2.2 Matter to Energy Conversion.
In (2.6) above, if the applied temporal force is in the opposite temporal
direction, i.e. if
so as to encourage temporal motion into the future, then via an identical
analysis to that above, (2.14) becomes
This result infers that energy is being extracted from the mass by the
applied temporal force.
Converting (2.16) to energy gives
and the rate of extraction is then
Consequently it would appear that the power available from such a conversion
could, from the application of a very small force, be substantial. For
instance, with Ft =1gm, in (2.18)
|
P = 300,000Kgm.m/sec = 2.94MW |
| (2.19) |
Also in (2.17) when E = 0
and therefore a 1gm.sec2/m mass of matter, subjected to the 1gm temporal force
would last for
delivering the above power. This is sufficient to provide all the power
requirements of about 1,960 typically British homes for nearly 10 years,
assuming losses are small by comparison.
Such a source of power, if it could be harnessed, would be virtually
limitless and extremely inexpensive. The main problem to be solved however,
which is obviously a formidable one, is the development of a means of
generating and applying a temporal force.
2.3 Spatial Travel at the Terminal Velocity of D0, ( ~ the Velocity of Light).
It is well known that Einstein's Special Theory of Relativity shows that it
is not possible to accelerate a mass to the velocity of light. To do so
would require an infinite amount of energy, and take an infinite amount of
time. However, this result assumes that the applied force is all purely
spatial.
To investigate the effect of a combined spatial/temporal force, consider the
time differential of (2.1), i.e.
|
F = |
dM
dt
|
= |
æ
è
|
m |
dv
dt
|
+ v |
dm
dt
|
ö
ø
|
r + j | ì
÷
í
÷
î
|
c |
dm
dt
|
|
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
- |
vm
|
|
dv
dt
|
ü
÷
ý
÷
þ |
|
| (2.22) |
where the subscript r has been dropped from the spatial velocity as no longer
necessary.
If the applied force is a spatial/temporal one such that
|
F = F0 |
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
r - j F0 |
v
c
|
|
| (2.23) |
where F0 is a constant, then substitution of (2.23) into (2.22) gives
Spatial
|
m |
dv
dt
|
+ v |
dm
dt
|
= F0 |
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
|
| (2.24) |
Temporal
|
c |
dm
dt
|
|
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
- |
vm
|
|
dv
dt
|
= - F0 |
v
c
|
|
| (2.25) |
and as will be shown below, (2.24) and (2.25) represent the "zero mass rate"
equations of motion of the mass m. In this case the spatial-temporal
Force/Velocity diagram is as shown in Fig.2.2 below.
Fig. 2.2 - Zero Mass Rate Force/Velocity Diagram
and the force/reaction diagram as in Fig. 2.3 below
Fig. 2.3 - Zero Mass Rate Force/Reaction Diagram.
These diagrams can be compared with [1], Figs. 3.1 and 3.2 to show the
difference to the application of a purely spatial force.
It is clear from (2.23) that the spatial/temporal force vector F is applied
precisely at right angles to the Existence Velocity vector and then
continuously adjusted to remain so. It is this feature of the applied force
vector that ensures that the mass rate remains non-existent. Hence the
absence of the mass rate terms in the diagrams. The ensuing analysis will
confirm this and derive the other characteristics of the motion.
With F0 being constant, (2.25) can be integrated by inspection to give
|
mc |
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
= - F0 |
r
c
|
+ k |
| (2.26) |
where although this has been derived from a temporal equation, r is the
spatial distance moved in the direction of travel. Invoking the initial
conditions viz. v = 0, r = 0, and m = m0 when t = 0 gives
and substitution of this into (2.26) then gives
Clearly the application of the negative temporal component of the applied
force is extracting energy from the mass. This energy could then be used to
supply the spatial part of the impressed force. The mass is thereby acting
as "matter fuel" to maintain its own drive. Also it would appear from (2.28)
that m = 0 when
and, it will shown later that this result still represents an anomaly, even
though at this point v=c so that in fact (2.28) then becomes indeterminate,.
Now differentiating (2.28) with respect to the time gives the time rate of
change of mass thus
|
|
dm
dt
|
= |
|
( m0 c2 - F0 r ) |
v
c2
|
|
dv
dt
|
- F0 v |
æ
è
|
1 - |
v2
c2
|
ö
ø
|
|
|
| (2.30) |
Eqs (2.28) and (2.30) may now be substituted into (2.24) to give
|
|
m0 c - F0 r
|
|
dv
dt
|
+
|
|
( m0 c2 - F0 r ) |
v2
c2
|
|
dv
dt
|
- F0 v2 |
æ
è
|
1 - |
v2
c2
|
ö
ø
|
|
= F0 |
æ
è
|
1 - |
v2
c2
|
ö
ø
|
1/2
|
|
| (2.31) |
which reduces to
|
|
m0 c2 - F0 r
|
|
dv
dt
|
= F0 |
æ
è
|
1 - |
v2
c2
|
ö
ø
|
1/2
|
|
| (2.32) |
and comparison of (2.32) with (2.24) and (2.28) clearly shows that
i.e. the mass rate is non-existent.
The interpretation of this is that the relativistic increase in mass due to
the application of the spatial part of the impressed force, is exactly
offset by the reduction in mass due to the extraction of energy by the
temporal part of the impressed force. Consequently not only are there no
inertial effects, but effectively there is no increase in mass due to the
storage of kinetic energy, i.e. energy mass is maintained at the level of
rest mass during the whole period that the impressed force obeys (2.23).
Accordingly (2.30) can be equated to zero to yield after minor reduction
where
and is the spatial acceleration at t = 0.
Also, in (2.28) this means that the mass m must be constant at the value of
rest mass , thus
This permits the spatial velocity v to be expressed as a function of the
spatial distance travelled thus
|
v = c |
ì
í
î
|
1 - |
æ
è
|
1 - |
a0 r
c2
|
ö
ø
|
2
|
ü
ý
þ
|
1/2
|
|
| (2.37) |
and it can easily be shown that the time differential of (2.37) equates to
(2.34). Converting (2.37) to a simple differential equation in r gives
cdt =
|
dr
|
|
ì
í
î
|
1 - |
æ
è
|
1 - |
a0 r
c2
|
ö
ø
|
2
|
ü
ý
þ
|
1/2
|
|
|
| (2.38) |
This is a standard integral easily solved by putting
|
1 - |
a0 r
c2
|
= cosf so that dr = |
c2
a0
|
sinf df |
| (2.39) |
The result is
|
r = |
c2
a0
|
|
æ
è
|
1 - cos |
a0 t
c
|
ö
ø
|
|
| (2.40) |
Consequently
and therefore finally
|
a = |
dv
dt
|
= a0 cos |
a0 t
c
|
|
| (2.42) |
From (2.41) v = c when
i.e. when
and clearly this time may be made as small as required by making a0 as
large as necessary. Also insertion of (2.40) and (2.41) into (2.28) confirms
that the mass remains constant at m0.
Thus it would appear that "light" velocity has been achieved in a finite
time whilst maintaining the mass constant at m0, the rest mass. However,
unfortunately, all is not as simple as it seems, as was indicated by (2.29).
The problem is revealed by determining the energy levels involved.
From [1], Eq.(3.19) the kinetic energy associated with spatial motion at
relativistic velocities is
expanding this to
|
Ek = mc2 |
ì
í
î
|
1- |
æ
è
|
1- |
v2
c2
|
ö
ø
|
1/2
|
ü
ý
þ
|
|
| (2.46) |
enables it to be sensibly applied to the current process. Inserting (2.41)
for v and m0 for m gives
|
Ek = m0 c2 |
æ
è
|
1 - cos |
a0 t
c
|
ö
ø
|
|
| (2.47) |
so that the matter energy is then
and therefore when t reaches the value in (2.44)
and
What appears to have happened is that the matter energy of the rest mass has
all been converted to kinetic energy. The precise manner in which this
kinetic energy exists in D0 is unclear, but may be addressed in a
future paper.
Accordingly, if the above process were proposed for the drive of a
futuristic interstellar vehicle, it would be necessary to apply the temporal
force in (2.23) only to that part of the vehicle mass identified as "matter
fuel". This would mean that the final velocity attained would be less that
that of light and therefore to maximise this velocity and thereby minimise
the relativistic mass increase of the non- matter fuel component of the
vehicle, its infra-structure would have to be as small and as light as
possible compared to the mass of the "matter fuel".
Also in this application, the ratio of the spatial and temporal components
of the impressed force, from (2.23) shows that while the spatial part of
(2.23) was larger than the temporal part, the energy extracted from the
matter fuel would be insufficient to supply all of the spatial drive.
Consequently some form of additional power would be needed during this
period. However, once the ratio had reversed and more energy was being
extracted than was needed to supply the spatial drive, the subsidiary fuel
could thereby be replenished.
R3 Version 1.0.0
Ó
P.G.Bass, December 2004
|