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:

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

where

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

is a known source term. By duality we have:

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

where

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

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

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

$$\left( \nabla^2 - \varepsilon_0 \mu_0 \frac{\partial^2}{\partial ... ...erline{\mathcal{E}} \\ \underline{\mathcal{H}} \end{array} =0 \space , \right.$$ (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 $c_0$ 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 $x$ and $y$ Cartesian coordinates. Then (1.64 becomes

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

whose general solution

$$\mathcal{E}_x(z,t)=f_+(z-c_0t)+f_-(z+c_0t)$$ (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 $c_0$ 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

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

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

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

where

$$Z_0=Y_0^{-1}=\sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 120 \pi \Omega \approx 377 \Omega$$ (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 $t$: (i) $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ are perpendicular to each other; (ii) the wave propagates in the direction of the vector

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

called Poynting's vector, which represents the power flow density associated with the wave at point $\underline{r}$ and time $t$; (iii) the ratio $\vert \underline{\mathcal{E}} \vert / \vert \underline{\mathcal{H}} \vert$ is independent of $\underline{r}$ and $t$.

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.

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 $S_s$ which bounds the sources volume $v_s$. We want the fields $\underline{\mathcal{E}}$ and $\underline{\mathcal{H}}$ produced by the sources in $v_s$ at a point $P$ and at a time $t$. The observation point $P$ is determined by the position vector $\underline{r}$ with respect to a fixed origin $O$; $P$ may be located inside or outside $v_s$. The closed surface $S_{\mathrm{ext}}$ surrounds both $v_s$ and the region of space where $P$ is located; it is assumed that no sources exist in the space between $S_{\mathrm{ext}}$ and $S_s$. The general solution of the homogeneous equations (1.64) yields the field produced at $P$ 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 $P$ by the sources in $v_s$, and are given by:

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

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

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

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

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

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

is the retarded time. Physically, this means that an electromagnetic disturbance generated by the sources at $\underline{r}'$ at a time $t'$, travels with the velocity of light $c_0$ in free space to reach the observation point $\underline{r}$ at a later time $t$. Thus, the disturbances reaching $P$ at a time t have originated at different times $t'$ from the various elements of $v_s$.

Another possible solution of (1.59) and (1.61) is obtained by replacing $t'$ in (1.72-1.73) with the advanced time $t+R/c_0$. 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:

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

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

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

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

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

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

where $\nabla'$ operates on $\underline{r}'$ with $t'$ constant.

Finally, note that some authors use the symbol:

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

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

where

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

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

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

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

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

so that (1.76-1.77) become:

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

where

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