Accelerating rocket/light beam question
David Rutherford
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"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
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"Horizon Problem Resolution"
Eric Gisse
> 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
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
>
>
>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)
David Rutherford
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
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
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.
Read about difference between acceleration and gravitation in A.A.
Logunov’s articles and books on relativistic theory of gravity (RTG)
in arXiv, for example.
Bob.
Eric Gisse
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
>
>
>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
>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.
>Read about difference between acceleration and gravitation in A.A.
>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
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
both. Please choose one.
Eric Gisse
>
> >>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
> both. Please choose one.
YOU CAN NOT KNOW.
That is the entire *point* of the equivalence principle.
Oh No
>
>
>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 both. Please choose one.
>
choose one without the other. That is what equivalence means.
David Rutherford
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
>>both. Please choose one.
>
> 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
>> 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
>>>both. Please choose one.
>> 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.
>
propulsion system.
Igor
>
> >>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
> both. Please choose one.
>
Six of one, half dozen of the other. Please choose one.
harry
"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:
-
http://groups.google.com/group/sci.physics.foundations/msg/68cd1c181f8191d2
Consistent with my analysis, Einstein had already shifted his position in
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
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
>>>>both. Please choose one.
>>>
>>> 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
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
> both. Please choose one.
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,
http://en.wikipedia.org/wiki/Inertial_navigation_system
Regards
Ken S. Tucker
Eric Gisse
> 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
> >>both. Please choose one.
>
> > 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
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
>>>>both. Please choose one.
>>
>>>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
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
>>both. Please choose one.
>
> 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
>
>
> 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
>>>>>both. Please choose one.
>>>
>>>>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
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
it with energy in the form of electromagnetic radiation. Think about
an antenna on a radio transmitter. The supplied field carried by the
conducting antenna causes the electron to accelerate, which in turn
causes the electron to radiate.
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^
(-29) J] which is radiated as electromagnetic radiation [5.5*10^(-52)
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
> radiates.
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
> 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
>> 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
>>>both. Please choose one.
>> 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..
>
Oh No
>> 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
>>>>>both. Please choose one.
>>>
>>>>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.
>
Einstein based general relativity.
Oh No
>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
radiation of observable photons.
Ken S. Tucker
> 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
> >>both. Please choose one.
>
> > Well there are alternatives.
>
> Any alternative (including GR) is more complicated.
That's a broad and subjective statement.
>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..
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
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
"=>" 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
> 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
> 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
> >>both. Please choose one.
>
> > 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
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
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
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
>>>>both. Please choose one.
>>>
>>> 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
>
> >>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
>
>
> 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
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.
>
>>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..
>
> 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
>
>
>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.
>
gravitational field is accelerating relative to inertial matter. This is
sometimes described as absolute or proper acceleration.
Oh No
>
>
>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
>
>
> 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
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news: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. Or do you claim that those masses
are also "relative" and can be transformed away?
Regards,
Harald
Ken S. Tucker
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.
>
> >>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..
>
> > 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
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"
instead of gravitational "curvature"?
Regards
Ken S. Tucker
David Rutherford
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
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
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
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"
> instead of gravitational "curvature"?
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
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"
"Action-reaction Paradox 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
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
>
>
> 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
>
>
> 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
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
>
>
>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?
>
(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
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
>
>> 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?
>
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.
Pleas see also the preceding 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
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
news:FjSJePAF...@charlesfrancis.wanadoo.co.uk...
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.
I'm now really curious about his answer! :-)
> 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
eric gisse
<druth...@softcom.net> wrote:
>
>
>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?
Acceleration is absolute, regardless.
The situations aren't even the same - a body at rest in a
gravitational field is being _HELD_ there, which requires acceleration
away from geodesic motion. Apples and oranges, yo.
Oh No
>On Thu, 8 Jan 2009 07:07:22 CST, David Rutherford
><druth...@softcom.net> wrote:
>
>>
>>
>>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?
>
>Acceleration is absolute, regardless.
No. Acceleration, like all motion, is relative. Absolute acceleration
refers to acceleration relative to inertial matter, and is better called
Oh No
>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.
>
I do not see the metaphysics disagreeing with Einstein's equivalence
principle in any way. It is true that there is a certain ambiguity in
what is meant by "gravitation" in gtr, but this concerns the fact that
gravitational fields are not, in general, uniform. Usually we see
gravitational fields as being non-local, and as referring to curvature
of space time. However, since he is only considering a local experiment
at this point, that does not apply. He might like to know that it took
Einstein nine years to get from the equivalence principle to gtr. He is
not likely to proceed any faster if he doesn't follow what is done in
the text books.
David Rutherford
Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>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,
It's not ambiguous to me. If the object `feels' a force, it is
undergoing an absolute acceleration. If it doesn't, it isn't. The only
possible ambiguity, I think, is that acceleration in time is not obvious.
Acceleration in time, as per gravity, doesn't have the same appearance
as acceleration in space. Objects on the surface of the earth are
accelerating/decelerating in time radially from the center of the earth.
However, they are all accelerating in the same direction in time
(backward). The distance between them doesn't increase, as it would for
a radial spatial acceleration (except to a free-falling observer). Their
distance from the center doesn't increase, as it would for a radial
spatial acceleration (except to a free-falling observer).
The effect of this acceleration in time is that free-falling objects and
light follow paths _as_if_ the objects `at rest' on the surface of the
earth were accelerating in space.
Oh No
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>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,
>
>It's not ambiguous to me. If the object `feels' a force, it is
>undergoing an absolute acceleration. If it doesn't, it isn't.
It is ambiguous because it it suggestive of acceleration with respect to
Newtonian absolute space, a concept which is known not to be valid.
>The only possible ambiguity, I think, is that acceleration in time is
>not obvious.
>
>Acceleration in time, as per gravity, doesn't have the same appearance
>as acceleration in space. Objects on the surface of the earth are
>accelerating/decelerating in time radially from the center of the
>earth. However, they are all accelerating in the same direction in time
>(backward). The distance between them doesn't increase, as it would for
>a radial spatial acceleration (except to a free-falling observer).
>Their distance from the center doesn't increase, as it would for a
>radial spatial acceleration (except to a free-falling observer).
>
>The effect of this acceleration in time is that free-falling objects
>and light follow paths _as_if_ the objects `at rest' on the surface of
>the earth were accelerating in space.
>
Acceleration in time is not an easy concept to put properly into words.
However this was how Einstein first approached gtr, and I have used
changes in the rate time as the basis for the development on the
website.
David Rutherford
Oh No wrote:
> Thus spake David Rutherford <druth...@softcom.net>
>
>>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.
Can you direct me to the experiment that was done that shows this? Does
it test for deviations from straight line motion in spacetime, or just
in space?
David Rutherford
It's not being _HELD_ there in time.
David Rutherford
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,
I was referring to the article at
http://einstein.stanford.edu/content/faqs/gpa_vessot.html
> it was done with succes in 1976, and another test
> (Gravity-probe B) is being done nowadays.
> http://en.wikipedia.org/wiki/Gravity_Probe
Both Gravity probe A and B are outside the setup of my thought
experiment. In my setup, the gravitational field is uniform, meaning
that the strength of gravity (the apparent acceleration imparted to free
falling objects) is the same everywhere in the rocket. The
`acceleration' at the top of the rocket is the same as at the bottom,
and the `acceleration' is perpendicular to the floor at all points
inside the rocket.
>>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.
The link you provided
http://einstein.stanford.edu/content/faqs/gpa_vessot.html
describes neither Gravity probe A nor B. It's a separate _proposed_
experiment that hasn't been done yet.
>>"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.
You're referring to a rocket moving in a gravitational field. That isn't
what my thought experiment is about.
David Rutherford
harry wrote:
> "David Rutherford" <druth...@softcom.net> wrote
>
>>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.
According to whom? You need to ask the free-falling observers, or
another free-falling observer, what their motion is relative to each
other. In the accelerating rocket in empty space example, take two
rockets accelerating in opposite directions. Two observers at rest in
space along the line of motion of the rockets, ahead of and outside each
rocket, will _seem_ to be accelerating toward each other, according to
observers `at rest' inside the rockets, after they compare results and
assume they are in a gravitational field. But according to the observers
outside the rockets, there is no change in the distance between them.
> - 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,
It/he isn't _actually_ accelerating. It just _looks_like_ it's/he's
accelerating from the viewpoint of an observer `at rest' in the rocket
on the surface of the earth. In the accelerating rocket in empty space,
a free-falling object _looks_like_ it's accelerating from the viewpoint
of an observer `at rest' on the floor of the accelerating rocket. In
both cases, it's the acceleration of the rocket that causes the
_illusion_ that the free-falling object is accelerating.
> and why does this physically
> affect the accelerating object?
It doesn't. The object isn't accelerating. It's an illusion.
Ken S. Tucker
Ok, how would you determine the direction of rotation?
> >> 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.
This thread is about light deflection, a GR effect.
> > 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.
In GR either A or B must be able to be considered to be
at rest, since an Observer at rest in either A or B must
find the Physical Laws of Nature apply in that FoR,
that's the meaning of GC.
> > 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"
> > instead of gravitational "curvature"?
>
> 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.
Ah, bingo!, thanks Harald, you are helping me understand the
problem. Now you are measuring the relative rotational difference
between the said aircraft and it's gyro. How would you determine
which is "absolutely rotating" and which is relatively rotating?
If you choose one or the other to be absolutely rotating you are
physically creating a "preferred" FoR, violating the PoR.
> - 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. :-)
Well we are edging toward EP + "Mach's Principle".
I think a new thread based on,
http://www.teleconnection.info/rqg/TheEquivalencePrinciple
would be a good start.
> Cheers,
> Harald
Cheers to you Harald.
You got me thinking :-).
Ken S. Tucker
Ken S. Tucker
In Maxwell's theory of EMR propagation the vector product
ExB *result* points in the vector direction of propagation "c",
which I should have written,
ExB => c.
Sorry for the confusion.
On Jan 8, 2:26 am, David Rutherford <drutherf...@softcom.net> wrote:
> 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 ...
It's a handy short-cut, by setting E and B to unit
vectors, reproduces a unit vector in direction c, so
If one then specifies c changes direction - deflects
as a function of time - then we obtain,
&c/&t => &(ExB)/&t =/= 0.
> > 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.
There have been numerous debates about the meaning
of "fictitious force" , that is why I encourage definitions,
of a mathematical nature.
> > How is that described mathematically?
>
> It depends on what the observers in K' think the force is that's causing
> c' to vary.
David, your over-all "space acceleration" theory is well
known, (I studied it in late 1960's) and it is very good.
In the end you will end up with something like the SS,
(Schwarz. Solution).
> Dave Rutherford
> ======================================= 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
I (wrongly) thought everyone knew ExB = direction of
propagation for Maxwell's EMR. Anyway I think it's to
much detail for this thread, so I'll drop the subject :-).
Regards
Ken S. Tucker
Oh No
>
>
>Oh No wrote:
>> Thus spake David Rutherford <druth...@softcom.net>
>>
>>>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.
>
>Can you direct me to the experiment that was done that shows this? Does
>it test for deviations from straight line motion in spacetime, or just
>in space?
>
really just a straightforward application of Newtonian gravity, which
still applies in approximation in gtr. So many measurements of that have
been done over the years that more are hardly necessary, and I don't
have a reference. Gravity probe B is a more sophisticated measurement,
based on giroscopes. The motion is geodesic in space-time - that is to
say locally straight in a small enough region, where the region is
defined with bounds in time as well as space.
harry
> harry wrote:
>> "David Rutherford" <druth...@softcom.net> wrote
>>
>>> 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.
>
> According to whom? You need to ask the free-falling observers, or
> another free-falling observer, what their motion is relative to each
> other. In the accelerating rocket in empty space example, take two
> rockets accelerating in opposite directions.
Note: I thought that the example we discussed was of free-falling observers,
not accelerating rockets.
> Two observers at rest in
> space along the line of motion of the rockets, ahead of and outside
> each rocket, will _seem_ to be accelerating toward each other,
> according to observers `at rest' inside the rockets, after they
> compare results and assume they are in a gravitational field. But
> according to the observers outside the rockets, there is no change in
> the distance between them.
All free-falling observers will measure the distance between them reducing
until they even crash onto earth - destruction is a very physical event, and
all observers everywhere will agree on that. :-)
>> - 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,
>
> It/he isn't _actually_ accelerating. It just _looks_like_ it's/he's
> accelerating from the viewpoint of an observer `at rest' in the rocket
> on the surface of the earth.
Ah - that negates your earlier claim that "gravitation is acceleration" -
thus perhaps (I added "perhaps" after reading what you wrote next!) you *do*
distinguish between acceleration and gravitation? And "looks like" can just
as much be said by the observer in the rocket, it's standard knowledge for
students who learn about accelerometers. To get back to your original
question: If an observer has enough information, he/she will draw the
correct conclusion.
> In the accelerating rocket in empty
> space, a free-falling object _looks_like_ it's accelerating from the
> viewpoint of an observer `at rest' on the floor of the accelerating
> rocket. In both cases, it's the acceleration of the rocket that
> causes the _illusion_ that the free-falling object is accelerating.
Hmm that's a little too complex for me, as you combine an accelerating
rocket with a free-falling object. Now you appear to claim that acceleration
is an illusion; however as you can see when an object crashes on the ground,
there has been real acceleration (change of velocity) of at least one of the
objects involved, such that they were heading for each other and ending in a
crash. For the physics, it's important to figure out which of the objects
changed state (and how much), that is, which were subject to a physical
change, and what was the cause, and by what means was the effect
transmitted.
That was how the Principia came to be; if we only define relative
acccelerations without considering the cause, or calll everything an
illusion, we won't be succesful in establishing a predictive physics.
>> and why does this physically
>> affect the accelerating object?
>
> It doesn't. The object isn't accelerating. It's an illusion.
Ok. I asked: If a rocket is keeping itself at constant height above the
earth, you say that it is accelerating relative
to what? Now you answer that it's "an illusion". In fact it is accelerating
relative to an observer who undergoes physical acceleration (change of
state), while the rocket isn't physically accelerated - indeed an "illusion"
in the sense that the acceleration did not involve a change of state of the
rocket.
I think we have progressed somewhat. :-)
Best regards,
Harald
harry
Yes.
>>> 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.
>
> In GR either A or B must be able to be considered to be
> at rest, since an Observer at rest in either A or B must
> find the Physical Laws of Nature apply in that FoR,
> that's the meaning of GC.
Indeed - and that partly fails for rotation. Measurement space becomes
distorted in an unnatural way, both locally and globally - very different
from "rest". Even Einstein admitted in 1920 that rotation is, after all,
"absolute". See also below.
Not necessarily. I had in mind an optical (Sagnac) gyroscope, which rotates
with the rotating object. It works because the light does not participate in
the rotation, and that is also much simpler for the physical description.
Note that that device got me to re-think "relativity" some years ago.
> How would you determine
> which is "absolutely rotating" and which is relatively rotating?
All the matter is here rotating relative to physical "space" as indicated by
the gyroscope reading.
> If you choose one or the other to be absolutely rotating you are
> physically creating a "preferred" FoR, violating the PoR.
I cannot physically create anything. :-)
But if you claim that the Sagnac gyroscope violates the "general PoR" of
Einstein, I agree. Probably most physicists don't adhere to the general PoR
nowadays ("modern usage demotes the uniform "gravitational" field back to
its old status as a pseudo-field") -
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html;
and most people mean with "PoR" the PoR of Newton and Poincare - which
Einstein called the "special" PoR.
>> - 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. :-)
>
> Well we are edging toward EP + "Mach's Principle".
> I think a new thread based on,
>
> http://www.teleconnection.info/rqg/TheEquivalencePrinciple
>
> would be a good start.
That could be nice - but note that I won't be able to participate for the
coming weeks.
>> Cheers,
>> Harald
>
> Cheers to you Harald.
> You got me thinking :-).
> Ken S. Tucker
Cheers to you too!
Harald
======================================= MODERATOR'S COMMENT:
Doesn't the mass energy distribution in the universe favourize cs the symmetries of whose account for its ones?
harry
> 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,
>
> I was referring to the article at
>
> http://einstein.stanford.edu/content/faqs/gpa_vessot.html
The experiment has been done, over 30 years ago.
>> it was done with succes in 1976, and another test
>> (Gravity-probe B) is being done nowadays.
>> http://en.wikipedia.org/wiki/Gravity_Probe
>
> Both Gravity probe A and B are outside the setup of my thought
> experiment.
A thought experiment is just one among many thought experimetns and real
experiments. If one experiments gives way to a certain theory, that theory
can be disproved with another experiment and it's no use to close our eyes
to that information.
> In my setup, the gravitational field is uniform, meaning
> that the strength of gravity (the apparent acceleration imparted to
> free falling objects) is the same everywhere in the rocket. The
> `acceleration' at the top of the rocket is the same as at the bottom,
> and the `acceleration' is perpendicular to the floor at all points
> inside the rocket.
>
>>> 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.
>
> The link you provided
>
> http://einstein.stanford.edu/content/faqs/gpa_vessot.html
>
> describes neither Gravity probe A nor B. It's a separate _proposed_
> experiment that hasn't been done yet.
As I pointed out to you, that's a 1976 article about Gravity probe A, and I
also gave you the Wikipedia link about it here above about the experiment's
historical success (according to Wikipedia it was done on June 18, 1976). I
even gave you specific information of the results!
>>> "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.
>
> You're referring to a rocket moving in a gravitational field. That
> isn't what my thought experiment is about.
Indeed. Next you discussed your physics with the suggestion that it differs
from that of Einstein (contrary to Einstein you claimed that "the reason for
the curvature of the beam must be the same in both cases"). If so,Gravity
probe A provides an interesting test to see if your model is consistent with
GRT predictions and measurements. But if not, then I wonder what you try to
discuss.
Regards,
Harald
Chalky
> All free-falling observers will measure the distance between them reducing
> until they even crash onto earth - destruction is a very physical event, and
> all observers everywhere will agree on that. :-)
Tell that to the Man in the Moon : -)
Ken S. Tucker
On Jan 9, 3:39 am, "harry" <harald.NOTTHISvanlin...@epfl.ch> wrote:
[doing some snips]
Ken:
> > In GR either A or B must be able to be considered to be
> > at rest, since an Observer at rest in either A or B must
> > find the Physical Laws of Nature apply in that FoR,
> > that's the meaning of GC.
Harald:
> Indeed - and that partly fails for rotation. Measurement space becomes
> distorted in an unnatural way, both locally and globally - very different
> from "rest". Even Einstein admitted in 1920 that rotation is, after all,
> "absolute". See also below.
Ah, where's a ref where AE published "Absolute Rotation (AR)"?
...
> > Ah, bingo!, thanks Harald, you are helping me understand the
> > problem. Now you are measuring the relative rotational difference
> > between the said aircraft and it's gyro.
>
> Not necessarily. I had in mind an optical (Sagnac) gyroscope, which rotates
> with the rotating object. It works because the light does not participate in
> the rotation, and that is also much simpler for the physical description.
> Note that that device got me to re-think "relativity" some years ago.
Sagnac gyro measures *changes* in rotation, not AR,
that's like a "ring laser gyro".
Consider a rocket to be launched from the North pole,
(Earth rotates every 24 hours), how do you calibrate the
gyro?
> > How would you determine
> > which is "absolutely rotating" and which is relatively rotating?
>
> All the matter is here rotating relative to physical "space" as indicated by
> the gyroscope reading.
That "physical space" is actually distant stars?
> > If you choose one or the other to be absolutely rotating you are
> > physically creating a "preferred" FoR, violating the PoR.
> Probably most physicists don't adhere to the general PoR
> nowadays
Did the GPoR get voted out, and I didn't get a memo?
> ("modern usage demotes the uniform "gravitational" field back to
> its old status as a pseudo-field") -http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_...
> and most people mean with "PoR" the PoR of Newton and Poincare - which
> Einstein called the "special" PoR.
"most physicists" and now "most people", who can
argue with "THEM".
...
> > Well we are edging toward EP + "Mach's Principle".
> > I think a new thread based on,
> >http://www.teleconnection.info/rqg/TheEquivalencePrinciple
> > would be a good start.
>
> That could be nice - but note that I won't be able to participate for the
> coming weeks.
Well, I'd like to go more mathematical too.
See you later.
Ken S. Tucker
======================================= MODERATOR'S COMMENT:
Mathematically, 'most' means 'all, but a finite amount of them'
harry
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:c610f2e1-ca76-4b08...@r15g2000prh.googlegroups.com...
> Hi Harald.
> On Jan 9, 3:39 am, "harry" <harald.NOTTHISvanlin...@epfl.ch> wrote:
> [doing some snips]
>
> Ken:
>> > In GR either A or B must be able to be considered to be
>> > at rest, since an Observer at rest in either A or B must
>> > find the Physical Laws of Nature apply in that FoR,
>> > that's the meaning of GC.
>
> Harald:
>> Indeed - and that partly fails for rotation. Measurement space becomes
>> distorted in an unnatural way, both locally and globally - very different
>> from "rest". Even Einstein admitted in 1920 that rotation is, after all,
>> "absolute". See also below.
>
> Ah, where's a ref where AE published "Absolute Rotation (AR)"?
> ...
I think that he didn't publish a paper on "Absolute Rotation", and he even
wrapped it up inside two long paragraphs so that only careful readers would
notice - haha a bit like the subprime packages:
"For the mechanical behaviour of a corporeal system hovering freely in empty
space depends not only on relative positions (distances) and relative
velocities, but also on its state of rotation, which physically may be taken
as a characteristic not appertaining to the system in itself. In order to be
able to look upon the rotation of the system, at least formally, as
something real, Newton objectivises space. Since he classes his absolute
space together with real things, for him rotation relative to an absolute
space is also something real. [...] 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 [...] 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
>> > Ah, bingo!, thanks Harald, you are helping me understand the
>> > problem. Now you are measuring the relative rotational difference
>> > between the said aircraft and it's gyro.
>>
>> Not necessarily. I had in mind an optical (Sagnac) gyroscope, which
>> rotates
>> with the rotating object. It works because the light does not participate
>> in
>> the rotation, and that is also much simpler for the physical description.
>> Note that that device got me to re-think "relativity" some years ago.
>
> Sagnac gyro measures *changes* in rotation, not AR,
> that's like a "ring laser gyro".
Indeed a ring laser gyro is similar but I was thinking of the orginal
Michelson-Gale device as well as the modern fibre "gyroscope":
- http://en.wikipedia.org/wiki/Fiber_optic_gyroscope
There is in principle no reason for it to only measure changes in rotation
although it is certainly much easier to measure changes, it's a matter of
calibration. Michelson measured the rotation speed of the earth thanks to a
smart calibration trick:
- http://en.wikipedia.org/wiki/Sagnac_effect
> Consider a rocket to be launched from the North pole,
> (Earth rotates every 24 hours), how do you calibrate the
> gyro?
See above.
>> > How would you determine
>> > which is "absolutely rotating" and which is relatively rotating?
>>
>> All the matter is here rotating relative to physical "space" as indicated
>> by
>> the gyroscope reading.
>
> That "physical space" is actually distant stars?
No certainly not - that would imply an occult and instantaneous "action at a
distance". Distant stars can be used as a measure for rotation, just as
mercury can be used as a measure for temperature. As Einstein said, Newton
called it "space" (maybe to hide that it implies some obscure metaphysics?)
and Lorentz called it "ether"; it's whatever is required to accomodate
Newtonian frames as well as field and wave models.
- http://www.tu-harburg.de/rzt/rzt/it/Ether.html
>> > If you choose one or the other to be absolutely rotating you are
>> > physically creating a "preferred" FoR, violating the PoR.
>
>> Probably most physicists don't adhere to the general PoR
>> nowadays
>
> Did the GPoR get voted out, and I didn't get a memo?
That's quite the impression I have - and I didn't get a memo either!
>> ("modern usage demotes the uniform "gravitational" field back to
>> its old status as a pseudo-field")
- http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_...
>> and most people mean with "PoR" the PoR of Newton and Poincare - which
>> Einstein called the "special" PoR.
>
> "most physicists" and now "most people", who can
> argue with "THEM".
> ...
Sorry I meant most textbooks and papers that I've seen about it as well as
of course the abovementioned physics FAQ.
>> > Well we are edging toward EP + "Mach's Principle".
>> > I think a new thread based on,
>> >http://www.teleconnection.info/rqg/TheEquivalencePrinciple
>> > would be a good start.
>>
>> That could be nice - but note that I won't be able to participate for the
>> coming weeks.
>
> Well, I'd like to go more mathematical too.
> See you later.
> Ken S. Tucker
Until next month!
Cheers,
Harald
======================================= MODERATOR'S COMMENT:
Sciama's 1953: Inertia is the effect of the current local gravito-elm field of all masses -> no action-at-distance necessary
harry
See above.
I wonder about the purpose of that MODERATOR'S COMMENT; it appears that some
more clarification may be useful. With field theory there is no
action-at-a-distance as the action is locally brought about by the field;
thus the "field" is assumed to physically exist. The (local) "field" is not
identical to "the distant stars". For an elaborate discussion, see
Einstein's 1920 presentation that is linked above.
Regards,
Harald
Peter
...
> ======================================= MODERATOR'S COMMENT:
> Sciama's 1953: Inertia is the effect of the current local gravito-elm field
> of all masses -> no action-at-distance necessary
>
> I wonder about the purpose of that MODERATOR'S COMMENT; it appears that some
> more clarification may be useful. With field theory there is no
> action-at-a-distance as the action is locally brought about by the field;
The gravito-elm field is a field like Maxwell's elm field
> thus the "field" is assumed to physically exist.
yes, why not?
the point is,
(i), that Einstein didn't know it (nobody at his time did, although
the analogy to elm is obvious;
(ii) Maxwell's elm field can exist without sources, while Newton's
field cannot
> The (local) "field" is not
> identical to "the distant stars".
of course, it is, see elementary electromagnetism
> For an elaborate discussion, see
> Einstein's 1920 presentation that is linked above.
here, gravito-electromagnetism is completely ignored
Peter
sels...@gmail.com
Machs Principle
***************
Mach's principle is defined as inertia originates in a kind of
interaction between bodies. (Wikipedia describing Einsteins view of
inertia recuperated the 18/02/2009)
Now most scientists agree with Mach's principle but few have a simple
explication of how it works.The words "kind of" are also not very
scientific.
Many experiments have proven this to many decimal values. Einstein's
general theory of relativity is a response to this question.
I will propose this formula over here without proof for simplicity.
F_acc=-G/c^2*M*m/r*a^2
G is the the gravitational constant
c the speed of light
M a distant mass
m the mass before us
and a the acceleration of mass m
This equation describes force on a far a way mass due to acceleration
and is what is responsible for inertia.
If we plug in some values we will see that the force in normal
circumstances is very nearly zero.
The first thing to note is the equation has a 1/r and not a 1/r^2.As
in general we will assume that the density is about equal in the whole
universe far away masses will have a much larger effect than those
near by. In order to understand this better one could look up any good
explication of Olber's paradox which applies if effect = 1/r^2.
If we now consider the acceleration of a mass m there will be an equal
and opposite force due to all the forces in the whole universe .
The sum of these forces is
f= m*integrate(G/c^2*(4*%pi*Density_Universe*r^2)/r,r,
0,Radius_Of_Universe)*a
Taking
radius of universe as 7e26m
and density of universe 9.9 e-27 kg/m^3
We get
G/c^2*integrate((4*%pi*Density_Universe*r^2)/r,r,0,Radius_Of_Universe)
==23
The true value is exactly 1 so that
f=m*a
Thus
2*G/c^2*%pi*Density_Universe*Radius_Of_Universe^2==1
This gives us a correspondence between the radius of the universe and
its density.
When one mass accelerates it causes all other masses in the whole
universe to accelerate in the opposite direction.
===============>
Our equation is f=m*a as we chose our units that this would be so.
We can see from this that if the ratio of the mass and / or radius of
the universe changes either c or G should also change.
Thus Mach's inertia is not just a philosophical question but a real
force that could be measured with extreme accelerations and very ,very
accurate measurements.
brad
> Thus spake David Rutherford <drutherf...@softcom.net>
>
>
>
>
>
> >Oh No wrote:
> >> Thus spake David Rutherford <drutherf...@softcom.net>
>
> >>>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.
> >> 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,
True!
> > but the observer at rest in the
> >gravitational field `feels' a force.
Not true! He only feels the force if he moves.
IOW he provides the acceleration that *illuminates* the field.
Aren't both instances considered inertial reference
frames, so long as the observer doesn't *move*?
Isn't modern cosmology based on the consideration that
Earth can be considered a fixed reference frame compared to
the distant stars? Isn't this equivalent to calling Earth's
reference frame an inertial one? Yet Earth resides in Sun's
G Field and *feels* its gravity at every point in its orbit.
In every case it is the observer who defines the nature
of his confinement.
> >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).
This is the basis of my *misunderstanding* also, so I sympathize
with his statement. Perhaps, the equivalence principle needs updating?
Or, perhaps, I need *illumination*.
Thanks, Brad