June 20, 2026. What is the connection between bit flips and the rate at which stars explode?

Introduction

Classical computers are very reliable. But there is a rather exotic form of error the designers probably didn’t expect: bombardment by high-energy particles from out of space. These are called cosmic rays, and they are mostly generated by exploding stars. We’ll do a quick Fermi estimate connecting how the rate at which cosmic rays cause bit flips in computers predicts the rate at which stars explode in the Milky Way.

Supernova

Supernovas occur when big stars run out of fuel, collapse in on themselves, and “bounce” explosively outward again. They tend to release huge amounts of energy when this happens, about $10^{44} \text{ J}$, or $10^{51} \text{ ergs}$ in older units. This leads to the name for this unit of energy, the “foe” for “fifty one ergs”. Steven Weinberg suggests the more august “bethe”. Cosmic rays are mostly from supernova, and tend to have energy around $4\times 10^{9} \text{ eV} \approx 6.4 \times 10^{-10} \text{ J}$. Finally, we know the most ($99\%$) of the energy of the supernova is carried away in neutrinos, which don’t really interact with matter. That leaves a total of

\[N_\text{super} = \frac{0.01 \text{ foe}}{6.4 \times 10^{-10} \text{ J}} \approx 1.6 \times 10^{51}\]

cosmic rays per supernova. Now, cosmic rays tend to bounce around the galaxy, bent by magnetic fields, before they’re either absorbed or leak out into intergalactic space. How long does it take for that to happen? As they bounce around, they collide with elements in the interstellar medium in a process called spallation, knock out protons, and create certain new elements such as beryllium that are rarely produced in stars. We know from the ratio of ${}^{10}B$ to (decayed) ${}^{9}B$ how long that beryllium has been travelling when it floats into our detectors, and it suggests that the average age of a cosmic ray is

\[\tau = 15 \text{ million years} \approx 4.7 \times 10^{14} \text{ s}.\]

So we know how many rays get produced per explosion and how long they last. What can we do with that information?

Soft errors

Here is where computers come in. A laptop at sea-level experiences the order of several bitflips a month. We can Fermi estimate the soft error rate (SER) of cosmic-ray induced bit flip errors as

\[\text{SER} = \Phi_n \cdot \sigma \cdot N_{\text{bits}}\]

where $\Phi_n$ is the flux of cosmic rays at sea level, $\sigma$ is the cross-sectional area, and $N_{\text{bits}}$ is the number of bits. A laptop with $16\text{ GB}$ of RAM has $N_{\text{bits}} \approx 1.28 \times 10^{11}$ bits, and the observed SER at sea level is around $10^{-6}$ errors per second. Finally, a modern transistor is on the order of a nanometer, so $\sigma \approx 10^{-18} \text{ m}^2$. Rearranging, we find a flux

\[\Phi_n = \frac{\text{SER}}{\sigma \cdot N_{\text{bits}}} \approx \frac{10^{-6}}{10^{-18} \cdot 1.28 \times 10^{11}\text{ m}^2{s}} \approx \frac{8}{\text{m}^2\text{s}}.\]

Now, cosmic rays hitting the top of the atmosphere tend to cascade into a shower of neutrons, and only one in $500$ actually makes it down to sea level. So the actual flux of cosmic rays is $500$ times higher! That is,

\[\Phi \approx \frac{4000}{\text{m}^2\text{s}}.\]

We can get a density by assuming the cosmic rays form an isotropic gas, i.e. travelling equally in every direction, at close to the speed of light. The basic intuition is that if you take a unit area of flux and drag it along at the speed of light, you should get the density. This actually underestimates by a factor of $4$ because it neglects the fact that there are travelling in all directions, so we have

\[\rho = \frac{4\Phi}{c} \approx \frac{4 \cdot 4000/\text{m}^2\text{s}}{3 \times 10^{8} \text{ m/s}} \approx 5.3 \times 10^{-5} \text{ m}^{-3}.\]

Thus, from errors on a microchip we can deduce the density of cosmic rays!

Conclusion

Each supernova produces around $N_\text{super}$ cosmic rays. Each cosmic ray has a lifetime of $\tau$. For a total of $N$ cosmic rays in the galaxy, the time between supernova, call it $T$, obeys

\[T = \frac{\tau N_\text{super}}{N} =\frac{\tau N_\text{super}}{\rho V},\]

where $\rho$ is the density we just computed and $V$ the volume of the milky way. If you model the Milky Way as a thick disk of $10^{5}$ light years in diameter and $1000$ light years thick, its volume is

\[V = \pi r^2 h = \pi (5\times 10^4)^{2} 10^3 \text{ ly}^3 \approx 8 \times 10^{60} \text{ m}^3.\]

Thus, the time between supernovas is

\[T = \frac{(4.7 \times 10^{14} \text{ s})(1.6 \times 10^{51})}{(5.3 \times 10^{-5})( 8 \times 10^{60})} \approx 3 \times 10^{7}\text{ s},\]

or around $56$ years. This is very close to observation, which records $1$-$3$ per century! So, the next time your data is mysteriously corrupted, take a moment to reflect on the death of a massive, ancient star, somewhere in the depths of the Milky Way.