# Converting Jy/beam to Jy?

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 ?

• Please add a little more info - define J and Jy, and what your beam source is -- for example is it a point source? Jun 22 '18 at 14:08
• Jun 22 '18 at 19:37

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

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.

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.

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). Note that Condon et al provide multiple equations that depend on how resolved thee source is. Based on the referenced paper it looks like what they're doing is minimizing the variance of their "corrected" values.