C Gravitational (Clock) Compass
223
∂ 2 p (α)
∂x∂y
= −P
−1
0
∂P 0
∂y
∂p (α)
∂x
− P
−1
0
∂P 0
∂x
∂p (α)
∂y
,
(C.26)
which can be verified by direct substitution from (C.16) or otherwise.
Now (C.2) gives X i in terms of x, y, r, u. To obtain the Minkowskian line
element in coordinates x, y, r, u we first have from (C.2)
dX
i
=
v
i
+ r
∂p i
∂u
du + p
i dr + r
∂p i
∂x
dx + r
∂p i
∂y
dy .
(C.27)
Now using the scalar products (C.19)–(C.22) the Minkowskian line element (C.1)
becomes
ds
2
= η ij dX
i dX
j
= −r
2 P
−2
0 (dx
2
+ dy
2 ) + 2 r
2 ω (α)(β)
∂p (α)
∂x
p
(β) du dx
+2 r
2 ω (α)(β)
∂p (α)
∂y
p
(β) du dy − dr
2
+{(1 − h 0 r)
2
− r
2 ω (σ )(α) ω (σ )(β) p
(α) p
(β)
}du
2 ,
(C.28)
with
h 0 = a i p
i
= a (α) p
(α) ,
(C.29)
by (C.15). We can rewrite (C.28) in the neat form
ds
2
= −r
2 P
−2
0 {(dx + a 0 du)
2
+ (dy + b 0 du)
2
}
−dr
2
+ (1 − h 0 r)
2 du
2 ,
(C.30)
with
a 0 = −P
2
0 ω (α)(β)
∂p (α)
∂x
p
(β) ,
b 0 = −P
2
0 ω (α)(β)
∂p (α)
∂y
p
(β) ,
(C.31)
since
P
2
0
ω (α)(β)
∂p (α)
∂x
p
(β)
2
+ P
2
0
ω (α)(β)
∂p (α)
∂x
p
(β)
2
= δ
σρ ω (σ )(α) ω (ρ)(β) p
(α) p
(β) .
(C.32)