### Equidistant curve coordinate system - viXra.org

```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
****************************************************************************************

(3 次元 (3)、反転 (2)、擬球)

アブストラクト:

1. 3 次元 (3)
1.1. kr

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

√
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

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

5
2. 反転 (2)
2.1. 座標変換

√
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 =

(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

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

√
)
(
√
a + a2 − r2 √ 2
2
2
2
z = a log
r = x + y , x = r cos φ, y = r sin φ
− a −r
r

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。

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。

[1] 菊池盛雄、”等距離線座標 (反転)” (vixra:1201.0090, 2012)
8
```