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Conditional work bound for finite memory reset

Physical scope

A conditional endpoint work bound for a finite memory reset with fixed retained-register marginal, a positive-temperature ideal bath, the stated entropy identification, and no endpoint energy change; no universal cost per bit or measured work is asserted.

Statement

For the reset model of Finite memory and retained-register reset model, use the exact SI value and conventions of SI quantities and unit conventions. Assume (i) the dimensionless Shannon entropies in nats from Conditional entropy of finite memory given side information give memory-register thermodynamic entropy $S_{m e m o r y + r e g i s t e r} = k_{B} H (M, R)$ in joules per kelvin at both endpoints, with $k_{B}$ in joules per kelvin, (ii) a single ideal reservoir remains at fixed temperature $T_{b} > 0$ kelvin and receives heat $Q_{b}$ in joules, (iii) the memory-register endpoint energy is unchanged so the adopted first-law convention gives work on the system $W = Q_{b}$ in joules, and (iv) the total entropy change obeys $Δ S_{m e m o r y + r e g i s t e r} + Q_{b} / T_{b} \geq 0$ in joules per kelvin. Then $W \geq k_{B} T_{b} H (M ∣ R)$ , with both sides in joules. Equality is not asserted without a separately specified reversible implementation.

Facts & Assumptions

Given: Finite sets $M, R$ , a joint law, the reset map, $T_{b} > 0$ , work positive on the system, and heat $Q_{b}$ positive into the bath.

[F1]

Finite memory and retained-register reset model: the reset makes $M^{'} = m_{0}$ and preserves the register marginal $a_{r}$ .

[F2]

Conditional entropy of finite memory given side information: actual current carrier status is draft; its inspected statement for finite joint laws gives $H (M, R) = H (R) + H (M ∣ R)$ and $H (M ∣ R) \geq 0$ . Its proof directly expands $p (m, r) = a_{r} q (m ∣ r)$ and treats zero marginals separately.

[A1]

Adopt the finite-memory entropy identification $S_{m e m o r y + r e g i s t e r} = k_{B} H (M, R)$ at each endpoint.

[A2]

Adopt an ideal fixed-temperature bath with $T_{b} > 0$ and entropy change $Q_{b} / T_{b}$ for heat delivered to it.

[A3]

Adopt unchanged endpoint energy and the convention $W = Q_{b}$ for work on the memory-register system and heat delivered to the bath.

[A4]

Adopt total entropy nondecrease for this reset process: $Δ S_{m e m o r y + r e g i s t e r} + Q_{b} / T_{b} \geq 0$ .

Proof

Given: [F1], [F2], and physical assumptions [A1]--[A4].

1.1F1F2A1algebra

Initially the memory-register entropy is $k_{B} [H (R) + H (M ∣ R)]$ by [F2] and [A1]. After reset, the memory is deterministic conditional on each register value, so $H (M^{'} ∣ R) = 0$ ; applying the same finite entropy chain rule gives final entropy $k_{B} H (R)$ .

2.1step 1.1A4algebra

Therefore $Δ S_{m e m o r y + r e g i s t e r} = - k_{B} H (M ∣ R)$ . Substitution into [A4] gives $Q_{b} \geq k_{B} T_{b} H (M ∣ R)$ because $T_{b} > 0$ .

3.1A1A2A3A4step 2.1∎

By the adopted endpoint first-law convention [A3], $W = Q_{b}$ , so $W \geq k_{B} T_{b} H (M ∣ R)$ . The conclusion is conditional on [A1]--[A4]; the entropy bound alone does not assert that the reset can be implemented reversibly or that every bit erasure costs $k_{B} T_{b} \log 2$ .

Conditional work bound for finite memory reset

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources