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 exercise 2.2 
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Joined: 01 May 2008, 09:18
Posts: 1
Post exercise 2.2
Hello,
I am a math novice.
In the solution for exericse 2.2, initially it says : distance is lambda x
Later on however 1/lamba is used. Why is this?
Thanks for your help.


01 May 2008, 09:31
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: exercise 2.2
This is because the formula for the hyperbolic distance between two points A and B is derived by assuming that units have been adjusted so that C=1 (this is not stated very clearly in Penrose's book and of course the derivations of the formulae are not presented either). So this means that the distances given by the formula are all effectively multiplied by $\lambda$. In order therefore to get back to distances measured in the previous units we have to divide by $\lambda$ or multiply by $\frac{1}{\lambda}$.
I hope this is clear, if not please let me know.
To summarise: lengths were multiplied by an appropriate $\lambda$ so that C could be 1 and so to get distances back to the original units we must divide by $\lambda$.


11 Jun 2008, 14:55

Joined: 12 Oct 2008, 03:19
Posts: 1
Post Re: exercise 2.2
I'm not exactly a novice (I've done well in calculus and differential equations), but I am still confused after working at this for hours. There is something basic I'm not getting. It has nothing to do with the formulae, really, but more to do with the idea of scaling units. I think I need real life examples perhaps with actual measurements in order to see what the purpose of "C" and "lambda" are, as well as how they are related. Can someone point me in the right direction? I'm happy to attempt to explain my difficulty more clearly if necessary. Thanks.


13 Oct 2008, 03:03
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: exercise 2.2
Jerry
In the formula in Penrose's book, C is a constant of proportionality and the formula is saying that the area of the triangle is proportional to \pi-(\alpha+\beta+\gamma).
Please see also attached file:
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Hooke.pdf [16.98 KiB]
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13 Oct 2008, 08:07
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