(b) to — t B = (1 — y1— (v/c) 2 ) / 0 /v or tB—tA= (1+V 1— (v/c) 2 ) 10 /v.1.353. (a) t (B)=1 0 1v, t (B') = (1 (^0) /v) -111 —(v/c) 2 ; (b) t (A) =
(1 0 Iv)1 y1— (vIc) 2 , t (A') = 10 /v.
1.354. See Fig. 12 showing the positions of hands "in terms of K
clocks".
Af
A
H
Fig. 12.
1.355. x = (1— y1 — [3^2 ) c/I3, where p = V/c.
1.356. It should be shown first that if At = t 2 — tl > 0, then
At' =-- to — t; > 0.
1.357. (a) 13 ns; (b) 4.0 m. Instruction. Employ the invariance
of the interval.
ii(v,— V) 1 +1(1 — V 2 1c 2 )
1.358. v' = (^1) —v,V/c2 •
1.359. (a) v = vi + v 2 = 1.25c; (b) v = (v1 + v 2 )/(1 viv^2 /c^2 ).
= 0.91c.
1.360. 1 = 1 0 (1 — 2 )/(1 p2), where p = v/c.
1.361. v = v 2 ,—(v 1 v 2 /c 2 ).
1.362. s = Ato (1
2
_ 02) _v , where 0. V/c.
1.363. tan 0'
cos 0
—02 sin 0 , where
— vic^3 =
1.364. tan 0 = vie y1— (V/c)2.
1.365. (a) w' = w (1 — 13 2 ) 3 / 2 /(1 — 13v/c) 3 ; (b) w' = w (1 — 32).
Here 13 = V/c.
1.366. Let us make use of the relation hetween the acceleration
w' and the acceleration w in the reference frame fixed to the Earth:
= (1 — v2Ic2)- 3/2 dv dt •This formula is given in the solution of the foregoing problem
(item (a)) where it is necessary to assume V = v. Integrating the
given equation (for w' = const), we obtain v = w't/V1 (w't/c) 2.
The sought distance is 1 = (y1 + (w't/c) 2 — 1) c 2 I w' = 0.91 light-
year; (c — v)/c = 1/2 (c/w't) 2 = 0.47%.