I Derivative consisting Levi-Civita

[vahdaneh]

I've got here so far, but first of all I'm not sure if i did it right till the last line and second, if I've been right, i do not know what to do with the rest.

should i consider each of levi-civita parentheses in the last line zero?

and one additional question about the term in the first line parentheses, i wrote it to replace E.B, are the indices in their right place? i mean, should both of them be down indices?

i'll be really grateful

 

[haushofer]

View attachment 253290
I've got here so far, but first of all I'm not sure if i did it right till the last line and second, if I've been right, i do not know what to do with the rest.

should i consider each of levi-civita parentheses in the last line zero?

and one additional question about the term in the first line parentheses, i wrote it to replace E.B, are the indices in their right place? i mean, should both of them be down indices?

i'll be really grateful

I'd write

$\epsilon^{\alpha\beta\gamma\sigma} F_{\alpha\beta}F_{\gamma\sigma} = \epsilon^{\alpha\beta\gamma\sigma} \partial_{\alpha}A_{\beta} \partial_{\gamma}A_{\sigma}$ because the contraction with epsilon already takes into account the antisymmetrization. Doing the differentiation then leads to two terms,

$\epsilon^{\alpha\beta\gamma\sigma} \delta_{\alpha}^{\mu} \delta^{\theta}_{\beta} \partial_{\gamma}A_{\sigma} + \epsilon^{\alpha\beta\gamma\sigma} \partial_{\alpha}A_{\beta} \delta_{\gamma}^{\mu} \delta^{\theta}_{\beta}$

From there on you can simplify.

 

[Orodruin]

I'd write

$\epsilon^{\alpha\beta\gamma\sigma} F_{\alpha\beta}F_{\gamma\sigma} = \epsilon^{\alpha\beta\gamma\sigma} \partial_{\alpha}A_{\beta} \partial_{\gamma}A_{\sigma}$

Just to mention that there should be a factor of 4 here since $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ leads to two terms per insertion of $F$.

 

[vahdaneh]

I'd write

$\epsilon^{\alpha\beta\gamma\sigma} F_{\alpha\beta}F_{\gamma\sigma} = \epsilon^{\alpha\beta\gamma\sigma} \partial_{\alpha}A_{\beta} \partial_{\gamma}A_{\sigma}$ because the contraction with epsilon already takes into account the antisymmetrization. Doing the differentiation then leads to two terms,

$\epsilon^{\alpha\beta\gamma\sigma} \delta_{\alpha}^{\mu} \delta^{\theta}_{\beta} \partial_{\gamma}A_{\sigma} + \epsilon^{\alpha\beta\gamma\sigma} \partial_{\alpha}A_{\beta} \delta_{\gamma}^{\mu} \delta^{\theta}_{\beta}$

From there on you can simplify.

yet i do not know in what other forms i can use the last line...

 

[samalkhaiat]

\frac{\partial}{\partial (\partial_{\alpha}A_{\beta})} \left( \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} \right) = 2 \frac{\partial}{\partial F_{\alpha \beta}} \left( \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} \right) ,
\frac{\partial}{\partial (\partial_{\alpha}A_{\beta})} \left( \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} \right) = 2 \epsilon^{\mu\nu\rho\sigma} \left( \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu} F_{\rho\sigma} + \delta^{\alpha}_{\rho} \delta^{\beta}_{\sigma} F_{\mu\nu}\right) = 4 \epsilon^{\alpha\beta\mu\nu}F_{\mu\nu}.

 

[samalkhaiat]

View attachment 253294
yet i do not know in what other forms i can use the last line...

In the last line use \epsilon^{\alpha\beta\mu\theta} = \epsilon^{\mu\theta\alpha\beta} , then use \epsilon^{\mu\theta\gamma\sigma}\partial_{\gamma}A_{\sigma} = \epsilon^{\mu\theta\beta\alpha}\partial_{\beta}A_{\alpha} = - \epsilon^{\mu\theta\alpha\beta}\partial_{\beta}A_{\alpha} .

 

[haushofer]

Just to mention that there should be a factor of 4 here since $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ leads to two terms per insertion of $F$.

Ah, yes, those pesky numerical factors became a bit rusty since I left academia :P Thanks!