2
$\begingroup$

In some papers (such as http://arxiv.org/abs/hep-th/9910184 and http://arxiv.org/find/all/1/all:+AND+kapustin+AND+topological+disorder/0/1/0/all/0/1) I am reading it is always referred at "the dual particle of a photon" in three (2+1) dimension.

This particle is a scalar $\sigma$, compared with the photon which is a vector field $A^{\mu} \quad \mu=0,1,2$.

What is meant with that? Is it just because the photon in $d=3$ as $1$ degree of freedom, just as the scalar, or there is a deeper reasoning behind that? Would a theory formulated in terms of the "dual photon" $\sigma$ be the same theory as the one formulated in terms of the original photon $A^{\mu}$? How can one express the original photon in terms of its dual?

$\endgroup$
  • 3
    $\begingroup$ It's not very profound. Look at the equation above, $J^i = \epsilon^{ijk} F_{jk}.$ Now calculate $\epsilon_{ijk} \partial^k \sigma$, substituting $\partial^k \sigma \mapsto \epsilon^{klm} F_{lm}.$ $\endgroup$ – Vibert Jan 17 '14 at 17:50
  • $\begingroup$ Im sorry but I still don't understand. Is $\partial^i\sigma=J^i$? Why? $\endgroup$ – Federico Carta Jan 18 '14 at 15:19

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.