1
$\begingroup$

I'm reading some ALMA proposals and I am often seeing a conversion from Jy to Kelvin when, for example, people quote noise levels or source flux.

For example on their sensitivity calculator (https://almascience.nrao.edu/proposing/sensitivity-calculator) 1 mJy is equivalent to around 1 K when the angular resolution is 1". I can tell it somehow depends on the resolution because it won't give me a conversion unless I enter a resolution.

What's happening under the hood here? And why does their documentation treat sensitivities entered in Kelvin differently from when they're entered in Jy? Is this an interferometry thing?

Thanks

$\endgroup$
6
$\begingroup$

This is, indeed, a result of how we measure things in radio astronomy. It's not just interferometry, but radio astronomy in general. The thing they're referring to is a concept called "brightness temperature". In the low frequency limit (valid for radio astronomy), we can use the Rayleigh-Jeans approximation, which gives us the expression $$ T_B = \frac{S_\nu \lambda^2}{2k \theta_s^2} $$ where $S_\nu$ is the flux density, $\lambda$ is the wavelength, $k$ is Boltzmann's constant, and $\theta_s$ is the size of the source (or the resolution in this case). As long as your source and wavelength are constant, brightness temperature and flux density can be interchanged.

The reason astronomers talk about temperatures instead of fluxes sometimes is twofold. First, a large number of objects examined by radio astronomers are not blackbodies. Therefore we cannot calculate their temperatures like we normally would. Instead, we calculate what their temperature would be if they were blackbodies. This at least gives us a value to compare them to other things.

Secondly, when discussing radio telescopes and antennae, we usually say that various parts of the antenna contribute different 'temperatures'. We can quantify system temperatures to help us understand noise.

$\endgroup$
1
  • 1
    $\begingroup$ Suggestion for improvement: Say what you're actually approximating with the formula given. As it stands now, a person without background in radio astronomy doesn't know what you're talking about. $\endgroup$ Apr 12 '17 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.