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In electronics, one of the most significant sources of noise is heat, usually in the form of infrared radiation. Everything gives off infrared radiation, and the hotter something is, the more radiation it produces. This radiation is undoubtedly one component of the noise experienced by camera sensors.

At colder temperatures, objects give off less infrared radiation. Therefore, it would stand to reason that when a camera is used on the cold, say at -10F Farenheit, the sensor will be receiving less ambient infrared radiation from the camera internals than it receives when operating on a hot summer day at 85F. Is this effect noticeable in images?

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    \$\begingroup\$ Your camera may not be designed for such low temperatures. The manual for mine says that it should only be used in temperatures between 32F and 104F. \$\endgroup\$ Aug 16, 2017 at 22:43
  • \$\begingroup\$ @PeterTaylor The question is not about the environmental operating capabilities of particular cameras. The question is whether low temperatures reduce sensor noise noticeably. \$\endgroup\$ Aug 16, 2017 at 22:53

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You can set an exposure such as 1/30 of a second at f/11 and you can duplicate this exposure by using an aperture of f/16 at 1/15 of a second or 1/8 of a second at f/22. This practice is called reciprocity (my definition equal work for equal pay). The reciprocity equation is e=it (exposure = intensity of light multiplied by the allotted time of the exposure. The law of reciprocity is violated when exposure times range into minutes and hours. It will take considerably more exposure time for film base photography when super long exposures are utilized. We have learned to increase time more than the equation tells us, this is called reciprocity failure.

By accident, astronomers working at high altitudes on mountain tops in the cold of winter, discovered that films better adhere to the equation when the film is cold. This evolved into “cold cameras” whereby dry ice film holders cool the film and thus circumvent reciprocity failure. The same techniques are commonly in use when digital imaging replaced astronomical film. The digital camera does not suffer much from reciprocity failure but noise is greatly reduced and noise is especially devastating when high ISO's and long exposure times are employed.

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  • \$\begingroup\$ Okay, to distill these two paragraphs down a little bit, your answer is "yes"? \$\endgroup\$ Aug 16, 2017 at 20:17
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    \$\begingroup\$ @ Tyler Durden -- Pushing 80 I can use all the help I can get. \$\endgroup\$ Aug 16, 2017 at 20:20
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Most of the noise in a digital picture is due to thermal noise. This is due to the random motion of electrons at the operating temperature of the camera, this is called Johnson–Nyquist noise and this is proportional to the square root of the absolute temperature. This means that changing the operating temperature from 20 C to -20 C will reduce the noise by slightly less than 8%. So, you have to cool the camera to extremely low temperatures to eliminate the thermal noise. In practice one uses liquid nitrogen cooled sensors for applications where extremely low noise levels are needed, see here for an example of such a camera sensor.

Another source of noise is due to the finite number of photons that will lead to some finite number of electrons in a photo-site. This leads to two sources of noise, the number of photons per unit time that hits a photo-site emitted from a constant power light source will fluctuate. If on average N photons will hit the photo-site during the exposure, then the fluctuation will be of the order of sqrt(N), this means that the relative fluctuation scales as 1/sqrt(N). So, the lower the intensity of light, the more important this source of noise is.

Another effect leading to noise is due to the stochastic nature of the interaction of photons with electrons. The same number of photons hitting a photo-site will not always lead to the same number of electrons being moved into the photo-site. The net effect of the finite number of photons releasing a finite number of electrons then leads to a net amount of noise, which is called shot noise.

Shot noise is the dominant noise source in pictures taken by professional liquid nitrogen cooled cameras like those of the Hubble space telescope, it doesn't play a role in ordinary digital photography where it is always swamped by thermal noise.

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    \$\begingroup\$ I wouldn't say shot noise is "always swamped" by read noise. It all depends on the sensor design and the camera's imaging pipeline as well as the length of exposure and the intensity of the light source. photo.stackexchange.com/a/40192/15871 \$\endgroup\$
    – Michael C
    Aug 17, 2017 at 5:33
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    \$\begingroup\$ See also: Is Poisson Noise (“Shot Noise”) a significant source of noise for typical photography? \$\endgroup\$
    – Michael C
    Aug 17, 2017 at 5:35
  • \$\begingroup\$ @MichaelClark The shot noise you get in normal cameras at low light conditions is itself a result of thermal processes where small numbers of electrons are mobilized randomly. Shot noise that would truly derive from photon shot noise which is then not thermal, is always going to be swamped by thermal processes. If you take a picture of the night sky at very short exposure time, you see only shot noise, but with the lens cap on you see the same thing, obviously the latter has nothing to do with photons. Yes, it's Poisson noise, but it is also thermal. \$\endgroup\$ Aug 17, 2017 at 21:25
  • \$\begingroup\$ Or, put differently, any claim of having observed photon shot noise can be proved by showing a noise free dark frame picture. \$\endgroup\$ Aug 17, 2017 at 21:27
  • \$\begingroup\$ If you see it with a lens cap shot, it is not shot noise. When no light strikes the sensor it is all dark current noise. When you take a normal daytime exposure the amount of thermal noise will be much the same as the lens cap shot (assuming similar temperatures). The shot noise will be orders of magnitude greater. It just won't be as noticeable because the increase in signal will be the square of the the increase in shot noise. But that does not mean the shot noise isn't there. It's just overwhelmed by the actual signal. \$\endgroup\$
    – Michael C
    Aug 18, 2017 at 2:53

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