Step 2:Isotropy and homogeneity of space and Step 5:Inversion i=-,i=-y,andi=-', homogeneity of time 动=-y ..imply that (i)x'is independent of y and z,(ii)y is independent of z,x,and t;(iii)z'is independent of x, y,and t;(iv)t'is independent of y and z,so P(民剪，，) r'=a11(v)x+a14(v)t, (5) =a22(v)y, (6) (,,, 2=a33(v)z, (7) t=a41(v)z+a44(v)t. (8) NOTE:The fact that y'and z'are independent of x and t follows from the requirement that the x'-axis (the line y'=2'=0)always coincides with the x-axis (the line y=z=0);this would not be possible if y'and 2' depended on z and t. IMPORTANT:Eq.(8)indicates that it is possible to have two spacially separated events A and B that are 1 simultaneous in frame K and,yet,non-simultaneous in frame K,that is FIG.3:(Color online.)"Inverted"frames in relative motion. △tAB=0,△xAB≠0：△tAB=a41△xAB+0.(9) ..which is just a relabeling of coordinate marks,pre- This is not as obvious as might seem:for example,be- serves the right-handedness of the coordinate systems fore Einstein it was assumed that whenever AtAB is zero, At'AB must also be zero.So keeping a41(v)in (8)is a and is physically equivalent to (inverted)frame K'mov- significant departure from classical Newtonian mechan- ing with velocity i=-v relative to (inverted)frame K ics. (see Fig.3),so that Once the standard method of clock synchronization is =a(-v)(位-vt) (18) adopted,it is,however,relatively easy to give an example of two events satisfying(9).Try that on your own! 到=k(-v), (19) 2'=k(-v)2, (20) t=6(-v):+y(-v)t, (21) Step 3:Isotropy of space or, ..also implies that y'and z'are physically equivalent, so that a22(v）=a33(v)≡k(v),and thus -x'=a(-v)(-x-t), (22) x'=a11(v)z+a14(v)t, (10) -=-k(-v)y, (23) y=k(v)y, (11) 2'=k(-v)z, (24) 2'=k(v)2, (12) =-6(-v)x+y(-v)t, (25) t=a41(v)x+a44(v)t. (13) which gives a(-v)=a(v) (26) Step 4:Motion of O'(the origin of frame K) k(-U)=k(v), (27) ..as seen from K gives xo uto',or o-vto =0. 6(-v)=-6(w), (28) On the other hand,as seen from K,o=0.For this y(-v)=Y(w) (29) to be possible,we must have 'o (x-vt),and thus I'=a(v)(x-ut), (14 Step 6:Relativity principle and isotropy of space y=k(v)4, (15) 2'=k(v)z, (16) ..tell us that the velocity of K relative to K',as mea- t'=8(v)I+y(v)t, (17) sured by K'using primed coordinates (r',t'),is equal to -v.REMINDER:the velocity of K'relative to K,as where we have re-labeled o41≡6anda44≡Y, measured by K using unprimed coordinates (z,t),is v. NOTE:The y just introduced will soon become the I justify this by considering two local observers co- celebrated gamma factor. moving with O and O,respectively,and firing identical3 Step 2: Isotropy and homogeneity of space and homogeneity of time . . . imply that (i) x 0 is independent of y and z, (ii) y 0 is independent of z, x, and t; (iii) z 0 is independent of x, y, and t; (iv) t 0 is independent of y and z, so x 0 = α11(v)x + α14(v)t, (5) y 0 = α22(v)y, (6) z 0 = α33(v)z, (7) t 0 = α41(v)x + α44(v)t. (8) NOTE: The fact that y 0 and z 0 are independent of x and t follows from the requirement that the x 0 -axis (the line y 0 = z 0 = 0) always coincides with the x-axis (the line y = z = 0); this would not be possible if y 0 and z 0 depended on x and t. IMPORTANT: Eq. (8) indicates that it is possible to have two spacially separated events A and B that are simultaneous in frame K and, yet, non-simultaneous in frame K0 , that is ∆tAB = 0, ∆xAB 6= 0 : ∆t 0 AB = α41∆xAB 6= 0. (9) This is not as obvious as might seem: for example, be￾fore Einstein it was assumed that whenever ∆tAB is zero, ∆t 0 AB must also be zero. So keeping α41(v) in (8) is a significant departure from classical Newtonian mechan￾ics. Once the standard method of clock synchronization is adopted, it is, however, relatively easy to give an example of two events satisfying (9). Try that on your own! Step 3: Isotropy of space . . . also implies that y 0 and z 0 are physically equivalent, so that α22(v) = α33(v) ≡ k(v), and thus x 0 = α11(v)x + α14(v)t, (10) y 0 = k(v)y, (11) z 0 = k(v)z, (12) t 0 = α41(v)x + α44(v)t. (13) Step 4: Motion of O 0 (the origin of frame K0 ) . . . as seen from K gives xO0 = vtO0 , or xO0 −vtO0 = 0. On the other hand, as seen from K0 , x 0 O0 = 0. For this to be possible, we must have x 0 ∝ (x − vt), and thus x 0 = α(v)(x − vt), (14) y 0 = k(v)y, (15) z 0 = k(v)z, (16) t 0 = δ(v)x + γ(v)t, (17) where we have re-labeled α41 ≡ δ and α44 ≡ γ. NOTE: The γ just introduced will soon become the celebrated gamma factor. Step 5: Inversion x˜ = −x, y˜ = −y, and x˜ 0 = −x 0 , y˜ 0 = −y 0 K K’ (x ’ , y ’ , z ’ , t’) P (x, y, z, t) x ’ O O’ v = − v y ’ z ’ y x z ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ FIG. 3: (Color online.) “Inverted” frames in relative motion. . . . which is just a relabeling of coordinate marks, pre￾serves the right-handedness of the coordinate systems and is physically equivalent to (inverted) frame K˜ 0 mov￾ing with velocity ˜v = −v relative to (inverted) frame K˜ (see Fig. 3), so that x˜ 0 = α(−v)(˜x − vt), (18) y˜ 0 = k(−v)˜y, (19) z 0 = k(−v)z, (20) t 0 = δ(−v)˜x + γ(−v)t, (21) or, −x 0 = α(−v)(−x − vt), (22) −y 0 = −k(−v)y, (23) z 0 = k(−v)z, (24) t 0 = −δ(−v)x + γ(−v)t, (25) which gives α(−v) = α(v), (26) k(−v) = k(v), (27) δ(−v) = −δ(v), (28) γ(−v) = γ(v). (29) Step 6: Relativity principle and isotropy of space . . . tell us that the velocity of K relative to K0 , as mea￾sured by K0 using primed coordinates (x 0 , t0 ), is equal to −v. REMINDER: the velocity of K0 relative to K, as measured by K using unprimed coordinates (x, t), is v. I justify this by considering two local observers co￾moving with O and O0 , respectively, and firing identical