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Author Topic: Why Light folows the space-Time curve?Not the the shorthest Line?  (Read 12395 times)

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.
 

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
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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

 

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?
 

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.
Quote
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 »
 

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?
 

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:)..
 

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.

 

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!
 

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 :)
 

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
 

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. :)
 

lean bean

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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.
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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''

 

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.
 

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 »
 

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
 

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 »
 

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?
 

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'.
 

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.
 

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.
 

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. 
 

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 »
 

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 »
 

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 »
 

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").
 

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