Landau & Lifshitz "Classical theory of fields" section 6 p. 19 define: $$ df^{ik} = dx^i dx'^k - dx^k dx'^i $$ and $$ df^{*ik}=\frac{1}{2} \; \epsilon^{iklm}df_{lm} \tag{6.11} $$ and states:

"It is obvious that $df^{ik} df^*_{ik}=0$."

I am unable to see that this is true. Can anyone help me?


1 Answer 1


It might help to forget about the d's.

Instead, let $F^{ik}=(v^i w^k -v^k w^i)$.

So, $F^{*}_{ik}= \frac{1}{2}\epsilon_{iklm} (v^l w^m -v^m w^l)$.

Altogether, we have $F^{ik}F^{*}_{ik}=(v^i w^k -v^k w^i) \frac{1}{2}\epsilon_{iklm} (v^l w^m -v^m w^l)$.

Remember that $\epsilon_{iklm}$ is totally-antisymmetric.

Can you finish?

  • 1
    $\begingroup$ It's still not obvious to me, since $\epsilon_{iklm}=\epsilon_{lmik}$... Doesn't it? $\endgroup$
    – hft
    Jul 6 at 21:11
  • 3
    $\begingroup$ @hft Expand out the right-hand-side. Then, look at each term. In each term, what multiplies the $\epsilon_{iklm}$? $\endgroup$
    – robphy
    Jul 6 at 21:15

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