"Phil" wrote in message
...
Phil wrote:

"Phil" wrote in message
...

Phil wrote:


"Phil" wrote in message


Well, I'm not certain I follow you, so I'll just state that the
results
can always be correctly predicted, using relativity's laws, if we use
a
single inertial reference frames throughout the experiment. However,
in
order to have a physics that is completely independent of "absolute
references," it MUST be possible to ignore any changes in absolute
velocity. In cases of constant acceleration, we actually can do that
by
pretending we are in a gravitational field of constant strength, and
assuming we are motionless in space, which means using the velocity of
all other objects/observers relative to ourselves in order to
calculate
the results of their and our experiments.


I was not aware that GR is consistent with this view. Namely, being

under

constant acceleration, is being motionless in space. Are you sure of

this?

Actually, no, but I have not yet seen a situation where an observer
undergoing constant acceleration cannot correctly predict the outcome of
events around him (including himself) by *assuming* he is motionless in
a gravitational field of constant strength everywhere, and then using
the laws of GR for gravitational effects, and SR for velocity effects,
and using the velocity of objects relative to himself in his
calculations.

snip


This is the case. But to assume that one (in that state of motion) is
motionless, is to assume that freely falling frames are accelerating.
So
you should understand why I said you were thinking along those lines. I
do
recognize, however, that you have been completely oblivious to this in
your
practice and application.

I have no idea what you are talking about. In general, my intuition says
that we largely agree, but I am almost amazed at how many times I
"wonder what planet you are referring to," meaning I simply don't know
what you are saying. I think I understand enough math and theory, but
for some reason, the premises behind your comments are very unclear to
me. Kind of a side issue, and it doesn't necessarily mean anything bad
about you, but I find it interesting...

Anyway, yes, someone motionless in a gravitational field sees freely
falling frames as accelerating.

You are not getting the point I'm trying to make. In GR, the accelerating
frame is the one which was referred to as "motionless in a gravitational
field". The accelerated frame is following a curved path in space-time
according to all the inertial frames. When this frame is described as
motionless wrt to the gravity field, gravitation is distinguished separate
from its (motionless frame's) acceleration (which is, IMHO, a pracitical and
thoughtwise violation of equivalence) and it is equivalent (in both thought
and practice) to assuming that freely falling frames are accelerated.

As for "So you should understand why I
said you were thinking along those lines," I haven't a clue what this
means or is referring to. Also, for "I do recognize, however, that you
have been completely oblivious to this in your practice and
application," well, oblivious to what, first, and could you perhaps give
an example of a "practice and application" that I've "been oblivious
to?" Mind you, if you see something I am currently unaware of, I don't
mind having that pointed out to me here (publicly), as it could well
give me a better understanding, but at present, I literally haven't the
faintest idea what the subject even is that you are talking about. You
have me curious!

The difference between inertial freefall and accelerated freefall is
anchored in conservation. If a freeling falling object is accelerated, then
something must exchange momentum with it, call it sea of momentum if you can
not stomach an object.

If momentum in the locality of the accelerated object is depleted by
exchanging momentum with the object, then momentum flows into the area local
to the accelerating object in a radial manner from places more distant.
Under constant acceleration, the flow of momentum is constant, which means
there is an acceleration gradient which is directly equivalent to our
present understanding of gravity. In other words, accelerated free-fall
necessarily is the cause of gravity, if _accelerated free-fall_ is assumed.

Gravity, under the assumption of accelerated freefall is what I have called,
the momentum transport theory of gravity, (which happens to be my pet
theory), where gravity is the manifestion of a field transporting momentum
to the locality of an object in accelerated free-fall.

Accelerated free-fall is testable and physically distinguishable from GR.
For example, a test of the gravitation constant while falling or rising in
gravitational potential would show variance in field strength. Also
consider two asteroids in highly elliptical orbits, the period of their
orbits should vary a function of the rate of momentum flux (which is
directly proportional to the system's change in momentum).

The point I've been trying to make is, that _it matters_ how one views
free-fall in both the logical and practical sense of mattering. If free
fall is accelerated, it is significally different physically than inertial
free fall. Particularly, the notion that gravity is a function of
acceleration, _because_, it provides a model for a closed universe where
conservation laws are preserved.

Phil

"Phil" wrote in message
...
Phil wrote:


I am saying that frame which is constantly accelerating on the surface
of
the non-rotating planet can easily calculate the clock effects in all
free
falling frames. (Also in any other accelerating frame.

Phil ... how can you describe a motionless frame in a gravitational
field as "accelerating?" Now, it has an *equivalence* on an accelerating
spaceship, but it is not accelerating. Remember -- unless you already
know this -- two identical experiments, one in a motionless frame in a
gravitational field, and the other in an accelerating frame, have the
same *results*, but gravity and acceleration do not have the same
*effects* on the experiment. Accelerating frames do not bend light, but
they move the experiment relative to the light, thereby giving the same
results as gravity, which actually bends the light. Also, frames that
are accelerating in gravitational fields are accelerating, and often are
free falling. Frames that are motionless "feel" a force, one which is
extremely similar to the force felt on an accelerating spaceship. I
realize I'm being simple here, but we need to have common definitions
for terms if we are to successfully communicate, and I am unaware of
anyone who describes a motionless frame in a gravitational field as
accelerating.


You can't be talking GR if you are distinguishing between accelerations

and

considering a frame with a "proper acceleration" as inertial.

Your example is different in that the orbiting clock has a "constant
acceleration" that is completely different from the "acceleration" of
the motionless observer.


The orbiting clock "is not" accelerating in GR. It is moving
inertially.

Only from one point of view.

It is assumed from the principle of equivalence. All we need is an
accelerometer co-moving in the orbit and we have "evidence" we can share
with all "points of view" that the orbit is unaccelerated.

Mutual time dilation is when two frames each find the other's time
dilated,
but the same pair of clocks are never compared twice. When two frames
have
the opportunity to determine the dilation by comparing their clocks
readings
(a second time) then time dilation can not be mutual.

I *think*, that by "comparing clocks a second time,"

I mean two clocks measuring the time of two unique events. Like that of
"contact", so they compare the times of the two contact events.

Either one is slower,
or the same time has elasped.

As far as I know, I agree completely.

Acceleration, is described as an explanation
for the phenomenon.

By many people, suggesting that acceleration in and of itself has
relativistic effects, which is incorrect.

Basically, the "accelerated" frame's path is through
more space and less time, "explaining" the non-mutuality of the time
dilation.

I think I understand you here ... well, maybe. If observer B goes on a
round-trip journey with a constant velocity of 0.866c relative to
inertial observer A (as seen by either or both of them), simple geometry
*guarantees* that B has an average velocity of at least 0.866c relative
to the medium of space, so it would be very weird indeed if B did *not*
have a time-rate half as fast as A's time-rate when they meet again.

Still not getting it. Ask yourself why the dilation is not mutual. The
answer will be that one had underwent more acceleration than the other.


The example I gave, is the opposite,

Okay Phil, I am not going to deny that here, but please give that
example again, this time with more detail, and be prepared for me to ask
you to clarify certain details. I promise I will not try to be stupid or
dishonest, but I honestly do not understand your counter example.

and it is not true that the time of
the accelerating frame is dilated. The "inertial" orbiting clock ticks
less
in a revolution than the "accelerating" motionless (as you described it)
clock.

Sorry, but here I just do not understand the (1) velocity and position
over time of the two clocks as seen from an IRF, or (2) the extant to
which either of them is affected by a gravitational field.

The problem is that things are being inferred that I never said nor had any
intention of saying. The point is that the gravitational field doesn't
change the physical situation (the clock comparisons) one iota except in
this regard. The "orbiting" frame has a zero reading on his accelerometer
while the "motionless" frame has a positive reading. In this regard, they
are opposites. Since you have found it possible to calculate the time of
any clock from the motionless frame, then you should have no problem
conceptualizing the physical situation.

Which is, a clock circles about a point at a given radius. Another clock
is at rest with this point at the same radius. Every revolution, the clocks
"meet" to compare clock readings at a cycling sequence of unique events,
(let's call them "clock comparing events").

Simply view it two ways. One where the point, which the one clock circles,
is in a place far from any source of gravity. The other, where the point
corresponds to the center of gravity of a massive object. Ask yourself if
you can distinguish the physical situations by comparing the clock readings
at the clock comparing events. Then ask yourself which clock would have to
be accelerating to be consistent with SR.

Phil (the other one)

"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:

Does that mean that you can consider an observer at space ship one to be
at rest ?
I doubt that.

Of course you can consider any massive body to be "at rest".

That is the question.
Apparently if I understand you correctly
you can consider the earth to be at rest
but a moving space ship which moves around a star
(Our Sun) in one year not.
If this is correct what makes this difference ?

The most obvious candidate to be at rest will be an observer
situated at the one star in this scenario.

For such an observer you can draw a spacetime diagram consisting
of just one vertical line at x = 0

For an observer at space ship 1 this will be a sinus curve with a maximum
at x = r with r being the distance to the star.

Space time diagrams in curved space time aren't' easy to draw.

The question is what will happen if you replace spaceship 1 with our
Earth.

In a universe with one star much more massive than the Earth,
the difference would be negligible.

But what was your point?
Did you have one?

The above question boils down to the following question:
What is the physical representation of the vertical line at x = 0
in the following url: (The line ER)
http://users.telenet.be/vdmoortel/di...insEvents.html

Apparently this can be our Earth but not a spaceship
following the same trajetory (path)

Nicolaas Vroom
http://users.pandora.be/nicvroom/

Nicolaas Vroom wrote:
"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:

Does that mean that you can consider an observer at space ship one to be
at rest ?
I doubt that.
Of course you can consider any massive body to be "at rest".

That is the question.
Apparently if I understand you correctly
you can consider the earth to be at rest
but a moving space ship which moves around a star
(Our Sun) in one year not.
If this is correct what makes this difference ?

Is there any particular part of "any massive body" you don't understand?
That is "massive" as in "having mass" or "ponderable".
A feather is a massive body, a photon is not.

The most obvious candidate to be at rest will be an observer
situated at the one star in this scenario.

For such an observer you can draw a spacetime diagram consisting
of just one vertical line at x = 0

For an observer at space ship 1 this will be a sinus curve with a maximum
at x = r with r being the distance to the star.
Space time diagrams in curved space time aren't' easy to draw.

The question is what will happen if you replace spaceship 1 with our
Earth.
In a universe with one star much more massive than the Earth,
the difference would be negligible.

But what was your point?
Did you have one?

The above question boils down to the following question:
What is the physical representation of the vertical line at x = 0
in the following url: (The line ER)
http://users.telenet.be/vdmoortel/di...insEvents.html

Apparently this can be our Earth but not a spaceship
following the same trajetory (path)

That url describes the twin paradox in flat space-time
where we can use SR.

You specifically described a scenario where SR is not applicable.
"In that case we have one mass (i.e. one star) and three space ships."

But I understand that you don't know what your point was, so forget it.

Paul

"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:
"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:

Does that mean that you can consider an observer at space ship one to
be at rest ?
I doubt that.
Of course you can consider any massive body to be "at rest".

That is the question.
Apparently if I understand you correctly
you can consider the earth to be at rest
but a moving space ship which moves around a star
(Our Sun) in one year not.
If this is correct what makes this difference ?

Is there any particular part of "any massive body" you don't understand?
That is "massive" as in "having mass" or "ponderable".
A feather is a massive body, a photon is not.

The most obvious candidate to be at rest will be an observer
situated at the one star in this scenario.

For such an observer you can draw a spacetime diagram consisting
of just one vertical line at x = 0

For an observer at space ship 1 this will be a sinus curve with a
maximum
at x = r with r being the distance to the star.
Space time diagrams in curved space time aren't' easy to draw.

The question is what will happen if you replace spaceship 1 with our
Earth.
In a universe with one star much more massive than the Earth,
the difference would be negligible.

But what was your point?
Did you have one?

The above question boils down to the following question:
What is the physical representation of the vertical line at x = 0
in the following url: (The line ER)
http://users.telenet.be/vdmoortel/di...insEvents.html

Apparently this can be our Earth but not a spaceship
following the same trajetory (path)

That url describes the twin paradox in flat space-time
where we can use SR.

Exactly.
But how does that work out in practice.

You specifically described a scenario where SR is not applicable.
"In that case we have one mass (i.e. one star) and three space ships."

Correct. That is no flat space-time.
This is called Example 1 (See below)

Now I increase the mass of the space ship 1
(The one that moves in one year around the star)
such that it becomes equal to our Earth.
Or to say it different:
I replace space ship 1 with our Earth

And we have now:
one mass (i.e. one star), one earth and two space ships
This is called Example 2 (See below)

And my question is:
Do I now have a situation where we can speak of flat space-time ?

If the answer is Yes
Than in example 2 can we draw a vertical line parallel
to the t axis to represent an Observer at rest on Earth ?

While in example 1 this is not possible ?
(i.e. an Observer being at rest in space ship 1)

Nicolaas Vroom
http://users.pandora.be/nicvroom/

Nicolaas Vroom wrote:
"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:
"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:
Does that mean that you can consider an observer at space ship one to
be at rest ?
I doubt that.
Of course you can consider any massive body to be "at rest".
That is the question.
Apparently if I understand you correctly
you can consider the earth to be at rest
but a moving space ship which moves around a star
(Our Sun) in one year not.
If this is correct what makes this difference ?
Is there any particular part of "any massive body" you don't understand?
That is "massive" as in "having mass" or "ponderable".
A feather is a massive body, a photon is not.

The most obvious candidate to be at rest will be an observer
situated at the one star in this scenario.

For such an observer you can draw a spacetime diagram consisting
of just one vertical line at x = 0

For an observer at space ship 1 this will be a sinus curve with a
maximum
at x = r with r being the distance to the star.
Space time diagrams in curved space time aren't' easy to draw.

The question is what will happen if you replace spaceship 1 with our
Earth.
In a universe with one star much more massive than the Earth,
the difference would be negligible.

But what was your point?
Did you have one?
The above question boils down to the following question:
What is the physical representation of the vertical line at x = 0
in the following url: (The line ER)
http://users.telenet.be/vdmoortel/di...insEvents.html

Apparently this can be our Earth but not a spaceship
following the same trajetory (path)
That url describes the twin paradox in flat space-time
where we can use SR.

Exactly.
But how does that work out in practice.

It doesn't.

You specifically described a scenario where SR is not applicable.
"In that case we have one mass (i.e. one star) and three space ships."

Correct. That is no flat space-time.
This is called Example 1 (See below)

Now I increase the mass of the space ship 1
(The one that moves in one year around the star)
such that it becomes equal to our Earth.
Or to say it different:
I replace space ship 1 with our Earth

And we have now:
one mass (i.e. one star), one earth and two space ships
This is called Example 2 (See below)

And my question is:
Do I now have a situation where we can speak of flat space-time ?

No.

If the answer is Yes
Than in example 2 can we draw a vertical line parallel
to the t axis to represent an Observer at rest on Earth ?

While in example 1 this is not possible ?
(i.e. an Observer being at rest in space ship 1)

Nicolaas Vroom
http://users.pandora.be/nicvroom/

Paul

"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:
"Paul B. Andersen" schreef in bericht
...
Nicolaas Vroom wrote:


The above question boils down to the following question:
What is the physical representation of the vertical line at x = 0
in the following url: (The line ER)
http://users.telenet.be/vdmoortel/di...insEvents.html

Apparently this can be our Earth but not a spaceship
following the same trajetory (path)
That url describes the twin paradox in flat space-time
where we can use SR.

Exactly.
But how does that work out in practice.

It doesn't.

Apperently you do not understand my question..
The issue is how do you apply SR in real.
There must be a situation where SR applies
or is SR only something that exists in theory ?

See also below.

You specifically described a scenario where SR is not applicable.
"In that case we have one mass (i.e. one star) and three space ships."

Correct. That is no flat space-time.
This is called Example 1 (See below)

Now I increase the mass of the space ship 1
(The one that moves in one year around the star)
such that it becomes equal to our Earth.
Or to say it different:
I replace space ship 1 with our Earth

And we have now:
one mass (i.e. one star), one earth and two space ships
This is called Example 2 (See below)

And my question is:
Do I now have a situation where we can speak of flat space-time ?

No.

Consider there is someone who claims that he or she is at rest ?
implying that his world-line is represented by a vertical straight line.

Where is this person ? (can this person be).
At a spaceship ?
At the Moon ?
At our Earth ?
At our Sun ?
At the Centre of the Milky Way ?

And if he or she is at rest is that not identical as flat space-time ?

Nicolaas Vroom
http://users.pandora.be/nicvroom/