Astrophotography redux times two   Leave a comment

It started as a simple question: “What is the faintest star we can image with our equipment?” Figuring out the answer turned out to be delightfully complicated.

The basics are straightforward: since the apparent magnitude of a star is proportional to the amount of light striking the camera sensor, all we need to do is figure out how much light we need to generate barely-detectable signal.

Sky & Telescope magazine published a few articles in 1989 and 1994 under the title “What’s the faintest star you can see?”, and while much of that discussion is still valid, the question was posed pre-digital imaging, and so the results reflect the natural variability in human vision. It would seem that digital sensors, with their well-defined specifications, would easily provide an unambiguous way to answer that question.

Answering the question seems simple enough- begin by relating the apparent magnitudes of two stars to the ratio of “brightness” of the two stars: m1 – m2 = -2.5 log(b1/b2). So, if you have two stars (or other celestial objects), for example the noonday sun (m1 = -27) and full moon (m2 = -13), the difference in magnitudes tells us that the noonday sun is 400,000 times brighter than the full moon. Going further, the full moon is 3,000,000 times as bright as a magnitude 6 star (the typical limit of unaided human vision), which is 2.5 times as bright as a magnitude 7 star, etc. etc. But that doesn’t really tell us how much light is hitting the sensor, only an amount of light that is relative to some other amount.

Unfortunately, astronomers have their own words for optical things. What astronomers call ‘brightness’, physicists call ‘irradiance’. In radiometry (actually, photometry), ‘brightness’ means something completely different- it is a subjective judgment about photometric (eye-response weighted) qualities of luminous objects.

In any case, we have a way to make the ‘relative brightness’ scale into a real, measureable, quantity. What we do is ‘standardize’ the scale to a convenient standard irradiance, something that stays constant over repeated measurements, is easily replicated, and commonly available- for example, the irradiance of the noonday sun = 1 kW/m2.

So now we have our apparent magnitude scale (say: sun, moon, mag 6, mag 12, mag 16) = (-27, -13, 6, 12, 16) and corresponding relative brightness scale (sun: moon, moon: mag. 6, mag 6: mag 12, mag12: mag 16) that we make absolute by standardizing to the solar irradiance: 1 kW/m2, 3*10-3 W/m2. In table form:

magnitude rel. brightness irradiance [W/m2]
-26.74         1.00E+00           1.00E+03
-12.74         2.51E-06             2.51E-03
6                  8.02E-14             8.02E-11
12                3.19E-16              3.19E-13
16                8.02E-18             8.02E-15
21                8.02E-20            8.02E-17

By specifying the entrance pupil of our telephoto lens (maximum aperture = 140mm; diameter = 1.5 *10-2 m2), for any given star we can calculate how many watts of optical power is incident onto the sensor. But we have to be careful: not all the light emitted by the sun (or any luminous object) is detected by the sensor.

In addition to not being a perfect detector (the ‘efficiency’ or ‘responsivity’ of a detector is always < 1), not all the colors of light can be detected by the sensor: for example, the sun emits radio waves, but our camera sensor is not able to detect those. Of the 1 kW/m2 of light incident on the earth, how much of that light is in the visible region- or equivalently, within the spectral sensitivity of the sensor?

Stars are blackbodies, and the blackbody spectrum depends on temperature- different stars have different temperatures, and so appear differently colored. For ‘typical’ temperatures ranging from 5700K (our sun) to 8000K, the fraction of light in the visible waveband is about 40%. So visible sunlight (V-band light), accessible to the camera sensor, provides an irradiance of about 600 W/m2 on the earth’s surface.

So much for the sources, now the detector- how much light is needed to generate a detectable signal? There are several pieces to this answer.

First is the average ‘quantum efficiency’ of the sensor over the visible waveband. Manufacturers of scientific CCDs and CMOS imagers generally provide this information, but the sensor in our camera (Sony Exmor R) is a consumer product, and technical datasheets aren’t readily available. Basing the responsivity of the Exmor R based on other CMOS imagers on the market, we estimate responsivity as about 0.7 (70%) This means that on average, 1 absorbed photon will produce 0.7 e-. That’s pretty good- our cooled EMCCD camera costs 100 times as much and only has a slightly higher quantum efficiency- 0.9 (90%).

So if we know how many photons are hitting the sensor, we know how many electrons are being generated. And since we know (based on other CMOS sensors) that the Exmor has a ‘full well capacity’ of about 21000 e- and a dark current level of about 0.2 e-/s, if we know the number of photons incident during an exposure, we can calculate how may electrons accumulate in the well, compare that to the noise level and full-well capacity, and determine if we can detect light from the star. How many photons are incident onto the sensor?

We can calculate the incident optical power, in Watts. If we can convert Watts into photons/second, we can connect the magnitude of the star with the number of electrons generated during an exposure. Can we convert Watts into photons/second?

Yes, but we have to follow the rules of blackbody radiation- lots of different colors, lots of different photon energies, blah blah blah. Online calculators came in handy for this. Skipping a few steps, our table looks like this:

magnitude rel. brightness irradiance [W/m2] power incident on sensor [W] 6000K blackbody V-band phot/sec             e-/s
-26.74           1.00E+00         1.00E+03                 1.54E+01                                        6.83E+18                                 4.78E+18
-12.74            2.51E-06           2.51E-03                 3.87E-05                                          1.72E+13                                   1.20E+13
6                     8.02E-14          8.02E-11                  1.23E-12                                           5.48E+05                                  3.83E+05
12                   3.19E-16            3.19E-13                   4.91E-15                                          2.18E+03                                   1.53E+03
16                   8.02E-18           8.02E-15                1.23E-16                                            5.48E+01                                   3.83E+01
21                   8.02E-20          8.02E-17                   1.23E-18                                         5.48E-01                                     3.83E-01
18                    1.27E-18           1.27E-15                   1.96E-17                                          8.68E+00                                  6.08E+00

You may have noticed a slight ‘cheat’- we kept the temperature of the blackbody constant, when in fact different stars are at different temperatures. Fortunately, the narrow waveband of interest means the variability is small enough to get away with 1 or 2 digits of accuracy. As long as we keep that in mind, we can proceed. Now, let’s check our answers:

Moon: we have 1.2E+13 electrons generated every second, which would fill the well (saturated image) after 2 nanoseconds. This agrees poorly with experience- we typically expose for 1/250s. What’s wrong?

We didn’t account for the number of pixels over which the image is spread: the full moon covers 2.1*106 pixels, so we actually generate 5.8E+06 electrons per pixel per second- and the time to saturation is now 1/280 second, much better agreement!

Now, for the stars: a mag. 6 star (covering 20 pixels) saturates the pixels after 1 second, a mag. 12 star after 280 seconds, and a magnitude 16 star requires 10,000 seconds- about 2.5 hours- of observation to reach saturation.

But we don’t need to saturate the pixel in order to claim detection- we just need to be higher than the noise level. A star of magnitude 18.5 produces (roughly) 0.2 e-/s and thus the SNR =1. In practice, we never get to this magnitude limit due to the infinitely-long integration time required, but image stacking does get us closer: our roughly 1-hour image stacks start to reveal stars and deep sky objects as faint as 15 magnitude, according to the database SIMBAD. Calculations show that observing a mag. 15 object requires 4300 seconds of time to saturate the detector, in reasonable agreement with our images.

This is pretty good- our light-polluted night sky is roughly mag. 3!

Why did we even think to ask this question? Wide field astrophotography is becoming more common due to the ready supply of full-frame digital cameras and free/nearly free post-processing software. In fact, the (relatively) unskilled amateur (us, for example) can now generate images on par with many professional observatories, even though we live in a relatively light-polluted area.

What we have recently tried out is panoramic stitching of stacked images to generate an image with a relatively large field of view. This time of year, Cygnus is in a good observing position, so we have had direct views of the galactic plane in all it’s nebulous glory:


This image, and those that follow, are best viewed at full resolution (or even larger- 200% still looks great) on a large monitor to fully appreciate what nighttime at Atacama or Antarctica are probably like.

Some details about how we make these images- we acquire a few hundred images at each field of view, separately stack each field of view with Deep Sky Stacker and then fuse the images with Hugin. It may be surprising to know that each field of view must be slightly ‘tuned’ to compensate for the differences in viewing direction, as opposed to simply stacking all of the images at once and choosing “maximum field of view” in DSS.

Here are two subfields within Cygnus- the edge of the North American Nebula, and the Veil Nebula:




Posted July 31, 2014 by resnicklab in Physics, pic of the moment, Science

Tagged with

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: