Covector fields - did I get them wrong?


by Kontilera
Tags: covector, fields
Kontilera
Kontilera is offline
#1
Nov6-12, 05:07 PM
P: 102
Hello!
I am a bit confused about how I can use covector fields on a differentiable manifold.
John M. Lee writes that they can be integrated in a coordinate independent way so I thought that the covector fields could give me a coordinate independent way of calculating distance over a manifold.

Lets say we are working in R^3. This means that if I have a curve [tex]\gamma: I \rightarrow \mathbb{R}^3[/tex] I can measure how far it stretches in the y-direction by doing the integral,
[tex] \int_\gamma dy .[/tex]
If we change coordinates my covector field, [tex]\omega = dy[/tex] gets pullbacked to [tex]\omega' = dy/dy' dy'[/tex] and we get,
[tex] \int_\gamma \frac{dy}{dy'} dy' .[/tex]
It seems coordinate independent in this sense but what if we would have started with the coordinates dy' form the beginning?
Then we would have arrived at:
[tex] \int_\gamma dy' .[/tex]
Which gives another value right?
What have I missed in this subject? :/
Thanks so much,
All the best!
/ Kontilera
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quasar987
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#2
Nov7-12, 07:07 AM
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The point is that the integral of the form w is independant of the coordinate system used to compute its integral.

Here, you start with the form w = dy. Then you change the cordinates and remark that it is written (dy/dy')dy' in the new coordinates. Then you are confused by the fact that the integral of dy' will give something different. But it's not the same form! In the y' cordinates, the form w is (dy/dy')dy', not dy'.

Hope this helps.
Kontilera
Kontilera is offline
#3
Nov7-12, 05:15 PM
P: 102
Thanks quasar! It helped. Could you agree with me on this:
When doing an integral over a certain covector field and path it doesnt matter in which coordinate system you start as long as the path and the field are constructed with regard to this coordinatesystem.
For example, if you start to draw R^2 by two orthogonal lines with out any scale, it doesnt matter how big you make your coordinate grid, the result of the integral is, trivially, invariant of your size.

The transformation matrix is needed when you change coordinates in that sense that the calculation is taking place in one coordinate system and the geometrical objects such as the path (for example the unit circle) and the covector field are constructed from another coordinate system.

This makes sense for me at least. :)
All the best!

quasar987
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#4
Nov8-12, 08:58 AM
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Covector fields - did I get them wrong?


This sounds correct. :)
Vargo
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#5
Nov8-12, 10:13 AM
P: 350
Howdy,

It looks like your question about changing coordinates was answered, but your comment about lengths was unclear, so I thought I'd add something.

You won't get lengths from covector fields, rather you'll get net changes. E.g. the integral of dy over your curve returns the net change in the y direction. If you wanted total change in the y direction, you would need to integrate |dy|, which is not strictly speaking a differential form.

To measure length, you need a formula for ds. For example to calculate euclidean length there are many possible formulas:
[itex] ds = \sqrt{dx^2+dy^2} [/itex]
[itex] ds=\sqrt{dx^2+x^2dy^2}[/itex]
The specific formula depends on the coordinates you choose. The first corresponds to cartesian coordinates of the plane. The second to polar coords. The formula for the element of arclength is called a riemannian metric.

|dy| is like a degenerate metric: [itex] |dy|=\sqrt{0x^2+y^2}[/itex]
Kontilera
Kontilera is offline
#6
Nov11-12, 04:53 AM
P: 102
Thanks! Yeah, net change is a better way to put it. :)
Another question that popped up was if it is misleading that we write integration as:
[tex]\int_S f(x,y)\,dxdy \quad ?[/tex]
After all, we are integrating over differential forms, ie:
[tex]\int_S f(x,y)\,dxdy = \int_S f(x,y)\,dx\wedge dy[/tex]
but juxtaposition of dxdy is used for the symmetric product so we should have:
[tex]dxdy = dx \wedge dy[/tex]
Which is not true since the left hand side is symmetric and the right hand side is alternating.
Vargo
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#7
Nov12-12, 10:10 AM
P: 350
Both notations make sense and are used for good reason.

The dxdy notation is measure theoretic and is the more fundamental. It has nothing to do with tensors. It is shorthand for the product measure dA and by Fubini's theorem it makes sense to write it as dxdy or dydx.

The differential form is a completely different object. By definition, its integrals are the same up to a sign that indicates orientation. You could define this integral directly if you were doing a Riemann type integral, but if you want it to be defined Lebesgue style, it just makes more sense to essentially define it as a signed version of the dxdy (measure theoretic) integral.

You just have to tell from context whether dx or dy is referring to Lebesgue measure or a differential form. And often it doesn't really matter anyway.


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