Accelerating rocket/light beam question

[David Rutherford]

An observer in a uniformly accelerating rocket observes a light beam
from a laser projected horizontally from one wall of the rocket to the
other (perpendicular to the direction of acceleration). The observer
cannot see outside, so he cannot determine his state of motion. For all
he knows, he could be at rest in a uniform gravitational field. The
light beam will follow a curved path. To what will the observer
attribute the curvature of the beam?

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Chalky]

On Jan 3, 6:06*pm, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket observes a light beam
> from a laser projected horizontally from one wall of the rocket to the
> other (perpendicular to the direction of acceleration). The observer
> cannot see outside, so he cannot determine his state of motion. For all
> he knows, he could be at rest in a uniform gravitational field. The
> light beam will follow a curved path. To what will the observer
> attribute the curvature of the beam?

Gravity. This is Einstein's General Postulate, virtually word for
word, except that Einstein used a large box, not a rocket, and did not
specify the type of light beam.

[Phillip Helbig---remove CLOTHES to reply]

In article <K7mdnVqL57c408PUnZ2dnUVZ_v_inZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> An observer in a uniformly accelerating rocket observes a light beam
> from a laser projected horizontally from one wall of the rocket to the
> other (perpendicular to the direction of acceleration). The observer
> cannot see outside, so he cannot determine his state of motion. For all
> he knows, he could be at rest in a uniform gravitational field. The
> light beam will follow a curved path. To what will the observer
> attribute the curvature of the beam?

It depends on the physics knowledge the observer has. Perhaps he will
attribute it to angels pushing on the photons. If the observer is aware
of the principle of equivalence, he will realise that the best
explanations, and the only plausible ones, are a) acceleration and b)
being at rest in a gravitational field. Following the principle of
equivalence, the two cases cannot be distinguished in the experiment
described. However, the observer could arrange for another light beam
slightly below the first. If it is bent more, then he would conclude he
is in a gravitational field; if not, he will conclude that he is
accelerating far from any massive object. Alternatively, he could
arrange for two light beams to be emitted parallelly from the ceiling
and check if they are still parallel when reaching the floor (if not,
gravitational field; if so, acceleration). Or, alternatively, drop two
balls and see if the distance between them diminishes more than one
would expect from their mutual gravitational attraction (which would be
very small).

[Rich L.]

On Jan 3, 12:06=A0pm, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket observes a light beam
> from a laser projected horizontally from one wall of the rocket to the
> other (perpendicular to the direction of acceleration). The observer
> cannot see outside, so he cannot determine his state of motion. For all
> he knows, he could be at rest in a uniform gravitational field. The
> light beam will follow a curved path. To what will the observer
> attribute the curvature of the beam?
>
> --
> Dave Rutherford
> "New Transformation Equations and the Electric Field Four-vector"http://w=
ww.softcom.net/users/der555
>
> Applications:
> "4/3 Problem Resolution"
> "Action-reaction Paradox Resolution"
> "Energy Density Correction"
> "Proposed Quantum Mechanical Connection"
> "Biot-Savart's Companion"
> "Horizon Problem Resolution"

The acceleration of the rocket. If the rocket was not accelerating, the
beam would not curve. This is covered by the Equivalence Principle.
Note that the constancy of the speed of light does not hold in
accelerating frames! Light always propagates at c in the observers
local frame, but if the light is being observed from a frame higher in a
gravitational field, or higher in an accelerating frame, the light will
appear to propagate at a slower speed. And visa versa.

Rich L.

[kushal]

On Jan 3, 11:06=A0pm, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket observes a light beam
> from a laser projected horizontally from one wall of the rocket to the
> other (perpendicular to the direction of acceleration). The observer
> cannot see outside, so he cannot determine his state of motion. For all
> he knows, he could be at rest in a uniform gravitational field. The
> light beam will follow a curved path. To what will the observer
> attribute the curvature of the beam?
>
> --
> Dave Rutherford
> "New Transformation Equations and the Electric Field Four-vector"http://w=
ww.softcom.net/users/der555
>
> Applications:
> "4/3 Problem Resolution"
> "Action-reaction Paradox Resolution"
> "Energy Density Correction"
> "Proposed Quantum Mechanical Connection"
> "Biot-Savart's Companion"
> "Horizon Problem Resolution"

The curvature of the beam can be attributed either to the
gravitational field or the acceleration of the rocket. Both these
possibilities are equally likely and hence cannot be differentiated
from within the rocket. Light bends in a gravitational field due to
space-time curvature.

Kushal.

[David Rutherford]

Rich L. wrote:
> On Jan 3, 12:06=A0pm, David Rutherford <drutherf...@softcom.net> wrote:
>
>>An observer in a uniformly accelerating rocket observes a light beam
>>from a laser projected horizontally from one wall of the rocket to the
>>other (perpendicular to the direction of acceleration). The observer
>>cannot see outside, so he cannot determine his state of motion. For all
>>he knows, he could be at rest in a uniform gravitational field. The
>>light beam will follow a curved path. To what will the observer
>>attribute the curvature of the beam?
>
> The acceleration of the rocket. If the rocket was not accelerating, the
> beam would not curve. This is covered by the Equivalence Principle.

The Equivalence Principle says that the beam does curve in a
gravitational field. Please see

http://en.wikipedia.org/wiki/Equivalence_principle

But congratulations! You got the right answer. Put an accelerometer in
the rocket and, no matter which scenario the observer is in, it'll tell
him he's accelerating.

Now for the tricky part. In a gravitational field, it isn't apparent
that he's accelerating (he looks like he's at rest), so how do you
justify claiming that he's accelerating? Hint: we can accelerate (or
decelerate) in both space _and_ time.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Chalky]

On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> On Jan 3, 12:06 pm, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket observes a light beam
> > from a laser projected horizontally from one wall of the rocket to the
> > other (perpendicular to the direction of acceleration). The observer
> > cannot see outside, so he cannot determine his state of motion. For all
> > he knows, he could be at rest in a uniform gravitational field. The
> > light beam will follow a curved path. To what will the observer
> > attribute the curvature of the beam?

> The acceleration of the rocket. If the rocket was not accelerating, the
> beam would not curve.

Wrong. When the rocket is just resting on the launch pad, the light
still bends.

> Note that the constancy of the speed of light does not hold in
> accelerating frames!

Except that I think you will find it does, if you drop the rocket down
a mine shaft instead of igniting the fuel. - That is, until it impacts
with the bottom of the shaft.

[torre@cc.usu.edu]

>
> The curvature of the beam can be attributed either to the
> gravitational field or the acceleration of the rocket.

I always feel very pedantic when this thought experiment comes up.
Strictly speaking, there isn't any gravitational field unless the
spacetime
is curved - something that could be checked with *two* light beams.

charlie torre

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
> In article <K7mdnVqL57c408PUnZ2dnUVZ_v_inZ2d@posted.docknet>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>An observer in a uniformly accelerating rocket observes a light beam
>>from a laser projected horizontally from one wall of the rocket to the
>>other (perpendicular to the direction of acceleration). The observer
>>cannot see outside, so he cannot determine his state of motion. For all
>>he knows, he could be at rest in a uniform gravitational field. The
>>light beam will follow a curved path. To what will the observer
>>attribute the curvature of the beam?
>
> It depends on the physics knowledge the observer has. Perhaps he will
> attribute it to angels pushing on the photons. If the observer is aware
> of the principle of equivalence, he will realise that the best
> explanations, and the only plausible ones, are a) acceleration and b)
> being at rest in a gravitational field. Following the principle of
> equivalence, the two cases cannot be distinguished in the experiment
> described. However, the observer could arrange for another light beam
> slightly below the first. If it is bent more, then he would conclude he
> is in a gravitational field; if not, he will conclude that he is
> accelerating far from any massive object.

Maybe I should have defined what I meant by uniform gravitational field
a little better. By a uniform gravitational field I meant a
gravitational field whose effects are indistinguishable from the effects
of the acceleration of the rocket, everywhere in the rocket. Maybe it
could be created by a large, thick, flat slab of gravitating material
whose gravitational field creates the same curvature of the light beam
as in the accelerating rocket, anywhere in the rocket.

In that case, both beams would bend the same, in each scenario, wouldn't
they? If not, please point me to the evidence that shows that they
wouldn't.

> Alternatively, he could
> arrange for two light beams to be emitted parallelly from the ceiling
> and check if they are still parallel when reaching the floor (if not,
> gravitational field; if so, acceleration).

In the uniform gravitational field I described, above, they would still
be parallel, in either scenario.

> Or, alternatively, drop two
> balls and see if the distance between them diminishes more than one
> would expect from their mutual gravitational attraction (which would be
> very small).

Again, the distance between them would remain constant except for what
one would expect from their mutual gravitational attraction, in both
scenarios.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[David Rutherford]

kushal wrote:
> On Jan 3, 11:06=A0pm, David Rutherford <drutherf...@softcom.net> wrote:
>
>>An observer in a uniformly accelerating rocket observes a light beam
>>from a laser projected horizontally from one wall of the rocket to the
>>other (perpendicular to the direction of acceleration). The observer
>>cannot see outside, so he cannot determine his state of motion. For all
>>he knows, he could be at rest in a uniform gravitational field. The
>>light beam will follow a curved path. To what will the observer
>>attribute the curvature of the beam?
>
> The curvature of the beam can be attributed either to the
> gravitational field or the acceleration of the rocket. Both these
> possibilities are equally likely and hence cannot be differentiated
> from within the rocket. Light bends in a gravitational field due to
> space-time curvature.

The observer should be able to determine the cause of the curvature of
the beam without knowing whether the rocket is accelerating or at rest
in a uniform gravitational field.

[[Mod. note -- The whole point of the equivalence principle is that
(assuming the gravitational field is indeed *uniform*), the observer
can't distinguish between these two situations -- any local observation
(i.e. not looking out the windows) she makes will give the *same*
results in both scenarios. -- jt]]

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Rich L.]

On Jan 6, 11:05Â*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
...
> > Note that the constancy of the speed of light does not hold in
> > accelerating frames! Â*
>
> Except that I think you will find it does, if you drop the rocket down
> a mine shaft instead of igniting the fuel. - That is, until it impacts
> with the bottom of the shaft.

You are mistaken here, or else misunderstood the comment. The speed
of light local to an observer is postulated to be constant, "c", for
any observer in any location, in General Relativity. Once you
consider accelerating frames, however, it is easy to show that the
speed of light observed non-locally cannot be a constant. This is a
consequence of the equivalence principle and the gravitational red
shift. To see this, consider a "clock" consisting of a meter stick
with two mirrors at each end, and a pulse of light that bounces back
and forth between the mirrors. Each time the pulse is reflected from
one of the mirrors a light flashes, which an observer can monitor
remotely. If the observer is right next to the clock, he will observe
flashes of light corresponding to the pulse propagating a distance of
1 meter (2 meters for a round trip) at the speed of light, "c". As
this observere lowers this clock down into a gravitational field, the
gravitational red shift will make the clock run slower. There is no
gravitational length contraction (certainly not along an isopotential
anyway), so the length of the meter stick remains 1 meter. The
frequency of flashes reduces as the clock is lowered, so the observer
must conclude that the speed of light, AS HE OBSERVES IT REMOTELY is
reduced.

If the observer had an assistant that stayed with the clock as it was
lowered, the assistant would be subject to the gravitational red shift
and would perceive that the speed of light remained constant. If the
assistant looked up to observe a duplicate clock in the upper
laboratory frame, he would see that that clock has sped up, and thus
conclude that the speed of light in THAT frame has increased.

This argument is used to make the point that the constancy of the
speed of light in General Relativity only makes sense LOCALLY. This
point is made quite clearly in every book on GR that I have read.

Rich L.

[Phillip Helbig---remove CLOTHES to reply]

In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdnZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> Maybe I should have defined what I meant by uniform gravitational field
> a little better. By a uniform gravitational field I meant a
> gravitational field whose effects are indistinguishable from the effects
> of the acceleration of the rocket, everywhere in the rocket.

In that case, by definition, there is no way to distinguish the two.

> Maybe it
> could be created by a large, thick, flat slab of gravitating material
> whose gravitational field creates the same curvature of the light beam
> as in the accelerating rocket, anywhere in the rocket.

No. IIRC, there is NO WAY to produce a uniform gravitational field.
Approximately uniform to some accuracy within a certain region, yes, but
not precisely uniform over a finite region, not even in principle.

[Chalky]

On Jan 3, 6:06*pm, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket observes a light beam
> from a laser projected horizontally from one wall of the rocket to the
> other (perpendicular to the direction of acceleration). The observer
> cannot see outside, so he cannot determine his state of motion. For all
> he knows, he could be at rest in a uniform gravitational field. The
> light beam will follow a curved path. To what will the observer
> attribute the curvature of the beam?

Gravity. This is Einstein's General Postulate, virtually word for
word, except that Einstein used a large box, not a rocket, and did not
specify the type of light beam.

[Chalky]

On Jan 7, 11:09*pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
remove CLOTHES to reply) wrote:
> In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdn...@posted.docknet>, David
>
> Rutherford <drutherf...@softcom.net> writes:
> > Maybe I should have defined what I meant by uniform gravitational field
> > a little better. By a uniform gravitational field I meant a
> > gravitational field whose effects are indistinguishable from the effects
> > of the acceleration of the rocket, everywhere in the rocket.
>
> In that case, by definition, there is no way to distinguish the two.
>
> > Maybe it
> > could be created by a large, thick, flat slab of gravitating material
> > whose gravitational field creates the same curvature of the light beam
> > as in the accelerating rocket, anywhere in the rocket.
>
> No. *IIRC, there is NO WAY to produce a uniform gravitational field. *
> Approximately uniform to some accuracy within a certain region, yes, but
> not precisely uniform over a finite region, not even in principle.

If we are going to be pedantic about this, I think you will find the
most significant source of non uniformity is not in the Earth's
gravitational field, but in the turbulence of the thrust of the rocket
propulsion system.

[Chalky]

On Jan 7, 11:09 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
remove CLOTHES to reply) wrote:
> In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdn...@posted.docknet>, David
>
> Rutherford <drutherf...@softcom.net> writes:
> > Maybe I should have defined what I meant by uniform gravitational field
> > a little better. By a uniform gravitational field I meant a
> > gravitational field whose effects are indistinguishable from the effects
> > of the acceleration of the rocket, everywhere in the rocket.
>
> In that case, by definition, there is no way to distinguish the two.
>
> > Maybe it
> > could be created by a large, thick, flat slab of gravitating material
> > whose gravitational field creates the same curvature of the light beam
> > as in the accelerating rocket, anywhere in the rocket.
>
> No. IIRC, there is NO WAY to produce a uniform gravitational field.
> Approximately uniform to some accuracy within a certain region, yes, but
> not precisely uniform over a finite region, not even in principle.

This would seem to be a theoretical as opposed to practical objection.
(If not, please advise where we too can purchase the clocks and rulers
which are capable of telling the difference between the Earth's g field
and the effects of acceleration, within the boundary walls of a rocket).

As a theoretical argument, it seems to me that it too can be
theoretically countered by postulating a sufficiently large distribution
of masses with a centre of gravity at a sufficiently large distance that
any theoretical differences in gravitational field over the volume of
the rocket would be swamped by the limits set by the uncertainty
principle.

[David Rutherford]

Rich L. wrote:
> On Jan 3, 12:06=A0pm, David Rutherford <drutherf...@softcom.net> wrote:
>
>>An observer in a uniformly accelerating rocket observes a light beam
>>from a laser projected horizontally from one wall of the rocket to the
>>other (perpendicular to the direction of acceleration). The observer
>>cannot see outside, so he cannot determine his state of motion. For all
>>he knows, he could be at rest in a uniform gravitational field. The
>>light beam will follow a curved path. To what will the observer
>>attribute the curvature of the beam?
>
> The acceleration of the rocket. If the rocket was not accelerating, the
> beam would not curve. This is covered by the Equivalence Principle.

The Equivalence Principle says that the beam does curve in a
gravitational field. Please see

http://en.wikipedia.org/wiki/Equivalence_principle

But congratulations! You got the right answer. Put an accelerometer in
the rocket and, no matter which scenario the observer is in, it'll tell
him he's accelerating.

Now for the tricky part. In a gravitational field, it isn't apparent
that he's accelerating (he looks like he's at rest), so how do you
justify claiming that he's accelerating? Hint: we can accelerate (or
decelerate) in both space _and_ time.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[David Rutherford]

David Rutherford wrote:

> kushal wrote:
>
>>The curvature of the beam can be attributed either to the
>>gravitational field or the acceleration of the rocket. Both these
>>possibilities are equally likely and hence cannot be differentiated
>>from within the rocket. Light bends in a gravitational field due to
>>space-time curvature.
>
> The observer should be able to determine the cause of the curvature of
> the beam without knowing whether the rocket is accelerating or at rest
> in a uniform gravitational field.
>
> [[Mod. note -- The whole point of the equivalence principle is that
> (assuming the gravitational field is indeed *uniform*), the observer
> can't distinguish between these two situations -- any local observation
> (i.e. not looking out the windows) she makes will give the *same*
> results in both scenarios. -- jt]]

According to GR, the observation will give the same results, in both
scenarios, but for _different_ reasons, in each scenario. My proposal
(that the rocket/observer are accelerating in both scenarios) makes the
results _and_ the reasons for the results the same, in both scenarios.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Chalky]

On Jan 7, 9:33Â*pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> On Jan 6, 11:05Â*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> ..
> > > Note that the constancy of the speed of light does not hold in
> > > accelerating frames! Â*
>
> > Except that I think you will find it does, if you drop the rocket down
> > a mine shaft instead of igniting the fuel. - That is, until it impacts
> > with the bottom of the shaft.
>
> You are mistaken here, or else misunderstood the comment.

I don't think so.

>Â*The speed
> of light local to an observer is postulated to be constant, "c", for
> any observer in any location, in General Relativity. Â*Once you
> consider accelerating frames, however, it is easy to show that the
> speed of light observed non-locally cannot be a constant.

Ditto for non constancy of the speed of light non locally in a
gravitational field.

>Â*This is a
> consequence of the equivalence principle and the gravitational red
> shift. Â*

You seem to have forgotten that the gravitational redshift means red
in the direction earthwards whereas, for the accelerating rocket
observer, the shift is blue in the direction of acceleration. These
two effects cancel out if the rocket is in free fall earthwards.


[[Mod. note -- The sign of gravitational redshift is that if you shine
light *up* in a static gravitational field (e.g. the Earth's gravitational
field), it redshifts (i.e. receiver measures a lower frequency than
transmitter). If you shine light *down* it blueshifts (i.e. receiver
measures a higher freqency than transmitter). Confusingly, the term
"gravitational redshift" is often used in both cases, just with a minus
sign stuck in somewhere.
-- jt]]

[Phillip Helbig---remove CLOTHES to reply]

In article
<daed09c1-41ea-4145-a319-a813e00c9eff@i18g2000prf.googlegroups.com>,
Chalky <chalkyspam@bleachboys.co.uk> writes:

> On Jan 7, 11:09?pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
> remove CLOTHES to reply) wrote:
> > In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdn...@posted.docknet>, David
> >
> > Rutherford <drutherf...@softcom.net> writes:
> > > Maybe I should have defined what I meant by uniform gravitational field
> > > a little better. By a uniform gravitational field I meant a
> > > gravitational field whose effects are indistinguishable from the effects
> > > of the acceleration of the rocket, everywhere in the rocket.
> >
> > In that case, by definition, there is no way to distinguish the two.
> >
> > > Maybe it
> > > could be created by a large, thick, flat slab of gravitating material
> > > whose gravitational field creates the same curvature of the light beam
> > > as in the accelerating rocket, anywhere in the rocket.
> >
> > No. IIRC, there is NO WAY to produce a uniform gravitational field.
> > Approximately uniform to some accuracy within a certain region, yes, but
> > not precisely uniform over a finite region, not even in principle.
>
> If we are going to be pedantic about this, I think you will find the
> most significant source of non uniformity is not in the Earth's
> gravitational field, but in the turbulence of the thrust of the rocket
> propulsion system.

You're missing the point. Yes, in practice, there are all sorts of
disturbances. However, I can, in principle, have a uniform
acceleration. It doesn't have to be a rocket; it could be done
electromagnetically or whatever. However, EVEN IN PRINCIPLE, I cannot
have a uniform gravitational field. Yes, I can make it arbitrarily
uniform over a small enough region, but that's not the same---and it
isn't what you meant by "indistinguishable". With acceleration, it's
more of a practical matter. For instance, I could imagine the
disturbances vanishing by chance or whatever and have a perfectly
uniform acceleration; with gravity I can't do that.

[Chalky]

On Jan 10, 4:58*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Jan 7, 9:33*pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
>
> > On Jan 6, 11:05*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> > ..
> > > > Note that the constancy of the speed of light does not hold in
> > > > accelerating frames! *
>
> > > Except that I think you will find it does, if you drop the rocket down
> > > a mine shaft instead of igniting the fuel. - That is, until it impacts
> > > with the bottom of the shaft.
>
> > You are mistaken here, or else misunderstood the comment.
>
> I don't think so.
>
> >*The speed
> > of light local to an observer is postulated to be constant, "c", for
> > any observer in any location, in General Relativity. *Once you
> > consider accelerating frames, however, it is easy to show that the
> > speed of light observed non-locally cannot be a constant.
>
> Ditto for non constancy of the speed of light non locally in a
> gravitational field.
>
> >*This is a
> > consequence of the equivalence principle and the gravitational red
> > shift. *
>
> You seem to have forgotten that the gravitational redshift means red
> in the direction earthwards whereas, for the accelerating rocket
> observer, the shift is blue in the direction of acceleration. These
> two effects cancel out if the rocket is in free fall earthwards.
>
> [[Mod. note -- The sign of gravitational redshift is that if you shine
> light *up* in a static gravitational field (e.g. the Earth's gravitational
> field), it redshifts (i.e. receiver measures a lower frequency than
> transmitter). *If you shine light *down* it blueshifts (i.e. receiver
> measures a higher freqency than transmitter). *Confusingly, the term
> "gravitational redshift" is often used in both cases, just with a minus
> sign stuck in somewhere.
> -- jt]]

Sorry about that. I was looking at this from the point of view of the
observer (which is the only thing we can unambiguously verify first
hand).

>From the point of view of that observer, lower sources are redshifted
whereas higher sources are blueshifted. However, whichever way you
look at it, there is no such thing as "higher" or "lower", for an
observer in free fall, which is another way of confirming that these 2
shifts then cancel out.

[Phillip Helbig---remove CLOTHES to reply]

In article
<4330a1e6-db8b-4193-8ce7-885541e17548@i18g2000prf.googlegroups.com>,
Chalky <chalkyspam@bleachboys.co.uk> writes:

> On Jan 7, 11:09 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
> remove CLOTHES to reply) wrote:
> > In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdn...@posted.docknet>, David
> >
> > Rutherford <drutherf...@softcom.net> writes:
> > > Maybe I should have defined what I meant by uniform gravitational field
> > > a little better. By a uniform gravitational field I meant a
> > > gravitational field whose effects are indistinguishable from the effects
> > > of the acceleration of the rocket, everywhere in the rocket.
> >
> > In that case, by definition, there is no way to distinguish the two.
> >
> > > Maybe it
> > > could be created by a large, thick, flat slab of gravitating material
> > > whose gravitational field creates the same curvature of the light beam
> > > as in the accelerating rocket, anywhere in the rocket.
> >
> > No. IIRC, there is NO WAY to produce a uniform gravitational field.
> > Approximately uniform to some accuracy within a certain region, yes, but
> > not precisely uniform over a finite region, not even in principle.
>
> This would seem to be a theoretical as opposed to practical objection.

Yes, though I would say an objection as a matter of principle as opposed
to merely a practical one.

> (If not, please advise where we too can purchase the clocks and rulers
> which are capable of telling the difference between the Earth's g field
> and the effects of acceleration, within the boundary walls of a rocket).

This is a merely practical objection.

> As a theoretical argument, it seems to me that it too can be
> theoretically countered by postulating a sufficiently large distribution
> of masses with a centre of gravity at a sufficiently large distance that
> any theoretical differences in gravitational field over the volume of
> the rocket would be swamped by the limits set by the uncertainty
> principle.

Yes. You can make it arbitrarily uniform over a small enough distance.
Then you can't distinguish the two. If your measurement apparatus isn't
precise, you can't distinguish the two. However, IN PRINCIPLE I can
have measuring instruments which can tell the difference, but not even
in principle can I have a perfectly uniform field.

Your criterion was an INDISTINGUISHABLE field. In practice, it doesn't
have to be perfectly uniform if the instruments are bad enough. In
principle, a perfectly uniform field is impossible.

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:

> In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdnZ2d@posted.docknet>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>Maybe I should have defined what I meant by uniform gravitational field
>>a little better. By a uniform gravitational field I meant a
>>gravitational field whose effects are indistinguishable from the effects
>>of the acceleration of the rocket, everywhere in the rocket.
>
> In that case, by definition, there is no way to distinguish the two.
>
>>Maybe it
>>could be created by a large, thick, flat slab of gravitating material
>>whose gravitational field creates the same curvature of the light beam
>>as in the accelerating rocket, anywhere in the rocket.
>
> No. IIRC, there is NO WAY to produce a uniform gravitational field.
> Approximately uniform to some accuracy within a certain region, yes, but
> not precisely uniform over a finite region, not even in principle.

This is a thought experiment. Just imagine it's possible.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Tom Roberts]

torre@cc.usu.edu wrote:
> someone else wrote:
>> The curvature of the beam can be attributed either to the
>> gravitational field or the acceleration of the rocket.

Please note this statement is about the curvature OF THE BEAM, and not
about any curvature of the spacetime manifold. So in mentioning
"gravitational field" you change the subject:

> I always feel very pedantic when this thought experiment comes up.
> Strictly speaking, there isn't any gravitational field unless the
> spacetime is curved

Hmmmm. Like nearly every attempt to make a general statement in GR, this
one has counterexamples. So it's not quite "strict":

The domain wall solution to the field equation has no curvature in the
vacuum regions outside the infinitesimally thin wall (i.e. the Riemann
tensor is identically zero throughout these regions), but there is
gravity there in the sense that all non-spacelike geodesics are
attracted toward the wall.

Yes, you said "gravitational field", not "gravity". But it's strange to
have what could only be described as "gravity" without any
"gravitational field", especially in a field theory like GR....
The puns and oddities pile up -- which is my point:

The concept of "gravity" is quite slippery in GR. I don't think there is
a simple statement that fully captures the concept "gravity" in a useful
way. Each and every attempt I have seen has difficulties.


Tom Roberts

[Phillip Helbig---remove CLOTHES to reply]

In article
<dbf414db-c977-4afd-8e99-6599abe7322e@a26g2000prf.googlegroups.com>,
Chalky <chalkyspam@bleachboys.co.uk> writes:

> > > If we are going to be pedantic about this, I think you will find the
> > > most significant source of non uniformity is not in the Earth's
> > > gravitational field, but in the turbulence of the thrust of the rocket
> > > propulsion system.
> >
> > You're missing the point.
>
> I don't think I am. The OP is posing a variant on Einstein's argument
> for the general postulate (aka the strong principle of equivalence)
>
> By denying that it is true, you are denying an axiomatic foundation of
> GR. It seems to me that this renders your faith in the accuracy of the
> geometrical interpretation of GR that results, precarious at best.

I don't see how denying VARIANT on Einstein's argument is denying an
axiomatic foundation of GR. I'm also not denying the EP itself.

The EP states that the effects of a gravitational field and acceleration
are indistinguishable provided that the space is small enough so that
the gravitational field can be regarded as uniform. Several folks
pointed out that they CAN be distinguished if we make use of the
non-uniformity (which implies that the space in which we study these
effects is large enough). Then the claim was made that they could still
not be distinguished if the gravitational field was uniform within the
entire rocket. I claimed that this is not physically possible, that is,
one cannot have a uniform gravitational field over an arbitrarily large
region. The argument that in practice the non-uniform acceleration of
the rocket would be a bigger distraction is a red-herring: This is a
thought-experiment, and as such everything is allowed which is allowed
IN PRINCIPLE. If one cannot detect the non-uniformity, then we are back
to the case that the space is "small enough" and we have the
conventional EP. My claim is that on a large enough scale, we can
always detect the non-uniformity. One can't just postulate that we
can't without describing how such an arbitrarily large non-uniform
gravitational field can be constructed. I don't see how this discussion
questions the EP at all.

Maybe the original poster is saying "EP would be even stronger if we
have a uniform gravitational field in an arbitrarily large volume".
However, unless it is clear how to construct such a field, there is

[David Rutherford]

David Rutherford wrote:

> kushal wrote:
>
>>The curvature of the beam can be attributed either to the
>>gravitational field or the acceleration of the rocket. Both these
>>possibilities are equally likely and hence cannot be differentiated
>>from within the rocket. Light bends in a gravitational field due to
>>space-time curvature.
>
> The observer should be able to determine the cause of the curvature of
> the beam without knowing whether the rocket is accelerating or at rest
> in a uniform gravitational field.
>
> [[Mod. note -- The whole point of the equivalence principle is that
> (assuming the gravitational field is indeed *uniform*), the observer
> can't distinguish between these two situations -- any local observation
> (i.e. not looking out the windows) she makes will give the *same*
> results in both scenarios. -- jt]]

According to GR, the observation will give the same results, in both
scenarios, but for _different_ reasons, in each scenario. My proposal
(that the rocket/observer are accelerating in both scenarios) makes the
results _and_ the reasons for the results the same, in both scenarios.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Chalky]

On Jan 7, 9:33Â*pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> On Jan 6, 11:05Â*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> ..
> > > Note that the constancy of the speed of light does not hold in
> > > accelerating frames! Â*
>
> > Except that I think you will find it does, if you drop the rocket down
> > a mine shaft instead of igniting the fuel. - That is, until it impacts
> > with the bottom of the shaft.
>
> You are mistaken here, or else misunderstood the comment.

I don't think so.

>Â*The speed
> of light local to an observer is postulated to be constant, "c", for
> any observer in any location, in General Relativity. Â*Once you
> consider accelerating frames, however, it is easy to show that the
> speed of light observed non-locally cannot be a constant.

Ditto for non constancy of the speed of light non locally in a
gravitational field.

>Â*This is a
> consequence of the equivalence principle and the gravitational red
> shift. Â*

You seem to have forgotten that the gravitational redshift means red
in the direction earthwards whereas, for the accelerating rocket
observer, the shift is blue in the direction of acceleration. These
two effects cancel out if the rocket is in free fall earthwards.


[[Mod. note -- The sign of gravitational redshift is that if you shine
light *up* in a static gravitational field (e.g. the Earth's gravitational
field), it redshifts (i.e. receiver measures a lower frequency than
transmitter). If you shine light *down* it blueshifts (i.e. receiver
measures a higher freqency than transmitter). Confusingly, the term
"gravitational redshift" is often used in both cases, just with a minus
sign stuck in somewhere.
-- jt]]

[Chalky]

On Jan 10, 4:58*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Jan 7, 9:33*pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
>
> > On Jan 6, 11:05*am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > On Jan 5, 8:09 pm, "Rich L." <ralivings...@sbcglobal.net> wrote:
> > ..
> > > > Note that the constancy of the speed of light does not hold in
> > > > accelerating frames! *
>
> > > Except that I think you will find it does, if you drop the rocket down
> > > a mine shaft instead of igniting the fuel. - That is, until it impacts
> > > with the bottom of the shaft.
>
> > You are mistaken here, or else misunderstood the comment.
>
> I don't think so.
>
> >*The speed
> > of light local to an observer is postulated to be constant, "c", for
> > any observer in any location, in General Relativity. *Once you
> > consider accelerating frames, however, it is easy to show that the
> > speed of light observed non-locally cannot be a constant.
>
> Ditto for non constancy of the speed of light non locally in a
> gravitational field.
>
> >*This is a
> > consequence of the equivalence principle and the gravitational red
> > shift. *
>
> You seem to have forgotten that the gravitational redshift means red
> in the direction earthwards whereas, for the accelerating rocket
> observer, the shift is blue in the direction of acceleration. These
> two effects cancel out if the rocket is in free fall earthwards.
>
> [[Mod. note -- The sign of gravitational redshift is that if you shine
> light *up* in a static gravitational field (e.g. the Earth's gravitational
> field), it redshifts (i.e. receiver measures a lower frequency than
> transmitter). *If you shine light *down* it blueshifts (i.e. receiver
> measures a higher freqency than transmitter). *Confusingly, the term
> "gravitational redshift" is often used in both cases, just with a minus
> sign stuck in somewhere.
> -- jt]]

Sorry about that. I was looking at this from the point of view of the
observer (which is the only thing we can unambiguously verify first
hand).

>From the point of view of that observer, lower sources are redshifted
whereas higher sources are blueshifted. However, whichever way you
look at it, there is no such thing as "higher" or "lower", for an
observer in free fall, which is another way of confirming that these 2
shifts then cancel out.

[Phillip Helbig---remove CLOTHES to reply]

In article
<daed09c1-41ea-4145-a319-a813e00c9eff@i18g2000prf.googlegroups.com>,
Chalky <chalkyspam@bleachboys.co.uk> writes:

> On Jan 7, 11:09?pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
> remove CLOTHES to reply) wrote:
> > In article <k7KdnZ_Onqb4L__UnZ2dnUVZ_gKdn...@posted.docknet>, David
> >
> > Rutherford <drutherf...@softcom.net> writes:
> > > Maybe I should have defined what I meant by uniform gravitational field
> > > a little better. By a uniform gravitational field I meant a
> > > gravitational field whose effects are indistinguishable from the effects
> > > of the acceleration of the rocket, everywhere in the rocket.
> >
> > In that case, by definition, there is no way to distinguish the two.
> >
> > > Maybe it
> > > could be created by a large, thick, flat slab of gravitating material
> > > whose gravitational field creates the same curvature of the light beam
> > > as in the accelerating rocket, anywhere in the rocket.
> >
> > No. IIRC, there is NO WAY to produce a uniform gravitational field.
> > Approximately uniform to some accuracy within a certain region, yes, but
> > not precisely uniform over a finite region, not even in principle.
>
> If we are going to be pedantic about this, I think you will find the
> most significant source of non uniformity is not in the Earth's
> gravitational field, but in the turbulence of the thrust of the rocket
> propulsion system.

You're missing the point. Yes, in practice, there are all sorts of
disturbances. However, I can, in principle, have a uniform
acceleration. It doesn't have to be a rocket; it could be done
electromagnetically or whatever. However, EVEN IN PRINCIPLE, I cannot
have a uniform gravitational field. Yes, I can make it arbitrarily
uniform over a small enough region, but that's not the same---and it
isn't what you meant by "indistinguishable". With acceleration, it's
more of a practical matter. For instance, I could imagine the
disturbances vanishing by chance or whatever and have a perfectly
uniform acceleration; with gravity I can't do that.

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
>
> The EP states that the effects of a gravitational field and acceleration
> are indistinguishable provided that the space is small enough so that
> the gravitational field can be regarded as uniform. Several folks
> pointed out that they CAN be distinguished if we make use of the
> non-uniformity (which implies that the space in which we study these
> effects is large enough). Then the claim was made that they could still
> not be distinguished if the gravitational field was uniform within the
> entire rocket. I claimed that this is not physically possible, that is,
> one cannot have a uniform gravitational field over an arbitrarily large
> region. The argument that in practice the non-uniform acceleration of
> the rocket would be a bigger distraction is a red-herring: This is a
> thought-experiment, and as such everything is allowed which is allowed
> IN PRINCIPLE.

In most elementary EM books, there's a problem involving the electric
field of an infinite uniform sheet of charge. An infinite sheet of
charge isn't allowed in principle, either.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Tom Roberts]

torre@cc.usu.edu wrote:
> someone else wrote:
>> The curvature of the beam can be attributed either to the
>> gravitational field or the acceleration of the rocket.

Please note this statement is about the curvature OF THE BEAM, and not
about any curvature of the spacetime manifold. So in mentioning
"gravitational field" you change the subject:

> I always feel very pedantic when this thought experiment comes up.
> Strictly speaking, there isn't any gravitational field unless the
> spacetime is curved

Hmmmm. Like nearly every attempt to make a general statement in GR, this
one has counterexamples. So it's not quite "strict":

The domain wall solution to the field equation has no curvature in the
vacuum regions outside the infinitesimally thin wall (i.e. the Riemann
tensor is identically zero throughout these regions), but there is
gravity there in the sense that all non-spacelike geodesics are
attracted toward the wall.

Yes, you said "gravitational field", not "gravity". But it's strange to
have what could only be described as "gravity" without any
"gravitational field", especially in a field theory like GR....
The puns and oddities pile up -- which is my point:

The concept of "gravity" is quite slippery in GR. I don't think there is
a simple statement that fully captures the concept "gravity" in a useful
way. Each and every attempt I have seen has difficulties.


Tom Roberts

[Chalky]

[[Mod. note -- Attributions moved around to make it clearer who wrote what.
-- jt]]

> Phillip Helbig---remove CLOTHES to reply wrote:
> > The EP states that the effects of a gravitational field and acceleration
> > are indistinguishable provided that the space is small enough so that
> > the gravitational field can be regarded as uniform. Â*Several folks
> > pointed out that they CAN be distinguished if we make use of the
> > non-uniformity (which implies that the space in which we study these
> > effects is large enough). Â*Then the claim was made that they could still
> > not be distinguished if the gravitational field was uniform within the
> > entire rocket. Â*I claimed that this is not physically possible, that is,
> > one cannot have a uniform gravitational field over an arbitrarily large
> > region. Â*The argument that in practice the non-uniform acceleration of
> > the rocket would be a bigger distraction is a red-herring: This is a
> > thought-experiment, and as such everything is allowed which is allowed
> > IN PRINCIPLE.

But, the salient principle here IS the equivalence principle, and, in
practice, rockets are small enough for the principle to hold true, to
within current limits of observational evidence. Correct me if I am
wrong, but afaik, tidal forces are only detectable on the scale of
orbiting space stations where
1) Dimensions are larger
2) Achievable durations are longer
3) What is observed in practice is the combined effect of a non
uniform centrifugal force (which is demonstrable on a scale of
centimetres in laboratory centrifuges), re-enforced by the tidal
gravity of the Earth.


On Jan 14, 1:26Â*am, David Rutherford <drutherf...@softcom.net> wrote:
> In most elementary EM books, there's a problem involving the electric
> field of an infinite uniform sheet of charge. An infinite sheet of
> charge isn't allowed in principle, either.

[[Mod. note -- Quoted signature deleted -- jt]]

[Uncle Al]

Chalky wrote:
>
> [[Mod. note -- Attributions moved around to make it clearer who wrote what.
> -- jt]]
>
> > Phillip Helbig---remove CLOTHES to reply wrote:
> > > The EP states that the effects of a gravitational field and acceleration
> > > are indistinguishable provided that the space is small enough so that
> > > the gravitational field can be regarded as uniform. Â Several folks
> > > pointed out that they CAN be distinguished if we make use of the
> > > non-uniformity (which implies that the space in which we study these
> > > effects is large enough). Â Then the claim was made that they could still
> > > not be distinguished if the gravitational field was uniform within the
> > > entire rocket. Â I claimed that this is not physically possible, that is,
> > > one cannot have a uniform gravitational field over an arbitrarily large
> > > region. Â The argument that in practice the non-uniform acceleration of
> > > the rocket would be a bigger distraction is a red-herring: This is a
> > > thought-experiment, and as such everything is allowed which is allowed
> > > IN PRINCIPLE.
>
> But, the salient principle here IS the equivalence principle, and, in
> practice, rockets are small enough for the principle to hold true, to
> within current limits of observational evidence. Correct me if I am
> wrong, but afaik, tidal forces are only detectable on the scale of
> orbiting space stations where
> 1) Dimensions are larger
> 2) Achievable durations are longer
> 3) What is observed in practice is the combined effect of a non
> uniform centrifugal force (which is demonstrable on a scale of
> centimetres in laboratory centrifuges), re-enforced by the tidal
> gravity of the Earth.
[snip]

Tidal forces are detectable over less than a meter diameter. Eotvos
balance rotors are no larger than your fist.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

[Dirk Bruere at NeoPax]

Chalky wrote:
> [[Mod. note -- Attributions moved around to make it clearer who wrote what.
> -- jt]]
>
>> Phillip Helbig---remove CLOTHES to reply wrote:
>>> The EP states that the effects of a gravitational field and acceleration
>>> are indistinguishable provided that the space is small enough so that
>>> the gravitational field can be regarded as uniform. Â Several folks
>>> pointed out that they CAN be distinguished if we make use of the
>>> non-uniformity (which implies that the space in which we study these
>>> effects is large enough). Â Then the claim was made that they could still
>>> not be distinguished if the gravitational field was uniform within the
>>> entire rocket. Â I claimed that this is not physically possible, that is,
>>> one cannot have a uniform gravitational field over an arbitrarily large
>>> region. Â The argument that in practice the non-uniform acceleration of
>>> the rocket would be a bigger distraction is a red-herring: This is a
>>> thought-experiment, and as such everything is allowed which is allowed
>>> IN PRINCIPLE.
>
> But, the salient principle here IS the equivalence principle, and, in
> practice, rockets are small enough for the principle to hold true, to
> within current limits of observational evidence.
...

A couple of points:

a) "...small enough for the principle to hold true..." is not the same
as "holding true".

b) Dismissing an effect because it can (apparently) be made arbitrarily
small didn't work so well for QM in the early days when the Uncertainty
Principle was being examined. It turned out that the problem did not lie
with experimental accuracy (or lack of) but was fundamental to the whole
notion of uncertainty in measurement. Might it be conceivable that the
fact that it is in principle (if not experiment) possible to distinguish
between an accelerating frame and a gravitational field due to
uniformity considerations is the undoing of the Equivalence Principle?

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff

[Phillip Helbig---remove CLOTHES to reply]

In article <mt2.1-1713-1231896408@oxygen.astro.indiana.edu>, David
Rutherford <drutherford@softcom.net> writes:

> In most elementary EM books, there's a problem involving the electric
> field of an infinite uniform sheet of charge. An infinite sheet of
> charge isn't allowed in principle, either.

Yes, but it can be a useful approximation. However, the question "What
happens if the sheet is REALLY infinite?" is not meaningful. Similarly,
a uniform gravitational field can be a good approximation in some
circumstances, but we know that on a large enough scale a real
gravitational field is not uniform, so the question "But what if there
WERE a real gravitational field which is uniform over a large region?"
is not meaningful. In particular, it's not of any use in trying to
understand the EP or GR.

[Chalky]

=======IMPORTANT moderator's note to the poster=====================

Dear Chalky,

You should know that sci.physics.research is a moderated news group,
i.e., each of your posting to this news group is checked by a team
of moderators whether it is on topic and appropriate for this news
group. That's why it takes some time, until your posting appears.
If a posting is rejected, we'd like to send you an e-mail, but your
e-mail address is not working, so that we cannot do that. For that reason,
I write this note here in the news group: Please send each posting only
once, because otherwise our mail boxes get swamped with douplicate postings.
Also a working e-mail address would be really a good idea!

Hendrik van Hees.
================================================== =================

On Jan 17, 2:21*am, Dirk Bruere at NeoPax <dirk.bru...@gmail.com>
wrote:

> A couple of points:
>
> a) "...small enough for the principle to hold true..." is not the same
> as "holding true".

It is, to the extent that physics is an observationally driven
science.

> b) Dismissing an effect because it can (apparently) be made arbitrarily
> small didn't work so well for QM in the early days when the Uncertainty
> Principle was being examined.


That is completely different. With the advent of the Bohr model of the
atom, it became obvious that classical expectation (that such an
electron would continuously emit radiation, until it collapsed into
the nucleus), was clearly at variance with empirical observational
evidence

>. Might it be conceivable that the
> fact that it is in principle (if not experiment) possible to distinguish


It is not a fact, it is a classical expectation which has not been
confirmed observationally on the small scale, just as the classical
expectation of the collapse of the atom has failed to be confirmed on
the small scale.

> between an accelerating frame and a gravitational field due to
> uniformity considerations is the undoing of the Equivalence Principle?


If so, it would also be the undoing of GR, as I think I have mentioned
before.

[Uncle Al]

Chalky wrote:
>
> [[Mod. note -- Attributions moved around to make it clearer who wrote what.
> -- jt]]
>
> > Phillip Helbig---remove CLOTHES to reply wrote:
> > > The EP states that the effects of a gravitational field and acceleration
> > > are indistinguishable provided that the space is small enough so that
> > > the gravitational field can be regarded as uniform. Â Several folks
> > > pointed out that they CAN be distinguished if we make use of the
> > > non-uniformity (which implies that the space in which we study these
> > > effects is large enough). Â Then the claim was made that they could still
> > > not be distinguished if the gravitational field was uniform within the
> > > entire rocket. Â I claimed that this is not physically possible, that is,
> > > one cannot have a uniform gravitational field over an arbitrarily large
> > > region. Â The argument that in practice the non-uniform acceleration of
> > > the rocket would be a bigger distraction is a red-herring: This is a
> > > thought-experiment, and as such everything is allowed which is allowed
> > > IN PRINCIPLE.
>
> But, the salient principle here IS the equivalence principle, and, in
> practice, rockets are small enough for the principle to hold true, to
> within current limits of observational evidence. Correct me if I am
> wrong, but afaik, tidal forces are only detectable on the scale of
> orbiting space stations where
> 1) Dimensions are larger
> 2) Achievable durations are longer
> 3) What is observed in practice is the combined effect of a non
> uniform centrifugal force (which is demonstrable on a scale of
> centimetres in laboratory centrifuges), re-enforced by the tidal
> gravity of the Earth.
[snip]

Tidal forces are detectable over less than a meter diameter. Eotvos
balance rotors are no larger than your fist.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

[Phillip Helbig---remove CLOTHES to reply]

In article <mt2.1-10670-1232158890@oxygen.astro.indiana.edu>, Dirk
Bruere at NeoPax <dirk.bruere@gmail.com> writes:

> > But, the salient principle here IS the equivalence principle, and, in
> > practice, rockets are small enough for the principle to hold true, to
> > within current limits of observational evidence.

Yes.

> A couple of points:
>
> a) "...small enough for the principle to hold true..." is not the same
> as "holding true".
>
> b) Dismissing an effect because it can (apparently) be made arbitrarily
> small didn't work so well for QM in the early days when the Uncertainty
> Principle was being examined.

The equivalence principle is a good example of a limit.

> It turned out that the problem did not lie
> with experimental accuracy (or lack of) but was fundamental to the whole
> notion of uncertainty in measurement.

Just because it is bad to neglect some things in some cases doesn't mean
that the entire concept of a limit---well established mathematically and
physically---needs to be questioned.

> Might it be conceivable that the
> fact that it is in principle (if not experiment) possible to distinguish
> between an accelerating frame and a gravitational field due to
> uniformity considerations is the undoing of the Equivalence Principle?

Not unless you define the Equivalence Principle in a way in which it is
normally not defined.

Suppose that we can distinguish a gravitational field from acceleration,
due to tidal effects. This has been known since the EP was formulated.
What consequences of the "undoing" do you envisage? Unless you can come
up with different predictions than in the standard theory, all of this
is merely playing with words.

[Phillip Helbig---remove CLOTHES to reply]

> =======IMPORTANT moderator's note to the poster=====================
>
> Dear Chalky,
>
> You should know that sci.physics.research is a moderated news group,
> i.e., each of your posting to this news group is checked by a team
> of moderators whether it is on topic and appropriate for this news
> group. That's why it takes some time, until your posting appears.
> If a posting is rejected, we'd like to send you an e-mail, but your
> e-mail address is not working, so that we cannot do that. For that reason,
> I write this note here in the news group: Please send each posting only
> once, because otherwise our mail boxes get swamped with douplicate postings.
> Also a working e-mail address would be really a good idea!

In this particular case, the problem might be with Google Groups, rather
than with Chalky. As you say, as a regular poster he should be aware of
the moderation delay. We had a similar problem a couple of years ago
(i.e. we got several copies of an article even though it was submitted
only once). After a while, Google finally solved the problem.

Nevertheless, posts are either accepted and appear in the newsgroup, or
rejected. If they are rejected, then we send a rejection notice if you
supply a valid email address or a munged version which is easily
decipherable (see mine for an example). If you munge, then it is better
to munge the domain part rather than the username part (otherwise the
SMTP server for that domain might get flooded with spam addressed to an
unknown user and bounce that, generating backscatter spam (which is
sometimes, but not always, what the spammers actually want to happen)).
While there is no harm in obvious (to humans) munging, it is probably
not very necessary anymore. These days, most spammers don't harvest
email addresses from usenet and the web (there are far too few), but
rather use viruses to get email addresses stored in "address books" on
PCs or just use a dictionary attack. Any anti-spam solution which
handles such spam will also handle spam sent to your (perhaps de-munged)

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
> In article <mt2.1-1713-1231896408@oxygen.astro.indiana.edu>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>In most elementary EM books, there's a problem involving the electric
>>field of an infinite uniform sheet of charge. An infinite sheet of
>>charge isn't allowed in principle, either.
>
> Yes, but it can be a useful approximation. However, the question "What
> happens if the sheet is REALLY infinite?" is not meaningful. Similarly,
> a uniform gravitational field can be a good approximation in some
> circumstances, but we know that on a large enough scale a real
> gravitational field is not uniform, so the question "But what if there
> WERE a real gravitational field which is uniform over a large region?"
> is not meaningful. In particular, it's not of any use in trying to
> understand the EP or GR.

Taking the EP to the extreme, that is on a global scale, tests its
correspondence with GR, on a local scale. If the EP and GR are logically
inconsistent on a global scale, then they are inconsistent on a local
scale. A uniform gravitational field created by an infinite, uniform
sheet of mass can be used, in a thought experiment, to test the
correspondence between the EP and GR on both global _and_ local scales.

According to the EP, light would curve in the uniform gravitational
field created by an infinite, uniform sheet of mass. But, according to
GR, since there's no spacetime curvature associated with a uniform
gravitational field, light would not curve.

Consequently, we have two different predictions for the behavior of
light in a uniform gravitational field, therefore, GR is logically
inconsistent with the EP on both global and local scales. The question
is, which one gives the correct prediction (if either)? I say it's the EP.

But if it is the EP, we have to find a new cause for the curvature of
the path of light (and the path of any free-falling body).
Unfortunately, if I told you what I think it is, I would violate the
charter of sci.physics.research (if I haven't, already :)), so I won't.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Phillip Helbig---remove CLOTHES to reply]

In article <gkuq6c$3og$1@fb07-hees.theo.physik.uni-giessen.de>, Chalky
<chalkyspam@bleachboys.co.uk> writes:

> >. Might it be conceivable that the
> > fact that it is in principle (if not experiment) possible to distinguish
>
> It is not a fact, it is a classical expectation

One can of course measure tidal effects. It depends on what you mean by
"small scale". There is absolutely no reason not to expect that these
tidal effects can be detected on a smaller and smaller scale as the
precision of the measurement becomes better and better.

> which has not been
> confirmed observationally on the small scale, just as the classical
> expectation of the collapse of the atom has failed to be confirmed on
> the small scale.

Completely different case. The classical expectation of the collapse of
the atom was not confirmed because it does not happen, since atoms
aren't classical (but rather quantum) systems. The classical
expectation mentioned above (where "classical" here includes GR, by the
way) is that it WILL be confirmed since it is only a problem of
sensitivity. Thus, "just as" seems out of place here.

[Uncle Al]

Phillip Helbig---remove CLOTHES to reply wrote:
>
> In article <mt2.1-10670-1232158890@oxygen.astro.indiana.edu>, Dirk
> Bruere at NeoPax <dirk.bruere@gmail.com> writes:
>
> > > But, the salient principle here IS the equivalence principle, and, in
> > > practice, rockets are small enough for the principle to hold true, to
> > > within current limits of observational evidence.
>
> Yes.
>
> > A couple of points:
> >
> > a) "...small enough for the principle to hold true..." is not the same
> > as "holding true".
> >
> > b) Dismissing an effect because it can (apparently) be made arbitrarily
> > small didn't work so well for QM in the early days when the Uncertainty
> > Principle was being examined.
>
> The equivalence principle is a good example of a limit.
>
> > It turned out that the problem did not lie
> > with experimental accuracy (or lack of) but was fundamental to the whole
> > notion of uncertainty in measurement.
>
> Just because it is bad to neglect some things in some cases doesn't mean
> that the entire concept of a limit---well established mathematically and
> physically---needs to be questioned.
>
> > Might it be conceivable that the
> > fact that it is in principle (if not experiment) possible to distinguish
> > between an accelerating frame and a gravitational field due to
> > uniformity considerations is the undoing of the Equivalence Principle?
>
> Not unless you define the Equivalence Principle in a way in which it is
> normally not defined.
>
> Suppose that we can distinguish a gravitational field from acceleration,
> due to tidal effects. This has been known since the EP was formulated.
> What consequences of the "undoing" do you envisage? Unless you can come
> up with different predictions than in the standard theory, all of this
> is merely playing with words.

Einstein's elevator renders the The Equivalence Principle definitive
*IF* the vacuum is isotropic and inert. If the vacuum is selectively
interactive there can be counterexamples. Goose down and lead marbles
obviously obey the EP in vacuum. If the background is interactive -
air at one atmosphere - there is divergence.

The vacuum is spectacularly isotropic in the massless sector. Photons
behave in the lab (arxiv/0706.2031), in astronomy (PSR J1903+0327,
arxiv/0805.2396), and at cosmic distances (Sean Carroll has commented
at length).

Nobody knows if the vacuum is isotropic in the massed sector. A
chiral vacuum background is selective. All prior physics slips
through. A parity Eotvos experiment opposing enantiomorphic space
groups P3(1)21 an P3(2)21 single crystal qaurtz test masses is the
definitive observation.

Since theory and prior observation demand the vacuum is isotropic,
there is no reason to look at a novel test. Theory also demanded the
Weak interaction was parity conserving. Yang and Lee wasted funding
and equipment being pariahs, for there was no reason to look.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

[Phillip Helbig---remove CLOTHES to reply]

In article <6pKdnWthItsGfe7UnZ2dnUVZ_j-dnZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> Taking the EP to the extreme, that is on a global scale, tests its
> correspondence with GR, on a local scale. If the EP and GR are logically
> inconsistent on a global scale, then they are inconsistent on a local
> scale. A uniform gravitational field created by an infinite, uniform
> sheet of mass can be used, in a thought experiment, to test the
> correspondence between the EP and GR on both global _and_ local scales.
>
> According to the EP, light would curve in the uniform gravitational
> field created by an infinite, uniform sheet of mass. But, according to
> GR, since there's no spacetime curvature associated with a uniform
> gravitational field, light would not curve.

This is wrong. It would not "curve" in either case, but it would be
"deflected" in both. The two cases are indistinguishable. You seem to
agree they are indistinguishable in a region small enough so that tidal
effects are not detected. In this case, they would be indistinguishable
over a larger (in fact infinite) space.

You need to look at the DIFFERENCE in the behaviour of two initially
parallel light beams. In your example, even in this case, there could
be no way of distinguishing. This isn't strange, it's simply the
limiting case. Whether I approach the limit by making the distance
between the initially parallel light beams arbitrarily small, or by
making the uniform gravitational field arbitrarily large, doesn't
matter. If I can't distinguish, then BY DEFINITION the gravitational
field is uniform to within the precision of my measurement.

> Consequently, we have two different predictions for the behavior of
> light in a uniform gravitational field,

Wrong.

> therefore, GR is logically
> inconsistent with the EP on both global and local scales.

Wrong conclusion due to wrong premise.

> The question
> is, which one gives the correct prediction (if either)? I say it's the EP.

The question is moot, since the conclusion is wrong.

> But if it is the EP, we have to find a new cause for the curvature of
> the path of light (and the path of any free-falling body).
> Unfortunately, if I told you what I think it is, I would violate the
> charter of sci.physics.research (if I haven't, already :)), so I won't.

I think you are coming dangerously close. Unless you can point to
falsifiable predictions of your theory, it isn't worth anything.

[Igor Khavkine]

On Jan 19, 3:13 am, David Rutherford <drutherf...@softcom.net> wrote:

> Taking the EP to the extreme, that is on a global scale, tests its
> correspondence with GR, on a local scale. If the EP and GR are logically
> inconsistent on a global scale, then they are inconsistent on a local
> scale.

Not true. You've just defined two principles, EP-local and EP-global.
It is well known that EP-global implies an absence of curvature: the
acceleration can be removed with a coordinate transformation and the
result is a flat space-time. On the other hand, it is also well known
that EP-local, with non-uniform accelerations (for example like the
acceleration due to gravity on the Earth's surface) is inconsistent
with absence of curvature, cf. Schild's argument from section 7.3 in
Misner, Thorne, and Wheeler.

To summarize, you've defined EP-local and EP-global as inequivalent
observed principles (which one is observed depends on the situation),
which I've argued are both consistent with GR. EP-global implies that
the region investigated by the experimenters is a flat space-time,
which is contained as in GR as a special case with no curvature. EP-
local, together with variations in the acceleration detected by
experimenters in the investigated region, is consistent with GR + non-
zero curvature. The latter claim is substantiated by all the joint
theoretical/experimental tests that GR has endured over the past
century.

> According to the EP, light would curve in the uniform gravitational
> field created by an infinite, uniform sheet of mass. But, according to
> GR, since there's no spacetime curvature associated with a uniform
> gravitational field, light would not curve.
>
> Consequently, we have two different predictions for the behavior of
> light in a uniform gravitational field, therefore, GR is logically
> inconsistent with the EP on both global and local scales. The question
> is, which one gives the correct prediction (if either)? I say it's the EP.

Again, nothing logically inconsistent. In a "uniform gravitational
field", which refers to the space-time solution generated by the
presence of an infinite, uniform sheet of mass, light rays curve only
for an accelerated observer. Similarly, light rays curve uniformly in
the laboratory only if the laboratory is uniformly accelerated with
respect to the flat space-time (which is precisely your what-if
scenario). Conclusion: consistency with GR. So it's not a dichotomy,
either EP or GR; it's both.

Igor

[Jonathan Thornburg]

There's a lot of confusion floating around this thread about just
what the equivalence principle (EP) means, and the extent to which
it can be tested experimentally. The usual EP wording is something
along the lines of:
[1] No local experiment can distinguish between a uniform gravitational
field and a uniformly-acclerating reference frame in flat (Minkowski)
spacetime.

Given the confusion in this thread about "uniform" gravitational
fields, it's instructive to try to state [1] more precisely.

Operationally (at least in a gedanken sense), we can measure the
spacetime metric $g_{ab}$ at each event. Let's Taylor-expand this
in (spacetime) position about some reference point (event). By
making a suitable coordinate choice, we can ensure that
(a) $g_{ab} = \eta_{ab}$ at the reference point,
where $\eta_{ab} = diag(-1, +1, +1, +1)$ is the flat-spacetime
(Minkowski) metric in Cartesian coordinates, and
(b) $\partial g_{ab} / \partial x^c = 0$ at the reference point
for all indices $c$ (indices range from 0 to 3).
This is basically the definition of a local inertial frame, a.k.a
a local Lorentz frame. Notice that in general we can NOT ensure
that
(c) $\partial^2 g_{ab} / \partial x^c x^d = 0$ for all indices
$c$ and $d$
(If that were true, then the Riemann curvature tensor would vanish
at the reference point, i.e. spacetime would truly be "flat" there.)

[What I've said in the previous paragraph is is discussed in more
detail in section 6.2 of Bernard F. Schutz, "A First Course in
General Relativity" (Cambridge U.P., 1985).]

Now let's pose the question: Is the gravitational field in this
local inertial frame "uniform"? Well, that depends on what we
mean by "uniform". By choosing a coordinate system (reference
frame) in which (a) and (b) above hold, we've ensured that the
Newtonian "little g" will vanish. But what about the gradient
of "little g"? That's encoded in the 2nd derivatives of the
metric (with respect to spacetime position), which (c) above says
don't vanish in general.

So what the EP really says is that
[2] local experiments done in a local Lorentz frame where the
2nd and higher derivatives of the metric (with respect to spacetime
position) are small enough
[i.e., the ambient gravitational field is "uniform enough"]
not to influence the experiment, can't distinguish between
this local Lorentz frame and truly flat (Minkowski) spacetime.


Actually, "the" EP is neither [1] nor [2]. It's useful to split up
"the" EP into several constutient principles:
* The "weak EP" (WEP) is the univerality of free-fall, i.e. it asserts
that all test bodies will free-fall at the same acceleration in a
gravitational field, regardless of their internal structure. More
generally, it says that the laws of mechanics are the same in any
local Lorentz frame. This is tested by, for example, Eotvos experiments.
[Note that the modern interpretation of the words "free-fall"
here is that we're implicitly also assuming that the test
body has zero charge and spin, and that the local environment
is free of significant electromagnetic fields. (In other
words, rediscovering Coulomb's law, the Lorentz force law,
and/or the well-known gravitomagnetic effects on spinning
particles, aren't considered to be WEP violations.)]
* The "Einstein EP" (EEP) strengthens the WEP to assert that *all* the
non-gravitational laws of physics, i.e. including electromagnetism
and the weak & strong nuclear forces, are the same in any local
Lorentz frame. For example, the EEP asserts that test masses of
different chemical composition (or chirality, as Uncle Al would
remind us) should free-fall at the same rate in an external
gravitational field, despite having significantly different
[That is, "significantly different" relative to the
experimental accuracy, which is now somewhere down
below 1 part in 10^13.]
different fractions of their rest mass comprised of
electromagnetic vs weak-nuclear binding energies within atoms.
* The "strong EP" (which is much closer to [1] and [2]) further
generalizes the EEP to include gravitational interactions.
For example, the SEP asserts that the Earth and the Moon should
free-fall at the same rate in the Sun's gravitational field,
despite having significantly different fractions of their
rest mass made up of self-gravitational binding energy.

>>From a modern perspective, the EEP essentialy says that gravitation
is a curved-spacetime (metric) phenomenon, while the SEP essentially
gives general relativity. This is discussed in sections 2.1 and 3.1
of Will's paper cited below.

For a more detailed discussion of the various forms of the EP, the
experimental evidence that supports them, and their implications for
what we can (& can't) say about gravitation, see
* Clifford Will's (superb) review paper
http://relativity.livingreviews.org/Articles/lrr-2006-3/index.html
(freely available online)
* Clifford Will's classic book "Theory and Experiment in Gravitational
Physics" (Cambridge U.P., 1st Edition 1981, revised Edition
some-more-recent-date-which-I-don't-have-handy).
* Haugan and Laemmerzahl, gr-qc/0103067
* L?merzahl, C., "The Einstein equivalence principle and the search
for new physics", in Giulini, D.J.W., Kiefer, C., and L?merzahl,
C., eds., Quantum Gravity: From Theory to Experimental Search,
Lecture Notes in Physics, vol. 631, pp. 367-394, (Springer, Berlin,
Germany; New York, U.S.A., 2003).

-- From: "Jonathan Thornburg [remove -animal to reply]" <jthorn@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[J. J. Lodder]

Chalky <chalkyspam@bleachboys.co.uk> wrote:

> On Jan 17, 2:21 am, Dirk Bruere at NeoPax <dirk.bru...@gmail.com>
> wrote:
>
> > A couple of points:
> >
> > a) "...small enough for the principle to hold true..." is not the same
> > as "holding true".
>
> It is, to the extent that physics is an observationally driven
> science.
>
> > b) Dismissing an effect because it can (apparently) be made arbitrarily
> > small didn't work so well for QM in the early days when the Uncertainty
> > Principle was being examined.
>
>
> That is completely different. With the advent of the Bohr model of the
> atom, it became obvious that classical expectation (that such an
> electron would continuously emit radiation, until it collapsed into
> the nucleus), was clearly at variance with empirical observational
> evidence
>
> >. Might it be conceivable that the
> > fact that it is in principle (if not experiment) possible to distinguish
>
>
> It is not a fact, it is a classical expectation which has not been
> confirmed observationally on the small scale, just as the classical
> expectation of the collapse of the atom has failed to be confirmed on
> the small scale.

There is nothing to confirm.
Classical electrodynamics predicts a collapse of the atom
in a finite time. (and a very short one on the human scale)
The complete calculation can be found in Landau and Lifschitz.

The mere existence of anything disproves it,

Jan

[Chalky]

On Jan 19, 6:31Â*pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
remove CLOTHES to reply) wrote:
> In article <gkuq6c$3o...@fb07-hees.theo.physik.uni-giessen.de>, Chalky
>
> <chalkys...@bleachboys.co.uk> writes:
> > >. Might it be conceivable that the
> > > fact that it is in principle (if not experiment) possible to distinguish
>
> > It is not a fact, it is a classical expectation
>
> One can of course measure tidal effects. Â*

Yes, of course. One can also verify that an EM radiator radiates
classically on much smaller scales than one can currently confirm such
tidal gravitational effects.

> It depends on what you mean by
> "small scale". Â*There is absolutely no reason not to expect that these
> tidal effects can be detected on a smaller and smaller scale as the
> precision of the measurement becomes better and better.

There is, by analogy, given historical context, and prior experience
WRT classical expectations for EM radiation at the quantum scale.

> > which has not been
> > confirmed observationally on the small scale, just as the classical
> > expectation of the collapse of the atom has failed to be confirmed on
> > the small scale.
>
> Completely different case. Â*The classical expectation of the collapse of
> the atom was not confirmed because it does not happen, since atoms
> aren't classical (but rather quantum) systems. Â*The classical
> expectation mentioned above (where "classical" here includes GR, by the
> way)

Yes, I know, hence the Journal title "Classical & Quantum Gravity"

> is that it WILL be confirmed since it is only a problem of
> sensitivity. Â*

That is merely your classical expectation.

I have no doubt whatsoever that you would have been equally
enthusiastic about expounding the, in principle, correspondingly
limitless accuracy of classical EM theory, if you had been around in
the 19th century.

The point is, you don't know, and assertions (by some) that quantum
gravity must necessarily obey classical gravitational field theory
with limitless accuracy are, imo, highly suspect, given our historical
experience in the more accurately understood (and tested) EM field.

[Igor Khavkine]

On Jan 20, 7:55 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Jan 19, 6:31 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
> remove CLOTHES to reply) wrote:

> > It depends on what you mean by
> > "small scale". There is absolutely no reason not to expect that these
> > tidal effects can be detected on a smaller and smaller scale as the
> > precision of the measurement becomes better and better.

> > [...] The classical expectation of the collapse of
> > the atom was not confirmed because it does not happen, since atoms
> > aren't classical (but rather quantum) systems. The classical
> > expectation mentioned above (where "classical" here includes GR, by the
> > way)
>
> Yes, I know, hence the Journal title "Classical & Quantum Gravity"
>
> > is that it WILL be confirmed since it is only a problem of
> > sensitivity.
>
> That is merely your classical expectation.

Whether the prediction of the measurement of tidal effects on smaller
and smaller scales with improving instrumentation is "classical" or even
an "expectation" is beside the point. It is a testable, falsifiable
prediction that is consistent with all available data, collected both
before and after it was first made. That is the best information that a
scientific theory can offer. Any alternative must do at least as well to
be considered seriously. And such serious alternatives are conspicuously
lacking in this thread.

Igor

[David Rutherford]

Igor Khavkine wrote:
> On Jan 19, 3:13 am, David Rutherford <drutherf...@softcom.net> wrote:
>
>>Taking the EP to the extreme, that is on a global scale, tests its
>>correspondence with GR, on a local scale. If the EP and GR are logically
>>inconsistent on a global scale, then they are inconsistent on a local
>>scale.
>
> Not true. You've just defined two principles, EP-local and EP-global.
> It is well known that EP-global implies an absence of curvature: the
> acceleration can be removed with a coordinate transformation and the
> result is a flat space-time.

The acceleration of what?

> On the other hand, it is also well known
> that EP-local, with non-uniform accelerations (for example like the
> acceleration due to gravity on the Earth's surface) is inconsistent
> with absence of curvature, cf. Schild's argument from section 7.3 in
> Misner, Thorne, and Wheeler.

Schild's argument sites the redshift of an electromagnetic signal
emitted from an observer on the surface of the earth to a second
observer located above the first observer. According to the EP, we could
repeat the experiment in an accelerating rocket in empty space and get
the same result. Interestingly, there's an absence of curvature in the
accelerating rocket, yet there's still a redshift of the signal. How
does Schild explain that?

>>According to the EP, light would curve in the uniform gravitational
>>field created by an infinite, uniform sheet of mass. But, according to
>>GR, since there's no spacetime curvature associated with a uniform
>>gravitational field, light would not curve.
>>
>>Consequently, we have two different predictions for the behavior of
>>light in a uniform gravitational field, therefore, GR is logically
>>inconsistent with the EP on both global and local scales. The question
>>is, which one gives the correct prediction (if either)? I say it's the EP.
>
> Again, nothing logically inconsistent. In a "uniform gravitational
> field", which refers to the space-time solution generated by the
> presence of an infinite, uniform sheet of mass, light rays curve only
> for an accelerated observer.

That statement is consistent with GR, but inconsistent with the EP.
According to the EP, light rays would curve for an observer in a rocket
at rest on the infinite sheet of mass, just as they would curve for an
observer at rest in a rocket accelerating in empty space. That's why I
claim that GR is logically inconsistent with the EP.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Phillip Helbig---remove CLOTHES to reply]

In article
<da31f85d-42b9-4e41-ba49-24c477d0097e@g39g2000pri.googlegroups.com>,
Chalky <chalkyspam@bleachboys.co.uk> writes:

> > One can of course measure tidal effects.
>
> Yes, of course. One can also verify that an EM radiator radiates
> classically on much smaller scales than one can currently confirm such
> tidal gravitational effects.

Because gravitation is weaker than electromagnetism. So what?

> > It depends on what you mean by
> > "small scale". There is absolutely no reason not to expect that these
> > tidal effects can be detected on a smaller and smaller scale as the
> > precision of the measurement becomes better and better.
>
> There is, by analogy, given historical context, and prior experience
> WRT classical expectations for EM radiation at the quantum scale.

I could come up with analogies and examples which would indicate the
opposite. In science, one cannot conclusively argue from analogy.

> > is that it WILL be confirmed since it is only a problem of
> > sensitivity.
>
> That is merely your classical expectation.

The burden of proof is on the person (you) who expects something strange
to happen, not on the person who points out that, barring evidence to
the contrary, there is no reason to think that the EP says anything that
what it is normally understood to say.

> I have no doubt whatsoever that you would have been equally
> enthusiastic about expounding the, in principle, correspondingly
> limitless accuracy of classical EM theory, if you had been around in
> the 19th century.

But classical EM theory didn't even appear to be limitlessly accurate.
There were known problems. Their investigation led to QM and
relativity.

> The point is, you don't know, and assertions (by some) that quantum
> gravity must necessarily obey classical gravitational field theory
> with limitless accuracy are, imo, highly suspect, given our historical
> experience in the more accurately understood (and tested) EM field.

Again, arguing from analogy is dangerous. That's not the way science
works. We know that at some level QM and GR are not completely
compatible. But there is no reason to think that this has any relation
to understanding the EP. If you have such evidence, please present it.

[Phillip Helbig---remove CLOTHES to reply]

In article <p8mdnRIMfoS-AuvUnZ2dnUVZ_qjinZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> Schild's argument sites the redshift of an electromagnetic signal
> emitted from an observer on the surface of the earth to a second
> observer located above the first observer. According to the EP, we could
> repeat the experiment in an accelerating rocket in empty space and get
> the same result. Interestingly, there's an absence of curvature in the
> accelerating rocket, yet there's still a redshift of the signal. How
> does Schild explain that?

That's the whole POINT of the equivalence principle, that the effects of
gravitation ("curvature") can mimick those of acceleration ("absence of
curvature") and vice versa WITHIN A SMALL ENOUGH SPACE. Yes, I can
distinguish them if my measurements are precise enough. If we couldn't
distinguish them AT ALL, then there probably wouldn't be an EP since,
again if we couldn't distinguish them AT ALL, then we would think of
them as the same phenomenon.

> > Again, nothing logically inconsistent. In a "uniform gravitational
> > field", which refers to the space-time solution generated by the
> > presence of an infinite, uniform sheet of mass, light rays curve only
> > for an accelerated observer.
>
> That statement is consistent with GR, but inconsistent with the EP.
> According to the EP, light rays would curve for an observer in a rocket
> at rest on the infinite sheet of mass, just as they would curve for an
> observer at rest in a rocket accelerating in empty space. That's why I
> claim that GR is logically inconsistent with the EP.

Your claim is based on a serious misunderstanding. You seem to be
confusing the words "curve" and "deflect". The example mentioned above
is just the standard example of the EP with the difference that the area
over which the gravitational field can be regarded as uniform is
arbitrarily large. You need to understand the concept of a limit.

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
> In article <p8mdnRIMfoS-AuvUnZ2dnUVZ_qjinZ2d@posted.docknet>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>Schild's argument sites the redshift of an electromagnetic signal
>>emitted from an observer on the surface of the earth to a second
>>observer located above the first observer. According to the EP, we could
>>repeat the experiment in an accelerating rocket in empty space and get
>>the same result. Interestingly, there's an absence of curvature in the
>>accelerating rocket, yet there's still a redshift of the signal. How
>>does Schild explain that?
>
> That's the whole POINT of the equivalence principle, that the effects of
> gravitation ("curvature") can mimick those of acceleration ("absence of
> curvature") and vice versa WITHIN A SMALL ENOUGH SPACE.

Take that "SMALL ENOUGH SPACE", in which the EP holds true exactly, and
blow it up to infinity, so that it holds true _exactly_. To do this (in
theory), you can use an infinite uniform sheet of mass, to create the
uniform gravitational field.

Now take a rocket and place it at rest on the sheet of mass. That will
be the gravitational part of the EP (Part 1). Then, for the acceleration
part (part 2), place a uniformly accelerating rocket in empty space so
that its acceleration exactly matches the (uniform) acceleration of
gravity inside the rocket in part 1. Project a beam of light across the
interior of the rocket in an initially horizontal direction, in both
part 1 and part 2.

You would apparently agree that in the "SMALL ENOUGH SPACE", the EP
holds true. Then you must also agree that it should hold true in the
infinite case, since we still have the same conditions within the
rocket, as in the "SMALL ENOUGH SPACE" case.

The behavior of the beam within the rocket, according to the EP, should
be the same in part 1 and part 2. That is, the beam will curve in both
parts 1 and 2, according to the EP (assume that the interior of the
rocket is wide enough that the curvature of the beam is measurable).

If the EP and GR are consistent, since the conditions within the rocket
in the infinite case are identical with the conditions within the rocket
in the "SMALL ENOUGH SPACE" case, then GR should predict that the beam
will curve in part 1. It doesn't. Thus, the EP and GR are logically
inconsistent.

> Yes, I can
> distinguish them if my measurements are precise enough. If we couldn't
> distinguish them AT ALL, then there probably wouldn't be an EP since,
> again if we couldn't distinguish them AT ALL, then we would think of
> them as the same phenomenon.

That's my point. They _are_ the same phenomenon. To prove it, put an
accelerometer on the floor of the rocket. It'll tell you that the rocket
is accelerating in both parts 1 and 2. Like a pilot flying in the fog,
trust your instruments!

>>>Again, nothing logically inconsistent. In a "uniform gravitational
>>>field", which refers to the space-time solution generated by the
>>>presence of an infinite, uniform sheet of mass, light rays curve only
>>>for an accelerated observer.
>>
>>That statement is consistent with GR, but inconsistent with the EP.
>>According to the EP, light rays would curve for an observer in a rocket
>>at rest on the infinite sheet of mass, just as they would curve for an
>>observer at rest in a rocket accelerating in empty space. That's why I
>>claim that GR is logically inconsistent with the EP.
>
> Your claim is based on a serious misunderstanding. You seem to be
> confusing the words "curve" and "deflect".

The term "deflect" implies (to me) a force acting on the beam. I don't
believe there are any forces acting on the beam in either part 1 or part 2.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Phillip Helbig---remove CLOTHES to reply]

In article <_cydnWofr9p3u-TUnZ2dnUVZ_jSdnZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> Take that "SMALL ENOUGH SPACE", in which the EP holds true exactly, and
> blow it up to infinity, so that it holds true _exactly_. To do this (in
> theory), you can use an infinite uniform sheet of mass, to create the
> uniform gravitational field.

The last part is OK (except that in practice an infinite uniform sheet
of mass does not exist, nor can one make an arbitrarily large uniform
sheet of mass). As for the first part, the EP holds EXACTLY only in an
INFINITELY SMALL space---that's what is meant by a limit. It's not
clear what happens when you blow up something infinitely small to
infinity.

> Now take a rocket and place it at rest on the sheet of mass. That will
> be the gravitational part of the EP (Part 1). Then, for the acceleration
> part (part 2), place a uniformly accelerating rocket in empty space so
> that its acceleration exactly matches the (uniform) acceleration of
> gravity inside the rocket in part 1. Project a beam of light across the
> interior of the rocket in an initially horizontal direction, in both
> part 1 and part 2.

OK. Again, this is no different than the standard EP explanation,
except that normally the region we study is "sufficiently small".

> You would apparently agree that in the "SMALL ENOUGH SPACE", the EP
> holds true. Then you must also agree that it should hold true in the
> infinite case, since we still have the same conditions within the
> rocket, as in the "SMALL ENOUGH SPACE" case.

Yes.

> The behavior of the beam within the rocket, according to the EP, should
> be the same in part 1 and part 2.

Yes.

> That is, the beam will curve in both
> parts 1 and 2, according to the EP (assume that the interior of the
> rocket is wide enough that the curvature of the beam is measurable).
>
> If the EP and GR are consistent, since the conditions within the rocket
> in the infinite case are identical with the conditions within the rocket
> in the "SMALL ENOUGH SPACE" case, then GR should predict that the beam
> will curve in part 1. It doesn't. Thus, the EP and GR are logically
> inconsistent.

To be clear, "curve" means that the beam is not just deflected downward,
but follows a curved path. In other words, nothing about the curvature
of space or space-time here, just what we can see the beam doing
(imagine the rocket filled with dusty air or smoke).

You're saying that in case 1, the beam will not be curved, merely be
deflected downward, while in case 2 it will be curved?

I just want to make sure we agree on the descriptions before continuing.

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
> In article <_cydnWofr9p3u-TUnZ2dnUVZ_jSdnZ2d@posted.docknet>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>That is, the beam will curve in both
>>parts 1 and 2, according to the EP (assume that the interior of the
>>rocket is wide enough that the curvature of the beam is measurable).
>>
>>If the EP and GR are consistent, since the conditions within the rocket
>>in the infinite case are identical with the conditions within the rocket
>>in the "SMALL ENOUGH SPACE" case, then GR should predict that the beam
>>will curve in part 1. It doesn't. Thus, the EP and GR are logically
>>inconsistent.
>
> To be clear, "curve" means that the beam is not just deflected downward,
> but follows a curved path. In other words, nothing about the curvature
> of space or space-time here, just what we can see the beam doing
> (imagine the rocket filled with dusty air or smoke).

Right.

> You're saying that in case 1, the beam will not be curved, merely be
> deflected downward, while in case 2 it will be curved?

No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
the EP. But according to GR, the beam will neither be curved nor
deflected downward, in part 1.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Phillip Helbig---remove CLOTHES to reply]

In article <SsednZq3kdk2bObUnZ2dnUVZ_hqdnZ2d@posted.docknet>, David
Rutherford <drutherford@softcom.net> writes:

> Phillip Helbig---remove CLOTHES to reply wrote:
> > In article <_cydnWofr9p3u-TUnZ2dnUVZ_jSdnZ2d@posted.docknet>, David
> > Rutherford <drutherford@softcom.net> writes:
> >
> >>That is, the beam will curve in both
> >>parts 1 and 2, according to the EP (assume that the interior of the
> >>rocket is wide enough that the curvature of the beam is measurable).
> >>
> >>If the EP and GR are consistent, since the conditions within the rocket
> >>in the infinite case are identical with the conditions within the rocket
> >>in the "SMALL ENOUGH SPACE" case, then GR should predict that the beam
> >>will curve in part 1. It doesn't. Thus, the EP and GR are logically
> >>inconsistent.
> >
> > To be clear, "curve" means that the beam is not just deflected downward,
> > but follows a curved path. In other words, nothing about the curvature
> > of space or space-time here, just what we can see the beam doing
> > (imagine the rocket filled with dusty air or smoke).
>
> Right.
>
> > You're saying that in case 1, the beam will not be curved, merely be
> > deflected downward, while in case 2 it will be curved?
>
> No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
> the EP. But according to GR, the beam will neither be curved nor
> deflected downward, in part 1.

Of course, since a uniform gravitational field doesn't exist in nature,
we can't solve this problem by observation.

Repeating a portion of the quoted text above:

> No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
> the EP.

I think we all agree on this point.

> But according to GR, the beam will neither be curved nor
> deflected downward, in part 1.

Can you give some evidence to back this up? You say "according to GR",
so you should present your evidence in the language of GR.

Most texts discuss in detail the opposite case, e.g. a rocket in free
fall in a gravitational field is the same as one "far enough away" from
any mass. But it should be possible to compare the accelerating rocket
and the rocket resting on the surface of the Earth, or within a
gravitational field which is uniform at least over the extent of the
rocket.

We can think of the uniform gravitational field as arising by making a
mass bigger and bigger but at the same time increasing our distance from
it, so that the acceleration remains constant. Observational
gravitational lensing is concerned with the deflection of beams of light
on scales from the radius of a star (the classic eclipse observation) to
deflection by galaxy clusters, several orders of magnitude larger. The
angle of deflection is that predicted by GR, using the mass of the
deflector and the impact parameter. If we imagine increasing the mass
and distance to approximate the uniform gravitational field, at what
point, in your view, does the deflection cease to exist? It doesn't
appear to do so gradually since we don't see this effect over the range
of scale mentioned above.

I think I see what you are saying: GR attributes the deflection to the
curvature of spacetime, but this curvature of spacetime doesn't exist in
a uniform gravitational field, since we don't observe tidal effects.
Right?

I'm not sure to what extent the concept of a limit is essential to this
discussion. The uniform gravitational field exists only as a limit.
Perhaps it is uniform in this limit in the sense of producing no tidal
effects, but not in the sense of not producing deflection. Think of the
standard way to calculate the slope of a curve. On the one hand, we
assume that our differential quantity is zero (or take that limit) in
order to calculate the derivative at a particular point. On the other
hand, it appears in the denominator and division by zero is not
meaningful. (I hear the ghost of Bishop Berkeley approaching, muttering
something about the ghosts of departed quantities.) The limit of
sin(x)/x is 1 as x approaches zero, even though both quantities are zero
at that point.

I hope I understand your claim. My claim is that, although it doesn't
appear to do so at first (or even second) glance, your claim that GR
predicts NO DEFLECTION in the case of a uniform gravitational field is
wrong. I suspect it has something to do with subtleties involving
limits. Maybe some GR expert (I am not one) can chime in here (at least
a few are regular readers of the newsgroup).

[David Rutherford]

Phillip Helbig---remove CLOTHES to reply wrote:
> In article <SsednZq3kdk2bObUnZ2dnUVZ_hqdnZ2d@posted.docknet>, David
> Rutherford <drutherford@softcom.net> writes:
>
>>>You're saying that in case 1, the beam will not be curved, merely be
>>>deflected downward, while in case 2 it will be curved?
>>
>>No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
>>the EP. But according to GR, the beam will neither be curved nor
>>deflected downward, in part 1.
>
> Of course, since a uniform gravitational field doesn't exist in nature,
> we can't solve this problem by observation.

We don't need observation to solve this problem. All we need is logic to
show that the EP and GR are _logically_ inconsistent. It doesn't matter
which one, if either, makes the correct predictions, for the problem I'm
posing, here.

> Repeating a portion of the quoted text above:
>
>>No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
>>the EP.
>
> I think we all agree on this point.
>
>>But according to GR, the beam will neither be curved nor
>>deflected downward, in part 1.
>
> Can you give some evidence to back this up? You say "according to GR",
> so you should present your evidence in the language of GR.

For a uniform gravitational field, as in this case, there is no
curvature of spacetime. And, according to GR, there is no curvature of
the paths of free-falling bodies or light without the curvature of
spacetime.

> I think I see what you are saying: GR attributes the deflection to the
> curvature of spacetime, but this curvature of spacetime doesn't exist in
> a uniform gravitational field, since we don't observe tidal effects.
> Right?

Right, there are no tidal effects, in a uniform gravitational field.
Therefore, no spacetime curvature. Therefore, no `deflection' (I would
say no curvature of the path).

> I'm not sure to what extent the concept of a limit is essential to this
> discussion. The uniform gravitational field exists only as a limit.
> Perhaps it is uniform in this limit in the sense of producing no tidal
> effects, but not in the sense of not producing deflection.

What produces the `deflection', in the limit according to GR, if there
are no tidal effects and no spacetime curvature?

> I hope I understand your claim. My claim is that, although it doesn't
> appear to do so at first (or even second) glance, your claim that GR
> predicts NO DEFLECTION in the case of a uniform gravitational field is
> wrong. I suspect it has something to do with subtleties involving
> limits. Maybe some GR expert (I am not one) can chime in here (at least
> a few are regular readers of the newsgroup).

You really don't expect them to agree with _me_, do you :)?

[Moderator's note: I think Dave's claim is now clear (at least to me).
Either it is right or wrong. If right, experts, then state your
agreement. If wrong, explain. To some extent, the situation might be
comparable to some issues involving the redshift, expansion, distances
etc in cosmology---cosmologists get the stuff right, but popular
expositions, even by cosmologists, are sometimes at best confusing and
at worst wrong. -P.H.]

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Tom Roberts]

David Rutherford wrote:
> No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
> the EP. But according to GR, the beam will neither be curved nor
> deflected downward, in part 1.

You have a rather strange notion of what GR predicts for this case [#].
If your notion were correct, then GR would easily be refuted by throwing
baseballs here on earth -- a baseball park is small enough that the
variation in gravity within it is negligible (for measurement accuracy
~feet) -- equivalently the earth can be considered to be an infinite
flat plate to such accuracy. Yet baseballs quite clearly follow curved
paths. GR predicts precisely that, and is not refuted so easily.

The same applies to your light beams.

[#] The deflection of geodesics depends on the connection,
not the curvature. The vacuum regions of the domain wall
manifolds have zero curvature, but a non-Euclidean
connection that causes geodesics to be deflected.


You said:
> Take that "SMALL ENOUGH SPACE", in which the EP holds true exactly, and
> blow it up to infinity, so that it holds true _exactly_.

This cannot be done, because there is no finite region in which the EP
"holds true _exactly_".

Bottom line: The EP is a statement about the equivalence of acceleration
and gravity in a region whose size depends on one's measurement accuracy
-- it claims YOU WILL NOT MEASURE A DIFFERENCE, not that there is no
difference. As others have pointed out, there quite clearly is a
difference (e.g. when one is allowed to look outside the elevator).


Tom Roberts

[Phillip Helbig---remove CLOTHES to reply]

In article <glijik$vdm$1@online.de>, helbig@astro.multiCLOTHESvax.de
(Phillip Helbig---remove CLOTHES to reply) writes:

> I hope I understand your claim. My claim is that, although it doesn't
> appear to do so at first (or even second) glance, your claim that GR
> predicts NO DEFLECTION in the case of a uniform gravitational field is
> wrong. I suspect it has something to do with subtleties involving
> limits. Maybe some GR expert (I am not one) can chime in here (at least
> a few are regular readers of the newsgroup).

One more thing before the experts chime in. At

http://www.mathpages.com/rr/s5-06/5-06.htm

there is a nice text on the EP which also addresses some common claims
that it is wrong.

There is a lot of interesting stuff, much relevant to physics (there is
an entire physics section as well) at

[Phillip Helbig---remove CLOTHES to reply]

[[Mod. note -- This is the correct version of this article. I mistakenly
posted a truncated version of it yesterday as article
<glmvep$nv6$1@online.de>
with the final url missing. I apologise to all concerned for the mixup.
-- jt]]

In article <glijik$vdm$1@online.de>, helbig@astro.multiCLOTHESvax.de
(Phillip Helbig---remove CLOTHES to reply) writes:

> I hope I understand your claim. My claim is that, although it doesn't
> appear to do so at first (or even second) glance, your claim that GR
> predicts NO DEFLECTION in the case of a uniform gravitational field is
> wrong. I suspect it has something to do with subtleties involving
> limits. Maybe some GR expert (I am not one) can chime in here (at least
> a few are regular readers of the newsgroup).

One more thing before the experts chime in. At

http://www.mathpages.com/rr/s5-06/5-06.htm

there is a nice text on the EP which also addresses some common claims
that it is wrong.

There is a lot of interesting stuff, much relevant to physics (there is
an entire physics section as well) at

http://www.mathpages.com/

[David Rutherford]

Tom Roberts wrote:
> David Rutherford wrote:
>
>>No, the beam _will_ be curved, in _both_ part 1 and part 2, according to
>>the EP. But according to GR, the beam will neither be curved nor
>>deflected downward, in part 1.
>
> You have a rather strange notion of what GR predicts for this case [#].
> If your notion were correct, then GR would easily be refuted by throwing
> baseballs here on earth -- a baseball park is small enough that the
> variation in gravity within it is negligible (for measurement accuracy
> ~feet) -- equivalently the earth can be considered to be an infinite
> flat plate to such accuracy. Yet baseballs quite clearly follow curved
> paths. GR predicts precisely that, and is not refuted so easily.
>
> The same applies to your light beams.
>
> [#] The deflection of geodesics depends on the connection,
> not the curvature. The vacuum regions of the domain wall
> manifolds have zero curvature, but a non-Euclidean
> connection that causes geodesics to be deflected.

Outside (and inside) the accelerating rocket, in empty space (part 2),
spacetime is Euclidean (you would probably say Minkowskian). Hence, all
of the Christoffel symbols (connections) are zero, since the g_uv are
all constants (the derivative of a constant is zero). Therefore, all of
the components of the curvature tensor will, also, be zero. But, if the
components of a tensor in any given coordinate system are all zero, then
its components in any other coordinate system will also be zero. This
includes whatever coordinate system is used by observers `at rest'
within the rocket, whether they think they are in an accelerating rocket
or at rest in a gravitational field. Thus, by the EP (and tensor
analysis), all of the Christoffel symbols (and the curvature tensor) in
part 1 must, also, vanish.

[ Mod. note: The above argument works for the cuvature tensor, but not
for Christoffel symbols. The latter are not tensors and can be non-zero
in one coordinate system, while they are identically zero in another
one. This is immediately evidenced by Minkowski space and the difference
in the Christoffel symbols associated to inertial and rotating,
or uniformly accelerating, coordinate systems. And it is the
Christoffel symbols that determine the trajectories of light rays
and baseballs in a given coordinate system. -ik ]

Therefore, according to [#], since "The deflection of geodesics depends
on the connection", there will be no curvature (or deflection) of the
paths of baseballs or light beams for an infinite uniform sheet of mass
(part 1), according to GR. But, the EP requires that the results
_must_be_ the same (curvature of the paths of baseballs and light beams)
for any region in which the EP holds true exactly, as in parts 1 and 2.

The description of the paths of baseballs and light beams in the EP and
GR, over any region in which the EP holds true exactly, must agree. But,
since in GR, the paths are _straight_ for vanishing connections and
vanishing curvature tensor, in part 1, and the paths are _curved_
(deflected), in the EP, the EP and GR give different descriptions and
are, thus, logically inconsistent.

> You said:
>
>>Take that "SMALL ENOUGH SPACE", in which the EP holds true exactly, and
>>blow it up to infinity, so that it holds true _exactly_.
>
>
> This cannot be done, because there is no finite region in which the EP
> "holds true _exactly_".

It holds true exactly in my example.

> Bottom line: The EP is a statement about the equivalence of acceleration
> and gravity in a region whose size depends on one's measurement accuracy
> -- it claims YOU WILL NOT MEASURE A DIFFERENCE, not that there is no
> difference. As others have pointed out, there quite clearly is a
> difference (e.g. when one is allowed to look outside the elevator).

It isn't necessary to look outside the elevator to show that the EP and
GR predict different results.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Tom Roberts]

David Rutherford wrote:
> Outside (and inside) the accelerating rocket, in empty space (part 2),
> spacetime is Euclidean (you would probably say Minkowskian). Hence, all
> of the Christoffel symbols (connections) are zero,

Only in an inertial frame. In coordinates fixed to the accelerating
rocket they are nonzero. Indeed, they are nonzero in precisely the
manner required to have the light beam bend (just like it does in a
gravitational field). As the moderator said, the connection is not a
tensor, and its components are explicitly coordinate dependent.


> [...] for any region in which the EP holds true exactly, [...]

I repeat: the EP is never exact in the real world, it is a statement
about a certain APPROXIMATION in which gravitation and acceleration are
indistinguishable within measurement resolutions. Only in an artificial
situation can there be an "exactly" flat gravitational field that can
"exactly" mimic an acceleration. But physics is not about such
unphysical situations, and is never "exact" in the real world.


> It isn't necessary to look outside the elevator to show that the EP and
> GR predict different results.

Well, this is not really "EP vs GR" -- the EP is an inherent part of GR.

The EP is really a statement about first order effects, and if you can
measure higher orders (e.g. tidal effects) then you can distinguish
gravity from acceleration inside the elevator. But in a small enough
elevator over a small enough time interval (related to your measurement
accuracy and the local geometry of the manifold), you won't be able to
MEASURE those higher-order effects, and won't MEASURE any difference --
THAT is what the equivalence principle says.


Tom Roberts

[David Rutherford]

David Rutherford wrote:
> Outside (and inside) the accelerating rocket, in empty space (part 2),
> spacetime is Euclidean (you would probably say Minkowskian). Hence, all
> of the Christoffel symbols (connections) are zero, since the g_uv are
> all constants (the derivative of a constant is zero). Therefore, all of
> the components of the curvature tensor will, also, be zero. But, if the
> components of a tensor in any given coordinate system are all zero, then
> its components in any other coordinate system will also be zero. This
> includes whatever coordinate system is used by observers `at rest'
> within the rocket, whether they think they are in an accelerating rocket
> or at rest in a gravitational field. Thus, by the EP (and tensor
> analysis), all of the Christoffel symbols (and the curvature tensor) in
> part 1 must, also, vanish.
>
> [ Mod. note: The above argument works for the cuvature tensor, but not
> for Christoffel symbols. The latter are not tensors and can be non-zero
> in one coordinate system, while they are identically zero in another
> one. This is immediately evidenced by Minkowski space and the difference
> in the Christoffel symbols associated to inertial and rotating,
> or uniformly accelerating, coordinate systems. And it is the
> Christoffel symbols that determine the trajectories of light rays
> and baseballs in a given coordinate system. -ik ]

If the components of the curvature tensor vanish in a given coordinate
system, then they must vanish in all coordinate systems, even
accelerating ones.

The components of the curvature tensor in the coordinate system of the
observers at rest inside the accelerating rocket of part 2 must vanish,
since they vanish in the (flat) coordinate system of the observers
outside the rocket. Therefore, observers inside the rocket determine
their spacetime to be flat, not curved, whether they think they are in
an accelerating rocket or at rest in a uniform gravitational field.

What I don't understand is how, in GR, the trajectories of baseballs or
light rays can be curved in the _flat_ spacetime of the rocket, even if
the Christoffel symbols are nonzero. I've always thought that, according
to GR, the trajectories of free-falling bodies and light rays don't
curve in flat spacetime.

[[Mod. note -- I think the problem here comes from mixing up two
different senses of the word "curve":

If you thought that, according to GR, the coordinate trajectories
(i.e., x = x(t), y = y(t), z = z(t)) of free-falling bodies and light
rays don't curve in flat spacetime, then I'm afraid you were mistaken.
As Igor Khavkine said, in GR coordinate trajectories are determined
by the Christoffel symbols, not by the curvature tensor, and the
Christoffel symbols (which are *not* tensors) can be (are) nonzero
in noninertial reference frames in flat (Minkowski) spacetime. The
result is that coordinate trajectories *do* curve in noninertial
frames (e.g., frames which are rotating and/or accelerating with
respect to an inertial frame) in *flat* spacetime.

On the other hand, maybe you were thinking of whether trajectories
"look" curved. The problem there is that we need a precise operational
definition of when a trajectory "looks" curved. Our usual notion
of this is to compare the trajectory with something that looks straight
-- but then what do we mean by "looks straight"? The classic "sight
along it and see if it looks straight" amounts to *defining* the path
of a light beam as "looks straight". But then (by the definition
we've just made) the path of a light beam can never "look curved",
even right next to a black hole! It's easy to get tied up in knots
this way!

The usual solution in GR is avoid the "looks curved" approach,
precisely because that leads to all sorts of trouble, and instead
stick to coordinate trajectories.

I'll expand on this a bit in a longer posting tomorrow...
-- jt]]

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"

[Phillip Helbig---remove CLOTHES to reply]

In article <Hq5hl.9842$8_3.4013@flpi147.ffdc.sbc.com>, Tom Roberts
<tjroberts137@sbcglobal.net> writes:

> David Rutherford wrote:
> > Outside (and inside) the accelerating rocket, in empty space (part 2),
> > spacetime is Euclidean (you would probably say Minkowskian). Hence, all
> > of the Christoffel symbols (connections) are zero,
>
> Only in an inertial frame. In coordinates fixed to the accelerating
> rocket they are nonzero. Indeed, they are nonzero in precisely the
> manner required to have the light beam bend (just like it does in a
> gravitational field). As the moderator said, the connection is not a
> tensor, and its components are explicitly coordinate dependent.

I think this essentially addresses the original claim. In other words,
EP and GR do not predict different things.

>
> > [...] for any region in which the EP holds true exactly, [...]
>
> I repeat: the EP is never exact in the real world, it is a statement
> about a certain APPROXIMATION in which gravitation and acceleration are
> indistinguishable within measurement resolutions. Only in an artificial
> situation can there be an "exactly" flat gravitational field that can
> "exactly" mimic an acceleration. But physics is not about such
> unphysical situations, and is never "exact" in the real world.

This is a red herring. We can make the gravitational field as uniform
as we like. We can even imagine a completely uniform one, as we can
imagine other idealisations.

I think Rutherford's claim is that GR states that the bending of the
light in a gravitational field REQUIRES there to be MEASURABLE tidal
effects, thus there should be no bending in a uniform gravitational
field. To counter his claim, one needs to show explicitly that it is
wrong.

[Tom Roberts]

David Rutherford wrote:
> What I don't understand is how, in GR, the trajectories of baseballs or
> light rays can be curved in the _flat_ spacetime of the rocket, even if
> the Christoffel symbols are nonzero.

In Minkowski spacetime: Consider a light pulse emitted from the side of
the accelerating elevator. To an inertial observer this light pulse must
travel in a uniform straight line. To the accelerated observer inside
the elevator it must therefore "curve", because the elevator observer's
coordinate system is accelerated. This is required for consistency,
because the choice of coordinate system cannot possibly affect the
physical trajectory of the light pulse; the different coordinates are
merely different descriptions of the same light pulse's trajectory.

The Christoffel symbols (connection components) in the inertial
coordinates are zero, and the light pulse travels in a uniform straight
line relative to those coordinates. The Christoffel symbols in the
accelerated elevator coordinates are nonzero, and the light pulse
travels in a curved line relative to those coordinates. For both
coordinate systems the geodesic equation is satisfied, but the values of
all the components and terms in that equation are different for the two
coordinate systems (there are lots of zeros in the equation expressed in
inertial coordinates; not so for the accelerated ones).

Note that the components of the Riemann curvature tensor are zero in
both coordinate systems. In the inertial coordinates this is so because
the connection coefficients are zero (and Riemann's components are a
certain combination of their derivatives). In the accelerating
coordinates this is so because the combinations of derivatives of the
connection components cancel each other.

In general, one cannot unambiguously say "this trajectory is curved" --
that is a coordinate-dependent statement. IOW there is no tensor that
specifies the "curvature" of a trajectory. All one can say is that the
light pulse follows a trajectory that is a null geodesic (this is a
coordinate-independent statement).

But there is a tensor that describes the curvature of the manifold, so
one can unambiguously discuss that.

In an N-dimensional manifold the Riemann curvature tensor
has N^2(N^2−1)/12 independent components. For N=1 that is
zero. So "curvature" has no definitive meaning for a 1-d
manifold, such as a trajectory.


Tom Roberts

[Igor Khavkine]

On Feb 2, 10:50 am, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
remove CLOTHES to reply) wrote:
> [...] We can make the gravitational field as uniform
> as we like. We can even imagine a completely uniform one, as we can
> imagine other idealisations.

I wonder if one can substantiate this statement with something
approaching mathematical precision.

A quick search of the literature revealed that domain-wall vacuum
solutions of Einsten's equations obtained by symmetry reduction have
non-trivial curvature (the Kretschmann scalar is nowhere zero) [1]. I
think the presence of smooth curvature is incompatible with any
reasonable definition of "uniform gravitational field".

Is the limit of ever thinner and denser sheets of mass described by
one of the metrics from [1] or will it be different? Otherwise, the
hypothetical construction of a uniform gravitational field used in
this thread doesn't work

[1] B. Linet (1985) IJTP v.24, p.1159
Static planar domain wall in general relativity with a
cosmological constant
http://dx.doi.org/10.1007/BF00670330

Igor

[Cephalobus_alienus@comcast.net]

On Jan 19, 4:07*pm, Uncle Al <Uncle...@hate.spam.net> wrote:

> Since theory and prior observation demand the vacuum is isotropic,
> there is no reason to look at a novel test. *Theory also demanded the
> Weak interaction was parity conserving. *Yangand Lee wasted funding
> and equipment being pariahs, for there was no reason to look.

You have been corrected on this point numerous times. Yang and
Lee were not "pariahs" as you state; nor did they waste equipment,
being theorists, not experimentalists.

Doubt concerning the principle of parity conservation began with
the "theta-tau puzzle". In 1949, C.F. Powell discovered a
particle, the tau meson, which decayed into three pions. At about
the same time, he observed another particle, the theta meson,
which disintegrated into two pions. By every test, the tau meson
and the theta meson had identical characteristics (mass, lifetime)
yet the products had different total parity. Hence tau and theta
could not be the same particle...or could they?

Yang and Lee were only two of many physicists attending the 1956
ICHEC conference who were intensely interested in the theta-tau
puzzle. The idea that they brought to the conference was that
for certain particles, parity might be a state rather than an
intrinsic property. According to this notion, which they called
"parity doubling", the theta and tau mesons would be in fact the
same particle, but in different parity states.

After Yang and Lee's talk, Richard Feynman presented a notion
that Martin Block had suggested to him the day before: Perhaps
the solution to the puzzle was that parity was simply not a
conserved quantity. This notion was seconded by Eugene Wigner.
Wigner's concurrence was especially significant, because more
than two decades earlier, Wigner had been the key player in the
proof that parity is conserved in atomic transitions.

Yang and Lee returned to the University of Rochester excited by
the discussions they had at the ICHEC conference, and decided to
make a systematic study of all known experiments involving the
weak and strong interactions from the standpoint of whether the
experiments were supportive of parity conservation. Although they
found ample evidence for parity conservation in the strong
interaction, they were startled to discover that there was no
such evidence for the weak interaction. They wrote up their
conclusions in a paper appearing in the Oct. 1, 1956 issue of
Physical Review, and proposed several experiments.

Yang and Lee had close contact with many of the most important
experimental physicists of the time, among whom were C.S. Wu.
Long before the appearance of Yang and Lee's paper in print, Wu
had already assembled a team of cryogenics, vacuum, radiation
detection and electronics experts at the Cryogenics Physics Lab
of the NBS. Less than three months after the publication of Yang
and Lee's paper, the team observed the sought-after effect.

Trigg, G.L., "Disproof of a Conservation Law", in Landmark
Experiments in Twentieth Century Physics, pp 155-178 (1975)
http://focus.aps.org/story/v22/st19
http://nobelprize.org/nobel_prizes/physics/laureates/1957/yang-lecture.html

In post after post, you have repeatedly overemphasized the role
of the lone wolf experimentalist in scientific advancement, not
hesitating to distort history to buttress your fanciful
imaginings. Like the work that led to Special Relativity five
decades earlier, the thoughts that led to the notion of parity
violation were part of the intellectual Zeitgeist of the time.
This does not diminish Yang and Lee's contribution. But you do
them, and yourself, no honor by falsifying history.

Yes, "somebody should look" at possible geometric parity effects
in gravity. But unlike Madame Wu, you have -absolutely no-
theoretical framework to back up the experiment that you propose.

Jerry

[Daryl McCullough]

In article <gm7mo7$spt$1@online.de>, Phillip Helbig---remove CLOTHES to reply
says...

>What about other limits? We can make a spherical mass (or a point mass)
>more and more massive, and move ourselves correspondingly farther away.
>We thus approach a uniform field in a given region with a well-studied
>gravitational field.
>
>Is the physical possibility of such a uniform field necessary for the
>discussion, though?

I wrote a post making the following point, but it never appeared, for
some reason.

The "gravitational field" inside an accelerating rocket is *not*
uniform. It falls off linearly with the distance from the rear of
the rocket.

If the acceleration is g at the rear of the rocket, then
the acceleration at the front of the rocket will be only
g/(1+gL/c^2), where L is the length of the rocket.

[ Mod. note: As this claim does not appear to be self-evident,
I hope some justification will follow in another post. -ik ]

--
Daryl McCullough
Ithaca, NY

[David Rutherford]

Tom Roberts wrote:
> David Rutherford wrote:
>
>>What I don't understand is how, in GR, the trajectories of baseballs or
>>light rays can be curved in the _flat_ spacetime of the rocket, even if
>>the Christoffel symbols are nonzero.
>
>
> In Minkowski spacetime: Consider a light pulse emitted from the side of
> the accelerating elevator. To an inertial observer this light pulse must
> travel in a uniform straight line. To the accelerated observer inside
> the elevator it must therefore "curve", because the elevator observer's
> coordinate system is accelerated. This is required for consistency,
> because the choice of coordinate system cannot possibly affect the
> physical trajectory of the light pulse; the different coordinates are
> merely different descriptions of the same light pulse's trajectory.
>
> The Christoffel symbols (connection components) in the inertial
> coordinates are zero, and the light pulse travels in a uniform straight
> line relative to those coordinates. The Christoffel symbols in the
> accelerated elevator coordinates are nonzero,

Can you please post the calculations that show this? What are the g_uv
for the observers in the accelerated elevator?

> and the light pulse
> travels in a curved line relative to those coordinates.

Again, can you show me your calculations?

> For both
> coordinate systems the geodesic equation is satisfied, but the values of
> all the components and terms in that equation are different for the two
> coordinate systems (there are lots of zeros in the equation expressed in
> inertial coordinates; not so for the accelerated ones).
>
> Note that the components of the Riemann curvature tensor are zero in
> both coordinate systems. In the inertial coordinates this is so because
> the connection coefficients are zero (and Riemann's components are a
> certain combination of their derivatives). In the accelerating
> coordinates this is so because the combinations of derivatives of the
> connection components cancel each other.

Can you please post your calculations?

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[Daryl McCullough]

Daryl McCullough says...

>If the acceleration is g at the rear of the rocket, then
>the acceleration at the front of the rocket will be only
>g/(1+gL/c^2), where L is the length of the rocket.
>
>[ Mod. note: As this claim does not appear to be self-evident,
> I hope some justification will follow in another post. -ik ]

Okay, sorry.

I'm going to give the details using an inertial coordinate
system in which the rocket is at rest initially. Let the
rocket have length L, and let's label various points along
the rocket by the label s. s runs from 0 (the rear of the
rocket) to L (the front of the rocket).

Let x(s,t) = the position of point s on the rocket at time t.
(So x(s,0) = s)

Then if every point on the rocket is undergoing constant
proper acceleration, then x(s,t) will be given by the following
formula: (I'm skipping the derivation, but I could give it in another
post if there is any interest) (where the dependence on s and t
of B and x is implicit):

1. x = s - B + square-root(B^2 + t^2)

where I'm using units where c=1, and where B(s)
is 1/g(s), where g(s) is the proper acceleration
of point s.

Without seeing a
derivation, you can easily see that at t=0, x=l,
and for t << B, we have, approximately

x = s - B + B (1 + 1/2 (t/B)^2)
= s + 1/2 t^2/B

We can see that for t << B, this approximately
agrees with the nonrelativistic expression

x = s + 1/2 g t^2

Note we're allowing B (and therefore g)
to depend on s, but not t. The goal is to find
the dependence of B (and therefore g) on s. So
we want to compute @B/@s.

Now, using equation 1, we derive the following
expressions for the partial derivatives of x:

2. @x/@s = 1 + @B/@s (-1 + B/square-root(B^2 + t^2))
= 1 + @B/@s (-1 + B/(x-s+B))

(where I used equation 1 to rewrite the square-root
in the denominator)

3. @x/@t = t/square-root(B^2 + t^2)

The expression @x/@t is just the velocity of point s. In
terms of this velocity, we can derive the length contraction
factor 1/gamma = square-root(1-(@x/@t)^2):

4. 1/gamma = square-root(1 - t^2/(B^2 + t^2))
= B/square-root(B^2 + t^2)
= B/(x-s+B)

(where again I used equation 1 to rewrite the square-root)

The significance of the time-dilation factor is this:
Let s and s+ds be two different points on the rocket,
infinitesimally removed. The distance between these
points, as measured in the rocket's instantaneous
inertial reference frame will just be ds. To compute
the distance between these points, as measured in our
initial inertial reference system, we use the Lorentz
transformations. The result is the well-known length
contraction formula: (again, a more detailed derivation
is available, if anyone wants it):

(distance in initial rest frame)
= (distance in instantaneous rest frame)/gamma

In the initial inertial coordinate system, the
quantity on the left is just

x(s+ds,t) - x(s,t)

So we have:

x(s+ds,t) - x(s,t) = ds/gamma

or

5. @x/@s = 1/gamma = B/(x-s+B)

(where I used equation 4 to rewrite gamma).

Comparing equations 2 and 5 gives us the following:

6. B/(x-s+B) = 1 + @B/@s (-1 + B/(x-s+B))

This implies that

B/(x-s+B) - 1 = @B/@l (-1 + B/(x-s+B))

Dividing both sides by (-1 + B/(x-l+B)) gives:

7. @B/@s = 1

which can be immediately solved by

8. B = s + B_0

where B_0 is the value of B at s=0. Remembering that
B = 1/g gives us the following:

9. 1/g = s + 1/g_0

or

10. g = g_0/(1+g_0 s)

So, for the particular case s=L, we get

11. g = g_0/(1+ g_0 L)

An alternative derivation of the same result uses
the accelerated "Rindler" coordinates (X,T) which
are related to the inertial coordinates (x,t) through

x = X cosh(gT)
t = X sinh(gT)

The coordinates X,T have nonzero connection
coefficients, which can be used to compute
the proper acceleration of an object at "rest"
in the X-T coordinate system.

--
Daryl McCullough
Ithaca, NY

[Igor Khavkine]

On Feb 2, 10:50 am, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---
remove CLOTHES to reply) wrote:
> [...] We can make the gravitational field as uniform
> as we like. We can even imagine a completely uniform one, as we can
> imagine other idealisations.

I wonder if one can substantiate this statement with something
approaching mathematical precision.

A quick search of the literature revealed that domain-wall vacuum
solutions of Einsten's equations obtained by symmetry reduction have
non-trivial curvature (the Kretschmann scalar is nowhere zero) [1]. I
think the presence of smooth curvature is incompatible with any
reasonable definition of "uniform gravitational field".

Is the limit of ever thinner and denser sheets of mass described by
one of the metrics from [1] or will it be different? Otherwise, the
hypothetical construction of a uniform gravitational field used in
this thread doesn't work

[1] B. Linet (1985) IJTP v.24, p.1159
Static planar domain wall in general relativity with a
cosmological constant
http://dx.doi.org/10.1007/BF00670330

Igor

[David Rutherford]

Daryl McCullough wrote:

> Daryl McCullough says...
>
>
>>If the acceleration is g at the rear of the rocket, then
>>the acceleration at the front of the rocket will be only
>>g/(1+gL/c^2), where L is the length of the rocket.
>>
>>[ Mod. note: As this claim does not appear to be self-evident,
>> I hope some justification will follow in another post. -ik ]
>
>
> Okay, sorry.
>
> I'm going to give the details using an inertial coordinate
> system in which the rocket is at rest initially. Let the
> rocket have length L, and let's label various points along
> the rocket by the label s. s runs from 0 (the rear of the
> rocket) to L (the front of the rocket).
>
> Let x(s,t) = the position of point s on the rocket at time t.
> (So x(s,0) = s)
>
> Then if every point on the rocket is undergoing constant
> proper acceleration, then x(s,t) will be given by the following
> formula: (I'm skipping the derivation, but I could give it in another
> post if there is any interest) (where the dependence on s and t
> of B and x is implicit):
>
> 1. x = s - B + square-root(B^2 + t^2)

Yes, could you please post your derivation of this formula. Also, do you
know of anywhere (online, hopefully) that I can find Rindler's
derivation of the Rindler coordinates for a uniformly accelerated
observer in Minkowski spacetime? I did a Google search, but couldn't
find it. Thanks, Daryl.

[[Mod. note -- Wikipedia has a overview of Rindler coordinates,
http://en.wikipedia.org/wiki/Rindler_coordinates
For more details, chapter 6 of Misner, Thorne, and Wheeler is a good
place to start.
-- jt]]

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

[David Rutherford]

David Rutherford wrote:
>
> Yes, could you please post your derivation of this formula. Also, do you
> know of anywhere (online, hopefully) that I can find Rindler's
> derivation of the Rindler coordinates for a uniformly accelerated
> observer in Minkowski spacetime? I did a Google search, but couldn't
> find it. Thanks, Daryl.
>
> [[Mod. note -- Wikipedia has a overview of Rindler coordinates,
> http://en.wikipedia.org/wiki/Rindler_coordinates
> For more details, chapter 6 of Misner, Thorne, and Wheeler is a good
> place to start.
> -- jt]]

Thanks for the link, jt, but I'm looking for the _derivation_ of the
Rindler coordinates, not the coordinates themselves. Neither Wikipedia
nor MTW (or anywhere else I've looked) has Rindler's derivation of the
coordinates.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555

Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"