# The dual of a surface element in 4-space

+ 5 like - 0 dislike
378 views

In reading the classic text, "The Classical Theory of Fields", Third Edition, by Landau and Lifschitz, I found an "obvious" statement not so obvious to me. It is on p.19, the statement of the normality of the dual of a surface element, $df^{*ik}=\frac{1}{2}e^{iklm}df_{lm}$ to the element $df^{ik}$. Yes, the contraction is zero, as one can see if he lists the 24 terms of the sum and takes account of the alternations of the sign of the completely antisymmetric tensor coefficient and the sign changes of the surface elements. That is a bit of tedium that I found necessary, because I did not find it obvious. Maybe that is because I was not clever about the way I listed the terms.

Question: Is there some way of listing the terms that would have quickly made clear that for every positive term there would be a negative one? One thought that suggested itself to me, after I did the work (!) was that if the terms were not all of the same sign, there would have to be an equal number of positive and negative terms because of the symmetry of the form and, thus, normality of the two surface elements. Is that the obvious quality that I first missed?

This post imported from StackExchange Physics at 2015-11-28 17:02 (UTC), posted by SE-user Albert D Horowitz
You don't need to write the terms in an explicit way at all. Observe that $(\star f)^{ik} f_{ik} = \epsilon^{iklm} f_{lm} f_{ik}$, where the $\epsilon$ is antisymmetric under exchange of the double indices $(ik)$ and $(lm)$, while $f_{lm} f_{ik}$ is symmetric. The contraction of a symmetric and an antisymmetric object is zero.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.