1. I have a question about whether a line can be intrinsically curved.

Is it a nonsense ? Is intrinsic curvature only a property of 2 dimensional (or higher) surfaces?

Also is a mathematical surface the set of points where a n-dimensional object intersects with a n+1 dimensional object ?(object ="set of points"?)

2.

3. Originally Posted by geordief
I have a question about whether a line can be intrinsically curved.

Is it a nonsense ? Is intrinsic curvature only a property of 2 dimensional (or higher) surfaces?
A line cannot be intrinsically curved. Intrinsic curvature only exists for spaces of two or more dimensions. Also, two-dimensional spaces can only possess Ricci scalar curvature. Three-dimensional spaces can only possess Ricci tensor curvature (including Ricci scalar curvature). Spaces of at least four dimensions are required for the space to possess Weyl tensor curvature. Spaces of four or more dimensions can possess the complete Riemann curvature and there are no more types of curvature for spaces of greater than four dimensions.

4. Is it more correct(or just different) to talk of an n-dimensional space rather than a n-dimensional surface?

If they are different concepts,then is there a general relationship between the space and the surface?

If the surface of a ball is 2-dimensional does that mean it has to be completely flat ?(as in an Earth with no hills)

Does including hills turn this surface into a 3-dimensional surface?

5. Originally Posted by geordief
Is it more correct(or just different) to talk of an n-dimensional space rather than a n-dimensional surface?

If they are different concepts,then is there a general relationship between the space and the surface?
One would typically use the term "surface" only for spaces that are embedded in a higher-dimensional space.

Originally Posted by geordief
If the surface of a ball is 2-dimensional does that mean it has to be completely flat ?(as in an Earth with no hills)

Does including hills turn this surface into a 3-dimensional surface?
No, the surface is still 2-dimensional.

6. Originally Posted by KJW
Originally Posted by geordief
Is it more correct(or just different) to talk of an n-dimensional space rather than a n-dimensional surface?

If they are different concepts,then is there a general relationship between the space and the surface?
One would typically use the term "surface" only for spaces that are embedded in a higher-dimensional space.

You say "typically". Can surfaces also not be embedded in a higher-dimensional space?

Originally Posted by geordief
If the surface of a ball is 2-dimensional does that mean it has to be completely flat ?(as in an Earth with no hills)

Does including hills turn this surface into a 3-dimensional surface?
No, the surface is still 2-dimensional.
Is a 3-dimensional surface embedded in a 4-dimensional space and as a result not something we would encounter in our normal experience? (except perhaps as a mathematical representation of an aspect of Spacetime)

7. Originally Posted by geordief
You say "typically". Can surfaces also not be embedded in a higher-dimensional space?
I think it should be noted that people who are using the term "surface" for anything other than a two-dimensional surface do not have strictness in mind, and are using the term as an analogy to a two-dimensional surface. I say "typically" because I do not claim exhaustive knowledge of all the possible ways in which a person could use a two-dimensional surface as an analogy. For example, one might speak of a coordinate system being overlayed onto a space, thus using the language of a surface even though the space is not embedded in a higher-dimensional space. In this case, it is a different analogy.

Originally Posted by geordief
Is a 3-dimensional surface embedded in a 4-dimensional space and as a result not something we would encounter in our normal experience?
A three-dimensional surface is not something that we would encounter in our normal experience. Also, some people would use the term "three-dimensional hypersurface" instead to describe the same thing.

8. Originally Posted by geordief
I have a question about whether a line can be intrinsically curved.
Yes it can (in spite of what others have said). According to the so-called Riemann geometry, which he himself called "intrinsic geometry", there is no need of an embedding space. That is, curvature is a property of the "object" itself, not relative to anything else. If you want the mathematics, it's free here.

Is it a nonsense ? Is intrinsic curvature only a property of 2 dimensional (or higher) surfaces?
No it's not nonsense, but it is incorrect. Worryingly, but correctly, a manifold of any dimension has the exact same properties as a space as one of any other dimension. This includes a 1-dimensional manifold, one of which is a line (the circle is a less trivial example). These have intrinsic curvature.

Again, I can serve up the meat and two veg if you want, but the mathematics is challenging (to say the least)

PS I admire your curiosity about science, but I would advise you to restrict yourself to an in-depth study of a single area.

9. Originally Posted by Guitarist
Originally Posted by geordief
I have a question about whether a line can be intrinsically curved.
Yes it can (in spite of what others have said). According to the so-called Riemann geometry, which he himself called "intrinsic geometry", there is no need of an embedding space. That is, curvature is a property of the "object" itself, not relative to anything else. If you want the mathematics, it's free here.
How can a 1-dimensional space have intrinsic curvature, given that the Riemann tensor is identically zero everywhere on it? The curvature of a line that is given by the covariant derivative of the tangent vector is not actually intrinsic because it does depend on the embedding space to provide the connection.

10. I just want to say that this thread is getting pretty bloody cool, and that there are fascinated lurkers.

11. Originally Posted by KJW
How can a 1-dimensional space have intrinsic curvature,
So what? Either the circle is not a manifold, or that it is, but "flat"
given that the Riemann tensor is identically zero everywhere on it {i.e the circle}?
This is an assertion in desperate search of a proof
The curvature of a line that is given by the covariant derivative of the tangent vector is not actually intrinsic because it does depend on the embedding space to provide the connection.
This is not the definition that I am used to. For example spacetime is a 4-manifold with a connection, but you will struggle to find the embedding space.

12. Originally Posted by Guitarist
Originally Posted by KJW
How can a 1-dimensional space have intrinsic curvature,
So what? Either the circle is not a manifold, or that it is, but "flat"
A circle is flat, just as a cylinder is a flat 2-dimensional surface.

Originally Posted by Guitarist
Originally Posted by KJW
given that the Riemann tensor is identically zero everywhere on it {i.e the circle}?
This is an assertion in desperate search of a proof
It's a consequence of the algebraic properties of where . Clearly, if then which for is the entire Riemann tensor.

Originally Posted by Guitarist
Originally Posted by KJW
The curvature of a line that is given by the covariant derivative of the tangent vector is not actually intrinsic because it does depend on the embedding space to provide the connection.
So where does it end? Spacetime has a connection, but I struggle to see the embedding space.

I don't follow. I said that one-dimensional spaces can't have intrinsic curvature. I didn't say that higher-dimensional spaces can't have intrinsic curvature. Indeed, I discussed intrinsic curvature for spaces of two or more dimensions.

However, if one is considering a line in spacetime, then it is the spacetime that is the embedding space.

13. Originally Posted by KJW

A circle is flat, just as a cylinder is a flat 2-dimensional surface.
What?? The circle is flat in the topological sense? That is, it is globally homeomorphic to the 1-plane? I really do not think so. Check your texts

14. Originally Posted by Guitarist
Originally Posted by KJW
A circle is flat, just as a cylinder is a flat 2-dimensional surface.
What?? The circle is flat in the topological sense? That is, it is globally homeomorphic to the 1-plane? I really do not think so. Check your texts
No one mentioned topology. I was discussing curvature, which is a local concept.

15. Originally Posted by KJW
No one mentioned topology.
They didn't need to - curvature is a property of manifolds which are of course topological spaces
curvature, which is a local concept.
No it is not. Any manifold known to man, beast or bacterium is locally indistinguishable from some and is, of course, locally "flat".

But the curvature of a manifold (I repeat this is a topological space with an additional property) need not be globally flat. It usually isn't

Try this.......

Suppose a line (do not assume it is a straight line in the Euclidean sense). Take a vector at some arbitrary point . Take another distinct, but arbitrary point and a tangent vector there. Then if and only if the tangent vectors at and coincide i.e are parallel, then one says the curvature is zero.

16. Originally Posted by Guitarist
Originally Posted by KJW
No one mentioned topology.
They didn't need to - curvature is a property of manifolds which are of course topological spaces

curvature, which is a local concept.
No it is not. Any manifold known to man, beast or bacterium is locally indistinguishable from some and is, of course, locally "flat".

But the curvature of a manifold (I repeat this is a topological space with an additional property) need not be globally flat. It usually isn't
No, because a general topological manifold is not equipped with curvature. That requires additional structure, specifically a connection. Therefore, homeomorphisms are not the appropriate mappings that define the equivalence classes relevant to the OP's question.

Originally Posted by Guitarist
Suppose a line (do not assume it is a straight line in the Euclidean sense). Take a vector at some arbitrary point . Take another distinct, but arbitrary point and a tangent vector there. Then if and only if the tangent vectors at and coincide i.e are parallel, then one says the curvature is zero.
Since we discussing the line from an intrinsic perspective, the two tangent vectors do coincide. Thus, they are parallel and the curvature is zero. The implicit notion of parallel transport to which you refer is a property of the embedding space, not the line itself.

17. Originally Posted by KJW
Originally Posted by geordief
I have a question about whether a line can be intrinsically curved.

Is it a nonsense ? Is intrinsic curvature only a property of 2 dimensional (or higher) surfaces?
A line cannot be intrinsically curved. Intrinsic curvature only exists for spaces of two or more dimensions. Also, two-dimensional spaces can only possess Ricci scalar curvature. Three-dimensional spaces can only possess Ricci tensor curvature (including Ricci scalar curvature). Spaces of at least four dimensions are required for the space to possess Weyl tensor curvature. Spaces of four or more dimensions can possess the complete Riemann curvature and there are no more types of curvature for spaces of greater than four dimensions.
Am I right to assume that all those types of(intrinsic?)curvature are mathematical definitions and only apply to physical scenarios to the extent that they can model them successfully?

When it comes to physical scenarios ,would I also be right to say that ,without exception every physical object actually possesses intrinsic curvature? (even in the hypothetical absence of gravitational fields)

To reassure myself that I am getting my ducks in a row ,when discussing intrinsic curvature are we here modeling physical /and or mathematical objects as sets of points**?

**(that's a manifold,I think)

18. I agree with GiantEvil. However it is a pity that DrRocket is not present for this fascinating exchange.

19. Originally Posted by geordief
Am I right to assume that all those types of(intrinsic?)curvature are mathematical definitions and only apply to physical scenarios to the extent that they can model them successfully?
It should be noted that the curvature of spacetime is a fact and not just an abstract theoretical notion that only applies to mathematical models of reality rather than physical reality itself. Being mathematically defined doesn't restrict a notion to mathematical models if one can define a mapping from the mathematics to the physics. For example, distances in spacetime can be physically measured (and hence defined) by rulers and clocks. Then using the mathematical expressions for curvature in terms of the metric, one can calculate the values of the curvature of physical spacetime. But one does need to establish a correspondence between the mathematical realm and the physical realm.

Originally Posted by geordief
When it comes to physical scenarios ,would I also be right to say that ,without exception every physical object actually possesses intrinsic curvature? (even in the hypothetical absence of gravitational fields)
Curvature is a property of spaces. But the Einstein equation does say that the mathematical quantity called the Einstein tensor field is equivalent to the physical quantity called the energy-momentum density tensor field. Even though energy-momentum is associated with physical objects, the Einstein tensor field is still a property of spacetime.

Originally Posted by geordief
To reassure myself that I am getting my ducks in a row ,when discussing intrinsic curvature are we here modeling physical /and or mathematical objects as sets of points**?

**(that's a manifold,I think)
The notion of intrinsic curvature goes way beyond the notion of "sets of points". It should be noted that the debate between Guitarist and myself was a debate about the appropriate level of generality to apply to the discussion. Mathematical notions are defined in terms of satisfying a given set of axioms. However, if one has the definition of a particular mathematical notion, then one can define specialised forms of that notion by adding more axioms to the original definition. Even a notion as general as topological spaces goes beyond mere sets of points. Manifolds are more special than topological spaces, but even these do not have the axioms necessary to equip the notion with curvature. Manifolds require additional axioms that define a connection in order for them to be equipped with the notion of curvature. However, I've gone a little bit further and considered not just manifolds equipped with a connection but manifolds equipped with a metric, as it is these that correspond to macroscopic physical spacetime.

20. Originally Posted by KJW

Curvature is a property of spaces. But the Einstein equation does say that the mathematical quantity called the Einstein tensor field is equivalent to the physical quantity called the energy-momentum density tensor field. Even though energy-momentum is associated with physical objects, the Einstein tensor field is still a property of spacetime.

What about objects like fluids?Can they be modeled using the concept of curvature? Can the motion of an object through a variably dense medium also be modeled like this?

Originally Posted by geordief
To reassure myself that I am getting my ducks in a row ,when discussing intrinsic curvature are we here modeling physical /and or mathematical objects as sets of points**?

**(that's a manifold,I think)

The notion of intrinsic curvature goes way beyond the notion of "sets of points".
It should be noted that the debate between Guitarist and myself was a debate about the appropriate level of generality to apply to the discussion. Mathematical notions are defined in terms of satisfying a given set of axioms. However, if one has the definition of a particular mathematical notion, then one can define specialised forms of that notion by adding more axioms to the original definition. Even a notion as general as topological spaces goes beyond mere sets of points. Manifolds are more special than topological spaces, but even these do not have the axioms necessary to equip the notion with curvature. Manifolds require additional axioms that define a connection in order for them to be equipped with the notion of curvature. However, I've gone a little bit further and considered not just manifolds equipped with a connection but manifolds equipped with a metric, as it is these that correspond to macroscopic physical spacetime.
A bit over my head (interesting nonetheless) ,but do you think you could expand a tiny bit on the part I have bolded? Can intrinsic curvature be modeled in ways that are unrelated to "sets of points"?

21. Originally Posted by geordief
What about objects like fluids? Can they be modelled using the concept of curvature? Can the motion of an object through a variably dense medium also be modelled like this?
Fluids can be modelled by specifying the energy-momentum field, then applying the Einstein equation.

Originally Posted by geordief
A bit over my head (interesting nonetheless) ,but do you think you could expand a tiny bit on the part I have bolded?
I actually did expand on that. But if you would like further details, have a look at topological spaces.

Originally Posted by geordief
Can intrinsic curvature be modelled in ways that are unrelated to "sets of points"?
I'm not sure of what you are expecting, but I prefer to downplay the geometrical aspect of curvature in favour of the analytical¹ aspect. The Riemann curvature tensor is the obstruction to an arbitrary connection object being coordinate-transformed to zero. This can be regarded as the necessary and sufficient condition for the existence of a solution to a particular system of differential equations.

¹ Calculus, as distinct from algebra.

22. Originally Posted by KJW
Originally Posted by geordief
What about objects like fluids? Can they be modelled using the concept of curvature? Can the motion of an object through a variably dense medium also be modelled like this?
Fluids can be modelled by specifying the energy-momentum field, then applying the Einstein equation.
And do those models incorporate intrinsic curvature?(it sounds like they might but I don't understand the Einstein equation or how it is arrived at or applied ...)

The Einstein equation is applicable to seemingly mundane** problems like modeling fluids and the passage of objects through non homogeneous fluids ?

**I mean mundane compared to modeling gravitational fields which I assume is what it is most known for.

23. Originally Posted by geordief
And do those models incorporate intrinsic curvature?(it sounds like they might but I don't understand the Einstein equation or how it is arrived at or applied ...)
Yes, general relativity is based on intrinsic curvature. And given that we have no access to outside of spacetime, spacetime curvature has to be intrinsic¹. The Einstein equation is a statement of the equivalence between physical energy-momentum and mathematical Einstein curvature (which I can't explain in non-mathematical terms). Given the energy-momentum, the Einstein curvature is obtained directly. But to obtain the spacetime metric, from which the total (Riemann) curvature is calculated, one needs to solve the equation.

Originally Posted by geordief
The Einstein equation is applicable to seemingly mundane** problems like modelling fluids and the passage of objects through non homogeneous fluids ?

**I mean mundane compared to modelling gravitational fields which I assume is what it is most known for.
Once one has a specification of the energy-momentum, then the Einstein equation can be applied, though solving the equation might not be so easy.

¹ Two interesting question are: (1) Given the metric of a space of some dimension, what higher-dimensional spaces can this space be embedded in? (2) Given the metric of a space of some dimension, what lower-dimensional spaces can be embedded in this space? In the case where the two spaces have the same dimension, embedding one into the other is the same as a coordinate transformation and can only be done if the Riemann curvature is the same (in the appropriate sense) for both spaces.

24. Originally Posted by KJW
Yes, general relativity is based on intrinsic curvature. And given that we have no access to outside of spacetime, spacetime curvature has to be intrinsic¹. The Einstein equation is a statement of the equivalence between physical energy-momentum and mathematical Einstein curvature (which I can't explain in non-mathematical terms).
Is it the equivalence that is hard to explain or the way curvature is described in the spacetime model?

Does the measurement process bear any comparison with,say how one might measure the curvature of a simple sphere?

Does one measure distances in various directions in spacetime and define curvature as some kind of a function of them?(with sources of energy-momentum causing particular changes to the distances)

25. Originally Posted by geordief
Is it the equivalence that is hard to explain or the way curvature is described in the spacetime model?
I apologise for the ambiguity. What is hard to explain in non-mathematical terms is "Einstein curvature" (and distinguishing between the different curvature tensors).

Originally Posted by geordief
Does the measurement process bear any comparison with, say how one might measure the curvature of a simple sphere?
Apart from the differences associated with the increased number of dimensions, the curvature of spacetime is essentially the same as the curvature of a sphere. However, the curvature of a sphere can be viewed extrinsically because the embedding in three-dimensional space is easy to visualise, whereas we don't visualise spacetime as embedded in anything. But note that a sphere only has simple scalar curvature, whereas spacetime has the full curvature that only exists in four or more dimensions.

Originally Posted by geordief
Does one measure distances in various directions in spacetime and define curvature as some kind of a function of them?
Yes. The metric is a generalised Pythagorean expression that specifies the distance between infinitesimally separated points specified by their coordinates. The coefficients of the dxudxv terms¹ collectively form the metric tensor and are functions of the coordinates. The curvature tensors contain second-order partial derivatives of the metric tensor.

¹ The superscripts simply specify the components of a vector and are not exponents.

26. Originally Posted by KJW

Originally Posted by geordief
Does the measurement process bear any comparison with, say how one might measure the curvature of a simple sphere?
Apart from the differences associated with the increased number of dimensions, the curvature of spacetime is essentially the same as the curvature of a sphere. However, the curvature of a sphere can be viewed extrinsically because the embedding in three-dimensional space is easy to visualise, whereas we don't visualise spacetime as embedded in anything. But note that a sphere only has simple scalar curvature, whereas spacetime has the full curvature that only exists in four or more dimensions.
If we choose a distance in spacetime such as 1 lightyear and make a map of all the events in space that would have been judged(ie potentially observed) by someone on (as an example) the Earth as having occured then **,would that set of events constitute a surface in spacetime?

And if the original spacetime distance (1 ly) was reduced or increased by an infinitesimal amount ,would the change in that surface be a method of calculating the curvature in spacetime produced by a massive object in the vicinity ?

**the observer would make an inventory of all events that had taken place and only select those which had apparently occurred "simultaneously" with the arrival of a beam of light originating 1 year in the past(a hypersurface of simultaneity?)

EDIT: How many dimensions are needed to model/describe that surface? (the set of all events appearing to an observer as having occurred at the same time)

27. Originally Posted by geordief
I have a question about whether a line can be intrinsically curved.
there exist two approaches in my mind

1) we can express almost all functions with fourier series. so ,the curvature.
2) in complex anlysis ,we can claim that a circle (means here a curve) is a line , or a line is a circle.

28. A little off topic ,but one of the tests for intrinsic curvature is that the value for pi of the sum of the angles of a triangle are different from a flat space.

If the curvature is present but constant how can one know that the value measured is not "flat"?

If one measured pi as (say) 5 what is to say that the space is not curved at all but that pi=5 is the correct number?

Does this test for intrinsic curvature only work when the intrinsic curvature varies from one point to another?

Might this imply that our universe may actually be intrinsically curved despite measurements apparently concluding it is flat ?

Just that the curvature is the same everywhere.....

29. 2) in complex anlysis ,we can claim that a circle (means here a curve) is a line , or a line is a circle.
Funny thing is, that the curvature is affected by dimensional scale. If we keep increasing the radius of a circle, the curvature of line will at some point disappear for a stationary observer - or surface of a sphere will become flat. Observer on the surface of Earth is not able to notice the curvature and you need to change the perspective by:
a) moving further away from the planet
b) changing the scale

This is probably why you can't observe the effects of gravitational time dilation until you won't compare the flow of time in different frames - just as you can't observe the curvature of a planet, while you're standing on it's surface...

30. Originally Posted by geordief
A little off topic ,but one of the tests for intrinsic curvature is that the value for pi of the sum of the angles of a triangle are different from a flat space.

If the curvature is present but constant how can one know that the value measured is not "flat"?

If one measured pi as (say) 5 what is to say that the space is not curved at all but that pi=5 is the correct number?

Does this test for intrinsic curvature only work when the intrinsic curvature varies from one point to another?

Might this imply that our universe may actually be intrinsically curved despite measurements apparently concluding it is flat ?

Just that the curvature is the same everywhere.....
I’m not sure if I understand your question, but I’m going to take a stab at it. My gut tells me the answer is no. My understanding is that the value of pi (the constant) is a relationship between the diameter & circumference of a circle drawn exclusively on a flat surface, or, like you said, in the case of drawn triangles, a relationship between their angles (the sum of each adding to 180). In other words, the constants in both cases are intrinsic to circles or triangles drawn on a flat surface (as opposed to, say, a sphere). Of course we’ll have to wait for people on the forum with real knowledge, the big guns if you will, to pipe in.

PS. As a cop out, I think I can safely say that the answer to your last question would be no, since intrinsic curvature is by definition detectable through measurements made by the “inhabitants” of its surface.

31. Originally Posted by Vexspits
.

PS. As a cop out, I think I can safely say that the answer to your last question would be no, since intrinsic curvature is by definition detectable through measurements made by the “inhabitants” of its surface.
I think you are probably right.

On reflection intrinsic non curvature should be as demonstrable as intrinsic curvature (using the same techniques) and so pi can only lie on the "sweet spot" between positive and negative curvature.

32. On reflection intrinsic non curvature should be as demonstrable as intrinsic curvature (using the same techniques) and so pi can only lie on the "sweet spot" between positive and negative curvature.
The definition of intrinsic curvature is not done via the numerical value of pi, but rather through what happens when you parallel-transport a tangent vector along a small closed curve on the surface in question. If there is intrinsic curvature, then, once the vector arrives back at the starting point, it will not coincide with the original vector you started with. The difference between the two is encoded by the Riemann curvature tensor.

33. Originally Posted by Markus Hanke

The definition of intrinsic curvature is not done via the numerical value of pi, but rather through what happens when you parallel-transport a tangent vector along a small closed curve on the surface in question. If there is intrinsic curvature, then, once the vector arrives back at the starting point, it will not coincide with the original vector you started with. The difference between the two is encoded by the Riemann curvature tensor.
I now think I was wrong to assume that there could be a negative intrinsic curvature in spacetime.It is either positive or zero.Correct?

Although intrinsic curvature is evaluated by parallel transport ,is not the value of pi a direct consequence?

34. Originally Posted by geordief
I now think I was wrong to assume that there could be a negative intrinsic curvature in spacetime.It is either positive or zero.Correct?
No. A saddle is an example of a surface with negative curvature.

35. Originally Posted by KJW
Originally Posted by geordief
I now think I was wrong to assume that there could be a negative intrinsic curvature in spacetime.It is either positive or zero.Correct?
No. A saddle is an example of a surface with negative curvature.
Are there any physical regions in spacetime that can be modeled geometrically as a saddle? Wouldn't that be a region where gravity is negative?(which is why I decided in my own mind they couldn't exist physically)

36. <removed>

37. Originally Posted by Markus Hanke

The definition of intrinsic curvature is not done via the numerical value of pi, but rather through what happens when you parallel-transport a tangent vector along a small closed curve on the surface in question. If there is intrinsic curvature, then, once the vector arrives back at the starting point, it will not coincide with the original vector you started with. The difference between the two is encoded by the Riemann curvature tensor.
Any lurkers for whom this is all new (like me) might find this link useful https://www.coursera.org/learn/gener...llel-transport It provides a decent visual of a parallel transport of a vector on a flat surface vs a spherical one.

38. If we draw a simple circle on a flat plane , does the curvature get greater the smaller we draw the circles?(and "obviously" vice versa)

If we can measure the curvature of this simple circle ,is it an intrinsic curvature?* (maybe we need to draw concentric circles to show it?)

Apologies for my (to be kind) simple intellect

** perhaps KJW and Guitarist disagree on this.....

39. Originally Posted by geordief
If we draw a simple circle on a flat plane , does the curvature get greater the smaller we draw the circles?(and "obviously" vice versa)
I think so, since the circle’s curvature is defined by the reciprocal of its radius, 1/r. Put another way, the curvature is the rate of change of direction of a curve with respect to distance along the curve.1 So if you were to take two circles of different size, examine the change of direction over a distance of one foot along the circumference of each, the change in direction of the curve on the smaller would obviously be greater.

1 https://www.britannica.com/science/curvature

40. Originally Posted by Vexspits
Originally Posted by geordief
If we draw a simple circle on a flat plane , does the curvature get greater the smaller we draw the circles?(and "obviously" vice versa)
I think so, since the circle’s curvature is defined by the reciprocal of its radius, 1/r. Put another way, the curvature is the rate of change of direction of a curve with respect to distance along the curve.1 So if you were to take two circles of different size, examine the change of direction over a distance of one foot along the circumference of each, the change in direction of the curve on the smaller would obviously be greater.

1 https://www.britannica.com/science/curvature
Looks like extrinsic curvature to me.

Seems to be no way of knowing how big or small the circle is by simply taking measurements along the circle.

If you are allowed to take a measurement of the radius I wonder if that would make any difference? (would it again be an extrinsic measurement?)

41. Originally Posted by geordief
Originally Posted by Vexspits
Originally Posted by geordief
If we draw a simple circle on a flat plane , does the curvature get greater the smaller we draw the circles?(and "obviously" vice versa)
I think so, since the circle’s curvature is defined by the reciprocal of its radius, 1/r. Put another way, the curvature is the rate of change of direction of a curve with respect to distance along the curve.1 So if you were to take two circles of different size, examine the change of direction over a distance of one foot along the circumference of each, the change in direction of the curve on the smaller would obviously be greater.

1 https://www.britannica.com/science/curvature
Looks like extrinsic curvature to me.
I’m not sure. I think an example of extrinsic curvature would be if you, say, folded that flat plane surface into a cylinder—that curvature would be extrinsic, because it relies on an additional dimension (resigned to the two dimensional flat plane, one would be at a loss to detect it?). But I still think the circle on the two dimensional plane has intrinsic curvature. Unfortunately, despite having re-read the exchange between Guitarist an KJW, and having checked other sources, I still can’t tell you why I believe that to be true (Lol).

42. Originally Posted by Vexspits
Originally Posted by geordief
Originally Posted by Vexspits
Originally Posted by geordief
If we draw a simple circle on a flat plane , does the curvature get greater the smaller we draw the circles?(and "obviously" vice versa)
I think so, since the circle’s curvature is defined by the reciprocal of its radius, 1/r. Put another way, the curvature is the rate of change of direction of a curve with respect to distance along the curve.1 So if you were to take two circles of different size, examine the change of direction over a distance of one foot along the circumference of each, the change in direction of the curve on the smaller would obviously be greater.

1 https://www.britannica.com/science/curvature
Looks like extrinsic curvature to me.
I’m not sure. I think an example of extrinsic curvature would be if you, say, folded that flat plane surface into a cylinder—that curvature would be extrinsic, because it relies on an additional dimension (resigned to the two dimensional flat plane, one would be at a loss to detect it?). But I still think the circle on the two dimensional plane has intrinsic curvature. Unfortunately, despite having re-read the exchange between Guitarist an KJW, and having checked other sources, I still can’t tell you why I believe that to be true (Lol).
Imagine you are lost in space,alone in your spacesuit as a result of a space walk gone wrong.

Amazingly you come across an apparently straight length of (magical) railway line that stretches before endlessly you as far as the eye can see .

How can you tell whether this railway line is straight or curved simply by taking measurements of distances from one part of the line to another ?

I think there is no way and so any actual curvature can only be measured by means of a bird's eye view and so any curvature would ,in my raw opinion be extrinsic.

I know Guitarist promised me "meat and vegetables" to show that a line can have intrinsic curvature.

Sadly I am fairly confident sure I would be unable to follow his maths .....

43. Originally Posted by geordief
How can you tell whether this railway line is straight or curved simply by taking measurements of distances from one part of the line to another ?
To test for extrinsic curvature, you carry an accelerometer with you while moving at a constant speed; if it reads anything other than zero, with speed remaining constant, then you know there is extrinsic curvature. For intrinsic curvature, you carry a watch, and measure the physical distance you travel in a set period of time without accelerating or decelerating along the direction of the train tracks. You do this in different places along the rail line. If you are travelling different distances in different sections of the rail line within the same amount of time as measured by the watch, and with speed having remained constant at all times, then the tracks have intrinsic curvature.

In practice though we live in a 3+1 dimensional world, so any test for curvature would be done in different ways.

44. Originally Posted by Markus Hanke
Originally Posted by geordief
How can you tell whether this railway line is straight or curved simply by taking measurements of distances from one part of the line to another ?
To test for extrinsic curvature, you carry an accelerometer with you while moving at a constant speed; if it reads anything other than zero, with speed remaining constant, then you know there is extrinsic curvature. For intrinsic curvature, you carry a watch, and measure the physical distance you travel in a set period of time without accelerating or decelerating along the direction of the train tracks. You do this in different places along the rail line. If you are travelling different distances in different sections of the rail line within the same amount of time as measured by the watch, and with speed having remained constant at all times, then the tracks have intrinsic curvature.

In practice though we live in a 3+1 dimensional world, so any test for curvature would be done in different ways.
Even if the curvature is constant (eg a circle)? (my bold)

45. I feel it necessary to point out that the term "curvature" applied to 1-dimensional curves is a different notion to the term "curvature" applied to spaces of 2 or more dimensions. Under the assumption that the 1-dimensional curve is embedded in a higher-dimensional space, "curvature" is defined as the absolute derivative of the unit tangent vector with respect to the arc-length parameter along the curve:

where is the unit tangent vector ( is the arc-length parameter along the curve)

All values, , , and are defined in terms of the embedding space. By contrast, for spaces of 2 or more dimensions, the term "curvature", when used without qualification, is defined as the Riemann curvature tensor field:

(It should be noted that different texts define the Riemann curvature tensor differently, resulting in a different sign, particularly in the Einstein equation)

46. Originally Posted by KJW
I feel it necessary to point out that the term "curvature" applied to 1-dimensional curves is a different notion to the term "curvature" applied to spaces of 2 or more dimensions. Under the assumption that the 1-dimensional curve is embedded in a higher-dimensional space, "curvature" is defined as the absolute derivative of the unit tangent vector with respect to the arc-length parameter along the curve:

where is the unit tangent vector ( is the arc-length parameter along the curve)

All values, , , and are defined in terms of the embedding space. By contrast, for spaces of 2 or more dimensions, the term "curvature", when used without qualification, is defined as the Riemann curvature tensor field:

(It should be noted that different texts define the Riemann curvature tensor differently, resulting in a different sign, particularly in the Einstein equation)
As I said above ,I am not able to comment on/follow the maths ,but feel that I am (trying to) take your side in that my physical example (idealized) cannot be determined (curved or not) without recourse to an external space (by taking tangents which seem to require an extra dimension)

I am happy that Time should be allowed in my "thought experiment"-so it is ID+T but I feel only extrinsic curvature can be determined .

I am now unclear after Markus' post whether this intrinsic curvature is possible if the curvature is not constant.

47. Even if the curvature is constant (eg a circle)? (my bold)
A circle does not have intrinsic curvature, only extrinsic one.

48. my physical example (idealized) cannot be determined (curved or not) without recourse to an external space (by taking tangents which seem to require an extra dimension)
Extrinsic curvature is a meaningful concept only when something is embedded in a higher dimensional space, as in the example of your rail line. If that is not the case, then there can only be intrinsic curvature. If there is intrinsic curvature, then there are always ways to detect this, but note that any meaningful notion of “measurement” requires the presence of time. So physically, and in practice, I think the simplest meaningful scenario would be 1+1 dimensions, since no measurements can take place in 1D only.

49. Originally Posted by geordief
looks like extrinsic curvature to me.
Said you with reference to the curvature of a circle. And then we have:

Originally Posted by markus hanke
a circle does not have intrinsic curvature, only extrinsic one.

There it is! And I’ve found at least one other corroborating reference: "Gaussian curvature is the most commonly studied intrinsic measure of curvature....This is a reasonably advanced notion in differential geometry, applying also for manifolds (extensions of surfaces with dimension greater than two). Note that for a 1-dimensional manifold (e.g. a line, curve, circle) there is no intrinsic curvature, only extrinsic curvature" (my emphasis). https://www.maths.ox.ac.uk/about-us/...ntial-geometry (under the section entitled "Intrinsic Curvature")

50. Yes I think only Guitarist in this thread maintains that a line can have intrinsic curvature .

I think his definition of intrinsic curvature may be different from the others'.

I wonder what it is (and whether I could understand it as he has warned that it might be daunting).

He said in post#14 "do not assume it is a straight line in the Euclidean sense"-which I don't understand (naturally enough)

51. Yes I think only Guitarist in this thread maintains that a line can have intrinsic curvature .
I must admit I also thought a 1D manifold can - at least in principle - have intrinsic curvature (though a circle quite obviously doesn’t), since the Riemann tensor in 1D still has one component. But I just remembered that the tensor is also skew-symmetric, so this component must be identically zero. Therefore there cannot be intrinsic curvature in 1D only. But there can be extrinsic curvature, if the line is embedded in a higher dimensional manifold.

52. I am still wondering if spacetime can have negative curvature...

It seems to me that it should (in practical applications) only have positive values and that negative values would be used potentially to describe anti gravity scenarios.

53. Originally Posted by geordief
I am still wondering if spacetime can have negative curvature...
Yes it can. This is in fact one of the options available to us in standard cosmology - the shape of the global universe can be that of a “saddle”, i.e. an open universe with negative curvature, if a quantity called the “density parameter” has a value of less than unity.

54. Originally Posted by Markus Hanke
Originally Posted by geordief
I am still wondering if spacetime can have negative curvature...
Yes it can. This is in fact one of the options available to us in standard cosmology - the shape of the global universe can be that of a “saddle”, i.e. an open universe with negative curvature, if a quantity called the “density parameter” has a value of less than unity.
Would that cause an ongoing global expansion? It seems like an anti gravity (or negative gravity) effect to me.

55. I should be grateful if somebody could explain what is meant by "negative curvature". As far as my learning goes, a manifold is either curved or it is not. And since the curvature tensor field is the second derivative of the metric tensor field, a non-zero curvature field implies a non-constant metric field. This what one might expect of a spacetime equipped with gravitating bodies.

56. I should be grateful if somebody could explain what is meant by "negative curvature".
I don’t know where exactly the nomenclature comes from, but this seems to be used in the context of geodesic deviation. If initially parallel geodesics converge as they extend into the future, it’s said that the region of spacetime in question is “positively curved”. Likewise, if the geodesics diverge, the curvature is said to be negative (hence geordief’s connecting this to anti-gravity).

I have seen this used in various physics texts over the years, but I guess it’s really just a convention.

57. Originally Posted by Guitarist
I should be grateful if somebody could explain what is meant by "negative curvature".
I think that it is used for "time" parameter. time is positively defined only and is first order polynomial function. either when the more than first order polynomial function ((polynom ,quadratic,qubic or more order polynoms)) is defined or when the negative curvature is possible for time parameter , these mean that we could visit the past times. at first one, when we use more than first order function for time,the currency is not same at everywhere. the second one is already clear.

this subject has close relation with black holes (or some (new) special forms)

note please at above,some descriptions are defined about "differential geometry" but I cannot interfere to all related subjects ,here I approach to the subject via using these common nomenclatures / key words.

1) convergence or divergence 2) ¢[a,b] 3)L[a,b] 5) completeness , 6) c ,7) c0 8) cauchy sequences 9)compactness

and like these

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