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Write $\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}$ as a total divergence $\partial_\mu G^\mu$

+ 3 like - 0 dislike
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I have the following homework problem in theoretical electrodynamics:

Show that the gauge invariant Lagrange density $\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}$ can be written as a total divergence of a four-vector.

This total divergence should be something like $\partial_\mu G^\mu$, right?

With $$ \hat F^{\mu\nu} = \frac 12 \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} $$ I figured that I write this Lagrange density (I'll refer to it as $L_2$) could be written as: $$ L_2 = 2 \hat F^{\mu\nu} F_{\mu\nu} $$

I simplified this by using the matrix representations of each $F$ and got it down to: $$ L_2 = -\frac 4c B_i E^i $$

$B$ and $E$ can be expressed in Terms of $A$ like so: $$ B_i = \epsilon_{ijk} \partial_j A^j ,\quad E_i = -\partial_i A^0 -\partial_0 A^i $$

How would I continue to find that four-vector $G$?

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user queueoverflow
asked Jan 20, 2013 in Theoretical Physics by queueoverflow (25 points) [ no revision ]
retagged Mar 22, 2014

Just a short comment. You can also write this lagrangian density as total divergence even for non-Abelian gauge fields. 

2 Answers

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If you write $$F_{\mu\nu} = \partial_{[\mu}A_{\nu ]} $$ Then your Lagrange density is $${ \mathcal{L}} = \epsilon^{\mu\nu\alpha\beta}\partial_{[\mu}A_{\nu ]}\partial_{[\alpha}A_{\beta ]}$$

Now we want a vector whose divergence is ${ \mathcal{L}}$. The $\partial_{\mu}$ looks promising, so we ask if we can bring that outside so it acts on the remaining vector, i.e. $${ \mathcal{L}} = \partial_{\mu}(\epsilon^{\mu\nu\alpha\beta}A_{\nu }\partial_{[\alpha}A_{\beta ]}) \ \ (1)$$ We don't need to worry that we've lost the antisymmetrization brackets on $\mu$ and $\nu$ because the epsilon symbol forces this.

As you pointed out, this isn't quite what we started with because we have an additional term of the form $$(\epsilon^{\mu\nu\alpha\beta}A_{\nu }\partial_{\mu}\partial_{[\alpha}A_{\beta ]}) $$ However the total antisymmetry of the epsilon symbol means we can treat the $\mu$ $\alpha$ $\beta$ contribution as $$\partial_{[\mu}F_{\alpha\beta]} $$ which vanishes due to the Maxwell equations. Hence the anzatz (1) holds. (The vector whose divergence we take looks like the abelian Chern Simons current).

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user twistor59
answered Jan 20, 2013 by twistor59 (2,490 points) [ no revision ]
Most voted comments show all comments
You mean $\partial_\mu G^\mu = \partial_{[\mu} \epsilon^{\mu\nu\alpha\beta} A_{\nu]} \partial_{[\alpha} A_{\beta]}$ and that is it? But then that $\partial_\mu$ acts onto the $A_\beta$ as well. Don't I need to write $\epsilon^{\mu\nu\alpha\beta} (\partial_{[\mu} A_{\nu]}) (\partial_{[\alpha}A_{\beta]})$ which then prevents this pulling out?

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user queueoverflow
If you write $G^{\mu} = \epsilon^{\mu\nu\alpha\beta}A_{\nu}\partial_{[\alpha}A_{\beta]}$ then $\partial_{\mu}G^{\mu}$ is what you want. The epsilon automatically antisymmetrizes $\mu$ and $\nu$ for you.

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user twistor59
In your definition of $F$, the $\partial$ only acts on $A$. But with that $G$, the $\partial_\mu$ will act on both $A_\nu$ and $A\beta$ with the product rule. Or does some (anti-)symmetry remove that part?

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user queueoverflow
Right, so you will get some terms like $\partial_{[\mu}F_{\alpha]\beta}$, but the epsilon will fully antisymmetrize over $\mu$, $\alpha$ and $\beta$, and this vanishes by Maxwell equations.

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user twistor59
That last point ought to be incorporated in your answer.

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user Emilio Pisanty
@EmilioPisanty OK as it was a homework question I was trying to keep the answer as lean as possible, but now it's pretty much there I should maybe beef it up a bit...

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user twistor59
Okay, this is where I have to use the Bianchi-Identity given on the problem set. That makes totally sense now!

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user queueoverflow
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Same answer as the one given by twistor59, but written in a differential form language. By definition, $F=dA$, so

$d(A \wedge F)=dA \wedge F + A \wedge dF= F \wedge F$

where we used $dF=0$ (because $d^2=0$).

Same thing in the non-abelian case, as suggested in a comment by ruifeng14. By definition, $F=dA + A \wedge A$, so

$d(Tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A) = Tr(dA \wedge dA + 2 dA \wedge A \wedge A )$

$=Tr(F \wedge F)$

where we used the cyclicity of $Tr$ to rearrange the terms like $Tr(A \wedge dA \wedge A)$ and to obtain the vanishing $Tr(A \wedge A \wedge A \wedge A)=0$.

The expression $Tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A) = Tr(A \wedge F -\frac{1}{3} A \wedge A \wedge A)$, which reduces to $A \wedge dA= A \wedge F$ is the abelian case, is called the Chern-Simons Lagrangian.

answered Jun 11 by 40227 (4,175 points) [ revision history ]

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