Curvature: Intrinsic vs. Extrinsic - What's the Difference?

[vibhuav]

In trying to explain the concept of curved space, many books use the example of the surface of a sphere, which can be considered as a curved 2D space embedded in a higher dimensional, 3D space. I could derive, starting from $a^2=x^2+y^2+z^2$, that the metric, or the line element, on the surface of the sphere - which we now consider to be 2D - is $ds^2=\frac{a^2 {d\rho}^2}{a^2-\rho^2}+\rho^2 {d\phi}^2$. But here’s my question: in the equation for the metric, what are $x(=\rho\cos\phi)$ and $y(=\rho\sin\phi)$? It makes sense that they are the original $x$ and $y$ from the 3D space, which are measurements in the external, third dimension, but I would rather have coordinates $\it{on}$ the 2D surface of the sphere. (Note the had we used $\theta$ and $\phi$, they too are in the external 3D space.)

Some books use a tangent plane at the point in question on the surface of the sphere, and define a 2D $x$-$y$ Cartesian coordinate system on it, but - and this is my main question - I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent space. What am I missing?

While we are at it, does the tangent plane move around on the surface of the sphere for different points? And the line element on such a tangent plane ought to be projected on to the curved, 2D space. What is the math for this?

 

[Nugatory]

Some books use a tangent plane at the point in question on the surface of the sphere, and define a 2D $x$-$y$ Cartesian coordinate system on it, but - and this is my main question - I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent space. What am I missing?

You are missing nothing, because there is no such relationship. The points making up the two-dimensional sphere are a subset of the points that make up the three-dimensional space, but the tangent space is a completely different mathematical object. It's a vector space and its elements are vectors, not points in the 3D space.

This might also be a good time to mention that we don't need the 3D space at all. Mathematically what we have is a "manifold", which is a set of points, a mapping from those points to pairs (and "pairs" is why say it's a two-dimensional space) of numbers $\theta, \phi$, and a rule for calculating a number $ds^2$ from differences between these $\theta$ and $\phi$ values. However, this is all very abstract and hard to visualize, so the textbooks draw a picture of a three-dimensional globe in three-dimensional space, then tell us to think about how the points on its two-dimensional surface behave.

While we are at it, does the tangent plane move around on the surface of the sphere for different points? And the line element on such a tangent plane ought to be projected on to the curved, 2D space. What is the math for this?

Every point in the manifold has its own tangent space. That tangent space is a mathematical tool that we need if we're going to reason about the behavior of vectors at that point; it doesn't have distances, let alone distances that need to be projected back onto the surface of the sphere.

Some other poster may be able to recommend a good introduction to differential geometry. That's really the only way to make sense of this stuff.

 

[fresh_42]

Some other poster may be able to recommend a good introduction to differential geometry. That's really the only way to make sense of this stuff.

I like this one very much, because it is written in a manner, which doesn't give you the feeling that it is unreachable math. It also has plenty of graphics, examples, exercises and diagrams which are especially in the field of differential geometry very helpful and in my opinion necessary. It might not contain all algebraic or topological aspects of the theory, but it provides a very good foundation, which enables its readers to continue in the field if needed.
https://www.amazon.com/dp/0387903577/?tag=pfamazon01-20

 

[vibhuav]

Thanks for the replies. I will have to think through them carefully to understand completely.

At this stage I would like to clarify my OP, quoted below:

Some books use a tangent plane at the point in question on the surface of the sphere, and define a 2D xxx-yyy Cartesian coordinate system on it, but - and this is my main question - I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent space. What am I missing?

...my main question is, I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent $\it{plane}$ and obtain a equation for the metric, $ds$, which still extols the non-flatness as the original metric equation did.

 

[Nugatory]

...my main question is, I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent $\it{plane}$ and obtain a equation for the metric, $ds$, which still extols the non-flatness as the original metric equation did.

You can calculate the metric for the two dimensional curved manifold (the sphere) from the metric for the three-dimensional space in which ithe sphere is embedded. Google for "induced metric" to see how this works - the Wikipedia article has the relevant formula. In this particular problem the calculation is straightforward (although you'll need some calculus and comfort with the Einstein summation convention) and it's fun watching it come out the way you expect.

However, you can't do something similar with the tangent space, because the tangent space at a given point isn't a submanifold of the three-dimensional space. Indeed, the term "tangent $\it{plane}$" is something of a misnomer - the tangent space at a given point can only be represented as a plane if you're drawing a picture in which your curved manifold is embedded in a higher-dimensional flat space. And in that case the relationship between coordinates on the plane and coordinates in the embedding space is pretty much unrelated to the geometry of the curved manifold.

 

[pervect]

Thanks for the replies. I will have to think through them carefully to understand completely.

At this stage I would like to clarify my OP, quoted below:...my main question is, I am not able to mathematically relate the original coordinates in 3D space to the coordinates of the tangent $\it{plane}$ and obtain a equation for the metric, $ds$, which still extols the non-flatness as the original metric equation did.

What you've found is a set of coordinates for the 2d surface of the sphere. It's possible to transform those coordinates by any invertible function (which is also usually required to be smooth, but I don't want to try to define smooth rigorously) to a different set of coordinates. So there are many different choices of coordinates.

Tensor methods tell you how the metric tensor transforms when you choose different coordinates. So using the tensor techniques, you can (if you know how to go through the mathematical manipulations) transform the metric tensor from any set of coordinates to any other set. It will probably take some study before you can actually carry this process out, but at this point I'm just pointing out that it's possible to do, that there is sufficient information.

There is a particularly nice set of coordinates called Riemann normal coordinates. You can google for more information, Wiki discusses it as a special case of normal coordinates <<link>>. There are probably better articles out there on the topic than wiki, this is just to give you some reference on this important keyword.

Riemann normal coordinates are very nice, and I started to do a writeup on them, but it got rather long, too long. So I won't focus on them, though they might well be what you're looking for.

What I'll focus on instead is something more modest. What you are doing has the mathematical name of "projection". Riemann normal coordinates are one way of projecting the surface of the sphere onto a flat map with an associated metric, but the details will be very messy. It would be much easier to consider a different projection, the usualattitude and longitude system, or even something like the Mercator projection. Something that is in common used to make 2d maps of the Earth, in other words.

If you understand that projections are just mappings, the only other detail you need to know is how the metric tensor transforms when you do such a coordinate change. It follows standard tensor transformation rules, which I'm not going to attempt to describe in detail in this post. If you need more details on that I suppose we can discuss it, the only point I want to make is that given a metric, and given a projection, we can find the induced metric of the projected geometry when we know the original metric and the mapping function that defines the projection process.

 

[Dale]

I would rather have coordinates onon\it{on} the 2D surface of the sphere

You do. $\rho$ and $\phi$ are coordinates in the sphere.

 

[vibhuav]

Thanks for all the replies; I am still working it all out.

In an attempt to understand how the metric of a 2D surface (even the tangent plane) embedded in 3D space still extols the non-flatness as the original metric equation did, I tried to consider the 1D “space” (basically a line) embedded on a 2D circle. So such a line is indeed curved, or so I thought.

First of all, how can an ant residing on such a 1D curved line determine intrinsically that its space is curved, just like how an ant on the 2D surface of a sphere can do so by measuring total angle of a large triangle and compare it to $180^\circ$? I could not think of any ways. Even a parallel transport of a vector does not seem to show any difference that exposes non-flatness. Is there a physical method that an ant on the 1D “space” can determine if it is curved or not?

Then I thought of the Riemannian curvature tensor, $R^\rho_{\;\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\;\nu\sigma} - \partial_\nu\Gamma^\rho_{\;\mu\sigma} + \Gamma^\rho_{\;\mu\lambda}\Gamma^\lambda_{\;\nu\sigma} - \Gamma^\rho_{\;\nu\lambda}\Gamma^\lambda_{\;\mu\sigma}$. For a 1D space, all the variables, $\rho, \sigma, \mu, \nu$ and $\lambda$ are just 1, and so there will be only one Riemannian curvature component, $R^1_{\;111} = \frac{\partial\Gamma^1_{\;11}}{\partial x} - \frac{\partial\Gamma^1_{\;11}}{\partial x} + \Gamma^1_{\;11}\Gamma^1_{\;11} - \Gamma^1_{\;11}\Gamma^1_{\;11}$ which evaluates to $0$.

In other words, the curvature of a 1D line embedded on a 2D circle is zero. Does that match with my understanding that there appears to be no practical way for an ant on a 1D space to determine if its space is curved or not? But we, the 2D and 3D people, can easily see that the line segment is curved. What gives?

 

[Dale]

First of all, how can an ant residing on such a 1D curved line determine intrinsically that its space is curved, just like how an ant on the 2D surface of a sphere can do so by measuring total angle of a large triangle and compare it to 180∘180∘180^\circ?

A 1D manifold cannot have any intrinsic curvature.

 

[pervect]

In other words, the curvature of a 1D line embedded on a 2D circle is zero. Does that match with my understanding that there appears to be no practical way for an ant on a 1D space to determine if its space is curved or not? But we, the 2D and 3D people, can easily see that the line segment is curved. What gives?

What appears to be happening is that you are not distinguishing intrinsic curvature from extrinsic curvature. The computation of the Riemann you did is an illustration that a line has no intrinsic curvature. When you look at the line and say that it is curved, you are talking about something that is not the intrinsic curvature measured by the Riemann tensor, but a different sort of curvature, extrinsic curvature.

A short summary of the difference is that intrinsic curvature can be computed by measurements made only on the manifold, without regard to any embedding of the manifold.

You appear to be only considering curvature caused by an emgbedding. However, the Riemann tensor and intrinsic curvature is defined in such a manner that it depends only on the geometry, it does not need an embedding to measure. For instance, a hypotehtical "flatlander" living on a 2 dimensional surface could determine if that surface was curved or flat, by summing the angles of triangles of various areas. If the surface was flat, the sum of the triangles would always be 180 degrees, if the surface was not flat, the sum of the angles of the triangles would depend on their area.