Can the Dot Product of Electric and Magnetic Fields be Proven as Invariant?

[turin]

Originally posted by selfAdjoint
within that frame{/i] you move the vector parallel to itself in a small (differentially small) closed curve and its direction changes. From the amount of turning you can deduce the connection components and derive the covariant derivative, in that coordinate system. Then you have to prove that unlike the connection coefficients, the covariant derivative is, uh, covariant.

Well, I follow what you're saying to the same extent I've followed it when anyone else has said it. The main thing with which I am uncomfortable is parallel transport. Why would you want to parallel transport a vector when this obviously distorts the vector? It seems like a parallel transported vector isn't really the same vector. Maybe I don't understand what a vector is.

The most popular simplified example that I've seen is the sphere. So parallel transport of a vector is the same as moving an arrow along a great circle while always maintaining the dot product of the arrow with an arrow tangent to the great circle. But this obviously shows that the transported arrow isn't really the same vector as I see it. The arrow obviously changes direction as it moves along a great circle in this manner.

 

[turin]

Originally posted by chroot
... if you slide the vector in Africa around the equator, you can transform it into the vector on the other side of the Earth -- proving they are, in fact, parallel.

Can I not transform any vector into any other vector, proving that every vector is parallel to every other vector?

"Sliding" a vector along some path (a geodesic, for instance) seems to have a different meaning than simply "translating" or "moving" a vector along that same path. It isn't at all clear to me why would want to knowingly change the direction of our vector.

 

[turin]

Originally posted by lethe
no, i don t agree with this picture at all. in general, there isn t even any such notion as perpendicularity with things that are not in the tangent space, and sometimes even with things that are int the tangent space.

I guess I just got confused by the diagrams. Is not the dual of a vector a set of surfaces, though, or is this flirting too much with the pancake analogy?

Originally posted by lethe
remember how a derivative works: you have to subtract the value of the thing you are taking the derivative of at two neighbouring points, and then take the limit as these points approach each other.

only thing is, there is no way to subtract one vector from another if they do not live in the same vector space.

Aha. I think things are starting to come together. In the sphere analogy, there is no 3-D space, really? It is useful to visualize, but really we're just talking about a bunch of 2-D tangent spaces, and the 3-D space does not exist? This still makes me a little uncomfortable, do to the readily available 3-D space. I guess there are situations, like space-time, that don't have such a readily available vector space?

Originally posted by lethe
what the rule looks like is completely arbitrary, you can define comparison between neighboring vectors any way you want, as long as you adhere to a few simple axioms: linearity and the Leibniz law for products.

such a linear derivation is called a connection.

no matter what it looks like, i must be able to express it in local coordinates, and that is just what i have done on the right hand side of that equation.

Why must it adhere to these axioms? What is the Leibniz law? What are local coordinates (as opposed to "global?" coordinates)?

 

[chroot]

Originally posted by turin
But this obviously shows that the transported arrow isn't really the same vector as I see it.

That's absolutely correct -- and very important! The simple fact is that a vector exists within a vector space. The tangent space at a point P in a manifold is such a vector space. However, another point Q in the manifold has a completely different tangent space, and thus completely different vectors live in it. A vector defined at one point in a manifold doesn't inherently have anything to do with any other vector defined at any other point in the manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.

To facilitate the comparison of vectors at P with vectors at Q, you have to introduce some valid mechanism to identify a vector in one tangent space with another (different!) vector in another tangent space. I assume there are lots of ways you can do this -- but the one most often done is to establish the mechanism of parallel transport.

When you parallelly transport a vector from point P to point Q, you are certainly changing the vector -- in fact, you're completely changing the space it lives in. In fact, that's not really possible. What you're doing is destroying the original vector and making an entirely new one in the new space at Q, and just labelling the new one "parallel" to the old one. It's only parellel in the sense that it was found from this operation called parallel transport. Yes, you can see the circularity inherent in the definition.

- Warren

 

[turin]

Originally posted by lethe
the notion of a connection is not specific to Riemannian geometry, so let's just say differential geometry.

Well, I'll try to remember to respect your wishes on this issue, but neither title means anything to be at all. Apologies for any disrespect to Mr. Riemann or Mr. differential.

Originally posted by lethe
the way vectors are often introduced, one writes their components, and decrees how they must transform under coordinate transformations.

this totally obscures the fact that they are geometric objects completely independent of what their components may be in some basis! i really dislike this definition of a vector.

I first really understood (I think I understood) what a vector was when I took QM (chapter 1 of Shankar). Are the vectors in QM and the vectors we're talking about now fundamentally different? I suppose the first concrete example of vectors in QM had a coordinatization (the x-basis), but I don't recall ever dealing with parallel transport or anything like that. Is this related to the problem of finding a QM theory of gravity?

Originally posted by lethe
<br /> \begin{align*}<br /> D_{a\mathbf{v}+b\mathbf{w}}s&amp;=aD_{\mathbf{v}}s+bD_{\mathbf{w}}s\\<br /> D_{\mathbf{v}}(s+t)&amp;=D_{\mathbf{v}}s+D_{\mathbf{v}}t\\<br /> D_{\mathbf{v}}as&amp;=\mathbf{v}(a)s+aD_{\mathbf{v}}s<br /> \end{align*}<br />
these are just the linearity requirements and the Leibniz rule. this approach is perhaps more abstract, but i much prefer it.

I don't quite follow your notation. Apparently, the a and b in the subscript are scalar multiples, the bold v and w are two vectors, the s and t superscripts in the subscript are parameterizations? What is D?

Originally posted by lethe
define parallel.

We could go around and around. Whatever I say, I'm pretty sure you will find some term that I use inadequate. Then, I'll have to define that, and inevitably use another term in that definition that you will find inadequate, and on and on, until I get tired of it. But what the hell, I'll give it a shot:

Two vectors are parallel if they point in the same direction.

I have no idea how to define direction, but for some strange reason, I think I know what it means.

Come to think of it, I really can't think of a good definition for "vector" or "point," either.

 

[turin]

Originally posted by Ambitwistor
Tell me: if I have a vector at the equator pointing north, then which direction should it be pointing at the north pole, in order to "not change direction" from how it was pointing when it was at the equator?

Towards Polaris.

 

[chroot]

Originally posted by turin
Come to think of it, I really can't think of a good definition for "vector" or "point," either.

Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.

However, it's true that you need to learn to walk (to understand the broad nature of what the mathematics does) before you can fly (to understand the rigorous definitions of each of the words). All of this stuff boils down to the simple concept of vector parallelism you learned about in grade school when the connection coefficients go to zero and the space is flat. As a result, I think it's best to learn this stuff by constantly reminding yourself of how this differential geometry can be boiled down to the stuff you already know.

In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them!

- Warren

 

[turin]

Originally posted by chroot
A vector defined at one point in a manifold doesn't inherently have anything to do with any other vector defined at any other point in the manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.

To facilitate the comparison of vectors at P with vectors at Q, you have to introduce some valid mechanism to identify a vector in one tangent space with another (different!) vector in another tangent space. I assume there are lots of ways you can do this -- but the one most often done is to establish the mechanism of parallel transport.

Good supplement to lethe's contribution. Why is it that they hammer a connection between vectors into your head in high school? Or was that just my experience?

Originally posted by chroot
When you parallelly transport a vector from point P to point Q, ... you're ... destroying the original vector and making an entirely new one in the new space at Q, and just labelling the new one "parallel" to the old one.

This is an excellent clarification! Thank you chroot!

 

[chroot]

Originally posted by turin
Towards Polaris.

You're still stuck in three-space! If you're an ant living on this two-dimensional surface, you can't look up or down. There's no such thing as Polaris. All you can do with your vectors is to push them around the surface. Imagine starting at some point P and pushing a vector along a great circle. After you've walked all the way around the sphere, the vector ends up being the same at P as it was when you started. That's parallel transport.

If you pushed your vector along another path that was not a great circle, you'll find that, much to your dismay, the vector you end up with at P is not the same as the one you started with at P.

The reason all this mathematical machinery is needed is because we need a way to define curvature without reference to some outside, higher-dimensional space. In other words, we need some way to define what the curvature is at a point on the Earth without any reference at all to any points not also on the surface of the Earth. This is the manifold's intrinsic curvature. Parallel transport allows us to determine this intrinsic curvature, without having to look up!

- Warren

 

[turin]

Originally posted by Ambitwistor
If you change the connection, then the transported vector can be different: a connection is what defines whether two vectors at different points are parallel, given a path between those points. That's why we say that parallel transport doesn't change the vector: there is no way, other than a connection, to define whether a vector has been changed or not.

I think this is becoming much more clear now. Thanks.

Originally posted by Ambitwistor
Can you define what those words mean? ("Sliding", "translating", and "moving".)

No. That's why I made the comment. The term, "sliding" seemed to be used ambiguously. In retrospect, I would say it means "parallelly transporting," or "mapping a vector from the vector space tangent to point P to a vector in the vector space tangent to point Q according to the transformation that corresponds to parallel transport (connection rule)." "Translating" and "moving" I intended to be ambiguous as dramatic contrast.

 

[turin]

Originally posted by chroot
Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.

In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them!

Actually I rather enjoy the rigor. I aspire for such in my own treatments. I was somewhat interested to see where the definition of "parallel" winds up. But, judging from the most recent posts, I see that it might not be all that interesting.

 

[turin]

Originally posted by Ambitwistor
It can't: there is no 3D embedding space. It has to lie within the surface of the sphere.

And now I think I appreciate that a lot more.

 

[chroot]

Originally posted by Ambitwistor
It can't: there is no 3D embedding space. It has to lie within the surface of the sphere.

To add to this, note that a vector, as we've been discussing, lives in the tangent space at a point in the manifold. In other words, if you take a sphere like the Earth, the tangent space at the north pole is a plane that just touches the Earth at that one point -- like resting a piece of cardboard on the top of a basketball. The tangent space to a 2D manifold is also 2D. In other words, a vector pointing from the north pole towards Polaris is not a member of the tangent space at the north pole. That vector can't exist!

- Warren

 

[turin]

Thanks to lethe, chroot and ambitwistor. Y'all are many-fold (pun intended) better at helping me get up to speed than my major professor.

I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)

So, the first thing my prof wants to do is define a global coordinate system using the usual polar and azimuthal angles. I didn't like this at the time, but couldn't put my finger on why I didn't like it. It seemed preferential in a way to something. Now, I think I can put my finger on it (almost):

1) How can you have this kind of a coordinate system, (&theta;,&phi;), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?

2) Why would you use &phi; as a coordinate, when taking the partial derivative of position with respect to &phi; clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?

3) At the poles, what is the &theta; direction? I mentioned that there were singularities at the poles, but he said there weren't.

4) Also, near the poles, it doesn't make sense why you would want to use &phi; as a coordinate, since going in that direction obviously has a gross difference in ds (close to the poles so sin&theta; << 1) for a given d&phi;.

 

[turin]

Originally posted by chroot
... a vector pointing from the north pole towards Polaris is not a member of the tangent space at the north pole. That vector can't exist!

Right. Like saying you have a vector pointing in the w direction in 3-D space, when the only directions you have are x, y, and z. It just doesn't make sense.

 

[chroot]

Originally posted by turin
I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)

The metric tensor is just a machine into which you can plug two vectors. It enables you to measure the lengths of things in the manifold. It allows you to determine the circumference of a large circle of given radius drawn in a curved space, for example.

1) How can you have this kind of a coordinate system, (&theta;,&phi;), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?

You can pick any sort of coordinates you like, actually. People use the spherical polar coordinates simply because they appeal to their knowledge of 2-spheres that are embedded in 3-space. It doesn't actually matter.

2) Why would you use &phi; as a coordinate, when taking the partial derivative of position with respect to &phi; clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?

This really feeds into your next question...

3) At the poles, what is the &theta; direction? I mentioned that there were singularities at the poles, but he said there weren't.

There absolutely are singularities in the spherical polar coordinate system on a 2-sphere. In fact, there is no way to cover the entire 2-sphere with just one coordinate system without having singularities.

However, you can cover the 2-sphere with two or more coordinate systems -- one that applies to only the northern hemisphere, for example, and a different one that applies only to the southern hemisphere. Where the two coordinate systems overlap, there are functions defined to convert from one to the other.

The resulting "atlas", composed of two different coordinate systems ("charts") stitched together, does not have any singularities. In other words, the singularities that result when you cover the 2-sphere with spherical polar coordinates are coordinate singularities. By picking a better (set) of coordinate system(s) (a better atlas), you can remove them. Some spaces are not so friendly, however. Some spaces contain singularities that cannot be removed by any clever use of charts stitched together.

- Warren

 

[turin]

Originally posted by chroot
The metric tensor is just a machine into which you can plug two vectors. It enables you to measure the lengths of things in the manifold. It allows you to determine the circumference of a large circle of given radius drawn in a curved space, for example.

From classical mechanics, I gathered that the metric tensor was the machine that calculates dot products. The reason it was introduced, as I understood it, was for use in coordinate systems that were not generally orthonormal, like polar coordinates in 3-D. But this just seemed to be like a using it for different coordinate systems in 3-D (or higher, depending on the number of particles) flat space, and that the coordinate system was curved artificially. In other words, it was allowed to define a vector as an arrow that points from the source to the observation point. Then, the length of that arrow could be found by:

|u|²=g_mnu^muⁿ.

But I have gotten the impression here today that this kind of a vector doesn't make the same kind of sense. It seems like, before you can construct such a vector, you have to define a connection rule, but the connection rule, as I understand it, must be defined in terms of the metric tensor.

Does a particular connection rule have a physical significance? It was said earlier that the connection rule is arbitrary, but, are there certain rules chosen because they have a certain physical significance? Then, do you have to figure out what the metric tensor is from the connection rule. This does seem circular.

Maybe I'll try to reword my confusion with this point in another way:

The metric tensor takes tangent vectors as inputs? These vectors must both be at the same point in the manifold? The vectors don't exist in the manifold, but rather in a space tangent to the manifold at a particular point where all this is goin' down? So then, the metric tensor takes these two objects that do not exist in the manifold, and spits out an object that does exist in the manifold? How can two vectors that are not in the manifold have anything to do with the length of something in the manifold?

 

[HallsofIvy]

Remember that vectors are derivatives. Two vectors that are not in the manifold can give, for example, a length in the manifold in the same way that, in basic calculus, integrating the tangent vector to a curve can tell you arc length.

 

[lethe]

Originally posted by turin
Well, I follow what you're saying to the same extent I've followed it when anyone else has said it. The main thing with which I am uncomfortable is parallel transport. Why would you want to parallel transport a vector when this obviously distorts the vector? It seems like a parallel transported vector isn't really the same vector. Maybe I don't understand what a vector is.

its not the same vector. its a parallel transported vector, the rule for parallel transport being whatever i want it to be.

Originally posted by turin
Can I not transform any vector into any other vector, proving that every vector is parallel to every other vector?

you can transform one vector into any other vector you want, but you must single out one of these transformations (per path between points) and call it the connection. then, only those vectors are parallel transported into one another, and others are not.

 

[lethe]

Originally posted by turin
I guess I just got confused by the diagrams. Is not the dual of a vector a set of surfaces, though, or is this flirting too much with the pancake analogy?

you can call it a "net of surfaces" if you want, as long as i don t hear you use the word perpendicular. there is no canonical notion of perpendicularity

Aha. I think things are starting to come together. In the sphere analogy, there is no 3-D space, really? It is useful to visualize, but really we're just talking about a bunch of 2-D tangent spaces, and the 3-D space does not exist? This still makes me a little uncomfortable, do to the readily available 3-D space. I guess there are situations, like space-time, that don't have such a readily available vector space?

well... there are theorems stating that any n dimensional differentiable manifold can be embedded in R²ⁿ. and i think any Riemannian manifold can be embedded isometrically in R^n(n+3)/2 (im not sure if i have that last one correct, but its something like that). that is a theorem proved by russell crowe... err... i mean john nash.

embeddings are hard to find, for some surfaces. they may live in spaces with very high dimension. also, there are many different ways to embed the same manifold.

so defining things in terms of the ambient space can be awkward. also, it makes it hard to tell which properties of your manifold depend only on the manifold, and which properties are particular to the arbitrary way you chose to embed it (intrinsic versus extrinsic properties). for a physicist, this is terrible. for example, physicists are mostly concerned with the manifold that is our spacetime, and it doesn t make physical sense to consider our spacetime as embedded in some higher dimensional space (although sometimes people do do this, see Hawking and Ellis, for example)

so the point is, whether or not there is an ambient space available to our manifold, it is highly desirable to develop all our machinery under the assumption that there is not.

Why must it adhere to these axioms? What is the Leibniz law? What are local coordinates (as opposed to "global?" coordinates)?

well, somewhere along the line, someone thought up the properties he thought a derivative of vectors on a smooth manifold ought to have, and wrote them down. these axioms seem very reasonable to me.

the Leibniz law may be more familiar to you in this form:

<br /> \frac{d}{dx}(f(x)g(x))=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}<br />

the connection must satisfy something like that (if its not obvious how what i wrote above about the connection is the same thing as this rule, that s OK, you may have to study some differential geometry for a while to see that)

and local coordinates are coordinates in some small neighborhood, that don t necessarily extend to the whole manifold (like the polar coordinates on a sphere)

but its not so useful to distinguish local coordinates from global coordinates (except for some trivial examples, no manifolds admit global coordinates)

 

[lethe]

Originally posted by chroot
manifold. The vectors are unconnected. There's no way at all to inherently compare a vector at one point P with another at point Q. None at all.

none at all? don t forget, the whole point of this discussion is the connection, which is the way to "connect" vectors in different vector spaces. i m sure you knew that, so i don t know why you say "none at all". in fact, there are infinitely many different ways to compare vectors in different tangent spaces.

perhaps you mean "no canonical way" which is certainly true.

Yes, you can see the circularity inherent in the definition.

i can t. and earnestly hope there isn t any.

 

[lethe]

Originally posted by turin
Well, I'll try to remember to respect your wishes on this issue, but neither title means anything to be at all. Apologies for any disrespect to Mr. Riemann or Mr. differential.

differential geometry is the study of smooth manifolds. Riemannian geometry is the study of smooth manifolds that also have metrics on them.

I first really understood (I think I understood) what a vector was when I took QM (chapter 1 of Shankar). Are the vectors in QM and the vectors we're talking about now fundamentally different? I suppose the first concrete example of vectors in QM had a coordinatization (the x-basis), but I don't recall ever dealing with parallel transport or anything like that. Is this related to the problem of finding a QM theory of gravity?

not really related, except in the sense that both concepts are vector spaces.

but in both cases, there is the issue of coordinate dependence.

I don't quite follow your notation. Apparently, the a and b in the subscript are scalar multiples, the bold v and w are two vectors, the s and t superscripts in the subscript are parameterizations? What is D?

the s and t are vectors (not necessarily tangent vectors, but you may assume that they are tangent vectors if you like).

D is the connection. a connection is any linear operator satisfying the rules i listed. these are the natural rules that characterise the notion of comparing vectors in neighboring vector spaces on a smooth manifold.

Come to think of it, I really can't think of a good definition for "vector" or "point," either.

actually, the definition of vector is a little tricky on a smooth manifold. there are many equivalent choices, my favorite of which is "derivation on the algebra of smooth functions on the manifold". but it will probably take quite some time to see why this definition has anything to do with what you normally think of a vector.

a point is just an element of the underlying set of the manifold.

 

[lethe]

Originally posted by chroot
Well, at least we're making you think! lethe is a big stickler on saying things with mathematic precision, and that's valuable, if sometimes a little grating.

think I am grating eh?

However, it's true that you need to learn to walk (to understand the broad nature of what the mathematics does) before you can fly (to understand the rigorous definitions of each of the words).

perhaps this is true for you. it is not true for everyone

In the end, it really is just a bunch of circular definitions, so try to keep lethe from leading you around the nose by them!

time for my mathematical sticklerism to show up: why do you keep saying the definition is circular? a circular definition is actually not a definition at all, and i claim the connection can be consistently defined in terms of the mathematical objects on the set (a linear derivation on sections and functions)

 

[lethe]

Originally posted by turin
I've got this confusion with the sphere still. I'm trying to understand, really understand, what the metric tensor is, and so I'm trying to understand it on the simplest manifold I can think of that would be non-trivial: the sphere. (if anyone knows of a simpler example, then please share!)

you remember the dot product from vector algebra? well, the metric tensor is just this same old dot product, defined pointwise, in a smooth way, on the manifold for vectors in the tangent space at each point.

this allows you do have concepts on your manifold like perpendicular, area, length, angle, parallel. it also allows you to have an isomorphism between the tangent vectors and their dual vectors. with a notion of length on hand, you can also have a notion of shortest length. i.e. the geodesic.

1) How can you have this kind of a coordinate system, (&theta;,&phi;), which seems to suggest that you know the surface is a sphere in 3-D space, when we aren't supposed to appeal to this higher D space?

"aren t supposed to" is not the same thing as "not allowed to"

but um... i don t think he used the 3D space. if he specifies a point by two points (&theta; and &phi;), then he is parametrizing the sphere with 2D space.

2) Why would you use &phi; as a coordinate, when taking the partial derivative of position with respect to &phi; clearly does not demonstrate a geodesic (except the equator)? Does it really make sense to use such a coordinate? Does this issue detract from application of the analogy to GR?

who cares if its not a geodesic? or, alternatively, maybe i can produce a funny metric on the sphere so that this vector does follow a geodesic.

3) At the poles, what is the &theta; direction? I mentioned that there were singularities at the poles, but he said there weren't.

there are no intrinsic singularities on the sphere. it is smooth anywhere. you just made a bad choice of coordinates (but remember, there is no such thing as global coordinates, so you can hardly be blamed for your bad choice)

here is an example of the importance of distinguishing properties that are intrinsic from properties that depend on your choice of coordinates or your embedding.

 

[lethe]

Originally posted by turin
From classical mechanics, I gathered that the metric tensor was the machine that calculates dot products. The reason it was introduced, as I understood it, was for use in coordinate systems that were not generally orthonormal, like polar coordinates in 3-D. But this just seemed to be like a using it for different coordinate systems in 3-D (or higher, depending on the number of particles) flat space, and that the coordinate system was curved artificially.

you can t really tell whether the space is curved until you calculate the curvature. even in spherical coordinates, flat space has zero curvature.

however, you are right, the metric tensor allows you to deal with non orthonormal bases on flat space. it also allows you to deal with nonflat space.

In other words, it was allowed to define a vector as an arrow that points from the source to the observation point. Then, the length of that arrow could be found by:

|u|²=g_mnu^muⁿ.

yeah.

But I have gotten the impression here today that this kind of a vector doesn't make the same kind of sense. It seems like, before you can construct such a vector, you have to define a connection rule, but the connection rule, as I understand it, must be defined in terms of the metric tensor.

you don t need a connection to define the length of a vector. the length of a vector is given by the inner product of that vector with itself.

the connection allows you to compare neighboring vectors. i don t see any relationship between the two.

Does a particular connection rule have a physical significance? It was said earlier that the connection rule is arbitrary, but, are there certain rules chosen because they have a certain physical significance?

no. choice of coordinates is completely arbitrary

Then, do you have to figure out what the metric tensor is from the connection rule. This does seem circular.

there is nothing circular going on here, depsite chroots warnings.

the connection and the metric are completely independent, although if you want the connection s concept of parallel transport to agree with the metrics idea of parallel (inner product = 0), along with preserve length, then there is some consistency relationship they must satisfy. a connection that satisfies this consistency relationship is called a Levi-Civita connection.

Maybe I'll try to reword my confusion with this point in another way:

The metric tensor takes tangent vectors as inputs? These vectors must both be at the same point in the manifold?

yep.

The vectors don't exist in the manifold, but rather in a space tangent to the manifold at a particular point where all this is goin' down?

yep

So then, the metric tensor takes these two objects that do not exist in the manifold, and spits out an object that does exist in the manifold?

no, the metric spits out a number in the real line. this is not in the manifold

How can two vectors that are not in the manifold have anything to do with the length of something in the manifold?

you get vectors in the tangent space by taking the derivative. you can get distances between points on the manifold by reversing that process, i.e. integrating the inner product of vectors.

 

[turin]

I have just hit a major learning curve.

Going back to the metric in classical mechanics. So, we only dealt with flat space, so the position vector could make sense as the "arrow" that starts at the origin and terminates at the point at that position. Generally, in flat space, vectors can make sense as starting at some point and terminating at another different point in the flat space, because a flat space is itself a vector space.

That's why I emphasised the coordinate system being curved, while, in actuality, the space is flat. If the space is curved, then the arrow that starts at some point will literally disappear from the manifold before it ever makes it out the door - it doesn't seem to make sense, from what I have gathered, to even have a position vector, or any vector to define the distance between two points in the manifold, no matter how small. No matter how close you choose two points to be, it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.

Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?

 

[chroot]

Originally posted by turin
I have just hit a major learning curve.

Hopefully you mean you've reached the summit!

it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.

Bingo! That's exactly right. Vectors don't exist between points in manifolds -- they exist only at one point. To connect a vector at one point with another vector at another point, you need a set of connection coefficients.

Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?

Yes. In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible. This means the metric of the spacetime can be expressed as

g_{\mu\nu} \equiv \eta_{\mu\nu} + h_{\mu\nu}

where the h_{\mu\nu} are small.

How exactly you define "small" depends upon the problem at hand!

- Warren

 

[turin]

Originally posted by chroot
Hopefully you mean you've reached the summit!

Huh. I wish. I feel like I fell over the cliff. Y'all are a tremendous help, though!

Originally posted by chroot
In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible.

This is my source of confusion. In light of this, I wonder how the metric tensor could ever be fundamentally different from that of flat space. For some curve, at everypoint on that curve, there is a tangent space where the vectors live. If, at every one of these points on the curve, the space can be assumed locally flat, (and apparently approximated as the tangent space?), then I don't understand why one even talks about the metric.

For instance, I'm sitting here on the surface of the earth. It certainly does look flat to me. I can go forward/backward or right/left. I have two degrees of freedom. If I move to the right 1 meter, then I move along the surface of the Earth 1 meter. If I move forward 1 meter, then I move along the surface of the Earth 1 meter. The metric seems to be the Kronecker-Delta. I can go to the other side of the world, but the situation seems the same to me. I will still have access to forward/backward and right/left, with the same consequence to my displacement along the surface of the earth. In mathematical notation, I can have a tangent vector of the general form:

dxe_right + dye_forward,

where the basis is:

e_right,e_forward.

The metric is:

(ds)² = (dx)² + (dy)².

From this, I infer that the metric tensor is &delta;_mn.

I don't see where the metric tensor for a sphere differs from the metric tensor for a flat plane.

 

[chroot]

turin,

Sure -- at every point on the earth, the space around you is approximately flat, and it's metric is approximately the Kronecker delta. That's what you're talking about -- you're walking around and measuring the local space around each point, and it's always locally flat.

Now, instead of considering the space at just individual points, try considering it over larger areas. For example, try drawing some circles on the surface of the Earth. To make the idea concrete, stand on the north pole as you draw these circles.

If you draw a small circle, the relationship between its circumference and diameter is approximately your good friend \pi. If you draw a much larger circle around yourself -- say you draw a huge one, the equator -- you'll notice that the ratio of circumference to radius is no longer \pi! This deviation is caused by the metric not being the Kronecker delta. The larger the circle you draw, the more the curvature matters.

- Warren

 

[lethe]

Originally posted by turin
I have just hit a major learning curve.

Going back to the metric in classical mechanics. So, we only dealt with flat space, so the position vector could make sense as the "arrow" that starts at the origin and terminates at the point at that position. Generally, in flat space, vectors can make sense as starting at some point and terminating at another different point in the flat space, because a flat space is itself a vector space.

you don t have to do classical mechanics in R³, or in a linear space at all. for example, you may like to do classical mechanics on the sphere. in which case, you cannot call the position a vector.

it just seems like an arrow starting at one of them cannot possibly point to the other, because the other isn't even in the same vector space.

that s right.

Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?

only in the limit that the two points are actually the same point.

in other words, no.

 

[lethe]

Originally posted by chroot

Originally posted by turin
Is there some notion that two tangent spaces, given that the points in the manifold to which they are tangent are sufficiently close together, can be said to be sufficiently the same tangent space?

Yes. In many manifolds, such as the one used in general relativity, the space can be assumed locally flat. For two points close together, the space between them is approximately flat, and the effect of curvature is negligible. This means the metric of the spacetime can be expressed as

g_{\mu\nu} \equiv \eta_{\mu\nu} + h_{\mu\nu}

where the h_{\mu\nu} are small.

How exactly you define "small" depends upon the problem at hand!

what on Earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)?

 

[lethe]

Originally posted by Ambitwistor
I'm not sure that I followed your argument, but it looks to me like you demonstrated that you can find an orthonormal basis at every point in the manifold (a frame field). That's true, but you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.

i think i already knew this, i mean, its clear, that if the space has nonzero curvature, then there are no coordinates in which the metric is flat.

but... i think i must not understand this idea as well as i should. why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space? perhaps these integral curves don t exist, but why not?

 

[chroot]

Originally posted by lethe
what on Earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)?

Oh, give me a break -- you understood what he was asking and you understood my answer, too.

But let's clarify it a bit, since I was speaking without much precision. What I want to know is this: in flat space, you can slide a vector around all you want without changing its direction. In other words, the connection coefficients go to zero. In a flat space, can the tangent spaces at points P and Q be said to be "the same" in some way? Or does each point maintain its own independent tangent space even in flat space?

- Warren

 

[lethe]

Originally posted by chroot
Oh, give me a break -- you understood what he was asking and you understood my answer, too.

i certainly understood what he was asking, and i believe i also understood your answer, but i fail completely to see what your answer has to do with his question.

In a flat space, can the tangent spaces at points P and Q be said to be "the same" in some way? Or does each point maintain its own independent tangent space even in flat space?

- Warren

here is what you are saying: in flat space, since the transport of the vectors is independent of path, there is a canonical isomorphism between any two tangent spaces, and thus they can be regarded as the same.

fine.

in flat space, everything is dandy, if we were only dealing with flat space, then the intuition we learned about R³ would be sufficient.

but turin was asking a question about smooth manifolds. not manifolds with connections... he wants to know if you can consider neighboring tangent spaces the same, and the answer is simply no! in general not!

insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal.

 

[chroot]

Originally posted by lethe
insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal.

Isn't that exactly what I said? lol

- Warren

 

[turin]

Originally posted by chroot
Now, instead of considering the space at just individual points, try considering it over larger areas. For example, try drawing some circles on the surface of the Earth. To make the idea concrete, stand on the north pole as you draw these circles.

If you draw a small circle, the relationship between its circumference and diameter is approximately your good friend \pi. If you draw a much larger circle around yourself -- say you draw a huge one, the equator -- you'll notice that the ratio of circumference to radius is no longer \pi! This deviation is caused by the metric not being the Kronecker delta. The larger the circle you draw, the more the curvature matters.

I don't interpret what you say to mean I should consider literally drawing circles, but that I should consider mathematical circles in R³ that coincide with the sphere S² which is imbedded in R³. Otherwise, I can't make sense of doing all of this from the north pole.

1) Isn't suggesting that the equator is a circle to be suggesting that a straight line is a circle? I'm uncomfortable with that.

2) I suppose we can define a circle as the collection of evey point that is the same geodesic (straight line) proper distance away from some given point (i.e. the north pole). In order to "draw" (construct) such a thing, I would have to walk along all possible geodesics that pass through the north pole for a certain distance, r, and stick a flag in the ground or something. I don't understand how I can do that while standing at the north pole.

3) Once I've stuck all my flags in the ground, I could walk around this path of flags, pull them out of the ground as I go, and keep track of how far I walked. When I was done pulling out all the flags, I would compare the distance I walked to r, and find that it is less than 2&pi;r. (The first thing I would think to myself is that I miscounted something)

4. I still don't see how this helps me understand the metric tensor, or, more importantly for my purposes, how to determine its components. No matter what I do, I take the metric tensor with me. Every step I take, I see Kronecker Delta componentes of my metric tensor. Drawing circles seems to instead be demonstrating an R^&alpha;_{&mu;&nu;&beta;}

 

[turin]

Originally posted by Ambitwistor
... you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.

Maybe that's my problem. I don't think I understand a coordinate system on a curved space, at least not without viewing it from the higher D Euclidean space. I am thinking of the coordinate system in terms of the tangent space, because that is what seems the most physical to me. I don't see signs on the highway that say, "Houston, 5 degrees lattidude;" I see signs that say, "Houston, 300 miles." These 300 miles I experience in my own personal tangent plane that I take with me.

 

[turin]

Originally posted by lethe
what on Earth does a nearly flat metric have to do with the tangent spaces nearly coinciding (which is what turin asked about)?

I thought that you could make the metric exactlty flat at a point. I thought that that might mean there is some neighborhood of that point that exists in the tangent space at that point.

 

[turin]

Originally posted by lethe
why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space?

Can you explain to us laymen what you are asking? This looks suspiciously like how I'm trying to picture it except for that "vielbein" thing.

 

[turin]

Originally posted by lethe
insofar as i can approximate the space by a trivial space, i can treat the tangent spaces as approximately equal.

What does it mean for a space to be trivial?

 

[lethe]

Originally posted by turin
I thought that you could make the metric exactlty flat at a point.

the word flat doesn t apply to a single point. what you can do is choose an orthonormal basis for the tangent space which makes the metric look like a kronecker delta at a point. but you cannot in general do this in any neighborhood.

I thought that that might mean there is some neighborhood of that point that exists in the tangent space at that point.

the points in the manifold are never in the tangent space. ever.

 

[lethe]

Originally posted by chroot
Isn't that exactly what I said? lol

make careful note of my usage of the phrase "insofar as". this means that in some cases, such an approximation might apply, only when we have added additional structure to our manifold

in general, the answer is a definite "no", which is a stark contrast to your answer, which was "yes"

you add things to the problem so that you can answer a different question than that which was asked, which makes it irrelevant. why do you introduce a metric when turin asks about tangent spaces?

 

[turin]

Originally posted by lethe
the points in the manifold are never in the tangent space. ever.

I think a light bulb just came on, though not the one I wanted to.

Points in a manifold are not vectors, period, so it doesn't make sense to say that they would be in any vector space, whether it be a tangent space to the manifold or otherwise? The point of the manifold to which some vector space is tangent is not even in the vector space?

 

[lethe]

Originally posted by turin
Can you explain to us laymen what you are asking? This looks suspiciously like how I'm trying to picture it except for that "vielbein" thing.

i think it is the same thing that you are thinking about. a vielbein, or frame or vierbein or tetrad in 4d, is an orthonormal basis for the tangent space.

you can use the gram-schmidt process to choose such an orthonormal basis at every point in the manifold. Ambitwistor says that this does not imply that you can choose coordinates that will make the space flat, which is a perfectly reasonable statement, since flatness is a coordinate independent property.

but my question is, if i can choose a basis for the tangent bundle that is orthonormal, cannot i just integrate the vectors in this basis to obtain some curves on the manifold, whose tangent vectors are this basis, and therefore coordinates for which the metric is flat?

clearly there is something wrong with my reasoning, since i know that flatness is coordinate independent, but what?

 

[lethe]

Originally posted by turin
What does it mean for a space to be trivial?

this means different things in different contexts. in this context, i mean euclidean or flat.

 

[lethe]

Originally posted by turin
Points in a manifold are not vectors, period, so it doesn't make sense to say that they would be in any vector space, whether it be a tangent space to the manifold or otherwise?

that is correct.

The point of the manifold to which some vector space is tangent is not even in the vector space?

correct.

points in a manifold are simply not vectors. you cannot add them or multiply them by numbers, which are the properties that characterize vectors.

just try to imagine a way to add or multiply the points in a sphere. it cannot be done, let alone in the linear fashion that a vector space would require.

so there is associated to each point in the manifold a vector space, but the point is in on sense in the vector space.

 

[turin]

Originally posted by Ambitwistor
What's so hard about visualizing lines of latitude and longitude? Why do you need a higher dimensional space?

They are not hard for me to visualize, even on a "chart" (I think that's the term I'm looking for, a piece of the manifold that is presented as a flat map). The issue I have with lat. & long. is that, if I look at a map of Scandanavia, for instance, I see this lat. & lon. coordinate system dramatically curved, but I see no good reason for it in the context of the chart. If I were to extend myself into 3-D and look at Scandanavia on the globe, then I would definitely have no problem understanding why these lines were curved, but that entails considering Scandanavia as residing on a globe in R³, which is a higher D space than what I should be considering.

Originally posted by Ambitwistor
What do tangent spaces have to do with coordinate systems?

I thought that a coordinate system had a set of basis vectors. I must be very confused about what a coordinate system is.

Originally posted by Ambitwistor
A sign that says "Houston, 300 miles" doesn't specify the location of Houston, relative to you or anything else. So what does this have to do with coordinate systems?

It specifies to me that Houston is 300 miles directly in front of me. How much more specific can one be about location? Specifying the distance in degrees of lattitude I find less specific, in the sense that I have a standard for the mile that will tell me how many times my tires will have to rotate before I get there, whereas I don't have a standard for a degree of lattitude that will tell me anything at all about the duration of my trip in terms of tire rotation or anything else. I would have to know what the radius of the Earth was, and that, again, seems to suggest required knowledge about the 3-D situation.

Originally posted by Ambitwistor
You don't "take a tangent plane with you". A tangent plane is associated with a point in space; it doesn't go anywhere.

So do I jump from one to another? Where am I in between tangent planes?

 

[turin]

Originally posted by lethe
but my question is, if i can choose a basis for the tangent bundle that is orthonormal, ...

Does "orthonormal" mean something as opposed to "perpendicular?"

Originally posted by lethe
... cannot i just integrate the vectors in this basis to obtain some curves on the manifold, ...

What does it mean to integrate a vector?

Originally posted by lethe
... whose tangent vectors are this basis, and therefore coordinates for which the metric is flat?

How did you go go from vectors to coordinates? I tried to suggest going from vectors to coordinates, but apparently I was doing it incorrectly or at least wording it inappropriately.

 

[selfAdjoint]

orthonormal means "mutually perpendicular and of magnitude one". Like a set of unit vectors spanning the space. v_i v_j = \delta_j^i

 

[turin]

Originally posted by selfAdjoint
orthonormal means "mutually perpendicular and of magnitude one". Like a set of unit vectors spanning the space. v_i v_j = \delta_j^i

I'm assuming this is supposed to be a dot product? Don't you need to raise one of the indices to do that? I don't mean to be picky, but I was under the impression that you must contract covariant components with contravariant components, but that lethe found the notion of "perpendicular" distasteful. I just wanted to hear how "orthonormal" does not require a notion of "perpendicular."