The cost of forgetting

Leans on Entropy is missing information.

Why would forgetting cost anything?

Start with the simplest memory there is: a ball in a box with a wall down the middle. Ball on the left means 0. Ball on the right means 1. That is one bit.

Reading it is easy: you look. Changing it is easy too, as long as you know where the ball is. You nudge it across the wall, and if you nudge gently enough, the nudge can cost as little energy as you like.

Now try to wipe the memory: set it to 0 whatever it holds, without looking first. Nudging won't do, because you don't know which way to push. What does work is this: pull the wall out, then slide a piston in from the right and push the ball's whole world into the left half. Wherever the ball was, it ends up on the left. The memory reads 0.

Wiping a bit you haven't looked at. Two possible states get squeezed into one, and the squeeze has to push heat out into the surroundings.

Pushing a piston against the moving ball takes work, and that work ends up as heat in the box's surroundings. Resetting either initial state to 0 reduces the memory's entropy. Under the second law of thermodynamics, the surroundings must gain at least as much entropy, carried by the heat.

Rolf Landauer, a physicist at IBM, worked this out in 1961. Erasing unknown information has a minimum heat cost at a given temperature. Adding, comparing, copying and moving bits can, in principle, use arbitrarily little energy when performed reversibly and slowly enough.

Guess first Which of these has a cost no cleverness can avoid: copying a bit onto a blank memory, or wiping that copy afterwards?

Only the wiping. Copying onto a memory you know is blank can be done, in principle, for free; Charles Bennett showed how in the 1970s. Wiping a memory whose contents you don't know throws away which state it was in, and that is what costs.

How much heat is one bit?

The price is simple to write down:

energy to forget one bit = k × T × ln 2

T is the temperature of the surroundings the heat goes into, in kelvin.
k is Boltzmann's constant, the exchange rate between temperature and energy: 1.380649 × 10⁻²³ joules per kelvin.
ln 2, about 0.69, is there because one bit is a choice between two.

At room temperature, 300 K, one bit costs 2.9 × 10⁻²¹ joules. A single joule, about what it takes to lift an apple onto a table, pays for forgetting 3.5 × 10²⁰ bits.

Wiping a whole mind, a state of 1 × 10¹⁷ bits, costs only 2.9 × 10⁻⁴ joules at room temperature. Repeated erasures across a civilization add up. The Vigil's first petawatt, if it did nothing but erase at 300 K, could wipe 3.5 × 10³⁵ bits every second, and every one of those erasures is heat that must be radiated away.

Is colder cheaper?

The temperature sits right in the formula, so the colder the surroundings, the cheaper it is to forget:

Where the heat goes	Temperature	Cost to forget one bit
A room on Earth, or the Vigil's computers	300 K	2.9 × 10⁻²¹ J
The Vigil's radiators	30 K	2.9 × 10⁻²² J
Deep space in 2540 (the cosmic microwave background)	2.7 K	2.6 × 10⁻²³ J
Vesper's own faint glow	4.3 billionths of a kelvin	4.1 × 10⁻³² J

Every tenfold drop in temperature allows ten times as many erasures per joule. Deep space in 2540 is 110 times cheaper to forget in than a warm room. Cold surroundings improve computing efficiency, and the universe's continued cooling could lower this cost further.

Vesper's low temperature does not make it a practical heat sink. Its small absorption area limits how much heat it can take, as The black hole is not your freezer explains.

How close did the computers of the 2020s come?

Nowhere near the floor. A 2023 survey of chip physics put the energy to switch one logic gate in a processor of that decade somewhere between 1 × 10⁻¹⁶ and 1 × 10⁻¹³ joules, depending on the chip and on what you count. That is 35,000 to 3.5 × 10⁷ times the Landauer cost at the same temperature.

The survey judged that charging and discharging tiny wires likely accounted for most of the switching energy, so those chips operated far above the erasure limit. In 2540 the Vigil's computers operate much closer to it.

Open everything

Where does k T ln 2 come from?A one-molecule engine and a little calculus

Make the ball a molecule

In 1929 Leo Szilard imagined the box holding a single gas molecule, bouncing around in contact with surroundings at temperature $T$ . One molecule obeys the ideal gas law on average, just as a whole gas does:

$pV = kT$

To wipe the bit, take the wall out and push a piston in from the right until the molecule is confined to the left half. Do it slowly, so the molecule stays at the temperature of its surroundings. The work you do on it is

$W = -\int_{V}^{V/2} p \, dV' = -\int_{V}^{V/2} \frac{kT}{V'} \, dV' = kT \ln 2$

The molecule's energy depends only on its temperature, which has not changed. So every joule of that work leaves as heat, $Q = kT\ln 2$ , into the surroundings.

The same answer from entropy

Entropy counts possibilities: $S = k \ln W$ , where $W$ is the number of states the system could be in. Before the wipe the memory could be in either of two states, so it holds $k \ln 2$ of entropy. Afterwards it is certainly in one, with entropy zero.

The second law says the total entropy of the memory and its surroundings can never fall. The surroundings must therefore gain at least $k \ln 2$ , and at temperature $T$ that takes heat of at least

$Q \ge T \, \Delta S = kT \ln 2$

The result applies to different memory media, including molecules, magnets and charges on capacitors.

A bit you can predict is cheaper

If the bit is not a fifty-fifty unknown, the minimum heat is the temperature times the bit's Shannon entropy, written in physical units:

$Q \ge -kT \sum_i p_i \ln p_i$

A bit you can mostly predict costs less to erase. Resetting a known bit has no minimum heat cost: if you know it holds 1, you can flip it reversibly. The bound depends on the uncertainty removed by the reset.

What the principle does and doesn't say

It is about steps you can't run backwards. Clearing, overwriting, and any gate that maps several inputs to one output (AND, OR) lose information and carry the floor. A step that keeps enough to undo itself (NOT, a controlled swap) has no floor at all.
Measurement can be reversible. Maxwell's demon can record and sort molecules without a minimum heat cost for the measurement. Its memory fills up, and resetting that memory releases at least the heat the sorting seemed to save. Bennett made this argument in 1982.
It is a limit for going infinitely slowly. Real erasures happen in finite time and cost more. How much more is one of the questions in the next fold.

Has anyone measured it, and can a clever machine dodge it?The experiments, the loopholes and the arguments

Measured, fifty years later

For half a century the principle was an argument on paper. Then in 2012 Antoine Bérut, Eric Lutz and their colleagues trapped a single glass bead in a pair of wells made of laser light: bead on the left was 0, bead on the right was 1. They erased the bit over and over and measured the heat each time. The slower they erased, the closer the heat came to $kT \ln 2$ , and on average it never went below.

Two years later John Bechhoefer's group repeated the test with a feedback trap and much higher precision. Single runs sometimes came in under the floor, exactly as the statistics of very small systems predict, while the average held. In 2016 Jeongmin Hong, Jeffrey Bokor and their colleagues at Berkeley did it with nanoscale magnets, the kind of bit real memory is made of, and measured an energy within the error bars of the limit.

Faster costs more

The floor is for erasing infinitely slowly. Real erasure takes finite time, and finite time costs extra. In 2020 Karel Proesmans, Jannik Ehrich and John Bechhoefer worked out the best anyone can do: when an erasure takes a time $\tau$ and is reasonably slow, the extra cost above $kT \ln 2$ falls off as $1/\tau$ . Erase twice as slowly and you halve the excess.

Slower erasure lets the Vigil's computers operate closer to the minimum heat cost. Large distributed minds also need time for signals to cross them, so their communication delays can make a slower operating pace useful for both reasons.

Ways around it, and what they cost

Don't erase. Charles Bennett showed in 1973 that any computation can be done reversibly: keep a record of every step, copy out the answer, then run the whole computation backwards to clean up. Nothing is forgotten and nothing has to be paid for. The price moves instead into memory and time, and that trade is the subject of computing without forgetting.
Erase somewhere colder. The cost scales with the temperature of the surroundings, so cold surroundings help. You still have to move the heat there, and the coldest object around, a black hole, takes heat only through a tiny window.
Pay in a different currency frontier. Joan Vaccaro and Stephen Barnett showed in 2011 that erasure can be paid for in angular momentum instead of energy, using a reservoir of spins rather than a heat bath. The entropy cost stays, but it no longer has to be heat. The proposal is theoretical.
Small and quantum surroundings. David Reeb and Michael Wolf gave in 2014 a tighter version of the bound for a finite, quantum heat bath. It is stricter than $kT \ln 2$ , with corrections that matter when the surroundings are small.

Still argued in 2026

Most physicists treat Landauer's principle as settled, and the experiments agree with it. The philosopher of physics John Norton has argued that the usual derivations assume what they set out to prove and need a better foundation. His objection concerns the derivations; the measured heat remains consistent with $kT \ln 2$ .

The open engineering questions of 2026 concern erasure that must be fast, reliable and error-corrected, and how closely working hardware can approach the bound. The Vigil's design minimizes erasures and performs them as slowly as its workload permits. frontier