A Maxwell theory invariant under dual field strength tensor application

[Mark99]

TL;DR

Dual field strength tensor and EOM

Hello everybody! I know in classical field theory adding in the Lagrangian density a term of the form F_αβ (*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/∂∂_μA_υ)-∂L/∂A_υ=0, I do not find this to be true.
In particular I get ∂(F_αβ (*F)^αβ)/∂∂_μA_ν=4(*F^μν), 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.

 

[vanhees71]

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 $A_{\mu}$.

 

[Mark99]

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

 

[vanhees71]

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.