i.e. m.Uy(v) = m.Uy(0) but Uy(v) = Uy(0).L(v)
Hence to fix the contradiction, we follow the precedent for time and suppose m depends on v. That is let:
m(v) = m(0)/L(v) *EQ12
For v < c the mass of an object is greater as measured by a stationary observer than the measure made by an observer travelling with the object, or as measured afterwards when the object is stationary relative to an observer. Note that as speed builds towards c, the objects mass increases without bound, hence no object with non zero mass can be given sufficient energy to reach the speed of light.
substituting for L(v)
Which is modified to
m(v)2c2 = m(0)2c2 + m(v)2v2
where m(v).v is the momentum, commonly called P
m(v)2c2 = m(0)2c2 + p2 *EQ13
Expanding 1/L(v) by Taylor series, EQ12 becomes
m(v) = m(0) [1 + v2/2c2 + 3v4/8c4 + ...]
multiplying by c2 and expanding
m(v)c2 = m(0)c2 + m(0)v2/2 + 3m(0)v4/8c4 + ... *EQ14
The units of both sides of EQ14 are those of energy and the familiar Newtonian Kinetic energy term appears second on the right hand side of the equation. The terms involving v can be equated to p2/m(v) using EQ13. Einstein proposed that EQ14 was in fact an energy equation and that the total energy of a particle is m(v)c2.
E = m(v)c2 *EQ15
To suppose that the intrinsic energy of a mass is m(0)c2 requires that an object of zero rest mass and velocity of less than c (since 1/L(c) is undefined) must have zero energy. Perhaps more justification for this conclusion is required. Consider two objects approaching each other at the same speed at an angle of A degrees to the X direction and colliding inelastically, that is they coalesce into one. The total momentum before the collision is 2m(v).vsin(A) in the Y direction, hence this is the total momentum afterwards as well.
If the angle A is small then the resultant velocity is small and so the relativistic mass increase is negligible, the mass being close to 2m(0). However this is not the case if it is accepted that mass is conserved in the collision. The resultant mass is M=2m(v), if v was large then this is noticeably greater than 2m(0), the combined rest mass. Hence the mass of the resultant object depends on the speed or kinetic energy of its precursors. Moreover, should a particle of mass M fly apart into two fast parts, as in nuclear fission, each will dissipate its kinetic energy, hence velocity, hence relativistic mass as it slows through interactions with neighbouring particles. The energy released (KE) is a function of the mass deficit of the initial particle and its now stationary products. Taking a clue from EQ15 this function is simply the multiplicative constant c2.
It would seem that energy can be stored as mass and so this is evidence to support the claim that the intrinsic energy of a mass could be m(0)c2. There are other implications of mass depending on creation velocity. The insides of a particle consist of the intrinsic masses of Protons etc. minus energy used to stick them together plus energy left over from the velocity of the creation of the particle, so what does a masson look like? Or is it that the insides of particles may emit protons etc. but that they do not actually exist inside the particle. As Richard Feynman put it, one can emit the word cat without having it ready made inside you, there is no finite supply of this word, only the energy used to make it. So it is for the contents of particles, the emission products do not necessarily exist internally.
A more useful and thought provoking version of EQ15 is obtained by squaring and dividing by c2 ...
then using *EQ13
In the preceding development, momentum was defined as a non zero mass multiplied by velocity. Although this is valid it does not exclude other forms of momentum. Consider an object with zero rest mass but with velocity c, calculating m(c) we have an ambivalent situation, both L(c) and m(0) are zero. Is their quotient zero, finite or infinite? If the answer is finite then there may be objects which have momentum and energy but only at the speed of light. Photons seem to fit this description. For example consider the momentum transferred from one particle to another by light, the source particle recoils on emission thus losing momentum, there is a slight travel time for the photon between particles, then the target particle gains momentum when hit by the photon. So if momentum is always conserved the photon must carry it. Hence let *EQ16 describe these as well.
This document was created on 23 August 1995
last modified on 23 August 1995
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