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?

Albertito wrote in message

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?

Warning:
http://i263.photobucket.com/albums/i...ne/troll5d.jpg

Dirk Vdm

On May 11, 5:38*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?

"Hey, ****ing moron, you can't even trolling!
I'll teach you how to troll in this NG. Find
some flaws in a well-established theory.
Once, you have convinced all of us those
flaws are real inconsistencies, then we'll let
you propose a new theory which can repair
those flaws. Then, and only then, we will
discuss about your 'wonderful theory'. And then,
all we'll see the great troll you can be. Until
then, **** off, moron! "

http://groups.google.com/group/sci.p...c?dmode=source

One troll trolls the other troll.

Idiot

Mike

On May 11, 11:51 am, Mike wrote:
On May 11, 5:38 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?

"Hey, ****ing moron, you can't even trolling!
I'll teach you how to troll in this NG. Find
some flaws in a well-established theory.
Once, you have convinced all of us those
flaws are real inconsistencies, then we'll let
you propose a new theory which can repair
those flaws. Then, and only then, we will
discuss about your 'wonderful theory'. And then,
all we'll see the great troll you can be. Until
then, **** off, moron! "

http://groups.google.com/group/sci.p...sg/d1133f915ff...

One troll trolls the other troll.

Idiot

Mike

Deer Mike:
Are you one of those shrunken heads needing to call how
idiotic are the others in order to forget the irreparable
idiocy that resides in you?

On May 11, 10:38 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?

In the gravitational field of the Earth, a test body located
at a distance equal to R =1 AU, would feel a escape
speed of v_e=sqrt(2GM_e/R), with M_e the mass of the
Earth. In the gravitational field of the Sun, a test body located
at a distance equal to R =1 AU, would feel a escape speed
of v_s=sqrt(2GM_s/R), with M_s the mass of the Sun.
This means both bodies, the Earth and the Sun, are
free falling toward each other with different relative speeds.
If the mass of the Earth were equal to the mass of the Sun,
then it would be v_s = - v_e. But, since M_s M_e, then
|v_s| |v_e|. If you could stop the orbital motion of the Earth
around the Sun, it would free fall towards the Sun along a
straight line, and would impact it after a time of t_e = R/v_e,
measured with a local clock in the earth. The same collision,
measured by a local clock at the center of the Sun, would
occur after a time of t_s = R/v_s.

On May 11, 5:38*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?


Just one question. What exists in empty spacetime that would make
them not be symmetric?

On May 11, 7:25*am, Albertito wrote:
On May 11, 11:51 am, Mike wrote:





On May 11, 5:38 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?

"Hey, ****ing moron, you can't even trolling!
I'll teach you how to troll in this NG. Find
some flaws in a well-established theory.
Once, you have convinced all of us those
flaws are real inconsistencies, then we'll let
you propose a new theory which can repair
those flaws. Then, and only then, we will
discuss about your 'wonderful theory'. And then,
all we'll see the great troll you can be. Until
then, **** off, moron! "

http://groups.google.com/group/sci.p...sg/d1133f915ff...

One troll trolls the other troll.

Idiot

Mike

Deer Mike:
Are you one of those shrunken heads needing to call how
idiotic are the others in order to forget the irreparable
idiocy that resides in you?- Hide quoted text -

- Show quoted text -

I just quoted you troll. Don't you recognize your own statements?

Mike

On May 11, 7:04 pm, Igor wrote:
On May 11, 5:38 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?

Just one question. What exists in empty spacetime that would make
them not be symmetric?

Energy, a lot of energy. Empty spacetime is not
actually empty.

On May 11, 7:28 pm, Mike wrote:
On May 11, 7:25 am, Albertito wrote:



On May 11, 11:51 am, Mike wrote:

On May 11, 5:38 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?

"Hey, ****ing moron, you can't even trolling!
I'll teach you how to troll in this NG. Find
some flaws in a well-established theory.
Once, you have convinced all of us those
flaws are real inconsistencies, then we'll let
you propose a new theory which can repair
those flaws. Then, and only then, we will
discuss about your 'wonderful theory'. And then,
all we'll see the great troll you can be. Until
then, **** off, moron! "

http://groups.google.com/group/sci.p...sg/d1133f915ff...

One troll trolls the other troll.

Idiot

Mike

Deer Mike:
Are you one of those shrunken heads needing to call how
idiotic are the others in order to forget the irreparable
idiocy that resides in you?- Hide quoted text -

- Show quoted text -

I just quoted you troll. Don't you recognize your own statements?

Mike

Then, stop quoting me. Quote your own statements, troll.

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