4
$\begingroup$

I know what a jansky is, but I don't understand what beam (in $\frac{1 Jy}{beam}$) is. If jansky is a unit of flux density, what kind of unit is $\frac{1 Jy}{beam}$?

$\endgroup$
2
  • $\begingroup$ Welcome to Astronomy Stack Exchange! Can you edit your question to include an example of such a map, along with the terminology you mention? That would make it much easier to answer. $\endgroup$
    – HDE 226868
    Commented Dec 14, 2016 at 18:37
  • $\begingroup$ When talking about fluxes here, there are two areas involved. One is the area through which the source is emitting and that is wrapped up in the Janksy. The beam is the area of the sky the radar is observing (more or less). So you're basically saying the energy per emitted area per frequency per measured area. I believe, that's the high-level answer, but my background is not in radio astronomy so perhaps someone with more technical knowledge can provide a better answer. $\endgroup$
    – zephyr
    Commented Dec 14, 2016 at 20:06

1 Answer 1

5
$\begingroup$

Jansky is defined as $1 Jy= 10^{-23}erg/s/cm^2/Hz$. So all the energy coming from a given solid angle per time, per frequency bin and per detector area.

This is convenient for sources of small angular extent (smaller than observing beam), e.g. for point sources, because the flux remains constant for varying beam sizes. For extended sources, the surface brightness is often described with units of Jy per solid angle. Radio contour maps, e.g. of interstellar clouds, usually show objects that are larger than the beam, so that the observed fluxes are a function of the beam size. To clean the beam influence from the date Jy/beam (beam area in steradian) is used.

So it is a surface brightness. For a flux of 1 Jy measured with a beam of 1 sr solid angle we find: $1 Jy/beam= 10^{-23} erg/s/cm^2/Hz/sr$.

Note, that 1 sr is already a huge portion of the sky. Beams are much smaller, on the order of arcsec to arcmin diameters.

$\endgroup$
6
  • 1
    $\begingroup$ I think the OP's confusion stems from not knowing what the "beam" is. Perhaps you could include an explanation of this concept (and its relation to field of view). Also, I think the last sentence might be a bit confusing, since what you say there is essentially "So it is a surface brightness Jy/sr", which is not really the case, except for a beam of exactly 1 sr. $\endgroup$
    – pela
    Commented Mar 14, 2017 at 11:13
  • $\begingroup$ @pela Correct me if I am mistaken, but if we dived a flux by a solid angle, we get a surface brightness, i.e. the mean flux per sr, If we observe a point source having a flux of 10Jy with a 1 sr beam it has a s.b. of 10 Jy/beam.(or Jy/sr) Observing with a 2 sr beam gives a s.b of 5 Jy/beam. Right? $\endgroup$ Commented Mar 14, 2017 at 11:45
  • 1
    $\begingroup$ Yes, exactly. That's what I mean. So SB can be expressed in Jy/sr or in Jy/beam (or Jy/arcsec² or whatever), but the two numbers will differ by a constant. Sorry, I also see now that my comment was a bit confusing, since "erg/s/cm²/Hz/sr" is not equal to "Jy/sr", but to "1e23 Jy/sr". Anyway, my point was that your last sentence might be interpreted as "1 Jy/beam is 1 erg/s/cm²/Hz/sr", which is not really correct. $\endgroup$
    – pela
    Commented Mar 14, 2017 at 12:36
  • $\begingroup$ @pela Thank you. I updated the text. Hope it is clearer now :) $\endgroup$ Commented Mar 14, 2017 at 13:22
  • $\begingroup$ Yes, excellent! $\endgroup$
    – pela
    Commented Mar 14, 2017 at 20:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .