Momentum Versus Kinetic Energy/Energy in Kinematics

Momentum and kinetic energy are present in Newtonian mechanics as fundamental ideas. Here, we try to consider how they arise in a practical sense. We consider momentum as being linked with a force interaction which occurs at specific x,t in a tiny dt, and is associated with a specific interaction arrangement in space, i.e the force vector. We note that given the specific orientation, momentum is a vector (without knowing its explicit functional form) and it depends on the particular co-ordinate system used. One would expect dp/dt = F without any of these being formally defined except that dt is tiny time interval. If on the other hand, there is a starting and ending point in space with many interaction points in between (in fact so many that each (x,y,z) point is seen as an interaction point), then there is a different px,py,pz at each point as well as an initial px0, py0, pz0 and pxf, pyf, pzf (0=initial, f=final). Given that one is not dealing with a specific interaction at t which takes dt, one is interested in a path in space between the two points with continuous interaction. In such a case, one has F(x,y,z) instead of F(t) so dp = F dt becomes v dot dp = F dot dr. If the integral is path independent, i.e. F = -grad V, then one may evaluate a scalar V(x,y,z final) - V(x,y,z initial) and there should exist a rotationally invariant function associated with the particle which may also be evaluated at the endpoints. If p is p proportional to v as an example, then a rotational scalar is constant v dot v. Thus, one has both the vector p for specific point interactions and the scalar E for path independent continuous interactions, and the two are related due to the notion of vector and rotational scalar. In Newtonian mechanics, one may note that giving a particle a hit (force) in a certain direction creates a velocity vector in that direction. Now momentum (which is of yet not defined) is also a vector. One might guess that it is linked to the vector v, but if that is the case, one needs a full functional form for momentum. (Incidentally, if momentum were proportional to v vector, then a rotational invariant is v dot v and measure which is independent of a the co-ordinate system should be a function of v dot v. ) In order to find a full expression for momentum and even the co-ordinate system independent scalar, one may consider viewing a particle from frames moving at different speeds, i.e. special relativity. A Lorentz transformation changes the vector r and number t, but if one has a rest mass mo ( a number), then in a moving frame it has a velocity. If one applies a transformation for r,t to 0, mo one may create an explicit vector p and number E by assuming that there is an invariant modulus, i.e. a link between mo and p,E. This necessarily involves a (1,-1) metric instead of the usual (1,1) for physical reasons. Given p and E, these are not simply math creations as the fact that a particle is seen as moving means that it was given a hit. In the Newtonian world, this may be described by dp1/dt = force. Here p1 is the Newtonian momentum (which might be different from the special relativity one), dt is tiny and implies a specific interaction at t and a position in space (together with its spatial geometrical force situation). One also wishes to have a co-ordinate independent scalar linked with p1. Given the explicit forms for p= mo v / sqrt(1-vv/cc) and E=mocc/ sqrt(1-vv/cc), one may see that in the v/c