There is a beautiful demonstration, available in the text Robert S. Elliot, Antenna theory and Design, Wiley-IEEE Press, page 17 (Stratton-Chu solution), which shows how the electromagnetic field at each point $ \mathbf { r} $ of a volume $ V $, with boundary $S_1, ..., S_N $:

enter image description here

can be found from the following integral:

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi}\int_V \left( \frac{\rho}{\epsilon_0}\nabla_S\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right)\mathrm{d}V+\frac{1}{4\pi}\int_{S_1,...,S_N}\left[ \left(\mathbf{1}_n\cdot\mathbf{E}\right)\nabla_S\psi+\left(\mathbf{1}_n\times\mathbf{E}\right)\times\nabla_S\psi -j\omega\psi\left(\mathbf{1}_n\times\mathbf{B}\right)\right]\mathrm{d}S$$


  1. the point $ P $ shown in the figure corresponds to what I have called $ \mathbf {r} $ in the formula, while the surface $ \Sigma $ is a comfortable surface that only serves to demonstrate the validity of the above formula, in fact at the end it no longer appears in it;

  2. in the formula $\rho=\rho(\mathbf{r}')$ is the charge density in $V$ (all, both impressed and induced);

  3. $\mathbf{J}=\mathbf{J}(\mathbf{r}')$ is the current density in $V$ (both impressed and induced);

  4. $\psi=\psi(\mathbf{r},\mathbf{r}')=\frac{e^{-j k |\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}$ is the Green function, the only one which is both a function of $\mathbf{r}'$ and $\mathbf{r}$.

  5. the integration is done with respect to $\mathbf{r}'$ and also $\nabla_S$ operates on $\mathbf{r'}$;

  6. $\mathbf{1}_n=\mathbf{1}_n(\mathbf{r}')$ is the unit normal vector, in each point, to the surfaces $S_1,...,S_N$;

  7. $\omega$ it is the angular frequency of the field, which is supposed to be fixed.

This formula is very useful because, for antenna problems, it allows to distinguish two types of approaches to the solution very well:

  1. there are antennas for which the distribution of the currents on the metal that composes them can be assumed to be known with a good degree of approximation (eg: dipole), therefore for such problems the volume $V$ will be chosen as the entire space $ \mathbb {R} ^ 3 $, without any exclusion surface, so in those cases the formula is reduced to:

$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi}\int_V \left( \frac{\rho}{\epsilon_0}\nabla_S\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right)\mathrm{d}V$

  1. for other types of antennas, for which instead the distribution of the field is known with a good degree of approximation (eg: horn), one can choose a volume $ V $ delimited by a surface $ S_1 $ (on which the field is known with good approximation) such that inside $ V $ there will be neither currents nor charges (they are all inside $ S_1 $). The formula in this case becomes:

$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi}\int_{S_1}\left[ \left(\mathbf{1}_n\cdot\mathbf{E}\right)\nabla_S\psi+\left(\mathbf{1}_n\times\mathbf{E}\right)\times\nabla_S\psi -j\omega\psi\left(\mathbf{1}_n\times\mathbf{B}\right)\right]\mathrm{d}S$

For a dipole antenna, using the continuity equation for the current $ \nabla \cdot \mathbf {J} = - j \omega \rho $, you will have:

$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi}\int_V \left( -\frac{\nabla_S\cdot\mathbf{J}}{j\omega\epsilon_0}\nabla_S\psi-j\omega\psi\frac{\mathbf{J}}{\mu_0^{-1}}\right)\mathrm{d}V$

where $ \mathbf{J} $ is all the current present in the whole space ($V = \mathbb {R} ^ 3 $), that is the surface currents distributed on the two plates of the dipole.

Let us now take another example, which falls into the second category of antennas: a slot in an infinite ground plane.

The book cited above, pag. 86, analyze this situation:

enter image description here

taking the volume $ V $ as the one bounded by the surface $ S_1 $: infinite hemisphere described by the plane $ x = 0 $ and all the space in front of the slot ($ x> 0 $). In this case, I would say that the field cannot be determined only with the surface integral $ S_1 $, since that of volume does not give a null contribution (there are all the currents that are established on the ground plane). The book instead states exactly the opposite, assumes that the field on the slot is done in a certain way and proceeds to calculate only the surface integral and only on the surface of the slot.

Isn't that a mistake? Or, in the case of an implicit approximation, is it not a too strong approximation? It is like saying that we are leaving out fundamental contributions that would lead to having a field different from the one expected.


2 Answers 2


The explanation is on page 80, where the source, transmission line, etc. of Figure 3.1 are shown to be enclosed by a hemi-sphere over which the contribution of the fields can be neglected; this figure is referenced on page 87.

  • $\begingroup$ Thank you. However, my doubt in fact was not on the hemisphere $S_3$, but on the fact that the currents induced on the ground plane are also neglected, which instead I think must be taken into account since $S_2$ is inside the metal. $\endgroup$
    – Nameless
    Aug 22, 2020 at 16:16
  • $\begingroup$ Exactly; in the figure 3.1 the surface $S_2$ is inside within, and not on, the metal hence no current can flow on it. So the only source of the field is in the "mouth", i.e., in the slot itself. There is surface current on the outside of the metal but that is handled with "image processing", pun intended. The diffracted field from the mouth does not satisfy the boundary conditions by itself but together with its image it will. For details of that derivation read the original Stratton-Chu "Diffraction Theory of Electromagnetic Waves," Phys. Rev. vol56, in section 6. $\endgroup$
    – hyportnex
    Aug 22, 2020 at 17:56
  • $\begingroup$ On p. 81 actually says to use the image theorem, but he uses it in a way that I don't understand. Having taken his S as the union of S1 (mouth of the slot, a little more "out" of the ground plane), S2 (inside the ground plane) and S3 (infinite hemisphere), he says that since the field is zero in the internal points of the volume enclosed by S, then I can pretend that the ground plane has no holes. But why? (continues...) $\endgroup$
    – Nameless
    Aug 24, 2020 at 19:32
  • $\begingroup$ The fields imposed on S as boundary conditions of the problem must be the true ones, i.e. those that would come out solving the problem in all space and not circumscribing the rest of the world (transmitter, wave fluid, and in particular the ground plane with the hole) in the comfort surface S. Otherwise if we reason as the book says, then in any electromagnetic problem, I build a comfortable surface, and outside of it I can place ground planes anywhere, just because the field is zero ... $\endgroup$
    – Nameless
    Aug 24, 2020 at 19:33
  • $\begingroup$ see imgur.com/u8WkEQ2 and imgur.com/6KRXy5e $\endgroup$
    – hyportnex
    Aug 25, 2020 at 11:52

I guess author implicitly used Babinet's principle where you can replace slot in an infinite ground with a dipole antenna.

  • $\begingroup$ He deduces this aposteriori, after having already made the integral on the slot surface only, without including the currents on the ground plane. $\endgroup$
    – Nameless
    Aug 22, 2020 at 16:18

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