Counting Photons in a Fractal Box

Transcription

1 Counting Photons in a Fractal Box Joe P. Chen Cornell University November 9, 2010

2 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

3 The experiment (circa 1900) A blackbody is a box which absorbs and emits light (EM radiation) of all frequencies f [0, ). Suppose we place the blackbody in a room of temperature T. Poke a tiny hole on the surface of the box, and measure the intensity of the radiation (total intensity, intensity as a function of freq f, etc.)

4 Experimental result

5 Classical physics fails to explain the result After 1862 (Maxwell) and before 1900, light was understood to consist of EM waves travelling at the constant speed c. Rayleigh and Jeans applied Maxwell s classical theory of light to the blackbody problem, and produced a result which works for small f, but blows up as f : ultraviolet catastrophe! Wien took a different approach, and reproduced the higher f portion of the curve, but not the low f regime.

6 Planck to the rescue Max Planck (1901) postulated that the light energy is distributed over the modes of charged oscillators in the blackbody. Specifically, he assumed that the energies of the oscillator with freq f can only be integer multiples of the fundamental energy E = hf = ω. h is Planck s constant. ( h/(2π), ω 2πf.) Using these hypotheses, Planck was able to reproduce the experimental result: I (ω, T ) = 1 ω 3 π 2 c 3 e ω/k BT 1. This marked the beginning of quantum mechanics. Through the work of Einstein, Bose, etc., we now understand that these oscillators are really quanta of light energy.

7 Quantum theory of light Light is made up of quanta of EM radiation, called photons. Photons are massless. Photons in vacuum travel at the speed of light c, and satisfy the wave equation ψ tt = c 2 ψ. Each photon has energy E = hf = ω (ω = 2πf ) and momentum of magnitude p = E/c = h/λ = k (λ = wavelength, k = wavevector, k = 2π/λ). Photons have spin-1: each photon carries polarization which can be left-circular, right-circular, or a linear combination thereof.

8 Photons as structural probes (a) Spiral galaxy (b) DNA (c) Quantum computer

9 What about using photons to probe fractals? Okay, maybe not the Julia set.

10 What about using photons to probe fractals? More realistic: A 3D fractal box which can be manufactured using, say, business cards.

11 Thermodynamics of quantum system with 1 particle (Ω,, ) = C Hilb. space for a single-particle sys. H : Ω Ω, Hamiltonian (energy) operator. (H 0) σ(h) : Spectrum of H = {E n } n=0. β = 1/(k B T ) > 0 : inverse temperature. Rule: Each particle occupies an energy level σ(h). At thermal equilibrium, the probability that a particle occupies energy level E n is given by the Boltzmann distribution P(n) = e βe n Tr Ω e βh = 1 Z 1 e βen. Z 1 = Tr Ω e βh is the one-particle partition function. F = β 1 ln Z 1 (equiv. Z 1 = e βf ) is the free energy.

12 Thermodynamics of qu. sys. with N identical particles W = Ω Ω = Hilb. space for N-particle system }{{} n tuple H : W W defined by H(ψ 1 ψ N ) = (Hψ 1 ) (Hψ N ). However, if the N particles are identical, then one cannot distinguish ψ i ψ j from ψ j ψ i For photons, P(ψ 1 ψ N ) = ψ 1 ψ N for any permutation P. In other words, the actual Hilbert space for photons is W = Sym(Ω Ω). Partition function is Z = Tr W e βh.

13 Photon statistics For convenience, assume that the system contains any number of photons. Example: A single-mode system with σ(h) = {E}. Z = e jβe 1 = 1 e βe. j=0 Example: A three-mode system with σ(h) = {E 1, E 2, E 3 }. 3 3 ( ) Z = e jβe n 1 =. 1 e βen n=1 j=0 n=1

14 Partition function of thermal photons If the Hamiltonian for a single photon is H = (i t ) = c, }{{}}{{} ω = k then the partition function for the ensemble of photons under the same Hamiltonian is ln Z = ( ) 1 ln 1 e βe = Tr Ω ln(1 e βh ). E σ(h) (Ω is the single-particle Hilbert space.)

15 Thermodynamic quantities For an extended system, the partition function depends on volume (V ) in addition to temperature (T ): Z = Z(T, V ) = e βf = Tre βh Many equilibrium thermodynamic quantities can be derived from Z. Examples: Mean energy Ē = TrHe βh Tre βh = 1 Z Z β = ln Z. β Pressure P = F V = 1 ln Z. β V

16 Counting states inside a cubic box U = [0, a] 3 R 3 : Cubic box of side a, a large. Each photon has wavefunction satisfying { ψtt (x, t) = c 2 ψ(x, t) in U {t 0} Periodic boundary condition on U {t 0}. Fourier transform shows that ˆψ(k, ω) = 0 ψ(x, t)e i(k x ωt) dt k where k = (2π/a)(n 1, n 2, n 3 ), n i N. The photons states are the Fourier modes of the box, specified by k. The wave equation implies that ω = c k, or E = ω = c k.

17 Counting states inside a cubic box Question: How many eigenmodes ρ(e)de are contained in an energy interval [E, E + de]? ρ(e)de = ρ( k )d k = [ ( a ) 3 2 4π k 2] d k 2π = ( a ) 3 8πE 2 2π ( c) }{{ 3 de } density of states 2 comes from the two possible polarizations of the photon.

18 Mean energy density of photons in a cubic box ln Z = E σ(h) ln(1 e βe ). We just saw that the mean energy density of the photons in the box is given by Ē V = 1 V = 1 V β ln Z ( E E σ(h) 1 e βe 1 ). Approximation: If the box side length a is large, one can replace so E σ(h) Ē V = 1 π 2 ( c) 3 by 0 0 ρ(e)de, ( ) E 3 1 e βe de. 1

19 Mean energy density of photons in a cubic box Ē V = 1 π 2 ( c) 3 = 1 π 2 ( c) 3 β 4 Two immediate observations: 0 0 ( E 3 1 e βe 1 η 3 e η 1 dη ) de (η βe) Each energy mode E = ω contributes an energy density I (ω, T ) = Get Planck s radiation law. 1 E 3 π 2 ( c) 3 e βe 1 = 1 ω 3 π 2 c 3 e β ω 1. The total energy density scales with β 4, or T 4. (Stefan-Boltzmann law)

20 Using a series of change of variables, one finds (for Re(n) > 1) So 0 η n 1 e η 1 dη = 1 k n k=1 0 u n 1 e u du = ζ(n)γ(n). Ē V = 1 π 2 ( c) 3 β 4 ζ(4)γ(4) = 1 π 2 ( c) 3 β 4 π4 90 3! = π 2 15( c) 3 β 4

21 Generalize the calculation to different geometries The photon state-counting argument for the cubic box relies on Fourier decomposition. However, Fourier decomposition is not a priori known on other geometries, especially fractals. Goal: Find an expression for the partition function Z which depends on the Laplacian endowed upon the given geometry U. Spectrum of photons radiated from U Laplacian on U Many techniques on the Laplacian on fractals are well-developed (see later), so we can apply them to give information about radiation from a fractal blackbody.

22 Manipulation (Akkermans, Dunne & Teplyaev, ArXiv: ) Accounting for zero-point energy, the partition function for a single mode E = ω is Z(ω) = e (j+1/2)β ω = e β ω/2 1 e β ω or j=0 ln Z(ω) = 1 2 β ω ln(1 e β ω ) = 1 2 β ω + }{{} vacuum term (Technical) Use the following identities 1 n e nβ ω = β 2 dτ 2τ e (nβ )2 /(4τ). π Poisson sum. relation: to rewrite ln Z(ω) as... 0 e ω τ 3/2 π t n= e π2 n 2 /t = n= n=1 e tn2. 1 n e nβ ω.

23 ln Z(ω) = dτ 2 τ e ω τ n= exp ( [ ] 2 2πn τ) β The full partition function is obtained by tracing over all modes ω: ln Z = 1 ( dτ [ ] 2 e ω2 τ 2πn exp τ) 2 τ β 0 ω n= Since ω 2 is the eigenvalue of tt = c 2, we can express the first sum as a trace of e c2τ over U: ln Z = 1 dτ 2 0 τ Tr U (e ) ( [ ] 2 c2 τ 2πn exp τ) β n= ω n := 2πn β (n Z) are known as the Matsubara frequencies. {iω n } are the poles of the function n B (ω) = 1 e β ω 1.

24 Partition function for general geometry Let s try to make everything dimensionless: L S : Characteristic length of the box (L S ) 2 : Dimensionless Laplacian L β βc : Thermal photon wavelength Upon a redefinition of variables, one gets ln Z = 1 ( dτ 2 τ 0 n= e (2πn)2 τ ) Tr U (e (L β/l S) 2 τ ). The partition function is connected to the heat kernel trace on U. State counting is encapsulated in the HKT.

25 Heat kernel Heat equation has solution u t (x, t) = u(x, t) in U {t > 0} u(x, 0) = f (x) in U boundary condition on U {t > 0} u(x, t) = e t f (x) = U p t (x, y)f (y)dy.. p t (x, y) is the heat kernel, the fundamental soln to the heat eqn. If U is compact, has pure point spectrum with eigensolutions {λ j, ϕ j }. The heat kernel can be expanded in the eigenbasis as p t (x, y) = Heat kernel trace is Tr(e t ) = U e tλ j ϕ j (x)ϕ j (y). j=1 p t (x, x)dx = e tλ j. j=1

26 Spectral asymptotics & Heat kernel trace Eigenvalue counting function N(s) = #{λ σ( ) : λ < s}. Heat kernel trace Tr U (e t ) = j=1 e tλ j The two expressions are related via a Laplace-Stieltjes transform. e tλ j = j=1 0 e ts dn(s). What we re interested in is the asymptotic scaling of the HKT as t 0 (correspondingly, the asymptotic scaling of N(s) as s ).

27 Revisiting the cubic box Weyl asymptotic formula for a bounded Riemannian manifold M in R d : N(s) = B dvol(m) (2π) d s d/2 + O(s (d 1)/2 ) as s. B d = πd/2 Γ((d/2)+1) : Volume of unit ball in Rd. This translates into the HKT asymptotics [ ( ) ] Tr M (e t ) = e ts Bn Vol(M) d 0 (2π) d s (d/2) 1 + O(s (d 3)/2 ) ds 2 = B n Vol(M) (2π) d t d/2 ( ) d 2 0 e u u (d/2) 1 du + O(t (d 1)/2 ) } {{ } =Γ(d/2) = Vol(M) (2 π) d t d/2 + O(t (d 1)/2 ) as t 0.

28 Take d = 3, M = cube of side L S. From the HKT asymptotics Tr(e t ) = Plug into the partition function: ln Z = 1 ( dτ 2 τ = 0 t 3/2 (2 π) 3 + O(t 1 ). n= = (vacuum term) + Check mean energy density Ē V = 1 L 3 S e (2πn)2 τ ( LS L β ) 3 c ζ(4)γ(2) ln Z = β L 4 β π 2 = Tr M ( e (L β/l S) 2 τ ) ) 3 ζ(4)γ(2) π 2 +. π 2 30( c) 3 β 4. Off by a factor of 2: But that s because I forgot to account for polarizations! So we re okay!

29 Fractal boxes What I have in mind: Subdivide a cube into p 3 equal subcubes. Remove k of these subcubes, with the restriction that any neighboring cells must adjoin at a common face. (For simplicity, also demand that full cubic symmetry remains intact.) Contract the said map by scale 1/p, then apply it to each of the remaining cells. Iterate. Example: Menger Sponge (p = 3, k = 7)

30 Spectral asymptotics on fractals 10 4 Eigenvalue counting function, Menger sponge L N(s) s N(s) scales as s ds/2, but d S d H! (d H = log 3 20 = ; d S = 2.51 ± 0.01 from numerical estimate.) There is extra structure on top of the scaling. To see this, plot W (s) := N(s)s d S/2, the Weyl ratio.

31 Spectral asymptotics on fractals Menger Sponge Weyl ratio (α=1.145) M1 M2 M3 W(s)=N(s)/s α s In the fractal limit, N(s) = s d S/2 G(ln s) + o(s d S/2 ) as s, where G is periodic and zig-zag.

32 Heat kernel trace on fractals Menger sponge Heat Kernel Trace (α=1.145) 0.32 M2 M3 0.3 t α p t (x,x) dx t In the fractal limit, Tr(e t ) = t d S/2 g( ln t) + o(t d S/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-periodic oscillations in the short-time regime.

33 Spectral dimension (d S ) Brownian motion r(t) := E x [(X t ) 2 ] t 1/d W, where d W is the walk dimension (Gaussian diffusion corresponds to d W = 2.) Heuristic derivation: Given a Brownian motion X t, the amount of fractal substrate explored by the process after (a short) time t is r(t) dh. So the probability of X t returning to O scales as r(t) d H t d H/d W. But this is the same as the trace of the heat kernel t ds/2. Relationship between spectral, Hausdorff, and walk dims d S 2 = d H d W For fractals, d H > d S, so d W > 2: sub-gaussian diffusion.

34 Partition function for a fractal box The leading term in the HKT is Tr U (e t ) = t d s/2 g( ln t). Convert into dimensionless units: Tr U (e ) ( (L β/l S) 2 τ L 2 S = L 2 β τ ) ds /2 ( [ ]) L 2 β τ g ln. L 2 S Plug into the partition function, get ln Z = y g( ln y) where y ( ) ds LS. β c Mean energy density: The appropriate volume is the spectral volume, not Hausdorff volume. L d S S Ē V = 1 L ds S β ln Z = d S β ds+1 ( c) ds (g + yg ). So d S can be experimentally probed by varying the temp. of the box!

35 Blackbody at T = 0: Vacuum energy The vacuum term in the partition function is ln Z 0 = 1 2 β ω ω = 1 β c 2 = 1 2 L S ( Lβ L S λ σ( ) λ 1/2 ) ( ζ M 1 ), 2 where ζ M (s) is the spectral zeta function. It contributes an energy E 0 = β ln Z 0 = 1 ( c ζ M 1 ), 2 L S 2 called the Casimir energy: It is the only energy present at absolute zero temperature. It has been measured for simple geometries.