Chapter I Special relativity and spacetime Dividing the upper and lower expressions on the right-hand side of this equation by At,and cancelling the Lorentz factors,gives Ax! (△x/△t-V) △P=Q-(Ax/△t)V7E阿 Now,if we suppose that the two events that we are considering are very close together一indeed,if we consider the limit as△tand△go to zero一then the quantities△x/△tand△x'/△'will become the instantaneous velocity components vz and v of a moving object that passes through the events 1 and 2. Extending these arguments to three dimensions by considering events that are not confined to the z-axis leads to the following velocity transformation rules: 以= Ur-V 1-vzV/c2 (1.43) = Uy y(V)1-.V/c2 (1.44) 2= Uz y(V)(1-vV/c2) (1.45) These equations may look rather odd at first sight but they make good sense in the context of special relativity.When v and V are small compared to the speed of light c,the term vV/c2 is very small and the denominator is approximately 1.In such cases,the Galilean velocity transformation rule,=v-V,is recovered as a low-speed approximation to the special relativistic result.At high speeds the situation is even more interesting,as the following question will show. An observer has established that two objects are receding in opposite directions.Object 1 has speed c,and object 2 has speed V.Using the velocity transformation,compute the velocity with which object 1 recedes as measured by an observer travelling on object 2. O Let the line along which the objects are travelling be the r-axis of the original observer's frame,S.We can then suppose that a frame of reference S'that has its origin on object 2 is in standard configuration with frame S,and apply the velocity transformation to the velocity components of object 1 with v=(-c,0,0)(see Figure 1.14).The velocity transformation predicts that as observed in S',the velocity of object 2 is v'=(v,0,0),where Ur-V -c-V 4=1-V7=1-(-e7=-6 So,as observed from object 2,object 1 is travelling in the -'-direction at the speed of light,c.This was inevitable,since the second postulate of special relativity (which was used in the derivation of the Lorentz transformations) tells us that all observers agree about the speed of light.It is nonetheless pleasing to see how the velocity transformation delivers the required result in this case.It is worth noting that this result does not depend on the value of V. Exercise 1.7 According to an observer on a spacestation,two spacecraft are moving away,travelling in the same direction at different speeds.The nearer spacecraft is moving at speed c/2,the further at speed 3c/4.What is the speed of one of the spacecraft as observed from the other? 30Chapter 1 Special relativity and spacetime Dividing the upper and lower expressions on the right-hand side of this equation by Δt, and cancelling the Lorentz factors, gives Δx % Δt % = (Δx/Δt − V ) (1 − (Δx/Δt)V /c 2) . Now, if we suppose that the two events that we are considering are very close together — indeed, if we consider the limit as Δt and Δx go to zero — then the quantities Δx/Δt and Δx %/Δt % will become the instantaneous velocity components vx and v % x of a moving object that passes through the events 1 and 2. Extending these arguments to three dimensions by considering events that are not confined to the x-axis leads to the following velocity transformation rules: v % x = vx − V 1 − vxV /c 2 , (1.43) v % y = vy γ(V )(1 − vxV /c 2 ) , (1.44) v % z = vz γ(V )(1 − vxV /c 2) . (1.45) These equations may look rather odd at first sight but they make good sense in the context of special relativity. When vx and V are small compared to the speed of light c, the term vxV /c 2 is very small and the denominator is approximately 1. In such cases, the Galilean velocity transformation rule, v % x = vx − V , is recovered as a low-speed approximation to the special relativistic result. At high speeds the situation is even more interesting, as the following question will show. ● An observer has established that two objects are receding in opposite directions. Object 1 has speed c, and object 2 has speed V . Using the velocity transformation, compute the velocity with which object 1 recedes as measured by an observer travelling on object 2. ❍ Let the line along which the objects are travelling be the x-axis of the original observer’s frame, S. We can then suppose that a frame of reference S% that has its origin on object 2 is in standard configuration with frame S, and apply the velocity transformation to the velocity components of object 1 with v = (−c, 0, 0) (see Figure 1.14). The velocity transformation predicts that as observed in S% , the velocity of object 2 is v % = (v % x , 0, 0), where v % x = vx − V 1 − vxV /c 2 = −c − V 1 − (−c)V /c 2 = −c. So, as observed from object 2, object 1 is travelling in the −x % -direction at the speed of light, c. This was inevitable, since the second postulate of special relativity (which was used in the derivation of the Lorentz transformations) tells us that all observers agree about the speed of light. It is nonetheless pleasing to see how the velocity transformation delivers the required result in this case. It is worth noting that this result does not depend on the value of V . Exercise 1.7 According to an observer on a spacestation, two spacecraft are moving away, travelling in the same direction at different speeds. The nearer spacecraft is moving at speed c/2, the further at speed 3c/4. What is the speed of one of the spacecraft as observed from the other? ■ 30