Is it possible to scale a Szilard engine up to the point of being able to power everyday things like cars, and if so what might a practical version look like?

My understanding is that Szilard engine essentially consists of a particle, say an electron, in a box, and in the thought experiment version a "daemon" which looks to see which side of the box the particle is in, and using this bit of information splits the box in half by inserting a slide, and places a piston on one side to extract energy from the particle. (http://www.quantumcomplexity.org/tutorials/knowledge-is-power-the-energy-content-of-bits/)

Crucially, the measurements (classically copying the state of the system to memory) can be done elastically and adiabaticly and so require no energy, and the system is at thermal equilibrium with its environment. However, by Landauer's principle erasing a bit on the hard drive will cost $kT\log(2)$, so the $kT\log(2)$ gained from the particle dumping its energy into the piston is balanced unless there is a temperature difference between the hard drive and the electron-box system.

Experimentally, this setup has been realized for a single bit (https://www.pnas.org/content/111/38/13786) by using a high speed camera of some sort (an SET electrometer, and other electronics).

A gallon of gas is $\approx 20kWh = 72 MJ$. If each particle carries $kT\log(2) J$, then at $200f=366.5 K$ we need $2.06\cdot 10^{28}$ particles or $\approx 34,000$ moles worth, or $\approx 1,000$ yottabytes.

Presumably you'd need a large hard drive connected to some sort of ultra-high resolution, ultra-fast camera, which in turns controls the necessary gates and many "pistons". I'm imagining a gas in a chamber which I suppose would cool off as the "pistons" engage while heat pours off the hard drive as bits written into it. Is this picture accurate?


4 Answers 4


Imagine, for the sake of argument, your proposal was physically feasible, and we successfully extracted the 72MJ from the gas using an enormously complex Szilard piston.

If we make a generous assumption that your hard drive is cooled with LHe and is at roughly $4K$ during the first erase cycle, then writing your thousand yottabytes of data will generate $10^{28} \times 4 \text{K} \times 1.3 \times 10^{-23} \text{J/K} \approx 382 \text{KJ} $. Which doesn't sound like a lot of waste heat compared to the 72MJ we got out, but it's the start of a problem.

Because dumping 382KJ of heat into an extremely cold area means it's not going to stay cold for long. 382KJ is enough to raise 1 mol of material to somewhere in the realm of $40000 \text{K}$. Even if you have a ten thousand mol reservoir to absorb that, then your ambient heat increases by another few Kelvin, but since your ambient temperature was only a few Kelvin to begin with, this multiplies the waste heat by a significant factor. You effectively have a positive feedback loop where any waste heat left over by the time the engine "cycles" makes it operate less efficiently next time, and so on until the system breaks down from the exponentially growing energies.

This is going to massively limit the power output of your engine, given that you need to evacuate at least 380KJ per cycle, and out of an environment much colder than the surroundings rather than a conventional engine being much hotter. You get your 72MJ out, but not necessarily at a useful rate.

  • $\begingroup$ This sort of makes sense. Follow up question is what makes this system essentially different from a conventional engine? 1) why does the hard drive need to be at 4K? To keep waste heat energy as low as posssible? Why can’t we just keep the HDD at room temp? 2) Is the factor 10^28, i.e. keeping track of every bit, the essential difference? $\endgroup$ Feb 1, 2021 at 3:05
  • 1
    $\begingroup$ @JacksonWalters 1) the min. entropy generated by consuming information is proportional to temperature - $4K$ is a semi-arbitrary value for how cold you could plausibly make bulk matter, as opposed to, e.g. laser cooling a gas. Room temp would multiply waste heat by ~100x 2) The $10^{28}$ factor is why conventional harddrives don't explode immediately and 3) conventional engines are hotter than their surroundings, and naturally lose energy rather than gain it, so they do not have the same exponential inefficiency $\endgroup$
    – redroid
    Feb 1, 2021 at 16:30
  • $\begingroup$ 1) Makes sense. 2) Could you elaborate on don't explode? Not sure I follow. I meant more why doesn't a factor of 10^28 appear when analyzing other heat engines. 3) Hm, right I choose 200f as the rough operating temperature of a conventional engine. Here, if the gas is at $200f=367K$ and hard drive is at $4K$, wouldn't we say the Szilard engine is operating at $367K$? Also, in your analysis why don't we see the efficiency depending on the difference $367K-4K$ and only on the temp of the cool hard drive? $\endgroup$ Feb 1, 2021 at 21:05

I have no idea on how to implement such an engine so I can not answer your question, but I can give you some context and theoretical limits on how it will look like.

The idea seems sound: we just need a big hard drive! Can we make one...?

If your calculation is correct, we need a 1000 yottabites HD. That's... a lot! Consider the total amount of data on Earth is estimated at $\approx 50$ zettabytes. 1 YB = 1000 ZB so we would need an HD capable of containing 20000 times the amount of data on Earth now! So you already see that it would be a big effort..

But ok, it is a huge HD but maybe we might have some day the technology - right? After all, the human race got to 50ZB, we can go much higher in the future!

But there is another problem: how big would this HD be?

There is a funny result called Berkenstein bound saying that you have a system of mass $M$ with radius $R$ (assume is a spherical hard-drive) the maximum amount of information you can put inside it is $\approx 10^{43}MR$ bits, that is $\approx 10^{42}MR$ bytes (with kgs and m as units).

This means that if you have a total information $I$ then

$$I< 10^{42} MR$$

which sets a limit on the size of your hard drive $$MR > I/10^42$$

However, it is not a very stringent limit: if you plug in the number, you get someting that is definitely feasible.

So if you want something that can stay inside a car, let's say $R=10cm=0.1m$ the hard drive will have at least a weight of

$$M>10^{27} bytes/ 10^{42} /0.1m = 10^{-16} Kg$$ which is definitely a weight one can afford to bring in a car!

However unfortunately at the moment the best we can do to store information (the information density is called areal density) is much worse then this theoretical limit and is around 10 TB per inches squared (for magnetic drives) or, interestingly, 1 PB per gram in the case of DNA so with today's technology to store $10^{27}$ bytes for a car engine you would need something with an area of $10^{15} in^2$ which is 645 160 000 000 $m^2$ or, using DNA, $10^{12}$ grams, that's 1000000000 Kg! That's how it would look like today: big! (And that's just for one hour of travel)


Presumably you'd need a large hard drive connected to some sort of ultra-high resolution, ultra-fast camera, which in turns controls the necessary gates and many "pistons". I'm imagining a gas in a chamber which I suppose would cool off as the "pistons" engage while heat pours off the hard drive as bits written into it. Is this picture accurate?

Yes, except that the hard drive serves as an information sink, I mean it serves as sink for waste information, which is also known as entropy.

The hard drive must be such that writing entropy on it requires just a very small amount of energy. Amount of entropy written per unit of energy is a temperature.

The hard drive is a 'cool' hard drive, which is getting less 'cool' as the engine is running.

Actually, to agree with what Landauer says, let's say that entropy density of our hard drive is low, so that we can say that entropy is written on a surface whose entropy is zero, which means that zero energy is required to write the entropy, just as Landauer says. "Erasing information requites energy, writing doesn't". There is another answer talking about Bekenstein bound. This "entropy density" thing here is related to that bound.

  • $\begingroup$ I'm a little confused, as the hard drive is storing bits which are then used to decide which side to place the piston. Isn't the waste entropy the information/energy which is released when the useful bit (the bit storing the molecule position) is written onto the hard drive? $\endgroup$ Jan 26, 2021 at 23:39

So the demon has a yotta-byte hard disk as its memory, which allows it to newer overwrite any information in its memory, which means that no erasing, or zeroing, of memory is necessary.

So the demon just opens and closes some engine valves, the gas, or one gas molecule, pushes piston, the engine runs and makes the car move. The molecule sucks heat energy from the cylinder walls.

And no heat is generated anywhere.

Except if the hard disk has filled up, and we want to erase it, then we must use energy to do that and heat is generated.

Of course making new hard disks must require such amount of energy that the laws of thermodynamics are not violated when we use those hard disks as parts of Szilard engines.

  • $\begingroup$ You already submitted an answer. Perhaps it would make more sense to merge the two? $\endgroup$ Jan 28, 2021 at 1:38

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