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Proof that E could Never Equal mc².
Date: 28th March 2004
Author: Greg Alexander
If you have already viewed my webpage “Are the ‘Laws of Physics’ Wrong?” listed in the links below, go straight to SECTION TWO.
SECTION ONE
Consider a rocket travelling in the frictionless environment of space. According to the equation for kinetic energy, K.E. = ½ mv², when the rocket doubles its velocity (by increasing v to 2v for example), four times as much kinetic energy would be obtained. However considering that the law of conservation of energy must always apply would it really take four times the amount of fuel to increase the velocity from v to 2v as it took to increase the velocity from 0 to v? Surely the energy input supplied by the burning fuel would at all times equal the kinetic energy obtained by the space rocket? Such a situation would surely suggest that the formula would actually be K.E. = mv, where the relation between kinetic energy and velocity is a linear one.
Such an equation simply states 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 law of conservation would clearly dictate that the energy supplied to the craft by its power source would at all times equal the resulting kinetic energy, 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.
Q.E.D.
You may also be interested to view the link listed below on the same subject and by the same author, "Is the Special Theory of Relativity Wrong?" which questions the very basis of the theory.
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Is the Special Theory of Relativity Wrong?
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