# Gauge Invariant terms of Lagrangian for Electromagnetism

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:

$$\textbf{Attempt}$$: $$$$\int d^{4}x \mathcal{L'} = -8 \int d^{4}x \vec{E} \cdot \vec{B}$$$$ Recall that $$\vec{B} = \nabla \times \vec{A}$$, where $$\vec{A}$$ is the vector potential. In order to evaluate the integration, we use the following identity: $$$$\nabla \cdot ( F \times G) = G\cdot (\nabla \times F) - F \cdot(\nabla \times G)$$$$ In here, we replace $$F \rightarrow E$$ and $$G \rightarrow A$$, we can obtain the following: $$$$\nabla \cdot( \vec{E} \times \vec{A}) = \vec{A} \cdot ( \nabla \times \vec{E}) - E \cdot (\nabla \times \vec{A})$$$$ By applying the integration on both sides: $$$$\int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\Omega} d^{4}x \Big( \vec{A} \cdot ( \nabla \times \vec{E}) \Big) - \int_{\Omega} d^{4}x \Big( E \cdot (\nabla \times \vec{A}) \Big)$$$$ By Divergence theorem, we can reduce the left-handed side: $$$$\int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\partial \Omega} dS ~\hat{n} \cdot ( \vec{E} \times \vec{A})$$$$ in here, I assume that the boundary $$\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.

• You can check whether it's right yourself by re-doing it covariantly, it's not only a lot easier (you just need to re-write it as a total derivative up to boundary terms and use the anti-symmetry vs. symmetry of second order partial derivatives and the same boundary conditions at infinity) but you can then check the terms in 3D notation to be sure. Sep 18, 2021 at 13:06
• Thanks for your comment, @bolbteppa. I got your point. Suppose I can transform $\epsilon FF$ to $\partial_{\mu} ( ....)$. By Divergence theorem, then I can evaluate the stuff (....) at boundary and (....) vanishes at infinity. Therefore, the contribution of $\int d^{4}x \mathcal{L'} = 0$. Sep 18, 2021 at 13:12

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 $$\partial_\mu F^{\mu\nu}=0$$.

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 $$$$A^0\boldsymbol=\phi\,, \quad A^1\boldsymbol=A_{\rm x}\,, \quad A^2\boldsymbol=A_{\rm y}\,, \quad A^3\boldsymbol=A_{\rm z} \tag{01}\label{01}$$$$ since these are the $$''$$general coordinates$$''$$.

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

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

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

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

From equation \eqref{03} $$$$\boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\underbrace{\boldsymbol\nabla\phi\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}}_{\boxed{\,1\,}}\boldsymbol+\underbrace{\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)}_{\boxed{\,2\,}} \tag{A-01}\label{A-01}$$$$ Because of the first vector formula \eqref{06} and due to the fact that $$\,\boldsymbol{\nabla\cdot}\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=0\,$$ we have $$$$\boxed{1\,}\boldsymbol=\boldsymbol\nabla\phi\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\boldsymbol=\boldsymbol\nabla\boldsymbol\cdot\bigl[\phi \left(\boldsymbol{\nabla\times}\mathbf A\right)\bigr] \tag{A-02}\label{A-02}$$$$ so expressing the term $$\,\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 $$$$\boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\mathbf A\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol+\boldsymbol{\nabla\cdot} \left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-03}\label{A-03}$$$$ But from the differentiation of a product $$$$\boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol-\mathbf A\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-04}\label{A-04}$$$$ Adding equations \eqref{A-03} and \eqref{A-04} side by side yields $$$$2\cdot\boxed{2\,}\boldsymbol=2\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol+\boldsymbol{\nabla\cdot} \left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right) \tag{A-05}\label{A-05}$$$$ so $$$$\boxed{2\,}\boldsymbol=\dfrac{\partial\mathbf A }{\partial t}\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol=\dfrac{\partial}{\partial t}\left[\tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{\partial\mathbf A }{\partial t}}\right]\boldsymbol+\boldsymbol{\nabla\cdot} \left[\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right)\right] \tag{A-06}\label{A-06}$$$$

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

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

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

The same holds also for the 3 equations with respect to the components of the vector potential $$\,\left(A^1,A^2,A^3\right)\boldsymbol=\mathbf A$$ $$$$\underbrace{\frac{\partial }{\partial t}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial t}\right)}\right]}^{\left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}\boldsymbol+\overbrace{\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial x_\nu}\right)}\right]}^{\boldsymbol-\tfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}{\partial t}}}_{0}\boldsymbol-\underbrace{\frac{\partial \mathcal L'}{\partial A^\mu}\vphantom{\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial t}\right)}}}_{0}\boldsymbol=0\:, \quad \mu\boldsymbol=1,2,3 \tag{B-07}\label{B-07}$$$$ where $$\,\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 $$$$\begin{split} &\frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^\mu}{\partial x^\nu}\right)}\right]\boldsymbol=\boldsymbol-\dfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^\mu}{\partial t}\\ &^{\left(\nu\boldsymbol=1,2,3\right)}\\ \end{split} \tag{B-08}\label{B-08}$$$$ would be clear if \eqref{B-02} is expressed with all details $$$$\begin{split} \mathcal L'\left(A^\mu, \partial_k A^\mu\right)& \boldsymbol=\boldsymbol-\mathbf E\boldsymbol\cdot \mathbf B\boldsymbol=\left(\boldsymbol\nabla\phi\boldsymbol+\dfrac{\partial\mathbf A }{\partial t}\right)\boldsymbol\cdot \left(\boldsymbol{\nabla\times}\mathbf A\right)\\ & \boldsymbol=\left(\dfrac{\partial\phi }{\partial x^1}\boldsymbol+\dfrac{\partial A^1}{\partial t}\right)\left(\dfrac{\partial A^3 }{\partial x^2}\boldsymbol-\dfrac{\partial A^2}{\partial x^3}\right)\boldsymbol+\left(\dfrac{\partial\phi }{\partial x^2}\boldsymbol+\dfrac{\partial A^2}{\partial t}\right)\left(\dfrac{\partial A^1 }{\partial x^3}\boldsymbol-\dfrac{\partial A^3}{\partial x^1}\right)\\ & \hphantom{\boldsymbol=}\boldsymbol+\left(\dfrac{\partial\phi }{\partial x^3}\boldsymbol+\dfrac{\partial A^3}{\partial t}\right)\left(\dfrac{\partial A^2 }{\partial x^1}\boldsymbol-\dfrac{\partial A^1}{\partial x^2}\right)\\ \end{split} \tag{B-09}\label{B-09}$$$$ For example $$$$\begin{split} \frac{\partial }{\partial x^\nu}\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^\nu}\right)}\right]&\boldsymbol=\frac{\partial }{\partial x^1}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^1}\right)}\right]}^{0}\boldsymbol+\frac{\partial }{\partial x^2}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^2}\right)}\right]}^{\boldsymbol-\left(\tfrac{\partial\phi }{\partial x^3}\boldsymbol+\tfrac{\partial A^3}{\partial t}\right)}\boldsymbol+\frac{\partial }{\partial x^3}\overbrace{\left[\frac{\partial \mathcal L'}{\partial \left(\dfrac{\partial A^1}{\partial x^3}\right)}\right]}^{\boldsymbol+\left(\tfrac{\partial\phi }{\partial x^2}\boldsymbol+\tfrac{\partial A^2}{\partial t}\right)}\\ & \boldsymbol=\boldsymbol-\frac{\partial }{\partial x^2}\left(\dfrac{\partial\phi }{\partial x^3}\boldsymbol+\dfrac{\partial A^3}{\partial t}\right)\boldsymbol+\frac{\partial }{\partial x^3}\left(\dfrac{\partial\phi }{\partial x^2}\boldsymbol+\dfrac{\partial A^2}{\partial t}\right)\\ & \boldsymbol=\boldsymbol-\frac{\partial }{\partial t}\left(\dfrac{\partial A^3}{\partial x^2}\boldsymbol-\dfrac{\partial A^2}{\partial x^3}\right)\boldsymbol=\boldsymbol-\dfrac{\partial \left(\boldsymbol{\nabla\times}\mathbf A\right)^1}{\partial t}\\ \end{split} \tag{B-10}\label{B-10}$$$$

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$$\texttt{ Is the 4-vector }\left(h^0,\mathbf h\right)\texttt{ of equation \eqref{05} a Lorentz 4-vector?}$$
So it would be reasonable to ask if the 4-vector function of \eqref{05} is a Lorentz one $$$$\mathbf H\boldsymbol= \begin{bmatrix} h^0\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \mathbf h\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol= \begin{bmatrix} \tfrac{1}{2}\mathbf A\boldsymbol\cdot\left(\boldsymbol{\nabla\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \phi\left(\boldsymbol{\nabla\times}\mathbf A\right)\boldsymbol+\tfrac{1}{2}\left(\mathbf A\boldsymbol\times\dfrac{\partial\mathbf A }{\partial t}\right)\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{C-01}\label{C-01}$$$$ 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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