<p>Actually to convert from Jy/beam to Jy/pixel you need to divide by the beam size.</p>

<p>Let's say you have a quantity of 1 Jy/beam, then</p>

<p><span class="math-container">$\frac{Jy}{beam} \frac{beam}{\Omega}$</span>, then to go from Jy/beam to Jy/pixel you would need to divide by <span class="math-container">$\Omega$</span>.</p>

<p>The values of the beam major and minor axis must be in pixels.</p>

<p>Source: <a href="https://science.nrao.edu/facilities/vla/proposing/TBconv" rel="nofollow noreferrer">NRAO</a></p>

CC BY-SA 4.0
2019-11-27T20:05:03.603Z
2019-11-27T22:33:23.617Z
2019-11-27T22:33:23.617Z
Miguel Cárcamo
<p>So long as you accurately know the beam size, then yes multiplying your Jy/beam measurement (effectively flux density) by the beam size (effectively area of flux) will give you the total Jy (effectively the flux).</p>

<p><a href="http://tsrice-research.blogspot.com/2016/05/how-to-manually-extract-flux-in-jy-from.html" rel="nofollow noreferrer">See this source as an example.</a></p>

CC BY-SA 4.0
2018-06-22T15:38:39.367Z
2018-06-22T15:38:39.367Z
zephyr
<p>It depends on what you mean by &quot;just Jy&quot;. Usually, what is meant is the surface brightness of a source, in some unit like <span class="math-container">$\operatorname{Jy}\operatorname{sr}^{-1}$</span> or <span class="math-container">$\operatorname{Jy}\operatorname{arcsec}^{-2}$</span>, integrated over solid angle to get the source's total flux. What a measurement of <span class="math-container">$I \operatorname{Jy}\operatorname{beam}^{-1}$</span> is telling you is, roughly, &quot;An unresolved source that is the nominal beam size with this peak flux will have total flux <span class="math-container">$I$</span> in Jy.&quot; So, if you want surface brightness, take your quantity in Jy/beam and divide by <span class="math-container">$\Omega$</span>/beam (see the relationship between <span class="math-container">$S$</span> [flux] and <span class="math-container">$I$</span> [surface brightness] in <A HREF="https://science.nrao.edu/facilities/vla/proposing/TBconv" rel="nofollow noreferrer">this brightness temperature NRAO tutorial</A>.</p>
<p>In most situations with radio astronomy, the clean beam will be a Gaussian, so
<span class="math-container">\begin{align}
   I &amp;= I_{\text{Jy/beam}} \frac{\text{beam}}{\Omega_{\text{beam}}} \\
     &amp;= I_{\text{Jy/beam}} \frac{1}{\pi \sigma_{\text{beam}}^2} \\
    &amp;= I_{\text{Jy/beam}} \frac{4\ln(2)}{\pi \theta_{\text{beam}}^2},
\end{align}</span>
with <span class="math-container">$\theta_{\text{beam}}$</span> the beams full width at half power and <span class="math-container">$\sigma_{\text{beam}}$</span> the standard deviation of the beam.</p>
<p>Once you have <span class="math-container">$I$</span>, getting Jy / pixel is as easy as multiplying by <span class="math-container">$\Omega_{\text{pixel}}$</span>. If you have some surface brightness profile you fit, like
<span class="math-container">\begin{align}
  I_{\text{model}} &amp;= A \exp\left(-\frac{(x-x_0)^2\sigma_y^2- 2\rho\sigma_x\sigma_y(x-x_0)(y-y_0) + (y-y_0)^2}{2\sigma_x^2\sigma_y^2(1-\rho^2)}\right),
\end{align}</span>
where <span class="math-container">$\sigma_x$</span> and <span class="math-container">$\sigma_y$</span> are normal standard deviations and <span class="math-container">$\rho$</span> is your normal correlation coefficient. For more normal astronomer usage, you'd use
<span class="math-container">\begin{align}
   \theta_M &amp;= \sqrt{8\ln(2)\left[\frac{\sigma_x^2 + \sigma_y^2}{2} + \sqrt{\left(\frac{\sigma_x^2 - \sigma_y^2}{2}\right)^2 + \rho^2\sigma_x^2\sigma_y^2}\right]} \\
   \theta_m &amp;=\sqrt{8\ln(2)\left[\frac{\sigma_x^2 + \sigma_y^2}{2} - \sqrt{\left(\frac{\sigma_x^2 - \sigma_y^2}{2}\right)^2 + \rho^2\sigma_x^2\sigma_y^2}\right]} \\
    \phi &amp;= \left\{\begin{array}{ll}
        0 &amp; \text{if }\rho=0,\ \sigma_y &gt; \sigma_x \\
        90^\circ &amp; \text{if }\rho=0,\ \sigma_x &gt; \sigma_y \\
        \operatorname{atan2}\left(\theta_M^2 - 8\ln(2)\sigma_x^2, \sigma_y^2\right) &amp; \text{otherwise}
    \end{array} \right.
\end{align}</span>
Then you can convert the peak surface brightness <span class="math-container">$A$</span> to total brightness by integrating <span class="math-container">$I$</span> over all <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, yielding
<span class="math-container">\begin{align}
   S &amp;= A 2\pi \sigma_x \sigma_y \sqrt{1-\rho^2} \\
     &amp;= A \frac{\pi \theta_m \theta_M}{4\ln(2)}.
\end{align}</span></p>
<p>Note what happens if you combine the conversions from Jy/beam to surface brightness to Jy. You get:
<span class="math-container">\begin{align}
   S &amp;= A \frac{\theta_m \theta_M}{\theta_{\mathrm{beam}}^2}.
\end{align}</span>
C.f. Equation 35 of <A HREF="https://ui.adsabs.harvard.edu/abs/1998AJ....115.1693C" rel="nofollow noreferrer">Condon et al (1998)</A>. Note that Condon et al provide multiple equations that depend on how resolved thee source is. Based on the <A HREF="https://ui.adsabs.harvard.edu/abs/1997PASP..109..166C/abstract" rel="nofollow noreferrer">referenced paper</A> it looks like what they're doing is minimizing the variance of their &quot;corrected&quot; values.</p>

CC BY-SA 4.0
2020-09-01T12:18:44.770Z
2020-09-01T12:18:44.770Z
Sean Lake
<p>Maybe its a dumb question, but to convert Jy/beam to Jy, I just have to multiply it by the beam size in sr right?</p>

<p>Being $\Omega$ the beam size: 
$\Omega = \frac{\pi \theta_{maj} ~~ \theta_{min}}{4 \ln{2}}$</p>

<p>Jy/beam $\cdot \ \Omega$ = Jy   ?</p>

CC BY-SA 4.0
2018-06-22T14:08:01.223Z
Please add a little more info - define J and Jy, and what your beam source is -- for example is it a point source?
Carl Witthoft
CC BY-SA 4.0
2018-06-22T19:37:41.390Z
See https://en.wikipedia.org/wiki/Jansky
Mike G
<radio-astronomy><units><interferometry>
Converting Jy/beam to Jy?
igreen21
# Converting Jy/beam to Jy?

(Asked by: username_0 on 2018-06-22)

Maybe its a dumb question, but to convert Jy/beam to Jy, I just have to multiply it by the beam size in sr right?

Being $\Omega$ the beam size:
$\Omega = \frac{\pi \theta\_{maj} ~~ \theta\_{min}}{4 \ln{2}}$

Jy/beam $\cdot \ \Omega$ = Jy ?

**Tags:** radio-astronomy, units, interferometry
> **Comments:**
> **username_1**: Please add a little more info - define J and Jy, and what your beam source is -- for example is it a point source?  
> **username_2**: See https://en.wikipedia.org/wiki/Jansky  

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## 3 answers

### Answer by username_3 on 2019-11-27. Score: 5

Actually to convert from Jy/beam to Jy/pixel you need to divide by the beam size.

Let's say you have a quantity of 1 Jy/beam, then

$\frac{Jy}{beam} \frac{beam}{\Omega}$, then to go from Jy/beam to Jy/pixel you would need to divide by $\Omega$.

The values of the beam major and minor axis must be in pixels.

Source: [NRAO](https://science.nrao.edu/facilities/vla/proposing/TBconv)


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### Answer by username_4 on 2018-06-22. Score: 3

So long as you accurately know the beam size, then yes multiplying your Jy/beam measurement (effectively flux density) by the beam size (effectively area of flux) will give you the total Jy (effectively the flux).

[See this source as an example.](http://tsrice-research.blogspot.com/2016/05/how-to-manually-extract-flux-in-jy-from.html)


---

### Answer by username_5 on 2020-09-01. Score: 2

It depends on what you mean by "just Jy". Usually, what is meant is the surface brightness of a source, in some unit like $\operatorname{Jy}\operatorname{sr}^{-1}$ or $\operatorname{Jy}\operatorname{arcsec}^{-2}$, integrated over solid angle to get the source's total flux. What a measurement of $I \operatorname{Jy}\operatorname{beam}^{-1}$ is telling you is, roughly, "An unresolved source that is the nominal beam size with this peak flux will have total flux $I$ in Jy." So, if you want surface brightness, take your quantity in Jy/beam and divide by $\Omega$/beam (see the relationship between $S$ [flux] and $I$ [surface brightness] in [this brightness temperature NRAO tutorial](https://science.nrao.edu/facilities/vla/proposing/TBconv).

In most situations with radio astronomy, the clean beam will be a Gaussian, so
\begin{align}
I &= I\_{\text{Jy/beam}} \frac{\text{beam}}{\Omega\_{\text{beam}}} \\
&= I\_{\text{Jy/beam}} \frac{1}{\pi \sigma\_{\text{beam}}^2} \\
&= I\_{\text{Jy/beam}} \frac{4\ln(2)}{\pi \theta\_{\text{beam}}^2},
\end{align}
with $\theta\_{\text{beam}}$ the beams full width at half power and $\sigma\_{\text{beam}}$ the standard deviation of the beam.

Once you have $I$, getting Jy / pixel is as easy as multiplying by $\Omega\_{\text{pixel}}$. If you have some surface brightness profile you fit, like
\begin{align}
I\_{\text{model}} &= A \exp\left(-\frac{(x-x\_0)^2\sigma\_y^2- 2\rho\sigma\_x\sigma\_y(x-x\_0)(y-y\_0) + (y-y\_0)^2}{2\sigma\_x^2\sigma\_y^2(1-\rho^2)}\right),
\end{align}
where $\sigma\_x$ and $\sigma\_y$ are normal standard deviations and $\rho$ is your normal correlation coefficient. For more normal astronomer usage, you'd use
\begin{align}
\theta\_M &= \sqrt{8\ln(2)\left[\frac{\sigma\_x^2 + \sigma\_y^2}{2} + \sqrt{\left(\frac{\sigma\_x^2 - \sigma\_y^2}{2}\right)^2 + \rho^2\sigma\_x^2\sigma\_y^2}\right]} \\
\theta\_m &=\sqrt{8\ln(2)\left[\frac{\sigma\_x^2 + \sigma\_y^2}{2} - \sqrt{\left(\frac{\sigma\_x^2 - \sigma\_y^2}{2}\right)^2 + \rho^2\sigma\_x^2\sigma\_y^2}\right]} \\
\phi &= \left\{\begin{array}{ll}
0 & \text{if }\rho=0,\ \sigma\_y > \sigma\_x \\
90^\circ & \text{if }\rho=0,\ \sigma\_x > \sigma\_y \\
\operatorname{atan2}\left(\theta\_M^2 - 8\ln(2)\sigma\_x^2, \sigma\_y^2\right) & \text{otherwise}
\end{array} \right.
\end{align}
Then you can convert the peak surface brightness $A$ to total brightness by integrating $I$ over all $x$ and $y$, yielding
\begin{align}
S &= A 2\pi \sigma\_x \sigma\_y \sqrt{1-\rho^2} \\
&= A \frac{\pi \theta\_m \theta\_M}{4\ln(2)}.
\end{align}

Note what happens if you combine the conversions from Jy/beam to surface brightness to Jy. You get:
\begin{align}
S &= A \frac{\theta\_m \theta\_M}{\theta\_{\mathrm{beam}}^2}.
\end{align}
C.f. Equation 35 of [Condon et al (1998)](https://ui.adsabs.harvard.edu/abs/1998AJ....115.1693C). Note that Condon et al provide multiple equations that depend on how resolved thee source is. Based on the [referenced paper](https://ui.adsabs.harvard.edu/abs/1997PASP..109..166C/abstract) it looks like what they're doing is minimizing the variance of their "corrected" values.


---

