# Maxwell-Chern-Simons Theory

1. Oct 12, 2014

### Maybe_Memorie

1. The problem statement, all variables and given/known data

$(a)$ Consider the following action in three space-time dimensions
$S_{CS}=\displaystyle \int d^3x\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho$
It is known as the Chern-Simons action.

Demonstrate that it is gauge-invariant.

Show that any solution to the equations of motion for the above action is gauge-equivalent to a trivial solution.

$(b)$ Find the equations of motion for the following Maxwell-Chern-Simons action:
$S=-\frac{1}{4e^2}\displaystyle \int d^3x F_{\mu\nu}F^{\mu\nu} + \displaystyle \int d^3x\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho$

Show that each component of $F_{\mu\nu}$ satisfies the Klein-Gordon equation.

2. Relevant equations

3. The attempt at a solution

$(a)$ The action is gauge invariant if it is invariant under the transformation $A_\mu \rightarrow A_\mu + \partial_\mu \Lambda$, where $\Lambda$ is a function of the space-time coordinates. Under this transformation, $S_{CS} \rightarrow S_{CS}' = k \displaystyle \int d^3x \epsilon^{\mu\nu\rho} (A_\mu + \partial_\mu \Lambda) \partial_\nu (A_\rho + \partial_\rho \Lambda)$

$= k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}(A_\mu \partial_\nu A_\rho + A_\mu\partial_\nu \partial_rho \Lambda + \partial_\mu \Lambda\partial_\nu A_\rho + \partial_\mu \Lambda \partial_\nu \partial_\rho \Lambda)$

$=S_{CS} + k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}(A_\mu\partial_\nu \partial_rho \Lambda + \partial_\mu \Lambda\partial_\nu A_\rho + \partial_\mu \Lambda \partial_\nu \partial_\rho \Lambda)$

$(*)$ Note that $\epsilon^{\mu\nu\rho}A_\mu\partial_\nu \partial_\rho \Lambda = - \epsilon^{\mu\rho\nu}A_\mu\partial_\nu \partial_\rho \Lambda$ by anti-symmetry of $\epsilon$
$=- \epsilon^{\mu\nu\rho}A_\mu\partial_\rho \partial_\nu \Lambda$ after renaming indicies
$= - \epsilon^{\mu\nu\rho}A_\mu\partial_\nu \partial_\rho \Lambda$ due to the symmetry $\partial_\nu \partial_\rho=\partial_\rho \partial_\nu$
$=0$.

Same argument holds for the last term in the integrand and we are left with $S_{CS}'=S_{CS} + k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}\partial_\mu \Lambda \partial_\nu A_\rho$

Applying the product rule backwards and using the same argument as above in $(*)$ we find that $\displaystyle \int d^3x \epsilon^{\mu\nu\rho}\partial_\mu \Lambda \partial_\nu A_\rho = \displaystyle \int d^3x \partial_\mu (\epsilon^{\mu\nu\rho} \Lambda \partial_\nu A_\rho)$

Applying the divergence theorem this becomes $\displaystyle \int dS n_\mu \epsilon^{\mu\nu\rho} \Lambda \partial_\nu A_\rho$, where $n_\mu$ is a normal vector pointing outward from the surface, which is a surface integral equal to $0$ under the assumption that the fields fall off sufficiently fast at infinity. Hence the action is gauge-invariant.

The Lagrangian for this action is $\mathcal{L}_{CS}=\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho$. I'm not going to write out the full procedure for getting the equations of motion, but we find that they are
$k\epsilon^{\rho\mu\nu}F_{\mu\nu}=0$

I am unsure what the phrase "gauge equivalent to a trivial solution" means. I know that the equations of motion are gauge-invariant so nothing will change, but I don't really know what it's asking me to do.

$(b)$ Again, not going through finding the e.o.m, but we find that they are $\partial_\mu F^{\mu\nu} + ke^2\epsilon^{\nu\alpha\beta}F_{\alpha\beta}=0$

The Klein-Gordon equation is $(\partial_\mu \partial^\mu + m^2)\phi=0$. I was reading a paper on this theory and it said this can be done by writing $\tilde{F^\mu}=\epsilon^{\mu\nu\rho}F_{\nu\rho}$ and then the e.o.m. can be written as $(\partial^\nu \partial_\nu + (ke^2)^2)\tilde{F^\mu}=0$. I have no idea why this is justified.

Any help on the two issues above would be very appreciated. Thank you.

2. Oct 13, 2014

### Maybe_Memorie

Bump

3. Oct 20, 2014

### Maybe_Memorie

Bump

4. Oct 25, 2014

### Maybe_Memorie

Bump

5. Oct 27, 2014

### Maybe_Memorie

Bump, but I'm guessing there's no point. This is the last time I'll do it.