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Exercise [08.07] b
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Joined: 07 Jun 2008, 08:21
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Exercise [08.07] b
There has been a quite a bit of discussion on this forum regarding what Roger Penrose is asking for in this exercise. See:

and

In my attached solution I have stated at the beginning what I think Penrose is asking for. If anyone finds any errors or would like any clarifications, please contact me via this forum.
Vasco

Attachments:
RTRex8.07.pdf [621.81 KiB]
19 Mar 2009, 09:52

Joined: 03 Jul 2011, 14:43
Posts: 8
Re: Exercise [08.07] b
Hi Vasco,

My problem is when you decided to rotate the unit circle by 90 degrees before shifting it to the left in order to get the correct answer. Why couldn't one just shift the unit disc up without rotating it at all?

Also, I think I've found a rather slick way of doing the problem:
First, we notice that t is in above the real axis iff. |t-i| = |-t+i| < |t+i| (this can be easily proved with a direct calculation).
Then, this means that t is above the real axis iff. (|(-t+i)/(t+i)|) < 1.
Letting z = (-t+i)/(t+i), we then have |z| < 1, the unit disk in the z-plane, iff. t is above the real axis in the t-plane.
Thus, z is the transformation with the properties we seek.

Regards,
cwjian

Last edited by cwjian on 30 Jul 2011, 18:55, edited 1 time in total.

28 Jul 2011, 19:14
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Re: Exercise [08.07] b
Hi
This is a good question. The answer is that what you are proposing - to just move the plane up by +i and then multiply the whole plane by 2 - achieves the same result, in the sense that the upper half-plane bounded by the real axis is mapped to the unit disc, and the lower plane to the rest of the plane.
In fact your transformation =(1+it)/(t+i)=-i(i-t)/(t+i). So your T=-i times Penrose's T or in other words:
your T= Penrose's T rotated 90 degrees clockwise.
If you look at Fig. 8.8 in RTR on page 145 and label the 3 points along the real axis of the t-plane as B, C, D going from left to right then Penrose's transformation means that B transforms to the South point of the circle in the z-plane, C to the east point and D to the north point, If you also label -infinity to the left in the t-plane as A and + infinity also as A (the point at infinity on the Riemann sphere) then the west point can be labelled A. As you travel from -infinity to +infinity along the horizontal axis of the t-plane you travel round the circle in the z-plane starting at the west point A and anticlockwise through B, C, D and back to A.
If you use your transformation instead then A in the t-plane corresponds to the north point in the z-plane and so on. I guess that this is probably why Penrose chose to put in the anticlockwise rotation as well, to make the points correspond in a more pleasing way from t-plane to z-plane.
I will have a look at the second part of your post and post my thoughts on that as soon as I can.
Thanks
Vasco

30 Jul 2011, 16:05
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Joined: 07 Jun 2008, 08:21
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Re: Exercise [08.07] b
Hi
As to your second point about the transformation, I agree with what you say. Your argument shows that the given transformation does map the upper half-plane to the unit circle and the lower half-plane to the rest of the plane. However this is not what Penrose is asking us to do.
He is saying:
"Show that the particular rotation that we are concerned with can be exhibited explicitly as a relation between the Riemann spheres of the complex parameters z and t given by the bilinear correspondence:
t=(z-1)/(iz+i), z=(-t+i)/(t+i)."

So you have to show that the Riemann sphere for the t-plane and the z-plane are equivalent to rotating the t-plane sphere through 90 degrees to exchange the real axis and the unit circle.
You have to decide first what the rotation is equivalent to in terms of a bilinear mapping and then you can apply your argument to the given mapping. However it seems to me to be more interesting and instructive to find the mapping itself rather than show that a given mapping solves the problem.
In fact by approaching the problem this way you have found a slightly different mapping which achieves almost the same result. In fact your mapping is exactly equivalent to the rotation of the Riemann sphere whereas Penrose's mapping includes an extra rotation of the plane through 90 degrees anticlockwise.
Vasco

30 Jul 2011, 16:48

Joined: 03 Jul 2011, 14:43
Posts: 8
Re: Exercise [08.07] b
Hi Vasco,

Thank you for clearing that up. I was concerned that it seemed a little bit arbitrary when you decided to rotate the t-plane instead of merely shifting it, but I do see the logic behind it now. I also agree, after having read the question again, your solution seems to be more instructive and closer to the requirements. I do find it somewhat unusual that Penrose decided to rotate the plane.

Also, you may want to add that the transformation given is a slightly modified Cayley transform.

cwjian

P.S. Is there a solution to 8.8 available on the board? I have my own solution, but am not quite sure of it, being as inexperienced with Riemann surfaces as I am...

30 Jul 2011, 18:53
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Joined: 07 Jun 2008, 08:21
Posts: 235
Re: Exercise [08.07] b
Hi
No one has posted a solution to 8.8 yet. I did a lot of work on it myself about a year ago but I did not manage to solve it. I don't think I was very far away from a solution and I would like to go back to it sometime.
Penrose has marked it as "not to be undertaken lightly" - so it's supposed to be very difficult.
You might as well post your solution and get some feedback.
Vasco

31 Jul 2011, 08:00
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