Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

As far as I know, today most of the computers are made from semiconductor devices, so the energy consumed all turns into the heat emitted into space.

But I wonder, is it necessary to consume energy to perform computation?

  • If so, is there a theoretical numerical lower bound of the energy usage ? (I even have no idea about how to measure the amount of "computation")

  • If not, is there a physically-practical Turing-complete model that needs no energy?

edit: Thank @Nathaniel for rapidly answering the question and pointing out it's actually Landauer's principle. Also thank @horchler for referring to the Nature News and the related article. There are lots of useful information in the comments; thank every one! This whole stuff is really interesting!

share|improve this question
Suggested title change: "Is it necessary to consume energy to perform computation?" At the moment the title looks like you want to know how cooling fans work, but the body text is asking a much deeper physical question. –  Chris White Jul 7 '13 at 16:00
You may be interested in the Applied Cryptography snippet about half way into the post on schneier.com/blog/archives/2009/09/the_doghouse_cr.html –  Michael Kjörling Jul 7 '13 at 19:06
Your question made me think of Maxwell's demon. See this Nature News article and the related Nature article by Bérut, et al. on the experimental verification of Landauer's principle. –  horchler Jul 8 '13 at 1:12
See also link"The thermodynamics of Computation - A Review" by Charles Bennet, Int J. Theor. Phys. Vol 21, No 12 1982 for a most excellent review of Landauer's principle discussed in Nathaniel's answer below. –  WetSavannaAnimal aka Rod Vance Jul 8 '13 at 8:32
In contrast, a cell's building of a protein forgets about 1.6 bits per amino acid (64 DNA codons code for 20 amino acids) and does it with the expenditure of about between $10 k T$ and $40 k T$, thus only one order of magnitude worse than the Landauer limit. See linkJ Avery "Information Theory and Evolution" 2003. I guess this is one of the reasons DNA computing might hold some promise. –  WetSavannaAnimal aka Rod Vance Jul 8 '13 at 8:52
show 2 more comments

1 Answer

up vote 84 down vote accepted

What you're looking for is Landauer's principle. You should be able to find plenty of information about it now that you know its name, but briefly, there is a thermodynamic limit that says you have to use $k_BT \ln 2$ joules of energy (where $k_B$ is Boltzmann's constant and $T$ is the ambient temperature) every time you erase one bit of computer memory. With a bit of trickery, all the other operations that a computer does can be performed without using any energy at all.

This set of tricks is called reversible computing. It turns out that you can make any computation reversible, thus avoiding the need to erase bits and therefore use energy, but you end up having to store all sorts of junk data in memory because you're not allowed to erase it. However, there are tricks for dealing with that as well. It's quite a well developed area of mathematical theory, partly because the theory of quantum computing builds upon it.

The energy consumed by erasing a bit is given off as heat. When you erase a bit of memory you reduce the information entropy of your computer by one bit, and to do this you have to increase the thermodynamic entropy of its environment by one bit, which is equal to $k_B \ln 2$ joules per kelvin. The easiest way to do this is to add heat to the environment, which gives the $k_BT \ln 2$ figure above. (In principle there's nothing special about heat, and the entropy of the environment could also be increased by changing its volume or driving a chemical reaction, but people pretty much universally think of Landauer's limit in terms of heat and energy rather than those other things.)

Of course, all of this is in theory only. Any practical computer that we've constructed so far uses many orders of magnitude more energy than landauer's limit.

share|improve this answer
+1: I love learning! (especially from well-written posts) –  joshphysics Jul 7 '13 at 16:55
@Nathaniel : What is the increasing of the energy for the environment ? For instance is it correct to state that the environnement gains energy $\frac{kT}{2}$, so that the Gibbs Free energy of the environnement decreases : $\Delta F = \Delta U - T\Delta S =kT(\frac{1}{2} - ln 2) < 0$ –  Trimok Jul 7 '13 at 18:15
The Feynman Lectures on Computation is a slim little volume that includes a presentation of the state of this art a few of decades ago. Very accessible to most physicists. –  dmckee Jul 7 '13 at 18:43
See the paper "Thermodynamics of prediction" by Still et al., available on [threeplusone.com](Gavin Crooks' website) (sorry, I kind of feel bad about deep-linking) for a refinement of Landauer's principle for general computation devices retaining memory and responding to some temporally correlated input signals. Such a system approaches Landauer's limit when it is maximally predictive, that is, the information in the system's memory contains a maximum amount of information about future input signals. –  Simeon Carstens Jul 7 '13 at 23:07
@Trimok it's simpler than that. You add $Q=k_BT\ln 2$ joules of heat, so the environment's entropy changes by $Q/T=k_B\ln 2 = 1\,\text{bit}$. The environment is typically assumed to be a heat bath, so we don't worry about its free energy, just its entropy. The total entropy change is $-1\,\text{bit}+1\,\text{bit} = 0$, satisfying the second law. –  Nathaniel Jul 8 '13 at 1:31
show 1 more comment

protected by Qmechanic Jul 8 '13 at 14:59

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.