thermodynamics of information part icsc.ucdavis.edu/~chaos/courses/poci/lectures/lecture37a...energy...
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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
Thermodynamics treats physical systems many elements.
Relevant quantities:1) Energy2) Work
Thermodynamics
Th
Tc
W
Qh
Qc
Thermodynamics treats physical systems many elements.
Relevant quantities:1) Energy2) Work3) Heat
Thermodynamics
Th
Tc
W
Qh
Qc
Thermodynamics treats physical systems many elements.
Relevant quantities:1) Energy2) Work3) Heat4) Temperature
Thermodynamics
Th
Tc
W
Qh
Qc
Thermodynamics treats physical systems many elements.
Relevant quantities:1) Energy2) Work3) Heat4) Temperature5) Entropy
Thermodynamics
Th
Tc
W
Qh
Qc
Physical system states:
Thermodynamic Entropy
X = {x}
Physical system states:
Probability of physical state at time :
Thermodynamic Entropy
X = {x}
t
Pr(Xt = x)
Physical system states:
Probability of physical state at time :
Thermodynamic entropy (Gibbs) of physical system:
Thermodynamic Entropy
X = {x}
t
Pr(Xt = x)
S[Xt
] = �k
B
X
x2XPr(X
t
= x) lnPr(Xt
= x)
Physical system states:
Probability of physical state at time :
Thermodynamic entropy (Gibbs) of physical system:
Shannon entropy of physical system:
Thermodynamic Entropy
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)
Physical system states:
Probability of physical state at time :
Thermodynamic entropy (Gibbs) of physical system:
Shannon entropy of physical system:
Thermodynamic Entropy
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
Thermodynamic entropy as a measure of disorder.
Disordered (Hot Gas) Ordered (Cold Solid)
Shannon entropy as a measure of uncertainty or unpredictability.
Entropy Comparison
Thermodynamic entropy as a measure of disorder.
Disordered (Hot Gas) Ordered (Cold Solid)
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 )}
Physical states described by positions and their conjugate momenta :
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)
Physical states described by positions and their conjugate momenta :
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)
First Law of Thermodynamics
Energy of each state specified by Hamiltonian function of state
Position changes as velocity and momentum changes as the force:
E(x) = H(~q, ~p)
dqidt
= @piH(~q, ~p)dpidt
= �@qiH(~q, ~p)
First Law of Thermodynamics
Energy of each state specified by Hamiltonian function of state
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
Entropy increases in isolated systems:
Second Law of Thermodynamics
dS[Xt]
dt� 0
dH[Xt]
dt� 0
Ordered (Hot Solid)
Entropy increases in isolated systems:
Second Law of Thermodynamics
dS[Xt]
dt� 0
dH[Xt]
dt� 0
Disordered (Hot Gas) Ordered (Hot Solid)
Entropy increases in isolated systems:
Example another and another and another
Second Law of Thermodynamics
dS[Xt]
dt� 0
dH[Xt]
dt� 0
Disordered (Hot Gas) Ordered (Hot Solid)
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
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:
where is the number of states with energy .
E
Pr(Xeq = x) = �
E(x),E/⌦(E)
E
⌦(E)
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:
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?
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
Energy/Entropy vs. Temperature
What is temperature?-Average energy?
Temperature of a system in equilibrium:
System Q
1
T
=@S[Xeq|E(x) = E
0]
@E
0
�Ssys =Q
Tsys
Energy/Entropy vs. Temperature
What is temperature?-Average energy?
Temperature of a system in equilibrium:
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
Entropy explains why heat diffuses from hot to cold:
Total entropy increases by heat diffusion:
Q THTC
�Stotal
= Q/TC �Q/TH > 0
TH > TC
Heat Flow
Entropy explains why heat diffuses from hot to cold:
Total entropy increases by heat diffusion:
The reverse is impossible:
Q THTC
Q THTC
�Stotal
= Q/TC �Q/TH > 0
TH > TC
Heat Flow
Entropy explains why heat diffuses from hot to cold:
Total entropy increases by heat diffusion:
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
Some energy can be redirected to produce work
Th
Tc
W
Qh
Qc
Heat Engine
Some energy can be redirected to produce work
Th
Tc
W
Qh
Qc Qc + W = Qh
Heat Engine
Some energy can be redirected to produce work
Entropy increase constrains efficiency
Th
Tc
W
Qh
Qc Qc + W = Qh
4Stotal
= Qc
/Tc
�Qh
/Th
� 0
Heat Engine
Some energy can be redirected to produce work
Entropy increase constrains efficiency
Th
Tc
W
Qh
Qc Qc + W = Qh
4Stotal
= Qc
/Tc
�Qh
/Th
� 0Qc
Qh� Tc
Th
Heat Engine
Some energy can be redirected to produce work
Entropy increase constrains efficiency
Th
Tc
W
Qh
Qc Qc + W = Qh
e =output
input
=
W
Qh= 1� Qc
Qh
Heat Engine
Some energy can be redirected to produce work
Entropy increase constrains efficiency
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?
Thermodynamics of Control
How do we extract work from a thermodynamic system?
!Thermal
Reservoir MassWQ
Feedback/Control
X
Thermodynamics of Control
How do we extract work from a thermodynamic system?
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
Thermodynamics of Control
How do we extract work from a thermodynamic system?
Through control of the energy landscape
dhEi =X
x
Pr(Xt
= x)dE(x) +X
x
E(x)dPr(Xt
= x)
= hdW i+ hdQi
!Thermal
Reservoir MassWQ
Feedback/Control
X
Thermodynamics of Control
How do we extract work from a thermodynamic system?
Through control of the energy landscape
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 total entropy production of the joint reservoir system
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
The total entropy production of the joint reservoir system
The fact that the reservoir is ideal means it has no memory of the ratchet, and so is uncorrelated with it:
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
The total entropy production of the joint reservoir system
The fact that the reservoir is ideal means it has no memory of the ratchet, and so is uncorrelated with it:
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 re-expressed with work
This can be interpreted as dissipated work
!Thermal
Reservoir Mass
Feedback/Control
X WQ
hWdissi = hW i ��FNEQ
T�Stotal
= hW i ��hE[X]i+ T�S[X]
Entropy of Control
This can be re-expressed with work
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
= hW i ��hE[X]i+ T�S[X]
Stepping Back
What does entropy do?
Stepping Back
What does entropy do?
1) Bounds function of heat engine
e 1� Tc
Th
e =output
input
=
W
Qh= 1� Qc
Qh
Stepping Back
What does entropy do?
1) Bounds function of heat engine with Carnot efficiency
e 1� Tc
Th
e =output
input
=
W
Qh= 1� Qc
Qh
Stepping Back
What does entropy do?
1) Bounds function of heat engine with Carnot efficiency
2) Bounds work and heat production in controlled thermodynamic systems
e 1� Tc
Th
e =output
input
=
W
Qh= 1� Qc
Qh
Stepping Back
What does entropy do?
1) Bounds function of heat engine with Carnot efficiency
2) Bounds work and heat production in controlled thermodynamic systems
hQi � �kBT ln 2�H[X]
e 1� Tc
Th
e =output
input
=
W
Qh= 1� Qc
Qh
Stepping Back
What does entropy do?
1) Bounds function of heat engine with Carnot efficiency
2) Bounds work and heat production in controlled thermodynamic systems
hQi � �kBT ln 2�H[X]
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
hQi � �kBT ln 2�H[X]
hW i � h�E[X]i � kBT ln 2�H[X]
Control Example
Initial vs. final energies and entropies:
Control Example
Initial vs. final energies and entropies:
Energy of each position is the same, so average energy of positions is the same: hE~qiinitial = hE~qifinal
Control Example
Initial vs. final energies and entropies:
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
Initial vs. final energies and entropies:
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
Initial vs. final energies and entropies:
Vinitial
Vfinal
Control Example
Initial vs. final energies and entropies:
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
Initial vs. final energies and entropies:
Probability of each position is uniform and inversely proportional to the available volume:
At equilibrium, the probability of momentum is the same initially and finally
Pr( ~Qfinal = ~q) =c
VfinalPr( ~Qinitial = ~q) =
c
Vinitial
Vinitial
Vfinal
Pr(~Pinitial = ~p) = Pr(~Pfinal = ~p) = Pr(~P eq = ~p)
Control Example
Initial vs. final energies and entropies:
Probability of each position is uniform and inversely proportional to the available volume:
At equilibrium, the probability of momentum is the same initially and finally, and independent of position distribution.
Pr( ~Qfinal = ~q) =c
VfinalPr( ~Qinitial = ~q) =
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
Initial vs. final energies and entropies:
Thus, the change in Shannon entropy is
Vinitial
Vfinal
�H[X] = H[Xfinal]�H[Xinitial]
Control Example
Initial vs. final energies and entropies:
Thus, the change in Shannon entropy is
Vinitial
Vfinal
�H[X] = H[Xfinal]�H[Xinitial]
= H[
~P eq]�
X
x2Vfinal
c
Vfinallog2
c
Vfinal�H[
~P eq] +
X
x2Vinitial
c
Vinitiallog2
c
Vinitial
Control Example
Initial vs. final energies and entropies:
Thus, the change in Shannon entropy is
Vinitial
Vfinal
�H[X] = H[Xfinal]�H[Xinitial]
= H[
~P eq]�
X
x2Vfinal
c
Vfinallog2
c
Vfinal�H[
~P eq] +
X
x2Vinitial
c
Vinitiallog2
c
Vinitial
= log2Vfinal
Vinitial
Control Example
Initial vs. final energies and entropies:
Thus, the change in Shannon entropy is
And the work and heat are bounded by the log ratio of the volumes (achievable if done very slowly):
Vinitial
Vfinal
�H[X] = H[Xfinal]�H[Xinitial]
= 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:
Szilard’s Information Engine
Take a particle in a box:
Insert a barrier at halfway:
Szilard’s Information Engine
Take a particle in a box:
Insert a barrier at halfway:
Measure which side the particle lands on:
left right
Szilard’s Information Engine
Take a particle in a box:
Insert a barrier at halfway:
Measure which side the particle lands on:
Use measurement to slowly slide barrier away from particle, extracting work:
left right
Szilard’s Information Engine
Take a particle in a box:
Insert a barrier at halfway:
Measure which side the particle lands on:
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
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
Maxwell’s Demon
87
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
hW i = hQi = �kBT ln 2
Maxwell’s Demon
88
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
hW i = hQi = �kBT ln 2
�H[X] = 0
Maxwell’s Demon
89
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
hW i = hQi = �kBT ln 2
�H[X] = 0
�Stotal
=hQiT
+ kB ln 2�H[X]
Maxwell’s Demon
90
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
hW i = hQi = �kBT ln 2
�H[X] = 0
�Stotal
=hQiT
+ kB ln 2�H[X]
= �kB ln 2 < 0
Maxwell’s Demon
91
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
hW i = hQi = �kBT ln 2
�H[X] = 0
�Stotal
=hQiT
+ kB ln 2�H[X]
= �kB ln 2 < 0 ?
Maxwell’s Demon
92
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
�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
Szilard Engine Cycle
A
A
BA
A
95
measure
measure
Szilard Engine Cycle
A
A
B
B
A
A
A
96
measure
measure
control
Szilard Engine Cycle
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
Szilard Engine Cycle
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
Szilard Engine Cycle
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
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
�Stotal
< 0?
Information Bearing Degrees of Freedom
102
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
Erasure
?Thermal
Reservoir MassWQ
�Stotal
< 0?
Information Bearing Degrees of Freedom
103
http://www.eoht.info/page/Maxwell's+demon
!Thermal
Reservoir MassWQ
Maxwell’s Demon Feedback/Control
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
Szilard Engine Dissipation
The average dissipated heat is exactly calculable as the work done to slide the barriers:
110
Q = W = �Z
PdV
Szilard Engine Dissipation
The average dissipated heat is exactly calculable as the work done to slide the barriers:
111
Q = W = �Z
PdV
= �T�S[X]
Szilard Engine Dissipation
The average dissipated heat is exactly calculable as the work done to slide the barriers:
112
Q = W = �Z
PdV
= �T�S[X]
�Stotal
= 0!
113
hQerasei hQmeasurei
hQcontrol
i
hQdissi[k
BT]
�
Szilard Engine Dissipation
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