7
$\begingroup$

An electromagnetic duality is a duality that maps electric to magnetic degrees of freedom of two distinct theories. Apart from source-less Maxwell electrodynamics, other theories require magnetic monopoles. I am not an expert in ($N=1$) Seiberg duality but as far as I know there is no magnetic charges in it. So why is it called an electromagnetic duality? Where are the "magnetic" degree of freedom of the theories? My guess is just because it is also an S-duality such as the Montonen-Olive or GNO dualities. Yet their are much different. The EM duality in the sense of Montonen-Olive is exact in the sense that it is (conjectured) valid at any scale. The Seiberg duality however is a map valid only at particular regimes. The Seiberg duality maps the IR regime of the "electric" theory to an IR free regime of a "magnetic" theory. This map is not valid along the RG flow.

Note: I now that Seiberg assume magnetic monopoles in the theory. But how do these monopoles appear? If they are topological solutions, where is the spontaneous symmetry breaking pattern? Where is the homotopy condition for stable monopoles? Where are the field confiturations of these solitons? Where is the magnetic charge quantization? What are the topological charge or topological sectors of these monopoles?

$\endgroup$

1 Answer 1

2
$\begingroup$

They're not different at all! Seiberg has used the term electric-magnetic duality in the title of his famous paper, too.

His duality is an electric-magnetic duality because the duality relates two descriptions and objects that are electrically charged under the gauge group on one side (e.g. quarks and gluons) are mapped to solitons (forms of magnetic monopoles) carrying the magnetic monopole charges under the gauge group in the dual description! So just like there is an "$SU(3)$-like" electric field $F_{0i}$ around a quark or gluon in the electric description, the same object looks like a magnetic field $F_{jk}$ in the dual description – and under the other description's gauge group.

Because this duality avoids an immediate contradiction, it must simultaneously be an S-duality, too. It couldn't be a weak-weak duality because the weak-coupling physics of any gauge theory is basically unique. "Easy to construct" excitations (quarks and gluons) may only be mapped to "complicated objects" (solitons) if the coupling is strong on one side.

Seiberg duality is just a generalization of the usual $F_{\mu\nu}\to *F_{\mu\nu}$ electric-magnetic duality of the Maxwell's theory, a generalization with different and more complex gauge groups on both sides, supersymmetry, and some quark matter. In Maxwell's theory, one can arguably "ban" magnetic monopoles but in non-Abelian gauge theories with complicated enough spectrum, they're basically unavoidable because one may construct them as classical solitonic solutions and these objects have to exist in the quantum theory, too.

In $d=4$, electric-magnetic duality and S-duality are basically synonyms. In different spacetime dimensionalities, these two notions become different because the electric-magnetic duality exchanges point-like charges with some branes of a different dimension while an S-duality should preserve the dimensionality (and location) of objects. To allow a more general symmetry that mixes the charges of different dimensions etc., one has to call it a U-duality.

$\endgroup$
1
  • 2
    $\begingroup$ Thanks for your answer but I'm afraid I disagree. There is a fundamental difference between them. The EM duality in the sense of Montonen-Olive is exact in the sense that it is (conjectured) valid at any scale. The Seiberg duality however is a map valid only at particular regimes. It maps the IR regime of the "electric" theory to an IR free regime of a "magnetic" theory. This map is not valid along the RG flow. Also your second paragraph describes an EM duality in the sense of Montonen-Olive. I do not see any solitons in Seiberg duality. That is exactly my point! $\endgroup$
    – Diracology
    Commented May 13, 2016 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.