# Tensor manipulations in Landau & Lifschitz "Classical theory of fields"

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?

• When I was a student, we sometimes spent hours on a Landau's "it is obvious that......" ! Jul 6 at 20:11
• Jul 7 at 20:34

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?

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