<p>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 <span class="math-container">$T(\theta,\phi)$$^{\dagger}$</span>, and assuming <span class="math-container">$h\nu\ll k_BT$</span>:
<span class="math-container">$$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}$$</span>
where we have made the approximation
<span class="math-container">$$e^{h\nu/k_BT}\approx1+\frac{h\nu}{k_BT}$$</span>
and I drop the explicit <span class="math-container">$(\theta,\phi)$</span> dependence. The flux density is then, by definition (and assuming the source is small, so <span class="math-container">$\cos\theta\approx1$</span>),
<span class="math-container">$$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}$$</span>
If we calculate the mean of the function <span class="math-container">$T(\theta,\phi)$</span> over the source, we get
<span class="math-container">$$T_{\text{mean}}=\frac{\int T(\theta,\phi)\;d\Omega}{\int d\Omega}$$</span>
and some algebra and noting that <span class="math-container">$\int d\Omega=\Omega$</span> 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.</p>

<hr>

<p><span class="math-container">$^{\dagger}$</span> The brightness temperature <span class="math-container">$T_b$</span> is typically <em>defined</em> as the temperature of a black body which would yield the same specific intensity.</p>

CC BY-SA 4.0
2020-05-27T08:07:40.823Z
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
CC BY-SA 4.0
2020-05-27T14:02:57.753Z
@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
CC BY-SA 4.0
2020-05-26T17:10:17.860Z
2020-05-27T14:01:50.263Z
2020-05-27T14:01:50.263Z
HDE 226868
<p>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 (<span class="math-container">$S_{\nu}$</span>) from this region equal to
<span class="math-container">$$ S_{\nu} = \frac{2\nu^2 k_B T_{\text{mean}}}{c^2} \times A$$</span>
where <span class="math-container">$\nu$</span> is the frequency of my map, <span class="math-container">$k_B$</span> is the Boltzmann constant, <span class="math-container">$T_{\text{mean}}$</span> is the mean value of the temperature inside my region, <span class="math-container">$c$</span> is the speed of light and <span class="math-container">$A$</span> is the area of my region in steradians?</p>

<observational-astronomy><radio-astronomy><units><ds9><flux>
Flux density from brightness map
Tutan
# Flux density from brightness map

(Asked by: username_0 on 2020-05-26)

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?

**Tags:** observational-astronomy, radio-astronomy, units, ds9, flux


---

## 1 answers

### Answer by username_1 on 2020-05-26. Score: 4 | Accepted 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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: @username_0 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.  

---

