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Exercise [08.07] c
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Joined: 19 Oct 2011, 16:47
Posts: 2
Exercise [08.07] c
Vasco's solution clearly shows that the bilinear correspondence of exercise 8.7 maps the plane on and above the real axis onto the unit disk, but I share tonychinnery's misgivings about the correspondence being a rotation. Please see the attached for an alternate solution.

Attachments:
Road to Reality, exercise [8.7].pdf [53.4 KiB]

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-Tony Dean

Last edited by TonyKDean on 31 Oct 2011, 16:15, edited 3 times in total.
24 Oct 2011, 17:26
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Joined: 07 Jun 2008, 08:21
Posts: 235
Re: Exercise [8.7]
Hi Tony
I have downloaded your document and I am currently working through it. There seem to be a lot of brackets missing in the pdf file - they have been replaced by spaces. Download it and you will see.

25 Oct 2011, 06:18
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Joined: 07 Jun 2008, 08:21
Posts: 235
Re: Exercise [08.07] c
Tony
I have looked at your document and I understand it. However both you and Tony Chinnery have missed the point here and that is why you are confused.
What Penrose is asking us to do in this exercise is to find the transformation of the Complex plane which is equivalent to a rotation of the Riemann sphere.
The transformation given by Penrose is NOT a transformation of the Riemann sphere but of the Complex plane - that is why it is not a rotation about the x-axis.
The idea here is that we have the Complex t-plane and an associated Riemann sphere obtained by stereographic projection. If we now apply a transformation to this Complex t-plane and effectively produce a new Complex plane called the z-plane and then generate the Riemann sphere associated with this new z-plane, we can then ask ourselves what is the transformation of the t-plane Riemann sphere which transforms it into the z-plane Riemann sphere.
So what we are saying is that if we apply a transformation to a Complex plane and get a new Complex plane, what is the transformation between the associated Riemann spheres that results from this transformation of the complex plane.
For example the transformation z=1/t (not a rotation of the complex plane) is equivalent to a rotation of the Riemann sphere about the real axis through an angle of \pi
The first part of my original solution to this exercise tries to explain this using a diagram. You might find it useful to read this.
Please let me know if this clarifies things for you. If not please ask me for further clarification.
Vasco

26 Oct 2011, 08:37

Joined: 19 Oct 2011, 16:47
Posts: 2
Re: Exercise [08.07] c
Vasco,

I respect your difference of opinion, but I don't believe I'm confused about Exercise 8.7. Many will find "Exercise [08.07] c" a valid way of addressing it, so I will let it stand.

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-Tony Dean

27 Oct 2011, 15:44
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Joined: 07 Jun 2008, 08:21
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Re: Exercise [08.07] c
Hi Tony
That's fine by me. When I first read your document I thought you were confusing rotations of the complex plane and rotations of the Riemann sphere, but now, re-reading your document, I can see that you aren't.
Vasco

28 Oct 2011, 06:43
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