Equidistant curve coordinate system Morio Kiuchi Abstract: An isometry is realized between Poincaré dis of which radius is not limited to 1 and upper half-plane. Poincaré metrics are the same in both regions when equidistant curve coordinate system is used. 1. Poincaré dis We try to give an interior point of Poincaré dis coordinates. However, a radius of Poincaré dis is not limited to 1 but arbitrary. As orthogonal coordinates is finite in it, a device to use full coordinates to infinity is needed. Figure 1 shows two equidistant curves which are parts of two circles. They are called an equidistant curve in direction of x and an equidistant curve in direction of y on the grounds that their centres are on x-axis and y-axis. An intersection of two equidistant curves represents an interior point of Poincaré dis, and its coordinates are non-euclidean distances on axes from origin. At the beginning, we obtain a centre a and a radius r a of an equidistant curve in direction of x which passes through x a on x-axis of Figure 1. Assuming a radius of Poincaré dis to be R R + a (x a a) is realized. Therefore a (x a R )/(x a ) r a x a a (x a + R )/( x a ) Similarly b (y b R )/(y b ) r b (y b + R )/( y b ) The equations of two circles are (x a) + y r a x + (y b) r b Assuming f(l, m) (x a R ) l (y b R ) m, the solutions of this are x f(1, )x a f(0, 1)x a f(, ) + 4R {f(0, )x a + f(, 0)y b } {f(0, )x a + f(, 0)y b } y f(, 1)y b f(1, 0)y b f(, ) + 4R {f(0, )x a + f(, 0)y b } {f(0, )x a + f(, 0)y b } 1
Next, we obtain a relation between a coordinate of an equidistant curve, namely an equidistant curve coordinate x and an orthogonal coordinate x a of Figure 1. Assuming z 1 0, z x a, z 3 R, z 4 R z 1 z 3 z z 3 z z 4 z 1 z 4 R + x a R x a Therefore x log{(r + x a )/(R x a )} From this x a R tanh{x /()} Similarly y b R tanh{y /()} Substituting them into x, y x R sinh x 1 + sinh x + sinh + 1, y R sinh 1 + sinh x + sinh + 1 We express a Poincaré metric in equidistant curve coordinates. dx + dy ( x x dx + x ) ( y y d + x dx + y ) y d R ( 1 + sinh x + sinh + 1) x cosh cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh 4 R (1 (R x y ) + sinh x R + sinh + 1) Therefore ds 4 R (dx + dy ) (R x y ) cosh x cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh
. Upper half-plane.1. Isometry A point w x + iy on Poincaré dis and a point z u + iv on upper half-plane have a relation of one-to-one correspondence through the linear transformation z ir R w R + w R y + ir(r x y ) (x + R) + y Using this linear transformation, we show that an isometry is realized between Poincaré dis and upper half-plane. u R y (x + R) + y, v R(R x y ) (x + R) + y ( du + dv dx + x ( ) + x + ) y dy + ( v x ) v dx + y dy ) ( ) dx + + y ) v dxdy x y ( v x ( x y + v ( ) v y dy Because partial derivatives of u, v satisfy the following expressions x v y, y v x ( ) + x ( ) v x (dx + dy ) 4R 4 {(x + R) + y } (dx + dy ) 3
Therefore ds (du + dv ) 4 R (dx + dy ) v (R x y ) The equidistant curves x > 0, y > 0 of Figure are mapped to the equidistant curves x > 0, y > 0 of Figure 3 respectively by the linear transformation. Therefore, coordinate axes of equidistant curve coordinate system of upper half-plane are a semicircle of which centre is origin and which passes through z ir and positive imaginary axis... Poincaré metrics Because equidistant curves stretch lie Figure 3 in upper half-plane, we express a Poincaré metric in equidistant curve coordinates. We obtain a relation between an equidistant curve coordinate y and an orthogonal coordinate y b in Figure 4. Assuming z 1 R, z y b, z 3 0, z 4 z 1 z 3 z z 3 z z 4 z 1 z 4 R y b Therefore y log(r/y b ) From this y b Re / The equation of a circle which passes through y b of Figure 4 is x + (y b) R + b namely x + y by R 0 4
We obtain a relation between θ e and x in Figure 4. The length of NA of Figure 5 is s log(av/au) Using sine theorem AV/ sin{π/ (α/ β)} AU/ sin(α/) namely AV/AU cos(α/ β)/ sin(α/) cos β cot(α/) + sin β s log{cos β cot(α/) + sin β} cos β cot(α/) + sin β e s/ tan(α/) cos β/(e s/ sin β) In Figure 6 tan θ e r b sin θ b + r b cos θ sin θ sin β + cos θ sin{π/ (α β)} sin β + cos{π/ (α β)} cos(α β) sin β + sin(α β) {1 tan (α/)} cos β + tan(α/) sin β tan(α/) cos β + tan (α/) sin β Substituting tan(α/) into it es/ e s/ cos β sinh(s/) cos β Because β is 0 in Figure 4 tan θ e sinh(x /) namely x/y sinh(x /) x y sinh(x /) Furthermore b (y b R )/(y b ) R sinh(y /) Substituting x, b into the equation of the circle {1 + sinh (x /)}y + R sinh(y /) y R 0 5
Solving this y R 1 + sinh x 1 + sinh + sinh, x R 1 + sinh x sinh x + sinh + sinh We express a Poincaré metric in equidistant curve coordinates. dx + dy ( x x dx + x ) ( y y d + x dx + y ) y d ( 1 + sinh x R + sinh + sinh x cosh cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh ) Therefore ds (dx + dy ) y cosh x cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh 6
References: [1] Morio Kiuchi, On coordinate systems by use of spheres (the 1st,..., the 4th) (009)(in Japanese) ******************************************************************************** : 1 1. 1 1 x y x y 1 x x a x a r a R R + a (x a a) a (x a R )/(x a ) r a x a a (x a + R )/( x a ) b (y b R )/(y b ) r b (y b + R )/( y b ) (x a) + y r a x + (y b) r b f(l, m) (x a R ) l (y b R ) m 7
x f(1, )x a f(0, 1)x a f(, ) + 4R {f(0, )x a + f(, 0)y b } {f(0, )x a + f(, 0)y b } y f(, 1)y b f(1, 0)y b f(, ) + 4R {f(0, )x a + f(, 0)y b } {f(0, )x a + f(, 0)y b } 1 x x a z 1 0, z x a, z 3 R, z 4 R z 1 z 3 z z 3 z z 4 z 1 z 4 R + x a R x a x log{(r + x a )/(R x a )} x a R tanh{x /()} y b R tanh{y /()} x, y x R sinh x 1 + sinh x + sinh + 1, y R sinh 1 + sinh x + sinh + 1 dx + dy ( x x dx + x ) ( y y d + x dx + y ) y d R ( 1 + sinh x + sinh + 1) x cosh cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh 4 R (1 (R x y ) + sinh x R + sinh + 1) 8
ds 4 R (dx + dy ) (R x y ) cosh x cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh..1. w x + iy z u + iv z ir R w R + w R y + ir(r x y ) (x + R) + y 1 1 u R y (x + R) + y, v R(R x y ) (x + R) + y ( du + dv dx + x ( ) + x + ) y dy + ( v x ) v dx + y dy ) ( ) dx + + y ) v dxdy x y ( v x ( x y + v ( ) v y dy u, v x v y, y v x 9
( ) + x ( ) v x (dx + dy ) 4R 4 {(x + R) + y } (dx + dy ) ds (du + dv ) 4 R (dx + dy ) v (R x y ) x > 0, y > 0 3 x > 0, y > 0 z ir.. 3 4 y y b z 1 R, z y b, z 3 0, z 4 z 1 z 3 z z 3 z z 4 z 1 z 4 R y b y log(r/y b ) y b Re / 4 y b 10
x + (y b) R + b x + y by R 0 4 θ e x 5 NA s log(av/au) AV/ sin{π/ (α/ β)} AU/ sin(α/) AV/AU cos(α/ β)/ sin(α/) cos β cot(α/) + sin β s log{cos β cot(α/) + sin β} cos β cot(α/) + sin β e s/ tan(α/) cos β/(e s/ sin β) 6 tan θ e r b sin θ b + r b cos θ sin θ sin β + cos θ sin{π/ (α β)} sin β + cos{π/ (α β)} cos(α β) sin β + sin(α β) {1 tan (α/)} cos β + tan(α/) sin β tan(α/) cos β + tan (α/) sin β tan(α/) es/ e s/ cos β sinh(s/) cos β 4 β 0 tan θ e sinh(x /) x/y sinh(x /) x y sinh(x /) 11
b (y b R )/(y b ) R sinh(y /) x, b {1 + sinh (x /)}y + R sinh(y /) y R 0 y R 1 + sinh x 1 + sinh + sinh, x R 1 + sinh x sinh x + sinh + sinh dx + dy ( x x dx + x ) ( y y d + x dx + y ) y d ( 1 + sinh x R + sinh + sinh x cosh cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh ) ds (dx + dy ) y cosh x cosh (dx + dy ) cosh x x dx dy 1 + sinh x + sinh 1
: [1] ( 1... 4 ) (009) 13