TransWikia.com

The dual of a surface element in 4-space

Physics Asked on December 26, 2020

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?

One Answer

It is OK to use an explicit form in a local orthonormal coordinate system (in the Minkowski sense). The dual for each component would then be just a possible flip of signs (see, e.g., https://en.wikipedia.org/wiki/Hodge_star_operator#Four_dimensions). Raising the indices would amount to doing nothing in that coordinate system. This works even if the space-time curvature is nonzero.

Answered by C Tong on December 26, 2020

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP