e NOT EQUAL TO mc2

From: vignesh (vignesh28_at_gmail.com)
Date: 02/16/05 

Date: 16 Feb 2005 01:20:50 -0800

SECTION ONE

Consider the problem of powered space flight. The equation giving the
kinetic energy obtained by any space vehicle which has been accelerated
to a velocity v is of course, K.E. = ½ mv². However observe that to
double the velocity of any space vehicle (by increasing v to 2v for
example), four times as much energy would be required each time if the
laws of the conservation of energy are to be maintained. This results
from the relation between velocity and kinetic energy being non-linear.
However surely this would go against common sense as the latter would
tend to suggest that the formula were as follows, K.E. = mv.

Such an equation would simply state that to increase your velocity by a
set amount in the frictionless environment of space the same proportion
of energy would be required each time no matter what speed you were
travelling at to begin with. The laws of conservation of energy would
suggest that the energy supplied to the craft by its power source would
at all times equal the kinetic energy obtained, taking into account the
various efficiency ratios involved of course. If the kinetic energy did
equal ½ mv² then this would result in increasingly larger amounts of
fuel being needed to accelerate the craft which in turn would produce a
limiting velocity requiring its own separate equation.

Of course we have all been taught since day one that it is actually
momentum which equals mv and not kinetic energy. But is it really that
straight forward? For example is momentum a force or is it actually a
form of energy? Indeed the units it is measured in are rather peculiar,
for example kilograms per metre per second. But which do these denote,
force or energy?

Having discovered that there is a possible problem with the current
equation for kinetic energy and that the exact nature of the physical
property of momentum is open to question, it is now necessary to
enquire where the underlying fault would appear to lie. In my opinion
the best place to start is with the definition of the joule itself
which we are told equals a force of one Newton applied through a
distance of 1 metre. An objection which can immediately be raised to
this apparently straight forward definition is that the work done by
one Newton through one metre is surely dependant upon the mass it is
acting upon? For example would it not take more work for one Newton to
move a 2 kg mass through one metre than it would the same force to move
1 kg through the same distance? The unit mass is entirely absent from
the definition.

To further illustrate my point, using F = ma and s = ½ at², 1 Newton
acting on a 1 kg mass would take 1.41 seconds to reach one metre (i),
whereas 1 Newton acting on a 2 kg mass would take 2 seconds to reach
one metre (ii). Surely it would be the case that if one Newton were
applied for a longer duration of time as with case (ii) more work would
have been done as a consequence? However if the equations v = at and
K.E. = ½ mv² are applied to the accelerations and velocities obtained
in (i) and (ii), we do in fact end up with 1 Joule of energy in each
case as is predicted. In fact this is the very basis of the definition.

But surely it has to be the case that 1 Newton applied for 2 seconds in
case (ii) would have carried out an extra 40% more work than in case
(i) where 1 Newton was only applied for 1.41 seconds? This point is
surely indisputable? If it is then a large part of classical physics
has to be in error and possibly as a consequence many other areas of
the same science would also be undermined?

The above demonstration would clearly suggest therefore, that the
formula Work Done = Force x Distance, would appear to be not at all
valid. Indeed it is this equation that the definition of the Joule is
based upon. Instead the correct equation would perhaps be Work Done =
Force x Time. Since the formula for kinetic energy is also derived from
the former apparently incorrect equation, it too must be unsound.
Indeed it is this formula which is actually responsible for calculating
the erroneous 1 Joule of work done for Cases (i) and (ii).
Interestingly if it is instead assumed that K.E. is proportional to mv,
then we obtain the energies of 1.41 Joules and 2 Joules respectively
for cases (i) and (ii). This is entirely consistent with a 40%
difference in work done between the two cases.

Indeed it would even appear possible to prove by integration that
kinetic energy equals mv if it is assumed that Work Done = Force x
Time, as follows:

K.E. = ∫ F.dt = ∫ m dv/dt . dt = ∫ m dv = mv.

or alternatively :

K.E. = Work Done = Force x Time = F x t = m a x t (and because v = at)
= m v

Further evidence that would suggest that kinetic energy were equal to
mv is the observation that when two objects of known masses collide, it
is the conservation of momentum mv which would apply and not the
conservation of kinetic energy according to the currently accepted
equation which covers it, ½mv². Indeed the two cases produce quite
different results : Consider two bowls each with a mass of 1.5 kg. The
first bowl has a velocity of 2 m/sec and impacts at a slight angle with
the second which is stationary. If the velocity of the first bowl after
impact is 0.5 m/sec, the velocity of the second bowl based upon the
conservation of momentum would be 1.5 m/sec (i.e. 2 – 0.5 = 1.5
m/sec). But if it were assumed that the conservation of kinetic energy
applied instead, the velocity of the second bowl would be as follows :

V = √(2² - 0.5²) = 1.94 m/sec - A different value altogether.

Since conservation would only occur experimentally where mv is used,
perhaps it can be assumed therefore that the kinetic energy also equals
mv?

SECTION TWO

According to the Special Theory of Relativity the equation for kinetic
energy is as follows:

K.E. = mc² / √(1 - v²/c²) - mc²

It is apparent that this equation is entirely based upon E = mc².
However in this case m happens to be the mass increase which is equal
to the relativistic mass minus the rest mass or to formulate it
algebraically, m / √(1 - v²/c²) – m. According to the Special
Theory the relativistic mass is obtained by dividing the rest mass by
the Lorentz factor √(1 - v²/c²) which in turn is obtained from the
Lorentz transformation. The mathematical properties of this factor are
as follows: when v is small compared with c, the speed of light, √(1
- v²/c²) = 1 and when v approaches c, √(1 - v²/c²) = 0 making the
relativistic mass tend towards infinity. Multiplying the mass
differential quoted above, m / √(1 - v²/c²) – m, by c² gives the
stated equation for relativistic kinetic energy. An alternative way of
looking at it is that the mass-energy of the relativistic particle
minus the rest energy of the particle equals the relativistic kinetic
energy of the particle as a result of v, the velocity.

If you put all the relevant numbers into the equation for relativistic
kinetic energy it does actually produce the same results as ½ mv² but
only when v is small compared with c, the speed of light, of course.
You can try this yourself at home with an appropriate software package
that can handle large numbers of decimal places. Most spreadsheets can
actually do this and you can divide the equation into a number of parts
which are then calculated in adjacent cells (or alternatively you can
do it all in the same cell if you are clever enough!). Knowing the
speed of light to be 3 x 10e8 m/sec (such that c² then equals 9 x
10e16) and that v, the velocity is also calculated in m/sec, the maths
is quite straightforward. The mass m can be left out of the equation
because it cancels on both sides or simply assume it to have a value of
1 kg which adds up to much the same thing.

Having proved in the first section that the kinetic energy could never
equal ½ mv² and that the classical equation is therefore wrong, this
brings into question the relativistic equation for kinetic energy since
it produces the equivalent results as ½ mv² at low velocities.
However altering the relativistic equation for kinetic energy
completely undermines the logic supporting the conservation of
mass-energy and hence also the equation of mass-energy equivalence
itself, E = mc².