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

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• Shirish
Shirish
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)?

So I'll finally try to frame the above in a more concrete way: given a force field ##F(x,y,z,t)##, does there exist a coordinate transformation ##(x,y,z,t)\to (x',y',z',t')## that removes the effects of the field, i.e. ##F(x',y',z',t')=0## [or is it supposed to be ##F'(x',y',z',t')=0##? I'm not sure]?

Fine so far (except the ##F## vs ##F'## confusion). The book then mentions that this question is similar to the question of deciding whether a geometry is flat. It goes into a long discussion, finally concluding that a manifold is flat if it's locally flat everywhere. And it's locally flat at a point ##p## if the metric tensor at ##p##, via a coordinate transformation ##(x,y,z,t)\to (x',y',z',t')##, gets transformed into the identity matrix (actually should be the Minkowski metric tensor but for simplicity the book is discussing Riemannian geometry at this point).

Individually I understand the logic on both sides, but I'm not sure about the connection (see the bold italic statement). I see the following similarity:

1. The effects of a "fake" gravitational field (as seen in ##R##) can be eliminated globally by switching to a reference frame ##R'## with a specific acceleration ##\vec{a}(t)## w.r.t. ##R## (via a global coordinate transformation). But the effects of any arbitrary gravitational field can be eliminated at any point locally via a similar procedure of switching to some other local reference frame (via a local coordinate transformation)
2. This is similar to how a manifold is globally flat if there exists a coordinate transformation that diagonalizes the metric tensor everywhere, vs. how for a general smooth manifold, we can still diagonalize the metric tensor at a point via a local coordinate transformation

But this is still just a vague similarity to me. Consider an extended region of spacetime in which, w.r.t. some coordinate system (i.e. reference frame) ##(x,y,z,t)##, we notice a gravitational field ##F##. We also notice that the transformation to a different coordinate system ##(x',y',z',t')## "gets rid of" ##F##.

If the above is true, is there some mathematical result that guarantees existence of a coordinate transformation diagonalizing the metric tensor in that particular region of spacetime? Or conversely, is there a result that if a diagonalizing coordinate transformation exists in a spacetime region, then there also exists a transformation that "gets rid of" ##F##?

If these kind of results don't exist, how can we say that the above two scenarios (points 1 and 2) are mathematically equivalent?

Mario Beaulieu
The connection between the two is tidal gravity. Spacetime curvature cannot be removed by a coordinate transformation and tidal effects are modeled by spacetime curvature. So tidal effects are the “real” gravity in the sense that it cannot be transformed away by choice of coordinates.

jbergman, Mario Beaulieu, vanhees71 and 3 others
Dale said:
The connection between the two is tidal gravity. Spacetime curvature cannot be removed by a coordinate transformation and tidal effects are modeled by spacetime curvature. So tidal effects are the “real” gravity in the sense that it cannot be transformed away by choice of coordinates.
That makes perfect sense, but on the mathematical side... is there a concrete result that shows any equivalence b/w a coordinate transformation removing the tidal force and a coordinate transformation diagonalizing the metric tensor? The broad idea is clear to me but I'm looking for details (see 3rd last and 2nd last paragraphs in my OP)

Or do you know of any reference book or paper that may clarify this?

Carroll approaches it in chapter 3 of his GR lecture notes by observing that he has enough degrees of freedom to set all first derivatives of the metric to zero at an event, but not enough to set second derivatives to zero. That tells him he can always pick coordinates so the metric is constant over "small enough" regions of spacetime but not over larger regions unless the curvature is actually zero.

Shirish, vanhees71 and topsquark
Shirish said:
is there a concrete result that shows any equivalence b/w a coordinate transformation removing the tidal force
There is no such coordinate transformation. So there can be no equivalence between that non-existent transform and anything else. Coordinate transform cannot remove tidal effects. That is why they are represented by a coordinate independent geometric feature of the manifold, curvature.

I am not certain what you are looking for. As @Ibix mentioned, there are mathematical proofs that show that at each event in spacetime there exists a coordinate system where the metric is the Minkowski metric to first order. The second and higher order deviations cannot be eliminated by coordinate transforms. Those are precisely the curvature/tidal effects associated with real gravity.

vanhees71
Dale said:
There is no such coordinate transformation. So there can be no equivalence between that non-existent transform and anything else. Coordinate transform cannot remove tidal effects. That is why they are represented by a coordinate independent geometric feature of the manifold, curvature.

I am not certain what you are looking for. As @Ibix mentioned, there are mathematical proofs that show that at each event in spacetime there exists a coordinate system where the metric is the Minkowski metric to first order. The second and higher order deviations cannot be eliminated by coordinate transforms. Those are precisely the curvature/tidal effects associated with real gravity.
Sorry my bad. I meant to say, "equivalence b/w a coordinate transformation removing the effect of tidal force locally at some point and a coordinate transformation diagonalizing the metric tensor at that point" - hopefully that's more precise.

I'm not saying that I'm absolutely sure such an equivalence exists - but if it does, I want to know its mathematical details. The reason I suspect an equivalence exists is because it's mentioned in the book:
The analogy between tidal forces and curvature actually is not an analogy, it is a very precise equivalence.

You already mentioned that tidal effects are actually modeled using curvature. With that in mind, I wanted to understand mathematically how, according to such a model, a coordinate transformation that locally removes tidal effects is related to (or is the same as?) a coordinate transformation that locally diagonalizes the metric tensor.

Shirish said:
I meant to say, "equivalence b/w a coordinate transformation removing the effect of tidal force locally at some point
There is no such thing.

Shirish said:
a coordinate transformation that locally removes tidal effects
There is no such thing. I don't know how to be more clear on that point.

Shirish said:
I'm not saying that I'm absolutely sure such an equivalence exists
How can an equivalence exist between anything and something that doesn't exist?

Dale said:
How can an equivalence exist between anything and something that doesn't exist?
I'm new to the subject and didn't know that even locally the tidal force effects cannot be removed, till you just told me. And since that's the case, I understand such an equivalence cannot exist.

Thanks for clarifying and clearing up my misconception.

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Dale
Dale said:
There is no such thing. I don't know how to be more clear on that point.
So I think the misconception arose because I tried to take the analogy between diagonalizing metric tensor and diagnosing a "fake" gravitational field a bit too far. I gave it some thought after your response and here's my updated understanding (sorry in advance for the long read but I want to be as explicit as possible):

In the book the introductory way of identifying curvature is: let's say I have some metric tensor field on a manifold. If the manifold is flat, I can change the metric tensor field representation into a Kronecker delta (i.e. diagonalize metric tensor everywhere) via a global coordinate transformation. If I can't do that, then that's a fundamental characteristic of a curved manifold (this is where curvature comes in).

But it's also the case that transforming from a regular coordinate system to a curvilinear one in a flat manifold also changes the metric. So given a metric corresponding to a curvilinear coordinate system in flat manifold, changing back to a regular coordinate system in the same manifold is equivalent to diagonalizing the metric.

But a specific change in coordinate system can also remove the effect of a fake gravitational field (i.e. gravitational field existed in normal coordinate system but went away in the curvilinear one). So finally, the very fact that such a transformation exists (that gets rid of the fake gravitational field) means that the metric gets diagonalized under this same transformation (gets diagonalized since we changed from curvilinear to normal coordinates) - i.e. the manifold is flat.

If the gravitational field is "real" (i.e. tidal effects exist), it's a physical phenomenon that doesn't care about coordinate transformations - whether local or global. Let's say we're again given some metric on the manifold. No curvilinear to regular coordinate transformation exists that removes the field effects.

I can now think of two ways to conclude that in the above case, the manifold is curved (i.e. no coordinate transformation exists that can diagonalize the metric):

1. I make the assumption: tidal effects are equivalent to curvature. With this assumption, I can now say: no transformation that removes tidal effect => curved manifold => no transformation that diagonalizes metric
2. Some result exists in differential geometry like: (loosely stating) for a smooth manifold, if a force field exists that doesn't vanish under any change of coordinates, then no change of coordinates can diagonalize the metric
If part 2 is true, then the statement "tidal effects are equivalent to curvature" follows as a consequence rather than being an assumption. Do you know if such a theorem (as described in part 2) exists in differential geometry (or any other theorem that finally lets me conclude that the manifold is curved)?

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Dale said:
There is no such thing.
Another source for the confusion is: on page 20 of the book it's written:

What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.

I took this to mean, in a small patch of spacetime (let's say a small neighborhood of point ##p##), the effects of a gravitational field (even one which is non-uniform on a global level) are indistinguishable from that of an accelerated reference frame. This is what made me think that in such a small, infinitesimal patch (i.e. locally), even the effects of a real gravitational field can be negated by switching to a (local) accelerated frame.

So it came as a natural thought that there would be a (local) coordinate transformation (i.e. switching to an accelerated frame) removing the effect of tidal force (negates effects of a real gravitational field) locally at some point (##p##). Maybe I'm missing a subtlety.

Shirish said:
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)?
By computing the Riemann tensor. If it's zero, it's a "fake gravitational field". If it's nonzero, it's a real gravitational field.

Shirish said:
The book then mentions that this question is similar to the question of deciding whether a geometry is flat. It goes into a long discussion, finally concluding that a manifold is flat if it's locally flat everywhere.
I don't think this is a good way to put it. As most other textbooks use the term "locally flat", it means that you an always find a coordinate chart that makes the metric take the Minkowski form at a single chosen event. And that can be done for any spacetime geometry, including curved ones.

In fact you can go further and find a coordinate chart that makes the metric take the Minkowski form, and sets all of the first derivatives of the metric to zero, at a single chosen event. This can also be done for any spacetime geometry. Such coordinates are called "Riemann normal coordinates" and you will find extensive discussion of them in, for example, Misner, Thorne & Wheeler. The usual usage of the term "locally flat" includes this as well--it basically means that you can find Riemann normal coordinates at any single chosen event. So any spacetime geometry in GR is locally flat in thise sense.

What you cannot do in a curved spacetime geometry is find coordinates that make the metric take the Minkowski form, set all of the first derivatives of the metric to zero, and set all of the second derivatives of the metric to zero, at a single chosen event. The second derivatives of the metric contain the information about spacetime curvature (in Riemann normal coordinates they are the nonzero components of the Riemann tensor), so there is no way to make them go away in a curved spacetime no matter what coordinates you choose.

PeterDonis said:
I don't think this is a good way to put it. As most other textbooks use the term "locally flat", it means that you an always find a coordinate chart that makes the metric take the Minkowski form at a single chosen event. And that can be done for any spacetime geometry, including curved ones.

In fact you can go further and find a coordinate chart that makes the metric take the Minkowski form, and sets all of the first derivatives of the metric to zero, at a single chosen event. This can also be done for any spacetime geometry. Such coordinates are called "Riemann normal coordinates" and you will find extensive discussion of them in, for example, Misner, Thorne & Wheeler. The usual usage of the term "locally flat" includes this as well--it basically means that you can find Riemann normal coordinates at any single chosen event. So any spacetime geometry in GR is locally flat in thise sense.

What you cannot do in a curved spacetime geometry is find coordinates that make the metric take the Minkowski form, set all of the first derivatives of the metric to zero, and set all of the second derivatives of the metric to zero, at a single chosen event. The second derivatives of the metric contain the information about spacetime curvature (in Riemann normal coordinates they are the nonzero components of the Riemann tensor), so there is no way to make them go away in a curved spacetime no matter what coordinates you choose.
Thanks! So in an infinitesimal neighborhood of an event in curved spacetime, we can diagonalize the metric and set its first derivatives to zero. But we can't make its second derivatives zero. In other words, "locally flat" (only metric is diagonalized) isn't the same as "absence of curvature" (metric diagonalized and 1st and 2nd derivatives of metric vanish). Sounds like unfortunate terminology since locally flat sounds like implying absence of curvature.

Another thing is,
1. If you refer to post #10, based on what's said in the book - seems like the local effect of a gravitational field (i.e. effect in an infinitesimal patch of spacetime) can be removed by changing (locally) to an accelerated reference frame.
2. On the other hand, based on the replies I've also realized that these real gravitational field effects are associated with higher order derivatives of metric tensor. Since higher order derivatives can't be set to zero using any coordinate transformation, even the field effects can't be removed (even locally)
The above two sound like a contradiction, so does that mean the book statement "What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference." is not precisely worded? Or have I misinterpreted its meaning in post #10?

Shirish said:
Sounds like unfortunate terminology since locally flat sounds like implying absence of curvature.
You are not the only one to make this comment.

Shirish said:
in an infinitesimal neighborhood of an event in curved spacetime, we can diagonalize the metric and set its first derivatives to zero.
No, you can only do that at a single chosen event. Even in an infinitesimal region around that event, the metric and its first derivatives won't be exactly zero. However, any changes will only be second order, because they will be produced by the second derivatives of the metric at the chosen event, the ones that can't be made to go away by any choice of coordinates. So for many purposes it is a good enough approximation to treat the metric as Minkowski and the first derivatives of the metric as zero within a sufficiently small patch of spacetime around the chosen event. (How small "sufficiently small" is depends on how large the second derivatives of the metric are at the chosen event, i.e., how large the spacetime curvature is.)

Shirish said:
In other words, "locally flat" (only metric is diagonalized) isn't the same as "absence of curvature"
Yes.

Shirish said:
Sounds like unfortunate terminology since locally flat sounds like implying absence of curvature.
That's why textbooks like MTW are very careful about how they define such terminology. Unfortunately, it doesn't sound like the Susskind book you refer to was careful enough.

Shirish said:
seems like the local effect of a gravitational field (i.e. effect in an infinitesimal patch of spacetime) can be removed by changing (locally) to an accelerated reference frame.
No, you have it backwards. In an accelerated frame, i.e., a frame in which an observer with nonzero proper acceleration is at rest, there is a "gravitational field"--freely falling objects accelerate "downwards" (in the opposite direction from the proper acceleration). You make the gravitational field vanish by switching to an inertial frame, in which freely falling objects have constant velocity.

Shirish said:
these real gravitational field effects
Are spacetime curvature, which is not the same as a "gravitational field" in the sense of freely falling objects accelerating downward. Even in an inertial frame in a curved spacetime, the effects of spacetime curvature don't go away--the Riemann tensor is still nonzero. But if the inertial frame is small enough, the curvature effects are small enough that they can be ignored for many purposes.

Shirish said:
does that mean the book statement "What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference." is not precisely worded? Or have I misinterpreted its meaning in post #10?
You've misinterpreted the meaning. What Einstein meant was what I described for an accelerated frame above. His example was a person standing in an "elevator" pulled by a rope (nowadays we would say accelerated by a rocket engine) so that it has a 1 g proper acceleration, compared to a person standing in an identical room at rest on the surface of the Earth. If the elevator and the room are small enough, the observers won't be able to tell which one they are in: both of them will see a dropped rock accelerate downwards with the same acceleration, so the "gravitational field" will be the same for both of them.

Shirish
First of all thanks a lot for the above post - it's very helpful.

PeterDonis said:
Are spacetime curvature, which is not the same as a "gravitational field" in the sense of freely falling objects accelerating downward. Even in an inertial frame in a curved spacetime, the effects of spacetime curvature don't go away--the Riemann tensor is still nonzero. But if the inertial frame is small enough, the curvature effects are small enough that they can be ignored for many purposes.
So even in a tiny region of spacetime, the metric tensor itself and its first derivative won't vanish completely - they'll just be so small that they can be practically ignored.
I'm picturing a "local reference frame at some event ##p##" like this: a 4D spacetime coordinate grid with origin ##p## and the axes extending up to a very small value ##\epsilon## about the origin. Something like this (sorry for the crude drawing):

For this, if ##\epsilon## is small enough, then the curvature effects up to first order at least are small enough to be ignored. Is that right?
PeterDonis said:
You've misinterpreted the meaning. What Einstein meant was what I described for an accelerated frame above. His example was a person standing in an "elevator" pulled by a rope (nowadays we would say accelerated by a rocket engine) so that it has a 1 g proper acceleration, compared to a person standing in an identical room at rest on the surface of the Earth. If the elevator and the room are small enough, the observers won't be able to tell which one they are in: both of them will see a dropped rock accelerate downwards with the same acceleration, so the "gravitational field" will be the same for both of them.
I'm really sorry I should've given more context behind that statement. Basically Susskind is describing the situation of a gravitational field around the earth (radially inward, varying with distance from earth center). Let me quote
Being stretched or shrunk, or both, by the Earth's gravitational field - if you are big enough - is an invariant fact. In summary, it is not quite true that gravity is equivalent to going to an accelerated reference frame.
...
He (Einstein) just had to qualify his statement and make it a bit more precise. What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.

Again, sorry for the confusion I may have caused. So the catch is, that last statement is supposed to apply to a tiny region in spacetime in earth's vicinity. Even in that region, as said, no difference between the field and an accelerated frame. Referring to my diagram above, let's say that tiny ##\epsilon##-sized spacetime region exists in earth's gravitational field. Even in the limit ##\epsilon\to 0##, based on what I've learned in this thread so far, only up to the first-order effects of curvature can be removed by coordinate transformation (switching to tiny inertial frame).

So should the last statement in the above quote be modified like so? : infinitesimally small objects, for an infinitesimally small length of time, cannot tell the difference between (up to) first order effects of a gravitational field and an accelerated frame of reference, but they can still tell the difference between second (and higher) order effects of the field and any accelerated reference frame.

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Shirish said:
So even in a tiny region of spacetime, the metric tensor itself and its first derivative won't vanish completely - they'll just be so small that they can be practically ignored.
The metric tensor never vanishes. In Riemann normal coordinates it will be the Minkowski metric at the chosen event, and the first derivatives will be zero at that event. As you move away from that event, differences of the metric from Minkowski, and of the first derivatives from zero, will be second order in the coordinate differential from the chosen event.

Shirish said:
For this, if is small enough, then the curvature effects up to first order at least are small enough to be ignored. Is that right?
Curvature effects are never first order. They are second order. Differences in the metric from Minkowski, and its first derivatives from zero, are second order, so up to first order they are zero.

Shirish said:
So the catch is, that last statement is supposed to apply to a tiny region in spacetime in earth's vicinity.
No, it applies to a small region of spacetime around an accelerated observer. The point is that, in that small region of spacetime, the observer cannot tell whether they are in Earth's vicinity (standing at rest on the Earth's surface) or in a rocket accelerating at 1 g in free space.

Shirish said:
So should the last statement in the above quote be modified like so?
Sort of. See above.

Shirish said:
Sounds like unfortunate terminology since locally flat sounds like implying absence of curvature.
This confusion is not specific to relativity, but is a more general issue about "curvature" vs. "effect of curvature".

When you consider a region of a spherical surface that gets smaller and smaller, the effects of curvature (deviation from the Euclidean circumference/radius-ratio or triangle angle sum) tend towards zero.

But the curvature stays constant in this process, because of the way curvature is defined: The "effect of curvature" is divided by the size of the considered region.

"Locally flat" doesn't mean that curvature goes to zero, but that, for a fixed curvature, deviations from (pseudo)Euclidean geometry go to zero, due to locality.

jbergman
PeterDonis said:
The metric tensor never vanishes. In Riemann normal coordinates it will be the Minkowski metric at the chosen event, and the first derivatives will be zero at that event. As you move away from that event, differences of the metric from Minkowski, and of the first derivatives from zero, will be second order in the coordinate differential from the chosen event.
My bad, I'm being a bit sloppy with my statements.

PeterDonis said:
As you move away from that event, differences of the metric from Minkowski, and of the first derivatives from zero, will be second order in the coordinate differential from the chosen event.
Is there a way to physically measure these second order effects of difference between the metric and Minkowski, and its first derivative? If so, then in this scenario:
PeterDonis said:
in that small region of spacetime, the observer cannot tell whether they are in Earth's vicinity (standing at rest on the Earth's surface) or in a rocket accelerating at 1 g in free space.
The observer can measure those second order effects, and be able to deduce based on this measurement whether they're in Earth vicinity or accelerating in free space.

Is this fine or am I wrong in saying the above?

Shirish said:
Is there a way to physically measure these second order effects of difference between the metric and Minkowski, and its first derivative?
Yes, but you might not be able to with measurements confined to a small region of spacetime. (Remember that spacetime includes time, so a small region of spacetime means a small distance and a short time around the chosen event.)

Shirish said:
The observer can measure those second order effects
If the region of spacetime over which the measurements extend is large enough, yes. But measurements always have a finite accuracy, so there will always be some region of spacetime that is small enough that no measurements can detect curvature effects.

Shirish said:
Is there a way to physically measure these second order effects of difference between the metric and Minkowski, and its first derivative?
Drop a ball in one corner of the room and another in the other corner. The failure of the "vertical" paths to be parallel is what you're looking for. On Earth, my mental arithmetic says that if the balls are about 6.4m apart (##R_E/10^6##) and 2m off the floor they'll be 2##\mu##m closer together when they hit the ground.

Ibix said:
Drop a ball in one corner of the room and another in the other corner. The failure of the "vertical" paths to be parallel is what you're looking for. On Earth, my mental arithmetic says that if the balls are about 6.4m apart (##R_E/10^6##) and 2m off the floor they'll be 2##\mu##m closer together when they hit the ground.
Thank you! And I see, so practically as Peter said, in an infinitesimally small region of spacetime this experiment will not yield any discernable deviation in the earth gravitation field vs. accelerating in free space cases.

But as a purely thought experiment, if a tiny observer in a tiny spacetime region (laboratory) does the measurement in the way prescribed above, then they should be able to figure out whether they're in earth's field or accelerating in free space? Because
1. In the limit of the size of the lab going to zero, they won't detect up to first order effects of difference between the metric and Minkowski, and its first derivative (because mathematically they vanish in that limit), BUT
2. They'll still detect second order effects of difference between the metric and Minkowski, and its first derivative
So this becomes a prescription for the observer to tell the difference between whether they're on earth or in an accelerated frame in free space, right?

Dale said:
There is no such coordinate transformation. So there can be no equivalence between that non-existent transform and anything else. Coordinate transform cannot remove tidal effects. That is why they are represented by a coordinate independent geometric feature of the manifold, curvature.
And it's very easy to explain, as soon as you have understood the power of the manifestly covariant description in terms of tensors. Tensors are invariant quantities, and curvature of spacetime is described by the Riemann curvature tensor. There's a gravitational field, if this tensor is different from 0, and that holds for its components wrt. any basis, i.e., when using "coordinate bases" (holonomous bases) as usually done in introductory texts, you cannot transform all components of the Riemann tensor away if there's a gravitational field.

As described in #21 what you always can do at a given point of spacetime is to transform away the connection components (Christoffel symbols) and thus the 1st-order partial derivatives of the metric components at this point. If the spacetime is not flat you cannot transform away the 2nd-order derivatives, and there's a "true" gravitational field.

Physically that means if there's a true gravitational field an observer at rest wrt. such a local inertial reference frame, i.e., a freely falling non-rotating frame, will observe "tidal forces", if he is able to resolve them within the region he is observing. That's why nowadays one talks about "micro gravity" rather then "weightlessness".
Dale said:
I am not certain what you are looking for. As @Ibix mentioned, there are mathematical proofs that show that at each event in spacetime there exists a coordinate system where the metric is the Minkowski metric to first order. The second and higher order deviations cannot be eliminated by coordinate transforms. Those are precisely the curvature/tidal effects associated with real gravity.
That's the simple and mathematically precise form of the "Einstein equivalence principle". It defines GR spacetime as a pseudo-Riemannian (Lorentzian) spacetime (which implies also the assumption of the connection being the unique torsion-free, metric-compatible connection).

Dale
vanhees71 said:
As described in #21 what you always can do at a given point of spacetime is to transform away the connection components (Christoffel symbols) and thus the 1st-order partial derivatives of the metric components at this point. If the spacetime is not flat you cannot transform away the 2nd-order derivatives, and there's a "true" gravitational field.
Right, and that's why I got tripped up by the statement that an observer in an infinitesimally small, closed laboratory won't be able to tell the difference whether they're on earth or in an accelerated room in free space.

If they have a way of physically measuring the 2nd-order derivatives, then that should enable them to figure it out - since those derivatives (at least as I understand at this point) will give different measurements depending on whether the observer is on earth ("true" grav field) or in an accelerated room in free space (no grav field)

vanhees71
Shirish said:
1. In the limit of the size of the lab going to zero, they won't detect up to first order effects of difference between the metric and Minkowski, and its first derivative (because mathematically they vanish in that limit), BUT
2. They'll still detect second order effects of difference between the metric and Minkowski, and its first derivative
1. is right, 2. is wrong. In the limit of the size of the lab going to zero, you can't detect the effects of any derivatives at all. Even defining derivatives requires an open neighborhood around a point; taking the limit of the size of the lab going to zero means you're just looking at a point, not its neighborhood.

vanhees71 and Shirish
PeterDonis said:
1. is right, 2. is wrong. In the limit of the size of the lab going to zero, you can't detect the effects of any derivatives at all. Even defining derivatives requires an open neighborhood around a point; taking the limit of the size of the lab going to zero means you're just looking at a point, not its neighborhood.
Makes sense. So that means the statement: What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.

is exactly, mathematically correct.

Shirish said:
So that means the statement: What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.

is exactly, mathematically correct.
No. Einstein's statement was not about the limit in which the size of the lab goes to zero. It was about a finite sized lab operating over a finite time that is small enough that the effects of spacetime curvature are not measurable with whatever finite measuring accuracy is available. So it doesn't even make sense to ask whether Einstein's statement was "exactly, mathematically correct", because it isn't a statement about math, it's a statement about practical physical measurements.

You are making this a lot more difficult than it needs to be. I don't understand why you keep insisting on making wrong restatements of what I and others tell you.

vanhees71
Shirish said:
Makes sense. So that means the statement: What Einstein really meant was that small objects, for a small length of time, cannot tell the difference between a gravitational field and an accelerated frame of reference.

is exactly, mathematically correct.
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.

vanhees71 and Shirish
PeterDonis said:
You are making this a lot more difficult than it needs to be. I don't understand why you keep insisting on making wrong restatements of what I and others tell you.
It's clear to me at this point. But you have to understand there's a considerable difference between our knowledge levels, and maybe even aptitude since I probably have less of it. Whenever I'm restating something in a different way - it's not because I want to contradict you and others.

It's only because I want to be absolutely sure about a new concept, and whenever I think of an interpretation or doubt regarding a concept it, I will try to clarify that in as much detail as possible. If you and others give responses - they'll obviously raise follow-up questions for me since I'm a beginner, and I won't hesitate in asking them.

In this case I understood that that statement holds physically, but I was trying to figure out how to translate it mathematically. And I was trying to get its precise, mathematical interpretation since GR physics is modeled using differential geometry. If it's not a statement about math as you say, I'll believe you.

vanhees71
Shirish said:
Whenever I'm restating something in a different way - it's not because I want to contradict you and others.
I understand that; but so far every restatement you have made has been wrong and has required correction.

Shirish said:
In this case I understood that that statement holds physically, but I was trying to figure out how to translate it mathematically.
But which statement do you want to "translate mathematically"? We have discussed multiple different statements in this thread.

Shirish
PeterDonis said:
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.

jbergman
Shirish said:
obviously I'm not wrong on purpose.
Yes, that's understood.

Shirish said:
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.

vanhees71
Ibix said:
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")

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.

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