From
Transformation
Result
(3-8)
Φ μ ν = [ C 00 exp ( i ( ∓ E t ± P r ) ) 0 0 C 11 exp ( i ( ∓ E t ± P r ) ) ] = [ C 00 exp ( i ( ∓ E t ± P r + 2 π ) ) 0 0 C 11 exp ( i ( ∓ E t ± P r + 2 π ) ) ]
2 π symmetry
Table 5
h μ ν 3 = ( a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b ) = ( − 1 0 0 0 0 cos ϕ − sin ϕ 0 0 sin ϕ cos ϕ 0 0 0 0 1 ) ( a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b ) ( − 1 0 0 0 0 cos ϕ sin ϕ 0 0 − sin ϕ cos ϕ 0 0 0 0 1 )
where a = C 00 exp ( i ( ∓ E t ± P r ) )
where b = C 11 exp ( i ( ∓ E t ± P r ) )
ϕ independent