Relativity Lite

Appendix | 95 shows that the lower triangle in figure 1 is similar to the upper triangle. Finally, the τ deriv- ative of (A3b) gives 0 = − = V A V A V A t t r r µ µ . (A3c) In the instantaneous rest frame of the uniformly accelerating rocket, V r = 0 and V t = c so that the angle θ in figure 1 goes to 0. In this case, (A3c) implies A t = 0 . Then the spatial part of the acceleration four-vector in the instantaneous rest frame of the rocket (with θ = 0 in figure 2, the triangle equivalent to figure 1 for acceleration) gives the size of the invariant acceleration, A A r 2 2 = , just as the temporal part of the momentum four-vector in the instan- taneous rest frame of the rocket gives the invariant mass, E mc 0 2 = . A vector with the former properties is called space-like , and those with the latter (more conventional) properties are called time-like . This means that for θ > 0, figure 2 must have A r on the hypotenuse. Then this triangle is similar to the one for velocity only if the three-acceleration is parallel to the three-velocity. All the following assumes this restriction . We take both three-acceleration and three-velocity parallel to the x-direction for simplicity; then, from (A3c), AV A V t t x x = . From figures 1 and 2, sin / / / θ = = = v c V V A A x t t x , (A4a) cos / / / θ γ = = = 1 c V A A t x , (A4b) tan / / / θ γ = = = v c V c A A x t . (A4c) A A A θ t r A r Figure 2. Equations (A4a–c). From sec tan 2 2 1 θ θ − = , one may show that A A V c x x = + 1 2 ( / ) (A5a) or dV V c A d x x 1 2 + = ( / ) τ . (A5b)