- #1
parton
- 83
- 1
Homework Statement
Hi,
I have to calculate the invariant: [tex] \tilde{F}^{\mu \nu} \, F_{\mu \nu} [/tex]
where F is the electromagnetic field tensor and [tex] \tilde{F} [/tex] the dual one.
Homework Equations
First, the contravariant components of the electromagnetic field tensor are given by:
[tex]F^{\mu\nu} = \begin{bmatrix} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & -B_3 & B_2 \\ E_2 & B_3 & 0 & -B_1 \\ E_3 & -B_2 & B_1 & 0 \end{bmatrix}[/tex]
and the covariant by:
[tex]F_{\mu\nu} = \begin{bmatrix} 0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & -B_3 & B_2 \\ -E_2 & B_3 & 0 & -B_1 \\ -E_3 & -B_2 & B_1 & 0 \end{bmatrix}[/tex]
And last but not least, the contravariant components of the dual elm. field tensor:
[tex]\tilde{F}^{\mu\nu} = \begin{bmatrix} 0 & -B_1 & -B_2 & -B_3 \\ B_1 & 0 & E_3 & -E_2 \\ B_2 & -E_3 & 0 & E_1 \\ B_3 & E_2 & -E_1 & 0 \end{bmatrix}
[/tex]
The Attempt at a Solution
First I note:
[tex]\tilde{F}^{a0} = B^{a} [/tex]
[tex]\tilde{F}^{ab} = \epsilon^{abi} E_{i}[/tex]
[tex]F_{a0} = - E_{a} [/tex]
[tex]F_{ab} = \epsilon_{abc} B^{c}[/tex]
Now, if I use these relations I obtain the wrong solution:
[tex] \tilde{F}^{\mu \nu} \, F_{\mu \nu} = 2 \tilde{F}^{a0} F_{a0} + \tilde{F}^{ab} F_{ab} = - 2 B^{a} E_{a} + \epsilon^{abi} E_{i} \epsilon_{abc} B^{c} = 0 [/tex]
where I used the relation: [tex] \epsilon^{abi} \epsilon_{abc} = 2 \delta^{i}_{c} [/tex]
Of course, if i simply insert the components explicitly (the "matrix elements") I get the result [tex]\tilde{F}^{\mu \nu} \, F_{\mu \nu} = - 4 \vec{B} \cdot \vec{E} [/tex] and everything is fine.
I used the following convention: [tex] \epsilon^{0123} = 1 [/tex], latin indices: {1,2,3}, greek indices: {0,1,2,3}
Further I calculated: [tex] \epsilon_{123} = \eta_{\alpha 1} \, \eta_{\beta 2} \, \eta_{\gamma 3} \, \epsilon^{\alpha \beta \gamma} = - 1 [/tex] where [tex] \eta = diag(1, -1, -1, -1) [/tex] is the metric tensor.
I think there has to be a wrong sign, but I don't find it. Does anyone have an idea?