Gauge Invariant terms of Lagrangian for Electromagnetism

Question

Besides the usual EM Lagrangian $\mathcal{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$, we can add an additional term $\mathcal{L'} = \epsilon_{\mu \nu \rho \sigma }F^{\mu \nu}F^{\rho \sigma} = -8 \vec{E} \cdot \vec{B} $. Then, I want to show that adding $\mathcal{L'}$ does not affect the Maxwell's equations. In order to prove it, I wanna show that $ \int d^{4}x \mathcal{L'} = C$, where $C$ is a constant. If $ \int d^{4}x \mathcal{L'} = C$ by least action principle, the Maxwell equations are unchanged. By Attempt is as follow:

$\textbf{Attempt}$: \begin{equation} \int d^{4}x \mathcal{L'} = -8 \int d^{4}x \vec{E} \cdot \vec{B} \end{equation} Recall that $\vec{B} = \nabla \times \vec{A}$, where $\vec{A}$ is the vector potential. In order to evaluate the integration, we use the following identity: \begin{equation} \nabla \cdot ( F \times G) = G\cdot (\nabla \times F) - F \cdot(\nabla \times G) \end{equation} In here, we replace $F \rightarrow E$ and $G \rightarrow A$, we can obtain the following: \begin{equation} \nabla \cdot( \vec{E} \times \vec{A}) = \vec{A} \cdot ( \nabla \times \vec{E}) - E \cdot (\nabla \times \vec{A}) \end{equation} By applying the integration on both sides: \begin{equation} \int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\Omega} d^{4}x \Big( \vec{A} \cdot ( \nabla \times \vec{E}) \Big) - \int_{\Omega} d^{4}x \Big( E \cdot (\nabla \times \vec{A}) \Big) \end{equation} By Divergence theorem, we can reduce the left-handed side: \begin{equation} \int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\partial \Omega} dS ~\hat{n} \cdot ( \vec{E} \times \vec{A}) \end{equation} in here, I assume that the boundary $\partial \Omega$ is very far away from the origin and both $\vec{E} =\vec{A} = 0 $ at the boundary. Besides, $\nabla \times \vec{E} = 0$ therefore $\int_{\Omega} d^{4}x \Big( E \cdot (\nabla \times \vec{A}) \Big) = 0 \rightarrow \int d^{4}x \mathcal{L'} = 0 $. Hence, the Maxwell equations are unchanged. This is my attempted proof. However, I am not sure whether I evaluate the integration $\int_{\partial \Omega} dS ~\hat{n} \cdot ( \vec{E} \times \vec{A}) $ and $\nabla \times \vec{E} =0$ are correct. Could anyone point out that whether my proof is correct? I appreciate any comment.

Answer 1

You start approaching the problem by the right way but at an intermediate step what you missed was to express the electric field $\,\mathbf E\,$ in terms of the potentials \begin{equation} A^0\boldsymbol=\phi\,, \quad A^1\boldsymbol=A_{\rm x}\,, \quad A^2\boldsymbol=A_{\rm y}\,, \quad A^3\boldsymbol=A_{\rm z} \tag{01}\label{01} \end{equation} since these are the $''$general coordinates$''$.

The problem is if this Lorentz invariant scalar of the electromagnetic field \begin{equation} \boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B \tag{02}\label{02} \end{equation} could be expressed as the 4-divergence of a 4-dimensional vector so its addition to the well-known Lagrangian density doesn't affect the Equations Of Motion, that is the Maxwell equations. Expressing the fields in terms of the potentials \begin{equation} \boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\left(\boldsymbol\nabla\phi\boldsymbol+\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right) \tag{03}\label{03} \end{equation} our target would be to check if it's possible to express it as follows \begin{equation} \boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\left(\boldsymbol\nabla\phi\boldsymbol+\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial h^0 }{\partial t}\boldsymbol+\boldsymbol{\nabla\cdot}\mathbf h \tag{04}\label{04} \end{equation} where $\,h^0\,$ and $\,\mathbf h\,$ scalar and 3-vector functions of the potentials respectively.

Now, we could verify that the following 4-dimensional vector \begin{equation} \begin{split} h^0 & \boldsymbol= \tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\\ \mathbf h & \boldsymbol=\phi\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol+\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right) \end{split} \tag{05}\label{05} \end{equation} satisfies equation \eqref{04}.

The following vector formulas would be useful \begin{equation} \begin{split} \boldsymbol{\nabla\cdot}\left(\psi\,\mathbf a\right) & \boldsymbol= \mathbf a\boldsymbol\cdot\boldsymbol\nabla \psi\boldsymbol+\psi\boldsymbol{\nabla\cdot}\mathbf a\\ \boldsymbol{\nabla\cdot}\left(\mathbf a\boldsymbol\times\mathbf b\right) & \boldsymbol= \mathbf b\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf a\right)\boldsymbol-\mathbf a\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf b\right)\\ \end{split} \tag{06}\label{06} \end{equation}

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ADDENDUM A :

$\texttt{ Derivation of the 4-dimensional vector }$ \eqref{05} $\texttt{ that satisfies equation }$ \eqref{04}

From equation \eqref{03} \begin{equation} \boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\underbrace{\boldsymbol\nabla\phi\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}}_{\boxed{\,1\,}}\boldsymbol+\underbrace{\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)}_{\boxed{\,2\,}} \tag{A-01}\label{A-01} \end{equation} Because of the first vector formula \eqref{06} and due to the fact that $\,\boldsymbol{\nabla\cdot}\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=0\,$ we have \begin{equation} \boxed{1\,}\boldsymbol=\boldsymbol\nabla\phi\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\boldsymbol=\boldsymbol\nabla\boldsymbol\cdot\bigl[\phi \left(\boldsymbol{\nabla\times}\mathbf A\right)\bigr] \tag{A-02}\label{A-02} \end{equation} so expressing the term $\,\boxed{1\,}\,$ as the divergence of a 3-vector function of the potentials.

The second vector formula \eqref{06} with $\,\mathbf a \boldsymbol=\mathbf A\,$ and $\,\mathbf b \boldsymbol=\partial\mathbf A /\partial t\,$ yields \begin{equation} \boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\mathbf A\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol+\boldsymbol{\nabla\cdot} \left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-03}\label{A-03} \end{equation} But from the differentiation of a product \begin{equation} \boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol-\mathbf A\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-04}\label{A-04} \end{equation} Adding equations \eqref{A-03} and \eqref{A-04} side by side yields \begin{equation} 2\cdot\boxed{2\,}\boldsymbol=2\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol+\boldsymbol{\nabla\cdot} \left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-05}\label{A-05} \end{equation} so \begin{equation} \boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol+\boldsymbol{\nabla\cdot} \left[\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right)\right] \tag{A-06}\label{A-06} \end{equation}

Finally adding equations \eqref{A-02} and \eqref{A-06} we have \begin{equation} \begin{split} \boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B & \boldsymbol=\boxed{1\,}\boldsymbol+\boxed{2\,}\\ & \boldsymbol=\dfrac{\partial}{\partial t}\underbrace{\left[\tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]}_{h^0}\boldsymbol+\boldsymbol{\nabla\cdot} \underbrace{\left[\phi\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol+\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right)\right]}_{\mathbf h} \\ \end{split} \tag{A-07}\label{A-07} \end{equation} qed.

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ADDENDUM B :

$\texttt{ The unaffected Euler-Lagrange (Maxwell's) Equations}$

That the addition of the term $\,\boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\,$ doesn't affect the Equations Of Motion, that is the Maxwell equations, could be proved directly in these Euler-Lagrange Equations.

So let the contravariant space-time position 4-vector, the contravariant potential 4-vector and the covariant 4-gradient \begin{equation} \begin{split} x^0 & \boldsymbol=t\,, \quad x^1\boldsymbol=\mathrm x\,, \quad x^2\boldsymbol=\mathrm y\,, \quad x^3\boldsymbol=\mathrm z\\ A^0 & \boldsymbol=\phi\,, \quad A^1\boldsymbol=A_{\rm x}\,, \quad A^2\boldsymbol=A_{\rm y}\,, \quad A^3\boldsymbol=A_{\rm z}\\ \partial_k & \boldsymbol\equiv\dfrac{\partial}{\partial x^k}\,, \qquad k=0,1,2,3 \end{split} \tag{B-01}\label{B-01} \end{equation} Mixing tensor calculus and 3-vector notations consider the Lagrangian \begin{equation} \mathcal L'\left(A^\mu, \partial_k A^\mu\right) \boldsymbol=\boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\left(\boldsymbol\nabla\phi\boldsymbol+\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right) \tag{B-02}\label{B-02} \end{equation} Using the summation convention for $k\boldsymbol=0,1,2,3$ we have the following 4 scalar Euler-Lagrange Equations \begin{equation} \begin{split} &\partial_k\left[\dfrac{\partial \mathcal L' }{\partial \left( \partial_k A^\mu\right)}\right]\boldsymbol- \frac{\partial \mathcal L'}{\partial A^\mu}\boldsymbol=0\:, \quad \mu\boldsymbol=0,1,2,3 \\ &^{\left(k\boldsymbol=0,1,2,3\right)}\\ \end{split} \tag{B-03}\label{B-03} \end{equation} Separating the time from the space partial derivatives and using the summation convention for the space indices $\nu\boldsymbol=1,2,3$ above equations are expressed more explicitly as follows \begin{equation} \begin{split} \frac{\partial }{\partial t}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial t}\right)}\right]\boldsymbol+&\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial x^\nu}\right)}\right]\boldsymbol- \frac{\partial \mathcal L'}{\partial A^\mu}\boldsymbol=0\:, \quad \mu\boldsymbol=0,1,2,3\\ &^{\left(\nu\boldsymbol=1,2,3\right)}\\ \end{split} \tag{B-04}\label{B-04} \end{equation} For $\mu\boldsymbol=0$, that is for the equation with respect to the scalar potential $\,A^0\boldsymbol=\phi$ \begin{equation} \begin{split} \frac{\partial }{\partial t}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^0}{\partial t}\right)}\right]\boldsymbol+&\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^0}{\partial x^\nu}\right)}\right]- \frac{\partial \mathcal L'}{\partial A^0}\boldsymbol=0\\ &^{\left(\nu\boldsymbol=1,2,3\right)}\\ \end{split} \tag{B-05}\label{B-05} \end{equation} or \begin{equation} \frac{\partial }{\partial t}\underbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial \phi}{\partial t}\right)}\right]}_{0}\boldsymbol+\underbrace{\boldsymbol\nabla\boldsymbol\cdot\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\boldsymbol\nabla\phi\right)}\right]}^{\boldsymbol{\nabla\times}\mathbf A}\vphantom{\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial \phi}{\partial t}\right)}}}_{0}\boldsymbol-\underbrace{\frac{\partial \mathcal L'}{\partial \phi}\vphantom{\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial \phi}{\partial t}\right)}}}_{0}\boldsymbol=0 \tag{B-06}\label{B-06} \end{equation} that is a null equation $0\boldsymbol=0$.

The same holds also for the 3 equations with respect to the components of the vector potential $\,\left(A^1,A^2,A^3\right)\boldsymbol=\mathbf A$ \begin{equation} \underbrace{\frac{\partial }{\partial t}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial t}\right)}\right]}^{\left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}\boldsymbol+\overbrace{\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial x_\nu}\right)}\right]}^{\boldsymbol-\tfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}{\partial t}}}_{0}\boldsymbol-\underbrace{\frac{\partial \mathcal L'}{\partial A^\mu}\vphantom{\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial t}\right)}}}_{0}\boldsymbol=0\:, \quad \mu\boldsymbol=1,2,3 \tag{B-07}\label{B-07} \end{equation} where $\,\left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu\,$ the $\mu\boldsymbol-$ component of the 3-vector $\,\left(\boldsymbol{\nabla\times}\mathbf A\right)$.

In above equation the proof that \begin{equation} \begin{split} &\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial x^\nu}\right)}\right]\boldsymbol=\boldsymbol-\dfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}{\partial t}\\ &^{\left(\nu\boldsymbol=1,2,3\right)}\\ \end{split} \tag{B-08}\label{B-08} \end{equation} would be clear if \eqref{B-02} is expressed with all details ^{\begin{equation} \begin{split} \mathcal L'\left(A^\mu, \partial_k A^\mu\right)& \boldsymbol=\boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\left(\boldsymbol\nabla\phi\boldsymbol+\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\\ & \boldsymbol=\left(\dfrac{\partial\phi }{\partial x^1}\boldsymbol+\dfrac{\partial A^1}{\partial t}\right)\left(\dfrac{\partial A^3 }{\partial x^2}\boldsymbol-\dfrac{\partial A^2}{\partial x^3}\right)\boldsymbol+\left(\dfrac{\partial\phi }{\partial x^2}\boldsymbol+\dfrac{\partial A^2}{\partial t}\right)\left(\dfrac{\partial A^1 }{\partial x^3}\boldsymbol-\dfrac{\partial A^3}{\partial x^1}\right)\\ & \hphantom{\boldsymbol=}\boldsymbol+\left(\dfrac{\partial\phi }{\partial x^3}\boldsymbol+\dfrac{\partial A^3}{\partial t}\right)\left(\dfrac{\partial A^2 }{\partial x^1}\boldsymbol-\dfrac{\partial A^1}{\partial x^2}\right)\\ \end{split} \tag{B-09}\label{B-09} \end{equation}} For example ^{\begin{equation} \begin{split} \frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^\nu}\right)}\right]&\boldsymbol=\frac{\partial }{\partial x^1}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^1}\right)}\right]}^{0}\boldsymbol+\frac{\partial }{\partial x^2}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^2}\right)}\right]}^{\boldsymbol-\left(\tfrac{\partial\phi }{\partial x^3}\boldsymbol+\tfrac{\partial A^3}{\partial t}\right)}\boldsymbol+\frac{\partial }{\partial x^3}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^3}\right)}\right]}^{\boldsymbol+\left(\tfrac{\partial\phi }{\partial x^2}\boldsymbol+\tfrac{\partial A^2}{\partial t}\right)}\\ & \boldsymbol=\boldsymbol-\frac{\partial }{\partial x^2}\left(\dfrac{\partial\phi }{\partial x^3}\boldsymbol+\dfrac{\partial A^3}{\partial t}\right)\boldsymbol+\frac{\partial }{\partial x^3}\left(\dfrac{\partial\phi }{\partial x^2}\boldsymbol+\dfrac{\partial A^2}{\partial t}\right)\\ & \boldsymbol=\boldsymbol-\frac{\partial }{\partial t}\left(\dfrac{\partial A^3}{\partial x^2}\boldsymbol-\dfrac{\partial A^2}{\partial x^3}\right)\boldsymbol=\boldsymbol-\dfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^1}{\partial t}\\ \end{split} \tag{B-10}\label{B-10} \end{equation}}

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ADDENDUM C :

$\texttt{ Is the 4-vector }\left(h^0,\mathbf h\right)\texttt{ of equation \eqref{05} a Lorentz 4-vector?}$

The 4-divergence of a Lorentz 4-vector function of space-time coordinates is a Lorentz invariant. The inverse is not valid in general. That is, if the 4-divergence of a 4-vector function of space-time coordinates is a Lorentz invariant this doesn't imply that the 4-vector function is a Lorentz one.

So it would be reasonable to ask if the 4-vector function of \eqref{05} is a Lorentz one \begin{equation} \mathbf H\boldsymbol= \begin{bmatrix} h^0\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \mathbf h\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol= \begin{bmatrix} \tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \phi\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol+\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right)\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{C-01}\label{C-01} \end{equation} I found that the above 4-vector IS NOT a Lorentz 4-vector, but I omit the proof as being out of the main subject of the post.

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Related 1 : Deriving Lagrangian density for electromagnetic field.

Related 2 : Squaring the E&M (Maxwell) field strength tensor.

Answer 2

Your approach works, but as @bolbteppa notes it's easier to work covariantly. Either way, it's a "because boundary terms vanish" argument.

Let's repeatedly use antisymmetries. Since $\partial_\mu(\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma)=\epsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\rho A_\sigma=0$,$$\partial_\mu(\epsilon^{\mu\nu\rho\sigma}A_\nu\partial_\rho A_\sigma)=\epsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu\partial_\rho A_\sigma=\tfrac14\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma},$$so $\int\mathcal{L}^\prime d^4x\propto\int\partial_\mu(\epsilon^{\mu\nu\rho\sigma}A_\nu\partial_\rho A_\sigma) d^4x$ is a boundary term. It's also worth noting how direct work with Euler-Lagrange equations looks:$$\frac{\partial\mathcal{L^\prime}}{\partial\partial_\mu A_\nu}\propto\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma\implies\frac{\partial\mathcal{L^\prime}}{\partial A_\nu}\stackrel{\text{on-shell}}{\propto}\partial_\mu(\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma)=0,$$i.e. the ELE is still $\partial_\mu F^{\mu\nu}=0$.