Galaxy Zoo Talk

Green AGN

  • Ghost_Sheep_SWR by Ghost_Sheep_SWR

    This edge-on galaxy looks very green. All bands in DECaLS DR2 have been imaged at least 3 times, and it is also visible in SDSS.

    The galaxy is classified as starforming and narrow-line AGN (NLAGN)

    Is the green halo caused by the AGN somehow?

    http://skyserver.sdss.org/dr12/en/tools/explore/Summary.aspx?id=1237655468061753517

    http://ned.ipac.caltech.edu/cgi-bin/objsearch?search_type=Obj_id&objid=6631201&objname=1&img_stamp=YES&hconst=73.0&omegam=0.27&omegav=0.73&corr_z=1

    Posted

  • JeanTate by JeanTate in response to Ghost_Sheep_SWR's comment.

    Hmm, from the looks of it in SDSS:

    enter image description here

    green on 'the top'; red 'underneath', and a Dec of only 1°, I'd check whether it's differential refraction, and/or bad seeing. For SDSS, PhotoObj gives an airmass of 1.25, which doesn't seem too bad. I can't quickly find out what the estimated seeing was ...

    Any idea on how to find the airmass and seeing for a DECaLS image?

    Hope this helps, and happy hunting! 😃

    Posted

  • Ghost_Sheep_SWR by Ghost_Sheep_SWR

    Any idea on how to find the airmass and seeing for a DECaLS image?

    Absolutely not! 😃 but i think your explanation already is sufficient enough i guess. Would be necessary for both telescopes though to see it in SDSS and DECaLS (roughly same location)

    • Apache Point-Sloan Digital Sky Survey
      254°10'45.9''E, 32°46'49.8''N, 2791.2m
    • Cerro Tololo-DECam 289°11'36.9''E, 30°10'10.5''S, 2202.7 m

    Well seems to be the case,

    Thanks!

    AND seeing same greenish colors on nearby galaxies too, which i ofcourse should have checked first..... 8)

    Posted

  • JeanTate by JeanTate in response to Ghost_Sheep_SWR's comment.

    Um, Cerro Tololo is located in Chile, so the latitude is -30°, making it rather a long way from Apache Point! 😮

    Shouldn't be hard to work out what the airmass is, if this galaxy were observed as it crossed the meridian, right? I think SDSS, in drift-scan mode, tried to make its observations close to the meridian, but I have no idea about DECaLS.

    Posted

  • Ghost_Sheep_SWR by Ghost_Sheep_SWR

    !haha whoops. Now i see the ssd uses N (+30) for north and S (-30) for south, missed that bit.

    http://ssd.jpl.nasa.gov/sbfind.cgi

    I have no clue to working out airmasses at a certain time at certain locations looking at certain points, being a citizen scientist with sadly not the right papers, and still trying to catch up on other stuff.

    So for now i feel comfortable knowing it's likely the viewpoint + airmass / viewing conditions. + other galaxies nearby in DECaLS exhibit the same greenish shade, adding to the evidence this is the correct answer.

    On a side note i think i've read somewhere DECaLS is aiming at 3 imaging moments, from which at least one is with good viewing conditions. So a certain area having all bands imaged three times should contain at least one time in good viewing conditions

    Thanks for now, and if you have a short-hand explanation on how to retrieve the airmass that would be very informative, i'll likely dive into it on a later time

    Posted

  • JeanTate by JeanTate in response to Ghost_Sheep_SWR's comment.

    Airmass (WP) is a fairly straight-forward concept, and the approximation sec(altitude) good enough for anything we're likely to use.

    The altitude of an object - angular distance above horizon - on the meridian is also easy to calculate, if you know the telescope's latitude and the object's Dec.

    Try this: the celestial pole is always at the same position in the sky (to a good approximation). If you're on the equator, both celestial poles are on the horizon (altitude 0°); if at either (geographic) pole, straight up (altitude 90°); at 45° latitude, 45° altitude. And so on.

    When it's on the meridian, what will be the altitude of an object with Dec = +1°, viewed from two locations, one with latitude +30°, the other -30°?

    Posted

  • Ghost_Sheep_SWR by Ghost_Sheep_SWR

    When i'm on the equator looking at the meridian it is straight up , altitude 90°

    When at latitude 45° looking at meridian its altitude 45°

    When at latitude +30° looking at meridian I'm guessing 60°

    DEC +1° viewing from latitude +30° would then be 61°, and from latitude -30° Altitude 59° (?)

    Not taking no doubt many factors like the hight of the telescopes (+2.7km / +2.2km above sea-level) etc.

    Thinking about it, it says nothing without the Right Ascension due to the rotation of the earth? The sun would get two times maximum and one time minimum airmass a day (standing on the equator)

    Posted