Distance formula in hyperbolic metric

  • Thread starter elffry
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  • #1
elffry
2
0
Hi!

I'm trying to derive the hyperbolic distance formula for the upper-half plane model.
It is given here: http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model" [Broken]

I have the first formula, (ds)2= ...
But I can't figure out how they got the distance formula below it.

I understand the integral of ds is the distance, but I dont know how they got to arccosh. I've never dealt with hyperbolic trig functions before, so I wonder if that's where I'm running into trouble. I've looked up the hyperbolic trig identities, but I'm unable to get something that when integrated, would result in arccosh. I've also tried taking the derivative of the arccosh function, but I'm not completely sure which variable I should be deriving with respect to.

Please a little advice to get me pointed in the right direction!
:smile:
Thank you!
 
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Answers and Replies

  • #2
lawsofform
15
0
Look up the trig identitites again. arccosh identifies with a rather complicated log function. The log comes from the y^2 in the demoninator of the path length. If you take the derivative of that log function you should be able to work backwards and see how it works out, but it looks like it will be extremely messy.

I have to say that you can get a much better feel for hyperbolic geometry, and also avoid ugly calculus, if you treat the upper half plane as a complex plane and use Mobius transformations to find lengths, but this might be tangential to what you're trying to do.
 
  • #3
elffry
2
0
[SOLVED]

Thanks,

turns out, I needed to take the hyperbolic cosine of a completely different formula and work from there. Thanks though.
 

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