Unity tensor

1. Jan 11, 2009

Rory9

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

Imagine a spatially 2d world. The electromagnetic field could be richer here, because you could add to the Lagrangian L an additional term (known as the Chern-Simons Lagrangian)

$$L_{CS} = \epsilon_{0}\frac{\kappa}{2}\epsilon^{\alpha \beta \gamma}(\partial_{\alpha}A_{\beta})A_{\gamma}$$​

where

$$\epsilon^{\alpha \beta \gamma}$$

denotes the completely antisymmetric unity tensor in a world with 2 spatial dimensions (and time) and $$\kappa[\tex] is a coupling constant. In order to be suitable as part of the total Lagrangian, [tex]L_{CS}[\tex] must be a Lorentz scalar. Explain why the Chern-Simons expression is indeed a scalar. Why is the action generated by [tex]L_{CS}[\tex] guage-invariant? 3. The attempt at a solution I must admit, I'm rather confused by this one, and haven't done much work with the unity tensor before (I've only just begun playing with tensors, really). I was hoping to learn something by trying this problem, but haven't got anywhere with it yet, and any help would be greatly appreciated. 2. Jan 11, 2009 tim_lou that sounds like an interesting problem. Firstly, it's obvious that L is either a scaler or pseudo-scaler. You just need to rule out that last possibility. Parity transformation maps [tex]u_0 \rightarrow u_0$$
$$u_1 \rightarrow -u_1$$
$$u_2 \rightarrow -u_2$$

it suffices to show that L is invariant under this transformation. As for gauge invariance, this is pretty much the transformation:
$$A_\alpha \rightarrow A_\alpha + \partial_\alpha V$$
for scaler V. If L changes by a divergences under this transformation, you get gauge-invariance.

3. Jan 12, 2009

Rory9

Thank you :-)

Would I be right in thinking that a covariant dot product can be formed in the above (and these are invariant) - another way of looking at it (?)

Regarding the guage invariance, are you proposing simply to substitute that transformation into the above expression and see if I can kill off all the $$\partial_\alpha V$$ terms, somehow (?)

Cheers.