference to velocity, results in completely symmetrical transformation equations (Equations 26 to 29 ).

4 Development of a Symmetrical Spacetime Diagram of Relative Motion Figure 5 Given the symmetry of relative motion shown in Figures 1 & 2, and the symmetry of the transformation equations (26 to 29), it should be possible to develop a spacetime diagram in a symmetrical model, to represent the symmetry of the relative motion. Essentially the concept of symmetry about a vertical axis is transferred from Figure 2 into a spacetime diagram (Figure 5). It can be clearly seen that rays of light are not at 45 degrees to either P s or Q s t axis. However, every ray of light is at 45 degrees to the axis of symmetry. The relationship of the axis of symmetry to P and Q is clarified later. It will also be shown later (Figures 10,11,12) that by reference to the hyperbola t² x² = 1, if the t axis is at angle θ to the vertical of the axis of symmetry, the x axis is at the same angle θ to the horizontal. It is not necessary to refer to a ray of light leaving the t axis, reflecting off the x axis and returning to the t axis to derive the position of the x axis. This is important in avoiding a circular argument in relation to time measurement, discussed later (Figures 14 and 15). This gives the full symmetrical spacetime diagram for relative motion (Figure 6). Correlation with the mathematics of special relativity will show that this diagram is a true representation of relative motion.

5 Figure 6 The Mathematics of Relativity It is now necessary to look at the mathematics of relativity and see how this compares with the model of symmetrical relative motion. The mathematics will be developed without reference to velocity. It is, otherwise, a standard derivation of the equations. Einstein begins his derivation in this way (Ref 2). He considers Inertial Frames K and K 1 in Standard Configuration, considering only events which are localized on the x-axis, with the convention that the speed of light is c. "Any event is represented with respect to the co-ordinate system K by the abscissa x and the time t, and with respect to the system K 1 by the abscissa x 1 and the time t 1. We require to find x 1 and t 1 when x and t are given. A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation x = ct or x ct = 0.(1) Since the same light-signal has to be transmitted relative to K 1 with the velocity c, the propagation relative to the system K 1 will be represented by the analogous formula x 1 ct 1 = 0 (2) Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation

6 (x 1 ct 1 ) = λ(x ct) (3) is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (x ct) involves the disappearance of (x 1 ct 1 ). If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition (x 1 +ct 1 ) = µ(x+ct) (4) By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants a and b in place of the constants λ and µ where a = ½ (λ +µ) (4a) and b = ½ (λ µ) (4b) we obtain the equations x 1 = ax bct ct 1 = act bx (5) We should thus have the solution of our problem, if the constants a and b were known." Einstein at this point chooses to define velocity and continue the analysis in terms of velocity. A different approach is taken here to determine the relationship between a and b. This avoids making any assumptions about velocity. From this point the convention used will be that c=1. Apart from this, Einstein s notation is continued. Thus the equations at (5) become x 1 = ax bt (5a) t 1 = at bx (5b) An observer at the origin of Inertial Frame K will find the expanding wavefront of a pulse of light from his location forms the surface of a sphere represented by the equation. t² = x²+ y²+ z².....(7a) and the observer at the origin of K 1 will find (t 1 )² = (x 1 )² + (y 1 )² +(z 1 )².....(7b) For light along the x axes this is simplified to t² - x² = (8) (t 1 )² - (x 1 )² = 0....(9) From (8) and (9) is derived t² - x² = (t 1 )² - (x 1 )².....(10)

7 From (10), when x = 0, t² = (t 1 )²- (x 1 )².....(11) From the equations (5a), when x = 0, x 1 = - bt (12) From (11) and (12) t² = (t 1 )²- (-bt) ²...(13) t² = (t 1 )²- b²t² (14) (t 1 )² = t²(1+ b²)..(15) From equations (5b), when x= 0, t 1 = at...(16) From (15) and (16) a² t² = t²(1+ b²)...(17) a² = 1+ b².....(18) b² = a² (19) It has been shown that x 1 = ax bt (5a) t 1 = at bx. (5b) From (5a) x = 1/a (x 1 + bt). (20) From (5b) and (20) t 1 = at b/a (x 1 + bt) at 1 = a²t bx 1 b²t at 1 = t(a² b²) bx 1... (21) From (19) a² b² = 1. (22) From (21) and (22) t = at 1 + bx 1.. (23) From (5b) and (23) t 1 = a(at 1 + bx 1 ) bx bx = t 1 (a²- 1) + ab x 1 (24) From (19) and (24) bx = t 1 (b²) + ab x 1 x = a x 1 + bt 1. (25)

8 Thus, the full transformation equations are x 1 = ax bt. (5a) t 1 = at bx.. (5b) x = ax 1 + bt 1... (25) t = at 1 + bx 1 (23) To clarify the meaning of the transformation equations, it is worthwhile quoting Einstein again. "An event wherever it may have taken place would be fixed in space with respect to K by the three perpendiculars x, y, z and with respect to time by the time value t. Relative to K 1, the same event would be fixed in respect of space and time by corresponding values x 1, y 1, z 1, t 1." A change is now made from Einstein s notation. Coordinate systems P and Q are considered. P has x P,t P axes, Q has x Q, t Q axes. The transformation equations become x Q = ax P bt P. (26) t Q = at P bx P... (27) x P = ax Q + bt Q... (28) t P = at Q + bx Q (29) Consider the point in the coordinate system P that has coordinates x P =0, t P =1..(30) The corresponding coordinates in the system Q are given by From (26 ) and (30) x Q = a(0) b(1) x Q = b (31) From (27 ) and (30) t Q = a(1) b(0) t Q = a... (32) Consider the point in the coordinate system Q that has coordinates x Q =0, t Q =1. (33) The corresponding coordinates in the system P are given by From (33) and (28) x P = a(0) + b(1) x P = b..(34) From (33) and (29) t P = a(1) + b(0)

9 t P = a. (35) Similarly, if a point in the system P has coordinates x P =0, t P =1/a..(36) The corresponding coordinates in the system Q are From (26), (27) and (36) x Q = b/a, t Q = 1.. (37) Figure 7 In Figure 7, it is clear that whatever angle θ is, angles GOL and HOM remain 90. Similarly, angles FOM and JOL remain 90. If NM is parallel to GJ, the angle ONM is 90. If LM is parallel to FH, OML is 90. Thus the triangles ONM and OML are right angle triangles. All these right angles are in relation to the axis of symmetry. They are not right angles in relation to the P coordinate system (x P,t P axes) or the Q coordinate system (x Q, t Q axes).

10 Figure 8 Consider now the point N on the t P axis of P (Figure 8). The length of ON is 1/a, giving the coordinates of N (according to P) as, x P =0, t P =1/a From (36) and (37), if the coordinates of N are x P =0, t P =1/a, according to P, the coordinates of N are x Q = b/a, t Q =1 according to Q. Point M is a distance 1 along the t Q axis, giving this point the coordinates x Q =0, t Q =1 (according to Q). The line MN is parallel with the x Q axis, this length giving the x Q coordinate of b/a for the point N (as judged by Q). According to Q, the point M has coordinates x Q =0, t Q =1 (33) Thus the coordinates of M, according to P are x P = b, t P = a....(34),(35) The line OL has length a, giving t P =a. The line LM is parallel to the x P axis, and is of length b, so that x P =b. Consider now the triangle OLM. It has sides OL, length a (according to P); LM length b (according to P); OM, length 1 (according to Q). This is a right angle triangle (according to the orthogonal coordinate system of the axis of symmetry). These lengths are consistent with equation (18), where a² = 1+ b². The diagram is reassuring that the symmetrical model of relative motion is consistent with the mathematics of relativity. However, there is the problem that the diagram is comparing lengths and times according to coordinate system P with those of system Q using the mathematics of a third coordinate system, the orthogonal axis of symmetry. It is like comparing an apple with a pear using an orange measurer. To resolve this problem, it is necessary to give some thought to the curve t² x² = 1.. (38)

11 This hyperbola can be written as t = ±(x² + 1) 1/2...(39) Figure 9: Consider the curve in the region t > 0, t = + (x² + 1) (40) Figure 10 It is clear, in figure 10, that the line OM meets the curve at M with coordinates x,t. The x value of M is HM and the t value OH. OM subtends an angle θ with OH, angle HOM.

12 Tan θ (angle HOM) = HM/OH = x/t (41) From Differential Calculus, we know that if t = (x² + 1).(40) dt/dx = x/ (x² + 1).(42) from (40) and (42) dt/dx = x/t..(43) The slope of the curve (dt/dx) at any point on the curve is the slope of the tangent drawn at that point. Thus the tangent at M, the line GM, has the slope x/t (43). GM forms the angle OGM with the x axis where tan (angle OGM) = x/t..(44) Thus, OM subtends an angle θ with the y axis where tan θ = x/t (41); the tangent at M, GM, subtends the same angle θ with the x axis (44). At this point it is helpful to quote from Minkowski (Ref 3). He uses the convention that the speed of light is c. "We draw.the upper branch of the hyperbola c²t² x² = 1, with its asymptotes. From the origin O we draw any radius vector OA 1 of this branch of the hyperbola; draw a tangent to the hyperbola at A 1 to cut the asymptote on the right at B 1 ; complete the parallelogram OA 1 B 1 C 1.Now if we take OC 1 and OA 1 as axes of oblique coordinates x 1,t 1, with measures OC 1 = 1, OA 1 = 1/c, then that branch of the hyperbola again acquires the expression c²(t 1 )² (x 1 )² = 1, t > 0, and the transition from x,..t to x 1,..t 1 is one of the transformations in question." Figure 11

13 From Minkowski s diagram (Figure 11), using the convention c=1, in the x 1,t 1 coordinate system (call it Q), OA 1 =1, OC 1 = 1 (42) In the x,t (call it R) coordinate system, OA 1 = OC 1 = (x R ² + t R ²). (43) Thus the value, 1 "Q unit", along the t Q axis ( line OA 1 ) is equivalent to the value (x R 2 + t R 2 ) "R units" (the same line OA 1 ) as judged by the x R and t R axes (R coordinate system). Figure 12 For any value of t on the curve t = (x² + 1) there is a positive value (point M) and a negative value (point F) of x (Figure 12). The line OF is symmetrical with OM about the t axis. The tangent at F makes the same angle θ with the x axis as the tangent at M. The line (OF) can be chosen to be the (t P ) axis (of coordinate system P). It satisfies the curve t P ² x P ² = 1. Thus the length OF = 1 "P unit" on the t P axis. According to the R coordinate system, (x R and t R axes), the length of OF² = ( x R ) ²+ t R 2 OF = (x R 2 + t R 2 ) "R units" Thus the t P axis along OF (system P) is on a different scale to the t R axis (system R) but will have the same scale as the t Q axis along OM (system Q). The coordinates according to the t Q and t P axes will have a linear, proportional relationship with the coordinates according to the orthogonal (x R and t R ) axes. As the t Q, t P, x Q and x P axes are on the same scale, lengths from these axes will be proportionally represented in the x R,t R orthogonal axis. The mathematics of the orthogonal coordinate system R can thus be applied.

14 Thus, using a symmetrical representation of relative motion, comparing values in system P with those of system Q using the mathematics of the orthogonal axis of symmetry, coordinate system R, is like comparing apples with apples using a small apple measurer. Figure 13 Superimposing the hyperbola t R ² x R ² = 1 on the diagram showing the lengths as derived from the mathematics of the transformation equations, the point M lies in the curve (Figure 13). At M, the coordinates according to P are t P =a (line OL), x P =b (line LM). Thus, as the point lies on the curve t P ² x P ² = 1, it follows that a² b² = 1 in relation to the P coordinate system. In relation to the axis of symmetry, OLM is a right angle triangle with hypotenuse OL. The lengths of the sides are in the ratio a,b,1 with a as the hypotenuse. Thus a² = 1+ b² a² b² = 1 Thus, in this model, both according to the P coordinate system and the axis of symmetry, a² b² = 1, as shown in the mathematics (22) It is concluded that the symmetrical model represents the equations of relativity.

15 Time measurement Figure 14 In an orthogonal reference frame, light emitted from the t axis at point D, x=0,t=t b, will reach the x axis at point H, x=b, t=t. If reflected there, it returns to the t axis at point G, x=0,t=t+b (Figure 14). Thus in any given inertial frame the time of the event at H is defined as t=t. Quoting from Einsteins 1905 paper (Ref 4). "If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighbourhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far only defined an "A time" and a "B time". We have not defined a common "time" for A and B, for the latter cannot be defined at all unless we establish by definition that the "time" required by light to travel from A to B equals the time" it requires to travel from B to A." Hence the definition of time at point H (Figure 14) depends on the assumption that the ray of light takes the same time to get from D to H as from H to G. Figure 15 examines the situation where the point H (frame Q) is moving relative to the inertial frame with the clock (frame P).

16 Figure 15 In Figure 15, Q, at point H, with his clock, measures the time at H as t Q =1, the line OH. According to P the time at H is t P =a, the line OF. Ps conclusion that t P =a at H is based on the assumption that the time it takes light to get from P to Q (DH) is the same time it takes for light to get back from Q to P (HG). It can be seen from the diagram that these two distances (and therefore times) appear not to be equal. This is confirmed by calculation. Because DEH is a right angle triangle, DH² = ED² + EH² (44) HE = b/a (37) OD = OF FD = a b ED = OE OD = 1/a (a b) ED = 1/a (1 a² + ab) (45) From (45 ) and (19) ED = 1/a (ab b²) =b/a (a b) (46) From (44) and (46) DH² = (b/a) ² [ 1 + (a b) ²] =(b/a) ² [ 1 + a² 2ab + b²]

17 =(b/a) ² [ 2a² 2ab ] =2b²/a (a b) (47) HG² = HE²+ EG² (48) EG = OG OE = (a + b) 1/a = 1/a (a² + ab 1) =1/a (b² + ab) = b/a (a + b)..(49) From (48), (49) & (37) HG ² = (b/a) ² [ 1 + (a + b) ²] =(b/a) ² [ 1 + a² +2ab + b²] =(b/a) ² [ 2a² +2ab ] =2b²/a (a + b)..(50) Thus DH ² + HG ² = 2b²/a (a b) + 2b²/a (a + b) = 2b²/a (2a ) = 4b² This gives the length DG as (4b² ) = 2b This is the length expected as light left P at t=a-b and returned at t=a+b, a difference of 2b. HG ² = 2b²/a (a + b)..(50) HG = (2b²/a) 1/2 (a + b) 1/2.(51) DH² = 2b²/a (a b) (47) DH = (2b²/a) 1/2 (a b) 1/2 (52) The proportion of the total light path that has elapsed when the light reaches H is DH/(DH + HG)..(53) From (51) and (52) DH + HG = (2b²/a) 1/2 [(a + b) ½ + (a b) ½ ]..(54) DH/(DH + HG) = (2b²/a) 1/2 (a b) 1/2 /(2b²/a) 1/2 [(a + b) ½ + (a b) ½ ] = (a b) 1/2 / [(a + b) ½ + (a b) ½ ] Multiplying numerator and denominator by (a + b) ½ DH/(DH + HG) = [(a b) (a + b)] 1/2 / {[(a + b)(a + b)] ½ + [(a + b) (a b)] ½ } = [a² b²] 1/2 / {[a + b] +[a² b²] 1/2 }

18 = 1/(a + b + 1).(55) The time change, according to P, between D and G is 2b. Ps timeline between D and G will have a linear relationship with the progress of the ray of light going first to H and then to G. Time taken for P to get from O to D is (a b) Therefore the time elapsed, according to P, when the light (and Q) arrive at H is (a b) + 2b[DH/(DH + HG)] = (a b) + 2b/(a + b + 1) = (a² + ab + a ab b² b +2b)/ (a + b + 1) = (a + b + 1)/ (a + b + 1) = 1.(56) It was shown at Figure 12 that the time scales of P and Q are the same relative to the axis of symmetry. If P and Q have clocks that are synchronised at O, when Q arrives at H his clock will read t Q =1. It would be expected from the symmetry of their time scales that Ps clock would also read 1 as Q (and the light from D) arrive at H. It has also been shown that Ps clock will read 1 when the ray of light from point D arrives at H, as judged by the proportion of the light path DHG that has elapsed at H (56). It was shown in equations (30) and (31) that when x P =0, (ie at the origin of P) and t P =1 (ie the time on Ps clock is 1, the clock being at the origin), x Q = b. That is the distance P has moved along Qs negative x Q axis is b (P and Q being in Standard Configuration). By the symmetry of relative motion, when the origin of P is at x Q = b, the origin of Q is at x P =b. Thus P can reckon the time on Qs clock when Q is at x P =b, by looking at his clock when he,p, is at x Q = b. It reads 1. Thus, via these three different methods, the time on Ps clock when Q gets to H is 1. If P reckons time to be t P =a, using the flawed approach of halving the time elapsed on the light path DHG, he overestimates time when Q gets to H by a factor of a. Einstein s definition of Velocity To continue the quote from Einstein, which ended at the equations 5 (Ref 2) " we obtain the equations x 1 = ax bct ct 1 = act bx (5) We should thus have the solution of our problem, if the constants a and b were known. These result from the

19 following discussion. For the origin of K 1 we have permanently x 1 = 0, and hence according to the first of the equations (5) x = (bc/a)t..(5c) If we call v the velocity with which the origin of K 1 is moving relative to K, we then have v = (bc/a) (6)" Einstein s definition of velocity therefore is From (5c) and (6) v = x/t =bc/a If c is defined as 1 v = x/t =b/a..(57) The significance of "Proper Velocity" The above discussion in relation to time shows that Ps clock reads 1 as Q gets to point H, x P =b (Figure 15). Thus distance of Q along Ps x P axis / The time on Ps clock = b.(58) This is not the same as the value x P / t P,which is the same as Einstein s definition (57) and has value b/a Recalling the transformation equations From x P = ax Q + bt Q... (28) When x Q =0, ie at the origin of Q, x P /t Q = + b...(59) (The distance of Q along Ps positive x P axis) / (The time on Qs clock) = +b From x Q = ax P bt P. (26) When x P =0, ie at the origin of P x Q /t P = b....(60) (The distance of P along Qs negative x Q axis) / (The time on Ps clock) = +b The difference in sign in equations (59) and (60) represents the opposite directions of movement of P and Q. The distance/time relationships x P /t Q and x Q /t P have been called "Proper velocity". These ratios give the same value as the calculation of velocity based on the time on Ps clock (57). The discussion started with P moving along Q and Q moving along P, a pair of interlinked and interdependent aspects of the symmetrical relative motion of P and Q. The relationships x P /t Q and x Q /t P are symmetrical because they both have the same value and each contain an element of P and an element of Q. This compares with the classical definition of velocity which is asymmetrical as it only has measures from one reference frame x P /t P or x Q /t Q.

20 Does it then follow that the true measure of velocity is, as the name "Proper Velocity" suggests, the relationships x P /t Q and x Q /t P? They are symmetrical and have the same value as velocity based on the time on Ps clock (57). The theoretical limit of velocity Whether velocity is b or b/a is of huge importance. a = (b 2 + 1). This value is always more than b. Thus b/a is always less than 1. If the speed of light is defined as 1, then b/a is always less than the speed of light. The value b has no such restriction. k Calculus By adding Einstein s equations (4a) and (4b), (a + b) = λ (which we can call k instead of λ) (59) It follows that 1/k = 1/(a + b)..(60) = (a b) / (a + b)(a b) = (a b) / (a² b²) From (22) and (60) 1/k = (a b) )..(61) From Figure 14, light that leaves P at t P = (a b), is reflected from Q at t Q = 1, returns to P at t P = (a +b). This can alternatively be stated that light leaves P at t P = 1/k, is reflected from Q at t Q = 1, returns to P at t P =k. Thus if light leaves P at t P = T (P units of time) it arrives at Q at t Q = kt(q units of time). Light leaving Q at t Q = kt(q units of time) arrives at P at t P = k 2 T (P units of time). Thus, in a symmetrical diagram, the "k Calculus" can be a useful way of looking at time values in a spacetime diagram using rays of light. In this, the P units of time are equal to the Q units of time and a direct comparison can be made. In a model (Figures 4, 10 & 15) where one coordinate frame is orthogonal and the other oblique, the scale of the oblique axes is not the same as the orthogonal axes, and direct comparison is invalid. The Twin Paradox The twin paradox raises concerns that all may not be well with the way the mathematics of special relativity has been interpreted. A paradox is essentially nonsense. If sense could be made, there would be no paradox. Any mathematical analysis that ends in nonsense has to be reconsidered. The twin paradox is often demonstrated using the k-calculus and an asymmetrical spacetime diagram. (Figure 16). It has been shown that the algebra of the k-calculus is that of the transformation equations 26 to 29. It follows that inertial frames Alex and Barbara can only be directly compared in a symmetrical model. Any conclusions drawn from direct comparison in an asymmetrical model (Figure 16) will be invalid.

21 Figure 16 A symmetrical model (Figure 17) shows Alex and Barbara ageing the same Figure 17 The Symmetrical Spacetime Model of Special Relativity has been shown to represent the transformation equations. Interesting questions have been raised about the measurement of time. It has been shown that there may not be a limit to achievable velocity. It has solved a longstanding paradox, the twin paradox. References 1. B.F. Schutz, A First Course in General Relativity. (Cambridge University Press, Cambridge, 1985)

22 2. A.Einstein, Relativity The Special and the General Theory. (Methuen and Co Ltd, London, 1920) 3. H. Minkowski, "Space and Time". Address delivered at the 80 th Assembly of German Natural Scientists and Physicians, Cologne, In "The Principle of Relativity," Dover Publications, New York, A.Einstein, "On The Electrodynamics of Moving Bodies," Ann. Phys. (Leipzig) 17, (1905) 5. R.D Inverno, Introducing Einstein s Relativity. (Oxford University Press, Oxford, 1992)