Chapter I Special relativity and spacetime The remaining task is to determine the coefficients a1,a2,b and b2. To do this,use is made of known relations between coordinates in both frames of reference.The first step is to use the fact that at any time t,the origin of S'(which is always at '=0 in S')will be at x Vt in S.It follows from Equation 1.21 that 0=b1Vt+b2t, from which we see that 62 =-b1V. (1.22) Dividing Equation 1.21 by Equation 1.20,and using Equation 1.22 to replace b2 by -b1V,leads to x'bix-b1Vt (1.23) at a2x Now,as a second step we can use the fact that at any time t',the origin of frame S (which is always at x=0 in S)will be at x'=-Vt'in S'.Substituting these values for z and x'into Equation 1.23 gives -Vt' -b1Vt (1.24) from which it follows that b1=a1. If we now substitute a =b into Equation 1.23 and divide the numerator and denominator on the right-hand side by t,then x'b1(x/t)-Vb1 (1.25) b1+a2(x/t) As a third step,the coefficient a2 can be found using the principle of the constancy of the speed of light.A pulse of light emitted in the positive z-direction from (ct =0,x=0)has speed c =x/t'and also c=x/t.Substituting these values into Equation 1.25 gives b1c-V61 c= b1+a2c which can be rearranged to give a2=-Vb1/c2=-Va1/c2. (1.26) Now that a2,b and b2 are known in terms of a1,the coordinate transformations between the two frames can be written as t=a1(t-Vz/c2) (1.27) r'=a1(x-Vt). (1.28) All that remains for the fourth step is to find an expression for a.To do this,we first write down the inverse transformations to Equations 1.27 and 1.28,which are found by exchanging primes and replacing V by-V.(We are implicitly assuming that a depends only on some even power of V.)This gives t=a(t'+Vx'/c2), (1.29) x=a1(x'+Vt'). (1.30) 22Chapter 1 Special relativity and spacetime The remaining task is to determine the coefficients a1, a2, b1 and b2. To do this, use is made of known relations between coordinates in both frames of reference. The first step is to use the fact that at any time t, the origin of S% (which is always at x % = 0 in S % ) will be at x = V t in S. It follows from Equation 1.21 that 0 = b1V t + b2t, from which we see that b2 = −b1V. (1.22) Dividing Equation 1.21 by Equation 1.20, and using Equation 1.22 to replace b2 by −b1V , leads to x % t % = b1x − b1V t a1t + a2x . (1.23) Now, as a second step we can use the fact that at any time t % , the origin of frame S (which is always at x = 0 in S) will be at x % = −V t% in S % . Substituting these values for x and x % into Equation 1.23 gives −V t% t % = −b1V t a1t , (1.24) from which it follows that b1 = a1. If we now substitute a1 = b1 into Equation 1.23 and divide the numerator and denominator on the right-hand side by t, then x % t % = b1(x/t) − V b1 b1 + a2(x/t) . (1.25) As a third step, the coefficient a2 can be found using the principle of the constancy of the speed of light. A pulse of light emitted in the positive x-direction from (ct = 0, x = 0) has speed c = x %/t% and also c = x/t. Substituting these values into Equation 1.25 gives c = b1c − V b1 b1 + a2c , which can be rearranged to give a2 = −V b1/c2 = −V a1/c2 . (1.26) Now that a2, b1 and b2 are known in terms of a1, the coordinate transformations between the two frames can be written as t % = a1(t − V x/c2 ), (1.27) x % = a1(x − V t). (1.28) All that remains for the fourth step is to find an expression for a1. To do this, we first write down the inverse transformations to Equations 1.27 and 1.28, which are found by exchanging primes and replacing V by −V . (We are implicitly assuming that a1 depends only on some even power of V .) This gives t = a1(t % + V x% /c2 ), (1.29) x = a1(x % + V t% ). (1.30) 22