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# 15.11: Maxwell’s Equations in Potential Form

In their usual form, Maxwell’s equations for an isotropic medium, written in terms of the fields, are

$\text{div} \,\textbf{D} =\rho \tag{15.11.1} \label{15.11.1}$

$\text{div} \,\textbf{B} = 0 \tag{15.11.2} \label{15.11.2}$

$\textbf{curl} \,\textbf{H} =\dot{\bf{D}} + \textbf{J} \tag{15.11.3} \label{15.11.3}$

$\textbf{curl} \,\textbf{E} = -\dot{\bf{B}} \tag{15.11.4} \label{15.11.4}$

If we write the fields in terms of the potentials:

$\textbf{E} = -\dot{\textbf{A}} - \textbf{grad}V \tag{15.11.5} \label{15.11.5}$

and $\textbf{B} = \textbf{curl A}, \tag{15.11.6} \label{15.11.6}$

together with $$\textbf{D} = \epsilon \textbf{E}$$ and $$\textbf{B} = \mu \textbf{H}$$, we obtain for the first Maxwell equation, after some vector calculus and algebra,

$\bigstar \quad \nabla^2V + \frac{\partial}{\partial t}(\text{div} \ \textbf{A}) = -\frac{\rho}{\epsilon}. \tag{15.11.7} \label{15.11.7}$

For the second equation, we merely verify that zero is equal to zero. ($$\text{div} \ \textbf{curl A}=0$$.)

For the third equation, which requires a little more vector calculus and algebra, we obtain

$\bigstar \quad \nabla^2 \textbf{A} - \epsilon \mu \frac{\partial^2 \textbf{A}}{\partial t^2} = \textbf{grad} \left( \text{div} \ \textbf{A} + \epsilon \mu \frac{\partial V}{\partial t} \right) - \mu \textbf{J}. \tag{15.11.8} \label{15.11.8}$

The speed of electromagnetic waves in the medium is $$1/\sqrt{\epsilon \mu}$$ and, in a vacuum, equation 15.11.8 becomes

$\bigstar \quad \nabla^2 \textbf{A} - \frac{1}{c^2}\frac{\partial^2 \textbf{A}}{\partial t^2} = \textbf{grad} \left(\text{div} \ \textbf{A} + \frac{1}{c^2} \frac{\partial V}{\partial t} \right) - \mu_0 \textbf{J}, \tag{15.11.9} \label{15.11.9}$

where $$c$$ is the speed of electromagnetic waves in a vacuum.

The fourth Maxwell equation, when written in terms of the potentials, tells us nothing new (try it), so equations $$\ref{15.11.7}$$ and $$\ref{15.11.8}$$ (or $$\ref{15.11.9}$$ in vacuo) are Maxwell’s equations in potential form.

These equations look awfully difficult – but perhaps we can find a gauge transformation, using some form for $$\chi$$, and subtracting $$\textbf{grad} \ \xi$$ from $$\textbf{A}$$ and adding $$\xi$$ to $$V$$, which will make the equations much easier and which will still give the right answers for $$\textbf{E}$$ and for $$\textbf{B}$$.

One of the things that make equations $$\ref{15.11.7}$$ and $$\ref{15.11.9}$$ look particularly difficult is that each equation contains both $$\textbf{A}$$ and $$V$$; that is, we have two simultaneous differential equations to solve for the two potentials. It would be nice if we had one equation for $$\textbf{A}$$ and one equation for $$V$$. This can be achieved, as we shall shortly see, if we can find a gauge transformation such that the potentials are related by

$\text{div} \ \textbf{A} = - \frac{1}{c^2} \frac{\partial V}{\partial t}. \tag{15.11.10} \label{15.11.10}$

You should check that the two sides of this equation are dimensionally similar. What would be the SI units?

You’ll see that this is chosen so as to make the “difficult” part of equation $$\ref{15.11.9}$$ zero.

If we make a gauge transformation and take the divergence of equation 15.10.5 and the time derivative of equation 15.10.6, we then see that condition $$\ref{15.11.10}$$ will be satisfied by a function $$\chi$$ that satisfies

$\nabla^2 \xi - \frac{1}{c^2} \frac{\partial^2 \xi}{\partial t^2} = - \text{div} \textbf{A}^{\prime} - \frac{1}{c^2} \frac{\partial V^{\prime}}{\partial t}.\tag{15.11.11} \label{15.11.11}$

Don’t worry – you don’t have to solve this equation and find the function $$\chi$$; you just have to be assured that some such function exists such that, when applied to the potentials, the potentials will be related by equation $$\ref{15.11.10}$$. Then, if you substitute equation $$\ref{15.11.10}$$ into Maxwell’s equations in potential form (equations $$\ref{15.11.7}$$ and $$\ref{15.11.9}$$), you obtain the following forms for Maxwell’s equations in vacuo in potential form, and the $$\textbf{A}$$ and $$V$$ are now separated:

$\bigstar \quad \nabla^2 V - \frac{1}{c^2} \frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon_0} \tag{15.11.12} \label{15.11.12}$

and $\bigstar \quad \nabla^2 \textbf{A} - \frac{1}{c^2}\frac{\partial^2\textbf{A}}{\partial t^2} = - \mu_0 \textbf{J}. \tag{15.11.13} \label{15.11.13}$

And, since these equations were arrived at by a gauge transformation, their solutions, when differentiated, will give the right answers for the fields.