Electromagnetic potentials

In the previous section we have obtained the fields by direct integration over the sources in free space, leading to the solutions (1.76-1.77) in the time domain and to the corresponding solutions (1.85) in the frequency (or phasor) domain. An alternative approach which utilizes electromagnetic potentials is often useful.

Let us rewrite Maxwell's equations (1.1-1.4) by splitting all field quantities $\underline{\mathcal{E}}$, $\underline{\mathcal{D}}$, $\underline{\mathcal{H}}$ and $\underline{\mathcal{B}}$ into the sum of two quantities:

$$\left\{ \begin{aligned}[l]\underline{\mathcal{E}}=\underline{\mat... ...rline{\mathcal{B}}_e+ \underline{\mathcal{B}}_m \space , \end{aligned} \right .$$

where the quantities with the ``e'' subscript depend only on the electric sources $\rho_e$ and $\underline{\mathcal{J}}_e$, while the quantities with the ``m'' subscript depend only on the magnetic sources $\rho_m$ and $\underline{\mathcal{J}}_m$:

$$\left\{ \begin{aligned}[l]\nabla \times \underline{\mathcal{H}}_e... ...\\ \nabla \cdot \underline{\mathcal{B}}_e &= 0 \space ; \end{aligned} \right .$$

$$\left\{ \begin{aligned}[l]\nabla \times \underline{\mathcal{H}}_m... ...nabla \cdot \underline{\mathcal{B}}_m &= \rho_m \space . \end{aligned} \right .$$

It is seen at once that by adding each of (1.88) to the corresponding (1.89) and using (1.87), we again obtain (1.1-1.4).

Because of the identity $\nabla \cdot \nabla \times \, \equiv 0$, the fourth of (1.88) and the third of (1.89) are always satisfied if:

$$\begin{array}{cc} \underline{\mathcal{B}}_e=\nabla \times \un... ...m=-\nabla \times \underline{\mathcal{A}}_m \space , \end{array}$$ (1.90)

where $\underline{\mathcal{A}}_e$ and $\underline{\mathcal{A}}_m$ are arbitrary vector functions of position and time, called the electric vector potential and the magnetic vector potential, respectively. Substitution of (1.90) into the second of (1.88) and the first of (1.89) yields:

$$\begin{array}{c c} \nabla \times \left( \underline{\mathcal{E... ...ne{\mathcal{A}}_m}{\partial t} \right) = 0 \end{array} \space .$$ (1.91)

On the strength of the identity $\nabla \times \nabla \equiv 0$, eqs. (1.91) are always satisfied if:

$$\begin{array}{c c} \underline{\mathcal{E}}_e=-\nabla \Phi_e -... ...ial \underline{\mathcal{A}}_m}{\partial t} \space , \end{array}$$ (1.92)

where $\Phi_e$ and $\Phi_m$ are arbitrary scalar functions of position and time, called the electric scalar potential and the magnetic scalar potential, respectively.

For given field quantities $\underline{\mathcal{E}}_e$ and $\underline{\mathcal{B}}_e$, the potentials $\Phi_e$ and $\underline{\mathcal{A}}_e$ are not uniquely defined; if $\Phi_e$ and $\underline{\mathcal{A}}_e$ satisfy the first of (1.90) and (1.92), the same is true of all potentials $\Phi_{eo}$ and $\underline{\mathcal{A}}_{eo}$ related to $\Phi_e$ and $\underline{\mathcal{A}}_e$ by the gauge transformation:

$$\Phi_{eo}=\Phi_e + \frac{\partial f_e}{\partial t} \space , \underline{\mathcal{A}}_{eo}=\underline{\mathcal{A}}_e - \nabla f_e \space ,$$ (1.93)

where $f_e(\underline{r},t)$ is an arbitrary scalar function of position and time. Similarly, if $\Phi_m$ and $\underline{\mathcal{A}}_m$ generate the fields $\underline{\mathcal{D}}_m$ and $\underline{\mathcal{H}}_m$ via the second of (1.90) and (1.92), the same is true of the potentials

$$\Phi_{mo}=\Phi_m + \frac{\partial f_m}{\partial t} \space , \underline{\mathcal{A}}_{mo}=\underline{\mathcal{A}}_m - \nabla f_m \space ,$$ (1.94)

where $f_m(\underline{r},t)$ is an arbitrary scalar function of position and time.

In order to obtain the remaining four field quantities $\underline{\mathcal{E}}_m$, $\underline{\mathcal{D}}_m$, $\underline{\mathcal{H}}_m$ and $\underline{\mathcal{B}}_m$ as functions of the potentials, it is necessary to specify the constitutive relations of the medium. If we assume free space:

$$\underline{\mathcal{E}}=\underline{\mathcal{E}}_e+ \frac{1}{\varepsilon_0} \underline{\mathcal{D}}_m \space , \underline{\mathcal{H}}_m= \frac{1}{\mu_0} \underline{\mathcal{B}}_e + \underline{\mathcal{H}}_m \space ,$$ (1.95)

and use of (1.90) and (1.92) yields:

$$\left\{ \begin{aligned}[l]\underline{\mathcal{E}}&=-\nabla \Phi_e... ...}{\mu_0} \nabla \times \underline{\mathcal{A}}_e \space . \end{aligned} \right.$$

The relations of the potentials to the sources is obtained by satisfying the remaining equations (1.88-1.89). Use of (1.90) and (1.92) in the first of (1.88), and in the second and fourth of (1.89), yields respectively:

$$\left\{ \begin{aligned}[l]\nabla^2 \underline{\mathcal{A}}_e - \v... ...\mu_0 \frac{\partial \Phi_e}{\partial t} \right) \space ; \end{aligned} \right.$$

$$\left\{ \begin{aligned}[l]\nabla^2 \underline{\mathcal{A}}_m - \v... ...\mu_0 \frac{\partial \Phi_m}{\partial t} \right) \space . \end{aligned} \right.$$

Eqs. (1.97) for the unknowns $\Phi_e$ and $\underline{\mathcal{A}}_e$ become decoupled if

$$\nabla \cdot \underline{\mathcal{A}}_e + \varepsilon_0 \mu_0 \frac{\partial \Phi_e}{\partial t}=0 \space ;$$ (1.99)

similarly, (1.98) decouple if

$$\nabla \cdot \underline{\mathcal{A}}_m + \varepsilon_0 \mu_0 \frac{\partial \Phi_m}{\partial t}=0 \space .$$ (1.100)

The Lorentz conditions (1.99) and (1.100) are sufficient but not necessary to the decoupling of (1.97) and (1.98); we could simply require that the left-hand-sides (LHS) of (1.99) and (1.100) be constant. The choice of zero for these two constants is suggested by the fact that (1.99) and (1.100) are covariant under a coordinate transformation in special relativity; thus, (1.99) and (1.100) have the appearance of true physical laws.

Potentials related to the above ones by the gauge transformation (1.93) and (1.94) also satisfy the Lorentz conditions, provided that $f_e$ and $f_m$ be solutions of the homogeneous wave equations:

$$\Box^2 f_{ \begin{array}{l} e \\ m \end{array} } =0$$ (1.101)

Under conditions (1.99) and (1.100), the potentials satisfy the inhomogeneous wave equations

$$\begin{array}{ll} \Box^2 \Phi_e = -\dfrac{\rho_e}{\varepsilon... ...= -\varepsilon_0 \underline{\mathcal{J}}_m \space , \end{array}$$ (1.102)

whose particular solutions are (the geometry of Fig. 1.3 applies):

$$\Phi_e(\underline{r},t) =\frac{1}{4\pi \varepsilon_0} \int_{v_s} \frac{\rho_e(\underline{r}',t')}{R} d \underline{r}' \space ,$$ (1.103)

$$\Phi_m(\underline{r},t) =\frac{1}{4\pi \mu_0} \int_{v_s} \frac{\rho_m(\underline{r}',t')}{R} d \underline{r}' \space ,$$ (1.104)

$$\underline{\mathcal{A}}_e(\underline{r},t) =\frac{\mu_0}{4\pi} \int_{v_s} \frac{\underline{\mathcal{J}}_e(\underline{r}',t')}{R} d \underline{r}' \space ,$$ (1.105)

$$\underline{\mathcal{A}}_m(\underline{r},t) =\frac{\varepsilon_0}{4\pi} \int_{v_s} \frac{\underline{\mathcal{J}}_m(\underline{r}',t')}{R} d \underline{r}'\space ,$$ (1.106)

with $R$ and $t'$ given by (1.74) and (1.75). Substitution of (1.103-1.106) into (1.96) yields:

$$\underline{\mathcal{E}}(\underline{r},t)=-\frac{1}{4\pi} \left\{ \frac{1}{\varepsilon_0} \nabla \int_{v_s} \frac{\rho_e (\... ...frac{\underline{\mathcal{J}}_e(\underline{r}',t')}{R} d\underline{r'} + \right.$$

$$\left. \nabla \times \int_{v_s} \frac{\underline{\mathcal{J}}_m(\underline{r}',t')}{R} d\underline{r}' \right\} \space ,$$ (1.107)

$$\underline{\mathcal{H}}(\underline{r},t)=-\frac{1}{4\pi} \left\{ \frac{1}{\mu_0} \nabla \int_{v_s} \frac{\rho_m(\underline... ...rac{\underline{\mathcal{J}}_m (\underline{r}',t')}{R} d\underline{r'} - \right.$$

$$\left. \nabla \times \int_{v_s} \frac{\underline{\mathcal{J}}_e(\underline{r}',t')}{R} d\underline{r}' \right\} \space ,$$ (1.108)

where $\nabla$ operates on $\underline{r}$ with t constant. Obviously, (1.107-1.108) must be equivalent to (1.76-1.77); the non-trivial proof of this equivalence is left as an exercise.

In the frequency (or phasor) domain, eqs. (1.96), (1.99-1.100), (1.103-1.104) become:

$$\begin{array}{l} \displaystyle{\underline{E}=-\nabla \Phi_e -... ...c{1}{\mu_0} \nabla \times \underline{A}_e} \space , \end{array}$$ (1.109)

$$\nabla \cdot \underline{\mathcal{A}}_e + j\omega \varepsilon_0 \mu_0 \Phi_e = 0 \space , \nabla \cdot \underline{\mathcal{A}}_m + j\omega \varepsilon_0 \mu_0 \Phi_m = 0 \space ,$$ (1.110)

$$\Phi_e(\underline{r}) = \frac{1}{\varepsilon_0} \int_{v_s} \rho_e(\underline{r}') G \, d\underline{r}' \space ,$$ (1.111)

$$\Phi_m(\underline{r}) = \frac{1}{\mu_0} \int_{v_s} \rho_m(\underline{r}') G \, d\underline{r}' \space ,$$ (1.112)

$$\underline{\mathcal{A}}_e(\underline{r}) = \mu_0 \int_{v_s} \underline{\mathcal{J}}_e(\underline{r}') G \, d\underline{r}' \space ,$$ (1.113)

$$\underline{\mathcal{A}}_m(\underline{r}) = \varepsilon_0 \int_{v_s} \underline{\mathcal{J}}_m(\underline{r}') G \, d\underline{r}'\space ,$$ (1.114)

where $G$ is given by (1.86). Substitution of (1.111-1.114) into (1.109) leads to expressions which must be equivalent to (1.85); the proof of this equivalence is left as an exercise. In the time-harmonic case, $\Phi_e$ and $\Phi_m$ are found to depend on $\underline{\mathcal{A}}_e$ and $\underline{\mathcal{A}}_m$, respectively, via (1.110). Substitution of (1.110) into (1.109) yields:

$$\left\{ \begin{aligned}[l]\underline{E}&=-j\omega\left( 1 + \frac... ... - \frac{1}{\mu_0} \nabla \times \underline{A}_e \space , \end{aligned} \right.$$

that is, in the time-harmonic case $\underline{E}$ and $\underline{H}$ are functions of the vector potentials only; this is because $\rho_e$ and $\rho_m$ are functions of $\underline{\mathcal{J}}_e$ and $\underline{\mathcal{J}}_m$, as noted previously.

The use of scalar and vector potentials has some advantages with respect to direct integration of the fields:

i

Each potential depends on one type of source only, and is therefore easier to calculate than the fields; once the potentials are known, the fields are easily obtained by differentiation.

ii

The arbitrarity of choice embodied in the gauge transformation allows for simplified calculations in many cases.

The potentials are ``more regular'' than the electric and magnetic fields, and this regularity can be enhanced further by introducing superpotentials, from which the fields are obtained by higher order differentiation. For example, conditions (1.99-1.100) are identically satisfied if we choose:

$$\underline{\mathcal{A}}_e =\varepsilon_0 \mu_0 \frac{\partial \underline{\mathcal{\Pi}}_e}{\partial t} \space , \qquad \Phi_e = - \nabla \cdot \underline{\mathcal{\Pi}}_e \space ,$$ (1.116)

$$\underline{\mathcal{A}}_m =\varepsilon_0 \mu_0 \frac{\partial \underline{\mathcal{\Pi}}_m}{\partial t} \space , \qquad \Phi_m = - \nabla \cdot \underline{\mathcal{\Pi}}_m \space ,$$ (1.117)

in which case (1.96) become:

$$\left\{ \begin{aligned}[l]\underline{\mathcal{E}}&=\nabla \nabla ... ... } \nabla \times \underline{\mathcal{\Pi}}_e \space . \\ \end{aligned} \right.$$

Use of the first of (1.116-1.117) into the wave equations (1.102) for $\underline{\mathcal{A}}_e$ and $\underline{\mathcal{A}}_m$ yields:

$$\begin{align}\Box^2 \underline{\mathcal{\Pi}}_e &= -\frac{1}{... ...line{\mathcal{J}}_m(\underline{r},t) \tau, d\tau \; \end{align}$$

the wave equations for $\Phi_e$ and $\Phi_m$ are satisfied if (1.7) are, on the strength of the continuity eqs. (1.5-1.6). Eqs. (1.7) and the identity $\nabla \times \nabla \times = \nabla \nabla \cdot - \nabla^2$ allow us to rewrite (1.118) as:

$$\left\{ \begin{aligned}[l]\underline{\mathcal{E}}&=\nabla \times ... ...ial}{\partial t} \nabla \times \underline{\Pi}_e \space . \end{aligned} \right.$$

Observe that $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ in (1.121) are unchanged if the gradients of arbitrary functions of space and time are added t $\underline{\Pi}_e$ and $\underline{\Pi}_m$; thus, for a given physical situation, a great amount of arbitrarity exists in the choice of $\underline{\Pi}_e$ and $\underline{\Pi}_m$.

The superpotentials $\underline{\Pi}_e$ and $\underline{\Pi}_m$ are called the electric Hertz vector and the magnetic Hertz vector, respectively. Instead of using two scalar and two vector potentials to drive $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$, we may use the two Hertz vectors $\underline{\Pi}_e$ and $\underline{\Pi}_m$.

In the frequency domain, from (1.116-1.117):

$$\underline{A}_e=j \omega \varepsilon_0 \mu_0 \underline{\Pi}_e \space , \underline{A}_mj \omega \varepsilon_0 \mu_0 \underline{\Pi}_m \space ,$$ (1.120)

where

$$\left( \nabla^2 + k_0^2 \right) \underline{\Pi}_e= \frac{j}{\omega \varepsilon_0} \underline{\mathcal{J}}_e \space , \left( \nabla^2 + k_0^2 \right) \underline{\Pi}_m \dfrac{j}{\omega \mu} \underline{\mathcal{J}}_m$$ (1.121)

hence, no significant difference exists between using $\underline{A}_e$, $\underline{A}_m$ and using $\underline{\Pi}_e$, $\underline{\Pi}_m$ in frequency domain analysis. Eqs. (1.121) become:

$$\left\{ \begin{aligned}[l]\underline{E}&=\nabla \times \nabla \ti... ...\frac{j}{\omega \mu_0} \underline{\mathcal{J}}_m \space , \end{aligned} \right.$$

which, with the aid of (1.122), are easily seen to be equivalent to (1.115).