Download 1 The Relationship between E = Mc and F = ma Copyright © 2007

The Relationship between E = Mc2 and F = ma
Copyright © 2007 Joseph A. Rybczyk
Abstract
An analysis is presented showing the complete relationship between Einstein’s E = Mc2
energy equation and Newton’s F = ma inertial force equation. During the analysis the reasons
why this relationship had not been established sooner is made clear. The present work takes the
relationships between energy and force developed over a series of previous works by this author
and finalizes them in a comprehensive treatment that includes new discoveries not previously
shown.
1
The Relationship between E = Mc2 and F = ma
Copyright © 2007 Joseph A. Rybczyk
1. Introduction
When finally accomplished the relationship between Einstein’s E = Mc2 energy equation
and Newton’s F = ma inertial force equation is not as simple and straightforward as might be
expected. As will be seen, the problems lie first in clarifying the true meanings of the two
different equations and then in the need to relativize Newton’s inertial force equation to make it
compatible with Einstein’s energy equation. Fortunately, this latter requirement was already
accomplished in earlier works by this author as will be shown during the analysis. An
unexpected final discovery is that there are actually two different forms of energy. More
specifically, it is found that energy has both a constant and relative value consistent with the
overall principles of relativity.
2. Clarifying Einstein’s Energy Equation
Before we can relate Einstein’s energy equation to Newton’s inertial force equation a
quick review of both is needed. Beginning then with Einstein’s energy equation it is important
to understand that there are four basic forms of the equation. These are
eo  mo c 2
(1) The Internal Energy of a Particle
Eo  M o c 2
(2) The Internal Energy of a System
e  mc 2
(3) The Total Energy of a Particle
E  Mc 2
(4) The Total Energy of a System
where the author took the liberty of extending the use of the o subscript from its customary use in
identifying the rest masses mo and Mo of a particle and system respectively to also show that eo
and Eo are the respective rest energies associated with such rest masses. This seems appropriate
for two different reasons. First, according to the textbooks, the internal energy of matter is
alternately referred to as its rest energy. Second, it will later be shown that eo is nothing more
than a special case of e since by definition e = eo for a particle at rest. The same relationship is
true of Eo and E in that E = Eo for a system at rest. What else is not clear from the above
equations is that Equations (3) and (4) are actually abbreviated forms of much more complicated
equations that include not only the rest masses of the respective particle or system, but also the
associated kinetic energies, if relative motion is involved. In the latter case there is also the
consideration of the relativistic transformation factor to allow for speeds that are a significant
fraction of light speed c. Let us now review the full versions of Equations (3) and (4) to develop
a better understanding of the mentioned relationships.
Where e is the total energy of a particle and E is the total energy of a system, we have
2
e  mo c 2
1
 mc 2
2
v
1 2
c
(5) Total Energy of a Particle
and
E  M oc 2
1
2
v
1 2
c
 Mc 2
(6) Total Energy of a System
where mo and Mo are the rest masses of the particle and system respectively, and m and M are the
total masses (rest mass + increased mass from motion) for the particle and system respectively.
In other words
m  mo
1
(7) Total Mass of a Particle
v2
1 2
c
and
M  Mo
1
(8) Total Mass of a System
v2
1 2
c
where the Lorentz/Einstein transformation factor
1
v2
1 2
c
(9) Lorentz/Einstein Transformation Factor
can be replaced by the equivalent millennium relativity1 transformation factor
c
c  v2
2
(10) Millennium Relativity Transformation Factor
and either can be replaced by the Greek gamma symbol γ for simplicity. Using the gamma
symbol for the relativistic transformation factor from this point forward gives
k  mo c 2  mo c 2
(11) Kinetic Energy of a Particle
where k is the kinetic energy of a particle and
3
K  M o c 2  M o c 2
(12) Kinetic Energy of a System
where K is the kinetic energy of a system. In summary then, the total energy for a particle or
system can be written as
e  mc 2
(13)
or
e  mo c 2
E  Mc 2
(15)
or
E  M o c 2
(14) Total Energy of a Particle
and
(16) Total Energy of a System
respectively where e is the total energy for a particle and E is the total energy for a system. In a
similar manner
k  e  mo c 2
(17)
or
k  e  eo
K  E  M oc 2
(19)
or
K  E  Eo
(18) Kinetic Energy of a Particle
and
(20) Kinetic Energy of a System
where k is the kinetic energy of a particle and K is the kinetic energy of a system given in their
respective abbreviated forms.
3. Clarifying Newton’s Force Equation
In the author’s earlier work, Time and Energy2, it was shown that Newton’s second law
of motion expressed mathematically as
a
F
m
or
(21)
F  ma
(22) Newton’s Second Law of Motion
where a is the acceleration of a particle, m is the inertial mass of the particle and ∑F is the vector
sum of the forces acting on the particle is true in the relativistic sense even if m does not increase
in value as a result of acceleration. In accordance with the millennium theory it is the effect that
time dilation has on velocity that results in the relativistic increase in energy that is observed as
the particle (or system when m is replaced by M) accelerates toward the speed of light. This
finding is expressed mathematically by the relationship
ac 
v
t
(23) The Law of Constant Acceleration
where ac is the constant rate of acceleration experienced in the acceleration frame (AF), v is the
instantaneous speed relative to the stationary frame (SF) and t is the time interval in the uniform
motion frame (UF) transitioned to during the stationary frame time interval T in which the
4
acceleration takes place. The relationship of the transitioned to uniform motion frame time
interval t to stationary frame time interval T is given by
t
T

(24) UF v Time Interval t Relationship to SF Time Interval T
that is the accepted relativistic formula for time dilation. Since time interval t rises
asymptotically along with speed v, the rate of acceleration ac given by Equation (23) is constant
by definition.
In the subsequent millennium relativity work involving the Laws of Acceleration3, the
relationship between time, speed and acceleration was expanded upon to include
ar 
v
T
(25) The Law of Relative Acceleration
where ar is the relative rate of acceleration experienced in the stationary frame during SF time
interval T. Since v rises asymptotically relative to the stationary frame during SF time interval T
that increases at a linear rate, the acceleration rate ar experienced in the stationary frame will
decrease toward zero as the instantaneous speed v increases toward light speed c. This, of
course, is the behavior observed in the field of particle physics as a particle is accelerated to near
the speed of light. In the work just mentioned the relationship of ac to ar is shown to be
ar 
ac

(26) Relationship of Acceleration Rate ac in AF to Acceleration Rate ar in SF
that is consistent with the evidence that supports relativistic principles. In the follow-up work,
Time and Energy, Inertia and Gravity4, these relationships are further clarified, but not until the
later work, The Four Principle Kinetic States of Material Bodies5, is it finally shown that there
are two definitions of inertial force. These are shown to be
f c  mac
(27) Constant Force in Acceleration Frame
which follows directly from the relationships of Equation (23) (except that Fc of the previous
work has been replaced by fc in this present work for reasons that will be explained shortly) and.
f r  mar
(28) Relative Force in Stationary Frame
which follows directly from the relationships of Equation (25) (except for the change from Fr in
the previous work to fr in the present work, again, as will be explained shortly). In brief, if mass
m is constant (as given in the millennium theory of relativity) and acceleration rate ac is constant,
then inertial force fc as shown in Equation (27) must also be constant. Conversely, if mass m is
constant but acceleration rate ar decreases toward zero as the instantaneous speed v approaches
light speed c, then inertial force fr must also simultaneously decrease toward zero.
And now for the explanation of the change from upper case F to lower case f in the above
treatment. Even though textbooks normally make no distinction between the mass of a particle
5
and the mass of a system regarding Newton’s second law of motion, in this present work for
purposes of consistency with the relativistic treatment of masses, Equations (27) and (28) will be
limited to use only in reference to particles with mass m. For systems we then have
Fc  Mac
(29) Constant Force in Acceleration Frame
where Fc is the constant inertial force involved in a system of mass M undergoing acceleration at
the constant rate ac. The corresponding formula for relative force Fr is
Fr  Mar
(30) Relative Force in Stationary Frame
where M is the mass of the system and ar is the relative rate of acceleration. In the previous
work5 it was then found that the relationship of relative inertial force fr to constant inertial force
fc (again replacing the original upper case F with the present lower case f) is given by
fc  f r
(31)
and
fr 
fc

(32) Relationship of fr to fc
where fc and fr are the respective constant and relative forces acting on a particle. For a system
we then have
Fc  Fr 
(33)
and
Fr 
Fc

(34) Relationship of Fr to Fc
where Fc and Fr are the respective constant and relative forces acting on a system.
Somewhat unexpectedly it is then found that the interpretation of the force quantities fc
and fr or Fc and Fr can be a bit confusing because it depends on which frame supplies the energy
used for the acceleration. If the energy is supplied by the stationary frame (in particle
acceleration, for example) then the force quantity fc is a direct function of the SF energy used
whereas force quantity fr is the resultant force applied to the particle, i.e. the force detected in the
acceleration frame of the particle. Alternately, if the energy is supplied by the acceleration
frame, in the case of a rocket or similar space object for example, then the force quantity Fc is
still a direct function of the energy used, but this time as determined in the acceleration frame
and the force quantity Fr is the resultant force experienced in the stationary frame. All of these
assumptions involving Newton’s second law of motion are consistent with the evidence that
supports relativity.
4. Converting to Similar Terms
Before establishing the relationship between energy and inertial force there is one
remaining area of confusion to resolve. Whereas Einstein’s theory of relativity distinguishes
between two different kinds of masses, i.e. rest masses mo and Mo and masses m and M that have
increased as a result of motion, both, Newtonian physics and millennium relativity physics treat
all masses as rest masses, i.e. as being unaffected by motion. It should be understood here that in
the case of the millennium theory treatment including that which has just been applied to
Newton’s second law involving force, the results are not affected by this consideration. In the
6
millennium theory treatment all of the changes are directly attributed to time dilation which in
turn gives the same results as would be obtained in Einstein’s theory. Thus, for consistency in
the combining of the two different theoretical approaches, we can either use the variables mo and
Mo for masses, or the variables m and M with the understanding that they all refer to rest masses
and not the two different conditions given in Einstein’s theory. For simplicity, the latter
variables, m and M, will be used with the understanding that mass is unaffected by motion.
5. The e to f Relationships
Now that the meanings of the two different quantities, e and f have been clarified it is a
relatively simple matter to determine the relationship of one to the other. Since energy Equation
(14) and force Equations (27) and (28) all have rest mass in common we can solve each for that
variable and equate the resulting expressions to each other. Rewriting energy Equation (14) as
e  mc 2
(35) Total Energy of a Particle
and solving for mass m gives
m
e
c 2
(36)
whereas force Equation (27)
f c  mac
(27) Constant Force in Energy Frame
when solved for mass m gives
m
fc
ac
(37)
and force Equation (28)
f r  mar
(28) Relative Force in Stationary Frame
when solved for mass m gives
m
fr
ar
(38)
where in each case the masses are assumed to be the same. Equating the right sides of Equations
(37) and (36) then gives
fc
e
 2
ac c 
7
(39)
that when solved for fc gives
fc  e
ac
c 2
(40) Relationship of e to fc
where fc is the force acting on a particle with a total energy e. Solving Equation (40) for e gives
the alternate relationship
c 2
e  fc
ac
(41) Relationship of fc to e
where e is the total energy of a particle acted on by force fc.
Doing the same thing with the right sides of Equations (38) and (36) then gives
fr
e
 2
ar c 
(42)
that when solved for fr gives
fr  e
ar
c 2
(43) Relationship of e to fr
where fr is the force acting on a particle with a total energy e. Solving Equation (43) for e gives
the alternate relationship
e  fr
c 2
ar
(44) Relationship of fr to e
where e is the total energy of a particle acted on by force fr.
Going back to Equation (40) and replacing γ with the special relativity Factor (9), and
simplifying then gives
fc  e
ac 1 
v2
c2
c2
(45) Relationship of e to fc
for the complete formula for fc whose alternate form is
e  fc
c2
v2
ac 1  2
c
(46) Relationship of fc to e
8
for e. Doing the same thing using the equivalent millennium relativity Factor (10) in Equation
(40) and simplifying gives the more concise yet equivalent complete formula for fc
ac c 2  v 2
fc  e
c3
(47) Relationship of e to fc
whose alternate form is
e  fc
c3
ac c2  v2
(48) Relationship of fc to e
for e.
Repeating the procedure for Equation (43) while again replacing γ with the special
relativity Factor (9), and simplifying gives
fr  e
ar 1 
v2
c2
c2
(49) Relationship of e to fr
for the complete formula for fr whose alternate form is
e  fr
c2
v2
ar 1  2
c
(50) Relationship of fr to e
for e. Similar to before, doing the same thing using the equivalent millennium relativity Factor
(10) in Equation (43) and simplifying gives the more concise
fr  e
ar c 2  v 2
c3
(51) Relationship of e to fr
yet equivalent complete formula for fr whose alternate form is
e  fr
c3
ar c2  v2
(52) Relationship of fr to e
for e.
6. The E to F Relationships
Repeating the procedure of Section 5 using energy Equation (16) for the total energy of a
system rewritten as
9
E  Mc 2
(53) Total Energy of a System
we can solve for mass M to obtain
M 
E
c 2
(54)
for use with constant force Equation (29)
Fc  Mac
(29) Constant Force in Acceleration Frame
also solved for M to give
M
Fc
ac
(55)
along with relative force Equation (30)
Fr  Mar
(30) Relative Force in Stationary Frame
solved for M to give
M
Fr
ar
(56)
needed for our next steps. Equating the right sides of Equations (55) and (54) then gives
Fc
E
 2
ac c 
(57)
that when solved for Fc gives
Fc  E
ac
c 2
(58) Relationship of E to Fc
for the force acting on a system with total energy E. Solving Equation (58) for E gives the
alternate relationship
E  Fc
c 2
ac
(59) Relationship of Fc to E
where E is the total energy of a system acted on by force Fc.
As before, doing the same thing with the right sides of Equations (56) and (54) then gives
10
Fr
E
 2
ar c 
(60)
that when solved for Fr gives
Fr  E
ar
c 2
(61) Relationship of E to Fr
where Fr is the force acting on a system with a total energy E. Solving Equation (61) for E gives
the alternate relationship
c 2
E  Fr
ar
(62) Relationship of Fr to E
where E is the total energy of a system acted on by force Fr.
Going back to Equation (58) and replacing γ with the special relativity Factor (9), and
simplifying then gives
Fc  E
ac 1 
v2
c2
c2
(63) Relationship of E to Fc
and
E  Fc
c2
v2
ac 1  2
c
(64) Relationship of Fc to E
for the complete special relativity versions of the formulas whereas using the equivalent
millennium relativity Factor (10) instead gives the more concise
Fc  E
ac c 2  v 2
c3
(65) Relationship of E to Fc
and
E  Fc
c3
ac c2  v2
(66) Relationship of Fc to E
millennium relativity versions.
Repeating the procedure for Equation (61) while again replacing γ with special relativity
Factor (9), and simplifying gives
11
Fr  E
ar 1 
v2
c2
(67) Relationship of E to Fr
c2
and
E  Fr
c2
(68) Relationship of Fr to E
v2
ar 1  2
c
for the complete special relativity versions of the formulas whereas using the equivalent
millennium relativity Factor (10) instead gives the more concise
ar c 2  v 2
Fr  E
c3
(69) Relationship of E to Fr
and
E  Fr
c3
ar c2  v2
(70) Relationship of Fr to E
millennium relativity versions.
7. The k to f Relationships
Rewriting the kinetic energy formula, Equation (11), using m in place of mo for mass that is
unaffected by motion we have
k  mc 2  mc 2
(71) Kinetic Energy of a Particle
k  mc 2 (  1)
(72) Kinetic Energy of a Particle
that can be simplified to
where k is the kinetic energy of a particle. Then, substituting the right side of constant force
Equation (37) for m in Equation (72) we obtain
k
fc 2
c (  1)
ac
(73) Kinetic Energy of a Particle
that can be rewritten as
12
k  fc
c 2 (  1)
ac
(74) Kinetic Energy of a Particle
to show the relationship of a constant force fc on a particle to its kinetic energy k. The alternate
form of the formula is then
f c k
ac
c (  1)
2
(75) Inertial Force of a Particle
giving the relationship of the kinetic energy k of a particle to its constant inertial force fc. Using
the millennium relativity Factor (10) in place of γ in Equation (74) then gives


c
c 2 
 1
2
2
c v

k  fc 
ac
(76) Kinetic Energy of a Particle
for the kinetic energy of a particle acted on by constant force fc. This same result can be
obtained using the special relativity Factor (9) but requires many more steps. The constant force
formula is then
fc  k
ac
c


 1
c 
2
2
 c v

2
(77) Inertial Force of a Particle
where fc is the constant inertial force of a particle with kinetic energy k.
Repeating the entire procedure beginning with
k  mc 2 (  1)
(72) Kinetic Energy of a Particle
where k is the kinetic energy of a particle, but this time substituting the right side of relative
force Equation (38) for m in Equation (72) we obtain
k
fr 2
c (  1)
ar
(78) Kinetic Energy of a Particle
c 2 (  1)
ar
(79) Kinetic Energy of a Particle
and
k  fr
for the relationship of a relative force fr on a particle to its kinetic energy k. The alternate form
of the formula is then
13
f r k
ar
c (  1)
2
(80) Inertial Force of a Particle
giving the relationship of the kinetic energy k of a particle to its relative inertial force fr. Using
the millennium relativity Factor (10) in place of γ in Equation (79) then gives


c
c 2 
 1
c2  v2

k  fr 
ar
(81) Kinetic Energy of a Particle
for the kinetic energy of a particle acted on by relative force fr. As before, this same result can
be obtained using the special relativity Factor (9) but requires many more steps. The relative
force formula is then
fr  k
ar
c


 1
c 
2
2
 c v

2
(82) Inertial Force of a Particle
where fr is the relative inertial force of a particle with kinetic energy k.
In analyzing kinetic energy Equation (76) it is easily seen that as the instantaneous speed
v from acceleration ac approaches c, the kinetic energy of the particle approaches infinity.
Conversely, as v approaches 0 the kinetic energy of the particle also approaches 0 as expected.
The same is true of Equation (81) where the smaller value of relative force fr is compensated by
the smaller value of relative acceleration rate ar. Although Equations (76) and (77) along with
Equations (81) and (82) can be simplified further, experimentation in a computer math program
shows these forms to be the most accurate at low Newtonian speeds.
8. The K to F Relationships
Using the procedure of Section 7 on the kinetic energy formula, Equation (12) for the
kinetic energy of a system gives
K  Mc 2  Mc 2
(83) Kinetic Energy of a System
where the total mass Mo of the system has been replaced by mass M. Simplification then gives
K  Mc 2 (  1)
(84) Kinetic Energy of a System
where K is the kinetic energy of a System. Then similar to before, substituting the right side of
constant force Equation (55) for M in Equation (84) gives
K
Fc 2
c (  1)
ac
(85) Kinetic Energy of a System
14
that can be restated as
K  Fc
c 2 (  1)
ac
(86) Kinetic Energy of a System
to show the relationship of a constant force Fc on a system to its kinetic energy K. For the
alternate form of the formula we then have
F c K
ac
c (  1)
2
(87) Inertial Force of a System
giving the relationship of the kinetic energy K of the system to its constant inertial force Fc.
Again using the millennium relativity Factor (10) in place of γ but this time in Equation (86)
gives


c
c 2 
 1
2
2
c v

K  Fc 
ac
(88) Kinetic Energy of a System
which again is the same result that could be obtained using the special relativity Factor (9) but
with far fewer steps. The system force formula is then
Fc  K
ac
c


 1
c 2 
2
2
 c v

(89) Inertial Force of a System
where Fc is the constant inertial force of a system with kinetic energy K.
As before, repeating the entire procedure beginning with
K  Mc 2 (  1)
(84) Kinetic Energy of a System
where K is the kinetic energy of a System, but this time substituting the right side of relative
force Equation (56) for M in Equation (84) we obtain
K
Fr 2
c (  1)
ar
(90) Kinetic Energy of a System
c 2 (  1)
ar
(91) Kinetic Energy of a System
and
K  Fr
15
for the relationship of a relative force Fr on a system to its kinetic energy K. For the alternate
form of the formula we then have
F r K
ar
c (  1)
2
(92) Inertial Force of a System
giving the relationship of the kinetic energy K of the system to its relative inertial force Fr.
Again using the millennium relativity Factor (10) in place of γ but this time in Equation (91)
gives


c
c 2 
 1
c2  v2

K  Fr 
ar
(93) Kinetic Energy of a System
which again is the same result that could be obtained using the special relativity Factor (9) but
with far fewer steps. The system relative force formula is then
Fr  K
ar
c


 1
c 
2
2
 c v

2
(94) Inertial Force of a System
where Fr is the relative inertial force of a system with kinetic energy K.
As before, in analyzing kinetic energy Equation (88) it is easily seen that as the
instantaneous speed v from constant acceleration ac approaches c, the kinetic energy of the
system approaches infinity while conversely, as v approaches 0 the kinetic energy approaches 0
as is expected. And also as before, the same is true of Equation (93) where the smaller value of
relative force Fr is compensated by the smaller value of relative acceleration rate ar. Although
Equations (88) and (89) along with Equations (93) and (94) can be simplified further,
experimentation in a computer math program shows these forms to be the most accurate at low
Newtonian speeds.
9. The er to ec Relationship
With the groundwork that has been laid in the previous Sections it is now possible to
show that the relativistic concept of a relative value and constant value also extends to energy.
Under the evidence supported assumption that a relativistic increase in energy approaching an
infinite amount of energy is required to continue the acceleration of a particle as it approaches
the speed of light we can claim that the constant force applied to the particle in the acceleration
frame equals the relative force determined in the stationary frame, i.e. the energy source frame.
In other words, it takes disproportionately greater amounts of stationary frame energy to
maintain a constant force on the particle as it approaches the speed of light. This relationship can
be expressed mathematically by equating the constant force fc of the acceleration frame to the
relative force fr of the stationary frame giving
16
fc  f r
(95) fr to fc Relationship of an Accelerated Particle
for the equality of these forces involving a particle under acceleration by energy applied
remotely from the stationary frame. Substituting the right sides of previously given Equation
(40) rewritten as
f c  ec
ac
c 2
(96) Relationship of ec to fc
to define the related energy ec as constant, and Equation (43) rewritten as
f r  er
ar
c 2
(97) Relationship of er to fr
to define the related energy er as relative in place of fc and fr respectively in Equation (95) we
obtain
ec
ac
a
 er 2r
2
c
c
(98)
that when solved for ec gives
ec  er
ar
ac
(99) Relationship of er to ec
for the relationship of relative energy er to constant energy ec. The alternate relationship is then
given by
er  ec
ac
ar
(100) Relationship of ec to er
where ec is the constant energy in the acceleration frame and er is the corresponding relative
energy in the stationary frame, i.e. the frame providing the energy for the acceleration of the
particle.
Given the previously stated relationship of constant acceleration rate ac to relative
acceleration rate ar
ar 
ac

(26) Relationship of Acceleration Rate ac in AF to Acceleration Rate ar in SF
we obtain
ac  ar
(101) Relationship of Acceleration Rate ar in SF to Acceleration Rate ac in AF
17
for the relationship of ar to ac. Substituting the right side of Equation (101) for ac in Equation
(99) gives
ec  er
ar
ar 
(99) Relationship of er to ec
that when simplified gives
ec 
er

(102) Relationship of er to ec
for the final relationship of relative energy er to constant energy ec. The alternate relationship is
then given by
er  ec
(103) Relationship of ec to er
where er is the relative energy of the stationary frame.
Since γ increases toward infinity as v increases toward c it is easily seen that Equation
(103) agrees with the evidence regarding the relative stationary frame energy er required to
accelerate a particle to the speed of light. From the opposite perspective, since γ decreases
toward 1 as v decreases toward 0, the stationary frame energy used in the acceleration of a
particle at Newtonian speeds is equal to the energy experienced in the acceleration frame of the
particle.
10. The Er to Ec Relationship
Repeating the procedure of Section 9 we can now derive the energy relationships that
apply to a system. As before, we begin by expressing the mathematical relationship between the
two different kinds of force. Thus by equating the constant force Fc of the acceleration frame to
the relative force Fr of the stationary frame we obtain
Fc  Fr
(104) Fr to Fc Relationship of an Accelerated Particle
for the equality of these forces involving a system under acceleration by energy applied remotely
from the stationary frame. Substituting the right sides of previously given Equation (58)
rewritten as
Fc  Ec
ac
c 2
(105) Relationship of E to Fc
to define the related energy Ec as constant, and Equation (61) rewritten as
Fr  Er
ar
c 2
(106) Relationship of E to Fr
18
to define the related energy Er as relative in place of Fc and Fr respectively in Equation (104) we
obtain
Ec
ac
a
 Er 2r
2
c
c
(107)
that when solved for Ec gives
Ec  Er
ar
ac
(108) Relationship of Er to Ec
for the relationship of relative energy Er to constant energy Ec. Similar to before the alternate
relationship is then given by
Er  Ec
ac
ar
(109) Relationship of Ec to Er
where Ec is the constant energy in the acceleration frame and Er is the corresponding relative
energy in the stationary frame, i.e. the frame providing the energy for the acceleration of the
system.
From the previously derived Equation (101)
ac  ar
(101) Relationship of Acceleration Rate ar in SF to Acceleration Rate ac in AF
we again obtain the relationship of ar to ac. Substituting the right side of Equation (101) for ac in
Equation (108) gives
Ec  Er
ar
ar
(110) Relationship of Er to Ec
that when simplified gives
Ec 
Er

(111) Relationship of Er to Ec
for the final relationship of relative energy Er to constant energy Ec. The alternate relationship is
then given by
Er  Ec
(112) Relationship of Ec to Er
where Er is the relative energy of the stationary frame.
As discussed previously, since γ increases toward infinity as v increases toward c it is
easily seen that Equation (112) agrees with the evidence regarding the relative stationary frame
19
energy Er required to accelerate a system to the speed of light. Again, from the opposite
perspective, since γ decreases toward 1 as v decreases toward 0, the stationary frame energy used
in the acceleration of a system at Newtonian speeds is equal to the energy experienced in the
acceleration frame of the system.
11. Conclusion
Obviously, many more relationships can be derived from those given in this work by
simply substituting other equivalent forms of the terms and variables used. Whereas most of
what was shown in this present work had first been discovered in preceding works, the discovery
of the two different forms of energy, i.e. constant and relative, was completely unexpected.
Thus, only future investigations involving this newest finding will determine its overall
importance in helping us to understand the physical universe in which we exist.
Appendix
The following Appendix is included as part of this paper:
Appendix A – Constant and Relative Force and Energy
REFERENCES
1
Joseph A. Rybczyk, Millennium Theory of Relativity, (2001), www.mrelativity.net
2
Joseph A. Rybczyk, Time and Energy, (2001), www.mrelativity.net
3
Joseph A. Rybczyk, The Laws of Acceleration, (2001), www.mrelativity.net
4
Joseph A. Rybczyk, Time and Energy, Inertia and Gravity, (2002), www.mrelativity.net
5
Joseph A. Rybczyk, The Four Principal Kinetic States of Material Bodies, (2005), www.mrelativity.net
20
Appendix A
Constant and Relative Force and Energy – The Relationship between E = Mc2 and F = ma
Constant and Relative Force and Energy.mcd
June 7, 2007
Joseph A. Rybczyk
page 1
Note: The formulas used in this analysis are identified by the numbers used in the associated paper,
"The Relationship between E = Mc
c  299792458
g  9.80665
M  1000
2
(53)
2
c 
E1  Fc
ac
(59)
2
c 
E2  Fr
ar
(62)
v  c .0000000000000001
ar 
(10)
2
c v
E  M  c  
and F = ma "
ac  g  1
c
 
2
2
Fc  M  ac (29)
ac

Fr  M  ar
ac
Fc1  E
2
c 
(58)
ar
Fr1  E
2
c 
(61)
c in m/s, g in m/s, a
c
in terms of g, v in terms of c
M in any terms,  in Millennium version, ar is relative
acceleration in terms of constant acceleration a c
(26)
E is energy of system, F c is constant force, F r is
(30)
relative force
E1 = E and F c1 = F c (Numerical subscripts for
comparison purposes only.)
E2 = E and F r1 = F r (Numerical subscripts for
comparison purposes only.)
Comparison of results from above formulas ---------------------------------------------------------------------------------------------------------------------19
E  8.988  10
19
E1  8.988  10
3
3
Fc  9.807  10
Fr  9.807  10
3
3
Fc1  9.807  10
Fr1  9.807  10
19
E2  8.988  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------In comparing the following results it is seen that as the above value of v changes, the value of Ec and E c1 remain constant.
Only the values of Er and Er1 are affected by changes in the value of v. (Numerical subscripts for comparison purposes only.)
If the Force is the same as in
ar
Ec  Er
ac
(108)
19
Ec  8.988  10
Ec1 
Fc
Er

Fr
and
Er  E
ac
Er  Ec
ar
(109)
(104)
(111)
19
then we have
Er1  Ec 
19
Ec1  8.988  10
(112)
19
Er  8.988  10
Er1  8.988  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Verification that both forms of kinetic energy equations (based on F c and ac or Fr and ar) give the same results. (Numerical
subscripts for comparison purposes only)
c 
2
c     1
2
K1  Fc
ac
(86)
K2  Fc
2
c     1
K3  Fr
ar
(91)
K4  Fr
 1


ac
c 
2
c
 2 2
 c v
c
 2 2
 c v
ar
(88)
K1  0
K2  0
(93)
K3  0
K4  0
 1


21
Appendix A
Constant and Relative Force and Energy – The Relationship between E = Mc2 and F = ma
Constant and Relative Force and Energy.mcd
June 7, 2007
Joseph A. Rybczyk
page 2
Note: The formulas used in this analysis are identified by the numbers used in the associated paper,
"The Relationship between E = Mc
c  299792458
g  9.80665
M  1000
2
(53)
2
c 
E1  Fc
ac
(59)
2
c 
E2  Fr
ar
(62)
v  c .5
ar 
(10)
2
c v
E  M  c  
and F = ma "
ac  g  1
c
 
2
2
Fc  M  ac (29)
ac

Fr  M  ar
ac
Fc1  E
2
c 
(58)
ar
Fr1  E
2
c 
(61)
c in m/s, g in m/s, a
c
in terms of g, v in terms of c
M in any terms,  in Millennium version, ar is relative
acceleration in terms of constant acceleration a c
(26)
E is energy of system, F c is constant force, F r is
(30)
relative force
E1 = E and F c1 = F c (Numerical subscripts for
comparison purposes only.)
E2 = E and F r1 = F r (Numerical subscripts for
comparison purposes only.)
Comparison of results from above formulas ---------------------------------------------------------------------------------------------------------------------20
E  1.038  10
20
E1  1.038  10
3
3
Fc  9.807  10
Fr  8.493  10
3
3
Fc1  9.807  10
Fr1  8.493  10
20
E2  1.038  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------In comparing the following results it is seen that as the above value of v changes, the value of Ec and E c1 remain constant.
Only the values of Er and Er1 are affected by changes in the value of v. (Numerical subscripts for comparison purposes only.)
If the Force is the same as in
ar
Ec  Er
ac
(108)
19
Ec  8.988  10
Ec1 
Fc
Er

Fr
and
Er  E
ac
Er  Ec
ar
(109)
(104)
(111)
19
then we have
Er1  Ec 
20
Ec1  8.988  10
(112)
20
Er  1.038  10
Er1  1.038  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Verification that both forms of kinetic energy equations (based on F c and ac or Fr and ar) give the same results. (Numerical
subscripts for comparison purposes only)
c 
2
c     1
2
K1  Fc
ac
(86)
K2  Fc
2
c     1
K3  Fr
ar
(91)
K4  Fr
 1


ac
c 
2
c
 2 2
 c v
c
 2 2
 c v
ar
19
K2  1.39  10
19
K4  1.39  10
(88)
K1  1.39  10
(93)
K3  1.39  10
19
 1


22
19
Appendix A
Constant and Relative Force and Energy – The Relationship between E = Mc2 and F = ma
Constant and Relative Force and Energy.mcd
June 7, 2007
Joseph A. Rybczyk
page 3
Note: The formulas used in this analysis are identified by the numbers used in the associated paper,
"The Relationship between E = Mc
c  299792458
g  9.80665
M  1000
2
(53)
2
c 
E1  Fc
ac
(59)
2
c 
E2  Fr
ar
(62)
v  c .9999999999999999
ar 
(10)
2
c v
E  M  c  
and F = ma "
ac  g  1
c
 
2
2
Fc  M  ac (29)
ac

Fr  M  ar
ac
Fc1  E
2
c 
(58)
ar
Fr1  E
2
c 
(61)
c in m/s, g in m/s, a
c
in terms of g, v in terms of c
M in any terms,  in Millennium version, ar is relative
acceleration in terms of constant acceleration a c
(26)
E is energy of system, F c is constant force, F r is
(30)
relative force
E1 = E and F c1 = F c (Numerical subscripts for
comparison purposes only.)
E2 = E and F r1 = F r (Numerical subscripts for
comparison purposes only.)
Comparison of results from above formulas ---------------------------------------------------------------------------------------------------------------------27
E  4.763  10
27
E1  4.763  10
4
3
Fc  9.807  10
Fr  1.85  10
4
3
Fc1  9.807  10
Fr1  1.85  10
27
E2  4.763  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------In comparing the following results it is seen that as the above value of v changes, the value of Ec and E c1 remain constant.
Only the values of Er and Er1 are affected by changes in the value of v. (Numerical subscripts for comparison purposes only.)
If the Force is the same as in
ar
Ec  Er
ac
(108)
19
Ec  8.988  10
Ec1 
Fc
Er

Fr
and
Er  E
ac
Er  Ec
ar
(109)
(104)
(111)
19
then we have
Er1  Ec 
27
Ec1  8.988  10
(112)
27
Er  4.763  10
Er1  4.763  10
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Verification that both forms of kinetic energy equations (based on F c and ac or Fr and ar) give the same results. (Numerical
subscripts for comparison purposes only)
c 
2
c     1
2
K1  Fc
ac
(86)
K2  Fc
2
c     1
K3  Fr
ar
(91)
K4  Fr
 1


ac
c 
2
c
 2 2
 c v
c
 2 2
 c v
ar
27
K2  4.763  10
27
K4  4.763  10
(88)
K1  4.763  10
(93)
K3  4.763  10
27
 1


23
27
The Relationship between E = Mc2 and F = ma
Copyright © 2007 Joseph A. Rybczyk
All rights reserved
including the right of reproduction
in whole or in part in any form
without permission.
Note: If this document was accessed directly during a search, you can visit the Millennium Relativity web site by clicking on the Home link
below:
Home
24