hwl

No - my question is as follows. There is acceleration in this field - not gravitational acceleration but acceleration of separation from

the inertial frame . Hence the acceleration vector must be NON-ZERO. The geodesic - metric equation defines the acceleration

vector in every field. But according to this equation the acceleration vector is ZERO because the metric is constant and so all its

partial derivatives are zero. There is obviously something to be explained here !

Is there anyone out there who can explain this ?

I am starting with an inertial frame with Minkowski coordinates. I could choose to do a variable non-linear Jacobian transformation of it such as the

conversion to spherical polar coordinates. This creates the auxiliary apparent field of the real Schwarzschild field. But instead I am

doing a constant linear Jacobian transformation which is not Lorentz by using four linear equations each of which expresses one of

the variables of the apparent field in terms of the Minkowski variables. This apparent field has no gravitational acceleration (due to the

matter source) but it does have apparent acceleration of separation from the inertial frame. Hence its acceleration vector must be

non-zero. My question is how do we reconcile this with the value defined by the geodesic - metric equation ?

Yes, the field is flat and so the acceleration is purely apparent as compared with a real field ( created by a non-Jacobian

transformation) where the acceleration is a combination of apparent acceleration (due to the choice of auxiliary apparent field) plus

gravitational (Riemann) acceleration (due to the matter source). But Einstein said that motion in all fields is correspondingly

equivalent - there are no exceptional fields to be excluded from consideration. Hence, the geodesic - metric equation must apply in

this field. This equation has the form :

(acceleration vector) = (metric connection)(velocity vector)(velocity vector) where the metric connection has the form :

g(pg + pg - pg) where each pg is a different partial derivative of the constant metric g. Thus if the acceleration vector is to be

non-zero then at least some of these partial derivatives must be non-zero.

The acceleration in this field cannot be ignored simply because it is apparent ( or fictional as you call it). Apparent acceleration makes

a definite contribution to the total acceleration in a real field which shares the same space-time variables with its auxiliary

apparent field.

There are only two possibilities to consider :

either (a) there is no acceleration in this field in which case it is an inertial frame with the wrong metric !

or (b) there is acceleration in this field in which case the acceleration vector is non-zero, even though the metric is constant ?????

Either we say that (a) is true or we must explain how (b) can be true.

Can anyone shed light on this puzzle ?

Suppose we do a constant Jacobian transformation (but not Lorentz) of a SR (inertial) frame by using four linear change of

variables (coordinates) equations. This defines an apparent gravity field with a constant metric (but not the SR metric) in which there is

apparent relative acceleration of separation and where the field clocks record unsynchronized time. From the geodesic metric equation

we see that the acceleration vector depends on the first partial derivatives of this constant metric and so these derivatives must therefore be non-zero ????

Also, the velocity vector along a geodesic is constant but it defines some sort of accelerated motion. How can this be ?? and what

sort of acceleration ?? constant acceleration along a straight line ??