One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. I suspect that I'm not
wording that quite right, but let me explain what I mean.

Given a lab frame resting on a non-rotating earth. Is this frame
equivalent, (equally relevent to make predictions and conduct experiments),
as a frame accelerating at 9.80 m/s^2 located in space far from any gravity
source. The answer I keep getting is _no_. Let me explain why .

Let both frames have identical systems of inertially moving objects. In the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the objects
all accelerate by identical quantity wrt the lab and all move inertially wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions, one
must know the actual physical situation of his lab, ie, whether he resting
on the launch pad or exiting the solar system with rockets propelling him.

I have taken this to mean that there is a fundamental difference between the
motions of the objects and the frames. In fact, it would seem, physically,
that they are precisely the opposite of each other, a mirror image so to
speak. Welcome any and all comments.

Phil

Phil wrote:

One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. [...]

There has been no experimental proof to indicate a centripetal time
dilation. So much for Einstein's Equivalence Principle!

Koobee Wublee wrote:
Phil wrote:

One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. [...]

There has been no experimental proof to indicate a centripetal time
dilation. So much for Einstein's Equivalence Principle!

Psst...Nobody expects centripetal time dilation.

****, this is even worked out in MTW's _Gravitation_. Specifically, it
is shown there is *no* redshift and thus no time dilation.

Why is it you can't even get the simple stuff right? The equivalence
principle has no bearing on this.

Phil wrote:
One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. I suspect that I'm not
wording that quite right, but let me explain what I mean.

Given a lab frame resting on a non-rotating earth. Is this frame
equivalent, (equally relevent to make predictions and conduct experiments),
as a frame accelerating at 9.80 m/s^2 located in space far from any gravity
source. The answer I keep getting is _no_. Let me explain why .

I don't see why it would be different, barring obvious objections like
locality and tidal effects. This is the equivalence principle in
action, imho.


Let both frames have identical systems of inertially moving objects. In the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the objects
all accelerate by identical quantity wrt the lab and all move inertially wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions, one
must know the actual physical situation of his lab, ie, whether he resting
on the launch pad or exiting the solar system with rockets propelling him.

I think you are right in this sense. If you break the locality
condition, you won't get an inertial frame of reference and everything
goes to hell.


I have taken this to mean that there is a fundamental difference between the
motions of the objects and the frames. In fact, it would seem, physically,
that they are precisely the opposite of each other, a mirror image so to
speak. Welcome any and all comments.

I don't see how you can conclude there is a difference between the
object's motion and the object's frame of reference.


Phil

"Phil" wrote in message
ink.net...
One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. I suspect that I'm not
wording that quite right, but let me explain what I mean.

Try reading this. http://xxx.lanl.gov/abs/physics/0204044
It is a paper I wrote a few years back. The intent of the paper is to
directly answer your question.

Given a lab frame resting on a non-rotating earth. Is this frame
equivalent, (equally relevent to make predictions and conduct
experiments),
as a frame accelerating at 9.80 m/s^2 located in space far from any
gravity
source. The answer I keep getting is _no_. Let me explain why .

Let both frames have identical systems of inertially moving objects. In
the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the objects
all accelerate by identical quantity wrt the lab and all move inertially
wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions, one
must know the actual physical situation of his lab, ie, whether he
resting
on the launch pad or exiting the solar system with rockets propelling him.

This is because you're thinking of a field with tidal accelerations. The
Equivalence Principle, as stated by Einstein, is this: A uniformly
accelerated frame of reference is identical to a *uniform gravitational
field*. For a field which has tidal forces it is to be applied *locally*
which means your measurement equipment does not have the precision to tell
the difference. E.g. drop a ball in your room and measure its acceleration.
The measurement is done with finite precision. Measure its instantanious
acceleration at the location of release. Now do the same thing at a point
closer to the floor. You'll see that with certain measurement devices that
there is no difference in the acceleration. Bringing in more precise
measurement devices only means that you're doing the experiments more
locally, i.e. within a smaller region of spacetime.

Best wishes

Pete

"Eric Gisse" wrote in message
oups.com...

Phil wrote:

Let both frames have identical systems of inertially moving objects. In
the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the
objects
all accelerate by identical quantity wrt the lab and all move inertially
wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions,
one
must know the actual physical situation of his lab, ie, whether he
resting
on the launch pad or exiting the solar system with rockets propelling
him.

I think you are right in this sense. If you break the locality
condition, you won't get an inertial frame of reference and everything
goes to hell.

Well the labs are not inertial by definition. But then, I have the sense
that "locality" isn't meaningful anyway. By this I mean that I don't think
the difference between the two accelerated frames disappear in an interval
of infinitesimal domain. Here is why. The geodesics are the integrations
of the contributions of each infinitesmal interval. If the geodesics
differ, there must be some difference in the infinitesimal contributions.
In other words, to be different in larger domains is to be different
infinitesimally, or vice versa.

Another way of saying this is that if they were identical infinetesimally,
they would integrate identically, resulting in identical motions wrt to the
accelerating lab frames. But then, we know that the motions are
distinguishable experimentally.

I have taken this to mean that there is a fundamental difference between
the
motions of the objects and the frames. In fact, it would seem,
physically,
that they are precisely the opposite of each other, a mirror image so to
speak. Welcome any and all comments.

I don't see how you can conclude there is a difference between the
object's motion and the object's frame of reference.

In re-reading what I posted, I can understand the confusion you are having
with what I said. I meant something entirely different than what you
inferred. I'll try to do a better job.

The the "inertial" objects are "all moving inertially wrt each other" and
"accelerating identically" wrt to the accelerating lab far from a gravity
source.

While in the accelerating lab resting on the surface of the planet, the
"inertial objects are all "accelerating with respect to each other" and
"accelerating differently" wrt the lab frame. Physically, they seem to be
opposites, that is, in terms of the qualities of their necessary motions.
Is this clearer?

Phil

"Pete" wrote in message
...

"Phil" wrote in message
ink.net...
One of the problems I have never resolved is the difference between
curvature and accelerations in flat space-time. I suspect that I'm not
wording that quite right, but let me explain what I mean.

Try reading this. http://xxx.lanl.gov/abs/physics/0204044
It is a paper I wrote a few years back. The intent of the paper is to
directly answer your question.

Given a lab frame resting on a non-rotating earth. Is this frame
equivalent, (equally relevent to make predictions and conduct
experiments),
as a frame accelerating at 9.80 m/s^2 located in space far from any
gravity
source. The answer I keep getting is _no_. Let me explain why .

Let both frames have identical systems of inertially moving objects. In
the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the
objects
all accelerate by identical quantity wrt the lab and all move inertially
wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions,
one
must know the actual physical situation of his lab, ie, whether he
resting
on the launch pad or exiting the solar system with rockets propelling
him.

This is because you're thinking of a field with tidal accelerations. The
Equivalence Principle, as stated by Einstein, is this: A uniformly
accelerated frame of reference is identical to a *uniform gravitational
field*. For a field which has tidal forces it is to be applied *locally*
which means your measurement equipment does not have the precision to tell
the difference. E.g. drop a ball in your room and measure its
acceleration.
The measurement is done with finite precision. Measure its instantanious
acceleration at the location of release. Now do the same thing at a point
closer to the floor. You'll see that with certain measurement devices that
there is no difference in the acceleration. Bringing in more precise
measurement devices only means that you're doing the experiments more
locally, i.e. within a smaller region of spacetime.

Best wishes

Pete

Thanks Pete for the link, I have downloaded it, and will be reading it
forthwith. I fully agree that the difference (the curvature) is a
consequence of the non-uniformity of the gravity field (tidal accerations).
Since the tidal accelerations are not insignificant, how are they
incorporated for predicting paths of motion? It would not be sufficient to
integrate the effects under a strict equivalence principle.

Phil

"Pete" wrote in message
...

Try reading this. http://xxx.lanl.gov/abs/physics/0204044
It is a paper I wrote a few years back. The intent of the paper is to
directly answer your question.

And I think it does. I found it a very satisfying read. I would like to
add that while MGR does seem to add the concept of curvature to Einstein's
definition of gravity, it is reasonable from this perspective. Gravity, (as
a field of matter/energy), is never devoid of curvature, in as much that the
concept of "uniform field" is an idealization for which we have no evidence
to exist anywhere in nature, though it may certainly be conceived to be
approachable, in that the effects of said curvature, become immeasurable.

Reading the historical references of Einstein's papers, I can not help but
to be reminded of the wonderful, tounge in cheek way, Einstein approached
his physics. Refreshing indeed.

Phil

Phil wrote:
"Eric Gisse" wrote in message
oups.com...

Phil wrote:

Let both frames have identical systems of inertially moving objects. In
the
frame located on the non-rotating earth, the objects all accelerate by
varying quantity wrt to the lab and all accelerate with respect to each
other. In the frame located away from any source of gravity, the
objects
all accelerate by identical quantity wrt the lab and all move inertially
wrt
to each other. So the labs are not identical nor do the paths of their
geodesics match in both frames, meaning that, for precise predictions,
one
must know the actual physical situation of his lab, ie, whether he
resting
on the launch pad or exiting the solar system with rockets propelling
him.

I think you are right in this sense. If you break the locality
condition, you won't get an inertial frame of reference and everything
goes to hell.

Well the labs are not inertial by definition. But then, I have the sense
that "locality" isn't meaningful anyway. By this I mean that I don't think
the difference between the two accelerated frames disappear in an interval
of infinitesimal domain. Here is why. The geodesics are the integrations
of the contributions of each infinitesmal interval. If the geodesics
differ, there must be some difference in the infinitesimal contributions.
In other words, to be different in larger domains is to be different
infinitesimally, or vice versa.

Geodesics aren't defined in terms of integrations, rather they are
defined in terms of how the metric changes. Path length is defined in
such a manner though.

I'm not sure how to respond to this in general. Yes the frames are
different at large scale. The basic point is that at *small* scale you
cannot tell the difference if you look at an area so small that it you
can construct an inertial coordinate system for the lab frame and so
small that you cannot observe tidal effects due to gravity.

If those two conditions are satisfied, you can't tell the difference
between that an an accelerated frame of reference. That is how I
understand the equivalene principle.


Another way of saying this is that if they were identical infinetesimally,
they would integrate identically, resulting in identical motions wrt to the
accelerating lab frames. But then, we know that the motions are
distinguishable experimentally.

Not arguing with you on this because you are right.


I have taken this to mean that there is a fundamental difference between
the
motions of the objects and the frames. In fact, it would seem,
physically,
that they are precisely the opposite of each other, a mirror image so to
speak. Welcome any and all comments.

I don't see how you can conclude there is a difference between the
object's motion and the object's frame of reference.

In re-reading what I posted, I can understand the confusion you are having
with what I said. I meant something entirely different than what you
inferred. I'll try to do a better job.

The the "inertial" objects are "all moving inertially wrt each other" and
"accelerating identically" wrt to the accelerating lab far from a gravity
source.

While in the accelerating lab resting on the surface of the planet, the
"inertial objects are all "accelerating with respect to each other" and
"accelerating differently" wrt the lab frame. Physically, they seem to be
opposites, that is, in terms of the qualities of their necessary motions.
Is this clearer?

Perfectly.


Phil

Pete wrote:

Try reading this. http://xxx.lanl.gov/abs/physics/0204044
It is a paper I wrote a few years back. The intent of the paper is to
directly answer your question.

Your historical account on the development of GR is very incomplete and
misleading.

We all know how Newton devised the law of gravity by watching a falling
apple under gravitational influence. Einstein's breakthrough tried
to outdo Newton by fantasizing himself as that apple falling under the
influence of gravity. Hey, as doctor Roberts ingeniously pointed out.
The outcome of that would eventually intercept the ground in a tragic
end. So, Einstein co-operated with Grossmann trying to devise a theory
of gravity based on his version of Equivalence Principle. It turned
out to be a failure. Disgusted with Grossmann, he started looking for
help in Goettingen.

At Goettingen, Klein, Hilbert, and Schwarzschild were very good friends
with Minkowski who had since passed away but not before suggesting the
postulate on the existence of spacetime after abandoning the Aether by
the scientific communities. Thus, Hilbert was interested in
Einstein's work. However, instead of co-operation, Einstein and
Hilbert became rivals with each one trying to beat the other one in
coming up with the field equations.

A spark came to Hilbert after Einstein bragged about derivation of
Mercury's orbital anomaly. This derivation without using GR and thus
without the necessary Schwarzschild metric was an entire plagiarism of
Gerber's work. Of course, Hilbert did not know Einstein's work was
total BS. Thinking he had lost the race, he basically pulled out a
Hail-Mary throw. He pulled out this Hilbert Langrangian out of his
*ss, and the field equations were born. Thus, GR was actually
Hilbert's work.

Somehow, you have just blacked out this part of the history, is it
because you don't understand the field equations?