 The radiometer equation is one of the fundamental equations of radio astronomy. It defines the ratio of signal to noise. It will determine whether you can see a source given your telescope set-up and external conditions. It is given by, $$\frac{\mbox{Signal}}{\mbox{Noise}} = \frac{\mbox{Source Temperature}}{\mbox{System Temperature}} \sqrt{\mbox{Bandwidth} \times \mbox{Integreation Time}}$$ This post will look at the components of the equation.

### The source temperature

A telescope can be thought of as a resistor. When power (In our case the source signal) is passed through a resistor it will heat up. As such we start with the Rayleigh-Jeans description of the radiation emitted by a black body to find the specific intensity, $$I$$, of radiation of frequency, $$\nu$$, at a given source temperature,

$$I = \frac{2k \cdot \mbox{Source Temperature}}{\lambda^2} \left[ \frac{\mbox{Watts}}{m^2 \cdot \mbox{Hz} \cdot \mbox{Rad}^2} \right]$$

$$k$$ is the Boltzmann constant and $$\lambda[\latex] is the wavelength at which the intensity is measured. We can find the total power received by the antenna by multiplying the specific intensity by both the solid angle spanned and the effective area of the telescope. The solid angle is a measure of how large an object appears to an observer looking from another point in 3d space.  \mbox{Power} = \frac{2k \cdot \mbox{Source Temperature} \cdot \mbox{Area} \cdot \mbox{Solid Angle}}{\lambda^2} \left[ \frac{\mbox{Watts}}{\mbox{Hz}}\right] Half the total power received by a telescope must be equal to the Boltzmann constant multiplied by the source temperature. Via some algebra this means that  \mbox{Source Temperature} = \frac{\mbox{Area}}{ 2k} \times I \times \mbox{Solid Angle} [\mbox{Kelvin}] The area divided by [latex]2k$$ is called the forward gain of a telescope and has units of Kelvin per Jansky. A Jansky is the unit of flux density where

$$1 [\mbox{Jansky}] = 10^{-26} \left[ \frac{\mbox{Watts}}{m^2 \cdot \mbox{Hz}}\right]$$

Written by Ed Muir – Science Team