5.4 Gravitational Clock Compass
117
K 2 :=
3
4
c
−1
11
−
(5,1) B + R (2)(0)(2)(0) + 2R 3020 + R 3030
−
1
3
(R (1)(2)(1)(2) + 2R (1)(2)(1)(3) + R (1)(3)(1)(3) )c
2
11
+
4
3
(R (0)(2)(1)(2) + R (0)(3)(1)(3) )c 11
,
(5.111)
K 3 :=
3
4
c
−1
22
−
(6,2) B + R (1)(0)(1)(0) + 2R 3010 + R (3)(0)(3)(0)
−
1
3
(R (1)(2)(1)(2) + 2R (3)(2)(1)(2) + R (2)(3)(2)(3) )c
2
22
−
4
3
(R (2)(1)(1)(0) + R (0)(3)(2)(3) )c 22
,
(5.112)
the three remaining curvature components can be expressed as:
19 : R (1)(0)(2)(3) =
1
3
(K 3 − K 1 ) ,
(5.113)
20 : R (2)(0)(1)(3) =
1
3
(K 2 − K 1 ) ,
(5.114)
21 : R (3)(0)(2)(1) =
1
3
(K 3 − K 2 ) .
(5.115)
See Fig. 5.5 for a symbolical sketch of the solution. Note that the sketches of
the clock configurations make use of a notation analogous to the one in [17].
The observer is indicated by a black circle, the prepared clocks are indicated by
hollow circles. Furthermore, we note that the sketches were introduced in [17] to
give a two dimensional visual representation of the solution. In particular they are
designed for counting the number of clocks/measurements at a glance, they do not
directly represent the three dimensional geometry of the measurement (we order
hollow circles, corresponding to different positions (n), starting at the three o’clock
position, advancing counter clockwise in 45 degree angles depending on the position
index n). Note that we order arrows, corresponding to different velocities (m),
starting at the twelve o’clock position, advancing clockwise in 45 degree angles
depending on the velocity index m.
Vacuum Solution
In vacuum the number of independent components of the curvature is reduced to the
10 components of the Weyl tensor C abcd . Replacing R abcd in the compass solution
117
K 2 :=
3
4
c
−1
11
−
(5,1) B + R (2)(0)(2)(0) + 2R 3020 + R 3030
−
1
3
(R (1)(2)(1)(2) + 2R (1)(2)(1)(3) + R (1)(3)(1)(3) )c
2
11
+
4
3
(R (0)(2)(1)(2) + R (0)(3)(1)(3) )c 11
,
(5.111)
K 3 :=
3
4
c
−1
22
−
(6,2) B + R (1)(0)(1)(0) + 2R 3010 + R (3)(0)(3)(0)
−
1
3
(R (1)(2)(1)(2) + 2R (3)(2)(1)(2) + R (2)(3)(2)(3) )c
2
22
−
4
3
(R (2)(1)(1)(0) + R (0)(3)(2)(3) )c 22
,
(5.112)
the three remaining curvature components can be expressed as:
19 : R (1)(0)(2)(3) =
1
3
(K 3 − K 1 ) ,
(5.113)
20 : R (2)(0)(1)(3) =
1
3
(K 2 − K 1 ) ,
(5.114)
21 : R (3)(0)(2)(1) =
1
3
(K 3 − K 2 ) .
(5.115)
See Fig. 5.5 for a symbolical sketch of the solution. Note that the sketches of
the clock configurations make use of a notation analogous to the one in [17].
The observer is indicated by a black circle, the prepared clocks are indicated by
hollow circles. Furthermore, we note that the sketches were introduced in [17] to
give a two dimensional visual representation of the solution. In particular they are
designed for counting the number of clocks/measurements at a glance, they do not
directly represent the three dimensional geometry of the measurement (we order
hollow circles, corresponding to different positions (n), starting at the three o’clock
position, advancing counter clockwise in 45 degree angles depending on the position
index n). Note that we order arrows, corresponding to different velocities (m),
starting at the twelve o’clock position, advancing clockwise in 45 degree angles
depending on the velocity index m.
Vacuum Solution
In vacuum the number of independent components of the curvature is reduced to the
10 components of the Weyl tensor C abcd . Replacing R abcd in the compass solution