====== Antenna temperature ======
It is not the physical temperature of an antenna, but the temperature of a matched resistor whose thermally generated power per unit frequency is equal to that produced by the antenna in the Nyquist approximation. So antenna temperature

$$ T_A = \frac{P_\nu}{k} $$

where $k$ is the [[Boltzmann constant]]. So a 1 K antenna temperature amounts to a power $P_\nu = kT = 1.38\times 10^{-23}$ W/Hz. It can be calibrated easilty using matched resistors or **loads**. And it is convenient because the receiver noise is also measured in K. Its relation with flux density can be found following the equations given in the article [[beam]]:

$$ T_A = \frac{A_e S}{2k} $$

where $S$ is measured in [[jansky]]. The point-source sensitivity of a radio telescope is often given in K/Jy as

$$ A_e = 2k\frac{T_A}{S} $$

which is $2761$ m$^2$ if we put $T_A=1$ K and $S=1$ Jy. Now let us put this antenna in a radiation field $I_\nu$. So the power received by the antenna as given in [[beam]]

$$ P_\nu = \frac{1}{2} \int A_e I_\nu d\Omega \Rightarrow kT_A = \frac{1}{2} \int A_e \frac{2k}{\lambda^2} T_b d\Omega $$

where $T_b$ is the [[brightness temperature]]; if this is constant the integral simplifies to

$$ T_A = \frac{T_b}{2} \int A_e d\Omega = T_b $$

which means //the antenna temperature produced by a diffuse smooth source much larger than the beam is equal to the brightness temperature of the source//.

On the other hand, if we have a compact source of uniform brightness temperature and a solid angle $\Omega_s < \Omega_A$ then its

$$ T_A = \frac{A_0 T_b \Omega_s}{\lambda^2} $$

where $A_0$ is the **on-axis** collecting area, the peak. Now if put $A_0\lambda^2 = \Omega_A$ then

$$ \frac{T_A}{T_b} = \frac{\Omega_s}{\Omega_A} = \Omega_{ff} $$


where $\Omega_{ff}$ is the beam filling factor. The **main beam** of an antenna is defined as

$$ \Omega_{MB} = \frac{1}{G_0} \int_{MB} G(\theta,\phi) d\Omega $$

and the beam efficiency

$$ \eta = \frac{\Omega_{MB}}{\Omega_A}. $$