- #1
demonelite123
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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 (x_0, y_0) and (x_1, y_1) on a vertical line is ln(y_1/y_0). 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 \frac{1}{2}ln(\frac{b+r}{b-r}).
I want to find the circumference of the circle x^2 + (y-b)^2 = b^2 - 1 so i parametrize x = \sqrt{b^2-1}cos(\theta), y = b + \sqrt{b^2-1}sin(\theta). I then use the arclength formula and get C = \int^{2\pi}_0 \frac{\sqrt{b^2-1}}{b+\sqrt{b^2-1}sin(\theta)} d\theta. I have the extra factor of y = b + \sqrt{b^2-1}sin(\theta) on the bottom since I am using the Poincare metric. This evaluates to 2\pi\sqrt{b^2-1} which i know shouldn't be right since the circumference of this circle using the regular euclidean metric is also just 2\pi\sqrt{b^2-1}. can someone help me figure out what went wrong here?
I want to find the circumference of the circle x^2 + (y-b)^2 = b^2 - 1 so i parametrize x = \sqrt{b^2-1}cos(\theta), y = b + \sqrt{b^2-1}sin(\theta). I then use the arclength formula and get C = \int^{2\pi}_0 \frac{\sqrt{b^2-1}}{b+\sqrt{b^2-1}sin(\theta)} d\theta. I have the extra factor of y = b + \sqrt{b^2-1}sin(\theta) on the bottom since I am using the Poincare metric. This evaluates to 2\pi\sqrt{b^2-1} which i know shouldn't be right since the circumference of this circle using the regular euclidean metric is also just 2\pi\sqrt{b^2-1}. can someone help me figure out what went wrong here?
