30 On the Causality Problem: Particle–Tachyon Collisions
333
to not get caught up in contradictions, also see the discussion based on the examples
shown in Chap. 28. Let us now more closely examine the observations made by Dr
Watson.
With t 2 = 0, x 2 = L and again v = v
= −
4
5
c o , Dr Watson observes for t
2 ,
the time of release of Mr Wacker,
t
2 =
t 2 − v
x 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
.
(480)
He also observes for x
2 ,
x
2 =
x 2 − v
t 2
1 − v 2 /c 2
o
=
5
3
2L ,
thus
x
2 =
10
3
L .
(481)
The following observations were made for the event E 2 , see also Fig. 30.2,
E 2 :
o : x 2 = 2L , t 2 = 0 ,
: x
2 =
10
3
L , t
2 = −
8
3
L
c o
,
: x
2 =
10
3
L , t
2 =
8
3
L
c o
.
⎫
⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎭
(482)
We see t
1 < t
2 from (478) and (482). Observed from the reference system
the
event E 1 occurs before the event E 2 . The more spatially right-hand side events do
occur, from the
point of view, at a later time point than in o due to the opposite
direction of motion of the Hermes Courier. This can cause the reversal of the order
of events in time, which itself is reversed from the viewpoint of the Mercury
Express. Here, we once again refer the reader to our theorem on the twin paradox in Chap. 17:
“The time dilatation of a moving clock is solely dependent on the square of its velocity.
The synchronisation regulation of a clock changes its sign when the direction of velocity
changes.”
According to the measurements made by Dr Watson, Dr Fast’s assistant was still
in prison after Mr Stus had already died. Thus, Mr Wacker also had an alibi and had
to be set free. Although the measurements made by Dr Watson showed Dr Fast to
be outside of prison before Mr Stus was killed, the measurements made by Holmes
exonerated him. Both suspects had to be set free because of their respective alibis.
333
to not get caught up in contradictions, also see the discussion based on the examples
shown in Chap. 28. Let us now more closely examine the observations made by Dr
Watson.
With t 2 = 0, x 2 = L and again v = v
= −
4
5
c o , Dr Watson observes for t
2 ,
the time of release of Mr Wacker,
t
2 =
t 2 − v
x 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
.
(480)
He also observes for x
2 ,
x
2 =
x 2 − v
t 2
1 − v 2 /c 2
o
=
5
3
2L ,
thus
x
2 =
10
3
L .
(481)
The following observations were made for the event E 2 , see also Fig. 30.2,
E 2 :
o : x 2 = 2L , t 2 = 0 ,
: x
2 =
10
3
L , t
2 = −
8
3
L
c o
,
: x
2 =
10
3
L , t
2 =
8
3
L
c o
.
⎫
⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎭
(482)
We see t
1 < t
2 from (478) and (482). Observed from the reference system
the
event E 1 occurs before the event E 2 . The more spatially right-hand side events do
occur, from the
point of view, at a later time point than in o due to the opposite
direction of motion of the Hermes Courier. This can cause the reversal of the order
of events in time, which itself is reversed from the viewpoint of the Mercury
Express. Here, we once again refer the reader to our theorem on the twin paradox in Chap. 17:
“The time dilatation of a moving clock is solely dependent on the square of its velocity.
The synchronisation regulation of a clock changes its sign when the direction of velocity
changes.”
According to the measurements made by Dr Watson, Dr Fast’s assistant was still
in prison after Mr Stus had already died. Thus, Mr Wacker also had an alibi and had
to be set free. Although the measurements made by Dr Watson showed Dr Fast to
be outside of prison before Mr Stus was killed, the measurements made by Holmes
exonerated him. Both suspects had to be set free because of their respective alibis.