Thermodynamics of Information Part Icsc.ucdavis.edu/~chaos/courses/poci/Lectures/Lecture37a...Energy and Entropy According to the ﬁrst law and second law of thermodynamics together,

Thermodynamics of InformationPart IAlec Boyd

1

Thermal Reservoir

Information Engine

Q MassW

Information Reservoir

I

What does information do?

H(X|Y ) I(X : Y ) H(Y |X)

H(X) H(Y ) How do we use the stuffin these bubbles?

Thermodynamics treats physical systems many elements.

Relevant quantities:1) Energy

Thermodynamics

Th

Tc

W

Qh

Qc


Relevant quantities:1) Energy2) Work

Thermodynamics

Th

Tc

W

Qh

Qc


Relevant quantities:1) Energy2) Work3) Heat

Thermodynamics

Th

Tc

W

Qh

Qc


Relevant quantities:1) Energy2) Work3) Heat4) Temperature

Thermodynamics

Th

Tc

W

Qh

Qc


Relevant quantities:1) Energy2) Work3) Heat4) Temperature5) Entropy

Thermodynamics

Th

Tc

W

Qh

Qc

Physical system states:

Thermodynamic Entropy

X = {x}


Probability of physical state at time :


X = {x}

t

Pr(Xt = x)



Thermodynamic entropy (Gibbs) of physical system:


X = {x}

t

Pr(Xt = x)

S[Xt

] = �k

B

X

x2XPr(X

t

= x) lnPr(Xt

= x)




Shannon entropy of physical system:


X = {x}

t

Pr(Xt = x)

S[Xt

] = �k

B

X

x2XPr(X

t

= x) lnPr(Xt

= x)

H[Xt

] = �X

x2XPr(X

t

= x) log2 Pr(Xt

= x)




Shannon entropy of physical system:


X = {x}

t

Pr(Xt = x)

S[Xt

] = �k

B

X

x2XPr(X

t

= x) lnPr(Xt

= x)

H[Xt

] = �X

x2XPr(X

t

= x) log2 Pr(Xt

= x)

S[Xt] = kB ln 2H[Xt]

Entropy Comparison

Thermodynamic entropy as a measure of disorder.

Disordered (Hot Gas) Ordered (Cold Solid)

Entropy Comparison



Shannon entropy as a measure of uncertainty or unpredictability.

Entropy Comparison



Shannon entropy as a measure of uncertainty or unpredictability.

String of measurements of unpredictable variable

String of measurements of predictable variable

...110111011001011110000000001... ...1010101010101010101010101...

Physical states described by positions and their conjugate momenta :

Classical Physics

qipi

X = {x} = {(q1, q2, ..., qN , p1, p2, ..., pN )}


Can be thought of as a collection of particles:

Classical Physics

qipi

X = {x} = {(q1, q2, ..., qN , p1, p2, ..., pN )}

(qi, qi+1, qi+2)

(pi, pi+1, pi+2)


Can be thought of as a collection of particles:

How does this change with time?

Classical Physics

qipi

X = {x} = {(q1, q2, ..., qN , p1, p2, ..., pN )}

(qi, qi+1, qi+2)

(pi, pi+1, pi+2)

First Law of Thermodynamics

Energy of each state specified by Hamiltonian function of state

E(x) = H(~q, ~p)



Position changes as velocity and momentum changes as the force:

E(x) = H(~q, ~p)

dqidt

= @piH(~q, ~p)dpidt

= �@qiH(~q, ~p)



Position changes as velocity and momentum changes as the force:

Equations of motion conserve energy

E(x) = H(~q, ~p)

dqidt

= @piH(~q, ~p)dpidt

= �@qiH(~q, ~p)

�t!t+⌧E = E(x(t+ ⌧))� E(x(t)) = 0

dE(x(t))

dt

= 0

Entropy increases in isolated systems:

Second Law of Thermodynamics

dS[Xt]

dt� 0

dH[Xt]

dt� 0



dS[Xt]

dt� 0

dH[Xt]

dt� 0

Ordered (Hot Solid)



dS[Xt]

dt� 0

dH[Xt]

dt� 0

Disordered (Hot Gas) Ordered (Hot Solid)


Example another and another and another


dS[Xt]

dt� 0

dH[Xt]

dt� 0

Disordered (Hot Gas) Ordered (Hot Solid)

https://www.youtube.com/watch?v=v30b5IAgwQw

https://gfycat.com/gifs/detail/scarcefemininelcont

https://youtu.be/cJyGoGPXTj4?t=159

https://i.imgur.com/Q8jSySp.gifv

Energy and Entropy

According to the first law and second law of thermodynamics together, the microstates evolve towards the maximum entropy distribution over a fixed energy , called the microcanonical ensemble:

E

Energy and Entropy


where is the number of states with energy .

E

Pr(Xeq = x) = �

E(x),E/⌦(E)

E

⌦(E)

Energy and Entropy


where is the number of states with energy .

This leads to an equilibrium estimate of entropy:

E

Pr(Xeq = x) = �

E(x),E/⌦(E)

E

⌦(E)

S[Xeq|E(x) = E

0] = kB ln⌦(E0)

Energy/Entropy vs. Temperature

What is temperature?-Average energy?



Temperature of a system in equilibrium:

1

T

=@S[Xeq|E(x) = E

0]

@E

0




System Q

1

T

=@S[Xeq|E(x) = E

0]

@E

0

�Ssys =Q

Tsys




In most systems, every degree of freedom stores energy proportional to the temperature:

System Q

1

T

=@S[Xeq|E(x) = E

0]

@E

0

hEqii ⇡ hEpii ⇡kBT

2

�Ssys =Q

Tsys

Heat Flow

Entropy explains why heat diffuses from hot to cold:

Q THTC

TH > TC

Heat Flow


Total entropy increases by heat diffusion:

Q THTC

�Stotal

= Q/TC �Q/TH > 0

TH > TC

Heat Flow



The reverse is impossible:

Q THTC

Q THTC

�Stotal

= Q/TC �Q/TH > 0

TH > TC

Heat Flow



The reverse is impossible:

Q THTC

Q THTC

�Stotal

= Q/TC �Q/TH > 0

�Stotal

= Q/TH �Q/TC < 0

TH > TC

Heat Engine

Some energy can be redirected to produce work

Heat Engine


Th

Tc

W

Qh

Qc

Heat Engine


Th

Tc

W

Qh

Qc Qc + W = Qh

Heat Engine


Entropy increase constrains efficiency

Th

Tc

W

Qh

Qc Qc + W = Qh

4Stotal

= Qc

/Tc

�Qh

/Th

� 0

Heat Engine



Th

Tc

W

Qh

Qc Qc + W = Qh

4Stotal

= Qc

/Tc

�Qh

/Th

� 0Qc

Qh� Tc

Th

Heat Engine



Th

Tc

W

Qh

Qc Qc + W = Qh

e =output

input

=

W

Qh= 1� Qc

Qh

Heat Engine



Th

Tc

W

Qh

Qc Qc + W = Qh

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Thermodynamics of Control

How do we extract work from a thermodynamic system?



!Thermal

Reservoir MassWQ

Feedback/Control

X



Through control of the energy landscape

dhEi =X

x

Pr(Xt

= x)dE(x) +X

x

E(x)dPr(Xt

= x)

!Thermal

Reservoir MassWQ

Feedback/Control

X




dhEi =X

x

Pr(Xt

= x)dE(x) +X

x

E(x)dPr(Xt

= x)

= hdW i+ hdQi

!Thermal

Reservoir MassWQ

Feedback/Control

X




We consider control of a system which interacts with an ideal thermal reservoir at temperature to produce work

dhEi =X

x

Pr(Xt

= x)dE(x) +X

x

E(x)dPr(Xt

= x)

= hdW i+ hdQi

T

!Thermal

Reservoir MassWQ

Feedback/Control

X

X

X

R

Entropy of Control

The total entropy production of the joint reservoir system

!Thermal

Reservoir Mass

Feedback/Control

X

�Stotal

= �S[R] +�S[X]��I[X;R]

WQ

Entropy of Control


The fact that the reservoir is ideal means it has no memory of the ratchet, and so is uncorrelated with it:

!Thermal

Reservoir Mass

Feedback/Control

X

�Stotal

= �S[R] +�S[X]��I[X;R]

I[X;R] = 0

WQ

Entropy of Control



Heat production in control is bounded by change:

!Thermal

Reservoir Mass

Feedback/Control

X

�Stotal

= �S[R] +�S[X]��I[X;R]

I[X;R] = 0

hQiT

+�S[X] = �Stotal

� 0

WQ

Entropy of Control



Heat production in control is bounded by change:

!Thermal

Reservoir Mass

Feedback/Control

X

�Stotal

= �S[R] +�S[X]��I[X;R]

I[X;R] = 0

hQiT

+�S[X] = �Stotal

� 0

WQ

hQi � �T�S[X]

Entropy of Control

This can be re-expressed with work

!Thermal

Reservoir Mass

Feedback/Control

X WQ

T�Stotal

= hW i ��hE[X]i+ T�S[X]

Entropy of Control


This can be interpreted as dissipated work

!Thermal

Reservoir Mass

Feedback/Control

X WQ

hWdissi = hW i ��FNEQ

T�Stotal


Entropy of Control


This can be interpreted as dissipated work

This includes a new quantity: the non-equilibrium free energy, which specifies the amount of work that could have been harvested:

!Thermal

Reservoir Mass

Feedback/Control

X WQ

hWdissi = hW i ��FNEQ

FNEQ = hE[X]i � TS[X]

T�Stotal


Stepping Back

What does entropy do?

Stepping Back


1) Bounds function of heat engine

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Stepping Back


1) Bounds function of heat engine with Carnot efficiency

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Stepping Back



2) Bounds work and heat production in controlled thermodynamic systems

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Stepping Back




hQi � �kBT ln 2�H[X]

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Stepping Back





hW i � h�E[X]i � kBT ln 2�H[X]

e 1� Tc

Th

e =output

input

=

W

Qh= 1� Qc

Qh

Control Example

Single particle in a box

Control Example

Single particle in a box, insert barrier

Control Example

Single particle in a box, insert barrier, slide barrier away from particle

Control Example

Single particle in a box, insert barrier, slide barrier away from particle, while in contact with heat reservoir at temperature T

Thermal Reservoir Q

Control Example

Single particle in a box, insert barrier, slide barrier away from particle, while in contact with heat reservoir at temperature , and extract work with the piston.T

Thermal Reservoir Q MassW

Control Example

Single particle in a box, insert barrier, slide barrier away from particle, while in contact with heat reservoir at temperature , and extract work with the piston.

How much much work and heat flow?

T

Thermal Reservoir Q MassW


hW i � h�E[X]i � kBT ln 2�H[X]

Control Example

Initial vs. final energies and entropies:

Control Example


Energy of each position is the same, so average energy of positions is the same: hE~qiinitial = hE~qifinal

Control Example


Energy of each position is the same, so average energy of positions is the same:If in equilibrium, average kinetic energy of momentum variable is the same:

hE~qiinitial = hE~qifinal

hE~piinitial = hE~pifinal =kBT

2= hKinetic Energyi

Control Example


Energy of each position is the same, so average energy of positions is the same:If in equilibrium, average kinetic energy of momentum variable is the same:

Average energy doesn’t change:

hE~qiinitial = hE~qifinal

hE~piinitial = hE~pifinal =kBT

2= hKinetic Energyih�E[X]i = 0

Control Example


Vinitial

Vfinal

Control Example


Probability of each position is uniform and inversely proportional to the available volume:

Pr( ~Qfinal = ~q) =c

VfinalPr( ~Qinitial = ~q) =

c

Vinitial

Vinitial

Vfinal

Control Example



At equilibrium, the probability of momentum is the same initially and finally



c

Vinitial

Vinitial

Vfinal

Pr(~Pinitial = ~p) = Pr(~Pfinal = ~p) = Pr(~P eq = ~p)

Control Example



At equilibrium, the probability of momentum is the same initially and finally, and independent of position distribution.



c

Vinitial

Vinitial

Vfinal

Pr(~Pinitial = ~p) = Pr(~Pfinal = ~p)

Pr(Xinitial = (~q, ~p)) =cPr(~Pinitial = ~p)

VinitialPr(Xfinal = (~q, ~p)) =

cPr(~Pfinal = ~p)

Vfinal

= Pr(~P eq = ~p)

Control Example


Thus, the change in Shannon entropy is

Vinitial

Vfinal

�H[X] = H[Xfinal]�H[Xinitial]

Control Example



Vinitial

Vfinal


= H[

~P eq]�

X

x2Vfinal

c

Vfinallog2

c

Vfinal�H[

~P eq] +

X

x2Vinitial

c

Vinitiallog2

c

Vinitial

Control Example



Vinitial

Vfinal


= H[

~P eq]�

X

x2Vfinal

c

Vfinallog2

c

Vfinal�H[

~P eq] +

X

x2Vinitial

c

Vinitiallog2

c

Vinitial

= log2Vfinal

Vinitial

Control Example



And the work and heat are bounded by the log ratio of the volumes (achievable if done very slowly):

Vinitial

Vfinal


= H[

~P eq]�

X

x2Vfinal

c

Vfinallog2

c

Vfinal�H[

~P eq] +

X

x2Vinitial

c

Vinitiallog2

c

Vinitial

hW i � �kBT ln 2 log2Vfinal

VinitialhQi � �kBT ln 2 log2

Vfinal

Vinitial

= log2Vfinal

Vinitial

Szilard’s Information Engine

Take a particle in a box:



Insert a barrier at halfway:




Measure which side the particle lands on:

left right





Use measurement to slowly slide barrier away from particle, extracting work:

left right





Use measurement to slowly slide barrier away from particle, extracting work:

Store work, repeat:

left right

hW i = �kBT ln 2

Maxwell’s Demon

85

http://www.eoht.info/page/Maxwell's+demon

Maxwell’s Demon


Maxwell’s Demon

86


!Thermal

Reservoir MassWQ

Maxwell’s Demon Feedback/Control


Maxwell’s Demon

87


!Thermal

Reservoir MassWQ


hW i = hQi = �kBT ln 2


Maxwell’s Demon

88


!Thermal

Reservoir MassWQ



�H[X] = 0


Maxwell’s Demon

89


!Thermal

Reservoir MassWQ



�H[X] = 0

�Stotal

=hQiT

+ kB ln 2�H[X]


Maxwell’s Demon

90


!Thermal

Reservoir MassWQ



�H[X] = 0

�Stotal

=hQiT

+ kB ln 2�H[X]

= �kB ln 2 < 0


Maxwell’s Demon

91


!Thermal

Reservoir MassWQ



�H[X] = 0

�Stotal

=hQiT

+ kB ln 2�H[X]

= �kB ln 2 < 0 ?


Maxwell’s Demon

92


!Thermal

Reservoir MassWQ


�Stotal

< 0?


Maxwellian Demons

Measure

Erase

Control repeat

93

Demon

Heat Bath Temp=T

System Under Study (SuS)

Closed System

EE

E

Imeasure

Icontrol

Szilard Engine Cycle

A

AA

94

measure


A

A

BA

A

95

measure

measure


A

A

B

B

A

A

A

96

measure

measure

control


A

A

B

B

A

BA

A

A

97

measure

control

measure

Demon is able to extract and store work every cycle.

kBT ln2

control


A

A

B

B

A

A

BA

A

A

98

measure

control

erase

measure

Demon is able to extract and store work every cycle.

kBT ln2

control


A

A

B

B

A

A

BA

A

A Demon is able to extract and store work every cycle.

Violates the second law?

99

measure

control

erase

measure

kBT ln2

control

Energetic Costs

100

Qerase � kBT ln 2Landauer’s principle:R. Landauer, “Irreversibility and heat generation in the computing process”, (1961).

0 1 0 1

Energy

0 1

ErasureErasure

4E > kBT

Energy Energy

Erasure:(resetting memory)

Measurement:(writing to memory)

L. Brillouin “Maxwell’s Demon Cannot Operate”, (1951).

� ⇡ hc

kBT

Information Bearing Degrees of Freedom

101


!Thermal

Reservoir MassWQ


�Stotal

< 0?



102


!Thermal

Reservoir MassWQ


Erasure

?Thermal

Reservoir MassWQ

�Stotal

< 0?



103


!Thermal

Reservoir MassWQ


Erasure

?Thermal

Reservoir MassWQ

Landauer’s Principle: Werase

� kBT ln 2 ! �Stotal

� 0

�Stotal

< 0?


Benefit of Information Processing

Information processing allows systems to leverage structure of environment

104

http://spectrum.ieee.org/aerospace/robotic-exploration/the-nearly-effortless-flight-of-the-albatross



Cost of Information Processing

105

http://pooh.wikia.com/wiki/Think,_Think,_Thinkhttp://spectrum.ieee.org/computing/hardware/new-tech-keeps-data-centers-cool-in-warm-climates

Processing information costs work and dissipates heat.

Szilard Example

The Szilard system can be reformulated as a 2D box of gas.

SuS Memory Macrostates

Demon Memory Macrostates

RightLeft

A

B

L0

L

106

�L

C. Bennet, “Thermodynamics of Computation-A Review”, (1981)

�L0

http://arxiv.org/abs/1506.04327

Demon Dynamics: Deterministic Chaos, the Szilard Map, and the Intelligence of Thermodynamic Systems

http://arxiv.org/abs/1506.04327

Szilard Engine Dissipation

The average dissipated heat is exactly calculable as the work done to slide the barriers:

110

Q = W = �Z

PdV



111

Q = W = �Z

PdV

= �T�S[X]



112

Q = W = �Z

PdV

= �T�S[X]

�Stotal

= 0!

113

hQerasei hQmeasurei

hQcontrol

i

hQdissi[k

BT]

�


A. Boyd, J. Crutchfield, “Maxwell Demon Dynamics: Deterministic Chaos, the Szilard Map, and the Intelligence of Thermodynamic Systems”, PRL, (2016)

Information is a Thermodynamic Fuel

114

I

Thermal Reservoir MassWQ

Instead of erasing, write to a hard drive

Documents

Thermodynamics of Information Part Icsc.ucdavis.edu/~chaos/courses/poci/Lectures/Lecture37a...Energy and Entropy According to the ﬁrst law and second law of thermodynamics together,