When I look at pictures of the sun shield on the James Webb Space telescope (JWST), I see something that looks highly reflective (and hence must have a very low emissivity). My intuition tells me that this is no accident.

On the NASA website (https://webb.nasa.gov/content/about/innovations/coating.html) it is stated that the solar shield has a coating consisting of highly reflective (low emissivity) aluminum but also that the coating includes doped-silicon which has a high emissivity that "emits the most heat and light and acts to block the sun's heat from reaching the infrared instruments that will be located underneath it". My first issue is that if the emissivity of the solar shield is something that is needed to be optimized (that is, it is necessary for it to have either a very high value or a very low value) then why use a mixture of high and low emissivity materials? Surely this would lead to a middle-of-the-range emissivity?

My second issue is that when I apply the Stefan-Boltzmann law to determine the equilibrium temperature of the shield, I get a result that is independent of emissivity

$$P_{in} =P_{out}$$ $$\Rightarrow solarConstant * A*\epsilon =A*\epsilon*\sigma*T^4$$ $$\Rightarrow 1370 W/m^2 * 294m^2 = 294m^2* \sigma *T_{equilibrium}^4$$ $$T=121.26\deg C$$ Clearly the emissivity's, denoted $\epsilon$, in the above derivation cancel leading to an equilibrium temperature independent of emissive values. This leads me to conclude that the emissivity of the coating is completely inconsequential and that the solar shield could have been coated with a hypothetical black-body paint with perfect absorptivity and it would still function as well as it does now with aluminum. But if this is the case, then why does the NASA website even mention emissivity? Is it just an accident that the shield appears highly reflective but it doesn't actually need to be (since the final equilibrium temperature is the same regardless of the reflective/emissive properties of the shield)?


1 Answer 1


Emissivity is equal to absorptivity at a given wavelength. But the emissivity/absorptivity of any surface does vary with the wavelength of the light in question. In this case, the Webb telescope is absorbing light in the near infrared & visible range of the spectrum and emitting it (hopefully) in the far infrared. So the calculation should be $$ S A \epsilon_\text{vis} = A \epsilon_\text{far IR} \sigma T^4 $$ and since $\epsilon_\text{vis} \neq \epsilon_\text{far IR}$, the two factors don't cancel out.

To minimize the temperature $T$, we can see that we want $\epsilon_\text{vis}$ to be relatively low and $\epsilon_\text{far IR}$ to be relatively high. So when the above paragraph refers to "low-emissivity aluminum", it is presumably referring to the emissivity in the near IR & visible spectrum; when it refers to "high-emissivity doped silicon", it is presumably referring to the far infrared.

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    $\begingroup$ Yep. In spacecraft engineering, α generally refers to mean absorptivity of a ~6000 K blackbody spectrum, while ε refers to mean emissivity of a ~300 K blackbody spectrum. These can be very different. $\endgroup$
    – John Doty
    Commented Aug 18, 2022 at 14:17
  • $\begingroup$ @Michael Thanks for the excellent answer. Explains pretty much all my issues. Am I correct in assuming that this explanation also works for the simple and often-used earth-equilibrium temperature model which yields $P_{in}=P_{out} \rightarrow S\,\pi r_E^2\,(1-A)=4\pi r_E^2 \,\epsilon \,\sigma T_{eq}^4$ where A is the albedo of the earth. In this model we assume that (1-A) is the emissivity of the earth in the visible spectrum while $\epsilon$ is the emissivity in the infrared spectrum right? $\endgroup$ Commented Aug 19, 2022 at 7:46
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    $\begingroup$ @SalahTheGoat: More or less. I'd actually prefer to call $(1-A)$ the "absorptivity" in the visible to emphasize the role it plays in the equation. But that's equal to the emissivity, so it's more of a semantic distinction than a physical one. $\endgroup$ Commented Aug 19, 2022 at 12:01

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