Equidistant curve coordinate system (3 dimensions(3), inversions(2), pseudo-sphere) Morio Kikuchi Abstract: A constant of length in an orthogonal sphere agrees with a constant of length in a plane which passes through origin. 1. Three dimensions(3) 1.1. kr We obtain the constant of length in the case that a sphere intersects the inﬁnity sphere in non-great circle. From Figure 1 2kRdx 2kr Rr dx kr Rr kR = 2 = 2 2, 2 2 2 2 z −R z −R zr + R r zr + R r 2 R zr 2 + R r 2 kr = k Rr z 2 − R 2 2 2 √ 2 x1 + z1 = R , z1 = R2 − x1 = √ 2 R2 − Rr , z = zr + z1 = zr + 2 √ R 2 − Rr 2 Assuming the radius of the sphere to be r √ zr = r + r 2 − Rr 2 Therefore kr = k r 1 R R √ √ √ =k √ Rr r 2 − R 2 + R 2 − R 2 Rr 1 − (Rr /r)2 + (R2 − Rr 2 )/r2 r r In the case of the arrangement like Figure 2 z = zr − z1 = zr + √ R 2 − Rr 2 Therefore kr = k r 1 R R √ √ √ =k √ Rr r 2 − R 2 − R 2 − R 2 Rr 1 − (Rr /r)2 − (R2 − Rr 2 )/r2 r r In the case that a sphere is at right angles to the inﬁnity sphere, the expression becomes kr = k. 1 2. Inversions(2) 2.1. Coordinate transformations We consider the coordinate transformation between A1 (x, y, z) and A2 (u, v, w) on z0 in Figure 6 of the reference [1]. Because A1 (x, y, z) and A2 (u, v, w) are in the relation of inversion on the sphere √ of which center and radius are z0 and z0 2 + R2 respectively √ √ x2 + y 2 + (z − z0 )2 u2 + v 2 + (w − z0 )2 = z0 2 + R2 The line which passes through the two points (0, 0, z0 ), (x, y, z) is u−0 v−0 w − z0 = = x−0 y−0 z − z0 (w − z0 )x (w − z0 )y u= , v= z − z0 z − z0 Substituting u, v into it { } { } ( ) x2 y2 2 2 2 2 2 2 2 x + y + (z − z0 ) + + 1 (w − z ) = z + R 0 0 (z − z0 )2 (z − z0 )2 2 (z0 2 + R2 ) (z − z0 )2 (w − z0 ) = 2 {x + y 2 + (z − z0 )2 }2 Assuming that w − z0 and z − z0 agree as to whether the values of them are positive or negative (z0 2 + R2 ) (z − z0 ) w − z0 = 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) (z − z0 ) + z0 w= 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) x u= 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) y v= 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) (w − z0 ) + z0 = z u2 + v 2 + (w − z0 )2 (z0 2 + R2 ) u =x u2 + v 2 + (w − z0 )2 (z0 2 + R2 ) v =y u2 + v 2 + (w − z0 )2 The coordinate transformation between A1 (x, y, z) and A2 (u, v, w) on x0 in Figure 8 of the reference [1] is similarly (x0 2 − R2 ) (x − x0 ) + x0 u= (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) y v= (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) z w= (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) (u − x0 ) + x0 = x (u − x0 )2 + v 2 + w2 (x0 2 − R2 ) v =y (u − x0 )2 + v 2 + w2 (x0 2 − R2 ) w =z (u − x0 )2 + v 2 + w2 2 2 2.2. Concrete example If the other sphere is a plane in an inversion, the inversion agrees with a stereographic projection. In Figure 3 and 4, the stereographic projections in three dimensions are expressed two-dimensionally being simpliﬁed. Assuming that b = 0, c = +∞, R = 1, it agrees with Figure 3. √ √ b2 + R2 c2 + R2 + bc − R2 z0 = =R=1 b√ +c √ 2 b2 + R 2 c2 + R 2 2 2 2 r z = z0 + R = z0 = 2R2 = 2 b+c 2x 2R2 x = 2 u= 2 2 2 x +y +R x + y2 + 1 2R2 y 2y v= 2 = 2 2 2 x +y +R x + y2 + 1 R(x2 + y 2 − R2 ) x2 + y 2 − 1 w= = x2 + y 2 + R 2 x2 + y 2 + 1 Assuming that b = 1/2, c = +∞, R = 0, it agrees with Figure 4. √ √ b2 + R2 c2 + R2 + bc − R2 z0 = =1 b√ +c √ 2 b2 + R 2 c2 + R 2 2 2 2 r z = z0 + R = z0 = 1 b+c x u= 2 x + y2 + 1 y v= 2 x + y2 + 1 x2 + y 2 w= 2 x + y2 + 1 3. Pseudo-sphere The surface which extends in the direction of z in Figure 5 is a pseudo-sphere. The curve of the vertical section of the pseudo-sphere is √ ) ( √ a + a2 − r2 √ 2 2 2 2 z = a log r = x + y , x = r cos φ, y = r sin φ − a −r r The metric on the surface is ds2 = a2 dφ2 + dψ 2 ψ2 3 Because r = a/ψ ≤ a, we have ψ ≥ 1. In Figure 5, xy plane and φψ plane being drawn as one plane, a line and a circle which represent equidistant curve coordinates are drawn on φψ plane, and two curves which represent corresponding equidistant curve coordinates are drawn on the pseudo-sphere. Pseudo-sphere can be understood to be a kind of window which shows φψ plane in three-dimensional space as well. References: [1] Morio Kikuchi, ”Equidistant curve coordinate system (inversions)” (vixra:1201.0090, 2012) 4 **************************************************************************************** 等距離線座標 (3 次元 (3)、反転 (2)、擬球) 菊池盛雄アブストラクト: 直交球面における長さの定数は原点を通る平面における長さの定数に一致します。 1. 3 次元 (3) 1.1. kr 球面が無限遠球面と非大円で交わる場合の球面上の長さの定数を求めます。図 1 より 2kRdx kR 2kr Rr dx kr Rr = 2 = 2 2, 2 2 2 2 z −R z −R zr + R r zr + R r 2 R zr 2 + R r 2 kr = k Rr z 2 − R 2 x1 2 + z1 2 = R2 , z1 = √ R 2 − x1 2 = √ R2 − Rr 2 , z = zr + z1 = zr + √ R 2 − Rr 2 球面の半径を r として √ zr = r + r 2 − Rr 2 したがって kr = k R r R 1 √ √ √ =k √ Rr r 2 − R 2 + R 2 − R 2 Rr 1 − (Rr /r)2 + (R2 − Rr 2 )/r2 r r 図 2 のようになっている場合には z = zr − z1 = zr + √ R 2 − Rr 2 したがって kr = k R r 1 R √ √ √ =k √ 2 2 Rr r 2 − R − R 2 − R Rr 1 − (Rr /r)2 − (R2 − Rr 2 )/r2 r r 球面が無限遠球面と直交している場合は kr = k となります。 5 2. 反転 (2) 2.1. 座標変換参考文献 [1] の図 6 の A1 (x, y, z), A2 (u, v, w) の z0 に関する座標変換を考えます。 √ A1 (x, y, z), A2 (u, v, w) は中心 z0 、半径 z0 2 + R2 の球面に関して反転の関係にあるので √ x2 + y2 + (z − z0 √ )2 u2 + v 2 + (w − z0 )2 = z0 2 + R2 2 点 (0, 0, z0 ), (x, y, z) を通る直線は u−0 v−0 w − z0 = = x−0 y−0 z − z0 (w − z0 )x (w − z0 )y u= , v= z − z0 z − z0 u, v を代入して { x + y + (z − z0 ) 2 2 2 } { } ( ) x2 y2 2 2 2 2 + + 1 (w − z ) = z + R 0 0 (z − z0 )2 (z − z0 )2 2 (w − z0 )2 = (z0 2 + R2 ) (z − z0 )2 {x2 + y 2 + (z − z0 )2 }2 w − z0 と z − z0 が正負が同じであるとして (z0 2 + R2 ) (z − z0 ) x2 + y 2 + (z − z0 )2 (z0 2 + R2 ) (z − z0 ) + z0 w= 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) x u= 2 x + y 2 + (z − z0 )2 (z0 2 + R2 ) y v= 2 x + y 2 + (z − z0 )2 2 (z0 + R2 ) (w − z0 ) + z0 = z u2 + v 2 + (w − z0 )2 (z0 2 + R2 ) u =x u2 + v 2 + (w − z0 )2 (z0 2 + R2 ) v =y u2 + v 2 + (w − z0 )2 w − z0 = 参考文献 [1] の図 8 の A1 (x, y, z), A2 (u, v, w) の x0 に関する座標変換は、同様に (x0 2 − R2 ) (x − x0 ) u= + x0 (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) y v= (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) z w= (x − x0 )2 + y 2 + z 2 (x0 2 − R2 ) (u − x0 ) + x0 = x (u − x0 )2 + v 2 + w2 (x0 2 − R2 ) v =y (u − x0 )2 + v 2 + w2 (x0 2 − R2 ) w =z (u − x0 )2 + v 2 + w2 6 2.2. 具体例反転において片方の球面が平面である場合は、その反転は立体射影と一致します。図 3、図 4 では 3 次元における立体射影を簡略化して 2 次元的に表現しています。b = 0, c = +∞, R = 1 とすれば図 3 の場合となります。 √ √ b2 + R2 c2 + R2 + bc − R2 z0 = =R=1 b√ +c √ 2 b2 + R 2 c2 + R 2 2 2 2 z0 = 2R2 = 2 r z = z0 + R = b+c 2x 2R2 x = 2 u= 2 2 2 x +y +R x + y2 + 1 2R2 y 2y v= 2 = 2 2 2 x +y +R x + y2 + 1 R(x2 + y 2 − R2 ) x2 + y 2 − 1 w= = x2 + y 2 + R 2 x2 + y 2 + 1 b = 1/2, c = +∞, R = 0 とすれば図 4 の場合となります。 √ √ b2 + R2 c2 + R2 + bc − R2 z0 = =1 b√ +c √ 2 b2 + R 2 c2 + R 2 2 2 2 r z = z0 + R = z0 = 1 b+c x u= 2 x + y2 + 1 y v= 2 x + y2 + 1 x2 + y 2 w= 2 x + y2 + 1 3. 擬球図 5 の z 方向に伸びる曲面が擬球です。擬球の断面の曲線は √ ) ( √ a + a2 − r2 √ 2 2 2 2 z = a log r = x + y , x = r cos φ, y = r sin φ − a −r r 曲面上の計量は ds2 = a2 dφ2 + dψ 2 ψ2 7 r = a/ψ ≤ a となっているので ψ ≥ 1 となります。図 5 では、xy 平面に φψ 平面が重ねてあり、φψ 平面には等距離線座標を示す直線と円が、擬球には対応する等距離線座標を示す二つの曲線が描かれています。擬球は φψ 平面を 3 次元空間内に現わす一種のウィンドウと考えることもできます。参考文献: [1] 菊池盛雄、”等距離線座標 (反転)” (vixra:1201.0090, 2012) 8