Page 1 of 1 [ 6 posts ]
 Print view Previous topic | Next topic
Exercise [12.10]
Author Message

Joined: 07 May 2009, 16:45
Posts: 62
Exercise [12.10]
Let . Explain why and evaluate this by chaning to polar coordinates .

Hence prove .
-----------------------------------------------------------------------------------------------------------------------
This, , is a 1-form.

So is this .

If , then also because the or is a dummy variable that disappears with integration.

So multiplying these two ways of expressing together would give us

This part is a little confusing to me. We don't multiply 1-forms, we "wedge product" them. Is that justification enough to go to the next step as follows,

?

is a 2-form, appropriate for integrating over a plane, as Penrose tells us (See page 231 and note 12.9 on page 245). A little confusion remains on my part.

Changing to polar coordinates:

The change in integration limits justified because for

(OK, I mean this is the limit as approaches )

( for going to )

and for

(for x going to )

What about the lower limit of corresponding to when ? We are integrating along the half line from a point where to in the integration. I think integrating along any line to infinity (except along the -axis in which case would not change) would do the trick, just so long as the angle remains constant in this integration. If we started at any other point than the origin on the -axis, at the start of this integration, but the angle would have to change as increased towards . That is about as good a reason as I can come up with as to why for a lower limit means for a lower limit with the variable change.

Continuing, and substituting

(I am a little uncertain about this step as well. I mean losing the wedge. Is it justified?)

So

,
and

Last edited by DimBulb on 30 Jul 2009, 06:02, edited 5 times in total.

29 Jul 2009, 19:14

Joined: 07 May 2009, 16:45
Posts: 62
Re: Exercise [12.10]
Also a little confusing is this notion that "The wedge product in Cartan's differential-form notation takes care of everything if we choose to change our coordinates" Seems to me we are still doing quite a bit of work with coordinate change, and while fairly simple in this case, we had to add an to the wedge product when we went from to .

And then there is this notion of a 2-form. I mean, a 1-form spans all the dimensions in the space except one. In our case the space is two dimensional so a 1-form represents a line or a "line density". A 2-form spans all the dimensions in a space except two. Does that means in our case a 2-form represents a point, or a "point density", which I guess is the same as an "area density"?

30 Jul 2009, 03:48

Joined: 07 May 2009, 16:45
Posts: 62
Re: Exercise [12.10]
The breaking up of

into

was necessitated by the fact that the transfer function

is not one-to-one, that is both and will give the same . So and which would make the starting and ending point of the integration the same, making the integration , which would be wrong.

30 Jul 2009, 19:56

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [12.10]
Dimbulb, I don't think you need to split your integrals up the way you have.

It's sufficient to note that:

and

both represent area integrals over the same region (manifold), namely the plane.

Also, you seem to be trying to change x into r and y into , which isn't kosher - you can't (generally) change coordinates pairwise like this.

If I recall my calculus correctly (and that's definitely an if!), to change coordinates in a regular area (or volume, etc.) integral, you need to (a) ensure that the integration is occurring over the same exact region, (b) change the variables in the term being integrated, and (c) insert det(J) into the integral, where J is the Jacobian matrix of the coordinate transformation.

i.e.

where

Intuitively can be understood as the ratio of the infinitesimal areas/volumes dx : dx'. For example, in the case of changing from cartesian coordinates dx dy into polar coordinates , dr has the same scale as dx or dy, but scales proportionally to r, so at any given point .

It appears that this is what the Grassman product notation provides "for free", in that the rules for coordinate transforms of 1-forms automatically involve calculation of the elements of J, and then calculating out the wedge-product automatically yeilds its determinant, so when changing coordinates in an integral involving p-forms, you don't need to worry about step (c).

Like Dimbulb, though, I'll confess I'm a little unclear about the step involving the combining two regular integrals to get a "wedge" integral, and then the later step of integrating out from the wedge product to get a regular integral again... it seems OK, and obviously works in this case, but it's not clear if there are caveats in more general cases...

17 Jul 2010, 19:14

Joined: 23 Aug 2010, 13:12
Posts: 33
Re: Exercise [12.10]
A comment on how the "r" in r dr is automatically generated by the wedge product. See PDF file.

Attachments:
Ex12.10_comment.pdf [42.34 KiB]