The following para is excerpted from Principles of Optics by Born, Wolf:

[...] In other words $\mathbf B$ anfd $\mathbf E$ are invariant under the transformation \begin{align}\mathbf A^\prime & = \mathbf A + \textrm{grad} ~\chi~,\tag{14a}\\ \phi^\prime &= \phi- \frac1c~\ddot{\chi}~.\tag{14b} \end{align} Now from $$\textrm{div}~\mathbf A+ \frac1c\dot \phi ~=~0\tag{10}$$ and $(14)$ $$\textrm{div}~\mathbf A^\prime + \frac1c\dot \phi +\left(\boldsymbol\nabla^2 \chi -\frac1{c^2}~\ddot\chi\right)~=~0\tag{15}$$ Hence $\mathbf A^\prime$ and $\phi^\prime$ will satisfy the Lorentz relation, if one imposes on $\chi$ the condition $$\left(\boldsymbol\nabla^2 \chi -\frac1{c^2}~\ddot\chi\right)~=~0\:.\tag{16}$$ ... In a region where the charge density $\rho$ is zero, $\phi$ satisfies the homogeneous wave equation $$\boldsymbol\nabla^2 \phi -\frac1{c^2}~\ddot\phi~=~0\tag{17}$$ and now, $\chi$ may then be so chosen, that the scalar potential vanishes. According to $(14\mathrm b)$ and $(16)$, it is only necessary to take $$\chi ~=~ c\int \phi ~\mathrm dt\,.\tag{18}$$

I've some queries on this above excerpt:

I have understood for the scalar potential $\phi^\prime$ to get vanished, $\chi$ was chosen to be equal to $c\displaystyle \int \phi ~\mathrm dt$ which can indeed be verified by looking at $(14\mathrm b)$ solely; how does this value of $\chi$ satisfy $(16)$?

Also, won't there be $\textrm{div}~\mathbf A$ instead of $\textrm{div}~\mathbf A^\prime$ in $(15)$?

  • $\begingroup$ Can you please verify the double time-derivative in Eq. (14b)? I think it should be $\phi \to \phi - \partial_t \chi$ $\endgroup$ – caverac Dec 11 '16 at 10:26

Note that

\begin{eqnarray} \nabla\cdot \mathbf{A}' &\stackrel{(14a)}{=}& \nabla \cdot \mathbf{A} + \nabla^2 \chi \\ &\stackrel{(10)}{=}& \left(-\frac{1}{c}\dot{\phi} \right) + \nabla^2\chi \\ &\stackrel{(14b)}{=}& -\frac{1}{c}\left(\dot{\phi}' + \frac{1}{c}\ddot\chi \right) + \nabla^2\chi \end{eqnarray}

Then you get

$$ \nabla \cdot \mathbf{A}' + \frac{1}{c}\dot{\phi}' = \nabla^2\chi - \frac{1}{c^2}\ddot\chi $$

From this you can conclude that the new potential follows the expression $\nabla \cdot \mathbf{A}' + \dot{\phi}'/c = 0$ only if

$$ \nabla^2\chi - \frac{1}{c^2}\ddot\chi = 0 $$

That should answer your first question. As for the second one

\begin{eqnarray} \nabla^2\chi &=& c \int{\rm d}t~\nabla^2 \phi \stackrel{(17)}{=} c\int{\rm d}t~\frac{1}{c^2}\ddot{\phi} = \frac{1}{c} \dot{\phi} \\ &\stackrel{(14b)}{=}& -\frac{1}{c^2}\ddot{\chi} \end{eqnarray}

where I have used the fact that the scalar potential vanishes

  • $\begingroup$ Yes, that was what I meant to say; it should be $\textrm{div}~\mathbf A^\prime + \frac1c\dot \phi \color{red}{-}\left(\boldsymbol\nabla^2 \chi -\frac1{c^2}~\ddot\chi\right)=0$ and not $+$ as they stated, isn't it? $\endgroup$ – user142971 Dec 11 '16 at 11:55
  • $\begingroup$ @MAFIA36790 I think it should be a '-' sign in there $\endgroup$ – caverac Dec 11 '16 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy