Cooling is the economy

Where does the heat go when there is no air?

On Earth you get rid of heat without thinking about it. A hot cup loses heat to the air touching it. A laptop blows it away with a fan. A body sweats. All of these hand the heat to some matter that carries it off.

In space there is no such matter. A spacecraft is hot metal in a near-perfect vacuum, and a vacuum conducts nothing and carries nothing. There is exactly one way left for heat to leave: light. Everything warm glows, in infrared, and every bit of glow carries energy away. A radiator in space is just a big surface whose job is to glow.

Radiated power depends strongly on temperature. The hotter a surface, the more light it gives off, and the dependence is steep: double the temperature and the glow per square metre goes up sixteen times. Cooling the surface reduces its emission by the same relation. Halve the temperature and each square metre sheds sixteen times less heat, so you need sixteen times as many of them.

Guess first The Vigil rejects one petawatt of heat. Roughly how much surface must glow to carry it away at 300 K, room temperature: a city, a large country, or a planet's worth? And at 30 K?

At 300 K it takes a large country's worth, 0.43 percent of the Earth's surface. At 30 K, ten times colder, the glow per square metre falls ten thousand times, and the surface grows ten thousand times with it: tens of Earths.

That is the Vigil's problem. Its computers run warm and its radiators run cold, because a cold end lets its engines harvest more of the heat on the way out, and makes forgetting cheaper (see Recycling heat and The cost of forgetting). The price is area. One petawatt of heat at 30 K needs 43 Earths of glowing surface, counting one face, which is 2.2 × 10¹⁶ square metres.

It does not sit in one place. The Vigil lays the surface out as a swarm of thin, kilometre-wide films spread through the space around Vesper, every one turned edge-on to Ember, the star Vesper circles with. Seen in ordinary light they are nearly black. Seen in infrared they glow faintly, a cold haze.

The required radiator mass dominates the Vigil's construction needs. The fuel for a petawatt is 0.11 kilograms a second, a trickle. The radiators require a much larger stock of material. A one-tonne seed machine that copies itself needs 37 doublings to make all that film, which is the story of Doubling. And the trade runs both ways: colder buys more thought from every joule and costs more film for every watt. That trade sits under the Vigil's whole case for growing inward rather than outward, in Inward, not outward.

How big is the radiator, in numbers?

The rule has a name and a formula. A perfectly black surface at temperature T sheds, per square metre,

power per m² = σ × T⁴

where σ (sigma, the Stefan-Boltzmann constant) is 5.670 × 10⁻⁸ watts per square metre per kelvin to the fourth. Real surfaces fall a little short of perfect, which a factor ε (emissivity, between 0 and 1) takes care of. The ledger here assumes ε = 1.

Space is not at absolute zero. The afterglow of the Big Bang, at 2.7 K in 2540, shines back a little, so a radiator really sheds σ(T⁴ − T_sky⁴). For the temperatures here it hardly matters.

Ember matters a great deal. Near Vesper its light delivers 99 watts on each square metre that faces it, 2,152 times what a 30 K film sheds (0.046 W/m²). A black ball there, turning so that it warms evenly, would sit near 144 K. Turned edge-on it catches almost nothing, but only just: a tilt of 0.46 milliradians lets in as much starlight as the film radiates. The same ball would cool to 30 K only beyond 35 AU from Ember, and a thin film held face-on to the star, which sheds from both faces but catches on one, only beyond 49 AU, far from the Engine and hours away by light. So the Vigil's radiators are pointed as carefully as its lasers design.

The cold-film heat balance assumes Ember-light absorptivity 0.05 and pointing error 0.50 mrad. These unsourced engineering targets and film flutter determine whether the temperature margin holds challenge.

Dividing the heat by what a square metre sheds gives the area:

area = heat power ÷ (σ × T⁴)

Radiator at	Each m² sheds	Area for 1 PW	In Earths
300 K	459 W	2.2 × 10¹² m²	0.0043
30 K	0.0459 W	2.2 × 10¹⁶ m²	43

Ten times colder costs 10,001 times the area. That is the T⁴ at work: ten to the fourth is ten thousand.

Two details turn the area into something you can build.

Two faces. A thin film glows from both sides. The glowing surface for the Vigil is 2.2 × 10¹⁶ m², but the film itself needs only half that, 1.1 × 10¹⁶ m², because each square metre of film counts twice.
Mass. The model assumes the Vigil's films weigh 0.01 kilograms per square metre, 10 grams, lighter than kitchen foil. That makes the 30 K radiator for one petawatt weigh 1.1 × 10¹⁴ kilograms. At 300 K the same job is 1.1 × 10¹⁰ kilograms: ten thousand times lighter. This unsourced film mass per area is an engineering target challenge.

Compare that with the energy itself. A petawatt for a year is the mass-energy (E = mc²) of 3.5 × 10⁵ kilograms of matter, and the cold radiator weighs 3.1 × 10⁸ times that. Heat rejection accounts for most of the Vigil's manufactured mass.

The calculator shows how increasing temperature reduces the required area, while increasing rejected power increases it proportionally.

The area a radiator needs for a fixed power, against its temperature, on log axes. Radiated power goes as the fourth power of temperature, so every step colder costs the fourth power more surface.

With JavaScript on, this is a calculator: choose the heat, the radiator temperature and one face or two, and it gives the area in Earths and the mass of film. The table above gives the same numbers for 30 K and 300 K.

What can you do about it?

Three levers exist, and the Vigil uses all of them.

Run the radiators hotter. Area falls as T⁴. The cost is that the engine has less temperature drop to work with, so less heat is recovered, and forgetting gets dearer.
Make the film thinner. The mass bill falls with the weight per square metre. 10 grams per square metre is the Vigil's unsourced engineering target challenge.
Make more of the heat useful before it leaves. As Recycling heat explains, engines recover part of the heat as work and return it to the computers. The radiator still rejects the full external supply, while supporting a larger internal computing load.

Each option changes the balance between radiator mass, engine recovery and erasure cost. An ideal engine fed from 300 K gets the most work out of each square metre of radiator with the radiator at 225 K, and there it gets 117 times as much per square metre as with a 30 K radiator. The next fold works out why that best case sits at three-quarters of the hot temperature, and why the Vigil stays far from it.

Open everything

Why a fourth power, and where does three-quarters come from?The Planck integral, view factors and a derivative

Where does σT⁴ come from?

A hot surface emits light of every wavelength. Planck's law gives the spectral radiance, the power per unit area per unit solid angle per unit wavelength, of a black surface at temperature $T$ :

$B_\lambda = \frac{2hc^2}{\lambda^5}\,\frac{1}{e^{hc/\lambda kT}-1}$

A flat surface radiates equally bright in every direction (Lambert's law), and integrating the cosine of the angle over the outward hemisphere gives a factor $\pi$ . The total power per unit area, the exitance, is therefore

$M = \pi \int_0^\infty B_\lambda \, d\lambda$

Substitute $x = hc/\lambda kT$ , so that $\lambda = hc/(xkT)$ and $d\lambda = -\,hc\,dx/(kT\,x^2)$ . The powers of $h$ , $c$ and $kT$ collect and leave

$M = \frac{2\pi (kT)^4}{h^3c^2}\int_0^\infty \frac{x^3}{e^x-1}\,dx$

The whole temperature dependence has come out as $T^4$ , because the integral over $x$ is a pure number. Expand $1/(e^x-1) = \sum_{n\ge1} e^{-nx}$ and integrate term by term, using $\int_0^\infty x^3 e^{-nx}dx = 6/n^4$ :

$\int_0^\infty \frac{x^3}{e^x-1}\,dx = 6\sum_{n\ge1}\frac{1}{n^4} = 6\,\zeta(4) = \frac{\pi^4}{15}$

That is about 6.494.

$M = \sigma T^4, \qquad \sigma = \frac{2\pi^5 k^4}{15\,h^3 c^2}$

The fourth power follows from the spectrum: the number of available light modes grows as $\omega^2$ , each carries energy of order $kT$ and the typical frequency is itself proportional to $T$ : three powers from the spectrum, one from the energy.

What does the surface see?

$\sigma T^4$ is the power leaving a surface. What matters for cooling is the net exchange with whatever the surface looks at. For a gray surface of area $A$ and emissivity $\varepsilon$ that sees only a sky at $T_s$ , through a view factor $F = 1$ :

$P_{\text{net}} = \varepsilon \sigma A\,(T^4 - T_s^4)$

The view factor $F_{ij}$ is the fraction of what leaves surface $i$ that lands on surface $j$ . It depends on geometry alone, and reciprocity ties the two directions together: $A_i F_{ij} = A_j F_{ji}$ . Three cases matter for a swarm.

A flat panel in empty space has $F = 1$ to the sky on each face, so both faces glow freely and the total is $2\varepsilon\sigma A T^4$ .
A panel next to another panel at the same temperature gets back part of what it sends out. Two parallel plates that see only each other exchange nothing, so a swarm must be spread thin or turned edge-on to its neighbours.
A panel facing something hot, a star or the minidisk where Vesper is fed, also absorbs $a\,S A_\perp$ , where $a$ is its absorptivity, $S$ the incoming flux and $A_\perp$ its area seen from the source, so it must be kept edge-on or shaded.

All the area numbers on this page are the ideal case: $F = 1$ , $\varepsilon = 1$ , no incoming light. Any real swarm needs more.

Why would a radiator-limited engine run its radiator at three-quarters?

Here is the question exactly. You have a hot source at a fixed temperature $T_h$ that will hand over as much heat as you ask for, so heat is free. You have a radiator of area $A$ , and area is the one thing that is scarce. Between them sits an ideal engine. At what radiator temperature $T_c$ does each square metre of radiator give you the most work? Four assumptions keep it simple: the engine is a perfect Carnot engine, the radiator is all at one temperature, it is black ( $\varepsilon = 1$ ), and the sky is cold enough to ignore. Heat leaves the radiator at

$Q_c = \sigma A T_c^4$

A Carnot engine delivers work $W$ while rejecting $Q_c$ , with $W = Q_c\,(T_h/T_c - 1)$ . Substitute:

$W = \sigma A\,T_c^3\,(T_h - T_c)$

Write $\theta = T_c/T_h$ and the power per unit of radiator area is

$\frac{W}{A} = \sigma T_h^4\;\theta^3(1-\theta)$

There is a trade. Colder (small $\theta$ ) makes the engine more efficient but the radiator glows feebly. Hotter ( $\theta$ near 1) makes the radiator glow brightly but the engine has almost no temperature drop to work with. The best compromise is where the derivative vanishes:

$\frac{d}{d\theta}\left[\theta^3 - \theta^4\right] = 3\theta^2 - 4\theta^3 = 0 \;\Rightarrow\; \theta = \tfrac{3}{4}$

The second derivative is negative there, so this is a maximum. The engine's efficiency at that point is $1 - 3/4 = 1/4$ , and the work per square metre of radiator is

$\frac{W}{A}\Big|_{\max} = \frac{27}{256}\,\sigma T_h^4$

The fraction 27/256 is 0.105. A gray radiator ( $\varepsilon < 1$ ) multiplies $W/A$ by $\varepsilon$ and leaves the three-quarters where it is.

For a 300 K source the best radiator temperature is 225 K, and each square metre of radiator then delivers 48.4 watts of work. The same ideal engine with its radiator at 30 K, $\theta = 1/10$ , delivers 0.41 watts per square metre, 117 times less. The Vigil uses a lower temperature and more radiator area to reduce the cost of erasure.

This is not the efficiency at maximum power of finite-time thermodynamics (Curzon and Ahlborn's result, on Recycling heat), which comes from heat having to cross finite conductances into and out of the engine. Here the engine is ideal and the only bottleneck is the glow of the radiator.

What is the lightest radiator, and could anyone see a swarm?Mass-optimal temperatures and the infrared sky

The lightest radiator is not the smallest

The temperature that minimizes radiator area can differ from the one that minimizes total plant mass. The following model includes both heat-source and radiator mass. Keep the ideal engine and the fixed $T_h$ of the last fold, but let the hot side, the reactor or collector, weigh $\beta$ kilograms per watt of heat it delivers, and the radiator $\mu$ kilograms per square metre. An engine delivering work $W$ takes in heat $W/\eta$ , with $\eta = 1-\theta$ , and rejects $W\theta/(1-\theta)$ , so its radiator area is $A = W/\big((1-\theta)\,\sigma T_h^4\theta^3\big)$ . The mass per watt of work is

$\frac{M}{W} = \frac{\beta}{1-\theta} + \frac{\mu}{\sigma T_h^4\,\theta^3(1-\theta)}$

With $\beta = 0$ the radiator is all that weighs anything, and minimizing the second term gives the same $\theta = 3/4$ as before. Give the hot side any weight and the first term, which rises with $\theta$ , tilts the slope at $3/4$ uphill, so the optimum moves to a colder radiator and a more efficient engine. Where it lands depends on the ratio of the two specific masses, which is why real designs differ. A published cousin is the study of an endoreversible space Carnot plant by Wang, Ge, Chen, Feng and Yu in 2021: they fix the total heat-exchanger area, maximize power, and find an optimal temperature for the low-temperature heat sink. That is closer to the three-quarters question of the last fold than to this mass question.

The Vigil's case is extreme. The fuel is a trickle, and the ledger counts the radiator as the mass of the plant, so $\beta$ is nearly zero and the lightest plant would run its radiators near 225 K. The Vigil chooses 30 K anyway, a tenth of the hot temperature, because the cold also makes forgetting ten times cheaper (see The cost of forgetting). This operating temperature is a design assumption.

Could anyone see it?

Every watt a civilization uses ends as heat, and in a steady state every watt of that heat must leave as light. This follows from energy conservation. Freeman Dyson pointed out in 1960 that a civilization capturing most of its star's light would therefore re-radiate it in the infrared, and that the infrared glow is how to look for it.

Searches have been made.

Carrigan (2009) went through the IRAS satellite's whole-sky catalogue looking for sources whose spectra fit a blackbody between 100 and 600 K, found only a few ambiguous candidates, and set upper limits on Dyson spheres.
The Ĝ search. Wright and colleagues set out the case for looking in mid-infrared light for the waste heat of civilizations with large energy supplies, and in 2014 reported a first, null result from the WISE all-sky catalogue: Kardashev Type III civilizations, ones using the power of a whole galaxy, are very rare in the local universe. In 2015 Griffith and colleagues vetted the reddest extended sources in WISE by eye, 563 of them, and found no galaxy hosting a civilization that turns more than 85 percent of its starlight into mid-infrared heat.
Project Hephaistos. In 2024 Suazo and colleagues screened 5,000,000 objects from the Gaia, 2MASS and WISE catalogues and picked out seven M-dwarf stars with an infrared excess that known astrophysics could not easily explain. Ren, Garrett and Siemion answered the same year: three of the seven sit close to radio sources, most likely dust-obscured galaxies far behind the stars, and such galaxies are common enough to account for all seven.

Now put the Vigil into that picture. These searches use mid-infrared light, with WISE reaching 22 micrometres at its longest. By Wien's law the peak of a glowing surface sits at $\lambda_\text{peak} = b/T$ with $b$ = 0.002898 metre-kelvins: 9.66 micrometres for 300 K, but 96.6 micrometres for 30 K, in the far infrared. At 22 micrometres each square metre of a 30 K surface is only 2.7 × 10⁻⁹ times as bright as a square metre at 300 K. The cold swarm has 10,001 times the area for the same power, and in that band it still comes out only 2.7 × 10⁻⁵ times as bright as a warm swarm shedding the same heat. These ratios are calculated from Planck's law; they do not constitute a published detection limit. It says a swarm like the Vigil's would slip past the mid-infrared searches above, and past Carrigan's 100 to 600 K window too. It would glow in the far infrared, where cold dust between the stars glows as well.

The waste heat produces an infrared signature, with lower temperatures shifting the emission toward longer wavelengths. In the 2020s no one could say whether any such swarm existed. The Hephaistos candidates remained unconfirmed, with background contamination offering a plausible explanation.