# Electric Magnetic duality

In this paper http://arxiv.org/abs/hep-th/9705122 Section 2

We have $$S_A = \frac{1}{4g^2} \int{d^4x F_{\mu\nu}(A)F^{\mu\nu}(A)}$$

where $F_{\mu\nu}(A) = \partial_{[\mu A\nu]}$. Its Bianchi Identity is $\partial_\mu *F^{\mu\nu}$ (Note that $*$ represents hodge dual)

Great. Now the author went to parent action:

$$S_{F,\Lambda} = \int{d^4x(\frac{1}{4g^2} F_{\mu\nu} F^{\mu\nu} +a \Lambda_\mu \partial_\nu *F^{\nu\mu}} )$$

He first varied it w.r.t $\Lambda_\mu$ and then w.r.t $F_{\mu\nu}$.

1)He got in the first case, $\partial_\mu *F^{\mu\nu} = 0$ and thus he mentioned that our parent action reduces to $S_A$.

2)He got in the second case, $$\frac{1}{2g^2} F^{\mu\nu} = \frac{a}{2} \partial_\rho \Lambda_\sigma \epsilon^{\rho \sigma \mu \nu}= \frac{a}{2} *G^{\mu\nu}$$ and thus he mentioned that now plugging this back into the action: $$S_{F,\Lambda} \rightarrow S_{\Lambda} = \frac{-g^2a^2}{4} \int{d^4x *G_{\mu\nu} *G^{\mu\nu}}$$ He then said knowing that $*G_{\mu\nu} *G^{\mu\nu}=-2G_{\mu\nu} G^{\mu\nu}$ We obtain perfectly $$S_\Lambda = \frac{g^2}{4} \int{d^4x G_{\mu\nu}(\Lambda)G^{\mu\nu}(\Lambda)}$$

And so this is duality with the coupling constants inversed. Perfect.

My questions:

A) When he said plugging this back into the action above (in italic). He plugged it in the first term of the parent action. What about the second term? Did he throw it away?

B) $\frac{1}{2g^2} F^{\mu\nu} = \frac{a}{2} \partial_\rho \Lambda_\sigma \epsilon^{\rho \sigma \mu \nu}= \frac{a}{2} *G^{\mu\nu}$ Where did this relation come from (The first and the second equality)?

• I should say that I found the article you linked very enlightening. Thanks! – Prahar Mitra Nov 30 '14 at 18:48

We start with the action $$S_{\Lambda,F} = \int d^4 x \left[ \frac{1}{4g^2} F_{\mu\nu} F^{\mu\nu} + a \Lambda_\mu \partial_\nu \ast F^{\mu\nu}\right]$$ This action is equivalent to $S_A$ in the sense that the equation of motion for $\Lambda_\mu$ when plugged into $S_{\Lambda,F}$ gives $S_A$. This is legit since $\Lambda_\mu$ appears linearly and without derivatives in the action.
On the other hand, we can find the equations of motion for $F_{\mu\nu}$. For this, we note $$\ast F^{\mu\nu} = \frac{1}{2} \varepsilon^{\mu\nu\rho\sigma} F_{\rho\sigma}$$ Varying the action, we then get $$\begin{split} \delta S_{\Lambda,F} &= \int d^4 x \left[ \frac{1}{2g^2} \delta F_{\mu\nu} F^{\mu\nu} + a \frac{1}{2} \Lambda_\mu \varepsilon^{\mu\nu\rho\sigma} \partial_\nu \delta F_{\rho\sigma}\right] \\ &= \frac{1}{2} \int d^4 x \left[ \frac{1}{g^2} F^{\rho\sigma} - a \varepsilon^{\mu\nu\rho\sigma}\partial_\nu \Lambda_\mu \right] \delta F_{\rho\sigma} \end{split}$$ Thus, the equations of motion are $$\frac{1}{g^2} F^{\rho\sigma} = a \varepsilon^{\mu\nu\rho\sigma}\partial_\nu \Lambda_\mu = \frac{a}{2} \varepsilon^{\mu\nu\rho\sigma} \left( \partial_\nu \Lambda_\mu - \partial_\mu \Lambda_\nu \right) \\ = -\frac{a}{2} \varepsilon^{\rho\sigma\nu\mu} \left( \partial_\nu \Lambda_\mu - \partial_\mu \Lambda_\nu \right)$$ We now define $$G_{\nu\mu} \equiv \partial_\nu \Lambda_\mu - \partial_\mu \Lambda_\nu$$ Using the definition of hodge star, we find $$F^{\rho\sigma} = - a g^2 \ast G^{\rho\sigma}$$ We can now plug this answer back into the action to get $$\begin{split} S_{\Lambda,F} &= \int d^4 x \left[ \frac{1}{4g^2} (-ag^2)^2 (\ast G)_{\mu\nu} (\ast G)^{\mu\nu} - \frac{1}{2} a (-ag^2) G_{\nu\mu} \ast \ast G^{\mu\nu}\right] \\ &= \int d^4 x \left[ \frac{a^2 g^2}{4} (\ast G)_{\mu\nu} (\ast G)^{\mu\nu} - \frac{1}{2} a^2 g^2 G_{\mu\nu} \ast \ast G^{\mu\nu}\right] \end{split}$$ Now, we use the fact that $$(\ast G)_{\mu\nu} (\ast G)^{\mu\nu} = - 2 G_{\mu\nu} G^{\mu\nu},~~~~ \ast \ast G^{\mu\nu} = - G^{\mu\nu}$$ we find $$S_{\Lambda,F} = \int d^4 x \left[ - \frac{a^2 g^2}{4} G_{\mu\nu} G^{\mu\nu} + \frac{1}{2} a^2 g^2 G_{\mu\nu} G^{\mu\nu}\right] = \int d^4 x \frac{a^2 g^2}{4} G_{\mu\nu} G^{\mu\nu}$$
• Thank you @prahar for your excellent answer, there is just one minus sign I didn't understand where it came from: $$1/g^2 F^{\rho\sigma} =-a/2 \epsilon^{\rho\sigma\nu\mu}(\partial_\nu \Lambda_\mu "-"\partial_\mu\Lambda_\nu)$$ – Fluctuations Nov 30 '14 at 11:29
• We can write $\varepsilon ^{\mu\nu\rho\sigma} M_{\rho\sigma} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma} M_{\rho\sigma} + \frac{1}{2} \varepsilon ^{\mu\nu\sigma\rho} M_{\sigma\rho}$. Here I have split the sum into two pieces and interchanged the summation variables $\sigma$ and $\rho$. Next I use antisymmetry of $\varepsilon$ to write $\varepsilon^{\mu\nu\sigma\rho}=-\varepsilon^{\mu\nu\rho\sigma}$. Putting it altogether, I get $\varepsilon^{\mu\nu\rho\sigma} M_{\rho\sigma}= \varepsilon^{\mu\nu\rho\sigma} (M_{\rho\sigma}-M_{\sigma\rho})$. – Prahar Mitra Nov 30 '14 at 13:57
• Excuse me, now as I was revising this, I noticed in the final answer something. In the original $S_{\Lambda,F}$ there is a $\partial_\nu$ where did this go in your final line when you replaced by $*F^{\mu\nu}$? – Fluctuations Nov 30 '14 at 17:54
• I integrated by parts and write $\Lambda_\mu \partial_\nu \ast F^{\mu\nu} = - \partial_\nu \Lambda_\mu \ast F^{\mu\nu}$. Then I used anti-symmetry of the epsilon tensor as indicated in my earlier comment to get something like $G_{\rho\sigma} \ast F^{\mu\nu}$ – Prahar Mitra Nov 30 '14 at 18:08
• Aha ok. But here there's no epsilon like in the previous case and we were not varying the action so that the term $\Lambda_\mu *F^{\mu\nu}$ (This term is the result of integration by parts) dies out and the one you just mentioned in your last comment stays. Am I mistaken? – Fluctuations Nov 30 '14 at 18:17