Topological Mass in Maxwell-Chern-Simons theory [closed]

Starting from the Maxwell-Chern-Simons lagrangian (in 2+1 dimensions):

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

I've derived equations of motion for the gauge field $A_\mu$.

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

In various texts on Chern-Simons theory [e.g. Dunne's notes] it is stated, that by using the 2+1 dimensional Hodge dual $\widetilde{F}^\mu=\frac{1}{2}\epsilon^{ \nu \alpha \beta}F_{\alpha \beta}$, these equations of motion can be rewritten as a massive Klein-Gordon equation

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

In these variables, we can now see that the massless vector field $A_\mu$ becomes massive due to the presence of the Chern-Simons term.

However, I cannot do this change of variables 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 $F_{\mu\nu}$.)

• 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). – ACuriousMind Jan 16 '15 at 18:40
• @ACuriousMind Thx for response, I've got it. :) – Oiale Jan 17 '15 at 11:15