Deriving Maxwell's Equations for the Potentials

We take Maxwell's equations and the fields written in terms of the potentials as input. In the left column the equations are given in the standard form while the right column gives the equivalent equation in terms of indexed components. The right column uses the totally antisymmetric tensor in 3D \bgroup\color{black}$\epsilon_{ijk}$\egroup and assumes summation over repeated indices (Einstein notaton). So in this notation, dot products can be simply written as \bgroup\color{black}$\vec{a}\cdot\vec{b}=a_ib_i$\egroup and any component of a cross product is written \bgroup\color{black}$(\vec{a}\times\vec{b})_k=a_ib_j\epsilon_{ijk}$\egroup.

\begin{eqnarray*}
\vec{\nabla}\cdot\vec{B} = 0 &\qquad & {\partial\over\partial ...
...al\over\partial x_k}\phi-{1\over c}{\partial A_k\over\partial t}
\end{eqnarray*}


If the fields are written in terms of potentials, then the first two Maxwell equations are automatically satisfied. Lets verify the first equation by plugging in the B field in terms of the potential and noticing that we can interchange the order of differentiation.

\begin{displaymath}\bgroup\color{black} \vec{\nabla}\cdot\vec{B}={\partial\over\...
...artial x_m}{\partial\over\partial x_i}A_n\epsilon_{mni} \egroup\end{displaymath}

We could also just interchange the index names \bgroup\color{black}$i$\egroup and \bgroup\color{black}$m$\egroup then switch those indices around in the antisymmetric tensor.

\begin{displaymath}\bgroup\color{black} \vec{\nabla}\cdot\vec{B}={\partial\over\...
...artial x_m}{\partial\over\partial x_i}A_n\epsilon_{mni} \egroup\end{displaymath}

We have the same expression except for a minus sign which means that \bgroup\color{black}$\vec{\nabla}\cdot\vec{B} = 0$\egroup.

For the second equation, we write it out in terms of the potentials and notice that the first term \bgroup\color{black}${\partial\over\partial x_i}{\partial\over\partial x_j}\phi\epsilon_{ijk}=0$\egroup for the same reason as above.

\begin{eqnarray*}
{\partial\over\partial x_i}E_j\epsilon_{ijk}+{1\over c}{\parti...
...ial x_i}{\partial A_j\over\partial t}\epsilon_{ijk}\right)=0 \\
\end{eqnarray*}


The last step was simply done by renaming dummy indices (that are summed over) so the two terms cancel.

Similarly we may work with the Gauss's law equation

\begin{eqnarray*}
\vec{\nabla}\cdot\vec{E} ={\partial\over\partial x_k}E_k
=-{\p...
...partial\over\partial t}(\vec{\nabla}\cdot\vec{A})&=&4\pi\rho \\
\end{eqnarray*}


For the fourth equation we have.

\begin{eqnarray*}
{\partial\over\partial x_i}B_j\epsilon_{ijk}-{1\over c}{\parti...
...+{1\over c}{\partial A_k\over\partial t}\right)={4\pi\over c}J_k
\end{eqnarray*}


Its easy to derive an identity for the product of two totally antisymmetric tensors \bgroup\color{black}$\epsilon_{mnj}\epsilon_{ijk}$\egroup as occurs above. All the indices of any tensor have to be different in order to get a nonzero result. Since the \bgroup\color{black}$j$\egroup occurs in both tensors (and is summed over) we can simplify things. Take the case that \bgroup\color{black}$i=1$\egroup and \bgroup\color{black}$k=2$\egroup. We only have a nonzero term if \bgroup\color{black}$j=3$\egroup so the other 2 terms in the sum are zero. But if \bgroup\color{black}$j=3$\egroup, we must have either \bgroup\color{black}$m=1$\egroup and \bgroup\color{black}$n=2$\egroup or vice versa. We also must not have \bgroup\color{black}$i=k$\egroup since all the indices have to be different on each epsilon. So we can write.

\begin{displaymath}\bgroup\color{black} \epsilon_{mnj}\epsilon_{ijk}=\epsilon_{m...
...ft(\delta_{km}\delta_{in}-\delta_{kn}\delta_{im}\right) \egroup\end{displaymath}

Applying this identity to the Maxwell equation above, we get.

\begin{eqnarray*}
{\partial\over\partial x_i}{\partial\over\partial x_k}A_i-{\pa...
... c}{\partial\phi\over\partial t}\right)={4\pi\over c}\vec{J} \\
\end{eqnarray*}


The last two equations derived are wave equations with source terms obeyed by the potentials. As discussed in the opening section of this chapter, they can be simplified with a choice of gauge.

Jim Branson 2013-04-22