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I’ve read that only cold-blooded animals develop IR vision to spot warm prey because a mammal would blind himself with his own heat and not be able to see anything.

But I’ve seen a FLIR camera, operating at room temperature, able to image warm and cool spots on walls, to find places where insullation is lacking. Clearly it’s not blinded by having its sensor, lenses, and housing all at room temperature, and is able to image cooler patches without any problem.

How does that work? Why is the popular explaination wrong?

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    $\begingroup$ Sensing heat is not the same thing as IR vision. Vision works near the quantum limit. Humans, for instance, can detect as few as maybe ten photons, some animals are claimed to do better than that by a factor of a few. At that photon count thermal noise is limiting. If we don't have to work at that limit we can average the thermal background and then the SNR can always be large enough to detect a signal. A heat sensing animal will probably be able to detect prey from a distance of a few meters, whereas vision, at the same source brightness, would allow it to see hundreds of times farther. $\endgroup$ – CuriousOne Jul 25 '16 at 2:41
  • $\begingroup$ Sensing heat via IR as with a pit viper; not, say, noticing the source of warmth through convection of the air so warmed. $\endgroup$ – JDługosz Jul 25 '16 at 5:38
  • $\begingroup$ Yes, that's a lot less sensitive than vision, which is, as you correctly observe, because the heat background is limiting the achievable SNR. One can probably compare our IR motion sensors with the sensory performance of animals that have these senses and it might turn out to be of similar performance. What I don't know is if that's near the physical limit, already, but it might be close. $\endgroup$ – CuriousOne Jul 25 '16 at 6:08
  • $\begingroup$ Just as a point of reference, the minimum temperature that our FLIR camera can measure is -40C, so ~60C below the sensor temperature. $\endgroup$ – Chris H Jul 25 '16 at 13:11
  • $\begingroup$ Remember that cold-blooded animals are also at room temperature, or actually rather above it because their bodies do produce heat. They just don't regulate their temperatures to a constant level, and they do operate cooler than warm-blooded creatures. So they are actually in a similar position to the FLIR camera in terms of internal heat, and perhaps worse off. Still a good question, though. $\endgroup$ – Mark Foskey Mar 21 at 17:47
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By comparing the signal to the background.

Suppose you get 10 IR photons from the camera and lens background but an extra 5 from the source then you can still detect the source.

There is a whole science of signal processing to detect signals much fainter than the background. Especially in IR astronomy.

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    $\begingroup$ That said, cameras used in IR astronomy are often kept quite cold. And when space-based IR telescopes run out of coolant, that generally signals the end of the mission. $\endgroup$ – user10851 Jul 25 '16 at 7:07
  • $\begingroup$ @ChrisWhite why not just use passive radiators like any decent KSP space station would? $\endgroup$ – John Dvorak Jul 25 '16 at 13:20
  • $\begingroup$ @JanDvorak It all depends on the particular needs of the mission. Sometimes passive radiators just can't cut it. I would not be surprised if there were cases where passive radiators were considered, but sufficient passive cooling to meet signal-to-noise requirements would have been too heavy. (And they also have limits as to how cold they can get, which may matter to some programs) $\endgroup$ – Cort Ammon Jul 25 '16 at 14:25
  • $\begingroup$ Some Earth-observing infrared instruments are not cooled at all, and most of the signal is self-emission. Calibration takes care of subtracting that. $\endgroup$ – gerrit Jul 25 '16 at 17:07
  • $\begingroup$ @Jan JWST will be mostly passive, but this depends on a highly complex unfolding Sun shield. This only gets it to 40 K, good enough for near-IR. The mid-IR instrument relies on active cooling down to 6 K -- better than bringing a supply of liquid helium, but very costly with power. It won't do far-IR science, because that requires colder temperatures. $\endgroup$ – user10851 Jul 25 '16 at 17:09
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In astronomy, the background from the camera itself is called "dark current" and is removed by first taking an exposure with the shutter closed for, say, half an hour, and then subtract those counts from the real observations, normalized to the exposure time of a given image.

Sometimes, if you're bored at the telescope due to bad weather, you can even take multiple dark exposures and add them, just to get better statistics.

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  • $\begingroup$ Maybe this should've been a comment… $\endgroup$ – pela Jul 25 '16 at 6:37
  • $\begingroup$ @pela "Dark Current" is generally used for the signal generated thermally in the detector. There is an additional thermal background from the instrument, optics, telescope and atmosphere. They are normally treated separately because their noise characterstics/statistics are different. $\endgroup$ – Martin Beckett Jul 25 '16 at 18:18
  • $\begingroup$ @MartinBeckett: It's been some years since I observed at telescopes myself, and I see you have an optics background, so I'm not going to argue with you, but as I remember my data reduction, we subtract dark, bias (small number added to each pixel to keep counts positive), and read-out noise. That is, we don't treat those backgrounds you mention separately, I think. Maybe it's because it's dominated by the detector noise? $\endgroup$ – pela Jul 25 '16 at 21:11
  • $\begingroup$ @pela it depends on the wavelength/detector technology. You subtract a dark frame to take out detector thermal background (classical Dark Current) and then perhaps chop on the sky, or a cold load to handle atmospheric and telescope thermal background. My background is in near IR where the detector dark current dominates so it's a lot easier $\endgroup$ – Martin Beckett Jul 26 '16 at 3:36
  • $\begingroup$ Okay, maybe the technology at the telescopes I used is more simple, since they're from the 70's and 80's (e.g. the 1.5m at La Silla and the 2.5m at La Palma). We simply set the final image equal to (raw – [bias+dark+read-out]) / (flat-field), where the flat-field accounts for the different sensitivity of the pixel. $\endgroup$ – pela Jul 26 '16 at 7:42
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I'm not sure whether it's best to resurrect this old question, or pose a new one, but here goes:

Is it not the case that the sensor emits IR toward the scene, just as the scene emits IR toward the sensor?

If the sensor is viewing a scene at a lower temperature, that cools the sensor. The sensor emits more energy toward the scene than the scene emits toward the sensor.

The camera body and lens have low emissivity, so they don't radiate enough thermal energy to drown out this effect. The sensor has high emissivity -- it must, because it needs to absorb incoming IR with high efficiency, instead of reflecting it away.

This same reasoning works if you consider an individual sensor pixel, and the corresponding small patch of the scene. That's how a thermal camera can image objects colder than itself.

This is also how the ground cools below air temperature on a clear night. The sky looks very cold, and as the ground radiates heat toward it, the sky doesn't radiate nearly as much heat back. The result: the ground can get cold enough to accumulate frost, even when the air temperature is above freezing, all without breaking a single law of thermodynamics!

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