B According to General Relativity, would the Earth be 'flat'?

[AstralSentient]

I thought about the geometry of Earth by definition and thought of the implications of 'curved space-time'. I understand a trajectory of a satellite and other objects to be straight geodesics through warped space-time. Would the Earth then therefore be able to described as a straight surface in this non homogeneous space-time continuum and in a sense, 'flat'?
I am presuming flat to mean a surface that is straight so that a tangent vector is always touching the surface.

 

[A.T.]

I understand a trajectory of a satellite and other objects to be straight geodesics through warped space-time. Would the Earth then therefore be able to described as a straight surface in this non homogeneous space-time continuum and in a sense, 'flat'?

No.

1) Different manifolds:
Space (Earth's surface) vs space-time (free fall worldline)

2) Different types of curvature (or lack of it)
Intrinsic curvature (Earth's surface) vs. (no) extrinsic curvature (free fall worldline)

 

[Dale]

I thought about the geometry of Earth by definition and thought of the implications of 'curved space-time'. I understand a trajectory of a satellite and other objects to be straight geodesics through warped space-time. Would the Earth then therefore be able to described as a straight surface in this non homogeneous space-time continuum and in a sense, 'flat'?
I am presuming flat to mean a surface that is straight so that a tangent vector is always touching the surface.

If you draw a triangle on the surface of the Earth you will find that the sum of the angles is greater than 180. This indicates that the surface of the Earth has positive curvature. It is measurably not flat.

 

[martinbn]

Your question isn't very clear and it is possible that you use the terminology in a non standard way, but here is how I understand your question. The world line of a satellite is a straight line in space-time (by which I mean a geodesic). The question then is: is the world sheet of the surface of the Earth a flat hypersurface in space-time? If the Schwartzschild solution is a good approximation, then the answer is yes, it is flat, in the sense that the induced metric has zero curvature. And no, it is not flat in the sense that the extrinsic curvature is not zero.

 

[AstralSentient]

No.

1) Different manifolds:
Space (Earth's surface) vs space-time (free fall worldline)

2) Different types of curvature (or lack of it)
Intrinsic curvature (Earth's surface) vs. (no) extrinsic curvature (free fall worldline)

But the Earth's gravitational field exists because of space-time curvature, so certainly, I can use space-time to define it and a transported vector touching the Earth's surface. Doesn't this imply the Earth's surface has no intrinsic curvature? I understand that you could plot out a 270 degree triangle by straight geodesics and call that intrinsic curvature, but in curved space-time, a tangent vector should be able to remain tangent to the surface by parallel transporting it, as if a line on the Earth's surface is a straight geodesic through curved space-time, defining the surface as flat.

If you draw a triangle on the surface of the Earth you will find that the sum of the angles is greater than 180. This indicates that the surface of the Earth has positive curvature. It is measurably not flat.

But what I was supposing is that the Earth's surface is a plane in non-euclidean space-time, and so this would be able to happen on a flat surface.

Your question isn't very clear and it is possible that you use the terminology in a non standard way, but here is how I understand your question. The world line of a satellite is a straight line in space-time (by which I mean a geodesic). The question then is: is the world sheet of the surface of the Earth a flat hypersurface in space-time? If the Schwartzschild solution is a good approximation, then the answer is yes, it is flat, in the sense that the induced metric has zero curvature. And no, it is not flat in the sense that the extrinsic curvature is not zero.

So, does this mean that it could be flat in that a transported tangent vector remains in the same position relative to the surface?

 

[Dale]

But what I was supposing is that the Earth's surface is a plane in non-euclidean space-time, and so this would be able to happen on a flat surface.

This cannot happen on a flat surface. It is one of the key distinctions between flat and curved surfaces.

but in curved space-time, a tangent vector should be able to remain tangent to the surface by parallel transporting it

This isn't the issue. The issue is that parallel transport along different closed paths yields different vectors. This is another key distinction between flat and curved surfaces, but it is harder to describe in a post so I usually go with the triangle.

 

[A.T.]

a line on the Earth's surface is a straight geodesic through curved space-time,

No it isn't. While a line on the Earth's surface can be the spatial projection of an geodesic through curved space-time, the same line can be the spatial projection of some other non-geodesic worldline through space-time. Neither implies anything about the intrinsic curvature of the Earth's surface for the 2 reasons stated in post #2.

 

[AstralSentient]

This cannot happen on a flat surface. It is one of the key distinctions between flat and curved surfaces.

This isn't the issue. The issue is that parallel transport along different closed paths yields different vectors. This is another key distinction between flat and curved surfaces, but it is harder to describe in a post so I usually go with the triangle.

But as long as the tangent vector remains tangent along the surface, it is a straight surface. Yes, it can be applied to a non-geodesic path and therefore different vector, but the moment you keep a consistent vector across, it remains a straight surface through curved space-time. I really don't think the triangle exceeding 180 degrees is mutually exclusive with a surface having a consistent tangent vector across and so being straight and flat, given curved space-time (since that kinda changes our defining geometry a lot). It's a unique non-euclidean 'plane' (not the best word choice, but it'll have to do for now) with general relativity (due to the property of space-time itself) is how I am seeing it here.

No it isn't. While a line on the Earth's surface can be the spatial projection of an geodesic through curved space-time, the same line can be the spatial projection of some other non-geodesic worldline through space-time.

Which doesn't go against the point since if you transport a tangent vector across earth, it will remain consistent with the Earth's surface in curved space-time, as a straight line is defined, implying a flat surface. You can do this from any point on Earths surface by simply following a vector, I don't see how being able to follow a non-geodesic path and therefore gradually have a non-straight vector would change this. This would be a unique property of curved space-time.

 

[Ibix]

In terms of extrinsic curvature, I thought that a flat surface would be one where all geodesics initially tangent to the surface remained in it. For example, a flat plane embedded in flat three space has zero extrinsic curvature, and all straight lines lying tangent to the plane at some point are tangent everywhere. But a cylinder (an extrinsically curved surface) in the same three space has both straight lines that are tangent at a point that do remain in the surface (parallel to the axis) and ones that don't (all others).

The Earth's surface is clearly an example of the latter case. There is a zero-altitude circular orbit, but there are an awful lot of initially-parallel-to-the-ground trajectories that don't stay that way.

 

[pervect]

I thought about the geometry of Earth by definition and thought of the implications of 'curved space-time'. I understand a trajectory of a satellite and other objects to be straight geodesics through warped space-time. Would the Earth then therefore be able to described as a straight surface in this non homogeneous space-time continuum and in a sense, 'flat'?
I am presuming flat to mean a surface that is straight so that a tangent vector is always touching the surface.

As others have mentioned, the spatial geometry of the Earth's surface is not a flat spatial geometry. A flat spatial geometry would satisfy Euclid's axioms, and there is a proof that the sum of the angles of a triangle on a Euclidean plane is always 180 degrees.

See for instance http://www.apronus.com/geometry/triangle.htm, or a textbook if you want a quality reference.

Geodesics on the surface of the Earth are great circles. Sometimes they are informally called "straight lines, but geodesics is a more precise term. A simple field that studies the geometry of the Earth is spherical trignometry, if we ignoree the fact that the Earth isn't quite spherical. It's useful for talking a bit about curvature in a way that is hopefully familiar and is easy to visualize and (cmopared to GR) relatively non-abstract.

We can draw triangles on a sphere, with geodesics forming the sides of the triangle. The sum of the angles of the traingle is always greater than 180 degrees, though it approaches 180 degrees for small triangles. You can look up more on this in spherical trignometry, but a simple example of a triangle on the surface of the Earth whose angles sum to 270 degrees (3*90) would be a spherical triangle formed by starting at the north pole, going south to the equator, making 1/4 a circuit around the equator, and then going back to the north pole.By this example, and others, we can therefore conclude and confirm that the Earth's surface is not a plane because it doesn't satisfy Euclid's axioms. We can even tell which axiom it doesn't follow - it doesn't follow the parallel postulate. Some posters have already mentioned that we replace the parallel postulate with "parallel transport", but learning how this works gets rather deep into Riemannian geometry.

There is a lot more to understanding curvature, but understanding the terminology well enough to know that the surface of the Earth is not flat is a good place to start. I'm not quite sure why you think it's flat, I hope I have convinced you that it can't be.

 

[Dale]

But as long as the tangent vector remains tangent along the surface, it is a straight surface.

No. This has nothing to do with flatness. I don't know what makes you think that it does. You have a wrong concept in mind about what defines flatness.

I really don't think the triangle exceeding 180 degrees is mutually exclusive with a surface having a consistent tangent vector across and so being straight and flat,

The exceeding 180 deg implies that the manifold is curved. The tangent vector remaining in the manifold does not imply that it is flat.

It's a unique non-euclidean 'plane'

Non-Euclidean is curved. Euclidean is flat. The fact that it is non-Euclidean is precisely what identifies it as curved.

You clearly have some basic wrong ideas about what it means for a manifold to be curved. I recommend studying the second chapter of Sean Carroll's lecture notes on general relativity:

https://arxiv.org/abs/gr-qc/9712019

 

[A.T.]

Which doesn't go against the point...

Then you have to be more precise about your reasoning, because so far I understand it like this:

- Low orbiting satellite has a worldline with no extrinsic curvature w.r.t space-time
- Low orbiting satellite follows Earth's surface in space
Therefore (you claim):
- Earth's surface has no intrinsic curvature w.r.t space-time

Now, replace the orbiting satellite with a truck following Earth's surface, which has a curved worldline w.r.t space-time, and the very same reasoning proves the exact opposite. So that reasoning is wrong

 

[davidge]

I recommend studying the second chapter of Sean Carroll's lecture notes on general relativity

It's probably too advanced for a beginner, no? Perhaps it's better to look for introductory books on differential geometry.

 

[Dale]

It's probably too advanced for a beginner, no? Perhaps it's better to look for introductory books on differential geometry.

By all means, if you have a better recommendation please make it!

 

[AstralSentient]

As others have mentioned, the spatial geometry of the Earth's surface is not a flat spatial geometry. A flat spatial geometry would satisfy Euclid's axioms, and there is a proof that the sum of the angles of a triangle on a Euclidean plane is always 180 degrees.

See for instance http://www.apronus.com/geometry/triangle.htm, or a textbook if you want a quality reference.

Geodesics on the surface of the Earth are great circles. Sometimes they are informally called "straight lines, but geodesics is a more precise term. A simple field that studies the geometry of the Earth is spherical trignometry, if we ignoree the fact that the Earth isn't quite spherical. It's useful for talking a bit about curvature in a way that is hopefully familiar and is easy to visualize and (cmopared to GR) relatively non-abstract.

We can draw triangles on a sphere, with geodesics forming the sides of the triangle. The sum of the angles of the traingle is always greater than 180 degrees, though it approaches 180 degrees for small triangles. You can look up more on this in spherical trignometry, but a simple example of a triangle on the surface of the Earth whose angles sum to 270 degrees (3*90) would be a spherical triangle formed by starting at the north pole, going south to the equator, making 1/4 a circuit around the equator, and then going back to the north pole.By this example, and others, we can therefore conclude and confirm that the Earth's surface is not a plane because it doesn't satisfy Euclid's axioms. We can even tell which axiom it doesn't follow - it doesn't follow the parallel postulate. Some posters have already mentioned that we replace the parallel postulate with "parallel transport", but learning how this works gets rather deep into Riemannian geometry.

There is a lot more to understanding curvature, but understanding the terminology well enough to know that the surface of the Earth is not flat is a good place to start. I'm not quite sure why you think it's flat, I hope I have convinced you that it can't be.

Not necessarily. You see, I'm talking about a curved space-time. We have a vector in space, it is tangent to a surface in space, space-time curves, but that vector remains tangent following a straight line through space, a straight and therefore flat surface. If you have someone walk a straight line through a warped space-time, relative to an independent observer, it appears to bend by following the warp but relative to the observer it's straight since it's non-homogeneous space.
There is a distinction to be made by curved space-time and a curved Earth in euclidean space, you can't say that they are the same or that a straight vector throughout non-homogeneous space is not a straight path because space is non-homogeneous. Precisely why I can refer to an orbit as inertial in curved space-time.

No. This has nothing to do with flatness. I don't know what makes you think that it does. You have a wrong concept in mind about what defines flatness.

The exceeding 180 deg implies that the manifold is curved. The tangent vector remaining in the manifold does not imply that it is flat.

Non-Euclidean is curved. Euclidean is flat. The fact that it is non-Euclidean is precisely what identifies it as curved.

You clearly have some basic wrong ideas about what it means for a manifold to be curved. I recommend studying the second chapter of Sean Carroll's lecture notes on general relativity:

https://arxiv.org/abs/gr-qc/9712019

That's precisely what I mean by a flat surface, a straight one, if I draw a line across a surface, with vectors defining it, and it's straight, that's a flat surface. Non-euclidean is curved indeed, but it's non-euclidean space-time, not a non-euclidean surface in flat homogeneous space-time.

Then you have to be more precise about your reasoning, because so far I understand it like this:

- Low orbiting satellite has a worldline with no extrinsic curvature w.r.t space-time
- Low orbiting satellite follows Earth's surface in space
Therefore (you claim):
- Earth's surface has no intrinsic curvature w.r.t space-time

Now, replace the orbiting satellite with a truck following Earth's surface, which has a curved worldline w.r.t space-time, and the very same reasoning proves the exact opposite. So that reasoning is wrong

I'm saying Earth's surface has no intrinsic curvature due to curved space-time, while space-time being curved allows a non-euclidean flat surface as it follows as a geodesic in curved space-time, essentially matching it's curvature, being straight and flat in curved space-time. Space-time is how we define surfaces and vectors here, without it, geometry isn't definable in the universe. If space-time is curved itself, it would be silly not to question our original understanding of the geometry of surface as assumed in basic flat space-time.
The Earth's surface is flat w.r.t to space-time with curved space-time in General Relativity, that's the point.

In terms of extrinsic curvature, I thought that a flat surface would be one where all geodesics initially tangent to the surface remained in it. For example, a flat plane embedded in flat three space has zero extrinsic curvature, and all straight lines lying tangent to the plane at some point are tangent everywhere. But a cylinder (an extrinsically curved surface) in the same three space has both straight lines that are tangent at a point that do remain in the surface (parallel to the axis) and ones that don't (all others).

The Earth's surface is clearly an example of the latter case. There is a zero-altitude circular orbit, but there are an awful lot of initially-parallel-to-the-ground trajectories that don't stay that way.

This isn't "flat" space, it's curved space, the surface is flat in that it's surface equals the curvature of the space-time manifold, so the vector remains tangent since it follows this curvature.

 

[Nugatory]

I am presuming flat to mean a surface that is straight so that a tangent vector is always touching the surface.

However, that's not what "flat" means. A flat spacetime is one in which the metric everywhere is Minkowski, meaning that the interval between two nearby events is given by $ds^2=-dt^2+dx^2+dy^2+dz^2$. A flat space is one in which the metric is everywhere Euclidean, meaning that the interval between two nearby points is given by $ds^2=dx^2+dy^2+dz^2$. By this definition, the two-dimensional space corresponding to the surface of the Earth is not flat, although the three-dimensional space around the Earth in which it is embedded is very close to flat. The four-dimensional spacetime in the vicinity of the Earth is not.

 

[pervect]

Not necessarily. You see, I'm talking about a curved space-time. We have a vector in space, it is tangent to a surface in space, space-time curves, but that vector remains tangent following a straight line through space, a straight and therefore flat surface. If you have someone walk a straight line through a warped space-time, relative to an independent observer, it appears to bend by following the warp but relative to the observer it's straight since it's non-homogeneous space.

Sorry, you lost me, and I strongly suspect you are using highly personal (and not generally accepted) defnitions of many important terms, such as "tangent", and curved.

To take such a simple case, let's talk about a tangent plane to a sphere. You seem to be thinking that the tangent plane touches the sphere at more than one point, when it doesn't do that, and the actual definition of a tangent space defines a different tangent space for every point on a manifold. In this context the surface of the Earth is a stand-in for the more abstract concept of a manifold.I don't wish to be impolite, but you're way, way, offtrack, and you seem to be more interested in pushing your own highly personal views than learning anything :(.

 

[FactChecker]

By your definition, every smooth surface would be "flat" no matter how many wiggles and warps it had. There are always geodesics on a surface. So that is not a useful definition of "flat". The useful (and correct) definition of "flat" is that the usual geometry of triangles and circles applies -- the sum of the angles of a triangle is 180° and the circumference of a circle is π times the diameter. The surface of the Earth is not flat.

 

[A.T.]

...as it follows as a geodesic in curved space-time...

The Earth's surface doesn't follow a geodesic in curved space-time

 

[Ibix]

This isn't "flat" space, it's curved space, the surface is flat in that it's surface equals the curvature of the space-time manifold, so the vector remains tangent since it follows this curvature.

This isn't the case for the surface of the Earth for any definition of curvature with which I am familiar. As A.T. points out, points on the surface of the Earth don't follow geodesics. The point of my flat space example was to show you that, although there are geodesics that do remain in the worldsheet of the Earth's surface, there are many that do not - so, like the cylinder, the Earth's surface is extrinsically curved.

 

[FactChecker]

I thought about the geometry of Earth by definition and thought of the implications of 'curved space-time'. I understand a trajectory of a satellite and other objects to be straight geodesics through warped space-time. Would the Earth then therefore be able to described as a straight surface in this non homogeneous space-time continuum and in a sense, 'flat'?
I am presuming flat to mean a surface that is straight so that a tangent vector is always touching the surface.

A satellite in orbit is following an unaccelerated straight space-time line. It is essentially in "free fall" under the influence of gravity. But objects on the Earth's surface are not in free fall. The Earth's surface is not following a straight line in space-time.

 

[AstralSentient]

However, that's not what "flat" means. A flat spacetime is one in which the metric everywhere is Minkowski, meaning that the interval between two nearby events is given by $ds^2=-dt^2+dx^2+dy^2+dz^2$. A flat space is one in which the metric is everywhere Euclidean, meaning that the interval between two nearby points is given by $ds^2=dx^2+dy^2+dz^2$. By this definition, the two-dimensional space corresponding to the surface of the Earth is not flat, although the three-dimensional space around the Earth in which it is embedded is very close to flat. The four-dimensional spacetime in the vicinity of the Earth is not.

I know the 4D space-time of the Earth isn't flat, that's my big point here. In General Relativity, we have curved space-time, and the vectors in space defining a flat surface are distorted relative to the surrounding homogeneous flat space. It would be silly to equate this to a sphere since space-time and surfaces are clearly distinct variables, the Earth's surface being flat in Non-Euclidean space-time (or relative to it for that matter).

Sorry, you lost me, and I strongly suspect you are using highly personal (and not generally accepted) defnitions of many important terms, such as "tangent", and curved.

To take such a simple case, let's talk about a tangent plane to a sphere. You seem to be thinking that the tangent plane touches the sphere at more than one point, when it doesn't do that, and the actual definition of a tangent space defines a different tangent space for every point on a manifold. In this context the surface of the Earth is a stand-in for the more abstract concept of a manifold.I don't wish to be impolite, but you're way, way, offtrack, and you seem to be more interested in pushing your own highly personal views than learning anything :(.

I didn't claim that at all, it clearly doesn't.

If I walk a straight line through warped space-time, such a straight path may appear curved. How if I have a flat surface in curved space-time, it may appear curved but is flat. This is because space-time itself, where we define our vectors and surfaces, is curved. Relative to space-time itself, the tangent plane is straight across, in curved space-time, it's touching across earth, as a flat surface.

I am actually learning what you people think of this, but I'm not going to give in and accept what you people are saying because you said so (if that's how you define "learning"), many of you are simply missing the point and trying to bring Euclidean spaces to this.

By your definition, every smooth surface would be "flat" no matter how many wiggles and warps it had. There are always geodesics on a surface. So that is not a useful definition of "flat". The useful (and correct) definition of "flat" is that the usual geometry of triangles and circles applies -- the sum of the angles of a triangle is 180° and the circumference of a circle is π times the diameter. The surface of the Earth is not flat.

3D vectors remaining tangent to a surface doesn't work on a sphere, it curves around in a circle in 3D space as a geodesic (if you look at that tangent plane above, it only touches at one point). But if space-time itself is curved, the defining spatial vectors of the tangent plane are equal to the non-homogeneous path of space-time, and so is touching the surface across as flat.

The Earth's surface doesn't follow a geodesic in curved space-time

Yes it does, it is a geodesic in curved space-time, that's how the Earth stays together, you know, we call it 'gravity'.

This isn't the case for the surface of the Earth for any definition of curvature with which I am familiar. As A.T. points out, points on the surface of the Earth don't follow geodesics. The point of my flat space example was to show you that, although there are geodesics that do remain in the worldsheet of the Earth's surface, there are many that do not - so, like the cylinder, the Earth's surface is extrinsically curved.

But not intrinsically curved, because it's in curved space-time. Curved space-time can make straight lines appear curved and curved lines appear straight, it's simply applying geodesics to non-homogeneous space-time itself. All I can conclude is that space-time is curved or the Earth's surface is curved in Euclidean space-time from realizing the failure of Euclid's postulates on Earth's surface. They clearly aren't the same.

By all of your reasoning, I could examine gravitational lensing and say the light is bending because straight lines don't curve, that's impossible, and then conclude light isn't traveling in a straight but is bending it's path, but of course, this is wrong, the light's path is straight, General Relativity has described it as the non-homogeneous warping of space-time itself (as one connected continuum).

Consider this, you travel straight through space-time, and eventually another straight path converges. This is a fundamental component of non-euclidean space-time, which is distinct from an non-euclidean surface.

 

[AstralSentient]

A satellite in orbit is following an unaccelerated straight space-time line. It is essentially in "free fall" under the influence of gravity. But objects on the Earth's surface are not in free fall. The Earth's surface is not following a straight line in space-time.

Yes, it can be described as being in an inertial frame, and you simply can't orbit a Non-Euclidean curved surface in an inertial frame.
Objects on Earth's surface experience acceleration because they are all in curved space-time, the surface is a distinct product of this curved space-time manifold.

 

[FactChecker]

Yes, it can be described as being in an inertial frame

You weren't just asking about any old inertial frame, you were asking about the curved space-time of GR. That gives a very specific definition of GR space-time geodesics that does not agree with a stationary object on the Earth's surface. If an object is traveling at orbital speed on the surface, then it would be following a GR space-time geodesic.

 

[Dale]

Relative to space-time itself, the tangent plane is straight across, in curved space-time,

This doesn't make sense. The intrinsic curvature that is used in GR is not "relative" to anything. The curvature of a manifold is a tensor, so it is a covariant object. It is not relative to a reference frame or another object or some other space.

I am actually learning what you people think of this, but I'm not going to give in and accept what you people are saying because you said so (if that's how you define "learning")

Given that you don't think that we are a credible resource then I don't think further discussion is productive. You will need to muddle through this material on your own. Please read chapter 2 of the Carroll lecture notes that I linked to earlier.

 

[PeterDonis]

I don't think further discussion is productive.

I agree. Thread closed.