# Overlooked Work and Heat of Intervention and the Fate of Information
Principles of Szilard and Landauer

P.D. Gujrati  1Department of Physics, 2School of Polymer Science and
Engineering, The University of Akron, Akron, OH 44325<EMAIL_ADDRESS>
###### Abstract

We show that any external intervention (insertion or removal of a partition)
that destroys the equilibrium or brings it in a system always requires work
and heat to ensure that the first law is obeyed, a fact that has been
completely overlooked in the literature. As a consequence, there is no second
law violation. We discuss the ramifications of our finding for information
principles of Szilard and Landauer and show that no information entropy is
needed. The relevance of this result for Maxwell’s demon is also considered.

††preprint:  ††preprint: UATP/2204

Szilard is said to be the first one to exorcise Maxwell’s demon Maxwell  by
introducing the concept of the ”intervention of intelligent beings” (the demon
D) in a milestone paper Szilard  dealing with what is conventionally called a
Maxwell’s pressure-demon system Rex . The system $\Sigma$ is an ideal gas with
$N$ particles in thermal contact with a heat bath
$\widetilde{\Sigma}_{\text{h}}$ at fixed temperature $T_{0}$ (see the red
arrow representing thermal contact in Fig. 1), which D separates into volumes
$V_{\text{1}}$ and $V_{\text{2}}$ by _inserting_ a partition P at a fixed
location NEQ (see the long dashed red arrow); the partition, which is _rigid_
and _impervious_ to particle flows across it, can also be used as a piston.
The demon and P shown by a broken red vertical line forms a part of a
mechanical work source $\widetilde{\Sigma}_{\text{w}}$ Note ; see the blue
arrow for mechanical contact. The idea is to distinguish four distinct
processes (i) _intervention_ during a period $\Delta\tau_{\text{I}}$ by
inserting or removing the partition _without_ requiring any work or heat, (ii)
_informational measurement_ of determining the state of the system during
$\Delta\tau_{\text{M}}$, (iii) _utilization of measurement_ by extracting work
from the gas at the expense of heat extraction from
$\widetilde{\Sigma}_{\text{h}}$ by isothermal expansion during
$\Delta\tau_{\text{E}}$, which is the only work considered, and (iv) removal
of P from EQ and return it into $\widetilde{\Sigma}_{\text{w}}$. The demon
also passes (removes) a rod R through $\Sigma$ and P as shown when inserting
(removing) P. The engine extracts a positive work in a cycle, and violates the
second law. Szilard _postulates_ that the information in (ii) is correlated
with ”… a certain definite average entropy production…” $\Delta S_{\text{SL}}$
(SL for Szilard-Landauer) to salvage the second law. Using this, Brillouin
Brillouin0  proclaims that every physical measurement requires minimum entropy
increase to be performed.

Figure 1: The system $\Sigma$ of ideal gas with $N$ particles in a volume $V$
with pressure $P$ and temperature $T_{0}$ kept constant by a heat medium
$\widetilde{\Sigma}_{\text{h}}$. The demon D is part of a work medium
$\widetilde{\Sigma}_{\text{w}}$ and manipulates the partition P for
intervention by inserting it at NEQ (which divides $\Sigma$ into $\Sigma_{1}$
on its left and $\Sigma_{2}$ on its right) and removing from EQ as described
in the text, the latter also defines an imaginary wall W; see the text. The
rod R and P move _together_ when connected. The rod R is used to identify the
state from outside the system after intervention.

Landauer argues that $\Delta S_{\text{SL}}$ is instead due to resetting
Landauer ; Plenio ; Frank ; Lent ; Klaers . Chambadal Chambadal  later argues
that entropy and information are not involved in (i) and (ii). Penrose Penrose
emphasizes fluctuations to generate work and create a perpetual machine. These
attempts have inspired a highly active informational approach to
thermodynamics that is still undergoing rapid growth Mallick ; Avery ; Wolpert
.

Although Szilard’s treatment is completely classical, there are recent quantum
mechanicanical treatments; see for example Zurek ; Ueda ; Serra . However,
there are still contentious issues in treating information as thermodynamic
entropy and not all agree with the significance of information for entropy and
work. A nice discussion of these issues and various suggestions is given by
Leff and Rex Rex . It has been pointed out Kish  that relocation of the
partition (not the insertion itself) must be carefully accounted to avoid any
violation of the second law or that a deeper connection exists between
thermodynamic efficiency and irrelevant information Still . Thus, the
controversy and confusion persist Norton ; Bub ; Bennett0 . Despite this,
information principles of Szilard and Landauer have been visionary in creating
the modern field of cybernetics and artificial intelligence in which
information is processed by machines.

Among all the confusion, the omission of the work and heat of intervention
until now seems most disturbing as they most probably contribute not only to
the net work and heat in a complete cycle but are also necessarily required by
the first law. Unfortunately, this law has not received any attention to date
as the entire field has been mired by the need to satisfy the second law
alone. Indeed to the best of our knowledge, almost all researchers including
Szilard and Landauer believe that classically, the intervention neither
involves heat nor work. This is quite surprising as any possible entropy
change $\Delta S_{\text{I}}$ due to intervention results in a directly
measurable heat transfer $\Delta Q_{\text{I}}$ that plays an important role,
and requires some accompanying work $\Delta W_{\text{I}}$ to satisfy the first
law. But $\Delta S_{\text{I}}$ must be _nonzero_ if the intervention desturbs
the state by destroying the equilibrium state $\mathfrak{M}_{\text{eq}}$ or
bringing it back. Then it must play a pivotal role in not only settling the
violation of the second law but also satisfying the first law. This also
suggests that almost all earlier attempts to satisfy the second law will need
reassessment.

Therefore, there is an urgent need to carefully look at the process of
intervention in (i) and the role of information in (ii). We discuss a many-
particle system to make a general argument about intervention for an arbitrary
system. Our main conclusion is in the form of the following theorem that
intervention by itself usually requires nonzero $\Delta S_{\text{I}}$, $\Delta
Q_{\text{I}}$, and $\Delta W_{\text{I}}$, which so far have been overlooked,
and have profound implications for the underpinnings of the entire field of
information theory.

Intervention Theorem: Any act of intervention such as insertion (Szilard) or
removal (Landauer) of a partition that destroys $M_{\text{eq}}$ or brings it
back in an interacting system always results in heat that can be externally
measured and satisfies the first law to determine work.

Proof: We consider a prefixed point NEQ that divides $\Sigma$ into fixed
volumes $V_{1}$ on its left and $V_{2}$ on its right. Initially, the gas is in
$\mathfrak{M}_{\text{eq}}$. At time $t=0$, D moves P from
$\widetilde{\Sigma}_{\text{w}}$ and inserts it at NEQ (different from EQ)
along with R. The partition divides $N$ _randomly_ into $N_{1}\geq 0$ and
$N_{2}=N-N_{1}\geq 0$. Different attempts to put P at NEQ will result in
different values of $N_{1}$ and $N_{2}$. The two gases, each in equilibrium at
some time $\tau_{\text{I}}$, are not in equilibrium with each other; this
state is denoted by $\mathfrak{M}_{\text{neq}}$. We give two independent
methods to find $\Delta W_{\text{I}}$.

(a) Let $\Delta
S_{\text{I}}=S(\mathfrak{M}_{\text{neq}})-S(\mathfrak{M}_{\text{eq}})$ be the
entropy change. Associated with this is the externally measured heat $\Delta
Q_{\text{I}}$. From the first law applied to the intervention in (i), the
corresponding work is $\Delta W_{\text{I}}=dQ_{\text{I}}-dE_{\text{I}}$.

(b) We consider the imaginary wall W at EQ in $\mathfrak{M}_{\text{eq}}$ so
that $N_{1}$ particles are in the volume $\bar{V}_{1}$ to its left and $N_{2}$
particles in $\bar{V}_{2}$ to its right, and $P$ is the pressure on both
sides. The entropy is $S(\mathfrak{M}_{\text{eq}})$. We treat W as an
imaginary piston P′ (R is not important in this process) that moves reversibly
from EQ to NEQ so that $\bar{V}_{1}\rightarrow V_{1}$, and simultaneously
$\bar{V}_{2}\rightarrow V_{2}$ to reproduce $\mathfrak{M}_{\text{neq}}$ in
(a). As $\mathfrak{M}_{\text{neq}}$ has been reproduced, it is irrelevant how
it is generated as $\Delta S_{\text{I}}$ and $\Delta E_{\text{I}}$ are the
same. If P′ is moved so that $\Delta Q_{\text{I}}$ is the same as in (a), then
$\Delta W_{\text{I}}$ is also the same as above. As the two gases in
$\bar{V}_{1}$ and $\bar{V}_{2}$ are in equilibrium with each other, D can
replace W by inserting P (see the reverse of the dotted short red arrow now
pointing to EQ) there. During this insertion at EQ, $\Delta
W_{\text{I}}^{\prime}=\Delta Q_{\text{I}}^{\prime}=0$. Now, D can use P in
place of P′ to move it to NEQ to yield the same $\Delta W_{\text{I}},\Delta
S_{\text{I}}$, and $\Delta Q_{\text{I}}$.

During removal of P from NEQ, D changes $\mathfrak{M}_{\text{neq}}$ to
$\mathfrak{M}_{\text{eq}}$ so that all $\Delta S_{\text{I}},\Delta
Q_{\text{I}}$, and $\Delta W_{\text{I}}$ are negative of the above quantities
for insertion. This proves the theorem.

Before proceeding further, we briefly use the above new understanding of the
insertion of P in a one-particle Szilard engine (Fig. 1 with
$N_{1}=N=1,N_{2}=0,V_{1}=V_{2}=V/2$). Before insertion, the probabilities are
$p_{1}=p_{2}=1/2$ of the particle being in $\Sigma_{1}$ and $\Sigma_{2}$,
respectively. After insertion, we have either $p_{1}=1,p_{2}=0$ (state $0$ or
$\mathfrak{M}_{10}$) or $p_{1}=0,p_{2}=1$ (state $1$ or $\mathfrak{M}_{01}$).
This results in $\Delta S_{\text{I}}=-\ln 2\Rightarrow\Delta
Q_{\text{I}}=-T_{0}\ln 2$ (externally measurable) due to insertion alone in
(i) so $\Delta W_{\text{I}}=-T_{0}\ln 2$ from $\Delta E_{\text{I}}=\Delta
Q_{\text{I}}-\Delta W_{\text{I}}=$ $0$. Thus, $\left|\Delta
Q_{\text{I}}\right|$ is given out to $\widetilde{\Sigma}_{\text{h}}$ and
$\left|\Delta W_{\text{I}}\right|$ to $\widetilde{\Sigma}_{\text{w}}$, i.e.,
D. All this is in accordance with the above theorem. This first law for the
intervention in (i) has never been used to date to draw this conclusion.

Work and Heat of Intervention: The above proof is for a general system but we
now determine these quantities for an ideal gas that is assumed to always have
the same temperature $T_{0}$. The pressure $P$ in $\mathfrak{M}_{\text{eq}}$
satisfies $PV=NT_{0}$. The pressure of the two gases in
$\mathfrak{M}_{\text{neq}}$ at time $\tau_{\text{I}}$ are
$P_{1}=N_{1}T_{0}/V_{1}$ and $P_{2}=N_{2}T_{0}/V_{2}$, respectively. We now
show that

$\Delta W_{\text{I}}=T_{0}[N\ln P-N_{1}\ln P_{1}-N_{2}\ln P_{2}]\leq 0.$ (1)

We first consider (a). Initially, the entropy is
$S(\mathfrak{M}_{\text{eq}})=N[g(T_{0})-\ln P]$, where $g(T_{0})$ is a
constant function of constant $T_{0}$ Landau  and plays no role in our
discussion so we will omit it from now on. Moreover, to simplify the
discussion, we will restrict ourselves to reversible processes. The entropy
after insertion is given by $S(\mathfrak{M}_{\text{neq}})=-N_{1}\ln
P_{1}-N_{2}\ln P_{2}$. Thus,

$\Delta S_{\text{I}}=N\ln P-N_{1}\ln P_{1}-N_{2}\ln P_{2}\leq 0,$ (2)

the equality (no entropy reduction) occurs if $\Sigma$ is in equilibrium
($\Delta P=P_{1}-P_{2}=0$), which happens when NEQ and EQ are the same point.
From $\Delta S_{\text{I}}$, we obtain $\Delta Q_{\text{I}}=T_{0}\Delta
S_{\text{I}}\leq 0$ so that $\Delta W_{\text{I}}=\Delta Q_{\text{I}}$ from the
first law, which proves Eq. (1). The negative values of $\Delta Q_{\text{I}}$
and $\Delta W_{\text{I}}$ mean that heat is ejected to
$\widetilde{\Sigma}_{\text{h}}$ and the work is done by D or
$\widetilde{\Sigma}_{\text{w}}$. During the removal of P by D from NEQ, gases
will come to equilibrium with a concomitant entropy gain $\left|\Delta
S_{\text{I}}\right|$ so heat $\left|\Delta Q_{\text{I}}\right|$ will be
absorbed and work $\left|\Delta W_{\text{I}}\right|$ done by the gas on D. We
obtain the same values if we follow (b).

The intervention does not require $\Delta S_{\text{I}},\Delta Q_{\text{I}}$,
and $\Delta W_{\text{I}}$ if it does not destroy $\mathfrak{M}_{\text{eq}}$
($\Delta P=0$); otherwise they are nonzero but always satisfy the first law.
We have only considered reversible processes as our goal is to merely justify
their existence. It should be noted that P is impermeable to particle flow so
once $N_{1}$ is determined, it cannot change. In this regard, our treatment is
different from that in the quantum version of Szilard’s engine Ueda . Our
approach is in keeping with Szilard-Landauer idea that $\mathfrak{M}_{10}$ and
$\mathfrak{M}_{01}$ for $N=1$ are distinct states.

Measurement and Information: We now follow the consequences of the new
understanding of the intervention for the measurement process. We treat P-R as
a movable piston inside $\Sigma$. Before the intervention, we know
$V_{1},V_{2},P,T_{0}$, and $N$. After intervention, we need to know internal
quantities $N_{1},N_{2}$ and $P_{1},P_{2}$ for a complete thermodynamic
description. We now show that thermodynamics alone is sufficient to determine
them so D does not need to _actively_ ”gather” any information. The
_spontaneous_ motion of P is always towards equilibrium at W during which
$\Delta P\rightarrow 0$, which D easily determines by simply observing the
direction R moves from the outside without making any mechanical contact with
$\Sigma$ so no work is involved: if moving to the right, $\Delta P>0$ and
$\bar{V}_{1}>V_{1},\bar{V}_{2}<V_{2}$; if to the left, $\Delta P<0$ and
$\bar{V}_{1}<V_{1},\bar{V}_{2}>V_{2}$. As soon as R moves, D may attach a
weight on its opposite end to perform external work. We will not consider any
weight as our interest is to show what unknown quantities are
thermodynamically determined. Observing W where P stops from the outside, D
determines $\bar{V}_{1}$ and $\bar{V}_{2}$, again without any work involved.
The ideal gas equation gives

$N_{1}=N\frac{\bar{V}_{1}}{V},N_{2}=N\frac{\bar{V}_{2}}{V},P_{1}=P\frac{\bar{V}_{1}}{V_{1}},P_{2}=P\frac{\bar{V}_{2}}{V_{2}}.$
(3)

Thus, we have verified that the state of the system is completely determined
thermodynamically; there is no need for D to obtain any internal information
that requires any work by mechanical contacts with $\Sigma$. As a consequence,
$\Delta W_{\text{M}}=\Delta Q_{\text{M}}=\Delta E_{\text{M}}=0$. The situation
with Szilard’s engine is even simpler; see below.

Isothermal Expansion: The positive work done by the gas in the isothermal
expansion in (iii) from NEQ (P) to EQ (W) after insertion is precisely the
magnitude $\Delta W_{\text{E}}=\left|\Delta W_{\text{I}}\right|$ on
$\widetilde{\Sigma}_{\text{w}}$, and the same amount of positive heat $\Delta
Q_{\text{E}}=\left|\Delta Q_{\text{I}}\right|$ is taken from
$\widetilde{\Sigma}_{\text{h}}$ (recall that $\Delta W_{\text{E}}$ is the only
work considered by Szilard-Landauer). They satisfy the first law with $\Delta
E_{\text{E}}=0$. As the system is at equilibrium at W, D now takes out P (by
separating from R) from EQ and brings it back into
$\widetilde{\Sigma}_{\text{w}}$ without involving any work or heat. This
completes the isothermal cycle during which the net change $\Delta E=\Delta
E_{\text{I}}+\Delta E_{\text{M}}+\Delta E_{\text{E}}=0$. The conclusion is the
same if P is removed; see later.

Information Entropy: However, remarkably, by recognizing nonzero $\Delta
W_{\text{I}}=\Delta Q_{\text{I}}$ during the intervention, not only the net
heat $\Delta Q$ taken from $\widetilde{\Sigma}_{\text{h}}$ but also the net
work $\Delta W$ done by the gas on $\widetilde{\Sigma}_{\text{w}}$ vanish
during an entire cycle so no change has occurred either in
$\Sigma,\widetilde{\Sigma}_{\text{h}}$, and $\widetilde{\Sigma}_{\text{w}}$
(including D). The second law is not challenged. As the entropy of the
universe has not changed, there is no need for Szilard-Landauer information
conjecture $\Delta S_{\text{SL}}>0$ due to the measurement. As we have assumed
a reversible cycle, it is not clear why this entropy generation is needed. It
is clear that the physical necessity of $\Delta S_{\text{I}},\Delta
Q_{\text{I}}$, and $\Delta W_{\text{I}}$ ensures that the second law is not
violated so the need for $\Delta S_{\text{SL}}>0$ loses any physical
significance from the way they have been justified so far.

As no net work is generated per cycle, the process cannot be considered as
forming an engine that can produce any work. It is merely performing a
conventional reversible isothermal (isoenergetic) cycle.

A word of caution. The new value of $N_{1}$ in the state
$\mathfrak{M}_{N_{1}N_{2}}$ of $\Sigma$ in the new cycle is again a random
number and need not be the previous value, just as in Szilard’s engine. We
follow the same set of processes as described above in the previous cycle.
Again, there will be no violation of the second law for the new state
$\mathfrak{M}_{N_{1}N_{2}}$.

The case $\mathfrak{M}_{N0}$ ($N_{1}=N,N_{2}=0$) with $P_{1}=PV/V_{1},P_{2}=0$
or $\mathfrak{M}_{0N}$ ($N_{1}=0,N_{2}=N$) with $P_{1}=0,P_{2}=PV/V_{2}$ is
very similar to the case considered by Szilard with the states
$\mathfrak{M}_{10}$ and $\mathfrak{M}_{01}$. We also see that the case is very
exceptional in that it is one of the most improbable fluctuations in $\Sigma$.
In this particular case, once D sees R move in a certain direction, the state
is immediately known to determine $P_{1}=0,P_{2}$. There is no need to follow
the entire motion of P to EQ. This step is also not needed in Szilard’s engine
that we now discuss.

Szilard’s Engine: For $N=1$, D lifts P from $\widetilde{\Sigma}_{\text{w}}$
and inserts it at NEQ to obtain the states $\mathfrak{M}_{10}$ and
$\mathfrak{M}_{01}$. For $\mathfrak{M}_{10}$ ($\mathfrak{M}_{01}$) with
$P_{1}=PV/V_{1},P_{2}=0$ (with $P_{1}=0,P_{2}=PV/V_{2}$), and W at the right
(left) end of $\Sigma$, we have $\Delta S_{\text{I}}=-\ln(V/V_{1}),\Delta
Q_{\text{I}}=-T_{0}\ln(V/V_{1})$ and $\Delta W_{\text{I}}=\Delta
Q_{\text{I}}$. The state is uniquely determined by how R moves; see the
discussion of $\mathfrak{M}_{N0}$ and $\mathfrak{M}_{0N}$ above. Thus, D knows
precisely whether the particle is in $\mathfrak{M}_{10}$ or
$\mathfrak{M}_{01}$ without having any mechanical contact (blue thick arrow)
with $\Sigma$ so $\Delta W_{\text{M}}=0$ due to this information. As D does
not have a thermal contact (red thick arrow), we also have $\Delta
S_{\text{M}}=0$ and $\Delta Q_{\text{M}}=0$, which refutes the conjecture
$\Delta S_{\text{SL}}=\ln(V/V_{1})$ . For the particle in $\mathfrak{M}_{10}$,
P stops at W on the right end of $\Sigma$ and we have $\Delta
S_{\text{E}}=\ln(V/V_{1}),\Delta Q_{\text{E}}=T_{0}\ln(V/V_{1})$ and $\Delta
W_{\text{E}}=\Delta Q_{\text{E}}$; same for the particle in
$\mathfrak{M}_{10}$. The system is in equilibrium once P arrives at W, where D
removes it and puts it back in $\widetilde{\Sigma}_{\text{w}}$ for the next
cycle without any work and heat. Obviously, no net work and heat are extracted
from the system during each cycle. Hence, there is neither any violation of
the second law nor $\Sigma$ performs as an engine.

Landauer’s Resetting: The distinct state $0\doteq\mathfrak{M}_{10}$ or
$1\doteq$ $\mathfrak{M}_{01}$ (identified by the motion of R without any
mechanical contact) interact with $\widetilde{\Sigma}_{\text{h}}$ and
$\widetilde{\Sigma}_{\text{w}}$. We take $V_{1}=V_{2}=V/2$. The demon starts
with $0$ and $1$ and always produces the same state, say $0$, by first
removing P from NEQ and putting in $\widetilde{\Sigma}_{\text{w}}$. As each
state is a nonequilibrium state before the intervention, the removal increases
the entropy $\Delta S_{\text{I}}=\ln 2$ so $\Sigma$ has positive heat $\Delta
Q_{\text{I}}=T_{0}\ln 2$ from $\widetilde{\Sigma}_{\text{h}}$ and positive
work $\Delta W_{\text{I}}=T_{0}\ln 2$ on $\widetilde{\Sigma}_{\text{w}}$ or D,
and the first law $\Delta E_{\text{I}}=0$ remains satisfied. These quantities
are never discussed for the intervention in the derivation of Landauer’s
principle. After the gas equilibrates in $\Sigma$, D picks P in
$\widetilde{\Sigma}_{\text{w}}$ and inserts it at the far right of $\Sigma$.
The insertion does not require any work or heat; see the Insertion Theorem so
$\Delta S_{\text{I}}^{\prime}=0$ and $\Delta Q_{\text{I}}^{\prime}=0=\Delta
W_{\text{I}}^{\prime}$. Now the gas is isothermally compressed (entropy loss
$\Delta S_{\text{E}}=-\ln 2$) to $V_{1}$ to create the state $0$ during which
D or $\widetilde{\Sigma}_{\text{w}}$ performs work so $\Delta
W_{\text{E}}=-T_{0}\ln 2$ and $\widetilde{\Sigma}_{\text{h}}$ receives heat so
$\Delta Q_{\text{E}}=-T_{0}\ln 2$. The particle always ends in the state $0$,
regardless of whether it was initially in the state $0$ or $1$. The entire
process is called ” _reset_ to $0$,” and is considered a logically
irreversible operation. However, net entropy change is $\Delta S=\Delta
S_{\text{I}}+\Delta S_{\text{I}}^{\prime}+\Delta S_{\text{E}}=0$ so the reset
is a thermodynamically reversible operation with no net work done $\Delta W$
and heat gained $\Delta Q$ by the gas as was the case for Szilard’s engine. As
the net heat loss by $\widetilde{\Sigma}_{\text{h}}$ is also zero, there is a
contradiction with the dissipation principle of Landauer. The entire logical
operation is thermodynamically reversible with zero reset cost. The same
reasoning also applies to the general set up with $N$ paritcles with the same
conclusion for inserting P above.

Maxwell’s Demon: While it is commonly asserted that Szilard’s information
approach is designed to exorcise Maxwell’s demon, there is really no
connection between the two. Szilard’s pressure demon always remains outside
the system as part of $\widetilde{\Sigma}_{\text{w}}$ and does work on the
system to reduce its entropy without any violation of the second law during
$\Delta\tau_{\text{I}}$. Maxwell’s demon is an internal part of $\Sigma$ that
is isolated Rex , and is claimed to internally create a temperature difference
$\Delta T$ and a corresponding entropy reduction $\Delta S_{\text{sort}}$ by
sorting and then allowing only slow ($v_{\text{s}}<\bar{v}$) and fast
($v_{\text{f}}<\bar{v}$) particles across a hole; here $\bar{v}$ is the
average speed of energy $3T_{0}/2$. However, $\Delta S_{\text{sort}}<0$ will
invalidate the second law. The sorting, i.e., observing fast and slow
particles relative to $\bar{v}$ occurs prior to the intervention of opening
the hole to let them through in different parts $\Sigma_{1}$ and $\Sigma_{2}$
followed by closing it in succession. This order is reverse of that considered
by Szilard and Landauer and causes a major difference. As D gains information
about $\bar{v},v_{\text{s}}$, and $v_{\text{f}}$ of incoming particles, there
is an information entropy production $\Delta s_{\text{SL}}$ per observation.
It is followed by interventions only for fast and slow particles. However, the
effects of the two successive parts (opening and closing) of each intervention
cancel each other out so the combined changes are $\Delta S_{\text{I}}=0$ and
$\Delta Q_{\text{I}}=0=\Delta W_{\text{I}}$ per intervention (particle
transfer). This scenario does not seem right as D observes particles with
$\bar{v}$ lot more often than those with $v_{\text{s}}$ and $v_{\text{f}}$ so
$\Delta S_{\text{SL}}$ is mostly determined by observing particles with
$\bar{v}$ and will far exceed the entropy reduction $\Delta S_{\text{sort}}$
due to particle transfers; the idea may seem similar to that proposed by Still
Still  but is not identical Note1 . Indeed, our recent investigation Gujrati-
NEQ-MD  shows that the Maxwell’s demon cannot destroy equilibrium so $\Delta
T=0\Longrightarrow\Delta S_{\text{sort}}=0$.

In summary, if anything, the current information theories are formulated by
not satisfying the first law during $\Delta\tau_{\text{I}}$, an observation
that has neither ever been made before nor its consequences discussed. Once
$\Delta S_{\text{I}},\Delta Q_{\text{I}}$, and $\Delta W_{\text{I}}$ due to
intervention are included to satisfy the first law, there is no support in the
setup considered for the conjectures by Szilard and Landauer for any
information entropy $\Delta S_{\text{SL}}$. Thermodynamics is sufficient to
yield all the information externally without any need to probe the interior of
the system.

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  * (26) Observing particles of speed $\bar{v}$ is crucial to not only identify $v_{\text{s}}$, and $v_{\text{f}}$ for passage but also to block particles of speed $\bar{v}$. Thus, this observation is not _irrelevant_ as defined by Still Still .
  * (27) P.D. Gujrati, arXiv:2112.12300, arXiv:2205.02313.