Stat Mech Flashcards
What does classical thermodynamics describe?
Macroscopic systems ~in equilibrium in terms of a few measurable variables
What are state variables?
Variables that describe the current state of a system. Importantly, they do not depend on the history of the system.
Includes temperature, pressure, volume, composition, internal energy, and entropy.
Cannot all be varied independently, equations linking them are equations of state.
What is quasi-equilibrium?
Changes to conditions (eg pressure) can be made sufficiently slowly that the system remains in equilibrium at all times.
What is a reversible process?
A process where an infinitesimal change in the external conditions is enough to reverse the direction of the process;
e.g. compression of a gas in a cylinder with a frictionless piston by exerting an external force only just sufficient to move the piston one way or the other.
What is a macrostate?
The state of a macroscopic system in equilibrium specified by a handful of macroscopically-manipulable variables. It completely ignores what is going on with the individual atoms that comprise the system.
What is a microstate?
A description of the positions and momenta of all atoms (or of their combined quantum state) of a system.
The Zeroth Law of Thermodymamics
If two bodies are separately in thermal equilibrium with a third body, they are also in thermal equilibriums with one another. All three are then said to be at the same temperature.
Absolute temperature
Measured in Kelvin, defined to be 273.16 K at the triple point of water:
T = lim(P→0) PV/(PV)ₜᵣᵢₚₗₑ × 273.16 K
Low pressure limit taken so real gases approach ideal behaviour.
Zero of the Kelvin scale is absolute zero. At this temperature, pressure of an ideal gas would vanish, because (classically) the motion of its molecules would cease.
First Law of Thermodynamics
Any change in internal energy, E, of a system is due to the amount of heat added to it and the work done on it.
∆E = Q + W
dE = đQ + đW
Energy and work in a reversible process.
dE = đQʳᵉᵛ + đWʳᵉᵛ
Adiabatic processes
No heat transfer
Q = 0
System in a cycle
Returns to its initial state
∆E = 0
Work done in reversible processes
Compression of a fluid:
đWʳᵉᵛ = −P dV
Stretching a wire of tension Γ by dl:
đWʳᵉᵛ = Γ dl
Increasing the area of a film of surface tension γ by dA:
đWʳᵉᵛ = γ dA
Increasing the magnetic field B imposed upon a paramagnetic sample:
đWʳᵉᵛ = −m · dB = −VM · dB
where M is magnetisation per unit volume, and m is total magnetic moment of the sample.
To calculate the internal energy change for irreversible processes (such as free expansion of a gas), it is necessary to find a reversible process linking the same initial and final states of the system.
Second law of thermodynamics
No process which would decrease the entropy of a system will happen spontaneously.
OR
If left alone, systems always head towards the same equilibrium as determined by external constraints.
OR
In an isolated system, entropy can only increase (or stay the same)
Kelvin and Planck statement
“It is impossible to construct an engine which, operating in a cycle, will produce no other effect than the extraction of heat from a reservoir and the performance of an equivalent amount of work”
Clausius statement
“It is impossible to construct an refrigerator which, operating in a cycle, will produce no other effect than the transfer of heat from a cooler body to a hotter one.”
Entropy
dS = đQʳᵉᵛ/T
For an isolated system, during any spontaneous change
dS ≥ 0
Fundamental thermodynamic
relation
dE = TdS − PdV
(dE = đQʳᵉᵛ + đWʳᵉᵛ)
E : energy
T : absolute temperature
S : entropy
P : pressure
V : volume
Carnot engine
Qₕ is removed from a hot reservoir and Q꜀ is discarded to a cold one in such a way that the combined entropy change is zero.
Qₕ/Tₕ = Q꜀/T꜀
The difference is available to do work
W = Qₕ − Q꜀
From the fundamental thermodynamic relation, derive functions for fixed properties.
dE = TdS − PdV
⇒E(S, V) with
(∂E/∂S)|ᵥ = T and (∂E/∂V)|ₛ = −P
dS = 1/T dE + P/T dV
⇒S(E, V) with
(∂S/∂E)|ᵥ = 1/T and (∂S/∂V)|ₑ = P/T
Helmholtz free energy
System in contact with heat bath, spontaneous change at constant volume, Q absorbed.
∆E = Q.
System starts and ends at temperature T. Total change in entropy has two parts, ∆S for system and −Q/T for surroundings.
∆Sₜₒₜ = ∆S − Q/T = 1/T (T∆S − ∆E) ≥ 0
⇒ ∆(TS − E) ≥ 0
In approach to equilibrium, quantity TS − E is maximised.
Helmholtz free energy, F, is negative of this, so it is minimised.
F = E − TS
⇒ dF = −SdT − PdV
(∂F/∂T)|ᵥ = −S and (∂F/∂V)|ₜ = −P
Gibbs free energy
Approach to equilibrium minimises Gibbs free energy, G, of a system.
G = E − TS + PV
⇒ dG = −SdT + VdP
(∂G/∂T)|ₚ = −S and (∂G/∂P)|ₜ = V
Gibbs free energy and chemical potential
µ = (∂G/∂N)|ₜ,ₚ
G = ΣμᵢNᵢ
μᵢ is chemical potential of each species
Nᵢ is number of molecules of each species
Entropy for an ideal gas as a function of temperature and pressure
S(T, P) = S(T₀, P₀) + Cₚ ln(T/T₀) − nRln(P/P₀)
Minimising chemical potential through pressure
Chemical potential is higher at higher pressures, and gas will diffuse from higher to lower pressure. Partial pressures of each species will minimise despite mechanical pressure. For example, If two ideal gases are at different concentrations on either side of a rigid membrane, but only one can pass through, the partial pressure of the mobile one will equalise even if that increases the mechanical total pressure on one side.
Comparing the chemical potential or Gibbs free energy per molecule at two different pressures but constant temperature.
µ(T₀, P₂) − µ(T₀, P₁) = kᴮT₀ln(P₂/P₁)
What is an ensemble?
The set of all allowed microstates.
Types of ensemble
Microcanonical (NVE)
Total energy and number of particles are fixed. The system must remain totally isolated (unable to exchange energy or particles with its environment) to stay in statistical equilibrium.
Canonical (NVT)
Energy is not known exactly but number of particles is fixed. In place of energy, the temperature is specified. It can describe a closed system which is/has been in weak thermal contact with a heat bath. The system must remain totally closed (unable to exchange particles with its environment) but may come into weak thermal contact with other systems that are described by ensembles with the same temperature.
Grand canonical (μVT)
Neither the energy nor particle number are fixed. Instead, the temperature and chemical potential are specified. It can describe an open system: one which is in, or has been in, weak contact with a reservoir (thermal contact, chemical contact, radiative contact, electrical contact, etc.). It will remain in statistical equilibrium if the system comes into weak contact with other systems that are described by ensembles with the same temperature and chemical potential.
What is an ensemble average?
If the value of some quantity X in the iₜₕ microstate is Xᵢ, and the probability that the system is in that
microstate is pᵢ, then the value of X in the macrostate is the ensemble average.
⟨X⟩ = Σ pᵢXᵢ
Number of ‘allowed’ or ‘accessible’ microstates
Allowed or accessible means having the same constraints, such as volume, particle number and total energy, as the macrostate.
pᵢ = 1/Ω
Σpᵢ = Ω 1/Ω = 1
For the example of blue/green counters
Ω = N!/n!(N − n)!
N!/n! (N − n)! are binomial coefficients, written ᴺCₙ
Entropy from microstates
More microstates exist with higher entropy, as a system evolves it goes to more likely microstates, hence entropy incerases.
S = kᴮ lnΩ
Total magnetic moment of an ideal paramagnet
Lattice of N sites, at each the spin points either up
or down. Total magnetic moment is the sum of the individual moments. If n↑ spins are pointing up and n↓ = N − n↑ are pointing down:
m = n↑µ + n↓(−µ) = µ(2n↑ − N)
where µ is magnetic moment.
Energy of a magnetic moment in an external magnetic field
−µ · B
Spin-up atoms will have energy −µB, and spin-down atoms will have energy µB
Total energy in paramagnet
E = n↑(−µB) + n↓(µB) = −Bm
Macrostate probability distribution for spin-1/2 paramagnet
For n↑ = 1/2 (N + m/µ)
Ω(n↑) = N!/n↑!(N − n↑)!
Plotting Ω(n) (normalised to 1 at the peak) as a function of n/N, get a gaussianesque shape with a sharper peak for higher N.
For large N, curve approximated by
Ω(n) ∝ exp(−(n−N/2)²/(N/2))
Mean N/2
Standard deviation σ = √N/2’
Tends to gaussian, central limit theorem
Temperature, pressure, and chemical potential from entropy
(∂S/∂E)|ᵥ,ₙ = 1/T
(∂S/∂V)|ₑ,ₙ = P/T
(∂S/∂N)|ₑ,ᵥ = −µ/T
They give
dS = 1/T dE + P/T dV − µ/T dN
which is the fundamental thermodynamic relation rearranged
Fundamental thermodynamic relation in spin-1/2 paramagnet
Instead of −PdV we have −mdB
dE = TdS − mdB + µdN
dS = 1/T dE + m/T dB − µ/T dN
(∂S/∂E)|ᴮ,ₙ = 1/T
(∂S/∂B)|ₑ,ₙ = m/T
(∂S/∂N)|ₑ,ᴮ = −µ/T
Temperature and magnetic moment from entropy for an isolated spin-half paramagnet
S = kᴮ ln Ω(E, B) = kᴮ ln(N!/n↑!(N − n↑)!)
Large N, Stirling’s approximation:
ln(n!) = nln(n)−n
S = kᴮ(Nln(N) − n↑ln(n↑) − (N−n↑)ln(N−n↑))
1/T = (∂S/∂E)ᴮ,ₙ
= (∂S/∂n↑)ₙ (∂n↑/∂E)ᴮ,ₙ
= kᴮ/2µB ln(n↑/n↓)
m/T = (∂S/∂B)ₑ,ₙ
= (∂S/∂n↑)ₙ (∂n↑/∂B)ₑ,ₙ
= −kᴮE/2µB² ln(n↑/n↓)
⇒ m = −E/B
Derive the ideal gas equation from first principles
Isolated system of N atoms in a box of volume V . Box is subdivided into many tiny cells of volume ∆V , so there are V/∆V cells total. Each atom can be in any cell, V/∆V microstates for each atom, and (V/∆V)N microstates for the gas as a whole.
S = Nkᴮln(V/∆V)
Change in entropy when volume changes (at fixed energy)
∆S = Nkᴮ ln(Vf/Vi)
From entropy,
P/T = (∂S/∂V)ₑ,ₙ=
⇒ P = NkᴮT/V
⇒ PV = NkᴮT
When can you treat atoms independently?
When they don’t interact AND they’re in a non-isolated system (energy isn’t limited)
Derive the Boltzmann distribution for a system in contact with a heat bath at temperature T.
System in contact with a heat bath at temperature T, probability it is in the ith microstate, energy εᵢ
System S in contact with heat reservoir R, the whole forming a single isolated system with energy E₀
Eᵣ = E₀ − ε
Eₛ = ε
Likelihood of a particular partition of energy depends on the number of microstates of whole system S + R that correspond to that partition. System and reservoir are independent, so total number of microstates factorises
Ω = ΩₛΩᵣ
Specify the ith microstate (with energy εᵢ). Probability pᵢ of finding the system in that microstate is proportional to the number of compatible microstates Ω(E₀, εᵢ) of the whole
system. Specified state of S, Ωₛ = 1, only the microstate of the reservoir is unspecified, Ω(E₀, εᵢ) = Ωᵣ(E₀ − εᵢ)
Using relation between Ω and entropy,
pᵢ ∝ Ωᵣ(E₀ − εᵢ) = exp{Sᵣ(E₀ − εᵢ)/kᴮ}
R ⟩⟩ S, so εᵢ ⟨⟨ E₀. Can expand Sᵣ about Sᵣ(E₀) and keep only the lowest terms:
Sᵣ(E₀ − εᵢ) = Sᵣ(E₀) − εᵢ(∂Sᵣ/∂E)ᵥ,ₙ + ½εᵢ²(∂²Sᵣ/∂E²)ᵥ,ₙ + . . .
(derivatives are evaluated at E0)
Derivative of S w.r.t. E is just the inverse of temperature. Dropping third term as negligibly small:
pᵢ ∝ exp{Sᵣ(E₀)/kᴮ − εᵢ/(kᴮT)}
∝ exp{−εᵢ/(kᴮT)}
(since Sᵣ(E₀) is a constant, independent of the microstate we are interested in)
Call constant of proportionality 1/Z, final result:
pᵢ = exp(−εᵢ/kᴮT) / Z
Where normalisation constant Z is found by saying total probability is one:
Σⱼ pⱼ = 1
Z = Σⱼ exp(−εⱼ/kᴮT)
This is also the partition function
Partition function
Z = Σⱼ exp(−εⱼ/kᴮT)
Energy from partition function
⟨E⟩ = −1/Z (∂Z/∂β)ₙ,ᵥ
or
⟨E⟩ = −∂lnZ/∂β
where
β ≡ 1/kᴮT
Heat capacity from energy
Cᵥ = (∂⟨E⟩/∂T)ᵥ,ₙ
Fluctuations in energy
(∆E)² = ⟨E⟩² − ⟨E⟩²
Since ⟨E⟩² = Σᵢ εᵢ² pᵢ,
⟨E⟩² = 1/Z (∂²Z/∂β²)ₙ,ᵥ
(∆E)² = 1/Z (∂²Z/∂β²) − (∂lnZ/∂β)² = ∂²lnZ/∂β²
= −∂T/∂β ∂⟨E⟩/∂T = (kᴮT)² Cᵥ/kᴮ
Normal macroscopic system, average energy is of order NkᴮT and heat capacity is of order Nkᴮ.
∆E/E ≈ 1/√N’
Derive Gibbs entropy
Gibbs entropy of system is entropy of ensemble divided by the number of copies, ν, in the ensemble.
⟨S⟩ = Sᵥ/ν
Ensemble has νᵢ copies in the ith microstate, number of ways of arranging these is
Ωᵥ = ν!/ν₁!ν₂!ν₃!…
Stirling’s approximation,
ln Ωᵥ = νlnν − ν − Σᵢ (νᵢ (lnνᵢ − νᵢ)
= Σᵢ νᵢ (lnν − lnνᵢ) (using ν = Σᵢνᵢ)
= −Σᵢ νᵢ (lnνᵢ/ν)
= −ν Σᵢ pᵢ lnpᵢ
Ensemble entropy is Sᵥ = kᴮlnΩᵥ and system entropy is
⟨S⟩ = −kᴮ Σᵢ pᵢ lnpᵢ
Values from differentiating Helmholtz free energy for system in contact with a heat bath
F = −kᴮTlnZ
F = E −T S, from fundamental thermodynamic relation,
dF = −SdT −PdV + µdN
S = −(∂F/∂T)ᵥ,ₙ
P = −(∂F/∂V)ₜ,ₙ
µ = (∂F/∂N)ₜ,ᵥ
For a magnetic system,
m = − (∂F/∂B)ₜ,ₙ
Partition fuction for single atom in spin-½ paramagnet
In an external field, two states have different energy; spin-up has energy −µB, and spin-down, µB.
Z₁ = exp(µB/kᴮT) + exp(−µB/kᴮT)
= 2cosh(µB/kᴮT)
= 2cosh (µBβ)
Energy of spin-½ paramagnet
Atoms non-interacting
⟨E⟩ = −N ∂lnZ₁/∂β = −NµB tanh(µBβ)
How does temperature affect spin alignmet in the spin-½ paramagnet
Low T, all spins aligned with the field and the energy per spin is close to −µB.
As T increases, thermal fluctuations start to flip some of the spins; becomes noticeable when kᴮT is of the order of µB.
T very large, energy tends to zero as the number of up and down spins become more nearly equal. (Remember, ⟨n↓⟩ / ⟨n↑⟩ = exp(−2µB/kᴮT), so it never exceeds one.
Heat capacity of spin-½ paramagnet
Cᵥ = ∂E/∂T = Nkᴮ (µBβ)² sech²(µBβ)
Heat capacity tends to zero both at high and low T.
Low T, heat capacity small because kᴮT is much smaller than the energy gap 2µB, thermal fluctuations which flip spins are rare and it is hard for the system to absorb heat. Quantisation means there is always a minimum excitation energy of a system and if the temperature is low enough, the system can no longer absorb heat.
High T, number of down-spins never exceeds number of up-spins, energy has a maximum of zero. As the temperature gets very high, limit is close to being reached, raising temperature makes very little difference. Only applies here because there is a finite number of energy levels (two). Most systems have an infinite tower of energy levels, there is no maximum energy and the heat capacity does not fall off.
Derive partition function for N non-interacting spins on a lattice (distinguishable)
System that has two single-particle energy levels,
ε₁ and ε₂. The single-particle partition function is
Z₁ = exp(−ε₁β) + exp(−ε₂β)
The partition function for two distinguishable particles is
Z₂ = exp(−2ε₁β + 2exp(−(ε₁+ε₂)β) + exp(−2ε₂β) = (Z₁)²
where the second state is multiplied by 2 because there are two ways that two distinguishable particles can be in different levels.
In general, for N particles, the energies are nε₁ + (N−n)ε₂, for 0 ≤ n ≤ N, and there are N!/n!(N−n)! separate microstate of this energy.
Z_N = Σᴺ N!/n!(N−n)! exp(−(nε₁+(N−n)ε₂)β)
=Σᴺ N!/n!(N−n)! exp(−nε₁β) exp(−(N−n)ε₂β)
=Σᴺ N!/n!(N−n)! (exp(−ε₁β))ⁿ (exp(−ε₂β))ᴺ⁻ⁿ
= (Z₁)ᴺ
(used the binomial expansion of (x+y)ᴺ)
Z_N = (Z₁)ᴺ
How do work and heat independently affect a spin-½ paramagnet
Doing work by adiabatically decreasing the field adds energy in the form of internal potential energy, bringing energy levels closer together and reducing the temperature.
Adding heat increases occupancy of the higher energy level, temperatre is higher.
Partition function w.r.t. degeneracy
Z = Σⱼ exp(−εⱼβ) = Σₙ (degeneracy) x exp(−εₙβ)
Using a paramagnet to reach low temperatures
See fig. 1 (pg 37)
Sample in a magnetic field B₁ at temperature T₁.
a→ b: Sample in contact with a heat bath at T₁, increase the magnetic field to B₂.
b→ c: Sample now isolated, slowly decrease the field to B₁ again. This adiabatic demagnetisation; because process is slow and adiabatic, entropy and magnetisation are unchanged.
Following these steps on a T − S plot, see that the second, constant entropy, step reduces temperature. Entropy is a function of B/T only, not B or T separately, so reducing B at constant S reduces T also.
In terms of spins (fig 2, pg 38):
First step increases level spacing while keeping temperature constant (isothermal), so population of upper level falls.
Second step reduces the level spacing again, but spins are isolated there is no change in level occupation. New, lower level occupation is now characteristic of a lower temperature.
By starting with a large sample, can repeat the process with a small sub-sample, the remaining material acting as a heat bath during the next magnetisation. This way, temperatures of a fraction of a Kelvin can be reached. However, after each step less and less is gained each time, as the curves come together as T → 0.
There is always a minimum excitation energy ε
of the system, and once kᴮT «_space;ε, there is no further way of lowering the temperature. The unattainability of absolute zero is the third law of thermodynamics.
Vibrational energy of a diatomic molecule
Energy levels of a quantum simple harmonic oscillator of frequency ω:
εₙ = (n + ½)ħω where n=0,1,2…
Z₁ = Σ∞ exp(−εₙβ)
= exp(−½ħωβ) (exp(0) + exp(−ħωβ) + exp(−2ħωβ). . .)
= exp(−½ħωβ)/1−exp(−ħωβ)
= (2sinh(½ħωβ))⁻¹
(used sum of geometric series, Σₙ xⁿ = (1−x)⁻¹, with x=exp(−ħωβ))
From this, obtain
⟨E₁⟩ = −∂lnZ₁/∂β = ½ħωcoth(½ħωβ)
Low temperature limit (kᴮT «_space;ħω; ħωβ → ∞):
⟨E₁⟩ → ½ħω,
Expected if only the ground state is populated.
High temperature limit (kᴮT»_space; ħω; ħωβ → 0):
⟨E₁⟩ → kᴮT
Typically only reached around 1000 K.
Rotational energy of a diatomic molecule
Energy levels of rigid rotor of moment of inertia I:
εₗ = l(l+1)ħ²/2I where l=0,1,2…
As well as the quantum number L, there is mₗ, −l ≤ mₗ ≤ l, and energy doesn’t depend on mₗ. Thus the lth energy level occurs 2l + 1 times in the partition function, giving
Z₁ = Σ(l=0→∞) Σ(mₗ=−l→l) exp(−l(l+1)ħ²β/2I)
= Σ(l=0→∞) (2l+1) exp(−l(l+1)ħ²β/2I)
2l + 1 is degeneracy factor since “degenerate” levels are levels with the same energy. For general β this cannot be further simplified.
Low temperatures, successive terms in Z₁ fall off quickly; only lowest levels will have any significant occupation probability, ⟨E₁⟩ → 0
High temperatures (kᴮT»_space; ħ²/2I), many accessible levels discrete becomes almost continuous; can replace the sum over l with integral dl; changing variables to x = l(l+1) gives
Z₁ = 2I/ħ²β
⟨E₁⟩ = kᴮT
Typically ħ²/2i ~ 10⁻³eV, so high-temperature limit reached well below room temperature.
Translational energy of a molecule in an ideal gas
Non-interacting atoms of gas in a cuboidal box of side lengths Lₓ, Lᵧ, and Lᶻ, and volume V ≡ LₓLᵧLᶻ. Wave function ψ satisfies the free Schrodinger equation:
−ħ²/2m ∇²ψ(x,y,z) = Eψ(x,y,z)
Equation and boundary conditions satisfied by:
ψ(x,y,z) = A sin(nₓπx/Lₓ) sin(nᵧπy/Lᵧ) sin(nᶻπz/Lᶻ)
with nₓ, nᵧ and nᶻ integers greater than zero.
Corresponding energy:
ε(nₓ,nᵧ,nᶻ) = ((nₓπ/Lₓ)² + (nᵧπ/Lᵧ)² + (nᶻπ/Lᶻ)²) ħ²/2m
≡ k²ħ²/2m
where k² = kₓ² + kᵧ² + kᶻ²
and kₓ = πnₓ/Lₓ etc.
One-particle partition function:
Z₁ = Σ{nₓ,nᵧ,nᶻ} exp(−ε(nₓ,nᵧ,nᶻ)β)
In general, this cannot be further simplified. However, it can be if kᴮT is much greater than the spacing between energy levels.
Volume of 1 litre, energy level spacing of order ħ²π²/2mL² ≈ 10⁻²⁰ eV. Tiny, but in high-temperature regime! Can replace sum over levels with an integral. (Continuum approximation).
Choose kₓ, kᵧ and kᶻ as the variables, and replace Σₙₓ with (Lₓ/π) ∫ dkₓ:
Z₁ = LₓLᵧLᶻ/π³ ∫∞ ∫∞ ∫∞ dkₓ dkᵧ dkᶻ exp(−ε(k)β)
= V/π³ ∫∞ ∫π/2 ∫π/2 k² sinθₖ dk dθₖ dφₖ exp(−ε(k)β)
(spherical polar)
= 1/8 4π V/π³ ∫∞ k² dk exp(−ε(k)β)
≡ ∫∞ g(k)exp(−ε(k)β) dk
where g(k) ≡ V k²/2π²
Factor of 1/8 comes from only integrating over positive values of kₓ etc, aka over the positive octant of k-space.
g(k) is the density of states in k-space; g(k)dk is the number of states within range of k → k + dk.
This depended on energy being independent of the direction of k.
Using actual form of ε(k) to complete calculation:
Z₁ = V/2π² ∫∞ k² exp(−ħ²k²β/2m) dk
= V(m/2πħ²β)³/²
≡ V n_Q
n_Q is the quantum concentration and is temperature-dependent.
From Z₁, can obtain average single particle energy:
⟨E₁⟩ = −∂lnZ₁/∂β = 3/2 kᴮT
Density of states for a particle in three dimensions
g(k) = gₛVk²/2π²
where gₛ = 2s+1
(spin factor)
Partition function using density of states
Z₁ = ∫∞ g(k) exp(-ε(k)β) dk
= gₛV(mkᴮT/2πħ²)³/² ≡ Vgₛn_Q.
ε(k) (3D non-relativistic)
ε(k) = ħ²k²/2m
g(ε) from g(k) for 3D non-relativistic particles
g(ε), ε(k) is energy of a particle with momentum ħk.
g(ε) defined so number of states with wave numbers between k and k+dk is the same as the number with energies between ε(k) and ε(k+dk) = ε(k)+dε.
g(k)dk = g(ε)dε
where dε = dε/dk dk
3D, non-relativistic,
ε(k) = ħ²k²/2m
g(ε) = gₛV(2m)³/² / 4π²ħ³ ε¹/²
Modification for relativistic density of states
ε(k) = ħck
Derive the Maxwell-Boltzmann distribution from the density of states
Speed v of a particle is related to the wave number k by mv = ħk.
Probability of a particle having k in range k → k + dk:
P(k → k+dk) = g(k)exp(−ε(k)β)/Z₁dk
where ε(k) = ħ²k²/2m
Corresponding probability of speed being in range v → dv:
P(v → v+dv) = V/2π² exp(−ε(v)β)/Z₁ (m/ħ)³ v² dv
where ε(v) = mv²/2
⇒ P(v) = √2/π’ (m/kᴮT)³’² v² exp(−mv²/2kᴮT)
This is the Maxwell-Boltzmann distribution.
Most probable speed (dP(v)/dv = 0):
vₚ = √2kᴮT/m’ ≈ 1.41√kᴮT/m’
Mean speed:
⟨v⟩ = √8kᴮT/πm’ ≈ 1.60√kᴮT/m’
rms speed:
vᵣₘₛ = √⟨v²⟩’ = √3kᴮT/m’ ≈ 1.73√kᴮT/m’
Show factorisation of partition functions when degrees of freedom are independent
Z = Σᵢ₁,ᵢ₂,…ᵢₙ exp(−Σ₁ᴺ εᵢₙ⁽ⁿ⁾β)
= Σᵢ₁,ᵢ₂,…ᵢₙ Π₁ᴺ exp(−εᵢₙ⁽ⁿ⁾β)
= Π₁ᴺ Σᵢₙ exp(−εᵢₙ⁽ⁿ⁾β)
= Π₁ᴺ Z⁽ⁿ⁾
What is the equipartition theorem?
Each quadratic term in the classical expression for the energy contributes ½kᴮT to the average energy and ½kᴮ to the heat capacity.
Oscillator:
E_vib = ½mẋ² + ½mω²x²
2 d.o.f, E → kᴮT
Rotor:
Eᵣₒₜ = ½Iω₁² + ½Iω₂²
2 d.o.f, E → kᴮT
Translational:
Eₜᵣ = ½mẋ² + ½mẏ² + ½mż²
3 d.o.f, E → ³⁄₂kᴮT
kᴮT must be much greater than the spacing between quantum energy levels, otherwise heat capacity will be reduced, dropping to zero at low temp. Corresponding degree of freedom is frozen out, like vibration at room temperature.
More temp, more translation/ rotation/ vibration, more ability to hold energy, higher heat capacity.
Avr. energy and heat capacity of a crystal, Einstein style
Equipartition, expect molar heat capacity to be 3R since each atom is free to vibrate in three directions. This is reproduced by the Einstein model for a crystal, which considers each atom linked to its neighbours by six springs (3N in total)
Algebra same as vibrations of a diatomic molecule:
⟨E⟩ = ³⁄₂ Nħωcoth(½ħωβ)
Cᵥ = ¾Nkᴮ(ħωβ)²(sinh½ħωβ)⁻²
Low temperature (β → ∞),
Cᵥ → 3Nkᴮ(ħωβ)² exp(−ħωβ)
Tends to zero, but doesn’t agree with observed low temperature behaviour, which is proportional to T³. Einstein model only works quantitatively at high temp.
Indistinguishability of particles
Quantum mechanics says that atoms of the same element are fundamentally indistinguishable, exactly the same. As a result all observables have to be unchanged when we interchange two identical atoms. We can’t give them labels and know which one is in which state.
N-particle partition function for indistinguishable particles
Differences between distinguishable and indistinguishable:
Distinguishable, four states, two of which have energy ε, two-particle partition function is
Z₂ = exp(0) + 2exp(−εβ) + exp(−2εβ) = (Z₁)²
Indistinguishable, can’t give labels so can’t know which is in which state. Only three states, partition function is
Z₂ = exp(0) + exp(−εβ) + exp(−2εβ) ≠ (Z₁)²
Using (Z₁)² over-counts the state where the particles are in different energy levels.
Note that (Z₁)ᴺ over-counts the states in which all N particles are in different energy levels by exactly N!
If there are many more accessible energy levels (levels with ε < a few kᴮT) than there are particles, probability of any two particles being in the same energy level is small, almost all states will have all particles in different levels. Hence a good approximation is:
Z_N = (Z₁)ᴺ/N!
Classical limit condition for partition function in an ideal gas
The one-particle translational partition function, at any attainable temperature, is
Z₁ = Vgₛn_Q
where n_Q ≡ (mkᴮT/2πħ²)³’²
Assuming atoms distinguishable yields non-extensive Helmholtz free energy, must treat them as indistinguishable. Want to use approximation
Z_N = (Z₁)ᴺ/N!
Can use if it is very unlikely that any two atoms are in the same energy level. Can calculate the number of levels below, for example, 2kᴮT, using
∫ₖₘₐₓ g(k) dk with ħ²kₘₐₓ²/2m = 2kᴮT
This gives 2.1n_QV
Can see that n_Q, the “quantum concentration”, is a measure of the number of states available.
Can use approximation Z_N = (Z₁)ᴺ/N! if
N ⟨⟨ n_QV
(or n ⟨⟨ n_Q)
This is the classical limit.
Also note that n_Q ≈ 1/λ³, where λ is thermal deBroglie wavelength (wavelength of a particle of energy order kᴮT). Condition n ⟨⟨ n_Q equivalent to saying separation of atoms ⟩⟩ wavelength, general condition for classical behaviour.
Helmholz free energy of an ideal gas
(Using Stirling’s approximation)
⟨F⟩ = −kᴮTlnZ_N
= −NkᴮT(lnZ₁ − lnN + 1)
= −NkᴮT [ln(V/N) + ln(gₛn_Q) + 1]
= NkᴮT [ln(n/gₛn_Q) − 1]
n_Q is composed only of constants and T, so it is intensive; number density n ≡ N/V is ratio of extensive quantities and so is also intensive. F is simply proportional
to N, and so is extensive (as required)
Pressure in an ideal gas
⟨F⟩ = −kᴮTlnZ_N
= NkᴮT [ln(n/gₛn_Q) − 1]
P = −(∂F/∂V)ₜ,ₙ
= NkᴮT/V
Entropy of an ideal gas, Sackur-Tetrode equation
⟨F⟩ = −kᴮTlnZ_N
= NkᴮT [ln(n/gₛn_Q) − 1]
S = −(∂F/∂T)ᵥ,ₙ
= −Nkᴮ [ln(n/gₛn_Q) − 1] + NkᴮT 1/n_Q dn_Q/dT
= Nkᴮ [ln(gₛn_Q/n) + ⁵⁄₂]
Since, n «_space;n_Q is required, S is positive as expected.
Eq can be unpicked to show dependence on V and T:
S = Nkᴮ(lnV + ³⁄₂lnT + const.)
Energies and partition functions for vibrations, rotations, and translations in ideal gas
Single particle
ε = εᵗʳ+ εʳᵒᵗ+ εᵛᶦᵇ (add)
Z₁ = Z₁ᵗʳ Z₁ʳᵒᵗ Z₁ᵛᶦᵇ (multiply)
Multiple particles
Z_N = (Z₁ᵗʳ)ᴺ (Z₁ʳᵒᵗ)ᴺ (Z₁ᵛᶦᵇ)ᴺ / N!
= Z_Nᵗʳ (Z₁ʳᵒᵗ)ᴺ (Z₁ᵛᶦᵇ)ᴺ
F = Fᵗʳ+ Fʳᵒᵗ+ Fᵛᶦᵇ
Energy and entropy also add.
Chemical potential of ideal gas
Can find the chemical potential from Gibbs free energy, given by G = E − TS + PV
Monatomic ideal gas, E = ³⁄₂NkᴮT, and PV = NkᴮT.
Sum of these exactly cancels constant term in Sackur-Tetrode entropy, giving
G = −TS + ⁵⁄₂NkᴮT = −NkᴮTln(gₛn_Q/n)
Chemical potential, Gibbs free energy per particle:
µ = −kᴮTln(gₛn_Q/n)
Chemical potential is large and negative, and is relative to lowest energy state of the system. Can be calculated if temperature and number density known.
Note we could also get the same result from ⟨F⟩ = NkᴮT [ln(n/gₛn_Q) − 1], using µ = (∂F/∂N)ₜ,ᵥ
Also note that the Sackur-Tetrode entropy and expression above only correct for non-relative particles.
Derive Gibbs distribution
System which is in diffusive contact with a particle reservoir at chemical potential µ. Gibbs distribution frees us from the constraint that total numbers of particles in each energy level has to add up to a fixed total, allowing us to treat each energy level independently.
Temperature is a measure of decrease in entropy of reservoir from giving heat to the system.
Chemical potential is a measure of energy decrease (entropy increase) of the reservoir from giving particles to the system.
Probability that system will be in microstate i with energy εᵢ and particle number Nᵢ:
(recall neither εᵢ nor Nᵢ will be small for a typical system)
Derivation follows that of the Boltzmann distribution closely.
* Probability of system being in the given microstate depends on number of microstates available to the reservoir with energy E₀−εᵢ and particle number N₀−Nᵢ.
* Express number of microstates as the exponential of the entropy
* Make a Taylor expansion of the entropy about Sᵣ(E₀, N₀)
* Express the derivatives of the entropy in terms of T and µ:
* (∂Sᵣ/∂E)ᵥ,ₙ = 1/T
* (∂Sᵣ/∂N)ₑ,ᵥ = −µ/T
pᵢ = exp((µNᵢ−εᵢ)/kᴮT)/ Ƶ
with
Ƶ = Σⱼ exp((µNⱼ−εⱼ)/kᴮT)
Normalisation constant Ƶ is the grand partition function. The ensemble is the grand canonical ensemble.
Grand partition function
Ƶ = Σⱼ exp((µNⱼ−εⱼ)/kᴮT)
Particle number, energy, entropy, and grand potential averages for grand canonical ensemble.
Ƶ = Σⱼ exp((µNⱼ−εⱼ)/kᴮT)
⟨N⟩ = kᴮT(∂lnZ/∂µ)|β
⟨E⟩ = −(∂lnZ/∂β)|µ + µ⟨N⟩
⟨S⟩ = −kᴮΣᵢ pᵢ lnpᵢ = 1/T (⟨E−µN⟩ + kᴮTlnƵ)
Φ_G ≡ −kᴮTlnƵ = ⟨E − TS − µN⟩
Absolute entropy, pressure, and particle number for grand canonical ensemble
Fundamental thermodynamic relation,
dΦ_G = −SdT − PdV − Ndµ
S = −(∂Φ_G/∂T)|V,µ
P = −(∂Φ_G/∂V)|T,µ
N = −(∂Φ_G/∂µ)|T,V
Prove grand potential is equal to -PV
Natural variables of grand potential are T, V and µ. Of these, T and µ are intensive. Φ_G itself is extensive, so must be proportional to V, the only extensive one:
Φ_G = VφG(T, µ)
φ_G(T, µ) = (∂ΦG/∂V)|T,µ = −P
⇒ Φ_G = −PV
Grand potential when more than one species is present
Φ_G ≡ E − TS − Σᵢ µᵢNᵢ
dΦ_G = −SdT − PdV − Σᵢ Nᵢdµᵢ
Can use this to prove µᵢ is Gibbs free energy per particle of species i.
G = E − TS + PV
= E − TS − (E − TS − Σᵢ µᵢNᵢ)
= Σᵢ µᵢNᵢ
Partition function and average occupancy for a system where energy levels can only have one particle in them.
Energy of empty site is 0, energy of occupied site is ε₀.
Surface is in contact with gas or solution of chemical potential µ (energy drop of the solution when it loses a molecule).
There are only two microstates:
* Unoccupied, N = 0 and ε = 0
* Occupied, N = 1 and ε = ε₀
Therefore only two terms in grand partition function:
Ƶ = exp(0) + exp((µ−ε₀)β) = 1 + exp((µ−ε₀)β)
Average occupancy:
⟨N⟩ = −(∂ΦG/∂µ)|β
= −(∂(−kBTlnƵ)/∂µ)|β
= 1/exp((ε₀−µ)β)+1
fig(56.1)
Plotting average occupancy as a function of ε₀, the energy of the level in question.
⟨N⟩ is always less than 1. If a level lies above the chemical potential, ε₀ > µ, it is less likely to be occupied, as it is energetically more favourable for the
molecule to remain in solution. If ε₀ < µ then it is more likely to be occupied, since that is the energetically favourable configuration. At zero temperature, distribution becomes a step function, with ⟨N⟩ = 1 if ε₀ < µ and ⟨N⟩ = 0 if ε₀ > µ.
Partition function and average occupancy for a system where energy levels can have many particles in them.
Energy of empty site is 0, energy of occupied site is Nε₀.
Surface is in contact with gas or solution of chemical potential µ (energy drop of the solution when it loses a molecule).
Infinitely many microstates, so infinite terms in grand partition function (geometric series):
Ƶ = exp(0) + exp((µ−ε₀)β) + exp(2(µ−ε₀)β) + …
= 1/1−exp((µ−ε₀)β)
Average occupancy:
⟨N⟩ = −(∂ΦG/∂µ)|β
= −(∂(−kBTlnƵ)/∂µ)|β
= 1/exp((ε₀−µ)β)−1
fig(56.2)
Plotting average occupancy as a function of ε₀, the energy of the level in question.
No limit to ⟨N⟩, it doesn’t make sense to consider states with ε₀ < µ, as occupancy will be infinite (formula for ⟨N⟩ no longer valid in that case.) For ε₀ close to µ, occupancy will be high, and it falls off as ε₀ increases. Rapidity of the drop depends on temperature; for T = 0 only a level with ε₀ = µ would have
non-zero occupancy.
What is a boson?
Bosons are particles with integer spin:
* Spin 0: ¹H and ⁴He in ground state, pion, Higgs boson
* Spin 1: ¹H and ⁴He in first excited state, ρ meson, photon, W and Z bosons, gluons
* Spin 2: ¹⁶O in ground state, graviton.
Force carriers (photon, gluon, W, Z) and the Higgs are bosons.
What is a fermion?
Fermions are particles with half-integer spin:
* Spin ½: ³He in ground state, proton, neutron, quark, electron, neutrino
* Spin ³⁄₂: ⁵He in ground state, ∆ baryons (excitations of proton and neutron)
The basic building blocks of atoms are all fermions.
What defines fundamental classes of composite particles?
Total “spin” of a composite particle, its total angular momentum (usually with symbol J rather than S). This is the (quantised, vector) sum of all spins and all orbital angular momenta of all constituents.
Orbital angular momentum is always an integer. Therefore, J is always half integer for an odd number of spin-½ particles and integer for an even number. Composite particles (nuclei, atoms, molecules) made of an odd number of protons, neutrons and electrons are also fermions, whereas those made of an even number are bosons.
Excitations of composite partcles (nuclei, atoms) can only change the spin by an integer amount, so don’t change its nature.
Wavefunction requirement for bosons
Wavefunction must be symmetric under exchange of any pair.
For spinless bosons:
Ψ(r₁, r₂, r₃…) = 1/√2’ φ(r₁)φ(r₂)φ(r₃)…
is a perfectly acceptable wave function and there is no Pauli exclusion principle. For spin-1 bosons, must ensure that the overall space-spin wave function is symmetric(details not important).
Bosons are free/”prefer” to crowd into the same quantum state.
This explains the spectrum of black-body radiation, the operation of lasers, the properties of liquid ⁴He, and superconductors.
Wavefunction requirement for fermions
Fermions obey the Pauli exclusion principle: no more than one fermion can occupy a single quantum state.
Note that spin quantum number mₛ is part of the state’s description; if ignored then two spin-½ or four spin-³⁄₂ particles can occupy the same spatial state.
This is the basis of atomic structure and the periodic table, it explains the properties of metals, white dwarves, and neutron stars.
“Grown-up” version of the Pauli exclusion principle:
Overall wave function of a system of identical fermions must be antisymmetric under exchange of any pair.
Two spin-½ particles in same spatial state (eg 1s state of helium), overall wave function must be:
Ψ(r₁, r₂, m₁, m₂) = 1/√2’ φ₁ₛ(r₁)φ₁ₛ(r₂)(↑↓ − ↓↑)
Spatial is symmetric, spin is antisymmetric. (Corresponds to overall
spin 0).
Can’t construct a state with three particles in the same spatial state, as there is no state (↑↓↑ − …) which changes sign when interchanging any pair.
It is possible to have a 2-fermion spinstate with ↑↑ (spin-1), but the particles have to be in different spatial states:
Ψ(r₁, r₂, m₁, m₂) = 1/√2’ (φ₁ₛ(r₁)φ₂ₛ(r₂) − φ₂ₛ(r₁)φ₁ₛ(r₂))↑↑
Mathematical meaning of “identical”
Nothing observable can change if we swap the particles.
|Ψ(r₁, r₂, m₁, m₂)|² = |Ψ(r₂, r₁, m₂, m₁)|²
This implies
Ψ(r₁, r₂, m₁, m₂) = exp(iα)Ψ(r₂, r₁, m₂, m₁)
for some real phase α.
Two swaps have to return us to the same state without any phase change at all, so 2α = 2nπ. Only α = π and α = 2π are possible.
Single-level partition functions and average occupancy when considering bosonic/ fermionic nature
Focussing on a single energy level of energy εᵣ. If it has n particles, total energy is nεᵣ.
Partition function of single energy level:
Ƶᵣ = 1 + exp((µ−εᵣ)β) + exp(2(µ−εᵣ)β)… ≡ Σₙ exp(n(µ−εᵣ)β)
(just first two terms for fermions)
Grand partition function of whole system:
Ƶ = Ƶ₁Ƶ₂Ƶ₃… ≡ Πᵣ Ƶᵣ
Note that we ignore interactions between particles, but we’re NOT treating particles as independent, because bosonic or fermionic nature influences the multi-particle system.
Log of a product is sum of logs of individual terms, so grand potential ΦG, energy, particle number, and entropy all consist of sums of contributions from each level:
⟨N⟩ = Σᵣ ⟨Nᵣ⟩
⟨E⟩ = Σᵣ ⟨Nᵣ⟩ εᵣ
etc.
New notation, ⟨Nᵣ⟩ occupancies of individual levels are microscopic, so use small letters and an overline,
⟨Nᵣ⟩ ≡ n̅ᵣ and
⟨N⟩= Σᵣ n̅ᵣ
etc.
Fermions:
Ƶᵣ = 1 + exp((µ−εᵣ)β)
n̅ᵣ = 1/exp((εᵣ−µ)β)+1
Bosons:
Ƶᵣ = 1/1−exp((µ−εᵣ)β)
n̅ᵣ = 1/exp((εᵣ−µ)β)−1
Average particle number and energy for ideal gas (boson or fermion)
⟨N⟩ = ∫∞ g(k)n̅(k)dk
⟨E⟩ = ∫∞ g(k)ε(k)n̅(k)dk
where ε(k) = ħ²k²/2m for non-relativistic particles
Fermi-Dirac and Bose-Einstein distributions
n̅(k) = 1/exp((ε(k)−µ)β)±1
where + is fermions
and − is bosons
Note that for bosons, µ must be less than the energy of the lowest level (normally zero) but for fermions µ can be (and often is) greater than 0.
Grand potential for bosons/fermions
ΦG ≡ −kᴮTlnƵ
= −kᴮT Σᵣ lnƵᵣ (using Ƶ = Πᵣ Ƶᵣ)
= ∓kᴮT Σᵣ ln(1±exp((µ−εᵣ)β))
= ∓kᴮT ∫ g(ε) ln(1±exp((µ−ε)β) dε
where r labels single particle energy levels, signs are for fermions and bosons respectively, and the last line switches to the continuum limit, valid for a gas.
Plot the Bose-Einstein, Boltzmann, and Fermi-Dirac distributions
Plot with β(ε−μ) on x-axis, and ⟨N⟩ on y-axis.
See fig.(62)
Occupation in fermi gas at zero temperature
Occupancy is given by:
n̅(ε) = 1/exp((ε−µ)β)+1
Zero temperature, ground state/ state of lowest possible energy must have all energy levels occupied up to a maximum, and all higher levels unoccupied. Occupation function n̅(ε) becomes a step function—one up to a certain value of ε and zero thereafter.
Limit T → 0, β → ∞:
* ε < µ, argument of the exponential is
very large and negative and the exponential itself can be ignored in the denominator, giving n̅(ε) = 1.
* ε > µ, argument of the exponential is very large and positive and the “+1” can be ignored in the denominator, so n̅(ε) → exp(−(µ−ε)β) → 0.
* n̅(ε) → 1 for ε < µ
* n̅(ε) → 0 for ε > µ
µ is the energy of the highest occupied state at zero temperature. Also known
as Fermi energy, εF.
A gas with this sort of occupancy is a ‘degenerate’ gas.
Other Fermi values
- εF: Fermi energy. Highest occupied energy level at zero temp.
- kF: Fermi momentum. Value of k corresponding to Fermi energy.
- Fermi sea. The filled levels.
- Fermi surface. Top of the Fermi sea.
Value of Fermi energy and momentum at zero temperature
N =∫∞ g(k)n̅(k) dk
= ∫kF g(k) dk
= gₛV/2π² ∫kF k² dk
= gₛV/2π² kF³/3
With gₛ = 2, εF = ħ²kF²/2m for non-relativistic electrons, and n = N/V:
kF = (3π²n)¹’³
and
εF = (3π²ħ³n)²’³/2m
Chemical potential changing w.r.t. Fermi energy as temperature rises
Electron density:
N/V = 1/V ∫∞ g(k)n(k)dk
= gₛ/2π² ∫∞ k²/exp((ħ²k²/2m−µ)β)+1 dk
= ½ (2m/ħ²β)³’² gₛ/2π² ∫∞ x¹’²/z̄eˣ+1 dx
≡ 2gₛnQ/√π’ F(z̄)
where x ≡ ε(k)β:
x = ħ²β/2m k²
k = (2m/ħ²β)¹’² x¹’²
dk = ½ (2m/ħ²β)¹’² x⁻¹’² dx
and z̄ = exp(−µβ)
dimensionless x-integral is F(z̄)
Rearranging:
F(z̄) = √π’/2 n/gₛnQ
F(z̄) becomes large as z̄ → 0, µ»_space; kᴮT
Thermal fluctuations only affect electrons within ~kᴮT of µ, so a very small fraction of them.
Electronic heat capacity, Fermi style
Thermal fluctuations affect only a small fraction of all electrons. Electronic heat capacity is much less than the ³⁄₂NkᴮT
predicted by equipartition. Thermal excitations only affect states with energies of order kᴮT below Fermi surface, roughly kᴮT/EF of the total, and their excess energy is about kᴮT. So the extra energy above zero temperature value of ⅗ NEF is of order
N(kᴮT)²/EF
and the electronic heat capacity is of order
Cᵥ ∝ NkᴮT/TF
A more careful calculation gives the constant of proportionality to be π²/2. This linear rise with temperature can be seen at very low temperatures; at higher temperatures the contribution of lattice vibrations dominates.
Zero temp. energy of Fermi gas in terms of Fermi energy
E = ∫∞ g(k)n̅(k)ε(k)dk
= ħ²/2m ∫kF k²g(k)dk
= gₛVħ²/4π²m kF⁵/5
= ⅗ NεF
Wiedemann-Franz law
κ/σ ∝ T
κ = thermal conductivity
σ = electrical conductivity
T = temperature
What can a white dwarf be treated as and why?
Main sequence star, cloud of hydrogen collapses till it is dense enough for nuclear fusion; equilibrium is obtained when outward pressure of hot plasma balances gravitational attraction. When hydrogen fuel mostly used up, equilibrium is lost. Further nuclear processes can raise central temperature and become a red giant, but when all processes are exhausted the cooling star will start to collapse. For a star the size of the sun, collapse continues until increasing density and decreasing temperature cause the electrons to cease to be a classical gas, better described by the Fermi-Dirac distribution. Though still very hot (“white hot”), density is such that µ»_space; kᴮT and electron gas is well described by a “zero-temperature” degenerate electron gas, with gravitational attraction balanced by degeneracy pressure.
Pressure of a degenerate Fermi gas
Zero temperature,
ΦG = E − µN = E − NεF
PV = −ΦG = −(⅗−1)NεF = ⅖NεF
⇒ P = ⅖nεF = ħ²(3π²)²’³/5mₑ n⁵’³
Differential pressure equation for white dwarf
Density, Fermi energy, and and pressure vary with radius. For equilibrium, pressure difference across a spherical shell at r must exactly balance the weight:
4πr² (P(r) − P(r+dr)) = 4πr² dr Gρ(r)M(r)/r²
⇒ dP/dr = −Gρ(r)/r² ∫ᵣ 4πr’²ρ(r’)dr’
Solution can be cast in terms of average density and a universal dimensionless function of the scaled variable r/R, where R is the radius of the star:
ρ(r) = ̄ρ̄ f(r/R)
mass-radius relation for white dwarf stars
dP/dr = −Gρ(r)/r² ∫ᵣ 4πr’²ρ(r’)dr’
Integrate for pressure difference from centre to surface r = R (at which P(R) = 0):
−∫R dP/dr dr = P(0) = ∫R Gρ(r)M(r)/r² dr
Using ρ(r) = ̄ρ̄ f(r/R), which implies function g for the mass:
M(r) = M g(r/R), where ⁴⁄₃ πR³ ̄ρ̄ = M
Pressure of Fermi gas,
P = ⅖nεF = ħ²(3π²)²’³/5mₑ n⁵’³
Central pressure is proportional to ρ⁻⁵’³:
ρ⁻⁵’³ ∝ GM ̄ρ̄ ∫R f(r/R)g(r/R)/r² dr
= GM ̄ρ̄ /R ∫¹ f(x)g(x)/x² dx
The integral is universal and dimensionless; i.e. a number which is the same for all white dwarfs of a similar composition.
Cancel one power of ρ from either side to get
(M/R³)²’³ ∝ M/R ⇒ MR³ = constant
Upper bound of white dwarf mass ~ 1.4 solar masses—the Chandrasekhar limit. Involves:
(ħc/G)³’²/(2mₚ)² ∼ Msolar
Radius ~ Earth radius
energy per unit volume per unit frequency
u(ω) = g(ω)/V ħω/exp(ħωβ−1)
= ħ/π²c³ ω³/exp(ħωβ−1)
low freq, ħω«kᴮT, denominator ~ħωβ, Rayleigh-Jeans result
What can cavity radiation be described by?
An ultrarelativistic Bose gas with chemical potential µ = 0.
The gas is made of massless photons of energy ħω. The two polarisation states translate to gₛ = 2
number of photons for frequency range ω → ω + dω
ω = ck and ε = ħω
⟨N(ε)⟩ dε = g(ε)n̅(ε)dε
energy density of photons for frequency range ω → ω + dω
u(ω)dω ≡ ⟨E(ε)⟩/V dε
= ε g(ε)/V n̅(ε) dε
= ε gₛε²/2π²(ħc)³ 1/exp(εβ)−1 dε
= ħ/π²c³ ω³/exp(ħωβ)−1 dω
Flux of blackbody radiation (power emitted per unit emitting area, per unit frequency) from energy density inside
F(ω) = c/4 u(ω)
Stefan’s law, and constant in terms of Planck and Boltzmann
L = AσT⁴
σ = π²kᴮ⁴/60ħ³c²
G(z̄)
G(z̄) ≡ ∫∞ x¹’²/z̄eˣ−1 dx
= 4π²/gₛ n (ħ²/2mkᴮT)³’²
= √π’/2 n/gₛnQ
Has a limit at high zbar
condensation temperature
T꜀ = 3.3125 ħ²/mkᴮ (n/gₛ)²’³
N below critical/condensation temperature
Treat ground state separately
N = N₀ + ∫∞ g(k)n̅(k)dk
= N₀ + gₛV/2π² ∫∞ k²/exp((ħ²k²/2m)β−1 dk
= N₀ + (2m/ħ²β)³’² Vgₛ/4π² ∫∞ x¹’²/eˣ−1 dx
= N₀ + 2.61238 gₛVnQ
= N₀ + N(T/T꜀)³’²
⇒ N₀ = N(1 − (T/T꜀)³’²)
T꜀ is itself a function of N
Bose-Einstein condensation
For any T even just a little below T꜀, N₀ is of order N. For macroscopic N, that is very much
bigger than, say, fluctuations of order √N’ that we ignore when using the grand canonical ensemble. As temperature drops below T꜀, a macroscopic number of particles “condense” into the ground state: Bose-Einstein condensation.
This is in comparison with a typical state of higher energy (say ε ∼ kBT) where the occupancy
is “a few”, of order 1.
As the temperature drops, the fraction of the particles in the condensate rises steadily, till as T → 0, N₀ → N. The condensate is a single, macroscopic,
collective quantum object which (for cold trapped atomic gases) can actually be seen by the naked eye.
