A Maxwell theory invariant under dual field strength tensor application

Mark99
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Dual field strength tensor and EOM
Hello everybody! I know in classical field theory adding in the Lagrangian density a term of the form Fαβ (*F)αβ (where by * we denote the dual of the field strength tensor) does not change the EOM, since this corresponds to adding a total derivative term to the action. However when computing the EOMs explicitly through ∂μ(∂L/∂∂μAυ)-∂L/∂Aυ=0, I do not find this to be true.
In particular I get ∂(Fαβ (*F)αβ)/∂∂μAν=4(*Fμν), when the result should be zero. I suppose I am not managing the Levi Civita tensor properly, but I do not understand my mistake. Is there someone who can do this derivation explicitly and show it is zero?
Thank you in advance.
 
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You have the additional term
$$L=F_{\alpha \beta} (^*F)^{\alpha \beta} = F_{\alpha \beta} F_{\gamma \delta} \epsilon^{\alpha \beta \gamma \delta}.$$
The variation is
$$\delta L = 2 \delta F_{\alpha \beta} F_{\gamma \delta} \epsilon^{\alpha \beta \gamma \delta} = 8 \partial_{\alpha} \delta A_{\beta} \partial_{\gamma} A_{\delta} \epsilon^{\alpha \beta \gamma \delta}.$$
Thus, integrating by parts
$$\delta S=\int \mathrm{d}^4 x \delta L = -8 \int \mathrm{d}^4 x \delta A_{\beta} \partial_{\alpha} \partial_{\gamma} A_{\delta} \epsilon^{\alpha \beta \gamma \delta} \equiv 0.$$
Thus ##\delta S=0## is identically fulfilled, and that's equivalent for the Euler-Lagrange equations being fullfilled for all ##A_{\mu}##.
 
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vanhees71 said:
You have the additional term
$$L=F_{\alpha \beta} (^*F)^{\alpha \beta} = F_{\alpha \beta} F_{\gamma \delta} \epsilon^{\alpha \beta \gamma \delta}.$$
The variation is
$$\delta L = 2 \delta F_{\alpha \beta} F_{\gamma \delta} \epsilon^{\alpha \beta \gamma \delta} = 8 \partial_{\alpha} \delta A_{\beta} \partial_{\gamma} A_{\delta} \epsilon^{\alpha \beta \gamma \delta}.$$
Thus, integrating by parts
$$\delta S=\int \mathrm{d}^4 x \delta L = -8 \int \mathrm{d}^4 x \delta A_{\beta} \partial_{\alpha} \partial_{\gamma} A_{\delta} \epsilon^{\alpha \beta \gamma \delta} \equiv 0.$$
Thus ##\delta S=0## is identically fulfilled, and that's equivalent for the Euler-Lagrange equations being fullfilled for all ##A_{\mu}##.
Thank you for your answer! I understand that. Is it possibile to get the same result showing that the term ∂(Fαβ (*F)αβ)/∂∂μAν in the equations of motion Is zero? Because I understand why your way Is correct but I do not understand why mine Is not
 
You can write it as
$$L=4 (\partial_{\alpha} A_{\beta})(\partial_{\gamma} A_{\delta}) \epsilon^{\alpha \beta \gamma \delta}.$$
Then
$$\frac{\partial L}{\partial (\partial_{\mu} A_{\nu})} =8 \delta_{\mu \alpha} \delta_{\nu \beta} (\partial_{\gamma} A_{\delta}) \epsilon^{\alpha \beta \gamma \delta}= 8 (\partial_{\gamma} A_{\delta}) \epsilon^{\mu \nu \gamma \delta}.$$
Then contracting with ##\partial_{\mu}## gives
$$\partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} A_{\nu})} = 8 \partial_{\mu} \partial_{\gamma} A_{\delta} \epsilon^{\mu \nu \gamma \delta}=0.$$
Since ##\partial L/\partial A^{\mu}=0##, that shows that the Euler-Lagrange equations are identically fulfilled, i.e., this term in the Lagrangian doesn't contribute to the equations of motion.
 
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