330
30 On the Causality Problem: Particle–Tachyon Collisions
Holmes in his reference system
. We calculate the time t
2 that Holmes registered
in his reference system
at which Wacker was released from prison.
Using the Lorentz transformation (151), Chap. 13, we calculate for t 2 = 0, x 2 =
2L and again with v =
4
5
c o ,
t
2 =
t 2 − vx 2 /c
2
o
1 − v 2 /c 2
o
=
0 −
4
5
c o 2L/c
2
o
1 −
16
25
c 2
o /c 2
o
= −
8
5
5
3
L
c o
,
thus
t
2 = −
8
3
L
c o
.
(473)
We also calculate
x
2 =
x 2 − vt 2
1 − v 2 /c 2
o
=
5
3
2L ,
thus
x
2 =
10
3
L .
(474)
This time, measurement (473) cannot exonerate Mr Wacker. We saw above, cf.
Eq. (253), that Holmes registered the time t
1 = −
1
2
L/c o , so that Holmes observes
t
2 < t
1 . This means that, from the reference system
, Wacker was already free
before Mr Stus was killed and could rightly be held suspect. The measurements
made by Mr Holmes can only exonerate Dr Fast. The occurring event E 2 , to the
right-hand side of E 1 , is seen to occur even earlier to the approaching Mercury
Express than the event E o , see Fig. 30.2. Holmes, anticipating this result, sent his
friend Dr Watson to travel with the Hermes Courier, a train travelling with the same
velocity as the Mercury express, but in the opposite direction, and asked him to take
several measurements using the equipment he had deposited there. Dr Watson was
thus in the reference system
, which seen from o had a velocity of v
= −
4
5
c o .
1
The instruments were set up in such a fashion that Dr Watson registered Dr Fast’s
release from prison at the coordinates x
o = 0, t
o = 0, thus
E o :
o : x o = 0 , t o = 0 ,
: x
o = 0 , t
o = 0 ,
: x
o = 0 , t
o = 0 .
⎫
⎬
⎭
(475)
1 This reference system is not identical to the introduced reference system in Chap. 17
dealing with the twin paradox.