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}{b-r}) [/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I want to find the circumference of the circle [itex] x^2 + (y-b)^2 = b^2 - 1 [/itex] so i parametrize [itex] x = \sqrt{b^2-1}cos(\theta), y = b + \sqrt{b^2-1}sin(\theta) [/itex]. I then use the arclength formula and get [itex] C = \int^{2\pi}_0 \frac{\sqrt{b^2-1}}{b+\sqrt{b^2-1}sin(\theta)} d\theta [/itex]. I have the extra factor of [itex] y = b + \sqrt{b^2-1}sin(\theta) [/itex] on the bottom since I am using the Poincare metric. This evaluates to [itex] 2\pi\sqrt{b^2-1} [/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^2-1} [/itex]. can someone help me figure out what went wrong here?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Circumference of a circle in Poincare Half Plane

**Physics Forums | Science Articles, Homework Help, Discussion**