# Effective aperture of isotropic antenna

I have always taken for granted that 'the aperture of a loss-less isotropic antenna is $\dfrac {\lambda^2} {4\pi}$'.

On a whim, I tried to look up how this expression was derived, but so far I have not been able to find out a reference that works it out. Can someone outline the proof, or help me with any pointers to some on-line source? Hopefully this can be done using 3D vector calculus?

Thanks...

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So here I am, answering my own question...

Long story short -- I found the answer here, and in page 149 of 'Tools of Radio Astronomy' [Rohlfs/Wilson/Huttemeister 5/e 2009], and page 24 of 'Radio Astronomy' [Pawsey/Bracewell 1955] and I will now express that answer in my own words to save the reader a click!

Interestingly, one part of this proof comes from Thermodynamics--specifically the Rayleigh-Jeans Law which says that, given a black body at temperature $T$, the spectral radiance $B_{\nu}$ (power radiated per unit solid-angle per unit area per unit frequency) is given by: $$B_{\nu} = \frac {2 {\nu}^2 k T }{c^2} = \frac {2 k T }{{\lambda}^2}$$

Another part is the power spectral density of thermal noise, which is: $$\frac {dP}{d\nu}=kT$$

Now we consider two spherical cavities made of perfectly absorbing material, at thermal equilibrium with temperature $T$. Now, we have an isotropic antenna of effective aperture (seen from any direction -- by definition) $A_e$ at the center of the first cavity, and a matched resistor $R$ at the center of the second. The antenna and the resistor are connected through a transmission line and a filter that passes frequencies between $\nu$ and $\nu + d\nu$. Like this:

The walls will emit blackbody radiation. The antenna will collect it, as will the resistor (as thermal boise.) Once collected, the power has nowhere else to go except down the transmission line towards the other cavity. Power outside the frequency band $\left[\nu, \nu+d\nu\right]$ meets infinite impedance at the filter, and is reflected right back to where it was received -- and is re-radiated back into the original cavity, electromagnetically in the left cavity, and thermally in the right.

Now the power 'seen' from the point in space where the antenna is located, as in the surface integral of the Poynting vector taken around the point, will be: $$\int_0^{\infty} \left(\int_{4\pi}A_e\;B_{\nu}\;d\Omega\right)d\nu$$ But this radiation is unpolarized, so the antenna can collect power from only one polarization -- only half of what is available. So in the left-hand cavity we have the following power collected by the antenna in the frequency band of our interest: $$d\nu\;\frac 12 \int_{4\pi} A_e\;B_{\nu}\;d\Omega = d\nu\;\frac {A_e kT} {\lambda^2} \int_{4\pi} 1\;d\Omega = d\nu\;\frac {4\pi A_ekT} {\lambda^2}$$

On the right hand side, the thermal noise power generated by the resistor in the frequency band of our interest will be $$kT\;d\nu$$

Since the two cavities are at the same temperature, by the second law of thermodynamics, no net power can flow beween the two cavities. Thus the power collected by the antenna must equal the noise power generated by the resistor. Thus: $$d\nu\;\frac {4\pi A_e kT} {\lambda^2} = kT\;d\nu$$ ... and therefore: $$A_e = \frac {\lambda^2} {4\pi}$$

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It's covered in this sample book chapter:

http://highered.mcgraw-hill.com/sites/dl/free/0072321032/62577/ch02_011_056.pdf

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Link-only answers are discouraged. Can you add some summary of the contents of that source? – Colin McFaul Jun 6 '13 at 14:11
Hi @BitBucket -- sorry, I couldn't locate any derivation of the $\frac{\lambda^2}{4\pi}$ expression in the chapter linked. Can you point to the page number please? – Avijit Jun 13 '13 at 9:21

## protected by Qmechanic♦Aug 8 '14 at 12:58

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