Thermodynamics - University of Edinburghgja/thermo/lectures/lecture12.pdf · Everything we need to...

...Thermodynamics

Adiabatic cooling

Joule-Kelvin process

Types of Fridge, exploiting ∂V∂T P

Inversion temperature: does expansion cool or heat

Cooling by stretching a rubber band

Graeme Ackland Lecture 12 October 30, 2017 1 / 24

Electromagnetic Thermodynamics

Internal energy in electromagnetic fields

Radiation pressure 〈S〉/c (Poynting vector S=E × H)

Volume (cavity)

Temperature (e.g. Black body)

The magnetic energy is −µ.B ≈ −µ0HM,

H is extensive (like P), M is intensive (like V).

Everything we need to apply Thermodynamics.


Defining Magnetic Central Equation

But defining work is trickier.Magnetic Energy = -BM ;Define increasing magnetisation as “Work”: d W = B.dMMagnetic Central Equation:

du = Tds + BdM

with associated Maxwell Relations such as(dBdS

)M

=(dTdM

)S

We could define magnetic equivalant of enthalpy etot = −BM + u.

detot = Tds −MdB

Similarly, equivalents of Gibbs and Helmholtz free energy; four magneticMaxwell relations, etc.


Summary

Can use thermodynamics for magnetic systems.

Definition of systems and surroundings still important.

Definition of process determines what to treat as “work”.

Magnetic cooling: Approximations

1 Small induced magnetisation B = µ0(H + M) ≈ µ0H2 scalar susceptibility; magnetisation parallel to field.

Generalisation to tensorial quantities is straightforward if tedious.

3 Negligible change in volume: no PdV work.

4 Use energy densities, intensive quantities (no “script” symbols).


Magnetic Cooling - Two types of entropy

1 Put paramagnet in contact withheat bath

2 Apply field isentropically.Magnetic entropy→Thermalentropy

3 Paramagnet loses heat to heatbath.

4 Remove paramagnet from heatbath

5 Remove field isentropically:Thermal entropy→ magneticentropy increases

6 Cold paramagnet can extractheat from fridge


Magnetic Cooling Cycle

a) Isothermal magnetisation processloses heat:(

dQ

dB

)T

= T

(dS

dB

)T

= T

(dM

dT

)B

where we use a magnetic Maxwellrelation.

b) Adiabatic/isenthalpicdemagnetisation reduces temperature(

dT

dB

)S

= −(

dT

dS

)B

(dS

dB

)T

= − T

cB

(dM

dT

)B

= −TB

cB

(dχ

dT

)B

Using Maxwell and product rules,susceptibility, heat capacity atconstant fieldχ = M/B; cB = T (∂S/∂T )B


Magnetic equation of state: Curie Weiss Law

Defined byM = aB/(T − Tc).cB = cB(B = 0) + aTB2/(T − Tc)3. ← Note cB is infinite at T = Tc

Above Tc , this gives(dTdB

)S

= (T − Tc)/B.

Magnetic cycle starting at T1, raising to B1 then dropping to B2 gives

T1 − Tc

T2 − Tc=

B1

B2

A cycle with complete demagnetisation would cool toTc

Work through details in handin, and note approximations like neglecting

cB(B = 0) and assuming B = µ0H mean that in practice the performance will be

less good.


Equipartition in Classical Mechanics

All degrees of freedom have same energy.All excitation of degrees of freedom contribute to entropy.


Equipartition in society

US Declaration of Independence (1776)We hold these truths to be self-evident, that all men are created equal,But when a long train of abuses and usurpations, pursuing invariably thesame Object evinces a design to reduce them under absolute Despotism, itis their right, it is their duty, to throw off such Government,(The King of England is an asshole because...)

Declaration of Arbroath (1320)It is in truth not for glory, nor riches, nor honours, that we are fighting,but for freedom.But if he should give us or our kingdom to the English or the king of theEnglish, we would immediately take steps to drive him out as the enemyand the subverter of his own rights and ours(The King of the English is an auld scunner because...)


Equipartition/entropy: All Degrees of Freedom

Not just “men”.


Equipartition in radiation: All degrees of freedom

UltravioletCatastropheAllowing equipartition ofenergy in all wavelentghsrequires infinite energy.


The energy spectrum of cavity radiation

The Planck distribution law, for energy density of cavity radiation as afunction of wavelength λ, and temperature T is:

uλ(λ,T ) ∝ 1

λ5

(1

ehc/λkBT − 1

)In what follows, we do not use this law.

We will show that classical thermodynamics...

shows u is a function of λ and T only.

shows uλ is a state variable

shows uλ can be treated as a separate system for each λ

enabled Planck to guess the right answer


Thermal radiation as a thermodynamic system

System: radiation inside a cavity,Surroundings :Cavity walls = heat bath temperature T .

State Variables for radiationFixed T , V . System energy, U(T ,V )

radiationin

cavity

cavity walls at T

AAK

BBBBBN

Equation of state from Electromagntism:Radiation pressure 〈S〉/c (Poynting vector S=E × H)

P = 13u


Equipartition in radiation

Consider an partitioned oven.

A and B: different materials, same T .

Same value of energy density atequilibrium.

Crossflow of radiation would make heatpass from A to B. if uA > uB :

Clausius violation unless higher u =⇒higher T , regardless of volumes.

u(T , v) = u(T )

walls at T

uA uB

material A material B I

R

Consider systems comprising radiation of a given wavelength.

Same argument: uλ = uλ(λ,T ). (n.b. λ not a state variable)

Heat flow between different λ until all Tλ are the same.


How Planck discovered Quantum Theory

I had no alternative but to tackle the problem again ...

from the side of thermodynamics. In fact, my previous

studies of the Second Law of Thermodynamics came to

stand me in good stead now, for at the very outset I hit

upon the idea of correlating not the temperature of the

oscillator but its entropy with the energy... While a host

of outstanding physicists worked on the problem of the

spectral energy distribution ... every one of them

directed his efforts solely towards exhibiting the

dependence of the intensity of radiation on the

temperature. On the other hand, I suspected thatthe fundamental connection lies in thedependence of entropy with the energy ...

Nobody paid any attention to the method which I

adopted and I could work out by calculations completely

at my leisure, with absolute thoroughness, without fear

of interference or competition.



What is the energy density of blackbody radiation in a cavity?

u =

∫ ∞0

uλ(λ,T )dλ

(“area under the curve”) is very strongly temperature-dependent, butcannot blow up for high or low λ.

Anne Elk’s Theory on BrontosaurusesAll brontosauruses are thin at oneend; much, much thicker in themiddle and then thin again at the farend.



What is the total energy of blackbody radiation?Consider all wavelengths separately

u =

∫ ∞0

uλ(λ,T )dλ

(“area under the curve”) is observed to be very stronglytemperature-dependent, but cannot blow up for high or low λFirst consider the explicit dependence of u on T . From the centralequation of thermodynamics, and one of the Maxwell’s relations(

∂U

∂V

)T

= T

(∂P

∂T

)V

− P


Stefan’s Law is Thermodynamics

(∂U

∂V

)T

= T

(∂P

∂T

)V

− P

Using P = 13u, U = uV and

u = u(T ) we get

u =1

3T

du

dT− 1

3u =⇒ 4

dT

T=

du

u

u =

(4σ

c

)T 4

energy density scales as T 4.The integration constant σ is“Stefan’s constant”.Black body radiation ∝ u(T).


Closing in on the Planck distribution

Integrating the Black Body equation of state gives Stefan Law’s

u(T ) =

∫ ∞0

uλ(λ,T )dλ =

(4σ

c

)T 4

and requires for uλ(λ,T )

A 5th power in λ (3rd power in ν) for long wavelengths.

A function which doesn’t blow up at high ν (short wavelength)

Gives the same T = (∂Hλ/∂Sλ)P for all λ.

Maximises the entropy for fixed energy.

... but how to calculate the entropy?


The reluctant Quantum Mechanic

I was desperate. I was compelled to considerstatistical version of the second lawS = k .logW that I rejected and hated themost...

“If E is considered to be a continuouslydivisible quantity, this distribution is possiblein infinitely many ways.”i.e. Boltzmann’s entropy would be infinite.

We consider, however - this is the mostessential point of the whole calculation - Eto be composed of a well-defined number ofequal parts.”i.e. make energy countable, to make

Boltzmann’s entropy finite.

“Much though I hate thee, Let me count the ways.”


Inventing Quantum Mechanics

uλ(λ,T ) =∝ λ−5(

1

ehc/λkBT − 1

)Nice guess, Max, but what does it mean...Planck later wrote in 1934: My maxim is always this: consider every stepcarefully in advance, but then, if you believe you can take responsibility forit, let nothing stop you.


Inventing Quantum Mechanics

Using Maxwell Relations, get expression for entropy Sλ(T )

Use Sλ = k ln Ωλ: count discrete states

Discrete amount of Uλ : photon!

Can only calculate ∆S = k∆lnΩ.

To count Ω, need to know the energy of one photon: Planck’sconstant.

Meanwhile, Boltzmann was still struggling to get the Germans to acceptatoms.And the ultraviolet catastophe still lurks in zero-point energy.


Other thermodynamic quantities

Specific heat capacity.

Cv =

(∂U

∂T

)V

= 4σoVT 3

with σo = 4σ/c , remembering that we’ve been using energy density, soU = uV .Entropy

S =

∫cvdT

T=

4

3σoVT 3

Enthalpy

H = U + PV =4

3σoVT 4 = TS

which we already knew from T =(∂H∂S

)P.


And finally

Gibbs Free Energy,

G = uV − TS + PV = σoVT 4 − 4

3σoVT 4 +

1

3σoVT 4 = 0


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