Comments below this answer suggest that there is a thin gold coating on the glass of spacesuit helmets that blocks thermal IR (say roughly about 10 to 30 microns) but is mostly transparent to visible light (about 0.4-0.7 microns)

How does this work exactly; what's the property of gold used by these helmet glass? Can it be best understood as:

  • related to the plasma frequency of the electrons in the special gold-based film being above that of thermal IR but below visible (it may be different than bulk gold)
  • due to discontinuities in the film of a size that's smaller than thermal IR wavelengths but larger than optical (like a wire mesh in the window of a microwave oven acting as electromagnetic shielding)
  • some combination of the two
  • something else entirely?

We know it's not simply skin effect because that goes the wrong way, a film of constant conductivity would be more transparent to the longer wavelengths, and blocking of the shorter wavelengths; the opposite direction of what is supposed to be happening here.

  • $\begingroup$ A spacesuit helmet should block UV light too. Very bad sunburns to the face would result in an EVA for some hours. $\endgroup$
    – Uwe
    Dec 3 '18 at 13:32
  • $\begingroup$ Might be a better question for physics.stackexchange? I imagine a proper explanation would have to invoke quantum mechanics. $\endgroup$ Dec 3 '18 at 14:06
  • 1
    $\begingroup$ There's some details on the visors in this pdf: core.ac.uk/download/pdf/81535913.pdf describing the deposition process and physical properties. This partially transparent film is about 500 A (0.000002 inches) in thickness. It is applied by thermal evaporation in vacuum to the visor, made of acrylic plastic $\endgroup$ Dec 3 '18 at 14:54
  • 1
    $\begingroup$ See this related question at physics stackexchange with a graph of the reflectance of aluminium, silver, copper and gold. $\endgroup$
    – Uwe
    Dec 3 '18 at 15:30
  • 1
    $\begingroup$ @uhoh no thanks, I don't understand the underlying mechanisms anywhere near well enough, just the basic properties. $\endgroup$ Dec 9 '18 at 12:58

If I Recall Correctly, The gold (and silver) was an early version of dichroic optical filters. Thin films of gold and silver, when applied in a specific order, create "thin-film interference" which essentially become an optical bandpass filter. The gold would create the hi-pass, and silver would create the low-pass (I may have it backwards.) Modern dichroics usually use some form of titanium oxide and chromium oxide since they are cheaper than gold and silver.)

more info on the wiki for dichroic filters, and dichroic glass.

  • 1
    $\begingroup$ This is a great lead, thanks for your post, and welcome to space! If you can add the link to the Wikipedia pages or even the subsections you find most relevant, that would be much better. Stack Exchange is a little different than other Q&A sites and adding verifying sources for information in answers is strongly encouraged. $\endgroup$
    – uhoh
    Dec 3 '18 at 23:23
  • $\begingroup$ I remember seeing something about hot mirrors and cold mirrors as well, but it was a long time ago. $\endgroup$
    – uhoh
    Dec 3 '18 at 23:24

@BlakeWalsh started to figure out the answer here and finished here but declined the invitation to post the answer, so I'll finish by posting it by proxy.

Why is gold "gold-colored"?

Several answers (1, 2, 3) to the Physics SE question Why are most metals gray/silver? explain this, and I'll summarize as follows:

In addition to the highly reflective property of the "sea" or "plasma" of conduction electrons of bulk or films of solid or liquid metals, including gold, gold also just happens to have very strong atomic absorption lines in the blue around 460 nm. If you could make gold gas and look at white light through it, the blue would be absorbed by individual gold atoms. This American Chemical Society communities post describes it nicely (clipped from a longer discussion):

If you Google for "NIST Atomic Spectra Database", click "Levels" and put in "Au I" (neutral gold atoms, Au II is for Au+ ions), you get a lot of spectral lines measured in cm-1. The ones you can see by eye are between violet = 25,000 cm-1 = 400 nm (just do 1/x on your calculator) and red = 14,000 cm-1 = 700 nm. You can see the line at 21,435 cm-1 = 466 nm = blue-violet, near the top. That wavelength of light has the right energy to move an electron from its lowest state, where there are 10 electrons in 5d orbitals and one in a 6s orbital (called 5d10 6s1) to a state with 5d9 6s2. If you were to shine visible white light through gold gas or vapor, everything except that blue-violet color would go through. The color that would pass through gold vapor would be reddish-yellow, much like solid gold.

enter image description here

Source click image for full size view

You can see at least a pair of strong transitions around 460 to 480 nm, and these would happen in bulk just as much as in a gas. Even though the skin depth of gold is quite shallow, atomic absorption is strong enough to absorb much of the blue light that the conduction electrons are working so diligently to reflect!

Why not other metals?

From this answer:

For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity). This shifts the light absorption (for gold primarily due to the 5d→6s transition) from the ultraviolet down to the lower frequency blue range. So gold tends to absorb blue light while it reflects the rest of the visible spectrum. This causes the yellowish hue we call 'golden'.

This answer also links to What Gives Gold that Mellow Glow?

enter image description here

Note that silver has a strong atomic absorption line as well, but it's in the UV were we can't notice it. This is one reason that aluminum is used instead of silver for telescopes that work in the UV as well as the visible, the other being that the native aluminum oxide that forms quickly on aluminum is not a problem compared to the thicker and darker oxides and sulphides that form on silver. (for more on that see When did “resilvering” large telescope mirrors actually refer to aluminization, and why was it necessary? as well as Do primary mirrors in large observatories undergo regular removal and re-coating of the aluminum? Why?)

How a modest difference in reflectivity can mean a huge difference in transmission

The answer has to do with comparing (1-x)-like terms, and with exponential attenuation.

Imagine I had two isotope samples, tenyearium with a half-life of 10 years, and oneyearium with a half-life of 1 years. If I wait 15 years, then most of each isotope sample is gone. 65% of tenyearium has decayed and 99+% of oneyearium has decayed. That's only a 35% difference in the decayed amount.

But this also means that while 35% of the tenyearium has remained only 0.003% of the oneyearium has remained. That means the oneyearium has been attenuated by a factor of over 11,000 more than the tenyearium.

Same with gold. While a think film of gold on an astronaut's visor may attenuate say 50% of the green and blue light, it may have attenuated 90 or 95% of the infrared light.

Differences in attenuation can be far bigger than differences in reflection. Thus the dip in gold's reflectivity probably understates the effectiveness as an attenuator when used in thin film transmission.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.