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A lot of information has been published about shielding the JWST from solar radiation and the need to maintain the detectors at a low temperature, but little has been said about the other sources of heat loading on the detectors. In particular, what is the heat load from the collected radiation of the main mirrors on the detector, and how does that vary depending on what objects or fields are being imaged?

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    $\begingroup$ Cool question (pun intended)! I've just asked in Astronomy SE What does the celestial sphere look like in thermal IR? answers of which may be helpful in working towards an answer here. $\endgroup$
    – uhoh
    Feb 3, 2022 at 8:21
  • $\begingroup$ The large mirror that collects and concentrates light received from the viewing segment of the sky and sends it to the detectors does not see the sun; the big thermal energy source. The sun shield keeps that from happening. Dark sky radiation likely has a very small thermal radiation component that would be too low to be absorbed and heat up the detectors. I do not have a specific reference for the previous statement. $\endgroup$
    – tckosvic
    Feb 3, 2022 at 17:06

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Most of the heat on the detectors and telescope components is from the spacecraft itself, not from illumination from astrophysical sources. I updated the "heat from astronomical sources" part at the end to quantify this.


This image shows the thermal regions of JWST (there are prettier versions around but they don't have the temperatures printed on them):

enter image description here

This design keeps the cold side at a pretty constant temperature. Here are a few more details about parts of the question, but if others have more technical info then please edit:

Mirrors: The mirrors and other parts of the telescope will produce thermal radiation depending on their own temperature, which on the cold side should be near 40 K. The low temperature is important because it means low background radiation. Also important is the thermal stability. The environment is very stable at L2, which is important because if the temperature of the telescope part (including mirrors) changes, then the background changes, so all the science corrections have to change with it.

The mirrors are exposed to space, so they are getting constantly irradiated by astronomical objects, the cold side of the heat shield, etc. In terms of heating, they are getting irradiated by the whole sky no matter what direction the telescope is pointed. So the 40 K temperature is the balance between all the heat they are receiving, and all the heat they are radiating. The advantage is that they get to radiate back into space:

enter image description here

Detectors: These are kept in a chilled box, the ISIM. They don't see the sky, except from the beam of light coming from the telescope. The electronics are further separated from the detectors themselves, to keep the detectors at a very constant temperature. All of the actual science observations are differential, an A - B. That is the reason why having a stable thermal system is as important as having a cold one. It's actually more complicated than A - B, more like a series of measurements B, A1, A2, A3, etc., and a slope is fitted to the "ramp" of the individual measurements. So a stable background should cancel out in the science data.

Conduction: This wasn't mentioned specifically in the question, but with the sunshield in place, a big heat concern for the telescope and instruments is conduction. All of that equipment is still attached to the components on the hot side, so the engineers put a lot of thought into managing the heat conducted back and forth (the instruments on the cold side are connected to radiators on the hot side).


Heat from astronomical sources:

The question talks about thermal loads changing depending on the source being observed, and @uhoh linked to a celestial sphere question. But JWST (including mirrors) is an open-design telescope, so the whole sky (well, the anti-Sun half of the sky) continues to shine on it all the time, no matter where it is pointed. So there is not a whole lot of change depending on where the telescope is pointing. For the detectors, active cooling keeps them at a constant temperature, so that should compensate for differences in source brightness. More than actual heating, "persistence" can be a bigger issue from bright sources. It's like an afterimage on the detector, where it takes a while for the detector to forget the bright source and go back to baseline.

A 1983 paper by Mathis et al. has been called the "benchmark" of local interstellar radiation field (ISRF) measurements. Here is Figure 1 from their paper, where I clumsily highlighted the relevant curves:

enter image description here

Here is their description of the yellow curve:

The mean intensity of radiation at a typical point in the Galaxy can be usefully thought of as being divisible into two distinct wavelength ranges. The ultraviolet-visible-near infrared region of the spectrum ( 𝜆 ≤ 8 μm) is produced practically entirely from stars of various sorts. The middle- and far-infrared region ( 𝜆 ≥ 8 μm) is produced entirely from dust, either warm (in circumstellar shells) or cold.

The green curve is the CMB spectrum (labeled 2.9 K in this paper from 1983, but which we now take to be 2.7 K).

Mathis et al. give integrated intensities for the stellar and dust components, which I have converted here to µW m-2 and equilibrium temperatures using the Stefan-Boltzmann law:

Source Integrated intensity (µW m-2) Equilibrium temperature (K)
Stars 21.7 4.4
Galactic dust 4.9 3.1
Cosmic background 3.0 2.7
Combined 29.7 4.8

There is also some radiation coming from zodiacal dust, but I didn't look that up. It would still contribute less heating than the starlight.

For comparison, winter time temperatures in permanently-shadowed lunar craters can reach about 20 K (Williams et al. 2019).

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    $\begingroup$ Hmmmm, OK. I am working on a detailed replacement of the "heat from astronomical sources" section. $\endgroup$
    – giardia
    Feb 4, 2022 at 21:10
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    $\begingroup$ outstanding answer! $\endgroup$
    – uhoh
    Feb 4, 2022 at 23:44
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Sky is 2.7K on average. Well, only if the Sun is out of your averaging area.

The cosmic microwave background is the dominant visible energy in our universe. Stars, quasars and other sources are all based on baryon matter that is pretty minor amount in regard to the CMB photons.

This is true for both the whole sky as well as for any possible field of view a telescope shaded from the Sun may have.

So, in short: the main source of heat on the cold side of JWST is the hot side. Any field of view the telescope may turn to is way colder (on average) than the telescope itself.

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    $\begingroup$ I think this needs to be supported quantitatively to be credible. According to sources cited in another answer stars and dust are pretty substantial sources of infrared light. Power scales as $T^4$ so when pointing near the galactic center it appears the heat load due to those sources will be larger than the microwave CMB. It's also not clear that CMB microwaves will make it all the way to the detectors. All metal coatings would have to be thicker than a skin depth just for example. $\endgroup$
    – uhoh
    Feb 4, 2022 at 23:59
  • $\begingroup$ Because of some thermodynamic limitations of our universe, it is not really easy to make a telescope that doesn't propagate either CMB or some hotter radiation from the telescope's own gear, to the detector. And, because no part of the telescope is colder than the CMB, propagating CMB and not something hotter to the detector is a good thing. $\endgroup$
    – fraxinus
    Feb 5, 2022 at 9:01
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    $\begingroup$ Your other considerations need some more thinking, e.g. do we have ~10 square arcminutes of sky that are considerably optically thick at any wavelength (spare for Solar system objects that are hidden behind the heatshield anyway). $\endgroup$
    – fraxinus
    Feb 5, 2022 at 9:10

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