Flux density from brightness map

Question

This is a naive question. I have a temperature map of radio emission at 1420 MHz, in Kelvin. I want to extract the flux density (in Janskys) from an extended region of this map. Is the flux density ($S_{\nu}$) from this region equal to $$ S_{\nu} = \frac{2\nu^2 k_B T_{\text{mean}}}{c^2} \times A$$ where $\nu$ is the frequency of my map, $k_B$ is the Boltzmann constant, $T_{\text{mean}}$ is the mean value of the temperature inside my region, $c$ is the speed of light and $A$ is the area of my region in steradians?

Answer 1

This expression is valid for low frequencies, including the case of the 1420 MHz hydrogen line. It arises from treating the source as a black body with temperature $T(\theta,\phi)$$^{\dagger}$, and assuming $h\nu\ll k_BT$: $$I_{\nu}=B_{\nu}=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1}\approx\frac{2h\nu^3}{c^2}\frac{1}{h\nu/k_BT}=\frac{2k_BT\nu^2}{c^2}$$ where we have made the approximation $$e^{h\nu/k_BT}\approx1+\frac{h\nu}{k_BT}$$ and I drop the explicit $(\theta,\phi)$ dependence. The flux density is then, by definition (and assuming the source is small, so $\cos\theta\approx1$), $$S_{\nu}=\int I_{\nu}\cos\theta\;d\Omega\approx\int I_{\nu}\;d\Omega=\frac{2k_B\nu^2}{c^2}\int T(\theta,\phi)\;d\Omega\tag{1}$$ If we calculate the mean of the function $T(\theta,\phi)$ over the source, we get $$T_{\text{mean}}=\frac{\int T(\theta,\phi)\;d\Omega}{\int d\Omega}$$ and some algebra and noting that $\int d\Omega=\Omega$ should convince you that we can make the desired substitution. We can essentially always take the mean of a well-behaved function when integrating over a patch of sky.

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$^{\dagger}$ The brightness temperature $T_b$ is typically defined as the temperature of a black body which would yield the same specific intensity.