Using Different Field Transformation

BACKGROUND STUFF TO THE KRONECKER PROPOSAL An Illustration of the Idea

Suppose a photon takes 2hrs to travel between A and B as measured by a clock on the photon’s world line. The clock, according to Observer A should arrive at B at time 5pm. If A observed the clock at some point xa = x*, where xa is the point occupied by the observer and x* the point occupied by the photon. Then suppose that for a tiny duration during which the observer holds the photon at xa, there is ∂ x* / ∂xa = 1. For infinitesimal amounts the observer is ‘holding the photon (i.e the photon is trapped in one state). This amounts to saying that observing a photon requires a tiny duration during which the path of the observer and the photon are synchronised. Such that; ξ (u) = 0. Given that ξ(u) = xa (u)- x* (u) , u being the time common to the photon and the observer. Some Assumptions 1. Considering that the photon is synchronised to events at Observer A (are

photons emitted from creation synchronised to the events at the beginning?), then the photon will have the same time as Observer A. Travelling along the world line the photon will arrive at B with the time ‘incorrect according to B. 2.

If there is a real physical process that governs A’s time, then a watch on the photon (when it arrives at B will record 5pm – 2hrs travelling time) will be incorrect according to B. It is like someone living in Australia receiving a watch that was set to British time. The crucial thing is that according to B: ∂ x* / ∂xb > 1, or possibly ∂ x* / ∂xb < 1 During the time that the Observer B reads the watch on the photon it is recording 5pm, whereas B’s own time is 8pm. Suppose then that B reads: ∂ x* / ∂xb = A. Based Upon This: Suppose, (dФ)a measures the rate of change of a field along the path xa , then one can relate (dФ)b (the rate of change in the same field as seen along xb ). These are essentially one-forms. The transform (i)

(dФ)a = (∂ x* / ∂xa )(dФ)* can be used.

But there is also (ii)

(dФ)b = (∂ x* / ∂xb )(dФ)*

One can solve by substitution (i) into (ii): (dФ)b = (∂ x* / ∂xb )[(dФ)a (∂xa / ∂ x* )] (dФ)b = (∂ x* / ∂xb )(∂xa / ∂ x* ) [(dФ)a ] (dФ)b = δa b (dФ)a

δa b - is the Kronecker delta according to Schultz (An Introduction To Genreal Relativity). This is partly the reason why I think it’s possible to connect this to other stuff shown on the my facebook page. Now, what is ∂ 2 x* / (∂xb )2 = ∂A /(∂xb ) ?