What Is the Physical Meaning of Curvature in General Relativity?

In summary: The geodesic is the shortest path between two points, but it doesn't have to be the shortest path between two nearby points. It can be the shortest path between any two points on the manifold.The path that is followed by a falling body is called a geodesic, like the shortest path on a sphere is a great circle.In summary, the conversation discusses the concept of curvature in the theory of general relativity. It is explained that space(time) does indeed curve, and this curvature is described mathematically using the Riemann tensor. It is also noted that there are two types of curvature involved in GR: intrinsic curvature caused by the presence of mass, and curvature of geodesic paths
  • #36
stevendaryl said:
If you didn't know that the world was round, and all you knew was the above map, then you might be puzzled as to why airplanes naturally take the curved path, instead of the straight path. The two possible explanations are: (1) the world is round, and the Great Circle route is the shortest path, or (2) there is some mysterious force that causes the paths of airplanes to tend to curve to the south when they travel east to west.

It's possible locally (just considering the region shown on the above map, not the whole world--if you consider the whole world, the fact that you can sail completely around the world proves that it's not flat) to make the second explanation work.

I don't think I can agree with this as written, depending on one's interpretation of "make it work".

One can detect the curvature of a sphere by measuring the distance between as few as four points (a total of 6 possible pairs, giving six distance measurements).

There was a thread on this a while back, the original idea can be traced back to Synge. One way of describing the procedure is to use plane geometry and the law of cosines to calculate the interior angles of all 4 triangles.

Code:
 ....A
......
...B...D
.......C

The 4 points are A,B,C,D, by omitting each of the points in turn, one creates 4 triangles, namely ABC, ABD, ACD, BCD. There are a total of six sides, AB, AC, AD, BC, BD, CD.

Using the law of cosines, one can solve for the interior angles of each of the 4 triangles: https://en.wikipedia.org/wiki/Law_of_cosines

$$\cos \gamma = \frac{a^2 + b^2 - c^2}{2 a b}$$

If the sum of the interior angles one calculates by assuming the figure is planar in this manner is not 180 degrees, one knows the figure can't possibly be planar.

If one repeats the process using spherical trig, one needs to modify the law of cosines - and one has the well known relationship that the sum of the angles of a triangle increases with the area of the triangle. For instance, one can draw a triangle on a sphere with three right angles whose area is 1/4 the area of the sphere the triangle is on.
 
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  • #37
PeterDonis said:
I haven't revised my view at all. You are not reading carefully.I also said that in GR, curvature is directly observable.
...assuming the actual universe is accurately described by GR". Which it is, to a good approximation (with usage of appropriate solutions of the EFE). That's why we can actually observe spacetime curvature, i.e., tidal gravity--because we live in a universe that is actually described by GR.
I don't think this has to do with reading carefully but maybe with writing carefully, A few posts above the post these quotes are taken from you also wrote

PeterDonis said:
nothing forces you to use Riemannian geometry to interpret the physical observation of tidal gravity.
I hope everyone understands that the curvature we are talking about as a direct observable and equivalent to tidal gravity in GR as per the quotes above is actually Riemannian curvature as defined in Riemannian geometry, so I'm simply suggesting that this might lead laymen to confusion. Since the main goal of this site according to the owner is educational I thought it was worth to call attention over a potentially confusing heuristic but I don't really want to drag this to any further discussion. I'd much rather move on to other points raised by posters on the key issue of curvature meaning.
 
  • #38
stevendaryl said:
[...]
If you didn't know that the world was round, and all you knew was the above map, then you might be puzzled as to why airplanes naturally take the curved path, instead of the straight path. The two possible explanations are: (1) the world is round, and the Great Circle route is the shortest path, or (2) there is some mysterious force that causes the paths of airplanes to tend to curve to the south when they travel east to west.

It's possible locally (just considering the region shown on the above map, not the whole world--if you consider the whole world, the fact that you can sail completely around the world proves that it's not flat) to make the second explanation work. The mysterious forces have to be of a very particular kind that affect all objects in the same way--so that whether you are talking about a plane traveling east to west, or throwing a ball east to west, or sailing a ship, they all would be subject to exactly the same sort of force. The force on an object must be proportional to the mass of the object. And the force has to be impossible to "screen".

In retrospect, the geometric explanation, in terms of curvature, is a lot simpler than the explanation in terms of forces. But the force explanation can be made to work.

The same two explanations work for gravity: It can be explained as spacetime curvature, or it can be explained in terms of forces. If we ignore quantum mechanics, the geometric explanation is a lot simpler. If we include quantum mechanics, then it becomes more complicated to say definitively which explanation is simpler.
I basically agree with what you are saying here with an important caveat. That GR is a theory not just about curvature, but about variable curvature. This qualifier is not covered by your round sphere example, since you only deal with constant curvature there. And the same problem applies to the interpretation of gravity as a <<1 force field perturbation to a flat background.It falls short for variable curvature even if it is able to model constant or quasi-constant curvature quite well.
 
  • #39
RockyMarciano said:
I hope everyone understands that the curvature we are talking about as a direct observable and equivalent to tidal gravity in GR as per the quotes above is actually Riemannian curvature as defined in Riemannian geometry

No, it isn't. That was my point. The direct observable is tidal gravity. Riemannian geometry is a mathematical model in which the Riemann curvature tensor (more precisely, various contractions of it with other vectors and tensors) generates predictions of tidal gravity observations. Saying that Riemann curvature "is" tidal gravity is confusing the model with the observations that the model's predictions are compared to.

RockyMarciano said:
I'm simply suggesting that this might lead laymen to confusion

Failing to precisely distinguish models from actual observations might indeed lead laymen to confusion.
 
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  • #40
RockyMarciano said:
I basically agree with what you are saying here with an important caveat. That GR is a theory not just about curvature, but about variable curvature. This qualifier is not covered by your round sphere example, since you only deal with constant curvature there. And the same problem applies to the interpretation of gravity as a <<1 force field perturbation to a flat background.It falls short for variable curvature even if it is able to model constant or quasi-constant curvature quite well.

Hmm. If you try to describe motion in curved space (or curved spacetime) then there will be extra terms appearing in the equations of motion (the connection coefficients). It's possible to locally interpret these terms as forces, rather than manifestations of curvature. I don't see how curvature being constant or not makes a difference.

For example, Newtonian gravity has a "force" of gravity that is variable, depending on the distribution of matter. The exact same theory can be described in terms of curvature (that's the Newton-Cartan theory).
 
  • #41
RockyMarciano said:
the same problem applies to the interpretation of gravity as a <<1 force field perturbation to a flat background.It falls short for variable curvature

Why do you think this? Varying the field strength in the force field model is equivalent to varying the curvature in the spacetime geometry model. The force field model certainly includes the possibility of varying field strength.
 
  • #42
PeterDonis said:
Saying that Riemann curvature "is" tidal gravity is confusing the model with the observations that the model's predictions are compared to.
Failing to precisely distinguish models from actual observations might indeed lead laymen to confusion.
Yeah, that's what I'm saying. Glad we agree i.e.:
PeterDonis said:
spacetime curvature is tidal gravity. So if you think about how you would observe tidal gravity, that is how you would observe spacetime curvature

geodesic deviation for test objects, i.e., tidal gravity.
 
  • #43
stevendaryl said:
Hmm. If you try to describe motion in curved space (or curved spacetime) then there will be extra terms appearing in the equations of motion (the connection coefficients). It's possible to locally interpret these terms as forces, rather than manifestations of curvature. I don't see how curvature being constant or not makes a difference.
Since this thread is labeled as Basic I'll try and address only the basic points leaving more concerned complexities for other more advanced thread somebody might want to start on curvature in GR.
So I guess that we all agree that GR is not based on constant curvature , and it was in thsi sense that I was saying that the Earth surface example is limited as a model of GR's curvature, in the sense that curvature in a sphere is contant in all its points and this is what makes possible that the parallelism between geometric curvature and forces be perfect. This was considered long time ago by Poincare that showed graphically how any geometric model of constant curvature is equivalent to a model of forces in a flat background.

For example, Newtonian gravity has a "force" of gravity that is variable, depending on the distribution of matter. The exact same theory can be described in terms of curvature (that's the Newton-Cartan theory).
Yes, you are right. Then again Newtonian gravity is actually more complex than a simple geometric model with constant curvature.

PeterDonis said:
Why do you think this? Varying the field strength in the force field model is equivalent to varying the curvature in the spacetime geometry model. The force field model certainly includes the possibility of varying field strength.
True, as long as the fied strength is treated perturbatively.
 
  • #44
Two questions:

1.) So the example everyone sees in introductory physics classes of the equivalence principle (light beam in an elevator) is NOT an example of curvature, right? Since you can use a transformation of coordinates to turn the "curved" space of the elevator into the flat space of the earth?

2.) Would this be a good intuitive way to think of curved spacetime? Curved space = two objects a given distance a part at time t0 which move parallel with respect to each other will be at a different distance apart at some later time t1. Curved time = two objects with synchronized clocks at a given distance from each other will no longer necessarily be synchronized if they move symmetrically and uniformly to another location and both turn around and return in the same amount of proper time? (say two clocks move at the same speed in opposite direction for the same distance, then turn around with the same acceleration, and return at the same velocity, but one has a trip that goes up a mountain and one has a trip that goes down a mountain)EDIT- wait... how could they have the same proper time if gravity affects time? Okay just make it so they move symmetrically, but one goes up, one goes down. Good intuitive picture or no?
 
  • #45
Battlemage! said:
1.) So the example everyone sees in introductory physics classes of the equivalence principle (light beam in an elevator) is NOT an example of curvature, right? Since you can use a transformation of coordinates to turn the "curved" space of the elevator into the flat space of the earth?

Yes, that's exactly right. That is one of the things that is counter-intuitive about General Relativity. The most obvious feature of gravity is the acceleration due to gravity. But in GR, that's not even an objective, physical quantity: The acceleration you observe is dependent on your frame of reference, and so there is nothing objective about it. Curvature is, roughly speaking, the variation of the force of gravity with location. The force itself is frame-dependent and can be made to vanish at a point by a suitable choice of coordinates, but the variation cannot be made to vanish. It's sort of like the relationship between force and potential energy in Newtonian physics.
 
  • #46
RockyMarciano said:
Glad we agree i.e...

Just to be clear on the context of the statement of mine that you quoted: I was responding to a question that basically amounted to: how do we observe spacetime curvature? My response was, to quote the paragraph in full (which you didn't):

PeterDonis said:
Physically, spacetime curvature is tidal gravity. So if you think about how you would observe tidal gravity, that is how you would observe spacetime curvature. (Pervect mentioned geodesic deviation, which is another way of saying tidal gravity.)

Note in particular the word that I bolded, which you did not include in your quote. In other words, here we are talking about the observable. I could have gone into more detail in that post about how "spacetime curvature" really refers to a model and "tidal gravity" refers to the actual observable, but I was trying to give a quick B level response. And in any case that very point came up later in the discussion; here's what I said in response to Ibix:

PeterDonis said:
"Spacetime curvature" is a way of describing tidal gravity, which is something we observe directly. But there's nothing that requires us to use that particular description. We use it because it works and because it has proven very fruitful.

And then a few posts later in response to tionis:

PeterDonis said:
Riemannian geometry is a mathematical model. A mathematical model can never force you to adopt it as an interpretation of physical observations.

PeterDonis said:
nothing forces you to use Riemannian geometry to interpret the physical observation of tidal gravity.

And there is where the discussion in this thread would have rested, exactly where you (now) say you agree it should have rested--with the distinction made between the model (Riemannian geometry--curved spacetime) and the observable (tidal gravity). Only then you came into the thread and said this:

RockyMarciano said:
I believe I have read you and others in this site remark that: a) curvature is an invariant therefore something observable and physical in that sense

Which just blurred the distinction again and led to the subthread we have been having in which I have been trying to unblur it.
 
  • #47
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Edit: The thread will remain closed
 
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