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

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