30 On the Causality Problem: Particle–Tachyon Collisions
335
or anyone. They would just go right ‘through’ Mr Stus without a trace, just as we
saw in Eq. (466) concerning the elastic collision process.
Dr Fast however also wanted revenge and calculated the following experiment:
The device in the possession of his assistant can only create tachyons with a negative
momentum parameter. He himself adjusts his device to create tachyons with a positive
momentum parameter. He now aims his device at Stus, who is in his office located at
x 1 = L, simultaneously together with Wacker’s, the later located at x 2 = 2L and his
device at x o = 0. What happens, or what could happen if the tachyons inelastically
collide at the location where Mr Stus is standing?
This situation is exactly the same situation that we had when we started to discuss
tachyons in Chap. 28. In both situations, we observe the inelastic collision of two
tachyons. However, we now assume that the rest frame o of Dr Fast is the system
in which the total momentum disappears and where the restmass M o is created after
the collision. The tachyon with a positive velocity u created by Dr Fast has the
momentum parameter m
T
∗ . The momentum parameter of the tachyon with the negative velocity −u created by Wacker is δ · m
T
∗ = −m
T
∗ , so that the total momentum
disappears. As we know, the momentum Eq. (418) and the energy equation (419) with
f (u) according to (419a) and |u| > c o are both fulfilled, whereby the Eq. (423a)
determines M o . This means,
M o =
2m
T
∗
u 2 /c 2
o − 1
.
(483)
If sufficiently ‘slow’ tachyons could be created, meaning tachyons whose velocities
are not much greater than c o , then a particle with an arbitrarily large mass M o could
be created by the collision. This particle would of course destroy anything at the
location of its creation.
In order to prohibit such an inelastic tachyon reaction resulting in the creation
of a restmass M o all the above-mentioned general objections of a non-linear field
theory can be applied. A particle with the mass M o can only be created if such
an ‘elementary particle’ actually exists. We have seen that, e.g. the restmass of the
‘elementary particle’ kink, in other words the solution (111) of the sine-Gordon
equation (88) can only assume a certain value m o according to (296). A kink with
an arbitrary restmass M o cannot be created in our crystal, because there is no such
thing as a kink with an arbitrary restmass in our crystal. In our depiction above, we
will have to trust the competent explanations made by the physicist Dr Fast and
accept that the described process, the creation of a particle with a certain restmass
M o from two tachyons, is possible.
In the framework of the sine-Gordon equation, it is a mathematically important question whether the two tachyons with opposing momentum parameters, i.e.
the tachyons belonging to the solutions (261) and (445), coming from t −→ −∞,
for t −→ +∞ can be composed into a particle solution, a kink of the form (111).
The answer to this question cannot be given here.
335
or anyone. They would just go right ‘through’ Mr Stus without a trace, just as we
saw in Eq. (466) concerning the elastic collision process.
Dr Fast however also wanted revenge and calculated the following experiment:
The device in the possession of his assistant can only create tachyons with a negative
momentum parameter. He himself adjusts his device to create tachyons with a positive
momentum parameter. He now aims his device at Stus, who is in his office located at
x 1 = L, simultaneously together with Wacker’s, the later located at x 2 = 2L and his
device at x o = 0. What happens, or what could happen if the tachyons inelastically
collide at the location where Mr Stus is standing?
This situation is exactly the same situation that we had when we started to discuss
tachyons in Chap. 28. In both situations, we observe the inelastic collision of two
tachyons. However, we now assume that the rest frame o of Dr Fast is the system
in which the total momentum disappears and where the restmass M o is created after
the collision. The tachyon with a positive velocity u created by Dr Fast has the
momentum parameter m
T
∗ . The momentum parameter of the tachyon with the negative velocity −u created by Wacker is δ · m
T
∗ = −m
T
∗ , so that the total momentum
disappears. As we know, the momentum Eq. (418) and the energy equation (419) with
f (u) according to (419a) and |u| > c o are both fulfilled, whereby the Eq. (423a)
determines M o . This means,
M o =
2m
T
∗
u 2 /c 2
o − 1
.
(483)
If sufficiently ‘slow’ tachyons could be created, meaning tachyons whose velocities
are not much greater than c o , then a particle with an arbitrarily large mass M o could
be created by the collision. This particle would of course destroy anything at the
location of its creation.
In order to prohibit such an inelastic tachyon reaction resulting in the creation
of a restmass M o all the above-mentioned general objections of a non-linear field
theory can be applied. A particle with the mass M o can only be created if such
an ‘elementary particle’ actually exists. We have seen that, e.g. the restmass of the
‘elementary particle’ kink, in other words the solution (111) of the sine-Gordon
equation (88) can only assume a certain value m o according to (296). A kink with
an arbitrary restmass M o cannot be created in our crystal, because there is no such
thing as a kink with an arbitrary restmass in our crystal. In our depiction above, we
will have to trust the competent explanations made by the physicist Dr Fast and
accept that the described process, the creation of a particle with a certain restmass
M o from two tachyons, is possible.
In the framework of the sine-Gordon equation, it is a mathematically important question whether the two tachyons with opposing momentum parameters, i.e.
the tachyons belonging to the solutions (261) and (445), coming from t −→ −∞,
for t −→ +∞ can be composed into a particle solution, a kink of the form (111).
The answer to this question cannot be given here.