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 Geometrical interpretation of Covector Fields 
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Joined: 28 Oct 2009, 19:06
Posts: 7
Post Geometrical interpretation of Covector Fields
I'm having some considerable difficulty in visualising the geometrical interpretation of a covector field.

If I understand it correctly, figure 10.8 shows how the exterior derivative of a scalar field phi (ie a covector field d(phi)) can be interpreted as the mapping of the value of phi (eg the three dimensional height above a reference level) to a 2 dimensional plane element. Then d(phi) is a covector field (or 1-form) which can be interpreted as either the directions along which d(phi) is zero (ie along the tangent to the contour line at any point in the plane element) or other directions for which d(phi) is non zero implying crossing contour lines? Is either a zero or a non-zero d(phi) (or both) to be interpreted as a covector?

Now I try to reconcile this notion with the text in chapter 12, para 12.3 on page 225 and figure 12.7 (Vintage books 2005 edition). OK, if we have an n-manifold our covector field (alpha) is defined within an n-1 dimensional "plane" element (fine, this agrees with the above example in fig 10.8 for a three dimensional space "projected" onto a 2 dimensional plane). But now the text states that "the directions lying within this (n-1) plane element are those determined by vectors ksi for which

(alpha).(ksi) = 0.

Why?? If alpha and ksi are non-zero, does this imply some sort of orthoganality in the (n-1) plane?

If we go back to figure 10.8 and say that the directions of the covector field lying within the 2 dimensional plane are those for which d(phi).ksi = ksi(phi) = 0, then we are saying that the covector field is in the direction of the contour lines. So if we have a conical hill, the covector field would be tangential to a set of concentric circles representing the contours.

HELP!!


03 Nov 2009, 16:12

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Geometrical interpretation of Covector Fields
Trying to cut the crap here. The way I first saw the Vector/Co-vector stuff!
If you want to find the instantaneous rate of change of a function in the space of x it is Sum(i) dF/dx(i)*dx(i) where i is the dimension. It simply is a statement of:
Find the rate and project it onto the constraining direction! Both dF/dx(i) and dx(i) are vectors, which is a co-vector is a matter of choice. But one must remember this:
The whole Vector/Co-VECTOR thing is just a way of saying: WE HAVE DEFINED A SCALAR PRODUCT IN THIS SPACE!!


04 Nov 2009, 03:33

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Geometrical interpretation of Covector Fields
Now to be a little more specific in regard to fig 10.8 , fig 12.7,,and the hyberbolic talk of mathematicians!
The "CONTOUR LINES" of 10.8 are simply F(x(i)) reduced by the constraint F(X(i))=0! thus these lines are in a space that is one less than the space x spans. This is our covector (aka tangent) space! (note that the only change in F happens in the co-vector space and the space depends on F). This space is therefore "tangent" to F ( ie,,the space where F changes), the lost dimension being the dimension where F does not change. Fig 12.7 is simply the space in which F has instantaneous change!
J-E-T-S!! JETS!! JETS!! JETS!!


04 Nov 2009, 04:02

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Geometrical interpretation of Covector Fields
So , the co-vector field in fig 10.8 is not in the horizontal plane you think ,,if you look at 10.8b more carefully you will see a tangent plane aside the hill,,that's your instantaneous covector space!
Did this work?


04 Nov 2009, 04:11

Joined: 28 Oct 2009, 19:06
Posts: 7
Post Re: Geometrical interpretation of Covector Fields
Aren't the contour lines of Fig 10.8 the F(x(i)) (ie the height of the hill at a point x(i)) reduced by the constraint dF(x(i)) = 0, rather than F(x(i)) = 0 ?

Surely F(x(i)) = 0 identifies the reference level for zero height?

Not sure about your comments re the tangent plane shown in fig 10.8 containing the covector field. I need to think about this more.

Thanks for your comments.


04 Nov 2009, 15:36
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Post Re: Geometrical interpretation of Covector Fields
dp4947 wrote:
I'm having some considerable difficulty in visualising the geometrical interpretation of a covector field.

If I understand it correctly, figure 10.8 shows how the exterior derivative of a scalar field phi (ie a covector field d(phi)) can be interpreted as the mapping of the value of phi (eg the three dimensional height above a reference level) to a 2 dimensional plane element. Then d(phi) is a covector field (or 1-form) which can be interpreted as either the directions along which d(phi) is zero (ie along the tangent to the contour line at any point in the plane element) or other directions for which d(phi) is non zero implying crossing contour lines? Is either a zero or a non-zero d(phi) (or both) to be interpreted as a covector?


It is not d\Phi which is zero when the \xi-arrow points along a contour line, but \xi(\Phi)=d\Phi\bullet\xi which is zero. The scalar product d\Phi\bullet\xi is zero because it represents the gradient in the direction of \xi.
Edited to correct an error 5th November 2009, Vasco.

I need to think a bit more before I answer your other questions.


Last edited by vasco on 05 Nov 2009, 11:05, edited 1 time in total.

04 Nov 2009, 15:48
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Post Re: Geometrical interpretation of Covector Fields
dp4947 wrote:
If we go back to figure 10.8 and say that the directions of the covector field lying within the 2 dimensional plane are those for which d(phi).ksi = ksi(phi) = 0, then we are saying that the covector field is in the direction of the contour lines. So if we have a conical hill, the covector field would be tangential to a set of concentric circles representing the contours.


Edited to correct an error 5th November 2009, Vasco
See my post below.


Last edited by vasco on 05 Nov 2009, 11:08, edited 1 time in total.

04 Nov 2009, 16:49

Joined: 28 Oct 2009, 19:06
Posts: 7
Post Re: Geometrical interpretation of Covector Fields
Re Vasco's first message.

Thanks. I must be missing something pretty fundamental here. You say d\Phi is not zero when \xi points along a contour line. Surely d\Phi must be zero along a contour line as the height \Phi is constant along a contour line and therefore has zero gradient?

Sorry if I'm being a bit thick! I really want to understand this.


04 Nov 2009, 18:48
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Post Re: Geometrical interpretation of Covector Fields
Quote:
Sorry if I'm being a bit thick! I really want to understand this.

I haven't got this sewn up by any means myself. I really want to understand this as well!. I think it could be very useful to us two and others if we post our thoughts on the forum.

So here goes. Carrying on from your last post:

\xi(\Phi) is "the gradient or slope of \Phiin the direction of \xi" as Penrose states on page 189 section 10.3 at the end of the first paragraph on that page, and illustrated in Fig 10.6.

So it is this gradient which must be zero when \xi is pointing along a contour of constant \Phi. In my solution to Exercise 10.9 (already posted) you will see that \xi(\Phi) can also be written as \displaystyle{\frac{d\Phi}{d\xi}} using the notation of infinitesimals. So in other words \displaystyle{\frac{d\Phi}{d\xi}} must be zero when \xi points along a contour.

The confusion arises because the notation d\Phi is being used, in this chapter of RTR, with a different meaning to the one that you and I are used to, and which I have used above. In this chapter, it does not represent an infinitesimal. Penrose is at pains to stress this on page 186 when he explains: "When I write 'd\Phi' in the displayed formula above, .... I mean a certain kind of geometrical entity that is called a 1-form .... A 1-form is not an 'infinitesimal'; it has a somewhat different interpretation...."

Notice that Penrose deliberately avoids using the notation \displaystyle{\frac{d\Phi}{d\xi}} and uses\xi(\Phi) instead. This is to try and avoid any confusion between d\Phi as an infinitesimal and d\Phi as a 1-form

When you say that because we are calculating the derivative in the direction of the contour then d\Phi =0, you are right, if you are thinking of d\Phi as an infinitesimal. But in the expression d\Phi\bullet\xi, d\Phi is not zero as it is not here being used as an infinitesimal.

As Penrose says on page 189, final paragraph: "We are now in a position to interpret d\Phi.....it carries the information about how \Phi is varying in all possible directions along S".

This is its interpretation as a 1-form.


05 Nov 2009, 10:54

Joined: 28 Oct 2009, 19:06
Posts: 7
Post Re: Geometrical interpretation of Covector Fields
Yes, I have given this some thought since my last posting on 4/11/09 and I think I'm beginning to see the light (maybe!).

I agree, it is the gradient that we're talking about when we refer to d(Phi) and not the infinitesimal. So in figure 10.8, the co-vector d(Phi) is in the tangent plane but points along the line of steepest ascent of the hillside at the point inquestion. So it is orthogonal to the vector ksi which points along the contour line, also in the tangent plane. So we have d(Phi).ksi = 0 although both d(Phi) and ksi are non-zero, so it's just like the scalar product of two orthogonal vectors, except that one of them is a co-vector in this case. Also, the stuff about being in a (n-1) dimensional plane element (para 12.3) is OK, seeing as we're looking at a three dimensional manifold (ie the hill in three dimensions) but the co-vector lies in a two dimensional plane.

In figure 12.7, in the general case I see now that the covector alpha lies in a (n-1) dimensional plane which also contains the vector ksi such that alpha.ksi = 0, just as for the 3 dimensional case of the hill.

Similarly in fig 12.9(b), back in the 3 dimensional case, the covector field alpha lies in a two dimensional plane which is the linear sum (or sort of weighted average) of those planes which are each orthogonal to the coordinate axes.

I think I'm getting there!

PS How do I embed Latex in a posting. I've tried doing it and then doing a preview but I just get the [itex] stuff with my actual code between the brackets.


06 Nov 2009, 16:34
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Post Re: Geometrical interpretation of Covector Fields
To use Latex, just type it in normally and then highlight it and click on the itex button at the top of the new post window.
If you click on the quote button at the top left of my post a new post window will open with my stuff in showing the latex.
Use the preview button (below) before you submit to check that everything looks OK.


07 Nov 2009, 10:39

Joined: 28 Dec 2009, 21:25
Posts: 3
Post Re: Geometrical interpretation of Covector Fields
Hello Everybody :)

I was having a lot of problems with this topic myself and I think I now figured out most of it, agreeing with dp4947's last post except for one thing.
You wrote "[...]seeing as we're looking at a three dimensional manifold (ie the hill in three dimensions) but the co-vector lies in a two dimensional plane." which refers to Fig. 10.8 I guess.
But as I understand it, the surface S in Fig. 10.8 is a 2-dimensional manifold we are referring to. So due to par. 12.3, p. 225, stating "[...] a covector alpha determines an (n-1)-dimensional plane element", the covector d\phi (the gradient of \phi) at a certain point determines an (2-1)=1-dimensional plane element, which is just a straight line.
This line gives the direction in which the gradient d\phi vanishes, i.e. d\phi\bullet\xi=0. So it is locally identical with the contour line through this point and represents a tangent to this contour line globally. In Fig. 10.8, this line is called the 'axis of tilt'.
So the covector d\phi indeed lies in a 2-dimensional space (the cotangent space) which has as many dimensions as the 2-manifold S, but defines a (2-1)-dimensional plane (the contour line).
Hope this made it clearer, please feel free to respond, I am willing to learn :)


28 Dec 2009, 22:03
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Post Re: Geometrical interpretation of Covector Fields
jan wrote:
.......the covector d\Phi (the gradient of \Phi) at a certain point......


and again

jan wrote:
This line gives the direction in which the gradient d\Phi vanishes, i.e. d\Phi\bullet\xi=0.


I disagree with you here.
As I said in my last post on this subject above, \xi(\Phi) is the gradient of \Phiin the direction of \xi. It is THIS quantity which is zero when \xi points along a contour of constant \Phi:
\xi(\Phi)=d\Phi\bullet\xi=0, NOT d\Phi.
\xi(\Phi) is zero not because d\Phi is zero, but because d\Phi and \xi are perpendicular.


30 Dec 2009, 17:12

Joined: 28 Dec 2009, 21:25
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Post Re: Geometrical interpretation of Covector Fields
I re-read this and according to Penrose you're right, vasco. The gradient is here defined, as you said, as the scalar product of d\phi and \xi. But this leaves me quite confused, since the result of a scalar product is just a number. Yet from vector calculus, I'm used to look at the gradient of a scalar function as a vector (field), that's why I mixed up the 1-form and the gradient. So am I missing something or is there a real difference between the gradient \xi(\phi) (scalar field) and the gradient grad(\phi)=\nabla\phi (vector field)? I hope somebody can lighten this up a bit.


30 Dec 2009, 18:30
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Post Re: Geometrical interpretation of Covector Fields
jan wrote:
.......So am I missing something or is there a real difference between the gradient \xi(\phi) (scalar field) and the gradient grad(\phi)=\nabla\phi (vector field)?


I can understand your confusion - I've been there too!

When I studied Vector Analysis at university, I never did all this covector and 1-form stuff that Penrose talks about. So I tried to relate it all to grad(\Phi)=\nabla\Phi, and came up with this:

Penrose's d\Phi, when he's using it to mean a 1-form and not an infinitesimal, is identical to (see below) what you and I and many others call grad(\Phi)=\nabla\Phi.

So you could write:

\xi(\Phi)=\nabla\Phi\bullet\xi

So the answer to your question is: yes, there is a difference between \xi(\Phi) and \nabla\Phi.

\nabla\Phi is a vector defined over the surface S, and has a magnitude equal to the gradient of \Phi at every point of the surface and in the direction of the maximum rate of increase of \Phi. In other words, at right angles to the contour lines.
So if we now take any vector \xi through a point in S, then \xi(\Phi)=\nabla\Phi\bullet\xi=d\Phi\bullet\xi is the magnitude of the gradient of \Phi in the direction of \xi. That is what the scalar or dot product means.

If the vector \xi is chosen to be pointing along a tangent to a contour of \Phi, then the magnitude of \xi(\Phi) is zero.

You probably recall that the scalar product of two vectors at right angles is zero, since the scalar product of two vectors a and b is defined as ab\cos\theta, and \cos\theta is zero when \theta=\pi /2.

This is how I have managed to make sense of all this. I hope it helps you to do the same. Looking forward to your next post.

Edited on 1st January 2010 by Vasco. I think it would be more accurate to replace the words 'identical to' highlighted in red above by 'functionally equivalent to'.


Last edited by vasco on 01 Jan 2010, 11:11, edited 2 times in total.

31 Dec 2009, 11:10
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