5 ms·

I remember reading Feynman's thought experiment and him deciding that it doesn't require any energy to flip a bit.[1] How does that square with Landauer's limi

I remember reading Feynman's thought experiment and him deciding that it doesn't require any energy to flip a bit.[1] How does that square with Landauer's limit? [1] Apologies for the lack of reference. I'll try and find it.

kobeya 9y ago
I’m not sure how it squares with physics. I’d recommend finding that reference.
- emmelaich 9y ago
  Well, perhaps it's only a theoretical minimum. It agrees with kT ln 2 when T=0 right?
  - kobeya 9y ago
    T is never 0.
    - emmelaich 9y ago
      Yeah I know :-) But OK here's a reference of sorts. Not by Feynman but by Bennet and Landauer. https://www.scientificamerican.com/article/the-fundamental-physical-limits-of-computation/ https://www.scientificamerican.com/article/the-fundamental-p... "there is no minimum amount of energy that must be expended in order to run a Brownian clockwork Turing machine." Also interesting but I didn't find the exact thing I was looking for: Simulating Physics with Computers Richard P. Feynman https://people.eecs.berkeley.edu/~christos/classics/Feynman.pdf https://people.eecs.berkeley.edu/~christos/classics/Feynman.... Richard Feynman and Computation https://cds.cern.ch/record/411350/files/p101.pdf https://cds.cern.ch/record/411350/files/p101.pdf
      - kobeya 9y ago
        Well for example, T cannot be less than the temperature of the cosmic microwave background. You can make regions with lower T, but only by pumping heat out in some way, which is more energy to do. Being on mobile and not able to explore in depth, that quote sounds like a variant of Maxwell’s demon. It is correct to say that the Landauer limit is not due to a single physical law that must hold true, but rather a lack of knowledge about the state of the universe and the fact that acquiring that knowledge to do a “free” bitflip requires at least equivalent energy expenditure as that bitflip. TANSTAAFL.
contravariant 9y ago
From what I understand flipping a bit isn't necessarily subject to this bound. However setting it to either 1 or 0 is. Of course if you can flip it without expending energy then reading it and flipping it if it's 1 necessarily requires the Landauer limit's worth of energy.
abecedarius 9y ago
Basically, it costs free energy to forget a bit. If you flip a bit, that doesn't mean you forget it, as long as you remember you'll have to flip it back (or reinterpret it). ("Basically" because there are whole books on the subtleties of physics and information. I haven't mastered them.)
- akvadrako 9y ago
  This is the correct answer. If you don't overwrite a bit, but instead move the old value to an auxiliary bit known to be zero, that can be theoretically done for free.

kobeya 9y ago

I’m not sure how it squares with physics. I’d recommend finding that reference.

emmelaich 9y ago
Well, perhaps it's only a theoretical minimum. It agrees with kT ln 2 when T=0 right?
- kobeya 9y ago
  T is never 0.
  - emmelaich 9y ago
    Yeah I know :-) But OK here's a reference of sorts. Not by Feynman but by Bennet and Landauer. https://www.scientificamerican.com/article/the-fundamental-physical-limits-of-computation/ https://www.scientificamerican.com/article/the-fundamental-p... "there is no minimum amount of energy that must be expended in order to run a Brownian clockwork Turing machine." Also interesting but I didn't find the exact thing I was looking for: Simulating Physics with Computers Richard P. Feynman https://people.eecs.berkeley.edu/~christos/classics/Feynman.pdf https://people.eecs.berkeley.edu/~christos/classics/Feynman.... Richard Feynman and Computation https://cds.cern.ch/record/411350/files/p101.pdf https://cds.cern.ch/record/411350/files/p101.pdf
    - kobeya 9y ago
      Well for example, T cannot be less than the temperature of the cosmic microwave background. You can make regions with lower T, but only by pumping heat out in some way, which is more energy to do. Being on mobile and not able to explore in depth, that quote sounds like a variant of Maxwell’s demon. It is correct to say that the Landauer limit is not due to a single physical law that must hold true, but rather a lack of knowledge about the state of the universe and the fact that acquiring that knowledge to do a “free” bitflip requires at least equivalent energy expenditure as that bitflip. TANSTAAFL.

contravariant 9y ago

From what I understand flipping a bit isn't necessarily subject to this bound. However setting it to either 1 or 0 is. Of course if you can flip it without expending energy then reading it and flipping it if it's 1 necessarily requires the Landauer limit's worth of energy.

abecedarius 9y ago

Basically, it costs free energy to forget a bit. If you flip a bit, that doesn't mean you forget it, as long as you remember you'll have to flip it back (or reinterpret it). ("Basically" because there are whole books on the subtleties of physics and information. I haven't mastered them.)

akvadrako 9y ago
This is the correct answer. If you don't overwrite a bit, but instead move the old value to an auxiliary bit known to be zero, that can be theoretically done for free.