380 18 From 3D to 4D: Lorentz Transformation, Maxwell Equations
This implies, e.g.A ̃^12 =(A 34 −A 43 )/ 2 =A 34 ,A ̃^13 =−(A 24 −A 42 )/ 2 =−A 24 ,
andA ̃^14 =−(A 23 −A 32 )/ 2 =−A 23.
As an example, consider a special antisymmetric tensor associated with two vec-
torsaandbaccording to
Aik=⎛
⎜
⎜
⎝
0 b 3 −b 2 a 1
−b 3 0 b 1 a 2
b 2 −b 1 0 a 3
−a 1 −a 2 −a 30⎞
⎟
⎟
⎠. (18.43)
The components of the contra-variant tensor have just the opposite sign in the fourth
row and column, viz.
Aik=⎛
⎜
⎜
⎝
0 b 3 −b 2 −a 1
−b 3 0 b 1 −a 2
b 2 −b 1 0 −a 3
a 1 a 2 a 3 0⎞
⎟
⎟
⎠. (18.44)
In the dual tensor the role of thea- andb-components are interchanged, in particular
A ̃ik=⎛
⎜
⎜
⎝
0 a 3 −a 2 −b 1
−a 3 0 a 1 −b 2
a 2 −a 1 0 −b 3
b 1 b 2 b 3 0⎞
⎟
⎟
⎠. (18.45)
The double contracted product of the tensor with its dual is a Lorentz scalar. For the
special case (18.43) and (18.45), the result is
A ̃ikAik=−(a 1 b 1 +a 2 b 2 +a 3 b 3 )=−a·b. (18.46)For comparison, the product of the contra-variant tensor with its co-variant version
is, in this special case,
AikAik= 2 (b·b−a·a). (18.47)
The determinant det(A)of the tensorAis equal to
det(A)=(A ̃ikAik)^2 =(a·b)^2. (18.48)The determinant is also determined by a fourfold product ofAaccording to
εk′′m′n′
Akk′A′Amm′Ann′=−det(A)εkmn, (18.49)or, equivalently
det(A)=1
24
εkmnεk
′′m′n′
Akk′A′Amm′Ann′. (18.50)