114
5 Gravitational (Clock) Compass
General Solution
Considering the non-vacuum case first one needs to measure 20 independent
components of the Riemann curvature tensor R abcd . Analogously to the analysis
of the gravitational compass in Sect. 5.2, we may now consider different setups of
clocks to measure as many curvature components as possible. Introducing different
initial values for the clocks:
(1) p
α
=
⎛
⎝
1
0
0
⎞
⎠ ,
(2) p
α
=
⎛
⎝
0
1
0
⎞
⎠ ,
(3) p
α
=
⎛
⎝
0
0
1
⎞
⎠ ,
(4) p
α
=
⎛
⎝
1
1
0
⎞
⎠ ,
(5) p
α
=
⎛
⎝
0
1
1
⎞
⎠ ,
(6) p
α
=
⎛
⎝
1
0
1
⎞
⎠ ,
(5.90)
and
(1) u
α
=
⎛
⎝
c 11
0
0
⎞
⎠ ,
(2) u
α
=
⎛
⎝
0
c 22
0
⎞
⎠ ,
(3) u
α
=
⎛
⎝
0
0
c 33
⎞
⎠ ,
(4) u
α
=
⎛
⎝
c 41
c 42
0
⎞
⎠ ,
(5) u
α
=
⎛
⎝
0
c 52
c 53
⎞
⎠ ,
(6) u
α
=
⎛
⎝
c 61
0
c 63
⎞
⎠ ,
(5.91)
we have again an algebraic system. It was shown in [17], that this system can be
used to determine all gravitational field components as follows:
01 : R (1)(0)(1)(0) =
(1,1) B,
(5.92)
02 : R (2)(1)(1)(0) =
3
4
c
−1
22 c
−1
42 (c 22 − c 42 )
−1
(1,1) Bc
2
22 −
(1,1) Bc
2
42
+
(1,2) Bc
2
42 −
(1,4) Bc
2
22
,
(5.93)
03 : R (1)(2)(1)(2) = −3c
−1
22 c
−1
42 (c 22 − c 42 )
−1
(1,1) Bc 22 −
(1,1) Bc 42
+
(1,2) Bc 42 −
(1,4) Bc 22
,
(5.94)
5 Gravitational (Clock) Compass
General Solution
Considering the non-vacuum case first one needs to measure 20 independent
components of the Riemann curvature tensor R abcd . Analogously to the analysis
of the gravitational compass in Sect. 5.2, we may now consider different setups of
clocks to measure as many curvature components as possible. Introducing different
initial values for the clocks:
(1) p
α
=
⎛
⎝
1
0
0
⎞
⎠ ,
(2) p
α
=
⎛
⎝
0
1
0
⎞
⎠ ,
(3) p
α
=
⎛
⎝
0
0
1
⎞
⎠ ,
(4) p
α
=
⎛
⎝
1
1
0
⎞
⎠ ,
(5) p
α
=
⎛
⎝
0
1
1
⎞
⎠ ,
(6) p
α
=
⎛
⎝
1
0
1
⎞
⎠ ,
(5.90)
and
(1) u
α
=
⎛
⎝
c 11
0
0
⎞
⎠ ,
(2) u
α
=
⎛
⎝
0
c 22
0
⎞
⎠ ,
(3) u
α
=
⎛
⎝
0
0
c 33
⎞
⎠ ,
(4) u
α
=
⎛
⎝
c 41
c 42
0
⎞
⎠ ,
(5) u
α
=
⎛
⎝
0
c 52
c 53
⎞
⎠ ,
(6) u
α
=
⎛
⎝
c 61
0
c 63
⎞
⎠ ,
(5.91)
we have again an algebraic system. It was shown in [17], that this system can be
used to determine all gravitational field components as follows:
01 : R (1)(0)(1)(0) =
(1,1) B,
(5.92)
02 : R (2)(1)(1)(0) =
3
4
c
−1
22 c
−1
42 (c 22 − c 42 )
−1
(1,1) Bc
2
22 −
(1,1) Bc
2
42
+
(1,2) Bc
2
42 −
(1,4) Bc
2
22
,
(5.93)
03 : R (1)(2)(1)(2) = −3c
−1
22 c
−1
42 (c 22 − c 42 )
−1
(1,1) Bc 22 −
(1,1) Bc 42
+
(1,2) Bc 42 −
(1,4) Bc 22
,
(5.94)
