# Beamsize in radio/microwave observations

In radio/microwave observations, I understand the beamsize is the response of the telescope to a point source - effectively, the telescope's resolution.

Now, the Planck satellite (par exemple) combines different frequency channels in a variance-weighted map. These different channels in general will have different sensitivities, and different beamsizes.

Suppose you coadd two (or any number really) of these channels. What will be the effective resolution of the resulting map? For instance, suppose we take channel 1 with a 5 arcminute beamsize (FWHM), and channel 2 with a 12 arcminute beamsize (and let's suppose they have the same sensitivity). What will be the resolution of the inverse-variance weighted map? How will this change if the sensitivities differ?

## 1 Answer

This is a fundamental problem in modern multi-color/frequency astronomy. You can not simply add two spectrally significant different channels because they contain information about different solid angles on the sky.

If you have two maps with different inherent beam sizes one common approach is to resample the map with the higher spatial resolution to the lower resolution of the other map by convolving with a Gaussian beam of size:

$\theta_{Gauss-kernel}=\sqrt{\theta_{low}^2-\theta_{high}^2}$

For example, if you have two maps with resolution (FWHM) 5" and 15" you need to smooth the 5" map with a Guassian kernel of size

$\theta_{Gauss-kernel}=\sqrt{15^2-5^2}=14.14"$

to obtain a comparable map.

Of course the beams are never true Gaussians and it depends whether you observe point sources or extended (w.r.t. the beam) source.

As an example you can take a look at the SPIRE Observers manual. SPIRE was an instrument onboard the Herschel Space Observatory. They discuss the involved problems in somewhat more detail.

• Yes, you cannot get a flux ratio (or colour) at a higher resolution than the lowest resolution channel. – ProfRob Oct 29 '15 at 13:50