A metal wire mesh can be nearly as reflective to an incident EM wave as a solid sheet of metal provided the wires are sufficiently conductive and the size of the mesh is much smaller than the incident wavelength.

Question: Suppose that for some reason one wants to make a solar sail reflector out of a mesh of extremely thin gold wires1. What physics dictates the minimum thickness of the wires to produce say at least half of the reflectivity of a uniform gold surface? Presumably the resistance of the wires is important (it is for RF reflectors) but 1) is the bulk conductivity for gold at say 500 THz (600 nm) a well-defined thing? Will skin effects dominate?

For specifics, assume a square grid, a wavelength of 600 nm and a grid spacing of 150 nm ($\lambda / 4$).

This question was inspired by the Space SE question How realistic is the 1 kg/km² solar sail in “Death's End”?

What is the relationship between Faraday cage mesh size and attenuation of cell phone reception signals? and it's answer don't really address this problem in more than a hand-waving way. Here I've asked specifically about the nature and minimum diameter of the wires and the physical processes that limit it.

1 and acknowledges that gold's optical properties don't allow it to reflect the shorter wavelengths, thus its color, see this answer to How does the thin gold film in the glass of spacesuit helmets block thermal IR but transmit visible? What's the property?.


A metal is reflective because it is conductive. Conductivity will limit reflectivity as it gets thin and narrow.

There are considerations beyond this of course if you want to make a solar sail. A metal wire mesh as thin as possible would be weak and fragile. Strength matters. Manufacturability of giant a sail matters. Ability to survive Earth's atmosphere before launch, launch, and deployment matter. Radiation damage matters. But I will focus on things that affect reflectivity.

Choice of material matters. One problem with gold is its reflectivity across the spectrum. You want it most reflective at wavelengths the Sun emits more strongly. This is the spectrum.

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This is the reflectivity.

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Gold is less reflective (less conductive) than silver or aluminum across an important part of the spectrum.

Reflection happens at the surface of a metal. Radiation makes conduction electrons vibrate. Vibrating electrons radiate. This is is the reflected light. The current decays exponentially with depth into the metal. The decay constant is depends on conductivity.

$$J = J_s e^{-(1+j)d/ \sigma}$$

where $\sigma$ is the skin depth

$$\sigma = \sqrt{\frac{2 \rho}{\omega \mu}}$$

If the layer of gold is thinner than a few $\sigma$, light will be transmitted though it. For gold, this is fairly thick. As the link in the post says, a $500$ angstrom layer of gold on astronaut's faceplates reflected essentially all of the infrared, but about half the visible. This is what you want for a faceplate more than a solar sail.

On the other hand, your link How realistic is the 1 kg/km² solar sail in “Death's End”? says that a 5 nm Al "foil" is optimal. At that thickness, Al is $50$% reflective. Then it suggests a grid of $5$ nm wires would also be effective.

The link is interested in getting the weight down as much as reflectivity up. It was not mentioned, but perhaps a grid of say $10$ nm wires would be closer to $100$% reflective, and not that heavy. A wire grid would be strengthened with other stronger wires.

Another consideration is that a metal is reflective because electrons spread out through the metal. They can easily move, which makes them conductive. Classically, the oscillating electric field in light vibrates the electrons, and vibrating electrons radiate light. It works out that the incoming light is absorbed and the radiated light is the reflection.

Electrons can move because neighboring atoms don't have individual separate orbitals, each with the same energy. Instead, they overlap and forming bands, states with many closely spaced energy levels.

Gold is such a solid. Gold is more conductive at lower frequencies. Red and infrared light is reflected well, but blue light is absorbed. So gold is gold colored.

For this to work, there have to be enough metal atoms to make a solid with a band structure. The solid must be big enough for the electrons to vibrate and radiate.

The thinnest sputtered gold films are typically $10$ nm thick. Below that, gold forms islands rather than a film. This paper gets it down to $5.4$ nm and seems to say that the optical properties are comparable to what you would expect in bulk gold. There is an effect from the substrate, but in a solar sail, people are talking about (perhaps wishing for) removable substrates. Still, it might be difficult to make gold wires this thin.

Given that a $54$ angstrom film behaves like bulk, skin depth is the limiting factor in a film.

It is hard to make wires as thin and narrow as a film. Sample records for thin gold wires has some links. One, Gold Wire-networks: Particle Array Guided Evaporation Lithograpy, is an abstract that mentions hopes to deposit wires in the micron to submicron range. I could not find the paper itself.

Wires must run in multiple directions. Light polarized parallel to the wires can excite electrons to vibrate along the length of the wire, and is reflected well. Light polarized perpendicular to the wires cannot, and is not reflected.

Wires must be spaced less than a wavelength of light apart or light will diffract between them.

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    $\begingroup$ I was mostly addressing the "how small" in your title. Some of the physics is addressed in the links you provided. A film must be conductive to be reflective. This requires enough atoms for the conduction electrons to form a conduction band. A film $5.4$ thick is thick enough. Likely a wire $5.4$ nm this is too. But you might have practical difficulties making such a wire. If there are islands that make a connected film with holes it works. That might be a problem in a wire. $\endgroup$ – mmesser314 Jun 8 at 17:41
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    $\begingroup$ Here is a little more. If we repeatedly divide a colorful solid in half, at what point will the color disappear? $\endgroup$ – mmesser314 Jun 8 at 17:43
  • $\begingroup$ Okay we need a sufficient number of free electrons in motion to generate a reflected wave, and if there is a thickness necessary to do so and it's related to the establishment of a conduction band, then that sounds like a great answer to "What physics dictates this?" which is indeed right there in the title. I think you could simply move the points in your comments back up to the beginning of your answer! $\endgroup$ – uhoh Jun 8 at 23:09
  • $\begingroup$ Thanks for the edit! We don't get notifications when answers are edited so it's always a good idea to ping the commenter and let them know the post has been improved. Yes I think you are right, gold is a poor choice for the wire mesh for optical reasons, though for mechanical reasons it's amenable to being made extremely thin; a self-supporting mesh made from nanometer-scale aluminum wires would be a lot harder to make. $\endgroup$ – uhoh Jun 10 at 1:41
  • $\begingroup$ In Optics Express, Yakubovsky et al. (2017) Optical constants and structural properties of thin gold films seems to be helpful. $\endgroup$ – uhoh Jun 10 at 1:42

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