642 J Gauss surface normal coordinates in geometry and gravity space
Continuation of Box.
First variation:
∂L
∂x µ (ˆ x
ν ) = 0 ∀ µ, ν ∈ {1, 2, 3, 4}
(J.9)
⇐⇒
(i) plane P
2 :
−(X − ˆ
x
1 ) + a 1 ˆ
x
4 = 0 , −(Y − ˆ
x
2 ) + a 2 ˆ
x
4 = 0 , −(Z − ˆ
x
3 ) + a 3 ˆ
x
4 = 0 ,
a 1 ˆ
x
1 + a 2 ˆ
x
2 + a 3 ˆ
x
3 + a 4 = 0 ;
(J.10)
(ii) sphere S
2
r :
−(X − ˆ
x
1 ) + ˆ
x
1 ˆ
x
4 = 0 , −(Y − ˆ
x
2 ) + ˆ
x
2 ˆ
x
4 = 0 , −(Z − ˆ
x
3 ) + ˆ
x
3 ˆ
x
4 = 0 ,
(ˆ x
1 )
2 + (ˆ x
2 )
2 + (ˆ x
3 )
2 − r
2 = 0 ;
(J.11)
(iii) ellipsoid-of-revolution E
2
A 1 ,A 2 :
−(X − ˆ
x
1 ) + A
2
2 ˆ
x
1 ˆ
x
4 = 0 , −(Y − ˆ
x
2 ) + A
2
2 ˆ
x
2 ˆ
x
4 = 0 , −(Z − ˆ
x
3 ) + A
2
1 ˆ
x
3 ˆ
x
4 = 0 ,
A
2
2 [(ˆ x
1 )
2 + (ˆ x
2 )
2 ] + A
2
1 (ˆ x
3 )
2 − A
2
1 A
2
2 = 0 ;
(J.12)
(iv) triaxial ellipsoid E
2
A 1 ,A 2 ,A 2 :
−(X − ˆ
x
1 ) + A
2
2 A
2
3 ˆ
x
1 ˆ
x
4 = 0 , −(Y − ˆ
x
2 ) + A
2
1 A
2
3 ˆ
x
2 ˆ
x
4 = 0 , −(Z − ˆ
x
3 ) + A
2
1 A
2
2 ˆ
x
3 ˆ
x
4 = 0 ,
A
2
2 A
2
3 (ˆ x
1 )
2 + A
2
1 A
2
3 (ˆ x
2 )
2 + A
2
1 A
2
2 (ˆ x
3 )
2 − A
2
1 A
2
2 A
2
3 = 0 .
(J.13)
Second variation.
The second variation decides about the solution of type “minimum” or “maximum” or “turning point”.
In our case
1
2
∂
2 L
∂x k ∂x l (ˆ x
γ ) > 0
(J.14)
⇐⇒
(i) plane P
2 :
2
4
+1 0 0
0 +1 0
0 0 +1
3
5 > 0 ;
(J.15)
(ii) sphere S
2
r :
2
4
+1 + ˆ
Λ
0
0
0
+1+ ˆ
Λ
0
0
0
+ 1+ ˆ
Λ
3
5 > 0 ;
(J.16)
(iii) ellipsoid-of-revolution E
2
A 1 ,A 2 :
2
4
+1 + A
2
2
ˆ
Λ
0
0
0
+1+A
2
2
ˆ
Λ
0
0
0
+ 1+A
2
1
ˆ
Λ
3
5 > 0 ;
(J.17)
(iv) triaxial ellipsoid E
2
A 1 ,A 2 ,A 2 :
2
4
+1 + A
2
2 A
2
3 12 ˆ
Λ
0
0
0
+ 1+A
2
1 A
2
3
ˆ
Λ
0
0
0
+ 1 + A
2
1 A
2
2
ˆ
Λ
3
5 > 0 .
(J.18)