5
$\begingroup$

I have found two different numbers for the area of the sky covered by the Hubble Ultra Deep Field (HUDF). According to this, the image is roughly 2.4 arcminutes wide. The image is also attached to the link with a scale showing it is around 2.4 arcminutes by 2.4 arcminutes.

I might be wrong, but I think this means that the area covered is 2.4 $\times$ 2.4 = 5.76 squared arcminutes.

But according to this, the image covers 11 square minute-of-arc region in the southern sky.

I actually doubt the method I used to calculate the area of 5.76 squared arcminutes. So am I wrong ? and if not, then which is the correct number ?

$\endgroup$
  • $\begingroup$ It may be useful to remember that the angular diameter of the moon is approx. 30 minutes of arc, i.e. more than twelve times the size of the Hubble Ultra Deep Field. Put in another way, the angle of this patch is 2.4/360*60=0.00011 of the full circle. It is a rather small area of the visible universe. $\endgroup$ – CuriousOne Dec 16 '14 at 3:36
4
$\begingroup$

Your method is correct. When the angle is "small" so that we can ignore curvature, then the rectangular solid angle is just the product of the two side angles. (This doesn't hold for "large" angles).

I cannot find a perfect explanation, but one source of this confusion may be that the UDF was imaged by two separate instruments, one optical, one infrared.

This page shows some details, including the fact that the optical survey is 3 arcminutes square and the IR is 2.4 arcminutes square. So I think the entire UDF is 11 square arcminutes (3.4 arcminutes square), but the image released does not cover the entire area.

An additional piece I found that supports this size is that the Hubble ACS wide-field channel (found here) is 202 arcseconds on a side, and this page (which has an image showing the relative size of the moon, the HUDF and some other Hubble surveys) states the HUDF is exactly the size of one wide-field image. 202 arcseconds squared is 11.3 square arcminutes.

$\endgroup$
  • $\begingroup$ So if for example I want to calculate how many HUDF images are needed to cover the entire sky, will it be correct if I divide the area of the sky which is 148,510,660 square arcminutes by 11 square arcminutes ? Or I will be ignoring the curvature by doing that ? $\endgroup$ – Abanob Ebrahim Dec 16 '14 at 1:42
  • $\begingroup$ That's correct. What you can't do is assume the sky is 180 degree * 360 degree = 233 million square arcminutes. $\endgroup$ – BowlOfRed Dec 16 '14 at 3:51
  • $\begingroup$ I should divide by Pi if I want to correct your number, right ? $\endgroup$ – Abanob Ebrahim Dec 16 '14 at 9:53
  • $\begingroup$ No. The actual error is about 1.5x. A sphere has $4 pi$ square radians of surface area. That's a bit more than 41000 square degrees. If we just assume 180 degrees of latitude and 360 degrees of longitude and mulitplied, we'd naively get almost 65000 square degrees. $\endgroup$ – BowlOfRed Dec 16 '14 at 10:05
  • $\begingroup$ For some reason I used 360*360 instead of 180*360. So that explains why we are off by a factor of two in error. Thank you. $\endgroup$ – Abanob Ebrahim Dec 16 '14 at 10:39
1
$\begingroup$

Its complicated. There wasn't just one Hubble Ultra Deep Field (UDF) observation, but several, taken at different times, with different instruments, but pointed in (almost) the same direction.

The first image you refer to was taken with the WFC Infra red camera in 2009, in near infrared bands (1-1.6 microns). The area covered by this camera is $2.4$ arminutes on a side. i.e. 5.76 square arcminutes.

The second Beckwith et al. (2005) paper you refer to, discusses a UDF taken with the Advanced Camera for Surveys (ACS) in 2004, an instrument that works in the visible part of the spectrum, edging just into the near infrared (300-900nm). The image is approximately square because it had to be taken using different roll angles of the spacecraft and the image was dithered. The final (cropped) square image is 11 square arcminutes.

There is then a 2012 Hubble UDF taken again with the WFC IR instrument, in the same direction, with increased exposure times, especially in the 1.05 micron band, and added a 1.4 micron deep image. This also covers 5.76 square arcminutes. It also added further exposure (another 128 orbits!) to the 800nm optical/near IR exposure with the ACS instrument (11 square arcminutes). http://adsabs.harvard.edu/abs/2013ApJS..209....3K

EDIT: The number of square degrees on the sky is $4\pi (180/\pi)^2 =$ 41253 square degrees, which gives your number.

$\endgroup$

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.