Isotropic radiator

In physics, an isotropic radiator is a point radiation or sound source. At a distance, the Sun is an isotropic radiator of electromagnetic radiation.

Antenna theory

In antenna theory, an isotropic antenna is a hypothetical antenna radiating the same intensity of radio waves in all directions. It thus is said to have a directivity of 0 dBi (dB relative to isotropic) in all directions. Since it is entirely non-directional, it serves as a hypothetical worst-case against which directional antennas may be compared.

In reality, a coherent isotropic radiator of linear polarization can be shown to be impossible.[lower-alpha 1] Its radiation field could not be consistent with the Helmholtz wave equation (derived from Maxwell's equations) in all directions simultaneously. Consider a large sphere surrounding the hypothetical point source, in the far field of the radiation pattern so that at that radius the wave over a reasonable area is essentially planar. In the far field the electric (and magnetic) field of a plane wave in free space is always perpendicular to the direction of propagation of the wave. So the electric field would have to be tangent to the surface of the sphere everywhere, and continuous along that surface. However the hairy ball theorem shows that a continuous vector field tangent to the surface of a sphere must fall to zero at one or more points on the sphere, which is inconsistent with the assumption of an isotropic radiator with linear polarization.

Incoherent isotropic antennas are possible and do not violate Maxwell's equations. In practice, small antennas of all types are approximately isotropic whenever their longest dimension is much less than one wavelength (say, ~1/ 10  wave or less): The smaller the antenna, the more nearly isotropic it becomes.[lower-alpha 2]

Even though an exactly isotropic antenna cannot exist in practice, it is used as a base of comparison to calculate the directivity of actual antennas. Antenna gain ${\displaystyle \scriptstyle \ G\ ,}$ which is equal to the antenna's directivity multiplied by the antenna efficiency, is defined as the ratio of the intensity ${\displaystyle \scriptstyle \ I\ }$ (power per unit area) of the radio power received at a given distance from the antenna (in the direction of maximum radiation) to the intensity ${\displaystyle \scriptstyle \ I_{\text{iso}}\ }$ received from a perfect lossless isotropic antenna at the same distance. This is called isotropic gain

$${\displaystyle G={\frac {I}{~\ I_{\text{iso}}\ }}~.}$$

Gain is often expressed in logarithmic units called decibels (dB). When gain is calculated with respect to an isotropic antenna, these are called decibels isotropic (dBi)

$${\displaystyle G{\text{(dBi)}}=10\ \log _{10}\left({\frac {I}{~\ I_{\text{iso}}\ }}\right)~.}$$

The gain of any perfectly efficient antenna averaged over all directions is unity, or 0 dBi.

Sound

A depiction of an isotropic radiator of sound, published in Popular Science Monthly in 1878. Note how the rings are even and of the same width all the way around each circle, though they fade as they move away from the source.

An isotropic sound radiator is a theoretical loudspeaker radiating equal sound volume in all directions. Since sound waves are longitudinal waves, a coherent isotropic sound radiator is feasible; an example is a pulsing spherical membrane or diaphragm, whose surface expands and contracts radially with time, pushing on the air.[1]

Diagram of antenna and resistor in cavity

The aperture of an isotropic antenna can be derived by a thermodynamic argument, which follows.[2][3][4]

Suppose an ideal (lossless) isotropic antenna A located within a thermal cavity CA is connected via a lossless transmission line through a band-pass filter F_ν to a matched resistor R in another thermal cavity CR (the characteristic impedance of the antenna, line and filter are all matched). Both cavities are at the same temperature ${\displaystyle \ T~.}$ The filter F_ν only allows through a narrow band of frequencies from ${\displaystyle \ \nu \ }$ to ${\displaystyle \ \nu +\Delta \nu ~.}$ Both cavities are filled with blackbody radiation in equilibrium with the antenna and resistor. Some of this radiation is received by the antenna.

The amount of this power ${\displaystyle \ P_{\text{A}}\ }$ within the band of frequencies ${\displaystyle \ \Delta \nu \ }$ passes through the transmission line and filter F_ν and is dissipated as heat in the resistor. The rest is reflected by the filter back to the antenna and is reradiated into the cavity. The resistor also produces Johnson–Nyquist noise current due to the random motion of its molecules at the temperature ${\displaystyle \ T~.}$ The amount of this power ${\displaystyle \ P_{\text{R}}\ }$ within the frequency band ${\displaystyle \ \Delta \nu \ }$ passes through the filter and is radiated by the antenna. Since the entire system is at the same temperature it is in thermodynamic equilibrium; there can be 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

$${\displaystyle P_{\text{A}}=P_{\text{R}}}$$

The radio noise in the cavity is unpolarized, containing an equal mixture of polarization states. However any antenna with a single output is polarized, and can only receive one of two orthogonal polarization states. For example, a linearly polarized antenna cannot receive components of radio waves with electric field perpendicular to the antenna's linear elements; similarly a right circularly polarized antenna cannot receive left circularly polarized waves. Therefore the antenna only receives the component of power density S in the cavity matched to its polarization, which is half of the total power density

$${\displaystyle S_{\text{matched}}={\frac {\ 1\ }{2}}S}$$

Suppose ${\displaystyle \ B_{\nu }\ }$ is the spectral radiance per hertz in the cavity; the power of black body radiation per unit area (m²) per unit solid angle (steradian) per unit frequency (hertz) at frequency ${\displaystyle \ \nu \ }$ and temperature ${\displaystyle \ T\ }$ in the cavity. If ${\displaystyle \ A_{\text{e}}(\theta ,\phi )\ }$ is the antenna's aperture, the amount of power in the frequency range ${\displaystyle \ \Delta \nu \ }$ the antenna receives from an increment of solid angle ${\displaystyle \ \mathrm {d} \Omega =\mathrm {d} \theta \;\mathrm {d} \phi \ }$ in the direction ${\displaystyle \ \theta ,\phi \ }$ is

$${\displaystyle \mathrm {d} P_{\text{A}}(\theta ,\phi )~=~A_{\text{e}}(\theta ,\phi )\ S_{\text{matched}}\ \Delta \nu \;{\text{d}}\Omega ~=~{\frac {\ 1\ }{2}}A_{\text{e}}(\theta ,\phi )\ B_{\nu }\ \Delta \nu \;\mathrm {d} \Omega }$$

To find the total power in the frequency range ${\displaystyle \ \Delta \nu \ }$ the antenna receives, this is integrated over all directions (a solid angle of ${\displaystyle \ 4\pi \ }$)

$${\displaystyle P_{\text{A}}={\frac {\ 1\ }{2}}\ \int \limits _{4\pi }A_{\text{e}}(\theta ,\phi )\ B_{\nu }\ \Delta \nu \;\mathrm {d} \Omega }$$

Since the antenna is isotropic, it has the same aperture ${\displaystyle \ A_{\text{e}}(\theta ,\phi )=A_{\text{e}}\ }$ in any direction. So the aperture can be moved outside the integral. Similarly the radiance ${\displaystyle \ B_{\nu }\ }$ in the cavity is the same in any direction

$${\displaystyle P_{\text{A}}={\frac {\ 1\ }{2}}A_{\text{e}}\ B_{\nu }\ \Delta \nu \ \int \limits _{4\pi }\mathrm {d} \Omega }$$

$${\displaystyle P_{\text{A}}=2\pi \ A_{\text{e}}\ B_{\nu }\ \Delta \nu }$$

Radio waves are low enough in frequency so the Rayleigh–Jeans formula gives a very close approximation of the blackbody spectral radiance[lower-alpha 3]

$${\displaystyle B_{\nu }={\frac {\ 2\nu ^{2}kT\ }{c^{2}}}={\frac {\ 2kT\ }{\lambda ^{2}}}}$$

Therefore

$${\displaystyle P_{\text{A}}={\frac {\ 4\pi \ A_{\text{e}}\ kT\ }{\lambda ^{2}}}\ \Delta \nu }$$

The Johnson–Nyquist noise power produced by a resistor at temperature ${\displaystyle \ T\ }$ over a frequency range ${\displaystyle \ \Delta \nu \ }$ is

$${\displaystyle P_{\text{R}}=kT\ \Delta \nu }$$

Since the cavities are in thermodynamic equilibrium ${\displaystyle \ P_{\text{A}}=P_{\text{R}}\ ,}$ so

$${\displaystyle {\frac {\ 4\pi A_{\text{e}}kT\ }{\lambda ^{2}}}\ \Delta \nu =kT\ \Delta \nu }$$

• Radiation pattern
• E-plane and H-plane

• Isotropic Radiators, Matzner and McDonald, arXiv Antennas
• Antennas D.Jefferies
• isotropic radiator AMS Glossary
• U.S. Patent 4,130,023 – Method and apparatus for testing and evaluating loudspeaker performance
• Non Lethal Concepts – Implications for Air Force Intelligence Archived 2007-04-30 at the Wayback Machine Published Aerospace Power Journal, Winter 1994
• Glossary
• Cosmic Microwave Background – Introduction
• Isotropic Radiators Archived 2014-08-19 at the Wayback Machine Holon Academic Institute of Technology