Cosmic rays and charge bleeding

Below is a small part of a image cropped and expanded from 1996's 09240300. It shows two artifacts worthy of mention.

One is charge bleeding. When a photon hits a CCD pixel, electronic charge is registered in that pixel. A single CCD pixel can hold only so much charge, much like a bucket can hold only so much water. When the bucket gets full, water spills out. When the pixel gets full, charge spills out. This spillage of charge is called "bleeding" because it tends to run down the columns (or up the columns depending on your perspective) like blood seeping from a cut. Sometimes the overloading of a pixel is so large that the bleeding goes both up and down. The bright star in the example image is bleeding charge upward along the rows of the CCD. The black-and-white streak is the bleed trail; the reason why it is both black and white (not just white) is that the electronics are overloaded and the data values become very unreliable in the bleed trail.

The second artifact is due to a cosmic ray interaction with the CCD. When a cosmic ray (a particle from outer space, such as a helium nulceus, for example) interacts with the CCD, it also creates charge in those pixels it passes through. The rate of cosmic ray events in a CCD on earth is approximately 1 per minute per square centimeter. Our CCD is approximately 0.3 square centimeters in area. So we expect a cosmic ray event to be detected approximately once very few minutes. The cosmic rays create charge in the CCD when they pass through, so we expect them to appear as random white spots (typically 1 or a few pixels only). But the cosmic ray event in this example is not white it is black; it is the black spot (or small diagonal streak) just below the two white stars to the left of the bright star. Why is it black?

To understand why the cosmic ray event could be black rather than white like stars, we have to know one important thing about how Stardial is taken:

Stardial data is taken by taking two images in succession. For the first image, the shutter is open so starlight can be detected. But for the second image, the shutter is kept closed, so we can measure the CCD and electronics response to other effects that are independent of the light falling on the CCD, which is of course what we want to know. The primary effect we wish to measure with the shutter closed is the response of the CCD to the jostling of atoms in the CCD itself. This jostling occurs in any substance that is warmer than absolute zero Kelvin, and all things are warmer than 0 K. So at the atomic level, the thermal jostling of the material (silicon) that makes up the CCD creates charge too, in addition to the charge created by light. To calibrate the amount of charge created by the thermal jostling, we take an image in the dark (with the shutter closed) and we take an image with the shutter open, and we subtract the two to eliminate the effect of the thermal jostling (which is equal in the two images and therefore cancels out in the subtraction).

But (!) if a cosmic ray hits the CCD during the exposure of the image made with the shutter closed (cosmic rays can pass right through the shutter), then the "dark image" has additional charge in those pixels hit by the cosmic ray, so the process of subtraction doesn't work properly for those particular pixels. A cosmic ray interacting with the CCD during the shutter-open image will create a white spot in the difference image (which is recorded and put on the Web). A comsic ray interacting with the CCD during the shutter-closed image will create a black spot in the difference image.