28 Particles and Tachyons
307
With a positive v
( and also positive u
), v and therefore also u become negative
if −V > v
is chosen, which is always possible for the reference system.
We will now illustrate the peculiarities of tachyons using two numerical examples.
An observer in o determines the velocity of a tachyon as u =
5
4
c o . He sends this
tachyon towards an observer moving with the velocity V =
1
2
c o . Using (415) one
can calculate the velocity u
that the observer (moving with V in o ) determines
for this tachyon, u
=
−
1
2 c o +
5
4 c o
1−
co 5co
2·4c 2
o
= 2c o . The observer running behind the tachyon
registers that the tachyon is in fact moving away from him at a faster speed than
bevor! We can find a tentative explanation in correspondence with Eq. (425). The
observer following the tachyon assumes a portion of the tachyon’s energy of motion.
According to (425), the amount of the energy of a tachyon becomes smaller if the
amount of its velocity increases. The tachyon loses energy when it gets faster. If the
velocity increases infinitely, the tachyon completely loses its energy.
The following situation is even more peculiar. In o the velocity u = 2c o is
measured for a tachyon. An observer follows this tachyon with the velocity V =
4
5
c o .
This observer then registers the velocity u
=
−
4
5 +2c o
1−
4co 2co
5c 2
o
= −2c o for the tachyon. Is
the tachyon in fact moving towards the observer, who was sent out to follow the
tachyon? This would not be anything extraordinary for velocities in the region of
|u| < c o , a simple overtaking procedure, e.g. one car overtaking the other. A tachyon
cannot however be overtaken! its velocity is always greater than c o , a velocity that
we can never achieve. Nevertheless, the velocity u
= x
//t
of the tachyon in
becomes negative, whilst its velocity u = x//t in o is positive. The sign of
the quantity x for the tachyon cannot change when the tachyon changes reference
systems, just as the tachyon’s momentum retains its sign as shown above (with
exactly the same argumentation). For a positive u and a positive x, u
can only
become negative if t
becomes negative! The tachyon therefore does not spatially
come closer to the observer in
. Its velocity becomes negative, because it moves
backwards through time! We will say that a tachyon moves backwards through time if
its velocity and momentum have opposite signs. If we uphold the notion of tachyon
velocity, we must then accept the fact that a tachyon, depending on the reference
system in which it is observed, can move with the attributed velocity even backwards
through time. This is where the causality problem connected to tachyons is anchored.
We will return to this during the next two chapters.
When the sign of the tachyon’s velocity changes, the sign of its energy E
T also
changes according to (425) and thus also per definition that of its mass m
T
= E
T
/c
2
o .
An explanation for all these peculiarities can be found in the following considerations. We remain in one and the same reference system o and realise an inversion (a
mirroring) of the spatial coordinates, see also Chap. 12. In other words, we describe
all positions using new space coordinates x, whilst retaining the time coordinates
according to
x = −x , t = t .
(431)
307
With a positive v
( and also positive u
), v and therefore also u become negative
if −V > v
is chosen, which is always possible for the reference system.
We will now illustrate the peculiarities of tachyons using two numerical examples.
An observer in o determines the velocity of a tachyon as u =
5
4
c o . He sends this
tachyon towards an observer moving with the velocity V =
1
2
c o . Using (415) one
can calculate the velocity u
that the observer (moving with V in o ) determines
for this tachyon, u
=
−
1
2 c o +
5
4 c o
1−
co 5co
2·4c 2
o
= 2c o . The observer running behind the tachyon
registers that the tachyon is in fact moving away from him at a faster speed than
bevor! We can find a tentative explanation in correspondence with Eq. (425). The
observer following the tachyon assumes a portion of the tachyon’s energy of motion.
According to (425), the amount of the energy of a tachyon becomes smaller if the
amount of its velocity increases. The tachyon loses energy when it gets faster. If the
velocity increases infinitely, the tachyon completely loses its energy.
The following situation is even more peculiar. In o the velocity u = 2c o is
measured for a tachyon. An observer follows this tachyon with the velocity V =
4
5
c o .
This observer then registers the velocity u
=
−
4
5 +2c o
1−
4co 2co
5c 2
o
= −2c o for the tachyon. Is
the tachyon in fact moving towards the observer, who was sent out to follow the
tachyon? This would not be anything extraordinary for velocities in the region of
|u| < c o , a simple overtaking procedure, e.g. one car overtaking the other. A tachyon
cannot however be overtaken! its velocity is always greater than c o , a velocity that
we can never achieve. Nevertheless, the velocity u
= x
//t
of the tachyon in
becomes negative, whilst its velocity u = x//t in o is positive. The sign of
the quantity x for the tachyon cannot change when the tachyon changes reference
systems, just as the tachyon’s momentum retains its sign as shown above (with
exactly the same argumentation). For a positive u and a positive x, u
can only
become negative if t
becomes negative! The tachyon therefore does not spatially
come closer to the observer in
. Its velocity becomes negative, because it moves
backwards through time! We will say that a tachyon moves backwards through time if
its velocity and momentum have opposite signs. If we uphold the notion of tachyon
velocity, we must then accept the fact that a tachyon, depending on the reference
system in which it is observed, can move with the attributed velocity even backwards
through time. This is where the causality problem connected to tachyons is anchored.
We will return to this during the next two chapters.
When the sign of the tachyon’s velocity changes, the sign of its energy E
T also
changes according to (425) and thus also per definition that of its mass m
T
= E
T
/c
2
o .
An explanation for all these peculiarities can be found in the following considerations. We remain in one and the same reference system o and realise an inversion (a
mirroring) of the spatial coordinates, see also Chap. 12. In other words, we describe
all positions using new space coordinates x, whilst retaining the time coordinates
according to
x = −x , t = t .
(431)