0

This question is extension of Electro magnetic duality, Strong weak duality and N=4 super Yangmils which i asked before.

Here what i want to know is compare of QED and N=4 SYM in terms of electro-magnetic duality.

As i heard, In QED, there is no such electro-magnetic duality but N=4 SYM theory has such duality. I want to know why this is true, and what is the distinction of two theories in the viewpoint of electro-magnetic duality.

I found proper(?) information for this question.

QED has a running coupling constant, but in $N=4$ SYM is conformal invaraint, $\beta=0$.

Community
  • 1
phy_math
  • 3,552

1 Answers1

0

relevant answer monopole might give the answer. $i.e$,

In the sense of electro-magnetic duality, we require magnetic monopoles (We assume $B= \frac{g}{2\pi} \frac{\mathbf{r}}{r^3}$ like $E$.) And see the duality $g$ and $e$, $B$ and $E$.

In QED we have no magnetic monopole. Considering Gerogi-Glashow model coupled with Higgs transforming adjoint representation for given gauge group. $i.e$ $SO(3)$, or $SU(2)$. In the process of BPS saturation, we obtain 'thooft polyakov monopole which is a kind of magnetic monopole but without any singular behaviour.

By computing $N=2$ SYM and $N=4$ SYM theories, computing its Hamiltonian and BPS saturation we obtain exactly same as 'thooft polyakov monopole, ($i.e$, Georgi-Glashow model coupled with Hiigs), thus magnetic monopole and associated $W$ boson exsits.
$N=4$ SYM, the magnetic monopole and $W$ boson is in same supersymmetry short multiplet. Thus in $N=4$ SYM we see exact Monoten-Olive duality (em duality). (Note $N=2$ case they are in different multiplet... spin ...)

I noticed from original papers of H. Osborn "TOPOLOGICAL CHARGES FOR N = 4 SUPERSYMMETRIC GAUGE THEORIES AND MONOPOLES OF SPIN 1"

phy_math
  • 3,552