1
$\begingroup$

Given $$G^{\mu\nu} = -2 \frac{\partial L}{\partial F_{\mu\nu}}$$ it is written in http://arxiv.org/abs/hep-th/9506035 that the field equations are (eq 1.8) $$\partial_{[\alpha} \star G_{\beta \gamma]}= 0$$

How is that? Why the dual and not the $G_{\mu\nu}$ itself? Here L is general form of Lagrangian which is a function of $F_{\mu\nu}$ and $g_{\mu\nu}$ and $\partial_{[\alpha} \star G_{\beta \gamma]}= 0$ is the anti-symmetrized way of writing the equation of motion. Please note that here we are considering duality.

$\endgroup$
  • 2
    $\begingroup$ Please incorporate the relevant content of the link into the post - in this case: What does $\partial_{[\alpha\star G_\beta\gamma]}$ mean? What kind of $L$ are we looking at? $\endgroup$ – ACuriousMind Dec 4 '14 at 16:53
  • $\begingroup$ I edited it, hope it is clearer now! $\endgroup$ – Fluctuations Dec 4 '14 at 17:26
  • $\begingroup$ I don't know what "the antisymmetrization way of writing the e.o.m." is, and which duality are we considering? Is the $\star$ supposed to be the Hodge dual? $\endgroup$ – ACuriousMind Dec 4 '14 at 17:32
  • $\begingroup$ @ACuriousMind: I have corrected his equation, using the arXiv reference. He just forgot to move the $G$ and $\star$ outside the subscript. $\endgroup$ – JamalS Dec 4 '14 at 17:36
  • 1
    $\begingroup$ @ACuriousMind: Yes, the exterior derivative is an anti-symmetrized differentiation operator, so it's exactly that. In other words, the codifferential $\delta G = 0$. The OP's choice of notation is unfortunate since he combines tensor calculus and exterior calculus :) $\endgroup$ – JamalS Dec 4 '14 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.