Hi Albert and all, I think this is very
well formulated posted question.

On May 11, 2:38 am, Albertito wrote:
Let two bodies, A and B, with equal masses, move inertially
along a straight line. Velocity of A wrt B is V_ab, and velocity
of B wrt A is V_ba. Since both bodies are inertially moving
along a straight line, we assume V_ab = - V_ba will always
hold, so we say both velocities are symmetrical. Suppose
now, body A accelerates during a time t at constant a_A along
the same straight line to yield a final velocity V_ab'. Can we
still claim the new velocity of B wrt A is V_ba' = -V_ab'? IOW,
isn't it reasonable to claim that the new V_ba' is actually not
that new, but V_ba' = - k*V_ab', for a real k 1? If it is true
that
V_ba' = - k*V_ab', for a real k 1, after the acceleration a_A
and V_ab' V_ab, then, can we conclusively say that
acceleration a_A has created an eventual gravitational field,
by claiming that both masses are no longer equal?

I'd like to examine that.
Ok, Albert and Ken are sitting on a park bench.
Albert and Ken will agree they are in relatively
opposite directions as they converse. (ok?).

In terms of Radius of separation we might use,
R(A,K) = - R(K,A) , which is vectorial, from
(Albert to Ken) and (Ken to Albert) respectively.
The magnitude of separation is "R", and is
agreed to by Albert and Ken.

Next Albert shifts on the bench away from Ken,
and that we may describe as a increment of the
radius R to be R' = R + dR.

While Albert shifted, both Albert and Ken can
agree that (dR/dt) was equal, and I suggest,
the Velocity of the Shift is,

V(A,K) = V(K,A), as a symmetry,

because both Albert and Ken agree that a
positive relative displacement occurred.

In addition, the reason I suggest that, is
because

V(A,K) = - V(K,A) is antisymmetrical,

and requires a rotation.

....K=
....|
....A

or

....K
....|
..=A

to make Ken and Albert move in relatively
opposite directions.

That's the way I try to analyse that problem.
Naturally criticism is welcome, I find that
stuff a bit challenging so be gentle :-).
Regards
Ken S. Tucker
kxsxt8