ELECTROMAGNETIC FIELD THEORY 2

Next: Phasors Up: Maxwell's equations Previous: Maxwell's equations in integral

CONTINUATION OF Maxwell's equations in transform domains

Maxwell's equations (1.1 to 1.4) are a system of eight first-order partial differential equations in four independent variables: three space coordinates and time, whose solution is often quite complicated. It may be advantageous to eliminate the dependence of the field quantities upon one or more of the independent variables by applying a Fourier (or Laplace) transform to (1.1 to 1.4), solving the resulting equations in the transform domain, and then obtaining the desired field quantities by an inverse transformation.
Obviously, the main advantage of a transform technique with respect to an independent variable is to change the dependence of the equations on that variable from a differential one to an algebraic one; thus, a four-fold Fourier transform can change the differential system (1.1-1.4) to an algebraic system in the transform domain.
The Fourier transform pair:

$\displaystyle \underline{\tilde{E}}(\underline{r},\omega)=$ $\displaystyle \int_{-\infty}^{\infty} \underline{\mathcal{ E}} (\underline{r}, t) e^{- j \omega t} \, dt \space ,$ (1.23)

$\displaystyle \underline{\mathcal{ E}}(\underline{r},t) =\frac{1}{2 \pi}$ $\displaystyle \int_{-\infty}^{\infty} \underline{\tilde{E}}(\underline{r},\omega) e^{j \omega t} \, d\omega \space ,$ (1.24)

allows us to transform the electric field from the time domain, where the appropriate field vector is $\underline{\mathcal{ E}}(\underline{r},t)$ , to the frequency domain, where the appropriate field vector is $\underline{\tilde{E}} (\underline{r},\omega)$ , and viceversa. Identical transformations can be applied to all field variables in (1.1-1.4); the appropriate symbols are listed in Table 1.1.

Table 1.1: Symbols

time-domain symbol $\underline{\mathcal{E}}$ $\underline{\mathcal{H}}$ $\underline{\mathcal{D}}$ $\underline{\mathcal{B}}$ $\rho_e$ $\rho_m$ $\underline{\mathcal{J}}_e$ $\underline{\mathcal{J}}_m$

frequency-domain symbol $\underline{\tilde E}$ $\underline{\tilde H}$ $\underline{\tilde D}$ $\underline{\tilde B}$ $\rho_e$ $\rho_m$ $\underline{\tilde J}_e$ $\underline{\tilde J}_m$

If equation (1.24) and similar formulas are used in (1.1-1.4) we obtain, on equating the integrands:

$\displaystyle \nabla \times \underline{\tilde{ H}}$ $\displaystyle =\underline{ \tilde J}_e + j \omega \underline{\tilde{D}} \space ,$ (1.25)

$\displaystyle \nabla \times \underline{\tilde{E}}$ $\displaystyle =-\underline{ \tilde{J}}_m - j \omega \underline{ \tilde{B}} \space ,$ (1.26)

$\displaystyle \nabla \cdot \underline{ \tilde{D}}$ $\displaystyle = \rho_e \space ,$ (1.27)

$\displaystyle \nabla \cdot \underline{ \tilde{B}}$ $\displaystyle = \rho_m \space ,$ (1.28)

similarly, the continuity equations (1.5) and (1.6) become:

$\displaystyle \nabla \cdot \underline{\tilde{J}}_e$ $\displaystyle + j \omega \tilde{\rho}_e=0 \space ,$ (1.29)

$\displaystyle \nabla \cdot \underline{\tilde{J}}_m$ $\displaystyle + j \omega \tilde{\rho}_m=0 \space .$ (1.30)

Observe that (1.25-1.30) are formally obtained from (1.1-1.6) by replacing the differential operator $\partial/\partial t$ with the multiplicative factor $j \omega$ , where $j=\sqrt{-1}$ and $\omega$ is measured in rad/s.
Consider, as an example, the sinusoidal (or time-harmonic) electric field

$\displaystyle \underline{\mathcal{ E}}(\underline{r},t)=\underline{E}_0 (\underline{r}) \cos (\omega_0 t + \phi ( \underline{r}) )$ (1.31)

with angular frequency $\omega_0$ and initial phase $\phi(\underline{r})$ . The corresponding field in the frequency domain is obtained by substituting (1.31) into (1.23) and using the integral representation

$\displaystyle \delta(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-j \omega t} \, dt$ (1.32)

for the one-dimensional delta-function; we find that

$\displaystyle \underline{\tilde{E}} (\underline{r},\omega)= \pi \underline{E}_0...
...ega_0) + e^{-j \varphi(\underline{r})} \delta(\omega+\omega_0) \right] \space .$ (1.33)

In order to express the spatial dependence of the field quantities in (1.1-1.4) in algebraic form, we introduce the three-dimensional Fourier transform pair:

$\displaystyle \underline{\dot{E}}(\underline{k},t)$ $\displaystyle = \int_{-\infty}^{\infty}\underline{\mathcal{ E}}(\underline{r},t) e^{j \underline{k} \cdot \underline{r}} \, d\underline{r} \space ,$ (1.34)

$\displaystyle \underline{\mathcal{ E}}(\underline{r},t)$ $\displaystyle = \frac{1}{( 2 \pi)^3} \int_{-\infty}^{\infty} \underline{\dot{E}...
...ine{k}, t) e^{- j \underline{k} \cdot \underline{r}} \, d\underline{k} \space ,$ (1.35)

where $\underline{k}=k_x \hat{x} + k_y \hat{y} + k_z \hat{z}$ is in $\mathrm{rad}/\mathrm{m}$ , $d \underline{r}=\, dx \, dy \, dz$ , $d\underline{k}=dk_x \, dk_y \, dk_z$ , and $\underline{k} \cdot \underline{r}=k_x x + k_y y +k_z z$ . With similar transformations for the other field quantities, Maxwell's equations in the wavenumber domain (or k-space) become:

$\displaystyle -j \underline{k} \times \underline{\dot{H}}$ $\displaystyle =\underline{\dot{J}}_e + \frac{\partial \underline{\dot{D}} }{\partial t} \space ,$ (1.36)

$\displaystyle j \underline{k} \times \underline{\dot{E}}$ $\displaystyle = \underline{\dot{J}}_m + \frac{\partial \underline{\dot{B}}}{\partial t} \space ,$ (1.37)

$\displaystyle - j \underline{k} \cdot \underline{\dot{D}}$ $\displaystyle = \dot{\rho}_e \space ,$ (1.38)

$\displaystyle - j \underline{k} \cdot \underline{\dot{B}}$ $\displaystyle =\dot{\rho}_m \space ,$ (1.39)

where all field quantities are functions of $\underline{k}$ and t. The continuity equations (1.5-1.6) become:

$\displaystyle \frac{\partial \dot{\rho}_e}{\partial t} - j \underline{k} \cdot \underline{\dot{J}}_e$ $\displaystyle =0 \space ,$ (1.40)

$\displaystyle \frac{\partial \dot{\rho}_m}{\partial t} - j \underline{k} \cdot \underline{\dot{J}}_m$ $\displaystyle =0 \space ,$ (1.41)

Finally, we may elect to work in $k-\omega$ space by subjecting all field quantities to a four-fold Fourier transform. For example, for the electric field we have the transform pair

$\displaystyle \underline{e}(\underline{k},\omega)$ $\displaystyle = \int_{-\infty}^{\infty} \underline{\mathcal{ E}}(\underline{r},...
...underline{k} \cdot \underline{r} - j \omega t } \,d\underline{r} \, dt \space ,$ (1.42)

$\displaystyle \underline{\mathcal{ E}}(\underline{k},t)$ $\displaystyle =\frac{1}{(2 \pi)^4} \int_{-\infty}^{\infty} \underline{e}(\under...
...line{k} \cdot \underline{r} + j \omega t } \,d\underline{k} \, d\omega \space ,$ (1.43)

According to (1.43), each field quantity may be thought of as the four-fold sum of infinitesimal plane waves propagating in different directions with different amplitudes, wavenumbers, and frequencies. Substitution of (1.43) and similar equations into (1.1-1.4) yields:

$\displaystyle - j \underline{k} \times \underline{h}$ $\displaystyle = \underline{j}_e + j \omega \underline{d} \space ,$ (1.44)

$\displaystyle j \underline{k} \times \underline{e}$ $\displaystyle = \underline{j}_m + j \omega \underline{b} \space ,$ (1.45)

$\displaystyle - j \underline{k} \cdot \underline{d}$ $\displaystyle = \rho_e \space ,$ (1.46)

$\displaystyle - j \underline{k} \cdot \underline{b}$ $\displaystyle = \rho_m \space ,$ (1.47)

where all field quantities are functions of $\underline{k}$ and $\omega$ . The continuity equations (1.5-1.6) become:

$\displaystyle \omega \rho_e$ $\displaystyle = \underline{k} \cdot \underline{j}_e \space ,$ (1.48)

$\displaystyle \omega \rho_m$ $\displaystyle = \underline{k} \cdot \underline{j}_m \space .$ (1.49)

Note that eqs. (1.44-1.49) are purely algebraic; the simplification thus achieved is, however, not as significant as might appear at first, because the initial and boundary conditions still must be satisfied.
Whenever a Fourier transform is effected to eliminate the variable t, the charge densities are obtained at once from the current densities, as seen from (1.29-1.30) and (1.48-1.49); it is then necessary to specify only the current densities as sources of the electromagnetic field.

Next: Phasors Up: Maxwell's equations Previous: Maxwell's equations in integral

1999-07-01

Next: Duality Up: Maxwell's equations Previous: Maxwell's equations in transform

Phasors
If all field quantities vary sinusoidally with time, with angular frequency $\omega_0$ , the electric field (1.31) may be written as:

$\displaystyle \underline{\mathcal{ E}}(\underline{r}, t)= \frac{1}{2} \left[ \u...
...hrm{Re} \left\{ \underline{E}(\underline{r}) e^{j \omega_0 t} \right\} \space ,$ (1.50)

where the asterisk means the complex conjugate, and

$\displaystyle \underline{E}(\underline{r})=\underline{E}_0 e^{j \varphi( \underline{r} ) }$ (1.51)

should be compared with the field $\underline{\tilde E}$ of (1.33) in the frequency domain. While $\underline{E}$ is measured in $\mathrm{V/m}$ , $\underline{\tilde E}$ is measured in $\mathrm{V \, s /m}$ as seen from (1.23). The electric field $\underline{\mathcal{E}}$ is obtained from $\underline{E}$ by multiplying $\underline{E}$ times the time-dependence factor $\exp(j \omega_0 t)$ and taking the real part of the product, and from $\underline{\tilde E}$ via the inverse transform (1.24).
When all field quantities are written as phasors, Maxwell's equations assume the form (1.25-1.30) and are thus indistinguishable from the equations in the frequency domain. Basically, phasor quantities and frequency-domain quantities lead to the same analytical derivations. Unless indicated otherwise, in the following we will use phasors and indicate the angular frequency with $\omega$ instead of $\omega_0$ . Then Maxwell's equations and the continuity equations in the phasor domain are:

$\displaystyle \nabla \times \underline{H}$ $\displaystyle =\underline{J}_e + j \omega \underline{D} \space ,$ (1.52)

$\displaystyle \nabla \times \underline{E}$ $\displaystyle =-\underline{J}_m - j \omega \underline{B} \space ,$ (1.53)

$\displaystyle \nabla \cdot \underline{D}$ $\displaystyle = \rho_e \space ,$ (1.54)

$\displaystyle \nabla \cdot \underline{B}$ $\displaystyle = \rho_m \space ,$ (1.55)

$\displaystyle \nabla \cdot \underline{J}_e + j \omega \rho_e$ $\displaystyle =0 \space ,$ (1.56)

$\displaystyle \nabla \cdot \underline{J}_m + j \omega \rho_m$ $\displaystyle = 0 \space .$ (1.57)

The scalar phasors $\rho_e$ and $\rho_m$ are complex scalar quantities. The vector phasors $\underline{E}$ , $\underline{H}$ , $\underline{D}$ , $\underline{B}$ , $\underline{J}_e$ , and $\underline{J}_m$ may be seen as vectors whose components are complex scalar quantities or, equivalently, as complex vectors whose real and imaginary parts are vectors in the ordinary sense.

Next: Duality Up: Maxwell's equations Previous: Maxwell's equations in transform

1999-07-01
Next: Electromagnetic field in free Up: Maxwell's equations Previous: Phasors

Duality
Maxwell's equations (1.1-1.4) are invariant under the substitutions:

$\displaystyle \left\{ \begin{array}{cc} \underline{\mathcal{E}} \rightarrow \un...
..._e \rightarrow \rho_m\space , & \rho_m \rightarrow -\rho_e \end{array} \right .$ (1.58)

The duality relation (1.58) may be restated by saying that Maxwell's equations are unchanged when each electric quantity is replaced by the corresponding magnetic quantity, and each magnetic quantity by the opposite of the corresponding electric quantity.

Next: Electromagnetic potentials Up: Maxwell's equations Previous: Duality

Electromagnetic field in free space

If the sources $\rho_e$ , $\rho_m$ , $\underline{\mathcal{J}}_e$ and $\underline{\mathcal{J}}_m$ are known everywhere in space and time, subject, of course, to the conditions (1.5) and (1.6), then the electric field $\underline{\mathcal{E}}$ and the magnetic field $\underline{\mathcal{H}}$ in free space can be found everywhere in space and time by solving the system (1.10-1.13).
By taking the curl of (1.11), using the relation $\nabla \times \nabla \times \equiv \nabla \nabla \cdot \, - \nabla^2$ , as well as (1.10) and (1.12), we obtain:

$\displaystyle \nabla^2 \underline{\mathcal{E}} - \varepsilon_0 \mu_0 \frac{\partial^2 \underline{\mathcal{E}}}{\partial t^2} = \underline{\Gamma}_e \space ,$ (1.59)

where

$\displaystyle \Gamma_e=\frac{1}{\varepsilon_0} \nabla \rho_e + \mu_0 \frac{\par...
...underline{\mathcal{J}}_e}{\partial t} + \nabla \times \underline{\mathcal{J}}_m$ (1.60)

is a known source term. By duality we have:

$\displaystyle \nabla^2 \underline{\mathcal{H}} - \varepsilon_0 \mu_0 \frac{\partial^2 \underline{\mathcal{H}}}{\partial t^2}=\Gamma_m \space ,$ (1.61)

where

$\displaystyle \Gamma_m=\frac{1}{\mu_0} \nabla \rho_m + \varepsilon_0 \frac{\par...
...{\mathcal{J}}_m}{\partial t} - \nabla \times \underline{\mathcal{J}}_e \space .$ (1.62)

In (1.59), the laplacian of $\underline{\mathcal{E}}$ means:

$\displaystyle \nabla^2 \underline{\mathcal{E}}= \hat{x} \nabla^2 \mathcal{E}_x + \hat{y} \nabla^2 \mathcal{E}_y + \hat{z} \nabla^2 \mathcal{E}_z \space ,$ (1.63)

where $\mathcal{E}_x$ , $\mathcal{E}_y$ , $\mathcal{E}_z$ are the Cartesian components of $\underline{\mathcal{E}}$ . A similar interpretation holds for $\nabla^2 \underline{\mathcal{H}}$ .
Before solving (1.59) and (1.61), let us consider the particular case when $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ are wanted in a region of space where no sources exist $(\Gamma_e=\Gamma_m=0)$ ; then (1.59) and (1.61) become

$\displaystyle \left( \nabla^2 - \varepsilon_0 \mu_0 \frac{\partial^2}{\partial ...
...erline{\mathcal{E}} \\ \underline{\mathcal{H}} \end{array} =0 \space , \right.$ (1.64)

which is a homogeneous wave equation in three dimensions for $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ . As previously stated in (1.9), $\varepsilon_0 \mu_0$ is the inverse of the square of the velocity with which electromagnetic waves (and therefore light) propagate in free space.
Let us consider a field $\underline{\mathcal{E}}=\hat{x} \mathcal{E}_x (z,t)$ parallel to the x-axis and independent of the and Cartesian coordinates. Then (1.64 becomes

$\displaystyle \left( \frac{\partial^2}{\partial z^2} - \frac{1}{c_0^2} \frac{\partial^2}{\partial t^2} \right) \mathcal{E}_x (z,t) = 0 \space ,$ (1.65)

whose general solution

$\displaystyle \mathcal{E}_x(z,t)=f_+(z-c_0t)+f_-(z+c_0t)$ (1.66)

is the superposition of two arbitrary functions $f_{\pm}$ of the independent variables $z \mp c_0 t$ , respectively; $f_{\pm}$ represents a plane wave propagating with velocity and without any shape or amplitude modification in the direction $\pm z$ respectively.
If $\underline{\mathcal{E}}=\hat{x} \mathcal{E}_x (z,t)$ is substituted into (1.10-1.13) with zero sources, it is found that

$\displaystyle \dfrac{\partial \mathcal{E}_x}{\partial z}=-\mu_0 \dfrac{\partial \mathcal{H}_y}{\partial t} \space ,$ $\displaystyle \dfrac{\partial \mathcal{E}_x}{\partial t}=-\dfrac{1}{\varepsilon_0} \dfrac{\partial \mathcal{H}_y}{\partial z} \,$ (1.67)

$\displaystyle \dfrac{\partial \mathcal{H}_x}{\partial t}=\dfrac{\partial \mathcal{H}_z}{\partial t}=0 \space ,$ $\displaystyle \dfrac{\partial \mathcal{H}_y}{\partial x}=\dfrac{\partial \mathcal{H}_y}{\partial y}=0 \space ;$ (1.68)

eqs. (1.68) mean that, neglecting magnetostatic fields, $\underline{\mathcal{H}}= \hat{y} \mathcal{H}_y(z,t)$ ; then (1.66) and (1.67) yield:

$\displaystyle \mathcal{H}_y(z,t)=Y_0 f_+(z-c_0t)-Y_0 f_-(z+c_0t) \space ,$ (1.69)

where

$\displaystyle Z_0=Y_0^{-1}=\sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 120 \pi \Omega \approx 377 \Omega$ (1.70)

is the intrinsic impedance of free space (in $\Omega$ ). From the above derivations it may be concluded that for a plane wave propagating in free space, at each point $\underline{r}$ and time : (i) $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ are perpendicular to each other; (ii) the wave propagates in the direction of the vector

$\displaystyle \underline{p}=\underline{\mathcal{E}} \times \underline{\mathcal{H}} \space ,$ $\displaystyle \mbox{(in $W/m^2$ )}$ $\displaystyle \space ,$ (1.71)

called Poynting's vector, which represents the power flow density associated with the wave at point $\underline{r}$ and time ; (iii) the ratio $\vert \underline{\mathcal{E}} \vert / \vert \underline{\mathcal{H}} \vert$ is independent of $\underline{r}$ and .
Let us now proceed with the solution of (1.59) and (1.61). From the theory of differential equations, it is known that the general solution of the inhomogeneous wave equation (1.59) and (1.61) is the sum of a particular solution of the equation plus the general solution of the corresponding homogeneous equation (1.64). Let us consider the geometry of Fig. (1.3).

Figure 1.3: Relation of field to sources.

$\begin{figure}
\begin{center}
\mbox{}
\centerline{\psfig{figure=chap1/FieldSources.eps,height=6cm}}
\end{center}\end{figure}$

The sources $\rho_e$ , $\rho_m$ , $\underline{\mathcal{J}}_e$ and $\underline{\mathcal{J}}_m$ which give rise to the known terms $\underline{\Gamma}_e$ and $\underline{\Gamma}_m$ in (1.59) and (1.61) are contained inside the closed fixed surface which bounds the sources volume . We want the fields $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ produced by the sources in at a point and at a time . The observation point is determined by the position vector $\underline{r}$ with respect to a fixed origin ; may be located inside or outside . The closed surface $S_{\mathrm{ext}}$ surrounds both and the region of space where is located; it is assumed that no sources exist in the space between $S_{\mathrm{ext}}$ and . The general solution of the homogeneous equations (1.64) yields the field produced at by all the sources located outside $S_{\mathrm{ext}}$ ; since we are not interested in these external sources, we neglect such solution. The particular solutions of (1.59) and (1.61) correspond to the fields produced at by the sources in , and are given by:

$\displaystyle \underline{\mathcal{ E}}(\underline{r},t)$ $\displaystyle =-\frac{1}{4 \pi} \int_{v_s} \frac{\underline{\Gamma}_e (\underline{r}', t')}{R} \, d\underline{r}' \space ,$ (1.72)

$\displaystyle \underline{\mathcal{H}}(\underline{r}, t)$ $\displaystyle =-\frac{1}{4 \pi} \int_{v_s} \frac{\underline{\Gamma}_m (\underline{r}', t')}{R} \, d\underline{r}' \space ,$ (1.73)

where $\underline{r}'$ is the position of the integration point inside , $d\underline{r}'=dx' \, dy' \, dz'$ is the elementary volume at $\underline{r}'$ ,

$\displaystyle R=\vert\underline{r}-\underline{r}'\vert$ (1.74)

is the distance between the observation point $\underline{r}$ and the integration point $\underline{r}'$ , and

$\displaystyle t'=t-\frac{R}{c_0}$ (1.75)

is the retarded time. Physically, this means that an electromagnetic disturbance generated by the sources at $\underline{r}'$ at a time , travels with the velocity of light in free space to reach the observation point $\underline{r}$ at a later time . Thus, the disturbances reaching at a time t have originated at different times from the various elements of .
Another possible solution of (1.59) and (1.61) is obtained by replacing in (1.72-1.73) with the advanced time . Physically, this would mean that the disturbance reaches the observation point P before manifesting itself at the source, i.e. violation of causality occurs. For this reason, the advanced solutions are neglected and only the retarded solutions corresponding to the choice (1.75) in (1.72-1.73) are considered as physically meaningful. With the use of (1.60) and (1.62), the retarded solutions (1.72) and (1.73) can be rewritten as:

$\displaystyle \underline{\mathcal{E}}(\underline{r},t)=$ $\displaystyle \displaystyle{-\frac{1}{4 \pi \varepsilon_0} \int_{v_s} \frac{1}{R} \nabla' \rho_e (\underline{r}',t') \, d\underline{r}', }$

$\displaystyle -\frac{\mu_0}{4\pi} \int_{v_s} \frac{1}{R} \frac{\partial \underline{J}_e (\underline{r}', t')}{\partial t'} \, d\underline{r}'$

$\displaystyle -\frac{1}{4 \pi} \int_{v_s} \frac{1}{R} \nabla' \times \underline{\mathcal{J}}_m (\underline{r}', t') d\underline{r}' \space ,$ (1.76)

$\displaystyle \underline{\mathcal{H}}(\underline{r},t)=$ $\displaystyle -\frac{1}{4 \pi \mu_0} \int_{v_s} \frac{1}{R} \nabla' \rho_m (\underline{r}',t') \, d\underline{r},$

$\displaystyle -\frac{\varepsilon_0}{4\pi} \int_{v_s} \frac{1}{R} \frac{\partial \underline{J}_m (\underline{r}', t')}{\partial t'} \, d\underline{r}'$

$\displaystyle -\frac{1}{4 \pi} \int_{v_s} \frac{1}{R} \nabla' \times \underline{\mathcal{J}}_e (\underline{r}', t') d\underline{r}' \space ,$ (1.77)

where $\nabla'$ operates on $\underline{r}'$ with constant.
Finally, note that some authors use the symbol:

$\displaystyle \Box^2 \, =\nabla^2 \, - \varepsilon_0 \mu_0 \frac{\partial^2}{\partial t^2}$ (1.78)

for the operator which appears in the wave equation; the operator $\Box^2$ is called the dalembertian operator.
In the frequency or phasor domains, the operator (1.78) becomes:

$\displaystyle \Box^2\, =\nabla^2\, + k_0^2 \space ,$ (1.79)

where

$\displaystyle k_0=\omega \sqrt{\varepsilon_0 \mu_0} = \frac{\omega}{c_0}= \frac{2\pi}{\lambda_0}$ (1.80)

is the wavenumber in free space, while $\lambda_0$ (in m) is the wavelength in free space.
In the frequency domain, eqs. (1.59) and (1.61) may be rewritten as

$\displaystyle \nabla^2 \underline{\tilde{E}} + k_0^2 \underline{\tilde{E}}=\und...
... \mu_0 \underline{\tilde{J}}_e + \nabla \times \underline{\tilde{J}}_m \space ,$ (1.81)

$\displaystyle \nabla^2 \underline{\tilde{H}} + k_0^2 \underline{\tilde{H}}=\und...
...ilon_0 \underline{\tilde{J}}_m + \nabla \times \underline{\tilde{J}}_e \space ,$ (1.82)

where the tilde may be dropped if all quantities are interpreted as phasors. With the aid of (1.23), solutions (1.72-1.73) become:

$\displaystyle \underline{\tilde{E}}(\underline{r}, \omega)$ $\displaystyle =-\frac{1}{4 \pi} \int_{v_s} \frac{\underline{\tilde{\Gamma}}_e (\underline{r}', \omega)}{R} e^{- j k_0 R} \, d\underline{r}' \space ,$ (1.83)

$\displaystyle \underline{\tilde {H}}(\underline{r}, \omega)$ $\displaystyle =-\frac{1}{4 \pi} \int_{v_s} \frac{\underline{\tilde{\Gamma}}_m (\underline{r}', \omega)}{R} e^{-j k_0 R} \, d\underline{r}' \space ,$ (1.84)

so that (1.76-1.77) become:

$\displaystyle \left\{ \begin{aligned}[l]\underline{E}(\underline{r})&= -\int_{v...
...rline{J}_e ( \underline{r}') \right] G \, d\underline{r} \end{aligned} \right .$

where

$\displaystyle G=G(\underline{r},\underline{r}')=\frac{e^{-j k_0 R}}{4 \pi R} \space ,$ (1.86)

and the dependence of all field quantities on the parameter $\omega$ has been omitted. The quantity G of (1.86) is called the free-space Green function.

Next: Electromagnetic potentials Up: Maxwell's equations Previous: Duality

1999-07-01