# Flux density from brightness map

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?

## 1 Answer

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.

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

• Thank you for your detailed explanation. I am aware of the connection between density flux and brightness temperature and I understand the Rayleigh-Jeans law. I was wondering if I could replace $\int I_{\nu} \cos\theta d\Omega$ with $I_{\nu} \times A$, where $A$ is the surface of my ds9 region and $T_{mean}$ is the mean temperature obtained from all the pixels inside my region. – Tutan May 27 '20 at 8:07
• @Tutan Ah, my bad - then the answer is certainly yes; we're treating $T$ as a function $T(\theta,\phi)$ and we can certainly take the mean of that function over the source - answer edited. – HDE 226868 May 27 '20 at 14:02