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