Why Light folows the space-Time curve?Not the the shorthest Line?

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Offline Spacetectonics

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Sounds like a stupid question !
But,
According to general relativity mass curves space-time giving the effect of gravity,
Why light follows a curved path as the shortest possible path between two points,when crossing ,say a huge mass ?and not a straight line ?
OK ,we may say it is because of the space-time "curved" and its intrinsic characteristic!now my question is what caused space to curve"bend"?
Lets be straight forward,HOW matter could "curve" the space-time?Under what physical regime,what formula?!

« Last Edit: 04/01/2013 14:16:29 by Spacetectonics »

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Offline Pmb

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Quote from: Spacetectonics
Sounds like a stupid question !
You asked
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Why Light follows the space-time curve?
It follows a spacetime curve by definition. That is to say that the name we give to the set of events through which light passes is defined as a worldline.

Quote from: Spacetectonics
But,
According to general relativity mass curves space-time giving the effect of gravity,
Its more accurate to say that mass can curve spacetime. It’s conceivable to have a distribution of matter and the spacetime region near the matter still be flat. It’s quite possible to have a gravitational field in flat spacetime.


Whether the spacetime is flat or curved, any particle, free or not, will my on a worldline by definition. A free particle moves on a geodesic.

Quote from: Spacetectonics
Why light follows a curved path as the shortest possible path between two points, when crossing ,say a huge mass ?and not a straight line ?
A particle follows a path on which the proper time is extremal. It can be shown from this that such a particle follows a geodesic. The derivation is pretty mathematical so if your mathematically inclined you can follow the derivation I placed on my website at
http://home.comcast.net/~peter.m.brown/math_phy/geodesics.htm

It shouldn’t call it the “shortest possible path” because that phrase refers to a spatial trajectory, not a spacetime trajectory. And that it does it a postulate so it can’t be derived from more fundamental facts.

Quote from: Spacetectonics
OK ,we may say it is because of the space-time "curved" and its intrinsic characteristic!now my question is what caused space to curve"bend"?
Lets be straight forward,HOW matter could "curve" the space-time?Under what physical regime,what formula?!
Nobody knows the “why”’s of your questions. We just know the “how”’s and they’re given by general relativity. Einstein’s field equations describes how spacetime curvature is related to matter. For more on Einstein’s field equations please see
http://home.comcast.net/~peter.m.brown/gr/einsteins_field_equations.htm

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Offline Spacetectonics

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Thanks BMP,

I have read about GP.B and have some inquires!!

Einstein's theory of General Relativity predicts that Earth, by rotating, twists space and time around with it, forming a mild vortex in the fabric of spacetime around our planet. Researchers call this "frame dragging." Most physicists believe the spacetime vortex is real, but no experiment to date has been sensitive enough to detect it unequivocally.

Until this has been done:   http://science.nasa.gov/science-news/scien.../26apr_gpbtech/

"Einstein’s field equations describes how space-time curvature is related to matter"

Does that mean everything defined as "mass ",could effect Space-T?!
-----------------------

When saying :  "by rotating, twists space and time around with it, forming a mild vortex"

And considering  that  "Sun rotates once every 27 days at the equator" and solar system rotates at the center of M.W galaxy  ,then should not we observer effect of a huge vortex at about our sun?and consequently on the light received from outside of the milky way?from other stars,including ones at these experiments?

In another world, have these vortexes affected the "light observation experiments"including the one you have mentioned?
What happen to the light crossing these vortexes?!(or by expanding this into "even" milky way, this effect is still negligible?)

Cheers,

« Last Edit: 05/01/2013 08:03:25 by Spacetectonics »

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lean bean

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It’s conceivable to have a distribution of matter and the spacetime region near the matter still be flat. It’s quite possible to have a gravitational field in flat spacetime.

Perhaps I'm being pernickety.
Some time ago on a NASA site there was a question... '' is spacetime a thing'' the answer given was... ''Yes, it's another name for the gravitational field of the universe''
Is it in this sense that you mean you can have a gravitational field even though spacetime may be flat?  i.e. a flat spacetime is still a gravitational field?
Or is it something to do with the evenness of the distribution of mass/energy in a region ?
« Last Edit: 05/01/2013 10:51:19 by lean bean »

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Offline yor_on

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To me it is energy dependent :)

Our universe is picky, and don't want to spend more energy than it must. So if the cheapest 'way' is bent, then that is the way it will follow. As for what is a straight line I'm not sure. I would call that observer dependent too.
=

Or geometry dependent?
Or all of the above possibly?
« Last Edit: 05/01/2013 19:17:32 by yor_on »
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Offline Pmb

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Quote from: lean bean
Perhaps I'm being pernickety.
Some time ago on a NASA site there was a question... '' is spacetime a thing'' the answer given was... ''Yes, it's another name for the gravitational field of the universe''
Then that NASA site is wrong. Another name for the gravitationall field is not the universe. That's just plain silly.

Quote from: lean bean
Is it in this sense that you mean you can have a gravitational field even though spacetime may be flat?  i.e. a flat spacetime is still a gravitational field?
Or is it something to do with the evenness of the distribution of mass/energy in a region ?
Spacetime curvature is the same thing as tidal forces. Tidal forces cannot be transformed away. For that reason some relativists like to think of spacetime curvature as being the true signature of the presense of a gravitational field. Einstein disagreed with that perspective and so don't I.

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Offline evan_au

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Re: mild vortex in the fabric of spacetime around our planet. Researchers call this "frame dragging."

The Earth, the Sun and the Galaxy have a relatively low density, and a fairly low rate of spin, so frame dragging was undetectable (until Gravity Probe B).

Neutrons stars have the have the mass of the sun in a diameter of about 10km, and can spin at up to 1000 revolutions per second. Frame dragging would be more easily detectable in this environment.

But the environment where it has the biggest impact is in the vicinity of a spinning black hole, which could have a mass much larger than the Sun in a smaller volume than a neutron star, with a greater angular velocity, with no risk of flying apart.

Some links to this story:
Frame Dragging due to Earth http://science.nasa.gov/missions/gravity-probe-b/
Frame Dragging due to a Black Hole http://science.nasa.gov/science-news/science-at-nasa/1997/ast06nov97_1/
« Last Edit: 06/01/2013 15:06:21 by evan_au »

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lean bean

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Spacetime curvature is the same thing as tidal forces. Tidal forces cannot be transformed away. For that reason some relativists like to think of spacetime curvature as being the true signature of the presense of a gravitational field. Einstein disagreed with that perspective and so don't I.

I think I can understand how comparing different points in a region leads to the idea of tidal forces and so to the idea of curvature. But, you say like Einstein you don’t agree with that idea?
If it’s not too mathematical could you explain the grist of what Einstein would agree with in defining gravity in a region if it’s not comparing different points of that region?

To me it is energy dependent :)
Our universe is picky, and don't want to spend more energy than it must. So if the cheapest 'way' is bent, then that is the way it will follow. As for what is a straight line I'm not sure. I would call that observer dependent too.

 Yes I can see something of where your coming from. A rock will take the path that will give the rocks wristwatch the maximum proper time in spacetime, that maybe the straightest path in spacetime but not necessarily a straight path in just space. Something like...in the rock's frame there's no energy expended.
I think!

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Offline Pmb

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Quote from: lean bean
I think I can understand how comparing different points in a region leads to the idea of tidal forces and so to the idea of curvature. But, you say like Einstein you don’t agree with that idea?
No. In essence, what Einstein said was that the presence of a gravitational field was not the non-vanishing of tidal forces (Reimann tensor) but the non-vanishing of gravitational acceleration (affine connection).

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lean bean

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Quote from: lean bean
I think I can understand how comparing different points in a region leads to the idea of tidal forces and so to the idea of curvature. But, you say like Einstein you don’t agree with that idea?
No. In essence, what Einstein said was that the presence of a gravitational field was not the non-vanishing of tidal forces (Reimann tensor) but the non-vanishing of gravitational acceleration (affine connection).
Thanks for getting back.
Now, you have made me really pernickety and allowing me to display my ignorance here...
I thought, when talking of GR, it was because of the differences at difference points we get acceleration?
What is causing a rock to accelerate, if all it is doing is moving in a spacetime which is the same at all points?
Not too mathematical please :)


« Last Edit: 06/01/2013 17:30:59 by lean bean »

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Offline Pmb

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Quote from: lean bean
I think I can understand how comparing different points in a region leads to the idea of tidal forces and so to the idea of curvature. But, you say like Einstein you don’t agree with that idea?
No. In essence, what Einstein said was that the presence of a gravitational field was not the non-vanishing of tidal forces (Reimann tensor) but the non-vanishing of gravitational acceleration (affine connection).
Thanks for getting back.
Now, you have made me really pernickety and allowing me to display my ignorance here...
I thought, when talking of GR, it was because of the differences at difference points we get acceleration?
What is causing a rock to accelerate, if all it is doing is moving in a spacetime which is the same at all points?
Not too mathematical please :)



There is no theory of what causes objects to accelerate.

Picture yourself in a uniform gravitational field. When standing in such a field you can't tell that you're in a uniform g-field or at rest in a uniformly accelerating frame of reference. in flat spacetime. However if there were tidal forces present then you'd b able to tell the difference. If tidal forces are present then you know its caused by a gravitational field. There is no way to simply change your frame of reference to mimic a frame of reference in which there are tidal forces.

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lean bean

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The concept of mass is only metaphysically tied to matter. The idea of "curved space-time" is gibberish. "Time", "space" and "space-time" are abstract coordinate systems invented to help describe reality, they are not somethingreal you can manipulate or "curve" in any fashion.

If from experiment and observation we create our reality and so our models (GR), then in what other way do you wish to explain reality if not by models?
 Observation, and so our models, suggests something about mass/energy alters spacetime in their vicinity.
Can you suggest a way of getting to the fundamental ''truth'' (whatever that means) and how we will know it is the fundamental ''truth'' and not just a very good model?
« Last Edit: 12/01/2013 16:14:04 by lean bean »

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Offline AndroidNeox

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Actually, light does follow a straight line, from its own perspective. Imagine you have a long string that doesn't weigh anything but is very strong pulled tightly. Shoot a light beam alongside it and the light and string will remain parallel, even as they pass through bent spacetime.

I was surprised to read that the NASA site says spacetime is a thing. Not only is it not a thing, but the universe appears to have no net gravity, when you add up all of the positive and negative energies the result seems to be zero.

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lean bean

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I was surprised to read that the NASA site says spacetime is a thing. Not only is it not a thing, but the universe appears to have no net gravity, when you add up all of the positive and negative energies the result seems to be zero.
I have spent a little time googling for that Nasa site where the question was asked,it was some time ago when I first saw it...

Bit more googling and Nasa site ...
Where does space come from?
Quote
Essentially, space is what we refer to as three of the four dimensions to a more comprehensive entity called the space-time continuum, and this continuum is itself just another name for the gravitational field of the universe. If you take away this gravitational field -- space-time itself vanishes!

http://www.nasa.gov/centers/kennedy/about/information/science_faq.html#46

These are interesting saying the same...
http://www.astronomycafe.net/qadir/BackTo283.html

http://www.astronomycafe.net/qadir/BackTo289.html

Don't shoot the messenger.




« Last Edit: 13/01/2013 19:12:24 by lean bean »

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Offline Pmb

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Quote from: AndroidNeox
Actually, light does follow a straight line, from its own perspective.
In what way does light does have a perspective? Light follows a geodesic if that’s what you mean but in a gravitational field its spatial trajectory is curved.

Quote from: AndroidNeox
Imagine you have a long string that doesn't weigh anything but is very strong pulled tightly. Shoot a light beam alongside it and the light and string will remain parallel, even as they pass through bent spacetime.
Not in general. E.g. in a uniform gravitational field a taught string will have the shape of a catenary. The more tension in the string the less it sags but regardless of how strong the tension is there will always be some sag.

Quote from: AndroidNeox
Shoot a light beam alongside it and the light and string will remain parallel, even as they pass through bent spacetime.
That’s incorrect. If you tied a string between two walls in a gravitational field the shape of the string will be a catenary. Now aim a laser beam starting off parallel with the string at one end. The string and laser beam will diverge. This happens because light falls in a gravitational field and thus the beam of light will be deflected downwards just as if you were throwing a ball parallel to the ground and it dropped as if moved and eventually hits the ground
« Last Edit: 13/01/2013 23:35:34 by Pmb »

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lean bean

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Quote
Actually, light does follow a straight line, from its own perspective. Imagine you have a long string that doesn't weigh anything but is very strong pulled tightly. Shoot a light beam alongside it and the light and string will remain parallel, even as they pass through bent spacetime.

Android
I think your find the idea is rocks, light and anything else unpowered follow a curved spacetime path in the vicinity of massive bodies such as the sun/ black holes. Powered things like rockets can travel in straight line paths ignoring curved spacetime.
Your ''pulled tightly'' straight rope has been pulled straight, that is, the work done in pulling, has stopped the rope from following a curved spacetime path. So light would not follow the rope parallel
I think :)


« Last Edit: 14/01/2013 18:21:16 by lean bean »

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Offline AndroidNeox

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Maybe I didn't make it clear that I was presenting a thought experiment in which a string of zero weight were under tension. Since it has no weight, it will not sag. It will be perfectly straight, locally, just as a light beam is. A light beam always follows a straight line, from its own perspective. I presume one can meaningfully speak about the perspective of a light beams since Einstein depended upon such thought experiments in forming Relativity. In thought experiments, one should ignore every inconvenience of real physics, so long as the logical basis of the argument isn't invalidated.

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Offline Pmb

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Maybe I didn't make it clear that I was presenting a thought experiment in which a string of zero weight were under tension. Since it has no weight, it will not sag. 
So you're not talking about a real string and thus not a real lsitiation but merely an approximation to the real thing. In anycase the string and beam of light will not remain parallel if there is a gravitational field present which is perpendicular to the beam of light.

In any case you still have given no meaning to the statement "A light beam always follows a straight line, from its own perspective." What does this mean? What is the perspective of light? What does it mean for light to have a perspective?.On what basis are you justifying this assertion? Are you speaking about a straightline in spacetime or in space? Is this spacetime curved? Is the space curved? If so then "straight line" has no meaning for curved surfaces.

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Offline AndroidNeox

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In any case you still have given no meaning to the statement "A light beam always follows a straight line, from its own perspective." What does this mean? What is the perspective of light? What does it mean for light to have a perspective?.On what basis are you justifying this assertion? Are you speaking about a straightline in spacetime or in space? Is this spacetime curved? Is the space curved? If so then "straight line" has no meaning for curved surfaces.

Here's a wikipedia reference that might help: http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

"In general relativity, a geodesic generalizes the notion of a "straight line" to curved spacetime."

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Offline Pmb

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In any case you still have given no meaning to the statement "A light beam always follows a straight line, from its own perspective." What does this mean? What is the perspective of light? What does it mean for light to have a perspective?.On what basis are you justifying this assertion? Are you speaking about a straightline in spacetime or in space? Is this spacetime curved? Is the space curved? If so then "straight line" has no meaning for curved surfaces.
That is true only for null worldliness, i.e. for particles moving at the speed of light.

The point is that only particles that move at the speed of light have speed c. The other person was talking about the speed of objects in spacetime where I was speaking about the speed of objects in space. The term “speed of an object in spacetime” is meaningless. There are a few people here and there who like to say that all objects move at the speed of light in spacetime but no relativist would ever make such a statement. In fact when they hear that they sigh and say that its wrong.
." What does this mean? What is the perspective of light? What does it mean for light to have a perspective?.On what basis are you justifying this assertion? Are you speaking about a straightline in spacetime or in space? Is this spacetime curved? Is the space curved? If so then "straight line" has no meaning for curved surfaces.
[/quote]

Here's a wikipedia reference that might help: http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

"In general relativity, a geodesic generalizes the notion of a "straight line" to curved spacetime."
[/quote]
I'm well aware of what a geodesic is and how its viewed as the 'straightest possible line in a curved space". The phrase "light beam always follows a straight line, from its own perspective." is not something that has been used to mean that by anybody else I've ever seen. If we were to use that kind of terminology then It can always be said that that all free particles move in straight lines from their own perspective. That holds true because all free particles move on geodesics

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lean bean

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Quote from: lean bean
Perhaps I'm being pernickety.
Some time ago on a NASA site there was a question... '' is spacetime a thing'' the answer given was... ''Yes, it's another name for the gravitational field of the universe''

Then that NASA site is wrong. Another name for the gravitationall field is not the universe. That's just plain silly.

 I have found another site where a Dr Sten Odenwald goes into more detail as to why he thinks...
Quote
Why is the gravitational field of the universe another name for space-time?

The development of any mathematical theory of natural phenomena such as gravity requires that the mathematical symbols defining the theory must be related to qualities of the phenomena such as the symbol T representing temperature, V representing velocity or M representing mass. In general relativity, a similar association had to be made by Einstein. We have seen how Einstein defined the gravitational field to be identical to the so-called metric tensor,

 g mu,nu

used by Riemann to describe the geometry of a space. This means that where Newtonian gravity dealt with one quantity to measure the gravitational field, Einstein's theory in the guise of "g-mu-nu" required a total of 10 unique quantities to more completely define how the gravitational field behaved. The force of gravity defined as changes in the gravitational field from place to place in Newtonian mechanics, was replaced by changes in the geometry of space from place to place in spacetime measured by the degree of curvature symbolized by "C-mu-nu" at each point. Einstein's minimalist adoption of "g-mu-nu" as the embodiment of the gravitational field was significant and has far-reaching ramifications. Before Einstein, the metric tensor "g-mu-nu" was a purely geometric quantity that expresses how to determine the distances between points in space. Geometers from the time of Gauss knew nothing about forces, mass and momentum, they did however use the metric tensor to uncover new and bizarre spaces resembling nothing that humans have ever experienced.

Einstein's appropriation of the metric tensor so that it also represented the gravitational field led to an inevitable, logical conclusion: If you took away the gravitational field, this meant that "g-mu-nu" would be everywhere and for all time equal to zero, but so too would the metric for spacetime. Spacetime would lose its metric, the distance between points in the manifold would vanish, and the manifold itself would disappear into nothingness. In Relativity: The Special and General Theory page 155, Einstein expressed this quality of spacetime as follows,
"Spacetime does not claim existence on its own but only as a structural quality of the [gravitational] field"
The article continues...
http://einstein.stanford.edu/content/relativity/q2442.html

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Offline Pmb

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Quote from: lean bean
I have found another site where a Dr Sten Odenwald goes into more detail as to why he thinks...
He’s quite wrong. I explain why below.

Quote
Why is the gravitational field of the universe another name for space-time?

The development of any mathematical theory of natural phenomena such as gravity requires that the mathematical symbols defining the theory must be related to qualities of the phenomena such as the symbol T representing temperature, V representing velocity or M representing mass. In general relativity, a similar association had to be made by Einstein. We have seen how Einstein defined the gravitational field to be identical to the so-called metric tensor,

 g mu,nu



Einstein's appropriation of the metric tensor so that it also represented the gravitational field led to an inevitable, logical conclusion: If you took away the gravitational field, this meant that "g-mu-nu" would be everywhere and for all time equal to zero, but so too would the metric for spacetime.
The reason why he’s wrong is because he’s misquoting Einstein. Einstein never said such things.

Einstein said in describing gravity the following (by “ordinary theory” he mean SR and the absence of gravity)
Quote
The case of the ordinary theory of relativity arises out of the here case considered if it is possible, by reason of the particular relations of the g-mu-nu in a finite region, to choose the system of reference in the finite region in such a way that the g-mu-nu assume the constant values diag(1, -1, -1, -1)
Its only when the metric tensor components are not constants and the coordinates Cartesian that when the system of reference leaves the components of the metric tensor as variables will there be a gravitational field. When g-mu-nu = diag(1, -1, -1, -1) there exists no gravitational field, just spacetime. So gravity and spacetime are quite different, but related, things. You can certainly have spacetime with no gravitational field.

He also said
Quote
The force of gravity defined as changes in the gravitational field from place to place in Newtonian mechanics, was replaced by changes in the geometry of space

Compare this with what Einstein said in a letter to Lincoln Barnett date June 19, 1948
Quote
I do not agree with the idea that the general theory of relativity is geometerzing physics or the gravitational field.  The concepts of physics have always been geometrical concepts and I cannot see why the g_ik field should be called more geometrical than f.i. the electromagnetic field or the distance of bodies in Newtonian mechanics. The notion probably comes from the fact that the origin of the g_ik field is the Gauss-Riemann theory of the metrical continuum which we are wont to look at as a part of geometry. I am convinced, however, that the distinction between geometrical and other kinds of fields is not logically founded.
« Last Edit: 20/01/2013 16:37:47 by Pmb »

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lean bean

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From "Relativity and the Problem of Space" Albert Einstein (1952)
http://www.relativitybook.com/resources/Einstein_space.html
Quote
Space-time does not claim existence on its own, but only as a structural quality of the field.
If we imagine the gravitational field, i.e. the functions gik, to be removed...

Quote
On the basis of the general theory of relativity, on the other hand, space as opposed to "what fills space", which is dependent on the co-ordinates, has no separate existence. Thus a pure gravitational field might have been described in terms of the gik (as functions of the co-ordinates), by solution of the gravitational equations. If we imagine the gravitational field, i.e. the functions gik, to be removed, there does not remain a space of the type (1), but absolutely nothing, and also no "topological space". For the functions gik describe not only the field, but at the same time also the topological and metrical structural properties of the manifold.
Quote
A space of the type (1), judged from the standpoint of the general theory of relativity, is not a space without field, but a special case of the gik field, for which – for the co-ordinate system used, which in itself has no objective significance – the functions gik have values that do not depend on the co-ordinates. There is no such thing as an empty space, i.e. a space without field.

  I take it that the space of type (1) here is Minkowski space.
If I'm misquoting or what may seem like selective quoting then it's unintentional...it's not my field(pun)
« Last Edit: 20/01/2013 18:50:19 by lean bean »

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Offline Pmb

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Quote from: lean bean
There is no such thing as an empty space, i.e. a space without field.
Frankly I find that to be just plain creepy. What could he possibly have had in mind for the definition of the term “field” when he wrote that.

In any case he did not mean that such a field was a gravitational field.

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lean bean

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In any case he did not mean that such a field was a gravitational field.
He seems to be  referring to a gravitational field.
Look at the first sentence here ( quote below)…
Quote
If we imagine the gravitational field, i.e. the functions gik, to be removed, there does not remain a space of the type (1), but absolutely nothing, and also no "topological space". For the functions gik describe not only the field, but at the same time also the topological and metrical structural properties of the manifold.
''If we imagine the gravitational field, i.e. the functions gik, to be removed,''
 So here he is associating the gravitational field with the functions gik.

Then he goes on to make a point about type one space
Quote
A space of the type (1), judged from the standpoint of the general theory of relativity, is not a space without field, but a special case of the gik field, for which – for the co-ordinate system used, which in itself has no objective significance – the functions gik have values that do not depend on the co-ordinates. There is no such thing as an empty space, i.e. a space without field.
So there his saying a space of type one is a special case of a gravitational field. And the very next sentence you have ''There is no such thing as an empty space, i.e. a space without field.''
So I think he is referring to a gravitational field.

And as he says, if you remove the gravitational field, your not even left with a type one space, your left with but absolutely nothing…
« Last Edit: 21/01/2013 18:05:18 by lean bean »

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Offline AndroidNeox

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I would presume that when Einstein said, "There is no such thing as an empty space, i.e. a space without field" he was referring to the fact that spacetime isn't a "thing". Spacetime is an artifact of the existence of stuff. Personally, I would say that spacetime is an artifact of the requirements for causal interactions.

When we get to the most fundamental questions, we need to use more care in our thinking. After all, science is not about "the universe". Science is about observation. Observation is not special... every physical event qualifies as an observation... when an electron absorbs a photon, that counts as an observation. Also, when I say "observation" I do not refer to "interpretation"... I refer to the totality of quanta by which the "observer" changes (absorbs or loses).

If one considers some of the deeper questions in those terms, they often become simpler. Then again, some questions are so freighted with bad assumptions that they should be discarded and replaced. For example, "What is time?" presumes time is a thing. A better question is, "When we measure time, what do we measure?" When we measure space, what do we measure?

The fact that quantum mechanics is entirely about what is "observable" and that Einstein based his thought experiments on the assumption that the appearance of the universe (light beams altered by motion and/or acceleration) to be, not the "appearance" of reality, but reality itself. It's not just that a rapidly moving object looks smaller... you can actually fit a 12 inch ruler through a 1 inch hole, if it's moving fast enough WRT the hole.

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Offline Pmb

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Quote from: lean bean
And as he says, if you remove the gravitational field, your not even left with a type one space, your left with but absolutely nothing…
In his text Einstein identifies the presence of a gravitational field with the spacetime variability of the copmponents of the metric tensor. Thus

ds2 = (1 + gz/c2)2 (cdt)2 - dx2 - dy2 - dz2

denotes the presence of a uniform gravitational field while

ds2 =  (cdt)2 - dx2 - dy2 - dz2

denotes the absence of a gravitational field. In section 2 of Einstein's review paper on GR (1916) he wrote
Quote
It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of coordinates.
That means that whether there is a gravitational field present or not depends on the choice of spacetime coordinates. Typically there is a gravitational field in non-inertial frames of reference. This comes pretty much as a shock to those not familiar with general relativity, but thes actually are Einstein's views.

Warning: Don't confuse this with those changes of coordinates from ain inertial frame of reference in flat spacetime using Cartesian coordinates to one in which curvilinear coordinates are used in an inertial frame of reference. The components will still vary with the position in spacetime. However there is still no gravitational field


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Offline yor_on

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A question about observer dependencies?

If I assumed that 'gravity' always need to be observed in some coordinate system to 'exist' as a global phenomena, including all observers description. Can we then assume a 'space' that no observers would be able to define a 'gravity' too?

If we can then 'space' clearly exist on its own, gravity not needed. If we can't?
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lean bean

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That's a good point yor_on.


In his text Einstein identifies the presence of a gravitational field with the spacetime variability of the copmponents of the metric tensor. Thus

ds2 = (1 + gz/c2)2 (cdt)2 - dx2 - dy2 - dz2

denotes the presence of a uniform gravitational field while

ds2 =  (cdt)2 - dx2 - dy2 - dz2

denotes the absence of a gravitational field.
Can you link to anywhere where that's shown. google's not helping me.

The concept of space in GR is mentioned earlier in my Einstein link…
http://www.relativitybook.com/resources/Einstein_space.html
Sorry for the great chunks of quotes, don’t go to sleep.

View of space in classical mechanics and special theory of relativity.
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In accordance with classical mechanics and according to the special theory of relativity, space (space-time) has an existence independent of matter or field.

Now what GR makes of space…
Quote
On the basis of the general theory of relativity, on the other hand, space as opposed to "what fills space", which is dependent on the co-ordinates, has no separate existence.

So, on the basis of GR, space has no separate existence.
Quote
Space-time does not claim existence on its own, but only as a structural quality of the field.

That’s the gravitational field. So what happens to space when you take that field away? we know what Dr Sten Odenwald thinks.



« Last Edit: 23/01/2013 18:01:09 by lean bean »

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Offline AndroidNeox

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A question about observer dependencies?

If I assumed that 'gravity' always need to be observed in some coordinate system to 'exist' as a global phenomena, including all observers description. Can we then assume a 'space' that no observers would be able to define a 'gravity' too?

If we can then 'space' clearly exist on its own, gravity not needed. If we can't?

I'm sorry, I didn't understand this post. Are you suggesting a model with a universe that has space but no matter or energy at all?

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Offline yor_on

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Gravity is observer dependent to me. It has to do with what coordinate system you use. But that's also because I think of it as local definitions. It all depend on how far you want to take the subject 'gravity'. Assuming a globally existing 'objective universe' becomes to me a theoretical exercise in where the observer dependencies either are ignored, or somehow described theoretically as, let's call it 'null' as they all somehow need to take each other out to form this theoretical universe.

That universe is not what you see though. Your universe as defined from your local experiments, and as those experiments can be done no other way practically, is 'observer dependent', and what you measure is what you get (WYMWYG:)..
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Offline Pmb

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Quote from: lean bean
Can you link to anywhere where that's shown. google's not helping me.
No. I'm sorry. I can't find a derivation. They're hard to follow. Here is a list of sources which provide derivations

[1] Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173/
[2] Radiation from a Uniformly Accelerated Charge, David G. Boulware, Ann. Phys., 124, (1980), page174.
[3] Relativistic solutions to the falling body in a uniform gravitational field, Carl G. Adler, Robert W. Brehme, Am. J. Phys. 59 (3), March 1991.
[4] Gravitation, Charles. W. Misner, Kip S. Thorne, John Archibald Wheeler, (1973), sect 6.6.
[5] The uniformly accelerated reference frame, J. Dwayne Hamilton, Am. J. Phys., 46(1), Jan. 1978.


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Offline Pmb

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Gravity is observer dependent to me. It has to do with what coordinate system you use.
Wonderful! You actually brought a tear to me eye. It's so wonderful when someone actually gets it! Bravo, sir. Bravo!

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Offline yor_on

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ahem :)

What I'm really proud over Pete is WYSWYG  (now © nota bene .. And by me, by God, and no other:)
A lasting contribution to the proud use of Acronyms.
May they fill our universe(s)..

(Yes, I hate acronyms..
Never remember what they stand for, and makes me feel like an idiot hearing other use them with such ease :)
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Offline Spacetectonics

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And  how electromagnetic field of the earth" behaved ;if there were no such a curve ( curved space-time )? could it be different in shape?(has it really curved ,is it detectable by instruments?!!)

cheers

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Offline Pmb

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I don't know whyh it took me so long to recall this (getting old? lol) but the quantum to classical limit is not defined by h->0 but by Bohr's Correspondence Principle, the behaviour of a quantum system must approach the classical system in the loimit of large quantum numbers. The only thing you get when you take h->0 is to wipe out the wave-particle duality and the uncertainty principle.

I'll state this warning yet one last time: Do NOT confuse a "classical photon" with real photons. They aren't much alike. That's been lightarrow's problem all along. I.e. he confuses classical photons with real photons. The former do not exist, l while the later does. Comes for a lack of paying close enough attention I suspect. :)

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Gravity is observer dependent to me. It has to do with what coordinate system you use. But that's also because I think of it as local definitions.

Albert Einstein.
Quote
Before proceeding farther, however, I must warn the reader against a misconception suggested by these considerations. A gravitational field exists for the man in the chest, despite the fact that there was no such field for the co-ordinate system first chosen. Now we might easily suppose that the existence of a gravitational field is always only an apparent one. We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes. 
http://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch20.htm

I'm wondering here, does Einstein mean a uniform gravitational field when he refers to special form? ''but only for those of quite special form''


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Offline Pmb

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Quote from: lean bean
I'm wondering here, does Einstein mean a uniform gravitational field when he refers to special form? ''but only for those of quite special form''
No. He means those gravitational fields in flat spacetime. Consider a rotating frame of reference. In that frame there will be two inertial forces, The Coriolis force and the centrifugal force. Since there is an inertial force in the rotating frame there are gravitational forces/gravitational field in such a frame. But you can transform the field away by the proper coordinate transformation. The "special kind" that Einstein refers to are gravitational fields in which the spacetime is flat.

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lean bean

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No. He means those gravitational fields in flat spacetime. Consider a rotating frame of reference. In that frame there will be two inertial forces, The Coriolis force and the centrifugal force. Since there is an inertial force in the rotating frame there are gravitational forces/gravitational field in such a frame. But you can transform the field away by the proper coordinate transformation. The "special kind" that Einstein refers to are gravitational fields in which the spacetime is flat.

In around about way I did mean that (my nutty way of asking the question)...
Reading what came before my selected quote (my link). A man in a large chest is in flat space being pulled by at an uniform accelerating rate, and so that man is experiencing a uniform gravitational field. If the chest stopped being pulled, that field in the chest vanishes and the chestman would then be in the coordinate frame as someone who had watched the chest being pulled.
-----------------------
So, for my understanding and no point to be made by me...
A gravitational field  'produced' by pulling or rotating in flat space can be transformed away.
-----------------------
Einstein same link
Quote
It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes.
Is this impossible because of tidal gradients or because the earth is a natural generator of a gravitational field?  something else?
« Last Edit: 30/01/2013 16:07:34 by lean bean »

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Offline imatfaal

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"Transformed away" in physics normally means that you perform a mathematical operation, or change coordinate systems but keep whats happening the same!  You have not transformed away the acceleration - you have changed the physical situation.  an accelerated frame of reference is not an inertial frame

  I can look at a rock travelling at a constant velocity from my "fixed" position or I can mathematically show what I look like from the rest frame of the rock (ie the rock is no longer moving).  What I cannot do is a mathematical operation that allows me to say that the one point of view the earth has a gravitational field - but from another frame of reference, or coordinate system describe the earth without that field
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lean bean

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"Transformed away" in physics normally means that you perform a mathematical operation, or change coordinate systems but keep whats happening the same!  You have not transformed away the acceleration - you have changed the physical situation.  an accelerated frame of reference is not an inertial frame
Thanks for pointing that out. 
  I can look at a rock travelling at a constant velocity from my "fixed" position or I can mathematically show what I look like from the rest frame of the rock (ie the rock is no longer moving).
I can understand that.


imatfaal 
Quote
What I cannot do is a mathematical operation that allows me to say that the one point of view the earth has a gravitational field - but from another frame of reference, or coordinate system describe the earth without that field
Very late edit:Been doing some googling.
 Apparently, it is to do with not being able to transform away the tidal gradients of the earth's gravitational field, no matter how small your frame the gradients still exist.
So your right imatfaal, there is no mathematical operation you can do to transform the gradients away. :)
« Last Edit: 31/01/2013 18:42:02 by lean bean »

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Offline LetoII

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i see really long and complicated answers here.
Isn't it as simple as this: the shortest path (a straight line) is not the path of least resistance?

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Offline yor_on

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Maybe, depends on how you define gravity and space-time. Gravity is just a preferred direction to me though, nothing 'touchable', and nothing definable to conglomerations of different bosons for example. But we might find it otherwise, considering if we can prove a Higgs boson. That seem to open for 'densities'.
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Offline Pmb

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"Transformed away" in physics normally means that you perform a mathematical operation, or change coordinate systems but keep whats happening the same!
When it comes to GR it means the following: Suppose there is a gravitational field in the current frame of reference, The presence of the gravitational field manifests itself by letting an object go free by dropping it. If the body accelerates with respect to the current frame of reference it means that there is a gravitational field present. Now invoke a change of the system of coordinates corresponding to a locally inertial frame of reference, If a body is let free and it remains at rest and doesn’t accelerate then there is no gravitational field present. That is what it means to “transform the gravitational field away.” At least according to Einstein. Obviously the reverse is true in that you can produce a gravitational field by an appropriate change in coordinate systems.

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Offline AndroidNeox

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"Transformed away" in physics normally means that you perform a mathematical operation, or change coordinate systems but keep whats happening the same!  You have not transformed away the acceleration - you have changed the physical situation.  an accelerated frame of reference is not an inertial frame

  I can look at a rock travelling at a constant velocity from my "fixed" position or I can mathematically show what I look like from the rest frame of the rock (ie the rock is no longer moving).  What I cannot do is a mathematical operation that allows me to say that the one point of view the earth has a gravitational field - but from another frame of reference, or coordinate system describe the earth without that field

Right.

Equivalence principle allows one to equate a gravitational field with linear acceleration only if the field is uniform... for example, for someone standing on an infinitely wide plane of mass in an otherwise empty universe. Real world gravitational fields only approximate this, locally.

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Offline imatfaal

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"Transformed away" in physics normally means that you perform a mathematical operation, or change coordinate systems but keep whats happening the same!
When it comes to GR it means the following: Suppose there is a gravitational field in the current frame of reference, The presence of the gravitational field manifests itself by letting an object go free by dropping it. If the body accelerates with respect to the current frame of reference it means that there is a gravitational field present. Now invoke a change of the system of coordinates corresponding to a locally inertial frame of reference, If a body is let free and it remains at rest and doesn’t accelerate then there is no gravitational field present. That is what it means to “transform the gravitational field away.” At least according to Einstein. Obviously the reverse is true in that you can produce a gravitational field by an appropriate change in coordinate systems.


Sorry Pete but can you run that again?  How can you produce a gravitational field by a coordinate transform - you can show that acceleration is indistinguishable (tidal aside) but after that I am flummoxed; it is the "elevator car" that is either in a gravitational field or accelerating - you cannot just transform that away. 
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Offline Spacetectonics

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@ Pmb:

 Obviously the reverse is true in that you can produce a gravitational field by an appropriate change in coordinate systems.
[/quote]

Thanks Pmb,

Would it it be possible for you ,to define "Appropriate"; where you have mentioned "an appropriate change in coordinate systems" please?

« Last Edit: 05/02/2013 18:29:16 by Spacetectonics »

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Offline Pmb

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Quote from: AndroidNeox
Equivalence principle allows one to equate a gravitational field with linear acceleration only if the field is uniform... for example, for someone standing on an infinitely wide plane of mass in an otherwise empty universe. Real world gravitational fields only approximate this, locally.
In Newtonian gravity it is quite possible set up a distribution of mass to get a perfectly uniform field, at least in principle. Nothing is ever perfect to a zillion decimal places, right? :)  I created a web page describing such an example. It’s a cavity inside a spherical body whose center is offset from the center of the spherical body (which otherwise has uniform mass density). See http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

In GR there are stresses to take into account and those stresses contribute to the gravitational field. However the field is still uniform to a large degree of accuracy.


Quote from: imatfaal
Sorry Pete but can you run that again?  How can you produce a gravitational field by a coordinate transform - you can show that acceleration is indistinguishable (tidal aside) but after that I am flummoxed; it is the "elevator car" that is either in a gravitational field or accelerating - you cannot just transform that away. 
First let’s look at where I got that notion from just so that the world can be sure that it’s not pmb who has been creating wild fantasies in his mind.

From The Foundations of the General Theory of Relativity by A. Einstein, Annalen der Physik, 49, 1916.
Quote
It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to “produce” a gravitational field merely by changing the system of co-ordinates.
If you were to invoke a spacetime coordinate transformation that is changes from an inertial frame of reference S in flat spacetime to a uniformly accelerating frame S’ then observers in S’ will observe that there is a uniform gravitational field in their frame of reference.

If you were in a frame of reference in which there was a gravitational field of the Earth’s gravitational field then you can only transform the gravitational field away locally (i.e. in a small region of spacetime). Please explain what your objection is and what the talk about the elevator has to do with it? I.e. please explain why it can’t be transformed away? You do understand, don’t you, that when the spacetime is curved then you can only transform the field away locally? What local means has to do with the precision of the instruments that you’re using to detect the tidal forces.

Quote from: Spacetectonics
Thanks Pmb,

Would it it be possible for you, to define "Appropriate"; where you have mentioned "an appropriate change in coordinate systems" please?
You’re most welcome Sir! :)

Appropriate means that not all changes of spacetime coordinates will work. Only those of a special kind will work. Obviously changing only the spatial coordinates from Cartesian coordinates to spherical to polar coordinates won’t be able to produce a gravitational field. But changing from one set of spacetime coordinates to a set of spacetime coordinates corresponding to a frame of reference which is accelerating relative to an inertial frame is an appropriate transformation. Mind you, these are changes from one set of spacetime coordinates to another set of spacetime coordinates. Not merely from one set of spatial coordinates to another.
« Last Edit: 06/02/2013 18:21:51 by Pmb »

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lean bean

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If you were in a frame of reference in which there was a gravitational field of the Earth’s gravitational field then you can only transform the gravitational field away locally (i.e. in a small region of spacetime). Please explain what your objection is and what the talk about the elevator has to do with it? I.e. please explain why it can’t be transformed away? You do understand, don’t you, that when the spacetime is curved then you can only transform the field away locally? What local means has to do with the precision of the instruments that you’re using to detect the tidal forces.
That’s answered something I was wondering about earlier.
Those tidal gradients are still there, it’s just a question of the precision of the instruments that you’re using to detect the tidal forces. Thanks. :)
« Last Edit: 06/02/2013 16:25:50 by lean bean »

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Offline Pmb

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Quote from: lean bean
That’s answered something I was wondering about earlier.
Those tidal gradients are still there, it’s just a question of the precision of the instruments that you’re using to detect the tidal forces. Thanks. :)
You’re welcome Sir! There are two opinions on this subject. One side says that the equivalence principle is wrong because you can detect the tidal forces while the other side says its right because you can ignore them. I’ve explained the “it’s right” side. Here is the “it’s wrong” side.

What is the principle of equivalence?, Hans C. Ohanian, Am. J. Phys. 45(10)), October 1977. The abstract reads
Quote
The strong principle of equivalence is usually formulated as an assertion that in a sufficiently small, freely falling laboratory the gravitational fields surrounding the laboratory cannot be detected. We show that this is false by presenting several simple examples of phenomena which may be used to detect the gravitational field through its tidal effects: we show that these effects are, in fact, local (observable in an arbitrarily small region). Alternative formulations of the strong principle are discussed and a new formulation of strong equivalence (the "Einstein principle") as an assertion about the field equations of physics, rather than an assertion about all laws or all experiments, is proposed. We also discuss the weak principle of equivalence and its two complimentary aspects: the uniqueness of free fall of a test particles in arbitrary gravitational fields ("Galileo principle") and the uniqueness of free fall of arbitrary systems in weak gravitational fields ("Newton's principle").