112
5 Gravitational (Clock) Compass
a 1 = −
2
3
P
2
0 R (α)(4)(β)(γ ) p
(α) ∂p (β)
∂x
p
(γ ) ,
(5.75)
b 1 = −
2
3
P
2
0 R (α)(4)(β)(γ ) p
(α) ∂p (β)
∂y
p
(γ ) .
(5.76)
When the functions are substituted into the 1-forms (5.49)–(5.52) the line element
(5.48) is given, in the coordinates x i = (x, y, r, u) with i = 1, 2, 3, 4 as ds 2 =
g ij dx i dx j with
g 11 = −r
2 P
−2
0 {1 + 2 (α 2 − q 2 ) r
2
+ O(r
3 )} ,
(5.77)
g 22 = −r
2 P
−2
0 {1 − 2 (α 2 + q 2 ) r
2
+ O(r
3 )} ,
(5.78)
g 12 = −r
2 P
−2
0 {2 β 2 r
2
+ O(r
3 )} ,
(5.79)
g 33 = −1 (g 31 = g 32 = g 34 = 0) ,
(5.80)
g 14 = −r
2 P
−2
0 {a 0 + a 1 r + O(r
2 )} ,
(5.81)
g 24 = −r
2 P
−2
0 {b 0 + b 1 r + O(r
2 )} ,
(5.82)
g 44 = {1 + (a · p) r}
2
− |ω × p|
2 r
2
+2 c 2 r
2
+ O(r
3 ) ,
(5.83)
with P 0 given by (C.17).
If x i = x i (s) with x i = (x, y, r, u) is an arbitrary time-like world line in the
neighbourhood of r = 0 with s proper time along it then, for small values of r and
using the line element (5.48) the formula (5.45) is modified to read
ds
du
2
= 1 − |u|
2
+ 2 {a · p − u · (ω × p)} r + {(a · p)
2
− |ω × p|
2
} r
2
+2 c 2 r
2
− 2 a 1 r
3 P
−2
0
dx
du
− 2 b 1 r
3 P
−2
0
dy
du
− 4 β 2 r
4 P
−2
0
dx
du
dy
du
−2 (α 2 − q 2 ) r
4 P
−2
0
dx
du
2
+ 2 (α 2 + q 2 )r
4 P
−2
0
dy
du
2
+ O(r
3 ) .
(5.84)
Here c 2 is given by (5.74). Using (5.75) and (5.76) we have
a 1 r
3 P
−2
0
dx
du
+ b 1 r
3 P
−2
0
dy
du
= −
2
3
r
2 R (α)(4)(β)(γ ) p
(α) u
(β) p
(γ ) , (5.85)