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Quantum and Relativity Theories UnifiedIn Concepts, Principles
and Laws
The New Relativity Theory
BY AZZAM K. I. ALMOSALLAMIP.O. BOX 1067
GAZA, PALESTINEEMAIL : [email protected]
Abstract
In our work (the new relativity theory), we will unify
relativity and quantumtheory (Copenhagen school) in concepts,
principles and laws. While this newtheory is in agreement with the
concepts, principles and laws of the quantumtheory (Copenhagen
school), it introduces changes from the abstract,undiscriptive and
unimaginative to descriptive, and imaginative. As
previously stated, quantum theory was applied to the micro world
while themacro world was controlled by the laws of classical
physics. We believed thatthe laws that control the micro and macro
worlds are the same. Because ofthis, we formulated a new theory to
unify micro and macro into one theorywith the same concepts,
principle and laws.
Introduction
When Einstein formulated his special relativity theory, he
believed in anobjective existence of phenomena, rather than in the
principles of continuity,determinism, and causality in the nature
laws. But Quantum theory(Copenhagen school) discovered that the
observer participates in determiningthe formulation of phenomena.
That is clear from Heisenberg's definition ofthe wave function,
(1958) it is a mixture of two things, the first is the realityand
the second is our knowledge of this reality. Thus, we get from
thisdefinition that phenomena does not exist without the observer
receiving it.
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Einstein was hardly refusing this concept for phenomena, as Pais
said in(1979) while I was walking with Einstein, he said, - look at
the moon do you believe it exists because we are looking at it? ]2[
.
Furthermore, quantum theory adopts the principles of
non-causality,indeterminism, and discontinuity in the nature
laws.
The mathematical formula for Einstein's relativity depends on
Riemans spaceof four dimensions , but in quantum Hilberts space
with infinite dimensions.Stapp said in (1972) the Copenhagen school
refused understanding theworld by the concepts of space-time, when
it considers relativity theory isinconsistent for understanding the
micro world, and where quantum theoryforms the basis for
understanding this world ]3[ .
Also in formulating his relativistic equations, Einstein depends
on the
possibility of simultaneously measuring the location of a
particle and itsmomentum. Heisenberg discovered this is impossible
to do (Heisenberguncertainty principle) ]6,5[ . Oppenheimer said
Einstein in the last days of hislife tried to prove the
inconsistency of quantum laws but failed. After that hesaid - I
dislike quantum theory, especially Heisenberg's
uncertaintyprinciple ]4[ .
Theory
1- The Hypotheses of the Theory
The First Hypothesis: The speed of light is constant and equal
to C in anyinertial frame of reference, where C is the speed of
light in a vacuum.
The Second Hypothesis: The speed of light in any frame moving
withconstant velocity V is equal to 'C for any inertial frame of
reference, where
22' V C C = , whereas 'C does not depend on the direction of the
velocity ofthe moving frame, but depends only on the absolute value
of the velocity.
To understand these two hypotheses, I assume a stationary
observer onthe earth's surface. In this case, the earth's surface
is considered a referenceframe. Now, if the stationary observer
made an experiment for measuring thespeed of light in his reference
frame, he would find it equals C , the speed oflight in vacuum.
Also, if there was a train moving with constant velocity V onthe
earth surface, and a stationary rider on the moving train made
anexperiment for measuring light speed inside this train, he would
find thespeed of light equals C . In this case, the moving train is
considered a
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stationary reference frame, and the result is the same as for
the stationaryobserver on the earth's surface. This is according to
the first hypothesis.
Now, assume the stationary observer on earth measured the speed
of lightinside the moving train. In this case he would find it
equals 22' V C C =according to the second hypothesiss. Now if C V ,
then from the secondhypothesis we get 0' C . This is equivalent to
quantum theory. For theobserver stationary on the earth's surface,
the solution of the wave function is 0 inside the moving train. So
the probability density (the probability ofgetting any information
inside the moving train for the earth observer) equalszero, where
0* , where * is the complex conjugate of .
2- Time in our Relativity Theory
(2.1) Consider a train at rest and fixed observer, (both on the
earth's surface)
where the length of the train is L . If one of the riders of the
train sends a rayof light along the length of the train, the time
required for the ray of light totraverse the length of the train
for both the fixed observer and the rider is 0t ,where
C L
t
=0 (2.1.1)
The above equation does not contradict Heisenberg's uncertainty
principle
since C and L
are not measured at the same time. The equation makes uspredict
the speed of light C when the displacement L is known.
Now, suppose the train moved with constant velocity V and the
rider sends aray of light along the length of the train during that
motion, and the fixedobserver sets his clock to compute the time
required for the ray to traverse thelength of the moving train.
According to the second hypothesis, the speed oflight inside the
moving train is 'C relative to the fixed observer, where
22' V C C = . Thus the time required for the ray of light to
traverse thelength of the moving train is t , where
22' V C
LC L
t
=
=
From the second hypothesis we propose, 'C does not depend on the
directionof the transmitted ray of light comparing with the
direction of the velocity ofthe train. Also, the above equation is
in contrast with the Lorentztransformation equations which are are
built on the concepts of continuity,
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causality, and determinism. In our work, we believe in
discontinuity, non-causality and indeterminism. The measurement
that is taken in the equationabove is taken from a wave function
and to get another measurement youshould use another wave function
and vise versa. Those wave functions areunrelated. Also, the
Lorentz transformation equations proposed that we canmeasure the
velocity of the train and its location at the same time. That is
in
contrast with the uncertainty principle of Heisenberg.
Therefore from equation (2.1.1), we get
0t C L =
Then
22
0
V C
t C t
=
Thus
2
2
0
1C
V t t
= (2.1.2)
Where 0t is the time required to the ray of light to pass the
length of the trainwhen it is at rest.
In the derivation of equation (2.1.2), we considered that the
fixed observer onthe earth's surface will measure the length of the
moving train to be equal to
L as it is at rest. That is in contrast with Einstein's length
contraction.
Equation (2.1.2) indicates that the time separation of any event
in any movingframe with constant velocity V is greater than the
rest time separation, (whenthe frame is at rest) for any inertial
frame of reference external to the movingframe.
(2.2) Now, suppose one of the riders of the moving train sets
his clock insidethe train to measure the time required for the
light beam to pass the length ofhis train during the motion.
According to equation (2.1.2), the time separationfor any event
happening in the train, is greater when it is moving than whenit is
at rest for the reference frame of the earth surface. Because the
clock'smotion on the train is considered as events occurring inside
the train, it will be slower than the clock of the fixed observer
in the reference frame of theearth surface. Now, if we assume both
the observer and the rider will agreeon the beginning of the event
and its end inside the moving train, then if theobserver computes
the time, via his clock as t for the light beam to pass the
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length of the moving train, then the rider will compute the time
't via hisclock, where
t Rt =1'
Where
2
21
1
C V
R
= . Since from equation (2.1.2)
0t Rt =
Thus we get0' t t =
From the above equation, we find that the time separation of the
event insidethe train that is measured by the rider via his clock
during the motion is equal
to the rest time separation of the same event (the measured time
when thetrain is at rest). From this, we can write equation (2.1.2)
as
't Rt = (2.2.1)
Accordingly, we can predict that the speed of light inside the
moving train forthe rider is equal to light speed in a vacuum. This
agrees with the first
hypothesis of the theory. As a result of the slowing of light
speed inside themoving train for the reference frame on the earth's
surface, it made time slow(movement of clocks) inside the train.
That made the measurement of lightspeed inside the train to be the
same as the speed of light in a vacuum for therider.
(2.3) Now suppose the stationary observer desires to compare the
motion ofthe clock of the moving rider with the motion of his
clock. According toequation (2.1.2), and, because the motion of the
clock of the rider is an eventinside the moving train, thus the
clock will be slower when the train ismoving than when it is at
rest for the observer in the reference frame of theearth's surface.
If the observer computes the time t at this instant, he willfind
that the clock of the rider will compute the time 't where
t Rt = 1'
(2.4) now, suppose the rider of the moving train desires to use
the clock of thestationary observer for computing the time required
to the light beam to pass
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the length of his train. The time which will be measured by the
stationaryobserver via his clock is t where
0t Rt =
If we consider the rider is moving with constant velocity
forward, then theclock of the observer should be moving with the
same velocity in the oppositedirection for the rider. In this case,
the riders frame is considered to be areference frame, and the
clock as a frame moving with constant velocity V forhim. According
to the preceding discussion, the clock will be slower for therider
than the observer for the reference frame of the earth surface.
Thus, ifthe observer computes the time t by his clock, at this
instant, the rider willcompute the time 't by the same clock {or by
his clock inside the train as wehave seen in (2.2), whereas
t Rt = 1
'
For more clarification, suppose the length of the train is m21 ,
and its speed isC 87.0 . If the clock computes by ns where sec101
9=ns ., then the time
required to the light beam to pass the length of the moving
train for the fixedobserver is t , where from equation (2.1.2) we
get
2
0
)87.0(1
=
t t
And
nst 70100.3
2180
=
= .
Where, we propose smC / 103 8= . Thus, we have
.1405.0
.70ns
nst ==
Thus, the fixed observer will compute ns140 via his clock for
the light beam topass the length of the moving train. For the
rider, the time is 't where from
equation (2.2.1) we get
.701' 22
nst C V
t ==
So, the rider will compute .70ns for the light beam to pass the
length of histrain. Both, the observer and the rider will agree on
the beginning and endingof the event, and when both used the same
clock to compute the time
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separation for this event, the clock was slower for the rider
than the observer.So, when the observer has ns140 for the time
separation, at that instant, therider receives only the first ns70
of the clock's motion that were experienced by the fixed observer
in the past. His present at this instant is at ns140 , whilethe
present of the rider is at ns70 . Subsequently, we can consider the
riderlives in the past of the observer on the earth surface during
his motion.
In this example we find that when both the rider and the
observer use thesame clock, each one creates his own clock to get
his reading. That is incontrast with the objective existence of the
phenomenon. In our example, wedetermine that the observer is the
main participant in the formulation of thephenomenon as in the
concepts of the Copenhagen School.
(2.5) Now suppose train ( ) is at rest and its length is L .
Also there is train( ) moving with constant velocity V and there is
a fixed observer on the
earth's surface. Both the fixed observer and the rider of train
will measurethe time required for the light beam to traverse the
length of the fixed train .For the observer, the measured time
according to his clock is 0t where
C L
t
=0
For the rider of train , since train is moving with constant
velocity V,thus the speed of light inside it compared with the
reference frame of the
fixed observer is22
' V C C =
, thus the rider should be computing the timer t for the event
where
022t R
V C
Lt r =
=
where, 0t is the time separation of the event when the rider's
train isstationary. Because the riders clock is slowed during the
motion for thereference frame of the earth surface {as we have seen
in (2.2)}, thus, the rider
will compute the time 't
, where
01' t t Rt r ==
(2.5.1)
Equation (2.5.1) indicates that, both the rider of the moving
train and thefixed observer will measure the same time separation
for the light beam topass the length of the fixed train . That
leads us to say that the measured
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speed of light is the same for each inside fixed train and is
equal to C (thespeed of light in a vacuum). Thus, we can write
equation (2.5.1) as
C L
t
= ' (2.5.2)
If both the fixed observer and the rider of the moving train
agree on thetime required for the light beam to pass the length of
the fixed train , then,they will differ as to the beginning and
ending of the event. We have seenpreviously that the rider of the
moving train was living in the past of theobserver on the earth's
surface during the motion.
To clarify, let us assume both the observer and the rider agree
on the beginning of the event, on the condition of
0=V at
0= t C V 87.0= at t >0
when the earth's clock points to zero at 0= t , (before
transmitting the light beam,) the velocity of train of the rider
was equal to zero, and at the firstmoment of transmitting the light
beam at t >0 , the velocity of the train wasequal to C 87.0 ( in
this case, for simplicity, we neglect the effect ofacceleration).
Subsequently, the fixed observer and the rider of the movingtrain
will agree on the beginning of transmitting the light beam inside
thefixed train , and will differ on its end.
If the length of fixed train is m21 , then, the time required
for the light beamto pass the length of fixed train for the fixed
observer is
.70100.3
218 nst =
=
For the rider of the moving train from equation (2.5.2),
.70100.3
21' 8 nst =
From that, we see that both the rider and the observer measure
the same timeseparation of the event. Bbut because the time (clock)
in the frame of themoving train is slower than the time (clock) of
the fixed observer for thereference frame of the earth's surface,
then the light beam will arrive at theend of the fixed train faster
for the observer than the rider. Thus, if the
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observer confirms that the light beam arrived at the end of the
train, at thisinstant, the rider confirms that the light beam
arrived at the middle of thetrain. If the observer confirms that
the light beam traversed distance x , atthis instant the rider will
confirm that the light beam traversed distance ' x ,where, x R x =
1' . Also we get, in this case, y R y = 1' , and z R z = 1' .Now,
if the observer looks at the clock of the rider, he will confirm
that theclock of the rider computes only ns35 at the moment that
his clockcomputes ns70 , where we get t Rt = 1' . But, if the rider
looks at the clock ofthe observer, he will confirm that the clock
of the observer computes only
ns35 , as in his clock, while the observer secures that his
clock computes .70ns
(2.6) Now, if the rider of the moving train observes the clock
of the fixedobserver at the condition of
0=V at 0= observer t
C V 87.0= at 0 < sec4 observer t .0=V at observer t > sec4
.
Where, observer t is the reading of the fixed observer from his
clock.
We can draw observer t versus rider t for the reference frame of
the earth surfaceas in figure (2.6.1), where, rider t is the
reading of the rider from the clock ofthe fixed observer
From figure (2.6.1), we find two straight lines; the first for 0
< .sec4 observer t
and its slope =0.5. The second is for observer t > .,sec4 and
its slope =1
We find from the figure, the seconds between 2 < .sec4
observer t would not be determined by the rider where the train of
the rider stopped at
observer t > sec4 . He would find that the observer was
reading the seconds at
observer t > .,sec4 while his last reading was equal to .sec2
That means the
events measured by the fixed observer between 2 < .sec4
observer t were not
received by the rider of the moving train.
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0
2
4
6
8
10
12
0 2 4 6 8 10
t (observer) sec.
t ( r i d e r ) s e c .
Figure (2.6.1): t (observer) versus t (rider).
From the figure we get, the observer is the main participant in
formulation ofthe phenomenon, where each one creates his own clock
during the motion.That is in contrast with the objective existence
of the phenomenon.
For greater clarification, let us study this example. Assume the
train started atrest to move with constant velocity V at the first
moment of sunrise at 5 AMas in figure (2.6.2). Now if both the
train rider and the earth observer observethe sun's motion, we know
from the preceding discussion, the moving trainrider will get the
events regarding the sun's motion slower than the earthobserver for
the reference frame of the earth's surface. For more
clarification,suppose the train's velocity is C 87.0 . Now if the
earth observer registers thesun's motion at time t=1 PM, the time
of the train rider will be
48)5.0(1' 22
=== t C V
t
Thus AM t 945' =+=
Thus, the train rider will observe the sun's motion at 9AM. That
is consideredthe past for the earth observer. Where the train rider
(in his present, lives inthe past of the earth observer at 9AM, and
views the events which happenedon earth at 9AM, while the present
of the earth is at 1PM.
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Figure (1): A illustrates the sun at 5 am for the earth observer
and the trainrider. At this instant, the train started at rest to
move with velocity V. B is theobservation of the earth observer for
the sun's motion at 1PM via his clock. C isthe observation of the
moving train rider for the sun's motion at 9Am via hisclock.
Figure (1): A At time 5 am forboth the earthobserver and
thetrain rider, thetrain started atrest to move
with velocity V.
BFigure (1): B
After 8 hours -viathe clock of the
earth observerfor starting of themoving train, theearth
observerregisters the time1 PM via hisoclock.
CFigure (1): C At the momentthat the earthobserver registersthe
time 1pm viahis clock, the trainrider registers thetime 9 am via
hisclock
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3- Velocity in the new relativity theory:
(3.1) Now let us go back to the moving train rider and the
stationary observeron the earth's surface, where both of do an
experiment to measure the velocityof the moving train. This can be
done with two pylons. The distance betweenthem is x . Now, the
measured time according to the earth observer for the
train to pass the distance x is t with respect to his clock, so
the measuredvelocity for him is observer V , whereas
V t
xV observer =
=
From the last equation we can observe that it is not
inconsistent withHeisenberg's uncertainty principle, since observer
V and x were not measuredsimultaneously. When the earth observer
determines the train location
precisely at x , he can predict by the last equation that the
train velocity wasequal to observer V .
Now, when the earth observer determines that the train reached
the end of its journey at x , at this instant, according to the
train rider (during his motion)the train did not reach the end of
the journey, and did not traverse the
distance x . The train reached the distance ' x where xC
V x = 2
2
1' , and
this distance was passed at a time separation equal to 't ,
where
t C V
t = 22
1'
Therefore the measured velocity with respect to the moving train
rider will berider V , whereas
V V t R
x Rt
xV observer rider ==
=
=
1
1
''
(3.1.1)
From that we find that both the earth observer and the rider on
the movingtrain will measure the same train velocity during the
motion, where both willmeasure the actual velocity. In the special
theory of relativity of Einstein, themeasured distance for the
moving train rider between the two pylons will be
' x where
x R x = 1'
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Therefore the distance between the two pylons will decrease
during the train'smotion according to the train rider. That is
because Einstein believed in theobjective existence of the
phenomenon. According to this concept, both theearth observer and
the moving train rider will be in agreement for the start ofmoving
the train from the first pylon and then will agree on the train
reaching
the end of its journey at the second pylon. Subsequently,
according to thereciprocity principle, the earth observer will also
judge the train's length will
be decreased according to the factor 22
1C V
in the direction of the velocity.
But in our new relativity theory both; the earth observer and
the moving trainrider will measure the same train length in the
velocity direction (say in
x direction) and also in any direction y and z , also both of
them will beagreed on the measured distance between the two pylons,
but the train'smotion makes the rider get his measurements at a
slower rate than the earth
observer. This is in agreement with Copenhagen School concepts,
where theobserver plays the major role in determining phenomena.
Both the earthobserver and the rider make their own determination
of the motion of thetrain and clocks.
(3.2) Now suppose, there is a stationary train on the earth's
surface whichcontains a clock. As we have seen in (2.4), the rider
of the moving train willdetermine the clock motion of train is
identical with his clock motion,where the time that is measured via
his clock is equal to the time that ismeasured via the clock of
train . Also the earth observer will determine thatthe motion of
the clock of train is identical with his clock motion. Now iftrain
is moving with constant velocity V between the two pylons, then
aswe have seen in (3.1), both the earth observer and the moving
train rider will agree on the actual measured velocity of the train
as V . Now if it sends alight beam along the length of train during
its motion, then the requiredtime for the light beam to pass train
's length with respect to the earthobserver is t where
2
2
0
1C
V
t t
=
But according to the train rider, the time separation of this
event is 't ,where
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2
2
0
1
''
C
V
t t
=
Where '0t is the time separation to the light beam to pass the
length of train with respect to the moving train rider, when train
is at rest. And, since
both the earth observer and the moving train rider will agree on
the timeseparation of this event when train is at rest as in
equation. (2.5.1), thus weget
00' t t =
Subsequently, we get
t t = 0'
From the last equation, we know that both the earth observer and
the movingtrain rider will be agreed on the time separation for the
light beam totraverse the length of moving train , but both of them
will disagree on the beginning of the event and its end. Also, both
will be agree, that the clockmotion on the moving train will be
slower than their clocks and they will be agreed on the slowing
rate.
We get from this example that the motion of train did not effect
the rider's
measurement, where his measurement was identical with the earth
observer'smeasurement during the motion, but the motion of train
made the rider getthis measurement at a slower rate than the earth
observer.
(3.3) Suppose a sphere is moving with constant velocity pV on
the earth'ssurface. As we have seen previously, both the earth
observer and the movingtrain rider will be agreed on the sphere's
velocity on the earth surface, where both of them will measure the
velocity to be equal to the actual velocity pV .Now, if this sphere
entered the moving train and traversed the length of thetrain, then
the time separation for the rider is 't via his clock and t for
theobserver via his clock where
't Rt =
====================In this case, both the earth observer and
the train rider should agree on the beginning of the event from the
initial sphere's motion inside the train and thefinal; when the
sphere traverses train length L , but they will differ on the
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measured time separation. Subsequently the measured velocity for
the sphereinside the moving train according to the train rider is
rider V , where
prider V t L
V =
=
'
The last equation agrees with Heisenberg's uncertainty principle
since rider V and L were not measured simultaneously. At the
instant that the riderexactly determines the sphere's position on L
inside the train, he predictsfrom the last equation that the sphere
was moving with velocity rider V insidehis train. According to the
earth observer, it is observer V , where
'1 2
2
t L
C V
t L
V observer
=
=
Therefore
pobserver V C V
V 22
1 = (3.3.1)
From equation (3.3.1) we find that the sphere's velocity inside
the movingtrain with respect to the earth observer is less than the
sphere velocity on the
earth's surface by the factor 22
1C
V . The sphere velocity inside the train
according to the earth observer is less than that of the train
rider.
(3.4) Now suppose that both the earth observer and the train
rider desireapplying this condition
0=V At 0= xC V 87.0= At 0< m x 100
0=V At 100= x
This condition illustrates the moving train's velocity in terms
of x , where x isthe train's traversed distance according to the
earth observer. Figure (3.4.1),illustrates the relationship between
x and ' x , where ' x is the train's traverseddistance according to
the train rider.
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0
20
40
60
80
100
120
0 20 40 60 80 100
x(m)
x ' ( m
)
Figure (3.4.1): illustrates the relationship between x and ' x
.
From figure (3.4.1) we find that the relationship between x and
' x is a straight
line. Its slope is 22
1C
V , and we find that when the earth observer judges
that the train covered m100 , then the rider will judge (during
the motion) thathis train passed only m50 . When the train is at
rest at m x 100= , and the riderleaves his train, he will be
surprised that the traversed distance is m100 , not
m50 . Subsequently he will avow that his train transferred from
m50 to m100in one instant, and the distance in the interval 50<
x 0, the train moved with constant
velocity V , and after the train traversed distance x
according to the earthobserver, the train stopped. In this case,
the earth observer records accordingto his clock, the time t for
the train to traverse distance x . Therefore hewill predict the
train was moving with velocity observer V , where
V t
xV observer =
=
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Then the measured train momentum according to him was equal to
observer Pwhere
mV t
xmV m p observer observer =
==
Where m is the measured mass according to the earth observer
during itsmotion. But according to the train rider, when the train
stopped, the traverseddistance is ' x where x x = ' as in figure
(3.4.1). On the other hand, the riderwill judge that this distance
was traversed in time separation 't according tohis clock,
where
t C
V t = 2
2
1'
where t is the measured time according to the earth observer.
Subsequentlythe rider will predict that the train velocity was
rider V where
2
2
2
2
2
2
111''
C
V
V
C
V
V
t C V
xt
xV observer rider
=
=
=
= (4.1.2)
We find that, equation (4.1.2) disagrees with equation (3.1.1),
where we findfrom equation (3.1.1) that V V V observer rider == ,
but equation (4.1.2) indicates to
2
2
2
2
11''
C
V
V
C
V
xt
xV rider
=
=
=
That is because equation (3.1.1) applied during the train motion
where the
traversed distance according to the rider is ' x wherein xC
V
x = 2
2
1' , where
x is the traversed distance according to the earth observer. But
x x = ' ,when the train is at rest as in figure (3.4.1). Therefore
the rider will judge thatthe train was moving with momentum rider P
, where
2
200
1C V
V mV mP rider rider
== (4.1.3)
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0m is the train rest mass, as we have assumed that the mass of
the movingtrain according to the train rider is the rest mass. Now
if we assume that boththe train rider and the earth observer agreed
on the momentum measurementof the moving train, subsequently after
the equivalence between the twoequations (4.1.1) and (4.1.3) we
get
2
20
1C V
V mmV
=
From the last equation, we can get
2
2
0
1C V
mm
= (4.1.4)
Equation (4.1.4) represents the relativistic mass of the moving
train according
to the earth observer, where the train mass increases by the
factor
2
2
1
1
C
V
during its motion.
(4.2) Suppose the rider on the moving train desires measuring
the mass ofthe stationary train on the earth's surface. As we have
seen he will judgethat the clock motion of train is symmetrical
with his clock motion, wherethe time that will be measured via the
train clock equals the time that will
be measured via his clock, which means that "' t t = , where 't
is the timeseparation that the rider will measure via his clock,
and "t is the timeseparation that the rider will measure via the
train clock. Now if train passed the distance ' x according to the
rider of , then we can consider forthe rider of train , that train
is moving with constant velocity V , but in theopposite direction
to the train velocity . Subsequently it may be consideredthat train
passed distance ' x with respect to train rider. Therefore
themeasured momentum of train with respect to the train rider is
rider Pwhere
''
t x
mPrider
= (4.2.1)
where m is the relativistic mass of train with respect to the
rider of themoving train . But according to the measured momentum
for the train
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with respect to itself according to the reference frame of the
moving train istrainP (according to the clock of train with respect
to rider of train ), where
''
"'
00 t x
mt x
mPrider
=
= (4.2.2)
where 0m is the rest mass of train .
Now when we equivalent the two equations (4.2.1) and (4.2.2) we
get
''
''
0 t x
mt
xm
=
Subsequently, we get
0mm = (4.2.3)
From the last equation we find that the rider of the moving
train willmeasure (during the motion) the mass of the stationary
train to be equal tothe rest mass according to the stationary
observer on the earth's surface.
(4.3) Suppose now train was moving with constant velocity V
between thetwo pylons, and after the train covered the distance ' x
between the two
pylons according to the rider of the moving train , then the
train stopped. Inthis case, the rider will judge that the train
passed this distance in timeseparation 't according to his clock,
and subsequently the rider will predictthat train was moving with
momentum rider P , whereas
''
t x
mmV P rider rider
== (4.3.1)
Where m is the relativistic mass of the moving train according
to the riderof the moving train . But according to the moving train
, the distance ' x
was covered in time separation ''t according to his clock with
respect to thereference frame of the moving train , where
'1''2
2
t C
V t =
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and subsequently the momentum of train can be predicted
according toitself with respect to the reference frame of the
moving train to be trainP ,where
'1
''''
2
200
t C
V
xm
t x
mPtrain
=
=
(4.3.2)
And by the equivalence of these two equations (4.3.1) and
(4.3.2), we get
2
2
0
1C
V
mm
= (4.3.3)
equation (4.3.3) represents the measured relativistic mass of
the moving train
with respect to the rider of the moving train
, whereas we find accordingto equation (4.3.3) that the train
mass will increase during the motion by the
factor
2
2
1
1
C
V
, according to the rider of the moving train , and also
according to the earth observer. Whereas we find that both the
earth observerand the rider of the moving train will agree on the
measured relativisticmass of moving train , but the motion of train
made the rider get hismeasurement at a slower rate than the earth
observer.
5- The Equivalence of Mass and Energy
The kinetic energy of the moving train , according to the earth
observer oraccording to the rider of moving train is defined
through calculating thework done by increasing train velocity from
zero to V . Suppose a force F affects train parallel to the
distance dx , subsequently the work done ontrain is dw , where
dxF dw .=
Newton's second law is defined as
)( mV dt d
F =
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Where m is the relativistic mass of train according to the earth
observer orthe rider of the moving train . From that we get
dxmV dt d
dw )( =
Sincedt dxV = , therefore we get
)( mV d V dw =
And as we saw previously the relativistic mass given by the
relation
2
2
0
1C
V
mm
= , where 0m is the rest mass, subsequently we can get
=
V
C
V
V d V C mw
0
2
2
20
1
And by integrating the last equation, we get
= 1
1
1
2
22
0
C
V C mw
And because of the work done makes an increase in the velocity
from zeroto V , this makes a change in the kinetic energy of train
from zero to k E .Subsequently we get
= 1
1
1
2
2
20
C
V C m E k
(5.1)
Equation (5.1) represents the kinetic energy of the moving train
accordingto both the earth observer and the rider of the moving
train , where both ofthem will agree on the measurement of k E
.
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(3) Stapp, H. "The Copenhagen School Interpretation and the
nature of space-time",(1972), American Journal of Physics, 40,
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(5) Robert, E., "Quantum Physics of Atoms, Melecules, Solids,
Nuclei, and Particles",(1985), John Wiley and Sons, 2 nd ed..
(6) Richard, L. Liboff, "Introductory Quantum Mechanics",
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(7) Einstein, A., "Autobiographical notes", in P. A. Schilpp,
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(8) H. J., Pain, " The Physics of Vibrations and Waves",
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ed..(9) Jerry, B. Marion, Stephen, T. Thornton, "Classical
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Systems", (1988), Harcourt Brace Jovanovich, 3 rd ed..(10)
Henry, Semat, "Introduction to Atomic and Nuclear Physics", (1962),
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