Counting Photons in a Fractal Box

Joe P. Chen

Cornell University

November 9, 2010

Outline

History of blackbody radiation

Radiation from a cubic box

Generalize the calculation to other geometries

Heat kernels on fractals

Radiation from a fractal box

The experiment (circa 1900)

I A blackbody is a box which absorbs and emits light (EM radiation)of all frequencies f ∈ [0,∞).

I Suppose we place the blackbody in a room of temperature T . Pokea tiny hole on the surface of the box, and measure the intensity ofthe radiation (total intensity, intensity as a function of freq f , etc.)

Experimental result

Classical physics fails to explain the result

I After 1862 (Maxwell) and before 1900, light was understood toconsist of EM waves travelling at the constant speed c .

I Rayleigh and Jeans applied Maxwell’s classical theory of light to theblackbody problem, and produced a result which works for small f ,but blows up as f → ∞: ultraviolet catastrophe!

I Wien took a different approach, and reproduced the higher f portionof the curve, but not the low f regime.

Planck to the rescue

I Max Planck (1901)postulated that the light energy is distributed overthe modes of charged oscillators in the blackbody.

I Specifically, he assumedthat the energies of the oscillator with freq f canonly be integer multiples of the fundamental energy

E = hf = ~ω.

I h is Planck’s constant. (~ ≡ h/(2π), ω ≡ 2πf .)

I Using these hypotheses, Planck was able toreproduce the experimental result:

I (ω,T ) =1

π2c3ω3

e~ω/kBT − 1.

I This marked the beginning of quantum mechanics.Through the work of Einstein, Bose, etc., we nowunderstand that these oscillators are really quanta oflight energy.

Quantum theory of light

Light is made up of quanta of EM radiation, called photons.

I Photons are massless.

I Photons in vacuum travel at the speed of light c , and satisfy thewave equation

ψtt = c2∆ψ.

I Each photon has energy E = hf = ~ω (ω = 2πf ) and momentum ofmagnitude |p| = E/c = h/λ = ~|k|(λ = wavelength, k = wavevector, |k| = 2π/λ).

I Photons have spin-1: each photon carries polarization which can beleft-circular, right-circular, or a linear combination thereof.

Photons as structural probes

(a) Spiral galaxy (b) DNA (c) Quantum computer

What about using photons to probe fractals?

Okay, maybe not the Julia set.

What about using photons to probe fractals?

More realistic: A 3D fractal box which can be manufactured using, say,business cards.

Thermodynamics of quantum system with 1 particle

I (Ω, ⟨ , ⟩) = C Hilb. space for a single-particle sys.

I H : Ω → Ω, Hamiltonian (energy) operator. (H ≥ 0)

I σ(H) : Spectrum of H = En∞n=0.

I β = 1/(kBT ) > 0 : inverse temperature.

Rule: Each particle occupies an energy level ∈ σ(H).At thermal equilibrium, the probability that a particle occupies energylevel En is given by the Boltzmann distribution

P(n) =e−βEn

TrΩe−βH=

1

Z1e−βEn .

I Z1 = TrΩe−βH is the one-particle partition function.

I F = −β−1 lnZ1 (equiv. Z1 = e−βF ) is the free energy.

Thermodynamics of qu. sys. with N identical particles

I W = Ω⊗ · · · ⊗ Ω︸ ︷︷ ︸n−tuple

= Hilb. space for N-particle system

I H : W → W defined by H(ψ1 ⊗ · · · ⊗ ψN) = (Hψ1)⊗ · · · ⊗ (HψN).

However, if the N particles are identical, then one cannot distinguish

· · · ⊗ ψi ⊗ · · · ⊗ ψj ⊗ · · ·

from· · · ⊗ ψj ⊗ · · · ⊗ ψi ⊗ · · ·

For photons, P(ψ1 ⊗ · · · ⊗ ψN) = ψ1 ⊗ · · · ⊗ ψN for any permutation P.In other words, the actual Hilbert space for photons isW = Sym(Ω⊗ · · · ⊗ Ω).

Partition function is Z = TrW e−βH .

Photon statistics

For convenience, assume that the system contains any number ofphotons.

I Example: A single-mode system with σ(H) = E.

Z =∞∑j=0

e−jβE =1

1− e−βE.

I Example: A three-mode system with σ(H) = E1,E2,E3.

Z =3∏

n=1

∞∑j=0

e−jβEn

=3∏

n=1

(1

1− e−βEn

).

Partition function of thermal photons

If the Hamiltonian for a single photon is

H = ~ (i∂t)︸︷︷︸ω

= ~c√−∆︸ ︷︷ ︸=|k|

,

then the partition function for the ensemble of photons under the sameHamiltonian is

lnZ =∑

E∈σ(H)

ln

(1

1− e−βE

)= −TrΩ ln(1− e−βH).

(Ω is the single-particle Hilbert space.)

Thermodynamic quantities

For an extended system, the partition function depends on volume (V ) inaddition to temperature (T ):

Z = Z (T ,V ) = e−βF = Tre−βH

Many equilibrium thermodynamic quantities can be derived from Z .Examples:

I Mean energy

E =TrHe−βH

Tre−βH= − 1

Z

∂Z

∂β= − ∂

∂βlnZ .

I Pressure

P = − ∂F

∂V=

1

β

∂

∂VlnZ .

Counting states inside a cubic box

I U = [0, a]3 ⊂ R3 : Cubic box of side a, a large.

I Each photon has ”wavefunction” satisfyingψtt(x, t) = c2∆ψ(x, t) in U × t ≥ 0Periodic boundary condition on ∂U × t ≥ 0 .

I Fourier transform shows that

ψ(k, ω) =

∫ ∞

0

∑k

ψ(x, t)e i(k·x−ωt)dt

where k = (2π/a)(n1, n2, n3), ni ∈ N. The photons states are theFourier modes of the box, specified by k.

I The wave equation implies that ω = c |k|, or E = ~ω = ~c |k|.

Counting states inside a cubic box

Question: How many eigenmodes ρ(E )dE are contained in an energyinterval [E ,E + dE ]?

ρ(E )dE = ρ(|k|)d |k| =

[2 ·( a

2π

)34π|k|2

]d |k|

=( a

2π

)3· 8πE

2

(~c)3︸ ︷︷ ︸density of states

dE

I 2 comes from the two possible polarizations of the photon.

Mean energy density of photons in a cubic box

lnZ = −∑

E∈σ(H)

ln(1− e−βE ).

We just saw that the mean energy density of the photons in the box isgiven by

E

V= − 1

V

∂

∂βlnZ

=1

V

∑E∈σ(H)

E

(1

eβE − 1

).

Approximation: If the box side length a is large, one can replace∑E∈σ(H)

by

∫ ∞

0

ρ(E )dE ,

so

E

V=

1

π2(~c)3

∫ ∞

0

E 3

(1

eβE − 1

)dE .

Mean energy density of photons in a cubic box

E

V=

1

π2(~c)3

∫ ∞

0

E 3

(1

eβE − 1

)dE

=1

π2(~c)3β4

∫ ∞

0

η3

eη − 1dη (η ≡ βE)

Two immediate observations:

I Each energy mode E = ~ω contributes an energy density

I (ω,T ) =1

π2(~c)3E 3

eβE − 1=

1

π2c3ω3

eβ~ω − 1.

Get Planck’s radiation law.

I The total energy density scales with β−4, or T 4. (Stefan-Boltzmannlaw)

Using a series of change of variables, one finds (for Re(n) > 1)∫ ∞

0

ηn−1

eη − 1dη =

∞∑k=1

1

kn

∫ ∞

0

un−1e−udu = ζ(n)Γ(n).

So

E

V=

1

π2(~c)3β4ζ(4)Γ(4) =

1

π2(~c)3β4· π

4

90· 3! = π2

15(~c)3β4

Generalize the calculation to different geometries

I The photon state-counting argument for the cubic box relies onFourier decomposition.

I However, Fourier decomposition is not a priori known on othergeometries, especially fractals.

I Goal: Find an expression for the partition function Z which dependson the Laplacian ∆ endowed upon the given geometry U.

Spectrum of photons radiated from U ↔ Laplacian on U

I Many techniques on the Laplacian on fractals are well-developed(see later), so we can apply them to give information aboutradiation from a fractal blackbody.

Manipulation (Akkermans, Dunne & Teplyaev, ArXiv:1010.1148)

Accounting for zero-point energy, the partition function for a single modeE = ~ω is

Z (ω) =∞∑j=0

e−(j+1/2)β~ω =e−β~ω/2

1− e−β~ω

or

lnZ (ω) = −1

2β~ω − ln(1− e−β~ω) = −1

2β~ω︸ ︷︷ ︸

vacuum term

+∞∑n=1

1

ne−nβ~ω.

(Technical) Use the following identities

I 1

ne−nβ~ω =

β~2√π

∫ ∞

0

dτ

τ 3/2e−ω2τe−(nβ~)2/(4τ).

I Poisson sum. relation:

√π

t

∞∑n=−∞

e−π2n2/t =∞∑

n=−∞e−tn2 .

to rewrite lnZ (ω) as...

lnZ (ω) =1

2

∫ ∞

0

dτ

τe−ω2τ

∞∑n=−∞

exp

(−[2πn

~β

]2τ

)The full partition function is obtained by tracing over all modes ω:

lnZ =1

2

∫ ∞

0

dτ

τ

∑ω

e−ω2τ∞∑

n=−∞exp

(−[2πn

~β

]2τ

)

Since ω2 is the eigenvalue of −∂tt = −c2∆, we can express the first sumas a trace of ec

2τ∆ over U:

lnZ =1

2

∫ ∞

0

dτ

τTrU

(ec

2τ∆) ∞∑

n=−∞exp

(−[2πn

~β

]2τ

)

I ωn := 2πn~β (n ∈ Z) are known as the Matsubara frequencies.

iωn are the poles of the function nB(ω) =1

eβ~ω−1.

Partition function for general geometry

Let’s try to make everything dimensionless:

I LS : Characteristic length of the box

I ∆ ≡ (LS)2∆ : Dimensionless Laplacian

I Lβ ≡ ~βc : Thermal photon ”wavelength”

Upon a redefinition of variables, one gets

lnZ =1

2

∫ ∞

0

dτ

τ

( ∞∑n=−∞

e−(2πn)2τ

)TrU

(e(Lβ/LS)

2τ∆).

I The partition function is connected to the heat kernel trace on U.

I State counting is encapsulated in the HKT.

Heat kernelHeat equation ut(x, t) = ∆u(x, t) in U × t > 0

u(x , 0) = f (x) in Uboundary condition on ∂U × t > 0

.

has solution

u(x , t) = et∆f (x) =

∫U

pt(x , y)f (y)dy .

I pt(x , y) is the heat kernel, the fundamental soln to the heat eqn.I If U is compact, −∆ has pure point spectrum with eigensolutions

λj , ϕj. The heat kernel can be expanded in the eigenbasis as

pt(x , y) =∞∑j=1

e−tλjϕj(x)ϕj(y).

I Heat kernel trace is

Tr(et∆) =

∫U

pt(x , x)dx =∞∑j=1

e−tλj .

Spectral asymptotics & Heat kernel trace

I Eigenvalue counting function

N(s) = #λ ∈ σ(−∆) : λ < s.

I Heat kernel trace

TrU(et∆) =

∞∑j=1

e−tλj

The two expressions are related via a Laplace-Stieltjes transform.

∞∑j=1

e−tλj =

∫ ∞

0

e−tsdN(s).

I What we’re interested in is the asymptotic scaling of the HKT ast ↓ 0 (correspondingly, the asymptotic scaling of N(s) as s → ∞).

Revisiting the cubic box

Weyl asymptotic formula for a bounded Riemannian manifold M in Rd :

N(s) =BdVol(M)

(2π)dsd/2 +O(s(d−1)/2) as s → ∞.

I Bd = πd/2

Γ((d/2)+1) : Volume of unit ball in Rd .

This translates into the HKT asymptotics

TrM(et∆) =

∫ ∞

0

e−ts

[BnVol(M)

(2π)d

(d

2

)s(d/2)−1 +O(s(d−3)/2)

]ds

=

BnVol(M)

(2π)dt−d/2

(d

2

)∫ ∞

0

e−uu(d/2)−1du︸ ︷︷ ︸=Γ(d/2)

+O(t−(d−1)/2)

=Vol(M)

(2√π)d

t−d/2 +O(t−(d−1)/2) as t ↓ 0.

Take d = 3, M = cube of side LS. From the HKT asymptotics

Tr(et∆) =t−3/2

(2√π)3

+O(t−1).

Plug into the partition function:

lnZ =1

2

∫ ∞

0

dτ

τ

( ∞∑n=−∞

e−(2πn)2τ

)TrM

(e(Lβ/LS)

2τ∆)

= · · ·

= (vacuum term) +

(LSLβ

)3ζ(4)Γ(2)

π2+ · · · .

Check mean energy density

E

V= − 1

L3S

∂

∂βlnZ =

3~cL4β

ζ(4)Γ(2)

π2=

π2

30(~c)3β4.

Off by a factor of 2: But that’s because I forgot to account forpolarizations! So we’re okay!

Fractal boxes

What I have in mind:

I Subdivide a cube into p3 equal subcubes.

I Remove k of these subcubes, with the restriction that anyneighboring cells must adjoin at a common face. (For simplicity, alsodemand that full cubic symmetry remains intact.)

I Contract the said map by scale 1/p, then apply it to each of theremaining cells. Iterate.

I Example: Menger Sponge (p = 3, k = 7)

Spectral asymptotics on fractals

100

101

102

103

104

100

101

102

103

104

s

N(s

)

Eigenvalue counting function, Menger sponge L3

I N(s) scales as sdS/2, but dS = dH!(dH = log3 20 = 2.7268...; dS = 2.51 ± 0.01 from numerical estimate.)

I There is extra structure on top of the scaling. To see this, plotW (s) := N(s)s−dS/2, the Weyl ratio.

Spectral asymptotics on fractals

100

101

102

103

100

s

W(s

)=N

(s)/

sα

Menger Sponge Weyl ratio (α=1.145)

M1M2M3

In the fractal limit,

N(s) = sdS/2G (ln s) + o(sdS/2) as s → ∞,

where G is periodic and zig-zag.

Heat kernel trace on fractals

10−3

10−2

10−1

100

0.22

0.24

0.26

0.28

0.3

0.32

t

tα ∫ p t(x

,x)

dxMenger sponge Heat Kernel Trace (α=1.145)

M2M3

In the fractal limit,

Tr(et∆) = t−dS/2g(− ln t) + o(t−dS/2) as t ↓ 0,

where g is periodic, bounded away from 0, and approximates a sinusoid.

HKT on fractals obeys a scaling with t modulated by log-periodicoscillations in the short-time regime.

Spectral dimension (dS)

Brownian motion

r(t) :=√Ex [(Xt)2] ∼ t1/dW , where dW is the walk dimension

(Gaussian diffusion corresponds to dW = 2.)

Heuristic derivation: Given a Brownian motion Xt , the amount of fractalsubstrate explored by the process after (a short) time t is ∼ r(t)dH . Sothe probability of Xt returning to O scales as r(t)−dH ∼ t−dH/dW . Butthis is the same as the trace of the heat kernel ∼ t−dS/2.

Relationship between spectral, Hausdorff, and walk dims

dS2

=dHdW

For fractals, dH > dS, so dW > 2: sub-Gaussian diffusion.

Partition function for a fractal boxI The leading term in the HKT is TrU(e

t∆) = t−ds/2g(− ln t).

I Convert into dimensionless units:

TrU

(e(Lβ/LS)

2τ∆)=

(L2SL2βτ

)dS/2

g

(− ln

[L2βτ

L2S

]).

I Plug into the partition function, get

lnZ = y · g(− ln y) where y ≡(

LSβ~c

)dS

.

I Mean energy density: The appropriate volume is the spectral volumeLdSS , not Hausdorff volume.

E

V= − 1

LdSS

∂

∂βlnZ =

dSβdS+1(~c)dS

(g + yg ′).

So dS can be experimentally probed by varying the temp. of the box!

Blackbody at T = 0: Vacuum energy

The vacuum term in the partition function is

lnZ0 = −1

2β~∑ω

ω = −1

2

β~cLS

∑λ∈σ(−∆)

λ1/2

= −1

2

(LβLS

)ζM

(−1

2

),

where ζM(s) is the spectral zeta function. It contributes an energy

E0 = − ∂

∂βlnZ0 =

1

2

~cLSζM

(−1

2

),

called the Casimir energy:

I It is the only energy present at absolute zero temperature.

I It has been measured for simple geometries.