As we know, from a surprising corollary to Kirchhoff's law of thermal radiation, just as darker* objects absorb more light (and therefore energy), darker objects also radiate more light (at lower temperatures, in the infrared) and therefore energy.

Observe the following photograph of the ISS:

International Space Station photographed by an STS-134 crew member of the Space Shuttle Endeavour on May 29, 2011

International Space Station photographed by an STS-134 crew member of the Space Shuttle Endeavour on May 29, 2011. Image Credit: NASA

I count ten radiators, all of them light-colored or metallic. The same sort of material appears on the Space Shuttle's radiators, which are on the inside of the payload doors:

The Space Shuttle Endeavour approaching the ISS during STS-118 on August 10, 2007

The Space Shuttle Endeavour approaching the ISS during STS-118 on August 10, 2007. Image Credit: NASA

What this says to me is that these radiators are inefficient.

From research for this question, I found that the consensus of the internet for terrestrial radiators (e.g. for heating) is that a dark coating isn't worth it: just use a larger radiator and, in the presence of an atmosphere, most heat transfer is by convection anyway.

However, in space, efficiency is paramount. Someone needs to check up on all those panels, fix them up, repair damage when possible, &c.--to say nothing of lugging all those precious grams into orbit in the first place. If you could, say, put a carbon nanotube layer on them and use one fewer, wouldn't you?

So my question is this: why, with the particular examples of the ISS and Space Shuttle, aren't radiators jet black?

* Note: darker here is not necessarily in the visible range. If these panels have low albedo in the nonvisible range, for example, then that might be better than some blacker material with a higher albedo in the nonvisible range. This would certainly explain these radiators.

  • $\begingroup$ Incidental comment: terrestrial heating radiators work on atmospheric convection of course, so that partly explains why radiative transfer is less important for them. $\endgroup$
    – Andy
    Apr 22, 2015 at 8:41
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    $\begingroup$ This is just a guess, I suspect that reflecting photons back into space is more efficient overall than absorbing the heat and then radiating it out again, therefore shiny is better. $\endgroup$
    – GdD
    Apr 22, 2015 at 10:14
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    $\begingroup$ The radiators should be black especially at IR frequencies; not necessarily on visible freqs. Also, special radiator coatings allow little solar absorptivity (mainly UV freqs) and good IR emissivity (I don't have a reference at hand now). $\endgroup$
    – Juancho
    Apr 22, 2015 at 11:29
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    $\begingroup$ Relevant: space.stackexchange.com/questions/5246/… $\endgroup$ Apr 22, 2015 at 14:49
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    $\begingroup$ @GdD: Not really; the average temperature of space is low enough (a few kelvins) that simply radiating efficiently is optimal. There are a few point sources of high-temperature radiation, but once you avoid the Sun and maybe Jupiter those diminish to insignificance against the expanse of nothing-in-particular. $\endgroup$ Apr 22, 2015 at 17:23

1 Answer 1


Kirchhoff's law is only valid for objects in radiative equilibrium. The emissivity and absorptivity of a material are the same for a given wavelength, but can vary dramatically for different wavelengths.

The radiators on a spacecraft are not in radiative equilibrium, since they lose heat to radiation. They emit heat in the longwave infrared spectrum, but receive heat energy (from the Sun) in the shortwave infrared, visible, and UV parts of the spectrum. (They do also receive longwave IR from Earth, but the amount is only around 250 W/m2 vs 1300 W/m2 for sunlight.) This means that the effective emissivity and absorptivity can differ while the radiators are in operation.

The radiators on the ISS are a high-emissivity white paint, meaning that they are dark in the infrared spectrum where the heat is emitted. They are white in the visible spectrum to reflect sunlight.

The radiators on the shuttle are have a two-layer coating: a silver reflective layer covered by a thin Teflon film. The Teflon layer is opaque to infrared light, so the high emissivity of Teflon dominates. Visible light passes through the Teflon layer and is reflected by the silver layer, so the solar absorbance is low.

The radiators on the Shuttle are exposed to more direct sunlight (the radiators on the ISS pivot so they are typically close to edge-on to the Sun), which is why they use the higher-performance but more expensive dual-layer design.

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    $\begingroup$ Thanks, but this is just restating my exact point of confusion. "Light colors are good emitters of radiation." is patently false. See e.g. this: "the emissivity is equal to the absorptivity". In your link, the inset figure on the right illustrates my point: the black side of the Leslie cube is radiating much more thermal radiation than the light side. In space, you can orient the radiators parallel to the solar radiation, so I suspect absorption is insignificant. $\endgroup$
    – imallett
    Apr 22, 2015 at 20:51
  • $\begingroup$ @imallett Kirchoff's law only holds for a body in equillibrium; the radiator is not in equilibrium with its environment since it's emitting heat. The white and black sides of the Leslie's cube appear practically the same under infrared light, where the energy is being emitted (the article says that paint, "including white" has an emissivity of around 0.9). Where solar energy is being absorbed (in the visible spectrum) we want to reflect as much incoming energy as possible, hence the white color. As for keeping edge-on to the Sun, what happens if the drive motors fail? $\endgroup$ Apr 22, 2015 at 23:47
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    $\begingroup$ White materials are not necessarily good emitters of visible light radiation. But since these materials are emitting in IR and not visible, it's not relevant. $\endgroup$
    – BowlOfRed
    Apr 23, 2015 at 21:26
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    $\begingroup$ @imallett regardless of what your and my intuition says, the white surface on the right side of the picture is just as IR-bright as the black surface on the left side $\endgroup$ Apr 23, 2015 at 21:56
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    $\begingroup$ @imallett You should probably edit your question to reflect the fact (no pun intended) that Kirchoff's law doesn't apply here: in fact the shiny radiator is the most efficient! $\endgroup$ Apr 24, 2015 at 23:16

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