circumference of a circle in Poincare Half Planeby demonelite123 Tags: circle, circumference, plane, poincare 

#1
Feb2312, 02:30 PM

P: 219

i am trying to figure out how to calculate the circumference of a circle in the Poincare Half Plane. I know that vertical lines are geodesics so using the arclength formula, the distance between 2 points [itex] (x_0, y_0) and (x_1, y_1) [/itex] on a vertical line is [itex] ln(y_1/y_0) [/itex]. Thus, if i have a circle centered at (0, b) on the y axis with radius r, then the radius of the circle using the Poincare metric will be [itex] \frac{1}{2}ln(\frac{b+r}{br}) [/itex].
I want to find the circumference of the circle [itex] x^2 + (yb)^2 = b^2  1 [/itex] so i parametrize [itex] x = \sqrt{b^21}cos(\theta), y = b + \sqrt{b^21}sin(\theta) [/itex]. I then use the arclength formula and get [itex] C = \int^{2\pi}_0 \frac{\sqrt{b^21}}{b+\sqrt{b^21}sin(\theta)} d\theta [/itex]. I have the extra factor of [itex] y = b + \sqrt{b^21}sin(\theta) [/itex] on the bottom since I am using the Poincare metric. This evaluates to [itex] 2\pi\sqrt{b^21} [/itex] which i know shouldn't be right since the circumference of this circle using the regular euclidean metric is also just [itex] 2\pi\sqrt{b^21} [/itex]. can someone help me figure out what went wrong here? 



#2
Feb2312, 02:54 PM

P: 1,412

Why do you think your answer is wrong? Can you say more? Justify it better?



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