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 Exercise [12.10]b 
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Joined: 07 May 2009, 16:45
Posts: 62
Post Exercise [12.10]b
Attempted a better version based on feedback I got from Exercise [12.10]

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Last edited by DimBulb on 02 Dec 2010, 19:26, edited 1 time in total.
29 Nov 2010, 18:38

Joined: 23 Aug 2010, 13:12
Posts: 33
Post Re: Exercise [12.10]b
You raise the issue of what happens if the y-axis is at an angle to the x-axis. The wedge product represents an (oriented) area. If the axes are at an angle the area is reduced by the cosine of that angle. The integral over the wedge product then picks up the cosine factor as you show. However, the product of the one-dimensional integrals is only equal to the 2-dimensional integral if the "angled" wedge product is divided by that cosine. The cosine factors cancel out.

Another interesting point relates to the orientation of the wedge product. We define the orientation to be positive for the chosen order dx^dy. The orientation of dr^d(theta) is then the same if theta is measured counter-clockwise from the r axis. The particular representation of x and y in terms of r and theta then follows. If we had chosen to define dy^dx as positive, then to preserve the orientation of r and theta as being positive, we would have defined theta to be measured in the clockwise sense. The roles of y and x are interchanged and the calculation goes through as before.

29 Nov 2010, 19:31

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]b
I meant sine instead of cosine. For a parallelogram the area is the length of the sides times the sine of the angle, duh.
I fixed it.

When Penrose introduces Grassmann algebras section 11.6, he says they do not need to "know" what "perpendicular" means. They are just concerned with a "plane element". But it seems to me that while they can cross at any angle, they need to "know" what that angle is, which implies they "know" what perpendicular means.

02 Dec 2010, 19:13
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