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

[turin]

Originally posted by lethe
i have never been able to once and for all put to memory exactly what the elements of the Lorentz transformation look like. if i don t know the elements of the Lorentz transformation, i can t figure out how the electric field and magnetic field transform.

I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation? Am I missing something here? It seems like, either way you do this, if you don't have the Faraday tensor memorized, then you're going to have to look it up. And, I don't understand how it would be so much more effort to look up the Lorentz transformation while you're at it.

But, like I said, your preference isn't my business. I'm just curious. Can you fill me in if I'm missing something here?

 

[lethe]

Originally posted by turin
I do recognize that your preference is your own and that I shouldn't really concern myself, but I just can't resist my curiosity.

Are you saying that you are willing to commit the components of the Faraday tensor to memory before you are willing to commit the elements of the Lorentz transformation?

yeah, you bet! but the Faraday tensor is really easy to remember. here it is: any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember. i ll never have to look that up in a book!

i think there are some plus or minus signs thrown in there for good measure, but who cares about them?

 

[turin]

Originally posted by lethe
any part of the Faraday field tensor with a 0 in it is electric field, and the other nonzero components are magnetic field. that s all i have to know, and its real easy to remember.

Every time I use it, I have to look up which magnetic field component is which Faraday component. What's your trick for remembering this?

 

[lethe]

Originally posted by turin
I got a hold of an undergrad math meth book by Arfken. In the chapter on tensors, it is said that the Levi-Civita symbol isn't really a tensor, but a pseudo-tensor. Any comments on this, and how it relates to the discussion?

this is exactly what i was saying with the above comment about changing sign in a mirror.

the Levi-Civita changes sign under parity transformation in any number of dimensions

Does this mean that the dot product in question isn't really a scalar, but a pseudo-scalar?

you bet! that s exactly what it means!

Is this only the case for the Levi-Civita symbol in 3-D, and in 4-D it is a true tensor?

any number of dimensions.

 

[lethe]

Originally posted by turin
Every time I use it, I have to look up which magnetic field component is which Faraday component. What's your trick for remembering this?

well, the component has two numbers in it, chosen from 1, 2 and 3. thus it doesn t have one of those three numbers. the number it doesn t have is the number for the component of the magnetic field.

for example, the F^{12} doesn t have a 3, so it must be the three component of B.

again, i think there may be some minus signs in there, but i m not sure, and don t really care anyway.

 

[lethe]

Originally posted by turin
That is the first I have heard of this. What about:

\eta_{\alpha\sigma}\eta_{\beta\tau}...\eta_{\kappa\epsilon}\eta_{\mu\eta}F^{\alpha\beta}F^{\gamma\delta}...F^{\kappa\lambda}F^{\mu\nu}

?

Is this not considered distinct from other contractions involving the metric

yeah, that s fine, but for this problem, i was only considering quadratic forms. usually, in any situation, there is some limit like that on the degree.

, or is it considered just another version of the same type of contraction. I don't understand how there can only be a few (or even a finite number of) valid manipulations.

i hope that s clear now. if you are only looking for quadratic forms, there are only 3 choices (one of which vanishes)

(For some reason, It won't let me edit that itex stuff. There are a few typos in it.)

it probably is letting you edit it, but you are just not seeing the edit show up. so after you edit, make sure you reload the page. the same thing happens to me.

What does it mean for a space to have an orientation or connection?

an orientation is a rule for choosing from two classes of coordinates. like, in Rⁿ, there are a bunch of coordinate systems you can choose, and you can change between any two coordinate systems with some matrix, and sometimes the determinant of that matrix will be positive and sometimes it will be negative. all the coordinate systems that are related by a positive determinant are said to have the same orientation, and the others have the opposite orientation.

to say a space has an orientation just means to choose a preferred class from all these coordinate systems. for example, in R³, the choice x,y,z (in that order) is called the right-handed orientation. if i wanted to change to a new coordinate system with y as my x-axis and x as my y-axis (and z the same), then this would be a left-handed orientation.

so the Levi-Civita tensor is sensitive to which orientation you choose (is it clear why?)

a connection is a rule for comparing tensors at different points. it is necessary for taking derivatives

 

[turin]

Originally posted by lethe
for example, the F^{12} doesn t have a 3, so it must be the three component of B.

again, i think there may be some minus signs in there, but i m not sure, and don t really care anyway.

Thanks, I think I figured it out:

You want ijk to be a cyclic permutation in the scheme cB_i = F^jk.

 

[turin]

Originally posted by lethe
i hope that s clear now. if you are only looking for quadratic forms, there are only 3 choices (one of which vanishes)

Very clear. Quadratic, I'm assuming, refers to the Faraday tensor appearing twice in the product.

Originally posted by lethe
... you can change between any two coordinate systems with some matrix, and sometimes the determinant of that matrix will be positive and sometimes it will be negative. all the coordinate systems that are related by a positive determinant are said to have the same orientation, and the others have the opposite orientation.

Very clear.

Originally posted by lethe
so the Levi-Civita tensor is sensitive to which orientation you choose (is it clear why?)

I'm assuming because the permutations have a sense, so, taking them in the other sense shows up as a minus sign, just like your example with flipping one of the axes.

Originally posted by lethe
a connection is a rule for comparing tensors at different points. it is necessary for taking derivatives

I'm still not quite getting this. Does this have anything to do with R^α_{μνβ} (or Γ^α_μν or g_μν)? Is this related to topology?

 

[chroot]

Originally posted by turin
I'm still not quite getting this. Does this have anything to do with R^α_{μνβ} (or Γ^α_μν or g_μν)? Is this related to topology?

The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points. This allows the identification of "parallel" vectors at two different points in the manifold.

The connection coefficients appear in the construction of the covariant derivative, which is basically a generalization of the partial derivative that transforms like a tensor (the partial derivative does not). In order for differentiation to obey tensor character, we have to use component differences at the same point in the manifold -- something not done by normal partial differentiation. The connection coefficients allow us to identify a vector at one point in the manifold with a parallel one at a neighboring point -- they define the operation of parallel transport. Thus the covariant derivative uses the component differences between a vector at one point and a vector at a neighboring point by first using the connection coefficients to parallely transport the neighboring vector so that both are defined at the same point.

The covariant derivative of a vector \lambda^a is thus defined as

\lambda^a_{;c} \equiv \frac{\partial \lambda^a}{\partial x^c} + \Gamma^{a}_{bc} \lambda^a

where x^c are the coordinates of the point at which this covariant derivative is taken.

- Warren

 

[lethe]

Originally posted by chroot
The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points.

not just the tangent space, you can have a connection for any vector bundle (including the cotangent bundle, and higher products of bundles)

The covariant derivative of a vector \lambda^a is thus defined as

\lambda^a_{;c} \equiv \frac{\partial \lambda^a}{\partial x^c} + \Gamma^{a}_{bc} \lambda^a

where x^c are the coordinates of the point at which this covariant derivative is taken.

this formula, as it stands, is a little confusing, since when we are doing gauge theory, we usually try to use different symbols for the gauge components of the connection than for the spacetime components.

so how about this formula instead:

<br /> D_\mu\lambda^a=\partial_\mu\lambda^a + A^a{}_{b\mu}\lambda^b<br />
when the connection is the metric connection from GR, then the gauge indices really ought to be the same as spacetime indices, (and we use a capital gamma, and call them Christoffel symbols), but when we are talking about electromagnetic gauge theories, the indices are "internal symmetry" indices, not the same as spacetime indices, and we use the symbol A and call these components the vector potential.

but its all really a matter of semantics or notation or whatever.

 

[lethe]

Originally posted by turin
For instance, I thought that raising and lowering (what he calls "index gymnastics") was defined by contraction with the covariant and contravriant components of the metric tensor. This is the first I've heard to the contrary, and now I am utterly confused.

um... yes, don t get confused here, you raise and lower indices with the metric, by definition.

but you can contract indices between any tensors you want.

I thought that I understood raising and lowering of the indices of a component to be changing that component between contravariant and covariant. I don't understand how the Levi-Civita symbol/tensor can do this, since it doesn't contain any dot products (or does it?).

no, the Levi-Civita doesn t contain the metric, only the orientation. so you can t quite raise and lower indices with the Levi-Civita. but like i said, you can contract. for example, F^{\mu\nu}\epsilon_{\mu\nu\rho\sigma} is not the same thing as F_{\rho\sigma}. it is a new tensor (called the dual tensor to the electromagnetic field tensor).

 

[lethe]

Originally posted by turin

I'm still not quite getting this. Does this have anything to do with R^&alpha;_{&mu;&nu;&beta;} (or &Gamma;^&alpha;_&mu;&nu; or g_&mu;&nu;)? Is this related to topology?

yeah, that &Gamma; thingy is a component of the connection (as you have no doubt read by now from other posts). this has to do with geometry, not topology, except inasmuch as topology and geometry are connected.

 

[turin]

Originally posted by lethe
not just the tangent space, you can have a connection for any vector bundle (including the cotangent bundle, and higher products of bundles)

I'm lost with this. We used that classical dynamics book by Jose & Salen this semester in my classical mechanics, and I think this stuff was in there. But I also think that my prof was a mathe-phobe, and he avoided appealing to the notation and termonolgy at every turn. So, this is about all I undertand:

The tangent bundle (called TQ in the book) of say a 1-D problem is the 1-D manifold of q (some curve), with a flat line tangent to the curve at every point to represent q_dot.

The cotangent bundle (calle T*Q in the book) of the same would be the same 1-D manifold but with flat lines attached perpendicular to the curve to represent p.

Do I have anything incorrect here? What is a higher product of a bundle?

Originally posted by lethe
... since when we are doing gauge theory, we usually try to use different symbols for the gauge components of the connection than for the spacetime components.

... when the connection is the metric connection from GR, then the gauge indices really ought to be the same as spacetime indices, ... but when we are talking about electromagnetic gauge theories, the indices are "internal symmetry" indices, not the same as spacetime indices, ...

I'm a little confused. It still looks like your taking the derivatives with respect to the spacetime coordinates. What is the capital D on the L.H.S.? Are you saying that this is a derivative with respect to the components of the potential?

 

[turin]

Originally posted by lethe
for example, F^{\mu\nu}\epsilon_{\mu\nu\rho\sigma} is not the same thing as F_{\rho\sigma}. it is a new tensor (called the dual tensor to the electromagnetic field tensor).

OK, I think I see what you're saying. The indices can literally be lowered by the Levi-Civita, but you don't get F_{\rho\sigma}. Instead you get *F_{\rho\sigma}?

 

[turin]

Originally posted by chroot
The \Gamma^a_{bc} are called "connection coefficients." As lethe said, they provide a means to compare vectors at two different points in a manifold. In other words, they "connect" the tanget space of one point in the manifold with the tangent spaces of neighoring points. This allows the identification of "parallel" vectors at two different points in the manifold.

This is something that confuses the hell out of me in this curved space/Riemannian geometry business. Lately, from classical mechanics, and my discussions with my major prof about GR, it has been hammered into my head that vectors are objects that exist independent of the coordinate system. But then these Christoffel symbols show up and yank the vectors around, stretching and twisting and bending and redirecting. It seems so artificial. If two vectors are parallel, then they are parallel, correct? Why should the components matter? Is it just that we need a way to find these components? In GR, are the vectors really more than 4-D and they only project onto 4-D?

 

[selfAdjoint]

A vector, as a rank 1 tensor shares the fact that tensorial equations are invariant under coordinate changes ("covariant" in a different sense of the overused word). So if a vector is zero in one coordinate system it is zero in every coordinate system. But obviously when you change coordinates the componentsof the vector will change, and there are specific formulas for doing that, one for covariant vectors and one for covariant vectors.

Connection components, aka Christoffel symbols in this case (but take notice of the generalizations mentioned above) are not tensors: they don't change right under new coodinates. So everything you do with them assumes you are for the moment in a fixed frame of reference with a specified coordinate frame. So 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.

 

[chroot]

Originally posted by turin
If two vectors are parallel, then they are parallel, correct?

Indubitably, but how do you determine whether or not they are parallel?

Think about a sphere, like the Earth -- a positively-curved two-dimensional Riemannian manifold.

Put a vector on the equator somewhere -- say in Africa -- that points due east. Now, put another on the opposite side of earth, also pointing due east. Are these two vectors parallel, or not? They both point in the same direction, east, in one sense -- yet, as viewed from above the north pole, they seem to point in exactly opposite directions!

The bottom line is that the notion of parallelism is non-trivial in curved space. You can't simply look at the vectors and declare them parallel or non-parallel like you can in flat space. You have to have some rigorous way of defining parallelism, and the connection coefficients provide the mechanism. For example, 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. This "sliding" action, called parallel transport, depends upon some numbers specific to the curvature of the space at each point along the path used for that sliding. These numbers are called connection coefficients. In flat space, the connection coefficients vanish -- they're all zero.

- Warren

 

[lethe]

Originally posted by turin
The tangent bundle (called TQ in the book) of say a 1-D problem is the 1-D manifold of q (some curve), with a flat line tangent to the curve at every point to represent q_dot.

to represent the vector space spanned by q-dot, yes.

The cotangent bundle (calle T*Q in the book) of the same would be the same 1-D manifold but with flat lines attached perpendicular to the curve to represent p.

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.

some GR books like to use a stack of pancakes to represent cotangent vectors, but i never really liked that. so i just say a cotangent vector is a dual vector to a tangent vector, and i don t need to draw a picture at all.

Do I have anything incorrect here? What is a higher product out of a bundle?

like, the tensor product, symmetrized tensor product, or antisymmetrized tensor product of any number of copies of the tangent bundle and the cotangent bundle.

also, you can just have an arbitrary vector bundle that is not built out of the tangent or cotangent bundle at all. just some arbitrary vector space bundle over some manifold. all these bundles admit connections.

I'm a little confused. It still looks like your taking the derivatives with respect to the spacetime coordinates.

I am

What is the capital D on the L.H.S.?

the (covariant) derivative

Are you saying that this is a derivative with respect to the components of the potential?

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. so a covariant derivative is just a rule that let's you do this. 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.

 

[lethe]

Originally posted by turin
OK, I think I see what you're saying. The indices can literally be lowered by the Levi-Civita, but you don't get F_{\rho\sigma}. Instead you get *F_{\rho\sigma}?

that s right. i can take any tensor A^{\mu\nu} and get a new tensor with lowered indices by contracting like this: B_{\mu\nu}=A^{\rho\sigma}t_{\rho\sigma\mu\nu} . if the tensor i use to lower those indices is the metric tensor, then i pretend that A and B are really the same tensor. if the tensor that i used to lower was the Levi-Civita, then i would say that A and B are dual to one another. in the general case, i wouldn t really say they have any important relationship.

in the problem at hand, there were only two natural tensors available in the problem, the metric tensor and the Levi-Civita tensor.

but if i were doing, say, symplectic geometry, then i would have another tensor available, the symplectic form, and i could have used that to construct a scalar as well.

 

[lethe]

Originally posted by turin
This is something that confuses the hell out of me in this curved space/Riemannian geometry business.

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

Lately, from classical mechanics, and my discussions with my major prof about GR, it has been hammered into my head that vectors are objects that exist independent of the coordinate system.

good! that s a good lesson to learn!

But then these Christoffel symbols show up and yank the vectors around, stretching and twisting and bending and redirecting.

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.

for the exact same reason, i really dislike defining a connection to be a set of Christoffel symbols. that is how i saw them the first time, and it was really confusing to me!

so now i much prefer to just think of a connection as a linear rule for comparing neighboring vectors in a derivative like way. in other words, forget the component formula you saw above, with the Christoffel symbols, and remember the following rules:

<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.

It seems so artificial. If two vectors are parallel, then they are parallel, correct?

define parallel..

Why should the components matter?

the components don t matter.

Is it just that we need a way to find these components? In GR, are the vectors really more than 4-D and they only project onto 4-D?

in GR, tangent vectors live in a 4D vector space (if you are doing 4D GR)

 

[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.