I Is a Gravitational Field Real or Just a Coordinate Transformation Artifact?

[Shirish]

I understand that; but so far every restatement you have made has been wrong and has required correction.

So be it - obviously I'm not wrong on purpose. Maybe I'm not too bright? It obviously is useful to me to be corrected so that I figure it out eventually (even if it takes me time). Not everyone is the same and what seems obvious to one may not be to another. At the end of the day, I gain understanding and a future viewer of the thread hopefully learns from it.

 

[PeterDonis]

obviously I'm not wrong on purpose.

Yes, that's understood.

 

[pervect]

I'm studying Susskind's GR TTM book, in which he gives a nice explanation of why differential geometry is needed for GR. But there is one gap that I want to fill.

The argument is: through a thought experiment, it seems that a uniform gravitation field can be seen as an artifact of going from an inertial frame $R$ to another one $R'$ that's uniformly accelerating w.r.t. $R$. More precisely, I can do a (curvilinear) coordinate transformation to $R'$ coordinate system. So the general question is: if there is a gravitational field $F$, how do we know if it's real or just the result of some weird coordinate transformation (i.e. a fake gravitational field)?

I'm not sure if specifying at force at every point will give us the information we really want to do an actual GR calculation. The sort of question a standard GR calculation can answer is "if we have the metric tensor at every point, is the space-time flat or curved?" Gravity, in GR, is fundamentally about the space-time metric, not about forces. It includes elements that can't be modeled by forces, such as effects on time (gravitational time dilation), effects on space (spatial curvature), and frame-dragging effects that do not neatly fit into the idea of a "force". Thinking of gravity "as a force" does not allow one to understand or even consider the existence of these additional GR effects.

There are various experiments we can do that don't involve the unsatisfactory notion of specifying forces at every point that can detect whether space-time is flat or not. It can also be insightful to consider the differences in geometry on a plane, which is flat, and geometry on the surface of a sphere, which is not flat. The idea of space-time curvature is rather abstract. The idea that a spherical surface is not a flat surface, while a plane is a flat surface, is probably one of the simplest and least abstract illustrations possible.

To carry out these tests we need at a minimum something equivalent to the notion of geodesic motion, the "natural" motion of a test particle that experiences no forces. An example of a suitable equivalent to geodesic motion would be the mathematical notion of a connection. A metric tensor would be best - this actually more than meets the requirements, so rather than being an equivalent notion it is a more complete notion.

For our simple example of the sphere, the notion of geodesic is simple - a geodesic on a sphere is a great circle. If we have additionally a metric tensor on our sphere, we can say that a geodesic - a great circle - is the shortest distance between two points that lies entirely on the sphere.

Testing to see if in initially parallel geodesics remain parallel, or whether they accelerate towards or away from each other, is one such test, and probably the simplest.

On the sphere, two great circles running through the poles would be initially parallel to each other at the equator - but they do not and cannot remain parallel, any pair of great circles must intersect at two points on a sphere.

You can apply this test in space-time, with a careful definition of "separation of geodesics". Basically, informallyh speaking, the separation should be measured in a co-moving inertial frame.

Other tests are possible, but I think it'd be better to keep things as simple as possible rather than attempting to describe them at this point.

 

[Shirish]

The correct statement would be that the differences between a gravitational field and an accelerating reference frame reduce over smaller 4d regions, and become undetectably small at some scale for any finite measurement precision. The manifestation of this in the maths is that the metric can always be diagonalised and first derivatives of it can be made to vanish at a point.

Thanks! If you don't mind, I want to clarify another thing.
I'm trying to understand the motivation of modeling gravitation using spacetime curvature. Here's my limited understanding:

1. "Fake" gravitational field can be made to vanish using a global coordinate transformation or can be introduced using one (regular to curvilinear or vice-versa)
2. Curvilinear coordinate system has a funky-looking metric (curved manifolds also have funky-looking metrics). So an inverse transformation back to the regular coordinate system diagonalizes the metric
3. In the book it's written that if on a manifold we can diagonalize the metric everywhere using a single coordinate transformation, then that manifold is flat
4. So for a flat manifold, I can switch from a metric to other weird-looking metrics using global coordinate transformations. And each of those transformations also induces a "fake" gravitational field
5. Thus there is a clear parallel between flat spacetime and fake gravitational field - both are linked by the existence of a global coordinate change that diagonalizes the metric in the former and removes the latter
6. A global coordinate change can't remove a "real" gravitational field, so conversely no global coordinate change can induce a real gravitational field
7. Similarly, in curved spacetime, no global coordinate change can diagonalize the metric everywhere
8. Thus there is a clear parallel between curved spacetime and real gravitational field - both are linked by the non-existence of a global coordinate change that can diagonalize the metric everywhere in the former, and can remove the latter

Does the above seem like a comprehensive enough justification for why we model gravity using curved spacetime? Or is there anything else I'm missing - maybe some result in differential geometry that provides a stronger justification?
(e.g. off the top of my head something like, "For a flat manifold, there exists no vector field that can be transformed away using a global coordinate change")

 

[Ibix]

I think that seems overcomplicated to me, although there may be points I have overlooked.

I'd say that Minkowski noted that the Lorentz transforms could be read as implying spacetime. Curved coordinate systems give gravity-like effects, so curved spacetime might too. And Newtonian gravity has an assumption that all test particles fall with the same acceleration independent of mass, so a new theory had to behave like that.

The obvious thing that would completely destroy a spacetime curvature model is two free falling rest masses with the same initial conditions falling on different paths. A force model can accomodate that quite easily - for example if we find an element called Unobtanium that accelerates twice as fast as anything else in free fall then Newtonian gravity becomes $F=kGMm/r^2$, where $k$ is a material parameter that is 1 for everything except Unobtanium, for which it is 2. But a spacetime curvature model cannot accommodate it at all.

 

[PeterDonis]

I'm trying to understand the motivation of modeling gravitation using spacetime curvature.

The motivation was simple: once you have the idea of spacetime, it is obvious that the spacetime of Special Relativity is flat. That means that two objects that are always in free fall, if they are at rest relative to each other at any given instant, will remain at rest relative to each other forever.

But in the presence of gravity--"real" gravity, produced by a real object like a planet or star, not the "fake" gravity produced by being at rest in an accelerating rocket--the above is no longer the case. Objects that are always in free fall, and are at rest relative to each other at some instant, in general will not stay at rest relative to each other. In Newtonian terms, this is called tidal gravity. And in GR terms, it is called spacetime curvature, because geometrically, freely falling objects move on geodesics, and if they are at rest relative to each other at some instant but don't stay at rest relative to each other, that is geodesic deviation, which is already known to be equivalent to curvature of the manifold.

Notice, btw, that in the above I didn't mention any of your 6 points at all. IMO all of the things you are looking at in those 6 points are distractions. You can't really understand this issue by looking at coordinates. You have to understand the physical and geometric reasons why GR models tidal gravity as spacetime curvature.

 

[Shirish]

The motivation was simple: once you have the idea of spacetime, it is obvious that the spacetime of Special Relativity is flat. That means that two objects that are always in free fall, if they are at rest relative to each other at any given instant, will remain at rest relative to each other forever.

But in the presence of gravity--"real" gravity, produced by a real object like a planet or star, not the "fake" gravity produced by being at rest in an accelerating rocket--the above is no longer the case. Objects that are always in free fall, and are at rest relative to each other at some instant, in general will not stay at rest relative to each other. In Newtonian terms, this is called tidal gravity. And in GR terms, it is called spacetime curvature, because geometrically, freely falling objects move on geodesics, and if they are at rest relative to each other at some instant but don't stay at rest relative to each other, that is geodesic deviation, which is already known to be equivalent to curvature of the manifold.

Notice, btw, that in the above I didn't mention any of your 6 points at all. IMO all of the things you are looking at in those 6 points are distractions. You can't really understand this issue by looking at coordinates. You have to understand the physical and geometric reasons why GR models tidal gravity as spacetime curvature.

Thank you. One follow-up thing though - you mentioned that objects in free fall, initially at rest w.r.t. each other, won't remain at rest w.r.t. each other in the presence of "real" gravity. In other words, real gravity causes geodesic deviation that is equivalent to curvature.

But then objects won't remain at rest w.r.t. each other in the presence of any force field, not just real gravity, right? So any force field would cause geodesic deviation, hence curvature?

I know my above argument is wrong but can't figure out how. I guess equivalence principle comes into the picture somewhere (since that is unique to gravity)?

 

[A.T.]

But then objects won't remain at rest w.r.t. each other in the presence of any force field, not just real gravity, right? So any force field would cause geodesic deviation, hence curvature?

It's not "geodesic deviation", if the object's world-lines aren't geodesics, because you have applied some unspecified force field.

 

[Ibix]

I know my above argument is wrong but can't figure out how. I guess equivalence principle comes into the picture somewhere (since that is unique to gravity)?

It's the point I made in my last post. You cannot model (e.g.) electromagnetism as spacetime curvature in the same way as gravity because two objects with the same postion and velocity can follow different paths due to having different charge-to-mass ratios. A geometric model requires every subsequent behaviour to depend solely on derivatives of the current position.

That is a form of the equivalence principle, yes. I can tell the difference between an accelerating box and one in a uniform electric field by releasing an electron and a neutral body and seeing if they move differently. I cannot do the same with a gravitational field (or so Einstein claims, and so far nobody has proved him wrong).

 

[PeterDonis]

you mentioned that objects in free fall, initially at rest w.r.t. each other, won't remain at rest w.r.t. each other in the presence of "real" gravity. In other words, real gravity causes geodesic deviation that is equivalent to curvature.

Yes.

objects won't remain at rest w.r.t. each other in the presence of any force field, not just real gravity, right?

Right. But objects subjected to non-gravitational forces do not have geodesic worldlines.

So any force field would cause geodesic deviation, hence curvature?

No. See above.

 

[pervect]

Thanks! If you don't mind, I want to clarify another thing.
I'm trying to understand the motivation of modeling gravitation using spacetime curvature. Here's my limited understanding:

1. "Fake" gravitational field can be made to vanish using a global coordinate transformation or can be introduced using one (regular to curvilinear or vice-versa)
2. Curvilinear coordinate system has a funky-looking metric (curved manifolds also have funky-looking metrics). So an inverse transformation back to the regular coordinate system diagonalizes the metric
3. In the book it's written that if on a manifold we can diagonalize the metric everywhere using a single coordinate transformation, then that manifold is flat
4. So for a flat manifold, I can switch from a metric to other weird-looking metrics using global coordinate transformations. And each of those transformations also induces a "fake" gravitational field
5. Thus there is a clear parallel between flat spacetime and fake gravitational field - both are linked by the existence of a global coordinate change that diagonalizes the metric in the former and removes the latter
6. A global coordinate change can't remove a "real" gravitational field, so conversely no global coordinate change can induce a real gravitational field
7. Similarly, in curved spacetime, no global coordinate change can diagonalize the metric everywhere
8. Thus there is a clear parallel between curved spacetime and real gravitational field - both are linked by the non-existence of a global coordinate change that can diagonalize the metric everywhere in the former, and can remove the latter

Does the above seem like a comprehensive enough justification for why we model gravity using curved spacetime? Or is there anything else I'm missing - maybe some result in differential geometry that provides a stronger justification?
(e.g. off the top of my head something like, "For a flat manifold, there exists no vector field that can be transformed away using a global coordinate change")

I don't see anything obviously objectionable here, but it seems a bit complicated. Also the terms "real" and "fake" are a bit judgmental as well as rather imprecise.

The way I'd describe the situation is this. Flat spaces - and flat space-time admits Cartesian coordinates, which have a lot of very nice properties and most people are introduced to physics using Cartesian coordinates. Curved spaces - and curved space-times - do not in general admit Cartesian coordinates. So if one's understanding of physics is based on the use of Cartesian coordinates, ones understanding is limited.

Studying the methods of using arbitrary coordinates on a flat space - or space-time - is helpful in developing the mathematical methods needed to study curved space-time. In general, it's not usually obvious how to assign coordinates in a curved space or space-time, so one does the extra work needed to be able to use arbitrary coordinates.

In studying the methods used to deal with arbitrary coordinates, it is helpful to introduce the idea of geometric objects, entities that are independent of coordinates. The theory of geometric objects is in the literature, on of my texts, MTW, provides a few references, though MTW does not discuss the subject in great depth.

To give a very crude example of a geometric approach, consider that with the geometric approach, a straight line is straight regardless of whether one uses Cartesian coordinates or polar coordinates. One example of a coordinate independent way of describing the notion of a straight line is that it's the shortest distance between two points (*). This doesn't even introduce the idea of coordinates, so it doesn't depend on them. And, one can use this definition, and the calculus of variations, to find the equations that represent a straight line in Cartesian coordinates, or in polar coordinates, or in any coordinates one likes. All one needs to use this definition is a metric, a way of defining the distance between nearby points, and the calculus of variations.

This definition also allows us to leverage the idea of "straight lines" to the sphere, where they become great circles, the curve of shortest length connecting two points that lies entirely in the sphere. However, we'd probably use some other term such as geodesic rather than straight line at this point.

Lagrangian methods in physics also lend themselves to a coordinate-free approach to physics, and illustrate the existence and power of coordinate-independent methods that handle arbitrary coordinates.

(*) Note that this definition of a straight line isn't the only possible one. At the elementary level, a geometric treatment of straight lines is given by Euclid's axiomatic formulation. This axiomatic approach shares the idea of a geometric - and it also illustrates why doing geometry doesn't require the adoption of coordinates at all. At a more advanced level, the idea of a straight line as the shortest distance between two points may be abandoned in favor of a notion based on a "connection". There are some definite advantages to using connections, which one will undoubtedly run into in the study of GR. One important reason for using connections relates to the important mathematical requirements of existence and uniqueness. Antipodal points (for example on the sphere) are a simple illustrate that this "shortest distance" definition has issues with uniqueness, issues which turn out to be more difficult to sidestep than one might hope for. However, in spite of its limits, the "shortrest distance" approach is a personal favorite of mine, and for better or worse has molded much of my own thinking.