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Matrioshka brains and IPv6: a thought experiment

Nich (one of my roommates) mentioned recently that discussion in his computer networking course this semester turned to IPv6 in a recent session, and we spent a short while coming up with interesting ways to consider the size of the IPv6 address pool.

Assuming 2128 available addresses (an overestimate since some number of them are reserved for certain uses and are not publicly routable), for example, there are more IPv6 addresses than there are (estimated) grains of sand on Earth by a factor of approximately \( 3 \times 10^{14} \) (Wolfram|Alpha says there are between 1020 and 1024 grains of sand on Earth).

A Matrioshka brain?

While Nich quickly lost interest in this diversion into math, I started venturing into cosmic scales to find numbers that compare to that very large address space. I eventually started attempting to do things with the total mass of the Solar System, at which point I made the connection to a Matrioshka brain.

“A what?” you might say. A Matrioshka brain is a megastructure composed of multiple nested Dyson spheres, themselves megastructures of orbiting solar-power satellites in density sufficient to capture most of a star’s energy output. A Matrioshka brain uses the captured energy to power computation at an incredible scale, probably to run an uploaded version of something evolved from contemporary civilization (compared to a more classical use of powering a laser death ray or something). Random note: a civilization capable of building a Dyson sphere would be at least Type II on the Kardashev scale. I find Charlie Stross’ novel Accelerando to be a particularly vivid example, beginning in a recognizable near-future sort of setting and eventually progressing into a Matrioshka brain-based civilization.

While the typical depiction of a Dyson sphere is a solid shell, it’s much more practical to build a swam of individual devices that together form a sort of soft shell, and this is how it’s approached in Accelerando, where the Solar System’s non-Solar mass is converted into “computronium”, effectively a Dyson swarm of processors with integrated thermal generators. By receiving energy from the sunward side and radiating waste heat to the next layer out, computation may be performed.

Let’s calculate

Okay, we’ve gotten definitions out of the way. Now, what I was actually pondering: how does the number of routable IPv6 addresses compare to an estimate of the number of computing devices there might be in a Matrioshka brain? That is, would IPv6 be sufficient as a routing protocol for such a network, and how many devices might that be?

A silicon wafer used for manufacturing electronics, looking into the near future, has a diameter of 450 millimeters and thickness of 925 micrometers (450mm wafers are not yet common, but mass-production processes for this size are being developed as the next standard). These wafers are effectively pure crystals of elemental (that is, monocrystalline) silicon, which are processed to become semiconductor integrated circuits. Our first target, then, will be to determine the mass of an ideal 450mm wafer.

First, we’ll need the volume of that wafer (since I was unable to find a precise number for a typical wafer’s mass):
$$ \pi \times \left( \frac{450 \;\mathrm{mm}}{2} \right)^2 \times 925 \;\mathrm{\mu m} = 147115 \;\mathrm{mm^3} $$
Given the wafer’s volume, we then need to find its density in order to calculate its mass. I’m no chemist, but I know enough to be dangerous in this instance. A little bit of research reveals that silicon crystals have the same structure as diamond, which is known as diamond cubic. It looks something like this:

Silicon crystal structure.

Now, this diagram is rather difficult to make sense of, and I struggled with a way to estimate the number of atoms in a given volume from that. A little more searching revealed a handy reference in a materials science textbook, however. The example I’ve linked here notes that there are 8 atoms per unit cell, which puts us in a useful position for further computation. Given that, the only remaining question is how large each unit cell is. That turns out to be provided by the crystal’s lattice constant.
According to the above reference, and supported by the same information from the ever-useful HyperPhysics, the lattice constant of silicon is 0.543 nanometers. With this knowledge in hand, we can compute the average volume per atom in a silicon crystal, since the crystal structure fits 8 atoms into a cube with sides 0.543 nanometers long.

$$ \frac{0.543^3 \mathrm{\frac{nm^3}{cell}}}{8 \mathrm{\frac{atoms}{cell}}} = .02001 \mathrm{\frac{nm^3}{atom}} $$

Now that we know the amount of space each atom (on average) takes up in this crystal, we can use the atomic mass of silicon to compute the density. Silicon’s atomic mass is 28.0855 atomic mass units, or about \( 4.66371 \times 10^{-23} \) grams.

$$ \frac{4.66371 \times 10^{-23} \mathrm{\frac{g}{atom}}}{.02001 \mathrm{\frac{nm^3}{atom}}} = 2.3307 \mathrm{\frac{g}{cm^3}} $$

Thus, we can easily compute the mass of a single wafer, given the volume we computed earlier.

$$ \frac{147115 \;\mathrm{mm}^3}{1000 \frac{mm^3}{cm^3}} \times 2.3307 \frac{g}{\mathrm{cm}^3} = 342.881 \;\mathrm{g} $$

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