324
30 On the Causality Problem: Particle–Tachyon Collisions
We now replace the tachyon velocities u and u
with their respective allocated
velocities v and v
from u = c
2
o /v and u
= c
2
o /v
. Furthermore, we replace the
corresponding square root factors according to the rule κ = γ u/c o . The conservation
laws (457) then have the following form:
m o w
1 − w 2 /c 2
o
+
m
T
∗ c o
1 − v 2 /c 2
o
=
m o w
1 − w 2 /c 2
o
+
m
T
∗ c o
1 − v 2 /c 2
o
, Momentum
m o c
2
o
1 − w 2 /c 2
o
+
m
T
∗ c o v
1 − v 2 /c 2
o
=
m o c
2
o
1 − w 2 /c 2
o
+
m
T
∗ c o v
1 − v 2 /c 2
o
. Energy
⎫
⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎭
(457a)
In order to simplify Eq. (457a), we use to our advantage the fact that we may freely
select the reference system in which we want to describe the occurring collision
process. Furthermore, we also have the positive-valued direction of the x-axis at our
disposition. We can therefore assume that the particle in the laboratory, in which we
describe its collision with the tachyon, is at rest before the collision, and furthermore,
that the momentum of the colliding tachyon points in the direction of the positive
x-axis, the momentum parameter m
T
∗ being therefore positive. We then introduce a
positive parameter μ according to
μ :=
m o
m T
∗
.
(458)
We again refer the reader to the curiosities of tachyon kinematics. Even though the
momentum of our colliding tachyon must always remain positive, its velocity can
change its sign. Equation (457a) becomes, by dividing by m
T
∗ ,
:
c o
1 − v 2 /c 2
o
=
μ w
1 − w 2 /c 2
o
+
c o
1 − v 2 /c 2
o
, Momentum
μ c
2
o +
c o v
1 − v 2 /c 2
o
=
μ c
2
o
1 − w 2 /c 2
o
+
c o v
1 − v 2 /c 2
o
. Energy
⎫
⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎭
(457b)
In order to get a clearer overview of the possible solutions of these equations, we
now introduce hyperbolic functions according to,
v
c o
= tanh α
,
v
c o
= tanh α ,
w
c o
= tanh β
.
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(459)
Because w = 0, we get with w/c o = tanh β = 0, also β = 0.
Under observance of 1/
1 − tanh
2 x = cosh x , tanh x/
1 − tanh
2 x = sinh x
and simple rearrangements, Eq. (457b) is followed by
30 On the Causality Problem: Particle–Tachyon Collisions
We now replace the tachyon velocities u and u
with their respective allocated
velocities v and v
from u = c
2
o /v and u
= c
2
o /v
. Furthermore, we replace the
corresponding square root factors according to the rule κ = γ u/c o . The conservation
laws (457) then have the following form:
m o w
1 − w 2 /c 2
o
+
m
T
∗ c o
1 − v 2 /c 2
o
=
m o w
1 − w 2 /c 2
o
+
m
T
∗ c o
1 − v 2 /c 2
o
, Momentum
m o c
2
o
1 − w 2 /c 2
o
+
m
T
∗ c o v
1 − v 2 /c 2
o
=
m o c
2
o
1 − w 2 /c 2
o
+
m
T
∗ c o v
1 − v 2 /c 2
o
. Energy
⎫
⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎭
(457a)
In order to simplify Eq. (457a), we use to our advantage the fact that we may freely
select the reference system in which we want to describe the occurring collision
process. Furthermore, we also have the positive-valued direction of the x-axis at our
disposition. We can therefore assume that the particle in the laboratory, in which we
describe its collision with the tachyon, is at rest before the collision, and furthermore,
that the momentum of the colliding tachyon points in the direction of the positive
x-axis, the momentum parameter m
T
∗ being therefore positive. We then introduce a
positive parameter μ according to
μ :=
m o
m T
∗
.
(458)
We again refer the reader to the curiosities of tachyon kinematics. Even though the
momentum of our colliding tachyon must always remain positive, its velocity can
change its sign. Equation (457a) becomes, by dividing by m
T
∗ ,
:
c o
1 − v 2 /c 2
o
=
μ w
1 − w 2 /c 2
o
+
c o
1 − v 2 /c 2
o
, Momentum
μ c
2
o +
c o v
1 − v 2 /c 2
o
=
μ c
2
o
1 − w 2 /c 2
o
+
c o v
1 − v 2 /c 2
o
. Energy
⎫
⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎭
(457b)
In order to get a clearer overview of the possible solutions of these equations, we
now introduce hyperbolic functions according to,
v
c o
= tanh α
,
v
c o
= tanh α ,
w
c o
= tanh β
.
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(459)
Because w = 0, we get with w/c o = tanh β = 0, also β = 0.
Under observance of 1/
1 − tanh
2 x = cosh x , tanh x/
1 − tanh
2 x = sinh x
and simple rearrangements, Eq. (457b) is followed by