# E-M Duality

1. Nov 29, 2014

### PhyAmateur

(This question is only long because I tried to give all the details (necessary and not) so that you don't refer to the paper every now and then)

In this paper http://arxiv.org/pdf/hep-th/9705122.pdf

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)?

Thank you for taking the time to read this.

2. Nov 30, 2014

### samalkhaiat

Let us start from the first order action
$$S [ F , B ] = \int d^{ 4 } x \left( \frac{ 1 }{ 4 g^{ 2 } } F^{ 2 } + a B_{ \sigma } \partial_{ \rho } *F^{ \rho \sigma } \right) .$$
Before varying with respect to $F_{ \mu \nu }$, we first integrate the second term by part to obtain:
$$S [ F , B ] = \int d^{ 4 } x \left( \frac{ 1 }{ 4 g^{ 2 } } F^{ 2 } - a ( \partial_{ \rho } B_{ \sigma } ) *F^{ \rho \sigma } \right) , \ \ \ \ (1)$$
and then substitute for
$$*F^{ \rho \sigma } = \frac{ 1 }{ 2 } \epsilon^{ \rho \sigma \mu \nu } \ F_{ \mu \nu } .$$
So, $\delta_{ F } S [ F , B ] = 0$, leads to
$$\int d^{ 4 } x \left( \frac{ 1 }{ 2 g^{ 2 } } F^{ \mu \nu } - \frac{ a }{ 2 } ( \partial_{ \rho } B_{ \sigma } ) \epsilon^{ \rho \sigma \mu \nu } \right) \delta F_{ \mu \nu } = 0 .$$
From this it follows that
$$F^{ \mu \nu } = a g^{ 2 } ( \partial_{ \rho } B_{ \sigma } ) \ \epsilon^{ \rho \sigma \mu \nu } . \ \ \ \ \ (2)$$
We can rewrite this as
$$F^{ \mu \nu } = a g^{ 2 } ( \frac{ 1 }{ 2 } ) \ \left( \partial_{ \rho } B_{ \sigma } - \partial_{ \sigma } B_{ \rho } \right) \epsilon^{ \rho \sigma \mu \nu } . \ \ \ (2a)$$
Introducing the field tensor $G_{ \rho \sigma } \equiv \partial_{ \rho } B_{ \sigma } - \partial_{ \sigma } B_{ \rho }$ and its dual $*G^{ \mu \nu } \equiv ( 1 / 2 ) G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu }$ into Eq(2a), we find
$$F^{ \mu \nu } = \frac{ a \ g^{ 2 } }{ 2 } \ G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu } = a g^{ 2 } \ *G^{ \mu \nu } .$$
For later use, let us lower the indices of the field F in the equation of motion Eq(2)
$$F_{ \mu \nu } = a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \epsilon^{ \rho \sigma \alpha \beta } \ \eta_{ \mu \alpha } \ \eta_{ \nu \beta } . \ \ \ \ (2b)$$
Now, contracting Eq(2) with $F_{ \mu \nu }$ gives
$$F^{ 2 } = 2 \ a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ ( \frac{ 1 }{ 2 } \epsilon^{ \rho \sigma \mu \nu } \ F_{ \mu \nu } ) = 2 \ a \ g^{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } .$$
Thus
$$\frac{ 1 }{ 4 \ g^{ 2 } } \ F^{ 2 } = \frac{ a }{ 2 } \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } . \ \ \ \ (3)$$
Now, inserting Eq(3) in Eq(1) gives
$$S [ F , B ] = - \frac{ a }{ 2 } \int d^{ 4 } x \ ( \partial_{ \rho } B_{ \sigma } ) \ \ *F^{ \rho \sigma } = - \frac{ a }{ 4 } \int d^{ 4 } x \left( \partial_{ \rho } B_{ \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \ F_{ \mu \nu } .$$
Using the equation of motion Eq(2b), we find (the dual theory action)
$$S [ B ] = - \frac{ a^{ 2 } g^{ 2 } }{ 4 } \int d^{ 4 } x \left( \partial_{ \rho } B_{ \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \left( \partial_{ \lambda } B_{ \tau } \ \epsilon^{ \lambda \tau \alpha \beta } \right) \eta_{ \mu \alpha } \ \eta_{ \nu \beta } .$$
In terms of the field tensor $G_{ \mu \nu }(B)$, the above action becomes
$$S [ B ] = - \frac{ a^{ 2 } g^{ 2 } }{ 4 } \int d^{ 4 } x \left( \frac{ 1 }{ 2 } G_{ \rho \sigma } \ \epsilon^{ \rho \sigma \mu \nu } \right) \left( \frac{ 1 }{ 2 } G_{ \lambda \tau } \ \epsilon^{ \lambda \tau \alpha \beta } \right) \ \eta_{ \mu \alpha } \ \eta_{ \nu \beta } .$$
Or
$$S [ B ] = - ( \frac{ a \ g }{ 4 } )^{ 2 } \int d^{ 4 } x \left( \epsilon^{ \rho \sigma \mu \nu } \ \epsilon^{ \lambda \tau }{}_{ \mu \nu } \right) G_{ \rho \sigma } \ G_{ \lambda \tau } .$$
Using the identity
$$\epsilon^{ \rho \sigma \mu \nu } \ \epsilon^{ \lambda \tau }{}_{ \mu \nu } = ( - 2 ) ( \eta^{ \rho \lambda } \ \eta^{ \sigma \tau } - \eta^{ \rho \tau } \ \eta^{ \sigma \lambda } ) ,$$
we find
$$S [ B ] = - ( \frac{ a \ g }{ 4 } )^{ 2 } \ ( - 2 ) \int d^{ 4 } x \ 2 \ G_{ \rho \sigma } \ G^{ \rho \sigma } .$$
Or
$$S [ B ] = \frac{ g^{ 2 } }{ 4 } \int d^{ 4 } x \ ( a G_{ \mu \nu } ) \ ( a G^{ \mu \nu } ) .$$
Finally, by rescaling the field $a G_{ \mu \nu } ( B ) \equiv \mathcal{ G }_{ \mu \nu } ( B )$, we obtain the action of the dual theory
$$S [ B ] = \frac{ g^{ 2 } }{ 4 } \int d^{ 4 } x \ \mathcal{ G }^{ 2 } ( B ) .$$

3. Dec 2, 2014

### PhyAmateur

That was brilliant explanation. Thank you a lot!

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