E-2 The differential equations of third order of a geodesic circle 545
{
M
KL } = {
M
LK }, “symmetry” of Christoffel symbols leads to
D
1 =
= [U
M + 3U
K U
L
{
M
KL } + U
K U
L U
N ({
M
KL }, N +{
Q
KL }{
M
NQ } − H KL H NQ G
QM )]G M +
+[3U
K U
L H KL + U
K U
L U
N ({
Q
KL }H NQ + H KL,N )]D 3 .
(E.12)
τ g then is obtained as
κ g τ g + κ
n = 3U
K U
L H KL + U
K U
L U
N ({
Q
KL }H NQ + H KL,N ) ,
κ
n = H KL,N U
K U
L U
N + 2H KL U
K U
L
=⇒
κ g τ g = H KL U
K U
L + {
Q
KL }H NQ U
K U
L U
N
=⇒
τ g =
H KL U
K U
L + {
Q
KL }H NQ U
K U
L U
N
G M 1 M 2 ({
M 1
K 1 L 1
} + U
M 1 )({
M 2
K 2 L 2
} + U
M 2 )
=⇒ (20.30) .
(E.13)
End of Proof.
E-2 The differential equations of third order of a geodesic circle
For the proof of Corollary 20.2, we depart from the Darboux derivational equations under the postulate
of geodesic circle κ g = const., κ n = const., and τ g = 0, namely
D
1 = −(κ
2
g + κ
2
n )D 1 = −(κ
2
g + κ
2
n )U
M G M .
(E.14)
Proof (Corollary 20.2).
D
1 = [U
M + 3U
K U
L
{
M
KL } + U
K U
L U
N ({
M
KL } ,N + {
Q
KL }{
M
NQ } − H KL H NQ G
QM )]G M
=⇒
(κ
2
g + κ
2
n )U
M = U
M + 3U
K U
L + U
K U
L U
N ({
M
KL }, N +{
Q
KL }{
M
NQ } − H KL H NQ G
QM ) ,
(E.15)
κ
2
g + κ
2
n = G M 1 M 2 (U
K 1 U
L 1 {
M 1
K 1 L 1
} + U
M 1 )(U
K 2 U
L 2 {
M 2
K 2 L 2
} + U
M 2 )+
+H K 1 L 1 H K 2 L 2 U
K 1 U
K 2 U
L 1 U
L 2
=⇒
U
M + 3U
K U
L
{
M
KL } + U
K U
L U
N ({
M
KL }, N +{
Q
KL }{
M
NQ }) − U
K U
L U
N H KL H NQ G
QM +
+G M 1 M 2 (U
K 1 U
L 1 {
M 1
K 1 L 1
} + U
M 1 )(U
K 2 U
L 2 {
M 2
K 2 L 2
}U
M 2 )U
M +
+H K 1 L 1 H K 2 L 2 U
K 1 U
K 2 U
L 1 U
L 2 U
M = 0
⇔
U
M + G KL U
K U
L U
M + 3{
M
KL }U
K U
L + 2G KL {
L
P Q }U
K U
L U
P U
M +
+({
M
KL } ,P + {
Q
KL }{
M
QP })U
K U
L U
P + G KL {
K
P Q }{
L
ST }U
P U
Q U
S U
T U
M = 0 .
(E.16)
Compare our result with W. O. Vogel (1970 p. 642, formula 3).
End of Proof.