[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...

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

[Pmb]

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.

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)

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

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

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.

[lean bean]

From "Relativity and the Problem of Space" Albert Einstein (1952)
http://www.relativitybook.com/resources/Einstein_space.html

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...

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.

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)

[Pmb]

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.

[lean bean]

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)…

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

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…

[AndroidNeox]

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.

[Pmb]

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

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

[yor_on]

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]

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.

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…

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.

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.

[AndroidNeox]

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?

[yor_on]

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:)..

[Pmb]

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.

[Pmb]

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!

[yor_on]

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

[Spacetectonics (OP)]

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

[Pmb]

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]

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.

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''

[Pmb]

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]

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

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?

[imatfaal]

"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