Maxwell’s Equations

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Maxwell’s Equations

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In a vaccum i.e. \rho=0, j=0, Maxwell’s Equations In SI and CGS system can be expressed as:

SI 

 
 (1)    \begin{equation*}  \begin{aligned}   \quad \nabla\times\mathcal{H} & =\frac{\partial\mathcal{D}}{\partial t} \quad,   & \quad \text{(Faraday's Law)} \\[5pt]  \quad \nabla\times\mathcal{E} & =-\frac{\partial\mathcal{B}}{\partial t}  \quad,  & \quad \text{(Ampere's Law)}   \\[5pt]  \nabla\cdot\mathcal{B}                \quad & = \quad 0,                        & \quad \text{( Gauss Law)}   \\[5pt]  \nabla\cdot\mathcal{D}                \quad & = \quad 0.                        & \quad \text{( Colomb's law)}\nonumber  \end{aligned}  \end{equation*}

CGS

 
 (2)    \begin{equation*}  \begin{aligned}  \quad \nabla\times\mathcal{H}\quad & =\quad\frac{1}{c} \frac{\partial\mathcal{D}}{\partial t} \quad,   & \quad \text{(Faraday's Law)} \\[5pt]  \quad \nabla\times\mathcal{E}\quad & =~ - \frac{1}{c}\frac{\partial\mathcal{B}}{\partial t}  \quad,  & \quad \text{(Ampere's Law)}   \\[5pt]  \nabla\cdot\mathcal{B}                \quad & = \qquad 0,                      & \quad \text{( Gauss Law)}   \\[5pt]  \nabla\cdot\mathcal{D}                \quad & =  \qquad0.                     & \quad \text{( Colomb's law)}\nonumber  \end{aligned}  \end{equation*}


Considering a Frequency dependency of these Fields \mathcal{H} and \mathcal{E}, i.e. \mathcal{H}, \mathcal{E} \propto e^{-i\omega t} and \kappa_0=\frac{\omega}{c},

 
 
 (3)    \begin{equation*}   \begin{aligned}    \quad \nabla\times\mathcal{H} & =- i \omega \epsilon \mathcal{E} \quad,   & \quad \text{(Faraday's Law)} \\[5pt]   \quad \nabla\times\mathcal{E} & = i \omega \mathcal{B}  \quad,  & \quad \text{(Ampere's Law)}   \\[5pt]   \nabla\cdot\mathcal{B}                \quad & = \quad 0,                        & \quad \text{( Gauss Law)}   \\[5pt]   \nabla\cdot\mathcal{D}                \quad & = \quad 0.                        & \quad \text{( Colomb's law)}\nonumber   \end{aligned}   \end{equation*}



  
  
 (4)    \begin{equation*}   \begin{aligned}    \quad \nabla\times\mathcal{H} & =-\frac{ i \omega \epsilon \mathcal{E}}{c}= - i \kappa_0 \epsilon \mathcal{E} \quad,   & \quad \text{(Faraday's Law)} \\[5pt]   \quad \nabla\times\mathcal{E} & =\frac{ i \omega \mathcal{B}}{c} =  i \kappa_0  \mathcal{B} \quad,  & \quad \text{(Ampere's Law)}   \\[5pt]   \nabla\cdot\mathcal{B}                \quad & = \quad 0,                        & \quad \text{( Gauss Law)}   \\[5pt]   \nabla\cdot\mathcal{D}                \quad & = \quad 0.                        & \quad \text{( Colomb's law)}\nonumber   \end{aligned}   \end{equation*}


These equations possess non-zero solutions which implies EM Field can exist even in the absence of any charges.
We can now choose the potentials of the EM wave such that the scalar potential is 0 i.e. \phi=0.

This can be satisfied by using the vector potential as a curl of the field \mathcal{H}. From eq.(2) : \quad \mathcal{H}= \nabla\times\mathcal{A} and consequently, \mathcal{E} =-\frac{1}{c^2}\frac{\partial\mathcal{A}}{\partial t} .

On substitution in the divergence equation, \nabla. \mathcal{A} =0.
Also, The choice of \mathcal{A} isn’t totally unique and it can expressed as a gradient of some function such that \phi =0 still holds. Using, these conditons we come to the d’Alembert equation or the wave equation:

 
 (5)    \begin{equation*}   \begin{aligned}   \quad \nabla^2\mathcal{A}\quad & -\quad\frac{1}{c^2} \frac{\partial^2\mathcal{A}}{\partial^2 t}=0 \quad  \\[5pt]   \end{aligned}   \end{equation*}
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