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Thermodynamics of Information Processing

20.1 Landauer Bound in Practice

The minimum energy dissipation per irreversible bit operation depends on the physical implementation:

  • CMOS transistor (2000s-era): 104kBT\sim 10^4\,k_BT per switch (vastly above the Landauer limit)
  • Modern CMOS (7 nm node): 102\sim 10^2103kBT10^3\,k_BT per switch
  • Adiabatic / reversible logic proposals: 1\sim 110kBT10\,k_BT per operation (approaching the limit)

The gap between theory (kBTln20.018k_BT\ln 2 \approx 0.018 eV at 300 K) and practice (1\sim 11010 fJ per switch) spans 5—6 orders of magnitude. Closing this gap requires fundamentally different computing paradigms.

20.2 Bennett”s Clock and Reversible Computing

Bennett (1982) showed that a computer can be made logically reversible at every step if it never erases information. Such a computer dissipates energy only during the initialisation of bits and during optional output, not during computation.

A logically reversible computation can be embedded in a thermodynamically reversible process by driving the system slowly enough that it remains near equilibrium at all times. The energy cost is then:

E=0τFλ(t)λ˙(t)dtE = \int_0^\tau \frac{\partial F}{\partial \lambda(t)}\,\dot{\lambda}(t)\,dt

For a quasi-static process: EΔFE \to \Delta F (minimum possible).

Fredkin and Toffoli gates are examples of logically reversible logic gates. Any computation can be made reversible by saving all intermediate results and running the computation in reverse to restore the input tape.

Worked Examples

Example 1: Carnot efficiency

Problem. A Carnot engine operates between 500K500 \mathrm{ K} and 300K300 \mathrm{ K}. Find the maximum efficiency.

Solution. η=1TC/TH=1300/500=10.6=0.4=40%\eta = 1 - T_C/T_H = 1 - 300/500 = 1 - 0.6 = 0.4 = 40\%.

\blacksquare

Example 2: Entropy change

Problem. Find the entropy change when 2mol2 \mathrm{ mol} of ice at 0°C0°\mathrm{C} melts (ΔHfus=6.01kJ/mol\Delta H_{\text{fus}} = 6.01 \mathrm{ kJ/mol}).

Solution. ΔS=QT=nΔHfusT=2×6010273=44.0J/K\Delta S = \frac{Q}{T} = \frac{n \Delta H_{\text{fus}}}{T} = \frac{2 \times 6010}{273} = 44.0 \mathrm{ J/K}.

\blacksquare

Summary

  • First law: ΔU=QW\Delta U = Q - W; conservation of energy.
  • Second law: ΔSuniverse0\Delta S_{\text{universe}} \geq 0; entropy of an isolated system never decreases.
  • Carnot efficiency: η=1TC/TH\eta = 1 - T_C/T_H (maximum possible for given temperatures).
  • Statistical mechanics: S=kBlnΩS = k_B \ln \Omega; Boltzmann distribution: pieEi/(kBT)p_i \propto e^{-E_i/(k_BT)}.

Cross-References

TopicSiteLink
[Thermal Physics]A-LevelView
[Thermal Physics]IBView
[Thermal Physics]DSEView
[Thermal Physics]UniversityView