I'm reading some introduction to antenna theory and I've often puzzled on the equation:

$$ A_{eff} = \frac{\lambda^2}{4\pi} $$

which relates the effective area by which an antenna captures radiation to the frequency at which that radiation is.

I have looked at this derivation of the formula and can understand the steps they take, but when trying to understand it on a higher level, I cannot reconcile it.

Going on the example used in the derivation above, I would assume that as the frequency that is selected to pass through the filter increases, the resistor would begin to give off more power, in line with Johnson-Nyquist Noise:

$$ dP = k_{b}Td\nu $$

This would mean, in order to keep thermal equilibrium, that the antenna on the other end would have to be more receptive to the blackbody radiation that the cavity would give off. So, I would assume, that would mean a larger Effective Aperture Area would be required in order to gather it.

But this seems to contradict the result, which says that higher frequencies need a smaller area.

Can anyone help me out and point out the flaws in my assumptions?


  • 1
    $\begingroup$ In the formula you want to derive there is no temperature or other thermodynamic quantities. It should be possible to derive it from EM theory only. $\endgroup$ – Ján Lalinský Mar 16 '15 at 6:04
  • $\begingroup$ I'm not even sure that the formula has a name that I could use to easily look it up in an EM textbook. :\ $\endgroup$ – Lawrence Trevor Mar 16 '15 at 21:23
  • $\begingroup$ You might have luck in the books on antenna theory for engineers. $\endgroup$ – Ján Lalinský Mar 16 '15 at 21:46
  • $\begingroup$ @JánLalinský Indeed one can do it. See my answer, and the more detailed on in the R. E. Collin work cited in my answer. $\endgroup$ – Selene Routley Sep 14 '17 at 4:00

Increasing frequency does not change the dv spectral interval of integration, and with an assumption of white noise there is no increase in dissipated power with frequency. Therefore with resistor power held constant there is a decrease of aperture area with wavelength squared.

| cite | improve this answer | |
  • $\begingroup$ Yep, I understand now. I was missing the point of that dv in the Johnson-Nyquist equation. A higher frequency filter will not give a higher dissipated power at the resistor, but will result in a higher energy being radiated from the black-body to the antenna. Hence, to conserve energy, a smaller effective area is needed. And in the way that blackbody energy increases with frequency^2, so too must the effective aperture area decrease with frequency^2. :D $\endgroup$ – Lawrence Trevor Jan 4 '16 at 4:23

Summary: The effecive antenna area arises essentially from a lowpass filtering imparted by the decay of evanescent waves near the antenna, or, alternatively, from discarding the antenna nearfield from any coupling calculation and using only the propagating farfield in the calculation.

Ah, the mysterious antenna effective area formula. I found it mysterious for a long time when I first met it too.

This linked Physics Stack EAxchange answer derives the formula from a thermodynamic argument, and it seems as though Sam Millicoat's answer has clarified that argument for you. But one can also derive the formula from EM theory without thermodynamics. R. E. Colin's ancient book[1] is a good albeit somewhat hard-to-read and complicated reference, but here is a sketch my own derivation. The actual details are a bit more messy (putting in the vector directions and so forth) but the argument is essentially the following.

Lorentz reciprocity shows that the power captured by the dipole lit by a plane wave is the same as the power launched into a plane wave. So it doesn't matter whether we derive the formula for transmitter or a receiver. It's more straightforward to derive the power captured by plane wave from an antenna, so let's avail ourselves of Lorentz and do it that way.

You can make a "real isotropic radiator" as a mixture of randomly real oriented dipoles. Let's first consider the dipole in a waveguide with its radiation maximum pointing along the waveguide. We assume the waveguide is much bigger than dipole, so the former's eigenfields can be taken to be constant in the latter's neighborhood. To calculate the amplitude of the field captured by a waveguide mode, we simply need to do an overlap integral:

$$\langle\hat{\psi}_w,\,\hat{\psi}_a\rangle=\int_A \mathbf{\hat{E}}_w \times \mathbf{\hat{H}}_a^\ast\cdot\mathbf{\hat{z}}\,\mathrm{d} A\tag{1}$$

where the fields are appropriately normalized (this will become clearer as we go) and the subscripts stand for waveguide and antenna fields. Now, we only include the propagating field: the antenna's evanescent field will not launch power.

Indeed it is essentially the lowpass filtering implicit in throwing away the evanescent / nearfield field that gives rise to effective area formula

So, we imagine propagating the antenna's fields far from the waveguide, stripping off the evanescent field, then running the propagation in reverse back to the position of the antenna and at last doing the overlap (1) with the propagating field. The antenna's farfield has only field components in the waveguide's transverse plane, so (1) reduces essentially to an inner product between scalar fields.

In wavevector (Fourier) space, the dipole's transverse fields have an amplitude that varies like $\cos\theta$ over the sphere of radius $k$, where $\theta$ is the spherical colatitude (zenith angle or elevation). Therefore the propagating transverse field (electric or magnetic) at the antenna is the following Fourier transform, calculated by summing up all the propagating plane waves of wavenumber $k$ corresponding to different points on the wavenumber space half sphere multiplied by the apodization factor $\cos\theta$ over the half sphere of radius $k$ in Fourier space. The result is proportional to:

$$\int_0^\frac{\pi}{2}\int_0^{2\pi} \exp\left(i\,k\,r\,\cos(\phi-\varphi)\right)\,\cos\theta\sin\theta\,\mathrm{d}\phi\,\mathrm{d}\theta \propto \frac{J_1(k\,r)}{r}\tag{2}$$

where $r$ and $\varphi$ are the radial and azimuthal co-ordinate in the waveguide. This is the wonted Airy disk beloved of microscopists and astronomers, so there are no surprises so far. We normalize (2) properly to get our scalar field that represents that propagating part of the antenna's field such that $\int_A\left|\hat{\psi}_a\right|^2\,\mathrm{d} A$ to find:

$$\hat{\psi}_a = \frac{J_1(k\,r)}{\sqrt{\pi}\,r}\tag{3}$$

Now we assume that the waveguide is very large so that its fields don't vary in the antenna's neighborhood. Therefore, the overlap (1) simply:

$$\begin{array}{lcl}\displaystyle{\langle\hat{\psi}_w,\,\hat{\psi}_a\rangle}&=&\displaystyle{\int_0^\infty\,r\,\int_0^{2\,\pi}\,\hat{\psi}_w(r,\,\varphi)\,\frac{J_1(k\,r)}{\sqrt{\pi}\,r}\,\mathrm{d}\varphi\,\mathrm{d} r = 2\,\pi\,\hat{\psi}_w(0,\,0)\,\int_0^\infty\,r\,\frac{J_1(k\,r)}{\sqrt{\pi}\,r}\,\mathrm{d} r}\\\\&=&\displaystyle{\frac{2\,\sqrt{\pi}}{k}\,\hat{\psi}_w(0,\,0)}\end{array}\tag{4}$$

This is our result. If the antenna launches unit power, this is the amplitude for the power unit to be captured by the mode in question. By Lorentz reciprocity, if the waveguide mode bears unit power, this is the amplitude of capture by the antenna, and so the fraction of power captured by the antenna is $\frac{4\,\pi}{k^2}\left|\hat{\psi}_w(0,\,0)\right|^2$. The intensity of the mode in the antenna's neighborhood is $\left|\hat{\psi}_w(0,\,0)\right|^2$. Therefore, we must ascribe an effective area to the dipole of $\frac{4\,\pi}{k^2}$.

But wait: this was for a perfectly aligned dipole, and we only integrated over half the sphere. When the antenna emits, half the power is transmitted away from the receiver. So we must apply a factor of $1/2$ to this effective area to account for half the power's being launched the "wrong" way. Moreover, we now consider a mixture of random dipoles - imagine a bunch of mutually incoherent dipoles all grouped together in a region much smaller than a wavelength - to simulate the isotropic antenna. This is a realistic scenario in system of fluorophores relaxing: there are thousands of randomly oriented dipoles relaxing at once from volumes of extents very small compared with a wavelength. A dipole whose axis points along the colatitude $\theta$ suffers a power loss factor of $\cos\theta$ in its coupling into the waveguide in question. When we average this $\cos\theta$ factor over the whole sphere, we find we must apply a further loss factor of $1/2$.

So, at last, our effective area is:

$$\frac{1}{4} \frac{4\,\pi}{k^2} = \frac{\pi}{k^2} = \frac{\lambda^2}{4\,\pi}\tag{5}$$

Mystery solved!

[1]: R. E. Collin, “The Receiving Antenna,” in Antenna Theory, Part I, R. E. Colling and F. J. Zucker, Eds., McGraww-Hill 1969

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.