超统一场与协变电磁学引论
2
1 12 13
0
0
0
0
21 22 23
31 32 32
b2
a
(5．79)
0
0
0
由此可求出肖军变换式的逆变换式是
1
x
a
0
0
0
a
0
0
0
a
ib
x
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
x
1
1
1
b2
a
x2
ib2 x2
x2
x
ib
a
x
x
3
3
3
x4
3
ib1 ib2 ib3
x4
x4
2
x
1 12 13 0
1
21 22 23
31 32 32
0
0
0
b2
a
x2
(5．80)
(5．81)
(5．82)
x
3
x4
0
0
0
由于
2
1
x
1 0 0 0 x
1
12 13 0
1
2
x2
0 1 0 0 x2
21 2
23
0
0
0
2
x
0 0 1 0 x
31 32 3
3
3
x4
0 0 0 0 x
0
0
0
4
所以，肖军逆变换式的最后形式是
x
a
0
0
0
a
0
0
0
a
ib
x
1
1
1
x2
ib2 x2
x
ib
a
x
3
3
3
x4
ib1 ib2 ib3
x4
肖军逆变换式也可写成
xj axj ibj x4
(5．83)
x4 ax4 ibj xj
若对上式两边分别平方
2
2
2
xj xj a xj xj 2iabj xj x4 b j x4x4
(5．84)
2
2
2
x x a x x 2iabj xj x4 b j xj xj
4
4
4
4
然后求和，可得到
2
2
2
2
xj xj x4 x4 a xj xj b x4x4 a x4x4 b xj xj
159