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  Equations of motion in Maxwell-Chern-Simons theory

+ 3 like - 0 dislike

I've started with the Maxwell-Chern-Simons lagrangian (in 2+1 dimensions):

$$L_{MCS}=-\frac{1}{4}F^{\mu \nu}F_{\mu\nu}+\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$

From this lagrangian I've derived equations of motion

$$\partial_\mu F^{\mu \nu}+\frac{g}{2}\epsilon^{ \nu \alpha \beta}F_{\alpha \beta}=0$$

I know statement that this equations could be rewrite in terms of $\widetilde{F}^\mu=\frac{1}{2}\epsilon^{ \nu \alpha \beta}F_{\alpha \beta}$ as

$$\left(\partial_\nu \partial^\nu+g^2\right)\widetilde{F}^\mu=0 $$

But I cannot do it explicitly. I've tried a lot of ways but only what I've found is $\partial_\nu \widetilde{F}^\nu=0$ (just differentiated initial equation of motion with respect to $x_\nu$ and used antisymmetric property of $F_{\mu\nu}$.)

This post imported from StackExchange Physics at 2015-01-18 13:43 (UTC), posted by SE-user Oiale

asked Jan 16, 2015 in Theoretical Physics by Oiale (15 points) [ revision history ]
Hint: Observe that your starting equation is $(\partial_\mu \epsilon^{\nu\mu\rho} + g\eta^{\mu\nu})\tilde{F}_\nu = 0$. Apply the operator $(\partial_\mu \epsilon^{\nu\mu\rho} + g\eta^{\mu\nu})$ once more to this to get the desired result ($\eta$ is the metric on your spacetime).

This post imported from StackExchange Physics at 2015-01-18 13:43 (UTC), posted by SE-user ACuriousMind
@ACuriousMind Thx for response, I've got it. :)

This post imported from StackExchange Physics at 2015-01-18 13:43 (UTC), posted by SE-user Oiale

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