Angle of parallelism

4 years ago by in Algebra, Algebra

In hyperbolic geometry, the angle of parallelism φ, also known as Π(p), is the angle at one vertex of a right hyperbolic triangle that has two asymptotic parallel sides. The angle depends on the segment length abetween the right angle and the vertex of the angle of parallelism φ. Given a point off of a line, if we drop a perpendicular to the line from the point, then a is the distance along this perpendicular segment, and φ is the least angle such that the line drawn through the point at that angle does not intersect the given line. Since two sides are asymptotic parallel,

 \lim_{a\to 0}\phi=\tfrac{1}{2}\pi\quad\text{ and }\quad\lim_{a\to\infty} \phi=0.

These five equivalent expressions relate φ and a:

 \sin\phi=\frac{1}{\cosh a}




 \tan\phi=\frac{1}{\sinh a}


 \cos\phi=\tanh a


 \phi=\tfrac{1}{2}\pi - \operatorname{gd}(a)

where gd is the Gudermannian function.


In the half-plane model of the hyperbolic plane (see hyperbolic motions) one can establish the relation of φ to a with Euclidean geometry. Let Q be the semicircle with diameter on the x-axis that passes through the points (1,0) and (0,y), where y > 1. Since Q is tangent to the unit semicircle centered at the origin, the two semicircles represent parallel hyperbolic lines. The y-axis crosses both semicircles, making a right angle with the unit semicircle and a variable angle φ with Q. The angle at the center of Q subtended by the radius to (0, y) is also φbecause the two angles have sides that are perpendicular, left side to left side, and right side to right side. The semicircle Q has its center at (x, 0), x < 0, so its radius is 1 − x. Thus, the radius squared of Q is

 x^2 + y^2=(1 - x)^2,


 x=\tfrac{1}{2}(1 - y^2).

The metric of the half-plane model of hyperbolic geometry parametrizes distance on the ray {(0, y) : y > 0 } with natural logarithm. Let log y =a, so y = ea. Then the relation between φ and a can be deduced from the triangle {(x, 0), (0, 0), (0, y)}, for example:

 \tan\phi=\frac{y}{-x}=\frac{2y}{y^2 - 1}=\frac{2e^a}{e^{2a} - 1}=\frac{1}{\sinh a}.

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