lattice inertia and gravity

[AntonioLao]

By the end of the mid 1900s inertial mass and gravity mass are well established to be equivalent. The former appeared in Newton’s 2nd law of motion. The latter appeared in his universal law of gravitation. In the 1st decade of 1900s Einstein stated this as the principle of equivalence. Subsequently he used it to formulate his general theory of relativity. Although a person free falling is weightless, the same person escaping a gravity field can experience twice or more increases in weights.

From Einstein’s field equations the side describing space-time curvatures relate to the lattice inertia. The side describing the energy-momentum tensors relate to lattice gravity. For these to become truly equivalent both sides must be multiplied by a factor equal to light speed showing that square of energy is equal to square of energy: E²=E². Furthermore, at the local infinitesimal domain of space-time, E² is the product of the local infinitesimal Hamiltonian and the local infinitesimal Lagrangian: (T+V)(T-V)=T²-V² where T is the local infinitesimal kinetic energy and V is the local infinitesimal potential energy. The form T²-V² suggests that T>V while the form V²-T² suggests that V>T. Both forms are applicable at the local infinitesimal domain of space-time lattices.

[Cubola Zaruka]

'both sides must be multiplied by a factor equal to light speed'
Any equation that is equivalent when both sides are multiplied by the same factor is also equivalent without the factor.

[AntonioLao]

The factor of light speed is to transform both sides to squares of energy.

[SteveA]

The factor of light speed is to transform both sides to squares of energy.

If you remap a random motion within a linear space to a parabolic space (squaring the displacement), the statistical deviation over time follows a linear growth.

If we have an object at position p(t) at some point in time and it moves to either p(t+1)=p(t)+1 or p(t+1)=p(t)-1 every unit of time, then the overall displacement in terms of statistical "energy" would be the average of these two (for a 50/50 "random" distribution):

p(t+1)^2 = ((p(t)+1)^2+(p(t)-1)^2)/2
p(t+1)^2 = (p(t)^2+2p(t)+1+p(t)^2-2p(t)+1)/2
p(t+1)^2 = (2p(t)^2+2)/2
p(t+1)^2 = p(t)^2+1

Hence there appears to be a statistically constant velocity associated with the squared energy term, which is most closely associated with a diffusion.

On the other hand, the standard deviation from p(t) over time grows inversely to this and is a square root and this non-linearity can construct a perception of spacial warping.

When we look at the average velocity over a larger period of time though, we find the object is ultimately stationary - 0=lim(sqrt(t)/t) as t->infinity.

So if an object is interpreted as diffusing within a space determined by the constant energy velocity space, but is interacted with via. the space warped by the square root of growth, then a warping of space appears between these two interpretations and if we look at it from the position of its long term velocity, then an acceleration can appear to be at work because the velocity approaches 0 and a dynamic acceleration would be required to both move an object, yet maintain an average velocity of zero.

We can find different mathematical forms that can fit the bill, including complex orbitals to describe a bell curve e^-(t^2), notice that if we replaced the t^2 term with a complex value, then we can describe an object with coherent features that still fits the form of a net influence of random displacements.

Also, notice that under relativity we can't actually measure the speed of light in any external sense, if the measurements are made relative to a mass and so NIST simply defines the speed of light to be a constant.

The velocity of a single photon isn't measurable because it takes at least 2 photons to determine the start and stop times for a measurement (and generally many more photons to determine the space or distance over which the measurement is being made - an summing these together to describe a distance can create properties following the e^-(t^2) form).

Also notice that at various positions along a Gaussian/Bell curve, we have slopes and accelerations that can be separated into approximately linear, constant, parabolic and exponential forms, so each form of interaction can appear to occur within different areas along this curves as projecting a phenomenon to an external space that's warped with inverse characteristics can invert this distortion and make the phenomenon appear undistorted, though at a distance and/or encompassing a section of space.

[AntonioLao]

Actually, a relativistic quantum mechanics must use the square of energy by this equation E²=c²p²+m²c. Both Dirac for fermions and Klein-Gordon for bosons equations were derived from it.

[SteveA]

Actually, a relativistic quantum mechanics must use the square of energy by this equation E²=c²p²+m²c. Both Dirac for fermions and Klein-Gordon for bosons equations were derived from it.

Just curious, did you intend the final c term to be to the fourth power? E^2=c^2*p^2+m^2*c^4?

[AntonioLao]

did you intend the final c term to be to the fourth power?

Yes. It is the square of E=mc². E²=c²p²+m²c is the currently accepted form of total relativistic energy.