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Metric:c^2dtau^2=c^2g_ttdt^2-g_xxdx^2-g_yydy^2-g_zzdz^2.For Schwarzschid Geometry,setting c=1,dtau^2=[1-2m/r]dt^2-[1-2m/r]^(-1)dr^2-r^2[(d theta)^2+sin^2 theta (dphi)^2] With the above metric dtau is the time interval in the local context while ‘r’ is measured from the origin.This is a concept typical of the metric.It is also to be noted that dr/d tau=0 can be valid in the strong gravity region (example:dropping of a stone)
(in continuation)Lorentz transformations as well as other transformations could cater to the requirement(discussed in the uploaded paper).Treating four potential as a four vector we can prove the invariance of Lorentz gauge. But starting from Lorentz gauge and using Lorentz gauge exclusively we cannot arrive at Lorentz transformations [transformation of four vectors]
Applying Lorentz gauge in any one inertial frame we obtain the wave equation with four potential :D Alembertian A^mu=4*pi j^mu,The quantity j^mu being a four vector,D Alembertian A^mu behaves like a four vector: D Alembertian A^mu=4*pi j^mu holds for all inertial frames,D Alembertian being an invariant operator.But this is conditioned by the fact that we are assuming the invariance of the Lorentz gauge.If it were not invariant the wave equation for four potential would not have appeared in the other frames leading to inconsistency.
Applying Lorentz gauge in any one inertial frame we obtain the wave equation with four potential : D Alembertian A^mu=4*pi j^mu,DAlembertian being an invariant operator and j^mu a four vector. But the solutions for A^mu may or may not be a four vector as discussed in the following paper:drive.google.com/file/d/0BymT8iD6LY1nS0wxcjRpTldyamc/…
The wave equation for four potential is deduced by applying Lorentz gauge. Four potential travels with a finite speed --at the speed of light.In that sense it is relativistically consistent.We might expect the four potential to be a four vector.But within the remit of Lorentz gauge we may have other types of gauges. One will be a four vector. Others may not be so.
Clarification on my last comment:E=0.B=0 does not necessarily imply phi=0,A=0;B=curl A=0 could be true for non zero variable A.[Aharonov Bohm effect].E=0 implies grad phi=-del A/del t without phi or A becoming constant/zero. Therefore four potential=0 in one frame does not necessarily imply such components are zero in all frames.Solutions other than 4 vectors seem to be possible.
There is nothing unreal/virtual about the amount of energy or momentum transferred by the virtual particles--especially if we look in to Feynman's diagrams.The only point is that they lie off the mass shell and hence remain unobserved except for the particle that receives the energy and momentum). The non conservation part is accounted for by Heisenberg's uncertainty principle[as stated in the first comment]
