Steps to Solve
:
\item Express four current density vector in covariant form
\item Express four potential vector in covariant form
\item Find equation of continuity in covariant form
\item Find Lorentz condition in covariant form
\item Find D' Alembertian Operator
\item Transformation equation for A
\item Transformation equation of J
\item Transformation equation of four momentum vector
\item Transformation of Electric field Components
\item Transformation of Magnetic field Components
\item Invariance of Maxwell's equation under Lorentz transformation
\item Transformation of Lorentz Force
\item Prove that $ \vec{E}\cdot\vec{B} $ is Lorentz invariant
\item Maxwell equations in terms of Electromagnetic Potentials are covariant under Lorentz transformation
\item Lorentz Condition is Invariant under Lorentz transformation
\item Lorentz Condition in covarinat form is Invariant under Lorentz transformation
\item Prove that $ 2 ( B^2-E^2/c^2) $ is Lorentz invariant
\item Prove that D' Alembertian Operator is Loretnz invariant
\item Electromagnetic field Tensor
\item Maxwells equation in terms of Electromagnetic field tensor
\item Lorentz force in terms of Electromagnetic field Tensor
\item Derive Lagrangian Density
\item Physical Meaning of 4th component of Lorentz force density four vector
\item Derive lagrange Equation for Charge Particle in Electromagnetic field
\item Derive Maxwell equations in terms of Electromagentic tensor and Dual Tensor
\item Derive Maxwell Equations from Euler Lagrange Equations of Charge particle in Electromagentic field
\end{enumerate}
\section{Preliminary four vectors:}
\subsection{Four Position Vector}
\[\begin{align}
& {{x}_{\mu }}=\left( {{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{0}} \right)=\left( x,y,z,ict \right) \\
& {{x}^{\mu }}=\left( {{x}^{0}},{{x}^{1}},{{x}^{2}},{{x}^{3}} \right)=\left( ict,x,y,z \right)\therefore {{x}_{1}}={{x}^{1}}=x,{{x}_{2}}={{x}^{2}}=y,{{x}^{3}}={{x}_{3}}=z,{{x}^{0}}={{x}_{0}}=ict \\
\end{align}\]
\subsection{Four Velocity Vector}
\[\begin{align}
{{u}_{\mu }} &=\left( {{u}_{1}},{{u}_{2}},{{u}_{3}},{{u}_{4}} \right) \\
& =\left( \frac{d{{x}_{1}}}{d\tau },\frac{d{{x}_{2}}}{d\tau },\frac{d{{x}_{3}}}{d\tau },\frac{d{{x}_{4}}}{d\tau } \right) \\
& =\left( \frac{d{{x}_{1}}}{dt}\frac{dt}{d\tau },\frac{d{{x}_{2}}}{dt}\frac{dt}{d\tau },\frac{d{{x}_{3}}}{dt}\frac{dt}{d\tau },\frac{d{{x}_{4}}}{dt}\frac{dt}{d\tau } \right) \\
& =\left( \frac{{{u}_{x}}}{\sqrt{1-\frac{{{u}^{2}}}{{{c}^{2}}}}},\frac{{{u}_{y}}}{\sqrt{1-\frac{{{u}^{2}}}{{{c}^{2}}}}},\frac{{{u}_{z}}}{\sqrt{1-\frac{{{u}^{2}}}{{{c}^{2}}}}},\frac{ic}{\sqrt{1-\frac{{{u}^{2}}}{{{c}^{2}}}}} \right) \\
\end{align}\]
\subsection{Four Momentum Vector}
\[\begin{align}
{{P}_{\mu }} &=\left( {{P}_{1}},{{P}_{2}},{{P}_{3}},{{P}_{4}} \right) \\
& =\left( {{P}_{x}},{{P}_{y}},{{P}_{z}},\frac{{{m}_{0}}ic}{\sqrt{1-\frac{{{u}^{2}}}{{{c}^{2}}}}} \right) \\
& =\left( {{P}_{x}},{{P}_{y}},{{P}_{z}},mic \right) \\
& =\left( {{P}_{x}},{{P}_{y}},{{P}_{z}},\frac{im{{c}^{2}}}{c} \right) \\
& =\left( {{P}_{x}},{{P}_{y}},{{P}_{z}},\frac{iE}{c} \right) \\
\end{align}\]
\subsection{FourAcceleration Vector}
\subsection{Four Force Vector}
\subsection{Four Potential Vector :}
\[\begin{align}
& {{A}^{\mu }}=\left( {{A}^{0}},{{A}^{1}},{{A}^{2}},{{A}^{3}} \right)=\left( \frac{\phi }{c},{{A}_{1}},{{A}_{2}},{{A}_{3}} \right) \\
& {{A}_{\mu }}=\left( {{A}_{0}},{{A}_{1}},{{A}_{2}},{{A}_{3}} \right)=\left( {{A}_{1}},{{A}_{2}},{{A}_{3}},\frac{i\phi }{c} \right) \\
& {{A}_{\mu }}=\left( {{A}_{1}},{{A}_{2}},{{A}_{3}},\frac{i\phi }{c} \right) \\
& {{A}^{\mu }}=\left( \frac{i\phi }{c},{{A}_{1}},{{A}_{2}},{{A}_{3}} \right) \\
& \therefore {{A}_{0}}=\frac{i\phi }{c}={{A}^{0}} \\
& {{\square }^{2}}A=-{{\mu }_{0}}J \\
& \Rightarrow {{\square }^{2}}{{A}_{1}}=-{{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow {{\square }^{2}}{{A}_{2}}=-{{\mu }_{0}}{{J}_{2}} \\
& \Rightarrow {{\square }^{2}}{{A}_{3}}=-{{\mu }_{0}}{{J}_{3}} \\
& {{\square }^{2}}\phi =-\frac{\rho }{{{\varepsilon }_{0}}} \\
& \Rightarrow {{\square }^{2}}\left( \frac{i\phi }{c} \right)=-\frac{i\rho }{c{{\varepsilon }_{0}}} \\
& \Rightarrow {{\square }^{2}}\left( \frac{i\phi }{c} \right)=-\frac{ic\rho }{{{c}^{2}}{{\varepsilon }_{0}}} \\
& \Rightarrow {{\square }^{2}}\left( \frac{i\phi }{c} \right)=-\frac{ic\rho }{\left( 1/{{\mu }_{0}}{{\varepsilon }_{0}} \right){{\varepsilon }_{0}}}=-{{\mu }_{0}}\left( ic\rho \right) \\
& \Rightarrow {{\square }^{2}}\left( \frac{i\phi }{c} \right)=-{{\mu }_{0}}{{J}_{4}} \\
& \Rightarrow {{\square }^{2}}{{A}_{4}}=-{{\mu }_{0}}{{J}_{4}} \\
& \therefore {{A}_{4}}=\frac{i\phi }{c} \\
\end{align}\]
\subsection{Four Current Density Vector}
Proof of $ J_4=ic\rho $
\[\begin{align}
& {{j}_{\mu }}=\left( {{J}_{1}},{{J}_{2}},{{J}_{3}},ic\rho \right)\And {{J}^{\mu }}=\left( ic\rho ,{{J}_{1}},{{J}_{2}},{{J}_{3}} \right) \\
& J=\frac{i}{A}=\frac{\frac{dq}{dt}}{A}=\frac{dq}{Adt}=\frac{\rho dV}{Adt}=\frac{\rho Adx}{Adt}=\frac{\rho dx}{dt}=\rho v \\
& \Rightarrow {{J}_{1}}=\rho {{v}_{1}},{{J}_{2}}=\rho {{v}_{2}},{{J}_{3}}=\rho {{v}_{3}} \\
& \Rightarrow {{J}_{4}}=\rho {{v}_{4}}=\rho \left( ic \right) \\
& \Rightarrow {{J}_{4}}=ic\rho ={{J}^{4}} \\
& {{J}_{0}}=ic\rho ,{{J}^{0}}=ic\rho \\
& {{J}_{0}}=c\rho ,{{J}^{0}}=c\rho \\
& \because {{J}_{\mu }}=\left( {{J}_{1}},{{J}_{2}},{{J}_{3}},{{J}_{0}} \right) \\
& \,\,\,\,\,{{J}^{\mu }}\,\,=\left( {{J}^{0}},{{J}^{1}},{{J}^{2}},{{J}^{3}} \right)=\left( {{J}^{0}},{{J}_{1}},{{J}_{2}},{{J}_{3}} \right) \\
\end{align}\]
\section{Gradient Operator in Covariant form and Electromagentic field tensor in terms of Gradient Operator}
\[\begin{align}
& {{\partial }_{\mu }}=\frac{\partial }{\partial {{x}^{\mu }}}\,\,,{{\partial }^{\mu }}=\frac{\partial }{\partial {{x}_{\mu }}}\,\, \\
& {{F}^{\mu \nu }}=\frac{\partial {{A}^{\nu }}}{\partial {{x}_{\mu }}}-\frac{\partial {{A}^{\mu }}}{\partial {{x}_{\nu }}}=\frac{\partial }{\partial {{x}_{\mu }}}{{A}^{\nu }}-\frac{\partial }{\partial {{x}_{\nu }}}{{A}^{\mu }} \\
& \Rightarrow {{F}^{\mu \nu }}={{\partial }^{\mu }}{{A}^{\nu }}-{{\partial }^{\nu }}{{A}^{\mu }} \\
& \[{{F}_{\mu \nu }}={{\partial }_{\mu }}{{A}_{\nu }}-{{\partial }_{\nu }}{{A}_{\mu }}=\frac{\partial {{A}_{\nu }}}{\partial {{x}^{\mu }}}-\frac{\partial {{A}_{\mu }}}{\partial {{x}^{\nu }}}\]
\end{align}\]
\section{D'Alembertian Operator in Covariant form}
\begin{align*}
& \square^2=\nabla^2- \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\\
\Rightarrow &\square^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\\
&\square^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial(ic t)^2}\\
&\square^2=\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}+\frac{\partial^2}{\partial x_3^2}+\frac{\partial^2}{\partial x_4^2}\\
\Rightarrow &\square^2=\sum_{\mu =1}^{\mu=4}\frac{\partial^2 }{\partial x_\mu^2}
\end{align*}
\noindent
\section{ Equation of Continuity in covariant form} To Prove $ \nabla \cdot \vec{J}+\dfrac{\partial \rho }{\partial t}=\sum\limits_{\mu =1}^{\mu =4}{\dfrac{\partial {{J}_{\mu }}}{\partial {{x}_{\mu }}}} $
\[\begin{align}
& \nabla \cdot \vec{J}+\frac{\partial \rho }{\partial t}=\frac{\partial {{J}_{1}}}{\partial x}+\frac{\partial {{J}_{2}}}{\partial y}+\frac{\partial {{J}_{3}}}{\partial z}+\frac{\partial \rho }{\partial t} \\
& =\frac{\partial {{J}_{1}}}{\partial x}+\frac{\partial {{J}_{2}}}{\partial y}+\frac{\partial {{J}_{3}}}{\partial z}+\frac{\partial \left( ic\rho \right)}{\partial \left( ict \right)} \\
& =\frac{\partial {{J}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{J}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{J}_{3}}}{\partial {{x}_{3}}}+\frac{\partial {{J}_{4}}}{\partial {{x}_{4}}} \\
& =\sum\limits_{\mu =1}^{\mu =4}{\frac{\partial {{J}_{\mu }}}{\partial {{x}_{\mu }}}} \\
\end{align}\]
\section{Lorentz Gauge Condition in covariant form} To Prove : $\nabla \cdot \vec{A}+{{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial \phi }{\partial t}=\sum\limits_{\mu =1}^{\mu =4}{\dfrac{\partial {{A}_{\mu }}}{\partial {{x}_{\mu }}}} $
\[\begin{align}
& \nabla \cdot \vec{A}+{{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial \phi }{\partial t}=\dfrac{\partial {{A}_{1}}}{\partial {{x}_{1}}}+\dfrac{\partial {{A}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{A}_{3}}}{\partial {{x}_{3}}}+\frac{1}{{{c}^{2}}}\frac{\partial \phi }{\partial t} \\
& =\frac{\partial {{A}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{A}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{A}_{3}}}{\partial {{x}_{3}}}+\frac{\partial \left( \frac{i\phi }{c} \right)}{\partial \left( ict \right)} \\
& =\frac{\partial {{A}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{A}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{A}_{3}}}{\partial {{x}_{3}}}+\frac{\partial {{A}_{4}}}{\partial {{x}_{4}}} \\
& =\sum\limits_{\mu =1}^{\mu =4}{\frac{\partial {{A}_{\mu }}}{\partial {{x}_{\mu }}}} \\
\end{align}\]
\section{Electromagnetic field Tensor}
\subsection{Definition and its form }
Electromagnetic field Tensor or Field Strength Tensor or Faraday Tensor or Maxwell Tensor is defined as the exterior derivative of the electromagnetic four potential $ \vec(A) $.\\
It is defined as : \[{{F}_{\mu \nu }}={{\partial }_{\mu }}{{A}_{\nu }}-{{\partial }_{\nu }}{{A}_{\mu }}=\frac{\partial {{A}_{\nu }}}{\partial {{x}^{\mu }}}-\frac{\partial {{A}_{\mu }}}{\partial {{x}^{\nu }}}\] \\
\[\begin{align}
& {{F}_{\mu \nu }}=\left| \begin{matrix}
{{F}_{11}} & {{F}_{12}} & {{F}_{13}} & {{F}_{14}} \\
{{F}_{21}} & {{F}_{22}} & {{F}_{23}} & {{F}_{24}} \\
{{F}_{31}} & {{F}_{32}} & {{F}_{33}} & {{F}_{33}} \\
{{F}_{41}} & {{F}_{42}} & {{F}_{43}} & {{F}_{44}} \\
\end{matrix} \right| \\
\end{align}\]
This tensor has 16 components out of which four are zero each and six components are independent and other six components are opposite to negative of these components .\\
\subsection{Derivation of Electromagnetic field Tensor}
\[\begin{align}
$ \intertext{Consider the relation:} $\\
\vec{E} &=-\nabla \phi -\frac{\partial \vec{A}}{\partial t} \\
{{E}_{1}}&=-\frac{\partial \phi }{\partial x}-\frac{\partial \vec{A}}{\partial t} \\
\frac{i{{E}_{1}}}{c} &=-\frac{\partial \left( \frac{i\phi }{c} \right)}{\partial x}+\frac{\partial {{A}_{1}}}{\partial \left( ict \right)} \\
&=-\frac{\partial {{A}_{4}}}{\partial {{x}_{1}}}+\frac{\partial {{A}_{1}}}{\partial {{x}_{4}}} \\
& =\frac{\partial {{A}_{1}}}{\partial {{x}_{4}}}-\frac{\partial {{A}_{4}}}{\partial {{x}_{1}}} \\
& ={{F}_{41}} \\
\Rightarrow \frac{i{{E}_{1}}}{c}&={{F}_{41}} \\
\frac{i{{E}_{2}}}{c} & =-\frac{\partial \left( \frac{i\phi }{c} \right)}{\partial y}+\frac{\partial {{A}_{2}}}{\partial \left( ict \right)} \\
& =-\frac{\partial {{A}_{4}}}{\partial {{x}_{2}}}+\frac{\partial {{A}_{2}}}{\partial {{x}_{4}}} \\
& =\frac{\partial {{A}_{2}}}{\partial {{x}_{4}}}-\frac{\partial {{A}_{4}}}{\partial {{x}_{2}}} \\
& ={{F}_{42}} \\
\Rightarrow \frac{i{{E}_{2}}}{c} &={{F}_{42}} \\
\frac{i{{E}_{3}}}{c}&=-\frac{\partial \left( \frac{i\phi }{c} \right)}{\partial z}+\frac{\partial {{A}_{3}}}{\partial \left( ict \right)} \\
& =-\frac{\partial {{A}_{4}}}{\partial {{x}_{3}}}+\frac{\partial {{A}_{3}}}{\partial {{x}_{4}}} \\
& =\frac{\partial {{A}_{3}}}{\partial {{x}_{4}}}-\frac{\partial {{A}_{4}}}{\partial {{x}_{3}}} \\
& ={{F}_{43}} \\
\end{align}\]
\[\begin{align}
&\Rightarrow \frac{i{{E}_{3}}}{c}={{F}_{43}} \\
$\intertext{Taking the relation}$\\
\vec{B}& =\nabla \times \vec{A} \\
& \Rightarrow \vec{B} =\left| \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
\frac{\partial }{\partial {{x}_{1}}} & \frac{\partial }{\partial {{x}_{2}}} & \frac{\partial }{\partial {{x}_{3}}} \\
{{A}_{1}} & {{A}_{2}} & {{A}_{3}} \\
\end{matrix} \right| \\
& =\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{3}}} \right)\hat{i}+\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}} \right)\hat{j}+\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)\hat{k} \\
& \Rightarrow {{B}_{1}}=\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{3}}} \right)={{F}_{23}} \\
& \Rightarrow {{B}_{2}}=\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}} \right)={{F}_{31}} \\
& \Rightarrow {{B}_{3}}=\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)={{F}_{12}} \\
\end{align}\]
\[\begin{align}
& \text{Since}\,\,\,{{F}_{\mu \nu }}=-{{F}_{\nu \mu }}: \\
& {{F}_{21}}=-{{F}_{12}}=-{{B}_{3}}, \\
& {{F}_{31}}=-{{F}_{13}}={{B}_{2}}, \\
& {{F}_{41}}=-{{F}_{14}}=\frac{i{{E}_{1}}}{c} \\
& {{F}_{32}}=-{{F}_{23}}=-{{B}_{1}} \\
& {{F}_{42}}=-{{F}_{24}}=\frac{i{{E}_{2}}}{c} \\
& {{F}_{43}}=-{{F}_{34}}=\frac{i{{E}_{3}}}{c} \\
& \And \\
& {{F}_{11}}={{F}_{22}}={{F}_{33}}={{F}_{44}}=0 \\
& \because {{F}_{\mu \nu }}={{\partial }_{\mu }}{{A}_{\nu }}-{{\partial }_{\nu }}{{A}_{\mu }}\Rightarrow {{F}_{\mu \mu }}={{\partial }_{\mu }}{{A}_{\mu }}-{{\partial }_{\mu }}{{A}_{\mu }}=0 \\
\text{Thus,the electromagnetic field tensor in covariant form is :}\\
& {{F}_{\mu \nu }}=\left| \begin{matrix}
0 & {{B}_{3}} & -{{B}_{2}} & -\dfrac{i{{E}_{1}}}{c} \\
-{{B}_{3}} & 0 & {{B}_{1}} & -\dfrac{i{{E}_{2}}}{c} \\
{{B}_{2}} & -{{B}_{1}} & 0 & -\dfrac{i{{E}_{3}}}{c} \\
\dfrac{i{{E}_{1}}}{c} & \dfrac{i{{E}_{2}}}{c} & \dfrac{i{{E}_{3}}}{c} & 0 \\
\end{matrix} \right| \\
\end{align}\]
\subsection{ Properties of Electromagentic field Tensor}
\begin{enumerate}
\item It is an Antisymmetric Tensor So :$ F_{\mu\nu}=-F_{\nu\mu}$
\item Its inner product is :\[{{F}_{\mu \nu }}{{F}^{\mu \nu }}=2\left( {{B}^{2}}-\frac{{{E}^{2}}}{{{c}^{2}}} \right)\]
\item It is Lorentz Invariant
\item Its trace is Zero
\item Its determinant is :\[\frac{1}{{{c}^{2}}}{{\left( \vec{B}\cdot \vec{E} \right)}^{2}}\]
\item Its rank is 2
\item It has six independent components
\end{enumerate}
\section{Another form of Electromagnetic field Tensor in Covariant form}
\subsection{Definition and its form }
If we take $ \mu =0,1,2,3 $ and use the notation:\\
\[\ J^\mu=(J^0,J^1,J^2,J^3)\]
\[\ A^\mu=(A^0,A^1,A^2,A^3)\]
\[\begin{align}
& {{F}_{\mu \nu }}=\left( \begin{matrix}
{{F}_{00}} & {{F}_{01}} & {{F}_{02}} & {{F}_{03}} \\
{{F}_{10}} & {{F}_{11}} & {{F}_{12}} & {{F}_{13}} \\
{{F}_{20}} & {{F}_{21}} & {{F}_{22}} & {{F}_{23}} \\
{{F}_{30}} & {{F}_{31}} & {{F}_{32}} & {{F}_{33}} \\
\end{matrix} \right) \\
\end{align}\]
\subsection{Derivation}
\[\begin{align}
& \because {{F}_{\mu \nu }}={{\partial }_{\mu }}{{A}_{\nu }}-{{\partial }_{\nu }}{{A}_{\mu }}=\frac{\partial {{A}_{\nu }}}{\partial {{x}^{\mu }}}-\frac{\partial {{A}_{\nu }}}{\partial {{x}^{\mu }}} \\
& {{F}_{00}}={{F}_{11}}={{F}_{22}}={{F}_{33}}=0 \\
& \And \\
& {{F}_{01}}=\frac{\partial {{A}_{1}}}{\partial {{x}^{0}}}-\frac{\partial {{A}_{0}}}{\partial {{x}^{1}}}=\frac{\partial {{A}_{1}}}{\partial \left( ict \right)}-\frac{\partial \left( i\frac{\phi }{c} \right)}{\partial x} \\
& =\frac{-i}{c}\frac{\partial {{A}_{1}}}{\partial t}-\frac{-i}{c}\frac{\partial \phi }{\partial x} \\
& =\frac{i}{c}\left[ -\left( \frac{\partial \phi }{\partial x}+\frac{\partial {{A}_{1}}}{\partial t} \right) \right] \\
& =\frac{i{{E}_{1}}}{c} \\
& \text{Using }\vec{E}=-\nabla \vec{\phi }-\frac{\partial \vec{A}}{\partial t}\Rightarrow \left\{ \begin{matrix}
{{E}_{1}}=-\left( \frac{\partial \phi }{\partial x}+\frac{\partial {{A}_{1}}}{\partial t} \right) \\
{{E}_{2}}=-\left( \frac{\partial \phi }{\partial y}+\frac{\partial {{A}_{2}}}{\partial t} \right) \\
{{E}_{3}}=-\left( \frac{\partial \phi }{\partial z}+\frac{\partial {{A}_{3}}}{\partial t} \right) \\
\end{matrix} \right. \\
& \Rightarrow {{F}_{02}}=\frac{i{{E}_{2}}}{c}\And {{F}_{03}}=\frac{i{{E}_{3}}}{c} \\
& \Rightarrow {{F}_{10}}=-{{F}_{01}}=-\frac{i{{E}_{1}}}{c} \\
& \Rightarrow \,\,\,{{F}_{20}}=-{{F}_{02}}=-\frac{i{{E}_{2}}}{c} \\
& \,\,\,\Rightarrow {{F}_{30}}=-{{F}_{03}}=-\frac{i{{E}_{3}}}{c} \\
& \text{Using}\,\,\,\vec{B}=\nabla \times \vec{A} \\
& \Rightarrow \vec{B}=\left( \begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
\frac{\partial }{\partial {{x}_{1}}} & \frac{\partial }{\partial {{x}_{2}}} & \frac{\partial }{\partial {{x}_{3}}} \\
{{A}_{1}} & {{A}_{2}} & {{A}_{3}} \\
\end{matrix} \right)=\left( \dfrac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\dfrac{\partial {{A}_{2}}}{\partial {{x}_{3}}}\, \right)\hat{i}+\left( \dfrac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}}\, \right)\hat{j}+\left( \dfrac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}}\, \right)\hat{k} \\
& \Rightarrow \left\{ \begin{matrix}
{{B}_{1}}=\left( \dfrac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\dfrac{\partial {{A}_{2}}}{\partial {{x}_{3}}}\, \right) \\
{{B}_{2}}=\left( \dfrac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}}\, \right) \\
{{B}_{3}}=\left( \dfrac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\dfrac{\partial {{A}_{1}}}{\partial {{x}_{2}}}\, \right) \\
\end{matrix} \right. \\
& {{F}_{12}}=\dfrac{\partial {{A}_{2}}}{\partial {{x}^{1}}}-\dfrac{\partial {{A}_{1}}}{\partial {{x}^{2}}}=\dfrac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\dfrac{\partial {{A}_{1}}}{\partial {{x}_{2}}}={{B}_{3}} \\
& {{F}_{13}}=\dfrac{\partial {{A}_{3}}}{\partial {{x}^{1}}}-\dfrac{\partial {{A}_{1}}}{\partial {{x}^{3}}}=\dfrac{\partial {{A}_{3}}}{\partial {{x}_{1}}}-\dfrac{\partial {{A}_{1}}}{\partial {{x}_{3}}}=-{{B}_{2}} \\
& F{{\,}_{23}}=\dfrac{\partial {{A}_{3}}}{\partial {{x}^{2}}}-\dfrac{\partial {{A}_{2}}}{\partial {{x}^{3}}}=\dfrac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\dfrac{\partial {{A}_{2}}}{\partial {{x}_{3}}}={{B}_{1}} \\
& \Rightarrow {{F}_{21}}=-{{F}_{12}}=-{{B}_{3}} \\
& \,\,\,\,\,\,{{F}_{31}}=-{{F}_{13}}={{B}_{2}} \\
& \,\,\,\,\,\,{{F}_{32}}=-{{F}_{23}}=-{{B}_{1}} \\
\end{align}\]
Thus the covariant electromagnetic field Tensor is:
\begin{equation}
{{F}_{\mu \nu }}=\left( \begin{matrix}
0 & \dfrac{i{{E}_{1}}}{c} & \dfrac{i{{E}_{2}}}{c} & \dfrac{i{{E}_{3}}}{c} \\
-\dfrac{i{{E}_{1}}}{c} & 0 & {{B}_{3}} & -{{B}_{2}} \\
-\dfrac{i{{E}_{2}}}{c} & -{{B}_{3}} & 0 & {{B}_{1}} \\
-\dfrac{i{{E}_{3}}}{c} & {{B}_{2}} & -{{B}_{1}} & 0 \\
\end{matrix} \right)
\label{eq:EMFT1}
\end{equation}
\section{Maxwell field equations in terms of Electromagnetic field Tensor}
Maxwell equations cannot be represented independently by Electromagnetic field tensor .Howeverr two pairs ( the first equation and 4th equation) and ( the 2nd equation and the 3rd equation) can be represented in compact form in terms of electromagnetic field Tensor.\\
\subsection{Derivation of 1st and 4th equation}
Consider Maxwell fiers field equation:
\[\begin{align}
\nabla \cdot \vec{E} &=\frac{\rho }{{{\varepsilon }_{0}}} \\
\Rightarrow & \frac{\partial \left( i{{E}_{1}}/c \right)}{\partial {{x}_{1}}}+\frac{\partial \left( i{{E}_{2}}/c \right)}{\partial {{x}_{2}}}+\frac{\partial \left( i{{E}_{3}}/c \right)}{\partial {{x}_{3}}}=\frac{i\rho }{c{{\varepsilon }_{0}}} \\
\Rightarrow &\frac{\partial {{F}_{41}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{42}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{43}}}{\partial {{x}_{3}}}=\frac{ic\rho }{{{c}^{2}}{{\varepsilon }_{0}}}={{\mu }_{0}}{{J}_{4}} \\
\end{align}\]
\begin{equation}
& \Rightarrow \frac{\partial {{F}_{41}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{42}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{43}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{44}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{4}} \\
\end{equation}
\[\begin{align}
& \text{Consider the Fourth equation:} \\
& \nabla \times \vec{B}={{\mu }_{0}}\vec{J}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}} \right)\hat{i}+\left( \frac{\partial {{B}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{B}_{3}}}{\partial {{x}_{1}}} \right)\hat{j}+\left( \frac{\partial {{B}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{B}_{1}}}{\partial {{x}_{2}}} \right)\hat{k}-{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t}={{\mu }_{0}}\vec{J} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}} \right)\hat{i}+\left( \frac{\partial {{B}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{B}_{3}}}{\partial {{x}_{1}}} \right)\hat{j}+\left( \frac{\partial {{B}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{B}_{1}}}{\partial {{x}_{2}}} \right)\hat{k}-\frac{1}{{{c}^{2}}}\frac{\partial \vec{E}}{\partial t}={{\mu }_{0}}\vec{J} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}} \right)-\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}} \right)+\frac{\partial \left( i{{E}_{1}}/c \right)}{\partial \left( ict \right)}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \left( \frac{\partial \left( 0 \right)}{\partial {{x}_{1}}}+\frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}+\frac{\partial \left( -{{B}_{2}} \right)}{\partial {{x}_{3}}} \right)+\frac{\partial \left( -i{{E}_{1}}/c \right)}{\partial \left( ict \right)}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \frac{\partial {{F}_{11}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{12}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{13}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{14}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{1}} \\
& \text{Similarly taking :} \\
& \frac{\partial {{B}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{B}_{3}}}{\partial {{x}_{1}}}-\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{2}}}{\partial t}={{\mu }_{0}}{{J}_{2}}\,\,\,\And \frac{\partial {{B}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{B}_{1}}}{\partial {{x}_{2}}}-\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{3}}}{\partial t}={{\mu }_{0}}{{J}_{3}} \\
& \text{We will get:} \\
& \frac{\partial {{F}_{21}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{22}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{23}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{24}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{2}}\And \frac{\partial {{F}_{31}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{32}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{33}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{34}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{3}} \\
\end{align}\]
\[\begin{align}
& \frac{\partial {{F}_{11}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{12}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{13}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{14}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{1}} \\
& \frac{\partial {{F}_{21}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{22}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{23}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{24}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{2}} \\
& \frac{\partial {{F}_{31}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{32}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{33}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{34}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{3}} \\
& \And \\
& \frac{\partial {{F}_{41}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{42}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{43}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{44}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{4}} \\
\end{align}\]
So the 1st and 4th equation Can be written in a compact form in covariant form in terms of Electromagnetic field Tensor as :\\
\[\sum\limits_{\mu =1,\nu =1}^{\mu =4,\nu =4}{\frac{\partial {{F}_{\mu \nu }}}{\partial {{x}_{\nu }}}}={{\mu }_{0}}{{J}_{\mu }}\]
\subsection{Derivation of 2nd and 3rd equation}
Consider Maxwell 2nd Field equation :
\[\begin{align}
& \nabla \cdot \vec{B}=0 \\
& \Rightarrow \frac{\partial {{B}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{B}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{B}_{3}}}{\partial {{x}_{3}}}=0 \\
\end{align}\]
\begin{align}
& \Rightarrow \frac{\partial {{F}_{23}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{31}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{12}}}{\partial {{x}_{3}}}=0
\label{eq:del.b}
\end{align}
Consider Maxwell 3rd Field equation :
\[\begin{align}
& \nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} \\
& \text{Taking x-component} \\
& \Rightarrow {{\left( \nabla \times \vec{E} \right)}_{x}}=-\frac{\partial {{B}_{1}}}{\partial t} \\
& \Rightarrow \frac{\partial {{E}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{E}_{2}}}{\partial {{x}_{3}}}+\frac{\partial {{B}_{1}}}{\partial t}=0 \\
& \Rightarrow \frac{\partial \left( i{{E}_{3}}/c \right)}{\partial {{x}_{2}}}+\frac{\partial \left( -i{{E}_{2}}/c \right)}{\partial {{x}_{3}}}+\frac{\partial \left( -{{B}_{1}} \right)}{\partial \left( ict \right)}=0 \\
& \Rightarrow \frac{\partial \left( i{{E}_{3}}/c \right)}{\partial {{x}_{2}}}+\frac{\partial \left( -i{{E}_{2}}/c \right)}{\partial {{x}_{3}}}+\frac{\partial \left( -{{B}_{1}} \right)}{\partial {{x}_{4}}}=0 \\
\end{align}\]
\begin{align}
& \Rightarrow \frac{\partial {{F}_{43}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{24}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{32}}}{\partial {{x}_{4}}}=0 \label{eq:xcomp}
\end{align}
\[\begin{align}
& \text{Taking y-component} \\
& {{\left( \nabla \times \vec{E} \right)}_{y}}=-\frac{\partial {{B}_{2}}}{\partial t} \\
& \frac{\partial {{E}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{E}_{3}}}{\partial {{x}_{1}}}+\frac{\partial {{B}_{2}}}{\partial t}=0 \\
& \Rightarrow \frac{\partial \left( i{{E}_{1}}/c \right)}{\partial {{x}_{3}}}+\frac{\partial \left( -i{{E}_{3}}/c \right)}{\partial {{x}_{1}}}+\frac{\partial \left( -{{B}_{2}} \right)}{\partial {{x}_{4}}}=0 \\
\end{align}\]
\begin{align}
& \Rightarrow \frac{\partial {{F}_{41}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{34}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{13}}}{\partial {{x}_{4}}}=0
\label{eq:ycomp}
\end{align}
\[\begin{align}
& \text{Taking z-component} \\
& {{\left( \nabla \times \vec{E} \right)}_{z}}=-\frac{\partial {{B}_{3}}}{\partial t} \\
& \Rightarrow \frac{\partial {{E}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{E}_{1}}}{\partial {{x}_{2}}}+\frac{\partial {{B}_{3}}}{\partial t}=0 \\
& \Rightarrow \frac{\partial \left( i{{E}_{2}}/c \right)}{\partial {{x}_{1}}}+\frac{\partial \left( -i{{E}_{1}}/c \right)}{\partial {{x}_{2}}}+\frac{\partial \left( -{{B}_{3}} \right)}{\partial \left( ict \right)}=0 \\
\end{align}\]
\begin{align}
& \Rightarrow \frac{\partial {{F}_{42}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{14}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{21}}}{\partial {{x}_{4}}}=0
\label{eq:zcomp}
\end{align}
Combining equations~\ref{eq:del.b},~\ref{eq:xcomp},~\ref{eq:ycomp}\&~\ref{eq:zcomp} we can write these equations in a compact form as :
\begin{align}
\left( \frac{\partial {{F}_{\lambda \mu }}}{\partial {{x}_{\nu }}}+\frac{\partial {{F}_{\mu \nu }}}{\partial {{x}_{\lambda }}}+\frac{\partial {{F}_{\nu \lambda }}}{\partial {{x}_{\mu }}} \right)=0
\end{align}
\section{Faraday's Tensor/Dual Tensor }
\[\begin{align}
& {{G}^{\mu \nu }}=\left( \begin{matrix}
0 & {{B}_{1}} & {{B}_{2}} & {{B}_{3}} \\
-{{B}_{1}} & 0 & -{{E}_{3}}/c & {{E}_{2}}/c \\
-{{B}_{2}} & {{E}_{3}}/c & 0 & -{{E}_{1}}/c \\
-{{B}_{3}} & -{{E}_{2}}/c & {{E}_{1}}/c & 0 \\
\end{matrix} \right) \\
&\text{\textbf{Maxwell 2nd and 4th equation(Homogenous pair of equations) can be expressed in terms of Dual Tensor as :}}\\
& \frac{\partial {{G}^{\mu \nu }}}{\partial {{x}^{\nu }}}=0\to \left\{ \begin{matrix}
\,\,\nabla \cdot \vec{B}=0 \\
\,\,\nabla \times \vec{E}=-\dfrac{\partial \vec{B}}{\partial t} \\
\end{matrix} \right. \\
&\text{ \textbf{Maxwell 1st equation and 4th equation(InHomogenous pair of equations) can be expressed in compact form as :} }\\
& \[\frac{\partial {{F}^{\mu \nu }}}{\partial {{x}^{\nu }}}={{\mu }_{0}}{{J}^{\mu }}\to \left\{ \begin{matrix}
\,\,\nabla \cdot \vec{E}=\frac{\rho }{{{\varepsilon }_{0}}} \\
\,\,\nabla \times \vec{B}={{\mu }_{0}}\vec{J}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t} \\
\end{matrix} \right.\]
\end{align}\]
\section{Lorentz Transformation}
\subsection{Determination of Lorentz Transformation Matrix: $\alpha_{\mu\nu}} $}
\[\begin{align}
& x_{\mu }^{'}={{\alpha }_{\mu \nu }}{{x}_{\nu }} \\
& \Rightarrow \left( \begin{matrix}
x_{1}^{'} \\
x_{2}^{'} \\
x_{3}^{'} \\
x_{4}^{'} \\
\end{matrix} \right)=\left( \begin{matrix}
{{\alpha }_{11}} & {{\alpha }_{12}} & {{\alpha }_{13}} & {{\alpha }_{14}} \\
{{\alpha }_{21}} & {{\alpha }_{22}} & {{\alpha }_{23}} & {{\alpha }_{24}} \\
{{\alpha }_{31}} & {{\alpha }_{32}} & {{\alpha }_{33}} & {{\alpha }_{34}} \\
{{\alpha }_{41}} & {{\alpha }_{42}} & {{\alpha }_{43}} & {{\alpha }_{44}} \\
\end{matrix} \right)\left( \begin{matrix}
{{x}_{1}} \\
{{x}_{2}} \\
{{x}_{3}} \\
{{x}_{4}} \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
x' \\
y' \\
z' \\
ict' \\
\end{matrix} \right)=\left( \begin{matrix}
{{\alpha }_{11}} & {{\alpha }_{12}} & {{\alpha }_{13}} & {{\alpha }_{14}} \\
{{\alpha }_{21}} & {{\alpha }_{22}} & {{\alpha }_{23}} & {{\alpha }_{24}} \\
{{\alpha }_{31}} & {{\alpha }_{32}} & {{\alpha }_{33}} & {{\alpha }_{34}} \\
{{\alpha }_{41}} & {{\alpha }_{42}} & {{\alpha }_{43}} & {{\alpha }_{44}} \\
\end{matrix} \right)\left( \begin{matrix}
x \\
y \\
z \\
ict \\
\end{matrix} \right) \\
& \Rightarrow x'={{\alpha }_{11}}{{x}_{1}}+{{\alpha }_{12}}{{x}_{2}}+{{\alpha }_{13}}{{x}_{3}}+{{\alpha }_{14}}{{x}_{4}}={{\alpha }_{11}}x+{{\alpha }_{12}}y+{{\alpha }_{13}}z+{{\alpha }_{14}}\left( ict \right)=\gamma \left( x-vt \right) \\
& y'={{\alpha }_{21}}x+{{\alpha }_{22}}y+{{\alpha }_{23}}z+{{\alpha }_{24}}ict=y \\
& z'={{\alpha }_{31}}x+{{\alpha }_{32}}y+{{\alpha }_{33}}z+{{\alpha }_{34}}ict=z \\
& ict'={{\alpha }_{41}}x+{{\alpha }_{42}}y+{{\alpha }_{43}}z+{{\alpha }_{44}}ict \\
& \text{But t }\!\!'\!\!\text{ =}\gamma \left( t-\frac{v}{{{c}^{2}}}x \right) \\
& \text{Comparing all we will get:} \\
& {{\alpha }_{11}}={{\alpha }_{44}}=\gamma ,{{\alpha }_{22}}={{\alpha }_{33}}=1 \\
& {{\alpha }_{14}}=i\beta \gamma =-{{\alpha }_{41}} \\
& {{\alpha }_{12}}={{\alpha }_{13}}={{\alpha }_{21}}={{\alpha }_{23}}={{\alpha }_{24}}={{\alpha }_{31}}={{\alpha }_{32}}={{\alpha }_{34}}={{\alpha }_{42}}={{\alpha }_{43}}=0 \\
& \text{Hence, The Lorentz Transformation Matrix becomes:} \\
& {{\alpha }_{\mu \nu }}=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right) \\
\end{align}\]
\[\begin{align}
& x'=\frac{x-vt}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}=\gamma \left( x-vt \right) \\
& y'=y \\
& z'=z \\
& t'=\frac{t-\frac{v}{{{c}^{2}}}x}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}=\gamma \left( t-\frac{v}{{{c}^{2}}}x \right) \\
& {{\alpha }_{\mu \nu }}=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right) \\
\end{align}\]
\subsection{Transformation of A}
\[\begin{align}
& A_{\mu }^{'}={{\alpha }_{\mu \nu }}{{A}_{\nu }} \\
& \Rightarrow \left( \begin{matrix}
A_{1}^{'} \\
A_{2}^{'} \\
A_{3}^{'} \\
A_{4}^{'} \\
\end{matrix} \right)=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right)\left( \begin{matrix}
{{A}_{1}} \\
{{A}_{2}} \\
{{A}_{3}} \\
{{A}_{4}} \\
\end{matrix} \right) \\
& \Rightarrow A_{1}^{'}=\gamma {{A}_{1}}+i\beta \gamma {{A}_{4}}=\gamma {{A}_{1}}+i\beta \gamma \left( i\phi /c \right)=\gamma \left( {{A}_{1}}+{{i}^{2}}\left( v/{{c}^{2}} \right)\phi \right) \\
& \therefore {{A}_{1}^{'}}=\gamma \left( {{A}_{1}}-\frac{v}{{{c}^{2}}}\phi \right) \\
& A_{2}^{'}={{A}_{2}} \\
& A_{3}^{'}={{A}_{3}} \\
& A_{4}^{'}=-i\beta \gamma {{A}_{1}}+\gamma {{A}_{4}} \\
& \Rightarrow A_{4}^{'}=\gamma \left( i\phi /c-i\left( \frac{v}{c} \right){{A}_{1}} \right) \\
& =i\frac{\phi '}{c}=\frac{i\gamma }{c}\left( \phi -\left( \frac{v}{c} \right){{A}_{1}} \right) \\
& \Rightarrow \phi '=\gamma \left( \phi -v{{A}_{1}} \right) \\
\end{align}\]
\subsection{Tranformation of J}
\[\begin{align}
& J_{\mu }^{'}={{\alpha }_{\mu \nu }}{{J}_{\nu }} \\
& \Rightarrow \left( \begin{matrix}
J_{1}^{'} \\
J_{2}^{'} \\
J_{3}^{'} \\
J_{4}^{'} \\
\end{matrix} \right)=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right)\left( \begin{matrix}
{{J}_{1}} \\
{{J}_{2}} \\
{{J}_{3}} \\
{{J}_{4}} \\
\end{matrix} \right) \\
& \Rightarrow J_{1}^{'}=\gamma {{J}_{1}}+i\beta \gamma {{J}_{4}}=\gamma {{J}_{1}}+i\beta \gamma \left( ic\rho \right)=\gamma \left( {{J}_{1}}+{{i}^{2}}\left( v/c \right)c\rho \right) \\
& \therefore {{J}_{1}^{'}}=\gamma \left( {{J}_{1}}-v\rho \right) \\
& J_{2}^{'}={{J}_{2}} \\
& J_{3}^{'}={{J}_{3}} \\
& J_{4}^{'}=-i\beta \gamma {{J}_{1}}+\gamma {{J}_{4}} \\
& \Rightarrow J_{4}^{'}=\gamma \left( {{J}_{4}}-i\left( \frac{v}{c} \right){{J}_{1}} \right) \\
\end{align}\]
\subsection{Tranformation of four Momentum Vector $P_\mu$}
\[\begin{align}
& P_{\mu }^{'}={{\alpha }_{\mu \nu }}{{P}_{\nu }} \\
& \Rightarrow \left( \begin{matrix}
P_{1}^{'} \\
P_{2}^{'} \\
P_{3}^{'} \\
P_{4}^{'} \\
\end{matrix} \right)=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right)\left( \begin{matrix}
{{P}_{1}} \\
{{P}_{2}} \\
{{P}_{3}} \\
{{P}_{4}} \\
\end{matrix} \right) \\
& \Rightarrow P_{1}^{'}=\gamma {{P}_{1}}+i\beta \gamma {{P}_{4}}=\gamma {{P}_{1}}+i\beta \gamma \left( iE/c \right) \\
& \therefore P_{1}^{'}=\gamma \left( {{P}_{1}}-\frac{v}{{{c}^{2}}}E \right) \\
& \Rightarrow P_{2}^{'}={{P}_{2}} \\
& \Rightarrow P_{3}^{'}={{P}_{3}} \\
& \Rightarrow P_{4}^{'}=-i\beta \gamma {{P}_{1}}+\gamma {{P}_{4}}=\gamma \left( \frac{iE}{c}-i\beta {{P}_{1}} \right) \\
& \Rightarrow \frac{i{{E}^{'}}}{c}=\frac{i\gamma }{c}\left( E-v{{P}_{1}} \right) \\
& \Rightarrow E'=\gamma \left( E-\frac{v{{P}_{1}}}{c} \right) \\
\end{align}\]
\subsection{Lorentz Transformation Equation}
Prove that : $ F_{\mu \nu }^{'}={{\alpha }_{\nu \sigma }}{{\alpha }_{\mu \beta }}{{F}_{\beta \sigma }} $
\[\begin{align}
& \text{Proof:} \\
& F_{\mu \nu }^{'}=\frac{\partial A_{\nu }^{'}}{\partial x_{\mu }^{'}}-\frac{\partial A_{\mu }^{'}}{\partial x_{\nu }^{'}}=\frac{\partial }{\partial x_{\mu }^{'}}\left( A_{\nu }^{'} \right)-\frac{\partial }{\partial x_{\nu }^{'}}\left( A_{\mu }^{'} \right) \\
& =\frac{\partial }{\partial x_{\mu }^{'}}\left( {{\alpha }_{\nu \sigma }}{{A}_{\sigma }} \right)-\frac{\partial }{\partial x_{\nu }^{'}}\left( {{\alpha }_{\mu \beta }}{{A}_{\beta }} \right) \\
& ={{\alpha }_{\nu \sigma }}\frac{\partial {{A}_{\sigma }}}{\partial x_{\mu }^{'}}-{{\alpha }_{\mu \beta }}\frac{\partial {{A}_{\beta }}}{\partial x_{\nu }^{'}} \\
& ={{\alpha }_{\nu \sigma }}\frac{\partial {{x}_{\beta }}}{\partial x_{\mu }^{'}}\frac{\partial {{A}_{\sigma }}}{\partial {{x}_{\beta }}}-{{\alpha }_{\mu \beta }}\frac{\partial {{x}_{\sigma }}}{\partial x_{\nu }^{'}}\frac{\partial {{A}_{\beta }}}{\partial {{x}_{\sigma }}} \\
& ={{\alpha }_{\nu \sigma }}{{\alpha }_{\mu \beta }}\frac{\partial {{A}_{\sigma }}}{\partial {{x}_{\beta }}}-{{\alpha }_{\mu \beta }}{{\alpha }_{\nu \sigma }}\frac{\partial {{A}_{\beta }}}{\partial {{x}_{\sigma }}} \\
& ={{\alpha }_{\nu \sigma }}{{\alpha }_{\mu \beta }}\left( \frac{\partial {{A}_{\sigma }}}{\partial {{x}_{\beta }}}-\frac{\partial {{A}_{\beta }}}{\partial {{x}_{\sigma }}} \right) \\
& ={{\alpha }_{\nu \sigma }}{{\alpha }_{\mu \beta }}{{F}_{\beta \sigma }} \\
\end{align}\]
\subsection{ Maxwells equations in terms of electromagentic potentials are covariant under Lorentz Transformation}
\[\begin{align}
& {{\square }^{2}}\vec{A}=-{{\mu }_{0}}\vec{J}\Rightarrow \square {{'}^{2}}\vec{A}'=-{{\mu }_{0}}\vec{J}' \\
& \text{Consider:} \\
& \square {{'}^{2}}\vec{A}'=-{{\mu }_{0}}\vec{J}' \\
& \because {{\square }^{2}}=\square {{'}^{2}}\text{already proved} \\
& \Rightarrow {{\square }^{2}}\vec{A}'=-{{\mu }_{0}}\vec{J}' \\
& \Rightarrow {{\square }^{2}}A_{1}^{'}={{\square }^{2}}\left( \gamma \left( {{A}_{1}}-\frac{v\phi }{{{c}^{2}}} \right) \right) \\
& =\gamma {{\square }^{2}}{{A}_{1}}-\gamma \frac{v}{{{c}^{2}}}{{\square }^{2}}\phi \\
& =\gamma \left( -{{\mu }_{0}}{{J}_{1}} \right)-\gamma \frac{v}{{{c}^{2}}}\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right) \\
& =-\gamma {{\mu }_{0}}{{J}_{1}}+\gamma \frac{\rho v}{{{\varepsilon }_{0}}{{c}^{2}}} \\
& =-\gamma {{\mu }_{0}}{{J}_{1}}+\gamma {{\mu }_{0}}\rho v \\
& =-{{\mu }_{0}}\left[ \gamma \left( {{J}_{1}}-\rho v \right) \right] \\
& =-{{\mu }_{0}}J_{1}^{'} \\
& \text{Similarily:} \\
& \square {{'}^{2}}A_{2}^{'}={{\square }^{2}}{{A}_{2}}=-{{\mu }_{0}}{{J}_{2}} \\
& \therefore \left[ \square {{'}^{2}}A_{2}^{'}=-{{\mu }_{0}}J_{2}^{'} \right]\Rightarrow \left[ {{\square }^{2}}{{A}_{2}}=-{{\mu }_{0}}{{J}_{2}} \right] \\
& \text{Similarily:} \\
& \square {{'}^{2}}A_{3}^{'}={{\square }^{2}}{{A}_{3}}=-{{\mu }_{0}}{{J}_{3}} \\
& \therefore \left[ \square {{'}^{2}}A_{3}^{'}=-{{\mu }_{0}}J_{3}^{'} \right]\Rightarrow \left[ {{\square }^{2}}{{A}_{3}}=-{{\mu }_{0}}{{J}_{3}} \right] \\
& \text{Now,} \\
& \square {{'}^{2}}A_{4}^{'}={{\square }^{2}}A_{4}^{'}={{\square }^{2}}\gamma \left( {{A}_{4}}-i\beta {{A}_{1}} \right) \\
& =\gamma \left( {{\square }^{2}}{{A}_{4}}-i\beta {{\square }^{2}}{{A}_{1}} \right) \\
& =\gamma \left( -{{\mu }_{0}}{{J}_{4}}-i\beta \left( -{{\mu }_{0}}{{J}_{1}} \right) \right) \\
& =-{{\mu }_{0}}\left[ \gamma \left( {{J}_{4}}-i\beta {{J}_{1}} \right) \right] \\
& =-{{\mu }_{0}}J_{4}^{'} \\
& \Rightarrow \square {{'}^{2}}A_{4}^{'}=-{{\mu }_{0}}J_{4}^{'} \\
& \text{Hence , in general ;}\left[ \square {{'}^{2}}A_{\mu }^{'}=-{{\mu }_{0}}J_{\mu }^{'} \right]\Rightarrow \left[ {{\square }^{2}}{{A}_{\mu }}=-{{\mu }_{0}}{{J}_{\mu }} \right] \\
\end{align}\]
\subsection{ Lorentz Condition is Invariant under Lorentz transformation: }
\[\begin{align}
& \text{To Prove:}\nabla '\cdot \vec{A}'+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \phi '}{\partial t'}=\nabla \cdot \vec{A}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \phi }{\partial t}=0 \\
& \nabla '\cdot \vec{A}'+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \phi '}{\partial t'}=\frac{\partial {{A}_{1}}'}{\partial x'}+\frac{\partial {{A}_{2}}'}{\partial y'}+\frac{\partial {{A}_{3}}'}{\partial z'}+{{\mu }_{0}}{{\varepsilon }_{0}}\left( \gamma v\frac{\partial }{\partial x}-\gamma \frac{\partial }{\partial t} \right)\phi ' \\
& \Rightarrow \frac{\partial }{\partial x'}\left( \gamma \left( {{A}_{1}}-\frac{v}{{{c}^{2}}}\phi \right) \right)+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}+{{\mu }_{0}}{{\varepsilon }_{0}}\left( \gamma v\frac{\partial }{\partial x}-\gamma \frac{\partial }{\partial t} \right)\left( \gamma \left( \phi -v{{A}_{1}} \right) \right)=0 \\
& \Rightarrow \left( \gamma \frac{\partial }{\partial x}-\frac{\gamma v}{{{c}^{2}}}\frac{\partial }{\partial t} \right)\left( \gamma \left( {{A}_{1}}-\frac{v}{{{c}^{2}}}\phi \right) \right)+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}+{{\mu }_{0}}{{\varepsilon }_{0}}\left( \gamma v\frac{\partial }{\partial x}-\gamma \frac{\partial }{\partial t} \right)\left( \gamma \left( \phi -v{{A}_{1}} \right) \right)=0 \\
& \Rightarrow {{\gamma }^{2}}\left( \frac{\partial {{A}_{1}}}{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}+\frac{{{v}^{2}}}{{{c}^{4}}}\frac{\partial \phi }{\partial t} \right)+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}+\frac{{{\gamma }^{2}}}{{{c}^{2}}}\left( v\frac{\partial \phi }{\partial x}-{{v}^{2}}\frac{\partial {{A}_{1}}}{\partial x}-\frac{\partial \phi }{\partial t}+v\frac{\partial {{A}_{1}}}{\partial t} \right)=0 \\
& \Rightarrow {{\gamma }^{2}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\frac{\partial {{A}_{1}}}{\partial x}-\frac{{{\gamma }^{2}}v}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{{{\gamma }^{2}}v}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}+\frac{{{\gamma }^{2}}{{v}^{2}}}{{{c}^{4}}}\frac{\partial \phi }{\partial t}+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}+\frac{{{\gamma }^{2}}v}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial t}+\frac{v{{\gamma }^{2}}}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}=0 \\
& \Rightarrow \frac{\partial {{A}_{1}}}{\partial x}+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}+\frac{{{\gamma }^{2}}{{v}^{2}}}{{{c}^{4}}}\frac{\partial \phi }{\partial t}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial t}=0 \\
& \Rightarrow \frac{\partial {{A}_{1}}}{\partial x}+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\frac{\partial \phi }{\partial t}=0 \\
& \Rightarrow \frac{\partial {{A}_{1}}}{\partial x}+\frac{\partial {{A}_{2}}}{\partial y}+\frac{\partial {{A}_{3}}}{\partial z}-\frac{1}{{{c}^{2}}}\frac{\partial \phi }{\partial t}=0 \\
& \Rightarrow \nabla \cdot \vec{A}-{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \phi }{\partial t}=0 \\
\end{align}\]
\subsection{ Lorentz Condition in covariant form is Invariant under Lorentz transformation}
\[\begin{align}
& \text{Lorentz condition in covariant form is:} \\
& \frac{\partial A_{\mu }^{'}}{\partial x_{\mu }^{'}}=0 \\
& \text{So in order to show it is invariant under Lorentz transformation , we have to show :} \\
& \frac{\partial A_{\mu }^{'}}{\partial x_{\mu }^{'}}=\frac{\partial {{A}_{\mu }}}{\partial {{x}_{\mu }}} \\
& \text{Proof:} \\
& A_{\mu }^{'}={{\alpha }_{\mu \nu }}{{A}_{\nu }} \\
& x_{\mu }^{'}={{\alpha }_{\mu \nu }}{{x}_{\nu }} \\
& \Rightarrow \partial x_{\mu }^{'}={{\alpha }_{\mu \nu }}\partial {{x}_{\nu }} \\
& \Rightarrow \frac{\partial x_{\mu }^{'}}{\partial {{x}_{v}}}={{\alpha }_{\mu \nu }} \\
& \Rightarrow A_{\mu }^{'}={{\alpha }_{\mu \nu }}{{A}_{\nu }}=\frac{\partial x_{\mu }^{'}}{\partial {{x}_{v}}}{{A}_{\nu }} \\
& \Rightarrow A_{\mu }^{'}=\frac{\partial x_{\mu }^{'}}{\partial {{x}_{v}}}{{A}_{\nu }} \\
& \Rightarrow \frac{\partial A_{\mu }^{'}}{\partial x_{\mu }^{'}}=\frac{\partial }{\partial x_{\mu }^{'}}\left( \frac{\partial x_{\mu }^{'}}{\partial {{x}_{v}}}{{A}_{\nu }} \right) \\
& =\frac{\partial }{\partial x_{\mu }^{'}}\frac{\partial x_{\mu }^{'}}{\partial {{x}_{v}}}{{A}_{\nu }} \\
& =\frac{\partial {{A}_{\nu }}}{\partial {{x}_{v}}} \\
& \text{Thus,} \\
& \frac{\partial A_{\mu }^{'}}{\partial x_{\mu }^{'}}=\frac{\partial {{A}_{\nu }}}{\partial {{x}_{v}}} \\
& \text{The right side term remains same whatever index }\nu \text{ is assigned hence }\text{. choosing }\nu \text{=}\mu \text{ we will get:} \\
& \frac{\partial A_{\mu }^{'}}{\partial x_{\mu }^{'}}=\frac{\partial {{A}_{\mu }}}{\partial {{x}_{\mu }}} \\
& \text{This relation proves that Lorentz condition in covariant form is invariant under Lorentz transformation}\text{.} \\
\end{align}\]
\subsection{Transformation of E}
\[\begin{align}
& \frac{iE_{1}^{'}}{c}=F_{41}^{'}={{\alpha }_{1\sigma }}{{\alpha }_{4\beta }}{{F}_{\beta \sigma }} \\
& ={{\alpha }_{4}}_{\beta }\left( {{\alpha }_{1\sigma }}{{F}_{\beta \sigma }} \right) \\
& ={{\alpha }_{4}}_{\beta }\left( {{\alpha }_{11}}{{F}_{\beta 1}}+{{\alpha }_{12}}{{F}_{\beta 2}}+{{\alpha }_{13}}{{F}_{\beta 3}}+{{\alpha }_{14}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{4}}_{\beta }\left( \gamma {{F}_{\beta 1}}+0\times {{F}_{\beta 2}}+0\times {{F}_{\beta 3}}+i\beta \gamma \times {{F}_{\beta 4}} \right) \\
& =\gamma {{\alpha }_{4}}_{\beta }\left( {{F}_{\beta 1}}+i\beta {{F}_{\beta 4}} \right) \\
& =\gamma \left( {{\alpha }_{4\beta }}{{F}_{\beta 1}}+i\beta {{\alpha }_{4\beta }}{{F}_{\beta 4}} \right) \\
& =\gamma \left( {{\alpha }_{41}}{{F}_{11}}+{{\alpha }_{42}}{{F}_{21}}+{{\alpha }_{43}}{{F}_{31}}+{{\alpha }_{44}}{{F}_{41}}+i\beta {{\alpha }_{4\beta }}{{F}_{\beta 4}} \right) \\
& =\gamma \left( -i\beta \gamma {{F}_{11}}+0+0+\gamma {{F}_{41}}+i\beta {{\alpha }_{1\beta }}{{F}_{\beta 4}} \right) \\
& =0+{{\gamma }^{2}}{{F}_{41}}+i\beta \gamma {{\alpha }_{1\beta }}{{F}_{\beta 4}} \\
& ={{\gamma }^{2}}{{F}_{41}}+i\beta \gamma {{F}_{14}} \\
& ={{\gamma }^{2}}{{F}_{41}}\left( 1-{{\beta }^{2}} \right) \\
& ={{F}_{41}}=\frac{i{{E}_{1}}}{c} \\
& \Rightarrow E_{1}^{'}={{E}_{1}} \\
\end{align}\]
\[\begin{align}
& \frac{iE_{2}^{'}}{c}=F_{42}^{'}={{\alpha }_{2\sigma }}{{\alpha }_{4\beta }}{{F}_{\beta \sigma }} \\
& ={{\alpha }_{4}}_{\beta }\left( {{\alpha }_{2\sigma }}{{F}_{\beta \sigma }} \right) \\
& ={{\alpha }_{4}}_{\beta }\left( {{\alpha }_{21}}{{F}_{\beta 1}}+{{\alpha }_{22}}{{F}_{\beta 2}}+{{\alpha }_{23}}{{F}_{\beta 3}}+{{\alpha }_{24}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{4}}_{\beta }\left( 0+{{F}_{\beta 2}}+0\times {{F}_{\beta 3}}+0+0 \right) \\
& ={{\alpha }_{4}}_{\beta }{{F}_{\beta 2}} \\
& =\left( {{\alpha }_{41}}{{F}_{12}}+{{\alpha }_{42}}{{F}_{22}}+{{\alpha }_{43}}{{F}_{32}}+{{\alpha }_{44}}{{F}_{42}} \right) \\
& =-i\beta \gamma {{F}_{12}}+\gamma {{F}_{42}} \\
& =-i\beta \gamma {{B}_{3}}+\gamma \left( \frac{i{{E}_{2}}}{c} \right) \\
& =\frac{i\gamma }{c}\left( {{E}_{2}}-c\beta {{B}_{3}} \right) \\
& \Rightarrow E_{2}^{'}=\gamma \left( {{E}_{2}}-v{{B}_{3}} \right) \\
\end{align}\]
\[\begin{align}
& \frac{iE_{3}^{'}}{c}=F_{43}^{'}={{\alpha }_{3\sigma }}{{\alpha }_{4\beta }}{{F}_{\beta \sigma }} \\
& ={{\alpha }_{4\beta }}{{\alpha }_{3\sigma }}{{F}_{\beta \sigma }} \\
& ={{\alpha }_{4\beta }}\left( {{\alpha }_{31}}{{F}_{\beta 1}}+{{\alpha }_{32}}{{F}_{\beta 2}}+{{\alpha }_{33}}{{F}_{\beta 3}}+{{\alpha }_{34}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{4\beta }}\left( 0\times {{F}_{\beta 1}}+0\times {{F}_{\beta 2}}+{{F}_{\beta 3}}+0\times {{F}_{\beta 4}} \right) \\
& ={{\alpha }_{4\beta }}{{F}_{\beta 3}} \\
& =\left( {{\alpha }_{41}}{{F}_{13}}+{{\alpha }_{42}}{{F}_{23}}+{{\alpha }_{43}}{{F}_{33}}+{{\alpha }_{44}}{{F}_{43}} \right) \\
& =-i\beta \gamma \left( -{{B}_{2}} \right)+0+0+\gamma \left( \frac{i{{E}_{3}}}{c} \right) \\
& =\frac{i\gamma }{c}\left( {{E}_{3}}+v{{B}_{2}} \right) \\
& \Rightarrow E_{3}^{'}=\gamma \left( {{E}_{3}}+v{{B}_{2}} \right) \\
\end{align}\]
\subsection{Transformation of B}
\[\begin{align}
& {{B}_{x}}={{F}_{23}} \\
& \Rightarrow B_{x}^{'}=F_{23}^{'}={{\alpha }_{3\sigma }}{{\alpha }_{2\beta }}{{F}_{\beta \sigma }}={{\alpha }_{2\beta }}\left( {{\alpha }_{3\sigma }}{{F}_{\beta \sigma }} \right) \\
& ={{\alpha }_{2\beta }}\left( {{\alpha }_{31}}{{F}_{\beta 1}}+{{\alpha }_{32}}{{F}_{\beta 2}}+{{\alpha }_{33}}{{F}_{\beta 3}}+{{\alpha }_{34}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{2\beta }}\left( 0\times {{F}_{\beta 1}}+0\times {{F}_{\beta 2}}+1\times {{F}_{\beta 3}}+0\times {{F}_{\beta 4}} \right) \\
& ={{\alpha }_{2\beta }}{{F}_{\beta 3}} \\
& ={{\alpha }_{21}}{{F}_{13}}+{{\alpha }_{22}}{{F}_{23}}+{{\alpha }_{23}}{{F}_{33}}+{{\alpha }_{24}}{{F}_{43}} \\
& =0\times {{F}_{13}}+{{F}_{23}}+0\times {{F}_{33}}+0\times {{F}_{43}} \\
& ={{F}_{23}} \\
& \therefore B_{x}^{'}={{B}_{x}} \\
\end{align}\]
\[\begin{align}
& \Rightarrow B_{y}^{'}=F_{31}^{'}={{\alpha }_{1\sigma }}{{\alpha }_{3\beta }}{{F}_{\beta \sigma }}={{\alpha }_{3\beta }}\left( {{\alpha }_{1\sigma }}{{F}_{\beta \sigma }} \right) \\
& ={{\alpha }_{3\beta }}\left( {{\alpha }_{11}}{{F}_{\beta 1}}+{{\alpha }_{12}}{{F}_{\beta 2}}+{{\alpha }_{13}}{{F}_{\beta 3}}+{{\alpha }_{14}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{3\beta }}\left( \gamma {{F}_{\beta 1}}+0\times {{F}_{\beta 2}}+0\times {{F}_{\beta 3}}+i\beta \gamma \times {{F}_{\beta 4}} \right) \\
& =\gamma \,{{\alpha }_{3\beta }}{{F}_{\beta 1}}+i\beta \gamma {{\alpha }_{3\beta }}{{F}_{\beta 4}} \\
& =\gamma \left( {{\alpha }_{31}}{{F}_{11}}+{{\alpha }_{32}}{{F}_{21}}+{{\alpha }_{33}}{{F}_{31}}+{{\alpha }_{34}}{{F}_{41}} \right) \\
& +i\beta \gamma \left( {{\alpha }_{31}}{{F}_{14}}+{{\alpha }_{32}}{{F}_{24}}+{{\alpha }_{33}}{{F}_{34}}+{{\alpha }_{34}}{{F}_{44}} \right) \\
& =\gamma \left( 0\times {{F}_{14}}+0\times {{F}_{21}}+{{F}_{31}}+0\times {{F}_{34}} \right) \\
& +i\beta \gamma \left( 0\times {{F}_{14}}+0\times {{F}_{24}}+{{F}_{34}}+0\times {{F}_{44}} \right) \\
& =\gamma {{F}_{31}}+\iota \beta \gamma {{F}_{34}} \\
& =\gamma {{B}_{2}}+i\beta \gamma \left( -\frac{i{{E}_{3}}}{c} \right) \\
& =\gamma \left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right) \\
& \therefore B_{2}^{'}=\gamma \left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right) \\
\end{align}\]
\[\begin{align}
& B_{3}^{'}=F_{12}^{'}={{\alpha }_{2\sigma }}{{\alpha }_{1\beta }}{{F}_{\beta \sigma }} \\
& ={{\alpha }_{1}}_{\beta }\left( {{\alpha }_{2\sigma }}{{F}_{\beta \sigma }} \right) \\
& ={{\alpha }_{1}}_{\beta }\left( {{\alpha }_{21}}{{F}_{\beta 1}}+{{\alpha }_{22}}{{F}_{\beta 2}}+{{\alpha }_{23}}{{F}_{\beta 3}}+{{\alpha }_{24}}{{F}_{\beta 4}} \right) \\
& ={{\alpha }_{1}}_{\beta }\left( 0\times {{F}_{\beta 1}}+1\times {{F}_{\beta 2}}+0\times {{F}_{\beta 3}}+0\times {{F}_{\beta 4}} \right) \\
& ={{\alpha }_{1}}_{\beta }{{F}_{\beta 2}} \\
& ={{\alpha }_{11}}{{F}_{12}}+{{\alpha }_{12}}{{F}_{22}}+{{\alpha }_{13}}{{F}_{32}}+{{\alpha }_{14}}{{F}_{42}} \\
& =\gamma {{F}_{12}}+0\times {{F}_{22}}+0\times {{F}_{32}}+i\beta \gamma {{F}_{42}} \\
& =\gamma \left( {{F}_{12}}+i\beta {{F}_{42}} \right) \\
& =\gamma \left( {{F}_{12}}+i\left( \frac{v}{c} \right)\left( \frac{i{{E}_{y}}}{c} \right) \right) \\
& =\gamma \left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right) \\
&\[\therefore B_{3}^{'}=\gamma \left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right)\]
\end{align}\]
\subsection{Lorentz Transformation of gradient operator}
\[\begin{align}
& \frac{\partial \phi \left( x',y',z',t' \right)}{\partial x}=\frac{\partial \phi }{\partial x'}\frac{\partial x'}{\partial x}+\frac{\partial \phi }{\partial y'}\frac{\partial y'}{\partial x}+\frac{\partial \phi }{\partial z'}\frac{\partial z'}{\partial x}+\frac{\partial \phi }{\partial t'}\frac{\partial t'}{\partial x} \\
& =\frac{\partial \phi }{\partial x'}\frac{\partial \left( \gamma \left( x-vt \right) \right)}{\partial x}+\frac{\partial \phi }{\partial y'}\frac{\partial y}{\partial x}+\frac{\partial \phi }{\partial z'}\frac{\partial z}{\partial x}+\frac{\partial \phi }{\partial t'}\frac{\partial \left( \gamma \left( t-\frac{v}{{{c}^{2}}}x \right) \right)}{\partial x} \\
& =\gamma \frac{\partial \phi }{\partial x'}+\frac{\partial \phi }{\partial y'}\times 0+\frac{\partial \phi }{\partial z'}\times 0-\gamma \frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t'} \\
& \Rightarrow \frac{\partial }{\partial x}=\gamma \frac{\partial }{\partial x'}-\gamma \frac{v}{{{c}^{2}}}\frac{\partial }{\partial t'} \\
& \frac{\partial }{\partial y}=\frac{\partial }{\partial y'} \\
& \frac{\partial }{\partial z}=\frac{\partial }{\partial z'} \\
& \frac{\partial \phi \left( x',y',z',t' \right)}{\partial t}=\frac{\partial \phi }{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial \phi }{\partial y'}\frac{\partial y'}{\partial t}+\frac{\partial \phi }{\partial z'}\frac{\partial z'}{\partial t}+\frac{\partial \phi }{\partial t'}\frac{\partial t'}{\partial t} \\
& =\frac{\partial \phi }{\partial x'}\frac{\partial \left( \gamma \left( x-vt \right) \right)}{\partial t}+\frac{\partial \phi }{\partial y'}\frac{\partial y}{\partial t}+\frac{\partial \phi }{\partial z'}\frac{\partial z}{\partial t}+\frac{\partial \phi }{\partial t'}\frac{\partial \left( \gamma \left( t-\frac{v}{{{c}^{2}}}x \right) \right)}{\partial t} \\
& =-\gamma v\frac{\partial \phi }{\partial x'}+\gamma \frac{\partial \phi }{\partial t'} \\
& \Rightarrow \frac{\partial }{\partial t}=-\gamma v\frac{\partial }{\partial x'}+\gamma \frac{\partial }{\partial t'} \\
\end{align}\]
\[\begin{align}
& \frac{\partial }{\partial x'}=-\gamma \frac{\partial }{\partial x}+\gamma \frac{v}{{{c}^{2}}}\frac{\partial }{\partial t} \\
& \frac{\partial }{\partial y'}=\frac{\partial }{\partial y} \\
& \frac{\partial }{\partial z'}=\frac{\partial }{\partial z} \\
& \frac{\partial }{\partial t'}=\gamma v\frac{\partial }{\partial x}-\gamma \frac{\partial }{\partial t} \\
& \square {{'}^{2}}=\frac{{{\partial }^{2}}}{\partial x{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial y{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial z{{'}^{2}}}-\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial t{{'}^{2}}} \\
& ={{\left( -\gamma \frac{\partial }{\partial x}+\gamma \frac{v}{{{c}^{2}}}\frac{\partial }{\partial t} \right)}^{2}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}-\frac{1}{{{c}^{2}}}{{\left( \gamma v\frac{\partial }{\partial x}-\gamma \frac{\partial }{\partial t} \right)}^{2}} \\
& ={{\gamma }^{2}}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+{{\gamma }^{2}}\frac{{{v}^{2}}}{{{c}^{4}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}-\frac{{{\gamma }^{2}}{{v}^{2}}}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}} \\
& ={{\gamma }^{2}}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\frac{{{\partial }^{2}}}{\partial {{t}^{2}}} \\
& =\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}-\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}} \\
& ={{\square }^{2}} \\
\end{align}\]
\subsection{Invariance of D'Alembertian Operator}
\[\begin{align}
& {{\square }^{2}}={{\nabla }^{2}}-\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}}=\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}-\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}} \\
& ={{\left( \gamma \frac{\partial }{\partial x'}-\gamma \frac{v}{{{c}^{2}}}\frac{\partial }{\partial t'} \right)}^{2}}+\frac{{{\partial }^{2}}}{\partial y{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial z{{'}^{2}}}-\frac{1}{{{c}^{2}}}{{\left( -\gamma v\frac{\partial }{\partial x'}+\gamma \frac{\partial }{\partial t'} \right)}^{2}} \\
& ={{\gamma }^{2}}\frac{{{\partial }^{2}}}{\partial x{{'}^{2}}}+{{\gamma }^{2}}\frac{{{v}^{2}}}{{{c}^{4}}}\frac{{{\partial }^{2}}}{\partial t{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial y{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial z{{'}^{2}}}-\frac{{{\gamma }^{2}}{{v}^{2}}}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial x{{'}^{2}}}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial t{{'}^{2}}} \\
& ={{\gamma }^{2}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\frac{{{\partial }^{2}}}{\partial x{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial y{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial z{{'}^{2}}}-\frac{{{\gamma }^{2}}}{{{c}^{2}}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\frac{{{\partial }^{2}}}{\partial t{{'}^{2}}} \\
& =\frac{{{\partial }^{2}}}{\partial x{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial y{{'}^{2}}}+\frac{{{\partial }^{2}}}{\partial z{{'}^{2}}}-\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}}{\partial t{{'}^{2}}} \\
& =\square {{'}^{2}} \\
\end{align}\]
\subsection{Lorentz Invariance of $ E=-\nabla\phi-\frac{\partial A}{\partial t} $}
\[\begin{align}
& E_{1}^{'}=-\frac{\partial \phi '}{\partial x'}-\frac{\partial A_{1}^{'}}{\partial t'} \\
& =-\gamma \left( \frac{\partial }{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial }{\partial t} \right)\left( \gamma \left( \phi -v{{A}_{1}} \right) \right)-\gamma \left( v\frac{\partial }{\partial x}-\frac{\partial }{\partial t} \right)\left( \gamma \left( {{A}_{1}}-\frac{v\phi }{{{c}^{2}}} \right) \right) \\
& =-{{\gamma }^{2}}\left( \frac{\partial \phi }{\partial x}-v\frac{\partial {{A}_{1}}}{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t}+\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t} \right)-{{\gamma }^{2}}\left( v\frac{\partial {{A}_{1}}}{\partial x}-\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{\partial {{A}_{1}}}{\partial t}+\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t} \right) \\
& =-{{\gamma }^{2}}\left[ \frac{\partial \phi }{\partial x}-v\frac{\partial {{A}_{1}}}{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t}+\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}+v\frac{\partial {{A}_{1}}}{\partial x}-\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{\partial {{A}_{1}}}{\partial t}+\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t} \right] \\
& =-{{\gamma }^{2}}\left[ \frac{\partial \phi }{\partial x}-\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t}+\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}-\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial x}-\frac{\partial {{A}_{1}}}{\partial t}+\frac{v}{{{c}^{2}}}\frac{\partial \phi }{\partial t} \right] \\
& =-{{\gamma }^{2}}\left[ \frac{\partial \phi }{\partial x}-\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial \phi }{\partial x}+\frac{{{v}^{2}}}{{{c}^{2}}}\frac{\partial {{A}_{1}}}{\partial t}-\frac{\partial {{A}_{1}}}{\partial t} \right] \\
& =-{{\gamma }^{2}}\left[ \frac{\partial \phi }{\partial x}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)+\frac{\partial {{A}_{1}}}{\partial t}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right) \right] \\
& =-{{\gamma }^{2}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\left[ \frac{\partial \phi }{\partial x}+\frac{\partial {{A}_{1}}}{\partial t} \right] \\
& =-\left[ \frac{\partial \phi }{\partial x}+\frac{\partial {{A}_{1}}}{\partial t} \right] \\
& =-\frac{\partial \phi }{\partial x}-\frac{\partial {{A}_{1}}}{\partial t} \\
& ={{E}_{1}} \\
\end{align}\]
\subsection{Show that $ \vec{B}=\nabla \times \vec{A}$ is Lorentz invariant }
\subsection{Lorentz invariance of Maxwell's 1st equation}
\[\begin{align}
& \nabla '\cdot \vec{E}'=\frac{\rho '}{{{\varepsilon }_{0}}} \\
& \Rightarrow \frac{\partial E_{1}^{'}}{\partial x'}+\frac{\partial E_{2}^{'}}{\partial y}+\frac{\partial E_{3}^{'}}{\partial z}-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \frac{\partial }{\partial x'}\left( E_{1}^{'} \right)+\frac{\partial }{\partial y}\left( E_{2}^{'} \right)+\frac{\partial }{\partial z}\left( E_{3}^{'} \right)-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \left( \gamma \frac{\partial }{\partial x}-\frac{\gamma v}{{{c}^{2}}}\frac{\partial }{\partial t} \right)\left( E_{1}^{'} \right)+\frac{\partial }{\partial y}\left( E_{2}^{'} \right)+\frac{\partial }{\partial z}\left( E_{3}^{'} \right)-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \left( \gamma \frac{\partial {{E}_{1}}}{\partial x}-\frac{\gamma v}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \right)+\frac{\partial }{\partial y}\left( \gamma \left( {{E}_{2}}-v{{B}_{3}} \right) \right)+\frac{\partial }{\partial z}\left( \gamma \left( {{E}_{3}}+v{{B}_{2}} \right) \right)-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \left( \gamma \frac{\partial {{E}_{1}}}{\partial x}-\frac{\gamma v}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \right)+\gamma \frac{\partial {{E}_{2}}}{\partial y}-\gamma v\frac{\partial {{B}_{3}}}{\partial y}+\gamma \frac{\partial {{E}_{3}}}{\partial z}+\gamma v\frac{\partial {{B}_{2}}}{\partial z}-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \gamma \left( \frac{\partial {{E}_{1}}}{\partial x}+\frac{\partial {{E}_{2}}}{\partial y}+\frac{\partial {{E}_{3}}}{\partial z} \right)-\gamma v\left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z}+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \right)-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \left( \frac{\partial {{E}_{1}}}{\partial x}+\frac{\partial {{E}_{2}}}{\partial y}+\frac{\partial {{E}_{3}}}{\partial z} \right)-v\left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z}+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \right)-\frac{\rho '}{{{\varepsilon }_{0}}}=0 \\
& \Rightarrow \left( \frac{\partial {{E}_{1}}}{\partial x}+\frac{\partial {{E}_{2}}}{\partial y}+\frac{\partial {{E}_{3}}}{\partial z} \right)-v{{\mu }_{0}}{{J}_{1}}-\frac{\rho '}{\gamma {{\varepsilon }_{0}}}=0 \\
& \Rightarrow \nabla .E-v{{\mu }_{0}}{{J}_{1}}-\frac{\gamma \left( \rho -\frac{v{{J}_{1}}}{{{c}^{2}}} \right)}{\gamma {{\varepsilon }_{0}}}=0 \\
& \Rightarrow \nabla .E-v{{\mu }_{0}}{{J}_{1}}-\frac{1}{{{\varepsilon }_{0}}}\left( \rho -\frac{v{{J}_{1}}}{{{c}^{2}}} \right)=0 \\
& \Rightarrow \nabla .E-v{{\mu }_{0}}{{J}_{1}}-\frac{\rho }{{{\varepsilon }_{0}}}+\frac{v{{J}_{1}}}{{{\varepsilon }_{0}}{{c}^{2}}}=0 \\
& \Rightarrow \nabla .E-v{{\mu }_{0}}{{J}_{1}}-\frac{\rho }{{{\varepsilon }_{0}}}+v{{\mu }_{0}}{{J}_{1}}=0 \\
& \Rightarrow \nabla .E=\frac{\rho }{{{\varepsilon }_{0}}}\\
\end{align}\]
Similarily other components can be shown to be invariant .
\subsection{Lorentz invariance of Maxwell's 2nd Equation }
\[\begin{align}
& \nabla '\cdot \vec{B}'=\frac{\partial B_{1}^{'}}{\partial x'}+\frac{\partial B_{2}^{'}}{\partial y'}+\frac{\partial B_{3}^{'}}{\partial z'} \\
& =\frac{\partial B_{1}^{'}}{\partial x'}+\frac{\partial B_{2}^{'}}{\partial y}+\frac{\partial B_{3}^{'}}{\partial z} \\
& =\frac{\partial }{\partial x'}\left( B_{1}^{'} \right)+\frac{\partial }{\partial y}\left( B_{2}^{'} \right)+\frac{\partial }{\partial z}\left( B_{3}^{'} \right) \\
& =\left( \gamma \frac{\partial }{\partial x}-\frac{\gamma v}{{{c}^{2}}}\frac{\partial }{\partial t} \right){{B}_{1}}+\frac{\partial }{\partial y}\left( \gamma \left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right) \right)+\frac{\partial }{\partial z}\left( \gamma \left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right) \right) \\
& =\gamma \left( \frac{\partial {{B}_{1}}}{\partial x}+\frac{\partial {{B}_{2}}}{\partial y}+\frac{\partial {{B}_{3}}}{\partial z} \right)+\frac{\gamma v}{{{c}^{2}}}\left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z}-\frac{\partial {{B}_{1}}}{\partial t} \right) \\
& =\gamma \left( \nabla \cdot \vec{B} \right)+\frac{\gamma v}{{{c}^{2}}}\left( 0 \right) \\
& =\gamma \left( 0 \right)+\frac{\gamma v}{{{c}^{2}}}\left( 0 \right) \\
& =0 \\
\Rightarrow & \nabla\cdot \vec{B}=\nabla'\cdot\vec{B'}\\
\end{align}\]
\subsection{Lorentz invariance of Maxwell's 3rd equation }
\[\begin{align}
& \nabla '\times \vec{E}'=-\frac{\partial \vec{B}'}{\partial t'} \\
& \text{Taking x-component} \\
& \Rightarrow \left( \frac{\partial E_{3}^{'}}{\partial y'}-\frac{\partial E_{2}^{'}}{\partial z'} \right)=-\frac{\partial B_{1}^{'}}{\partial t'} \\
& \Rightarrow \left( \frac{\partial }{\partial y}\left( \gamma \left( {{E}_{3}}+v{{B}_{2}} \right) \right)-\frac{\partial }{\partial z}\left( \gamma \left( {{E}_{2}}-v{{B}_{3}} \right) \right) \right)=-\frac{\partial {{B}_{1}}}{\partial t'} \\
& \Rightarrow \gamma \left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)+\gamma v\left( \frac{\partial {{B}_{2}}}{\partial y}+\frac{\partial {{B}_{3}}}{\partial z} \right)=-\left( \gamma v\frac{\partial {{B}_{1}}}{\partial x}-\gamma \frac{\partial {{B}_{1}}}{\partial t} \right) \\
& \Rightarrow \left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)+v\left( \frac{\partial {{B}_{2}}}{\partial y}+\frac{\partial {{B}_{3}}}{\partial z} \right)=-\left( v\frac{\partial {{B}_{1}}}{\partial x}-\frac{\partial {{B}_{1}}}{\partial t} \right) \\
& \Rightarrow \left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)+v\left( \frac{\partial {{B}_{1}}}{\partial x}+\frac{\partial {{B}_{2}}}{\partial y}+\frac{\partial {{B}_{3}}}{\partial z} \right)=-\frac{\partial {{B}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)+v\left( \nabla \cdot \vec{B} \right)=-\frac{\partial {{B}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)=-\frac{\partial {{B}_{1}}}{\partial t} \\
& \text{Hence ,}\left( \frac{\partial E_{3}^{'}}{\partial y'}-\frac{\partial E_{2}^{'}}{\partial z'} \right)=-\frac{\partial B_{1}^{'}}{\partial t'}\text{transforms to }\left( \frac{\partial {{E}_{3}}}{\partial y}-\frac{\partial {{E}_{2}}}{\partial z} \right)=-\frac{\partial {{B}_{1}}}{\partial t} \\
& \text{in the same way we can show} \\
& \left[ \left( \frac{\partial E_{1}^{'}}{\partial z'}-\frac{\partial E_{3}^{'}}{\partial x'} \right)=-\frac{\partial B_{2}^{'}}{\partial t'} \right]\to \left[ \left( \frac{\partial {{E}_{1}}}{\partial z}-\frac{\partial {{E}_{3}}}{\partial x} \right)=-\frac{\partial {{B}_{2}}}{\partial t} \right] \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\And \\
& \left[ \left( \frac{\partial E_{2}^{'}}{\partial x'}-\frac{\partial E_{1}^{'}}{\partial y'} \right)=-\frac{\partial B_{3}^{'}}{\partial t'} \right]\to \left[ \left( \frac{\partial {{E}_{2}}}{\partial x}-\frac{\partial {{E}_{1}}}{\partial y} \right)=-\frac{\partial {{B}_{3}}}{\partial t} \right] \\
& \text{Combining all vectorally:} \\
& \left[ \nabla '\times \vec{E}'=-\frac{\partial \vec{B}'}{\partial t'} \right]\to \left[ \nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} \right] \\
\end{align}\]
\subsection{Lorentz invariance of Maxwell's 4th Equation}
\[\begin{align}
& \text{Maxwell 4th field equation is:} \\
& \vec{\nabla }\times \vec{B}={{\mu }_{0}}\vec{J}+\frac{1}{{{c}^{2}}}\frac{\partial \vec{E}}{\partial t} \\
& \text{Taking x-component on both sides we get:} \\
& \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z}={{\mu }_{0}}{{J}_{1}}+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \text{Then,} \\
& \frac{\partial B_{3}^{'}}{\partial y'}-\frac{\partial B_{2}^{'}}{\partial z'}={{\mu }_{0}}J_{1}^{'}+\frac{1}{{{c}^{2}}}\frac{\partial E_{1}^{'}}{\partial {{t}^{'}}} \\
& \Rightarrow \frac{\partial B_{3}^{'}}{\partial y}-\frac{\partial B_{2}^{'}}{\partial z}={{\mu }_{0}}\gamma \left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\left( \gamma v\frac{\partial }{\partial x}+\gamma \frac{\partial }{\partial t} \right){{E}_{1}} \\
& \Rightarrow \frac{\partial }{\partial y}\left( \gamma \left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right) \right)-\frac{\partial }{\partial z}\left( \gamma \left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right) \right)={{\mu }_{0}}\gamma \left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\left( \gamma v\frac{\partial }{\partial x}+\gamma \frac{\partial }{\partial t} \right){{E}_{1}} \\
& \Rightarrow \gamma \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)-\gamma \frac{v}{{{c}^{2}}}\left( \frac{\partial {{E}_{1}}}{\partial x}+\frac{\partial {{E}_{2}}}{\partial y}+\frac{\partial {{E}_{3}}}{\partial z} \right)={{\mu }_{0}}\gamma \left( {{J}_{1}}-v\rho \right)+\gamma \frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)-\frac{v}{{{c}^{2}}}\left( \frac{\partial {{E}_{1}}}{\partial x}+\frac{\partial {{E}_{2}}}{\partial y}+\frac{\partial {{E}_{3}}}{\partial z} \right)={{\mu }_{0}}\left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)-\frac{v}{{{c}^{2}}}\nabla \cdot \vec{E}={{\mu }_{0}}\left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)-\frac{\rho v}{{{\varepsilon }_{0}}{{c}^{2}}}={{\mu }_{0}}\left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)-{{\mu }_{0}}\rho v={{\mu }_{0}}\left( {{J}_{1}}-v\rho \right)+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \left( \frac{\partial {{B}_{3}}}{\partial y}-\frac{\partial {{B}_{2}}}{\partial z} \right)={{\mu }_{0}}{{J}_{1}}+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \text{Similarly other components will satisfy Lorentz transforamation and in general:} \\
& \vec{\nabla }'\times \vec{B}'={{\mu }_{0}}\vec{J}'+\frac{1}{{{c}^{2}}}\frac{\partial \vec{E}'}{\partial t'}\Rightarrow \vec{\nabla }\times \vec{B}={{\mu }_{0}}\vec{J}+\frac{1}{{{c}^{2}}}\frac{\partial \vec{E}}{\partial t} \\
& \\
\end{align}\]
\subsection{Lorentz Invariance of $ \vec{E}\cdot \vec{B} $ }
\[\begin{align}
& \vec{E}'\cdot \vec{B}'=E_{1}^{'}B_{1}^{'}+E_{2}^{'}B_{2}^{'}+E_{3}^{'}B_{3}^{'} \\
& ={{E}_{1}}{{B}_{1}}+\gamma \left( {{E}_{2}}-v{{B}_{3}} \right)\gamma \left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right)+\gamma \left( {{E}_{3}}+v{{B}_{2}} \right)\gamma \left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right) \\
& ={{E}_{1}}{{B}_{1}}+{{\gamma }^{2}}\left( \left( {{E}_{2}}-v{{B}_{3}} \right)\left( {{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{3}} \right)+\left( {{E}_{3}}+v{{B}_{2}} \right)\left( {{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}} \right) \right) \\
& ={{E}_{1}}{{B}_{1}}+{{\gamma }^{2}}\left( {{E}_{2}}{{B}_{2}}+\frac{v}{{{c}^{2}}}{{E}_{2}}{{E}_{3}}-v{{B}_{2}}{{B}_{3}}-\frac{{{v}^{2}}}{{{c}^{2}}}{{B}_{2}}{{E}_{3}}+{{E}_{3}}{{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}}{{E}_{3}}+v{{B}_{2}}{{B}_{3}}-\frac{v}{{{c}^{2}}}{{E}_{2}}{{E}_{3}} \right) \\
& ={{E}_{1}}{{B}_{1}}+{{\gamma }^{2}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right){{E}_{2}}{{B}_{2}}+{{\gamma }^{2}}\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right){{E}_{3}}{{B}_{3}} \\
& ={{E}_{1}}{{B}_{1}}+{{E}_{2}}{{B}_{2}}+{{E}_{3}}{{B}_{3}} \\
\end{align}\]
\subsection{Invariance of $ B^2-E^2/c^2 $}
\subsection{Invariance of $ P^2-E^2/c^2 $ $ E$ =Energy}
\section{Derivation of Lagrangian Density}
\[\begin{align}
& \nabla \cdot \vec{E}=\frac{\rho }{{{\varepsilon }_{0}}} \\
& \Rightarrow \nabla \cdot \left( -\nabla \phi -\frac{\partial \vec{A}}{\partial t} \right)=\frac{\rho }{{{\varepsilon }_{0}}}=0 \\
\end{align}\]
\begin{equation}
& \Rightarrow \frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)+\nabla \cdot \nabla \phi -\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
\label{eq:lag1}
\end{equation}
\begin{equation}
& \nabla \times \vec{B}={{\mu }_{0}}\vec{J}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t} \\
\end{equation}
\[\begin{align}
& \Rightarrow \nabla \times \left( \nabla \times \vec{A} \right)={{\mu }_{0}}\vec{J}-{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial }{\partial t}\left( \nabla \phi +\frac{\partial \vec{A}}{\partial t} \right) \\
& \Rightarrow \frac{1}{{{c}^{2}}}\frac{\partial }{\partial t}\left( \nabla \phi +\frac{\partial \vec{A}}{\partial t} \right)+\nabla \cdot \left( \frac{\partial \vec{A}}{\partial {{x}_{k}}}-\nabla {{{\vec{A}}}_{k}} \right)-{{\mu }_{0}}\vec{J}=0 \\
& \Rightarrow \frac{\partial }{\partial t}\left( \nabla \phi +\frac{\partial \vec{A}}{\partial t} \right)+\nabla \cdot \left[ {{c}^{2}}\left( \frac{\partial \vec{A}}{\partial {{x}_{k}}}-\nabla {{{\vec{A}}}_{k}} \right) \right]-{{\mu }_{0}}{{c}^{2}}\vec{J}=0 \\
\end{align}\]
\begin{equation}
\Rightarrow \frac{\partial }{\partial t}\left( \frac{\partial \phi }{\partial {{x}_{k}}}+\frac{\partial {{{\vec{A}}}_{k}}}{\partial t} \right)+\nabla \cdot \left[ {{c}^{2}}\left( \frac{\partial \vec{A}}{\partial {{x}_{k}}}-\nabla {{{\vec{A}}}_{k}} \right) \right]-\frac{{{{\vec{J}}}_{k}}}{{{\varepsilon }_{0}}}=0
\end{equation}
\begin{equation}
\frac{\partial }{\partial t}\left( \frac{\partial L}{\partial {{{\dot{A}}}_{\mu }}} \right)+\nabla \cdot \frac{\partial L}{\partial \left( {{\partial }_{k}}{{A}_{\mu }} \right)}-\frac{\partial L}{\partial {{A}_{\mu }}}=0
\end{equation}
\[\begin{align}
& \mu =4 \\
& {{A}_{4}}=\frac{i\phi }{c},{{{\dot{A}}}_{\mu }}=\frac{i\dot{\phi }}{c} \\
& \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial \left( \frac{i\dot{\phi }}{c} \right)} \right)+\nabla \cdot \frac{\partial L}{\partial \left( {{\partial }_{k}}\left( i\phi /c \right) \right)}-\frac{\partial L}{\partial \left( i\phi /c \right)}=0 \\
& \Rightarrow \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial \dot{\phi }} \right)+\nabla \cdot \frac{\partial L}{\partial \left( {{\partial }_{k}}\phi \right)}-\frac{\partial L}{\partial \phi }=0 \\
\end{align}\]
\begin{equation}
& \Rightarrow \frac{\partial }{\partial t}\left( \frac{\partial L}{\partial \dot{\phi }} \right)+\nabla \cdot \frac{\partial L}{\partial \left( \nabla \phi \right)}-\frac{\partial L}{\partial \phi }=0 \\
\label{eq:lagrangefhi}
\end{equation}
Comparing equation 4.5 with 4.1 :
\[\begin{align}
& \frac{\partial L}{\partial \dot{\phi }}=\nabla \cdot \vec{A},\frac{\partial L}{\partial \left( \nabla \phi \right)}=\nabla \phi ,\frac{\partial L}{\partial \phi }=-\frac{\rho }{{{\varepsilon }_{0}}} \\
& {{L}_{{{\alpha }_{1}}}}=\left( \nabla \cdot \vec{A} \right)\dot{\phi } \\
& {{L}_{{{\alpha }_{2}}}}=\frac{{{\left| \nabla \phi \right|}^{2}}}{2} \\
& {{L}_{{{\alpha }_{3}}}}=-\frac{\rho }{{{\varepsilon }_{0}}}\phi \\
& {{L}_{\alpha }}={{L}_{{{\alpha }_{1}}}}+{{L}_{{{\alpha }_{2}}}}+{{L}_{{{\alpha }_{3}}}}=\left( \nabla \cdot \vec{A} \right)\dot{\phi }+\frac{{{\left| \nabla \phi \right|}^{2}}}{2}-\frac{\rho }{{{\varepsilon }_{0}}}\phi \\
\end{align}\]
\begin{equation}
{{L}_{\alpha }}=\left( \nabla \cdot \vec{A} \right)\dot{\phi }+\frac{{{\left| \nabla \phi \right|}^{2}}}{2}-\frac{\rho }{{{\varepsilon }_{0}}}\phi
\end{equation}
Comparing with 4.4 and 4.3 :
\[\begin{align}
& \frac{\partial L}{\partial {{{\dot{A}}}_{\mu }}}=\frac{\partial \phi }{\partial {{x}_{\mu }}}+\frac{\partial {{{\vec{A}}}_{\mu }}}{\partial t}=\nabla \phi +{{{\dot{A}}}_{\mu }}\Rightarrow {{L}_{{{\beta }_{1}}}}=\nabla \phi \cdot {{{\vec{\dot{A}}}}_{\mu }}+\frac{{{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}}{2} \\
& \frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{A}_{\mu }} \right)}={{c}^{2}}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}-\nabla {{{\vec{A}}}_{\mu }} \right) \\
& \Rightarrow \frac{\partial L}{\partial \left( \nabla {{A}_{\mu }} \right)}={{c}^{2}}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}-\nabla {{{\vec{A}}}_{\mu }} \right) \\
& \Rightarrow {{L}_{{{\beta }_{2}}}}=\frac{{{c}^{2}}}{2}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }}-\frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2} \right) \\
& \frac{\partial L}{\partial {{A}_{\mu }}}=\frac{{{{\vec{J}}}_{k}}}{{{\varepsilon }_{0}}} \\
& \Rightarrow {{L}_{{{\beta }_{3}}}}=\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}}{{{\varepsilon }_{0}}} \\
& {{L}_{\beta }}={{L}_{{{\beta }_{1}}}}+{{L}_{{{\beta }_{2}}}}+{{L}_{{{\beta }_{3}}}} \\
\end{align}\]
\begin{equation}
& {{L}_{\beta }}=\nabla \phi \cdot {{{\vec{\dot{A}}}}_{\mu }}+\frac{{{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}}{2}+\frac{{{c}^{2}}}{2}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }}-\frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2} \right)+\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}}{{{\varepsilon }_{0}}} \\
\end{equation}
\begin{equation}
L={{L}_{\alpha }}+{{L}_{\beta }}=\left( \nabla \cdot \vec{A} \right)\dot{\phi }+\frac{{{\left| \nabla \phi \right|}^{2}}}{2}-\frac{\rho }{{{\varepsilon }_{0}}}\phi +\nabla \phi \cdot {{{\vec{\dot{A}}}}_{\mu }}+\frac{{{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}}{2}+\frac{{{c}^{2}}}{2}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }}-\frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2} \right)+\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}}{{{\varepsilon }_{0}}}
\label{eq:Total Lagrangian}
\end{equation}
The Lagrangian density given by equation ~\eqref{eq:Total Lagrangian} is wrong as it does not fit into equation ~\ref{eq:lagrangefhi}
Comparing equation ~\ref{eq:lagrangefhi} with ~\ref{eq:lag1} we will get:
\[\begin{align}
& \frac{\partial L}{\partial \dot{\phi} }=\nabla \cdot \vec{A},\frac{\partial L}{\partial \left( \nabla \phi \right)}=\nabla \phi +\vec{\dot{A}},\frac{\partial L}{\partial \phi }=-\frac{\rho }{{{\varepsilon }_{0}}} \\
& \text{Putting these values in equation ~\ref{eq:lag1}}\\
& \frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)+\nabla \cdot \left( \nabla \phi \right)-\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
& \Rightarrow \frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)+\nabla \cdot \left( \nabla \phi +\vec{\dot{A}} \right)-\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
& \Rightarrow \frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)+\nabla \cdot \nabla \phi +\nabla \cdot \frac{\partial \vec{A}}{\partial t}-\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
& \Rightarrow \frac{\partial }{\partial t}\left( 2\left( \nabla \cdot \vec{A} \right) \right)+\nabla \cdot \nabla \phi -\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
\end{align}\]
We find extra term in $ \frac{\partial}{\partial t}(\nabla\cdot \vec{A})$ this extra term appears due to $(\nabla\cdot \vec{A})$ in $ L_{\beta_{1}} $. The equation ~\ref{eq:lag1}} can be written as :
\[\begin{align}
& \frac{\partial L}{\partial \phi }=\nabla \cdot \vec{A},\frac{\partial L}{\partial \left( \nabla \phi \right)}=\nabla \phi +\vec{\dot{A}},\frac{\partial L}{\partial \phi }=-\frac{\rho }{{{\varepsilon }_{0}}} \\
& \frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)+\nabla \cdot \nabla \phi -\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
& {{\nabla }^{2}}\phi +\frac{\partial }{\partial t}\left( \nabla \cdot \vec{A} \right)-\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
\end{align}\]
\begin{equation} & \nabla \cdot \left( \nabla \phi +\frac{\partial \vec{A}}{\partial t} \right)-\left( -\frac{\rho }{{{\varepsilon }_{0}}} \right)=0 \\
\end{equation}
\[\begin{align}
& \frac{\partial L}{\partial \dot{\phi }}=0\Rightarrow L_{{{\alpha }_{1}}}^{'}=0 \\
& \frac{\partial L}{\partial \left( \nabla \phi \right)}=\nabla \phi +\vec{\dot{A}}\Rightarrow L_{{{\alpha }_{2}}}^{'}=\frac{{{\left| \nabla \phi \right|}^{2}}}{2}+\left( \nabla \phi \cdot \vec{A} \right) \\
& \frac{\partial L}{\partial \phi }=-\frac{\rho }{{{\varepsilon }_{0}}}\Rightarrow L_{{{\alpha }_{3}}}^{'}={{L}_{{{\alpha }_{3}}}}=-\frac{\rho \phi }{{{\varepsilon }_{0}}} \\
& \Rightarrow L_{\alpha }^{'}=L_{{{\alpha }_{1}}}^{'}+L_{{{\alpha }_{2}}}^{'}+L_{{{\alpha }_{3}}}^{'}=\frac{{{\left| \nabla \phi \right|}^{2}}}{2}+\left( \nabla \phi \cdot \vec{A} \right)-\frac{\rho \phi }{{{\varepsilon }_{0}}} \\
& \text{it is necessary to omit }\nabla \phi \cdot \vec{\dot{A}}\,\text{from}\,{{L}_{{{\beta }_{1}}}} \\
& L_{{{\beta }_{1}}}^{'}\ne {{L}_{{{\beta }_{1}}}}=\frac{{{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}}{2} \\
& L_{{{\beta }_{2}}}^{'}={{L}_{{{\beta }_{2}}}}=\frac{{{c}^{2}}}{2}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }}-\frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2} \right) \\
& L_{{{\beta }_{3}}}^{'}={{L}_{{{\beta }_{3}}}}=\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}}{{{\varepsilon }_{0}}} \\
& \Rightarrow {{L}_{corrected}}=L_{\alpha }^{'}+{{L}_{\beta }} \\
& =\frac{{{\left| \nabla \phi \right|}^{2}}}{2}+\left( \nabla \phi \cdot \vec{A} \right)-\frac{\rho \phi }{{{\varepsilon }_{0}}}+\nabla \phi \cdot {{{\vec{\dot{A}}}}_{\mu }}+\frac{{{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}}{2}+\frac{{{c}^{2}}}{2}\left( \frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }}-\frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2} \right)+\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}}{{{\varepsilon }_{0}}} \\
& =\frac{1}{2}\left( {{\left| {{{\dot{A}}}_{\mu }} \right|}^{2}}+{{\left| \nabla \phi \right|}^{2}}+2\nabla \phi \cdot {{{\vec{\dot{A}}}}_{\mu }} \right)-\frac{{{c}^{2}}}{2}\left( \frac{{{\left| \nabla {{A}_{\mu }} \right|}^{2}}}{2}-\frac{\partial \vec{A}}{\partial {{x}_{\mu }}}\cdot \nabla {{A}_{\mu }} \right)+\frac{{{{\vec{J}}}_{\mu }}\cdot {{{\vec{A}}}_{\mu }}-\rho \phi }{{{\varepsilon }_{0}}} \\
& =\frac{1}{2}\left( {{\left| -\nabla \phi -\frac{\partial A}{\partial t} \right|}^{2}} \right)-\frac{{{c}^{2}}}{2}\left( {{\left| \nabla \times \vec{A} \right|}^{2}} \right)+\frac{{{J}_{1}}{{A}_{1}}+{{J}_{2}}{{A}_{2}}+{{J}_{3}}{{A}_{3}}+\left( ic\rho \right)\left( i\phi /c \right)}{{{\varepsilon }_{0}}} \\
& =\frac{1}{2}\left( {{E}^{2}}-{{c}^{2}}{{B}^{2}} \right)+\frac{{{J}_{1}}{{A}_{1}}+{{J}_{2}}{{A}_{2}}+{{J}_{3}}{{A}_{3}}+{{J}_{4}}{{A}_{4}}}{{{\varepsilon }_{0}}} \\
& =\frac{1}{2}\left( {{E}^{2}}-{{c}^{2}}{{B}^{2}} \right)+\frac{\vec{J}\cdot \vec{A}}{{{\varepsilon }_{0}}} \\
& =\frac{-{{c}^{2}}}{4}\left[ 2\left( {{B}^{2}}-{{\frac{E}{{{c}^{2}}}}^{2}} \right) \right]+\frac{\vec{J}\cdot \vec{A}}{{{\varepsilon }_{0}}} \\
& =\frac{-{{c}^{2}}}{4}{{F}_{\mu \nu }}{{F}^{\mu \nu }}+\frac{\vec{J}\cdot \vec{A}}{{{\varepsilon }_{0}}} \\
& L/V={{\varepsilon }_{o}}L=\frac{-{{c}^{2}}{{\varepsilon }_{0}}}{4}{{F}_{\mu \nu }}{{F}^{\mu \nu }}+\vec{J}\cdot \vec{A} \\
& =-\frac{{{F}_{\mu \nu }}{{F}^{\mu \nu }}}{4{{\mu }_{0}}}+\vec{J}\cdot \vec{A} \\
\end{align}\]
\subsection{Proof of $ \sum\limits_{\mu =1,\nu =1}^{\mu =4,\nu =4}{{{F}_{\mu \nu }}{{F}^{\mu \nu }}} = 2(B^2-E^2/c^2) $} We have:\\
\[\begin{align}
& 2\left( {{B}^{2}}-\frac{{{E}^{2}}}{{{c}^{2}}} \right)=2\left( B_{1}^{2}+B_{1}^{2}+B_{1}^{2}-\frac{B_{1}^{2}}{{{c}^{2}}}-\frac{B_{1}^{2}}{{{c}^{2}}}-\frac{B_{1}^{2}}{{{c}^{2}}} \right) \\
& =2\left( B_{1}^{2}+B_{2}^{2}+B_{3}^{2}-\frac{E_{1}^{2}}{{{c}^{2}}}-\frac{E_{2}^{2}}{{{c}^{2}}}-\frac{E_{3}^{2}}{{{c}^{2}}} \right) \\
& =\left( 2{{\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{3}}} \right)}^{2}}+2{{\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}} \right)}^{2}}+2{{\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)}^{2}}+2{{\left( \frac{i{{E}_{1}}}{c} \right)}^{2}}+2{{\left( \frac{i{{E}_{2}}}{c} \right)}^{2}}+2{{\left( \frac{i{{E}_{3}}}{c} \right)}^{2}} \right) \\
& 2{{\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{3}}} \right)}^{2}}={{\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{3}}} \right)}^{2}}+{{\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{2}}} \right)}^{2}}={{\left( {{F}_{23}} \right)}^{2}}+{{\left( {{F}_{32}} \right)}^{2}} \\
& 2{{\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}} \right)}^{2}}={{\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{A}_{3}}}{\partial {{x}_{1}}} \right)}^{2}}+{{\left( \frac{\partial {{A}_{3}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{3}}} \right)}^{2}}={{\left( {{F}_{31}} \right)}^{2}}+{{\left( {{F}_{13}} \right)}^{2}} \\
& 2{{\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)}^{2}}={{\left( \frac{\partial {{A}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)}^{2}}+{{\left( \frac{\partial {{A}_{1}}}{\partial {{x}_{2}}}-\frac{\partial {{A}_{2}}}{\partial {{x}_{1}}} \right)}^{2}}={{\left( {{F}_{12}} \right)}^{2}}+{{\left( {{F}_{21}} \right)}^{2}} \\
& 2{{\left( \frac{i{{E}_{1}}}{c} \right)}^{2}}={{\left( \frac{i{{E}_{1}}}{c} \right)}^{2}}+{{\left( -\frac{i{{E}_{1}}}{c} \right)}^{2}}={{\left( {{F}_{14}} \right)}^{2}}+{{\left( {{F}_{41}} \right)}^{2}} \\
& 2{{\left( \frac{i{{E}_{2}}}{c} \right)}^{2}}={{\left( \frac{i{{E}_{2}}}{c} \right)}^{2}}+{{\left( -\frac{i{{E}_{2}}}{c} \right)}^{2}}={{\left( {{F}_{24}} \right)}^{2}}+{{\left( {{F}_{42}} \right)}^{2}} \\
& 2{{\left( \frac{i{{E}_{3}}}{c} \right)}^{2}}={{\left( \frac{i{{E}_{3}}}{c} \right)}^{2}}+{{\left( -\frac{i{{E}_{3}}}{c} \right)}^{2}}={{\left( {{F}_{34}} \right)}^{2}}+{{\left( {{F}_{43}} \right)}^{2}} \\
\end{align}\]
\[\begin{align}
&\Rightarrow 2\left( {{B}^{2}}-\frac{{{E}^{2}}}{{{c}^{2}}} \right)=\left( 2{{B}^{2}}-2\frac{{{E}^{2}}}{{{c}^{2}}} \right)\\
\end{align}\]
\[=\left( \begin{align}
& {{\left( {{F}_{23}} \right)}^{2}}+{{\left( {{F}_{32}} \right)}^{2}}+{{\left( {{F}_{31}} \right)}^{2}}+{{\left( {{F}_{12}} \right)}^{2}}+{{\left( {{F}_{12}} \right)}^{2}}+{{\left( {{F}_{21}} \right)}^{2}} \\
& +{{\left( {{F}_{14}} \right)}^{2}}+{{\left( {{F}_{41}} \right)}^{2}}+{{\left( {{F}_{24}} \right)}^{2}}+{{\left( {{F}_{42}} \right)}^{2}}+{{\left( {{F}_{34}} \right)}^{2}}+{{\left( {{F}_{43}} \right)}^{2}} \\
\end{align} \right)\]
\[\begin{align}
& =\left\{ 0+{{\left( {{F}_{12}} \right)}^{2}}+{{\left( {{F}_{13}} \right)}^{2}}+{{\left( {{F}_{14}} \right)}^{2}}+{{\left( {{F}_{21}} \right)}^{2}}+0+{{\left( {{F}_{23}} \right)}^{2}}+{{\left( {{F}_{24}} \right)}^{2}} \right. \\
& \left. +{{\left( {{F}_{31}} \right)}^{2}}+{{\left( {{F}_{32}} \right)}^{2}}+0+{{\left( {{F}_{34}} \right)}^{2}}+{{\left( {{F}_{41}} \right)}^{2}}+{{\left( {{F}_{42}} \right)}^{2}}+{{\left( {{F}_{43}} \right)}^{2}}+0 \right\} \\
\end{align}\]
\[\begin{align}
& =\left\{ {{\left( {{F}_{11}} \right)}^{2}}+{{\left( {{F}_{12}} \right)}^{2}}+{{\left( {{F}_{13}} \right)}^{2}}+{{\left( {{F}_{14}} \right)}^{2}}+{{\left( {{F}_{21}} \right)}^{2}}+{{\left( {{F}_{22}} \right)}^{2}}+{{\left( {{F}_{23}} \right)}^{2}}+{{\left( {{F}_{24}} \right)}^{2}} \right. \\
& \left. +{{\left( {{F}_{31}} \right)}^{2}}+{{\left( {{F}_{32}} \right)}^{2}}+{{\left( {{F}_{33}} \right)}^{2}}+{{\left( {{F}_{34}} \right)}^{2}}+{{\left( {{F}_{41}} \right)}^{2}}+{{\left( {{F}_{42}} \right)}^{2}}+{{\left( {{F}_{43}} \right)}^{2}}+{{\left( {{F}_{44}} \right)}^{2}} \right\} \\
& \[=\sum\limits_{\mu =1,\nu =1}^{\mu =4,\nu =4}{{{F}_{\mu \nu }}{{F}^{\mu \nu }}}\]
\end{align}\]
\section{Euler-Lagrange Equation Derivation }
Euler Lagrange Equation for Charge Particle
\[\begin{align}
& {{\partial }_{\nu }}\left( \frac{\partial L}{\partial \left( {{\partial }_{\nu }}{{A}_{\mu }} \right)} \right)-\left( \frac{\partial L}{\partial {{A}_{\mu }}} \right)=0 \\
& \Rightarrow {{\partial }_{\nu }}\left( \frac{\partial L}{\partial \left( {{\partial }_{\nu }}{{A}_{\mu }} \right)} \right)-{{J}^{\mu }}=0 \\
& \Rightarrow {{\partial }_{\nu }}\left( -\frac{{{F}^{\nu \mu }}}{{{\mu }_{0}}} \right)-{{J}^{\mu }}=0 \\
& \Rightarrow {{\partial }_{\nu }}\left( {{F}^{\nu \mu }} \right)+{{\mu }_{0}}{{J}^{\mu }}=0 \\
& \mu =1 \\
& \\
\end{align}\]
\[\begin{align}
&$ \intertext{Lagrangian density of the charge particle is:}$\\
& L=2\left( F_{12}^{2}+F_{13}^{2}\,+F_{14}^{2}+F_{23}^{2}+F_{24}^{2}+F_{34}^{2} \right) \\
& =2\left( {{\left( \frac{\partial A}{\partial x}-\frac{\partial A}{\partial x} \right)}^{2}}+{{\left( \frac{\partial A}{\partial x}-\frac{\partial A}{\partial x} \right)}^{2}}+{{\left( \frac{\partial A}{\partial x}-\frac{\partial A}{\partial x} \right)}^{2}} \right) \\
&$ \intertext{The classical Euler-Lagrange equation is :}$\\
& \frac{d}{dt}\left( \frac{\partial L}{\partial {{{\dot{q}}}_{k}}} \right)-\frac{\partial L}{\partial {{q}_{k}}}=0 \\
& \frac{d}{dt}\left( \frac{\partial L}{\partial \frac{\partial {{q}_{k}}}{\partial t}} \right)-\frac{\partial L}{\partial {{q}_{k}}}=0 \\
&$ \intertext{The electrodynamic Euler-Lagrange equation is :}$\\
\end{align}\]
\begin{align}
& \frac{\partial }{\partial {{x}_{k}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{\mu }}}{\partial {{x}_{k}}} \right)} \right)-\frac{\partial }{\partial {{A}_{\mu }}}=0
\end{align}
\subsection{Derivation of Maxwell's Equations from Euler Lagrange Equation of Charge particle }
\[\begin{align}
& \text{Set }\mu \text{=1,k=1,2,3 }\!\!\And\!\!\text{ 4} \\
& \Rightarrow \frac{\partial }{\partial {{x}_{1}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{1}}}{\partial {{x}_{1}}} \right)} \right)-\frac{\partial }{\partial {{A}_{1}}}=0 \\
\end{align}\]
Expanding this equation for $ \mu =1 $ :
\[\begin{align}
& \frac{\partial }{\partial {{x}_{1}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{1}}}{\partial {{x}_{1}}} \right)} \right)+\frac{\partial }{\partial {{x}_{2}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{1}}}{\partial {{x}_{2}}} \right)} \right)+\frac{\partial }{\partial {{x}_{3}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{1}}}{\partial {{x}_{3}}} \right)} \right)+\frac{\partial }{\partial {{x}_{4}}}\left( \frac{\partial L}{\partial \left( \frac{\partial {{A}_{1}}}{\partial {{x}_{4}}} \right)} \right)-\frac{\partial L}{\partial {{A}_{1}}}=0 \\
& \Rightarrow \frac{\partial }{\partial {{x}_{1}}}\left( \frac{{{F}_{11}}}{{{\mu }_{0}}} \right)+\frac{\partial }{\partial {{x}_{2}}}\left( \frac{{{F}_{12}}}{{{\mu }_{0}}} \right)+\frac{\partial }{\partial {{x}_{3}}}\left( \frac{{{F}_{13}}}{{{\mu }_{0}}} \right)+\frac{\partial }{\partial {{x}_{4}}}\left( \frac{{{F}_{14}}}{{{\mu }_{0}}} \right)-{{J}_{1}}=0 \\
& \Rightarrow \frac{\partial {{F}_{11}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{12}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{13}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{14}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{1}} \\
& \text{For }\mu \text{=2:} \\
& \Rightarrow \frac{\partial {{F}_{21}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{22}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{23}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{24}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{2}} \\
& \text{For }\mu \text{=3:} \\
& \Rightarrow \frac{\partial {{F}_{31}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{32}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{33}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{34}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{3}} \\
& \text{For }\mu \text{=4:} \\
& \Rightarrow \frac{\partial {{F}_{41}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{42}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{43}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{44}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{4}} \\
\end{align}\]
\subsection{Maxwells first field equation: Gauss Law from Euler Lagrange Equation}
\[\begin{align}
& \text{Consider the case for }\mu \text{=4:} \\
& \frac{\partial {{F}_{41}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{42}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{43}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{44}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{4}} \\
& \Rightarrow \frac{\partial \left( i{{E}_{1}}/c \right)}{\partial {{x}_{1}}}+\frac{\partial \left( i{{E}_{2}}/c \right)}{\partial {{x}_{2}}}+\frac{\partial \left( i{{E}_{3}}/c \right)}{\partial {{x}_{3}}}+\frac{\partial \left( 0 \right)}{\partial {{x}_{4}}}={{\mu }_{0}}\left( ic\rho \right) \\
& \Rightarrow \frac{\partial {{E}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{E}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{E}_{3}}}{\partial {{x}_{3}}}={{\mu }_{0}}{{c}^{2}}\rho \\
& \Rightarrow \nabla \cdot \vec{E}=\frac{\rho }{{{\varepsilon }_{0}}} \\
\end{align}\]
\subsection{Maxwells fourth field equation: Modified Ampere's Law:from Euler Lagrange Equation}
Derivation of Maxwells 4th field equation : Modified Amperes Law: Considering the cases for $ \mu=1,2,3 $
\[\begin{align}
& \text{For }\mu \text{=1:} \\
& \frac{\partial {{F}_{11}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{12}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{13}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{14}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \frac{\partial \left( 0 \right)}{\partial {{x}_{1}}}+\frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}}-\frac{\partial \left( i{{E}_{1}}/c \right)}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}}-\frac{\partial \left( i{{E}_{1}}/c \right)}{\partial \left( ict \right)}={{\mu }_{0}}{{J}_{1}} \\
& \Rightarrow \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}}={{\mu }_{0}}{{J}_{1}}+\frac{1}{{{c}^{2}}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \Rightarrow \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}}={{\mu }_{0}}{{J}_{1}}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial {{E}_{1}}}{\partial t} \\
& \text{For }\mu \text{=2:} \\
& \frac{\partial {{F}_{21}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{22}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{23}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{24}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{2}} \\
& \text{Will yield :} \\
& \frac{\partial {{B}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{B}_{3}}}{\partial {{x}_{1}}}={{\mu }_{0}}{{J}_{2}}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial {{E}_{2}}}{\partial t} \\
& \text{Similarly}\,\text{for}\mu \text{=3 :} \\
& \frac{\partial {{F}_{31}}}{\partial {{x}_{1}}}+\frac{\partial {{F}_{32}}}{\partial {{x}_{2}}}+\frac{\partial {{F}_{33}}}{\partial {{x}_{3}}}+\frac{\partial {{F}_{34}}}{\partial {{x}_{4}}}={{\mu }_{0}}{{J}_{3}} \\
& \text{will give:} \\
& \frac{\partial {{B}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{B}_{1}}}{\partial {{x}_{2}}}={{\mu }_{0}}{{J}_{3}}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial {{E}_{3}}}{\partial t} \\
& \text{Combining these three equations in vector form:} \\
& \text{LHS is:} \\
& \left( \frac{\partial {{B}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{B}_{2}}}{\partial {{x}_{3}}} \right)\hat{i}+\left( \frac{\partial {{B}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{B}_{3}}}{\partial {{x}_{1}}} \right)\hat{j}+\left( \frac{\partial {{B}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{B}_{1}}}{\partial {{x}_{2}}} \right)\hat{k} \\
& =\nabla \times \vec{B} \\
& \text{RHS is:} \\
& {{\mu }_{0}}\left( {{J}_{1}}\hat{i}+{{J}_{2}}\hat{j}+{{J}_{3}}\hat{k} \right)+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial }{\partial t}\left( {{E}_{1}}\hat{i}+{{E}_{2}}\hat{j}+{{E}_{3}}\hat{k} \right) \\
& ={{\mu }_{0}}\vec{J}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t} \\
& \text{Thus we will get :} \\
& \nabla \times \vec{B}={{\mu }_{0}}\vec{J}+{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial \vec{E}}{\partial t} \\
\end{align}\]
\subsection{Maxwell's 2nd and 3rd equations from Euler Lagrange equation}
Maxwell's 2nd and 3rd equatiions are homogeneous differential equations So $J=0 $
These equations can not be derived from the simplified Euler-Lagrange equation in terms of covariant electromagnetic field Tensor:\[\frac{\partial {{F}_{\mu \nu }}}{\partial {{x}_{\nu }}}={{\mu }_{0}}{{J}_{\mu }}\]\\ For this, we have to define a \textbf{Dual Tensor G} as given by :\\
\[\begin{align}
& {{G}^{\mu \nu }}=\frac{1}{2}{{\varepsilon }^{\alpha \beta \gamma \delta }}{{F}_{\mu \nu }} \\
& \because {{F}_{\mu \nu }}=\left( \begin{matrix}
0 & {{E}_{1}}/c & {{E}_{2}}/c & {{E}_{3}}/c \\
-{{E}_{1}}/c & 0 & -{{B}_{3}} & {{B}_{2}} \\
-{{E}_{2}}/c & {{B}_{3}} & 0 & -{{B}_{1}} \\
-{{E}_{3}}/c & -{{B}_{2}} & {{B}_{1}} & 0 \\
\end{matrix} \right) \\
& \text{By Changing E}\to \text{B }\!\!\And\!\!\text{ B}\to \text{-E one can get Dual Tensor from }{{\text{F}}_{\mu \nu }} \\
& {{G}^{\mu \nu }}=\left( \begin{matrix}
0 & -{{B}_{1}} & -{{B}_{2}} & -{{B}_{3}} \\
{{B}_{1}} & 0 & {{E}_{3}}/c & -{{E}_{2}}/c \\
{{B}_{2}} & -{{E}_{3}}/c & 0 & {{E}_{1}}/c \\
{{B}_{3}} & {{E}_{2}}/c & -{{E}_{1}}/c & 0 \\
\end{matrix} \right) \\
& \text{Since}:{{\partial }_{\mu }}{{F}^{\mu \nu }}=0 \\
& \Rightarrow {{\partial }_{\mu }}{{G}^{\mu \nu }}=0 \\
& \Rightarrow \frac{\partial {{G}^{\mu \nu }}}{\partial {{x}^{\mu }}}=0 \\
& for\,\mu =1,2,3\And \nu =0 \\
& {{\partial }_{1}}{{G}^{10}}+{{\partial }_{2}}{{G}^{20}}+{{\partial }_{3}}{{G}^{30}}=0 \\
& \Rightarrow \frac{\partial {{G}^{10}}}{\partial {{x}_{1}}}+\frac{\partial {{G}^{20}}}{\partial {{x}_{2}}}+\frac{\partial {{G}^{30}}}{\partial {{x}_{3}}}=0 \\
& \Rightarrow \frac{\partial {{B}_{1}}}{\partial {{x}_{1}}}+\frac{\partial {{B}_{2}}}{\partial {{x}_{2}}}+\frac{\partial {{B}_{3}}}{\partial {{x}_{3}}}=0 \\
& \Rightarrow \nabla \cdot \vec{B}=0 \\
\end{align}\]
\subsection{Derivation of Maxwell's 3rd equation from Euler-Lagrange equation}
\[\begin{align}
& \mu =0,2,3\And \nu =1 \\
& \Rightarrow {{\partial }_{0}}{{G}^{01}}+{{\partial }_{2}}{{G}^{21}}+{{\partial }_{3}}{{G}^{31}}=0 \\
& \Rightarrow -{{\partial }_{0}}{{B}_{1}}+{{\partial }_{2}}\left( -i{{E}_{3}}/c \right)+{{\partial }_{3}}\left( i{{E}_{2}}/c \right)=0 \\
& \Rightarrow -\frac{\partial {{B}_{1}}}{\partial {{x}_{0}}}=\frac{i}{c}\left( \frac{\partial {{E}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{E}_{2}}}{\partial {{x}_{3}}} \right) \\
& \Rightarrow -\frac{1}{ic}\frac{\partial {{B}_{1}}}{\partial t}=\frac{i}{c}\left( \frac{\partial {{E}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{E}_{2}}}{\partial {{x}_{3}}} \right) \\
\end{align}\]
\begin{equation}
& \Rightarrow -\frac{\partial {{B}_{1}}}{\partial t}=\frac{\partial {{E}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{E}_{2}}}{\partial {{x}_{3}}} \\
\label{eq:homo1}
\end{equation}
\[\begin{align}
& \mu =0,1,3\And \nu =2 \\
& \Rightarrow {{\partial }_{0}}{{G}^{02}}+{{\partial }_{1}}{{G}^{12}}+{{\partial }_{3}}{{G}^{32}}=0 \\
& \Rightarrow -{{\partial }_{0}}{{B}_{2}}+{{\partial }_{1}}\left( i{{E}_{3}}/c \right)+{{\partial }_{3}}\left( -i{{E}_{1}}/c \right)=0 \\
& \Rightarrow -\frac{1}{ic}\frac{\partial {{B}_{2}}}{\partial t}=-\frac{i}{c}\left( \frac{\partial {{E}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{E}_{3}}}{\partial {{x}_{1}}} \right) \\
\end{align}\]
\begin{equation}
& \Rightarrow -\frac{\partial {{B}_{2}}}{\partial t}=\left( \frac{\partial {{E}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{E}_{3}}}{\partial {{x}_{1}}} \right) \\
\label{eq:homo2}
\end{equation}
\[\begin{align}
& \mu =0,1,2\And \nu =3 \\
& \Rightarrow {{\partial }_{0}}{{G}^{03}}+{{\partial }_{1}}{{G}^{13}}+{{\partial }_{2}}{{G}^{23}}=0 \\
& \Rightarrow -{{\partial }_{0}}{{B}_{3}}+{{\partial }_{1}}\left( -i{{E}_{2}}/c \right)+{{\partial }_{2}}\left( i{{E}_{1}}/c \right)=0 \\
& \Rightarrow -\frac{1}{ic}\frac{\partial {{B}_{3}}}{\partial t}=\frac{-i}{c}\left( \frac{\partial {{E}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{E}_{1}}}{\partial {{x}_{2}}} \right) \\
\end{align}\]
\begin{equation}
& \Rightarrow -\frac{\partial {{B}_{3}}}{\partial t}=\left( \frac{\partial {{E}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{E}_{1}}}{\partial {{x}_{2}}} \right) \\
\label{eq:homo3}
\end{equation}
Combining equations ~\ref{eq:homo1},~\ref{eq:homo2} \& ~\ref{eq:homo3} in vector form we will get :\\
\[\begin{align}
& \left( \frac{\partial {{E}_{3}}}{\partial {{x}_{2}}}-\frac{\partial {{E}_{2}}}{\partial {{x}_{3}}} \right)\hat{i}+\left( \frac{\partial {{E}_{1}}}{\partial {{x}_{3}}}-\frac{\partial {{E}_{3}}}{\partial {{x}_{1}}} \right)\hat{j}+\left( \frac{\partial {{E}_{2}}}{\partial {{x}_{1}}}-\frac{\partial {{E}_{1}}}{\partial {{x}_{2}}} \right)\hat{k}=-\left( \frac{\partial {{B}_{1}}}{\partial t}\hat{i}+\frac{\partial {{B}_{2}}}{\partial t}\hat{j}+\frac{\partial {{B}_{3}}}{\partial t}\hat{k} \right) \\
& \Rightarrow \nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} \\
\end{align}\]
\section{Lorentz Force in covariant form in terms of Electromagnetic field Tensor}
\[\begin{align}
& \vec{F}=q\vec{E}+q\left( \vec{v}\times \vec{B} \right) \\
& =\rho V\vec{E}+\rho V\left( \vec{v}\times \vec{B} \right) \\
& f=\rho \vec{E}+\left( \rho \vec{v}\times \vec{B} \right) \\
& =\rho \vec{E}+\left( \vec{J}\times \vec{B} \right) \\
& \text{The x-component of the Lorentz force per unit volume is :} \\
& {{f}_{1}}=\rho {{E}_{1}}+\left( {{J}_{2}}{{B}_{3}}-{{J}_{3}}{{B}_{2}} \right) \\
& =0{{J}_{1}}+{{B}_{3}}{{J}_{2}}+{{\left( -B \right)}_{2}}{{J}_{3}}+ic\rho \left( -\frac{i{{E}_{1}}}{c} \right) \\
& =0{{J}_{1}}+{{B}_{3}}{{J}_{2}}+\left( -{{B}_{2}} \right){{J}_{3}}+\left( -\frac{i{{E}_{1}}}{c} \right)ic\rho \\
& ={{F}_{11}}{{J}_{1}}+{{F}_{12}}{{J}_{2}}+{{F}_{13}}{{J}_{3}}+{{F}_{14}}{{J}_{4}} \\
& \text{Similarly y-component of the Lorentz force per unit volume is :} \\
& {{f}_{2}}={{F}_{21}}{{J}_{1}}+{{F}_{22}}{{J}_{2}}+{{F}_{23}}{{J}_{3}}+{{F}_{24}}{{J}_{4}} \\
& \text{Similarly z-component of the Lorentz force per unit volume is :} \\
& {{f}_{3}}={{F}_{31}}{{J}_{1}}+{{F}_{32}}{{J}_{2}}+{{F}_{33}}{{J}_{3}}+{{F}_{34}}{{J}_{4}} \\
& \text{Now , }{{f}_{1}},{{f}_{2}}\And {{f}_{3}}\text{ can be written as :Force-density four vector:} \\
& {{f}_{\mu }}={{F}_{\mu \nu }}{{J}_{\nu }} \\
& \text{since }\frac{\partial {{F}_{\mu \nu }}}{\partial {{x}_{\nu }}}={{\mu }_{0}}{{J}_{\mu }} \\
& {{f}_{\mu }}=\frac{1}{{{\mu }_{0}}}\sum{{{F}_{\mu \nu }}}\frac{\partial {{F}_{\nu \lambda }}}{\partial {{x}_{\lambda }}} \\
\end{align}\]
This is Tensor form of the Lorentz force density vector in covariant form.
\subsection{Significance of 4th Component of Lorentz force density vector}
\[\begin{align}
& {{f}_{\mu }}=\sum{{{F}_{\mu \nu }}}{{J}_{\nu }} \\
& \Rightarrow {{f}_{4}}=\sum{{{F}_{4\nu }}}{{J}_{\nu }} \\
& ={{F}_{41}}{{J}_{1}}{{+}_{42}}{{J}_{2}}{{+}_{43}}{{J}_{3}}{{+}_{44}}{{J}_{4}} \\
& =\left( \frac{i{{E}_{1}}}{c} \right){{J}_{1}}+\left( \frac{i{{E}_{2}}}{c} \right){{J}_{2}}+\left( \frac{i{{E}_{3}}}{c} \right){{J}_{3}}+\left( 0 \right){{J}_{4}} \\
& =\frac{i}{c}\left( {{E}_{1}}{{J}_{1}}+{{E}_{2}}{{J}_{2}}+{{E}_{3}}{{J}_{3}} \right) \\
& =\frac{i}{c}\left( \vec{E}\centerdot \vec{J} \right) \\
& =\frac{i\rho }{c}\left( \vec{E}\centerdot \frac{{\vec{J}}}{\rho } \right) \\
& =\frac{i\rho }{c}\left( \vec{E}\centerdot \vec{v} \right) \\
& =\frac{i}{c}\left( \frac{q\vec{E}}{V}\centerdot \vec{v} \right) \\
& =\frac{i}{c}\left( \frac{{\vec{F}}}{V}\centerdot \vec{v} \right) \\
& =\frac{i}{c}\left( \frac{\vec{F}\centerdot \vec{v}}{V} \right)=\frac{i}{c}\left( \frac{P}{V} \right) \\
\end{align}\]
Thus, the fourth component of Lorentz force density vector represents workdone per unit time per unit volume by the electric field on the charge \\
\subsection{Derivation of Lorentz transformation equation for Lorentz force}
\[\begin{align}
& f_{\mu }^{'}={{\alpha }_{\mu \nu }}{{f}_{\nu }} \\
& \Rightarrow \left( \begin{matrix}
f_{1}^{'} \\
f_{2}^{'} \\
f_{3}^{'} \\
f_{4}^{'} \\
\end{matrix} \right)=\left( \begin{matrix}
\gamma & 0 & 0 & i\beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i\beta \gamma & 0 & 0 & \gamma \\
\end{matrix} \right)\left( \begin{matrix}
{{f}_{1}} \\
{{f}_{2}} \\
{{f}_{3}} \\
{{f}_{4}} \\
\end{matrix} \right) \\
& \Rightarrow f_{1}^{'}=\gamma {{f}_{1}}+i\beta \gamma {{f}_{4}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,=\gamma {{f}_{1}}+i\beta \gamma \left( 0 \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,=\gamma {{f}_{1}}\,\,\,\,\, \\
& f_{2}^{'}={{f}_{2}}\,\,\,\,\,\,\, \\
& f_{3}^{'}={{f}_{3}} \\
& f_{4}^{'}={{f}_{4}}=0 \\
& \because \text{ in the rest frame no work is done on the moving charge} \\
& \text{But} \\
& f_{\mu }^{'}=\int{{{f}_{\mu }}}d\tau ' \\
& \Rightarrow f_{x}^{'}=\int{f_{_{1}}^{'}}d\tau '=\int{\gamma {{f}_{1}}}d\tau \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\
& =\int{{{f}_{1}}}d\tau ={{F}_{1}}={{F}_{x}} \\
& \Rightarrow F_{x}^{'}={{F}_{x}} \\
& F_{y}^{'}=\int{{{f}_{2}}}d\tau \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}={{F}_{y}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\
& F_{z}^{'}=\int{{{f}_{3}}}d\tau \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}={{F}_{z}}\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\
\end{align}\]