A numerical example of invariance example 12
Figure ae shows two frames of reference in motion relative to
one another atv = 3/5. (For this velocity, the stretching and
squishing of the main diagonals are both by a factor of 2.) Events
are marked at coordinates that in the frame represented by the
square are(t,x) = (0, 0) and
(t,x) = (13, 11).The interval between these events is 13^2 − 112 = 48. In the
frame represented by the parallelogram, the same two events lie
at coordinates(t′,x′) = (0, 0) and
(t′,x′) = (8, 4).Calculating the interval using these values, the result is
82 − 42 = 48, which comes out the same as in the other frame.424 Chapter 7 Relativity