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 Exercise [08.04] c 
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Joined: 07 Jun 2008, 08:21
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Post Exercise [08.04] c
Here is a development of the previous proof, which takes into account the special case when the circle in the z-plane passes through the origin and is transformed into a straight line (circle with infinite radius as Penrose calls it).
I noticed a small typing error in the attached .pdf file which I have now corrected.

"Equation (5) is of the same form as equation (1)" should be "Equation (5) is of the same form as equation (2)"


Thanks to Azrael's post below I have corrected a typo and added a clarification. See Azrael's post for details. The pdf file below has also been amemded to correct the typo and add the clarification. Edited by Vasco on 10/10/2011.
Vasco

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RTRex8.04v2.pdf [63.06 KiB]
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Last edited by vasco on 10 Oct 2011, 08:00, edited 1 time in total.

28 Feb 2009, 09:01

Joined: 03 Jul 2011, 14:43
Posts: 8
Post Re: Exercise [08.04] c
Vasco, I have a question. Why does shifting the plane by i upwards not work, whereas rotating the plane by pi/2 radians and then subtracting 1 does?


25 Jul 2011, 18:13
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Post Re: Exercise [08.04] c
Hi
I am a bit confused! Are you sure you are talking about this exercise? Don't you mean Exercise 8.7?
If you do mean this exercise, can you explain in a bit more detail what you mean?
Thanks
Vasco


25 Jul 2011, 21:32

Joined: 03 Jul 2011, 14:43
Posts: 8
Post Re: Exercise [08.04] c
Sorry Vasco, yes, I posted in the wrong thread...

Would you mind if I continued this discussion in the other thread? Or should I continue it here?


28 Jul 2011, 15:22
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Post Re: Exercise [08.04] c
Hi cwjian
I think it would be more useful to everybody if you transferred to the other thread. Presumably the one for exercise 8.7 - Is that right?
Also could you explain in a bit more detail what it is that you do not understand?
Thanks
Vasco


28 Jul 2011, 15:51

Joined: 03 Jul 2011, 14:43
Posts: 8
Post Re: Exercise [08.04] c
Hi Vasco,

I've posted in the Problem 8.07 thread.

cwjian


29 Jul 2011, 18:19

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [08.04] c
Nice solution. I believe there are few typos, equation (1) should really look like:
z=z_0+re^{it}, but this doesn't affect the proof obviously. Also on the last page I think you mean to take z_0=x_0+iy_0 (missing an i as z_0 is complex constant), again just some typos hopefully make it easier for those going through your proof.


09 Oct 2011, 15:05
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Post Re: Exercise [08.04] c
Azrael84 wrote:
Nice solution. I believe there are few typos, equation (1) should really look like:
z=z_0+re^{it}, but this doesn't affect the proof obviously. Also on the last page I think you mean to take z_0=x_0+iy_0 (missing an i as z_0 is complex constant), again just some typos hopefully make it easier for those going through your proof.

Thanks for reading my solution. I'm glad you like it!
I agree that on the last page it should be z_0=x_0+iy_0, but I don't agree with your first point.
r is a complex number here and so z=z_0+r is correct. That's why I called it a radius vector in line 1. I think that I should have added |r|=constant to make it clearer.
As z changes so does r, but only the argument of r changes not its modulus.
I will amend my first post above and upload a corrected and amended pdf file.


10 Oct 2011, 07:48

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [08.04] c
vasco wrote:
but I don't agree with your first point.
r is a complex number here and so z=z_0+r is correct. That's why I called it a radius vector in line 1. I think that I should have added |r|=constant to make it clearer.
As z changes so does r, but only the argument of r changes not its modulus.
I will amend my first post above and upload a corrected and amended pdf file.


Ah, OK. Yes I agree it makes sense if r is complex. I thought you had in mind 'r' as the radial type parameter of polar coords centered about the point z_0 and had accidentally missed off the e^{it} factor...


10 Oct 2011, 19:01
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