# Accelerating rocket/light beam question

5 views

### David Rutherford

Jan 3, 2009, 4:56:27 AM1/3/09
to
An observer in a uniformly accelerating rocket, far from any gravitating
bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
rocket in an unconscious state and wakes up after the rocket has
attained its uniform acceleration. Consequently, he doesn't know whether
he is accelerating or 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"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"

### Eric Gisse

Jan 3, 2009, 8:01:44 AM1/3/09
to
On Jan 3, 12:56 am, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket, far from any gravitating
> bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
> rocket in an unconscious state and wakes up after the rocket has
> attained its uniform acceleration. Consequently, he doesn't know whether
> he is accelerating or 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?

Either.

This is the fundamental meaning of the equivalence principle :
locally, you _can not_ tell the difference.

### David Rutherford

Jan 3, 2009, 1:31:18 PM1/3/09
to

Eric Gisse wrote:
> On Jan 3, 12:56 am, David Rutherford <drutherf...@softcom.net> wrote:
>
>>An observer in a uniformly accelerating rocket, far from any gravitating
>>bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
>>rocket in an unconscious state and wakes up after the rocket has
>>attained its uniform acceleration. Consequently, he doesn't know whether
>>he is accelerating or 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?
>
> Either.
>
> This is the fundamental meaning of the equivalence principle :
> locally, you _can not_ tell the difference.

What I meant by a uniform gravitational field was a field in which the
bending of the beam would match exactly the bending of the beam in the
accelerating rocket. Maybe it's created by a very large thick flat slab
of material so that the gravitational field is everywhere perpendicular
to the slab. In other words, the gravitational field would be such that
the bending of the beam is the same, in either case.

If both cases are equivalent, then shouldn't the reason for the bending
of the beam be the same, in both cases? That is, either the accelerating
rocket creates a gravitational field, which then causes the beam to
bend, or the rocket is accelerating as it sits `at rest' in the uniform
gravitational field.

This example is analogous to the problem in EM of the magnet and the
wire loop. In that problem, there are two scenarios, one in which the
magnet is moved through the loop, the other in which the loop is moved
over the magnet. These were seen to be two completely different
phenomenon until Einstein came along and said, essentially, that these
are the same thing, stupid (I'm not calling you stupid :))!

Why isn't this a similar situation.

### Oh No

Jan 3, 2009, 3:24:24 PM1/3/09
to
Thus spake David Rutherford <druth...@softcom.net>

>
>
>Eric Gisse wrote:
>> On Jan 3, 12:56 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>An observer in a uniformly accelerating rocket, far from any
>>>gravitating
>>>bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
>>>rocket in an unconscious state and wakes up after the rocket has
>>>attained its uniform acceleration. Consequently, he doesn't know whether
>>>he is accelerating or 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?
>> Either.
>> This is the fundamental meaning of the equivalence principle :
>> locally, you _can not_ tell the difference.
>
>What I meant by a uniform gravitational field was a field in which the
>bending of the beam would match exactly the bending of the beam in the
>accelerating rocket. Maybe it's created by a very large thick flat slab
>of material so that the gravitational field is everywhere perpendicular
>to the slab. In other words, the gravitational field would be such that
>the bending of the beam is the same, in either case.

>
>If both cases are equivalent, then shouldn't the reason for the bending
>of the beam be the same, in both cases?

It is. This is Einstein's equivalence principle.

>That is, either the accelerating rocket creates a gravitational field,
>which then causes the beam to bend, or the rocket is accelerating as it
>sits `at rest' in the uniform gravitational field.

>
>This example is analogous to the problem in EM of the magnet and the
>wire loop. In that problem, there are two scenarios, one in which the
>magnet is moved through the loop, the other in which the loop is moved
>over the magnet. These were seen to be two completely different
>phenomenon until Einstein came along and said, essentially, that these
>are the same thing, stupid (I'm not calling you stupid :))!

who exactly are you calling stupid? It would be stupid not to be able to
see it even after Einstein explained it.

>Why isn't this a similar situation.

It is. This insight lead eventually to the general theory of relativity,
and the geometrical structure of spacetime, through which we now
understand gravity.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

### David Rutherford

Jan 3, 2009, 4:25:02 PM1/3/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>

>>If both cases are equivalent, then shouldn't the reason for the bending
>>of the beam be the same, in both cases?
>
> It is. This is Einstein's equivalence principle.

Then what's the (one) reason in both cases?

>>This example is analogous to the problem in EM of the magnet and the
>>wire loop. In that problem, there are two scenarios, one in which the
>>magnet is moved through the loop, the other in which the loop is moved
>>over the magnet. These were seen to be two completely different
>>phenomenon until Einstein came along and said, essentially, that these
>>are the same thing, stupid (I'm not calling you stupid :))!
>
> who exactly are you calling stupid? It would be stupid not to be able to
> see it even after Einstein explained it.

I'm not calling anybody stupid. I'm saying that that's basically what
Einstein was saying about those who couldn't see it _before_ he
explained it.

>>Why isn't this a similar situation.
>
> It is. This insight lead eventually to the general theory of relativity,
> and the geometrical structure of spacetime, through which we now
> understand gravity.

I'm trying to show that we don't _really_ understand gravity, yet.

### Peter

Jan 3, 2009, 4:32:32 PM1/3/09
to

1) I would not call scenarios "analogous", where the one deals with
inertial systems and the other with non-inertial ones

2) According to Petry (http://siegfried-petry.de/page1.php), Einstein
has claimed asymmetries which may have been seen by others, but which
are not really in Maxwell's theory; moreover, the asymmetries claimed
by Einstein are not relativistic ones

Best wishes,
Peter

### Bob_for_short

Jan 3, 2009, 4:30:35 PM1/3/09
to
An accelerating reference frame is NOT equivalent to a gravitational
field, even locally.
In a gravitational field a falling charge radiates, in an accelerating
reference frame it does not. On the contrary, there a “resting” charge
radiates. If you look outside your rocket, you will see the whole
universe accelerating, so it is not at all equivalent to some “local”
gravity. You use a too poor experiment to make too general
conclusions.
Logunov’s articles and books on relativistic theory of gravity (RTG)
in arXiv, for example.
Bob.

### Eric Gisse

Jan 3, 2009, 5:32:16 PM1/3/09
to
wrote:

> An accelerating reference frame is NOT equivalent to a gravitational
> field, even locally.

Except it is. By explicit assumption.

> In a gravitational field a falling charge radiates, in an accelerating
> reference frame it does not. On the contrary, there a “resting” charge
> radiates. If you look outside your rocket, you will see the whole
> universe accelerating, so it is not at all equivalent to some “local”
> gravity. You use a too poor experiment to make too general
> conclusions.

What part of "locally" is not understood? You don't _GET_ to look
outside the rocket - the whole point is that locally, they are
indistinguishable. Naturally they are globally distinguishable.

### Oh No

Jan 3, 2009, 6:45:54 PM1/3/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>If both cases are equivalent, then shouldn't the reason for the
>>>bending
>>>of the beam be the same, in both cases?
>> It is. This is Einstein's equivalence principle.
>
>Then what's the (one) reason in both cases?

The reason is that a gravitational field is locally equivalent to an
accelerating reference frame.

>
>>>This example is analogous to the problem in EM of the magnet and the
>>>wire loop. In that problem, there are two scenarios, one in which the
>>>magnet is moved through the loop, the other in which the loop is moved
>>>over the magnet. These were seen to be two completely different
>>>phenomenon until Einstein came along and said, essentially, that these
>>>are the same thing, stupid (I'm not calling you stupid :))!
>> who exactly are you calling stupid? It would be stupid not to be
>>able to
>> see it even after Einstein explained it.
>
>I'm not calling anybody stupid. I'm saying that that's basically what
>Einstein was saying about those who couldn't see it _before_ he
>explained it.
>
>>>Why isn't this a similar situation.
>> It is. This insight lead eventually to the general theory of
>>relativity,
>> and the geometrical structure of spacetime, through which we now
>> understand gravity.
>
>I'm trying to show that we don't _really_ understand gravity, yet.
>

First you should master general relativity, then say what we do and
don't understand.

### Oh No

Jan 3, 2009, 6:53:38 PM1/3/09
to

>An accelerating reference frame is NOT equivalent to a gravitational
>field, even locally.
>In a gravitational field a falling charge radiates, in an accelerating
>reference frame it does not.

This isn't true.

>On the contrary, there a “resting” charge
>radiates. If you look outside your rocket, you will see the whole
>universe accelerating, so it is not at all equivalent to some “local”
>gravity. You use a too poor experiment to make too general
>conclusions.

You completely miss the point that in gtr physics is described with
local laws. Of course you can look out of the rocket, but what you see
is not local. That can also be described, but not in the same way.

>Logunov’s articles and books on relativistic theory of gravity (RTG)
>in arXiv, for example.

In so far as I can tell, Logunov is merely doing gtr in a particular
reference frame. The choice of frame appears to be Logunov's, not that
of nature.

### David Rutherford

Jan 4, 2009, 12:30:53 PM1/4/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>Oh No wrote:
>>
>>>Thus spake David Rutherford <druth...@softcom.net>
>>>
>>>>If both cases are equivalent, then shouldn't the reason for the
>>>>bending of the beam be the same, in both cases?
>>>
>>> It is. This is Einstein's equivalence principle.
>>
>>Then what's the (one) reason in both cases?
>
> The reason is that a gravitational field is locally equivalent to an
> accelerating reference frame.

If the two cases are equivalent, then the _reason_ for the curvature of
the beam must be the same in both cases. What is that _one_ reason?
Either it's the acceleration of the rocket for both, or it's gravity for

### Eric Gisse

Jan 4, 2009, 1:08:03 PM1/4/09
to
On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
> Oh No wrote:
> >  Thus spake David Rutherford <drutherf...@softcom.net>
>
> >>Oh No wrote:
>
> >>>Thus spake David Rutherford <drutherf...@softcom.net>

>
> >>>>If both cases are equivalent, then shouldn't the reason for the
> >>>>bending of the beam be the same, in both cases?
>
> >>> It is. This is Einstein's equivalence principle.
>
> >>Then what's the (one) reason in both cases?
>
> > The reason is that a gravitational field is locally equivalent to an
> > accelerating reference frame.
>
> If the two cases are equivalent, then the _reason_ for the curvature of
> the beam must be the same in both cases. What is that _one_ reason?
> Either it's the acceleration of the rocket for both, or it's gravity for

YOU CAN NOT KNOW.

That is the entire *point* of the equivalence principle.

### Oh No

Jan 4, 2009, 1:18:09 PM1/4/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>Oh No wrote:
>>>
>>>>Thus spake David Rutherford <druth...@softcom.net>
>>>>
>>>>>If both cases are equivalent, then shouldn't the reason for the
>>>>>bending of the beam be the same, in both cases?
>>>>
>>>> It is. This is Einstein's equivalence principle.
>>>
>>>Then what's the (one) reason in both cases?
>> The reason is that a gravitational field is locally equivalent to an
>> accelerating reference frame.
>
>If the two cases are equivalent, then the _reason_ for the curvature of
>the beam must be the same in both cases. What is that _one_ reason?
>Either it's the acceleration of the rocket for both, or it's gravity
>
As stated, the reason is that both are the same thing, so you can't
choose one without the other. That is what equivalence means.

### David Rutherford

Jan 4, 2009, 3:35:33 PM1/4/09
to

Eric Gisse wrote:
> On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>
>>Oh No wrote:
>>
>>>The reason is that a gravitational field is locally equivalent to an
>>>accelerating reference frame.
>>
>>If the two cases are equivalent, then the _reason_ for the curvature of
>>the beam must be the same in both cases. What is that _one_ reason?
>>Either it's the acceleration of the rocket for both, or it's gravity for
>
> YOU CAN NOT KNOW.

Sure you can. The reason is the acceleration of the rocket, in both
cases, which results in the illusion of the curvature of the beam, in
both cases.

### Oh No

Jan 4, 2009, 3:52:59 PM1/4/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Eric Gisse wrote:
>> On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>Oh No wrote:
>>>
>>>>The reason is that a gravitational field is locally equivalent to an
>>>>accelerating reference frame.
>>>
>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>the beam must be the same in both cases. What is that _one_ reason?
>>>Either it's the acceleration of the rocket for both, or it's gravity for
>> YOU CAN NOT KNOW.
>
>Sure you can. The reason is the acceleration of the rocket, in both
>cases, which results in the illusion of the curvature of the beam, in
>both cases.
>
The question assumed that you did not look outside the rocket at the
propulsion system.

### Igor

Jan 4, 2009, 9:38:26 PM1/4/09
to
On Jan 4, 12:30 pm, David Rutherford <drutherf...@softcom.net> wrote:
> Oh No wrote:
> >  Thus spake David Rutherford <drutherf...@softcom.net>
>
> >>Oh No wrote:
>
> >>>Thus spake David Rutherford <drutherf...@softcom.net>

>
> >>>>If both cases are equivalent, then shouldn't the reason for the
> >>>>bending of the beam be the same, in both cases?
>
> >>> It is. This is Einstein's equivalence principle.
>
> >>Then what's the (one) reason in both cases?
>
> > The reason is that a gravitational field is locally equivalent to an
> > accelerating reference frame.
>
> If the two cases are equivalent, then the _reason_ for the curvature of
> the beam must be the same in both cases. What is that _one_ reason?
> Either it's the acceleration of the rocket for both, or it's gravity for
>

Six of one, half dozen of the other. Please choose one.

### harry

Jan 5, 2009, 9:29:14 AM1/5/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:LNOdnen0V7_qAsLU...@posted.docknet...

>
>
> Eric Gisse wrote:
>> On Jan 3, 12:56 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>An observer in a uniformly accelerating rocket, far from any gravitating
>>>bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
>>>rocket in an unconscious state and wakes up after the rocket has
>>>attained its uniform acceleration. Consequently, he doesn't know whether
>>>he is accelerating or 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?
>>
>> Either.
>>
>> This is the fundamental meaning of the equivalence principle :
>> locally, you _can not_ tell the difference.
>
> What I meant by a uniform gravitational field was a field in which the
> bending of the beam would match exactly the bending of the beam in the
> accelerating rocket. Maybe it's created by a very large thick flat slab of
> material so that the gravitational field is everywhere perpendicular to
> the slab. In other words, the gravitational field would be such that the
> bending of the beam is the same, in either case.
>
> If both cases are equivalent, then shouldn't the reason for the bending of
> the beam be the same, in both cases?

It depends on what you mean with "equivalent". Dictionary.com:
"equal in value, measure, force, effect, significance, etc.: His silence is
equivalent to an admission of guilt."
Generally, "Equivalent" is not identical with "Identical" and different
causes can have identical effects. If you mean similar or analogue, then the
argument is false.

> That is, either the accelerating rocket creates a gravitational field,
> which then causes the beam to bend, or the rocket is accelerating as it
> sits `at rest' in the uniform gravitational field.

That was what Einstein thought (apart of a glitch in your formulation: one
cannot have both the acceleration and the field that replaces it!) and what
brought him to1916 GRT - according to which also acceleration is relative.
Consequently, acceleration would be free to choose so that instantaneous
homogeneous gravitational fields could be created by the rocket engine.
However, that claim that acceleration is physically relative resulted in the
clock paradox as described by him in 1918. I argued in this group that his
1918 solution does not really work, see:
-

1920 (but without frankly admitting so):
"It is true that Mach tried to avoid having to accept as real something
which is not observable by endeavouring to substitute in mechanics a mean
acceleration with reference to the totality of the masses in the universe in
place of an acceleration with reference to absolute space. But inertial
resistance opposed to relative acceleration of distant masses presupposes
[instantaneous] action at a distance; and [...] the modern physicist does
not believe that he may accept this action at a distance"
- http://www.tu-harburg.de/rzt/rzt/it/Ether.html

Nowadays GRT is usually not anymore presented as a theory of relative
acceleration but as a theory of gravitation - although in fact his theory of
gravitation was only *part* or a *consequence* of 1916 GRT:
"in pursuing the general theory of relativity we shall be led to a theory of
gravitation, since we are able to "produce" a gravitational firld merely by
changing the system of co-ordinates."
- http://www.alberteinstein.info/gallery/gtext3.html (p.150)

> This example is analogous to the problem in EM of the magnet and the wire
> loop. In that problem, there are two scenarios, one in which the magnet is
> moved through the loop, the other in which the loop is moved over the
> magnet. These were seen to be two completely different phenomenon until
> Einstein came along and said, essentially, that these are the same thing,
> stupid (I'm not calling you stupid :))!
>
> Why isn't this a similar situation.

Einstein thought that it was the same; and if the problem does not include
acceleration, the GRT debate doesn't even come into play. Otherwise it's a
tricky and interesting topic (but slightly different from the title of this
thread). For example if you discuss rotating magnets the literature is full
of contradicting claims, and acceleration of a charged particle is most
fascinating:
-
http://sci.tech-archive.net/Archive/sci.physics.relativity/2006-02/msg00767.html

Regards,
Harald

======================================= MODERATOR'S COMMENT:
One should also carefully discriminate between 'identical' and 'equal'

### David Rutherford

Jan 5, 2009, 9:24:28 AM1/5/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>Eric Gisse wrote:
>>
>>>On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>>
>>>>Oh No wrote:
>>>>
>>>>>The reason is that a gravitational field is locally equivalent to an
>>>>>accelerating reference frame.
>>>>
>>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>>the beam must be the same in both cases. What is that _one_ reason?
>>>>Either it's the acceleration of the rocket for both, or it's gravity for
>>>
>>> YOU CAN NOT KNOW.
>>
>>Sure you can. The reason is the acceleration of the rocket, in both
>>cases, which results in the illusion of the curvature of the beam, in
>>both cases.
>
> The question assumed that you did not look outside the rocket at the
> propulsion system.

You don't have to look outside the rocket. Just put an accelerometer in
the rocket, in both cases. It will tell you that the rocket is
accelerating, in both cases. As I've said before ... like a pilot flying
in the fog, trust your instruments.

### Ken S. Tucker

Jan 5, 2009, 5:53:16 PM1/5/09
to
Hi David and all.

On Jan 4, 9:30 am, David Rutherford <drutherf...@softcom.net> wrote:
> Oh No wrote:
...

> > The reason is that a gravitational field is locally equivalent to an
> > accelerating reference frame.

> If the two cases are equivalent, then the _reason_ for the curvature of
> the beam must be the same in both cases. What is that _one_ reason?
> Either it's the acceleration of the rocket for both, or it's gravity for

Well there are alternatives. I'm unable to find a text-book
ref, so we'll need to solve that ourselves, with a bit of GR
and EM, that you and most in this group know more about
than I do, so I expect corrections, because some initiative
is required.

I'll use the "geodesic equation" as a starting point, to look
at gravitational and inertial deflection of light. We only need,

http://en.wikipedia.org/wiki/Solving_the_geodesic_equations#The_geodesic_equation

To simplify, (U^u is a 4-vector), the speed will be non-
relativitistic (U^0~1, U^1~0) and the fields weak, in
direction of 0,1 (t,r), reducing to

dU^1 / dt = -GAMMA_1,00 = - 1/2 ( 2*g_01,0 - g_00,1).

= (1/2) g_00,1 - g_01,0 , (ken1),

sufficient for our purposes, is that ok?

CASE1: We are sitting on our chair's reading a screen,
subject to Earth's g-field. Relative to a CS attached to
Earth our "coordinate acceleration" dU^1/dt = 0, so
Eq.(ken1) becomes,

(1/2) g_00,1 = g_01,0 when dU^1/dt=0.

Conventionally, g_00 = 1 - 2m/r,
g_00,1 converts to Newtons - 2GM/r^2, giving

GM/r^2 = -g_01,0 , Eq.(Case1).

CASE2: Using the same CS, except with Earth's mass
set to zero so g_00,1=0, we accelerate, using thrusters,
so the "coordinate acceleration" is non-zero, yielding
from Eq.(ken1),

dU^1/dt = -g_01,0 , Eq.(Case2).

Evidently we need to define g_01 to proceed further,
and I'm certainly open to suggestions/initiatives.

I'll suggest, at this point, we substitute

g_01 = q*F_01 = q*E ,

(E=Electric field, q=charge), as an "antisymmetrical"
component of g_01, to provide,

g_01,0 = q*&E/&t , (Maxwells "displacement current").

When a light-ray deflects, it's E component changes
direction in time, and is the basis of some advanced
accelometers, see,

Regards
Ken S. Tucker

### Eric Gisse

Jan 5, 2009, 6:01:47 PM1/5/09
to
On Jan 4, 11:35 am, David Rutherford <drutherf...@softcom.net> wrote:
> Eric Gisse wrote:
> > On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>
> >>Oh No wrote:
>
> >>>The reason is that a gravitational field is locally equivalent to an
> >>>accelerating reference frame.
>
> >>If the two cases are equivalent, then the _reason_ for the curvature of
> >>the beam must be the same in both cases. What is that _one_ reason?
> >>Either it's the acceleration of the rocket for both, or it's gravity for
>
> > YOU CAN NOT KNOW.
>
> Sure you can. The reason is the acceleration of the rocket, in both
> cases, which results in the illusion of the curvature of the beam, in
> both cases.

Four hundred years of experimentation shows that you can not tell the
difference.

Look up Eotvos.

### David Rutherford

Jan 5, 2009, 9:55:14 PM1/5/09
to

Eric Gisse wrote:
> On Jan 4, 11:35 am, David Rutherford <drutherf...@softcom.net> wrote:
>
>>Eric Gisse wrote:
>>
>>>On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>>Oh No wrote:
>>
>>>>>The reason is that a gravitational field is locally equivalent to an
>>>>>accelerating reference frame.
>>
>>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>>the beam must be the same in both cases. What is that _one_ reason?
>>>>Either it's the acceleration of the rocket for both, or it's gravity for
>>
>>>YOU CAN NOT KNOW.
>>
>>Sure you can. The reason is the acceleration of the rocket, in both
>>cases, which results in the illusion of the curvature of the beam, in
>>both cases.
>
> Four hundred years of experimentation shows that you can not tell the
> difference.

I'm not saying you can tell the difference between gravitation and
acceleration. I'm saying that gravitation _is_ acceleration.

### David Rutherford

Jan 5, 2009, 9:55:38 PM1/5/09
to

Ken S. Tucker wrote:
> Hi David and all.
>
> On Jan 4, 9:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>
>>Oh No wrote:
>
> ...
>
>>>The reason is that a gravitational field is locally equivalent to an
>>>accelerating reference frame.
>
>>If the two cases are equivalent, then the _reason_ for the curvature of
>>the beam must be the same in both cases. What is that _one_ reason?
>>Either it's the acceleration of the rocket for both, or it's gravity for
>
> Well there are alternatives.

Any alternative (including GR) is more complicated. The simpler answer
of the two, above, is that it's the acceleration of the rocket, in both
cases. Using this assumption (which is simpler than GR) gives many, if
not all, of the effects of GR, including the (apparent) curvature of
light, the (apparent) curvature of spacetime, the slowing of clocks, etc..

### harry

Jan 6, 2009, 5:20:53 AM1/6/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:k7KdnZnOnqZhL__U...@posted.docknet...

>
>
> Eric Gisse wrote:
>> On Jan 4, 11:35 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>Eric Gisse wrote:
>>>
>>>>On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>>
>>>>>Oh No wrote:
>>>
>>>>>>The reason is that a gravitational field is locally equivalent to an
>>>>>>accelerating reference frame.
>>>
>>>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>>>the beam must be the same in both cases. What is that _one_ reason?
>>>>>Either it's the acceleration of the rocket for both, or it's gravity
>>>>>for
>>>
>>>>YOU CAN NOT KNOW.
>>>
>>>Sure you can. The reason is the acceleration of the rocket, in both
>>>cases, which results in the illusion of the curvature of the beam, in
>>>both cases.
>>
>> Four hundred years of experimentation shows that you can not tell the
>> difference.
>
> I'm not saying you can tell the difference between gravitation and
> acceleration. I'm saying that gravitation _is_ acceleration.

That's not clear enough for me to understand. If a rocket is keeping itself
at constant height above the earth, you say that it is accelerating relative
to what?

Regards,
Harald

======================================= MODERATOR'S COMMENT:
For geo-stationary orbits, even 'keeping itself' is not necessary

### nigelbr...@hotmail.com

Jan 6, 2009, 8:25:38 AM1/6/09
to
wrote:

> An accelerating reference frame is NOT equivalent to a gravitational
> field, even locally.
> In a gravitational field a falling charge radiates, in an accelerating
> reference frame it does not. On the contrary, there a “resting” charge
> radiates. If you look outside your rocket, you will see the whole
> universe accelerating, so it is not at all equivalent to some “local”
> gravity.

In any kind of quantum gravity theory, gravitational charge will be
exchanging gravitons, which are the field quanta. So you need to say
what kind of charge and what kind of radiation you refer to. I'm
assuming you are just thinking of classical electromagnetic radiation
from an accelerating electric charge like an electron. Actually, an
electron decelerates when it radiates energy as electromagnetic
waves. If you want an electron to accelerate, you also need to supply
an antenna on a radio transmitter. The supplied field carried by the
conducting antenna causes the electron to accelerate, which in turn

In the case of an electric charge being accelerated by a gravitational
field, gravitons supply energy and some of that energy is then
radiated as electromagnetic waves. Let's quantify this process to see
how the numbers go for non-relativistic electron fall velocities (i.e.
well below c).

A falling electron released in a uniform gravitational acceleration,
g, gains velocity v = gt.

Kinetic energy, E = (1/2)mv^2 = (1/2)m(gt)^2

Hence after 1 second, an electron in freefall gains a maximum
(ignoring electromagnetic radiation) of 4.4*10^(-29) Joules.

Electromagnetic energy is radiated whenever there is acceleration of
electric charge, as given by Larmor's non-relativistic formula for
radiated power: P = [(qg)^2]/[6*{Pi}*{Permittivity of free space}*c^3]
watts, where q is charge and g is the acceleration of that charge.

Hence, the freefalling electron in terrestrial gravity, g = 9.81 ms^
(-2), radiates 5.5*10^(-52) watts. Since a watt is 1 Joule per second,
over the first second of fall an electron radiates 5.5*10^(-52)
Joules.

Thus, the proportion of its kinetic energy after 1 second [4.4*10^
ms^(-2) J] is only [5.5*10^(-52)]/[4.4*10^(-29)] = 1.3*10^(-23). This
radiated electromagnetic energy is a trivial proportion of the kinetic
energy that the electron gains from the gravitational acceleration.

In quantum gravity, accelerations are due to the unequal exchange of
gravitons radiated by masses, while uniform motions occur where there
is an equilibrium of graviton exchange. This would apply to mainstream
spin-2 graviton conjecture by Pauli and Fietz, as well as any other
graviton spin (there is a severe problem with the spin-2 graviton
ideas, as it pretends that only two masses radiate gravitons in the
universe, and uses this falsehood to prove than spin-2 is needed to
get universal attraction between similar charge as Zee uses in his QFT
textbook; but if you accept the simple fact that there is a lot of
mass in the universe around us and no mechanism exists to stop
graviton exchanges with all that mass, then your two similar
gravitational charges are accelerated together by the converging
spin-1 graviton flux from the immense mass in the surrounding universe
which exceeds the repulsion due to spin-1 gravitons exchanged between
the two masses in question, see http://nige.wordpress.com/2008/01/30/book/
).

> In a gravitational field a falling charge radiates, in an accelerating
> reference frame it does not. On the contrary, there a “resting” charge

I disagree with this. A resting charge (relative to an observer) is
not going to radiate energy, or the principle of conservation of
energy will be violated. Where does the supposed radiated energy come
from? An accelerating charge in any gravitation or other accelerating
field (e.g. where an electron is accelerated by an electric field
carried by a conductor) will radiate. A resting charge will not
observably radiate in the frame of reference in which it is observed
to be resting.

======================================= MODERATOR'S COMMENT:
Thus, photon exchange occurs only between charges being accelerated against each another?

### nigelbr...@hotmail.com

Jan 6, 2009, 8:26:31 AM1/6/09
to
On Jan 3, 9:56 am, David Rutherford <drutherf...@softcom.net> wrote:
> An observer in a uniformly accelerating rocket, far from any gravitating
> bodies, observes a light beam projected from a laser 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. Assume that he has been somehow transported to the
> rocket in an unconscious state and wakes up after the rocket has
> attained its uniform acceleration. Consequently, he doesn't know whether
> he is accelerating or 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?

This is illustrated in Figure 24, "Light propagation in an accelerated
rocket" on page 121 of George Gamow's book, "Gravity" (Dover
Publications, Inc., N.Y., 2002 reprint of 1962 edition).

Gamow comments on p. 120: "The observer inside the [rocket] chamber,
considering all the phenomena he observes as due to gravity, will
conclude from his experiment that the light ray is bent when
propagating through a gravitational field. Thus, concluded, Einstein,
if the principle of equivalence [between gravitational and other
accelerations] is a general principle of physics, light rays from
distant stars should be bent if they pass close to the surface of the
Sun on the way to a terrestrial observer."

The curvature of the light beam tells the observer in the rocket that
the rocket is in some kind of accelerative field, whether due to
resting on the surface of a planet, or due to accelerating under
uniform rocket thrust. If the observer could tell it was rocket
thrust (rather than terrestrial gravitation) by glancing at the fuel
consumption dial or feeling the vibration from the rocket engine, then
she would know she was in motion. This is where general relativity
differs from restricted (aka "special") relativity, where non-
accelerating motions are relative. Accelerations are absolute:

‘The special theory of relativity … does not extend to non-uniform
motion … The laws of physics must be of such a nature that they apply
to systems of reference in any kind of motion. Along this road we
arrive at an extension of the postulate of relativity… The general
laws of nature are to be expressed by equations which hold good for
all systems of co-ordinates, that is, are co-variant with respect to
any substitutions whatever (generally co-variant). …’ – Albert
Einstein, ‘The Foundation of the General Theory of Relativity’,
Annalen der Physik, v49, 1916.

‘The great stumbing-block for a philosophy which denies absolute space
is the experimental detection of absolute rotation [and other forms of
acceleration].’ – Professor A.S. Eddington (who confirmed Einstein’s
general theory of relativity in 1919), Space Time and Gravitation: An
Outline of the General Relativity Theory, Cambridge University Press,
Cambridge, 1921, p. 152.

Only non-accelerating motion is relative, whereas accelerating motion
produces forces. In the real world, all motion requires some
acceleration to start and stop it, and gravitational fields exist
everywhere, so the restricted theory of relativity simply isn't
correct for understanding nature and is only useful for deriving the
Lorentz transformation and mass-energy equivalence. It's like the
difference between the Bohr atom and quantum mechanics. The Bohr atom
gives you the wrong idea about what happens in the atom, but it is
useful in helping to give you simple equations. An observer in a
spacecraft would always be able to ascertain the absolute motion of
that spacecraft by integrating the absolute accelerations measured
from the forces to gyroscopes, which is analogous (but in practice
easier) to watching light rays being deflected. Einstein's 1905
restricted relativity postulate is restricted to uniform, non-
accelerating motion, and therefore doesn't apply to any real physical
situations in this particular universe we inhabit where motions are
due to acceleration and where there are gravitational fields
(gravitons) causing acceleration.

### Oh No

Jan 6, 2009, 8:35:22 AM1/6/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Ken S. Tucker wrote:
>> Hi David and all.
>> On Jan 4, 9:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>Oh No wrote:
>> ...
>>
>>>>The reason is that a gravitational field is locally equivalent to an
>>>>accelerating reference frame.
>>
>>>If the two cases are equivalent, then the _reason_ for the curvature
>>>of
>>>the beam must be the same in both cases. What is that _one_ reason?
>>>Either it's the acceleration of the rocket for both, or it's gravity for
>> Well there are alternatives.
>
>Any alternative (including GR) is more complicated. The simpler answer
>of the two, above, is that it's the acceleration of the rocket, in both
>cases. Using this assumption (which is simpler than GR) gives many, if
>not all, of the effects of GR, including the (apparent) curvature of
>light, the (apparent) curvature of spacetime, the slowing of clocks,
>etc..
>
This is not simpler than GR. It is the basis of gr.

### Oh No

Jan 6, 2009, 8:31:31 AM1/6/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Eric Gisse wrote:
>> On Jan 4, 11:35 am, David Rutherford <drutherf...@softcom.net> wrote:
>>
>>>Eric Gisse wrote:
>>>
>>>>On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>>
>>>>>Oh No wrote:
>>>
>>>>>>The reason is that a gravitational field is locally equivalent to an
>>>>>>accelerating reference frame.
>>>
>>>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>>>the beam must be the same in both cases. What is that _one_ reason?
>>>>>Either it's the acceleration of the rocket for both, or it's gravity for
>>>
>>>>YOU CAN NOT KNOW.
>>>
>>>Sure you can. The reason is the acceleration of the rocket, in both
>>>cases, which results in the illusion of the curvature of the beam, in
>>>both cases.
>> Four hundred years of experimentation shows that you can not tell
>>the
>> difference.
>
>I'm not saying you can tell the difference between gravitation and
>acceleration. I'm saying that gravitation _is_ acceleration.
>
So, what's the big deal?. This is the principle of equivalence, on which
Einstein based general relativity.

### Oh No

Jan 6, 2009, 8:35:55 AM1/6/09
to
Thus spake nigelbr...@hotmail.com

>I disagree with this. A resting charge (relative to an observer) is not
>going to radiate energy, or the principle of conservation of energy
>will be violated. Where does the supposed radiated energy come from?
>An accelerating charge in any gravitation or other accelerating field
>(e.g. where an electron is accelerated by an electric field carried by
>a conductor) will radiate. A resting charge will not observably
>radiate in the frame of reference in which it is observed to be
>resting.
>
>
>======================================= MODERATOR'S COMMENT:
> Thus, photon exchange occurs only between charges being accelerated
>against each another?
>

There is a difference between photon exchange between charges and the

### Ken S. Tucker

Jan 6, 2009, 9:26:51 AM1/6/09
to
On Jan 5, 6:55 pm, David Rutherford <drutherf...@softcom.net> wrote:
> Ken S. Tucker wrote:
> > Hi David and all.
>
> > On Jan 4, 9:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>
> >>Oh No wrote:
>
> > ...
>
> >>>The reason is that a gravitational field is locally equivalent to an
> >>>accelerating reference frame.
>
> >>If the two cases are equivalent, then the _reason_ for the curvature of
> >>the beam must be the same in both cases. What is that _one_ reason?
> >>Either it's the acceleration of the rocket for both, or it's gravity for
>
> > Well there are alternatives.
>
> Any alternative (including GR) is more complicated.

That's a broad and subjective statement.

> of the two, above, is that it's the acceleration of the rocket, in both
> cases. Using this assumption (which is simpler than GR) gives many, if
> not all, of the effects of GR, including the (apparent) curvature of
> light, the (apparent) curvature of spacetime, the slowing of clocks, etc..

Yes, that cycles us back to Equivalence Principle.
What I found difficult to do is define an accelometer.
How do you intend to define it?

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

Cheers
Ken S. Tucker

### Ken S. Tucker

Jan 6, 2009, 5:05:32 PM1/6/09
to
Hi Nigel.

I understand the Principle of General Relativity differently.
In it's most simplest form, it allows any point or particle (#)
to be at rest, independent of motion, including someone
bumping along on a roller-coaster ride.
That also includes you and me sitting in our chairs.
(# an exception maybe a photon).

> ‘The special theory of relativity … does not extend to non-uniform
> motion … The laws of physics must be of such a nature that they apply
> to systems of reference in any kind of motion. Along this road we
> arrive at an extension of the postulate of relativity… The general
> laws of nature are to be expressed by equations which hold good for
> all systems of co-ordinates, that is, are co-variant with respect to
> any substitutions whatever (generally co-variant). …’ – Albert
> Einstein, ‘The Foundation of the General Theory of Relativity’,
> Annalen der Physik, v49, 1916.

That General Covariance, means the same as thing as I wrote,
any Frame may choosen to be at rest.

> ‘The great stumbing-block for a philosophy which denies absolute space
> is the experimental detection of absolute rotation [and other forms of
> acceleration].’ – Professor A.S. Eddington (who confirmed Einstein’s
> general theory of relativity in 1919), Space Time and Gravitation: An
> Outline of the General Relativity Theory, Cambridge University Press,
> Cambridge, 1921, p. 152.

I think Eddington is wrong. An "absolute rotation" requires
some apparatus like a Bar-Bell. Suppose we use this with
accelometers at A,o and B, rigidly connected, with arrows

<= A-----o-----B =>

1) The EP (Equivalence), does NOT apply, since there is
curvature from A to o and to B. The EP applies to small
regions where the curvature is zero (metric is constant).

2) The Bar-bell need NOT rotate to provide the same
accelometer readings, by placing an appropriate Mass
near the ends, using gravity.

M <= A-----o-----B => M

> Only non-accelerating motion is relative, whereas accelerating motion
> produces forces. In the real world, all motion requires some
> acceleration to start and stop it, and gravitational fields exist
> everywhere, so the restricted theory of relativity simply isn't
> correct for understanding nature and is only useful for deriving the
> Lorentz transformation and mass-energy equivalence. It's like the
> difference between the Bohr atom and quantum mechanics. The Bohr atom
> gives you the wrong idea about what happens in the atom, but it is
> useful in helping to give you simple equations. An observer in a
> spacecraft would always be able to ascertain the absolute motion of
> that spacecraft by integrating the absolute accelerations measured
> from the forces to gyroscopes, which is analogous (but in practice
> easier) to watching light rays being deflected.

I think that statement has a *logic bomb* ;-), for example,
it implies so-called inertial frames may be regared as
rest frames. What happens when I'm free-falling in Earths
g-field?
Regards
Ken S. Tucker

### Chalky

Jan 6, 2009, 9:35:49 PM1/6/09
to
On Jan 4, 8:35 pm, David Rutherford <drutherf...@softcom.net> wrote:
> Eric Gisse wrote:
> > On Jan 4, 8:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>
> >>Oh No wrote:
>
> >>>The reason is that a gravitational field is locally equivalent to an
> >>>accelerating reference frame.
>
> >>If the two cases are equivalent, then the _reason_ for the curvature of
> >>the beam must be the same in both cases. What is that _one_ reason?
> >>Either it's the acceleration of the rocket for both, or it's gravity for
>
> > YOU CAN NOT KNOW.
>
> Sure you can. The reason is the acceleration of the rocket, in both
> cases, which results in the illusion of the curvature of the beam, in
> both cases.

I am sure someone said this here before but, in 2 later passes through
the responses I could not find it, so, to repeat::- the common reason
is space-time curvature.

To clarify, and add something new in specific relation to my response
at SPR, the terms "gravity" and " space-time curvature" are
synonymous.

I don't know where your "the illusion of" comes in, unless you are
evangelising the prevailing philosophies and religions of the Indian
sub-continent. I would not want to argue with that point of view,
except to the extent that it adds nothing to the scientific
discussion.

In summary, and, hopefully, to put this matter to bed, you seem to
have re-invented the wheel, but left it a bit "square".

OK?

### David Rutherford

Jan 6, 2009, 9:34:11 PM1/6/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>

>>I'm not saying you can tell the difference between gravitation and
>>acceleration. I'm saying that gravitation _is_ acceleration.
>
> So, what's the big deal?. This is the principle of equivalence, on which
> Einstein based general relativity.

As far as I know, in GR, an observer `at rest' in a gravitational field
is _not_ accelerating. In my proposal, he _is_ accelerating.

### David Rutherford

Jan 6, 2009, 9:36:15 PM1/6/09
to

harry wrote:

> "David Rutherford" <druth...@softcom.net> wrote in message
> news:k7KdnZnOnqZhL__U...@posted.docknet...
>

>>I'm not saying you can tell the difference between gravitation and
>>acceleration. I'm saying that gravitation _is_ acceleration.
>
> That's not clear enough for me to understand. If a rocket is keeping itself
> at constant height above the earth, you say that it is accelerating relative
> to what?

It's accelerating (or rather, decelerating) in time, relative to an
inertial (free falling) observer.

### David Rutherford

Jan 6, 2009, 9:36:49 PM1/6/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>Ken S. Tucker wrote:
>>
>>>Hi David and all.
>>>
>>> On Jan 4, 9:30 am, David Rutherford <drutherf...@softcom.net> wrote:
>>>
>>>>Oh No wrote:
>>>
>>> ...
>>>
>>>>>The reason is that a gravitational field is locally equivalent to an
>>>>>accelerating reference frame.
>>>
>>>>If the two cases are equivalent, then the _reason_ for the curvature of
>>>>the beam must be the same in both cases. What is that _one_ reason?
>>>>Either it's the acceleration of the rocket for both, or it's gravity for
>>>
>>> Well there are alternatives.
>>
>>Any alternative (including GR) is more complicated. The simpler answer
>>of the two, above, is that it's the acceleration of the rocket, in both
>>cases. Using this assumption (which is simpler than GR) gives many, if
>>not all, of the effects of GR, including the (apparent) curvature of
>>light, the (apparent) curvature of spacetime, the slowing of clocks,
>>etc..
>
> This is not simpler than GR. It is the basis of gr.

I was under the impression that, in GR, light _actually_ (not
apparently) bends in a gravitational field and that spacetime is
_actually_ (not apparently) curved. What I'm saying is that, to an
observer in the rocket `at rest' in a gravitational field, the bending
of light is an illusion, and the curvature of spacetime is also an
illusion, attributable to the acceleration of the rocket/observer.

### Eric Gisse

Jan 7, 2009, 3:39:21 AM1/7/09
to
On Jan 6, 5:34 pm, David Rutherford <drutherf...@softcom.net> wrote:
> Oh No wrote:
> > Thus spake David Rutherford <drutherf...@softcom.net>

>
> >>I'm not saying you can tell the difference between gravitation and
> >>acceleration. I'm saying that gravitation _is_ acceleration.
>
> > So, what's the big deal?. This is the principle of equivalence, on which
> > Einstein based general relativity.
>
> As far as I know, in GR, an observer `at rest' in a gravitational field
> is _not_ accelerating. In my proposal, he _is_ accelerating.

You are sitting on your ass - are you accelerating?

Distinguish between 3-acceleration and 4-acceleration, as the answers
differ.

### harry

Jan 7, 2009, 3:41:12 AM1/7/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:zsWdnbAS-NNGifnU...@posted.docknet...

>
>
> harry wrote:
>
>> "David Rutherford" <druth...@softcom.net> wrote in message
>> news:k7KdnZnOnqZhL__U...@posted.docknet...
>>
>>>I'm not saying you can tell the difference between gravitation and
>>>acceleration. I'm saying that gravitation _is_ acceleration.
>>
>> That's not clear enough for me to understand. If a rocket is keeping
>> itself at constant height above the earth, you say that it is
>> accelerating relative to what?
>
> It's accelerating (or rather, decelerating) in time, relative to an
> inertial (free falling) observer.

Your metaphysics is completely obscure to me - according to you, the free
falling observer is in rest relative to what? He/she is not in general in
rest relative to other free falling observers, as they all fall with
different angles and speeds; moreover, observers don't have occult powers to
pull rockets with them.

Best regards,
Harald

### David Rutherford

Jan 7, 2009, 3:41:45 AM1/7/09
to

Ken S. Tucker wrote:
> On Jan 5, 6:55 pm, David Rutherford <drutherf...@softcom.net> wrote:
>
>>Any alternative (including GR) is more complicated.
>
> That's a broad and subjective statement.
>
>>of the two, above, is that it's the acceleration of the rocket, in both
>>cases. Using this assumption (which is simpler than GR) gives many, if
>>not all, of the effects of GR, including the (apparent) curvature of
>>light, the (apparent) curvature of spacetime, the slowing of clocks, etc..
>
> Yes, that cycles us back to Equivalence Principle.
> What I found difficult to do is define an accelometer.
> How do you intend to define it?

Accelerometer: A device for measuring acceleration.

--

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

Applications:

### Oh No

Jan 7, 2009, 4:56:11 AM1/7/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>I'm not saying you can tell the difference between gravitation and
>>>acceleration. I'm saying that gravitation _is_ acceleration.
>> So, what's the big deal?. This is the principle of equivalence, on
>>which
>> Einstein based general relativity.
>
>As far as I know, in GR, an observer `at rest' in a gravitational field
>is _not_ accelerating. In my proposal, he _is_ accelerating.
>
In gr acceleration is relative. In gr, an observer at rest in a
gravitational field is accelerating relative to inertial matter. This is
sometimes described as absolute or proper acceleration.

### Oh No

Jan 7, 2009, 5:05:15 AM1/7/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>

>>>Any alternative (including GR) is more complicated. The simpler answer

>>>of the two, above, is that it's the acceleration of the rocket, in both
>>>cases. Using this assumption (which is simpler than GR) gives many, if
>>>not all, of the effects of GR, including the (apparent) curvature of
>>>light, the (apparent) curvature of spacetime, the slowing of clocks,
>>>etc..

>> This is not simpler than GR. It is the basis of gr.
>
>I was under the impression that, in GR, light _actually_ (not
>apparently) bends in a gravitational field and that spacetime is
>_actually_ (not apparently) curved. What I'm saying is that, to an
>observer in the rocket `at rest' in a gravitational field, the bending
>of light is an illusion, and the curvature of spacetime is also an
>illusion, attributable to the acceleration of the rocket/observer.
>

It depends on the situation. Here you are assuming a local frame in
which you do your measurement of acceleration. In this case you can take
spacetime to be approximately Minkowski. In each part of its motion,
there is a frame in which the path of light is straight. The analogy is
that, in each part of the surface of the earth, you can make a flat map.

But consider a larger region, say one in which the rocket is orbiting
the earth. For simplicity turn the rocket engines off. The astronaut can
now do experiments showing the existence of a gravitational field. For
more details, see

http://www.teleconnection.info/rqg/TheEquivalencePrinciple

### harry

Jan 7, 2009, 5:35:14 AM1/7/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:zsWdnbYS-NNujvnU...@posted.docknet...

>
>
> Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>I'm not saying you can tell the difference between gravitation and
>>>acceleration. I'm saying that gravitation _is_ acceleration.
>>
>> So, what's the big deal?. This is the principle of equivalence, on which
>> Einstein based general relativity.
>
> As far as I know, in GR, an observer `at rest' in a gravitational field
> is _not_ accelerating. In my proposal, he _is_ accelerating.

Have you by any chance tested your theory with the results of Gravity probe
A?
http://archives.uah.edu/gpa/gpa.html
Although not the primary purpose of the mission, the frequency of the clock
was measured also while the rocket was rising and falling; and the results
appeared to support the hypothesis that a falling rocket is increasing in
speed.
"At low altitudes, where the probe moves rapidly, the frequency of the probe
clock will appear to be retarded by about two cycles per second due to the
second-order Doppler effect."
http://einstein.stanford.edu/content/faqs/gpa_vessot.html

Regards,
Harald

### harry

Jan 7, 2009, 5:34:43 AM1/7/09
to
Hi Ken,

"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
> Hi Nigel.
[....]

>> ‘The great stumbing-block for a philosophy which denies absolute space
>> is the experimental detection of absolute rotation [and other forms of
>> acceleration].’ – Professor A.S. Eddington (who confirmed Einstein’s
>> general theory of relativity in 1919), Space Time and Gravitation: An
>> Outline of the General Relativity Theory, Cambridge University Press,
>> Cambridge, 1921, p. 152.
>
> I think Eddington is wrong. An "absolute rotation" requires
> some apparatus like a Bar-Bell. Suppose we use this with
> accelometers at A,o and B, rigidly connected, with arrows
> "=>" indicating an acceleration reading,
>
> <= A-----o-----B =>
>
> 1) The EP (Equivalence), does NOT apply, since there is
> curvature from A to o and to B. The EP applies to small
> regions where the curvature is zero (metric is constant).
>
> 2) The Bar-bell need NOT rotate to provide the same
> accelometer readings, by placing an appropriate Mass
> near the ends, using gravity.
>
> M <= A-----o-----B => M

To me, claiming that rotation is relative, is equivalent to claiming that
those masses that you placed don't exist. Or do you claim that those masses
are also "relative" and can be transformed away?

Regards,
Harald

### Ken S. Tucker

Jan 7, 2009, 11:59:59 AM1/7/09
to
Hi David.
I respect your knowlege of EM.

On Jan 7, 12:41 am, David Rutherford <drutherf...@softcom.net> wrote:
> Ken S. Tucker wrote:
> > On Jan 5, 6:55 pm, David Rutherford <drutherf...@softcom.net> wrote:
>
> >>Any alternative (including GR) is more complicated.
>
> > That's a broad and subjective statement.
>
> >>of the two, above, is that it's the acceleration of the rocket, in both
> >>cases. Using this assumption (which is simpler than GR) gives many, if
> >>not all, of the effects of GR, including the (apparent) curvature of
> >>light, the (apparent) curvature of spacetime, the slowing of clocks, etc..
>
> > Yes, that cycles us back to Equivalence Principle.
> > What I found difficult to do is define an accelometer.
> > How do you intend to define it?
>
> Accelerometer: A device for measuring acceleration.

Consider unit vectors E and B then using Maxwell's
ExB=c = a constant velocity in an inertial frame K.

In an accelerating frame K' , c' is NOT a constant
velocity, do you see what I mean?
How is that described mathematically?
Regards
Ken S. Tucker

### Ken S. Tucker

Jan 7, 2009, 1:35:12 PM1/7/09
to
Hi Harald.

On Jan 7, 2:34 am, "harry" <harald.NOTTHISvanlin...@epfl.ch> wrote:
> Hi Ken,
>

> "Ken S. Tucker" <dynam...@vianet.on.ca> wrote in messagenews:78d23473-19af-4282...@u18g2000pro.googlegroups.com...

> > Hi Nigel.
> [....]
> >> ‘The great stumbing-block for a philosophy which denies absolute space
> >> is the experimental detection of absolute rotation [and other forms of
> >> acceleration].’ – Professor A.S. Eddington (who confirmed Einstein’s
> >> general theory of relativity in 1919), Space Time and Gravitation: An
> >> Outline of the General Relativity Theory, Cambridge University Press,
> >> Cambridge, 1921, p. 152.
>
> > I think Eddington is wrong. An "absolute rotation" requires
> > some apparatus like a Bar-Bell. Suppose we use this with
> > accelometers at A,o and B, rigidly connected, with arrows
> > "=>" indicating an acceleration reading,
>
> > <= A-----o-----B =>
>
> > 1) The EP (Equivalence), does NOT apply, since there is
> > curvature from A to o and to B. The EP applies to small
> > regions where the curvature is zero (metric is constant).
>
> > 2) The Bar-bell need NOT rotate to provide the same
> > accelometer readings, by placing an appropriate Mass
> > near the ends, using gravity.
>
> > M <= A-----o-----B => M
>
> To me, claiming that rotation is relative, is equivalent to claiming that
> those masses that you placed don't exist.

Ah, I don't understand that sentence.

>Or do you claim that those masses
> are also "relative" and can be transformed away?

Yes, I think I can claim that.
Let's do another gedanken.

3) Two CS's "A" and "B" are *unattached* in radial "free-fall"
in a g-field,

A=> B=> M
- +

B is always nearer to M and so accelerates faster than A,
relative to M, (nothing mysterious about that, just plain old
curvature, in Newton or GR), and it's interesting to note A
and B are inertial frames undergoing relative acceleration,
IOW's A and B are accelerating *away* from one another,
is either A or B absolutely accelerating?

4) Two CS's "A" and "B" are *rigidly attached* in radial
"free-fall" in a g-field,

A------B=> M (Fig.4.gravity)
+ -

with B pulling A (assuming they have some small mass),
causing a deacceleration of B and an acceleration of A.
The relative accelometer readings on A and B will look like,

<= A-----o-----B => (Fig.4.rotation)
+ -
which is a copy of the of the rotating Bar-Bell above (1).

On the basis of A and B's accelometer readings in Figs.(4),
how would you determine the effect is "absolute rotation"
Regards
Ken S. Tucker

### David Rutherford

Jan 8, 2009, 5:16:41 AM1/8/09
to

Eric Gisse wrote:
>
> You are sitting on your ass - are you accelerating?
>
> Distinguish between 3-acceleration and 4-acceleration, as the answers
> differ.

Me and my ass are both accelerating (or rather, decelerating) in time.

### David Rutherford

Jan 8, 2009, 5:17:51 AM1/8/09
to

Oh No wrote:

> But consider a larger region, say one in which the rocket is orbiting
> the earth. For simplicity turn the rocket engines off. The astronaut can
> now do experiments showing the existence of a gravitational field. For
> more details, see
>
> http://www.teleconnection.info/rqg/TheEquivalencePrinciple

Have experiments actually been done in rockets orbiting the earth,
showing non-inertial motion between objects in free fall inside the rocket?

### David Rutherford

Jan 8, 2009, 5:17:32 AM1/8/09
to

harry wrote:
>
> Your metaphysics is completely obscure to me - according to you, the free
> falling observer is in rest relative to what? He/she is not in general in
> rest relative to other free falling observers, as they all fall with
> different angles and speeds; moreover, observers don't have occult powers to
> pull rockets with them.

I didn't say the free falling observer is in/at rest. He's an inertial
observer. That means he can be moving with uniform velocity relative to
other inertial observers. He doesn't have to be at rest.

### harry

Jan 8, 2009, 5:19:55 AM1/8/09
to

Do you deny the existence of the masses that you placed? If you acknowledge
the masses, then there is no room for claiming rotation instead of
gravitation (apart of the fact that a simple test of motion inside the
system distinguishes between the two cases: a curved trjectory of any test
object or laser beam gives it away).

>> Or do you claim that those masses
>> are also "relative" and can be transformed away?
>
> Yes, I think I can claim that.
> Let's do another gedanken.
>
> 3) Two CS's "A" and "B" are *unattached* in radial "free-fall"
> in a g-field,
>
> A=> B=> M
> - +

OK, - and + are relative accelerator readings.

> B is always nearer to M and so accelerates faster than A,
> relative to M, (nothing mysterious about that, just plain old
> curvature, in Newton or GR), and it's interesting to note A
> and B are inertial frames

Note: you use "inertial" in GRT sense, not SRT sense. A and B are not SRT
reference frames.

> undergoing relative acceleration,
> IOW's A and B are accelerating *away* from one another,
> is either A or B absolutely accelerating?

That example is not the one you criticized if B is always nearer to M, this
means that the system is not rotating. Eddington's argument was that
*rotation* is very easy to detect, and thus absolute in every sense of the
word. Apart of that I would say that both are "absolutely" accelerating at
different rates (and thus also relatively), but that would be much more
difficult to prove.

> 4) Two CS's "A" and "B" are *rigidly attached* in radial
> "free-fall" in a g-field,
>
> A------B=> M (Fig.4.gravity)
> + -
>
> with B pulling A (assuming they have some small mass),
> causing a deacceleration of B and an acceleration of A.

I would say they accelerate at the same pace. But no doubt we mean the same
thing.

> The relative accelometer readings on A and B will look like,
>
> <= A-----o-----B => (Fig.4.rotation)
> + -
> which is a copy of the of the rotating Bar-Bell above (1).
>
> On the basis of A and B's accelometer readings in Figs.(4),
> how would you determine the effect is "absolute rotation"

I don't claim that with any limited set of observations we can distinguish
between different causes. If you sufficiently blindfold someone he will be
able to distinguish less and less phenomena while physicists from the time
of Newton set out to do quite the contrary - the same for airline pilots,
they prefer to have enough instruments in order to figure out what is really
happening. :-)
- I would put a gyroscope in the system in order to measure absolute
rotation.
- Your example seems to be not one of a rotating system. Then my gyroscope
reading will be zero, and thus even if I don't see any mass I can be pretty
sure that there must be matter M and/or an electric attractor nearby (two
other causes to distinguish by other means!), since I don't have a rocket
engine running. :-)

Cheers,
Harald

### David Rutherford

Jan 8, 2009, 5:26:44 AM1/8/09
to

Ken S. Tucker wrote:
>
> Consider unit vectors E and B then using Maxwell's
> ExB=c = a constant velocity in an inertial frame K.

ExB doesn't have the units of velocity. But anyway ...

> In an accelerating frame K' , c' is NOT a constant
> velocity, do you see what I mean?

Assuming c is a velocity (which it isn't), for our purposes, here.
Whatever force is causing c' to vary, according to observers in K', is a
fictitious force.

> How is that described mathematically?

It depends on what the observers in K' think the force is that's causing
c' to vary.

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

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

======================================= MODERATOR'S COMMENT:
I pass this posting because of its pointing to the fact that ExB is not a velocity; no more postings about such an absurdity should be approved

### David Rutherford

Jan 8, 2009, 5:35:44 AM1/8/09
to

harry wrote:

> "David Rutherford" <druth...@softcom.net> wrote:
>
>>As far as I know, in GR, an observer `at rest' in a gravitational field
>>is _not_ accelerating. In my proposal, he _is_ accelerating.
>
> Have you by any chance tested your theory with the results of Gravity probe
> A?
> http://archives.uah.edu/gpa/gpa.html
> Although not the primary purpose of the mission, the frequency of the clock
> was measured also while the rocket was rising and falling; and the results
> appeared to support the hypothesis that a falling rocket is increasing in
> speed.
>
> "At low altitudes, where the probe moves rapidly, the frequency of the probe
> clock will appear to be retarded by about two cycles per second due to the
> second-order Doppler effect."
> http://einstein.stanford.edu/content/faqs/gpa_vessot.html

This is a _proposed_ experiment. It hasn't been done yet. Also you have
things reversed. The excerpt, above, refers to the rising probe, not the
falling probe. You forgot to include that part (below):

"At low altitudes, where the probe moves rapidly, the frequency of the
probe clock will appear to be retarded by about two cycles per second

due to the second-order Doppler effect. As the probe gains altitude and
slows down, this effect diminishes and will be offset by the
gravitational shift, which makes the probe block appear to run faster,
eventually reaching one cycle per second at apogee."

### harry

Jan 8, 2009, 7:02:54 AM1/8/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:NYOdndFHSKeoB_jU...@posted.docknet...

>
>
> harry wrote:
>
>> "David Rutherford" <druth...@softcom.net> wrote:
>>
>>>As far as I know, in GR, an observer `at rest' in a gravitational field
>>>is _not_ accelerating. In my proposal, he _is_ accelerating.
>>
>> Have you by any chance tested your theory with the results of Gravity
>> probe A?
>> http://archives.uah.edu/gpa/gpa.html
>> Although not the primary purpose of the mission, the frequency of the
>> clock was measured also while the rocket was rising and falling; and the
>> results appeared to support the hypothesis that a falling rocket is
>> increasing in speed.
> >
>> "At low altitudes, where the probe moves rapidly, the frequency of the
>> probe clock will appear to be retarded by about two cycles per second due
>> to the second-order Doppler effect."
>> http://einstein.stanford.edu/content/faqs/gpa_vessot.html
>
> This is a _proposed_ experiment. It hasn't been done yet.

No, that's wrong, it was done with succes in 1976, and another test
(Gravity-probe B) is being done nowadays.
http://en.wikipedia.org/wiki/Gravity_Probe

> Also you have things reversed. The excerpt, above, refers to the rising
> probe, not the falling probe. You forgot to include that part (below):

The link was merely a quick ref. by means of Google, as it describes the
set-up.

> "At low altitudes, where the probe moves rapidly, the frequency of the
> probe clock will appear to be retarded by about two cycles per second due
> to the second-order Doppler effect. As the probe gains altitude and slows
> down, this effect diminishes and will be offset by the gravitational
> shift, which makes the probe block appear to run faster, eventually
> reaching one cycle per second at apogee."

I did not include that on purpose as it may distract from the issue here.
According to GRT and consistent with their results, the direction of the
rocket's velocity and acceleration does not matter for the clock rate.

Harald

### harry

Jan 8, 2009, 7:02:20 AM1/8/09
to

"David Rutherford" <druth...@softcom.net> wrote in message
news:NYOdnc1HSKc5B_jU...@posted.docknet...

>
>
> harry wrote:
>>
>> Your metaphysics is completely obscure to me - according to you, the free
>> falling observer is in rest relative to what? He/she is not in general in
>> rest relative to other free falling observers, as they all fall with
>> different angles and speeds; moreover, observers don't have occult powers
>> to pull rockets with them.
>
> I didn't say the free falling observer is in/at rest. He's an inertial
> observer. That means he can be moving with uniform velocity relative to
> other inertial observers. He doesn't have to be at rest.

- He is *not* moving with uniform velocity relatively to all other free
falling observers; for example, two free-falling observers on opposite sides
of the earth accelerate relative to each other.
- I meant if you have a metaphysical concept, that is, a physical model. If
you have one, relative to what is the free falling object (your "observer")
either physically in rest or accelerating, and why does this physically
affect the accelerating object?

Regards,
Harald

### David Rutherford

Jan 8, 2009, 8:07:22 AM1/8/09
to

Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>
>>Oh No wrote:
>>
>>>Thus spake David Rutherford <druth...@softcom.net>
>>>
>>>>I'm not saying you can tell the difference between gravitation and
>>>>acceleration. I'm saying that gravitation _is_ acceleration.
>>>
>>> So, what's the big deal?. This is the principle of equivalence, on
>>>which
>>>Einstein based general relativity.
>>
>>As far as I know, in GR, an observer `at rest' in a gravitational field
>>is _not_ accelerating. In my proposal, he _is_ accelerating.
>
> In gr acceleration is relative. In gr, an observer at rest in a
> gravitational field is accelerating relative to inertial matter. This is
> sometimes described as absolute or proper acceleration.

The inertial matter `feels' no force, but the observer at rest in the
gravitational field `feels' a force. That means the latter is
_absolutely_ accelerating and the former is not _absolutely_
accelerating, wouldn't you say?

### Oh No

Jan 8, 2009, 8:37:04 AM1/8/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>
>>>Oh No wrote:
>>>
>>>>Thus spake David Rutherford <druth...@softcom.net>
>>>>
>>>>>I'm not saying you can tell the difference between gravitation and
>>>>>acceleration. I'm saying that gravitation _is_ acceleration.
>>>>
>>>> So, what's the big deal?. This is the principle of equivalence, on
>>>>which
>>>>Einstein based general relativity.
>>>
>>>As far as I know, in GR, an observer `at rest' in a gravitational field
>>>is _not_ accelerating. In my proposal, he _is_ accelerating.
>> In gr acceleration is relative. In gr, an observer at rest in a
>> gravitational field is accelerating relative to inertial matter. This is
>> sometimes described as absolute or proper acceleration.
>
>The inertial matter `feels' no force, but the observer at rest in the
>gravitational field `feels' a force. That means the latter is
>_absolutely_ accelerating and the former is not _absolutely_
>accelerating, wouldn't you say?
>
Indeed, that is what I did say, and it is what general relativity says
(I personally prefer to call it proper acceleration, rather than
absolute acceleration which I find a somewhat ambiguous phrase, however
since Weinberg uses the phrase absolute acceleration the terminology is
accepted).

### Oh No

Jan 8, 2009, 8:34:28 AM1/8/09
to
Thus spake harry <harald.NOTT...@epfl.ch>

David was talking of local measurements, and he did not say *all*.
Inertial observers do move with uniform velocity relative to other
inertial observers locally. His "metaphysics" here is precisely the same
as Einstein's in gtr, so I do not see why it should trouble you.

### Oh No

Jan 8, 2009, 8:48:32 AM1/8/09
to
Thus spake David Rutherford <druth...@softcom.net>
>
>
>Oh No wrote:
>
>> But consider a larger region, say one in which the rocket is orbiting
>> the earth. For simplicity turn the rocket engines off. The astronaut can
>> now do experiments showing the existence of a gravitational field. For
>> more details, see
>> http://www.teleconnection.info/rqg/TheEquivalencePrinciple
>
>Have experiments actually been done in rockets orbiting the earth,
>showing non-inertial motion between objects in free fall inside the
>rocket?
>
Inertial motion and free fall motion are one and the same thing. That is
the definition of inertial. But yes, experiments have been done in
orbiting space craft showing that inertial motion is not straight line
motion over extended time periods. This is described as being due to
"tidal forces", as explained on the above web page.

http://www.teleconnection.info/rqg/BasicsOfCurvature

for a general explanation of how the mathematics of curved surfaces is
adapted in general relativity, such that we describe space time as
consisting of locally Minkowski regions such that inertial motion is
straight line motion in a small locality.

### harry

Jan 8, 2009, 11:08:17 AM1/8/09
to

"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message

That is irrelevant for my question to him. Once more:
I asked David to clarify his metaphysics, in particular relative to what he
defines the motion (rest/acceleration) of the "free falling observer"; he
should account for the fact that the free falling observer is not *in
general* in rest nor moving with uniform velocity relative to other free
falling observers, as they all fall with different angles and speeds; and
that observers don't have occult powers to cause physical effects such as
time dilation at a distance.

> Inertial observers do move with uniform velocity relative to other
> inertial observers locally. His "metaphysics" here is precisely the same
> as Einstein's in gtr, so I do not see why it should trouble you.

There is definitely a lack of clarity which doesn't trouble me - just
helping out here. Contrary to Einstein he is saying "that gravitation _is_
acceleration". Thus his metaphysics, as he pointed out to us, *disagrees*
considerably with that of Einstein, but it's far from clear what exact
physical model corresponds with his words.

Regards,
Harald