The aperture of an
isotropic antenna, the basis of the definition of gain above, can be derived on the basis of consistency with thermodynamics. Suppose that an ideal isotropic antenna
A with a
driving-point impedance of
R sits within a closed system CA in
thermodynamic equilibrium at temperature
T. We connect the antenna terminals to a resistor also of resistance
R inside a second closed system CR, also at temperature
T. In between may be inserted an arbitrary lossless electronic filter
Fν passing only some frequency components. Each cavity is in thermal equilibrium and thus filled with
black-body radiation due to temperature
T. The resistor, due to that temperature, will generate
Johnson–Nyquist noise with an open-circuit voltage whose mean-squared
spectral density is given by : \overline{v_n^2} = 4 k_\text{B} T R \, \eta(f), where \eta(f) is a quantum-mechanical factor applying to frequency
f; at normal temperatures and electronic frequencies \eta(f) = 1, but in general is given by : \eta(f) = \frac{hf/k_\text{B} T}{e^{hf/k_\text{B} T} - 1}. The amount of power supplied by an electrical source of impedance
R into a
matched load (that is, something with an impedance of
R, such as the antenna in CA) whose
rms open-circuit voltage is
vrms is given by : P = \frac{\text{v}_\text{rms}^2}{4\text{R}}. The mean-squared voltage \overline{v_n^2} = \text{v}_\text{rms}^2 can be found by integrating the above equation for the spectral density of mean-squared noise voltage over frequencies passed by the filter
Fν. For simplicity, let us just consider
Fν as a narrowband filter of bandwidth
B1 around central frequency
f1, in which case that integral simplifies as follows: : P_R = \frac{\int_0^\infty 4 k_\text{B} T R \, \eta(f) \, F_\nu(f) \, df}{4\text{R}} : \qquad = \frac{4 k_\text{B} T R \, \eta(f_1) \, B_1}{4\text{R}} = k_\text{B} T \, \eta(f_1) \, B_1. This power due to Johnson noise from the resistor is received by the antenna, which radiates it into the closed system CA. The same antenna, being bathed in black-body radiation of temperature
T, receives a spectral radiance (power per unit area per unit frequency per unit
solid angle) given by
Planck's law: : \text{P}_{f,A,\Omega}(f) = \frac{2hf^3}{c^2} \frac{1}{e^{hf / k_\text{B} T} - 1} = \frac{2f^2}{c^2} \, k_\text{B} T \, \eta(f), using the notation \eta(f) defined above. However, that radiation is unpolarized, whereas the antenna is only sensitive to one polarization, reducing it by a factor of 2. To find the total power from black-body radiation accepted by the antenna, we must integrate that quantity times the assumed cross-sectional area
Aeff of the antenna over all solid angles Ω and over all frequencies
f: : P_A = \int_0^\infty \int_{4\pi} \, \frac{P_{f,A,\Omega}(f)}{2} A_\text{eff}(\Omega, f) \, F_\nu(f) \, d\Omega \, df. Since we have assumed an isotropic radiator,
Aeff is independent of angle, so the integration over solid angles is trivial, introducing a factor of 4π. And again we can take the simple case of a narrowband
electronic filter function
Fν which only passes power of bandwidth
B1 around frequency
f1. The double integral then simplifies to : P_A = 2\pi P_{f,A,\Omega}(f) A_\text{eff} \, B_1 = \frac{4\pi \, k_\text{B} T \, \eta(f_1)}{\lambda_1^2} A_\text{eff} B_1, where \lambda_1 = c/f_1 is the free-space wavelength corresponding to the frequency
f1. Since each system is in
thermodynamic equilibrium at the same temperature, we expect no net transfer of power between the cavities. Otherwise one cavity would heat up and the other would cool down in violation of the
second law of thermodynamics. Therefore, the power flows in both directions must be equal: : P_A = P_R. We can then solve for
Aeff, the cross-sectional area intercepted by the
isotropic antenna: : \frac{4 \pi \, k_\text{B} T \, \eta(f_1)}{\lambda_1^2} A_\text{eff} B_1 = k_\text{B} T \, \eta(f_1) \, B_1, : A_\text{eff} = \frac{\lambda_1^2}{4\pi}. We thus find that for a hypothetical isotropic antenna, thermodynamics demands that the effective cross-section of the receiving antenna to have an area of λ2/4π. This result could be further generalized if we allow the integral over frequency to be more general. Then we find that
Aeff for the same antenna must vary with frequency according to that same formula, using λ =
c/
f. Moreover, the integral over solid angle can be generalized for an antenna that is
not isotropic (that is, any real antenna). Since the angle of arriving
electromagnetic radiation only enters into
Aeff in the above integral, we arrive at the simple but powerful result that the
average of the effective cross-section
Aeff over all angles at wavelength λ must also be given by Although the above is sufficient proof, we can note that the condition of the antenna's impedance being
R, the same as the resistor, can also be relaxed. In principle, any antenna impedance (that isn't totally reactive) can be
impedance-matched to the resistor
R by inserting a suitable (lossless)
matching network. Since that network is
lossless, the powers
PA and
PR will still flow in opposite directions, even though the voltage and currents seen at the antenna and resistor's terminals will differ. The spectral density of the power flow in either direction will still be given by k_\text{B} T \, \eta(f), and in fact this is the very thermal-noise power spectral density associated with one
electromagnetic mode, be it in free-space or transmitted electrically. Since there is only a single connection to the resistor, the resistor itself represents a single mode. And an antenna, also having a single electrical connection, couples to one mode of the
electromagnetic field according to its average effective cross-section of \lambda_1^2/(4\pi). ==See also==